{"text": "\\chapter{Dynamic Programming}\n\n\n\n\\section{Introduction}\nThe core philosophy of dp:\n\\begin{enumerate}\n\\item The definition of \\textbf{states} \n\\item The definition of the \\textbf{transition functions} among states \n\\end{enumerate} \n\nThe so called concept dp as memoization of recursion does not grasp the core philosophy of dp. \n\nThe formula in the following section are unimportant. Instead, what is important is the definition of dp array and transition function derivation.\n\\subsection{Common practice}\n\\runinhead{Dummy.} Use dummies to avoid using if-else conditional branch.\n\\begin{enumerate}\n\\item Use $n+1$ dp arrays to reserve space for dummies. \n\\item Iteration range is $[1, n+1)$.\n\\item $n+k$ for k dummies \n\\end{enumerate}\n\\runinhead{State definition.} Two general sets of state definitions - the state \n\\begin{enumerate}\n\\item ends \\textit{at} index $i$\n\\item ends \\textit{before} index $i$\n\\item ends \\textit{at} or \\textit{before} index $i$\n\\end{enumerate}\n\n\\runinhead{Space optimization.} To avoid MLE, we need to carry out space optimization. Let $o$ be other subscripts, $f$ be the transition function. \n\nFirstly,\n$$\nF_{i, o} = f\\big(F_{i-1, o'}\\big)\n$$\n\nshould be reduced to \n$$\nF_{o} = f\\big(F_{o'}\\big)\n$$\n\nSecondly,\n$$\nF_{i, o} = f\\big(F_{i-1, o'}, F_{i-2. o'}\\big)\n$$\n\nshould be reduced to \n$$\nF_{i, o} = f\\big(F_{(i-1)\\%2, o'}, F_{(i-2)\\%2. o'}\\big)\n$$\n\nMore generally, we can be $(i-b)\\%a$ to reduce the space down to $a$.\n\nNotice:\n\\begin{enumerate}\n\\item Must iterate $o$ \\textbf{backward} to un-updated value. \n\\end{enumerate}\n\n\n\n\\section{Sequence}\\label{dpSequence}\n\n\\subsection{Single-state dp}\n\\runinhead{Longest common subsequence.} Let $F_{i, j}$ be the LCS at string $a[:i]$ and $b[:j]$. We have two situations: $a[i]==b[j]$ or not.\n\\begin{eqnarray*}\nF_{i. j} = \\left\\{ \\begin{array}{rl}\n F_{i-1, j-1}+1 &\\mbox{// if $a[i]==b[j]$} \\\\\n \\max\\Big(F_{i-1, j},&\\mbox{// otherwise} \\\\\n F_{i,j-1}\\Big)\n \\end{array} \\right.\n\\end{eqnarray*}\n\n\n\\runinhead{Longest common substring.} Let $F_{i, j}$ be the LCS at string $a[:i]$ and\n$b[:j]$. We have two situations: $a[i]==b[j]$ or not.\n\\begin{eqnarray*}\nF_{i. j} = \\left\\{ \\begin{array}{rl}\n F_{i-1, j-1}+1 &\\mbox{// if $a[i]==b[j]$} \\\\\n 0 &\\mbox{// otherwise}\n \\end{array} \\right.\n\\end{eqnarray*}\n\nBecause it is not necessary that $F_{i,j}\\geq F_{i',j'}, \\forall i,j\\cdot i>i', j>j'$, the $gmax=max\\big(\\{{F_{i,j}\\}\\big)$.\n\\runinhead{Longest increasing subsequence}. Find the longest increasing subsequence of an array $A$.\n\nlet $F_i$ be the LIS length ends at $A_i$. \n\\begin{eqnarray*}\nF_i = \\max(F[j]+1 \\cdot\\forall j < i) \\text{ // if $A_i>A_j$}\n\\end{eqnarray*}\n\nThen the global $maxa$ is:\n$$\nmaxa = \\max(F_i\\cdot \\forall i)\n$$\n\nTime complexity: $O(n^2)$\n\nIn code, notice the \\pyinline{else}.\n\\begin{python}\nF[i] = max(\n F[j] + 1 if A[i] > A[j] else 1\n for j in xrange(i)\n)\n\\end{python}\n\nAlternative solution using binary search in $O(n \\log n)$ - Section \\ref{extremeValueProblem}.\n\n\\runinhead{Maximum subarray sum.} Find the maximum subarray sum of $A$. \n\nLet $F_i$ be the maximum subarray sum ending at $A_{i}$\n$$\nF_i = \\max(F_{i-1}+A_{i}, 0)\n$$\n\nThen the global $maxa$ is:\n$$\nmaxa = \\max(F_i\\cdot \\forall i)\n$$\n\n\n\\runinhead{Maximum sum of non-adjacent cells.} Get the maximum sum of non-adjacent\ncells of an array $A$.\n\nLet $F_i$ be the maximum sum of non-adjacent cells for $A[:i]$. You have tow options:\nchoose $A_{i-1}$ or not.\n\\begin{align*}\nF_{i} = \\max\\big(F_{i-1}, F_{i-2}+A_{i-1}\\big)\n\\end{align*}\n\\runinhead{Edit distance} Find the minimum number of steps required to convert words $A$ to $B$ using inserting, deleting, replacing. \n\nLet $F_{i, j}$ be the minimum number\nof steps required to convert $A[:i]$ to $B[:j]$.\n\\begin{eqnarray*}\nF_{i, j} = \\left\\{ \\begin{array}{rl}\n F_{i-1, j-1} &\\mbox{// if $a[i]==b[j]$} \\\\\n \\min\\Big(F_{i, j-1}+1, &\\mbox{// otherwise, insert}\\\\\n F_{i-1, j}+1, &\\mbox{// delete}\\\\\n F_{i-1, j-1}+1\\Big) &\\mbox{// replace}\\\\\n \\end{array} \\right.\n\\end{eqnarray*}\n\n\\runinhead{Maximal square.} Find the largest rectangle in the matrix:\n\\begin{lstlisting}\n1 0 1 0 0\n1 0 1 1 1\n1 1 1 1 1\n1 0 0 1 0\n\\end{lstlisting}\nLet $F_{i, j}$ represents the max square's length ended at $mat_{i, j}$ (lower right corner).\n\\begin{eqnarray*}\nF_{i, j} = \\left\\{ \\begin{array}{rl}\n \\min\\big(F_{i-1, j-1}, F_{i-1, j}, F_{i, j-1}\\big)+1 &\\mbox{// if $mat_{i, j}==1$} \\\\\n 0 &\\mbox{// otherwise}\n \\end{array} \\right.\n\\end{eqnarray*}\n\n\\subsection{Dual-state dp}\n\\runinhead{Maximal product subarray.} Find the subarray within an array $A$ which has the largest product. \n\\begin{itemize}\n\\item Let $small_i$ be the smallest product end with $A_i$. \n\\item Let $large_i$ be the largest product end with $A_i$.\n\\item The states can be negative. \n\\end{itemize}\n\\begin{eqnarray*}\n&& small_i = \\min\\big( A_i,\\ small_{i-1}\\cdot A_i,\\ large_{i-1}\\cdot A_i \\big)\n\\nonumber \\\\\n&& large_i = \\max\\big( A_i,\\ small_{i-1}\\cdot A_i,\\ large_{i-1}\\cdot A_i \\big)\n\\end{eqnarray*}\n\nIt can be optimized to use space $O(1)$. \n\n\\runinhead{Trapping Rain Water}\nGiven n non-negative integers representing an elevation map where the width of each\nbar is 1, compute how much water it is able to trap after raining.\n\\begin{figure}[]\n \\centerline{\\includegraphics[height = 1in]{rainwatertrap}}\n \\caption{Trapping Rain Water}\n \\label{fig:rainwatertrap}\n\\end{figure}\n\nLet $maxL_i$ be the $\\max(A[:i])$; let $maxR_i$ be the $\\max(A[i:n])$. The dp of obtaining max is trivial. \n\nThe the total volume $vol$:\n$$\nvol = \\sum_i(\\max\\big(0,\\min(maxL_i, maxR_{i+1})-A[i]\\big))\n$$\n\\runinhead{Zigzag subsequence.} Find the max length zigzag subsequence which goes up and down alternately within the array $A$.\n\nLet $U_i$ be the max length of zigzag subsequence end at $A_i \\wedge$ going up.\n\nLet $D_i$ be the max length of zigzag subsequence end at $A_i \\wedge$ going down.\n\\begin{align*}\nU_i &= max(D_j+1 \\cdot \\forall j < i) \\text{ // if $A_i > A_j$} \\\\ \nD_i &= max(U_j+1 \\cdot \\forall j < i) \\text{ // if $A_i < A_j$} \n\\end{align*}\n\nNotice in python implementation, don't use list comprehension since two states are interleaved and interdependent. \n\n\\section{String}\n\\runinhead{Word break.} Given a string $s$ and a dictionary of words $dict$, determine if $s$ can be segmented into a space-separated sequence of $dict$ words.\n\nLet $F_i$ be whether \\pyinline{s[:i]} can be segmented. \n\\begin{eqnarray*}\nF_{i} = \\left\\{ \\begin{array}{rl}\n F_{i-len(w)} &\\mbox{// if $\\exists w\\in dict$, \\pyinline{s[i-len(w):i]==w}}\n\\\\\n false &\\mbox{// otherwise}\n \\end{array} \\right.\n\\end{eqnarray*}\nReturn all such possible sentences. In original case, we use a bool array to record whether a dp could be segmented. Now we should use a vector for every dp to record how to construct that dp from another dp.\n\nLet $F_i$ be all possible segmented words ends at \\pyinline{s[i-1]}. $F_i$ is a list. $\\exists F_i$ means $F_i$ is not empty.\n\\begin{eqnarray*}\nF_{i} = \\left\\{ \\begin{array}{rl}\n F_{i}+[w] &\\mbox{// $\\forall w\\in dict\\cdot$} \\\\\n & \\mbox{ if \\pyinline{s[i-len(w):i]==w} $\\wedge\\ \\exists\nF_{i-len(w)$ } \\\\\n F_{i}} &\\mbox{// otherwise}\n \\end{array} \\right.\n\\end{eqnarray*}\n\nReconstruct the sentence from $F_i$. It is like building path for the tree. Using backtracking: \n\\begin{python}\ndef build(self, dp, i, cur, ret):\n if cur_index == 0:\n ret.append(\" \".join(list(cur)))\n return\n\n # backtracking\n for word in dp[i]:\n cur.appendleft(word)\n self.build(dp, i-len(word), cur, ret)\n cur.popleft()\n\n\\end{python}\n\n\\runinhead{Is palindrome.} Given a string $s$, use an array to determine whether $s[i:j]$.\n\nLet $P_{i,j}$ indicates whether $s[i:j]$ is palindrome. We have one condition - whether the head and the end letter are equal: \n\\begin{eqnarray*}\nP_{i. j} = P_{i-1, j+1}\\ \\wedge\\ s[i] = s[j-1]\n\\end{eqnarray*}\n\nThe code for palindrome dp is error-prone due to indexing. Notice that $i \\in [0, n), j \\in [i, n+1)$.\n\\begin{python}\nn = len(s)\npa = [[False for _ in xrange(n+1)] for _ in xrange(n)]\nfor i in xrange(n):\n pa[i][i] = True\n pa[i][i+1] = True\n\nfor i in xrange(n-2, -1, -1):\n for j in xrange(i+2, n+1):\n pa[i][j] = pa[i+1][j-1] and s[i] == s[j-1]\n\\end{python}\n\\runinhead{Minimum palindrome cut.} Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s.\n\nLet $C_i$ be the min cut for $s[:i]$. We have 1 more cut from previous state to make $S[:i]$ palindrome. \n\\begin{eqnarray*}\nC_{i} = \\left\\{ \\begin{array}{rl}\n \\min\\big(C[k]+1 \\cdot \\forall k 0$.\nThen\n\\begin{align*}\n M_i - m_i \\geq \\abs{f(0+) - f(0)} > \\epsilon\n\\end{align*}\nwhich is a contradiction, so $f(0+) = f(0)$.\n\n(b) Only difference from (a) is that we need $f(0-) = f(0)$ instead of $f(0+) = f(0)$. The only changes that\nneed to be made to the proof of (a) involve replacing $[0, x_i]$ with $[x_{i-1}, 0]$.\n\n(c) If $f$ is continuous at 0, we can consider the partition $P$ that contains $x_{i-1} < x_i = 0 < x_{i+1}$.\nThen $U(P, f, \\beta_2) = \\frac{M_{i} + M_{i+1}}{2}$ and $L(P, f, \\beta_2) = \\frac{m_i + m_{i+1}}{2}$.\nSince $f$ is continuous at 0, $M_i, M_{i+1}, m_i, m_{i+1} \\to f(0)$ as $x_{i-1} \\to 0$ and $x_{i+1} \\to 0$,\nso $f \\in \\mathscr{R}(\\beta_2)$.\nFor the other direction, we proceed similarly to the proof of (a) and see that\n\\begin{align*}\n \\frac{1}{2} (M_i + M_{i + 1}) - \\frac{1}{2} (m_i + m_{i + 1}) &= \\frac{1}{2} ((M_i - m_i) + (M_{i+1} - m_{i+1})) \\\\\n &\\geq \\frac{1}{2} (\\abs{f(0-) - f(0)} + \\abs{f(0+) - f(0)}) \\\\\n &> \\epsilon\n\\end{align*}\nfor some $\\epsilon > 0$ unless $f(0-) = f(0+) = f(0)$, so $f$ must be continuous at 0 if $f \\in \\mathscr{R}(\\beta_2)$.\n\n(d) If $f$ is continuous at 0 then we have $f(0+) = f(0-) = f(0)$, so we are done by parts (a)-(c).\n\n\\subsection{Exercise 4}\nFrom the density of the rationals and irrationals in the reals, we have that $M_i - m_i = 1 \\: \\forall i$, so\n$f \\notin \\mathscr{R}$.\n\n\\subsection{Exercise 5}\nIf we consider $f(x) = 1$ for all rational $x$ and $f(x) = -1$ for all irrational $x$, we have that\n$f^2 \\in \\mathscr{R}$ but $f \\notin \\mathscr{R}$. However, if $f^3 \\in \\mathscr{R}$, then $f \\in \\mathscr{R}$.\nThis is because $m \\leq f^3 \\leq M$ (since $f$ is bounded) and $x^{\\frac{1}{3}}$ is continuous on any\n$[m, M]$, so we can apply Theorem 6.11.\n\n\\subsection{Exercise 6}\n$P$ is compact, so the open cover consisting of neighborhoods around each point of $P$ has a finite subcover.\nThus, $P$ can be covered by finitely many segments. Additionally, the total length of these segments can be\nmade arbitrarily small, since $P$ contains no segments itself. We can then proceed exactly as in the proof of\nTheorem 6.10 to get the desired result.\n\n\\subsection{Exercise 7}\n(a) If $f \\in \\mathscr{R}$, then by Theorem 6.12 (c) we have\n\\begin{align*}\n \\abs{\\int_{c}^{1} f dx - \\int_{0}^{1} f dx} &= \\abs{\\int_{0}^{c} f dx} \\\\\n &\\leq cM < \\epsilon \\quad \\text{for} \\:\\: c < \\frac{\\epsilon}{M}\n\\end{align*}\nWhere $M = \\sup{f}$ on $[0, 1]$.\n\n(b) Consider $f$ such that\n\\begin{align*}\n f(x) =\n \\begin{cases}\n 0 & x = 0 \\\\\n \\frac{1}{x} & 0 < x < \\frac{1}{2} \\\\\n 0 & x = \\frac{1}{2} \\\\\n -\\frac{1}{x} & \\frac{1}{2} < x \\leq 1\n \\end{cases}\n\\end{align*}\nThen $\\int_{0}^{1} f dx = 0$, but $\\int_{0}^{1} \\abs{f} dx$ diverges.\n\n\\subsection{Exercise 8}\nWe see that the partition $P = {1, 2, ..., b + 1}$ of the interval $[1, b + 1]$ corresponds to the upper sum\n$U(P, f) = \\sum_{n = 1}^b f(n)$, and $\\int_{1}^{b + 1} f(x) dx \\leq U(P, f)$. \nThus, if $\\sum_{n = 1}^\\infty f(n)$ converges, then we have that \n$\\int_{1}^{b + 1} f(x) dx$ is bounded and monotonically increasing (since $f(x) \\geq 0$),\nso $\\int_{1}^{\\infty} f(x) dx $ converges. For the other direction, we can take the same partition $P$ \nand consider $L(P, f) = \\sum_{n = 1}^b f(n + 1)$. Since $L(P, f) \\leq \\int_{1}^{b + 1} f(x) dx$, if \n$\\int_{1}^{\\infty} f(x) dx$ converges then so does $\\sum_{n = 2}^\\infty f(n)$, and we are done (as\n$f(1) < \\infty$).\n\n\\subsection{Exercise 9}\nTheorem: Let $f$ and $g$ be differentiable functions on $[0, \\infty)$ such that \n$\\lim_{x \\to \\infty} f(x)g(x) = 0$ and $f', g' \\in \\mathscr{R}$ for every $[0, b]$. Then we have that\n\\begin{align*}\n \\int_{0}^{\\infty} f(x) g'(x) dx = -f(0)g(0) - \\int_{0}^{\\infty} f'(x) g(x) \n\\end{align*}\nif both improper integrals converge.\n\nProof: Take limits on both sides of the integration by parts formula.\n\nLetting $f(x) = \\frac{1}{1 + x}$ and $g'(x) = cos(x)$ in the above formula shows that \n\\begin{align*}\n \\int_{0}^{\\infty} \\frac{cos(x)}{1 + x} dx = \\int_{0}^{\\infty} \\frac{sin(x)}{(1 + x)^2} dx \n\\end{align*}\n\nWe have that $\\abs{cos(n)} + \\abs{cos(n + 1)} \\geq c$ for some\nconstant $c$ since $cos(n)$ and $cos(n + 1)$ cannot both be 0. As such, \n$\\sum_{n = 1}^\\infty \\frac{cos(n)}{1 + n}$ diverges since the harmonic series diverges and therefore\nthe integral on the left also diverges by the integral test.\n\n\\subsection{Exercise 10}\n(a) We can use the convexity of $e^x$ to show this. Let $\\lambda = \\frac{1}{p}$ and let $u^p = e^a, v^q = e^b$\nfor some $a, b$ (this is possible because $u, v \\geq 0$). Then we have\n\\begin{align*}\n e^{\\lambda a + (1- \\lambda) b} &\\leq \\lambda u^p + (1 - \\lambda) v^q \\\\\n e^{\\lambda a} e^{(1 - \\lambda) b} &\\leq \\frac{u^p}{p} + \\frac{v^q}{q} \\\\\n uv &\\leq \\frac{u^p}{p} + \\frac{v^q}{q}\n\\end{align*}\n\n(b) From part (a) we have that \n\\begin{align*}\n f(x) g(x) &\\leq \\frac{f^p(x)}{p} + \\frac{g^q(x)}{q} \\\\\n \\int_{a}^{b} fg d\\alpha &\\leq 1 \n\\end{align*}\n\n(c) We can use part (a) with $u = \\frac{\\abs{f(x)}}{\\bigg(\\int_{a}^{b} f^p d\\alpha\\bigg)^{\\frac{1}{p}}}$ \nand $v = \\frac{\\abs{g(x)}}{\\bigg(\\int_{a}^{b} g^q d\\alpha\\bigg)^{\\frac{1}{q}}}$ to get the desired result.\n\n(d) Assuming the limits exist, the inequality follows for impromper integrals by taking limits on both sides\nand then using limit rules.\n", "meta": {"hexsha": "21fa41d3160b116a000d743cb6011f89adbece34", "size": 7717, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_6.tex", "max_stars_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_stars_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-09-19T07:33:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-19T07:33:25.000Z", "max_issues_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_6.tex", "max_issues_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_issues_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_6.tex", "max_forks_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_forks_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.1214285714, "max_line_length": 132, "alphanum_fraction": 0.5776856291, "num_tokens": 3000, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812553, "lm_q2_score": 0.8688267626522814, "lm_q1q2_score": 0.699954806641055}} {"text": "\\documentclass[oneside]{book}\r\n\\usepackage[utf8]{inputenc}\r\n\\usepackage{authblk}\r\n\\usepackage{setspace} \r\n\\usepackage{amsmath}\r\n\\usepackage{textcomp}\r\n\\usepackage{amssymb}\r\n\\usepackage{geometry}\r\n\\usepackage{amsthm}\r\n\\usepackage{runic}\r\n\\usepackage{mathtools}\r\n\\usepackage{graphicx}\r\n\\usepackage[breaklinks=true,a4paper=true,pagebackref=true]{hyperref}\r\n\\graphicspath{ {figures/} }\r\n\\geometry{\r\n a4paper,\r\n total={170mm,257mm},\r\n left=20mm,\r\n top=20mm,\r\n}\r\n\\hypersetup{\r\n colorlinks=true,\r\n linktoc=true,\r\n linkcolor=blue,\r\n}\r\n\\title{Libraria Algebrae}\r\n\\author{Liam Gardner}\r\n\\date{\\today}\r\n%\\doublespacing\r\n\\newcommand\\tab[1][1cm]{\\hspace*{#1}}\r\n\\newcommand\\nextline{\\newline\\tab}\r\n\\newcommand\\nextquestion{\\newline\\newline}\r\n\\newcommand\\soln{$\\text{sol}^\\text{n}\\text{ }$}\r\n\\newcommand\\fs{\\mbox{\\large $\\mathrlap{f}s\\,$}\\,}\r\n\\newcommand\\thm[2]{\\section*{Theorem: #1}\\label{sec:#2}\\addcontentsline{toc}{section}{Theorem: #1}}\r\n\\newcommand\\propn[2]{\\section*{Proposition: #1}\\label{sec:#2}\\addcontentsline{toc}{section}{Proposition: #1}}\r\n\\newcommand\\defn{\\textbf{Definition}: }\r\n\r\n\\renewcommand\\mod[1]{\\text{ }\\left(\\text{mod }#1\\right)}\r\n\r\n\r\n\\begin{document}\r\n\\DeclarePairedDelimiter\\abs{\\lvert}{\\rvert}\r\n\r\n\\maketitle\r\n\\tableofcontents\r\n\\chapter{Linear Diophantine Equations in $\\mathbb{Z}^2$}\r\n\\tab\r\nNote that for the general equation $ax+by$, we assume that $ab\\neq0$, since if one is 0, then the equation is trivial. We wish to answer three fundamental problems\r\n\\begin{enumerate}\r\n\\item Does there exist an integer solution?\r\n\\item If the answer is yes, find an integer solution.\r\n\\item Can we find \\textit{all} solutions?\r\n\\end{enumerate}\r\n\\tab\r\nIt is common for existential theorems (those that say solutions exist) to not give a means of how to find said solutions.\r\n\\subsection{Example}\r\nSolve $506x + 391y = 23$. Notice that $\\gcd(506,391)=23$. Thus, by b\\'ezout's\\ lemma, a solution exists. We can use the EEA (Extended Euclidean Algorithm) to find a solution to the equation. \r\n\\newline\r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|c|}\r\n\\hline\r\n$x$ & $y$ & $r$ & $q$ \\\\\r\n\\hline\r\n\\hline\r\n1 & 0 & 506 & 0 \\\\\r\n0 & 1 & 391 & 0 \\\\\r\n1 & -1 & 115 & 1.0 \\\\\r\n-3 & 4 & 46 & 3.0 \\\\\r\n7 & -9 & 23 & 2.0 \\\\\r\n-17 & 22 & 0 & 2.0 \\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nThus, we know that $(7,-9)$ is a solution. Now, we can subtract the equation $506x+391y=23$ with $506x_0 + 391y_0=23$, which gives $506(x-x_0) + 391(y-y_0) = 0$. Thus, we get $506(x-x_0) = -391(y-y_0)$. We can divide by the GCD of 506 and 391 to get the equation $22(x-x_0) = -17(y-y_0)$. Since the GCD is a common divisor to 506 and 391, we get that both $\\frac{506}{23}$ and $\\frac{391}{23}$ are integers and are coprime to each other. Now, since we know that $-17\\lvert -17(y-y_0)$, and that $-17(y-y_0)=22(x-x_0)$, we get that $-17\\lvert 22(x-x_0)$. Thus, by CAD, we get $17\\lvert (x-x_0)$. Therefore, we find that $x-x_0 = 17n$ for some $n\\in\\mathbb{Z}$. Thus, a solution for $x$ can be found with $x_0 + 17n$, $\\forall n\\in\\mathbb{Z}$.\r\n\\nextline\r\nFollowing the same process, we get that $y=y_0+22n$. Therefore, all solutions to the Linear Diophantine Equation $506x + 391y = 23$ are given by the points $(7 + 17n, -9 - 22n)$.\r\n\\newline\r\n\\newline\r\nSolve $506x + 391y = 24$.\r\n\\nextline\r\nThere are no \\soln: Since $23=\\gcd(506,391)$, we get that $23\\lvert 506x+391y$ however, $23\\nmid 24$, therefore, we've run into a contradiction.\r\n\\newline\r\n\\newline\r\nSolve $506x + 391y = 46 = 2\\cdot23$. We know that $506\\cdot7 + 391\\cdot(-9) = 23$, thus if we multiply both sides by two, we get that $506(7\\cdot2) + 391(-9\\cdot2) = 2\\cdot23 = 46$. $506(14) + 391(-18) = 46$ gives the \\soln (14,-18).\r\n\r\n\\thm{LDET Part 1}{ldeta}\r\n\r\nSuppose $a,b,c\\in\\mathbb{Z}$ and $ab\\neq0$ then $ax+by=c$ has a \\soln in integers if and only if $\\gcd(a,b)\\lvert c$.\r\n\\subsection{Proof of forwards direction}\r\n\\tab\r\nAssume $ax+by=c$ has an integer \\soln $(x_0, y_0)$. Let $d=\\gcd(a,b)$. Since $d\\lvert a$ and $d\\lvert b$, and since $c=ax_0+by_0$, then by Divsibility of Integer Combinations, $d\\lvert c$.\r\n\\subsection{Proof of backwards direction}\r\n\\tab\r\nLet $d=\\gcd(a,b)$, and assume that $d\\lvert c$, thus $c=kd$ \\fs integer $k$. Then By B\\'ezout's\\ Lemma, we can find $x_0$ and $y_0$ $\\in\\mathbb{Z}$ such that $ax_0+by_0=d$, but then $k(ax_0+by_0)=kd$. Thus $a(kx_0) + b(ky_0) = c$\r\n\\subsection{Remark}\r\n\\tab\r\nif $d\\lvert c$ then the proof tells us how to find a \\soln.\r\n\\begin{enumerate}\r\n\\item solve $ax+by=d$ using EEA to get $(x,y)=(x_0,y_0)$.\r\n\\item take $x=kx_0$ and $y=ky_0$, where $k=\\frac{c}{d}$.\r\n\\end{enumerate}\r\n\r\n\\thm{LDET Part 2}{ldetb}\r\n\r\nSuppose that $(x_0, y_0)$ is a particular solution to the LDE $ax+by=c$.\r\n\\nextline\r\nThen the set of all \\soln $\\in\\mathbb{Z}$ is given by the following set:\r\n$$S = \\left\\{\r\n(x,y) \\Huge\\mid x=x_0 + \\frac{b\\cdot n}{\\gcd(a,b)},\\,\\,\\, y=y_0 - \\frac{a\\cdot n}{\\gcd(a,b)}\r\n\\right\\}$$\r\n\\subsection{Proof}\r\n\\tab\r\nLet $D$ be the set of all integer solutions to $ax+by=c$. I.E.\r\n$$D = \\left\\{(x,y) \\mid x,y,\\in\\mathbb{Z}, ax+by=c \\right\\}$$\r\n\\tab\r\nThis can be proven by showing $S \\subseteq D$ and $D \\subseteq S$\r\n\\newline $S\\subseteq D:$\r\n\\nextline\r\nLet $(x,y) \\in S$, thus $x=x_0 + \\frac{bn}{d}$ and $y=y_0-\\frac{an}{d}$.\r\n$$ax+by = a\\left(x_0+\\frac{bn}{d}\\right) + b\\left(y_0 - \\frac{an}{d}\\right)$$\r\n$$= ax_0 + by_0 = c$$\r\n\\tab\r\nsince by definition, we know that $x_0$ and $y_0$ are a particular solution to the equation. Therefore, $S\\subseteq D$\r\n\\newline $D\\subseteq S:$\r\n\\nextline\r\nLet $(x,y)\\in D$, thus $x,y\\in\\mathbb{Z}$ and $ax+by=c$. Since, $(x_0,y_0)\\in D$, we know that $ax_0 + by_0 = c$. We can subtract $ax_0+by_0=c$ from $ax + by = c$ to get $a(x-x_0) + b(y-y_0) = 0$. Dividing by the GCD of $a$ and $b$, we get that $\\frac{a(x-x_0)}{d} = \\frac{-b(y-y_0)}{d}$. Thus, since $\\frac{b}{d}\\lvert \\frac{a}{d}(x-x_0)$ then by CAD, we get that $\\frac{b}{d}\\lvert (x-x_0)$, since $\\frac{b}{d}$ and $\\frac{a}{d}$ are coprime. Then we get $x-x_0=n\\frac{b}{d}\\Rightarrow x=x_0+\\frac{bn}{d}$ \\fs $n\\in\\mathbb{Z}$. Similarly, we get $y=y_0-\\frac{an}{d}$. Therfore, $(x,y)\\in S$.\r\n\\section{More Examples}\r\n$12x+18y=13$.\r\n\\nextline\r\n$\\gcd(12,18) = 6$. Since $6\\nmid13$ the equation has no \\soln by \\hyperref[sec:ldeta]{LDET1}\r\n\\newline\r\n\\newline\r\n$14x-49y=28$\r\n\\nextline\r\n$\\gcd(14,-49) = 7$. Since $7\\lvert 28$ the equation has solutions by \\hyperref[sec:ldetb]{LDET2}\r\n\\nextline\r\nConsider $14x-49y=7\\Rightarrow 2x-7y=1$ which has \\soln $(4,1)$. If we multiply $14(4) - 49(1) = 7$ by 4, we get that $14(x_0\\cdot4) - 49(y_0\\cdot4) = 7$, and from this we can get all \\soln using LDET2.\r\n\\newline\r\n\\newline\r\n\\tab\r\nFind all \\soln to $15x+35=5$.\r\n\\newline\r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|c|}\r\n\\hline\r\n$x$ & $y$ & $r$ & $q$ \\\\\r\n\\hline\r\n\\hline\r\n1 & 0 & 15 & 0 \\\\\r\n0 & 1 & 35 & 0 \\\\\r\n1 & 0 & 15 & 0.0 \\\\\r\n-2 & 1 & 5 & 2.0 \\\\\r\n7 & -3 & 0 & 3.0 \\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nThus, we know that $\\gcd(15,35)=5$, and $5\\lvert 5$, there is a \\soln. By EEA, we find $x=-2, y=1$ is a \\soln. Then, we can find the general solution using\r\n\\hyperref[sec:ldetb]{LDET2}\r\n to be $x=-2+\\frac{35n}{5}\\Rightarrow 7n-2$, $y=1+\\frac{15n}{5}\\Rightarrow 1+3n$\r\n\\subsection{Geometric Understanding}\r\n\\tab\r\nGraphing the line $15x+35y=5$ or $3x+7y=1$, we can rearrange for $y$ to get $y=\\frac{-3}{7}x+\\frac{1}{7}$. Thus, picking any lattice point, we can construct a triangle of length 7 and height 3 from that point to find the next lattice point.\r\n\r\n\\begin{figure}[h]\r\n\\centering\r\n\\includegraphics[width=0.5\\textwidth]{l24f0}\r\n\\caption{Triangle formed from moving between two lattice points}\r\n\\end{figure}\r\n\r\n\\subsection{Nonnegative \\soln}\r\n\\tab\r\nThe solutions to $15x+35y=5$ is given by $(-2+7n, 1-3n)$, these will be nonnegative if $x\\geq0$ and $y\\geq0$.\r\n$$-2+7n\\geq 0 \\iff n\\geq \\frac{2}{7}$$\r\n$$1-3n \\geq 0 \\iff n \\leq \\frac{1}{3}$$\r\nThus, since there are no integer solutions in the range $\\frac{2}{7}\\leq n \\leq \\frac{1}{3}$, as $n\\in\\mathbb{Z}$ there are no nonnegative solutions.\r\n\\subsection{Find all integer \\soln to $15x+35y^2 = 5$}\r\n\\tab\r\nLet $Y=y^2$, then, since we have the \\soln to $15x+35Y=5$, given by $(-2+7n, 1-3n)$, all we have to do is see when $1-3n$ is a perfect square.\r\nOne way to solve this is to say let $z=y^2$ and solve $1-3n=z$. We can also solve this algebraically\r\n\\newline\r\n\\begin{center}\r\n\\begin{tabular}{c c c}\r\n$\\iff$ & $1-3n = y^2$ & \\\\\r\n$\\iff$ & $-3n=y^2-1$ & \\\\\r\n$\\iff$ & $3\\lvert y^2-1$ & by euclid's lemma \\\\\r\n$\\iff$ & $3\\lvert(y-1)(y+1)$ & \\\\\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nBy Euclid's Lemma, we know either $3\\lvert(y-1)$ or $3\\lvert(y+1)$, though not both. Suppose $3\\lvert(y-1) \\iff y-1=3k\\iff y=3k+1\\, \\fs k\\in\\mathbb{Z}$\r\nThen, if $y=3k+1$, we get that $y^2=(1+3k)^2=1-3(-2k-3k^2)$. Let $n=(-2k-3k^2)$, then we get $y^2=1-3n$. Thus, if we take $x=-2+7n=-2+7(-2-3k^2)$ and $y=1-3k$, we get a perfect square \\soln to the diophantine equation. \r\n\\nextline\r\nIf $3\\lvert y+1$, then we have $y=-1+3l\\, \\fs l \\in \\mathbb{Z}$. Then $y^2=(-1+3l)^2=1-3(2l+3l^2)$. Thus, $y=1-3l$ and $x=-2+7(2l+3l^2)$. Therfore, we can generate infinitely many perfect square solutions.\r\n\\chapter{Congruence and Modular Arithmetic}\r\n\\section{Clockwork Arithmetic Analogy}\r\n\\tab\r\nImagine a clock, we know that a clock has 12 spokes. If we look at it and see the hour hand at 2, and we know that 12 has already passed, then we also know that the clock really means it's 14. Thus, we can say $2\\approx14$\r\n\\section{Definition}\r\n\\tab\r\n$\\forall a,b\\in\\mathbb{Z}$, we say that ``$a$ is congruent to $b$ mod(ulo) 12'' if $m\\lvert(a-b)$.\r\n\\nextline\r\nNotation: $a\\equiv b\\mod{m}$\r\n\\section{Examples}\r\n\\tab\r\n$m=1$ then $a\\equiv b\\mod{1} \\iff 1\\lvert (a-b)$ which is true $\\forall a,b\\in\\mathbb{Z}$\r\n\\nextline\r\n$m=2$ then $a\\equiv b\\mod{2} \\iff 2\\lvert (a-b) \\iff a-b$ is even $\\iff a$ and $b$ are both even or both odd.\r\n\\nextline\r\n$2\\equiv-116\\mod{2}$, however $3\\not\\equiv 10024\\mod{2}$\r\n\\nextline\r\n$14\\equiv 2\\mod{12} \\iff 12\\lvert(14-2)$\r\n\\nextline\r\n$6\\equiv 26\\mod{10} \\iff 10\\lvert(6-26)$\r\n\\nextline\r\n$6\\not\\equiv -26\\mod{10} \\implies 10\\nmid(6+26)$\r\n\\propn{Congruence is an Equivalence Relation}{CER}\r\n\\begin{center}\r\n\\begin{tabular}{l|l}\r\n$\\forall m\\in\\mathbb{N}, \\forall a,b,c,\\in\\mathbb{Z}$ & Congruence is \\\\\r\n\\hline\r\n\\hline\r\n$a\\equiv a\\mod{m}$ & symmetric \\\\\r\nif $a\\equiv b\\mod{m}$ and $b\\equiv c\\mod{m}$ then $a\\equiv c\\mod{m}$ & transitive \\\\\r\n$a\\equiv b \\mod{m}\\implies b\\equiv a\\mod{m}$ & Reflexive\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nRemark: Any relation $a\\sim b$ that satisfies all above properties is called an equivalence relation.\r\n\\nextline\r\nIn calculus, we say two functions $f(x)\\sim g(x)$ are an equivalence relation if $f^\\prime(x) = g^\\prime(x)$\r\n\\section{Recap}\r\n\\tab\r\nFix $m\\in\\mathbb{N}$, $\\forall a,b\\in\\mathbb{Z}$\r\n\\nextline\r\n\\begin{tabular}{l c l}\r\n$a\\equiv b\\mod{m}$ & $\\iff$ & $m\\lvert (a-b)$ \\\\\r\n& $\\iff$ & $a-b=mk\\,\\, \\fs k\\in\\mathbb{Z}$ \\\\\r\n& $\\iff$ & $a=b+mk$\r\n\\end{tabular}\r\n\\newline\r\n\\propn{Arithmetic Rules of Congruence}{ARC}\r\nSuppose $a\\equiv a^\\prime \\mod{m}$ and $b\\equiv b^\\prime \\mod{m}$ then,\r\n\\begin{enumerate}\r\n\\item $a+b\\equiv a^\\prime + b^\\prime \\mod{m}$\r\n\\item $a-b\\equiv a^\\prime - b^\\prime \\mod{m}$\r\n\\item $ab\\equiv a^\\prime b^\\prime \\mod{m}$\r\n\\end{enumerate}\r\n\\subsection{Examples}\r\n$$2\\equiv 9\\mod{7}\\land3\\equiv 17\\mod{7}\\implies 2+3\\equiv9+17\\mod{7}\\implies 5=26$$\r\n\\newline\r\n$$56\\cdot30 \\mod{40}$$\r\n$$56=16+40\\equiv 16\\mod{40}$$\r\n$$30=-10\\equiv \\mod{40}$$\r\n$$56\\cdot30 \\equiv 16\\cdot(-10)\\mod{40}$$\r\n$$\\equiv-160\\mod{40}$$\r\n$$\\equiv-40\\cdot4\\mod{40}\\equiv 0\\mod{40}$$\r\n\\subsection{Proof of addition}\r\n\\tab\r\nSince $a\\equiv a^\\prime \\mod{m}$ and $b\\equiv b^\\prime \\mod{m}$ then $m\\lvert(a-a^\\prime)$ and $m\\lvert(b-b^\\prime)$\r\n\\nextline\r\nWe have $(a+b)-(a^\\prime-b^\\prime)=a-a^\\prime + b-b^\\prime$, thus by DIC we get $m\\lvert((a+b)-(a^\\prime+b^\\prime))$. Therefore $a+b\\equiv a^\\prime + b^\\prime \\mod{m}$\r\n\\subsection{Remark on Division}\r\n\\tab Care is needed with division\r\n$ab\\equiv ac\\mod{m} \\not\\implies b\\equiv c\\mod{m}$ even if $a\\not\\equiv 0 \\mod{m}$\r\n\\nextline\r\n$$10\\equiv 4\\mod{6}$$\r\n$$2\\cdot5 \\equiv 2\\cdot2 \\mod{6}$$\r\n$$5\\not\\equiv 2\\mod{6}$$\r\n\\propn{Congruent Division}{CD}\r\n\\tab\r\nIf $ab\\equiv ac\\mod{m}$ and $a$ is coprime to $m$, then $b\\equiv c\\mod{m}$\r\n\\propn{Congruent Powers}{CP}\r\n\\tab\r\n$a\\equiv b\\mod{m}\\implies a^n\\equiv b^n\\mod{m}$\r\n\\subsection{Proof}\r\n\\tab\r\n$$ab\\equiv bc\\mod{m}\\iff m\\lvert (ab-ac)\\iff m\\lvert a(b-c)$$\r\n\\nextline\r\nThen by CAD we get $m\\lvert(b-c)$ since $m$ and $a$ are coprime\\nextline\r\n$\\implies b\\equiv c \\mod{m}$\r\n\\nextline\r\nBy applying the above proposition repeatedly; if $a_1\\equiv a^\\prime_1 \\mod{m}, a_2\\equiv a^\\prime_2 \\mod{m}, \\cdots a_n\\equiv a^\\prime_n \\mod{m}$, then we get the following result\r\n\\begin{enumerate}\r\n\\item $a_1+\\cdots+a_n\\equiv a^\\prime_1 + \\cdots a^\\prime_n \\mod{m}$\r\n\\item $a_1-\\cdots-a_n\\equiv a^\\prime_1 - \\cdots a^\\prime_n \\mod{m}$\r\n\\item $a_1\\cdots a_n\\equiv a^\\prime_1 \\cdots a^\\prime_n \\mod{m}$\r\n\\item (special case) $\\forall q \\in \\mathbb{N}, a^q \\equiv \\left(a^\\prime\\right)^q \\mod{m}$\r\n\\end{enumerate}\r\n\\subsection{More Examples}\r\nSimplify $4^10 \\mod{18}$\\nextline\r\n$4^{10} = \\left(4^2\\right)^5 = 16^5=(18-2)^5$ \\nextline\r\n$(18-2)^5\\equiv-2^5\\mod{18}$ \\nextline\r\n$\\equiv-32\\mod{18}$ \\nextline\r\n$-32+2\\cdot 18\\mod{18}$ \\nextline\r\n$4\\mod{18}$\r\n\\newline\r\n\\newline\r\nIs $3^9 + 62^{2020} - 20$ divisible by 7?\r\n\\nextline\r\nLet $n=3^9+62^{2020}-20$. We know that $7\\lvert n \\iff 7\\lvert(n-0) \\iff n\\equiv0\\mod{7}$\r\n\\nextline\r\nWe can compute $3^9=\\left(3^3\\right)^3=27^3=(28-1)^3$ \\nextline\r\n$\\equiv (-1)^3\\mod{7}$\\nextline\r\n$\\equiv -1\\mod{7}$\\nextline\r\nWe also know that $62^{2020} = (63-1)^{2020} \\equiv (-1)^{2020}\\mod{7} \\equiv 1$ \\nextline\r\n$20=21-1\\equiv -1\\mod{7}$ \\nextline\r\nUsing \\hyperref[sec:ARC]{The arithmetic rules} we get that $n\\equiv -1+1-(-1)\\mod{7}\\equiv1\\mod{7}$ and thus $n$ is not divisible by 7.\r\n\\thm{Congruence and Remainders}{CR}\r\n\\subsection{Example}\r\n\\tab\r\nWhat day of the week is it going to be a year from now?\r\n\\nextline\r\nSince days cycle every 7, let's determine $365\\mod{7}$. We know that $350=50\\cdot7$ and thus \\nextline $365 = 350 + 15$ \\nextline\r\n$=7\\cdot50+14+1$ \\nextline\r\n$=7\\cdot50+7\\cdot2+1$\\nextline\r\n$=7(50+2)+1$\\nextline\r\n$\\equiv 1\\mod{7}$.\\nextline\r\n$365\\equiv1\\mod{7}$\\nextline\r\nTherefore, the day of the week one year from now is the same as the day of the week tomorrow.\r\n\\subsection{Observation}\r\n\\tab\r\nif $n\\in\\mathbb{Z}$ \\nextline\r\nAny block of consecutive numbers will cycle through the numbers 0-6 inclusive:\r\n$$\\{\\cdots,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,\\cdots\\}$$\r\n$$\\equiv\\{\\cdots,0,1,2,3,4,5,6,0,1,2,3,4,5,6,0,1,2,3,4,5,6,\\cdots\\}\\mod{7}$$\r\n\\section{Warning}\r\n\\tab\r\n$\\forall a,b,b^\\prime \\in \\mathbb{N}$, if $b\\equiv b^\\prime \\mod{m}$ then in general $a^b\\not\\equiv a^{b^\\prime}\\mod{m}$\r\n\\subsection{Example}\r\n\\tab\r\n$4\\equiv1\\mod{3}$\\nextline\r\n$2^4=16\\equiv1\\mod{3}$ however $2^1\\equiv 2\\mod{3}$ \\nextline\r\nThus $4\\equiv1\\mod{3}$ however $2^4\\not\\equiv2^1\\mod{3}$\r\n\\propn{Finite Integers}{FI}\r\n$\\forall a,b,\\in\\mathbb{Z} a\\equiv b \\mod{m}\\iff a$ and $b$ have the same remainder after division by $m$\r\n\\subsection{Proof}\r\n\\tab\r\nApplying the division algorithm, we get $a=qm+b$ and $b=q^\\prime m+r^\\prime$ where $0\\leq r,r^\\prime < m$.\\nextline\r\nNotice if $a\\equiv b\\mod{m}$, we get that $m\\lvert(a-b)$ and thus $m\\lvert(qm+r - q^\\prime m - r^\\prime)$\\nextline\r\n$\\implies m\\lvert(m(q+q^\\prime) + (r-r^\\prime))$ Then by DIC it follows that \\nextline\r\n$\\iff m\\lvert (r-r^\\prime)$ \\nextline\r\nNow since $0\\leq r < m$ and $0 \\leq r^\\prime < m$, we get that $-m\\leq r-r^\\prime < m$. Now, since $m\\lvert(r-r^\\prime)$, we get that by BBD, that $m\\leq \\abs{r-r^\\prime}$ and so for both inequalities to hold, $r=r^\\prime$.\r\n\\propn{Congruent if and only if same remainder}{CISR}\r\n\\tab\r\n$a\\equiv b\\mod{m} \\iff a$ and $b$ have the same remainder after division by $m$.\r\n\\propn{Congruent to Remainder}{CTR}\r\n$\\forall a,b\\in\\mathbb{Z}\\,, 0\\leq b \\leq m-1 : a\\equiv b\\mod{m}\\iff$ the remainder of $a$ after division by $m$ is $b$.\r\n\\nextline\r\nConsequently, every $a\\in\\mathbb{Z}$ is congruent to a unique integer in $[0,m-1]\\subseteq\\mathbb{Z}$ mod $m$.\r\n\\subsection{Examples}\r\n\\begin{tabular}{r l l}\r\n25 & $\\equiv 32$ & $\\mod{7}$ \\\\\r\n& $\\equiv 18$ & $\\mod{7}$ \\\\\r\n& $\\equiv 11$ & $\\mod{7}$ \\\\\r\n& $\\equiv \\mathbf{4}$ & $\\mod{7}$ \\\\\r\n& $\\equiv -3$ & $\\mod{7}$ \\\\\r\n& $\\equiv -10$ & $\\mod{7}$ \\\\\r\n& $\\cdots$ & $\\mod{7}$\r\n\\end{tabular}\r\n\\nextline\r\n4 is distinguished, as it is the remainder of 25 after division by 7.\r\n\\newline\r\n\\nextline\r\n\\textit{Find the remainder of $5^{10}$ after division by 7}\r\n\\nextline\r\nWe want to compute $5^{10}\\mod{7}$. Since $5^{10} = \\left(5^2\\right)^5 = 25^5$. We know $25\\equiv 4\\mod{7}$ and by\r\n\\hyperref[sec:CP]{Congruent Powers}\r\nwe get $\\equiv 4^5\\mod{7}$\r\n\\nextline\r\n$\\equiv 4^3\\cdot 4^2\\mod{7}$ \\nextline\r\n$\\equiv (63+1) \\cdot (14+2) \\mod{7}$ \\nextline\r\n$\\equiv 1\\cdot 2 \\mod{7}$ \\nextline\r\n$\\equiv 2\\mod{7}$ \\nextline\r\nTherefore, the remainder of $5^{10}$ after division by 7 is 2.\r\n\\newline\r\n\\nextline\r\n\\textit{Find the remainder of} $77^{100}\\cdot 999 - 6^{83} \\mod{4}$\r\n\\nextline\r\nWe know that $77=80-3\\equiv -3\\mod{4}\\equiv 1\\mod{4}$ by \\hyperref[sec:CP]{Congruent Powers}.\r\n$999=1000-1\\equiv -1\\mod{4}\\equiv 3\\mod{4}$. Now, notice that $6^{83}=6^2\\cdot6^{81}$ and since $6^2\\equiv 0\\mod{4}$ we get that $0\\cdot6^{81}\\equiv 0\\mod{4}$ by \\hyperref[sec:CP]{Congruent Powers}\r\n\\nextline\r\n$77^{100}\\cdot999 - 6^{83} \\equiv 1^{100}\\cdot3 - 0\\mod{4} \\equiv 3\\mod{4}$\r\n\\hyperref[sec:ARC]{Congruence and Multiplication}\r\n\\subsection{Sum of Factorial Example}\r\n\\tab\r\n\\textit{What is the last decimal of the following expression?}\r\n$$\\sum_{n=1}^{100} n!$$\r\n\\nextline\r\nNotice that the last digit of a number is the remainder mod 10. As a smaller example, notice that $7!\\equiv 0\\mod{10}$ because $7!$ contains a factor $2\\cdot5=10$ and thus is a multiple of 10. Therefore, we know that if $k\\geq 5$ we get that $k!\\equiv 0\\mod{10}$\r\n\\nextline\r\nGoing back to our original problem, we can notice that $\\forall n\\geq 5$ the sum of $n!\\mod{10}$ will be zero, and thus we only have to compute $1!+2!+3!+4!=1+2+6+24=33\\equiv 3\\mod{10}$\r\n\\section{Divisibility Rules}\r\n\\subsection{Divisibility by 3}\r\n\\tab\r\n$\\forall a \\in \\mathbb{Z} 3\\lvert a \\iff 3\\lvert$ the digit sum of $a$.\r\n\\nextline\r\n$3\\lvert 2046 \\iff 3\\lvert(2+0+4+6)$. $2+4+6=12$ and since $3\\lvert12$ we know $3\\lvert 2046$. \\nextline\r\n$3\\nmid271 \\iff 3\\nmid(2+7+1)$ $2+7+1=10$ and since we know $3\\nmid 10$ we know $3\\nmid 271$.\r\n\\subsection{Proof of divisibility by 3}\r\n\\tab\r\nIf the digits of $a$ are $d_k, d_{k-1}, d_{k-2}, \\cdots, d_1+d_0$ then $a=10^kd_k + 10^{k-1}d_{k-1}+10^{k-2}d_{k-2}, \\cdots, 10d_1, d_0$. This is called the decimal expansion of $a$. Since $10\\equiv 1\\mod{3}$ we get that \\newline$a\\equiv d_k + d_{k-1} + d_{k-2} + \\cdots + d_1 + d_0\\mod{3}$\r\n\\subsection{Divisibility by 11}\r\n\\tab\r\n$11\\lvert a \\iff 11\\lvert$ the alternatign sum of the digits of $a$\\nextline\r\n$11\\lvert108097\\iff 11\\lvert(1+8+9) - (0+0+7)$ and since $1+8+9-7=11$ and $11\\lvert 11$ we get that $11\\lvert 108097$\\nextline\r\n$11\\nmid133 \\iff 11\\nmid(1+3-3)$, thus $11\\nmid1\\implies 11\\nmid133$\r\n\\subsection{Proof of divisibility by 11}\r\n\\tab take\r\n$a=10^kd_k + 10^{k-1}d_{k-1}+10^{k-2}d_{k-2}, \\cdots, 10d_1+ d_0$ to be the decimal representation of $a$. Since $10\\equiv-1\\mod{11}\\equiv 10\\mod{11}$ thus\r\n$a\\equiv (-1)^kd_k + (-1)^{k-1}d_k-1 + \\cdots + (-1)^1d_1 + 1d_0$. Now, notice that this is the sum of the digits of $a$ indexed by even values of $k$ subtracted by the odd-indexed digits of $a$ mod 11.\r\n\\section{Linear Congruence Relations}\r\n\\subsection{Problem: does $x^3+x^2-x+1=0$ have an integer solution?}\r\n\\tab\r\nSuppse $x=a\\in\\mathbb{Z}$ is a \\soln. Thus, $a^3+a^2-a+1=0$. Thus, since both sides are integers, we know that $\\forall m \\in \\mathbb{N}$, $a^3+a^2-a+1\\equiv 0 \\mod{m}$.\r\n\\nextline\r\nConsider the equation in modulo 3. Notice now that $a$ can be either 0, 1, or 2 (mod 3). If $a\\equiv0\\mod{3}$ we get that $1\\equiv0\\mod{3}$ which is false. If $a\\equiv1\\mod{3}$ we get that $1+1=2\\equiv0\\mod{3}$ which is still false. If $a\\equiv2\\mod{3}$ then $a\\equiv-1\\mod{3}$ and thus we get $-1+1-(-1)+1=2\\equiv0\\mod{3}$ which is still false. Therefore, since there are no integer solutions moldulo 3, there are no integer solutions to the equation $a^3+a^2-a+1=0$.\r\n\\chapter{Linear Congruence}\r\n\\tab $y^2=4x+2$ can be solved over the integers by solving the relation mod m. If there are no solutions for a particular $m$, then there are no solutions over the integers. $y^2\\equiv4x+2\\mod{m}$. There's a finite process to check $y^2 \\equiv 4x+2\\mod{m}$ compared to the infinite process to check $y^2=4x+2$.\r\n\\nextline\r\n\\defn A Linear Congruence Equation is an equation of the form $ax\\equiv c\\mod{m}$ where $a,c\\in\\mathbb{Z}$ and $m\\in\\mathbb{N}$ are fixed and $a\\not\\equiv 0\\mod{m}$. We wish to find \\soln for $x$ over the integers.\r\n\\section{Methods of Solving}\r\n\\subsection{Brute Force}\r\n$5x\\equiv 2\\mod{3}$ \r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|c|}\r\n$x\\mod{3}$ & 0 & 1 & 2 \\\\\r\n\\hline\r\n$5x$ & 0 & 5 & 10 \\\\\r\n\\hline\r\n$5x\\mod{3}$ & 0 & 2 & 1 \\\\\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nWe see that the only solution to this is when $x\\equiv 1\\mod{3}$.\r\n\\subsection{LDE}\r\n$5x\\equiv 2\\mod{3} \\iff 5x=2+3k$ $\\fs k\\in\\mathbb{Z}$. \\nextline\r\n$\\iff 5x-3k=2$ \\nextline\r\n$\\iff 5x+3y=2$ for $y=-k$ \\nextline\r\nLet $d=\\gcd(5,3)$, then notice that $d\\lvert 2$ thus by \\hyperref[sec:ldeta]{LDET1} there are infinite solutions.\r\nSolve for $x,y$.\r\n\\nextline\r\nBy inspection we see that a particular \\soln is $(x_0,y_0) = (1,-1)$. Using \\hyperref[sec:ldetb]{LDET2} we can get the general solutions given by\r\n$$\\begin{cases}\r\nx=1 + 3n \\\\\r\ny=-1 - 5n\r\n\\end{cases}$$\r\n\\section{Examples}\r\n$2x\\equiv3\\mod{4}$\r\n\\nextline\r\nBy Method 1:\r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|c|c|}\r\n$x\\mod{4}$ & 0 & 1 & 2 & 3 \\\\\r\n\\hline\r\n$2x$ & 0 & 2 & 4 & 6 \\\\\r\n\\hline\r\n$2x\\mod{4}$ & 0 & 2 & 0 & 2 \\\\\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nSince there is no value of $3$ in the table, there is no \\soln to $2x\\equiv 3\\mod{4}$\r\n\\nextline\r\nBy Method 2:\r\n\\nextline\r\n$2x\\equiv3\\mod{4} \\iff 2x+4y=3$. Since $\\gcd(2,4)\\nmid3$, by \\hyperref[sec:ldeta]{LDET1} there is no \\soln to the equation.\r\n\\thm{Linear Congruence Theorem}{LC}\r\n\\tab\r\nConsider the Linear Congruence Equation $ax\\equiv c\\mod{m}$ where $a,c\\in\\mathbb{Z}$ and $m\\in\\mathbb{N}$ are fixed and $a\\not\\equiv 0\\mod{m}$. Let $d=\\gcd(a,m)$. Then, solutions exist if and only if $d\\lvert c$ by \\hyperref[sec:ldeta]{LDET1}. If $d\\lvert c$ and if $x_0$ is a \\soln then the general solution set is given by the following (using \\hyperref[sec:ldetb]{LDET2})\r\n$$\\left\\{\r\nx\\in\\mathbb{Z}\\mid x=x_0+\\frac{m}{d}n\\, \\fs n\\in\\mathbb{Z}\r\n\\right\\}$$\r\n$$\\left\\{\r\nx\\in\\mathbb{Z}\\mid x\\equiv x_0\\mod{\\frac{m}{d}}\r\n\\right\\}$$\r\n\\tab\r\nNotice for the second set, there are $d$ \\soln mod $m$\r\n\\subsection{Proof}\r\n$ax\\equiv c\\mod{m}\\iff ax+my=c$ so \\soln exist $\\iff d\\lvert c$ by \\hyperref[sec:ldeta]{LDET1}.\\nextline\r\nIf $x_0$ is a \\soln for x, then by \\hyperref[sec:ldetb]{LDET2} the general \\soln set is\r\n$$\\left\\{\r\nx\\in\\mathbb{Z} \\mid x=x_0+\\frac{m}{d}n\\,\\fs n\\in\\mathbb{Z}\r\n\\right\\}$$\r\n\\tab\r\nSince $x=x_0+\\frac{m}{d}n \\iff x\\equiv x_0\\mod{\\frac{m}{d}}$\r\n\\nextline\r\nFinally, if $x=x_0+\\frac{m}{d}n$ then by applying the Division Algorithm, we get $n=qd+r$, $0\\leq r < d$. Thus $x=x_0+\\frac{m}{d}n \\iff x=x_0+(qd+r)\\frac{m}{d} = x_0 + mq + r\\frac{m}{d}$\\nextline\r\n$\\iff x\\equiv x_0+r\\frac{m}{d}\\mod{m}$, $0\\leq r \\leq d-1$\r\n\\newline\r\n\\null\\hfill$\\mathcal{QED}$\r\n\\subsection{More Examples}\r\nSolve $12x\\equiv 9\\mod{15}$\\nextline\r\n\\soln: Step1: (gcd check)\r\n$d=\\gcd(12,15)=3$ and $3\\lvert9$ thus solutions exist\\newline\r\nStep2 (Particular \\soln)\\nextline\r\n$12x+15y=9$. By EEA we get $(x_0, y_0) = (-3,3)$.\\nextline\r\n$x\\equiv x_0\\mod{\\frac{m}{d}}$.\\nextline\r\n$\\equiv -3\\mod{5}$\\nextline\r\n$\\equiv 2\\mod{5}$\\nextline\r\n$x=2+5k$\\nextline\r\nSince the only unique solutions are in the integer range $[0,14]$, we know that the only solutions to $x=2+5k$ in that interval are $x\\equiv2,7,12\\mod{15}$.\r\n\\section{Nonlinear Congruence Equations}\r\n\\tab\r\nThere is no general/efficient method of finding solutions.\\nextline\r\n\\subsection{Examples}\r\n$x^2\\equiv 1\\mod{2}$.\r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|}\r\n$x$ & 0 & 1 \\\\\r\n\\hline\r\n$x^2$ & 0 & 1\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nThus $x\\equiv 1\\mod{2}$ is the only \\soln\r\n\\nextline\r\n$x^2\\equiv 1\\mod{4}$\r\n\\begin{center}\r\n\\begin{tabular}{|c|c|c|c|c|}\r\n$x$ & 0 & 1 & 2 & 3\\\\\r\n\\hline\r\n$x^2$ & 0 & 1 & 4 & 9 \\\\\r\n\\hline\r\n$x^2\\mod{4}$ & 0 & 1 & 0 & 1\r\n\\end{tabular}\r\n\\end{center}\r\n\\tab\r\nTherefore, the solutions are $x\\equiv 1,3\\mod{4}$.\r\n\\nextline\r\nSolving $x^2\\equiv 1\\mod{8}$ gives 4 solutions $\\left(x\\equiv1,3,5,7\\mod{8}\\right)$. Solving $x^2\\equiv 1\\mod{2^k}\\, \\fs k\\geq3$ will have only 4 solutions.\r\n\\end{document}", "meta": {"hexsha": "a6446c20a17b649aaf3fcae10765dd2e1c606137", "size": 24837, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Libraria Algebrae/Libraria Algebrae.tex", "max_stars_repo_name": "GardnerLiam/Libraria-Mathematica", "max_stars_repo_head_hexsha": "bfc2b3734230e883e439ea9bfec3b6c4103a01ed", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-11-23T20:16:39.000Z", "max_stars_repo_stars_event_max_datetime": "2020-11-23T20:16:39.000Z", "max_issues_repo_path": "Libraria Algebrae/Libraria Algebrae.tex", "max_issues_repo_name": "GardnerLiam/Libraria-Mathematica", "max_issues_repo_head_hexsha": "bfc2b3734230e883e439ea9bfec3b6c4103a01ed", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-11-17T06:21:41.000Z", "max_issues_repo_issues_event_max_datetime": "2020-11-17T06:21:41.000Z", "max_forks_repo_path": "Libraria Algebrae/Libraria Algebrae.tex", "max_forks_repo_name": "GardnerLiam/Libraria-Mathematica", "max_forks_repo_head_hexsha": "bfc2b3734230e883e439ea9bfec3b6c4103a01ed", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.5984990619, "max_line_length": 742, "alphanum_fraction": 0.6494745742, "num_tokens": 10165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478256, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.6999548053726969}} {"text": "\\subsection{Total variation}\\label{subsec:total_variation}\n\n\\begin{definition}\\label{def:riemann_stieltjes_integral}\n The most common generalization of the \\hyperref[def:riemann_integral]{Riemann integral} is the Riemann-Stieltjes integral. It is not as well-behaved, hence we will give the most general definition and not attempt to prove equivalences.\n\n Let \\( \\mscrX \\) be a real \\hyperref[def:separation_axioms/T2]{Hausdorff} \\hyperref[def:topological_vector_space]{topological vector space}. Fix two \\hyperref[def:function]{functions} \\( f, \\alpha: [a, b] \\to X \\).\n\n The \\term{Riemann sum} of \\( f \\) with respect to \\( \\alpha \\) corresponding to the \\hyperref[def:riemann_partition/tagged]{tagged partition} \\eqref{eq:def:riemann_partition/tagged} is defined as\n \\begin{equation*}\n S(f, \\alpha, \\Delta, \\Xi) \\coloneqq \\sum_{k=1}^n f(\\xi_k) (\\alpha(x_k) - \\alpha(x_{k-1})).\n \\end{equation*}\n\n The limit of the net\n \\begin{equation}\\label{eq:def:riemann_stieltjes_integral/net}\n \\{ S(f, \\alpha, \\Delta, \\Xi) \\}_{(\\Delta, \\Xi) \\in \\op{tpart}([a, b])},\n \\end{equation}\n if it exists, is called the \\term{Riemann-Stieltjes integral} of \\( f \\) with respect to \\( \\alpha \\) and is denoted by\n \\begin{equation*}\n \\int_a^b f(x) d \\alpha(x).\n \\end{equation*}\n\\end{definition}\n", "meta": {"hexsha": "90632af471eed3aef692741385a9e568115ce795", "size": 1294, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/total_variation.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/total_variation.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/total_variation.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.8181818182, "max_line_length": 237, "alphanum_fraction": 0.7117465224, "num_tokens": 431, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772417253256, "lm_q2_score": 0.8006920116079209, "lm_q1q2_score": 0.699866664977754}} {"text": "\\section{Linear hypothesis using linear regression}\n\n\\subsection{Data}\n\\begin{figure}[!ht]\n \\includegraphics[width=\\textwidth,height=0.4\\textheight,keepaspectratio]{scatter_plot.png}\n \\caption{Scatter plot of data}\n \\label{fig:scatter_plot}\n\\end{figure}\n\n\\subsection{Implementation}\nA linear function given by equation \\ref{eq:simple_linear_model} was selected as the hypothesis.\nLinear regression with learning learning rate($\\alpha$) = 0.01 and gradient descent approach was used for optimizing the parameters.\nThe training stopped after reaching the maximum allowed number of iterations which was 10000.\nFor this specific dataset we can use a threshold of 0.5 to classify the data. The predicting data\ncan be assigned a class 1 (fail) if $h_{\\theta}$(x) $<$ 0.5 and class 2 (pass) otherwise.\n\n\\begin{equation}\n\\label{eq:simple_linear_model}\nY = \\theta_0 + \\theta_1 * X\n\\end{equation}\n\n\\subsection{Observation}\nThe final values of $\\theta$s for a run are as follows:\n\n% Theta 0 final value\n\\begin{equation}\n\\theta_0 = -0.15184881886181856\n\\end{equation}\n\n% Theta 1 final value\n\\begin{equation}\n\\theta_1 = 0.23405702659773173\n\\end{equation}\n\n\\begin{figure}[!ht]\n \\includegraphics[width=\\textwidth,height=0.4\\textheight,keepaspectratio]{regression_line_0_01.png}\n \\caption{Linear regression line}\n \\label{fig:regression_line}\n\\end{figure}\n\nThe regression line corresponding to the above $\\theta$s is shown in figure \\ref{fig:regression_line}\n\n\\subsection{Source Code}\n\\lstinputlisting[language=python]{task_1.py}\n", "meta": {"hexsha": "066669b36bd1d0b2e71510a91694102dda5c0627", "size": 1519, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assignment_2/linear_classifier.tex", "max_stars_repo_name": "diwasblack/machine_learning", "max_stars_repo_head_hexsha": "83bf5af98a3db5e13f628f39d7519575c580497d", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assignment_2/linear_classifier.tex", "max_issues_repo_name": "diwasblack/machine_learning", "max_issues_repo_head_hexsha": "83bf5af98a3db5e13f628f39d7519575c580497d", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assignment_2/linear_classifier.tex", "max_forks_repo_name": "diwasblack/machine_learning", "max_forks_repo_head_hexsha": "83bf5af98a3db5e13f628f39d7519575c580497d", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.7555555556, "max_line_length": 132, "alphanum_fraction": 0.7774851876, "num_tokens": 430, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772482857833, "lm_q2_score": 0.8006920044739461, "lm_q1q2_score": 0.6998666639950148}} {"text": "% !TeX root = ./apxthy.tex\n\n\n\\section{Least Squares Methods}\n\n\\subsection{Motivation}\n%\nWe first describe least squares methods in abstract terms. Let\n$[a,b]$ be an interval and\n$b_1, \\dots, b_N \\in C([a,b])$ be $N$ linearly independent basis functions\nfor an approximation space\n\\[\n \\AA_N := {\\rm span}\\b\\{ b_1, \\dots, b_N \\b\\}.\n\\]\nGiven $w \\in C(a,b) \\cap L^1(a,b)$ (note the open interval!) we can\ndefine a weighted $L^2$-inner product\n\\[\n \\< f, g \\>_{L^2_w} := \\int_a^b w(x) f(x) g(x)^* \\,dx,\n\\]\nwith associated norm $\\|f\\|_{L^2_w} := \\_{L^2_w}^{1/2}$. The best\napproximation of a function $f \\in C([a,b])$ with respect to this weighted norm\nis then given by\n\\begin{equation} \\label{eq:lsq:ctslsq}\n p_N \\in \\min_{p \\in \\AA_N} \\b\\| f - p \\b\\|_{L^2_w}^2.\n\\end{equation}\nWe call this a continuous least squares problem.\n\nComputationally, we typically need to discretise \\eqref{eq:lsq:ctslsq}.\nTo that end, we choose points $x_1, \\dots, x_M \\in [a, b]$ and weights\n$w_1, \\dots w_M$ and define the discrete inner product\n\\begin{equation} \\label{eq:lsq:disip}\n \\< f, g \\>_{\\ell^2_w} := \\sum_{m = 1}^M w_m f(x_m) g(x_m)^*\n\\end{equation}\nwith associated norm $\\|f\\|_{\\ell^2_w} := \\_{\\ell^2_w}^{1/2}$. This\ngives the discrete least squares problem\n\\begin{equation}\n \\label{eq:lsq:dislsq}\n p_N \\in \\min_{p \\in \\AA_N} \\b\\| f - p \\b\\|_{\\ell_w^2}.\n\\end{equation}\nThis is the typical kind of least squares problem encountered in\nreal applications.\n\nWe distinguish two scenarios:\n\\begin{enumerate}\n\\item {\\bf User Chooses Data: } In this scenario the ``user'' is given a\nfunction $f$ to be approximated. She may then choose the points $x_m$, weights\n$w_m$ and evaluations $f(x_m)$ in order to fine-tune and optimise the fit $p_N$.\nFor example it is then feasible to start from \\eqref{eq:lsq:ctslsq} and design a\ndiscrete LSQ system \\eqref{eq:lsq:dislsq} that approximates\n\\eqref{eq:lsq:ctslsq} in a suitable sense. An arbtirary amount of data $(x_m,\nf(x_m))$ may then be generated to ensure a good fit.\n\n\\item {\\bf Data is provided: } Some data has been collected outside the control\nof the person (``user'') designing the fit. Given the data points $(x_m, f(x_m))$\n(possibly subject to noise, i.e. $y_m = f(x_m) + \\eta_m$ is then provided)\none then needs to choose an appropriate approximation space $\\AA_N$,\napproximation degree $N$ and weights $w_m$ to ensure a good fit in a sense\ndictated by the application.\n\\end{enumerate}\n\nWe will study both scenarios but note that the second one is the more\ntypical in applications.\n\n\\subsection{Solution methods}\n%\n\\label{sec:lsq:soln}\n%\nWe convert \\eqref{eq:lsq:dislsq} into a linear algebra problem.\nBy writing\n\\[\n Y_m := f(x_m), \\sqrt{W} := {\\rm diag}(\\sqrt{w_1}, \\dots, \\sqrt{w_m}) \\in \\R^{M \\times M}\n\\]\nand\n\\[\n p(x_m) = \\sum_{n = 1}^N c_n b_n(x_m) = A c,\n \\quad \\text{where } A_{mn} = b_n(x_m),\n\\]\nthen $A \\in \\R^{M \\times N}$ and we obtain\n\\[\n \\sum_{m = 1}^M w_m | p(w_m) - f(x_m)|^2\n = \\big\\| \\sqrt{W} A c - \\sqrt{W} Y \\big\\|^2,\n\\]\nwhere $\\|\\cdot\\|$ denotes the standard Euclidean norm in $\\R^M$.\nWe write $\\tilde{A} := \\sqrt{W} A, \\tilde{Y} := \\sqrt{W} Y$ and\nwrite the least squares functional equivalently as\n\\[\n \\Phi(c) := \\big\\| \\sqrt{W} A c - \\sqrt{W} Y \\big\\|^2\n = c^T \\tilde{A}^T \\tilde{A} c - 2 c^T \\tilde{A}^T \\tilde{Y} + \\|\\tilde{Y}\\|^2.\n\\]\nA minimiser must satisfy $\\nabla\\Phi(c) = 0$, which gives the linear system\n\\begin{equation} \\label{eq:lsq:normaleqns}\n \\tilde{A}^T\\tilde{A} c = \\tilde{A}^T \\tilde{Y}.\n\\end{equation}\nThese are called the normal equations, which can be solved using\nthe LU or Cholesky factorisation.\n\nIt turns out that they are often (though not always) ill-conditioned.\nAn alternative approach is therefore to perform the (numerically stable)\n{\\em thin QR factorisation}\n\\[\n \\tilde{A} = Q R,\n\\]\nwhere $R \\in \\R^{N \\times N}$ is upper triangular and $Q \\in \\R^{M \\times N}$\nhas ortho-normal columns, i.e., $Q^T Q = I \\in \\R^{N \\times N}$.\nWith the QR factorisation in hand the normal equations can be rewritten as\n\\begin{align*}\n \\tilde{A}^T\\tilde{A} c &= \\tilde{A}^T \\tilde{Y} \\\\\n \\Leftrightarrow \\qquad\n R^T Q^T Q R c &= R^T Q^T \\tilde{Y} \\\\\n \\Leftrightarrow \\qquad\n R c &= Q^T \\tilde{Y},\n\\end{align*}\nprovided that $R$ is invertible (which is equivalent to $A^T A$ being invertible\nand to $A$ having full rank). Thus, the solution of the least squares problem\nbecomes\n\\begin{equation}\n \\label{eq:lsq:qr}\n R c = Q^T \\sqrt{W} Y, \\qquad \\text{where} \\qquad\n \\sqrt{W} A = QR.\n\\end{equation}\n\nIt is worthwhile comparing the computational cost of the two approaches.\n\\begin{enumerate}\n\\item The assembly of thenormal equations requires the multiplication\n$A^T A$ which requires $O(M N^2)$ operations, followed by the\nCholesky factorisation of $A^T A$ which requires $O(N^3)$ operiations.\nThus the cost of solving \\eqref{eq:lsq:normaleqns} is $O(M N^2)$.\n\\item The cost of the QR factorisation in \\eqref{eq:lsq:qr} is\n$O(M N^2)$ as well, while the inversion of $R$ is only $O(N^2)$ and the\nmultiplication with $Q^T$ is $O(NM)$.\n\nThus both algorithms scale like $O(M N^2)$.\n\\end{enumerate}\n\n\n\\subsection{Orthogonal Polynomials}\n%\n\\label{sec:lsq:orthpolys}\n%\nWe have so far encountered orthogonal polynomials in the context of the\nChebyshev basis, which arise naturally due to their connection to trigonometric\npolynomials. More generally, we can consider orthogonal polynomials with respect\nto {\\em any} inner product $\\<\\cdot, \\cdot\\>$. For simplicity we will continue\nto work on the domain $[-1, 1]$. In the context of least squares problems, we\ncan think of \\eqref{eq:lsq:ctslsq} or \\eqref{eq:lsq:dislsq} and the inner\ncontinuous or discrete products associated with these least squares problems.\n\nThe main result we want to discuss here is that the three-point recursion\n\\eqref{eq:poly:chebrecursion} for the Chebyshev basis is not special, but that\nall families of orthogonal polynomials satisfy such a recursion. That is,\ngiven an inner product $\\< \\cdot, \\cdot\\>$ we will construct sequences of\ncoefficients, $A_k, B_k, C_k$ such that the sequence of polynomials given by\n%\n\\begin{equation} \\label{eq:lsq:general_3ptrec}\n \\phi_{k+1}(x) := (x - B_k) \\phi_k(x) - C_k \\phi_{k-1}(x)\n\\end{equation}\n%\nare orthogonal. By construction, we immediately see that the leading term in\n$\\phi_{k}$ is $x^k$; hence they also span the space of all polynomials.\n\nTaking the inner product of \\eqref{eq:lsq:general_3ptrec} with $\\phi_k$ and then\n$\\phi_{k-1}$ we obtain\n%\n\\begin{align*}\n 0 &= \\< x \\phi_k, \\phi_k \\> - B_k \\< \\phi_k, \\phi_k \\>, \\\\\n 0 &= \\< \\phi_k, x \\phi_{k-1} \\> - C_k \\< \\phi_{k-1}, \\phi_{k-1} \\>,\n\\end{align*}\n%\nwhich gives expressions for $B_k, C_k$,\n%\n\\begin{equation} \\label{eq:lsq:coeffs_3pt}\n \\begin{split}\n B_k &:= \\frac{\\< x \\phi_k, \\phi_k \\>}{\\< \\phi_k, \\phi_k \\>}, \\\\\n C_k &:= \\frac{\\< \\phi_k, x \\phi_{k-1} \\>}{\\< \\phi_{k-1}, \\phi_{k-1} \\>}.\n \\end{split}\n\\end{equation}\n%\nIt is worth noting that this construction is simply the Gram-Schmidt procedure,\nbut truncated at a three-term recursion rather than the full recursion to\n$\\phi_0$. In particular, by construction, we have that $\\phi_{k+1} \\perp \\phi_k,\n\\phi_{k-1}$ and it only remains to show that they it is also orthogonal to\n$\\phi_0, \\dots, \\phi_{k-2}$. Concretely we obtain the following result.\n\n\\begin{proposition}\n Suppose that $\\< \\cdot, \\cdot\\>$ is an inner product on the space of\n polynomials such that the operator $p \\mapsto x \\cdot p$ is self-adjoint,\n (i.e., $\\ = \\< p, xq\\>$ for all polynomials $p, q$). Suppose,\n moreover, that $\\phi_0 = 1, \\phi_1 = x - \\<1, x\\>$, and that $\\phi_k, k \\geq 2,$\n is given by the three-point recursion \\eqref{eq:lsq:general_3ptrec} with\n coefficients \\eqref{eq:lsq:coeffs_3pt}. Then $\\{ \\phi_k : k \\in \\N \\}$ is\n a basis of the space of polynomials which is orthogonal with respect to\n $\\< \\cdot, \\cdot \\>$.\n\\end{proposition}\n\\begin{proof}\n By construction we have that $\\phi_1 \\perp \\phi_0$ and that $\\phi_{k+1}\n \\perp \\phi_k, \\phi_{k-1}$ for $k \\geq 2$. Thus, it only remains to prove\n that, for $k \\geq 2$, $\\phi_{k+1} \\perp \\phi_j$ for $j = 0, \\dots, k-2$.\n\n By induction we may assume that $\\< \\phi_j, \\phi_i \\> = 0$ for $i \\neq j$\n and $i \\leq k$. Then, we have\n \\begin{align*}\n \\< \\phi_{k+1}, \\phi_j\\>\n &=\n \\< x \\phi_k, \\phi_j \\> - B_k \\< \\phi_k, \\phi_j\\> - C_k \\< \\phi_{k-1}, \\phi_j\\> \\\\\n &= \\< \\phi_k, x \\phi_j \\>,\n \\end{align*}\n where we also used self-adjointness of multiplication by $x$. Since\n the degree of $x \\phi_j$ is at most $k-1$ and, again by induction, $\\phi_k$\n is orthogonal to $\\phi_0, \\dots, \\phi_{k-1}$ it follows that\n $\\< \\phi_k, x \\phi_j \\> = 0$. This completes the proof.\n\\end{proof}\n\n\n\\begin{exercise}\n Derive a recursion for an {\\em orthonormal} basis of the form\n \\begin{align*}\n A_0 \\phi_0 &= 1, \\\\\n A_1 \\phi_1 &= x - B_1, \\\\\n A_k \\phi_k &= (x - B_k) \\phi_{k-1} - C_k \\phi_{k-2}.\n \\end{align*}\n Make sure to prove that all $A_k$ are non-zero.\n\\end{exercise}\n\n\\begin{exercise}\n Consider the inner product\n \\[\n \\< p, q\\> = \\int_{-1}^1 pq + p'q' \\, dx;\n \\]\n prove that the multiplication operator $p \\mapsto xp$ is not self-adjoint.\n\n If we were to construct a sequence of orthogonal polynomials by the\n Gram-Schmidt procedure, would we again obtain a three-term recursion?\n\n {\\it Hint: The involved calculations are somewhat boring. You may wish to use\n a computer algebra system to explore this question.}\n\\end{exercise}\n\n\\begin{remark}\n A discrete inner product of the form \\eqref{eq:lsq:disip} is not strictly\n an inner product on the space of all polynomials, but depending on the\n summation points it may be an inner product on a subspace $\\mathcal{P}_N$.\n In this case the recursion formula can simply be terminated at degree\n $k = N$ to obtain an orthogonal (or orthonormal) basis of $\\mathcal{P}_N$.\n\\end{remark}\n\n\n\n\\subsection{Accuracy and Stability I: Least Squares and Nodal Interpolation}\n%\nConsider fitting trigonometric polynomials $\\AA_{2N} = \\TT_N'$ with equispaced\ngrid points $x_n = \\pi n / N$ and uniform weights $w_n = 1$. Then the\nleast squares fit\n\\[\n \\min \\sum_{n = 0}^{2N-1} |f(x_n) - t(x_n)|^2\n\\]\nis equivalent to trigonometric interpolation, for which we have sharp and error\nestimates that predict a close to optimal rate of approximation.\n\nWe could leave it at this, but it is still interesting to observe what happens\nto the least squares system. The matrix $A$ is now given by\n\\[\n A_{nk} = e^{ikx_n}\n\\]\nand the entries in the normal equation by\n\\[\n [A^* A]_{kk'} = \\sum_{n} e^{-ikx_n} e^{ik' x_n} = 2 N \\delta_{kk'}\n\\]\naccording to Exercise~\\ref{exr:trig:trapezoidal rule}(i).\nThis is due to the fact that the discrete inner product\n\\eqref{eq:lsq:disip} (up to a constant factor) identical to\nthe $L^2(\\TT)$-inner product on the space $\\TT_N'$, that is,\n\\[\n \\< f, g \\>_{\\ell^2} = 2N \\mint_{-\\pi}^\\pi f g^* \\,dx \\qquad\n \\forall f, g \\in \\TT_N'.\n\\]\nNo QR factorisation is needed and the lsq fit is given by\n\\[\n c = (2 N)^{-1} A^* Y,\n\\]\nwhere the operation $(2N)^{-1} A^T Y$ can be performed at $O(N \\log N)$\ncomputational cost using the FFT.\n\nAnalogous observations are of course true for connecting least squares\nmethods and algebraic polynomials.\n\n\n\\subsection{Accuracy and Stability II: Random data}\n%\n\\label{sec:lsq:rand}\n%\nThe situation gets more interesting when we are not allowed to optimise the\npoints at which to fit the approximant. There is an infinite variety of\ndifferent situations that can occur when the provided data is application\ndriven, which goes far beyond the scope of this module. Here, we will assume\nthat the points $x_m$ are distributed according to some probability law. That\nis, they are random. This is in fact a rather common situation in applications\nas well. Note also that we are now in the Case-2 situtation where we are given a\nfixed amount of data $(x_m, f(x_m))_{m = 1}^M$ and should choose $N$ ensure the\nbest possible fit given the data we have. In particular this means that we\nshould not choose $N$ too large!\n\nSpecifically, we shall assume throughout this section that\n\\begin{equation}\n \\label{eq:lsw:wm_law}\n x_m \\sim w \\,dx, \\qquad \\text{are iid}, \\quad \\text{for } m = 1, \\dots, M.\n\\end{equation}\n(identically and independently distributed) and without loss of generality that\n$\\int_a^b w \\,dx = 1$ as this can always be achieved by rescaling. We also\nassume that $w \\in C(a,b) \\cap L^1(a,b)$ as before. In this case, we can\nconstruct a sequence of orthogonal polynomials as in \\S~\\ref{sec:lsq:orthpolys}\nwith respect to the $L^2_w$-inner product and we will target best approximation\nwith respect to the same inner product.\n\nWe will discuss two fundamental results due to Cohen, Davenport and Leviatan\n\\cite{Cohen2013-yj}, but we won't prove them. The first result concerns the\n{\\em stability} of the normal equations. Specifically, we will show that if we\nuse an $L^2_w$-orthogonal basis then $A^T A$ will be close to identity (and in\nparticular invertible) for a sufficiently large number of sample points $x_m$.\n\n\n\n\\begin{theorem}[Stability] \\label{th:lsq:randstab}\n Let $\\phi_1, \\dots, \\phi_N$ be $L^2_w$-orthonormal, $A_{mk} := \\phi_k(x_m)$,\n then\n \\[\n \\mathbb{P}\\B[ \\| A^* A - I \\|_{\\rm op} > \\delta\\B]\n \\leq 2 N \\exp\\B( - \\smfrac{C_\\delta M}{K(N)}\\B),\n \\]\n where $C_\\delta = (1+\\delta) \\log (1+\\delta) - \\delta$ and\n \\[\n K(N) = \\sup_{x \\in [a,b]} \\sum_{k = 1}^N |\\phi_k(x)|^2.\n \\]\n\\end{theorem}\n\nLet us specifically focus on the Chebyshev measure where $w(x) =\n(1-x^2)^{-1/2}$, the Chebyshev basis $T_k(x)$ on the interval $[a,b]= [-1,1]$.\nSince $|T_k(x)| \\leq 1$ it readily follows that $K(N) \\leq N$. Moreover, the\nrecursion formulat for $T_k$ implies that $T_k(1) = 1$ for all $k$, hence this\nbound is sharp, i.e., $K(N) = N$ in this case.\n\nTo make $N \\exp( - \\smfrac{C_\\delta M}{N})$ small, we therefore need\nto choose $N \\leq \\alpha M / \\log M$. With this choice,\n\\[\n N \\exp( - \\smfrac{C_\\delta M}{N})\n =\n N \\exp\\b( - \\alpha C_\\delta \\log M\\b)\n \\leq\n M^{1 -\\alpha C_\\delta} / \\log M\n\\]\nand by choosing $\\alpha$ sufficiently large we can ensure that this value tends\nto zero as $M \\to \\infty$. (the case of sufficiently large amounts of data).\nConversely, if $\\alpha$ is too small, then $M^{1 -\\alpha C_\\delta} / \\log M \\to\n\\infty$ as $M \\to \\infty$ which shows that the choice $N \\leq \\alpha M/\\log M$\nis sharp. This is a very mild restriction!\n\nThe next result we discuss concerns the approximation $p_{NM} = \\sum_n c_n\n\\phi_n$ we obtain by solving the least squares problem $\\| f - p_{NM}\n\\|_{\\ell^2(\\{x_m\\})}^2 \\to \\min$.\n\n\\begin{theorem} \\label{th:lsq:randerr}\n There exists a constant $c$ such that, if\n \\[\n K(N) \\leq \\frac{c}{1+r} \\frac{M}{\\log M},\n \\]\n then,\n \\[\n \\mathbb{E}\\b[ \\|f - p_{NM}\\|_{L^2_w}^2 \\b]\n \\leq\n (1+\\epsilon(M)) \\|f - \\Pi_N f\\|_{L^2_w}^2\n + 8 \\|f\\|_{L^\\infty(a,b)}^2 M^{-r},\n \\]\n where $\\epsilon(M) \\to 0$ as $M \\to \\infty$, and $\\Pi_N$ denotes the\n best-approximation operator with respect to the $L^2_w$-norm.\n\\end{theorem}\n\nAs a first comment, we observe that our restriction $N \\leq \\alpha M / \\log M$\nfor sufficiently small $\\alpha$ re-appears. (In fact, this prerequisite is\nrequired to be able to apply Theorem~\\ref{th:lsq:randstab}.)\n\nWe can now ask what consequence the choice $N = \\alpha M / \\log M$ has\non the error. In this case, $\\alpha = c / (1+r)$, or equivalently,\n$r = c/\\alpha - 1$ hence (sacrifycing just a log-factor)\n\\[\n M^{-r} \\leq N^{-r} = N^{1 - c/\\alpha} =: N^{-\\alpha'},\n\\]\nwhere $\\alpha' > 0$ provided that $\\alpha$ is chosen sufficiently small. In this\ncase, we can conclude that\n\\[\n \\mathbb{E}\\b[ \\|f - p_{NM}\\|_{L^2_w}^2 \\b]\n \\lesssim\n \\|f - \\Pi_N f\\|_{L^2_w}^2\n +\n N^{- \\alpha'}\n\\]\nfor some $\\alpha' > 0$. Thus, for differentiable functions $f$, such a choice is\nquasi-optimal.\n\nHowever, for analytic functions the rate in the error estimate is\nreduced. Let us assume that $f \\in A(E_\\rho)$, then\n$\\| f - \\Pi_N f \\|_{L^2_w} \\lesssim \\rho^{-N}$ hence we must balance the\ntwo constributions\n\\[\n \\rho^{-N} + M^{-r}.\n\\]\nWe have already seen that $N = \\alpha M/\\log M$ leads to $\\rho^{-N} \\ll M^{-r}$,\nhence we instead attempt to choose $N = a (M/\\log M)^{\\alpha}$ for some $0 < \\alpha < 1$,\nwhich gives\n\\[\n r = c' (M / \\log M)^{1-\\alpha}.\n\\]\nThus, we wish to balance\n\\begin{align*}\n \\exp\\B[ - (\\log \\rho) N\\B] &+ \\exp\\B[ - r \\log M \\B] \\\\\n = \\exp\\B[ - c'' M^\\alpha (\\log M)^{-\\alpha} \\B] &+ \\exp\\B[ - c M^{1-\\alpha} (\\log M)^{-\\alpha} \\B]\n\\end{align*}\nWe can now see that the two terms are balanced when $\\alpha = 1/2$, that is,\nthe quasi-optimal choice of $N$ appears to be\n\\[\n N = a \\b(M / \\log M\\b)^{1/2}.\n\\]\nThis is also somewhat consistent with the observation that in the $C^j$ case we\nneed to decrease $\\alpha$ for increasing $j$.\n\nThat said, we should remember that we have balanced an error estimate and not\nthe actual error. At this point, it is highly advisable to test these\npredictions numerically, which is done in \\nblsq, where we see --- for some\nlimited examples --- that the stability condition $N \\lesssim M/\\log M$ appears\nto be crucial but the stronger requirement $N \\lesssim (M/\\log M)^{1/2}$ seems\nto not be required.\n\nIn summary, the foregoing analysis is intended to demonstrate how different\ncompeting contributions to approximation by fitting from data can be balanced at\nleast in principle but also the limitations of analysis.\n\n\n\n\n\n\\subsection{Exercises}\n%\n\\label{sec:lsq:exercises}\n%\n", "meta": {"hexsha": "a07be35397b8f1339f0b366a14ab6d223495d854", "size": 17750, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/lsq.tex", "max_stars_repo_name": "cortner/MA3J8ApxThyApp", "max_stars_repo_head_hexsha": "9400c557187dbd82468df2dbd0a7da99d7f08f8f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-05-22T05:11:01.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-03T02:47:25.000Z", "max_issues_repo_path": "tex/lsq.tex", "max_issues_repo_name": "cortner/MA3J8ApxThyApp", "max_issues_repo_head_hexsha": "9400c557187dbd82468df2dbd0a7da99d7f08f8f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-01-03T22:23:55.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-21T01:58:58.000Z", "max_forks_repo_path": "tex/lsq.tex", "max_forks_repo_name": "cortner/ApxThyApp", "max_forks_repo_head_hexsha": "0b28c5c4370eb4d9c5a9063c2c5c1b938aa54a3d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-02T02:44:56.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-02T02:44:56.000Z", "avg_line_length": 40.0677200903, "max_line_length": 103, "alphanum_fraction": 0.6656901408, "num_tokens": 6000, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772351648677, "lm_q2_score": 0.800692004473946, "lm_q1q2_score": 0.6998666534892025}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\chapter{Regression}\n\\label{chap:regression}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Linear Regression (OLS)}\n\\label{regression:linear}\n\nLinear regression fits the best hyperplane, or line in 1D,\nto a collection of $m$ points $\\vb{x}_{i}, y_{i}$,\ntypically via the method of least squares.\nIf $\\vb{x}$ has $n$ features we can represent the\nlinear relationship between $\\vb{x}$ and $y$ as:\n\n\\begin{equation}\\label{eq:linear:one_point}\ny_{i} = \\beta_{0} + \\sum_{j=1}^{n}\\, \\beta_{j} x_{ij} + \\epsilon_{i}\\,,\n\\end{equation}\n\n\\noindent where $\\beta_{j}$ are the parameters of the regression\nand $\\epsilon$ represent random errors.\nTransitioning to matrix notation\\footnote{Note\nthat \\textit{linear} regression refers to the linearity in the model parameters\n$\\vb*{\\beta}$, not $\\mathbf{X}$.\nThe components of $\\mathbf{X}_{i}$ can be, and often are,\nnon-linear functions of other input features.}, this is simply:\n\n\\begin{equation}\\label{eq:linear:matrix}\n\\vb{y} = \\mathbf{X} \\vb*{\\beta} + \\vb*{\\epsilon}\\,,\n\\end{equation}\n\n\\noindent where we have set $X_{i0} =1$.\nHere $\\mathbf{X}$ is $m \\times n$,\n$\\vb*{\\beta}$ is $n \\times 1$,\nand $\\vb{y}$, $\\vb*{\\epsilon}$ are $m \\times 1$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Derivation}\n\\label{regression:linear:derivation}\n\nThe ordinary least squares (OLS) estimate\\footnote{When\nthe errors are assumed to be normal,\nas in \\cref{item:regression:linear:exogeneity,item:regression:linear:spherical,item:regression:linear:normality},\nOLS is equivalent to the maximum likelihood estimation (MLE) approach of \\cref{opt:MLE}.\nSee one derivation \\href{https://machinelearningmastery.com/linear-regression-with-maximum-likelihood-estimation/}{here}.} of\nthe parameters $\\hat{\\vb*{\\beta}}$\ncan be found by minimizing the squares of the errors,\n\\ie the objective function $S\\left(\\vb*{\\beta}\\right) = \\norm{\\vb*{\\epsilon}}^{2}$:\n\n\\begin{subequations} \\label{eq:linear:ols}\n\\begin{align}\n\\hat{\\vb*{\\beta}} &= \\argmin_{\\vb*{\\beta}} S\\left(\\vb*{\\beta}\\right)\\,, \\label{eq:linear:argmin} \\\\\nS\\left(\\vb*{\\beta}\\right)\n&= \\norm{\\vb{y} - \\mathbf{X} \\vb*{\\beta}}^{2} = \\left(\\vb{y} - \\mathbf{X} \\vb*{\\beta}\\right)\\transpose \\left(\\vb{y} - \\mathbf{X} \\vb*{\\beta}\\right) \\label{eq:linear:S_matrix} \\\\\n&= \\sum_{i=1}^{m} \\, \\abs{y_{i} - \\sum_{j=0}^{n} \\, \\beta_{j} x_{ij}}^{2}\\,. \\label{eq:linear:S_components}\n\\end{align}\n\\end{subequations}\n\nWe then find the minimum of $S\\left(\\vb*{\\beta}\\right)$ with respect to $\\vb*{\\beta}$\nby taking the gradient and setting it equal to zero:\n\n\\begin{subequations} \\label{eq:linear:ols_derivation}\n\\begin{align}\nS\\left(\\vb*{\\beta}\\right) &=\n \\vb{y}\\transpose \\vb{y}\n-\\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\vb{y}\n-\\vb{y}\\transpose \\mathbf{X} \\vb*{\\beta}\n+\\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta} \\label{eq:linear:S_expand} \\\\\n&= \\vb{y}\\transpose \\vb{y}\n-2\\, \\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\vb{y}\n+\\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta}\\,, \\label{eq:linear:S_expand_simplified} \\\\\n\\partial_{\\vb*{\\beta}} \\, S\\left(\\vb*{\\beta}\\right) &=\n0\n-2\\, \\partial_{\\vb*{\\beta}} \\, \\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\vb{y}\n+\\partial_{\\vb*{\\beta}} \\, \\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta} \\label{eq:linear:grad_S_expand} \\\\\n&= -2\\, \\mathbf{X}\\transpose \\vb{y} + 2\\, \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta} = 0. \\, \\implies \\label{eq:linear:grad_S} \\\\\n\\mathbf{X}\\transpose \\vb{y} &= \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta}\\,, \\label{eq:linear:penultimate}\n\\end{align}\n\\end{subequations}\n\n\\noindent where we have used\\footnote{See \\href{https://economictheoryblog.com/2015/02/19/ols_estimator/}{here}\nand \\href{https://economictheoryblog.com/2018/10/17/derivation-of-the-least-squares-estimator-for-beta-in-matrix-notation-proof-nr-1/}{here}\nfor the component-wise proof, but it is essentially the same as the 1D $\\partial_{x} b x = b$, $\\partial_{x} b x^{2} = 2 b x$.\nAlso note that \\cref{eq:linear:S_expand_simplified} results from $\\vb{y}\\transpose \\mathbf{X} \\vb*{\\beta}$ being scalar\nand thus equal to its own transpose, $\\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\vb{y}$.}:\n\n\\begin{subequations} \\label{eq:grad_relations}\n\\begin{align}\n\\partial_{\\vb*{\\beta}} \\, \\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\vb{y} &= \\mathbf{X}\\transpose \\vb{y}\\,, \\label{eq:grad_relations:1} \\\\\n\\partial_{\\vb*{\\beta}} \\, \\vb*{\\beta}\\transpose \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta} &= 2 \\, \\mathbf{X} \\transpose \\mathbf{X} \\vb*{\\beta}\\,. \\label{eq:grad_relations:2}\n\\end{align}\n\\end{subequations}\n\nConsequently, the optimal $\\hat{\\vb*{\\beta}}$ of \\cref{eq:linear:ols}\nhas a closed form solution\\footnote{The fitted prediction is $\\hat{\\vb{y}} = \\mathbf{X} \\hat{\\vb*{\\beta}}$, with residuals $\\hat{\\vb*{\\epsilon}} = \\vb{y} - \\hat{\\vb{y}}$.}:\n\n\\begin{equation}\\label{eq:linear:betahat_OLS}\n\\hat{\\vb*{\\beta}}^{\\text{OLS}} = \\left(\\mathbf{X}\\transpose\\mathbf{X}\\right)^{-1}\\mathbf{X}\\transpose \\vb{y}\\,,\n\\end{equation}\n\n\\noindent which is the best linear unbiased estimator (BLUE),\nas can be shown with the Gauss-Markov Theorem provided\nthe following assumptions hold\\footnote{For additional details see\n\\href{https://economictheoryblog.com/2015/04/01/ols_assumptions/}{here},\n\\href{http://people.duke.edu/~rnau/testing.htm}{here}, and\n\\href{https://economictheoryblog.com/2015/02/26/markov_theorem/}{here}.}:\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Assumptions}\n\\label{regression:linear:assumptions}\n\n\\begin{enumerate}[noitemsep]\n \\item The underlying relationship between $\\vb{x}$ and $y$ is linear, and there are no major outliers.\\label{item:regression:linear:linear}\n \\item The columns of $\\mathbf{X}$, \\ie features, are linearly independent, \\ie $\\mathrm{rank}\\left(\\mathbf{X}\\right) = n$ (no multicollinearity). This allows $\\mathbf{X}\\transpose\\mathbf{X}$ to be inverted.\\label{item:regression:linear:multicollinearity}\n \\item The errors $\\epsilon$ have conditional mean 0, $\\expvalE{\\epsilon \\mid \\mathbf{X}} = 0$ (exogeneity). The errors thus:\\label{item:regression:linear:exogeneity}\n \\begin{enumerate}[noitemsep]\n \\item Have a mean of zero, $\\expvalE{\\epsilon} = 0$.\n \\item Are not correlated with the input features, $\\expvalE{\\mathbf{X}\\transpose\\epsilon} = 0$.\n \\end{enumerate}\n \\item The errors are spherical, $\\mathrm{var}\\left(\\epsilon \\mid \\mathbf{X}\\right) = \\sigma^{2} \\identity$. Thus:\\label{item:regression:linear:spherical}\n \\begin{enumerate}[noitemsep]\n \\item Each observation $\\vb{x}_{i}$ has the same constant variance $\\sigma^{2}$ (homoscedasticity).\n \\item The errors are uncorrelated between observations, $\\expvalE{\\epsilon_{i}\\epsilon_{j \\neq i} \\mid \\mathbf{X}} = 0$ (no autocorrelation).\n \\end{enumerate}\n \\item The errors are normally distributed (multivariate normality)\\footnote{This is not required for OLS to be the BLUE, but hypothesis testing works if true.}.\\label{item:regression:linear:normality}\n\\end{enumerate}\n\nIf these assumptions are violated the following issues arise,\nnamely the model may be biased and or have a large or invalid estimated variance:\n\n\\begin{itemize}[noitemsep]\n \\item[\\cref{item:regression:linear:linear}.] If you are fitting nonlinear data the predictions will have large errors,\nparticularly when extrapolated beyond the range of the fitted data.\nThis will show up as systematic errors in the residuals plot,\nor may be obvious when comparing observed vs predicted values.\nPossible fixes include applying a nonlinear transformation to some of the features to linearize the data, \\eg take the log,\nadding more combinations of features, \\eg higher polynomial terms,\nor finding new independent features which may explain the nonlinearity.\n\n \\item[\\cref{item:regression:linear:multicollinearity}.] If some of the features are not linearly independent (multicollinearity),\nthey can be biasing the model and should be removed in turn until linear independence is restored.\nMulticollinearity can be spotted in the input feature correlation matrix,\nwith the variance inflation factor of \\cref{regression:linear:VIF},\nor if the residuals correlate to any of the features.\n\n \\item[\\cref{item:regression:linear:exogeneity}.] If something is wrong with the errors\nsuch that they have a non-zero mean or correlate to the input features\nthe OLS $\\hat{\\vb*{\\beta}}$ is biased and inconsistent.\nThis can happen if there are omitted variables or measurement errors.\nFurthermore, if $\\expvalE{\\epsilon \\mid \\mathbf{X}} = c$ only the $\\beta_{0}$ intercept is affected.\n\n \\item[\\cref{item:regression:linear:spherical}.] If something is wrong with the errors\nsuch that they have a changing variance or correlate across observations\\footnote{Thus the residuals correlate with row number, \\ie autocorrelation.}\nthe reported confidence intervals on the model parameters may be over or underestimated.\nThe OLS $\\hat{\\vb*{\\beta}}$ remains unbiased and has consistent coefficients, but will be biased for standard errors.\n\n \\item[\\cref{item:regression:linear:normality}.] If the errors are not normally distributed the confidence intervals are again suspect.\nThis can be diagnosed by comparing the errors to the normal distribution with a normal probability plot, or normal quantile plot,\nor through a statistical method like the Anderson-Darling and Kolmogorov-Smirnov tests.\nNote that violating normality in the errors is not as much of an issue compared to the other assumptions\nsince the fit will still give usable coefficients provided the assumed form of the model is correct.\nProblems of this kind can arise from nonlinear data or influential outliers.\nIf the errors really are non-normal, a generalized linear model (GLM) could be employed to model them correctly.\n\\end{itemize}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Goodness of Fit}\n\\label{regression:linear:goodness_of_fit}\n\n\\subsubsection{Coefficient of Determination ($R^{2}$)}\n\\label{regression:linear:goodness_of_fit:R2}\n\nThe coefficient of determination\\footnote{For additional details see\n\\href{https://economictheoryblog.com/2014/11/05/the-coefficient-of-determination-latex-r2/}{here},\n\\href{https://economictheoryblog.com/2014/11/05/proof/}{here}, and\n\\href{http://people.duke.edu/~rnau/rsquared.htm}{here}.}, $R^{2}$,\nmeasures how much of $\\vb{y}$'s variance is explained by the fitted model $\\hat{\\vb{y}}$,\nversus by random errors $\\vb*{\\epsilon}$ or other unknown sources,\nas $\\vb{y} = \\hat{\\vb{y}} + \\vb*{\\epsilon}$.\n$R^{2} \\approx 1$ is ideal and, depending on the particular definition,\nit can take values outside the typical $0 \\leq R^{2} \\leq 1$ range.\n\n\\begin{equation}\\label{eq:regression:linear:goodness_of_fit:R2}\nR^{2} = \\frac{\n\\sum_{i=1}^{m} \\left(\\yhat_{i} - \\expval{\\yhat}\\right)^{2}\n}{\n\\sum_{i=1}^{m} \\left(y_{i} - \\expval{y}\\right)^{2}\n} = 1 - \\frac{\n\\sum_{i=1}^{m} \\hat{\\epsilon}_{i}^{\\,2}\n}{\n\\sum_{i=1}^{m} \\left(y_{i} - \\expval{y}\\right)^{2}\n} = \\rho_{y,\\yhat}^{2}\n\\end{equation}\n\nThe adjusted $R^{2}$, $R^{2}_{\\text{adj}}$, accounts for the degrees of freedom in the\nfitted sample, $m-1$, and in the model, $m-n-1$, and is thus a better metric.\n\n\\begin{equation}\\label{eq:regression:linear:goodness_of_fit:R2_adj}\nR^{2}_{\\text{adj}} = 1 - \\frac{\n\\left(\\sum_{i=1}^{m} \\hat{\\epsilon}_{i}^{\\,2}\\right)/\\left(m-n-1\\right)\n}{\n\\left(\\sum_{i=1}^{m} \\left(y_{i} - \\expval{y}\\right)^{2}\\right)/\\left(m-1\\right)\n}\n= 1 - \\left(1-R^{2}\\right)\\frac{m-1}{m-n-1}\n\\end{equation}\n\n\\subsubsection{Standard Error of the Regression and $\\chi_{\\nu}^{2}$}\n\\label{regression:linear:goodness_of_fit:reduced_chi2}\n\nThe standard error of the regression, $s$,\nand reduced \\chiSqstat, $\\chi_{\\nu}^{2}$,\nare additional ways of quantifying goodness of fit.\n$s$ is a measure of the typical residual, in the units of $y$.\n$\\chi_{\\nu}^{2}$ is a convenient way of reporting this spread in a dimensionless manner,\nparticularly in more sophisticated regressions where each $y_{i}$ has an \\apriori estimated uncertainty, $\\sigma_{i}$.\n\n\\begin{subequations}\\label{eq:regression:linear:goodness_of_fit:s_red_chi2}\n\\begin{align}\ns^{2} &= \\frac{\\norm{\\hat{\\vb*{\\epsilon}}}^{2}}{m-n} = \\frac{1}{m-n} \\sum_{i=1}^{m} \\left(\\yhat_{i} - y_{i}\\right)^{2} \\label{eq:regression:linear:goodness_of_fit:s} \\\\\n\\chi_{\\nu}^{2} &= \\frac{1}{m-n} \\sum_{i=1}^{m} \\frac{\\left(\\yhat_{i} - y_{i}\\right)^{2}}{\\sigma_{i}^{2}} \\label{eq:regression:linear:goodness_of_fit:red_chi2}\n\\end{align}\n\\end{subequations}\n\n$\\chi_{\\nu}^{2}$ can be interpreted as follows:\n\n\\begin{table}[H]\n \\centering\n \\begin{tabular}{c | p{8cm}}\n$\\chi_{\\nu}^{2} \\gg 1$ & The true model may be different than the fitted model. \\\\\n$\\chi_{\\nu}^{2} > 1$ & Fit doesn't fully explain data (underfitting), or the \\apriori $\\sigma_{i}$ are underestimated. \\\\\n$\\chi_{\\nu}^{2} \\approx 1$ & Good fit! \\\\\n$\\chi_{\\nu}^{2} < 1$ & Fit overexplains data (overfitting), or the \\apriori $\\sigma_{i}$ are overestimated. \\\\\n$\\chi_{\\nu}^{2} \\ll 1$ & Extreme overfitting or overestimated $\\sigma_{i}$.\n \\end{tabular}\n% \\caption{}\n \\label{table:red_chi2_interp}\n\\end{table}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Coefficient Significance}\n\\label{regression:linear:coeff_significance}\n% TODO what is the significance of any particular \\beta_{i} coefficient, via \\ttest, \\chiSqtest, \\Ftest?? Should be able to get distribution of any \\beta_{i} analytically for OLS, but can use bootstrap sampling for more complex models.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Variance Inflation Factor (VIF)}\n\\label{regression:linear:VIF}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Ridge Regression (L2, or Tikhonov)}\n\\label{regression:linear:ridge}\n\nAdding ridge (L2, or Tikhonov) regularization to OLS is much more straight forward than LASSO,\nso we shall discuss it first\\footnote{See \\href{https://stats.stackexchange.com/a/164546}{here} for an interesting geometric explanation.}.\nWe wish to minimize the square errors $\\norm{\\vb*{\\epsilon}}^{2}$,\nnow subject to the condition that $\\norm{\\vb*{\\beta}}^{2} < t$ for some $t \\geq 0$.\nFor any $t \\geq 0$ there is a $\\lambda \\geq 0$ such that minimizing the following\nobjective function $S\\left(\\vb*{\\beta}\\right)$ without conditions is\nequivalent\\footnote{This is sometimes termed soft-thresholding, but is really an application of Lagrange multipliers, as described in \\cref{opt:lagrange_mult}.}:\n\n\\begin{equation} \\label{eq:linear:ridge}\nS\\left(\\vb*{\\beta}\\right) = \\norm{\\vb{y} - \\mathbf{X} \\vb*{\\beta}}^{2} + \\lambda \\norm{\\vb*{\\beta}}^{2}\\,.\n\\end{equation}\n\nAfter taking the gradient, we have the same terms as \\cref{eq:linear:grad_S},\nplus a new term\\footnote{Note the similarity to \\cref{eq:grad_relations:2}.} \\cref{eq:linear:ridge_derivation:new}.\nRearranging \\cref{eq:linear:ridge_derivation:penultimate} we can move the new term in with the old,\nand quickly arrive at a modified version of the OLS solution \\cref{eq:linear:ridge_derivation:betahat}.\n\n\\begin{subequations} \\label{eq:linear:ridge_derivation}\n\\begin{align}\n\\partial_{\\vb*{\\beta}} \\, S\\left(\\vb*{\\beta}\\right)\n&= -2\\, \\mathbf{X}\\transpose \\vb{y} + 2\\, \\mathbf{X}\\transpose \\mathbf{X} \\vb*{\\beta} \\label{eq:linear:ridge_derivation:grad_S_repeat} \\\\\n&+ 2\\,\\lambda \\vb*{\\beta} = 0 \\,\\, \\implies \\label{eq:linear:ridge_derivation:new} \\\\\n\\mathbf{X}\\transpose \\vb{y} &= \\left(\\mathbf{X}\\transpose \\mathbf{X} + \\lambda\\,\\identity\\right) \\vb*{\\beta} \\label{eq:linear:ridge_derivation:penultimate} \\\\\n\\hat{\\vb*{\\beta}}^{\\text{ridge}} &= \\left(\\mathbf{X}\\transpose\\mathbf{X} + \\lambda\\,\\identity\\right)^{-1}\\mathbf{X}\\transpose \\vb{y} \\label{eq:linear:ridge_derivation:betahat}\n\\end{align}\n\\end{subequations}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{LASSO Regression (L1)}\n\\label{regression:linear:lasso}\n\nAs promised, adding LASSO (L1) regularization to OLS is challenging\nas the condition, $\\norm{\\vb*{\\beta}} < t$ for some $t \\geq 0$,\nis not differentiable, see the corners of \\cref{fig:ml:l1l2:l1}, and must be dealt with carefully.\nIn the special case that $\\mathbf{X}$ is orthonormal, $\\mathbf{X}\\transpose \\mathbf{X} = \\identity$,\na closed form solution \\cref{eq:linear:lasso:betahat:1} can be derived on a case-by-case basis\nat the $\\beta_{i}$ coordinate-level\\footnote{See \\href{https://stats.stackexchange.com/questions/17781/derivation-of-closed-form-lasso-solution}{here},\n\\href{https://en.wikipedia.org/wiki/Lasso_(statistics)\\#Orthonormal_covariates}{here},\nand \\href{https://xavierbourretsicotte.github.io/lasso_derivation.html}{here}.}.\nIn the general case, more sophisticated methods can find numerical solutions.\n\n\\begin{subequations} \\label{eq:linear:lasso:betahat_all}\n\\begin{align}\n\\hat{\\vb*{\\beta}}^{\\text{LASSO}}_{j}\n&= \\hat{\\vb*{\\beta}}^{\\text{OLS}}_{j} \\max\\left(0, \\frac{\\lambda}{\\abs{\\hat{\\vb*{\\beta}}^{\\text{OLS}}_{j}}}\\right) \\label{eq:linear:lasso:betahat:1} \\\\\n&= \\sign{\\hat{\\vb*{\\beta}}^{\\text{OLS}}_{j}} \\left(\\abs{\\hat{\\vb*{\\beta}}^{\\text{OLS}}_{j}} - \\lambda\\right)^{+} \\label{eq:linear:lasso:betahat:2} \\\\\n\\hat{\\vb*{\\beta}}^{\\text{OLS}}\n&= \\left(\\mathbf{X}\\transpose\\mathbf{X}\\right)^{-1}\\mathbf{X}\\transpose \\vb{y}\n= \\mathbf{X}\\transpose \\vb{y} \\label{eq:lasso:modified_betahat_OLS}\n\\end{align}\n\\end{subequations}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{LASSO vs Ridge Regression}\n\\label{regression:linear:lasso_vs_ridge}\n\nAll the usual concepts from \\cref{ml_general:reg} apply, but specifically for linear regression\nboth LASSO and ridge allow us to relax the multicollinearity condition.\nLasso tends to use one feature per group of correlated $\\vb{x}_{j}$'s, setting the rest to $\\approx 0$,\nwhile ridge will keep all of the associated $\\beta_{j}$'s at a similar magnitude.\nAs one may expect from LASSO shrinking $\\beta_{j}$'s to zero,\nit does a better job when only some of the $n$ input features infuence \\yhat,\nwhile ridge does better if most of the features have similar importances.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{OLS Example}\n\\label{regression:linear:example}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Bayesian Linear Regression}\n\\label{regression:bayesian_linear}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{LASSO and Ridge as Priors}\n\\label{regression:bayesian_linear:lasso_vs_ridge}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Weighted Least Squares}\n\\label{regression:WLS}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Generalized Least Squares (GLS)}\n\\label{regression:GLS}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Generalized Linear Models (GLM)}\n\\label{regression:GLM}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Binomial Regression}\n\\label{regression:GLM:binomial}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Poisson Regression}\n\\label{regression:GLM:poisson}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Gaussian Process Regression (Kriging)}\n\\label{regression:kriging}\n% TODO\n\n% TODO see \\cite{Brochu2010} section 2.6 and the additional references it cites\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Difference in Differences (DID)}\n\\label{regression:DID}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Fixed-Effects Model}\n\\label{regression:fixed_effects}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Mixed-Effects Model}\n\\label{regression:mixed_effects}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Simpson's Paradox}\n\\label{regression:simpsons_paradox}\n% TODO\n", "meta": {"hexsha": "8c95f4a5049abe90ec1381151a46856fa1c7c4aa", "size": 20529, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/regression.tex", "max_stars_repo_name": "mepland/data_science_notes", "max_stars_repo_head_hexsha": "f529a86490110fc6a30d1af6d37c0add2517244f", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-05-30T15:15:50.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-23T01:01:08.000Z", "max_issues_repo_path": "sections/regression.tex", "max_issues_repo_name": "mepland/data_science_notes", "max_issues_repo_head_hexsha": "f529a86490110fc6a30d1af6d37c0add2517244f", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/regression.tex", "max_forks_repo_name": "mepland/data_science_notes", "max_forks_repo_head_hexsha": "f529a86490110fc6a30d1af6d37c0add2517244f", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.1040609137, "max_line_length": 256, "alphanum_fraction": 0.6538068099, "num_tokens": 6061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772253241802, "lm_q2_score": 0.800692004473946, "lm_q1q2_score": 0.6998666456098428}} {"text": "\\section{Direct Sums and Free Modules}\r\n\\begin{definition}\r\n If $M_1,\\ldots,M_n$ are $R$-modules, then their direct sum $M_1\\oplus\\cdots\\oplus M_n$ is the set $M_1\\times \\cdots\\times M_n$ with entry-wise addition and scalar multiplications.\r\n\\end{definition}\r\n\\begin{example}\r\n 1. $R^n$ is simply $R\\oplus\\cdots\\oplus R$ of $n$ copies of $R$.\\\\\r\n 2. If $M_1,M_2\\le M$, then the $R$-module homomorphism $M_1\\oplus M_2\\to M$ by $(m_1,m_2)\\mapsto m_1+m_2$ is an isomorphism iff $M_1\\cap M_2=\\varnothing$ and $M_1+M_2=M$.\r\n\\end{example}\r\n\\begin{lemma}\r\n If $M=\\bigoplus_{i=1}^nM_n$, and $N_1\\le M_i$.\r\n Take $N=\\bigoplus_{i=1}^nN_i$, then\r\n $$M/N\\cong\\bigoplus_{i=1}^nM_i/N_i$$\r\n\\end{lemma}\r\n\\begin{proof}\r\n Apply the first isomorphism theorem to the surjective $R$-module map $\\phi:M\\to\\bigoplus_{i=1}^nM_i/N_i$ by $(m_1,\\ldots,m_n)\\mapsto(m_1+N_1,\\ldots,m_n+N_n)$.\r\n\\end{proof}\r\n\\begin{example}\r\n Taking $R=\\mathbb Z$ then $\\mathbb Z^2=\\mathbb Z\\oplus \\mathbb Z$, then we have $(\\mathbb Z\\oplus\\mathbb Z)/(m\\mathbb Z\\oplus n\\mathbb Z)\\cong(\\mathbb Z/m\\mathbb Z)\\oplus(\\mathbb Z/n\\mathbb Z)$.\r\n\\end{example}\r\n\\begin{definition}\r\n Let $m_1,\\ldots,m_n\\in M$.\r\n The set $\\{m_1,\\ldots,m_n\\}$ is independent if $r_1m_1+\\cdots +r_nm_n=0\\implies \\forall i,r_i=0$.\r\n\\end{definition}\r\n\\begin{definition}\r\n A subset $S$ of an $R$-module $M$ generates $M$ freely if $S$ generates $M$ and any function $\\psi:S\\to N$ for another $R$-module $N$ extends to an $R$-module homomorphism $M\\to N$.\r\n\\end{definition}\r\nNote that if such an extension exists then it is necessarily unique.\r\n\\begin{definition}\r\n A freely-generated $R$-module is called a free $R$-module.\r\n The corresponding $S$ is called the free basis.\r\n\\end{definition}\r\n\\begin{proposition}\r\n For an $R$-module $M$ and a subset $S=\\{m_1,\\ldots,m_n\\}\\subset M$, the followings are equivalent:\\\\\r\n 1. $S$ generates $M$ freely.\\\\\r\n 2. $S$ generates $M$ and $S$ is independent.\\\\\r\n 3. Every $m\\in M$ can be written uniquely in the form $m=r_1m_1+\\cdots +r_nm_n$ for $r_1,\\ldots,r_n\\in R$.\\\\\r\n 4. The $R$-module homomorphism $R^n\\to M$ by $(r_1,\\ldots,r_n)\\mapsto r_1m_1+\\ldots r_nm_n$ is an isomorphism.\r\n\\end{proposition}\r\n\\begin{proof}\r\n $1\\implies 2$: We already knows that $S$ generates $M$, so it suffices to show that $S$ is independent.\r\n Suppose for sake of contradiction that $r_1m_1+\\ldots+r_nm_n=0$ for some $r_i\\in R$ and some $r_j$ is nonzero.\r\n Consider the function $\\psi:S\\to R$ by $m_j\\mapsto 1$ and $m_i\\mapsto 0$ for any $i\\neq j$.\r\n Suppose this extends to an $R$-module map $\\theta:M\\to R$, then $0=\\theta(0)=\\theta(r_1m_1+\\cdots +r_nm_n)=r_j$, contradiction.\\\\\r\n Remaining implications $2\\implies 3\\implies 1$ and $3\\iff 4$ are just as easy if not easier.\r\n\\end{proof}\r\nSadly not all $R$-modules are free.\r\nEven if it is, the free basis does not behave like what we expect from a vector space.\r\n\\begin{example}[non-example]\r\n 1. Suppose we have a nontrivial finite abelian group $A$, then $A$ is not free as a $\\mathbb Z$-module since it is not isomorphic to $\\mathbb Z^n$ which is infinite.\\\\\r\n 2. The set $\\{2,3\\}\\subset\\mathbb Z$ generates $\\mathbb Z$ as a $\\mathbb Z$-module, but it is not independent and no subset of it gives a free basis.\r\n\\end{example}\r\n\\begin{proposition}[Theorem on Invariant of Dimension]\r\n Let $R$ be a nonzero ring.\r\n If $R^m\\cong R^n$ as $R$-modules, then $m=n$.\r\n\\end{proposition}\r\nWe introduce the following general construction: Let $R$ be a ring and $I\\unlhd R$ and $M$ is an $R$-module.\r\nWe write $IM=\\{im:i\\in I,m\\in M\\}\\le M$.\r\nThen the quotient $M/(IM)$ is an $R/I$ module by $(r+I)(m+IM)=rm+IM$.\r\nAlso by Zorn's Lemma, for any proper ideal $I$ in a ring $R$, there is a maximal ideal containing $I$ (this is obvious when $R$ is Noetherian).\r\n\\footnote{I think we can prove the proposition without using AC (or equivalence)}\r\n\\begin{proof}\r\n Return to our proof, suppose $R^m\\cong R^n$.\r\n Choose $I\\unlhd R$ maximal, then we have\r\n $$(R/I)^m\\cong R^m/(IR^m)\\cong R^n/(IR^n)\\cong (R/I)^n$$\r\n But $R/I$ is a field, so $m=n$.\r\n\\end{proof}", "meta": {"hexsha": "78e19c0df07e54a65db936f9b636868ce354c6a9", "size": 4128, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "15/free.tex", "max_stars_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_stars_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "15/free.tex", "max_issues_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_issues_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "15/free.tex", "max_forks_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_forks_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.5454545455, "max_line_length": 199, "alphanum_fraction": 0.6712693798, "num_tokens": 1511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059609645723, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.699858802232432}} {"text": "\\documentclass{article}\n\\input{preamble}\n\n\\newcommand{\\version}{Draft: 02 April 2020}\n\n\\lhead{\\scriptsize}\n\\chead{\\scriptsize}\n\\rhead{\\scriptsize \\thepage}\n\n\\lfoot{\\scriptsize}\n\\cfoot{\\scriptsize}\n\\rfoot{\\scriptsize \\version}\n\n\n%==============================================================================\n\\begin{document}\n%==============================================================================\n\nWe define the Oneka-type regional flow model by the following discharge potential:\n%\n\\begin{equation}\\label{1}\n \\Phi(x,y) = A x^2 + B y^2 + C xy + Dx + Ey + F\n\\end{equation}\n%\nEquation~\\eqref{1} is a quadratic form in real variables. We define $z$ and $\\bar{z}$ as\n%\n\\begin{equation}\\label{2}\n z = x + i y \\qquad \\qquad \\bar{z} = x - i y\n\\end{equation}\n%\nInverting yields\n%\n\\begin{equation}\\label{3}\n x = \\frac{z + \\bar{z}}{2} \\qquad \\qquad y = \\frac{z - \\bar{z}}{2i}\n\\end{equation}\n%\nSubstituting \\eqref{3} into \\eqref{1} yields\n%\n\\begin{equation}\\label{4}\n \\Phi(z,\\bar{z})\n = A \\left(\\frac{z + \\bar{z}}{2}\\right)^2 + B \\left(\\frac{z - \\bar{z}}{2i}\\right)^2 +\n C \\left(\\frac{z + \\bar{z}}{2}\\right)\\left(\\frac{z - \\bar{z}}{2i}\\right) +\n D \\left(\\frac{z + \\bar{z}}{2}\\right) + E \\left(\\frac{z - \\bar{z}}{2i}\\right) + F\n\\end{equation}\n%\nRearranging \\eqref{4} yields\n%\n\\begin{equation}\\label{5}\n \\Phi(z, \\bar{z}) = a z^2 + b \\bar{z}^2 + c z \\bar{z} + d z + e \\bar{z} + f\n\\end{equation}\n%\nwhere\n%\n\\begin{align}\n a &= \\frac{A - B - iC}{4} \\\\\n b &= \\frac{A - B + iC}{4} \\\\\n c &= \\frac{A + B}{2} \\\\\n d &= \\frac{D - iE}{2} \\\\\n e &= \\frac{D + iE}{2} \\\\\n f &= F\n\\end{align}\n%\nAlternatively,\n%\n\\begin{equation}\n \\frac{1}{4}\n \\begin{bmatrix}\n 1& -1& -1i& 0& 0& 0 \\\\\n\t1& -1& 1i& 0& 0& 0 \\\\\n\t2& 2& 0& 0& 0& 0 \\\\\n\t0& 0& 0& 2& -2i& 0 \\\\\n\t0& 0& 0& 2& 2i& 0 \\\\\n\t0& 0& 0& 0& 0& 4\n \\end{bmatrix}\n \\begin{bmatrix}A \\\\ B \\\\ C \\\\ D \\\\ E \\\\ F \\end{bmatrix}\n =\n \\begin{bmatrix}a \\\\ b \\\\ c \\\\ d \\\\ e \\\\ f \\end{bmatrix}\n\\end{equation}\n%\nInverting yields\n%\n\\begin{equation}\n \\begin{bmatrix}\n\t1& 1& 1& 0& 0& 0 \\\\\n\t-1& -1& 1& 0& 0& 0 \\\\\n\t2i& -2i& 0& 0& 0& 0 \\\\\n\t0& 0& 0& 1& 1& 0 \\\\\n\t0& 0& 0& 1i& -1i& 0 \\\\\n\t0& 0& 0& 0& 0& 1\n \\end{bmatrix}\n \\begin{bmatrix}a \\\\ b \\\\ c \\\\ d \\\\ e \\\\ f \\end{bmatrix}\n =\n \\begin{bmatrix}A \\\\ B \\\\ C \\\\ D \\\\ E \\\\ F \\end{bmatrix}\n\\end{equation}\n\nNote that $b = \\bar{a}$ and $e = \\bar{d}$, so \\eqref{5} simplifies to\n%\n\\begin{align}\\label{7}\n \\Phi(z, \\bar{z}) \n &= a z^2 + \\bar{a} \\bar{z}^2 + c z \\bar{z} + d z + \\bar{d} \\bar{z} + f \\\\\n &= \\real{a z^2 + d z + f} + c z \\bar{z}\n\\end{align}\n%\nwhere $a$ and $d$ are complex, but $c$ and $f$ are real.\n\n%==============================================================================\n\\end{document}\n%==============================================================================\n\n\n\n\n\n\\newpage\n%------------------------------------------------\n\\subsection{Regional recharge}\n%------------------------------------------------\nThe regional recharge is given by\n%\n\\begin{equation}\\label{2.3}\n N = -2 \\left( A + B \\right)\n\\end{equation}\n%\n\\begin{equation}\\label{2.4}\n \\ev{N} = -2 \\left( \\ev{A} + \\ev{B} \\right)\n\\end{equation}\n%\n\\begin{equation}\\label{2.5}\n \\var{N} = 4 \\left( \\var{A} + \\var{B} + 2\\cov{A,B} \\right)\n\\end{equation}\n\n", "meta": {"hexsha": "20d0ace9818e8afd4e74aced90f4f1e995797862", "size": 3276, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/ComplexOneka.tex", "max_stars_repo_name": "RandalJBarnes/NagadanPy", "max_stars_repo_head_hexsha": "4fc3cf7e7650adb39d0e68ee9cddc243b85ef904", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "docs/ComplexOneka.tex", "max_issues_repo_name": "RandalJBarnes/NagadanPy", "max_issues_repo_head_hexsha": "4fc3cf7e7650adb39d0e68ee9cddc243b85ef904", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/ComplexOneka.tex", "max_forks_repo_name": "RandalJBarnes/NagadanPy", "max_forks_repo_head_hexsha": "4fc3cf7e7650adb39d0e68ee9cddc243b85ef904", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.8181818182, "max_line_length": 88, "alphanum_fraction": 0.4703907204, "num_tokens": 1319, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8824278757303677, "lm_q2_score": 0.7931059487389968, "lm_q1q2_score": 0.6998587975748708}} {"text": "\\section{Improper Integrals}\\label{sec:ImproperIntegrals}\r\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\nRecall that the Fundamental Theorem of Calculus says that if $f$ is a \\underline{{\\bf continuous}} function on the closed interval $[a,b]$, then\r\n$$\\ds{\\int_a^b f(x)~dx=F(x)\\bigg|_a^b=F(b)-F(a)},$$\r\nwhere $F$ is any antiderivative of $f$. \r\n\r\nBoth the {\\bf continuity} condition and {\\bf closed interval} must hold to use the Fundamental Theorem of Calculus, and in this case, $\\ds\\int_a^b f(x)\\,dx$ represents the net area under $f(x)$ from $a$ to $b$: \r\n$$\\includegraphics[width=2.25in]{images2/area-under}$$\r\n\r\nWe begin with an example where blindly applying the Fundamental Theorem of Calculus can give an incorrect result.\r\n\r\n\\begin{example}{Using FTC}{Using FTC}\r\nExplain why $\\ds\\int_{-1}^1\\frac{1}{x^2}\\,dx$ is not equal to $-2$.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\nHere is how one might proceed: \r\n$$\r\n\\int_{-1}^1\\frac{1}{x^2}\\,dx \r\n~=~ \\int_{-1}^1 x^{-2}\\,dx \r\n~=~ -x^{-1}\\bigg|_{-1}^1 \r\n~=~ -\\frac{1}{x}\\bigg|_{-1}^1 \r\n~=~ \\left(-\\frac{1}{1}\\right) - \\left(-\\frac{1}{(-1)}\\right) \r\n~=~ -2$$ \r\n However, the above answer is {\\bf WRONG!} \r\n Since $f(x)=1/x^2$ is not continuous on $[-1,~1]$, we cannot directly apply the Fundamental Theorem of Calculus.\r\nIntuitively, we can see why $-2$ is not the correct answer by looking at the graph of $f(x)=1/x^2$ on $[-1,~1]$.\r\nThe shaded area appears to grow without bound as seen in the figure below.\r\n$$\\includegraphics[width=2.5in]{images2/improper-integral-example-1}$$\r\n\\end{solution}\r\n\r\nFormalizing this example leads to the concept of an improper integral.\r\nThere are two ways to extend the Fundamental Theorem of Calculus.\r\nOne is to use an {\\bf infinite interval}, i.e., $[a,\\infty)$, $(-\\infty,b]$ or $(-\\infty,\\infty)$.\r\nThe second is to allow the interval $[a,b]$ to contain an infinite {\\bf discontinuity} of $f(x)$.\r\nIn either case, the integral is called an {\\bf improper integral}. \r\nOne of the most important applications of this concept is probability distributions. \r\n\r\nTo compute improper integrals, we use the concept of limits along with the Fundamental Theorem of Calculus.\r\n \r\n\\begin{definition}{Definitions for Improper Integrals}{Definitions for Improper Integrals}\r\nIf $f(x)$ is continuous on $[a,\\infty)$, then the improper integral of $f$ over $[a,\\infty)$ is:\r\n$$\\int_{a}^{\\infty} f(x)\\,dx:=\\lim_{R\\to\\infty}\\int_a^R f(x)\\,dx.$$ \r\nIf $f(x)$ is continuous on $(-\\infty,b]$, then the improper integral of $f$ over $(-\\infty,b]$ is:\r\n$$\\int_{-\\infty}^b f(x)\\,dx:=\\lim_{R\\to -\\infty}\\int_R^b f(x)\\,dx.$$\r\n\\end{definition}\r\n\r\nIf the limit exists and is a finite number, we say the improper integral {\\bf converges}. Otherwise, we say the improper integral {\\bf diverges}.\r\n\r\nTo get an intuitive (though not completely correct) interpretation of improper integrals, we attempt to analyze $\\ds\\int_a^\\infty f(x)\\,dx$ graphically. \r\n Here assume $f(x)$ is continuous on $[a,\\infty)$: \r\n$$\\includegraphics[width=5in]{images2/improper-integral-theory-2}$$ \r\n We let $R$ be a fixed number in $[a,\\infty)$. \r\n Then by taking the limit as $R$ approaches $\\infty$, we get the improper integral: \r\n$$\\int_a^\\infty f(x)\\,dx:=\\lim_{R\\to\\infty}\\int_a^R f(x)\\,dx.$$ \r\n We can then apply the Fundamental Theorem of Calculus to the last integral as $f(x)$ is continuous on the closed interval $[a,R]$.\r\n\r\nWe next define the improper integral for the interval $(-\\infty,~\\infty)$.\r\n\r\n\\begin{definition}{Definitions for Improper Integrals}{Definitions for Improper Integrals}\r\n If both $\\ds\\int_{-\\infty}^a f(x)\\,dx$ and $\\ds\\int_{a}^{\\infty} f(x)\\,dx$ are convergent, then the improper integral of $f$ over $(-\\infty,\\infty)$ is:\r\n$$\\int_{-\\infty}^{\\infty} f(x)\\,dx:=\\int_{-\\infty}^a f(x)\\,dx+\\int_{a}^{\\infty} f(x)\\,dx$$\r\n\\end{definition}\r\n\r\nThe above definition requires {\\bf both} of the integrals\r\n$$\\int_{-\\infty}^a f(x)\\,dx\\qquad\\mbox{and}\\qquad\\int_{a}^{\\infty} f(x)\\,dx$$\r\nto be convergent for $\\ds\\int_{-\\infty}^{\\infty} f(x)\\,dx$ to also be convergent. \r\n If {\\bf either} of $\\ds\\int_{-\\infty}^a f(x)\\,dx$ or $\\ds\\int_{a}^{\\infty} f(x)\\,dx$ is divergent, then so is $\\ds\\int_{-\\infty}^{\\infty} f(x)\\,dx$.\r\n\r\n\\begin{example}{Improper Integral}{Improper Integral}\r\nDetermine whether $\\ds\\int_1^\\infty\\frac{1}{x}\\,dx$ is convergent or divergent.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\n Using the definition for improper integrals we write this as:\r\n$$\t\\int_1^\\infty \\frac{1}{x}\\,dx= \\lim_{R\\to\\infty} \\int_1^R\\frac{1}{x}\\,dx\r\n\t= \\lim_{R\\to\\infty} \\ln|x|\\bigg|_1^R\r\n\t=\\lim_{R\\to\\infty} \\ln|R| - \\ln|1|\r\n\t= \\lim_{R\\to\\infty} \\ln|R|\r\n\t= +\\infty$$\r\n Therefore, the integral is {\\bf divergent}.\r\n\\end{solution}\r\n\r\n\\begin{example}{Improper Integral}{Improper Integral}\r\nDetermine whether $\\ds\\int_{-\\infty}^\\infty x\\sin(x^2)\\,dx$ is convergent or divergent.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\n We must compute both $\\ds\\int_0^\\infty x\\sin(x^2)\\,dx$ and $\\ds\\int_{-\\infty}^0 x\\sin(x^2)\\,dx$. \r\n Note that we don't have to split the integral up at $0$, any finite value $a$ will work. \r\n First we compute the indefinite integral. \r\n Let $u=x^2$, then $du=2x\\,dx$ and hence, \r\n$$\\int x\\sin(x^2)\\,dx=\\frac{1}{2} \\int \\sin(u)\\,du=-\\frac{1}{2}\\cos(x^2)+C$$\r\nUsing the definition of improper integral gives: \r\n$$\\int_0^\\infty x\\sin(x^2)\\,dx = \\lim_{R\\to\\infty} \\int_0^R x\\sin(x^2)\\,dx \r\n=\\lim_{R\\to\\infty} \\left[-\\frac{1}{2}\\cos(x^2)\\right] \\bigg|_0^R \r\n= -\\frac{1}{2} \\lim_{R\\to\\infty} \\cos(R^2) +\\frac{1}{2}$$\r\nThis limit does not exist since $\\cos x$ {\\bf oscillates} between $-1$ and $+1$. \r\n In particular, $\\cos x$ does not approach any particular value as $x$ gets larger and larger. \r\n Thus, $\\ds\\int_0^\\infty x\\sin(x^2)\\,dx$ diverges, and hence, $\\ds\\int_{-\\infty}^\\infty x\\sin(x^2)\\,dx$ diverges.\r\n\\end{solution}\r\n\r\nWhen there is a discontinuity in $[a,b]$ or at an endpoint, then the improper integral is as follows.\r\n\r\n\\begin{definition}{Definitions for Improper Integrals}{Definitions for Improper Integrals}\r\nIf $f(x)$ is continuous on $(a,b]$, then the improper integral of $f$ over $(a,b]$ is: \r\n$$\\int_a^b f(x)\\,dx:=\\lim_{R\\to a^+}\\int_R^b f(x)\\,dx.$$ \r\nIf $f(x)$ is continuous on $[a,b)$, then the improper integral of $f$ over $[a,b)$ is: \r\n$$\\int_a^b f(x)\\,dx:=\\lim_{R\\to b^-}\\int_a^R f(x)\\,dx.$$\r\n\\end{definition}\r\n\r\nIf the limit above exists and is a finite number, we say the improper integral {\\bf converges}.\r\nOtherwise, we say the improper integral {\\bf diverges}. \r\n\r\nWhen there is a discontinuity in the interior of $[a,b]$, we use the following definition.\r\n\r\n\\begin{definition}{Definitions for Improper Integrals}{Definitions for Improper Integrals}\r\nIf $f$ has a discontinuity at $x=c$ where $c\\in[a,b]$, and both\r\nboth $\\ds\\int_a^c f(x)\\,dx$ and $\\ds\\int_c^b f(x)\\,dx$ are convergent, then $f$ over $[a,b]$ is: \r\n$$\\int_a^b f(x)\\,dx:=\\int_a^c f(x)\\,dx+\\int_c^b f(x)\\,dx$$\r\n\\end{definition}\r\n\r\nAgain, we can get an intuitive sense of this concept by analyzing $\\ds\\int_a^b f(x)\\,dx$ graphically. \r\nHere assume $f(x)$ is continuous on $(a,b]$ but discontinuous at $x=a$: \r\n$$\\includegraphics[width=5in]{images2/improper-integral-theory-1}$$ \r\n We let $R$ be a fixed number in $(a,b)$. \r\n Then by taking the limit as $R$ approaches $a$ from the {\\bf right}, we get the improper integral: \r\n$$\\int_a^b f(x)\\,dx:=\\lim_{R\\to a^+}\\int_R^b f(x)\\,dx.$$ \r\n Now we can apply FTC to the last integral as $f(x)$ is continuous on $[R,b]$.\r\n\r\n\\begin{example}{A Divergent Integral}{A Divergent Integral}\r\nDetermine if $\\ds\\int_{-1}^1\\frac{1}{x^2}\\,dx$ is convergent or divergent.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\n The function $f(x)=1/x^2$ has a discontinuity at $x=0$, which lies in $[-1,1]$. \r\n We must compute $\\ds\\int_{-1}^0 \\frac{1}{x^2}\\,dx$ and $\\ds\\int_0^1 \\frac{1}{x^2}\\,dx$. Let's start with $\\ds\\int_0^1 \\frac{1}{x^2}\\,dx$: \r\n$$\\int_0^1 \\frac{1}{x^2}\\,dx = \\lim_{R\\to 0^+} \\int_R^1 \\frac{1}{x^2} \\,dx = \\lim_{R\\to 0^+} -\\frac{1}{x}\\bigg|_R^1\r\n= -1 + \\lim_{R\\to 0^+} \\frac{1}{R}$$\r\nwhich diverges to $+\\infty$.\r\n Therefore, $\\ds\\int_{-1}^1\\frac{1}{x^2}\\,dx$ is {\\bf divergent} since one of $\\ds\\int_{-1}^0 \\frac{1}{x^2}\\,dx$ and $\\ds\\int_0^1 \\frac{1}{x^2}\\,dx$ is divergent.\r\n\\end{solution}\r\n\r\n\r\n% % % % % % %\r\n% % The following example 'Integral of the Logarithm' can be replaced with the next example if Integration by Parts is not included in the text adaptation.\r\n\\begin{example}{Integral of the Logarithm}{Integral of the Logarithm}\r\nDetermine if $\\ds\\int_0^1 \\ln x \\,dx$ is convergent or divergent. Evaluate it if it is convergent.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\nNote that $f(x)=\\ln x$ is discontinuous at the endpoint $x=0$. \r\nWe first use integration by parts to compute $\\ds\\int\\ln x\\,dx$.\r\nWe let $u=\\ln x$ and $dv=dx$.\r\nThen $du=(1/x)dx$, $v=x$, giving:\r\n\\begin{eqnarray*}\r\n\\int \\ln x\\,dx &=& \\ds x\\ln x-\\int x\\cdot\\frac{1}{x}\\,dx\\\\\r\n&=& x\\ln x-\\int 1\\,dx\\\\\r\n&=& x\\ln x-x+C\\\\\r\n\\end{eqnarray*}\r\n Now using the definition of improper integral for $\\ds\\int_0^1 \\ln x \\,dx$: \r\n$$\r\n\t\\int_0^1 \\ln x \\,dx = \\lim_{R\\to 0^+} \\int_R^1 \\ln x\\,dx \r\n\t= \\lim_{R\\to 0^+} (x\\ln x-x)\\bigg|_R^1 \r\n\t%= \\lim_{R\\to 0^+} \\left(\\left[\\ln 1-1\\right] - \\left[R\\ln R-R\\right]\\right)\r\n\t= -1 - \\lim_{R\\to 0^+}(R\\ln R) + \\lim_{R\\to 0^+}R \r\n$$\r\nNote that $\\ds\\lim_{R\\to 0^+}R=0$. \r\nWe next compute $\\ds\\lim_{R\\to 0^+}(R\\ln R)$.\r\nFirst, we rewrite the expression as follows:\r\n$$\\lim_{x\\to0^+}(R\\ln R)=\\lim_{R\\to0^+}\\frac{\\ln R}{1/R}.$$\r\nNow the limit is of the indeterminate type $(-\\infty)/(\\infty)$ and l'H\\^opital's Rule can be applied. \r\n$$\\lim_{R\\to0^+}(R\\ln R) =\\lim_{R\\to0^+}\\frac{\\ln R}{1/R}\r\n=\\lim_{R\\to0^+}\\frac{1/R}{-1/R^2}\r\n=\\lim_{R\\to0^+}-\\frac{R^2}{R}\r\n=\\lim_{R\\to0^+}(-R)\r\n=0$$\r\nThus, $\\ds\\lim_{R\\to 0^+}(R\\ln R)=0$.\r\nThus\r\n$$\\int_0^1 \\ln x \\,dx = -1,$$ \r\nand the integral is convergent to $-1$. \r\n\r\nGraphically, one might interpret this to mean that the net area under $\\ln x$ on $[0,1]$ is $-1$ (the area in this case lies below the $x$-axis). \r\n$$\\includegraphics[width=2.5in]{images2/improper-integral-example-2}$$\r\n\\end{solution}\r\n\r\n\\begin{example}{Integral of a Square Root}{IntSquareRoot}\r\nDetermine if $\\ds\\int_0^4\\frac{dx}{\\sqrt{4-x}}$ is convergent or divergent. Evaluate it if it is convergent.\r\n\\end{example}\r\n\\begin{solution}\r\nNote that $\\frac{1}{\\sqrt{4-x}}$ is discontinuous at the endpoint $x=4$. We use a $u$-substitution to compute $\\int \\frac{dx}{\\sqrt{4-x}}$. We let $u=4-x$, then $du=-dx$, giving:\r\n\\begin{align*}\r\n\\ds\\int\\frac{dx}{\\sqrt{4-x}}&=\\int-\\frac{du}{u^{1/2}}\t\\\\\r\n&=\\int -u^{-1/2}\\,du\t\\\\\r\n&=-2(u)^{1/2}+C\t\\\\\r\n&=-2\\sqrt{4-x}+C\r\n\\end{align*}\r\nNow using the definition of improper integrals for $\\ds\\int_0^4\\frac{dx}{\\sqrt{4-x}}$:\r\n\\[\\ds\\int_0^4\\frac{dx}{\\sqrt{4-x}}=\\lim_{R\\to4^{-}}(-2\\sqrt{4-x})\\bigg|_0^R=\\lim_{R\\to4^{-}}-2\\sqrt{4-R}+2\\sqrt{4}=4\\]\r\n\\end{solution}\r\n\r\n\\begin{example}{Improper Integral}{ImpIntExample}\r\nDetermine if $\\ds\\int_{1}^{2}\\dfrac{dx}{\\left( x-1\\right) ^{1/3}}$ is\r\nconvergent or divergent. Evaluate it if it is convergent.\r\n\\end{example}\r\n\\begin{solution}\r\nNote that $f\\left( x\\right) =\\dfrac{1}{\\left( x-1\\right) ^{1/3}}$\r\nis discontinuous at the endpoint $x=1.$ We first use substitution to find\r\n$\\ds\\int \\dfrac{dx}{\\left( x-1\\right) ^{1/3}}.$ We let $u=x-1.$ Then $du=dx,$\r\ngiving%\r\n\\begin{equation*}\r\n\\int \\dfrac{dx}{\\left( x-1\\right) ^{1/3}}=\\int \\frac{du}{u^{1/3}}=\\int\r\nu^{-1/3}du=\\frac{3}{2}u^{2/3}+C=\\frac{3}{2}\\left( x-1\\right) ^{2/3}+C.\r\n\\end{equation*}%\r\nNow using the definition of improper integral for $\\ds\\int_{1}^{2}\\dfrac{dx}{%\r\n\t\\left( x-1\\right) ^{1/3}}:$%\r\n\\begin{equation*}\r\n\\int_{1}^{2}\\dfrac{dx}{\\left( x-1\\right) ^{1/3}}=\\lim_{R\\rightarrow\r\n\t1^{+}}\\int_{R}^{2}\\dfrac{dx}{\\left( x-1\\right) ^{1/3}}=\\left.\r\n\\lim_{R\\rightarrow 1^{+}}\\frac{3}{2}\\left( x-1\\right) ^{2/3}\\right\\vert\r\n_{R}^{2}=\\frac{3}{2}-\\lim_{R\\rightarrow 1^{+}}\\frac{3}{2}\\left( R-1\\right)\r\n^{2/3}=\\frac{3}{2},\r\n\\end{equation*}%\r\nand the integral is convergent to $\\frac{3}{2}.$ Graphically, one might\r\ninterpret this to mean that the net area under $\\dfrac{1}{\\left( x-1\\right)\r\n\t^{1/3}}$ on $\\left[ 1,2\\right] $ is $\\frac{3}{2}$.\r\n\r\n$$\\includegraphics[width=8cm]{images/improper-int-example}$$\r\n\\end{solution}\r\n\r\nThe following test allows us to determine convergence/divergence information about improper integrals that are hard to compute by comparing them to easier ones. \r\nWe state the test for $[a,\\infty)$, but similar versions hold for the other improper integrals.\r\n\r\n\\begin{formulabox}[The Comparison Test]\r\nAssume that $f(x)\\geq g(x)\\geq 0$ for $x\\geq a$.\r\n\\begin{enumerate}[(i)]\r\n\\item\tIf $\\ds\\int_a^\\infty f(x)\\,dx$ {\\bf converges}, then $\\ds\\int_a^\\infty g(x)\\,dx$ also {\\bf converges}.\r\n\\item\tIf $\\ds\\int_a^\\infty g(x)\\,dx$ {\\bf diverges}, then $\\ds\\int_a^\\infty f(x)\\,dx$ also {\\bf diverges}. \r\n\\end{enumerate}\r\n\\end{formulabox}\r\n\r\nInformally, (i) says that if $f(x)$ is larger than $g(x)$, and the area under $f(x)$ is finite (converges), then the area under $g(x)$ must also be finite (converges). \r\nInformally, (ii) says that if $f(x)$ is larger than $g(x)$, and the area under $g(x)$ is infinite (diverges), then the area under $f(x)$ must also be infinite (diverges). \r\n\r\n$$\\includegraphics[width=6in]{images2/improper-integral-theory-3}$$\r\n\r\n\\begin{example}{Comparison Test}{Comparison Test}\r\nShow that $\\ds\\int_2^\\infty \\frac{\\cos^2x}{x^2} \\,dx$ converges.\r\n\\end{example} \r\n\r\n\\begin{solution}\r\nWe use the comparison test to show that it converges. \r\nNote that $0\\leq \\cos^2x\\leq 1$ and hence \r\n$$0 \\leq\\frac{\\cos^2x}{x^2}\\leq\\frac{1}{x^2}.$$\r\nThus, taking $f(x)=1/x^2$ and $g(x)=\\cos^2x / x^2$ we have $f(x)\\geq g(x)\\geq 0$. \r\nOne can easily see that $\\ds\\int_2^\\infty \\frac{1}{x^2}\\,dx$ converges. \r\nTherefore, $\\ds\\int_2^\\infty \\frac{\\cos^2x}{x^2} \\,dx$ also converges.\r\n\\end{solution}\r\n\r\n\r\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\n\\Opensolutionfile{solutions}[ex]\r\n\\section*{Exercises for Section \\ref{sec:ImproperIntegrals}}\r\n\r\n\\begin{enumialphparenastyle}\r\n\r\n\\begin{ex}\r\n\tDetermine whether $\\ds\\int_1^{\\infty}\\frac{1}{x^2}\\,dx$ is convergent or divergent.\r\n\t\\begin{sol}\r\n\t\tConverges to 1.\r\n\t\\end{sol}\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tDetermine whether $\\ds\\int_e^{\\infty}\\frac{1}{x\\sqrt{\\ln x}}\\,dx$ is convergent or divergent.\r\n\t\\begin{sol}\r\n\t\tDiverges.\r\n\t\\end{sol}\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tEvaluate the improper integral $\\ds\\int_0^{\\infty}e^{-3x}\\,dx$.\r\n\t\\begin{sol}\r\n\t\t1/3\r\n\t\\end{sol}\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tDetermine if $\\ds\\int_1^e\\frac{1}{x(\\ln x)^2}\\,dx$ is convergent or divergent. Evaluate it if it is convergent.\r\n\t\\begin{sol}\r\n\t\tDivergent.\r\n\t\\end{sol}\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tShow that $\\ds\\int_0^{\\infty}e^{-x}\\sin^2\\left(\\frac{\\pi x}{2}\\right)\\,dx$ converges.\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tEvaluate $\\ds\\int_{-\\infty}^{\\infty}\\frac{1}{x^2+1}\\,dx$ and $\\ds\\int_{-\\infty}^{\\infty}\\frac{x}{x^2+1}\\,dx$.\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tDetermine whether the following improper integrals are convergent or divergent. Evaluate those that are convergent.\r\n\t\\begin{enumerate}\r\n\t\t\\item\t$\\int_{0}^{\\infty}\\dfrac{1}{x^2+1}\\,dx$\r\n\t\t\\item\t$\\int_{0}^{\\infty}\\dfrac{x}{x^2+1}\\,dx$\r\n\t\t\\item\t$\\int_{0}^{\\infty}e^{-x}(\\cos x+\\sin x)\\,dx$. [Hint: What is the derivative of $-e^{-x}\\cos x?$]\r\n\t\t\\item\t$\\int_{0}^{\\pi/2}\\sec^{2}x\\,dx$\r\n\t\t\\item\t$\\int_{0}^{4}\\dfrac{1}{(4-x)^{2/5}}\\,dx$\r\n\t\\end{enumerate}\r\n\t\\begin{sol}\r\n\t\t\\begin{enumerate}\r\n\t\t\t\\item\t$\\pi/2$\r\n\t\t\t\\item\tdivergent (to $\\infty$)\r\n\t\t\t\\item\t1\r\n\t\t\t\\item\tdivergent (to $\\infty$)\r\n\t\t\t\\item\t$\\frac{5}{3}(4^{3/5})$\r\n\t\t\\end{enumerate}\r\n\t\\end{sol}\r\n\\end{ex}\r\n\r\n\\begin{ex}\r\n\tProve that the integral $\\int_{1}^{\\infty}\\dfrac{1}{x^p}\\,dx$ is convergent if $p>1$ and divergent if $00$. Find all values of $p$ for which $\\int_{0}^{1}\\dfrac{1}{x^p}\\,dx$ converges.\r\n\t\\begin{sol}\r\n\t\t$0\n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section*{Question 7}\n\nSuppose that, instead of sorting an array, we just require that the elements increase on average.\nMore precisely, we call an $n$-element array $A$, \\textbf{k-sorted} if, for all $i = 1, 2, ..., n-k$ Equation \\ref{eq71} holds.\n\n\\begin{equation}\\label{eq71}\n\\frac{\\sum_{j=i}^{i + k - 1} A[j]}{k} \\leq \\frac{\\sum_{j=i+1}^{i + k} A[j]}{k}\n\\end{equation}\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item What does it mean for an array to be 1-sorted?\n\\item Give a permutation of the numbers $1$, $2$, $3$, ..., $10$, that is $2$-sorted but not sorted.\n\\item Prove that an $n$-element array is $k$-sorted if and only if $A[i] \\leq A[i+k]$ for all $i = 1, 2, ..., n-k$.\n\\item Give an algorithm that $k$-sorts an $n$-element array in $\\mathcal{O}(n \\log (n/k))$ time.\n\\end{enumerate}\n\n\\subsection*{Solution}\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item Using Equation \\ref{eq71} and substituting $k$ with 1 would give Equation \\ref{eq72}.\n\\begin{equation}\\label{eq72}\nA[i] \\leq A[i+1]\n\\end{equation}\nTherefore, any array that is 1-sorted as actually sorted in ascending order.\n\\item One possible permutation of numbers $1$ to $10$ to result in a 2-sorted array is $1$, $6$, $2$, $7$, $3$, $8$, $4$, $9$, $5$ and $10$.\nThis is due that for any $1 \\leq i < 9$, Equation \\ref{eq73} would be valid.\n\\begin{equation}\\label{eq73}\n\\frac{\\sum_{j=i}^{i + 1} A[j]}{2} \\leq \\frac{\\sum_{j=i+1}^{i + 2} A[j]}{2}\n\\end{equation}\n\\item Proof is given by induction on length of array $n$.\n\\begin{enumerate}\n\\item[] \\textit{Base case}: ($n = 2$)\\\\\nWhen length of array is 2, $k = 1$, Equation \\ref{eq71} dictates that $A[i] \\leq A[i+1]$ which confirms the hypothesis.\n\\item[] \\textit{Induction Step}\\\\\nWe form inductive hypothesis as for any array of length $n$, the array is $k$-sorted if and only if $A[i] \\leq A[i+k]$.\nUsing the inductive hypothesis, we will show that the statement would hold true for arrays of length $n + 1$.\n\nAssuming that the $(n+1)$-element array is $k$-sorted, Equation \\ref{eq71} would be available to use.\nSubtracting the inductive hypothesis from Equation \\ref{eq71} would directly lead to $A[j] \\leq A[j+k]$ which is the right-side of the equation.\n\nWe should as well show that using $A[i] \\leq A[i+1]$ and adding the inductive hypothesis, Equation \\ref{eq71} would directly be obtained which shows the array is $k$-sorted.\n\\end{enumerate}\n\\item Inspired by the proposed $2$-sorted array of numbers $1$ to $10$, the algorithm to $k$-sort an $n$-element array is given as follows.\n\nWe first divide the array into $k$ subarrays.\nThis clearly is done in constant time and has no effect on runtime of the algorithm.\nThen, we sort all $k$ subarrays using Merge Sort.\nSince each array has length $\\frac{n}{k}$, the runtime for sorting $k$ subarrays would be $\\mathcal{O}(k\\frac{n}{k}\\log(\\frac{n}{k}))$.\nAll that remains is to reconstruct the final array by picking one element from $k$ subarrays, one in turn.\nThis leading to a run time of $\\mathcal{O}(n)$.\nTherefore the total runtime of this algorithm would be $\\mathcal{O}(n \\log(\\frac{n}{k}) + n ) = \\mathcal{O}(n \\log(\\frac{n}{k}) )$.\n\\end{enumerate}\n", "meta": {"hexsha": "ff9602043f8201cbea0a3f1b7d8fe982c6cc857d", "size": 3472, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "umb-cs624-2015s/src/tex/hw02/hw02q07.tex", "max_stars_repo_name": "ghorbanzade/beacon", "max_stars_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-11-13T20:00:10.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-01T11:16:51.000Z", "max_issues_repo_path": "umb-cs624-2015s/src/tex/hw02/hw02q07.tex", "max_issues_repo_name": "ghorbanzade/beacon", "max_issues_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "umb-cs624-2015s/src/tex/hw02/hw02q07.tex", "max_forks_repo_name": "ghorbanzade/beacon", "max_forks_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-09-20T05:58:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T17:18:05.000Z", "avg_line_length": 57.8666666667, "max_line_length": 173, "alphanum_fraction": 0.6667626728, "num_tokens": 1094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835411997898, "lm_q2_score": 0.8376199572530448, "lm_q1q2_score": 0.6998176880653905}} {"text": "\\chapter{\\proj Wavelet and Multifractal Analysis}\n\\label{ch_fractal}\n\\index{fractal}\n% \\chapterhead{Programs}\n\\markright{Wavelet and Multifractal Analysis}\n\n\\section{Fractal}\n\\subsection{Introduction}\nThe word ``fractal'' was introduced by Mandelbrot \n(1977) \\cite{frac:mandel83}, and\ncomes from the Latin word {\\em fractus} which means ``to break''.\nAccording to Mandelbrot, a fractal is an object which has \na greater dimension than its topological dimension. A typical\nfractal is the Cantor set.\n \n\\subsubsection*{Example: the triadic Cantor set} \n\\label{triadic Cantor set}\nThe Cantor set is built in the following way: considering a\nsegment of dimension $L$, we separate it into three equal parts,\nand suppress the middle part. There remain two segments of size\n${L \\over 3}$. Repeating the process on both segments, we get \nfour segments of size ${3^{-2}L}$. After $n$ iterations, we have\n$2^n$ segments of size ${3^{-n}L}$. The Cantor set is obtained \nwhen $n$ tends to infinity. The set has the following properties:\n\\begin{enumerate}\n\\item It is self-similar.\n\\item It has a fine structure, i.e.\\ detail on arbitrary small\nscales.\n\\item It is too irregular to be described in traditional \ngeometrical language, both locally and globally.\n\\item It is obtained by successive iterations.\n\\item It has an infinite number of points but its length is null.\n\\end{enumerate}\nThese properties define in fact a fractal object. \n\nA real fractal does not exist in the nature, and we always need\nto indicate at which scales we are talking about fractality.\n\n\\subsection{The Hausdorff and Minkowski measure} \n\\label{Hausdorff}\n\\subsubsection*{Measure}\nAn object dimension describes how an object $F$ fills the space.\nA simple manner of measuring the length of\ncurves, the area of surfaces or the volume of objects is to divide the \nspace into small boxes (segment in one dimension, surface in 2D, and cubes\nin 3D) of diameter $\\delta$ as shown in Figure \\ref{fig_mes} \n\\cite{frac:falconer90}. These boxes\nare chosen so that their diameter is not greater than a given size $\\delta$,\nwhich corresponds to the measure resolution.\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\\psfig{figure=fig_frac1.ps,bbllx=5cm,bblly=12cm,bburx=16cm,bbury=18cm,height=6cm,width=12cm,clip=}\n}}\n\\caption{Measuring the ``size'' of curves (Feder 1988).}\n\\index{Mallat's multiresolution}\n\\label{fig_mes}\n\\end{figure}\n\nWe consider now the quantity:\n\\begin{eqnarray}\nL_{\\delta}^d(F) = \\sum diam(B_i)^d\n\\end{eqnarray}\nwhere $d$ is a real number, $diam(B_i)$ is the diameter of the box $i$.\n$L_{\\delta}^s(F)$ represents an estimation of the size of $F$ at the resolution\n$\\delta$. Depending on the choice of the boxes, the measure is more or less \ncorrect. Generally, it is easier to manipulate the Minkowski-Bouligand \nmeasure which fixes all the boxes to the same size $\\delta$. Using this\nmeasure, the size of $F$ is given by:\n\\begin{eqnarray}\nM^s(F) = \\lim_{\\delta \\rightarrow 0} \\sum diam(B_i)^s = \\delta^s N_B(\\delta)\n\\end{eqnarray}\nwhere $N_B$ is the number of boxes needed to cover the object $F$.\nThen the curves of length $L_\\ast$ in Figure~\\ref{fig_mes} \ncan be measured by finding the number $N_B(\\delta)$ of line segments\n (respectively squares and cubes for the second and third object)\nof length $\\delta$ needed to cover the object. The three sizes are:\n\\begin{eqnarray}\nL & = M^1(F) = & N_B(\\delta) \\delta^1 \\mathop\\rightarrow_{\\delta \\to 0} \n L_\\ast \\delta^0 \\nonumber \\\\\nA & = M^2(F) = & N_B(\\delta) \\delta^2 \\mathop\\rightarrow_{\\delta \\to 0}\n L_\\ast \\delta^1 \\nonumber \\\\\nV & = M^3(F) = & N_B(\\delta) \\delta^3 \\mathop\\rightarrow_{\\delta \\to 0}\n L_\\ast \\delta^2 \n\\end{eqnarray}\n\n\\subsection{The Hausdorff and Minkowski dimension}\n\nThe Hausdorff dimension $d_H$ of the set $F$ is the {\\it\ncritical dimension} for which the measure $H^d(F)$ jumps from infinity\nto zero:\n\\begin{eqnarray}\n H^d(F) = \\left\\{ \\begin{array}{ll}\n 0, & d > d_H, \\\\\n \\infty, & d < d_H.\n \\end{array}\n \\right.\n \\label{eq-ch2-1}\n\\end{eqnarray}\nBut $H^{d_H}(F)$ can be finite or infinite. For a simple set \n(segment, square, cube),\nHausdorff dimension is equal to the topological dimension (i.e.\\ 1, 2, \nor 3). This is not true for a more complex set, such as the Cantor set.\nMinkowski dimension $d_M$ (and $d_H (F) \\le d_M(F)$) is defined \nin a similar way using Minkowski measure.\n\nBy definition, we have:\n\\begin{eqnarray}\nM^{d_M} = \\lim_{\\delta \\rightarrow 0} \\delta^{d_M} N_B(\\delta)\n\\end{eqnarray}\nWhen $\\delta \\rightarrow 0$, we have $d_M \\ln M= d_M\\ln{\\delta} + \\ln{N_N(\\delta)}$. If\n$M$ is finite, the Minkowski dimension, also called box-counting,\n can be defined by\n\\begin{eqnarray}\n d_M = \\lim_{\\delta \\rightarrow 0} {\\ln{N_B(\\delta)} \\over - \\ln{\\delta}}\n\\end{eqnarray}\n\nIn the case of the Cantor set, at iteration $n$, we have $2^n$ segments\nof size $3^{-n}$ ($\\delta = 3^{-n}$). When $n \\rightarrow \\infty$, we have\n\\begin{eqnarray}\nd_M (Cantor) = {\\ln{2^n} \\over {- \\ln{3^{-n}}}} = {\\ln{2} \\over \\ln{3}}\n\\end{eqnarray}\n\n\\section{Multifractality}\nThe multifractal picture is a refinement and\ngeneralization of the fractal properties\nthat arise naturally in the case of self-similar distributions.\nThe singularity spectrum $f(\\alpha)$ can be introduced as a quantity\nwhich characterizes the degree of regularity and homogeneity of a \nfractal measure.\n\n\\subsection{Definition}\n\\subsubsection*{H\\\"older exponent}\nA multifractal measure describes a non-homogeneous set $A$. \nSuch a measure is called multifractal if it is everywhere self-similar, i.e.\\\nif the measure varies locally as a power law, at any point of A. \nDenoting $\\mu$ a measure, we call the H\\\"older exponent or singularity exponent\nat $x_0$ the limit\n\\be\n\\alpha(x_0) = \\lim_{\\delta \\rightarrow 0} { \\ln{\\mu(B_{x_0}(\\delta))} \n \\over \\ln{\\delta} }\n\\ee\nwhere $B_{r_0}$ is a box centered at $r_0$ of size $\\delta$.\nWe have:\n\\be\n\\mu(B_{x_0}(\\delta)) \\propto \\delta^{\\alpha(x_0)}\n\\ee\nThe smaller the value $\\alpha(x_0)$, the less the measure is regular around\n$x_0$. For example, if $\\mu$ corresponds to a Dirac distribution centered \nat $0$, \nthen $\\alpha(0) = 0$, and if $\\mu$ corresponds to a Gaussian distribution, then\n$\\alpha(0) = -1$.\n\n\\subsubsection*{Singularity spectrum}\n\nThe singularity spectrum, associated with a measure $\\mu$, is the function\nwhich associates with $\\alpha$ the fractal dimension of \nany point $x_0$ such that\n$\\alpha(x_0) = \\alpha$:\n\\be\nf(\\alpha) = d_F(\\{ x_0 \\in A \\mid \\alpha(x_0) = \\alpha \\})\n\\ee\n The function $f(\\alpha)$ is usually (\\cite{frac:paladin86})\na single-humped\nfunction with the maximum at $max_{\\alpha} f(\\alpha) = D$, \nwhere $D$ is the dimension of the support. In the case\nof a single fractal, the function $f(\\alpha)$ is reduced to\na single point: $f(\\alpha) = \\alpha = D$.\n\nThe singularity spectrum describes statistically the $\\alpha$ exponent\ndistribution on the measure support. For example, if we split the\nsupport into boxes of size $\\delta$, then the number of boxes with a\nmeasure varying as $\\delta^\\alpha$ for a given $\\alpha$ is\n\\be\nN_\\alpha(\\delta) \\propto \\delta^{-f(\\alpha)}\n\\ee\n$f(\\alpha)$ describes the histogram of $N_\\alpha(\\delta)$ when\n$\\delta$ is small. A measure is homogeneous if its singularity spectrum\nis concentrated in a single point. If $f(\\alpha)$ is large, the\nmeasure is multifractal.\n\n\\subsubsection*{Multifractal quantities}\nFrom a practical point of view one does not determine directly the\nspectrum of exponents $[f(\\alpha), \\alpha]$; it is more convenient to\ncompute its Legendre transformation $[\\tau(q),q]$ given by\n\\be\n\\left\\{\n\\begin{array}{l}\n\\label{mf12}\nf(\\alpha) = q \\cdot \\alpha - \\tau(q)\\\\\n\\alpha = \\frac{d\\tau(q)}{dq}\n\\end{array}\n\\right.\n\\ee\nIn the case of a simple fractal one has $\\alpha=f(\\alpha)=D$. In\nterms of the Legendre transformation this corresponds to\n\\be\n\\tau(q) = D(q-1)\n\\ee\ni.e.\\ the behavior of $\\tau(q)$ versus $q$ is a straight line with\ncoefficient given by the fractal dimension.\n\n\n\\subsection{Generalized fractal dimension}\n\n\\subsubsection*{Definition}\nThe generalized fractal dimension, also called Reyni dimension of\norder $q$, is given by:\n\\be\nD_q = { \\tau(q) \\over q -1 }\n\\ee\n$D_0$ is also called capacity dimension, and coincides with the\nHausdorff dimension. Dimensions $D_1$, and $D_2$ are respectively \ncalled information and correlation dimension.\n \n\\subsubsection*{Partition function}\nThe partition function $Z$ is defined by:\n\\be\nZ(q,\\delta) = \\sum_{i=1}^{N(\\delta)} \\mu_i^q(\\delta)\n\\ee\nwhere we denote $\\mu_i(\\delta) = \\mu(B_i(\\delta))$. If the measure\n$\\mu$ is multifractal, $Z$ follows a power law in the limit $\\delta\n\\rightarrow 0$.\n\\be\nZ(q,\\delta) \\propto \\delta^{\\tau(q)} \n% \\label{partition function}\n\\ee\n\nThe box-counting method consists of calculating the partition\nfunction, to derive from $Z$ $\\tau(q)$, and to obtain, the multifractal\nspectrum by a Legendre transform.\n\n\\section{Wavelets and Multifractality}\n\n\\subsection{Singularity analysis}\n\nLet $f(x)$ be the input signal, $x_0$ the singularity location,\n$\\alpha(x_0)$ the H\\\"older exponent at the singularity point $x_0$ and\n$n$ the degree of Taylor development such that\n$n\\leq\\alpha(x_0)n$ vanishing moments, then we have\nfor the wavelet transform of $f(x)$ at $x_0$ when the scale goes to 0\n($\\psi$ is orthogonal to polynomials up to order $n$):\n\n\\be\n\\lim_{scale s \\rightarrow 0^+} T_\\psi[f](x_0,s) \\sim {a^{\\alpha(x_0)}}\n% \\label{Mallat Hwang}\n\\ee\n\nOne can prove that if $f$ is $C^\\infty$, then we have\n\n\\be\n\\lim_{scale s \\rightarrow 0^+} T_\\psi[f](x_0,s) \\sim {a^{n_\\psi}}\n\\ee\n\nThus, we have \n\\begin{eqnarray}\n\\left\\{ \\begin{array}{ll}\n \tT_\\psi[f] \\sim s^{n_\\psi} & \\mbox{ where the signal f is regular} \\\\\n\tT_\\psi[f] \\sim s^{\\alpha} (>>s^{n_\\psi}) & \\mbox{ around a singular zone}\n\t\\end{array}\n\t\\right.\n\\end{eqnarray} \n\nFor a fixed scale s, $T_\\psi[f](.,s)$ will be greater when the signal is\nsingular. This local maximum is organized in maxima lines (function of s) \nwhich\nconverges, when s goes to 0, to a singularity of the signal.\nMallat and Hwang (1992) demonstrate that to recover the H\\\"older exponent \n$\\alpha(x_0)$ at $x_0$, one need only study the wavelet transform along these\nlines of maxima which converge (when the scale goes to 0) towards the singularity\npoint $x_0$.\n\nAlong this maxima line l we have\n\\be\nT_\\psi[f](b,s) \\sim a^{\\alpha(x_0)}, \\,\\,(b,a)\\in l, s\\rightarrow 0^+\n\\ee\nFigure~\\ref{fig_2} displays the function $f(x)=K(x-x_0)^{0.4}$ with a singular\npoint at $x_0$. The H\\\"older exponent at $x_0$ is equal to 0.4.\nIn Figure~\\ref{fig_3}, we display the wavelet transform of $f(x)$ with a wavelet\n$\\psi$ which is the first derivative of a Gaussian. In Figure~\\ref{fig_4}, we\ndisplay $log_2 {|T_\\psi[f](x,s)|}$ as a function of $\\log_2(s)$. \n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\psfig{figure=singul.ps,height=5cm,width=8cm}\n}}\n\\caption{Function $f(x)=K(x-x_0)^{0.4}$ \\cite{frac:arneodo90}.}\n\\label{fig_2}\n\\end{figure}\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\psfig{figure=WT.ps,height=5cm,width=8cm}\n}}\n\\caption{\nWavelet transform of $f(x)$ with a wavelet $\\psi$ which is the first \nderivative of\na Gaussian. The small scales are at the top. The maxima line converges to the\nsingularity point at $x_0$.}\n\\label{fig_3}\n\\end{figure}\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\psfig{figure=holder_expo.ps,height=5cm,width=8cm}\n}}\n\\caption{\nEstimation of the H\\\"older exponent $\\alpha(x_0)$ at \n$x_0$ by computing the slope of $\\log_2 {|T_\\psi[f](x,s)|}$ versus\n$\\log_2(s)$ along a maxima line which converges to $x_0$.}\n\\index{Mallat's multiresolution}\n\\label{fig_4}\n\\end{figure}\n\nWhen we compute the slope of the curve $log_2 {|T_\\psi[f](x,s)|}$\nversus $\\log_2(s)$ along a maxima line which converges at $x_0$, \nwe obtain an estimation of the H\\\"older exponent (in this case\n$\\alpha(x_0) \\approx 0.4$\nwhich corresponds to the theoretical value).\n\n\n\\subsection{Wavelet transform of multifractal signals }\n\nThe estimation of the H\\\"older exponent by this method \n becomes inaccurate\nin the case of multifractal signals \\cite{frac:arneodo90}. \nWe need to use a more global\nmethod. One can define the wavelet-based partition function by\n\n\\be\nZ(q,s) = \\sum_{b_i} |T_\\psi[\\mu](b_i,s)|^q\n\\ee\nwhere ${(b_i,s)}_i$ are all local maxima at scale $s$.\n\nLet $\\tau(q)$ be the scaling exponent. We can prove that we have\n\\be\nZ(q,s) \\sim a^{\\tau(q)}\n\\ee\n\nWe can then calculate the singularity spectrum $D(\\alpha)$ by its Legendre\ntransformation \n\\be\nD(\\alpha) = \\underset{q}{\\min} (q\\alpha-\\tau(q))\n\\ee\n \nThis method is called the {\\em Wavelet Transform Modulus Maxima} (WTMM).\n \n\n\\subsection{Numerical applications of WWTM method}\n\nThe calculation of the singularity spectrum of signal f proceeds as follows:\n\\begin{list}{--}{\\itemindent=-5mm \\itemsep=-1mm}\n\\item compute the wavelet transform and the modulus maxima $T_\\psi[f]$ for\nall $(s,q)$. We chain all maxima across scale\nlines of maxima\n\\item compute the partition \nfunction $Z(q,s) = \\sum_{b_i} |T_\\psi[f](b_i,s)|^q$\n\\item compute $\\tau(q)$ with \n$\\log_2{Z(q,s)}\\,\\approx\\,\\tau(q)\\log_2(s)+C(q)$\n\\item compute $D(\\alpha) = \\underset{q}{\\min} (q\\alpha-\\tau(q))$\n\\end{list}\n\n\\subsubsection*{The triadic Cantor set}\n\nDefinition : The measure associated with the triadic Cantor set is defined by \n$f(x) = \\int_0^x d\\mu$ where $\\mu$ is the uniform measure lying on the triadic \nCantor set described in section \\ref{triadic Cantor set}. To compute $f(x)$, we used the next recursive\nfunction called the Devil's Staircase function \n(which looks like a staircase whose steps are\nuncountable and infinitely small, see Figure~\\ref{fig_5}). \n\n\\begin{eqnarray}\nf(x)=\t\\left\\{ \\begin{array}{ll}\n\tp_1f(3x) & \\mbox{ if $x \\in [0,\\frac{1}{3}]$}\\\\\n\tp_1 & \\mbox{ if $x \\in [\\frac{1}{3}, \\frac{2}{3}]$ }\\\\\n\tp_1+p_2f(3x-2) & \\mbox{ if $x \\in [\\frac{2}{3}, 1]$}\n\t\\end{array}\n\t\\right.\n% \\label{recursive Cantor func}\n\\end{eqnarray}\n\nThis is a continuous function that increases from 0 to 1 \non [0,1]. The recursive\nconstruction of $f(x)$ implies that $f(x)$ is self-similar.\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\t\\psfig{figure=DSC3.ps,height=5cm,width=8cm}\n \\psfig{figure=DSC3ParFun.ps,height=5cm,width=8cm}\n}}\n\\caption{Classical Devil's Staircase function (associated with\ntriadic Cantor set) with $p_1=0.5$ and $p_2=0.5$\n(left) and partition function $Z(q,s)$ for several \nvalues of $q$ (right). The\nwavelet transform is calculated with $\\psi$ equal to the \nfirst derivative of a Gaussian.}\n\\label{fig_5}\n\\end{figure}\n\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\t\\psfig{figure=DSC3ScaExp.ps,height=5cm,width=8cm}\n \\psfig{figure=DSC3SinSpe.ps,height=5cm,width=8cm}\n}}\n\\caption{Scaling Exponent estimation $\\tau(q)$ (left) and Singularity\nSpectrum $D(\\alpha)$ (right). The Scaling Exponent curve corresponds\n to the theoretical curve\n$\\tau(q)=(q-1)\\log_2(2)/\\log_2(3)$. Points of the Singularity Spectrum\nwhere $q \\neq \\infty(\\max\\,q)$ and $q \\neq-\\infty(\\min\\,q)$ are reduced \nto a single point\n($\\alpha=\\log_2(2)/\\log(3), D(\\alpha)=\\log _2(2)/\\log _2(3))$. This point\ncorresponds to\nthe Hausdorff dimension of the triadic Cantor Set.}\n\\label{fig_6}\n\\end{figure}\n\n\n\n\\subsubsection*{The generalized Devil's Staircase with $p_1=0.4$ and $p_2=0.6$}\n\nOne can prove \\cite{frac:arneodo90} that the theoretical singularity spectrum $D(\\alpha)$\nof the generalized Devil's Staircase function $f(x)=\\int_0^x{d\\mu}$ verifies\nthe following: \n\\begin{itemize}\n\\item The singular spectrum is a convex curve with a maximum value\n$\\alpha_{max}$ which corresponds to the fractal dimension of the\nsupport of the measure ($\\mu$). \n\\item The theoretical support of $D(\\alpha)$ is reduced at the interval\n$[\\alpha_{min},\\alpha_{max}]$:\n\n\\begin{eqnarray}\n\\left\\{ \\begin{array}{l}\n \\alpha_{min}=\\min({\\frac{\\ln p_1}{\\ln (1/3)}, \\frac{\\ln p_2}{\\ln (1/3)}})\\\\\n \\alpha_{min}=\\min({\\frac{\\ln p_1}{\\ln (1/3)}, \\frac{\\ln p_2}{\\ln (1/3)}})\n\t\\end{array}\n\t\\right.\n\\end{eqnarray}\n\\end{itemize}\n\nFigure~\\ref{fig_7} displays the generalized Devil's Staircase and its \npartition \nfunction $Z(q,s)$. In Figure~\\ref{fig_8}, we can see the Scaling exponent and\nthe Singularity spectrum. This one is in perfect ``accord'' with \nthe theoretical\nvalues: bell curve, $D(\\alpha)_{max}=\\log_2(2)/\\log_2(3)$,\n$\\alpha_{min}\\approx 0.47$ and $\\alpha_{max}\\approx 0.83$\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n \\psfig{figure=DSC3b.ps,height=5cm,width=8cm}\n \\psfig{figure=DSC3bParFun.ps,height=5cm,width=8cm}\n}}\n\\caption{Devil's Staircase function with $p_1=0.4$ and $p_2=0.6$\n(left) and partition function $Z(q,s)$ for several values of $q$ (right). \nThe wavelet transform is calculated\nwith $\\psi$ equal to the first derivative of a Gaussian.}\n\\label{fig_7} \n\\end{figure}\n\n\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\t\\psfig{figure=DSC3bScaExp.ps,height=5cm,width=8cm}\n \\psfig{figure=DSC3bSinSpe.ps,height=5cm,width=8cm}\n}}\n\\caption{Scaling Exponent estimation $\\tau(q)$ (left) and Singularity\nSpectrum $D(\\alpha)$ (right). The theoretical maximum value of $D(\\alpha)$ is\nobtained for $\\alpha_{max D}=\\log_2(2)/\\log_2(3)$ and\n$D(\\alpha_{max D})=\\log_2(2)/\\log_2(3)$.}\n\\label{fig_8}\n\\end{figure}\n\n\n\\newpage\n\\section{Program}\n\\subsection{Devil's Staircase: mf1d\\_create\\_ds}\n\\index{mf1d\\_create\\_ds}\nProgram {\\em mf1d\\_create\\_ds} creates a Devil's Staircase function lying on the\nCantor sets measure. If the ``-b'' option is set, a border is added, which\nis equal to 0 on the left side, and to 1 on the right side. The size of the\nborder is ${N\\over 3}$ where $N$ is the number of points. Hence the output\nsignal size becomes $N + {2N\\over 3} = {5N\\over 3}$. It is recommended to\nchoose a number of points which can be divided by 3.\n\n{\\bf\n\\begin{center}\nUSAGE: mf1d\\_create\\_ds options image out\n\\end{center}}\nwhere options are \n\\begin{itemize}\n\\item {\\bf [-n number\\_of\\_points]} \\\\\nNumber of points of Devil's Staircase. Default is 243.\n\\item {\\bf [-p Prob1]} \\\\\nCoefficient $p_1$ of the first interval of the Devil's Staircase. \n$p_2$ is equal to $1 - p_1$. \\\\\nDefault value is $0.5$. \n\\item {\\bf [-b]} \\\\\nAdd a border. Default is no.\n\\item {\\bf [-v]} \\\\\nVerbose. Default is False.\n\\end{itemize}\n\\subsubsection*{Examples:}\n\\begin{itemize}\n\\item mf1d\\_create\\_ds NameOut \\\\\nCreate a Devil's Staircase with all default values (243 points,\nprobabilities equal to $0.5$, ...).\n\\item mf1d\\_create\\_ds -p 0.4 NameOut \\\\\nDitto, but probabilities $p_1$ and $p_2$ are respectively equal to $0.4$ and\n$0.6$.\n\\end{itemize}\n\n\n\n\\subsection{Chain maxima wavelet coefficients: mf1d\\_chain}\n\\index{mf1d\\_chain}\n\nProgram {\\em mf1d\\_chain} calculates the wavelet transform of a signal,\nfinds its maxima wavelet coefficients, and chains them. Finally, chains which\ndo not respect some criteria are deleted. Two files are created: the first\ncontains the maxima, and the second the skeleton.\n{\\bf\n\\begin{center}\n USAGE: mf1d\\_chain options signal\\_in\n\\end{center}}\nwhere options are \n\\begin{itemize}\n\\item {\\bf [-t type\\_of\\_transform]} \\\\\n0: Gaussian derivative wavelet transform \\\\\n1: Mexican hat wavelet transform \\\\\nDefault is 0.\n\\item {\\bf [-r min\\_length\\_chain]} \\\\\nRemove chains with length $<$ min\\_length. Default is 0.\n\\item {\\bf [-s dynamique(\\%)]} \\\\\nRemove points of the skeleton map \nif level $<$ dynamique $*$ max(current scale). \\\\\nDefault is 0.\n% \\item {\\bf [-c]} \\\\\n% Draw chains of max. \\\\\n% Default is false.\n\\item {\\bf [-v]} \\\\\nVerbose\n\\end{itemize}\nWavelet transform file name is {\\bf WTMM\\_signal\\_in.fits}. \\\\\nSkeleton file name is {\\bf WTMMskel\\_signal\\_in.fits}.\n\\subsubsection*{Example:}\n\\begin{itemize}\n\\item mf1d\\_chain Data01.fits \\\\\n Create the files{\\bf WTMM\\_Data01.fits} and {\\bf WTMMskel\\_Data01.fits}.\n\\end{itemize}\n\n\\subsection{Partition function: mf1d\\_repart}\n\\index{mf1d\\_repart}\nProgram {\\em mf1d\\_repart} computes the partition function. It needs the two\noutput files of the {\\em mf1d\\_chain} program (i.e.\\ WTMMxxx.fits \nand WTMMskelxxx.fits), and creates three files:\n\\begin{itemize}\n\\item {\\bf q\\_FileNameOut.fits}: which contains the one-dimensional array $Q$ \nof the $q$ values for which the\npartition function is calculated.\n\\item {\\bf s\\_FileNameOut.fits}: which contains the one-dimensional array $S$\nof the $\\log_2 s$ values for which the\npartition function is calculated.\n\\item {\\bf Z\\_FileNameOut.fits}: which contains the partition functions. It\nis a two dimensional array. $Z(i,j)$ is the partition function value \n for $q = Q(i)$ and $s = S(j)$ ($i$ indexing the x-axis, and $j$ the y-axis).\n\\end{itemize}\n{\\bf\n\\begin{center}\nUSAGE: mf1d\\_repart options MaxFileName MaxSupFileName FileNameOut\n\\end{center}}\nwhere options are \n\\begin{itemize}\n\\item {\\bf [-a q\\_minq]} \\\\\nValue min of vector $q$. Default is -10.00.\n\\item {\\bf [-b q\\_max]} \\\\\nValue max of vector $q$. Default is +10.00.\n\\item {\\bf [-c number\\_of\\_q]} \\\\\nNumber of values of vector $q$. Default is 20.\n\\end{itemize}\n$Z(q,s)$ file name is {\\bf Z\\_FileNameOut.fits} \\\\\n$q$ file name is {\\bf q\\_FileNameOut.fits} \\\\\n$s$ file name is {\\bf s\\_FileNameOut.fits}. \\\\\n\\subsubsection*{Example:}\n\\begin{itemize}\n\\item mf1d\\_chain Data01.fits \\\\\n mf1d\\_repart WTMM\\_Data01.fits WTMMskel\\_Data01.fits Repart.fits \\\\\n Create the files {\\bf Z\\_Data01.fits}, {\\bf q\\_Data01.fits}, and\n {\\bf s\\_Data0.fits}.\n\\end{itemize}\n\n\n\\subsection{Multifractal analysis: mf\\_analyse}\n\\index{mf\\_analyse}\nProgram {\\em mf\\_analyse} realizes the multifractal analysis. It\n computes the scaling\nexponent estimation $\\tau(q)$ and the singularity spectrum $\\alpha(q)$.\nIt needs the three files created by {\\em mf1d\\_repart}. Four files are \ncreated:\n\\begin{itemize}\n\\item {\\bf X\\_Tau.fits}: which contains the x-coordinates of the \n$\\tau(q)$ curve.\n\\item {\\bf Y\\_Tau.fits}: which contains the y-coordinates of the \n$\\tau(q)$ curve.\n\\item {\\bf X\\_FracSpectrum.fits}: which contains the x-coordinates of the \nsingularity spectrum curve (i.e. $\\alpha$ values).\n\\item {\\bf Y\\_FracSpectrum.fits}: which contains y-coordinates of the \nsingularity spectrum curve (i.e. $D(\\alpha)$ values).\n\\end{itemize}\n\n{\\bf \\begin{center}\n USAGE: mf\\_analyse options FileNameIn\n\\end{center}}\nwhere options are \n\\begin{itemize}\n\\item {\\bf [-d $\\alpha_{min}$]} \\\\\nValue min of vector $\\alpha$. Default is 0.00.\n\\item {\\bf [-e $\\alpha_{max}$]} \\\\\nValue max of vector $\\alpha$. Default is 1.00.\n\\item {\\bf [-f number\\_of\\_$\\alpha$]} \\\\\nNumber of values of vector $\\alpha$. Default is 20.\n\\item {\\bf [-m ind\\_scale\\_min]} \\\\\nind scale min for computing $\\tau$. Default is 30. \n\\item {\\bf [-M ind\\_scale\\_max]} \\\\\nind scale max forn computing $\\tau$. Default is scale\\_max-2.\n\\end{itemize}\n$Z(q,s)$ are read in {\\bf Z\\_FileNameIn.fits} file \\\\\n$q$ are read in {\\bf q\\_FileNameIn.fits} file\\\\\n$s$ are read in {\\bf s\\_FileNameIn.fits} file .\n\\subsubsection*{Example:}\n\\begin{itemize}\n\\item mf1d\\_chain Data01.fits \\\\\n mf1d\\_repart WTMM\\_Data01.fits WTMMskel\\_Data01.fits Repart \\\\\n mf\\_analyse Repart \\\\\n Create the files {\\bf X\\_Tau.fits},{\\bf Y\\_Tau.fits},{\\bf X\\_FracSpectrum.fits},{\\bf Y\\_FracSpectrum.fits},\n and {\\bf Y\\_TauTheo.fits}.\n\\end{itemize}\n\n\n\\subsection{Examples}\n\n\\subsubsection*{Fractal analysis of Devil's Staircase function}\nCreate Devil's Staircase function with $p_1=.6$ and $p_2=.4$.\n\\begin{verbatim}\nmf1d_create_ds -p .4 -b DSC3\n\\end{verbatim}\nCompute the wavelet transform WTMM\\_DSC3.fits, the skeleton WTMMskel\\_DSC3.fits\n(default transform: $\\psi$ is the first derivative of a Gaussian).\n\\begin{verbatim}\nmf1d_chain DSC3\n\\end{verbatim}\nCompute the partition function $Z(q,s)$ for $q\\in[-10.,10.]$ with 60 values,\nwrite the $Z(q,s)$ to file Z\\_DSC3.fits, the $s$(scale) file to s\\_DSC3.fits and the\n$q$ file to q\\_DSC3.fits.\n\\begin{verbatim}\nmf1d_repart -c 60 WTMM_DSC3 WTMMskel_DSC3 DSC3\n\\end{verbatim}\nCompute the scaling exponent $\\tau(q)$, write it in X\\_Tau.fits and Y\\_Tau.fits,\ncompute the singularity spectrum $\\alpha(q)$, write it in X\\_FracSpectrum.fits\nand Y\\_FracSpectrum.fits (100 values are used for $\\alpha$).\n\\begin{verbatim}\nmf_analyse -f 100 DSC3\n\\end{verbatim}\n\n\\subsubsection*{Fractal analysis}\n\nCompute the wavelet transform WTMM\\_DSC3.fits, the skeleton WTMMskel\\_DSC3.fits\n(use the Mexican hat for $\\psi$)\n\\begin{verbatim}\nmf1d_chain -t 1 signal\n\\end{verbatim}\nCompute the partition function $Z(q,s)$ for $q\\in[-5.,5.]$ with 100 values,\nwrite the $Z(q,s)$ in file Z\\_signal.fits, the $s$(scale) file in s\\_signal and the\n$q$ file in q\\_signal.\n\\begin{verbatim}\nmf1d_repart -a -5. -b 5. -c 100 WTMM_signal WTMMskel_signal sigout\n\\end{verbatim}\nCompute the scaling exponent $\\tau(q)$, write it in X\\_Tau.fits and Y\\_Tau.fits,\ncompute the singularity spectrum $\\alpha(q)$, write it in X\\_FracSpectrum.fits\nand Y\\_FracSpectrum.fits ($\\alpha \\in [0,2]$, 100 values are used).\n\\begin{verbatim}\nmf_analyse -d 0. -e 2. -f 100 sigout\n\\end{verbatim}\n \n\n\n\n\n\n", "meta": {"hexsha": "4f3205d1d413e16e07a25d2117ffa23812553444", "size": 25068, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr1/ch_fractal.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr1/ch_fractal.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr1/ch_fractal.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.6079545455, "max_line_length": 110, "alphanum_fraction": 0.709230892, "num_tokens": 8086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835371034368, "lm_q2_score": 0.8376199613065411, "lm_q1q2_score": 0.6998176880208329}} {"text": "\\documentclass[letterpaper]{article}\n\n\\usepackage{fullpage}\n\\usepackage{nopageno}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\n\\title{Notes}\n\\date{September 8, 2014}\n\\maketitle\n\\section*{excercise}\nif $a_n\\le b_n\\forall n$ then $\\lim_{n\\to\\infty}a_n\\le\\lim_{n\\to\\infty}b_n$\n\\begin{align*}\n a_n&\\le b_n\\\\\n 0&\\le b_n-a_n\\\\\n \\lim_{n\\to\\infty}0&\\le\\lim_{n\\to\\infty}(b_n-a_n)\\\\\n 0&\\le \\lim_{n\\to\\infty}b_n-\\lim_{n\\to\\infty}a_n\\\\\n \\lim_{n\\to\\infty}a_n&\\le\\lim_{n\\to\\infty}b_n\n\\end{align*}\n\n\\section*{monotone sequences}\ndefinition:\n\na sequence is increasing iff $a_{n+1}\\ge a_n$ for every $n\\in\\mathbb{N}$ strictly increasing ...decreasing if $a_{n+1}\\ge a_n$ for every $n\\in\\mathbb{N}$, strictly decreasing.... monotone if is it any of these types\n\n\\section*{theorem 2.6.1}\nan increasing sequence that is bounded above is convergent. a decreasing sequence that is bounded below is convergent.\n\\subsection*{proof}\nwe are given $\\{a_n\\}_{n=1}^\\infty$, increasing $a_n\\le a_n+1\\forall n\\in\\mathbb{N}$ and bounded above. since it is bdd above it has a supremum $L$. we prove that $\\lim_{n\\to\\infty}a_n=L$.\n\\begin{align*}\n L&=\\sup\\{a_n:n\\in\\mathbb{N}\\}\n\\end{align*}\n$L$ is the least upper bound. if $M0,\\exists N\\in\\mathbb{N}$ such that $|a_n-L|<\\epsilon\\forall n\\ge N$.\n\n$\\forall a_n,a_n\\le L$\n\n$\\forall \\epsilon>0\\exists N$\n$L-\\epsilonL-\\epsilon$ because $\\{a_n\\}$ is increasing. hence $a_n-L|<\\epsilon, \\forall n\\ge N\\to\\lim_{n\\to\\infty}=L\\Box$\n\\section*{example}\nlet $0 0 \\) if\n \\begin{equation*}\n \\rho_Y(f(x_1), f(x_2)) \\leq L \\rho_X(x_1, x_2)^\\alpha \\quad\\forall x_1, x_2 \\in X.\n \\end{equation*}\n\n We refer to the smallest such constant, if any, as \\enquote{the} H\\\"older constant.\n\n \\thmitem{def:lipschitz_continuity/locally_holder} We say that \\( f \\) is \\term{locally H\\\"older continuous} if every point has a neighborhood where \\( f \\) is H\\\"older continuous with the same exponent, but possibly with with a different constant.\n\n \\thmitem{def:lipschitz_continuity/lipschitz} If \\( \\alpha = 1 \\), we say that \\( f \\) is \\term{Lipschitz continuous}.\n\n \\thmitem{def:lipschitz_continuity/contraction} If \\( X = Y \\) and if \\( f \\) is Lipschitz with constant \\( L < 1 \\), we call \\( f \\) a \\term{contraction mapping}.\n\n \\thmitem{def:lipschitz_continuity/calm}\\cite[53]{DontchevRockafellar2014} We say that \\( f \\) is \\term{calm} at \\( x \\) if it satisfies the Lipschitz condition with one of the points fixed:\n \\begin{equation*}\n \\rho_Y(f(x), f(x')) \\leq L \\rho_X(x, x') \\quad\\forall x' \\in X.\n \\end{equation*}\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:holder_map_is_uniformly_continuous}\n A H\\\"older map is uniformly continuous.\n\\end{proposition}\n\\begin{proof}\n Let \\( f: X \\to Y \\) be a H\\\"older map with constant \\( L \\) and exponent \\( \\alpha \\).\n\n Fix \\( \\varepsilon > 0 \\). Then is enough to choose \\( \\delta < \\sqrt[\\alpha]{\\frac \\varepsilon L} \\), so that\n \\begin{equation*}\n \\rho_X(x_1, x_2) < \\delta \\implies \\rho_Y(f(x_1), f(x_2)) \\leq L \\rho_X(x_1, x_2)^\\alpha < L \\delta^\\alpha < \\varepsilon.\n \\end{equation*}\n\n This implies uniform continuity.\n\\end{proof}\n\n\\begin{corollary}\\label{thm:locally_holder_map_is_continuous}\n A locally H\\\"older map is continuous.\n\\end{corollary}\n\n\\begin{theorem}[Banach's fixed point theorem]\\label{thm:banach_fixed_point_theorem}\\mcite[exer. 4.3.J]{Engelking1989}\n A contraction \\hyperref[def:lipschitz_continuity/contraction]{mapping} in a \\hyperref[def:complete_metric_space]{complete metric space} has a unique fixed \\hyperref[def:fixed_point]{point}.\n\\end{theorem}\n\\begin{proof}\n Let \\( f: X \\to X \\) be a contraction mapping. Fix any point \\( x_0 \\in X \\) and inductively define the sequence\n \\begin{equation*}\n x_{k+1} \\coloneqq f(x_k), k = 1, 2, \\ldots\n \\end{equation*}\n\n Fix \\( \\varepsilon > 0 \\). Since \\( L < 1 \\), there exists an index \\( k_0 > \\log_L(\\varepsilon) \\) such that for positive integers \\( m \\) and \\( k > k_0 \\),\n \\begin{balign*}\n \\rho(x_k, x_{k+m})\n & =\n \\rho(f^k(x_0), f^{k+m}(x_0))\n \\leq \\\\ &\\leq\n L^k \\rho(x_0, x_m)\n < \\\\ &<\n \\varepsilon \\rho(x_0, x_m).\n \\end{balign*}\n\n Note that\n \\begin{balign*}\n \\rho(x_0, x_m)\n & \\leq\n \\sum_{i=1}^m \\rho(x_{i-1}, x_i)\n \\leq \\\\ &\\leq\n \\rho(x_0, x_1) \\sum_{i=1}^m L^{i-1}\n = \\\\ &=\n \\rho(x_0, x_1) \\frac {1 - L^m} {1 - L}\n \\leq \\\\ &\\leq\n \\rho(x_0, x_1) \\frac 1 {1 - L}.\n \\end{balign*}\n\n Thus,\n \\begin{equation*}\n \\rho(x_k, x_{k+m}) < \\frac {\\varepsilon \\rho(x_0, x_1)} {1 - L}.\n \\end{equation*}\n\n The constant on the right is linear in \\( \\varepsilon \\) and does not depend on \\( k \\) or \\( m \\), hence \\( \\{ x_k \\}_{k=0}^\\infty \\) is a fundamental sequence. Since \\( X \\) is complete, the sequence has a limit \\( x \\).\n\n Because of the continuity of \\( f \\) (see \\fullref{thm:holder_map_is_uniformly_continuous}),\n \\begin{equation*}\n f(x) = f(\\lim_{k \\to \\infty} x_k) = \\lim_{k \\to \\infty} f(x_k) = \\lim_{k \\to \\infty} x_{k+1} = x.\n \\end{equation*}\n\\end{proof}\n", "meta": {"hexsha": "bc41efb40df7cd897aa067894854b4f4011d4ab6", "size": 3935, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/lipschitz_continuity.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/lipschitz_continuity.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/lipschitz_continuity.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.7222222222, "max_line_length": 251, "alphanum_fraction": 0.6401524778, "num_tokens": 1447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245870332531, "lm_q2_score": 0.8397339716830605, "lm_q1q2_score": 0.6997709651705798}} {"text": "\\subsection{Particle Swarm Optimization (PSO)}\nPSO is an algorithm that simulates the sociological behavior of bird flocking. In the search space, we define a swarm $X$, which has many particles $x^p(t) \\in \\{x^1(t), x^2(t), \\ldots, \\gamma\\}$ where $\\gamma$ is a number of particles and $t$ is the $t$th iteration among $T$ total iterations. Each particle $x^p$ consists of many elements, $x^p_i$ where $i = 1,2,3, \\ldots, \\kappa$, with $\\kappa$ is the dimension of a particle.\n\nFirstly, the particles of the swarm are initialized and evaluated by a predefined fitness function. The objective of the PSO is to minimize these fitness function values. In order to move the particle in the search space, a velocity $v^p_j(t)$ is defined. It is the flight speed of the $j$th element of the $p$th particle at the $t$th iteration. Additionally, $x^p_j(t)$ is the current position of that element. The velocity and current position is then calculated by the following formulae:\n\\begin{equation}\nv_{j}^{p}(t) = 2 \\cdot rand() \\cdot (pbest_{j}^{p} - x_{j}^{p}(t-1))\n+ 2 \\cdot rand() \\cdot (gbest_{j} - x_{j}^{p}(t-1))\n\\end{equation}\n\n\\begin{equation}\nx^p_j(t) = x^p_j(t - 1) + v^p_j(t)\n\\end{equation}\n\nAnother improved version of PSO which is introduced in \\cite{eberhart2000comparing}, added two factors: the constriction and inertia which change the velocity calculation to:\n\n\\begin{equation}\nv_{j}^{p}(t) = k \\cdot \\{w \\cdot v^p_j(t-1) + \\varphi_1 \\cdot rand() \\cdot (pbest_{j}^{p} - x_{j}^{p}(t-1))\n+ \\varphi_2 \\cdot rand() \\cdot (gbest_{j} - x_{j}^{p}(t-1)) \\}\n\\end{equation}\n\nwhere $k$ is the constriction factor and $w$ is the inertia factor, $\\varphi_1$ and $\\varphi_2$ are constants.\n\nEven with the improved version of PSO, when the particle coincides with the global best position, the particle will move away from this point if the two factors above are different from zero. Furthermore, the particles will stop moving when their velocities are close to zero and they catch up with the global best position. This phenomena is called $stagination$ \\cite{eberhart1998comparison}. To overcome this phenomena, the mutation process of GA has been added to PSO to form up a Hybrid PSO with Mutation (HPSOM). We will discuss this mutation process among with the Wavelet Mutation in the next section.\n\n\\subsection{Hybrid PSO with Wavelet Mutation}\nThe initial version of PSO with mutation process works as follows: a random particle is selected to be moved to another position in the search space by using mutation under the following operation:\n\\begin{equation}\nmut(x_j) = x_j - \\omega, r < 0\n\\end{equation}\n\\begin{equation}\nmut(x_j) = x_j + \\omega, r \\geq 0\n\\end{equation}\nwhere $x_j$ is a randomly chosen element of the particle above, $\\omega$ is randomly generated in the range of $[0, 0.1 x (para^j_{max} - para^j_{min})]$, $r$ is a random number between +1 and -1, $para^j_{max}$ and $para^j_{min}$ are the upper and lower bounds of each particle element. Therefore, we can see that $\\omega$ represents $1/10$ of the search space.\n\nThis version of PSO can overcome the $stagination$ phenomena. However, it can easily be seen that the mutating space is fixed by $\\omega$. It is not always the best strategy for using the same size mutation space all the time. Therefore, in \\cite{ling2008hybrid}, the author proposed an improvement method to dynamically adjust the mutating size $\\omega$ based on the wavelet theory as follows.\n\n\\subsubsection{Hybrid PSO with Wavelet Mutation}\n\\\nEach particle element has a chance to mutate which is controlled by a probability of mutation $p_m \\in [0, 1]$. For each element, we generate a random number between 0 and 1 $r$, if $r \\leq p_m$, the mutation will take place on that element. In details, let $x^p(t) = [x^p_1(t), x^p_2(t), \\ldots, x^p_k(t)]$ be the current selected particle, with $x^p_j(t)$ is its randomly chosen element, $x^p_j(t) \\in [param^j_{min}, param^j_{min}]$. After the mutation process, the resulting particle is $\\bar{x}^p(t)$\n\n\\[ \\bar{x}^p(t) = \\left\\{ \\begin{array}{ll}\n x^p_j(t) + \\sigma \\times (para^j_{max} - x^p_j(t)) & \\mbox{if $\\sigma > 0$} \\\\\n x^p_j(t) + \\sigma \\times (x^p_j(t) - para^j_{min}) & \\mbox{if $\\sigma \\leq 0$}.\\end{array} \\right. \\] \n\nwhere $j \\in 1,2,3, \\ldots k$ is the dimension of the particle and:\n\\begin{equation}\n\t\\sigma = \\frac{1}{\\sqrt{a}}\\psi(\\frac{\\varphi}{a})\n\\end{equation}\n\nwhere $\\psi(x)$ is a wavelet - a continuous time function (see Figure \\ref{fig:fig1}) represents a certain seismic signal that satisfies two properties:\n\n\\begin{equation}\n\t\\int_{-\\infty}^{+\\infty}\\psi(x)dx = 0\n\\end{equation}\n\nand\n\n\\begin{equation}\n\t\\int_{-\\infty}^{+\\infty}|\\psi(x)|^2dx < 0\n\\end{equation}\n\n\\begin{figure} [H]\n\\centering\n\\includegraphics[width=60.72mm]{resources/morlet1}\n\\caption{The original form of the Morlet wavelet}\\label{fig:fig1}\n\\end{figure}\n\nIn this paper, we adapt this approach (HPSOWM) and transform the real-valued particles into binary-based ones. This will be discussed in the next section.", "meta": {"hexsha": "4260245e5083f7c7f548ae911d0aaae392be641b", "size": 5006, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/chapters/p5.tex", "max_stars_repo_name": "minhprg/binary-hpsowm", "max_stars_repo_head_hexsha": "c40a7c3575b9560205252b501b3a9b506a715260", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/chapters/p5.tex", "max_issues_repo_name": "minhprg/binary-hpsowm", "max_issues_repo_head_hexsha": "c40a7c3575b9560205252b501b3a9b506a715260", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/chapters/p5.tex", "max_forks_repo_name": "minhprg/binary-hpsowm", "max_forks_repo_head_hexsha": "c40a7c3575b9560205252b501b3a9b506a715260", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.6176470588, "max_line_length": 609, "alphanum_fraction": 0.7231322413, "num_tokens": 1520, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.863391624034103, "lm_q2_score": 0.8104789109591831, "lm_q1q2_score": 0.6997607031784402}} {"text": "\\chapter{Principal Components Analysis \\label{chapter:pca}}\n\nClustering (Chapter~\\ref{chapter:clustering}) is one approach to uncovering latent structure in data. Another are matrix decomposition methods, the most famous of which is \\textbf{principal components analysis (PCA)}. PCA is a powerful statistical tool for analyzing and visualizing datasets. It has been independently discovered several times over the course of history, and can be formulated mathematically a few different ways.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Principal Components: What are they?}\n\nYou can think of PCA as a rotation of the feature space onto a set of axes that most efficiently represent the data. The \\textbf{principal components} are these axes, or \\textbf{basis vectors}. Here are pictures of the first two principal components for a small dataset. These were borrowed from Andrew Ng's lecture notes for CS229 at Stanford University. \n\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/l04-ang-pc1.png}\n\\includegraphics[width=0.7\\textwidth]{img/l04-ang-pc2.png}\n\\end{center}\n\n\\begin{question}{}\nWhat do you notice about principal components 1 (top) and 2 (bottom)? Think in terms of (a) the variance (spread) of the data, and (b) the projection errors (dotted lines). \n\\end{question}\n\nMathematically, PCA is a linear projection of $p$-dimensional datapoints, $x^{(1)}, \\dots, x^{(n)}$ onto a $k$-dimensional space ($k \\leq p$) defined by basis vectors $u_1, \\dots, u_k$ such that:\n\\begin{itemize}\n\\item the projection maximizes the variance from the original dataset that is retained\n\\item the projection minimizes projection error (square loss)\n\\item the basis vectors, $\\left\\{u_1, \\dots, u_k\\right\\}$, are orthogonal (i.e., perpendicular)\n\\end{itemize}\n\nThe number of distinct principal components is the smaller of the number of original variables ($p$) or the number of observations ($n$) minus one.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Definitions}\n\n\\subsection{Feature Vectors}\n\nSay you have $n$ samples in a dataset, and each has dimensionality $p$. Let $x^{(i)}$ be the vector of $p$ features for the $i$th person. We write\n\n$$ x^{(i)} = \\begin{bmatrix}\n x_{1}^{(i)} \\\\\n x_{2}^{(i)} \\\\\n \\vdots \\\\\n x_{p}^{(i)}\n \\end{bmatrix} $$\n \n\\noindent and we can write our entire dataset as a $n \\times p$ matrix like this:\n\n$$ X = \\begin{bmatrix}\nx_1^{(1)} & x_2^{(1)} & \\dots & x_p^{(1)} \\\\\nx_1^{(2)} & x_2^{(2)} & \\dots & x_p^{(2)} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\nx_1^{(n)} & x_2^{(n)} & \\dots & x_p^{(n)} \\\\\n\\end{bmatrix} $$ \nsuch that the $i$th row of the matrix is the vector of features for the $i$th person. This is exactly the same notation that we have used in previous chapters. \n\n\\subsection{Variance and Covariance \\label{ssect:varcovar}}\n\nThe \\textbf{sample covariance} of two features $x_j$ and $x_k$ is given by\n$$ \\text{cov}(x_j, x_k) = S_{jk} = \\frac{1}{n-1} \\sum_{i=1}^{n}\\left(x_j^{(i)}-\\overline{x_j}\\right) \\left(x_k^{(i)}-\\overline{x_k}\\right) $$\nwhere the bar ($\\overline{x_j}$) refers to the mean of $x_j$ across the entire dataset. We can also write it in matrix format as\n$$ S = \\frac{1}{n-1} \\sum_{i=1}^n (x^{(i)} - \\overline{x})(x^{(i)} - \\overline{x})^T. $$\n\nThe covariance can be any real number between positive and negative infinity. One weird thing about it is that it has units - the product of the units of the two features. This means that the covariance depends on the scale that is chosen for each feature, which is somewhat arbitrary. The \\textbf{sample variance} of a given feature is just the covariance in cases where $j = k$ (the covariance of a feature with itself). \n\nThe \\textbf{correlation} (technically the \\textbf{Pearson correlation}) is a unitless, normalized version of the covariance and has the form\n$$ \\text{cor}(x_j, x_k)=\\frac{\\sum_{i=1}^{n}(x_j^{(i)}-\\overline{x_j})(x_k^{(i)}-\\overline{x_k})}{\\sqrt{\\sum_{i=1}^{n}(x_j^{(i)}-\\overline{x_j})^2}{\\sqrt{\\sum_{i=1}^{n}(x_k^{(i)}-\\overline{x_k})^2}}}. $$\n\n\\subsection{Orthogonality}\n\nTwo vectors are \\textbf{orthogonal} if their \\textbf{inner product}, or \\textbf{dot product}, is zero. This is defined as\n$$ \\langle x, y \\rangle = x \\cdot y = \\sum_{i=1}^n x_i y_i $$\nwhere the sum is over the vector components. If this mathematical notation makes you uncomfortable, just think of ``orthogonality'' as ``perpendicularity'' and you'll be able to understand it visually. \n\nIn PCA, the vectors whose inner products we are concerned with will \\emph{not} be the feature vectors for individual samples. They will be, instead, the \\emph{columns} of $X$ -- the values of a single feature for all $n$ samples. \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Example: Running PCA in R}\n\nLet's say you have measurements of weight, height, and IQ ($p=3$) for $n=100$ men. Here are the first $10$ rows of the $100$ row dataset:\n\n\\begin{center}\n\\begin{tabular}{cccc}\nPatient ID ($i$) & Height (in) & Weight (lbs) & IQ \\\\\n\\midrule\n1 & 74.9 & 155.6 & 109 \\\\\n2 & 66.5 & 171.6 & 117 \\\\\n3 & 74.4 & 175.2 & 91 \\\\\n4 & 65.8 & 151.3 & 84 \\\\\n5 & 70.2 & 162.7 & 107 \\\\\n6 & 78.8 & 206.4 & 117 \\\\\n7 & 69.1 & 162.7 & 96 \\\\\n8 & 70.6 & 150.3 & 110 \\\\\n9 & 65.2 & 156.2 & 121 \\\\\n10 & 66.8 & 153.3 & 93 \\\\\n\\end{tabular}\n\\end{center}\n\n\\noindent Here are some scatterplots showing pairwise comparisons of the three features:\n\n\\begin{center}\n\\includegraphics[width=0.32\\textwidth]{img/l04-example-1.png}\n\\includegraphics[width=0.32\\textwidth]{img/l04-example-2.png}\n\\includegraphics[width=0.32\\textwidth]{img/l04-example-3.png}\n\\end{center}\n\n\\noindent The \\textbf{correlation matrix} for these data looks like this:\n\n\\begin{verbatim}\n height weight iq\nheight 1.000 0.790 -0.103\nweight 0.790 1.000 -0.078\niq -0.103 -0.078 1.000\n\\end{verbatim}\n\n\\vspace{2mm}\n\n\\begin{question}{}\nLooking at the correlation matrix, which features are most tightly correlated? What does this imply about the direction of the first principal component, PC1? \n\\end{question}\n\n\\subsection{Centering and Scaling}\n\nPCA is sensitive to the relative scales of the different variables in the original dataset. For this reason, datasets are usually \\emph{scaled} and \\emph{centered} before PCA is performed. All this means is that for each column (feature) of the dataset, we subtract its mean and divide by its standard deviation. The transformed column will have mean 0 and standard deviation 1. \n\nIf the data are scaled and centered, we can rewrite $S$ (the sample covariance matrix from Section~\\ref{ssect:varcovar}) as $X^T X/(n-1)$.\n\n\\vspace{2mm}\n\n\\begin{question}{}\nWhat would happen to the principal components if you didn't center and scale the data?\n\\end{question}\n\n\\begin{question}{}\nWhy do you think the interpretation of the principal components becomes more difficult if you have features measured on lots of different scales (e.g., some categorical, some numeric/roughly normal, some numeric/highly skewed)? \n\\end{question}\n\n\\subsection{Input and Output}\n\nIn R, we can run PCA on our dataset, \\texttt{d}, using the following command:\n\\begin{center}\n\\verb|p <- prcomp(d[,2:4], center = TRUE, scale. = TRUE, rank. = 3)|\n\\end{center}\nwhere \\texttt{rank.} is an optional parameter indicating the number, $k$, of principal components desired. The output looks like this:\n\n\\begin{multicols}{2}\n{\\scriptsize\n\\begin{verbatim}\n> p$sdev\n[1] 1.3454074 0.9899765 0.4580672\n\n> p$rotation\n PC1 PC2 PC3\nheight 0.6996236 -0.09446119 -0.70823998\nweight 0.6971414 -0.12698944 0.70559726\niq -0.1565906 -0.98739595 -0.02299201\n\n> p$center\n height weight iq \n 70.74468 174.43346 101.12000 \n \n> p$scale\n height weight iq \n 3.914989 13.853272 14.183701 \n \n> p$x\n PC1 PC2 PC3\n [1,] -0.189268854 1.976180297 -0.416704298\n [2,] 1.794925031 0.710160037 0.575478783\n [3,] -0.404617344 0.283418147 0.151536071\n [4,] -0.471675025 -0.710146894 0.347006397\n [5,] -1.073877096 -0.234684815 -0.638243675\n [6,] -0.654663317 0.043481120 -0.107082277\n [7,] -1.705841441 0.368990046 -0.794047654\n [8,] 2.311353981 -0.501565781 0.033112309\n [9,] -0.006387264 -0.535422493 -1.126177513\n [10,] -1.557136911 1.399023167 -0.045281059\n [11,] 2.632974868 -1.046037442 -0.239934797\n [12,] -0.978337630 -0.193476727 0.007814938\n [13,] -0.625980903 -0.456274882 -0.305349616\n (continued)\n\\end{verbatim}\n}\n\\end{multicols}\n\n\\begin{question}{}\nDescribe/draw the directions of the three principal component vectors, PC1, PC2, and PC3, in the coordinate system of the original predictors, height, weight, and IQ.\n\\end{question}\n\n\\begin{question}{}\nHere is a picture of the flow cytometry dataset we first encountered in Chapter~\\ref{chapter:clustering}.\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/biomarker-data-big-no-labels.png}\n\\end{center}\nWhat would PC1 and PC2 look like for this dataset? (Why is there no PC3?) How could PCA help you separate the two clusters?\n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Applications of PCA}\n\n\\subsection{Eigenfaces}\n\nEigenfaces were an early application of PCA to computer vision, specifically image search and retrieval. Here's what you do to create a set of eigenfaces:\n\\begin{enumerate}\n\\item Prepare training set of images taken under the same lighting conditions, with mouths and eyes aligned, resampled to a common pixel resolution.\n\\item Generate a vector for each image by concatenating the pixel intensities across the rows. So if the image is of dimension $r \\times c$, the image vector will have $p = rc$ features.\n\\item Create the data matrix from the $n$ vectors.\n\\item Center the data (Why wouldn't we scale the data?).\n\\item Calculate the eigenvectors and eigenvalues of the sample covariance matrix. Each eigenvector corresponds to one eigenface. \n\\end{enumerate}\n\nSome real eigenfaces are shown below \\footnote{Figure details: (top) some of the original faces from the training set (there were 86 images total); (bottom) the eigenfaces corresponding to the 18 largest eigenvalues of the covariance matrix. From https://www.clear.rice.edu/elec301/Projects99/faces/images.html.}. \n\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{img/l04-eigenfaces-1.png}\\\\[5mm]\n\\includegraphics[width=0.8\\textwidth]{img/l04-eigenfaces-2.png}\n\\end{center}\n\n\\begin{question}{}\nWhat is $X$ for the eigenfaces problem? What are the principal components? How could you use the principal components to match a new face to an existing database of faces?\n\\end{question}\n\n\\subsection{Image Compression}\n\nPCA can also be used for image compression. In that case, the raw pixels for the image are the data matrix, $X$ (so each ``sample'' is one row of the image and each ``feature'' is one column of the image). Generally the data are centered but not scaled here (Why?). A subset of the principal components corresponding to the $r$ largest eigenvalues are used to reconstruct the image. The picture below illustrates the process\\footnote{Figure: Using PCA for image compression. The image is recomposed by adding together many gridlike images like that shown in (a). From \\texttt{https://www.projectrhea.org/rhea/index.php/PCA\\_Theory\\_Examples}.}. \n\n\\begin{center}\n\\includegraphics[width=\\textwidth]{img/l04-compression.png}\n\\end{center}\n\n\\begin{question}{}\nWhat is $X$ for the image compression problem? What are the principal components? How does using PCA help compress the image?\n\\end{question}\n\n\\subsection{Genetic Ancestry}\n\nThere are many awesome population genetics papers that use PCA to locate individuals within the ``space'' of possible genetic mutations. Here's some text from one, a figure from which is shown below\\footnote{From European Journal of Human Genetics (2016) 24, 931-936 (2016).}. Here's a quote from the original paper:\n\n\\begin{quote}\nHere, we analyse genome-wide variation in 173 Armenians and compare them with 78 other worldwide populations. We find that Armenians form a distinctive cluster linking the Near East, Europe, and the Caucasus. We show that Armenian diversity can be explained by several mixtures of Eurasian populations that occurred between ~3000 and ~2000 BCE, a period characterized by major population migrations after the domestication of the horse, appearance of chariots, and the rise of advanced civilizations in the Near East. However, genetic signals of population mixture cease after ~1200 BCE when Bronze Age civilizations in the Eastern Mediterranean world suddenly and violently collapsed. Armenians have since remained isolated and genetic structure within the population developed ~500 years ago when Armenia was divided between the Ottomans and the Safavid Empire in Iran. \n\\end{quote}\n\n\\begin{center}\n\\includegraphics[width=\\textwidth]{img/l04-genetics.jpg}\n\\end{center}\n\n\\begin{question}{}\nWhat is $X$ for the genetic ancestry problem? What are the principal components? How were the principal components used to produce the figure above?\n\\end{question}\n\n\\subsection{Topic Modeling}\n\nPCA also has many uses in natural language processing. It forms the basis for latent semantic indexing (see Deerwester 1990) and can also be used for topic modeling, a form of mixture model (see Section~\\ref{sect:mixturemodels})\\footnote{Figure: Topic modeling French crime fiction. From https://dragonfly.hypotheses.org/530.}.\n\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/l04-topics.png}\n\\end{center}\n\n\\begin{question}{}\nWhat is $X$ for the topic modeling problem? For the latent semantic indexing problem? What do the principal components represent?\n\\end{question}\n\n\\begin{question}{}\nThink of 2-3 different unsupervised learning problems from biology or medicine where PCA makes sense, conceptually at least, for modeling the data. How would you set up the data matrix in each case? What would the principal components correspond to in the data? \n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Technical Details (Advanced)}\n\n\\subsection{Eigenvalues and Eigenvectors \\label{ssect:eig}}\n\nOne interpretation of matrix multiplication, $Ax$, where $A$ is an $m \\times m$ square matrix and $x$ a vector of length $m$, is that $A$ linearly transforms $x$ by rotating it, changing its magnitude, or both. An \\textbf{eigenvector} of $A$ is a vector that, when multiplied by $A$, grows or shrinks in magnitude but does not rotate. The multiplier of its magnitude is given by the corresponding \\textbf{eigenvalue}. There are $m$ different eigenvalues and eigenvectors for an $m \\times m$ matrix, although the eigenvalues may not all be distinct. The set of all eigenvalues of $A$ is called the \\textbf{spectrum} of $A$. \n\nFor all eigenvalue-eigenvector pairs, the relationship $Av = \\lambda v$, where $v$ is an eigenvector and $\\lambda$ its corresponding eigenvalue, must hold. To find the $\\lambda$s and $v$s analytically, we can set the determinant $ | A - \\lambda I | $, called the \\textbf{characteristic polynomial}, equal to zero and solve for the eigenvalue and eigenvector corresponding to each root. Note that any scalar multiple of an eigenvector is itself an eigenvector; usually we restrict eigenvectors to have magnitude $1$ for this reason.\n\nIf a matrix is \\textbf{symmetric} and \\textbf{positive semidefinite} (all positive or zero eigenvalues), all of its eigenvectors will be orthogonal.\n\n\\subsection{PCA: Eigendecomposition Version} \n\nThe first way we can find the principal components is by performing an eigen-decomposition identical to the one we performed in Section~\\ref{ssect:eig} on the sample covariance matrix. The principal components will be the eigenvectors of this matrix, and the corresponding eigenvalues tell you how much of the dataset's overall variance is accounted for by each eigenvector.\n\nOne useful fact about the covariance matrix is that, because it is a symmetric matrix, it can be diagonalized:\n\\begin{equation} S = \\frac{1}{n-1} X^T X = V L V^T \\label{eqn:eigendiag} \\end{equation}\nwhere $V$ is a matrix of the $p$ eigenvectors of $S$ (each column is an eigenvector) and $L$ is a diagonal matrix with the eigenvalues of $S$ along the diagonal. This fact will be important later.\n\n\\subsection{PCA: Singular Value Decomposition (SVD) Version}\n\nAlthough the eigendecomposition of the covariance matrix is generally the way PCA is presented, in practice most software uses another matrix decomposition called the \\textbf{singular value decomposition (SVD)}, even though it is slower to compute, because it's more numerically stable.\n\nHere's why the two methods are equivalent. Any matrix, $M$, with real or complex values (we'll focus on real valued matrices here) has an SVD of the form\\footnote{This is in contrast with the eigendecomposition, which works only on positive semidefinite matrices.}:\n$$ M = U D V^T $$\nwhere $U$ is an orthogonal matrix, meaning that $U^TU = I$, $D$ is diagonal and all of its nonzero elements are non-negative real numbers called \\textbf{singular values}, and $V$ is also orthogonal. For a square matrix, the SVD can be viewed as ``breaking up'' the transformation described by a matrix into three separate transformations: a rotation/reflection, a scaling, and another rotation/reflection\\footnote{Figure: Graphical representation of the SVD of a square matrix, $M$. Note that the illustrator here uses $\\Sigma$ instead of $D$ for the middle, diagonal matrix. Attribution: By Georg-Johann (Own work), via Wikimedia Commons.}:\n\n\\begin{center}\n\\includegraphics[width=0.5\\textwidth]{img/l04-svd.pdf}\n\\end{center}\n\n\\noindent If we perform the SVD on $X$, we obtain a decomposition\n$$ X = U D V^T $$\nwhere $D$ is a diagonal matrix with singular values $d_1, \\dots, d_p$. Knowing this, we can rewrite our covariance matrix as\n\\begin{align*} S &= \\frac{1}{n-1} X^T X \\\\\n&= \\frac{1}{n-1} (UDV^T)^T U D V^T \\\\\n&= \\frac{1}{n-1} V D U^T U D V^T \\\\\n&= V \\frac{D^2}{n-1} V^T. \\end{align*}\n\nComparing this to the form from Equation~\\ref{eqn:eigendiag}, we see that the eigenvectors correspond to the right singular values of the SVD, and the eigenvalues are related to the singular values on the diagonal of $D$ by $\\lambda_i = d_i^2/(n-1)$.\n\nThere are a lot of interpretations of the SVD. The one I like best is that the SVD provides a ``nearest orthogonal matrix'' to the original matrix. It's like the data have a coordinate system of orthogonal axes in which they most like to live, and the SVD tells you what that system is.\n\n", "meta": {"hexsha": "dea260015174a6cdc6e53209f7431039696a65a8", "size": 18672, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/mcds-principal-components-analysis.tex", "max_stars_repo_name": "blpercha/mcds-notes", "max_stars_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-12-10T16:51:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-03T01:31:23.000Z", "max_issues_repo_path": "tex/mcds-principal-components-analysis.tex", "max_issues_repo_name": "blpercha/mcds-notes", "max_issues_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/mcds-principal-components-analysis.tex", "max_forks_repo_name": "blpercha/mcds-notes", "max_forks_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T17:16:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T17:16:44.000Z", "avg_line_length": 60.2322580645, "max_line_length": 872, "alphanum_fraction": 0.7261675236, "num_tokens": 5140, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703224, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.6997606940800105}} {"text": "\n\\subsection{Chapter 5}\n\n\\begin{p}{Show that any 2-form $F$ on $\\R\\times S$ can be uniquely expressed as $B+E\\wedge dt$ in \nsuch a way that for any local coordinates $x^i$ on $S$ we have \n$E=E_i dx^i$ and $B=\\frac{1}{2}B_{ij} dx^i\\wedge dx^j$.}\n\\end{p}\n\nFor some patch $U\\subset S$, we can use the local coordinates $x^i$ along with $t\\in\\R$ to define an\nelement of $\\R^{n+1}$. Any 2-form $F$ can be written as $F=F_{\\alpha\\beta} dx^\\alpha\\wedge dx^\\beta$ where the summation\nruns over $0\\dots n$ and $x^0=t$. We can also express this as $F=F_{0i}dt\\wedge dx^i+F_{i0}dx^i\\wedge dt+F_{ij}dx^i\\wedge dx^j$. \nBut then by \nantisymmetry of the wedge product, we might as well have $F_{i0}=-F_{0i}$. Then defining $E_i=2F_{0i}$ we have\n$F=E_i dx^i\\wedge dt+F_{ij}dx^i\\wedge dx^j$. With $B_{ij}=2F_{ij}$ we obtain $F=E\\wedge dt+B$. This decomposition is unique \nbecause if it were true using different functions $E'_i$ and $B'_{ij}$, then their difference would be zero.\n\n\\begin{p}{Show that for any form $\\om$ on $\\R\\times S$ there is a unique way to write $d\\om=dt\\wedge\\partial_t\\om+d_S\\om$\nsuch that for any local coordinates $x^i$ on $S$, writing $t=x^0$, we have $d_S\\om=\\partial_i \\om_I dx^i\\wedge dx^I$ and\n$dt\\wedge \\partial_t\\om=\\partial_0\\om_I dx^0\\wedge dx^I$.}\n\\end{p}\n\nFirst write $\\om=\\om_I dx^I$ for $I$ a multi-index of length $p$ (i.e. a $p$-tuple whose entries range over $0\\dots n$, where\n$n$ is the dimension of $S$). Then $d_S\\om=\\partial_i \\om_I dx^i\\wedge dx^I$ and $\\partial_t\\om=\\partial_0\\om_I$. Again, uniqueness\nfollows from linearity: if there were two possibilities, their difference would be zero by the linearity of $d$, implying their equality.\n\n\\begin{p}{Use the nondegeneracy of the metric to show that the map from $V$ to $V^*$ given by $v\\mapsto g(v,\\cdot)$ is \nan isomorphism, that is, 1-to-1 and onto.}\\end{p}\n\nNondegeneracy of the metric means that if $g(v,w)=0$ for all $w$, then $v=0$. Now, for the map to be an isomorphism, we have to \ncheck the two conditions. First 1-to-1: different inputs result in different outputs. So we check if the difference of the \noutputs is zero: $g(v,\\cdot)-g(w,\\cdot)=g(v-w,\\cdot)\\neq 0$ unless $v=w$. Next is the onto condition: is every element of $V^*$ the image\nof some $v\\in V$? Yes, consider $\\om\\in V^*$ and consider the action on a basis: $\\om_\\mu=\\om(e_\\mu)$. Setting $\\om_\\mu=g(v,e_\\mu)$, \nwe have to solve for $v$. Expanding $v=v^\\mu e_\\mu$ we obtain $\\om_\\mu=v^\\nu g(e_\\nu,e_\\mu)=v^\\nu g_{\\nu,\\mu}$. But\nthe nondegeneracy of $g$ implies that $g_{\\nu,\\mu}$ is invertible. Suppose we have $g(v,w)=g_{\\mu,\\nu}v^\\mu w^\\nu=0$ for all $w$. Then\n$g_{\\mu,\\nu}v^\\mu=0$ for all $\\nu$. But this only occurs when $v^\\mu=0$ for all $\\mu$, meaning that $g_{\\mu,\\nu}$ never annihilates\nan input vector; it has full rank. Thus, it's invertible. So we can solve the equation for $v$.\\\\\n\nThere's a more direct proof for the vector spaces we are considering here: by exercise 28, we know the dual space\nhas the same dimension as the original space. Thus, nothing in the dual can be outside the image of the original space, since if\nwere, it would consititute a new basis element.\n\n\\begin{p}{Let $v=v^\\mu e_\\mu$ be a vector field on a chart. Show that the corresponding 1-form $g(v,\\cdot)$ is equal to $v_\\nu f^\\nu$, where $f^\\nu$ is the dual basis of 1-forms and $v_\\nu=g_{\\mu,\\nu}v^\\mu$.}\\end{p}\n\nWe can follow the lines of the previous exercise to derive the form, or simply show that the equation is correct. Choosing an \narbitrary vector field $w$, $g(v,w)=g_{\\mu,\\nu}v^\\mu w^\\nu$ which is equal to $v_\\nu f^\\nu(w)=v_\\nu w^\\nu$.\n\n\\begin{p}{Let $\\om=\\om_\\mu f^\\mu$ be a 1-form on a chart. Show that the corresponding vector field is equal to\n$\\om^\\nu e_\\nu$ where $\\om^\\nu=g^{\\mu,\\nu}\\om_\\mu$.}\\end{p}\n\nFor arbitrary $v$, $\\om(v)=\\om_\\mu f^\\mu(v)=\\om_\\mu v^\\mu$. But there should be a $w$ such that $g(w,v)=\\om_\\mu v^\\mu$. \nUsing the series expression for $v$ and $w$, we obtain $\\om_\\nu v^\\nu=g_{\\mu,\\nu}w^\\mu v^\\nu$. This should hold for arbitrary $v$,\nso $\\om_\\nu=g_{\\mu,\\nu}w^\\mu$, or $w^\\mu=g^{\\mu,\\nu}\\om_\\nu$.\n\n\\begin{p}{Let $\\eta$ be the Minkowski metric on $\\R^4$. Show that its components in the standard basis\nare $\\eta_{\\mu,\\nu}=\\delta_{\\mu,\\nu}(1-2\\delta_{\\mu,0})$.}\\end{p}\n\nThe Minkowski metric is defined for two vectors $v$ and $w$ as $\\eta(v,w)=-v^0w^0+v^1w^1+v^2w^2+v^3w^3$. The\nform of $\\eta_{\\mu,\\nu}$ follows immediately. \n\n\\begin{p}{Show that $g^\\mu_\\nu=\\delta^\\mu_\\nu$.}\\end{p}\n\nIf we start with $g_{\\lambda,\\nu}$ we can raise the first index with the metric, obtaining $g^\\mu_\\nu=g^{\\mu,\\lambda}g_{\\lambda,\\nu}$.\nBut $g^{\\lambda,\\nu}$ is the matrix inverse of $g_{\\lambda,\\nu}$, so the result of the product is the Kronecker delta.\n\n\\begin{p}{Show that the inner product of $p$-forms is nondegenerate by supposing that $(e^1,\\dots,e^n)$ is any\northonormal basis of 1-forms in some chart, with $g(e^i,e^i)=\\epsilon(i)$ for $\\epsilon(i)=\\pm 1$. Show the $p$-fold\nwedge products $e^{i_1}\\wedge\\dots\\wedge e^{i_p}$ form an orthonormal basis of $p$-forms with \n$\\langle e^{i_1} \\wedge\\dots\\wedge e^{i_p},e^{i_1}\\wedge\\dots\\wedge e^{i_p}\\rangle=\\epsilon(i_1)\\cdots\\epsilon(i_p)$.}\\end{p}\n\nFrom exercise 52 we already know that the nondegeneracy of $g$ is equivalent to the invertibility of the matrix of $g$ applied to basis \nelements. The same will be true here. First recall that the terms in\nthe $p$-fold wedge products must be distinct so that the $p$-form doesn't vanish. \nSuppose we have two basis $p$-forms in which $m$ basis 1-forms\nappear in both. Reordering so that these come first, the matrix of $g(e^{i_j},e^{k_\\ell})$ will be block diagonal with the first\n$m\\times m$ block equal to the identity matrix, and zero everywhere else. This matrix obviously has zero determinant, thus\nthe basis $p$-forms with any distinct elements are orthogonal. Otherwise, the determinant yields the product of the individual\nnormalizations, establishing the last equation posed in the problem. This in turn means that the metric matrix \nfor the basis $p$-forms is diagonal with entries $\\pm 1$, whence it has full rank and is therefore nondegenerate.\n\n\\begin{p}{Let $E=E_j dx^j$ be a 1-form on $\\R^3$ with its Euclidean metric. Show that $\\langle E,E\\rangle=E_x^2+E_y^2+E_z^2$. Similarly, let $B=B_x dy\\wedge dz+B_y dz\\wedge dx+B_z dx\\wedge dy$ be a 2-form. \nShow that $\\langle B,B\\rangle = B_x^2+B_y^2+B_z^2$.}\\end{p}\n\n$\\langle E,E\\rangle=\\delta^{i,j}E_i E_j=E_x^2+E_y^2+E_z^2$. $\\langle B,B\\rangle=\\langle B_x dy\\wedge dz+B_y dz\\wedge dx+B_z dx\\wedge dy,\nB_x dy\\wedge dz+B_y dz\\wedge dx+B_z dx\\wedge dy\\rangle= B_x^2+B_y^2+B_z^2$ since the different wedge products are orthonormal.\\\\\n\n\\begin{p}{In $\\R^4$ let $F$ be the 2-form given by $F=B+E\\wedge dt$, where $E$ and $B$ are given by the formulas\nin the previous exercise. Using the Minkowski metric, calculate $-\\frac{1}{2}\\langle F,F\\rangle$.}\\end{p}\n\n$\\langle F,F\\rangle=\\langle B+E\\wedge dt,B+E\\wedge dt\\rangle=\\langle B,B\\rangle+\\langle B,E\\wedge dt\\rangle+\\langle E\\wedge dt,B\\rangle\n+\\langle E\\wedge dt,E\\wedge dt\\rangle$. The middle two terms are zero by orthogonality of different wedge products, leaving only the\nlast term to deal with. $\\langle E\\wedge dt,E\\wedge dt\\rangle={\\rm det}\\left(\\begin{array}{cc}\\langle E,E\\rangle & \\langle E,dt\\rangle\\\\\n\\langle dt,E\\rangle & \\langle dt,dt\\rangle\\end{array}\\right)={\\rm det}\\left(\\begin{array}{cc}\\langle E,E\\rangle & 0\\\\\n0 &-1\\end{array}\\right)=-\\langle E,E\\rangle$. Hence $-\\frac{1}{2}\\langle F,F\\rangle=\\frac{1}{2}(\\langle E,E\\rangle-\\langle B,B\\rangle)$,\nwhich is the Lagrangian of the source(charge)-free field.\\\\\n\n\\begin{p}{Show that any even permutation of a given basis has the same orientation, while any odd permutation has the\nopposite orientation.}\n\\end{p}\n\nTo check this we must examine the determinant. Since the determinant of a product is the product of determinants, we can\nbreak up the permutation of basis elements into a product of pairwise swaps, whose individual determinant is minus one. Thus,\nif the permutation has an even number of swaps, the new basis has the same orientation, while an odd number of swaps \nimplies the determinant will equal minus one.\n\n\\begin{p}{Let $M$ be an oriented manifold. Show that we can cover $M$ with oriented charts $\\phi_\\alpha:U_\\alpha\\rightarrow\\rn$, that is,\ncharts such that the basis $dx^\\mu$ of cotangent vectors on $\\rn$, pulled back to $U^\\alpha$ by $\\phi_\\alpha$, is positively\noriented.}\n\\end{p}\n\nThe point here is that the volume form on $\\rn$ can be pulled back to $U_\\alpha$ by $\\phi_\\alpha$. \nSince $M$ is orientable, we can define a volume form $\\om$ on all of $M$ to specify the standard orientation. In the chart\n$U_\\alpha$ we can also define a volume form by pulling back the standard volume form $\\bigwedge_\\mu dx^\\mu$ from $\\rn$. \nIf this specifies a different orientation, just reorder the basis of $\\rn$ so as to have the same orientation as given by $\\om$.\nSince $\\om$ exists in all charts $U_\\alpha$, we can always orient these appropriately.\n\n\\begin{p}{Given a diffeomorphism $\\phi:M\\rightarrow N$ from one oriented manifold to another, we say that $\\phi$ is\norientation-preserving if the pullback of any right-handed basis (standard orientation) of a cotangent space in $N$ is a \nright-handed basis of a cotangent space in $M$. Show that if we can cover $M$ with charts\nsuch that the transition functions $\\varphi_\\alpha\\circ \\varphi_\\beta^{-1}$ are orientation-preserving, we \ncan make $M$ into an oriented manifold by using the charts to transfer the \nstandard orientation on $\\rn$ to an orientation on $M$.}\n\\end{p}\n\nIn each chart pull the standard volume form on $\\rn$ back to $M$. Since\nany two charts are connected by an orientation-preserving transition function, the orientation is\nthe same in any chart and the manifold is oriented everywhere.\\\\\n\n\\begin{p}{Let $M$ be an oriented $n$-dimensional semi-Riemannian manifold and let $\\{e_\\mu\\}$ be an oriented orthonormal\nbasis of cotangent vectors at some point $p\\in M$. Show that $e_1\\wedge\\dots\\wedge e_n={\\rm vol}_p$ where \\emph{vol} is \nthe volume form associated to the metric on $M$, and \\emph{vol}$_p$ is its value at $p$.}\n\\end{p}\n\nThe oriented basis in question can be generated from the standard basis by applying a transformation $T$. Since \nthe input and output are orthonormal bases, $T$ is an orthogonal matrix, implying det$T=\\pm 1$. \nBecause it's orientation preserving, det$T=1$. Since in transforming $e_1\\wedge\\dots\\wedge e_n$ into the standard\nbasis at $p$, we pick up a factor equal to the determinant of $T$, $e_1\\wedge\\dots\\wedge e_n={\\rm vol}_p$.\n\n\\begin{p}{Show that if we define the Hodge star operator in a chart using the formula $\\star(e^{i_1}\\wedge \\cdots \\wedge\ne^{i_p})=\\pm e^{i_{p+1}}\\wedge \\cdots \\wedge e^{i_n}$, where the second set of indices consists of those not contained in the\nfirst and the $\\pm$ is given by ${\\rm sign}(i_1,\\dots,i_n)\\epsilon(i_1)\\cdots \\epsilon(i_p)$, it satisfies the property $\\om\\wedge\\star\\mu=\n\\langle\\om,\\mu\\rangle{\\rm vol}$.}\n\\end{p}\n\nFirst expand $\\om$ and $\\mu$: $\\om=\\om_\\alpha e^{\\alpha_1}\\wedge\\cdots\\wedge e^{\\alpha_p}$, \n$\\mu=\\mu_\\beta e^{\\beta_1}\\wedge\\cdots\\wedge e^{\\beta_p}$. Now compute their inner product:\n$\\langle \\om,\\mu\\rangle=\\om_\\alpha\\mu_\\beta\\langle e^{\\alpha_1}\\wedge\\cdots\\wedge e^{\\alpha_p}, \ne^{\\beta_1}\\wedge\\cdots\\wedge e^{\\beta_p}\\rangle=\\om_\\alpha\\mu_\\beta\\delta^{\\alpha,\\beta}\\epsilon(\\alpha_1)\\cdots\\epsilon(\\alpha_p)$. \nNow $\\star\\mu=\\pm\\mu_\\alpha e^{\\alpha_{p+1}}\\wedge\\cdots\\wedge \ne^{\\alpha_n}$, and taking the wedge product with $\\om$, one obtains $\\om\\wedge\\star\\mu=\\pm\\om_\\alpha\\mu_\\beta\\delta^{\\alpha,\\beta}\ne^{\\alpha_1}\\wedge\\cdots e^{\\alpha_n}$. Finally, $e^{\\alpha_1}\\wedge\\cdots e^{\\alpha_n}={\\rm sign}(\\alpha_1,\\dots,\\alpha_n){\\rm vol}$, \nso we obtain the desired result.\n\n\\begin{p}{Calculate $\\star d\\om$ when $\\om$ is a 1-form on $\\R^3$.}\\end{p}\n\n$\\star d\\om$ is essentially the curl of $\\om$. \nFirst expand $\\om$ in an orthonormal basis $e^j$: $\\om=\\om_je^j$. $d\\om=\\partial_k\\om_j dx^k\\wedge dx^j$. Finally, $\\star d\\om=\n\\partial_j\\om_k\\, {\\rm sign}(j,k,\\ell)dx^\\ell=\\partial_j\\om_k \\epsilon^{jk}_\\ell dx^\\ell.$ Note that $\\epsilon(i)=1$ for all $i$ in this case.\n\n\\begin{p}{Calculate $\\star d\\,{\\star}\\om$ when $\\om$ is a 1-form on $\\R^3$.}\\end{p}\n\nThis gives the divergence of $\\om$. Expanding $\\om$ in a basis as above, we obtain $\\star\\om=\\om_j\\epsilon^j_{k\\ell}dx^k\\wedge dx^\\ell$.\nApplying $d$ gives $d\\star\\om=\\partial_m\\om_j\\epsilon^j_{k\\ell}dx^m\\wedge dx^k\\wedge dx^\\ell$. Another $\\star$ yields the final\nresult: $\\star d\\star\\om=\\partial_m\\om_j\\epsilon^j_{k\\ell}\\epsilon^{mk\\ell}$ and since $\\epsilon^j_{k\\ell}\\epsilon^{mk\\ell}=\\delta^{jm}$, we\nhave $\\star d\\star\\om=\\partial_j\\om_k\\delta^{jk}$, the divergence.\n\n\\begin{p}{Give $\\R^4$ the Minkowski metric and the orientation in which\n$(dt,dx,dy,dz)$ is positively oriented. Calculate the Holdge star operator on all wedge products of\n$dx^\\mu$'s Show that on $p$-forms $\\star^2=(-1)^{p(4{-}p)+1}$.}\n\\end{p}\n\nFirst we have the definition vol$=dt\\wedge dx\\wedge dy\\wedge dz$. Using this we can make a table\nof the various starred wedge products. \n\n\n\\begin{tabular}{c|c}\n$\\omega$ & $\\star\\omega$\\\\\\hline\n$dt$ & $-dx\\wedge dy\\wedge dz$\\\\\n$dx$ & $-dt\\wedge dy\\wedge dz$ \\\\\n$dy$ & $dt\\wedge dx\\wedge dz$\\\\\n$dz$ & $-dt\\wedge dx\\wedge dy$\\\\\n$dt\\wedge dx$ & $-dy\\wedge dz$\\\\\n$dt\\wedge dy$ & $dx\\wedge dz$ \\\\\n$dt\\wedge dz$ &$-dx\\wedge dy$\\\\\n$dx \\wedge dy$ & $dt\\wedge dz$ \\\\\n$dx \\wedge dz$ & $-dt\\wedge dy$\\\\\n$dy \\wedge dz$ & $dt\\wedge dx$\\\\\n$dt\\wedge dx\\wedge dy$ & $-dz$ \\\\\n$dt\\wedge dx\\wedge dz$ & $dy$\\\\\n$dt\\wedge dy\\wedge dz$ & $-dx$\\\\\n$dx\\wedge dy\\wedge dz$ & $-dt$\n\\end{tabular}\\\\\n\nFrom the table we immediately see that the given condition is fulfilled.\n\n\\begin{p}{Let $M$ be an oriented semi-Riemannian manifold of dimension\n$n$ and signature $(s,n{-}s)$. Show that on $p$-forms $\\star^2=(-1)^{p(n-p)+s}$.}\n\\end{p}\n\n$\\om\\wedge \\star \\om=\\langle \\om,\\om\\rangle {\\rm vol}$ and $\\star\\om \\wedge \\star^2\\om=\\langle \\star\\om,\n\\star\\om\\rangle{\\rm vol}$, so $\\star\\om \\wedge \\star^2\\om=(-1)^{p(n{-}p)}\\star^2\\!\\om\\wedge \\star\\om=\n\\frac{\\langle \\star\\om, \\star\\om\\rangle}{\\langle \\om,\\om\\rangle}\\,\\om\\wedge\\star\\om$. Hence\n$\\star^2=(-1)^{p(n{-}p)}\\frac{\\langle \\star\\om, \\star\\om\\rangle}{\\langle \\om,\\om\\rangle}$. \nTo evaluate this expression,\nlet $\\om$ be a basis $p$-form $e^{i_1}\\wedge\\dots\\wedge e^{i_p}$. \nSetting $\\epsilon(\\mu)=\\langle e^\\mu,e^\\mu\\rangle$, we have \n$\\langle\\om,\\om\\rangle=\\prod_{j=1}^p \\epsilon(i_j)$. On the other hand, \n$\\star\\om=\\pm e^{i_{p+1}}\\wedge\\dots\\wedge e^{i_n}$, so $\\langle \\star\\om,\\star\\om\\rangle=\n\\prod_{j=p+1}^n \\epsilon(i_{p+j})$. Since all the terms in the denominator are $\\pm 1$, we might as\nwell multiply by $\\langle\\om,\\om\\rangle$ instead of dividing. We then obtain\n$\\star^2=(-1)^{p(n{-}p)}\\prod_{j=1}^n \\epsilon(i_j)$. The latter term equals $(-1)^s$ as $s$ is the \nnumber of basis elements with norm $-1$.\n\n\\begin{p}{Let $M$ be an oriented semi-Riemannian manifold of dimension $n$ and signature $(s,n{-}s)$. \nLet $e^\\mu$ be an orthonormal basis of 1-forms on some chart. Define the \nLevi-Civita sympbol for $1\\leq i_j\\leq n$ by \n\\[ \\epsilon_{i_1\\dots i_n}=\\left\\{\n\\begin{array}{ll}\n{\\rm sign}(i_1\\dots i_n) & \\textrm{all $i_j$ distinct}\\\\0 &{\\rm otherwise}\n\\end{array}\n\\right. \\]}\nShow that for any $p$-form $\\om=\\frac{1}{p!}\\om_{i_1\\dots i_p}e^{i_1}\\wedge\\dots\\wedge e^{i_p}$ we\nhave $(\\star\\om)_{j_1\\dots j_{n{-}p}}=\\frac{1}{p!}\\epsilon^{i_1\\dots i_p}_{j_1\\dots j_{n{-}p}}\\om_{i_1\\dots i_p}$.\n\\end{p}\n\nUsing the results of exercise 64, we have $\\star(e^{i_1}\\wedge \\cdots \\wedge\ne^{i_p})=\\epsilon(i_1)\\cdots \\epsilon(i_p)\\epsilon^{i_1\\dots i_p}_{i_{p+1}\\dots i_{n}}e^{i_{p+1}}\\wedge \\cdots \\wedge e^{i_n}$ \n(no sum). Now $(\\star\\om)_{j_1\\dots j_{n{-}p}}=\\epsilon(j_1)\\cdots \\epsilon(j_{n{-}p})\\langle e^{j_1}\\wedge\\dots \\wedge e^{j_{n{-}p}},\\star \\om\\rangle$, so $(\\star\\om)_{j_1\\dots j_{n{-}p}}=\\frac{(-1)^s}{p!}\\epsilon^{i_1\\dots i_p}_{j_1\\dots j_{n{-}p}}\\om_{i_1\\dots i_p}$. \n\n\\begin{p} {Show that the Maxwell equations $\\nabla\\cdot \\vec{E}=\\rho, \\nabla\\times \\vec{B}-\\frac{\\partial \\vec{E}}{\\partial t}=\\vec{j}$ can be rewritten as $\\star_S\\, d_S \\star_S\\! E=\\rho$ and $-\\partial_t E+\\star_S \\,d_S\\star_S\\! B=j$.}\n\\end{p}\n\nStart from $E=E_j dx^j$. For the remainder of the exercise, $\\star$ means $\\star_S$. So $\\star E=\\frac{1}{2}E_j \\epsilon^{j}_{k\\ell}dx^k\\wedge dx^\\ell$. Then $d\\star\\!E=\\frac{1}{2}\\epsilon^j_{k\\ell}\\partial_m E_j dx^m\\wedge dx^k\\wedge dx^\\ell$, and finally $\\star d\\star\\! E=\\frac{1}{2}\\epsilon^{j}_{k\\ell}\\epsilon^{mk\\ell}\\partial_m E_j=\\partial_j E_j=\\nabla\\cdot\\vec{E}$. Meanwhile $B=\\frac{1}{2}\\epsilon^j_{k\\ell} B_j dx^k\\wedge dx^\\ell$. Thus $\\star B=\\frac{1}{2}\\epsilon^j_{k\\ell}\\epsilon^{k\\ell}_m B_j dx^m=B_j dx^j$. Then $d\\star\\!B=\\partial_k B_j dx^k\\wedge dx^j$\nand $\\star d\\star\\!B=\\epsilon^{kj}_\\ell \\partial_k B_j dx^\\ell$, the components of which are simply\n$\\nabla\\times \\vec{B}$.\\\\\n\n\\begin{p}\n{Show that on a semi-Riemannian manifold $M$ which can\nbe decomposed into $\\R\\times S$, where $S$ is space, the general Maxwell \nequations $dF=0$ and $\\star d\\star F=J$ can be transformed into their ``usual'' \nappearance $d_S E=0, \\partial_t B+d_SE=0, \\star_Sd_S\\star_SE=\\rho,$ and\n$\\star_Sd_S\\star_SB-\\partial_tE=j$.}\n\\end{p}\n\nGiven the decomposition of $M$, we write $F=B+E\\wedge dt$ where $B$ is the\nportion of $F$ defined only on $S$. Similarly $J=j-\\rho dt$. \nThe first equation then reads $dB+dE\\wedge dt=0$. Observe that \n$dB=d_SB+\\partial_t B\\wedge dt$\nwhile $dE\\wedge dt=\\partial_t E dt\\wedge dt+d_S E\\wedge dt$. Thus we have \n$d_S B=0$ and $\\partial_t B+d_S E=0$; the first equation deals with three forms \nonly on space, while the second involves three forms on space and time. \nAssume further that\nthe metric can be decomposed into $g=-dt^2+{}^3g$ \n{\\small (can this always be done when the underlying space is a product? \nThe tangent space can be decomposed, it seems clear, \nfrom the picture of vectors as arrows pointing tangentially to the surface. In the case of\na product space, there are arrows for each manifold, and we take their formal product\nto get the tangent to the total manifold. But from the point of view of vectors \nas derivatives? I think it might be possible by showing that the value of the derivative \nis equal to making changes on each submanifold separately and then adding them. In\nother words, we have a curve $\\gamma(t)$ on $M$, but since $M=\\R\\times S$, we\ncan write this as $(\\gamma_1(t),\\gamma_2(t))$. Now $\\gamma'(t)[f]=\\frac{\\rm d}{{\\rm d}t}f(\\gamma(t))=\\frac{\\rm d}{{\\rm d}t}f(\\gamma_1(t),\\gamma_2(t))=\\frac{\\partial f}{\\partial \\gamma_1}\\frac{\\partial\\gamma_1}{\\partial t}+\\frac{\\partial f}{\\partial \\gamma_2}\\frac{\\partial\\gamma_2}{\\partial t}=\\gamma_1'(t)[f]|_{\\gamma_2(t)}+\\gamma_2'[f]|_{\\gamma_1(t)}\n\\equiv(\\gamma_1'(t),\\gamma_2'(t))[f].$ That's\nwhat we do in $\\R^n$, after all. I guess it's obvious once you map open sets of \nthe manifold to $\\R^n$. But this doesn't mean the metric has to be block diagonal and respect the split; after all that \\emph{doesn't} always happen in $\\R^n$.)}\nNow let $\\star_S$ be the Hodge dual on (differential forms on) $S$. Since $E$ is a one form\non $S$, $\\star E\\wedge dt=\\star_S E$ and similarly since $B$ is a two form on $S$, \n$\\star B=-\\star_S B\\wedge dt$ (using the chart from 67). So $\\star F=\\star_S E-\\star_S B\\wedge dt$ and then $d\\star F=d_S\\star_S E+\\star_S(\\partial_t E)\\wedge dt-d_S\\star_S B\\wedge dt$. Note that the first term is a three form on space, so $\\star d_S\\star_S E$ must be a one form on time, $\\star d_S\\star_S E=-(\\star_S d_S\\star_S E) dt$, where\nthe minus sign can be determined again by the chart from 67. The second term is a three form on space and time, and due to the ordering we'll have $\\star(\\star_S(\\partial_t E)\\wedge dt)=-\\partial_t E$. The last term is also a three form on space and time and\nwe end up with $\\star(d_S\\star_S B\\wedge dt)=-\\star_S d_S\\star_S B$. Thus $\\star_S d_S\\star_S E=\\rho$ and $-\\partial_t E+\\star_S d_S\\star_SB=j$.\n\n\n\\begin{p}{Show that in a Riemannian 4-dimensional manifold any 2-form $F$ can be written as a sum of self-dual and \nanti-self-dual parts, $F=F_++F_-$ for $\\star F_\\pm=\\pm F_\\pm$, if we take $F_\\pm=(F\\pm\\star F)/2$.}\n\\end{p}\n\nClearly $F=F_++F_-$. And \n$\\star F_\\pm=(\\star F\\pm\\star^2 F)/2=(\\pm F+\\star F)/2=\\pm(F\\pm\\star F)/2=\\pm F_\\pm$.\n\n\\begin{p}{Show that in a Lorentizan 4-manifold any 2-form $F$ can be written as a sum of self-dual and \nanti-self-dual parts, $F=F_++F_-$ for $\\star F_\\pm=\\pm i F_\\pm$.}\n\\end{p}\n\nTake $F_\\pm=(F\\mp \\star i F)/2$. Then $\\star F_\\pm=(\\star F\\mp\\star^2 i F)/2=(\\pm i F+\\star F)/2=\\pm i(F\\mp \\star i F)/2=\\pm i F_\\pm$.\n\n\\begin{p}{Show that the equations $\\star_S E=iB$ and $\\star_S B=-i E$ are \nequivalent and that they both hold if at every time $t$ we have $E=E_i dx^i$ and $B=-(i/2)\\varepsilon^j_{\\phantom{j}k\\ell}E_j dx^k\\wedge dx^\\ell$.}\n\\end{p}\n\nApplying $\\star_S$ turns one equation into the other, to they are equivalent. \nFrom exercise 69 we know that $\\star_S E=(1/2)\\varepsilon^j_{\\phantom{j}k\\ell}E_j dx^k\\wedge dx^\\ell$ which is clearly equal to $iB$.\n\n\\begin{p}{Show that the second Maxwell equation $\\partial_t B+d_SE=0$ leads to ${}^3k\\wedge E=k_0 B$.}\n\\end{p}\n\nStart with $d_S E=d_S(e^{ik_\\mu x^\\mu}E_j dx^j)=\\partial_k(e^{ik_\\mu x^\\mu}E_j)dx^j\\wedge dx^k=i k_k E_jdx^j\\wedge dx^k=-i{}^3k\\wedge E$. Meanwhile \n$\\partial_t B=ik_0 B$, so $\\partial_t B+d_SE=i(k_0 B-{}^3k\\wedge E)=0$, the desired \nresult.\n\n\\begin{p}%76\n{Show that ${}^3k\\wedge E=-ik_0 \\star_s E$ implies $k_\\mu k^\\mu=0$.}\n\\end{p}\n\nStart from $\\langle {}^3k\\wedge E,{}^3k\\wedge E\\rangle=k_0^2\\langle\\star_s E,\\star_s E\\rangle$. \nNote that the inner product should be antilinear in one of its inputs. Using the coordinate expressions for $E$\nand ${}^3k$ we have $\\langle {}^3k\\wedge E,{}^3k\\wedge E\\rangle=k_\\ell k_{\\ell'}^*E_j E_{j'}^*\\langle dx^\\ell\\wedge dx^j,dx^{\\ell'}\\wedge dx^{j'}\\rangle=k_\\ell k_{\\ell'}^*E_j E_{j'}^*(\\delta^{j j'}\\delta^{\\ell\\ell'}-\\delta^{j'\\ell}\\delta^{j\\ell'})=\\langle {}^3k,{}^3k\\rangle\\,\\langle E,E\\rangle-|\\langle {}^3k,E\\rangle|^2=\\langle {}^3k,{}^3k\\rangle\\,\\langle E,E\\rangle$, since $E$ is orthogonal to ${}^3k$.\nOn the other hand, $k_0^2\\langle\\star_s E,\\star_s E\\rangle=\\frac{k_0^2}{4}\\varepsilon^j_{k\\ell}\\varepsilon^{j'}_{k'\\ell'}E_jE_{j'}\\langle dx^k\\wedge dx^\\ell,dx^{k'}\\wedge dx^{\\ell'}\\rangle=\\frac{k_0^2}{4}\\varepsilon^j_{k\\ell}\\varepsilon^{j'}_{k'\\ell'}E_jE_{j'}(\\delta^{kk'}\\delta^{\\ell \\ell'}-\\delta^{k\\ell'}\\delta^{k'\\ell})=k_0^2\\langle E,E\\rangle$.\nThus, $\\langle {}^3k,{}^3k\\rangle=k_0^2$. Using the metric, we have $k_0^2=-k_0k^0$, and the \nother components are unchanged, so $k_\\mu k^\\mu=0$.\n\n\n\\begin{p}%77\n{Show $\\vec{E}=(0,e^{i(t-x)},-ie^{i(t-x)})$, $\\vec{B}=i\\vec{E}$ satisfy the vacuum Maxwell equations. [corrected from the book]}\n\\end{p}\n\nSince $\\vec{\\nabla}\\cdot \\vec{E}=0$ (by inspection), \n$\\vec{\\nabla}\\cdot \\vec{B}=0$, too. Now $\\vec{\\nabla}\\times \\vec{E}=\\vec{E}$ and $\\partial_t \\vec{B}=i\\partial_t \\vec{E}=-\\vec{E}$, so $\\vec{\\nabla}\\times \\vec{E}+\\partial_t \\vec{B}=0$. Finally,\n$ \\vec{\\nabla}\\times \\vec{B}=i\\vec{E}$, so $\\vec{\\nabla}\\times \\vec{B}-\\partial_t \\vec{E}=0$.\n\n\\begin{p}%{78}\n{Prove that all self-dual and anti-self-dual plane wave solutions are left and right circularly polarized, respectively.}\n\\end{p}\n\nWithout loss of generality we can take ${}^3k$ to point in the $x$ direction. Then the self-dual case is worked\nout in detail in the book. Left circular polarization can be recognized from the form of $E$, namely that the \npolarization vector is proportional to $(1,-i)$; right circular polarization is given by $(1,i)$. \nIn the anti-self-dual case $\\star_S E=-iB$ or, equivalently,\n$\\star_SB=iE$. The first Maxwell equation $d_SB=0$ reads as before, $B\\wedge {}^3k=0$, so\n$\\langle E,{}^3k\\rangle=0$ still holds. The second equation, $\\partial_t B+d_SE=0$, leads to ${}^3k\\wedge E=k_0B$ as before. Rewriting this in terms of $E$ we obtain ${}^3k\\wedge E=ik_0\\star_SE$, which\nagain leads to $\\langle {}^3k\\wedge E,{}^3k\\wedge E\\rangle=k_0^2\\langle\\star_s E,\\star_s E\\rangle$, as \nin exercise 76. Thus we know that $k$ must be lightlike, so we can assume without loss of generality \nthat $k=dt-dx$. $E$ must be orthogonal to ${}^3k$, so $E=a dy+bdz$. We now use the second \nMaxwell equation to determine $a$ and $b$. ${}^3k\\wedge E=-(a dx\\wedge dy+b dx\\wedge dz)$, \nwhile $ik_0\\star_SE=i(a dz\\wedge dx+b dx\\wedge dy)$, so $b=ia$, which is the condition for right circular\npolarization.\n\n\\begin{p}%{79}\n{Let $P:\\R^4\\rightarrow\\R^4$ be the parity transformation $P(t,x,y,z)=(t,-x,-y,-z)$. Show that if $F$ is a self-dual solution of Maxwell's equations, the pullback $P^*F$ is an anti-self-dual solution, and vice versa.}\n\\end{p}\n\nWe don't really need to show both, since the vice versa part follows from $P^*P^*F=F$. To see that $P^*F$ takes a self-dual solution to an anti-self-dual solution, note that $P^*E=-E$, while $P^*B=B$. (This is why\n$B$ is called an axial vector sometimes.) If $F=dE\\wedge dt+B$ was self-dual to begin with, meaning $\\star_SE=iB$, the new $F'=dE'\\wedge dt'+B'=-dE\\wedge dt+B$ is anti-self-dual: $\\star_SE'=-\\star_SE=-iB=-iB'$. \n\n", "meta": {"hexsha": "921ffcae5e9b240052726d8c01e8cbf55bdb0ac9", "size": 25213, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/I5.tex", "max_stars_repo_name": "joerenes/Baez-Muniain-solutions", "max_stars_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-04-13T12:10:03.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T18:18:34.000Z", "max_issues_repo_path": "src/I5.tex", "max_issues_repo_name": "joerenes/Baez-Muniain-solutions", "max_issues_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-04-13T12:15:30.000Z", "max_issues_repo_issues_event_max_datetime": "2017-04-13T20:19:44.000Z", "max_forks_repo_path": "src/I5.tex", "max_forks_repo_name": "joerenes/Baez-Muniain-solutions", "max_forks_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.5072886297, "max_line_length": 571, "alphanum_fraction": 0.6892872724, "num_tokens": 9011, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.810478913248044, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.6997606909101299}} {"text": "\\subsection{Graph embedding}\\label{subsec:graph_embedding}\n\n\\begin{definition}\\label{def:quiver_geometric_realization}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver}. Our goal is to construct a \\hyperref[def:topological_space]{topological space} that translates the connectivity properties of \\( Q \\) into their topological equivalents.\n\n Consider the \\hyperref[def:topological_sum]{topological sum}\n \\begin{equation*}\n \\mscrS \\coloneqq \\parens[\\Bigg]{ \\coprod_{a \\in A} [0, 1] } \\amalg \\parens[\\Bigg]{ \\coprod_{\\deg(v) = 0} \\set{ v } }.\n \\end{equation*}\n\n The space \\( \\mscrS \\) consists of disjoint unit intervals, one for each arc, and of disjoint points, one for each \\hyperref[def:hypergraph/degree]{isolated vertex}.\n\n We now want to glue common endpoints of arcs in \\( \\Sigma \\). We define the following function:\n \\begin{equation}\\label{eq:def:quiver_geometric_realization}\n \\begin{aligned}\n &R: V \\cup A \\to \\pow(\\mscrS), \\\\\n &R(v) \\coloneqq \\begin{cases}\n \\set[\\Big]{ (v, v) }, &\\deg(v) = 0 \\\\\n \\set[\\Big]{ (0, a) \\given h(a) = v } \\cup \\set[\\Big]{ (1, a) \\given t(a) = v }, &\\deg(v) > 0,\n \\end{cases} \\\\\n &R(a) \\coloneqq \\set[\\Big]{ \\set{ (x, a) } \\given 0 \\leq x \\leq 1 }.\n \\end{aligned}\n \\end{equation}\n\n The family\n \\begin{equation*}\n \\mscrX \\coloneqq \\set[\\Big]{ R(v) \\given* v \\in V } \\cup \\set[\\Big]{ \\Int R(a) \\given* a \\in A }.\n \\end{equation*}\n is a \\hyperref[def:set_partition]{partition} of \\( \\mscrS \\). For each vertex, there is a single point in \\( \\mscrX \\) (which is a set in \\( \\mscrS \\)) and for each arc, the interior of the arc is a subset of \\( \\mscrX \\).\n\n We can endow the partition \\( \\mscrX \\) it with a \\hyperref[def:topological_quotient]{quotient topology} \\( \\mscrT \\). We will call the topological space \\( (\\mscrX, \\mscrT, R) \\) endowed with \\( R \\) the \\term{geometric realization} of \\( G \\).\n\n \\begin{thmenum}\n \\thmitem{def:quiver_geometric_realization/undirected} For an \\hyperref[def:undirected_multigraph]{undirected multigraph} \\( G = (V, E, \\mscrE) \\), the geometric realization is any of the geometric realizations of its \\hyperref[def:multigraph_orientation]{orientations}. This construction is dependent on a choice function, but fortunately all the geometric realizations are homeomorphic as shown in \\fullref{thm:undirected_multigraph_geometric_realizations_homeomorphic}. Hence, for a lot of purposes, we can speak of \\enquote{the} geometric realization of an undirected multigraph.\n\n \\thmitem{def:quiver_geometric_realization/drawing} We will call any \\hyperref[def:global_continuity]{continuous function} with domain \\( (\\mscrX, \\mscrT) \\) a \\term{graph drawing}. The term \\enquote{graph drawing} is not standard terminology, but unfortunately non-injective continuous images of the realization have no established name.\n\n \\thmitem{def:quiver_geometric_realization/embedding} An injective graph drawing is called a \\term{graph embedding}.\n\n Every graph can be embedded into \\( \\BbbR^3 \\) as shown in \\fullref{thm:quiver_can_be_embedded_into_r3}.\n\n \\thmitem{def:quiver_geometric_realization/linear} If a graph can be embedded into \\( \\BbbR \\), we say that it is \\term{linear}.\n\n \\thmitem{def:quiver_geometric_realization/planar} If a graph can be embedded into \\( \\BbbR^2 \\), we say that it is \\term{planar}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:def:quiver_geometric_realization}\n We will give a few examples of \\hyperref[def:quiver_geometric_realization/undirected]{quiver geometric realizations}.\n\n \\begin{thmenum}\n \\thmitem{ex:def:quiver_geometric_realization/edgeless} The \\hyperref[def:quiver_geometric_realization/undirected]{geometric realization} of an edgeless quiver is the empty topological space.\n\n \\thmitem{ex:def:quiver_geometric_realization/positive_integers} Consider the reduced positive integer graph \\eqref{eq:ex:infinite_integer_graphs/positive}. We start with \\( \\aleph_0 \\) copies of \\( [0, 1] \\) and glue both ends of each of them except for the first. Thus, we obtain (a space homeomorphic to)\n \\begin{equation*}\n \\bigcup_{k \\geq 0} [k, k + 1] = [0, \\infty).\n \\end{equation*}\n\n Therefore, \\eqref{eq:ex:infinite_integer_graphs/positive} is a \\hyperref[def:quiver_geometric_realization/linear]{linear graph}.\n\n \\thmitem{ex:def:quiver_geometric_realization/k3} The graph with vertices \\( V = \\set{ a, b, c } \\) and arcs \\( \\set{ \\overbrace{a \\to b}^{e_1}, \\overbrace{b \\to c}^{e_2}, \\overbrace{c \\to a}^{e_3} } \\) is more subtle.\n\n We start with three copies of the interval \\( [0, 1] \\), depicted in \\eqref{eq:ex:def:quiver_geometric_realization/k3/relization} as upward-pointing arrows, and use dashed lines to connect the endpoints that we want to glue together.\n \\begin{equation}\\label{eq:ex:def:quiver_geometric_realization/k3/relization}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__def__graph_geometric_realization.pdf}\n \\end{aligned}\n \\end{equation}\n\n After contracting the dashed lines, we obtain a topological space that can easily be \\hyperref[def:quiver_geometric_realization/embedding]{embedded} into \\( \\BbbR^2 \\). An obvious embedding corresponds to \\enquote{pulling up} \\( e_2 \\) and \\( e_3 \\):\n \\begin{equation}\\label{eq:ex:def:quiver_geometric_realization/k3/embedding}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__def__graph_geometric_realization.pdf}\n \\end{aligned}\n \\end{equation}\n\n This is only one possible embedding of the geometric realization. It is sufficient, however, for proving that the graph is \\hyperref[def:quiver_geometric_realization/planar]{planar}. The underlying undirected graph is the \\hyperref[ex:complete_graph]{complete graph} \\( K_3 \\), hence we have shown that \\( K_3 \\) is also planar.\n\n \\thmitem{ex:def:quiver_geometric_realization/k4} \\Cref{fig:ex:complete_graph} shows that the complete graph \\( K_4 \\) is planar.\n\n This is not-at-all obvious from its geometric realization, however.\n \\begin{equation}\\label{eq:ex:def:quiver_geometric_realization/k4/realization}\n \\begin{aligned}\n \\includegraphics[page=3]{output/ex__def__graph_geometric_realization.pdf}\n \\end{aligned}\n \\end{equation}\n\n This example shows that constructing embeddings can be a tedious task.\n \\end{thmenum}\n\\end{example}\n\n\\begin{proposition}\\label{thm:linear_quiver_equivalence}\n If finite quiver is \\hyperref[def:quiver_geometric_realization/linear]{linear}, it has degree at most \\( 2 \\).\n\\end{proposition}\n\\begin{proof}\n Let \\( Q = (V, A, t, h) \\) be a quiver and let \\( (\\mscrX, \\mscrT, R) \\) be its geometric realization. Let \\( f: \\mscrX \\to \\BbbR \\) be an injective continuous function, i.e. a topological embedding.\n\n Suppose that the vertex \\( v \\) has degree larger than \\( 2 \\). It is sufficient to consider the case where \\( a \\), \\( b \\) and \\( c \\) are distinct arcs incident to \\( v \\).\n\n We have\n \\begin{equation*}\n R(v) \\in R(a) \\cap R(b) \\cap R(c)\n \\end{equation*}\n thus\n \\begin{equation*}\n f(R(v)) \\in f(R(a)) \\cap f(R(b)) \\cap f(R(c))\n \\end{equation*}\n\n Since \\( R(a) \\), \\( R(b) \\) and \\( R(c) \\) are \\hyperref[def:connected_space]{connected}, so are their images under \\( f \\). If \\( f(R(a)) \\) has a point to the right of \\( f(R(v)) \\), then \\( f(R(b)) \\) must be left of \\( R(v) \\) and there remains nowhere to place \\( R(c) \\).\n\n Therefore, \\( \\deg(v) \\leq 2 \\) and, since \\( v \\) was arbitrary, \\( \\deg(Q) \\leq 2 \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:quiver_can_be_embedded_into_r3}\n Every finite quiver can be embedded into \\( \\BbbR^3 \\).\n\\end{proposition}\n\\begin{proof}\n Let \\( Q = (V, A, t, h) \\) be a finite quiver of order \\( n \\). By definition of cardinality, there exists a bijection from \\( n \\) to \\( V \\).\n\n Place the vertices of \\( Q \\) along the \\hyperref[thm:moment_curve]{moment curve} by using \\( \\gamma(k) \\) as the position for the \\( k \\)-th vertex of \\( V \\). Then by \\fullref{thm:moment_curve}, no four of these points are \\hyperref[def:collinear_complanar]{complanar}. Hence, if we connect their vertices using a straight line where there is an arc, no two lines would intersect.\n\n Therefore, this is an embedding.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:quiver_geometric_realization/properties}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver} and let \\( (\\mscrX, \\mscrT, R) \\) be its \\hyperref[def:quiver_geometric_realization]{geometric realization}.\n\n \\begin{thmenum}\n \\thmitem{thm:def:quiver_geometric_realization/properties/isolated} A vertex \\( v \\in V \\) is isolated in \\( Q \\), i.e. has degree zero, if and only if \\( R(v) \\) is an isolated point of \\( \\mscrX \\).\n\n \\thmitem{thm:def:quiver_geometric_realization/properties/t1} If \\( Q \\) is \\hyperref[def:hypergraph/degree]{locally finite}, then the space \\( (\\mscrX, \\mscrT) \\) satisfies the \\ref{def:separation_axioms/T1} separation axiom.\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:quiver_geometric_realization/properties/isolated} For every isolated vertex \\( v \\), the point \\( R(v) = \\set{ (v, v) } \\) is isolated by definition.\n\n Now suppose that \\( v \\) is not an isolated vertex. Then \\( R(v) \\) is defined differently in \\eqref{eq:def:quiver_geometric_realization}. If there exists an arc \\( a \\) such that \\( v = h(a) \\), then \\( (a, 0) \\in R(v) \\) and hence there exists no neighborhood of \\( R(v) \\) disjoin from \\( R(a) \\), hence \\( R(v) \\) is not a disjoint point of \\( \\mscrX \\). The case \\( v = t(a) \\) is handled analogously.\n\n \\SubProofOf{thm:def:quiver_geometric_realization/properties/t1} Let \\( x \\in \\mscrX \\).\n\n If \\( x = \\set{ (t, a) } \\) for some arc \\( a \\) and \\( 0 < t < 1 \\), then \\( \\set{ x } \\) is closed because \\( [0, 1] \\) satisfies \\hyperref[def:separation_axioms/T1]{T1} and hence \\( \\set{ t } \\) is closed in \\( [0, 1] \\).\n\n If \\( x = R(v) = \\set{ (v, v) } \\) for some isolated vertex \\( v \\), then \\( R(v) \\) is clopen by definition.\n\n If \\( x = R(v) \\) for some vertex \\( v \\) of positive degree, then\n \\begin{equation*}\n R(v) = \\set[\\Big]{ (0, a) \\given h(a) = v } \\cup \\set[\\Big]{ (1, a) \\given t(a) = v }.\n \\end{equation*}\n\n Both \\( \\set{ 0 } \\) and \\( \\set{ 1 } \\) are closed in \\( [0, 1] \\), hence \\( \\set{ 0, 1 } \\) is also closed in \\( [0, 1] \\). Since \\( Q \\) is locally finite, \\( R(v) \\) is the union of finitely many closed sets and is thus itself closed.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:undirected_multigraph_geometric_realizations_homeomorphic}\n Let \\( G = (V, E, \\mscrE) \\) be an \\hyperref[def:undirected_multigraph]{undirected multigraph}.\n\n Let \\( (\\mscrX_c, \\mscrT_c, R_c) \\) and \\( (\\mscrX_d, \\mscrT_d, R_d) \\) be geometric realizations corresponding to the \\hyperref[def:multigraph_orientation]{orientations} \\( O_c(G) \\) and \\( O_d(G) \\) of \\( G \\).\n\n Then \\( (\\mscrX_c, \\mscrT_c) \\) and \\( (\\mscrX_c, \\mscrT_d) \\) are homeomorphic.\n\\end{proposition}\n\\begin{proof}\n Let \\( h_c \\) and \\( h_d \\) be the head functions from the \\hyperref[def:quiver]{quivers} \\( O_c(G) \\) and \\( O_d(G) \\).\n\n Define the function\n \\begin{equation*}\n \\begin{aligned}\n &f: \\mscrX_c \\to \\mscrX_d \\\\\n &f(x) \\coloneqq \\begin{cases}\n \\set[\\Big]{ (0, e) \\given h_d(e) = v } \\cup \\set[\\Big]{ (1, e) \\given t_d(e) = v }, &x = R_c(v) \\T{and} \\deg(v) > 0 \\\\\n \\set[\\Big]{ (e, 1 - t) }, &x = \\set{ (e, t) } \\T{and} h_c(e) \\neq h_d(e) \\\\\n x, &\\T{otherwise.}\n \\end{cases}\n \\end{aligned}\n \\end{equation*}\n\n This function \\enquote{reverses} the direction of some of the intervals in the construction of the realizations and fixes everything else in place. It is clearly bijective. It is also continuous because it satisfies \\fullref{def:global_continuity/closure}. Finally, it is a homeomorphism because the inverse function is defined in the same way by interchanging \\( c \\) and \\( d \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:quiver_geometric_realization_paths}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver} and let \\( (\\mscrX, \\mscrT, R) \\) be its \\hyperref[def:quiver_geometric_realization]{geometric realization}.\n\n \\begin{thmenum}\n \\thmitem{thm:quiver_geometric_realization_paths/quiver_to_realization} If there exists an \\hyperref[def:quiver_path/undirected]{undirected path} \\( p = (v, e_1, \\ldots, e_n) \\) from \\( s \\) to \\( f \\), then there exists a \\hyperref[def:parametric_curve]{continuous path} \\( \\gamma: [0, 1] \\to \\mscrX \\) from \\( R(s) \\) to \\( R(f) \\).\n\n \\thmitem{thm:quiver_geometric_realization_paths/realization_to_quiver} If \\( Q \\) is finite and if there exists a simple continuous path \\( \\gamma: [0, 1] \\to \\mscrX \\) from \\( R(s) \\) to \\( R(f) \\), then there exists a simple undirected path from \\( s \\) to \\( f \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n The case \\( s = f \\) is trivial, hence we assume that \\( s \\neq f \\).\n\n \\SubProofOf{thm:quiver_geometric_realization_paths/quiver_to_realization} Suppose that \\( p \\) is an undirected path from \\( s \\) to \\( f \\). We will use \\hyperref[rem:induction/well_founded]{strong induction} on the length of \\( p \\) to show that there exists a continuous path between the points \\( R(s) \\) and \\( R(f) \\) of \\( \\mscrX \\).\n\n Suppose that the statement holds for paths of length smaller than \\( n \\) and let\n \\begin{equation*}\n p = (v, e_1, \\ldots, e_{n-1}, e_n)\n \\end{equation*}\n be a path of length \\( n \\) from some vertex \\( s \\) to \\( f \\). The inductive hypothesis holds for the initial segment \\( (e_1, \\ldots, e_{n-1}) \\) of \\( p \\), hence there exists a continuous path \\( \\gamma: [0, 1] \\to \\mscrX \\) from \\( s \\) to an endpoint of \\( e_{n-1} \\).\n \\begin{itemize}\n \\item If both \\( e_{n-1} \\) and \\( e_n \\) are positively oriented, then \\( t(e_{n-1}) = h(e_n) \\) and thus \\( \\gamma \\) is a continuous path from \\( R(s) \\) to \\( R(h(e_n)) \\). We can then append to \\( \\gamma \\) the continuous path from \\( R(h(e_n)) \\) to \\( R(t(e_n)) = R(f) \\) to obtain a path from \\( R(s) \\) to \\( R(f) \\).\n\n \\item If \\( e_{n-1} \\) is positively oriented but \\( e_n \\) is not, then \\( \\gamma \\) is a continuous path from \\( R(s) \\) to \\( R(h(e_{n-1})) \\). We can then append to \\( \\gamma \\) the paths from \\( R(h(e_{n-1})) \\) to \\( R(t(e_{n-1})) = R(t(e_n)) \\) and from \\( R(t(e_n)) \\) to \\( R(h(e_n)) = R(f) \\).\n\n \\item Similarly, if \\( e_{n-1} \\) is negatively oriented but \\( e_n \\) is not, then \\( \\gamma \\) is a continuous path from \\( R(s) \\) to \\( R(t(e_{n-1})) \\), and we can append to it the paths from \\( R(t(e_{n-1})) \\) to \\( R(h(e_{n-1})) = R(h(e_n)) \\) and from \\( R(h(e_n)) \\) to \\( R(t(e_n)) \\).\n\n \\item Finally, if both \\( e_{n-1} \\) and \\( e_n \\) are negatively oriented, then \\( \\gamma \\) is a continuous path from \\( R(s) \\) to \\( R(t(e_{n-1})) \\), and we can append to it the paths from \\( R(t(e_{n-1})) \\) to \\( R(h(e_{n-1})) = R(t(e_n)) \\) and from \\( R(t(e_n)) \\) to \\( R(h(e_n)) = R(f) \\).\n \\end{itemize}\n\n We have shown that there exists a continuous path from \\( R(s) \\) to \\( R(f) \\).\n\n \\SubProofOf{thm:quiver_geometric_realization_paths/realization_to_quiver} Suppose that \\( Q \\) is finite and let \\( \\gamma: [0, 1] \\to \\mscrX \\) be a continuous path from \\( R(s) \\) to \\( R(f) \\). We will show that there is an undirected path from \\( s \\) to \\( f \\).\n\n \\SubProof*{\\( \\gamma \\) contains no isolated vertices} We have\n \\begin{equation}\\label{eq:thm:quiver_geometric_realization_paths/full_preimage}\n \\gamma^{-1}(\\mscrX) = \\bigcup_{a \\in A} \\gamma^{-1}(R(a)) \\cup \\bigcup_{\\mathclap{\\deg(v) = 0}} \\gamma^{-1}(R(v)).\n \\end{equation}\n\n For each arc \\( a \\), the set \\( R(a) \\) is closed as a homeomorphic image of the unit interval \\( [0, 1] \\). From \\fullref{thm:def:quiver_geometric_realization/properties/t1} it follows that \\( \\set{ R(v) } \\) is a closed set for every vertex \\( v \\in V \\). Therefore, \\( \\gamma^{-1}(\\mscrX) \\) is a union of disjoint closed sets.\n\n If we assume that \\( \\gamma \\) passes through an isolated vertex \\( v \\), then \\( \\gamma^{-1}(v) \\) would be a nonempty closed set. Then \\( \\img(\\gamma) = \\set{ v } \\) because otherwise \\( [0, 1] \\) would be the union of finitely many nonempty disjoint closed sets, which would contradict \\fullref{def:connected_space/closed_union} because \\( [0, 1] \\) is \\hyperref[def:connected_space]{connected}.\n\n But \\( \\gamma \\) passes through at least two vertices because \\( s \\neq t \\), and hence it doesn't pass through any isolated vertex.\n\n \\SubProof*{\\( \\gamma \\) contains the entirety of each arc it intersects} Suppose that \\( R(a) \\cap \\img(\\gamma) = \\varnothing \\) for some arc \\( a \\).\n\n Let \\( l \\coloneqq \\inf\\set{ 0 < t < 1 \\given \\gamma(t) \\in R(a) } \\) and \\( r \\coloneqq \\sup\\set{ 0 < t < 1 \\given \\gamma(t) \\in R(a) } \\). The closed interval \\( [l, r] \\) is compact, hence \\( \\gamma([l, r]) \\) is a continuous image of a compact set and hence is itself compact.\n\n Since \\( \\Int R(a) \\) is a subset of \\( \\gamma([l, r]) \\) and since \\( \\gamma([l, r]) \\) is closed, its closure \\( R(a) \\) is also a subset of \\( \\gamma([l, r]) \\).\n\n Therefore, if an internal point of an arc belongs to \\( \\img(\\gamma) \\), so does the entire arc.\n\n \\SubProof*{\\( \\gamma \\) contains an arc} From \\eqref{eq:thm:quiver_geometric_realization_paths/full_preimage} if follows that\n \\begin{equation*}\n \\gamma^{-1}(\\mscrX)\n =\n \\bigcup_{a \\in A} \\gamma^{-1}(R(a))\n \\reloset{\\ref{thm:topological_closure_operator_axioms/CO3}} =\n \\cl\\parens*{ \\bigcup_{a \\in A} \\gamma^{-1}(\\Int R(a)) }.\n \\end{equation*}\n\n Hence, \\( \\img(\\gamma) \\) contains at least one internal point of some arc and thus the entire arc.\n\n \\SubProof*{\\( \\gamma \\) induces an undirected path from \\( s \\) to \\( f \\)} We have that \\( \\gamma(0) = R(s) \\). By what we have already shown, \\( \\img(\\gamma) \\) contains no isolated points, hence \\( \\gamma \\) contains the set \\( R(a) \\), where \\( a \\) is some arc incident to \\( s \\).\n\n Suppose that \\( h(a) = s \\). Then \\( p = (s, a) \\) is an undirected path from \\( s \\) to \\( t(a) \\). If \\( t(a) = f \\), the proof is finished. Otherwise, define\n \\begin{equation*}\n x_0 \\coloneqq \\sup\\set{ x \\in [0, 1] \\given \\gamma(x) = R(t(a)) }\n \\end{equation*}\n and\n \\begin{equation*}\n \\begin{aligned}\n &\\delta: [0, 1] \\to \\mscrX, \\\\\n &\\delta(x) \\coloneqq \\gamma(x_0 + (1 - x_0) x).\n \\end{aligned}\n \\end{equation*}\n\n Then \\( \\delta \\) is a continuous path from \\( R(t(a)) \\) to \\( R(f) \\).\n\n We now proceed by \\fullref{thm:bounded_transfinite_induction} bounded by the number of arcs to define an undirected path \\( p \\) from \\( s \\) to \\( f \\). Since the path \\( \\gamma \\) is simple, it does not intersect itself and hence the image of the arc \\( a \\) at each step will not be in \\( \\delta \\).\n\\end{proof}\n\n\\begin{corollary}\\label{thm:quiver_geometric_realization_connectedness}\n A finite \\hyperref[def:quiver]{quiver} is \\hyperref[def:quiver_connectedness/weak]{weakly connected} if and only if its \\hyperref[def:quiver_geometric_realization/undirected]{geometric realization} is \\hyperref[def:path_connected]{path connected}.\n\\end{corollary}\n\\begin{proof}\n Follows from \\fullref{thm:quiver_geometric_realization_paths}.\n\\end{proof}\n\n\\begin{corollary}\\label{thm:undirected_multigraph_geometric_realization_connectedness}\n A finite \\hyperref[def:undirected_multigraph]{undirected multigraph} is \\hyperref[def:undirected_multigraph_connectedness]{connected} if and only if its \\hyperref[def:quiver_geometric_realization/undirected]{geometric realization} is a \\hyperref[def:path_connected]{path connected}.\n\\end{corollary}\n\\begin{proof}\n Follows from \\fullref{thm:quiver_geometric_realization_connectedness}.\n\\end{proof}\n", "meta": {"hexsha": "dfe0a2b3594cf085b08255071dc4c74ffdef7a1b", "size": 19928, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/graph_embedding.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/graph_embedding.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/graph_embedding.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 72.9963369963, "max_line_length": 586, "alphanum_fraction": 0.6605279004, "num_tokens": 6465, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430562234878, "lm_q2_score": 0.8418256472515683, "lm_q1q2_score": 0.6996775412639843}} {"text": "\\subsection{The Lanczos Tridiagonalization Step}\n\nGolub explains in \\cite{golub13} that one of the problems with the\nPower Method, is that it does not take advantage of the previously\ncalculated information. During the iterations of the Power Method, say\nuntil step $k$, we have calculated already the set of vectors\n$K(A,q_0,k) = \\{A\\vec{q_0},A\\vec{q_1},\\dots,A\\vec{q_k}\\}$; still, they\nare not used \nat all when looking for an estimate of the eigenvector. Such\nlimitation is addressed by the Lanczos Process, named after its\ncreator in 1950 (\\cite{lanczos1950}). The subspace spanned by the\n$K(A,q_0,k)$ is called Krylov Subspace of order $k$ \\footnote{The\n concept itself of Krylov Spaces is thanks for \n Krylov and dates back to 1931 (see \\cite{krylov1931})}, which is why\nthe Lanczos Process is usually cataloged as a Krylov Subspace\nmethod. \\\\\n\nGoing back to Lanczos, this subsection will only explain the iterative\nstep (called Lanczos Tridiagonalization Step). It works as follows; let us\nsuppose that we have an square symmetric matrix $A^{n \\times n}$, and\nthat we want a few of its biggest eigenvalues (as it is the case in LSI\napplications) \\footnote{The Lanczos Process can also calculate a few\n of the smallest eigenvalues, but we are not interested in such case\n for LSI applications.}. Each iteration $k$ of the algorithm generates a\ntridiagonal matrix $T_k \\in \\R{k \\times\n k}$ \\footnote{Matrices with the middle, upper and lower\n diagonals.}, and the whole sequence ${T_k}$ is progressively \napproximating the biggest eigenvalues of the original matrix\n$A$. \\\\\n\nThere are several ways of stating the algorithm for the Lanczos\nTridiagonalization Step, the following is taken from Golub \\cite{golub13};\nthough it is not the the most numerically stable. That honor\ncorresponds to the ones created by Paige\n(\\cite{paige71},\\cite{paige76}); we preferred Golub's one for our\nexposition, aiming to have an easier introduction to the procedure: \n\n\\begin{algorithm}\n \\label{alg:lanczos-step}\n \\caption{The Lanczos Tridiagonalization Step}\n%\n \\setstretch{1.5}\n \\SetKwInOut{Input}{Input}\n \\SetKwInOut{Output}{Output}\n \\DontPrintSemicolon\n%\n \\Input{A unit vector $\\vec{q_1} \\in \\R{n}$ and a symmetric matrix $A^{n\n \\times n}$}\n%\n \\Output{The sequences $\\{\\alpha_i\\}$, $\\{\\beta_i\\}$ and matrix $Q\n = [ \\vec{q_1} | \\vec{q_2} | \\cdots ]$ }\n%\n $k \\gets 0, \\beta_0 \\gets 1, \\vec{q_0} \\gets 0, r_0 \\gets \\vec{q_1}$ \\;\n \\While {$k = 0 \\lor \\beta_k \\ne 0$}\n {\n $\\vec{q_{k+1}} \\gets \\dfrac{\\vec{r_k}}{B_k}$ \\;\n $k \\gets k + 1$ \\;\n $\\alpha_k \\gets \\trans{\\vec{q_k}}A\\vec{q_k}$ \\;\n $\\vec{r_k} \\gets A\\vec{q_k} - \\alpha_kq_k - \\beta_{k-1}\\vec{q_{k-1}}$ \\;\n $\\beta_k \\gets \\norm{\\vec{r_k}}_2$ \\;\n }\n%\n return $(\\{\\alpha_i\\}, \\{\\beta_i\\}, \n Q = [ \\vec{q_1} | \\vec{q_2} | \\cdots ])$ \\;\n\\end{algorithm}\n\\hfill\n\nThe \\cref{alg:lanczos-step} is essentially applying Gram-Schmidt process,\nbut only against the last two vectors. Golub derives the algorithm\nfrom a relation between tridiagonalization, and the QR factorization of\nthe matrix formed by vectors $K(A,q_0,k)$; see \\cite{golub13} for\nfurther details. \\\\\n\nGolub goes even further in the cited book, and proves the following\nproperties about \\cref{alg:lanczos-step}. We will omit the theorem\nstatement, and just comment directly its results: \\\\\n\n\\begin{itemize}\n \\item The algorithm runs until $k = m = rank(K(A,q_0,k))$. This contrasts \n with the unknown number of steps of the Power Method\n (\\cref{alg:power-method}). \\\\\n\n \\item For $k = 1:m$ we have $AQ_k = Q_kT_k + \\vec{r_k}\\trans{e_k}$,\n where $Q = [\\vec{q_1} | \\cdots | \\vec{q_k} ]$ has orthonormal\n columns that span the Krylov subspace $K(A,\\vec{q_1},k)$, and $e_k\n = I_n(:,k)$ (the $k$ column of the identity matrix). This\n justifies the orthogonalization step of the algorithm (line 6),\n which only considers the last two vectors; whether that is enough to\n guarantee that all the $\\vec{q}$\\apos{s} will be orthogonal is\n certainly not evident, and gets proved on the same theorem. \\\\\n\n \\item The matrix $T_k$ has tridiagonal shape, that is: \\\\\n \\[\n \\begin{bmatrix}\n \\alpha_1 & \\beta_1 & \\cdots & 0 \\\\\n \\beta_1 & \\ddots & \\ddots & \\vdots \\\\\n \\vdots & \\ddots & \\ddots & \\beta_{k-1} \\\\\n 0 & \\cdots & \\beta_{k-1} & \\alpha_k\n \\end{bmatrix}\n \\]\n \\hfill\n\n This shape allows us to calculate its eigenvalues with much less\n effort than for original matrix $A$ (which was the whole\n motivation on the beginning). There are several options for such\n calculation, but we will consider only the (implicit) QL Algorithm\n (see \\cite{dubrulle71}), as that is the one used by Berry for his\n famous routines in the context of LSI (see further\n subsections). \n\\end{itemize}\n\\hfill\n\nIn addition to the above properties, which kind of guarantee the\n``correctness'' of the \\cref{alg:lanczos-step} (to some extent); Golub\nalso cites in \\cite{golub13} another theorem that establishes the\napproximation quality of matrix $T_k$ as a function of $k$. This is\nthe result that justifies our original claim that the sequence of\nmatrices $\\{T_k\\}$, approximates better the eigenvalues of $A$ as $k$\nincreases. \\\\\n\nFinally, Golub adds in \\cite{golub13} that not everything is flakes\nand honey with this \nalgorithm; the orthogonality that we expect on vectors \\vec{q}\\apos{s}\nis at jeopardy as $\\tilde{\\beta_k}$, the numerical approximation of\n$\\beta_k$, becomes really small; this is because that implies the\ncancellation of $\\vec{r_k}$). Main credit of this result goes again to\nPaige (\\cite{paige71},\\cite{paige76}), and we will come back to \nit on next subsection, when we show the full Lanczos Algorithm.\n\n\n\n", "meta": {"hexsha": "5c9fbe279c3788e76cc72b67c106ff29bf5c909b", "size": 5798, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "svd-lanczos-serial-step.tex", "max_stars_repo_name": "rzavalet/svd-lsi-project-master", "max_stars_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "svd-lanczos-serial-step.tex", "max_issues_repo_name": "rzavalet/svd-lsi-project-master", "max_issues_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "svd-lanczos-serial-step.tex", "max_forks_repo_name": "rzavalet/svd-lsi-project-master", "max_forks_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.9242424242, "max_line_length": 78, "alphanum_fraction": 0.699206623, "num_tokens": 1780, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430478583168, "lm_q2_score": 0.8418256432832333, "lm_q1q2_score": 0.6996775309237148}} {"text": "\\section*{Exercises}\n\n\\begin{ex}\nLet $V=\\R^{3}$ and let\n\\begin{equation*}\nW=\\sspan (S), \\mbox{ where } S=\\set{\\begin{mymatrix}{r}\n1 \\\\\n-1 \\\\\n1\n\\end{mymatrix} ,\\begin{mymatrix}{r}\n-2 \\\\\n2 \\\\\n-2\n\\end{mymatrix},\\begin{mymatrix}{r}\n-1 \\\\\n1 \\\\\n1\n\\end{mymatrix},\\begin{mymatrix}{r}\n1 \\\\\n-1 \\\\\n3\n\\end{mymatrix} }\n\\end{equation*}\nFind a basis of $W$ consisting of vectors in $S$.\n\n\\begin{sol}\nIn this case $\\dim (W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (non-zero) vector from $S$.\n\\end{sol}\n\\end{ex}\n\n\\begin{ex}\n Let $T$ be a linear transformation given by\n\\[\nT \\begin{mymatrix}{r}\nx\\\\\ny\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 &1 \\\\\n1 & 1\n\\end{mymatrix}\n\\begin{mymatrix}{r}\nx\\\\\ny\n\\end{mymatrix}\n\\]\nFind a basis for $\\ker (T)$ and $\\func{im} (T)$.\n\n\\begin{sol}\nA basis for $\\ker (T)$ is\n$\\set{\\begin{mymatrix}{r}\n1 \\\\\n-1\n\\end{mymatrix} }$\nand a basis for $\\func{im} (T)$ is\n$\\set{\\begin{mymatrix}{r}\n1 \\\\\n1\n\\end{mymatrix} }$. \\\\\nThere are many other possibilities for the specific bases, but in this case\n$\\dim (\\ker (T))=1 $ and $\\dim (\\func{im} (T))=1$.\n\\end{sol}\n\n\\end{ex}\n\n\\begin{ex}\n Let $T$ be a linear transformation given by\n\\[\nT \\begin{mymatrix}{r}\nx\\\\\ny\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 \\\\\n1 & 1\n\\end{mymatrix}\n\\begin{mymatrix}{r}\nx\\\\\ny\n\\end{mymatrix}\n\\]\nFind a basis for $\\ker (T)$ and $\\func{im}\n(T)$.\n\n\\begin{sol}\nIn this case $\\ker (T) =\\set{0}$\nand $\\func{im} (T) = \\R^2$ (pick any basis of $\\R^2$).\n\\end{sol}\n\n\\end{ex}\n\n\\begin{ex}\nLet $V=\\R^{3}$ and let\n\\begin{equation*}\nW=\\sspan\\set{\\begin{mymatrix}{r}\n1 \\\\\n1 \\\\\n1\n\\end{mymatrix} ,\\begin{mymatrix}{r}\n-1 \\\\\n2 \\\\\n-1\n\\end{mymatrix} }\n\\end{equation*}\nExtend this basis of $W$ to a basis of $V$.\n\n\\begin{sol}\nThere are many possible such extensions, one is (how do we know?):\n\\begin{equation*}\n\\set{\\begin{mymatrix}{r}\n1 \\\\\n1 \\\\\n1\n\\end{mymatrix} ,\\begin{mymatrix}{r}\n-1 \\\\\n2 \\\\\n-1\n\\end{mymatrix} ,\\begin{mymatrix}{r}\n0 \\\\\n0\\\\\n1\n\\end{mymatrix}\n}\n\\end{equation*}\n\\end{sol}\n\\end{ex}\n\n\\begin{ex}\n Let $T$ be a linear transformation given by\n\\[\nT \\begin{mymatrix}{r}\nx\\\\\ny \\\\\nz\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 1 & 1 \\\\\n1 & 1 & 1\n\\end{mymatrix}\n\\begin{mymatrix}{r}\nx\\\\\ny \\\\\nz\n\\end{mymatrix}\n\\]\nWhat is $\\dim ( \\ker (T) )$?\n\n\\begin{sol}\nWe can easily see that $\\dim ( \\func{im} (T) ) =1$, and thus\n$\\dim ( \\ker (T) ) = 3 - \\dim ( \\func{im} (T) ) = 3- 1 = 2$.\n\\end{sol}\n\\end{ex}\n\n", "meta": {"hexsha": "2257f4d89d45ad2c9996faac4fbbd3c4e1cf0678", "size": 2405, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/exercises/LinearTransformations-KernelImage.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "old/exercises/VectorSpaces-LinearTransformations-KernelImage.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/exercises/VectorSpaces-LinearTransformations-KernelImage.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 15.9271523179, "max_line_length": 132, "alphanum_fraction": 0.6012474012, "num_tokens": 1035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430353105599, "lm_q2_score": 0.8418256492357358, "lm_q1q2_score": 0.6996775253080721}} {"text": "\n\nThe description of \\textit{Fuzzy Logic} has been used by Lotfi A. Zadeh for the very first time within his famous published paper \"Fuzzy Sets\" .\n\nZadeh defined Fuzzy Logic ' a kind of logic using graded or qualified statements rather than ones that are strictly true or false . The results of fuzzy reasoning are not as definite as those derived by strict logic but they cover a larger field of discourse ' \\cite{lotfi} .\n\nFuzzy logic doesn't represents the logic that is fuzzy but the idea that all things can be a member of any cluster at least partly member. The idea behind fuzzy logic describes that all transmissions between clusters (memberhsips) are not so strict that you can change memberships just between 0 and 1. There are also soft and mid transitions between 0-1 .\n\nTo understand fuzzy logic we ought get into crisp sets too. In short comparision of \\textit{Crispt Sets} and \\textit{Fuzzy Sets} both sets are questioning the classical \"tall men\" problem's memberships according to height of men (See Fig. \\ref{fig:comparison}) .\n\n\\subsection{Crisp Sets and Fuzzy Sets}\n\nIn crisp sets theory a member can be in only one set. For example, for the relationship with the set “A”, the characteristic function gives us 1 (true) if an element named “u” is the member of the set “A” or 0 (false) if an element named “u” is not the member of the set “A” in the crisp logic \\cite{Kasabov:1996:FNN:525657}. Hence, this logic cannot represent vague and ambiguous solutions of any paradoxes.\n\nIn fuzzy set theory all elements can be definitely true or false and partly true or false. Fuzzy Logic uses the continuum of logical values between 0 and 1 . When you put your height(cm) in Fig.\\ref{fig:comparison} you can see that your degree of membership can be fractional and ambiguous .\n\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[scale=.3]{Images/crisp_and_fuzzy.png}\n \\caption{Crisp and Fuzzy Sets of \"\\textit{tall men}\"}\n \\label{fig:comparison}\n\\end{figure}\n\n\\subsection{Fundamentals of Fuzzy Logic} \n\nFundamentals of fuzzy decision making can be listed as : Fuzzification , inference and defuzzification.\n\n\n\n\\textit{1) Fuzzification}\n\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[scale=.55]{Images/fuzzification.png}\n \\caption{Example of nonlinear scaling of an input measurement. \\cite{jantzen} }\n \\label{fig:fig_fuzzification_jantzen}\n\\end{figure}\n\n\nFuzzification explains the process of converting every data to degrees of membership in one or more membership functions. The fuzzificaiton method thus matches the input data with the conditions of the rules . Every data has its own degree of memberships within a fuzzy set which can be explained with a linguistic term .\n\n\\textit{2) Inference}\n\nAfterwards of fuzzification an inference is made based on rules . The process which called 'inference' basicly combines all results that evaluated from each rule to obtain a final result. This output given after inference step is also called \\textit{fuzzy output set} .\n\n\n\\textit{3) Defuzzification}\n\nAfter the inference there will be fuzzy output sets. But , its not recognizable in real life . In order to make it usable and recognizable in real life those values should be translated into real values. This translation named \\textit{defuzzification} \\cite{defuzzification} .\n\n", "meta": {"hexsha": "1feb1b54d9437893904303d68af506ab5b86f145", "size": 3324, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Texes/FuzzyLogicAndFuzzySet.tex", "max_stars_repo_name": "ertanturan/Using-artificial-intelligence-for-modeling-of-the-realistic-animal-behaviors-in-a-virtual-island", "max_stars_repo_head_hexsha": "c0389aeeaa900e5c28bff98498f3c9f3637ab2a9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-06-27T06:21:48.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-28T11:05:04.000Z", "max_issues_repo_path": "Texes/FuzzyLogicAndFuzzySet.tex", "max_issues_repo_name": "ertanturan/Using-artificial-intelligence-for-modeling-of-the-realistic-animal-behaviors-in-a-virtual-island", "max_issues_repo_head_hexsha": "c0389aeeaa900e5c28bff98498f3c9f3637ab2a9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Texes/FuzzyLogicAndFuzzySet.tex", "max_forks_repo_name": "ertanturan/Using-artificial-intelligence-for-modeling-of-the-realistic-animal-behaviors-in-a-virtual-island", "max_forks_repo_head_hexsha": "c0389aeeaa900e5c28bff98498f3c9f3637ab2a9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.7169811321, "max_line_length": 408, "alphanum_fraction": 0.7782791817, "num_tokens": 794, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.841825635346563, "lm_q2_score": 0.8311430415844384, "lm_q1q2_score": 0.6996775190456946}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[fleqn]{amsmath}\n\\usepackage{textcomp}\n\\usepackage{gensymb}\n\\usepackage{amsfonts}\n\\usepackage{enumitem}\n%\\usepackage{tikz} % Include for figures.\n%\\usepackage{subfiles} % Include for subfiles.\n\n\\newcommand{\\HOMEWORKNUM}{5}\n\\newcommand{\\NAME}{D. Choi}\n\\newcommand{\\DATE}{2020-05-28}\n\n\\title{\\vspace{-4\\baselineskip}MATH 225 - Homework \\#\\HOMEWORKNUM}\n\\author{\\NAME}\n\\date{\\DATE}\n\n%\\pagenumbering{gobble} % Include for single-page document.\n\n\\begin{document}\n\\maketitle\n\n\\section*{1.}\n\\textit{Let $A$ be the projection matrix that projects onto the plane\n$z = \\sqrt{3}y$ in $\\mathbb{R}^3$. \\\\ Find $A$.}\n\\\\[\\baselineskip]\nNote that $z = \\sqrt{3}y = \\tan(60 \\degree)y$.\n\\begin{gather*}\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t\\overset{A}{\\longrightarrow}\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t\\\\\n\t\\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}\n\t\\overset{A}{\\longrightarrow}\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\cos(60 \\degree) \\cos(60 \\degree) \\\\\n\t\t\\cos(60 \\degree) \\sin(60 \\degree)\n\t\\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\frac{1}{4} \\\\\n\t\t\\frac{\\sqrt{3}}{4}\n\t\\end{pmatrix}\n\t\\\\\n\t\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\n\t\\overset{A}{\\longrightarrow}\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\cos(-30 \\degree) \\cos(60 \\degree) \\\\\n\t\t\\cos(-30 \\degree) \\sin(60 \\degree)\n\t\\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\frac{\\sqrt{3}}{4} \\\\\n\t\t\\frac{3}{4}\n\t\\end{pmatrix}\n\\end{gather*}\n\\begin{equation*}\n\tA =\n\t\\boxed{\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & 0 \\\\\n\t\t\t0 & \\frac{1}{4} & \\frac{\\sqrt{3}}{4} \\\\\n\t\t\t0 & \\frac{\\sqrt{3}}{4} & \\frac{3}{4}\n\t\t\\end{pmatrix}\n\t}\n\\end{equation*}\n\\newpage\n\n\\section*{2.}\n\\textit{Let $B$ be the reflection matrix over the plane\n$z = \\sqrt{3}y$ in $\\mathbb{R}^3$. \\\\ Find $B$.}\n\\\\[\\baselineskip]\nNote (still) that $z = \\sqrt{3}y = \\tan(60 \\degree)y$.\n\\begin{gather*}\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t\\overset{B}{\\longrightarrow}\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t\\\\\n\t\\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}\n\t\\overset{B}{\\longrightarrow}\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\cos(2 \\cdot 60 \\degree) \\\\\n\t\t\\sin(2 \\cdot 60 \\degree)\n\t\\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t-\\frac{1}{2} \\\\\n\t\t\\frac{\\sqrt{3}}{2}\n\t\\end{pmatrix}\n\t\\\\\n\t\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\n\t\\overset{B}{\\longrightarrow}\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\cos(90 - 2 \\cdot 30 \\degree) \\\\\n\t\t\\sin(90 - 2 \\cdot 30 \\degree)\n\t\\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t\\frac{\\sqrt{3}}{2} \\\\\n\t\t\\frac{1}{2}\n\t\\end{pmatrix}\n\\end{gather*}\n\\begin{equation*}\n\tB =\n\t\\boxed{\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & 0 \\\\\n\t\t\t0 & -\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\\n\t\t\t0 & \\frac{\\sqrt{3}}{2} & \\frac{1}{2}\n\t\t\\end{pmatrix}\n\t}\n\\end{equation*}\n\n\\section*{3.}\n\\textit{With $A$ and $B$ from \\textbf{(1)} and \\textbf{(2)},\nfind the following.}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item $A^{100} \\begin{pmatrix} 1 \\\\ 1 \\\\ \\sqrt{3} \\end{pmatrix} =\n\tA \\begin{pmatrix} 1 \\\\ 1 \\\\ \\sqrt{3} \\end{pmatrix} = \n\t\\boxed{\n\t\t\\begin{pmatrix} 1 \\\\ 1 \\\\ \\sqrt{3} \\end{pmatrix}\n\t}$\n\t\\item $B \\begin{pmatrix} 1 \\\\ 1 \\\\ \\sqrt{3} \\end{pmatrix} =\n\t\\boxed{\n\t\t\\begin{pmatrix} 1 \\\\ 1 \\\\ \\sqrt{3} \\end{pmatrix}\n\t}$\n\t\\item $B^3 \\begin{pmatrix} 10 \\\\ -\\sqrt{3} \\\\ 1 \\end{pmatrix} =\n\tB^2 B \\begin{pmatrix} 10 \\\\ -\\sqrt{3} \\\\ 1 \\end{pmatrix} =\n\tB \\begin{pmatrix} 10 \\\\ -\\sqrt{3} \\\\ 1 \\end{pmatrix} =\n\t\\boxed{\n\t\t\\begin{pmatrix} 10 \\\\ \\sqrt{3} \\\\ -1 \\end{pmatrix}\n\t}$\n\\end{enumerate}\n\n\\section*{4.}\n\\textit{Find the matrix that rotates space $30 \\degree$ about the y-axis in\n$\\mathbb{R}^3$.}\n\\\\[\\baselineskip]\nLet $\\text{Rot}_y(\\theta)$ be the rotation matrix where\n\\begin{equation*}\n\t\\text{Rot}_y(\\theta) =\n\t\\begin{pmatrix}\n\t\t\\cos \\theta & 0 & \\sin \\theta \\\\\n\t\t0 & 1 & 0 \\\\\n\t\t-\\sin \\theta & 0 & \\cos \\theta\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\n\\begin{equation*}\n\t\\text{Rot}_y(30 \\degree)\n\t=\n\t\\begin{pmatrix}\n\t\t\\cos(30 \\degree) & 0 & \\sin(30 \\degree) \\\\\n\t\t0 & 1 & 0 \\\\\n\t\t-\\sin(30 \\degree) & 0 & \\cos(30 \\degree)\n\t\\end{pmatrix}\n\t=\n\t\\boxed{\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{\\sqrt{3}}{2} & 0 & \\frac{1}{2} \\\\\n\t\t\t0 & 1 & 0 \\\\\n\t\t\t-\\frac{1}{2} & 0 & \\frac{\\sqrt{3}}{2}\n\t\t\\end{pmatrix}\n\t}\n\\end{equation*}\n\n\\end{document}", "meta": {"hexsha": "efa6d3e6213e20d288ad38b91e09b4f4acbbd1b3", "size": 4061, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "usc-20202-math-225-39425/hw05/main.tex", "max_stars_repo_name": "Floozutter/coursework", "max_stars_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "usc-20202-math-225-39425/hw05/main.tex", "max_issues_repo_name": "Floozutter/coursework", "max_issues_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "usc-20202-math-225-39425/hw05/main.tex", "max_forks_repo_name": "Floozutter/coursework", "max_forks_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.2057142857, "max_line_length": 75, "alphanum_fraction": 0.5900024624, "num_tokens": 1791, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951104066293, "lm_q2_score": 0.8289387998695209, "lm_q1q2_score": 0.6995374000362282}} {"text": "\\subsection{Trees}\\label{subsec:trees}\n\nIn this section, we will regard the edges of simple undirected graphs as sets of unordered tuples and edges of simple directed graphs as sets of ordered tuples.\n\n\\begin{definition}\\label{def:tree}\n The possibly infinite \\hyperref[def:undirected_multigraph]{simple undirected graph} \\( T= (E, V, \\mscrE) \\) is called a \\term{tree} if any of the following equivalent conditions hold:\n \\begin{thmenum}\n \\thmitem{def:tree/connected_acyclic} \\( T \\) is \\hyperref[def:undirected_multigraph_connectedness]{connected} and \\hyperref[def:undirected_multigraph_path/cycle]{acyclic}.\n \\thmitem{def:tree/maximally_acyclic} \\( T \\) is \\term{maximally acyclic}, meaning that adding an edge between existing vertices would create a \\hyperref[def:undirected_multigraph_path/cycle]{cycle}.\n \\thmitem{def:tree/minimally_connected} \\( T \\) is \\term{minimally connected}, meaning that removing an edge will make the graph \\hyperref[def:quiver_connectedness]{disconnected}.\n \\thmitem{def:tree/single_path} For every pair of vertices in \\( T \\), there exists a unique \\hyperref[def:undirected_multigraph_path]{path} connecting them.\n\n This condition motivates the definition of \\hyperref[def:arborescence]{arborescences}.\n \\end{thmenum}\n\n It is conventional to use the term \\term{node} for the vertices of a tree.\n\\end{definition}\n\\begin{proof}\n \\ImplicationSubProof{def:tree/connected_acyclic}{def:tree/maximally_acyclic} Suppose that \\( T \\) is connected and acyclic. There is always an edge not in \\( T \\) because the \\hyperref[ex:complete_graph]{complete graph} \\( K_V \\) contains cycles and \\( T \\) does not. Let \\( \\set{ u, v } \\) be an edge not in \\( T \\) and let \\( T' \\) be the supergraph that adjoins only this new edge.\n\n Since \\( T \\) is connected, there exists a path \\( p \\) connecting \\( u \\) and \\( v \\). Appending the edge \\( \\set{ u, v } \\) to this path creates a cycle in \\( T' \\).\n\n Therefore, \\( T' \\) is not acyclic.\n\n \\ImplicationSubProof{def:tree/maximally_acyclic}{def:tree/connected_acyclic} Suppose that \\( T \\) is maximally acyclic. We will show that it is connected.\n\n Suppose that \\( T \\) is not connected. Then there exist vertices \\( u \\) and \\( v \\) with no path between them. Let \\( T' \\) be the supergraph that adjoins only the edge \\( \\set{ u, v } \\).\n\n Since there is no path between \\( u \\) and \\( v \\) in \\( T \\), \\( T' \\) is also acyclic. But this contradicts the maximality of \\( T \\).\n\n Therefore, \\( T \\) is connected.\n\n \\ImplicationSubProof{def:tree/connected_acyclic}{def:tree/minimally_connected} Suppose that \\( T \\) is connected and acyclic. Let \\( \\set{ u, v } \\) be any edge of \\( T \\) and let \\( T' \\) be the subgraph that does not contain this edge.\n\n Suppose that \\( T' \\) is connected. Let \\( w \\) be any vertex and let \\( p \\) be a path in \\( T' \\) from \\( u \\) to \\( w \\). Then adding \\( \\set{ u, v } \\) creates a cycle in \\( T \\), which contradicts our assumption that \\( T \\) is acyclic.\n\n Therefore, \\( T' \\) is not connected.\n\n \\ImplicationSubProof{def:tree/minimally_connected}{def:tree/connected_acyclic} Suppose that \\( T \\) is minimally connected. We will show that it is acyclic.\n\n Suppose that \\( T \\) has a cycle \\( p \\). We can thus remove any edge of \\( p \\) from \\( T \\) and the resulting subgraph will also be connected. This contradicts the minimality of \\( T \\).\n\n Therefore, \\( T \\) is acyclic.\n\n \\ImplicationSubProof{def:tree/connected_acyclic}{def:tree/single_path} Suppose that \\( T \\) is connected and acyclic. Since \\( T \\) is connected, there exists at least one path between any two vertices. Since it is acyclic, this path must be unique because otherwise we could easily create a cycle by joining two such paths.\n\n \\ImplicationSubProof{def:tree/single_path}{def:tree/connected_acyclic} Suppose that, for every pair of vertices in \\( T \\), there exists a unique path connecting them. It is clear that \\( T \\) is connected.\n\n Suppose that \\( T \\) has a cycle \\( p \\) from \\( u \\) to \\( u \\) passing through \\( v \\). We thus obtain two paths from \\( u \\) to \\( v \\) --- one not containing the first edge in \\( p \\) and one not containing the last edge. But this contradicts our assumption that any two paths from \\( u \\) to \\( v \\) are equal.\n\n Therefore, \\( T \\) is acyclic.\n\\end{proof}\n\n\\begin{example}\\label{ex:def:tree}\n The reduced infinite integer graphs \\eqref{eq:ex:infinite_integer_graphs/positive}, \\eqref{eq:ex:infinite_integer_graphs/negative} and \\eqref{eq:ex:infinite_integer_graphs/two_sided} from \\fullref{ex:infinite_integer_graphs} are simple examples of infinite trees.\n\n Examples of finite trees are \\hyperref[def:proof_tree]{proof trees} from \\fullref{subsec:deductive_systems} and \\hyperref[def:concrete_syntax_tree]{concrete} and \\hyperref[def:abstract_syntax_tree]{abstract syntax trees} from \\hyperref[def:formal_grammar]{formal grammar}.\n\\end{example}\n\n\\begin{definition}\\label{def:arborescence}\\mcite[ch. 4, sec. 3.1]{GondranMinoux1984Graphs}\n Let \\( T = (V, A) \\) be a \\hyperref[def:quiver/simple]{simple directed graph}. Suppose that there exists a vertex \\( r \\in V \\) such that for every other vertex \\( v \\in V \\), there exists a unique \\hyperref[def:quiver_path/directed]{directed path} from \\( r \\) to \\( v \\). The pair \\( (G, r) \\) is called a (directed) \\term{arborescence}. It is often more convenient to consider the triple \\( T = (V, A, r) \\) instead. The vertex \\( r \\) is called the \\term{root} of the arborescence.\n\n \\begin{thmenum}\n \\thmitem{def:arborescence/undirected} A \\hyperref[def:tree]{tree} with a distinguished vertex \\( r \\) is called a \\term{rooted tree}.\n\n For every tree \\( T \\), as a consequence of \\fullref{def:tree/single_path}, every vertex \\( r \\) of \\( T \\) induces an \\hyperref[def:multigraph_orientation]{orientation} \\( O(T) \\) of \\( T \\) such that \\( O(T) \\) is an arborescence.\n\n For this reason, we identify rooted trees with their induced arborescences.\n\n \\thmitem{def:arborescence/depth} The \\term{depth} of a node \\( v \\) is the length of the path from \\( r \\) to \\( v \\). The term \\term{level} is also sometimes used but with a slightly different meaning --- the greater the depth, the lower the level.\n\n \\thmitem{def:arborescence/ancestry} If \\( v \\) has a strictly greater depth than \\( u \\), we say that \\( u \\) is an \\term{ancestor} of \\( v \\) and that \\( v \\) is a \\term{descendant} of \\( u \\).\n\n The ancestor of \\( v \\) at the lowest possible level is called the \\term{parent} of \\( v \\). If \\( u \\) is a parent of \\( v \\), \\( u \\) is a \\term{child} of \\( v \\). If a node has no children, we say that it is a \\term{leaf node}.\n\n Finally, if \\( u \\) and \\( v \\) are on the same level, we call them \\term{siblings}.\n\n \\thmitem{def:arborescence/height} The \\term{height} or \\term{depth} of the entire tree \\( T \\), if it exists, is the supremum of the depths of all nodes.\n\n \\thmitem{def:arborescence/width} The \\term{width} or \\term{breadth} of the entire tree \\( T \\), if it exists, is the supremum of the number of siblings among all vertices. It is equal to the degree of \\( T \\) minus \\( 1 \\).\n\n We use terminology similar to \\fullref{def:relation/arity}, e.g. \\enquote{binary tree}, \\enquote{ternary tree}, \\ldots.\n\n \\thmitem{def:arborescence/sub-arborescence} A \\hyperref[def:quiver/submodel]{subgraph} of an arborescence that is itself an arborescence is called a \\term{sub-arborescence}. The same holds for rooted subtrees.\n\n For every node \\( v \\) of \\( T \\), we define the \\term{induced sub-arborescence} of \\( T \\) with root \\( v \\) as the subgraph \\hyperref[def:quiver/submodel]{induced} by \\( V \\setminus \\dom(p) \\), where \\( p \\) is the unique path from \\( r \\) to \\( v \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{theorem}[K\\\"onig's lemma]\\label{thm:konigs_lemma}\n Every \\hyperref[def:hypergraph/degree]{locally finite} \\hyperref[def:arborescence]{arborescence} of \\hyperref[def:hypergraph/order]{infinite} \\hyperref[def:hypergraph/order]{order} contains a \\hyperref[def:undirected_multigraph_path/simple]{simple path} of infinite length.\n\\end{theorem}\n\\begin{proof}\n Let \\( T = (V, A, r) \\) be a locally finite infinite arborescence. We will build an infinite simple path\n \\begin{equation*}\n p = (p_1, p_2, \\cdots)\n \\end{equation*}\n using \\hyperref[rem:natural_number_recursion]{natural number recursion} starting at one. Let \\( c \\) be a \\hyperref[def:choice_function]{choice function} on the family\n \\begin{equation*}\n \\pow(V) \\setminus \\set{ \\varnothing }.\n \\end{equation*}\n\n Such a choice function exists by the \\hyperref[def:zfc/choice]{axiom of choice}.\n\n \\begin{itemize}\n \\item Since \\( T \\) is locally finite, there are finitely many children of \\( r \\). If we suppose that the \\hyperref[def:arborescence/sub-arborescence]{subarborescence} induced by \\( v \\) is finite for every child \\( v \\) of \\( r \\), then we would obtain that \\( T \\) itself is finite, which is a contradiction. Therefore, for at least one child, the induced sub-arborescence is infinite. Using the choice function \\( c \\), pick one such child and denote it by \\( v_1 \\).\n\n Define \\( p_1 \\) to be the arc \\( r \\to v_1 \\).\n\n \\item Fix \\( k > 1 \\). Using the same argument on the children of \\( t(e_{k-1}) \\), we can pick a child \\( v_k \\) that has an infinite induced sub-arborescence.\n\n Define \\( e_k \\) to be the arc \\( t(e_{k-1}) \\to v_k \\). The path\n \\begin{equation*}\n (e_1, e_2, \\ldots, e_{k-1}, e_k)\n \\end{equation*}\n is simple by construction because \\( T \\) is an arborescence and thus contains no directed cycles.\n \\end{itemize}\n\n Thus, we have constructed an infinite simple path.\n\\end{proof}\n", "meta": {"hexsha": "22984cb3113574bff0faa055be8e031eef5ab417", "size": 9690, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/trees.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/trees.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/trees.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 80.0826446281, "max_line_length": 487, "alphanum_fraction": 0.6991744066, "num_tokens": 2910, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289387998695209, "lm_q2_score": 0.8438951084436076, "lm_q1q2_score": 0.6995373984090033}} {"text": "\\chapter{Lagrange's Equations of the Second Kind}\\label{c2}\n\\section{Generalised coordinates}\\label{c2s1}\nIn the previous chapter we observed that in a system of $N$ particles and $r$\nconstraints, the $3N$ virtual displacements $\\delta x^{(n)}_i$ are not \nindependent. Any subset of $f = 3N - r$ virtual displacements is independent. \n$f$ is called the \\emph{number of degrees of freedom} of the system.\n\nConsider a system of two particles connected by a rigid rod. The two particles\neach need three Cartesian coordinates. But the constraint makes only five of\nthem independent. It is not clear which five to choose. Although the lack of\nindependence of virtual displacements in all coordinates is taken into account\nby the Lagrange's equations of the first kind, one still has to describe the \nsystem in terms of $6$ coordinates. This situation gets worse in the case of\na rigid body. If it made up of $N$ particles then we need $3N$ coordinates to\ndescribe its position when only $6$ of them are independent.\n\nGeneralised coordinates are any $f$ numbers that suffice to describe the\nposition of a system of $f$ degrees of freedom. For example, a rigid body\nwith $N$ particles can be described by three Cartesian coordinates of its\ncentre of mass and three Euler angles. We shall denote the $f$ generalised\ncoordiates by $\\{q_1, \\ldots, q_f\\}$.\n\nLet us therefore consider a system of $N$ particles and $f$ degrees of freedom\nand let $q_1, \\ldots, q_f$ be its generalised coordinates. Then the Cartesian\ncoordinates of its particles can be written as\n\\begin{eqnarray}\nx_i &=& x_i(q_1, \\ldots, q_f, t) \\\\\ny_i &=& y_i(q_1, \\ldots, q_f, t) \\\\\nz_i &=& z_i(q_1, \\ldots, q_f, t).\n\\end{eqnarray}\nNote that the Cartesian coordinates of a particle are allowed to depend on \n\\emph{all} generalised coordinates. Unlike their Cartesian counterparts, the\ngeneralised coordinates are not tied to a single particle. Further, note that\nthe Cartesian coordinates may depend on time as well. From these equations\nwe have\n\\begin{eqnarray}\ndx_i &=& \\pd{x_i}{q_j}dq_j + \\pd{x_i}{t}dt \\\\\ndy_i &=& \\pd{y_i}{q_j}dq_j + \\pd{y_i}{t}dt \\\\\ndz_i &=& \\pd{z_i}{q_j}dq_j + \\pd{z_i}{t}dt\n\\end{eqnarray}\nWe have used the summation convention in these three equations. Although\nthe functional forms of $x_i, y_i, z_i$ are arbitrary, the differentials\n$dx_i, dy_i, dz_i$ are linear functions of $dq_j$ and $dt$.\n\nThe three Cartesian velocities are\n\\begin{eqnarray}\n\\dot{x}_i &=& \\pd{x_i}{q_j}\\dot{q}_j + \\pd{x_i}{t} \\label{c2s1e7} \\\\\n\\dot{y}_i &=& \\pd{y_i}{q_j}\\dot{q}_j + \\pd{y_i}{t} \\label{c2s1e8} \\\\\n\\dot{z}_i &=& \\pd{z_i}{q_j}\\dot{q}_j + \\pd{z_i}{t} \\label{c2s1e9} \n\\end{eqnarray}\nNote that the symbol $\\dot{x}_i$ denotes the total time derivative of $x_i$ and\nit differs from $\\partial x_i/\\partial t$. The generalised coordinates, however,\ndo not depend explicitly on time. The kinetic energy of the system is\n\\begin{eqnarray}\nT &=& \\frac{1}{2}m_i(\\dot{x}_i^2 + \\dot{y}_i^2 + \\dot{z}_i^2) \\nonumber \\\\\n &=& \\frac{1}{2}m_i\\left(\\pd{x_i}{q_j}\\dot{q}_j + \\pd{x_i}{t}\\right)\n \\left(\\pd{x_i}{q_k}\\dot{q}_k + \\pd{x_i}{t}\\right) + \n \\nonumber \\\\\n & & \\frac{1}{2}m_i\\left(\\pd{y_i}{q_j}\\dot{q}_j + \\pd{y_i}{t}\\right)\n \\left(\\pd{y_i}{q_k}\\dot{q}_k + \\pd{y_i}{t}\\right) + \n\t \\nonumber \\\\\n & & \\frac{1}{2}m_i\\left(\\pd{z_i}{q_j}\\dot{q}_j + \\pd{z_i}{t}\\right)\n \\left(\\pd{z_i}{q_k}\\dot{q}_k + \\pd{z_i}{t}\\right) \n\t \\label{c1s1e10} \n\\end{eqnarray}\nWe now compute the derivative of $T$ with respect to the generalised \nvelocity $\\dot{q}_l$.\n\\begin{eqnarray}\n\\pd{T}{\\dot{q}_l} &=& \n m_i\\left(\\pd{x_i}{q_j}\\dot{q}_j + \\pd{x_i}{t}\\right)\\pd{x_i}{q_l} +\n m_i\\left(\\pd{y_i}{q_j}\\dot{q}_j + \\pd{y_i}{t}\\right)\\pd{y_i}{q_l} +\n \\nonumber \\\\\n& & m_i\\left(\\pd{z_i}{q_j}\\dot{q}_j + \\pd{z_i}{t}\\right)\\pd{z_i}{q_l} \n \\label{c2s1e11} \n\\end{eqnarray}\nUsing equations \\eqref{c2s1e7}, \\eqref{c2s1e8} and \\eqref{c2s1e9} we get\n\\begin{equation}\\label{c2s1e12}\n\\pd{T}{\\dot{q}_l} = m_i\\left(\\dot{x}_i\\pd{x_i}{q_l} + \\dot{y}_i\\pd{y_i}{q_l} + \n \\dot{z}_i\\pd{z_i}{q_l}\\right).\n\\end{equation}\nWe now take the total time derivative of this equation.\n\\begin{eqnarray}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) &=& \n m_i\\left(\\ddot{x}_i\\pd{x_i}{q_l} + \\dot{x}_i\\frac{d}{dt}\n \\left(\\pd{x_i}{q_l}\\right)\\right) + \n m_i\\left(\\ddot{y}_i\\pd{y_i}{q_l} + \\dot{y}_i\\frac{d}{dt}\n \\left(\\pd{y_i}{q_l}\\right)\\right) + \\nonumber \\\\\n& & m_i\\left(\\ddot{z}_i\\pd{z_i}{q_l} + \\dot{z}_i\\frac{d}{dt}\n \\left(\\pd{z_i}{q_l}\\right)\\right) \n\\end{eqnarray}\nEvaluating the total time derivatives on the right hand side,\n\\begin{eqnarray}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) &=& \n m_i\\left(\\ddot{x}_i\\pd{x_i}{q_l} + \n \\dot{x}_i\\dot{q}_k\\frac{\\partial^2 x_i}{\\partial q_k\\partial q_l} +\n \\dot{x_i}\\frac{\\partial}{\\partial q_l}\\pd{x_i}{t}\\right) \\nonumber \\\\\n & & m_i\\left(\\ddot{y}_i\\pd{y_i}{q_l} + \n \\dot{y}_i\\dot{q}_k\\frac{\\partial^2 y_i}{\\partial q_k\\partial q_l} +\n \\dot{y_i}\\frac{\\partial}{\\partial q_l}\\pd{y_i}{t}\\right) \\nonumber \\\\\n & & m_i\\left(\\ddot{z}_i\\pd{z_i}{q_l} + \n \\dot{z}_i\\dot{q}_k\\frac{\\partial^2 z_i}{\\partial q_k\\partial q_l} +\n \\dot{z_i}\\frac{\\partial}{\\partial q_l}\\pd{z_i}{t}\\right) \\label{c2s1e14}\n\\end{eqnarray}\nWe now observe that\n\\[\n\\pd{\\dot{q_1}}{q_k} = \\frac{\\partial}{\\partial t}\\delta_{lk} = 0\n\\]\nso that we can write\n\\begin{eqnarray}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) &=& \n m_i\\left(\\ddot{x}_i\\pd{x_i}{q_l} + \n \\dot{x}_i\\frac{\\partial}{\\partial q_l}\n \\left(\\dot{q}_k\\pd{x_i}{q_k} + \\pd{x_i}{t}\\right)\\right) \\nonumber \\\\\n & & m_i\\left(\\ddot{y}_i\\pd{y_i}{q_l} + \n \\dot{y}_i\\frac{\\partial}{\\partial q_l}\n \\left(\\dot{q}_k\\pd{y_i}{q_k} + \\pd{y_i}{t}\\right)\\right) \\nonumber \\\\\n & & m_i\\left(\\ddot{z}_i\\pd{z_i}{q_l} + \n \\dot{z}_i\\frac{\\partial}{\\partial q_l}\n \\left(\\dot{q}_k\\pd{z_i}{q_k} + \\pd{z_i}{t}\\right)\\right) \\label{c2s1e15}\n\\end{eqnarray}\nUsing equations \\eqref{c2s1e7}, \\eqref{c2s1e8} and \\eqref{c2s1e9} we get\n\\[\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) =\nm_i\\left[\\left(\\ddot{x}_i\\pd{x_i}{q_l} + \\dot{x}_i\\pd{\\dot{x}_i}{{q}_l}\\right) \n + \\nonumber \\\\\n\\left(\\ddot{y}_i\\pd{y_i}{q_l} + \\dot{y}_i\\pd{\\dot{y}_i}{{q}_l}\\right) \n + \\nonumber \\\\\n\\left(\\ddot{z}_i\\pd{z_i}{q_l} + \\dot{z}_i\\pd{\\dot{z}_i}{{q}_l}\\right)\\right].\n\\]\nRearranging it\n\\[\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) =\n\\pd{T}{q_l} + m_i\\left(\\ddot{x}_i\\pd{x_i}{q_l} +\n\\ddot{y}_i\\pd{y_i}{q_l} +\\ddot{z}_i\\pd{z_i}{q_l}\\right)\n\\]\nor\n\\begin{equation}\\label{c2s1e16}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) - \\pd{T}{q_l} =\n m_i\\left(\\ddot{x}_i\\pd{x_i}{q_l} +\n \\ddot{y}_i\\pd{y_i}{q_l} +\\ddot{z}_i\\pd{z_i}{q_l}\\right)\n\\end{equation}\nMultiplying both sides by $\\delta q_l$,\n\\begin{equation}\\label{c2s1e17}\n\\left(\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) - \\pd{T}{q_l}\\right)\\delta q_l\n= m_i\\left(\\ddot{x}_i\\delta x_i + \\ddot{y}_i\\delta y_i +\n \\ddot{z}_i\\delta z_i\\right)\n\\end{equation}\nComparing with equation \\eqref{c1s2e13} we surmise that the left hand side\nof the above equation is\n\\begin{equation}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) - \\pd{T}{q_l}\\delta q_l =\nX_i\\delta x_i + Y_i\\delta y_i + Z_i \\delta z_i,\n\\end{equation}\nwhere $(X_i, Y_i, Z_i)$ are components of the external force on the $i$th\nparticle. If this force is conservative, we can write it in terms of a \npotential $V$ as\n\\begin{eqnarray}\n\\frac{d}{dt}\\left(\\pd{T}{\\dot{q}_l}\\right) - \\pd{T}{q_l}\\delta q_l &=&\n\\pd{V}{x_i}\\delta x_i + \\pd{V}{z_i}\\delta y_i + \\pd{V}{z_i} \\delta z_i \\nonumber \\\\\n&=&\n\\pd{V}{x_i}\\pd{x_i}{q_l}\\delta q_l + \\pd{V}{z_i}\\pd{y_i}{q_l}\\delta q_l + \n\\pd{V}{z_i} \\pd{z_i}{q_l}\\delta q_l \\nonumber \\\\\n&=& \\pd{V}{q_1}\\delta q_l \\label{c1s1e20}\n\\end{eqnarray}\nNote that the last term here is not $3\\partial V/\\partial q_l$ because $V$\ndepends on all $x_1, \\ldots, z_N$ and each of these is a function of all \ngeneralised coordinates. If $V$ \\emph{does not} depend on generalised \nvelocities then we can write \\eqref{c1s1e20} as\n\\begin{equation}\\label{c2s1e21}\n\\frac{d}{dt}\\pd{L}{\\dot{q}_l} - \\pd{L}{q_l} = 0,\n\\end{equation}\nwhere the function\n\\begin{equation}\\label{c1s1e22}\nL(q_1, \\ldots, q_f, \\dot{q}_1, \\ldots, \\dot{q}_f) = T - V\n\\end{equation}\nis the Lagrangian of the system and equations \\eqref{c2s1e21} are called\nLagrange's equation of the second kind.\n\nA few remarks about \\eqref{c2s1e21}:\n\\begin{itemize}\n\\item Often times the phrase `second kind' is omitted and equations\nare called Lagrange's equations.\n\\item They equations are presently derived only for conservative forces.\n\\item They do not involve the forces of constraint. Neither do they insist on\nthe Cartesian coordinates. \n\\item They are valid in all frames of references. However $T$ depends on \ngeneralised velocities and will change across frames of reference. Therefore,\n$T$ should always be calculated with respect to an inertial frame of \nreference \\cite[p. 31]{akr}.\n\\end{itemize}\n\nProblems.\n\\begin{enumerate}\n\\item A uniform rod leans against a wall \\cite[Problem 1, chapter 4]{akr}.\nLet $l$ be its length. Its motion is confined to a plane. Its unconstrained\nmotion has three degrees of freedom, the position of its centre of mass and\nits orientation. In the present problem, the $x$ coordinate of its end\ntouching the wall is constrained to be $0$ and the $y$ coordinate of its \nend touching the floor is constrained to be $0$. Therefore, one generalised\ncoordinate suffices to describe the motion. \n\nWe choose $\\theta$ the (smaller) angle made by the rod with the $x$ axis. If\nthe rod is in the first quadrant then it makes and angle $\\pi -\\theta$ with\nthe positive $x$ axis.\n\nThe centre of the rod is at point $(0, l/2)$ when $\\theta = \\pi/2$ and it is\nat $(0, l/2)$ when $\\theta = 0$. It traces the arc of a circle of radius\n$l/2$ as the rod slides down the path.\n\nThe moment of inertia of the rod abouts its centre of mass is $ml^2/12$. \nTherefore, the rotational kinetic energy is\n\\[\nT_r = \\frac{I}{2}\\dot{\\theta}^2 = \\frac{m}{24}l^2\\dot{\\theta}^2\n\\]\nThe translational kinetic energy of its centre of mass is\n\\[\nT_c = \\frac{m}{2}v^2 = \\frac{m}{8}l^2\\dot{\\theta}^2\n\\]\nThe total kinetic energy is \n\\[\nT = \\frac{ml^2}{12}\\dot{\\theta}^2.\n\\]\n\nWe will now find the potential energy. If $dm$ is the mass element of the\nrod at a height $h$ then\n\\[\ndV = gy dm\n\\]\nIf $\\lambda$ be the linear mass density of the rod, $dm = \\lambda ds = mds/l$.\nThus,\n\\[\ndV = \\frac{mg}{l} yds = \\frac{mg}{l}y \\sqrt{(dx)^2 + (dy)^2} = \n\\frac{mg}{l}ydy\\sqrt{1 + \\left(\\frac{dx}{dy}\\right)^2}.\n\\]\nNow the equation of the straight line representing the rod is $y = y_0 -\n\\tan\\theta x$ so that \n\\[\n\\frac{dx}{dy} = \\cot\\theta\n\\]\nso that\n\\[\ndV = \\frac{mg}{l}\\csc\\theta ydy\n\\]\nor\n\\[\nV = \\int_{lsin\\theta}^0 dV = -\\frac{mg}{2}l\\sin\\theta.\n\\]\nTherefore, the Lagrangian is\n\\[\nL = \\frac{ml^2}{12}\\dot{\\theta}^2 + \\frac{mg}{2}l\\sin\\theta\n\\]\nand the equation of motion is\n\\[\n\\frac{ml^2}{6}\\ddot{\\theta} - \\frac{mgl}{2}\\cos\\theta = 0 \\Rightarrow\n\\frac{l}{3}\\ddot{\\theta} - g\\cos\\theta = 0 \\Rightarrow \\ddot{x} = \n\\frac{3g}{l}\\cos\\theta.\n\\]\n\n\\item An old model of the He atom consisted of a fixed nucleus and two\nelectrons on the opposite ends of the diameter of the circle centred at\nthe nucleus \\cite[Problem 3, chapter 4]{akr}. Without the constraints, the\nmotion of the two electrons in a plane would have required four coordinates.\nTheir constraints are:\n\\begin{itemize}\n\\item They are always at the opposite ends of a diameter. This means that\nwe can consider them to be end of a rigid, weightless rod.\n\\item Their centre of mass is fixed at the nucleus. This fixes the two \ncoordinates of their centre of mass.\n\\end{itemize}\nThis leaves the system with just one degree of freedom. If $\\theta$ is the\nangle made by one electron with the positive $x$ axis, the other electron\nmakes an angle $\\theta + \\pi/2$. If $m$ is the mass of the electrons, their\nkinetic energy is\n\\[\nT = \\frac{m}{2}r^2\\dot{\\theta}^2 + \\frac{m}{2}r^2\\dot{\\theta}^2 =\nmr^2\\dot{\\theta}^2.\n\\]\nTheir potential energy is\n\\[\nV = -\\frac{2e^2}{r} - \\frac{2e^2}{r} = -\\frac{4e^2}{r}\n\\]\nso that the Lagrangian is\n\\[\nL = mr^2\\dot{\\theta}^2 + \\frac{4e^2}{r^2}.\n\\]\nThe equation of motion is $\\ddot{\\theta} = 0$ or that $\\theta = \\alpha t + \n\\beta$, where $\\alpha$ and $\\beta$ are the initial conditions.\n\\end{enumerate}\n", "meta": {"hexsha": "0fee44a555c4e92180dde3b48f712a3d8122becf", "size": 12207, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cm/lm/c2.tex", "max_stars_repo_name": "amey-joshi/physics", "max_stars_repo_head_hexsha": "66ae9bf4a363bd32b09df22a049e281953adb39b", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "cm/lm/c2.tex", "max_issues_repo_name": "amey-joshi/physics", "max_issues_repo_head_hexsha": "66ae9bf4a363bd32b09df22a049e281953adb39b", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cm/lm/c2.tex", "max_forks_repo_name": "amey-joshi/physics", "max_forks_repo_head_hexsha": "66ae9bf4a363bd32b09df22a049e281953adb39b", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.2387543253, "max_line_length": 83, "alphanum_fraction": 0.6800196609, "num_tokens": 4617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951025545427, "lm_q2_score": 0.8289388019824946, "lm_q1q2_score": 0.699537395310457}} {"text": "\\section{Proposed method}\n\n\\begin{frame}{Proposed method}\n Some mathematics formula\n \\begin{equation}\n\t\tf(n) =\n \\begin{cases}\n n/2 & \\quad \\text{if } n \\text{ is even}\\\\\n -(n+1)/2 & \\quad \\text{if } n \\text{ is odd}\n \\end{cases}\n\t\\end{equation}\n\\end{frame}", "meta": {"hexsha": "715c168b4b138f0adb841360fc6981f49da64972", "size": 321, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "slide-presentation/proposed_method/proposed_method.tex", "max_stars_repo_name": "nhutnamhcmus/project-report-template", "max_stars_repo_head_hexsha": "8dec0d20f00959d1f29a476f6b52fbb3eb75eda3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "slide-presentation/proposed_method/proposed_method.tex", "max_issues_repo_name": "nhutnamhcmus/project-report-template", "max_issues_repo_head_hexsha": "8dec0d20f00959d1f29a476f6b52fbb3eb75eda3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "slide-presentation/proposed_method/proposed_method.tex", "max_forks_repo_name": "nhutnamhcmus/project-report-template", "max_forks_repo_head_hexsha": "8dec0d20f00959d1f29a476f6b52fbb3eb75eda3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.75, "max_line_length": 64, "alphanum_fraction": 0.5140186916, "num_tokens": 101, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8791467801752451, "lm_q2_score": 0.7956580976404297, "lm_q1q2_score": 0.6995002546609446}} {"text": "%\\columnbreak\n\\section{Model Selection for Clustering}\n\n% ===\nWhat is the appropriate \\#clusters $k$ for my data?\n\n\\emph{General approach:} Measure quality (neg. log-likelihood) for different $k$ \\enspace $\\to$ \\textbf{elbow}.\n\n% ===\n\\subsection{Complexity-based Model Selection}\n\n\\textbf{Strategy:} add a complexity term to neg. log-likelihood\n\n\\textbf{Attention:} MDL/BIC rely on likelihood optimisation $\\to$ \\underline{not generally applicable}\n\n\n\\emph{Ocam's razor:}\\enspace\nChoose the model that provides the shortest description of the data.\n\n% ---\n\\subsubsection{Min. Description Length\\quad (MDL)}\n\nMinimise \\textbf{descr. length}:\\enspace\n$-\\log p(\\bm X\\mid \\theta) - \\log p(\\theta)$\n\nApprox.:\\enspace\n$\\hat k \\in \\arg\\min_k \\highlight*{-\\log p(\\bm X\\mid \\hat\\theta) + \\frac{k'}{2} \\log n}$\n\n% ---\n\\subsubsection{Bayesian Information Crit.\\quad (BIC)}\n\nParametrise likelihood $p(\\bm X\\mid M)$ by $\\theta$:\\\\\n\\enspace $p(\\bm X\\mid M) = \\int_{\\Theta_M} \\! \\exp(\\log p(\\bm X\\mid M, \\theta)) \\cdot p(\\theta\\mid M) \\diff\\theta$\n\nAssume flat prior $p(\\theta\\vert M) \\approx \\mathit{const}$ and\\\\\nexpand log-likelihood by ML estimator $\\hat\\theta$:\\\\\n$\\overline\\ell (\\theta) = \\frac{\\ell(\\theta)}{n} = \\frac1n \\log p(\\bm X\\vert M,\\theta) \\overset{\\text{i.i.d.}}{=} \\frac1n \\sum_i \\ell(\\theta, X_i) \\overset{\\textrm{Taylor}}{\\approx} \\ldots$\n\n$\\implies p(\\bm X\\mid M) = \\mathit{const}_2 \\cdot \\exp( \\highlight*{\\ell (\\hat\\theta) - \\frac{k'}{2} \\log n} )$\\\\\n\\quad where $k'$ : dimension of (trainable) parameters\n\n% ===", "meta": {"hexsha": "99cd49dc39d906ae49df8e6eb03173ffa849e99d", "size": 1520, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/SLT21/sections/08_model_selection.tex", "max_stars_repo_name": "tstreule/eth-cheat-sheets", "max_stars_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-26T23:11:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T23:11:57.000Z", "max_issues_repo_path": "src/SLT21/sections/08_model_selection.tex", "max_issues_repo_name": "tstreule/eth-cheat-sheets", "max_issues_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/SLT21/sections/08_model_selection.tex", "max_forks_repo_name": "tstreule/eth-cheat-sheets", "max_forks_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.1904761905, "max_line_length": 189, "alphanum_fraction": 0.6776315789, "num_tokens": 514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.879146761176671, "lm_q2_score": 0.7956581073313275, "lm_q1q2_score": 0.6995002480642967}} {"text": "\\section{Delta Method}%\n\\label{sec:delta_method}\n\n\\begin{thm}\n If $\\sqrt{n}(T_n - \\theta) \\convl N(0, \\tau^2)$ and $f'(\\theta) \\ne 0$, then\n \\begin{equation*}\n \\sqrt{n}(f(T_n) - f(\\theta)) \\convl N(0, \\tau^2 [f'(\\theta)]^2).\n \\end{equation*}\n\\end{thm}\n", "meta": {"hexsha": "cf2f56d5cf6d9e70217e070ed66401c6eb9bd51e", "size": 268, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "statistics/inference/src/04-delta-method.tex", "max_stars_repo_name": "jems-lee/notes", "max_stars_repo_head_hexsha": "2e121f2131c4225776d3c820ac4372968e8248d3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "statistics/inference/src/04-delta-method.tex", "max_issues_repo_name": "jems-lee/notes", "max_issues_repo_head_hexsha": "2e121f2131c4225776d3c820ac4372968e8248d3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "statistics/inference/src/04-delta-method.tex", "max_forks_repo_name": "jems-lee/notes", "max_forks_repo_head_hexsha": "2e121f2131c4225776d3c820ac4372968e8248d3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.8, "max_line_length": 80, "alphanum_fraction": 0.5634328358, "num_tokens": 115, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9161096090086368, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6994348170442947}} {"text": "% !TEX root = hott_intro.tex\n\n\\section{Homotopy pullbacks}\n\nSuppose we are given a map $f:A\\to B$, and type families $P$ over $A$, and $Q$ over $B$.\nThen any family of maps\n\\begin{equation*}\ng:\\prd{x:A}P(x)\\to Q(f(x))\n\\end{equation*}\ngives rise to a commuting square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n\\sm{x:A}P(x) \\arrow[r,\"{\\total[f]{g}}\"] \\arrow[d,swap,\"\\proj 1\"] & \\sm{y:B}Q(y) \\arrow[d,\"\\proj 1\"] \\\\\nA \\arrow[r,swap,\"f\"] & B\n\\end{tikzcd}\n\\end{equation*}\nwhere $\\total[f]{g}$ is defined as $\\lam{(x,y)}(f(x),g(x,y))$. In the main theorem of this chapter we show that $g$ is a family of equivalences if and only if this square satisfies a certain universal property: the universal property of \\emph{pullback squares}.\n\nPullback squares are of interest because they appear in many situations. Cartesian products, fibers of maps, and substitutions can all be presented as pullbacks. Moreover, the fact that a family of maps $g:\\prd{x:A}P(x)\\to Q(f(x))$ is a family of equivalences if and only if it induces a pullback square has the very useful corollary that a square of the form\n\\begin{equation*}\n \\begin{tikzcd}\n C \\arrow[d,swap,\"p\"] \\arrow[r] & D \\arrow[d,\"q\"] \\\\\n A \\arrow[r,swap,\"f\"] & B\n \\end{tikzcd}\n\\end{equation*}\nis a pullback square if and only if the induced family of maps between the fibers\n\\begin{equation*}\n \\prd{x:A}\\fib{p}{x}\\to\\fib{q}{f(x)}\n\\end{equation*}\nis a family of equivalences. This connection between pullbacks and \\emph{fiberwise equivalences} has an important role in the descent theorem\\index{descent} in \\cref{chap:descent}.\n\nA second reason for studying pullback squares is that the dual notion of \\emph{pushouts} is an important tool to construct new types, including the $n$-spheres for arbitrary $n$. The duality of pullbacks and pushouts makes it possible to obtain proofs of many statements about pushouts from their dual statements about pullbacks.\n\n\\subsection{The universal property of pullbacks}\n\n\\begin{defn}\\label{defn:cospan}\n A \\define{cospan}\\index{cospan|textbf} consists of three types $A$, $X$, and $B$, and maps $f:A\\to X$ and $g:B\\to X$.\n\\end{defn}\n\n\\begin{defn}\n Consider a cospan\n \\begin{equation*}\n \\begin{tikzcd}\n A \\arrow[r,\"f\"] & X & B \\arrow[l,swap,\"g\"] \n \\end{tikzcd}\n \\end{equation*}\n and a type $C$. A \\define{cone}\\index{cone!on a cospan|textbf} on the cospan $A \\rightarrow X \\leftarrow B$ with \\define{vertex} $C$\\index{vertex!of a cone|textbf} consists of maps $p:C\\to A$, $q:C\\to B$ and a homotopy $H:f\\circ p\\htpy g\\circ q$ witnessing that the square\n \\begin{equation*}\n \\begin{tikzcd}\n C \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\n A \\arrow[r,swap,\"f\"] & X\n \\end{tikzcd}\n \\end{equation*}\n commutes. We write\\index{cone(C)@{$\\mathsf{cone}(\\blank)$}|textbf}\n\\begin{equation*}\n\\mathsf{cone}(C)\\defeq \\sm{p:C\\to A}{q:C\\to B}f\\circ p\\htpy g\\circ q\n\\end{equation*}\nfor the type of cones with vertex $C$.\n\\end{defn}\n\nIt is good practice to characterize the identity type of any type of importance. In the following lemma we give a characterization of the identity type of the type $\\mathsf{cone}(C)$ of cones on $A\\rightarrow X\\leftarrow B$ with vertex $C$. Such characterizations are entirely routine in homotopy type theory.\n\n\\begin{lem}\\label{lem:id_cone}%\n\\index{identity type!of cone@{of $\\mathsf{cone}(C)$}|textit}%\nLet $(p,q,H)$ and $(p',q',H')$ be cones on a cospan $f:A\\rightarrow X \\leftarrow B:g$, both with vertex $C$. Then the type $(p,q,H)=(p',q',H')$ is equivalent to the type of triples $(K,L,M)$ consisting of\n\\begin{align*}\nK & : p \\htpy p' \\\\\nL & : q \\htpy q'\n\\end{align*}\nand a homotopy $M : \\ct{H}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}$ witnessing that the square\n\\begin{equation*}\n\\begin{tikzcd}\nf\\circ p \\arrow[r,\"f\\cdot K\"] \\arrow[d,swap,\"H\"] & f\\circ p' \\arrow[d,\"{H'}\"] \\\\\ng\\circ q \\arrow[r,swap,\"g\\cdot L\"] & g\\circ q'\n\\end{tikzcd}\n\\end{equation*}\nof homotopies commutes.\n\\end{lem}\n\n\\begin{comment}\n\\begin{rmk}\nThe homotopy $M$ is a homotopy of homotopies, and for each $z:C$ the identification $M(z)$ witnesses that the square of identifications\n\\begin{equation*}\n\\begin{tikzcd}[column sep=huge]\nf(p(z)) \\arrow[r,equals,\"\\ap{f}{K(z)}\"] \\arrow[d,equals,swap,\"H(z)\"] & f(p'(z)) \\arrow[d,equals,\"{H'(z)}\"] \\\\\ng(q(z)) \\arrow[r,equals,swap,\"\\ap{g}{L(z)}\"] & g(q'(z))\n\\end{tikzcd}\n\\end{equation*}\ncommutes. \n\\end{rmk}\n\\end{comment}\n\n\\begin{proof}\nBy the fundamental theorem of identity types (\\cref{thm:id_fundamental}) it suffices to show that the type\n\\begin{equation*}\n \\sm{(p',q',H'):\\sm{p':C\\to A}{q':C\\to B}f\\circ p'\\htpy g \\circ q'}{K:p\\htpy p'}{L:q\\htpy q'} \\ct{H}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}\n\\end{equation*}\nis contractible. Using associativity of $\\Sigma$-types and commutativity of cartesian products, it is easy to show that this type is equivalent to the type\n\\begin{equation*}\n \\sm{(p',K):\\sm{p':C\\to A}p\\htpy p'}\\sm{(q',L):\\sm{q':C\\to B}q\\htpy q'}\\sm{H:f\\circ p'\\htpy g \\circ q'}\\ct{H}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}\n\\end{equation*}\nNow we observe that the types $\\sm{p':C\\to A}p\\htpy p'$ and $\\sm{q':C\\to B}q\\htpy q'$ are contractible, with centers of contraction\n\\begin{samepage}\n\\begin{align*}\n(p,\\mathsf{htpy\\usc{}refl}_p) & : \\sm{p':C'\\to A} p\\htpy p' \\\\\n(q,\\mathsf{htpy\\usc{}refl}_q) & : \\sm{q':C'\\to B} q\\htpy q'.\n\\end{align*}%\n\\end{samepage}%\nThus we apply \\cref{ex:contr_in_sigma} to see that the type of tuples $((p',K),(q',L),(H',M))$ is equivalent to the type\n\\begin{equation*}\n\\sm{H':f\\circ p'\\htpy g\\circ q'} \\ct{H}{\\mathsf{htpy\\usc{}refl}_{g\\circ q}}\\htpy \\ct{\\mathsf{htpy\\usc{}refl}_{f\\circ p}}{H'}.\n\\end{equation*}\nOf course, the type $\\ct{H}{\\mathsf{htpy\\usc{}refl}_{g\\circ q}}\\htpy \\ct{\\mathsf{htpy\\usc{}refl}_{f\\circ p}}{H'}$ is equivalent to the type $H\\htpy H'$, and $\\sm{H':f\\circ p\\htpy g\\circ q} H\\htpy H'$ is contractible.\n\\end{proof}\n\nGiven a cone with vertex $C$ on a span $A\\stackrel{f}{\\rightarrow} X \\stackrel{g}{\\leftarrow} B$ and a map $h:C'\\to C$, we construct a new cone with vertex $C'$ in the following definition.\n\n\\begin{defn}\nFor any cone $(p,q,H)$ with vertex $C$ and any type $C'$, we define a map\\index{cone map@{$\\mathsf{cone\\usc{}map}$}|textbf}\n\\begin{equation*}\n\\mathsf{cone\\usc{}map}(p,q,H):(C'\\to C)\\to\\mathsf{cone}(C')\n\\end{equation*}\nby $h\\mapsto (p\\circ h,q\\circ h,H\\circ h)$. \n\\end{defn}\n\n\\begin{defn}\nWe say that a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$ is a \\define{pullback square}\\index{pullback square|textbf}, or that it is \\define{cartesian}\\index{cartesian square|textbf}, if it satisfies the \\define{universal property} of pullbacks\\index{universal property!of pullbacks}, which asserts that the map\n\\begin{equation*}\n\\mathsf{cone\\usc{}map}(p,q,H):(C'\\to C)\\to\\mathsf{cone}(C')\n\\end{equation*}\nis an equivalence for every type $C'$. \n\\end{defn}\n\nWe often indicate the universal property with a diagram as follows:\n\\begin{equation*}\n\\begin{tikzcd}\nC' \\arrow[drr,bend left=15,\"{q'}\"] \\arrow[dr,densely dotted,\"h\"] \\arrow[ddr,bend right=15,swap,\"{p'}\"] \\\\\n& C \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\n& A \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nsince the universal property states that for every cone $(p',q',H')$ with vertex $C'$, the type of pairs $(h,\\alpha)$ consisting of $h:C'\\to C$ equipped with $\\alpha:\\mathsf{cone\\usc{}map}((p,q,H),h)=(p',q',H')$ is contractible by \\cref{thm:contr_equiv}.\n\nAs a corollary we obtain the following characterization of the universal property of pullbacks.\n\n\\begin{lem}\\label{thm:pullback_up}\nConsider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$\nThen the following are equivalent:\\index{universal property!of pullbacks (characterization)|textit}\n\\begin{enumerate}\n\\item The square is a pullback square.\n\\item For every type $C'$ and every cone $(p',q',H')$ with vertex $C'$, the type of quadruples $(h,K,L,M)$ consisting of a map $h:C'\\to C$, homotopies\n\\begin{align*}\nK & : p\\circ h \\htpy p' \\\\\nL & : q\\circ h \\htpy q',\n\\end{align*}\nand a homotopy $M : \\ct{(H\\cdot h)}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}$ witnessing that the square\n\\begin{equation*}\n\\begin{tikzcd}\nf\\circ p\\circ h \\arrow[r,\"f\\cdot K\"] \\arrow[d,swap,\"H\\cdot h\"] & f\\circ p' \\arrow[d,\"{H'}\"] \\\\\ng\\circ q\\circ h \\arrow[r,swap,\"g\\cdot L\"] & g\\circ q'\n\\end{tikzcd}\n\\end{equation*}\ncommutes, is contractible.\n\\end{enumerate}\n\\end{lem}\n\n\\begin{proof}\nThe map $\\mathsf{cone\\usc{}map}(p,q,H)$ is an equivalence if and only if its fibers are contractible. By \\cref{lem:id_cone} it follows that the fibers of $\\mathsf{cone\\usc{}map}(p,q,H)$ are equivalent the the described type of quadruples ($h,K,L,M)$.\n\\end{proof}\n\nIn the following lemma we establish the uniqueness of pullbacks up to equivalence via a \\emph{3-for-2 property} for pullbacks.\n\n\\begin{lem}\\label{lem:pb_3for2}\\index{pullback!3-for-2 property|textit}\\index{3-for-2 property!of pullbacks|textit}%\nConsider the squares\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] & {C'} \\arrow[r,\"{q'}\"] \\arrow[d,swap,\"{p'}\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X & A \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith homotopies $H:f\\circ p \\htpy g\\circ q$ and $H':f\\circ p'\\htpy g\\circ q'$.\nFurthermore, suppose we have a map $h:C'\\to C$ equipped with\n\\begin{align*}\nK & : p\\circ h \\htpy p' \\\\\nL & : q\\circ h \\htpy q' \\\\\nM & : \\ct{(H\\cdot h)}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}.\n\\end{align*}\nIf any two of the following three properties hold, so does the third:\n\\begin{samepage}%\n\\begin{enumerate}\n\\item $C$ is a pullback.\n\\item $C'$ is a pullback.\n\\item $h$ is an equivalence.\n\\end{enumerate}%\n\\end{samepage}%\n\\end{lem}\n\n\\begin{proof}\nBy the characterization of the identity type of $\\mathsf{cone}(C')$ given in \\cref{lem:id_cone} we obtain an identification\n\\begin{equation*}\n\\mathsf{cone\\usc{}map}((p,q,H),h)=(p',q',H')\n\\end{equation*}\nfrom the triple $(K,L,M)$. \nLet $D$ be a type, and let $k:D\\to C'$ be a map. We observe that\n\\begin{align*}\n\\mathsf{cone\\usc{}map}((p,q,H),(h\\circ k)) & \\jdeq (p\\circ (h\\circ k),q\\circ (h\\circ k),H\\circ (h\\circ k)) \\\\\n& \\jdeq ((p\\circ h)\\circ k,(q\\circ h)\\circ k, (H\\circ h)\\circ k) \\\\\n& \\jdeq \\mathsf{cone\\usc{}map}(\\mathsf{cone\\usc{}map}((p,q,H),h),k) \\\\\n& = \\mathsf{cone\\usc{}map}((p',q',H'),k).\n\\end{align*}\nThus we see that the triangle \n\\begin{equation*}\n\\begin{tikzcd}[column sep=-1em]\n(D\\to C') \\arrow[rr,\"{h\\circ \\blank}\"] \\arrow[dr,swap,\"{\\mathsf{cone\\usc{}map}(p',q',H')}\"] & & (D\\to C) \\arrow[dl,\"{\\mathsf{cone\\usc{}map}(p,q,H)}\"] \\\\\n& \\mathsf{cone}(D)\n\\end{tikzcd}\n\\end{equation*}\ncommutes. Therefore it follows from the 3-for-2 property of equivalences established in \\cref{ex:3_for_2}, that if any two of the maps in this triangle is an equivalence, then so is the third. Now the claim follows from the fact that $h$ is an equivalence if and only if $h\\circ\\blank : (D\\to C')\\to (D\\to C)$ is an equivalence for any type $D$, which was established in \\cref{lem:postcomp_equiv}.\n\\end{proof}\n\nPullbacks are not only unique in the sense that any two pullbacks of the same cospan are equivalent, they are \\emph{uniquely unique}\\index{uniquely uniqueness!of pullbacks} in the sense that the type of quadruples $(h,K,L,M)$ as in \\cref{lem:pb_3for2} is contractible.\n\n\\begin{cor}\nSuppose both commuting squares\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] & {C'} \\arrow[r,\"{q'}\"] \\arrow[d,swap,\"{p'}\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X & A \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith homotopies $H:f\\circ p \\htpy g\\circ q$ and $H':f\\circ p'\\htpy g\\circ q'$ are pullback squares.\nThen the type of quadruples $(e,K,L,M)$ consisting of an equivalence $e:\\eqv{C'}{C}$ equipped with\n\\begin{align*}\nK & : p\\circ e \\htpy p' \\\\\nL & : q\\circ e \\htpy q' \\\\\nM & : \\ct{(H\\cdot h)}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H'}.\n\\end{align*}\nis contractible.\n\\end{cor}\n\n\\begin{proof}\nWe have seen that the type of quadruples $(h,K,L,M)$ is equivalent to the fiber of $\\mathsf{cone\\usc{}map}(p,q,H)$ at $(p',q',H')$. By \\cref{lem:pb_3for2} it follows that $h$ is an equivalence. Since $\\isequiv(h)$ is a proposition by \\cref{ex:isprop_isequiv}, and hence contractible as soon as it is inhabited, it follows that the type of quadruples $(e,K,L,M)$ is contractible. \n\\end{proof}\n\n\\subsection{Canonical pullbacks}\n\nFor every cospan we can construct a \\emph{canonical pullback}.\n\n\\begin{defn}\nLet $f:A\\to X$ and $g:B\\to X$ be maps. Then we define\n\\begin{align*}\nA\\times_X B & \\defeq \\sm{x:A}{y:B}f(x)=g(y) \\\\\n\\pi_1 & \\defeq \\proj 1 & & : A\\times_X B\\to A \\\\\n\\pi_2 & \\defeq \\proj 1\\circ\\proj 2 & & : A\\times_X B\\to B\\\\\n\\pi_3 & \\defeq \\proj 2\\circ\\proj 2 & & : f\\circ \\pi_1 \\htpy g\\circ\\pi_2.\n\\end{align*}\nThe type $A\\times_X B$ is called the \\define{canonical pullback}\\index{canonical pullback|textbf} of $f$ and $g$.\n\\end{defn}\n\nNote that $A\\times_X B$ depends on $f$ and $g$, although this dependency is not visible in the notation.\n\n\\begin{rmk}\n Given $(x,y,p)$ and $(x',y',p')$ in the canonical pullback $A\\times_X B$, the identity type $(x,y,p)=(x',y',p')$ is equivalent to the type of triples $(\\alpha,\\beta,\\gamma)$ consisting of $\\alpha:x=x'$, $\\beta:y=y'$, and an identification $\\gamma:\\ct{p}{\\ap{g}{\\beta}}=\\ct{\\ap{f}{\\alpha}}{p'}$ witnessing that the square\n \\begin{equation*}\n \\begin{tikzcd}[column sep=large]\n f(x) \\arrow[r,equal,\"\\ap{f}{\\alpha}\"] \\arrow[d,swap,equal,\"p\"] & f(x') \\arrow[d,equal,\"{p'}\"] \\\\\n g(y) \\arrow[r,swap,equal,\"\\ap{g}{\\beta}\"] & g(y')\n \\end{tikzcd}\n \\end{equation*}\n commutes. The proof of this fact is similar to the proof of \\cref{lem:id_cone}.\n\\end{rmk}\n\n\\begin{thm}\\label{thm:canonical-pullback}\nGiven maps $f:A\\to X$ and $g:B\\to X$, the commuting square\\index{canonical pullback|textit}\n\\begin{equation*}\n\\begin{tikzcd}\nA\\times_X B \\arrow[r,\"\\pi_2\"] \\arrow[d,swap,\"\\pi_1\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X,\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{thm}\n\n\\begin{proof}\nLet $C$ be a type. Our goal is to show that the map\n\\begin{equation*}\n\\mathsf{cone\\usc{}map}(\\pi_1,\\pi_2,\\pi_3): (C\\to A\\times_X B)\\to \\mathsf{cone}(C)\n\\end{equation*}\nis an equivalence. Note that we have the commuting triangle\n\\begin{equation*}\n \\begin{tikzcd}[column sep=-4em]\n C\\to\\sm{x:A}{y:B}f(x)=g(y) \\arrow[dd,swap,\"\\mathsf{cone\\usc{}map}\"] \\arrow[dr,\"\\mathsf{choice}\"] \\\\\n & \\sm{p:C\\to A}\\prd{z:C}\\sm{y:B} f(p(z))= g(y) \\arrow[dl,\"\\mathsf{choice}\"] \\\\\n \\sm{p:C\\to A}{q:C\\to B} f\\circ p \\htpy g\\circ q.\n \\end{tikzcd}\n\\end{equation*}\nIn this triangle the functions $\\mathsf{choice}$ are equivalences by \\cref{thm:choice}. Therefore, their composite is an equivalence.\n\\end{proof}\n\n\\begin{defn}\nGiven a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p \\htpy g \\circ q$, we define the \\define{gap map}\\index{gap map|textbf}\\index{pullback!gap map|textbf}\n\\begin{equation*}\n\\mathsf{gap}(p,q,H):C \\to A\\times_X B\n\\end{equation*}\nby $\\lam{z}(p(z),q(z),H(z))$.\n\\end{defn}\n\nThe following theorem provides a useful characterization of pullback squares, because in many situations it is easier to show that the gap map is an equivalence.\n\n\\begin{thm}\\label{thm:is_pullback}\nConsider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p \\htpy g \\circ q$. The following are equivalent:\n\\begin{enumerate}\n\\item The square is a pullback square\n\\item There is a term of type\n\\begin{equation*}\n\\mathsf{is\\usc{}pullback}(p,q,H)\\defeq \\isequiv(\\mathsf{gap}(p,q,H)).\n\\end{equation*}\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\n Observe that we are in the situation of \\cref{lem:pb_3for2}. Indeed, we have two commuting squares\n \\begin{equation*}\n \\begin{tikzcd}\n A\\times_X B \\arrow[r,\"\\pi_2\"] \\arrow[d,swap,\"\\pi_2\"] & B \\arrow[d,\"g\"] &[2em] C \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\n A \\arrow[r,swap,\"f\"] & X & A \\arrow[r,swap,\"f\"] & X,\n \\end{tikzcd}\n \\end{equation*}\n and we have the gap map $\\mathsf{gap}:C\\to A\\times_X B$, which comes equipped with the homotopies\n\\begin{align*}\nK & : \\pi_1\\circ \\mathsf{gap} \\htpy p & K & \\defeq \\lam{z}\\refl{p(z)} \\\\\nL & : \\pi_2\\circ \\mathsf{gap} \\htpy q & L & \\defeq \\lam{z}\\refl{q(z)} \\\\\nM & : \\ct{(\\pi_3\\cdot \\mathsf{gap})}{(g\\cdot L)} \\htpy \\ct{(f\\cdot K)}{H} & M & \\defeq \\lam{z}\\mathsf{right\\usc{}unit}(H(z)).\n\\end{align*}\nSince $A\\times_X B$ is shown to be a pullback in \\cref{thm:canonical-pullback}, it follows from \\cref{lem:pb_3for2} that $C$ is a pullback if and only if the gap map is an equivalence.\n\\end{proof}\n\n\\subsection{Cartesian products and fiberwise products as pullbacks}\n\nAn important special case of pullbacks occurs when the cospan is of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[r] & \\unit & B. \\arrow[l]\n\\end{tikzcd}\n\\end{equation*}\nIn this case, the pullback is just the \\emph{cartesian product}.\n\n\\begin{lem}\\label{lem:prod_pb}\nLet $A$ and $B$ be types. Then the square\n\\begin{equation*}\n\\begin{tikzcd}\nA\\times B \\arrow[r,\"\\proj 2\"] \\arrow[d,swap,\"\\proj 1\"] & B \\arrow[d,\"\\mathsf{const}_{\\ttt}\"] \\\\\nA \\arrow[r,swap,\"\\mathsf{const}_{\\ttt}\"] & \\unit\n\\end{tikzcd}\n\\end{equation*}\nwhich commutes by the homotopy $\\mathsf{const}_{\\refl{\\ttt}}$ is a pullback square.\\index{cartesian product!as pullback}\n\\end{lem}\n\n\\begin{proof}\nBy \\cref{thm:is_pullback} it suffices to show that\n\\begin{equation*}\n\\mathsf{gap}(\\proj 1,\\proj2,\\lam{(a,b)}\\refl{\\ttt})\n\\end{equation*}\nis an equivalence. Its inverse is the map $\\lam{(a,b,p)}(a,b)$.\n\\end{proof}\n\nThe following generalization of \\cref{lem:prod_pb} is the reason why pullbacks are sometimes called \\define{fiber products}\\index{fiber product|textbf}.\n\n\\begin{thm}\nLet $P$ and $Q$ be families over a type $X$. Then the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=8em]\n\\sm{x:X}P(x)\\times Q(x) \\arrow[r,\"{\\lam{(x,(p,q))}(x,q)}\"] \\arrow[d,swap,\"{\\lam{(x,(p,q))}(x,p)}\"] & \\sm{x:X}Q(x) \\arrow[d,\"\\proj 1\"] \\\\\n\\sm{x:X}P(x) \\arrow[r,swap,\"\\proj 1\"] & X,\n\\end{tikzcd}\n\\end{equation*}\nwhich commutes by the homotopy\n\\begin{equation*}\nH\\defeq \\lam{(x,(p,q))}\\refl{x},\n\\end{equation*}\nis a pullback square.\n\\end{thm}\n\n\\begin{proof}\nBy \\cref{thm:is_pullback} it suffices to show that the gap map is an equivalence. The gap map is homotopic to the function\n\\begin{equation*}\n\\lam{(x,(p,q))}((x,p),(x,q),\\refl{x}).\n\\end{equation*}\nIt is easy to check that this function is an equivalence. I\nts inverse is the map \n\\begin{equation*}\n\\lam{((x,p),(y,q),\\alpha)}(y,(\\mathsf{tr}_P(\\alpha,p),q)).\\qedhere\n\\end{equation*}\n\\end{proof}\n\n\\begin{cor}\nFor any $f:A\\to X$ and $g:B\\to X$, the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=8em]\n\\sm{x:X}\\fib{f}{x}\\times\\fib{g}{x} \\arrow[r,\"{\\lam{(x,((a,p),(b,q)))}b}\"] \\arrow[d,swap,\"{\\lam{(x,((a,p),(b,q)))}a}\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{cor}\n\n\\subsection{Fibers of maps as pullbacks}\n\n\\begin{lem}\\label{lem:fib_pb}\nFor any function $f:A\\to B$, and any $b:B$, consider the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n\\fib{f}{b} \\arrow[r,\"\\mathsf{const}_\\ttt\"] \\arrow[d,swap,\"\\proj 1\"] & \\unit \\arrow[d,\"\\mathsf{const}_b\"] \\\\\nA \\arrow[r,swap,\"f\"] & B\n\\end{tikzcd}\n\\end{equation*}\nwhich commutes by $\\proj 2 : \\prd{t:\\fib{f}{b}} f(\\proj 1(t))=b$. This is a pullback square.\\index{fiber!as pullback|textit}\n\\end{lem}\n\n\\begin{proof}\nBy \\cref{thm:is_pullback} it suffices to show that the gap map is an equivalence. The gap map is homotopic to the function\n\\begin{equation*}\n\\mathsf{total}(\\lam{x}{p}(\\ttt,p))\n\\end{equation*}\nThe map $\\lam{x}{p}(\\ttt,p)$ is a family of equivalences by \\cref{ex:contr_in_sigma}, so it induces an equivalence on total spaces by \\cref{thm:fib_equiv}.\n\\end{proof}\n\n\\begin{cor}\nFor any type family $B$ over $A$ and any $a:A$ the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\nB(a) \\arrow[d,swap,\"{\\lam{y}(a,y)}\"] \\arrow[r,\"\\mathsf{const}_\\ttt\"] & \\unit \\arrow[d,\"\\lam{\\ttt}a\"] \\\\\n\\sm{x:A}B(x) \\arrow[r,swap,\"\\proj 1\"] & A\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{cor}\n\n\\begin{proof}\n To see this, note that the triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=0]\n B(a) \\arrow[rr,\"{\\lam{b}((a,b),\\refl{a})}\"] \\arrow[dr,swap,\"\\mathsf{gap}\"] & & \\fib{\\proj 1}{a} \\arrow[dl,\"\\mathsf{gap}\"] \\\\\n & \\Big(\\sm{x:A}B(x)\\Big)\\times_A\\unit.\n \\end{tikzcd}\n \\end{equation*}\n Since the top map is an equivalence by \\cref{ex:fib_replacement}, and the map on the right is an equivalence by \\cref{lem:fib_pb}, it follows that the map on the left is an equivalence. The claim follows.\n\\end{proof}\n\n\\subsection{Families of equivalences}\n\n\\begin{lem}\\label{lem:pb_subst}\nLet $f:A\\to B$, and let $Q$ be a type family over $B$. Then the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=6em]\n\\sm{x:A}Q(f(x)) \\arrow[r,\"{\\lam{(x,q)}(f(x),q)}\"] \\arrow[d,swap,\"\\proj 1\"] & \\sm{y:B}Q(b) \\arrow[d,\"\\proj 1\"] \\\\\nA \\arrow[r,swap,\"f\"] & B\n\\end{tikzcd}\n\\end{equation*}\ncommutes by $H\\defeq \\lam{(x,q)}\\refl{f(x)}$. This is a pullback square.\\index{substitution!as pullback|textit}\n\\end{lem}\n\n\\begin{proof}\nBy \\cref{thm:is_pullback} it suffices to show that the gap map is an equivalence. The gap map is homotopic to the function\n\\begin{equation*}\n\\lam{(x,q)}(x,(f(x),q),\\refl{f(x)}).\n\\end{equation*}\nThe inverse of this map is given by $\\lam{(x,((y,q),p))}(x,\\mathsf{tr}_Q(p^{-1},q))$, and it is straightforward to see that these maps are indeed mutual inverses.\n\\end{proof}\n\n\\begin{thm}\\label{thm:pb_fibequiv}\nLet $f:A\\to B$, and let $g:\\prd{a:A}P(a)\\to Q(f(a))$ be a family of maps\\index{family of maps|textit}. The following are equivalent:\n\\begin{enumerate}\n\\item The commuting square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n\\sm{a:A}P(a) \\arrow[r,\"{\\total[f]{g}}\"] \\arrow[d,->>] & \\sm{b:B}Q(b) \\arrow[d,->>] \\\\\nA \\arrow[r,swap,\"f\"] & B\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\item $g$ is a family of equivalences.\\index{family of equivalences|textit}\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nThe gap map is homotopic to the composite\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n\\sm{x:A}P(x) \\arrow[r,\"\\total{g}\"] & \\sm{x:A}Q(f(x)) \\arrow[r,\"{\\mathsf{gap}'}\"] & A \\times_B \\Big(\\sm{y:B}Q(y)\\Big)\n\\end{tikzcd}\n\\end{equation*}\nwhere $\\mathsf{gap}'$ is the gap map for the square in \\cref{lem:pb_subst}. Since $\\mathsf{gap}'$ is an equivalence, it follows by \\cref{ex:3_for_2,thm:fib_equiv} that the gap map is an equivalence if and only if $g$ is a family of equivalences.\n\\end{proof}\n\nOur goal is now to extend \\cref{thm:pb_fibequiv} to\narbitrary pullback squares. Note that every commuting\nsquare\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[d,swap,\"f\"] \\arrow[r,\"h\"] & B \\arrow[d,\"g\"] \\\\\nX \\arrow[r,swap,\"i\"] & Y\n\\end{tikzcd}\n\\end{equation*}\nwith $H: i\\circ f ~ g \\circ h$ induces a map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq} : \\prd{x:X} \\fib{f}{x} \\to \\fib{g}{f(x)}\n\\end{equation*}\non the fibers, by\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}(x,(a,p))\\defeq (h(a),\\ct{H(a)^{-1}}{\\ap{i}{p}}).\n\\end{equation*}\n\n\\begin{thm}\\label{cor:pb_fibequiv}\nConsider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[d,swap,\"f\"] \\arrow[r,\"h\"] & B \\arrow[d,\"g\"] \\\\\nX \\arrow[r,swap,\"i\"] & Y\n\\end{tikzcd}\n\\end{equation*}\nwith $H: i\\circ f ~ g \\circ h$. The following are equivalent:\n\\begin{enumerate}\n\\item The square is a pullback square.\\index{pullback square!characterized by families of equivalences|textit}\n\\item The induced map on fibers\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq} : \\prd{x:X} \\fib{f}{x} \\to \\fib{g}{f(x)}\n\\end{equation*}\nis a family of equivalences.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nFirst we observe that the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=huge]\n\\sm{x:X}\\fib{f}{x} \\arrow[d,swap,\"\\eqvsym\"] \\arrow[r,\"\\total{\\mathsf{fib\\usc{}sq}}\"] &\n\\sm{x:X}\\fib{g}{f(x)} \\arrow[d,\"\\total{\\total{\\mathsf{inv}}}\"] \\\\\nA \\arrow[r,swap,\"\\mathsf{gap}\"] & X \\times_Y B\n\\end{tikzcd}\n\\end{equation*}\ncommutes. To construct such a homotopy, we need to construct an identification\n\\begin{equation*}\n(f(a),h(a),H(a))=(x,h(a),(\\ct{H(a)^{-1}}{\\ap{i}{p}})^{-1})\n\\end{equation*}\nfor every $x : X$, $a : A$, and $p : f(a) = x$. This is shown by path induction on $p : f(a)=x$. Thus, it suffices to show that\n\\begin{equation*}\n(f(a),h(a),H(a))=(f(a),h(a),(\\ct{H(a)^{-1}}{\\refl{i(f(a))}})^{-1}),\n\\end{equation*}\nwhich is a routine exercise. \n\nNow we note that the left and right maps in this square are both equivalences. Therefore it follows that the top map is an equivalence if and only if the bottom map is. The claim now follows by \\cref{thm:fib_equiv}.\n\\end{proof}\n\n\\begin{cor}\\label{cor:pb_trunc}\nConsider a pullback square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X.\n\\end{tikzcd}\n\\end{equation*}\nIf $g$ is a $k$-truncated map, then so is $p$. In particular, if $g$ is an embedding then so is $p$.\\index{truncated!map!pullbacks of truncated maps|textit}\\index{embedding!pullbacks of embeddings|textit}\n\\end{cor}\n\n\\begin{proof}\nSince the square is assumed to be a pullback square, it follows from \\cref{cor:pb_fibequiv} that for each $x:A$, the fiber $\\fib{p}{x}$ is equivalent to the fiber $\\fib{g}{f(x)}$, which is $k$-truncated. Since $k$-truncated types are closed under equivalences by \\cref{thm:ktype_eqv}, it follows that $p$ is a $k$-truncated map.\n\\end{proof}\n\n\\begin{cor}\\label{cor:pb_equiv}\nConsider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X.\n\\end{tikzcd}\n\\end{equation*}\nand suppose that $g$ is an equivalence. Then the following are equivalent:\n\\begin{enumerate}\n\\item The square is a pullback square.\n\\item The map $p:C\\to A$ is an equivalence.\\index{equivalence!pullback of|textit}\n\\end{enumerate}\n\\end{cor}\n\n\\begin{proof}\nIf the square is a pullback square, then by \\cref{thm:pb_fibequiv} the fibers of $p$ are equivalent to the fibers of $g$, which are contractible by \\cref{thm:contr_equiv}. Thus it follows that $p$ is a contractible map, and hence that $p$ is an equivalence.\n\nIf $p$ is an equivalence, then by \\cref{thm:contr_equiv} both $\\fib{p}{x}$ and $\\fib{g}{f(x)}$ are contractible for any $x:X$. It follows by \\cref{ex:contr_equiv} that the induced map $\\fib{p}{x}\\to\\fib{g}{f(x)}$ is an equivalence. Thus we apply \\cref{cor:pb_fibequiv} to conclude that the square is a pullback.\n\\end{proof}\n\n\\begin{thm}\\label{thm:pb_fibequiv_complete}\nConsider a diagram of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[d,swap,\"f\"] & B \\arrow[d,\"g\"] \\\\\nX \\arrow[r,swap,\"h\"] & Y.\n\\end{tikzcd}\n\\end{equation*}\nThen the type of triples $(i,H,p)$ consisting of a map $i:A\\to B$, a homotopy $H:h\\circ f\\htpy g\\circ i$, and a term $p$ witnessing that the square\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[d,swap,\"f\"] \\arrow[r,\"i\"] & B \\arrow[d,\"g\"] \\\\\nX \\arrow[r,swap,\"h\"] & Y.\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square, is equivalent to the type of families of equivalences\n\\begin{equation*}\n\\prd{x:X}\\eqv{\\fib{f}{x}}{\\fib{g}{h(x)}}.\n\\end{equation*}\n\\end{thm}\n\n\\begin{cor}\\label{cor:pb_fibequiv_complete}\nLet $h:X\\to Y$ be a map, and let $P$ and $Q$ be families over $X$ and $Y$, respectively.\nThen the type of triples $(i,H,p)$ consisting of a map \n\\begin{equation*}\ni:\\Big(\\sm{x:X}P(x)\\Big)\\to \\Big(\\sm{y:Y}Q(y)\\Big),\n\\end{equation*}\na homotopy $H:h\\circ \\proj 1\\htpy \\proj 1\\circ i$, and a term $p$ witnessing that the square\n\\begin{equation*}\n\\begin{tikzcd}\n\\sm{x:X}P(x) \\arrow[d,swap,\"\\proj 1\"] \\arrow[r,\"i\"] & \\sm{y:Y}Q(y) \\arrow[d,\"\\proj 1\"] \\\\\nX \\arrow[r,swap,\"h\"] & Y.\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square, is equivalent to the type of families of equivalences\n\\begin{equation*}\n\\prd{x:X}\\eqv{P(x)}{Q(h(x))}.\n\\end{equation*}\n\\end{cor}\n\nOne useful application of the connection between pullbacks and families of equivalences is the following theorem, which is also called the \\define{pasting property} of pullbacks.\\index{pasting property!of pullbacks|textit}\n\n\\begin{thm}\\label{thm:pb_pasting}\nConsider a commuting diagram of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[r,\"k\"] \\arrow[d,swap,\"f\"] & B \\arrow[r,\"l\"] \\arrow[d,\"g\"] & C \\arrow[d,\"h\"] \\\\\nX \\arrow[r,swap,\"i\"] & Y \\arrow[r,swap,\"j\"] & Z\n\\end{tikzcd}\n\\end{equation*}\nwith homotopies $H:i\\circ f\\htpy g\\circ k$ and $K:j\\circ g\\htpy h\\circ l$, and the homotopy\n\\begin{equation*}\n\\ct{(j\\cdot H)}{(K\\cdot k)}:j\\circ i\\circ f\\htpy h\\circ l\\circ k\n\\end{equation*}\nwitnessing that the outer rectangle commutes. Furthermore, suppose that the square on the right is a pullback square. Then the following are equivalent:\n\\begin{samepage}%\n\\begin{enumerate}\n\\item The square on the left is a pullback square.\n\\item The outer rectangle is a pullback square.\n\\end{enumerate}%\n\\end{samepage}%\n\\end{thm}\n\n\\begin{proof}\nThe commutativity of the two squares and the outer rectangle induces a commuting triangle\n\\begin{equation*}\n\\begin{tikzcd}[column sep=tiny]\n\\fib{f}{x} \\arrow[rr,\"\\mathsf{fib\\usc{}sq}_{(f,k,H)}(x)\"] \\arrow[dr,swap,\"\\mathsf{fib\\usc{}sq}_{f,l\\circ k,\\ct{(j\\cdot H)}{(K\\cdot k)}}(x)\"] & & \\fib{g}{i(x)} \\arrow[dl,\"\\mathsf{fib\\usc{}sq}_{(g,l,K)}(i(x))\"] \\\\\n& \\fib{h}{j(i(x))}.\n\\end{tikzcd}\n\\end{equation*}\nA homotopy witnessing that the triangle commutes is constructed by a routine calculation.\n\nSince the triangle commutes, and since the map $\\mathsf{fib\\usc{}sq}_{(g,l,K)}(i(x))$ is an equivalence for each $x:X$ by \\cref{cor:pb_fibequiv}, it follows\nby the 3-for-2 property of equivalences that for each $x:X$ the top map in the triangle is an equivalence if and only if the left map is an equivalence.\nThe claim now follows by a second application of \\cref{cor:pb_fibequiv}.\n\\end{proof}\n\n\\subsection{Descent theorems for coproducts and \\texorpdfstring{$\\Sigma$}{Σ}-types}\n\n\\begin{thm}\\label{thm:descent-coprod}\nConsider maps $f:A'\\to A$ and $g:B'\\to B$, a map $h:X'\\to X$, and commuting squares of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA' \\arrow[r] \\arrow[d,swap,\"f\"] & X' \\arrow[d,\"h\"] & B' \\arrow[r] \\arrow[d,swap,\"g\"] & X' \\arrow[d,\"h\"] \\\\\nA \\arrow[r] & X & B \\arrow[r] & X.\n\\end{tikzcd}\n\\end{equation*}\nThen the following are equivalent:\n\\begin{enumerate}\n\\item Both squares are pullback squares.\n\\item The commuting square \n\\begin{equation*}\n\\begin{tikzcd}\nA'+B' \\arrow[d,swap,\"f+g\"] \\arrow[r] & X' \\arrow[d,\"h\"] \\\\\nA+B \\arrow[r] & X\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nBy \\cref{cor:pb_fibequiv} it suffices to show that the following are equivalent:\n\\begin{enumerate}\n\\item For each $x:A$ the map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}:\\fib{f}{x}\\to\\fib{h}{\\alpha_A(x)}\n\\end{equation*}\nis an equivalence, and for each $y:B$ the map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}:\\fib{g}{y}\\to\\fib{h}{\\alpha_B(y)}\n\\end{equation*}\nis an equivalence.\n\\item For each $t:A+B$ the map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}:\\fib{f+g}{t} \\to \\fib{h}{\\alpha(t)}\n\\end{equation*}\nis an equivalence.\n\\end{enumerate}\nBy the dependent universal property of coproducts, the second claim is equivalent to the claim that both for each $x:A$ the map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}:\\fib{f+g}{\\inl(x)}\\to \\fib{h}{\\alpha_A(x)}\n\\end{equation*}\nis an equivalence, and for each $y:B$, the map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq}:\\fib{f+g}{\\inr(y)}\\to \\fib{h}{\\alpha_B(y)} \n\\end{equation*}\nis an equivalence.\n\nWe claim that there is a commuting triangle\n\\begin{equation*}\n\\begin{tikzcd}[column sep=-1em]\n\\fib{f}{x} \\arrow[rr] \\arrow[dr] & & \\fib{f+g}{\\inl(x)} \\arrow[dl] \\\\\n\\phantom{\\fib{f+g}{\\inl(x)}} & \\fib{h}{\\alpha_A(x)}\n\\end{tikzcd}\n\\end{equation*}\nfor every $x:A$. To see that the triangle commutes, we need to construct an identification\n\n\nThe top map is given by\n\\begin{equation*}\n(a',p)\\mapsto (\\inl(a'),\\ap{\\inl}{p}).\n\\end{equation*}\nThe triangle then commutes by the homotopy\n\\begin{equation*}\n(a',p)\\mapsto \\mathsf{eq\\usc{}pair}(\\refl,\\ap{\\mathsf{concat}(H(a')^{-1})}{\\mathsf{ap\\usc{}comp}_{[\\alpha_A,\\alpha_B],inl}})\n\\end{equation*}\nWe note that the top map is an equivalence, so it follows by the 3-for-2 property of equivalences that the left map is an equivalence if and only if the right map is an equivalence. \n\nSimilarly, there is a commuting triangle\n\\begin{equation*}\n\\begin{tikzcd}[column sep=-1em]\n\\fib{g}{y} \\arrow[rr] \\arrow[dr] & & \\fib{f+g}{\\inr(y)} \\arrow[dl] \\\\\n\\phantom{\\fib{f+g}{\\inr(y)}} & \\fib{h}{\\alpha_B(y)}\n\\end{tikzcd}\n\\end{equation*}\nin which the top map is an equivalence, completing the proof.\n\\end{proof}\n\nIn the following corollary we conclude that coproducts distribute over pullbacks. \n\n\\begin{cor}\nConsider a cospan of the form\n\\begin{equation*}\n\\begin{tikzcd}\n& Y \\arrow[d] \\\\\nA+B \\arrow[r] & X.\n\\end{tikzcd}\n\\end{equation*}\nThen there is an equivalence\n\\begin{equation*}\n(A+B)\\times_X Y \\simeq (A\\times_X Y)+(B\\times_X Y).\n\\end{equation*}\n\\end{cor}\n\n\\begin{thm}\\label{thm:descent-Sigma}\nConsider a family of maps $f_i:A'_i\\to A_i$ indexed by a type $I$, a map $h:X'\\to X$, and a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nA'_i \\arrow[r] \\arrow[d,swap,\"f_i\"] & X' \\arrow[d,\"h\"] \\\\\nA_i \\arrow[r,swap,\"\\alpha_i\"] & X\n\\end{tikzcd}\n\\end{equation*}\nfor each $i:I$. Then the following are equivalent:\n\\begin{enumerate}\n\\item For each $i:I$ the square is a pullback square.\n\\item The commuting square\n\\begin{equation*}\n\\begin{tikzcd}\n\\sm{i:I}A'_i \\arrow[d,swap,\"\\total{f}\"] \\arrow[r] & X' \\arrow[d,\"h\"] \\\\\n\\sm{i:I}A_i \\arrow[r,\"\\alpha\"] & X\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nBy \\cref{cor:pb_fibequiv} it suffices to show that the following are equivalent for each $i:I$ and $a:A_i$:\n\\begin{enumerate}\n\\item The map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq} : \\fib{f_i}{a} \\to \\fib{g}{\\alpha_i(a)}\n\\end{equation*}\nis an equivalence.\n\\item The map\n\\begin{equation*}\n\\mathsf{fib\\usc{}sq} : \\fib{\\total{f}}{i,a}\\to \\fib{g}{\\alpha_i(a)}\n\\end{equation*}\nis an equivalence.\n\\end{enumerate}\nTo see this, note that we have a commuting triangle\n\\begin{equation*}\n\\begin{tikzcd}[column sep=-1em]\n\\fib{f_i}{a} \\arrow[rr] \\arrow[dr] & & \\fib{\\total{f}}{i,a} \\arrow[dl] \\\\\n\\phantom{\\fib{\\total{f}}{i,a}} & \\fib{g}{\\alpha_i(a)},\n\\end{tikzcd}\n\\end{equation*}\nwhere the top map is an equivalence by \\cref{lem:fib_total}. Therefore the claim follows by the 3-for-2 property of equivalences.\n\\end{proof}\n\nIn the following corollary we conclude that $\\Sigma$ distributes over coproducts.\n\n\\begin{cor}\nConsider a cospan of the form\n\\begin{equation*}\n\\begin{tikzcd}\n& Y \\arrow[d] \\\\\n\\sm{i:I}A_i \\arrow[r] & X.\n\\end{tikzcd}\n\\end{equation*}\nThen there is an equivalence\n\\begin{equation*}\n\\Big(\\sm{i:I}A_i\\Big)\\times_X Y \\simeq \\sm{i:I} (A_i\\times_X Y).\n\\end{equation*}\n\\end{cor}\n\n\n\n\\begin{exercises}\n\\item \\label{ex:id_pb}\\index{identity type!as pullback}\n\\begin{subexenum}\n\\item Show that the square\\index{identity type!as pullback}\n\\begin{equation*}\n\\begin{tikzcd}\n(x=y) \\arrow[r] \\arrow[d] & \\unit \\arrow[d,\"\\mathsf{const}_y\"] \\\\\n\\unit \\arrow[r,swap,\"\\mathsf{const}_x\"] & A\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\item Show that the square\\index{diagonal!of a type!fibers of}\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n(x=y) \\arrow[r,\"\\mathsf{const}_{x}\"] \\arrow[d,swap,\"\\mathsf{const}_\\ttt\"] & A \\arrow[d,\"\\delta_A\"] \\\\\n\\unit \\arrow[r,swap,\"{\\mathsf{const}_{(x,y)}}\"] & A\\times A\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square, where $\\delta_A:A\\to A\\times A$ is the diagonal of $A$, defined in \\cref{ex:diagonal}.\n\\end{subexenum}\n\\item \\label{ex:trunc_diagonal_map}In this exercise we give an alternative characterization of the notion of $k$-truncated map, compared to \\cref{thm:trunc_ap}. Given a map $f:A\\to X$ define the \\define{diagonal}\\index{diagonal!of a map} of $f$ to be the map $\\delta_f:A\\to A\\times_X A$ given by $x\\mapsto (x,x,\\refl{f(x)})$.\n\\begin{subexenum}\n\\item Construct an equivalence\n\\begin{equation*}\n\\eqv{\\fib{\\delta_f}{(x,y,p)}}{\\fib{\\apfunc{f}}{p}}\n\\end{equation*}\nto show that the square\\index{action on paths!fibers of}\\index{diagonal!of a map!fibers of}\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\n\\fib{\\apfunc{f}}{p} \\arrow[r,\"\\mathsf{const}_x\"] \\arrow[d,swap,\"\\mathsf{const}_\\ttt\"] & A \\arrow[d,\"\\delta_f\"] \\\\\n\\unit \\arrow[r,swap,\"{\\mathsf{const}_{(x,y,p)}}\"] & A\\times_X A\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square, for every $x,y:A$ and $p:f(x)=f(y)$.\n\\item Show that a map $f:A\\to X$ is $(k+1)$-truncated if and only if $\\delta_f$ is $k$-truncated.\\index{truncated!map!by truncatedness of diagonal}\n\\end{subexenum}\nConclude that $f$ is an embedding if and only if $\\delta_f$ is an equivalence.\\index{embedding!diagonal is an equivalence}\n\\item Consider a commuting square \n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X \n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$. Show that this square is a pullback square if and only if the square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"p\"] \\arrow[d,swap,\"q\"] & A \\arrow[d,\"f\"] \\\\\nB \\arrow[r,swap,\"g\"] & X \n\\end{tikzcd}\n\\end{equation*}\nwith $H^{-1}:g\\circ q\\htpy f\\circ p$ is a pullback square.\n\\item Show that any square of the form\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r] \\arrow[d] & B \\arrow[d] \\\\\n\\emptyt \\arrow[r] & X\n\\end{tikzcd}\n\\end{equation*}\ncommutes and is a pullback square. This is the \\emph{descent property} of the empty type.\\index{descent!empty type}\n\\item Consider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$. Show that the following are equivalent:\n\\begin{enumerate}\n\\item The square is a pullback square.\n\\item For every type $T$, the commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC^T \\arrow[r,\"q\\circ\\blank\"] \\arrow[d,swap,\"p\\circ\\blank\"] & B^T \\arrow[d,\"g\\circ\\blank\"] \\\\\nA^T \\arrow[r,swap,\"f\\circ\\blank\"] & X^T\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n\\end{enumerate}\nNote: property (ii) is really just a rephrasing of the universal property of pullbacks.\\index{pullback square!universal property}\n\\item \\label{ex:pb_diagonal}Consider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$. Show that the following are equivalent:\n\\begin{enumerate}\n\\item The square is a pullback square.\n\\item The square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"g\\circ q\"] \\arrow[d,swap,\"{\\lam{x}(p(x),q(x))}\"] & X \\arrow[d,\"\\delta_X\"] \\\\\nA\\times B \\arrow[r,swap,\"f\\times g\"] & X\\times X\n\\end{tikzcd}\n\\end{equation*}\nwhich commutes by $\\lam{z}\\mathsf{eq\\usc{}pair}(H(z),\\refl{g(q(z))})$ is a pullback square.\n\\end{enumerate}\n\\item \\label{ex:pb_prod}Consider two commuting squares\\index{pullback!cartesian products of pullbacks}\n\\begin{equation*}\n\\begin{tikzcd}\nC_1 \\arrow[r] \\arrow[d] & B_1 \\arrow[d] & C_2 \\arrow[r] \\arrow[d] & B_2 \\arrow[d] \\\\\nA_1 \\arrow[r] & X_1 & A_2 \\arrow[r] & X_2.\n\\end{tikzcd}\n\\end{equation*}\n\\begin{subexenum}\n\\item Show that if both squares are pullback squares, then the square\n\\begin{equation*}\n\\begin{tikzcd}\nC_1\\times C_2 \\arrow[r] \\arrow[d] & B_1\\times B_2 \\arrow[d] \\\\\nA_1 \\times A_2 \\arrow[r] & X_1\\times X_2. \n\\end{tikzcd}\n\\end{equation*}\nis also a pullback square.\n\\item Show that if there are terms $t_1:A_1\\times_{X_1}B_1$ and $t_2:A_2\\times_{X_2}B_2$, then the converse of (a) also holds.\n\\end{subexenum}\n\\item Consider for each $i:I$ a pullback square\n\\begin{equation*}\n\\begin{tikzcd}\nC_i \\arrow[r,\"q_i\"] \\arrow[d,swap,\"p_i\"] & B_i \\arrow[d,\"g_i\"] \\\\\nA_i \\arrow[r,swap,\"f_i\"] & X_i\n\\end{tikzcd}\n\\end{equation*}\nwith $H_i: f_i\\circ p_i\\htpy g_i\\circ q_i$.\\index{pullback!Pi-type of pullbacks@{$\\Pi$-type of pullbacks}}\\label{ex:pb_pi}\nShow that the commuting square\n\\begin{equation*}\n\\begin{tikzcd}\n\\prd{i:I}C_i \\arrow[r] \\arrow[d] & \\prd{i:I}B_i \\arrow[d] \\\\\n\\prd{i:I}A_i \\arrow[r] & \\prd{i:I}X_i\n\\end{tikzcd}\n\\end{equation*}\nis a pullback square.\n%\\item \n%\\begin{subexenum}\n%\\item Show that \\index{equivalence!type of equivalences!as pullback}\n%\\begin{equation*}\n%\\begin{tikzcd}[column sep=8em]\n%\\eqv{A}{B} \\arrow[r] \\arrow[d] & \\unit \\arrow[d,\"{(\\idfunc[A],\\idfunc[B])}\"] \\\\\n%A^B\\times B^A \\times A^B \\arrow[r,swap,\"{(h,f,g)\\mapsto (h\\circ f,f\\circ g)}\"] & A^A \\times B^B\n%\\end{tikzcd}\n%\\end{equation*}\n%is a pullback square.\n%\\item Show that \\index{contractible!type of contractibility!as pullback}\n%\\begin{equation*}\n%\\begin{tikzcd}[column sep=6em]\n%\\iscontr(A) \\arrow[r,\"\\mathsf{const}_{\\ttt}\"] \\arrow[d,swap,\"\\proj 1\"] & \\unit \\arrow[d,\"{\\lam{\\ttt}\\idfunc[A]}\"] \\\\\n%A \\arrow[r,swap,\"{\\lam{x}\\mathsf{const}_x}\"] & A^A\n%\\end{tikzcd}\n%\\end{equation*}\n%is a pullback square.\n%\\end{subexenum}\n%\\item Consider a commuting square\n%\\begin{equation*}\n%\\begin{tikzcd}\n%C \\arrow[r] \\arrow[d] & A \\arrow[d] \\\\\n%B \\arrow[r] & X.\n%\\end{tikzcd}\n%\\end{equation*}\n%Show that this square is cartesian if and only if the induced map $C\\to A\\times_X B$ has a retraction.\n%\\end{subexenum}\n%\\item Suppose that the squares\n%\\begin{equation*}\n%\\begin{tikzcd}\n%C \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] & {C'} \\arrow[r,\"{q'}\"] \\arrow[d,swap,\"{p'}\"] & B \\arrow[d,\"g\"] \\\\\n%A \\arrow[r,swap,\"f\"] & X & A \\arrow[r,swap,\"f\"] & X\n%\\end{tikzcd}\n%\\end{equation*}\n%with homotopies $H:f\\circ p \\htpy g\\circ q$ and $H':f\\circ p'\\htpy g\\circ q'$ are both pullback squares. Show that the type of equivalences $e:\\eqv{C'}{C}$ equipped with an identification\n%\\begin{equation*}\n%\\mathsf{cone\\usc{}map}((p,q,H),e)=(p',q',H')\n%\\end{equation*}\n%is contractible.\n\\begin{comment}\n\\item Consider a \\define{natural transformation of cospans}\\index{cospan!natural transformation of}, i.e., a commuting diagram of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[r,\"f\"] \\arrow[d,swap,\"i\"] & X \\arrow[d,swap,\"j\"] & B \\arrow[l,swap,\"g\"] \\arrow[d,\"k\"] \\\\\nA' \\arrow[r,swap,\"{f'}\"] & X' & B'. \\arrow[l,\"{g'}\"]\n\\end{tikzcd}\n\\end{equation*}\nShow that the map\n\\begin{equation*}\n(a,b,p)\\mapsto (i(a),j(b),\\mathsf{ap}_k(p)): A \\times_X B \\to A'\\times_{X'} B'\n\\end{equation*}\nis $k$-truncated if each of the vertical maps is.\n\\end{comment}\n\\begin{comment}\n\\item \\label{ex:pb_fib}Consider a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[d,swap,\"p\"] \\arrow[r,\"q\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X.\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$, and let $h:C\\to A\\times_X B$ be the map given by $h(z)\\defeq (p(c),q(c),H(c))$. \nShow that the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=6.5em]\n\\fib{\\mathsf{gap}(p,q,H)}{(a,b,\\alpha)} \\arrow[d,swap,\"\\mathsf{const}_{\\ttt}\"] \\arrow[r,\"{\\lam{(c,\\beta)}(c,\\ap{\\pi_1}{\\beta})}\"] & \\fib{p}{a} \\arrow[d,\"{\\fibf{(f,g,H)}}\"] \\\\\n\\unit \\arrow[r,swap,\"\\mathsf{const}_{(b,\\alpha^{-1})}\"] & \\fib{g}{f(a)}\n\\end{tikzcd}\n\\end{equation*}\n\\end{comment}\n\\item \\label{ex:pb_3by3}Consider a commuting diagram of the form\n\\begin{equation*}\n\\begin{tikzcd}\nA_0 \\arrow[r] \\arrow[d] & B_0 \\arrow[d] & C_0 \\arrow[l] \\arrow[d] \\\\\nA_1 \\arrow[r] & B_1 & C_1 \\arrow[l] \\\\\nA_2 \\arrow[u] \\arrow[r] & B_2 \\arrow[u] & C_2 \\arrow[u] \\arrow[l]\n\\end{tikzcd}\n\\end{equation*}\nwith homotopies filling the (small) squares. Construct an equivalence\n\\begin{align*}\n& (A_0\\times_{B_0} C_0) \\times_{(A_1\\times_{B_1} C_1)} (A_2\\times_{B_2} C_2) \\\\\n& \\qquad \\eqvsym (A_0\\times_{A_1} A_2) \\times_{(B_0\\times_{B_1} B_2)} (C_0\\times_{C_1} C_2).\n\\end{align*}\nThis is also known as the \\define{3-by-3 lemma}\\index{3-by-3 lemma!for pullbacks} for pullbacks.\n\\end{exercises}\n", "meta": {"hexsha": "990842cbdef88b5fb7faef73f2c3794191047dce", "size": 43970, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/pullback.tex", "max_stars_repo_name": "tadejpetric/HoTT-Intro", "max_stars_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Book/pullback.tex", "max_issues_repo_name": "tadejpetric/HoTT-Intro", "max_issues_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Book/pullback.tex", "max_forks_repo_name": "tadejpetric/HoTT-Intro", "max_forks_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.7884972171, "max_line_length": 397, "alphanum_fraction": 0.6737548328, "num_tokens": 16480, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267118111485245, "lm_q2_score": 0.8459424314825853, "lm_q1q2_score": 0.6993505996583548}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% CS624: Analysis of Algorithms\n% Copyright 2015 Pejman Ghorbanzade \n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section*{Question 1}\n\nA stack supports two operations: \\textsc{Push} and \\textsc{Pop}.\nImplementing the two operations using a linked-list takes $\\mathcal{O}(1)$ time per operation.\nSuppose we are given two stacks, $A$ and $B$ with $n$ and $m$ elements, respectively.\nWe want to implement the following operations:\n\\begin{description}\\itemsep=0pt\n\\item[\\textsc{Push}($A, x$)] Push $x$ elements into $A$.\n\\item[\\textsc{Push}($B, x$)] Push $x$ elements into $B$.\n\\item[\\textsc{MultipopA($k$)}] pop $\\text{min}\\{k,n\\}$ elements from $A$.\n\\item[\\textsc{MultipopB($k$)}] pop $\\text{min}\\{k,m\\}$ elements from $B$.\n\\item[\\textsc{Transfer($k$)}] repeatedly pop an element from $A$ and push it to $B$ until either $k$ elements have been transferred or $A$ is empty.\n\\end{description}\n\\begin{enumerate}[label=(\\alph*)]\n\\item What is the worst-case running time of \\textsc{MultipopA}, \\textsc{MultipopB} and \\textsc{Transfer}?\n\\item Show that the amortized running time per operation is $\\mathcal{O}(1)$ for a sequence of \\textsc{Push}($A,n$) and \\textsc{Transfer}($n$).\nYou may use any technique shown in class on a sequence on $n$ operations.\n\\end{enumerate}\n\n\\subsection*{Solution}\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item Operations \\textsc{Push}($A,x$) and \\textsc{Push}($B,x$) have a runtime cost of $\\mathcal{O}(x)$ as \\textsc{Push}() operation with runtime $\\mathcal{O}(1)$ should be performed for $x$ times.\n\nOperations \\textsc{MultipopA}($k$) and \\textsc{MultipopB}($k$) both repeat the \\textsc{Pop()} operation for $\\text{min}(k,n)$ and $\\text{min}(k,m)$ times, respectively.\nIn the worst-case where $n,m < k$, the operations are limited by the number of elements in the stacks.\nTherefore, their worst-case runtime cost are $\\mathcal{O}(n)$ and $\\mathcal{O}(m)$, respectively.\n\nOperation \\textsc{Transfer($k$)} has a worst-case runtime of $\\mathcal{O}(k)$, since worst scenario happens when $k < n$ and $k < m$ and there will be $k$ items to \\textsc{Pop()} from stack $A$ and $k$ items to \\textsc{Push()} to stack $B$.\n\n\\item For \\textsc{Push}($A,n$), amortized runtime would be constant $C$ where $C$ is the cost to push one element to a stack.\nThis can be shown easily using the aggregate method as shown in Eq. \\ref{eq11}.\n\n\\begin{equation}\n\\sum_{i=1}^{n} c_i = \\sum_{i=1}^{n} c = c \\sum_{i=1}^{n} 1 = nc \\Rightarrow \\text{Amortized Cost} = \\frac{nc}{n} = c\n\\label{eq11}\n\\end{equation}\n\nFor \\textsc{Transfer($k$)} operation, the worst case happens when there are at least $k$ elements in stack $k$, in which case amortized cost of the operation can be obtained as given in Eq. \\ref{eq12}.\n\n\\begin{equation}\n\\sum_{i=1}^{k} c_i = \\sum_{i=1}^{k}(c_1 + c_2) = (c_1 + c_2) \\sum_{i=1}^{k} 1 = (c_1 + c_2) k\n\\label{eq12}\n\\end{equation}\n\nSince cost of \\textsc{Pop()} ($c_1$) and \\textsc{Push()} ($c_2$) are $\\mathcal{O}(1)$, amortized cost of \\textsc{Transfer($k$)} is obtained from \\ref{eq12}.\n\n\\begin{equation}\n\\text{Amortized Cost} = \\frac{k(c_1 + c_2)}{k} = c_1 + c_2 = 2\n\\end{equation}\n\n\\end{enumerate}\n", "meta": {"hexsha": "c287954c177fb7f647c3b6a939696d61009d9ae1", "size": 3360, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "umb-cs624-2015s/src/tex/m02/m02q01.tex", "max_stars_repo_name": "ghorbanzade/beacon", "max_stars_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-11-13T20:00:10.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-01T11:16:51.000Z", "max_issues_repo_path": "umb-cs624-2015s/src/tex/m02/m02q01.tex", "max_issues_repo_name": "ghorbanzade/beacon", "max_issues_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "umb-cs624-2015s/src/tex/m02/m02q01.tex", "max_forks_repo_name": "ghorbanzade/beacon", "max_forks_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-09-20T05:58:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T17:18:05.000Z", "avg_line_length": 56.0, "max_line_length": 240, "alphanum_fraction": 0.6672619048, "num_tokens": 1099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.845942439250491, "lm_q1q2_score": 0.699350591633295}} {"text": "\\section{Mathematical basics}\nThis section is going to describe the mathematical concepts of an general computer algebra library which shall be implemented. Although a swath of computer algebra programs is available none of these programs comes with a library the an interested user may simply include in his/her own project.\\\\\n\\indent In addition, as anyone can imagine no single program can provide all objecs which are currently investigated. Worse, the user is forced to delve into each programs programming structures and restrictions and potentially has to change the program if desires and needs are not met.\n\\subsection{Categories and functor}\nAs we briefly outlined in the first chapter, we discuss the functorial approach. Firstly, we like to explain some the features needed.\n\\subsubsection{Basic concepts}\nWe will denote an arbitrary category with $\\mathcal{C}$.\n\\paragraph{Categories} Categories comprise of three concepts or data encoding it:\n\\begin{description}\n\\item[Objects] are the instances of a category. They are not necessarily sets in a set theoretic sense rather classes. They are denoted by $\\mathrm{Obj}_{\\mathcal{C}}$ or $\\objc$.\n\\item[Morphisms] are maps connecting objects in a category sometimes trivially called arrows. These maps are not arbitrary but preserve or respect the structure of each object. The class of morphisms are denoted by $\\morc$.\n\\item[Identity] is a unique map, actually a functor mapping each object $A \\in \\objc$ to its identity morphism $id_A : A \\longrightarrow A$.\n\\end{description}\nThe last point is of utmost importance: the identity map is always a morphism in a given category. Furthermore, given a triple $A, B, C \\in \\objc$ and morhisms $f : A \\longrightarrow B, g : \\longrightarrow C \\in \\morc$ the composition yields a new morphism:\n$$g \\circ f : A \\longrightarrow C \\in \\morc.$$\nSome authors require all five properties to be fulfilled to called a category others omit some.\n\\paragraph{Functors}\nFunctors can be thought of as maps between categories. To specify given two categories $\\mathcal{C}$ and $\\mathcal{D}$, a functor $F$ is a pairing $F : \\mathcal{C} \\longrightarrow \\mathcal{D}$ of these two such that:\n\\begin{description}\n\\item[Variance] For $A, B \\in \\objc$ and $f \\in \\morc$ either of the two diagrams commutes:\n$$\\begin{array}{cc}\n\\xymatrix{\nA \\ar[d]_F\\ar[r]^f& B\\ar[d]^F\\\\\nF(A) \\ar[r]_{(f)} &F(B)\\\\\n} & \\xymatrix{\nA \\ar[d]_F\\ar[r]^f& B\\ar[d]^F\\\\\nF(A) &F(B).\\ar[l]^{F(f)}\\\\\n} \n\\end{array}\n$$\nThe left diagram represents a so called covariant functor $F$ - the right one a so called contravariant functor.\n\\item[Morphism and identity] The map $F(f) : F(A) \\longrightarrow F(B)$ in the covariant case and the map $F(f) : F(B) \\longrightarrow F(A)$ in the contravariant case are morphisms in $\\mor{\\mathcal{D}}$. In particular, $F(id_A) = id_{F(A)}$ and $F(id_A) = id_{F(B)}$ are a morphisms.\n\\end{description}\nThere are categories consisting of categories or functors. We will leave it to the reader to explore those. In addition, we may get new categories \n\\subsubsection{Examples}\nAs any brief discussion, we will a some examples to illustrate these concepts.\n\\paragraph{The categories $\\mathrm{Set}$ and $\\mathrm{Top}$}\nThe most general category we are going to discuss is the category of all sets, $\\mathrm{Set}$, with morphisms $\\mor{\\mathrm{Set}}$ simply as maps\n$$f \\in \\mathrm{map}(A,B) = B^A \\subset \\mor{\\mathrm{Set}},$$\nfor all $A, B \\in \\obj{\\mathrm{Set}}$.\\\\\nWe call a category $\\calc$ small if the class of all its objects is a set in a set theoretic sense. The above example is not small, therefore appropiately called large (i.e. a large category).\\\\\n\\indent The next example is the category of all topological spaces, $\\mathrm{Top}$, with continuous maps as morphisms:\nFurthermore, we may define two pairings\n$$\\begin{array}{rrcl}\n\\call{O} :& \\mathrm{Top} &\\longrightarrow& \\mathrm{Set},\\\\\n&&&\\\\\n& (A, \\tau) &\\longmapsto &\\left\\{U \\in \\tau : U\\ \\mathrm{open}\\right\\},\\\\\n&&&\\\\\n\\call{F} :& \\mathrm{Top}&\\longrightarrow& \\mathrm{Set},\\\\\n&&&\\\\\n&(A, \\tau) &\\longmapsto& \\left\\{F \\in \\tau : F \\ \\mathrm{closed}\\right\\}\\\\\n\\end{array}\\\\\n$$\nthen we get a map $f \\in C(A,B) \\subset \\mor{\\mathrm{Top}}$ if and only if the following diagrams:\n$$\\begin{array}{ccc}\n\\xymatrix{\nA \\ar[r]^f\\ar[d]_{\\call{F}} & B\\ar[d]^{\\call{F}}\\\\\n\\call{F}(A) &\\call{F}(B)\\ar[l]^{\\call{F}(f)}\\\\\n}&\\mathrm{and}&\\xymatrix{\nA \\ar[r]^f\\ar[d]_{\\call{O}} & B\\ar[d]^{\\call{O}}\\\\\n\\call{O}(A) &\\call{O}(B)\\ar[l]^{\\call{O}(f)}\\\\\n}\\\\\n\\end{array}\n$$\ncommute. Moreover, what we simply called a pairing from topological spaces with their respetive sub topologies of open or closed sets, respectively, are in fact two contravariant functors.\n\\paragraph{Pointed spaces and partially ordered sets}\nGiven a set $X$ and an element $x_0 \\in X$, we call the pair $(X,x_0)$ a pointed space and $x_0$ its base point. The class of all pointed spaces forms a category $\\mathrm{Pt}$ wrt. to maps preserving base points as morphisms. By this defintion, $\\mathrm{Pt}$ is a proper subcategory of $\\mathrm{Set}$ as the empty set is not a member of its object class. However, the terminal objects are the singletons - objects with exactly one element.\\\\\n\\indent We call a category $\\calc$ with an object $\\cdot$ a category with terminal object $\\cdot$ if and only if for any pair of objects $A, B \\in \\objc$ with arrow $f \\in \\objc(A,B)$ there exist two unique morphisms $\\pi_A : A \\longrightarrow \\cdot, \\pi_B : B \\longrightarrow \\cdot \\in \\morc$ such that the following diagram commutes:\n$$\\xymatrix{\nA \\ar[r]^f \\ar[dr]_{\\pi_A} & B\\ar[d]^{\\pi_B}\\\\\n&\\cdot.\\\\\n}$$\nClearly, each pointed space $(X,x_0)$ is an object with terminal object.\\\\\n\\indent Reversing arrows, we call a category $\\calc$ with object $\\ast$ a category with initial object if for all objects $A, B \\in \\objc$ and arrow $f : B \\longrightarrow A$ there are two unique morphisms $\\iota_A : \\ast \\longrightarrow A, \\iota_B : \\ast \\longrightarrow B \\in \\morc$ such that\n$$\\xymatrix{\nA & B\\ar[l]_f\\\\\n&\\ast\\ar[u]_{\\iota_B} \\ar[ul]^{\\iota_A} .\\\\\n}$$\ncommutes.\\\\\n\\indent We call an object $\\ast \\in \\objc$ a null object if it is initial and terminal. The category of sets, $\\mathrm{Set}$, has the empty set as initial and the class of singletons as terminal object. The category of pointed spaces has its singletons as null object.\\\\\n\\indent A relation $R \\subset X^2$ is called a partial order on a set $X$ if for all $x, y, z \\in X$:\n\\begin{enumerate}\n\\item $x R x$ (reflexive),\n\\item $x R y$ and $y R x$ implies $x = y$ (antisymmetry) and\n\\item $x R y$ and $y R z$ implies $x R z$ (transitivity).\n\\end{enumerate}\nPartially ordered sets are of great importance in conjunction with algebraic categories which we will describe shortly.\n\\paragraph{Algebraic categories}\nIn its very beginning, category theory was mainly an algebraic endeavour. As it became apparent that certain theoremes had a wider application then initially anticipated.\\\\\nSince an algebraic library is the aim of this essay we shall spend some time on algebraic categories.\n\\subparagraph{Semigroups, monoids and groups}\nBefore we start a short definition. We call a category $\\calc$ a category with finite products if for all objects $A, B \\in \\objc$ the set theoretic product \n$$A \\times B = \\{(a,b) : a \\in A, b \\in B\\} \\in \\objc.$$\nIn particular, each canonical projection $\\pi_X : A \\times B \\longrightarrow X$ and their respective embeddings $\\iota_X : X \\longrightarrow A \\times B$, $X = A, B$ are morphisms in $\\calc$.\\\\\n\\begin{enumerate}\n\\item Let $S$ be a set and $m : S \\times S \\longrightarrow S$ be a binary operation. We call a pair $(S, m)$ a semi group if for all $(s, s')$ there is a $s'' \\in S$ such that $m(s,s') = s''$ and the following diagram commutes \n$$\\xymatrix{\nS \\times S \\times S \\ar[rr]^{id_S \\times m}\\ar[d]_{m \\times id_S} && S \\times S\\ar[d]^{m}\\\\\nS \\times S \\ar[rr]_{m} &&S.\\\\\n}$$\nWe call the first property closedness (under operation) and the second associativity. Similarily to the category of sets, the category of semi groups has empty set as initial and singletons as terminal object.\\\\\n\\item Let $(M, m)$ be a semi group. We call $M$ a monoid if there is a initial object $\\ast$ and an embedding $e : \\ast \\longrightarrow M$ such that the following diagram commutes:\n$$\\xymatrix{\n& M \\times M \\ar[dd]_{m}& \\\\\n\\ast \\times M \\ar[ur]^{e \\times id_M}\\ar[dr]_{\\simeq} && M \\times \\ast \\ar[ul]_{id_M \\times e}\\ar[dl]^{\\simeq} \\\\\n&M.&\\\\\n}$$\nThe map $e$ is a morphism in $\\mathrm{Mon}$ and called the unit (map). We denote a monoid with triple $(M, m, e)$. Note that $\\mathrm{Mon}$ is a category with null object (initial and terminal are isomorphic). Moreover, it is a subcategory of pointed spaces with trivial submonoid as null object.\n\\item We call a monoid $(G, m, e)$ a group if there is a map $S : G \\longrightarrow G$ such that the following diagram commutes:\n$$\\xymatrix{\n&G \\ar[ld]_{\\Delta} \\ar[ddd] \\ar[dr]^{\\Delta}&\\\\\nG \\times G \\ar[d]_{S \\times id_G} & & \\ar[d]^{id_G \\times S} G \\times G\\\\\nG \\times G \\ar[dr]_{m} & & \\ar[dl]^{m} G \\times G\\\\\n&\\ast.&\\\\\n}$$\nThe map $S$ is called the antipode or more popularly, the inverse/inversion. It induces a group (anti-) isomorphism:\n$$S : G \\stackrel{\\sim}{\\longrightarrow} G^{\\mathrm{op}},$$\n$$m^{\\mathrm{op}} = m \\circ (S \\times S) \\circ \\tau = \\left[(g, h) \\longmapsto m(h^{-1},g^{-1}) = h^{-1} g^{-1}\\right]$$\nIt is an anti isomorphism if considered as map $G \\longrightarrow G$ and an isomorphism $(G,m,e,S) \\longrightarrow (G^{\\mathrm{op}},m^{\\mathrm{op}},e^{\\mathrm{op}} = e, S^{op} = S)$ (we exchange the multiplication map $m$ with $m^{\\mathrm{op}}$). A group is called abelian if\n$$\\xymatrix{\nG \\times G \\ar[d]_{m^{\\mathrm{op}}} \\ar[r]^{m} &G\\ar[ld]^{S}\\\\\nG \n}$$\ncommutes. They form a proper subcategory of $\\mathrm{Grp}$, denoted $\\mathrm{Abel}$. Again, they have a null object (trivial subgroup). Furthermore, this category has finite products (at least two distinct in as far as):\n$$\\begin{array}{rrcl}\n\\times : &\\mathrm{Grp} \\times \\mathrm{Grp}& \\longrightarrow &\\mathrm{Grp},\\\\ &\\left((G, m_G, e_G, S_G), (H, m_H, e_H, S_H)\\right)& \\longmapsto &(G \\times H, m_{G} \\times m_H, e_{G} \\times e_{H},\\\\\n&&& S_G \\times S_H)\\\\\n&&&\\\\\n\\times_{\\alpha} : &\\mathrm{Grp} \\times \\mathrm{Grp} &\\longrightarrow& \\mathrm{Grp}\\\\\n&(G,m_G,e_G,S_G),(H,m_H,e_H,S_H))&\\longmapsto&(G \\times H, m^{\\alpha}_{G \\times H}, e_G \\times e_H, S_{G} \\times S_{H})\\\\\n\\end{array}$$\nwith\n$$\\begin{array}{rcl}\nm^\\alpha_{G \\times H} &=& (m_G \\times m_H) (id_G \\times \\alpha \\times id_H \\times id_G \\times id_H)\\\\&&(id_G \\times id_H \\times \\tau_{H\\times G}\\times id_H)(id_G \\times \\Delta_H \\times id_G \\times id_H),\\\\\n&&\\\\\n&=& [(g_1,h_1,g_2,h_2) \\longmapsto (g_1,h_1,h_1,g_2,h_2) \\longmapsto (g_1,h_1,g_2,h_1,h_2) \\longmapsto\\\\&&~ (g_1, \\alpha(h_1)(g_2),h_1,h_2) \\longmapsto (g_1 \\alpha(h_1)(g_2), h_1 h_2)] \\ \\mathrm{and}\\\\\n&&\\\\\n\\alpha &=& [h \\longmapsto \\alpha(h)] \\in \\mathrm{Grp}(H,\\mathrm{Aut}_{\\mathrm{Grp}}(G))\\\\\n\\end{array}$$\nThe former product is called direct (only the neutral element is in both groups and both are normal in $G \\times H$), the latter is called the semi direct product for a given group homomorphism $\\alpha : H \\longrightarrow \\mathrm{Aut}(G)$. They coincide if $\\alpha$ reduces to the trivial map\n$$\\alpha = e_{\\mathrm{Aut}(G)} \\circ \\pi_H = [ h \\longmapsto e_H \\longmapsto id_G].$$\n\\end{enumerate}\n", "meta": {"hexsha": "ad6fe184f3e36273cff6c647ab6714d116eca648", "size": 11352, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manifesto/impl2.tex", "max_stars_repo_name": "gmuel/texlib", "max_stars_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "manifesto/impl2.tex", "max_issues_repo_name": "gmuel/texlib", "max_issues_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "manifesto/impl2.tex", "max_forks_repo_name": "gmuel/texlib", "max_forks_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 79.3846153846, "max_line_length": 441, "alphanum_fraction": 0.7003171247, "num_tokens": 3578, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.905989822921759, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.6992823586853701}} {"text": "\\paragraph{Population} is a set, of which we would like to learn some properties.\n \n\\paragraph{Sample} is a subset of a population, on which we calculate the properties and then make\nstatements about the whole population.\n\n\\paragraph{Random sample} is a sample taken in an unpredictable way, i.e. the method of taking the\nsample involves an unpredictable component.\n\n\\paragraph{Self-weighting sample} is a sample taken in such manner that each element of population has equal\nchance of being included in the sample.\n\n\\paragraph{Simple random sample} is a kind of self-weighting sample selected so that all samples of\nsame size have equal chance of being selected.\n\n\\subsection*{Remark}\nFurther on we in most cases assume random sample of size $n$, $X = \\sample$.\n", "meta": {"hexsha": "2ee41e77755a6d5d76b8e2aee2a98a80fc4ccf66", "size": 764, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cs_2a_basics.tex", "max_stars_repo_name": "mbdevpl/wut-bsc-computer-statistics-formulas", "max_stars_repo_head_hexsha": "ce5febce6f6fc680c445257ca263962a0a76a688", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "cs_2a_basics.tex", "max_issues_repo_name": "mbdevpl/wut-bsc-computer-statistics-formulas", "max_issues_repo_head_hexsha": "ce5febce6f6fc680c445257ca263962a0a76a688", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cs_2a_basics.tex", "max_forks_repo_name": "mbdevpl/wut-bsc-computer-statistics-formulas", "max_forks_repo_head_hexsha": "ce5febce6f6fc680c445257ca263962a0a76a688", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.9411764706, "max_line_length": 108, "alphanum_fraction": 0.7866492147, "num_tokens": 168, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8577681122619883, "lm_q2_score": 0.8152324960856175, "lm_q1q2_score": 0.699280439221989}} {"text": "\n\\section{The conjugate gradient method}\n\n\\begin{intro}\n Relying on Hilbert space structure more than Richardson's iteration\n is the \\putindex{conjugate gradient method} (cg), since it uses\n orthogonal search directions. Nevertheless, it also relies on\n constructing search directions from residuals, such that a\n \\putindex{Riesz isomorphism} enters the same way as before and can\n then be used for preconditioning.\n \n The beauty of the conjugate gradient method is, that it is parameter\n and tuning free, and it converges considerably faster than a linear\n iteration method.\n\\end{intro}\n\n\\begin{Definition*}{steepest-descent}{Method of steepest descent}\n Let $a(.,.)$ be a symmetric, positive definite bilinear form on\n $V$. Then, the method of \\define{steepest descent} for the energy functional\n \\begin{gather}\n E(v) = \\tfrac12 a(v,v) - f(v),\n \\end{gather}\n reads: given an initial vector $u^{(0)}$,\n compute % $p^{(1)} = \\nabla E(u^{(0)}) \\in V^*$ and\n for $k\\ge 0$ iteratively\n \\begin{xalignat}2\n \\scal(p^{(k)},v) &= \\scal({-\\nabla E(u^{(k)})},v)_{V^*\\times V} &\\forall v&\\in V \\\\\n \\alpha_k &= \\operatorname*{argmin}_{\\alpha>0} E\\left(u^{(k)} + \\alpha p^{(k)}\\right)\\\\\n u^{(k+1)} &= u^{(k)} + \\alpha_k p^{(k)}\n \\end{xalignat}\n\\end{Definition*}\n\n\\begin{Lemma}{steepest-descent}\n Let $u\\in V$ be the unique minimizer of the energy functional $E(.)$. Then, there holds for all $k$\n \\begin{gather}\n a\\left(u^{(k+1)}-u, p^{(k)}\\right) = a(u^{(k+1)}, p^{(k)}) - f(p^{(k)}) = 0.\n \\end{gather}\n With $r^{(k)} = f-a(u^{(k)},.)\\in V^*$, there holds\n \\begin{gather}\n \\alpha_k = \\frac{\\scal(r^{(k)},p^{(k)})_{V^*\\times V}}{a\\left(p^{(k)},p^{(k)}\\right)}.\n \\end{gather}\n\\end{Lemma}\n\n\\begin{Definition}{cg-method}\n Let $V$ be a Hilbert space and $V^*$ its dual. The \\define{conjugate\n gradient method} for an iteration vector $u^{(k)} \\in V$ involves the\n residuals $r^{(k)} \\in V^*$ as well as the update direction $p^{(k)}\n \\in V$ and the auxiliary vector $w^{(k)} \\in V$. It consists of the\n steps\n \\begin{enumerate}\n \\item Initialization: for $f$ and $u^{(0)}$ given, compute $r^{(0)} = f- a(u^{(0)},.)$ and\n \\begin{xalignat*}{2}\n \\scal(p^{(0)}, v) = \\scal(w^{(0)}, v) &= \\scal(r^{(0)},v)_{V^*\\times V} & \\forall v &\\in V.\n \\end{xalignat*}\n \\item Iteration step: for $u^{(k)}$, $r^{(k)}$, $w^{(k)}$, and\n $p^{(k)}$ given, compute\n \\begin{xalignat*}2\n u^{(k+1)} &= u^{(k)} + \\alpha_k p^{(k)}\n &\n \\alpha_k &= \\frac{\\scal(r^{(k)},w^{(k)})_{V^*\\times V}}{a\\left(p^{(k)},p^{(k)}\\right)} \\\\\n r^{(k+1)} &= r^{(k)} - \\alpha_k a\\left(p^{(k)},.\\right) \\\\\n \\scal(w^{(k+1)}, v) &= \\scal(r^{(k+1)},v)_{V^*\\times V} & \\forall v &\\in V \\\\\n p^{(k+1)} &= w^{(k+1)} + \\beta_k p^{(k)}\n &\n \\beta_k &= \\frac{\\scal(r^{(k+1)},w^{(k+1)})_{V^*\\times V}}{\\scal(r^{(k)},w^{(k)})_{V^*\\times V}}\n \\end{xalignat*}\n \\end{enumerate}\n\\end{Definition}\n\n% \\begin{remark}\n% The results on orthogonality and minimization properties of the cg\n% method in~\\cite{GrossmannRoosStynes07} or~\\cite{Saad00} remain valid in this\n% context. Differences occur in the interpretation of these\n% properties. The conjugate gradient method does not necessarily\n% converge in a finite number of steps, and if the bilinear form is\n% unbounded, no convergence rate is guaranteed.\n% \\end{remark}\n\n% \\begin{lemma}\n% Let $a(.,.)$ be symmetric and elliptic. Then, either $u^{(k)}$ is a\n% solution, or $u^{(k+1)}$ can be computed by a step of the conjugate\n% gradient method. Furthermore, there are the \n% \\end{lemma}\n\n\\begin{Definition}{pcg-method}\n The \\define{preconditioned cg method} is obtained from\n above algorithm by reinterpreting the \\putindex{Riesz isomorphism}\n in the computation of $w^{(k+1)}$ as a preconditioning operation,\n much alike Definition~\\ref{definition:richardson:2} of the\n preconditioned Richardson iteration. Thus, the line defining\n $w^{(k+1)}$ is replaced by\n \\begin{xalignat*}2\n b(w^{(k+1)}, v) &= r^{(k+1)}(v) & \\forall v &\\in V .\n \\end{xalignat*}\n Here, like there, the preconditioner enters naturally from the weak\n form of the algorithm.\n\\end{Definition}\n\n\\begin{Definition}{krylov-space}\n The $n$th \\define{Krylov space} as subspace of the Hilbert space $V$\n with inner product $b(.,.)$ of the operator $A$ and seed vector\n $w \\in V$ is\n \\begin{gather}\n \\label{eq:cg:2}\n \\mathcal K_n = \\mathcal K_n(B^{-1}A, w)\n = \\operatorname{span}\\left\\{w, B^{-1}A w, (B^{-1}A)^2 w, \\dots, (B^{-1}A w)^{n-1}\\right\\}.\n \\end{gather}\n\\end{Definition}\n\n\\begin{Lemma}{cg-orthogonality}\n The vectors generated by the conjugate gradient iteration have the\n following properties: either $u^{(k)}$ is the solution to the\n linear system or\n \\begin{alignat}2\n r^{(k)} &= f - a(u^{(k)},.)\\\\\n \\scal(r^{(k)},w^{(k)})_{V^*\\times V} &= \\scal(r^{(k)},p^{(k)})_{V^*\\times V} \\\\\n \\scal(r^{(k)},p^{(j)}) &= 0 & j&1$. This time the algebraic integer ring is different from what you may have thought: $\\OK = \\Z[\\omega]$ where\n\t\t\t\t\\[\n\t\t\t\t\\omega = \\begin{cases}\n\t\t\t\t\t\\frac{1+\\sqrt{d}}{2}, &\\quad d = 4k+1, \\\\\n\t\t\t\t\t\\sqrt{d}, &\\quad \\text{otherwise}.\n\t\t\t\t\\end{cases}\n\t\t\t\t\\]\n\t\t\t\\end{example}\n\t\t\tIt turns out we are studying polynomials such as\n\t\t\t\\begin{itemize}\n\t\t\t\t\\item $x^2+1=0$.\n\t\t\t\t\\item $x^2-d=0$.\n\t\t\t\\end{itemize}\n\t\t\tIt also turns out that many properties are not restricted to $\\Z$, but to a specific class of rings. Hence we will investigate some properties in the sense of commutative ring theory. The next example deserves more discussion.\n\t\t\t\n\t\t\t\\subsubsection{The Cyclotomic Fields}\n\t\t\t\tLet $\\omega = e^{2\\pi i/m}$. This is the $m$-th root of $1$. If $\\lambda$ is a conjugate of $\\omega$, i.e. they both are roots of the same irreducible polynomial over $\\Q$, then $\\lambda$ is also an $m$-th root of $1$ and is not an $n$-th root of $1$ whenever $n 0$ or some \"if\" constraint that all makes our ILP non linear. The effective workaround to solve this issue is considered out of scope of this paper.\n\n\\subsection{Scalability}\nThe tier can be scalable or not. The not scalable tier can have only 1 or 0 containers. If the tier is absent in the plan, it will have 0 containers, otherwise it will be 1, whilst the scalable tier can have any number of containers $\\geq 0$ allocated on different VMs. In the ILP formulation not scalable containers are not considered, so the current workaround is too \n\n\\subsection{Limitation of the ILP formulation}\nThis ILP formulation has some limitations:\n\\begin{enumerate}\n \\item This formulation of the problem requires us to know the number of VMs beforehand, which we do not know. But we can make upper-bound estimation.\n \\item This formulation does not differentiate between \"container update\" and \"container use\" (do nothing with container). \n \\item Weights should be chosen wisely to satisfy NFR-0.\n \\item Bin packing problem is NP-hard\n \\item We have 1 optimal solution and can not choose between several different solutions\n\\end{enumerate}\n\n\\subsection{Hooks}\nHooks are scripts that trigger on the tier / container scaling. The goal is to adjust the application's configuration after the allocation is changed.\n\nWe tried to extract and describe use cases of adjusting required:\n\\begin{enumerate}\n \\item Jboss needs to adapt the number of threads considering the number of CPU allocated.\n \\item LoadBalancer needs to be provided with the list of Jboss containers and resources allocated to adapt it weights.\n \\item Jboss needs to be provided with the DB address on container creation / change.\n\\end{enumerate}\n\nThe adjustment 1 should be run for the container after it is created or updated. We call this adjustment \"scale-hook\" or \"on\\_node\\_scale\" hook. The script should be provided with 4 arguments: previous CPU cores allocated, previous RAM memory units allocated, and new CPU cores and RAM memory units allocated.\n\nIn the 2nd adjustment the LoadBalancer tier depends on the Jboss tier. So the dependencies should be specified. Also the dependent node (LoadBalancer) should wait until all the containers of the dependee are processed (JBOSS). After that the adjustment script should be run on all the containers of the dependent node (LoadBalancer). We call this adjustment \"tier-hook\" or \"on\\_dependency\\_scale\" hook.\n\nThe 3d adjustment should be run only on the container start. To simpliy the first version of the application we decided not to implement this adjustment for the dependecy 1-to-n. Only for the dependencies n-to-1 and 1-to-1 after the dependee tier is processed, the ip of the dependee tier is provided to the dependent node on the creation (parameter \"add-host\"). \n\nThis creates some limitations: the dependee container should be processed before dependent containers. For example, if the first plan will have only dependent containers, in the next plan it will be added dependee container, and in the last plan the dependence between them will be specified, then \"add-host\" feature will not be propagated and the dependee containers will not know about the dependant one.\n\nSimilar problem if we decided to remove the not-scalable tier from the plan, the dependee containers will have not valid \"add-host\" link and even if later we put the \"not-scalable\" tier back to the plan, it will have another IP address.\n\nSo considering these limitations we suggest one of the two workarounds.\n\nThe first is to manage the DB dependencies out of the topology and provide the IP address in the topology as \"docker parameter\".\n\nThe second is just use docker \"add-host\" feature as it. So if the dependent container is removed, or the dependency is added, the user should be aware that \"add-host\" will not be specified.\n\n\\subsection{Dependencies}\nConsidering the requirement FR-9 only 1-to-1, 1-to-n and n-to-1 dependencies are supported. The dependencies between scalable tiers are not supported. The dependencies are only used to manager tier-hooks and . \nThe flow for container allocation and hooks execution should be:\n\\begin{enumerate}\n \\item get tiers dependency graph (G)\n \\item get set of Tiers without outgoing dependencies (T)\n \\item run container create / update / delete for each tier in T\n \\item run scale hooks if needed for each tier container in T\n \\item run tier hooks if needed for each tier container in T\n \\item remove T from G\n \\item if G is not empty go to 2\n\\end{enumerate}\n\n\\section{Approaches}\nBesides the ILP approach to solve container allocation problem discussed above it was also tried another approache: Constraint Satisfaction Problem.\n\n\\subsection{Constraint Satisfaction Problem}\n\nCSP is defined as a triple \\textless X, D, C\\textgreater, where $X$ is a set of variables, $D$ is a set of the respective domains of values, $C$ is a set of constraints.\n\nIn our context we have 2 matrices of variables as $X$: \\\\\n$x_{ij}$ is a matrix that specify how many CPU is used by Tier[j] on VM[i],\\\\\n$y_{ij}$ is a matrix that specify how many RAM units are used by Tier[j] on VM[i].\n\nThe domain $D$ for us is all possible values of CPU cores and RAM units. For example if we have AWS t2.medium with 2 CPU cores and 8 RAM units, we have: \\\\\n$d_{0ij}$ is $\\{0, 1, 2\\}$ for CPU cores,\\\\\n$d_{1ij}$ is $\\{0, 1, 2, 3, 4, 5, 6, 7, 8\\}$ for RAM mem units.\n\nFor a set of constraints $C$ we should consider similar to ILP:\n\\begin{itemize}\n \\item availability constraints on each VM for CPU cores and for RAM mem units\n \\item demand constraint considering the plan\n \\item 0 CPU cores with RAM \\textgreater{} 0 or other way round\n\\end{itemize}\n\n\\subsection{Limitation of the CSP}\nThis CSP formulation has also some limitations:\n\\begin{enumerate}\n \\item Comparing to ILP, changing domain to continuous values may cause problems.\n \\item The solution of the CSP is a list of different solutions, that can differ only by a permutation. We can have a factorial number of different solutions which we need to look to found the optimal one considering the initial allocation.\n \\item The problem is NP-Hard and the complexity growth is much faster than the ILP.\n\\end{enumerate}\n\nThe CSP formulation has one insuperable problem that it does not scale. If we have 10 Tiers, 10 VMs, and 16 CPU cores and 64 RAM mem units, the CSP solution does not complete in appropriate time.\n\n\\section{Architecture}\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=350px,natwidth=688,natheight=617]{./pictures/architecture}\n \\caption{Architecture}\n\\end{figure}\n\nThe architecture consists of two main components: Executor Main Node and Executor Agent. The main node accepts a monolithic plan, orchestrates agents \nand has a whole picture of VMs allocated. The agent node can accept a distributed plan to execute, can run / start / stop docker containers, provides the allocation information and output of docker inspect command.\n\nThese executor nodes are \"Execute\" part of the autonomous system MAPE framework. The Monitor component measures different metrics of applications that are\nrunning in the containers on the Executor agents, and provide this information to the Analyze component. The Plan component together with the Analyze component produce a new plan (a resource demand for tiers) which is sent to Executor.\n\n\\clearpage\n\\subsection{Components}\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=350px,natwidth=553,natheight=303]{./pictures/main-classes}\n \\caption{Executor main node classes}\n\\end{figure}\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[height=150px,natwidth=113,natheight=203]{./pictures/agent-classes}\n \\caption{Executor agent node classes}\n\\end{figure}", "meta": {"hexsha": "098c9beff18bac556f39e5d34da3d2463c280fa3", "size": 15924, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "solutiondesign.tex", "max_stars_repo_name": "n43jl/thesis", "max_stars_repo_head_hexsha": "f7da3a5e8a01f6d9a83edabc6778683a5c225a5f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "solutiondesign.tex", "max_issues_repo_name": "n43jl/thesis", "max_issues_repo_head_hexsha": "f7da3a5e8a01f6d9a83edabc6778683a5c225a5f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "solutiondesign.tex", "max_forks_repo_name": "n43jl/thesis", "max_forks_repo_head_hexsha": "f7da3a5e8a01f6d9a83edabc6778683a5c225a5f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 84.7021276596, "max_line_length": 813, "alphanum_fraction": 0.7319140919, "num_tokens": 4202, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9124361533336451, "lm_q2_score": 0.766293653760418, "lm_q1q2_score": 0.69919403376114}} {"text": "\\subsection{ReLU Fourier representation }%-- Simple version}\n\nWe introduce the Taylor expansion of $e^{iz}$ with integral remainder as follows.\n\\begin{lemma}\nFor $|z|\\leq c$, \n\\begin{equation} \ne^{iz} - iz -1\n= \n- \\int_{0}^c\\left[(z - u)_+e^{iu} + (-z - u)_+e^{-iu} \\right]du.\n\\end{equation} \n\\end{lemma}\n\\begin{proof} \nBy the Taylor expansion with integral remainder,\n\\begin{equation} \ne^{iz} = 1 + iz - \\int_0^z e^{iu}(z-u)du.\n\\end{equation}\nLet $u_+=\\max (u, 0)$ and $u_-=\\min(u,0)$. Then, $u_-=-(-u)_+$ and \n$$\nz-u=(z-u)_+ + (z-u)_-=(z-u)_+ - (u-z)_+.\n$$\nIt follows that\n\\begin{equation}\n\\begin{split}\n\\int_{0}^z (z-u)e^{iu} du=&\\int_{0}^z (z-u)_+e^{iu} du + \\int_{0}^z -(u-z)_+e^{iu} du\n\\\\\n=&\\int_{0}^z (z-u)_+e^{iu} du + \\int_{0}^{-z} (-u-z)_+e^{-iu} du\n\\\\\n=&\\int_{0}^c (z-u)_+e^{iu} du + (-u-z)_+e^{-iu} du.\n\\end{split}\n\\end{equation}\nThus,\n\\begin{equation} \ne^{iz} - 1 - iz \n= \n-\\int_{0}^c\\left[(z - u)_+e^{iu} + (-z - u)_+e^{-iu} \\right]du,\n\\end{equation} \nwhich completes the proof.\n\\end{proof}\n\nLet $z=\\omega\\cdot x$, $u=\\|\\omega\\|_{\\ell_1}t$ and $\\bar \\omega={\\omega\\over \\|\\omega\\|_{\\ell_1}}$. If $\\Omega$ is bounded, say $|x|\\le T$, $|\\bar \\omega \\cdot x|\\le T$. There exists the following expansion for $e^{i\\omega\\cdot x}$.\n\\begin{lemma}\\label{lm:talorcomplex}\nIf $|x|\\le T$,\n\\begin{equation} \ne^{i\\omega\\cdot x} - 1 - i\\omega\\cdot x \n= \n- \\|\\omega\\|_{\\ell_1}^2\\int_{0}^T\\left[(\\bar \\omega\\cdot x - t)_+e^{i\\|\\omega\\|_{\\ell_1}t}\n+ (-\\bar \\omega\\cdot x - t)_+e^{-i\\|\\omega\\|_{\\ell_1}t} \\right]dt.\n\\end{equation} \nDenote \n$$\nD^\\alpha = \\partial_1^{\\alpha_1}\\partial_2^{\\alpha_2}\\cdots \\partial_d^{\\alpha_d},\\quad \\omega^\\alpha = \\omega_1^{\\alpha_1}\\omega_2^{\\alpha_2}\\cdots \\omega_d^{\\alpha_d},\\quad \\alpha!=\\alpha_1!\\alpha_2!\\cdots \\alpha_d!.\n$$\n\\end{lemma}\nIt follows the following Taylor expansion with an integral remainder.\n\\begin{lemma}\\label{lm:probabilityexpan}\nSuppose $|x|\\le T$. There exists\n\\begin{equation}\nf(x) = f(0) + \\nabla f(0)\\cdot x\n+ \\int_{\\{-1,1\\}\\times [0,T]\\times \\mathbb{R}^{d}} g(x, \\theta)\\lambda(\\theta)d\\theta \n\\end{equation} \nwith $g(x,\\theta)$ and $\\lambda(\\theta)$ defined in \\eqref{eq:straglam}.\n\\end{lemma}\n\\begin{proof}\nSince $\n f(x) = \\int_{\\mathbb{R}^d} e^{i\\omega\\cdot x}\\hat{f}(\\omega)d\\omega\n$\nand \n$\n\\nabla f(x)=\\int_{\\mathbb{R}^d} i^{|\\alpha|}\\omega e^{i\\omega\\cdot x}\\hat{f}(\\omega)d\\omega,\n$\n\\begin{eqnarray}\n\\nabla f(0)\\cdot x=\\int_{\\mathbb{R}^d} i\\omega\\cdot x\\hat{f}(\\omega)d\\omega.\n\\end{eqnarray} \n It follows that\n\\begin{equation}\n\\nabla f(0) \\cdot x=i \\int_{\\mathbb{R}^d} \\omega\\cdot x\\hat{f}(\\omega)d\\omega\n= \\int_{\\mathbb{R}^d} i\\omega\\cdot x\\hat{f}(\\omega)d\\omega.\n\\end{equation} \nLet $\\hat{f}(\\omega)=|\\hat{f}(\\omega)|e^{ib(\\omega)}$. Then, $e^{i\\|\\omega\\|_{\\ell_1}t}\\hat{f}(\\omega) = |\\hat{f}(\\omega)|e^{i(\\|\\omega\\|_{\\ell_1}t + b(\\omega))}$.\nBy Lemma \\ref{lm:talorcomplex},\n\\begin{equation}\\label{eq:fftaylor}\n\\begin{split}\n&f(x) - f(0) - \\nabla f(0) \\cdot x\n\\\\\n= &\\int_{\\mathbb{R}^d} \\big (e^{i\\omega\\cdot x} - 1 - i\\omega\\cdot x \\big )\\hat{f}(\\omega)d\\omega.\n\\\\\n=&{\\rm Re} \\bigg (-\\int_{\\mathbb{R}^d} \\int_{0}^T\\left[(\\bar \\omega\\cdot x - t)_+e^{i\\|\\omega\\|_{\\ell_1}t}\n+ (-\\bar \\omega\\cdot x - t)_+e^{-i\\|\\omega\\|_{\\ell_1}t} \\right]\\hat{f}(\\omega)\\|\\omega\\|_{\\ell_1}^{2}dt d\\omega\\bigg )\n\\\\\n=& \\int_{\\{-1,1\\}}\\int_{\\mathbb{R}^d} \\int_{0}^T (z\\bar \\omega\\cdot x - t)_+ s(zt,\\omega) |\\hat{f}(\\omega)|\\|\\omega\\|_{\\ell_1}^{2}dtd\\omega dz\n\\end{split}\n\\end{equation}\nwith $\\int_{\\{-1, 1\\}} r(z) dz = r(-1) + r(1)$ and\n\\begin{equation} \ns(zt,\\omega)= -\\cos(z\\|\\omega\\|_{\\ell_1}t + b(\\omega)) \n\\end{equation} \nDefine $G=\\{-1,1\\}\\times [0,T]\\times \\mathbb{R}^{d}$, $\\theta=(z, t, \\omega)\\in G$,\n\\begin{equation}\\label{eq:straglam}\ng(x,\\theta)= (z\\bar \\omega\\cdot x - t)_+ {\\rm sgn} s(zt,\\omega),\\qquad \\lambda(\\theta)={\\rho(\\theta)\\over \n\\int_{\\{-1,1\\}\\times [0,T]\\times \\mathbb{R}^{d}} \\rho(\\theta)d\\theta}.\n\\end{equation}\nwith $\\rho(\\theta) = |s(zt,\\omega)||\\hat{f}(\\omega)|\\|\\omega\\|_{\\ell_1}^{2}$. \n\nThen \\eqref{eq:fftaylor} can be written as \n\\begin{equation}\\label{eq:reluintegral}\nf(x) = f(0) + \\nabla f(0) \\cdot x\n+ \\int_{\\{-1,1\\}\\times [0,T]\\times \\mathbb{R}^{d}} g(x, \\theta)\\lambda(\\theta)d\\theta, \n\\end{equation} \nwhich completes the proof.\n\\end{proof}\n\nAn application of the Monte Carlo method in Lemma \\ref{MC} to the integral \\eqref{eq:reluintegral} gives the following estimate.\n\\begin{theorem} \nSuppose $|x|\\le T$ and \n$$\n \\int_{\\mathbb{R}^{d}} |\\hat{f}(\\omega)|\\|\\omega\\|_{\\ell_1}^{2} d\\omega<\\infty.\n$$\nThere exist $\\|\\bar \\omega_j\\|_{\\ell_1}=1$, $t\\in [0,T]$ such that \n$$\nf_n(x)= f(0) + \\nabla f(0) \\cdot x + {1\\over n}\\sum_{j=1}^{n} (\\bar \\omega_j\\cdot x - t_j)_+\n$$ \nsatisfies the following estimate \n\\begin{equation}\n\\|f - f_n \\|_{L^2(\\Omega)} \\leq C n^{-{1\\over 2}}.\n\\end{equation} \n%\\begin{equation}\n%\\|D^\\beta (f(x)- f_n(x))\\|_{L^2(\\Omega)}\\le \\sqrt{2^{m-k-2}(2m-k)\\over k!(m-k)!}|\\Omega|^{1/2} n^{-{1\\over 2}-{1\\over d}},\\quad |\\beta|=k\\le m.\n%\\end{equation}\n\\end{theorem}\n\n\\iffalse\n\\noindent\\textbf{A modified analysis using stratified sampling}\n\nAccording to \\eqref{eq:straglam}, the main ingredient $(z\\bar \\omega\\cdot x - t)_+$ of $g(x,\\theta)$ only includes the direction $\\bar\\omega$ of $\\omega$ which belongs to a bounded domain $\\mathbb{S}^{d-1}$. Thanks to the continuity of $(z\\bar \\omega\\cdot x - t)_+$ with respect to $(z, \\bar\\omega, t)$ and the boundedness of $\\mathbb{S}^{d-1}$,\nthe application of the stratified sampling to the residual term of the Taylor expansion leads to the following approximation property.\n\\begin{theorem}\\label{est:stratify}\nSuppose $|x|\\le T$ and \n$$\n \\int_{\\mathbb{R}^{d}} |\\hat{f}(\\omega)|\\|\\omega\\|_{\\ell_1}^{2} d\\omega<\\infty.\n$$\nThere exist $\\beta_j\\in [-2^d,2^d]$, $\\|\\bar \\omega_j\\|_{\\ell_1}=1$, $t\\in [0,T]$ such that \n$$\nf_n(x)= f(0) + \\nabla f(0) \\cdot x + {1\\over n}\\sum_{j=1}^{n}\\beta_j (\\bar \\omega_j\\cdot x - t_j)_+\n$$ \nsatisfies the following estimate \n\\begin{equation}\n\\|f - f_n \\|_{L^2(\\Omega)} \\leq C n^{-{1\\over 2}-{1\\over d}}.\n\\end{equation} \n%\\begin{equation}\n%\\|D^\\beta (f(x)- f_n(x))\\|_{L^2(\\Omega)}\\le \\sqrt{2^{m-k-2}(2m-k)\\over k!(m-k)!}|\\Omega|^{1/2} n^{-{1\\over 2}-{1\\over d}},\\quad |\\beta|=k\\le m.\n%\\end{equation}\n\\end{theorem}\n\\begin{proof}\nBy Lemma \\ref{lem:stratifiedapprox}, for any decomposition $G=\\cup_{i=1}^M G_i$, there exist $\\{\\theta_i\\}_{i=1}^n$ and $\\{\\beta_i\\}_{i=1}^n\\in [0,1]$ such that \n\\begin{equation}\n\\| f - f_n\\|_{L^2(\\Omega)} = \\| r - r_{n}\\|_{L^2(\\Omega)} \\leq {1\\over n^{1/2}}\\max_{1\\le j\\le M}\\sup_{\\theta_{j},\\theta_{j}'\\in G_j} \\| g(x,\\theta_j) - g(x,\\theta_j') \\|_{L^2(\\Omega)} \n\\end{equation}\nwith \n$$\nf_n(x)= f(0) + \\nabla f(0) \\cdot x + r_{n}(x), \\qquad r_{n}(x)={1\\over n}\\sum_{j=1}^{n}\\beta_j (\\bar \\omega_j\\cdot x - t_j)_+, \n$$\n$$\nr(x)=\\int_{\\{-1,1\\}\\times [0,T]\\times \\mathbb{R}^{d}} g(x, \\theta)\\lambda(\\theta)d\\theta.\n$$\nConsider a particular decomposition $G=\\cup_{i=1}^M G_i$ as follows. \nThe variable $z$ is in the set $\\{-1,1\\}$, which can be divided into two subsets $\\{-1\\}$ and $\\{1\\}$. \nGiven a positive integer $n$, for the random variable $t$, the interval $ [0,T]$ can be divided into $n_t$ subintervals $\\{G_i^t\\}_{i=1}^{n_t}$ such that \n$$\n|t-t'|<{1\\over 2}n^{-{1\\over d}}\\quad t,t'\\in G_i^t,\\quad 1\\leq i\\leq n_t\n$$ \nfor $n_t>2\\lceil T n^{1\\over d}\\rceil$. \nFor variable $\\bar \\omega=\\omega/\\|\\omega\\|_{\\ell_1}\\in \\mathbb{S}^{d-1}$ where $\\mathbb{S}^{d-1}=\\{\\bar \\omega\\in \\mathbb{R}^d: \\|\\bar \\omega\\|_{\\ell_1}=1\\}$. Note that $\\mathbb{S}^{d-1}$ can be divided into $n_\\alpha$ subdomains $\\{G_i^s \\}_{i=1}^{n_s}$ such that\n$$\n\\|\\bar \\omega- \\bar \\omega'\\|_{\\ell_1}\\leq {1\\over 2}n^{-{1\\over d}}\\qquad \\bar \\omega, \\bar \\omega' \\in G_i^s,\\quad 1\\leq i\\leq n_s\n$$\nfor $(2n^{1\\over d})^{d-1}\\leq n_s\\leq \\lceil (5n^{1\\over d})^{d-1}\\rceil$ \\cite{klusowski2016uniform}.\nThen \n$$\nG=\\displaystyle \\cup \\{G_{ijk\\ell}: 1\\leq i\\leq 2,\\ 1\\leq j\\leq n_t,\\ 1\\leq k\\leq n_s,\\ 1\\le \\ell\\le 2\\}\n$$\nwith \n\\begin{equation}\nG_{ijk\\ell} = \\{(z, t, \\omega): z=(-1)^i,\\ t\\in G_j^t, \\bar \\omega \\in G_k^s,\\ {\\rm sgn} s(zt,\\omega)=(-1)^\\ell\\}.\n\\end{equation}\nDenote this decomposition of $G$ by $G=\\cup_{i=1}^{M} G_i$ with $M=4n_sn_t\\le 2^{d}n$. For each $G_i$,\n\\begin{equation}\nz=z',\\ |t-t'|<{1\\over 2}n^{-{1\\over d}},\\ \\|\\bar \\omega - \\bar \\omega'\\|_{\\ell^1}<{1\\over 2}n^{-{1\\over d}}\\qquad \\forall \\theta=(z, t, \\omega),\\ \\theta'=(z', t', \\omega')\\in G_i.\n\\end{equation}\nFor any $\\theta_i, \\theta'_i\\in G_i$ and $|\\alpha |=1$, \n$$\n| g(x,\\theta_i) - g(x,\\theta_i') | = | \\bar\\omega - \\bar\\omega' |\\le n^{-{1\\over d}}.\n$$ \nThus, there exist $\\theta_{i,j}$ such that\n\\begin{equation}\n\\| f - f_n\\|_{L^2(\\Omega)} \\le C n^{-{1\\over 2}-{1\\over d}}.\n\\end{equation}\nwith\n$$\nf_n(x)= f(0) + \\nabla f(0) \\cdot x + {1\\over n}\\sum_{j=1}^{n}\\beta_j (\\bar \\omega_j\\cdot x - t_j)_+\n$$ \nwith $\\beta_j\\in [-2^d,2^d]$,\nwhich completes the proof.\n\\end{proof}\n\n\n\\fi\n \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "df7ae607d44d95437944fd614ce94db8924771b2", "size": 8910, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/ReLUFourierSimple.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/ReLUFourierSimple.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/ReLUFourierSimple.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.7767857143, "max_line_length": 347, "alphanum_fraction": 0.598989899, "num_tokens": 3947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.912436167620237, "lm_q2_score": 0.7662936377487305, "lm_q1q2_score": 0.6991940300992219}} {"text": "\\section{Useful Languages}\n\n\\paragraph{Dyck Language} $S \\rarr bSeS | \\epsilon$\n\\paragraph{Palindromes} $S \\rarr aSa | bSb | a | b | \\epsilon$.\n\\paragraph{Antisymmetric} For $a^nb^m$ with $0 \\le n \\le m \\le kn$: $S \\rarr aSb | \\ldots | aSb^k | \\epsilon$ (this is ambiguous, impose an order, from $aXb$ to $aYb^k$, on the rules to avoid that).\n\\paragraph{Math Expressions}\n\\begin{align*}\n E &\\rarr E + T | T \\\\\n T &\\rarr T * F | F \\\\\n F &\\rarr (E) | i\n\\end{align*}\nThe operators in this example are left-associative.\n", "meta": {"hexsha": "ab881be26ca92ba0db40f71380d0679d76bce040", "size": 525, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "grammars/useful-languages.tex", "max_stars_repo_name": "Kakasinho/FLC-cheatsheet", "max_stars_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-01-13T14:36:20.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-18T16:22:18.000Z", "max_issues_repo_path": "grammars/useful-languages.tex", "max_issues_repo_name": "Kakasinho/FLC-cheatsheet", "max_issues_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "grammars/useful-languages.tex", "max_forks_repo_name": "Kakasinho/FLC-cheatsheet", "max_forks_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-05-21T11:05:53.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-17T14:59:50.000Z", "avg_line_length": 40.3846153846, "max_line_length": 198, "alphanum_fraction": 0.6419047619, "num_tokens": 201, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8933094003735664, "lm_q2_score": 0.7826624738835051, "lm_q1q2_score": 0.699159745239766}} {"text": "\\part{Calculation commands reference}\\label{part:Rechenbefehle}\r\n\r\nWhen using calculation commands in Warteschlangensimulator,\r\nit is distinguished between \\textbf{expressions} and \\textbf{comparisons}.\r\nExpressions are used to calculate a numerical values, which are e.g.\\ used as periods of time.\r\nComparisons provide a yes/no decision (for example, whether a client should be directed\r\nin a particular direction). Unlike expressions, comparisons always contain at least one comparison operator.\r\n\r\nAll commands presented below are each recognized in \\textbf{any case}.\r\nThere is no distinction between different case types.\r\n\r\n\r\n\r\n\\chapter{Constants}\r\n\r\nThe following constants are available in all calculation commands:\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{e}: Returns the basis of the exponential function $e^x$. It is\r\n$e\\approx 2.718281828459$.\r\n\r\n\\item\r\n\\cmd{pi}: Returns the value of the circle constant $\\pi$. It is\r\n$\\pi\\approx 3.1415926535898$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{Variables}\r\n\r\nWhen calculating values in the context of a concrete client, the variables\r\n\\begin{itemize}\r\n\\item\r\n\\cmd{w} for the \\textbf{previous waiting time} of the client,\r\n\\item\r\n\\cmd{t} for the \\textbf{previous transfer time} of the client ant\r\n\\item\r\n\\cmd{p} for the \\textbf{previous operating time} of the client are always available.\r\n\\end{itemize}\r\n\r\nIf the calculation is done for getting a clients score, the variable \\cmd{w} does not contain\r\nthe total waiting time of the current client but the waiting time of the current client at the\r\ncurrent station.\r\n\r\nFurthermore, all variables that are defined by an assignment element are always available.\r\nBefore the first assignment of a value to a variable, it has the value 0.\r\n\r\n\r\n\r\n\\chapter{Basic arithmetic operations}\r\n\r\nSupported instructions for the basic arithmetic operations:\r\n\\begin{itemize}\r\n\\item Addition: \\cmd{$+$}\r\n\\item Subtraction: \\cmd{$-$}\r\n\\item Multiplication: \\cmd{$*$}\r\n\\item Division: \\cmd{$/$}\r\n\\item Potentiate: \\cmd{$\\hat~$}\r\n\\end{itemize}\r\n\r\nThe rule \\textbf{point before line calculation} is taken into account.\r\nTo enforce deviating evaluations, \\textbf{brackets} can be set.\r\n\r\n\r\n\r\n\\chapter{Trailing instructions}\r\n\r\nThe following expressions can be written directly behind a number:\r\n\r\n\\begin{itemize}\r\n\\item\r\n\\cmd{\\%}:\r\nThe numerical value left to this symbol is interpreted as a percent value,\r\nfor example $30\\%=0.3$.\r\n\r\n\\item\r\n\\cmd{$^2$}:\r\nExponentiate number by 2.\r\n\r\n\\item\r\n\\cmd{$^3$}:\r\nExponentiate number by 3.\r\n\r\n\\item\r\n\\cmd{!}:\r\nCalculate factorial of the number, for example\\ $4!=1\\cdot2\\cdot3\\cdot4=24$.\r\n\r\n\\item\r\n\\cmd{$^{\\circ}$}:\r\nConverts the value left to this symbol from grad to radian, for example $180^{\\circ}=3.1415\\ldots$.\\\\\r\n(See also section \\ref{sec:Winkelfunktionen} in which the supported\r\ntrigonometric functions are presented.)\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{General functions}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{abs(x)}:\r\nAbsolute value, for example \\cm{abs(-5)=5}.\r\n\r\n\\item\r\n\\cmd{ceil(x)}:\r\nRound to next bigger integer number, for example \\cm{ceil(2.1)=3}\r\n\r\n\\item\r\n\\cmd{exp(x)}:\r\nExponential function $e^x$.\r\n\r\n\\item\r\n\\cmd{factorial(x)}:\r\nFactorial, for example $4!=1\\cdot2\\cdot3\\cdot4=24$.\r\n\r\n\\item\r\n\\cmd{binom(n;k)}:\r\nBinomial coefficient\r\n\r\n\\item\r\n\\cmd{floor(x)}:\r\nRound to next smaller integer number, for example \\cm{floor(2.9)=2}\r\n\r\n\\item\r\n\\cmd{frac(x)}:\r\nFraction part, for example \\cm{frac(1.3)=0,3}\r\n\r\n\\item\r\n\\cmd{gamma(x)}:\r\nGamma funkcion, for example \\cm{gamma(5)=4!=24}\r\n\r\n\\item\r\n\\cmd{int(x)}:\r\nInteger part, for example \\cm{int(2.9)=2}\r\n\r\n\\item\r\n\\cmd{log(x)}:\r\nLogarithm to the base $e$.\r\n\r\n\\item\r\n\\cmd{log(x;b)}:\r\nLogarithm to the base $b$.\r\n\r\n\\item\r\n\\cmd{ld(x)}:\r\nLogarithm to the base $2$, for example \\cm{ld(256)=8}.\r\n\r\n\\item\r\n\\cmd{lg(x)}:\r\nLogarithm to the base $10$, for example \\cm{lg(100)=2}.\r\n\r\n\\item\r\n\\cmd{ln(x)}:\r\nLogarithm to the base $e$.\r\n\r\n\\item\r\n\\cmd{modulo(a;b)} oder \\cmd{mod(a;b)}:\r\nDivision reminder when dividing a/b\r\n\r\n\\item\r\n\\cmd{pow(x;y)}:\r\nExponentiate $x^y$.\r\n\r\n\\item\r\n\\cmd{round(x)}:\r\nRound, for example \\cm{round(4.4)=4} and \\cm{round(4.5)=5}.\r\n\r\n\\item\r\n\\cmd{sign(x)}:\r\nSign of a number, for example \\cm{sign(3)=1} and \\cm{sign(-3)=-1}.\r\n\r\n\\item\r\n\\cmd{sqrt(x)}:\r\nSquare root, for example \\cm{sqrt{81}=9}.\r\n\r\n\\item\r\n\\cmd{sqr(x)}:\r\nSquare the number, for example \\cm{sqr(4)=16}.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Random numbers}\r\n\r\nThe following commands can be used to generate random numbers\r\nthat are \\textbf{equally distributed} in a certain area.\r\nSection \\ref{sec:Wahrscheinlichkeitsverteilungen} introduces\r\nadditional functions for generating random numbers according\r\nto certain distribution functions.\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{random()}:\r\nRandom number between 0 (inclusive) and 1 (exclusive).\r\n\r\n\\item\r\n\\cmd{random(x)}:\r\nRandom number between 0 (inclusive) and x (exclusive).\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\r\n\r\n\\chapter{Trigonometric functions}\\label{sec:Winkelfunktionen}\r\n\r\nThe trigonometric functions always refer to $2\\pi$ as a full circle (radians).\r\nIf angles in degrees ($360^\\circ$ for the full circle) are to be specified\r\nin the elementary trigonometric functions,\r\nthese have to be converted to radians using the angle functions, for example\r\n\\cm{sin($90^\\circ$)=1}.\r\n\r\n\r\n\r\n\\section{Elementary trigonometric functions}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{sin(x)}:\r\nSine\r\n\r\n\\item\r\n\\cmd{cos(x)}:\r\nCosine\r\n\r\n\\item\r\n\\cmd{tan(x)}:\r\nTangent\r\n\r\n\\item\r\n\\cmd{cot(x)}:\r\nCotangent\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Hyperbolic trigonometric functions}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{sinh(x)}:\r\nSine hyperbolicus\r\n\r\n\\item\r\n\\cmd{cosh(x)}:\r\nCosine hyperbolicus\r\n\r\n\\item\r\n\\cmd{tanh(x)}:\r\nTangent hyperbolicus\r\n\r\n\\item\r\n\\cmd{coth(x)}:\r\nCotangent hyperbolicus\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Inverse of the elementary trigonometric functions}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{arcsin(x)}:\r\nArcus sine\r\n\r\n\\item\r\n\\cmd{arccos(x)}:\r\nArcus cosine\r\n\r\n\\item\r\n\\cmd{arctan(x)}:\r\nArcus tangent\r\n\r\n\\item\r\n\\cmd{arccot(x)}:\r\nArcus cotangent\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Inverse of the hyperbolic trigonometric functions}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{arcsinh(x)}:\r\nArcus sine hyperbolicus\r\n\r\n\\item\r\n\\cmd{arccosh(x)}:\r\nArcus cosine hyperbolicus\r\n\r\n\\item\r\n\\cmd{arctanh(x)}:\r\nArcus-Tangent hyperbolicus\r\n\r\n\\item\r\n\\cmd{arccoth(x)}:\r\nArcus-Cotangent hyperbolicus\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{Functions with multiple parameters}\r\n\r\nThe following functions can accept any number of parameters.\r\nThe individual parameters have to be specified separately by semicolon \";\".\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{Min(a;b;c;...)}:\r\nCalculates the minimum of the given numbers.\r\n\r\n\\item\r\n\\cmd{Max(a;b;c;...)}:\r\nCalculates the maximum of the given numbers.\r\n\r\n\\item\r\n\\cmd{Sum(a;b;c;...)}:\r\nCalculates the sum of the given numbers.\r\n\r\n\\item\r\n\\cmd{Mean(a;b;c;...)}:\r\nCalculates the mean value of the given numbers.\r\n\r\n\\item\r\n\\cmd{Median(a;b;c;...)}:\r\nCalculates the median of the given numbers.\r\n\r\n\\item\r\n\\cmd{Var(a;b;c;...)}:\r\nCalculates the sample variance of the given numbers.\r\n\r\n\\item\r\n\\cmd{SD(a;b;c;...)}:\r\nCalculates the sample standard deviation of the given numbers.\r\n\r\n\\item\r\n\\cmd{SCV(a;b;c;...)}:\r\nCalculates the squared coefficient of variation of the given numbers.\r\n\r\n\\item\r\n\\cmd{CV(a;b;c;...)}:\r\nCalculates the coefficient of variation of the given numbers.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{Probability distributions}\\label{sec:Wahrscheinlichkeitsverteilungen}\r\n\r\nBy using the following commands both values of the density and the cumulative distribution function\r\nof the following probability distributions can be calculated as well as random numbers are based on\r\none of these probability distributions:\r\n\r\n\r\n\r\n\\section{Hypergeometric distribution \\texorpdfstring{$Hg(N,K,n)$}{Hg(N,K,n)}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{HypergeometricDist(k;N;K;n)}:\r\nCalculates the counting probability density at $k$.\r\n\r\n\\item\r\n\\cmd{HypergeometricDist(N;K;n)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Binomial distribution \\texorpdfstring{$B(n,p)$}{B(n,p)}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{BinomialDist(k;n;p)}:\r\nCalculates the counting probability density at $k$.\r\n\r\n\\item\r\n\\cmd{BinomialDist(n;p)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Poisson distribution \\texorpdfstring{$P(l)$}{P(l)}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{PoissonDist(k;l)}:\r\nCalculates the counting probability density at $k$.\r\n\r\n\\item\r\n\\cmd{PoissonDist(l)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Exponential distribution with mean \\texorpdfstring{$a$}{a}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ExpDist(x;a;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{ExpDist(x;a;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{ExpDist(a)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{ExpDistRange(min;max;a)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Uniform distribution in the interval \\texorpdfstring{$[a;b]$}{[a;b]}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{UniformDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{UniformDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{UniformDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Normal distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{NormalDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{NormalDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{NormalDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{NormalDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Log-normal distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LogNormalDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LogNormalDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LogNormalDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{LogNormalDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Gamma distribution with parameters \\texorpdfstring{$\\alpha=a$}{a} and \\texorpdfstring{$\\beta=b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{GammaDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{GammaDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{GammaDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{GammaDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Gamma distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{GammaDistDirect(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{GammaDistDirect(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{GammaDistDirect(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{GammaDistDirectRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Erlang distribution with parameters \\texorpdfstring{$n$}{n} and \\texorpdfstring{$\\lambda=l$}{l}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ErlangDist(x;n;l;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{ErlangDist(x;n;l;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{ErlangDist(n;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{ErlangDistRange(min;max;n;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Beta distribution in the interval \\texorpdfstring{$[a;b]$}{[a;b]} and with parameters \\texorpdfstring{$\\alpha=c$}{c} and \\texorpdfstring{$\\beta=d$}{d}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{BetaDist(x;a;b;c;d;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{BetaDist(x;a;b;c;d;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{BetaDist(a;b;c;d)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Beta distribution in the interval \\texorpdfstring{$[a;b]$}{[a;b]} and with mean \\texorpdfstring{$c$}{c} and standard deviation \\texorpdfstring{$d$}{d}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{BetaDistDirect(x;a;b;c;d;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{BetaDistDirect(x;a;b;c;d;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{BetaDistDirect(a;b;c;d)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Weibull distribution with parameters Scale=\\texorpdfstring{$a$}{a} and Form=\\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{WeibullDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{WeibullDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{WeibullDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{WeibullDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Cauchy distribution with mean \\texorpdfstring{$a$}{a} and Scale=\\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{CauchyDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{CauchyDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{CauchyDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{CauchyDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{\\texorpdfstring{Chi$^2$}{Chi2} distribution with \\texorpdfstring{$n$}{n} degrees of freedom}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ChiSquareDist(x;n;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{ChiSquareDist(x;n;1)}:\r\nChi$^2$ distribution with $n$ degrees of freedom.\r\n\r\n\\item\r\n\\cmd{ChiSquareDist(n)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{ChiSquareDistRange(min;max;n)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Chi distribution with \\texorpdfstring{$n$}{n} degrees of freedom}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ChiDist(x;n;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{ChiDist(x;n;1)}:\r\nChi distribution with $n$ degrees of freedom.\r\n\r\n\\item\r\n\\cmd{ChiDist(n)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{ChiDistRange(min;max;n)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{F distribution with \\texorpdfstring{$a$}{a} degrees of freedom for the numerator and \\texorpdfstring{$b$}{b} degrees of freedom for the denominator}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{FDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{FDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{FDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{FDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Johnson SU distribution with parameters \\texorpdfstring{$\\gamma=a$}{a}, \\texorpdfstring{$\\xi=b$}{b}, \\texorpdfstring{$\\delta=c$}{c} and \\texorpdfstring{$\\lambda=d$}{d}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{JohnsonSUDist(x;a;b;c;d;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{JohnsonSUDist(x;a;b;c;d;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{JohnsonSUDist(a;b;c;d)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{JohnsonSUDistRange(min;max;a;b;c;d)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Triangular distribution over \\texorpdfstring{$[a;c]$}{[a;c]} with most likely value \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{TriangularDist(x;a;b;c;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{TriangularDist(x;a;b;c;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{TriangularDist(a;b;c)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Pert distribution over \\texorpdfstring{$[a;c]$}{[a;c]} with most likely value \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{PertDist(x;a;b;c;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{PertDist(x;a;b;c;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{PertDist(a;b;c)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Laplace distribution with mean \\texorpdfstring{$mu$}{mu} and scale factor \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LaplaceDist(x;mu;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LaplaceDist(x;mu;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LaplaceDist(mu;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{LaplaceDistRange(min;max;mu;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Pareto distribution with scale parameter \\texorpdfstring{$x_{\\rm min}=xmin$}{xmin} and shape parameter \\texorpdfstring{$\\alpha=a$}{a}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ParetoDist(x;xmin;a;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{ParetoDist(x;xmin;a;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{ParetoDist(xmin;a)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Logistic distribution with mean \\texorpdfstring{$\\mu=mu$}{mu} and scale parameter \\texorpdfstring{$s$}{s}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LogisticDist(x;mu;s;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LogisticDist(x;mu;s;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LogisticDist(mu;s)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{LogisticDistRange(min;max;mu;s)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\t\r\n\\section{Inverse gaussian distribution with \\texorpdfstring{$\\lambda=l$}{l} and mean \\texorpdfstring{$mu$}{mu}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{InverseGaussianDist(x;l;mu;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{InverseGaussianDist(x;l;mu;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{InverseGaussianDist(l;mu)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{InverseGaussianDist(min;max;l;mu)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Rayleigh distribution with mean \\texorpdfstring{$mu$}{mu}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{RayleighDist(x;mu;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{RayleighDist(x;mu;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{RayleighDist(mu)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{RayleighDistRange(min;max;mu)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Log-Logistic distribution with \\texorpdfstring{$\\alpha$}{alpha} and mean \\texorpdfstring{$\\beta$}{beta}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LogLogisticDist(x;alpha;beta;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LogLogisticDist(x;alpha;beta;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LogLogisticDist(alpha;beta)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{LogLogisticDistRange(min;max;alpha;beta)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Power distribution on \\texorpdfstring{$[a;b]$}{[a;b]} with exponent \\texorpdfstring{$c$}{c}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{PowerDist(x;a;b;c;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{PowerDist(x;a;b;c;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{PowerDist(a;b;c)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Gumbel distribution with expected value \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{GumbelDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{GumbelDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{GumbelDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{GumbelDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Fatigue life distribution with location parameter \\texorpdfstring{$\\mu$}{mu}, scale parameter \\texorpdfstring{$\\beta$}{beta} and form parameter \\texorpdfstring{$\\gamma$}{gamma}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{FatigueLifeDist(x;mu;beta;gamma;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{FatigueLifeDist(x;mu;beta;gamma;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{FatigueLifeDist(mu;beta;gamma)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{FatigueLifeDistRange(min;max;mu;beta;gamma)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Frechet distribution with location parameter \\texorpdfstring{$\\delta$}{delta}, scale parameter \\texorpdfstring{$\\beta$}{beta} and form parameter \\texorpdfstring{$\\alpha$}{alpha}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{FrechetDist(x;delta;beta;alpha;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{FrechetDist(x;delta;beta;alpha;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{FrechetDist(delta;beta;alpha)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{FrechetDistRange(min;max;delta;beta;alpha)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Hyperbolic secant distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{HyperbolicSecantDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{HyperbolicSecantDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{HyperbolicSecantDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\item\r\n\\cmd{HyperbolicSecantDistRange(min;max;a;b)}:\r\nGenerates a random number based on this distribution but limits the range of the value to $min\\ldots max$.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Left sawtooth distribution over \\texorpdfstring{$[a;b]$}{[a;b]}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Left sawtooth distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDistDirect(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDistDirect(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{LeftSawtoothDistDirect(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Right sawtooth distribution over \\texorpdfstring{$[a;b]$}{[a;b]}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{RightSawtoothDist(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{RightSawtoothDist(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{RightSawtoothDist(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Right sawtooth distribution with mean \\texorpdfstring{$a$}{a} and standard deviation \\texorpdfstring{$b$}{b}}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{RightSawtoothDistDirect(x;a;b;0)}:\r\nCalculates the probability density at $x$.\r\n\r\n\\item\r\n\\cmd{RightSawtoothDistDirect(x;a;b;1)}:\r\nCalculates the cumulative distribution function at $x$.\r\n\r\n\\item\r\n\\cmd{RightSawtoothDistDirect(a;b)}:\r\nGenerates a random number based on this distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\section{Distribution based on empirical values}\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{EmpiricalDensity(x;value1;value2;value3;...;max)}:\\\\\r\nCalculates the probability density at $x$.\r\nThe specified values will be used for the density in the range from 0 to $max$.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistribution(x;value1;value2;value3;...;max)}:\\\\\r\nCalculates the cumulative distribution function at $x$.\r\nThe specified values will be used for the density in the range from 0 to $max$.\r\n\r\n\\item\r\n\\cmd{EmpiricalRandom(value1;value2;value3;...;max)}:\\\\\r\nGenerates a random number based on this distribution.\r\nThe specified values will be used for the density in the range from 0 to $max$.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionMean(value1;value2;value3;...;max)}:\\\\\r\nCalculates the expected value of the distribution.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionMedian(value1;value2;value3;...;max)}:\\\\\r\nCalculates the median of the distribution.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionQuantil(value1;value2;value3;...;max;p)}:\\\\\r\nCalculates the quantil for the probability p of the distribution.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionSD(value1;value2;value3;...;max)}:\\\\\r\nCalculates the standard deviation of the distribution.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionVar(value1;value2;value3;...;max)}:\\\\\r\nCalculates the variance of the distribution.\r\n\r\n\\item\r\n\\cmd{EmpiricalDistributionCV(value1;value2;value3;...;max)}:\\\\\r\nCalculates the coefficient of variation of the distribution.\r\n\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{Erlang C calculator}\r\n\r\nBy using the following command some performance indicators can be calculated\r\nusing the extended Erlang C formula:\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;-1)}:\\\\\r\nCalculates the average queue length $\\E[N_Q]$. \r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;-2)}:\\\\\r\nCalculates the average number of clients in the system $\\E[N]$.\r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;-3)}:\\\\\r\nCalculates the average waiting time $\\E[W]$.\r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;-4)}:\\\\\r\nCalculates the average residence time $\\E[V]$.\r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;-5)}:\\\\\r\nCalculates the average accessibility $1-P(A)$.\r\n\r\n\\item\r\n\\cmd{ErlangC(lambda;mu;nu;c;K;t)}:\\\\\r\nCalculates the the probability for the service level at the $t$ seconds threshold $P(W\\le t)$.\r\n\r\n\\end{itemize}\r\n\r\nThe parameters have the following meanings:\r\n\\begin{itemize}\r\n\\item\r\n\\cm{lambda}:\\\\\r\nArrival rate $\\lambda$ (in clients per time unit), i.e.\\ inverse of the mean inter-arrival time.\r\n\\item\r\n\\cm{mu}:\\\\\r\nService rate $\\mu$ (in clients per time unit), i.e.\\ inverse of the mean service time.\r\n\\item\r\n\\cm{nu}:\\\\\r\nCancelation rate $\\nu$ (in clients per time unit), i.e.\\ inverse of the mean waiting time tolerance.\r\n\\item\r\n\\cm{c}:\\\\\r\nNumber of available parallel operating servers.\r\n\\item\r\n\\cm{K}:\\\\\r\nNumber of available places in the system (waiting and processing places together, i.e.\\ it is $K\\ge c$).\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\chapter{Allen-Cunneen approximation formula}\r\n\r\nBy using the following command some performance indicators can be calculated\r\nusing the Allen-Cunneen approximation formula:\r\n\r\n\\begin{itemize}\r\n\r\n\\item\r\n\\cmd{AllenCunneen(lambda;mu;cvI;cvS;c;-1)}:\\\\\r\nCalculates the average queue length $E[N_Q]$. \r\n\r\n\\item\r\n\\cmd{AllenCunneen(lambda;mu;cvI;cvS;c;-2)}:\\\\\r\nCalculates the average number of clients in the system $\\E[N]$.\r\n\r\n\\item\r\n\\cmd{AllenCunneen(lambda;mu;cvI;cvS;c;-3)}:\\\\\r\nCalculates the average waiting time $\\E[W]$.\r\n\r\n\\item\r\n\\cmd{AllenCunneen(lambda;mu;cvI;cvS;c;-4)}:\\\\\r\nCalculates the average residence time $\\E[V]$.\r\n\\end{itemize}\r\n\r\nThe parameters have the following meanings:\r\n\\begin{itemize}\r\n\\item\r\n\\cm{lambda}:\\\\\r\nArrival rate $\\lambda$ (in clients per time unit), i.e.\\ inverse of the mean inter-arrival time.\r\n\\item\r\n\\cm{mu}:\\\\\r\nService rate $\\mu$ (in clients per time unit), i.e.\\ inverse of the mean service time.\r\n\\item\r\n\\cm{cvI}:\\\\\r\nCoefficient of variation of the inter-arrival times $\\CV[I]$ (small values mean that the inter-arrival times are very homogeneous).\r\n\\item\r\n\\cm{cvS}:\\\\\r\nCoefficient of variation of the service times $\\CV[S]$ (small values mean that the operations are very homogeneous).\r\n\\item\r\n\\cm{c}:\\\\\r\nNumber of available parallel operating servers.\r\n\\end{itemize}", "meta": {"hexsha": "cdff58044a4cc5a30aaa6d55213bfb6681640e80", "size": 29973, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Simulator/build/Help/Scripting/en/Calculation1.tex", "max_stars_repo_name": "A-Herzog/Warteschlangensimulator", "max_stars_repo_head_hexsha": "fd83d400944a59147a465a4683b2f9258c3de5d0", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2020-07-14T05:36:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-30T14:24:12.000Z", "max_issues_repo_path": "Simulator/build/Help/Scripting/en/Calculation1.tex", "max_issues_repo_name": 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YES\n2. YES", "lm_q1_score": 0.8652240895276223, "lm_q2_score": 0.8080672112416737, "lm_q1q2_score": 0.699159217123702}} {"text": "\\documentclass[a4paper]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\n\\hbadness=99999\n\\voffset=-1in\n\\oddsidemargin=5pt\n\\textwidth=450pt\n\\textheight=700pt\n\n\\begin{document}\n\\title{Linear Algebra: Week 4 Notes and Exercises}\n\\author{Christopher Aytona}\n\\maketitle\n\n\\section{Notes}\nNotes\\\\\n\nVector Space\\\\\n1) $\\vec{0} \\in V$\\\\\n2) $\\vec{u}, \\vec{v} \\in V$ then must have $\\vec{u} + \\vec{v} \\in V$\\\\\n3) $c \\in \\mathbb{R}, \\vec{u} \\in V$ then must have $c \\vec{u} \\in V$\\\\\nExercise\\\\\n\n2) a) True because scaling $\\vec{u}$ with either a positive or negative value will still be in $W$.\\\\\n$\\vec{u} = \\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix}, x_1 \\geq 0, y_1 \\geq 0$\\\\\n$c \\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix} = \\begin{bmatrix}\ncx_1\\\\\ncy_1\n\\end{bmatrix}$\\\\\n$c \\geq 0 \\Rightarrow cx_1 \\geq 0 \\Rightarrow cy_1 \\geq 0$\\\\\n$c \\leq 0 \\Rightarrow cx_1 \\leq 0 \\Rightarrow cy_1 \\leq 0$\\\\\nb)Let $\\vec{u} = \\begin{bmatrix}\n4\\\\\n2\n\\end{bmatrix}$, $\\vec{v} = \\begin{bmatrix}\n-1\\\\\n-8\n\\end{bmatrix}$ $\\therefore \\vec{u} + \\vec{v} = \\begin{bmatrix}\n3\\\\\n-6\n\\end{bmatrix}$ which is not in any of the quadrants associated with $W \\therefore W$ is not a vector space .\\\\\n\n6) $p(t) = a + t^2$\\\\\n$V = \\begin{bmatrix}\nx_1&x_2&x_3\n\\end{bmatrix} \\in \\mathbb{R}^3 | x_1 \\in \\mathbb{R}, x_2 = 0, x_3 = 1$\\\\\n$\\begin{bmatrix}\na&0&1\n\\end{bmatrix}$\nIts not a vector space because scaling or adding vectors will change the $x_3$.\\\\\n\n\\section{Exercises}\n\n1. Let $V$ be the first quadrant in the $xy$-plane; that is let $V = \\{\n\\begin{bmatrix}\nx\\\\\ny\n\\end{bmatrix} : x \\geq 0, y \\geq 0 \\} $\\\\\na) If $u$ and $v$ are in $V$, is $u+v$ in $V$? Why?\\\\\nIf $\\vec{u}, \\vec{v} \\in V$ is $\\vec{u} + \\vec{v} \\in V$\\\\\n$\\vec{u} \\in V \\Rightarrow \\vec{u} = \\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix}, x_1 \\geq 0, y_1 \\geq 0$\\\\\n$\\vec{v} \\in V \\Rightarrow \\vec{v} = \\begin{bmatrix}\nx_2\\\\\ny_2\n\\end{bmatrix}, x_2 \\geq 0, y_2 \\geq 0$\\\\\n$\\vec{u} + \\vec{v} = \\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix} + \\begin{bmatrix}\nx_2\\\\\ny_2\n\\end{bmatrix} = \\begin{bmatrix}\nx_1 + x_2\\\\\ny_1 + y_2\n\\end{bmatrix} = x_1 + x_2 \\geq 0, y_1 + y_2 \\geq 0$\\\\\n$\\Rightarrow \\vec{u} + \\vec{v} \\in V$\\\\\nb) Find a specific vector $u$ in $V$ and a specific scalar $c$ such that $cu$ is not in $V$. (This is enough to show that $V$ is not a vector space.)\\\\\nLet $\\vec{u} = \\begin{bmatrix}\n1\\\\\n1\n\\end{bmatrix}$ then $\\vec{u} \\in V$,\\\\\nLet $c = -1$ then $c\\vec{u} = -1\\begin{bmatrix}\n1\\\\\n1\n\\end{bmatrix} = \\begin{bmatrix}\n-1\\\\\n-1\n\\end{bmatrix} \\notin V$\\\\\n2. Let $W$ be the union of the first and third quadrants in the $xy$-plane. That is, let $W = \\{\\begin{bmatrix}\nx\\\\\ny\n\\end{bmatrix} : xy \\geq 0 \\}$\\\\\na) If $u$ is in $W$ and $c$ is any scalar, is $cu$ in W? Why?\\\\\nIf $\\vec{u} \\in W, c \\in \\mathbb{R}$, is $c\\vec{u} \\in W$\\\\\nLet $\\vec{u} = \\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix}, x_1y_1 \\geq 0$, Let $c \\in \\mathbb{R}$\\\\\n$cu = c\\begin{bmatrix}\nx_1\\\\\ny_1\n\\end{bmatrix} = \\begin{bmatrix}\ncx_1\\\\\ncy_1\n\\end{bmatrix}$\\\\\n$(cx_1)(cy_1) = c^2x_1y_1$\\\\\nBut $c^2 \\geq 0, \\forall c \\in \\mathbb{R}$\\\\\n$c^2x_1y_1 \\geq 0 \\Rightarrow c\\vec{u}\\in W$\\\\\nb) Find specific vectors $u$ and $v$ in $W$ such that $u + v$ is not in $W$. This is enough to show that $W$ is not a vector space.\\\\\nLet $\\vec{u} = \\begin{bmatrix}\n1\\\\\n2\n\\end{bmatrix}$, let $\\vec{v} = \\begin{bmatrix}\n-2\\\\\n-1\n\\end{bmatrix}$\\\\\nThen $\\vec{u} + \\vec{v} = \\begin{bmatrix}\n1-2\\\\\n2-1\n\\end{bmatrix} = \\begin{bmatrix}\n-1\\\\\n1\n\\end{bmatrix} \\notin W, since (-1)(1) = -1$\\\\\n\nIn Exercise 5-8, determine if the given set is a subspace of $\\mathbb{P}_n$ for an appropriate value of $n$. Justify your answers.\\\\\nAll polynomials $p(t) = at^2$, $a \\in \\mathbb{R}$\\\\\n5. All polynomials of the form $p(t) = at^2$, where $a$ is in $\\mathbb{R}$\\\\\nIf $a = 0, p(t) = 0t^2 = \\vec{0}$ $\\therefore \\vec{0} \\in \\{p(t)\\}$\\\\\nIf $u(t) = a_1t^2$, $v(t) = a_2t^2$, then $\\vec{u} + \\vec{v} = a_1t^2+a_2t^2 = (a_1+a_2)t^2 \\in \\{ p(t) \\}$\\\\\nIf $u(t) = a_1t^2$ and $c \\in \\mathbb{R}$, $c\\vec{u} = ca_1t^2 = (ca_1)t^2 \\in \\{p(t)\\}$\\\\\n$\\therefore \\{p(t)\\}$ is a subspace\\\\\n6. All polynomials of the form $p(t) = a + t^2$, where $a$ is in $\\mathbb{R}$\\\\\n$\\vec{0} = 0+0t+t^2$ is not in $\\{p(t)\\}$\\\\\nIf $n(t) = a_1 + t^2$, $c = 0$\\\\\n$c\\vec{u} = 0(a_1 + t^2) = 0 \\notin \\{p(t)\\}$\\\\\n$\\therefore$ not in subspace\\\\\n7. All polynomials of degree at most 3, with integers of coefficients.\\\\\nIf the scalars are any number in $\\mathbb{R}$ then $c = \\frac{1}{10}$, $p(t) = 1+x+x^2 \\Rightarrow cp(t) = \\frac{1}{10} + \\frac{1}{10}x + \\frac{1}{10}x^2 \\notin$ the set.\\\\\nBut if we restrict scalars to the set of integers then it will be a subspace.\\\\\n8. All polynomials in $\\mathbb{P}_n$ such that $p(0) = 0$.\\\\\n$\\vec{0} = 0+0t+0t^2 \\cdots , \\vec{0} = 0$\\\\\nFor $u(t) = a_0 +a_1t+ \\cdots , u(0) = 0$\\\\\n$v(t) = b_0 + b_1t+ \\cdots, v(0) = 0$\\\\\n$u(t) + v(t) = a_0+b_0 + (a_1+b_1)t + \\cdots , (u + v) = 0$\\\\\nif $u(0) = 0$, $c \\in \\mathbb{R}$ then $cu(0) = c(0) = 0$\\\\\n$\\therefore$ is a subspace.\\\\\n9. Let $H$ be the set of all vectors of the form $\\begin{bmatrix}\n-2t\\\\\n5t\\\\\n3t\n\\end{bmatrix}$. Find a vector $v$ in $\\mathbb{R}^3$ such that $H = Span\\{v\\}$. Why does this show that $H$ is a subspace of $\\mathbb{R}^3$?\\\\\n$\\vec{v} = t\\begin{bmatrix}\n-2\\\\\n5\\\\\n3\n\\end{bmatrix}$\\\\\n$t = 0 \\Rightarrow \\begin{bmatrix}\n-2(0)\\\\\n5(0)\\\\\n3(0)\n\\end{bmatrix} = \\begin{bmatrix}\n0\\\\\n0\\\\\n0\n\\end{bmatrix} \\in H$, so $\\vec{0} \\in H$\\\\\n$\\vec{u} = \\begin{bmatrix}\n-2t_1\\\\\n5t_1\\\\\n3t_1\n\\end{bmatrix}$, $\\vec{v} = \\begin{bmatrix}\n-2t_2\\\\\n5t_2\\\\\n3t_2\n\\end{bmatrix} \\Rightarrow \\vec{u} + \\vec{v} = \\begin{bmatrix}\n-2(t_1+t_2)\\\\\n5(t_1+t_2\\\\\n3(t_1+t_2)\n\\end{bmatrix} \\in H$\\\\\n$\\forall c \\in \\mathbb{R}$ $c \\vec{u} = \\begin{bmatrix}\n-2(ct)\\\\\n5(ct)\\\\\n3(ct)\n\\end{bmatrix} \\in H$\\\\\n$\\therefore H$ is a subspace.\\\\\n\nWhich of the following sets are linearly independent?\\\\\nWhich form a basis for $\\mathbb{R}^3$\\\\\n$\\begin{bmatrix}\n1\\\\\n0\\\\\n0\n\\end{bmatrix}$, $\\begin{bmatrix}\n2\\\\\n3\\\\\n0\n\\end{bmatrix}$\n$\\begin{bmatrix}\n1&2&|&0\\\\\n0&3&|&0\\\\\n0&0&|&0\n\\end{bmatrix}$\n$\\frac{1}{2}R_2 = \\begin{bmatrix}\n1&2&|&0\\\\\n0&1&|&0\\\\\n0&0&|&0\n\\end{bmatrix}$\n$R_1 - R_2 = \\begin{bmatrix}\n1&0&|&0\\\\\n0&1&|&0\\\\\n0&0&|&0\n\\end{bmatrix}$\\\\\n$t$ has no free parameter.\\\\\n$\\therefore$ are linearly independent.\\\\\n$\\begin{bmatrix}\n1\\\\\n0\\\\\n0\n\\end{bmatrix}$, $\\begin{bmatrix}\n2\\\\\n3\\\\\n0\n\\end{bmatrix}$, $\\begin{bmatrix}\n4\\\\\n5\\\\\n6\n\\end{bmatrix}$\\\\\n$\\begin{bmatrix}\n1&2&4&|&0\\\\\n0&3&5&|&0\\\\\n0&0&6&|&0\n\\end{bmatrix}$\n$\\frac{1}{6}R_3 = \\begin{bmatrix}\n1&2&4&|&0\\\\\n0&3&5&|&0\\\\\n0&0&1&|&0\n\\end{bmatrix}$\n$R_1 - 4R_3 = \\begin{bmatrix}\n1&2&0&|&0\\\\\n0&3&5&|&0\\\\\n0&0&1&|&0\n\\end{bmatrix}$\n$R_2 - 5R_3 = \\begin{bmatrix}\n1&2&0&|&0\\\\\n0&3&0&|&0\\\\\n0&0&1&|&0\n\\end{bmatrix}$\n$\\frac{1}{3}R_2 = \\begin{bmatrix}\n1&2&0&|&0\\\\\n0&1&0&|&0\\\\\n0&0&1&|&0\n\\end{bmatrix}$\n$R_1 - 2R_2 = \\begin{bmatrix}\n1&0&0&|&0\\\\\n0&1&0&|&0\\\\\n0&0&1&|&0\n\\end{bmatrix}$\\\\\n$t$ has no free parameter, $\\therefore$ are linearly independent.\\\\\n$\\begin{bmatrix}\n1\\\\\n0\\\\\n0\n\\end{bmatrix}$, $\\begin{bmatrix}\n2\\\\\n3\\\\\n0\n\\end{bmatrix}$, $\\begin{bmatrix}\n4\\\\\n5\\\\\n6\n\\end{bmatrix}$, $\\begin{bmatrix}\n7\\\\\n8\\\\\n9\n\\end{bmatrix}$\\\\\n4 Vectors in $\\mathbb{R}^3$ must have at least 1 free parameter. $\\therefore$ linearly dependent.\\\\\n\\end{document}", "meta": {"hexsha": "b1c529b9b348af9921b8e05db711e79874c767d1", "size": 7107, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/Week4/Week4.tex", "max_stars_repo_name": "aytona/LinearAlgebra", "max_stars_repo_head_hexsha": "2a278b2957bc12456eb4bbc3f4d13b3c06d8d8d1", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Notes/Week4/Week4.tex", "max_issues_repo_name": "aytona/LinearAlgebra", "max_issues_repo_head_hexsha": "2a278b2957bc12456eb4bbc3f4d13b3c06d8d8d1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/Week4/Week4.tex", "max_forks_repo_name": "aytona/LinearAlgebra", "max_forks_repo_head_hexsha": "2a278b2957bc12456eb4bbc3f4d13b3c06d8d8d1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.7630662021, "max_line_length": 172, "alphanum_fraction": 0.5934993668, "num_tokens": 3464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240686758841, "lm_q2_score": 0.8080672089305841, "lm_q1q2_score": 0.6991591982744857}} {"text": "\\newpage\n\n\\section{Earley Method}\n\nIt works on any grammar (even nondeterministic ones) by using a vector of sets.\n\nIt builds a vector of size $n+1$ where $n=|x|$ where $E[i]$ is a set of $\\langle s,j \\rangle$ with $j0,i=1,2,\\dots,D; \\sum_{i=1}^D x_i=\\lambda \\right. \\right\\}. \\]\nWhere $\\lambda>0$ can be any positive real-valued constant. By definition, the compositions are the quantities denoted by $\\mathbf{x}$.\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.5\\textwidth]{appendix_c/Aitchison-simplex.jpg}\n \\caption[An illustration of the Aitchison simplex.]\n {An illustration of the Aitchison simplex. Here, there are 3 parts, $x_1, x_2, x_3$ represent values of different proportions. A, B, C, D and E are 5 different compositions within the simplex. A, B and C are all equivalent and D and E are equivalent.}\n \\label{figcS1}\n \\end{figure}\nThere are three core axioms that the Aitchison simplex, namely\n\n\\subsection{Scale invariance}\nWhether data is represented as proportions, percentages or probabilities, in the context of the Aitchison simplex all of these measurements are equivalent since they only differ by a constant scaling factor.\n\n\\subsection{Subcompositional coherence}\nObservations on shared species should be consistent. For example, supposed that there are 2 biologists that visited the exact same rainforest to count insects. One biologist observed 3 species of spiders and 10 species of ants whereas the other biologist only observed 2 species of spiders and 7 species of ants. If these biologists observed the same 2 spider species and the same 7 species of ants, their conclusions about those species should be the same. For instance, they should notice that the ratio of those two spider species are consistent between their observations. While the universal formal definition is still not clearly established, this concept can be formalized to distance metrics as follows\n\n \\[\nd(x_k, y_k) \\leq d(x, y) \\qquad\n\\forall x, y \\in S^D, \\; x_k, y_k \\in S^k, \\; S^k \\subset S^D\n\\]\n\n\\subsection{Permutation invariance}\nThe ordering of how the proportions or were measured or counted doesn't matter. This is analogous to how combinations are invariant to the order of selection.\n\n\n\\section{Vector Space Structure}\n\\subsection{Properties}\nThe Aitchison simplex has the following operators defined using the \\textbf{closure} operation as follows\n\n\\subsubsection{Perturbation}\n\\[ x \\oplus y = [\\frac{x_1 y_1}{\\sum_{i=1}^D x_i y_i},\\frac{x_2}{\\sum_{i=1}^D x_i y_i}, \\dots,\\frac{x_D y_D}{\\sum_{i=1}^D x_i y_i}] =\nC[x_1 y_1, ..., x_D y_D] \\qquad \\forall x, y \\in S^D\n\\]\n\n\\subsubsection{Powering}\n\n\\[\n\\alpha \\odot x = [\\frac{x_1^{\\alpha}}{\\sum_{i=1}^D x_i^{\\alpha}},\\frac{x_2^{\\alpha}}{\\sum_{i=1}^D x_i^{\\alpha}}, \\dots,\\frac{x_D^{\\alpha}}{\\sum_{i=1}^D x_i^{\\alpha}}] =\nC[x_1 y_1, ..., x_D y_D] \\qquad \\forall x \\in S^D, \\quad \\alpha \\in \\mathbb{R}\n\\]\n\n\\subsubsection{Inner product}\n\n\\[\n\\langle x, y \\rangle = \\frac{1}{2D}\n\\sum\\limits_{i=1}^{D}\n\\sum\\limits_{j=1}^{D}\n\\log \\frac{x_i}{x_j}\n\\log \\frac{y_i}{y_j}\n\\qquad \\forall x, y \\in S^D\n\\]\n\nUnder these these operations alone, it is sufficient to show that the Aitchison simplex forms a Euclidean vector space.\n\n\\subsection{Orthonormal bases}\nSince the Aitchison simplex forms a finite Hilbert space, it is possible to construct orthonormal bases in the simplex. Every composition can be decomposed as follows\n\n\\[ x = \\bigoplus_{i=1}^D x_i \\odot e_i \\]\n\nWhere $e_1, \\ldots e_{D-1} $ forms an orthonormal basis in the simplex \\cite{ilr}.\n\n\\section{Linear transformations}\n\nThere are 3 well-characterized isomorphisms that transform from the Aitchison simplex to real space. All of these transforms satisfy linearity and as given below\n\n\\subsection{Additive Logratio Transform}\nThe additive log ratio (alr) transform is an where $alr: S^D \\rightarrow \\mathbb{R}^{D-1} $. This is given by\n\n\\[ alr(x) = \\big[ \\log \\frac{x_1}{x_D} \\ldots \\log \\frac{x_{D-1}}{x_D} \\big]\\]\n\nThe choice of denominator component is arbituary, and could be any specified component.\nThis transform is commonly used in chemistry with measurements such as pH. In addition, this is the transform most commonly used for Multinomial logistic regression. The alr transform is not an isometry, meaning that distances on transformed values will not be equivalent to distances on the original compositions in the simplex.\n\n\\subsection{Center Logratio Transform}\nThe center log ratio (clr) tranform is both an isomorphism and an isometry where \\\\$clr: S^D \\rightarrow \\mathbb{U}, \\quad U \\subset \\mathbb{R}^{D} $\n\n\\[clr(x) = \\big[ \\log \\frac{x_1}{g(x)} \\ldots \\log \\frac{x_{D-1}}{g(x)} \\big] \\]\n\nThe inverse of this function is also known as the softmax function commonly used in neural networks.\n\n\\subsection{Isometric Logratio Transform}\nThe isometric log ratio (ilr) tranform is both an isomorphism and an isometry where $ilr: S^D \\rightarrow \\mathbb{R}^{D-1} $\n\n\\[\nilr(x) = \\big[ \\langle x, e_1 \\rangle, \\ldots \\langle x, e_{D-1} \\rangle]\n\\]\n\nThere are multiple ways to construct orthonormal bases, including using the Gram–Schmidt process Singular-value decomposition of clr transformed data.\nAnother alternative is to construct log contrasts from a bifurcating tree. If are given a bifurcating tree, we can construct a basis from the internal nodes in the tree.\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.5\\textwidth]{appendix_c/Orthogonal-tree-basis.jpg}\n \\caption[An illustration of the bifurcating trees as an orthonormal basis.]\n {A representation of a tree in terms of its orthogonal components. l represents an internal node, an element of the orthonormal basis. This is a precursor to using the tree as a scaffold for the ilr transform.}\n \\label{figcS1}\n \\end{figure}\n\nEach vector in the basis would be determined as follows\n\n\\[e_l = C[exp( \\underbrace{0,...0}_{k}, \\underbrace{a,...,a}_{r},\\underbrace{b,...,b}_s,\\underbrace{0,...0)}_t]\\]\n\nThe elements within each vector are given as follows\n\n\\[a = \\frac{\\sqrt{s}}{\\sqrt{r(r+s)}} \\quad \\textrm{and} \\quad b = \\frac{-\\sqrt{r}}{\\sqrt{s(r+s)}}\\]\n\nwhere $k, r, s, t$ are the respective number of tips in the corresponding subtrees shown in the figure. It can be shown that the resulting basis is orthonormal \\cite{groups_of_parts}.\n\nOnce the basis $\\Psi$ is built, the ilr transform can be calculated as follows\n\n\\[\nilr(x) = C[\\exp(clr(x) \\Psi)]\n\\]\n\n\nwhere each element in the ilr transformed data is of the following form\n\n\n\\[\nb_i = \\sqrt{\\frac{rs}{r+s}} \\log \\frac{g(x_R)}{g(x_S)}\n\\]\n\nwhere $ x_R$ and $ x_S$ are the set of values corresponding to the tips in the subtrees $R$ and $S$.\n", "meta": {"hexsha": "18ca8519c64ce717019f7983283d2ac9646b49fe", "size": 7792, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "appendixc.tex", "max_stars_repo_name": "mortonjt/phd-thesis", "max_stars_repo_head_hexsha": "f2b381322236b2591b51e4f9fca5899e0922654c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "appendixc.tex", "max_issues_repo_name": "mortonjt/phd-thesis", "max_issues_repo_head_hexsha": "f2b381322236b2591b51e4f9fca5899e0922654c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "appendixc.tex", "max_forks_repo_name": "mortonjt/phd-thesis", "max_forks_repo_head_hexsha": "f2b381322236b2591b51e4f9fca5899e0922654c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.1492537313, "max_line_length": 714, "alphanum_fraction": 0.7416581109, "num_tokens": 2167, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8872045996818986, "lm_q2_score": 0.7879311981328135, "lm_q1q2_score": 0.6990561832163016}} {"text": "\\section{Some classic CNN models}\\label{sec:CNNs}\nIn this section, we will use the notation introduced above to \ngive a brief description of some classic CNN models.\n\n\\subsection{LeNet-5, AlexNet and VGG}\nThe LeNet-5 \\cite{lecun1998gradient}, AlexNet \\cite{krizhevsky2012imagenet} and VGG \\cite{simonyan2014very}\ncan be written as:\n\t\\begin{equation}\n\t\\begin{cases}\n\tf^{1,0} &= \\theta^0(f), \\\\\n\tf^{\\ell,i} &= \\theta^{\\ell,i} \\circ \\sigma (f^{\\ell, j-1}), \\quad i = 1:\\nu_\\ell ~\\text{and}~ \\ell = 1:J,\\\\\n\tf^{\\ell+1,0} &= R_\\ell^{\\ell+1}( f^{\\ell,m+\\ell}). \\\\\n\t\\end{cases}\n\t\\end{equation}\n\twhere $R_\\ell^{\\ell+1}$ can be general pooling operators and $\\theta^{\\ell,i}$ can be convolution with stride 1, \n\tor fully connected operators. \n\tThen the CNN model will be defined by\n\t\\begin{equation}\\label{eq:cnndefine}\n\tH_0(f) = f^{L,\\nu_\\ell}.\n\t\\end{equation}\n\tIn these three classic CNN models, they still need some \n\textra fully connected layers in nonlinear mapping $H_0$ before the logistic regression as it contains \n\ta fully connected layer as in \\eqref{eq:log_reg}. \n\tThese fully connected layers are removed in ResNet to be described below.\n\\subsection{ResNet}\nThe ResNet \\cite{he2016deep} can be written as\n\t\\begin{equation}\\label{ori-ResNet}\n\t\\begin{cases}\n\tf^{1,0} &=R_{\\rm max}\\circ \\sigma \\circ \\theta^0(f), \\\\\n\tf^{\\ell,i} &= \\sigma \\left( f^{\\ell, i-1} + \\mathcal{F}^{\\ell, i} (f^{\\ell,i-1}) \\right), \\quad i = 1:\\nu_\\ell ~\\text{and}~ \\ell = 1:J ,\\\\\n\tf^{\\ell+1,0} &= \\sigma \\left( R_\\ell^{\\ell+1} (f^{\\ell, \\nu_\\ell} )+ \\mathcal{F}^{\\ell, 0} (f^{\\ell, \\nu_\\ell} ) \\right), \\quad \\ell = 1:J-1,\\\\\n\tH_0(f) &= R_{\\rm ave}(f^{L,\\nu_\\ell}). \\\\\n\t\\end{cases}\n\t\\end{equation}\n\tHere\n\t$$\n\t\\mathcal{F}^{\\ell,i} (f^{i-1}) = \\xi^{i} \\circ \\sigma \\circ \\eta^{i} (f^{i-1}).\n\t$$\n\tGenerally, $\\xi^{\\ell,i}$ and $\\eta^{\\ell,i}$ takes the form of \\label{eq:conv-1} with zero padding and stride 1,\n\texcept, $\\eta^{\\ell,0}$ is taken as convolution with stride 2 with the same output dimension of $R_\\ell^{\\ell+1}$.\n\t\n\\subsection{iResNet} \n\tThe iResNet\\cite{he2016identity} can be written as:\n\t\\begin{equation}\\label{eq:iResNet1}\n\t\\begin{cases}\n\tf^{1,0} &=R_{\\rm max}\\circ \\sigma \\circ \\theta^0(f), \\\\\n\tf^{\\ell,i} &= f^{\\ell, i-1} + \\mathcal{F}^{\\ell, i} (f^{\\ell,i-1}), \\quad i = 1:\\nu_\\ell ~\\text{and}~ \\ell = 1:J ,\\\\\n\tf^{\\ell+1,0} &= R_\\ell^{\\ell+1} (f^{\\ell, \\nu_\\ell} )+ \\mathcal{F}^{\\ell, 0} (f^{\\ell, \\nu_\\ell} ) , \\quad \\ell = 1:J-1,\\\\\n\tH_0(f) &= R_{\\rm ave}(f^{L,\\nu_\\ell}). \\\\\n\t\\end{cases}\n\t\\end{equation}\n\twhere\n\t$$\n\t\\mathcal{F}^{\\ell,i} (f^{\\ell,i -1}) = \\xi^{\\ell,i} \\circ \\sigma \\circ \\eta^{\\ell,i} \\sigma (f^{\\ell,i-1}).\n\t$$\n\tThe only difference between ResNet and iResNet can be viewed as \n\tputting a $\\sigma$ in different places. \n\tThe connection of those three models are often shown with next diagrams:\n\t\\begin{figure}[!htb]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=.6\\textwidth, height=.13\\textheight]{comparison-net} \n\t\t\\end{center}\n\t\t\\caption{Comparison of CNN Structures}\n\t\\end{figure}\n\t\n\nWithout loss of generality, we extract the key \nfeedforward steps on the same grid in different CNN models as follows.\n\\begin{description}\n\t\\item[Classic CNN] \n\t\\begin{equation}\\label{eq:cCNN}\n\tf^{\\ell,i} = \\xi^i \\circ \\sigma (f^{\\ell,i-1}) \\quad \\text{or} \\quad f^{\\ell,i} = \\sigma \\circ \\xi^{i} (f^{\\ell,i-1}) .\n\t\\end{equation}\n\t\\item[ResNet] \n\t\\begin{equation}\\label{eq:ResNet}\n\tf^{\\ell,i} = \\sigma( f^{\\ell,i-1} + \\xi^{\\ell,i} \\circ \\sigma \\circ \\eta^{\\ell,i}(f^{\\ell,i-1})).\n\t\\end{equation}\n\t\\item[iResNet]\n\t\\begin{equation}\\label{eq:iResNet}\n\tf^{\\ell,i} = f^{\\ell,i-1} + \\xi^{\\ell,i} \\circ \\sigma \\circ \\eta^{\\ell,i}\\circ \\sigma(f^{\\ell,i-1}).\n\t\\end{equation}\n\\end{description} \n\n", "meta": {"hexsha": "f3273461ddbc2271f3b37850f02ed5622348d1e5", "size": 3690, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/mgnet_classicalCNN.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/mgnet_classicalCNN.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": 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YES\n2. YES", "lm_q1_score": 0.8872045996818986, "lm_q2_score": 0.7879311956428946, "lm_q1q2_score": 0.699056181007234}} {"text": "% chapters/dp.tex\n\n\\chapter{Dynamic Programming}\t\\label{chapter:dp}\n\n\\input{algs/dp/binom-recursive}\n\n\\begin{figure}[h]\n \\includegraphics[width = 0.75\\textwidth]{figs/binom-4-2}\n \\caption{Calculate $\\binom{4}{2}$ recursively.}\n \\label{fig:binom-recursive}\n\\end{figure}\n\n\\begin{figure}[h]\n \\includegraphics[width = 0.65\\textwidth]{figs/pascal}\n \\caption{Pascal triangle for binomial coefficients.}\n \\label{fig:pascal}\n\\end{figure}\n\n\\input{algs/dp/binom-dp}\n\n\\[\n (n-k+1) + (k) + k (n-k) = nk - k^2 + n + 1\n\\]\n\n\\input{algs/dp/max-subarray-origin}\n\\input{algs/dp/max-subarray}\n", "meta": {"hexsha": "705059a688c88e627c88f57187ba67f09a913c1c", "size": 580, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/dp.tex", "max_stars_repo_name": "hengxin/algorithms-pseudocode", "max_stars_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-11-06T08:52:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-27T13:01:13.000Z", "max_issues_repo_path": "chapters/dp.tex", "max_issues_repo_name": "hengxin/algorithms-pseudocode", "max_issues_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/dp.tex", "max_forks_repo_name": "hengxin/algorithms-pseudocode", "max_forks_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.4814814815, "max_line_length": 58, "alphanum_fraction": 0.6879310345, "num_tokens": 216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396211, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.6990416046129629}} {"text": "%% Chapter 4 : Wind Energy Estimation\n\n\\section{Kinetic Energy of Air:}\n\\\n\\\n\\\n\\\nThe wind turbine converts kinetic energy of air into electrical energy. The kinetic energy possessed by the air is due to the flow of winds. The eq (\\ref{W1}) gives the kinetic energy of the air\n\n\\begin{equation}\n\\label{W1}\n\\text{K.E.(Joules)}=\\frac{1}{2}mV^2\n\\end{equation}\\\\\nwhere,\\\\\n$ m $ = Mass of Air $ (kg) $\\\\\n$ V $ = Velocity of air $ (m/s) $\\\\\n\n\n\\section{Power in Wind:}\n\\\n\\\n\\\n\\\nThe flow of the mass of air passing through the cross-section of the wind turbine blades is given by the eq (\\ref{W2})\n\n\\begin{equation}\n\\label{W2}\n\\text{mass flowrate} (kg/sec) = mass flow per sec = \\rho\\times A\\times V\n\\end{equation}\n\nThe volumetric flow rate of air is given by the eq (\\ref{W3})\n\n\\begin{equation}\n\\label{W3}\n\\text{volumetric flowrate} (m^3/sec) = A \\times V\n\\end{equation}\\\\\nwhere,\\\\\n$ \\rho $ = Air Density $ (kg/m^{3}) $\\\\\n$ A $ = Swept Area of Rotor $ (m^{2}) $\\\\\n\nThe mass flow rate of air through the cross-section of the wind turbine blades provides for the energy which is converted to electricity.\\\\\n\n\nThe eq (\\ref{W4}) gives us the relationship between power and energy.\n\n\\begin{equation}\n\\label{W4}\n\\text{Power} = \\frac{Energy}{Time}\n\\end{equation}\n\nThe power contained in wind (moving air) is given by eq (\\ref{W5}), which is the combination of eq (\\ref{W1},\\ref{W2},\\ref{W3},\\ref{W4}).\n\n\\begin{equation}\n\\label{W5}\n\\text{Power in moving air(Watts)} = \\left(\\frac{1}{2}\\right)\\times (mass flow per second)\\times V^2 \n\\end{equation}\n\nThe eq (\\ref{W6}) combines eq (\\ref{W5},\\ref{W3}) to give the mechanical power in wind in terms of air density, area and wind velocity; this is a more convenient equation for wind power computations.\n\n\\begin{equation}\n\\label{W6}\n\\text{Mechanical power in wind(Watts)} = \\left(\\frac{1}{2}\\right) \\times (\\rho \\times A ) \\times V^3\n\\end{equation}\n\nThe area in eq (\\ref{W6}) in context of wind turbine blades is the rotor-swept area (defined later in the text), but a more general equation of power in wind independent of the area variable is given in the eq (\\ref{W7}). It gives the power in wind per meter square of area.\n\n\\begin{equation}\n\\label{W7}\n\\text{Specific wind power of a site} (W/m^2) = \\left(\\frac{1}{2}\\right) \\times \\rho \\times V^3 \n\\end{equation}\\\\\n\n\n\n\\section{Power Extracted from Wind:}\n\\\n\\\n\\\n\\\nThe Fig (\\ref{figc5h1}) illustrates the different components of a typical Wind Turbine.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.75]{Wind11}\n\\caption{Wind Turbine [6]}\n\\label{figc5h1} %% to refer use, \\ref{}\n\\end{figure}\n\nAs the wind passes through the cross-section of the wind turbine blades, some portion of its kinetic energy is captured by the wind turbine blades and converted to electricity. In this process the wind loses some of its velocity due to loss of kinetic energy giving us two different wind velocities of upstream (wind velocity in front of the wind turbine) and downstream (wind velocity behind the wind turbine). Hence, the power extracted by the wind turbine from the wind is given by eq (\\ref{W8}).\n\n\\begin{equation}\n\\label{W8}\nP_0 = \\left(\\frac{1}{2}\\right) \\times \\text{(mass flow per sec)} \\times (V^2 - {V_0}^2)\n\\end{equation}\\\\\nwhere,\\\\\n$P_0$ = mechanical power extracted by the rotor (Watts)\\\\\n$ V $ = upstream wind velocity at the entrance of rotor\\\\\n$V_0$ = downstream wind velocity at the exit of rotor \\\\\n\nThe mass of the air passing through the rotor blades is given by the eq (\\ref{W9}).\n\n\\begin{equation}\n\\label{W9}\n\\text{Mass of air through the rotor blades} (kg/sec) = \\rho \\times A \\times \\left(\\frac{V + V_0}{2}\\right)\n\\end{equation}\n\nCombining the eq (\\ref{W8},\\ref{W9}) we get a detailed equation for the mechanical power extracted by the wind turbine rotor from the wind in terms of the eq (ref{W99}).\n\n\\begin{equation}\n\\label{W99}\nP_{0}= \\frac{1}{2}\\left[\\rho A \\frac{\\left(V+V_{0} \\right)}{2} \\right]\\left(V^{2}-V_{0}^{2} \\right)\n\\end{equation}\n\nOn further rearranging the eq (\\ref{W99}) we get the eq (\\ref{W88}).\n\n\\begin{equation}\n\\label{W88}\nP_{0} =\\frac{1}{2} \\rho A V^(3) \\frac{\\left(1 + \\frac{V_0}{V}\\right)\\times \\left[1 - \\left(\\frac{V_0}{V}\\right)^2\\right]}{2}\n\\end{equation}\\\\\n\nNow the power extracted by the wind turbine rotor blades is expressed as a fraction of upstream wind power in eq (\\ref{W10}).\n\n\\begin{equation}\n\\label{W10}\nP_0 = \\left(\\frac{1}{2}\\right) \\times \\rho \\times A \\times V^3 \\times C_p\n\\end{equation}\nwhere,\\\\\n$C_P$ = Wind Turbine Power Co-efficient\\\\\n\nThe wind Turbine Power Co-efficient is an important variable which depends on the mechanical design of the rotor blades and is given by the eq (\\ref{W11}).\n\n\\begin{equation}\n\\label{W11}\nC_p = \\frac{\\left(1 + \\frac{V_0}{V}\\right)\\times \\left[1 - \\left(\\frac{V_0}{V}\\right)^2\\right]}{2}\n\\end{equation}\n\nMaximum power extracted from the wind:\n\n\\begin{equation}\n\\label{W12}\nP_{max} = \\left(\\frac{1}{2}\\right) \\times \\rho \\times A \\times V^3 \\times 0.59\n\\end{equation}\\\\\n\nTheoretical max. value of $C_p = 0.59$ is called the Betz Limit.\n\nIf $C_p = 0.5$, maximum output power of the wind turbine is given by the eq (\\ref{W13}).\n\n\\begin{equation}\n\\label{W13}\n\\text{$P_{max} (Watts/m^2)$}= \\left(\\frac{1}{4}\\right) \\times \\rho \\times V^3 \n\\end{equation}\\\\\n\n\n\\section{Effect of Rotor-Swept Area on Wind Power}\n\\\n\\\n\\\n\\\nThe rotor-swept area is the cross-sectional area covered by the wind turbine rotor blades. For a horizontal axis wind turbine the rotor-swept area demarcates a circle, and is given by the eq (\\ref{W14}).\n\t\n\\begin{equation}\n\\label{W14}\n\t\\text{Rotor swept area}(m^2) = A_H = \\left(\\frac{\\Pi}{4}\\right) \\times D^2\n\\end{equation}\n\nFrom eq (\\ref{W10}) we observe that the amount of power extracted by the wind turbine rotor from the wind is directly proportional to the rotor-swept area. Hence, to increase power extraction capacity of a wind turbine the lengths of its rotor blades can be increased.\n\n\\section{Effect of Air Density on Wind Power}\n\\\n\\\n\\\n\\\nFrom eq (\\ref{W10}) we can see that the power extracted by a wind turbine is directly proportional to the air density. Hence, denser the air more will be the power extracted from the wind.\\\\\n\nHowever, the density of air depends on the on the pressure and temperature of the location as given by the ideal gas law in eq (\\ref{W16}).\n\n\\begin{equation}\n\\label{W16}\n\\rho = \\frac{P}{R \\times T}\n\\end{equation}\\\\\nwhere,\\\\\n$ P $ = Air pressure $ (atm) $\\\\\n$ T $ = Temperature $ (K) $\\\\\n$ R $ = Gas constant $ 8.205745 \\time 10^{-5} m^{3} atm K^{-1} mol^(-1) $\\\\\n\nThe air density at sea-level at $1$ atm $15 \\deg$Celsius is $1.2250 kg/m^3$. Using this as a reference $\\rho$ is corrected, for the site specific temperature and pressure.Hence, the density of air now is directly proportional to pressure and inversely proportional to temperature of the location.\\\\\n\nTemperature and pressure both vary with \"altitude\". Their combined effect is given by the eq (\\ref{W17}), which is valid upto $6000$m or $20000$ft of site elevation above sea-level. This equation is however approximate, and the detailed effect of both temperature and pressure on the air density is given in the subsequent sections of the text.\n\n\\begin{equation}\n\\label{W17}\n\\rho = \\rho_0 \\times e^{-{\\left(\\frac{0.297\\times H_m}{3048}\\right)}}\n\\end{equation}\\\\\nwhere,\\\\\n$ \\rho $ = Air Density at new height $(kg/m^3)$ \\\\ \n$ \\rho_{o} $ = Air Density at sea level $(kg/m^3)$ \\\\\n$ H_{m} $ = Altitude of the location $(m)$\\\\\n\n\n\\subsection{Correction for Pressure:}\n\\\n\\\n\\\n\\\n\nTo correct the air density at a given specific location, we first compute the air pressure at the location using its altitude in the eq (\\ref{W19}). this gives us the pressure at the specific location.\n\n\\begin{equation}\n\\label{W19}\nP = P_{o} \\times e^{(-1.185 \\times 10^{-4} \\times H)}\n\\end{equation}\\\\\nwhere,\\\\\n$ P $ = Pressure at new height $(atm)$ \\\\ \n$ P_{o} $ = Pressure at sea level $(atm)$ \\\\\n$ H_{m} $ = Altitude of the location $(m)$\\\\\n\n\n\\subsection{Correction for Temperature}\n\\\n\\\n\\\n\\\nAfter correcting the air pressure for altitude of the location, we use this pressure in the ideal gas law equation along with the temperature of the specific location as given in eq (\\ref{W20}).\n\n\\begin{equation}\n\\label{W20}\n\\rho = \\frac{P \\times M.W \\times 10^{-3}}{R \\times T}\\\\\n\\end{equation}\\\\\nwhere,\\\\\n$ M.W $ = Molecular weight of air $ $\\\\\n\nThe eq (\\ref{W20}) computes the air density at the specific location accurately as it takes into account both the correction in pressure and temperature.\n\n\n\\section{Effect of Hub-Height on Wind Power:}\n\\\n\\\n\\\n\\\n\nThe wind shear at a ground level surface causes the wind speed to increase with height in accordance with eq (\\ref{W36}).\n\n\\begin{equation}\n\\label{W36}\nV_2 = V_1 \\times \\left(\\frac{h_2}{h_1}\\right)^\\alpha\n\\end{equation}\\\\\nwhere,\\\\\n$V_1 = \\text{wind speed measured at the reference height} \\,h_1$\\\\\n$V_2 = \\text{wind speed estimated at height} \\,h_2$\\\\\n$\\alpha = \\text{ground surface friction coefficient}$\\\\\n\nThe Table (\\ref{c5Tab1}) gives different values of $\\alpha$ for different ground surface roughness class.\n\n\\begin{table}[H]\n \\centering\n \\caption{Ground Surface Friction Co-Efficient Claassification}\n \\begin{tabular}{|l|c|}\n \\hline\n \\textbf{Terrain Characteristics} & \\multicolumn{1}{l|}{\\textbf{Friction Coefficient (α)}} \\bigstrut\\\\\n \\hline\n \\textbf{Smooth hard ground, calm water } & 0.1 \\bigstrut\\\\\n \\hline\n \\textbf{Tall grass on level ground } & 0.15 \\bigstrut\\\\\n \\hline\n \\textbf{High crops, hedges and shrubs } & 0.2 \\bigstrut\\\\\n \\hline\n \\textbf{Wooded countryside, many trees } & 0.25 \\bigstrut\\\\\n \\hline\n \\textbf{Small town with trees and shrubs } & 0.3 \\bigstrut\\\\\n \\hline\n \\textbf{Large city with tall buildings } & 0.4 \\bigstrut\\\\\n \\hline\n \\end{tabular}%\n \\label{c5Tab1}%\n\\end{table}%\n\nFrom the eq (\\ref{W36}) we can observe that the velocity of wind increases with height. Moreover, as the power extracted from the wind by the wind turbine is directly proportional to the cube of the wind velocity, increasing the hub height has a huge improvement in the power extraction capacity of a wind turbine.\\\\\n\t \n\\section{Wake Effect Models:}\n\\\n\\\n\\\n\\\nThe wake effect is one of the biggest factors for reduction in power capture capacity of wind turbines in a wind farm. The wake effect has two main impacts on the wind farm; firstly there is reduction of wind speed as we move towards the down stream turbines, sometimes causing the downstream turbines to completely stop this causes a considerable amount of loss in power capturing capacity of the wind farm and secondly the the downstream wind is more turbulent causing structural fatigue and eventually lower lifespan of wind turbines. Modeling of Wake Effect helps in the computation of downstream wind velocity deficit (helps in calculating the reduced power capture) and the turbulence patterns so that the wind turbines in a wind farm are arranged spatially to optimize the the power capture. The Fig (\\ref{figc5h10}) shows a wake effect simulation in a wind farm.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{WakeEffect1}\n\\caption{Wake effect Simulation}\n\\label{figc5h10} %% to refer use, \\ref{}\n\\end{figure}\n\n\\subsection{Jensen's Model:}\n\\\n\\\n\\\n\\\n\nThe analytical wake model developed by Jensen is simple and requires less inputs as compared other complex wake models. It is used in mostly is optimizing position of the wind turbines within a wind farm. It is based on the global momentum conservation and on the assumption that the wake has a linearly expanding diameter.\\\\\n\nThe Fig (\\ref{figc5h11}) illustrates the principle of Jensen's wake model.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{WakeEffect2}\n\\caption{Jensen - Wake Model Principle [56]}\n\\label{figc5h11} %% to refer use, \\ref{}\n\\end{figure}\n\nThe eq (\\ref(W44),\\ref(W45),\\ref(W47),\\ref(W48)) describe the Jensen wake model.\n\n\\begin{equation}\n\\label{W44}\n\tD_{wake} = D(1 + 2ks)\\\\\n\\end{equation}\\\\\n \n\\begin{equation}\n\\label{W45}\n u_{def} = U_{\\infty}\\left[\\frac{1-\\sqrt{1-C_T}}{(1 + 2ks)^2}\\right]\\\\\n\\end{equation}\\\\ \n \n\\begin{equation}\n\\label{W46}\n u = U_{\\infty}\\left[1 - \\frac{1-\\sqrt{1-C_T}}{(1 + 2ks)^2}\\right]\\\\\n\\end{equation}\\\\\n \n\\begin{equation}\n\\label{W47}\n s = x/D\\\\\n\\end{equation}\\\\\n where,\\\\\n $ D_{wake} $ = Wake Diameter $ (m) $ \\\\\n $ D $ = Rotor Diameter $ (m) $ \\\\\n $ k $ = Wake Decay Constant (onshore=0.075 and offshore=0.05) \\\\\n $ s $ = Normalized downstream distance from turbine $ (m) $ \\\\\n $ x $ = Axial Distance between Turbines $ (m) $ \\\\ \n $ u_{def} $ = Velocity deficit in fully developed Wake $ (m/s) $ \\\\\n $ U_{\\infty} $ = Wind Speed $ (m/s) $ \\\\\n $ C_{T} $ = Induction Factor of Turbine \\\\\n $ u $ = Wind Speed after Wake effect $ (m/s) $ \\\\\n \n \t \t \n\\subsection{Frandsen's Model:}\n\\\n\\\n\\\n\\\nThe analytical wake model developed by Frandsen has been adopted in the Storpark Analytical Model (SAM). This model predicts the wind speed deficit in large offshore wind farms using a rectangular site area and straight rows of wind turbines with equidistant spacing between the wind turbines and rows. This model is more accurate than the Jensen model, but requires more input variables.\\\\\n\nThe eq (\\ref(W48),\\ref(W49),\\ref(W50),\\ref(W51)) describe the Frandsen wake model.\n\n\\begin{equation}\n\\label{W48}\n\tD_{wake} = D(\\beta ^{(k/2)} + \\alpha s)^{\\frac{1}{k}}\\\\\n\\end{equation}\\\\\n\t\n\\begin{equation}\n\\label{W49}\n\t\\beta = \\frac{1 + \\sqrt{1-C_T}}{2 \\sqrt{1-C_T}}\\\\\n\\end{equation}\\\\\n\t\n\\begin{equation}\n\\label{W50}\n\tu_{def} = \\frac{U_{\\infty}}{2} \\left(1 \\pm \\sqrt{1 - 2 \\frac{A}{A_{wake}} C_T}\\right)\\\\\n\\end{equation}\\\\\t\n \n\\begin{equation}\n\\label{W51}\n s = x/D\\\\\n\\end{equation}\\\\\n where,\\\\\n $ D_{wake} $ = Wake Diameter $ (m) $ \\\\\n $ D $ = Rotor Diameter $ (m) $ \\\\\n $ k $ = Wake Decay Constant $ k=2 \\text{, for Schlichting solution to the wake expansion; } k=3 \\text{, for square root shape chosen } $ \\\\\n $ \\alpha $ = Initial Wake expansion \\\\\n $ s $ = Normalized downstream distance from turbine $ (m) $ \\\\\n $ x $ = Axial Distance between Turbines $ (m) $ \\\\ \n $ u_{def} $ = Velocity deficit in fully developed Wake $ (m/s) $ \\\\\n $ U_{\\infty} $ = Wind Speed $ (m/s) $ \\\\\n $ C_{T} $ = Induction Factor of Turbine \\\\\n $ A $ = Swept area of rotor $ (m^{2}) $ \\\\ \n $ A_{wake} $ = Swept area of Wake $ (m^{2}) $ \\\\\n \n\\newpage \n \n\\section{Concept of TSR:}\n\\\n\\\n\\\n\\\nTip Speed Ratio (TSR) is defined as the ratio of the angular speed of the wind turbine rotor blade to the wind velocity, it is given by the eq (\\ref{W38}).\n\n\\begin{equation}\n\\label{W38}\nTSR = \\frac{\\text{Linear speed of the blade's outermost tip}}{\\text{Free upstream wind velocity}}\n\\end{equation}\\\\\n\nThe TSR in terms of rotor radius and angular speed in rads/s is given by the eq (\\ref{W39}).\n\n\\begin{equation}\n\\label{W39}\nTSR = \\frac{\\omega \\times R}{V}\n\\end{equation}\\\\\nwhere,\\\\\n$ R $ = rotor radius $ (m) $\\\\\n$\\omega$ = angular speed of the rotor $(rads/s)$\\\\\n\nThe TSR is an important quantity in determining the amount of power extracted from wind by the wind turbine, as the Wind Turbine Power Co-efficient $(C_{P})$ is a function of the TSR. The relationship between TSR and $C_{P}$ is dealt with in detail in the next section.\\\\ \n\n\n\\subsection{Power Coefficient Analysis:}\n\\\n\\\n\\\n\\\nThe power extracted from the wind by the wind turbine is directly proportional to the the Wind Turbine Power Co-efficient $(C_{P})$ as seen in the eq (\\ref{W10}). $C_{P}$ is experimentally measured by the Wind Turbine Manufactures and its graph is provided in the data-sheet.\\\\\n \n$C_{P}$ describes the aerodynamic behaviour of the wind turbine rotor blades. It is a highly non-linear function of TSR and the blade pitch angle $(\\theta)$. For a given $\\theta$ the $C_{P}$ is plotted for a range of TSR values. The relationship between $C_{P}$, TSR and $\\theta$ differs from turbine to turbine; however, its knowledge helps in designing maximum power tracking controllers which try to maintain $C_{P}$ at its maximum value under the given wind speed condition by adjusting the blade pitch angle (Pitch Control) and/or the gear ratio (Speed Control).\\\\\n\nThe eq (\\ref{W41},\\ref{W42}) gives a detailed analytical model for the relationship between $C_{P}$, TSR and $\\theta$, it is developed by (Anderson & Bose, 1983).\n\n\\begin{equation}\n\\label{W41}\nC_p(\\lambda , \\theta) = C_1\\times(C_2 \\frac{1}{\\beta} - C_3 \\beta \\theta - C_4 \\theta ^x - C_5)e^{-C_6 \\frac{1}{\\beta}}\\\\\n\\end{equation}\\\\\n\n\\begin{equation}\n\\label{W42}\n\\frac{1}{\\beta} = \\frac{1}{\\lambda + 0.08\\theta} - \\frac{0.035}{1 + \\theta ^3}\\\\\n\\end{equation}\\\\\nwhere,\\\\\n$ C_{P} $ = Wind Turbine Power Coefficient \\\\\n$ C_{1},C_{2},C_{3},C_{4},C_{5},C_{6} \\text{ and }x_ $ = Turbine rotor design dependent constants \\\\\n$ \\lambda $ = Tip speed ratio \\\\\n\nThe eq (\\ref{W43}) describes another simpler analytical model for C_{P}, which does not take into account physical design differences between the wind turbine rotor blades; it is also developed by (Anderson & Bose, 1983).\n\n\\begin{equation}\n\\label{W43}\n\tC_p = \\frac{1}{2}(\\lambda - 0.022\\theta ^2 - 5.6)e^{-0.17\\lambda}\\\\\n\\end{equation}\\\\\t\nwhere,\\\\ \t\n$\\lambda = \\frac{V_w \\text{ (mph) }}{\\omega _{b} (rads_-1) }$\\\\\nwhere,\\\\\n$ \\lambda $ = Tip Speed Ratio \\\\\n$ V_w $ = Wind speed $ (mph) $ \\\\\n$ \\omega_b $ = Angular speed of turbine rotor $ (rads/s) $ \\\\ \n \n\\section{Weibull Wind Speed Distribution:}\n\\\n\\\n\\\n\\\nThe variation in wind speed distribution is best described by the Weibull probability distribution function 'h' with two parameters:\n\n\\begin{enumerate}\n\\item \\blindtext The shape parameter 'k'\n\\item \\blindtext The scale parameter 'c'\n\\end{enumerate}\n\n\nThe probability of wind speed being $\\upsilon$ during any time interval is given by eq (\\ref{W21})\n\n\\begin{equation}\n\\label{W21}\nh(\\upsilon) = \\left(\\frac{k}{c}\\right) \\times \\left(\\frac{\\upsilon}{c}\\right)^{k-1} \\times \\exp{\\left(\\frac{\\upsilon}{c}\\right)}^k \\text{, for $0< \\upsilon < \\infty$}\n\\end{equation}\\\\\n\nThe eq (\\ref{W21}) tells us the probability of occurrence of a particular wind speed. As it is a probability distribution, hence its summation from infinity to infinity is 1. Therefore, we can multiply it with time parameter to get probability of wind speed in terms of hours in a year (if multiplied by 8760) or any other time parameter which fits best for the wind speed being described by the probability distribution.\\\\\n\nWeibull probability distribution changes its shape according to the shape parameter $k$ as follows:\n\n\\begin{enumerate}\n\\item \\blindtext \t$k=1$ makes it the exponential distribution, \\,$h = \\lambda \\times e^{-(\\lambda\\upsilon)},\\, where \\, \\lambda = \\frac{1}\t\t\t{c}$\n\\item \\blindtext $k=2$ makes it the Rayleigh distribution, \\,$h = 2 \\times \\lambda^2 \\times \\upsilon \\times e^{-(\\lambda\\upsilon)^2}$\n\\item \\blindtext $k>3$ makes it approach a normal bell-shaped distribution\n\\end{enumerate}\n\nThe eq (\\ref{W25}) gives the relationship between the shape parameter (k), scale parameter (c) and the mean wind speed for a give period of time for which the wind data was gathered. Knowing $V_{mean}$ and adjusting the shape parameter (k) so as to come close to the actual wind speed distribution the scale parameter (c) can be computed.\n\n\\begin{equation}\n\\label{W25}\nV_{mean} = c \\times \\sqrt{\\left(1 + \\frac{1}{k}\\right)}\n\\end{equation}\\\\\n\n\nRoot mean cube wind speed is defined in similar manner as $V_(rms)$ in AC electrical circuits, it is given by eq (\\ref{W28}).\n\n\\begin{equation}\n\\label{W28}\n\tV_(rmc) = 3 \\times \\sqrt{\\frac{1}{8760}\\int\\limits_0^\\infty h\\upsilon^3 \\, d\\upsilon}\n\\end{equation}\\\\\n\nUsing $V_{rmc}$ the 'annual average power generation' can be computed as given in eq (\\ref{W29}).\n\n\\begin{equation}\n\\label{W29}\n\tP_{rmc} = \\frac{1}{4}\\times \\rho \\times {V_{rmc}}^3\n\\end{equation}\\\\\n\nFurthermore, using eq (\\ref{W30}) we can compute the average annual energy production of the specific site.\n\t\n\\begin{equation}\n\\label{W30}\n\t\\text{Average annual energy production of the site} = P_{rmc} \\times \\text{total number of hours per year}\n\\end{equation}\\\\\n\nSo, using the concepts of Weibull Distribution and Root Mean Cube Wind Speed, we can estimate the potential of a site for setting up a wind farm with minimum amount of data (just annual mean wind speed can suffice for performing the analysis).\\\\\n\n\\newpage \n\t\t\n\\section{Power Flow through Wind Turbine Grid Connected System}\n\\\n\\\n\\\n\\\nThe Fig (\\ref{Windloss}) illustrates the power flow through a Wind Turbine grid connected system and the various losses incurred in the this system.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.75]{WindLoss1}\n\\caption{Flow Diagram - Power Flow and Power Losses in a Wind Turbine Grid Connected System}\n\\label{WindLoss} %% to refer use, \\ref{}\n\\end{figure}\n\n\\subsection{Wind Speed Correction}\n\\\n\\\n\\\n\\\nThe wind speed measured in a Wind Turbine Power Plant by a weather station at ground level. In order to know the exact wind speed at the height of the wind turbine rotors, correction has to be implemented as directed by the eq (\\ref{}).\n\nMoreover, wake effect has to be taken into account to incorporate the downstream wind velocity deficit as explained in the section (5.7), either using Jensen or Frandsen wake effect models.\n\n\\subsection{Power Curve Correction}\n\\\n\\\n\\\n\\ \nAs discussed in the section (5.5), the air density has a direct proportion to the power extracted by the wind turbine from the wind. The Power vs Wind Speed curves provided by the manufacturers are experimentally measured at sea level at a controlled temperature of $15^{\\circ}C$. However the the altitude and the temperature at the specific wind farm location is different, hence power curve correction based on the air density correction has to performed for more accurate power estimation from the wind turbines.\n\n\\subsection{Ohmic and Transformer Losses}\n\\\n\\\n\\\n\\\t\nAC cabling is done to connect wind turbines to the transformer. All these cables add to the resistance of the Wind Turbine system causing huge ohmic losses. Hence, in order to reduce these ohmic losses, the length of the cabling should be minimum and the cable material should be of low resistivity.\\\\\n\nTransformers are another indispensible part of the grid connected Wind Turbine systems. They step-up the voltage of the Wind Turbine system output, so that the power can be transferred to the transmission lines operating at higher voltages. But, transformers consist of copper coils which result in resistive losses, and iron cores which cause iron losses. Hence, the power output to the grid from the the PV system is reduced.\\\\ \n\n\\newpage\n \n\\section{MATLAB Model Algorithm} \n\nThe Fig (\\ref{figc5h4} ) depicts an simplified flowchart of the algorithm which is developed for the creation of the Wind Energy Estimation Application. \t \t \t\n \n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{WindScheme}\n\\caption{Wind Turbine Energy Estimation Schematic}\n\\label{figc5h4} %% to refer use, \\ref{}\n\\end{figure}\n\nBased on the theory discussed in previous sections of this chapter and the algorithm developed as illustrated in Fig (\\ref{WindLoss1},\\ref{figc5h4}); a GUI based application for energy estimation of grid connected wind turbine power plants is developed in MATLAB, the results of which are presented in the next section and the application GUIs can be found in the Appendix.\n\n\\newpage\n\n\\section{Results}\n\\\n\\\n\\\n\\\nThe Wind Energy Estimation App has been tested with a hypothetical wind turbine power plant, and the Weibull Distribution sub-module has been tested with random site information. The results obtained are doumented in the following sections.\n\n\\subsection{Wind Turbine Power Plant Site Potential Estimation using Weibull Distribution App}\n\\\n\\\n\\\n\\\nThe Fig (\\ref{WinResImg1}) shows the site potential estimation performed by the Weibull Distribution sub-module for monthly data.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{Weibull_Monthly}\n\\caption{Weibull Distribution-Monthly Simulation }\n\\label{WinResImg1} %% to refer use, \\ref{}\n\\end{figure}\n\nThe Fig (\\ref{WinResImg2}) shows the site potential estimation performed by the Weibull Distribution sub-module for yearly data.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{Weibull_Yearly}\n\\caption{ Weibull Distribution-Yearly Simulation }\n\\label{WinResImg2} %% to refer use, \\ref{}\n\\end{figure}\n\n\\subsection{Cp Curve Generator App}\n\\\n\\\n\\\n\\\nThe Table (\\ref{CpCurveTab1}) gives the rotor information used to generate C_{P} curves using equation (\\ref{W41}).\n\n\\begin{table}[htbp]\n \\centering\n \\caption{Cp Curve Generator App - Equation 1 Simulation Parameters}\n \\begin{tabular}{|l|l|}\n \\hline\n \\multicolumn{2}{|c|}{\\textbf{Cp Curve Equation 1}} \\bigstrut\\\\\n \\hline\n \\multicolumn{1}{|c|}{\\textbf{Parameters }} & \\multicolumn{1}{c|}{\\textbf{Value}} \\bigstrut\\\\\n \\hline\n \\multicolumn{1}{|c|}{Number of Rotor Types} & 3 \\bigstrut\\\\\n \\hline\n c1 & 0.5,0.7,0.3 \\bigstrut\\\\\n \\hline\n c2 & 116,150,85 \\bigstrut\\\\\n \\hline\n c3 & 0.4,0.6,0.2 \\bigstrut\\\\\n \\hline\n c4 & 0,0,0 \\bigstrut\\\\\n \\hline\n c5 & 5,10,3 \\bigstrut\\\\\n \\hline\n c6 & 21,31,11 \\bigstrut\\\\\n \\hline\n x & 0,0,0 \\bigstrut\\\\\n \\hline\n Theta & 1,2,3,4 \\bigstrut\\\\\n \\hline\n \\end{tabular}%\n \\label{CpCurveTab1}%\n\\end{table}%\n\nThe Fig (\\ref{WinResImg3},\\ref{WinResImg4},\\ref{WinResImg5}) shows the C_{P} curve graphs obtained from the Cp Curve Generator App for different values of the blade pitch angles.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{CpCurve_Eq1_1}\n\\caption{CpCurve Generator App-Equation 1 Rotor 1 Simulation}\n\\label{WinResImg3} %% to refer use, \\ref{}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{CpCurve_Eq1_2}\n\\caption{CpCurve Generator App-Equation 1 Rotor 2 Simulation}\n\\label{WinResImg4} %% to refer use, \\ref{}\n\\end{figure}\n\\begin{figure}[H]\n\n\\centering\n\\includegraphics[scale=0.5]{CpCurve_Eq1_3}\n\\caption{CpCurve Generator App-Equation 1 Rotor 3 Simulation}\n\\label{WinResImg5} %% to refer use, \\ref{}\n\\end{figure}\n\nThe Table (\\ref{CpCurveTab2}) gives the rotor information used to generate C_{P} curves using equation (\\ref{W43}).\n\n\\begin{table}[htbp]\n \\centering\n \\caption{Cp Curve Generator App - Equation 2 Simulation Parameters}\n \\begin{tabular}{|l|l|}\n \\hline\n \\multicolumn{2}{|c|}{\\textbf{Cp Curve Equation 2}} \\bigstrut\\\\\n \\hline\n \\multicolumn{1}{|c|}{\\textbf{Parameters }} & \\multicolumn{1}{c|}{\\textbf{Value}} \\bigstrut\\\\\n \\hline\n Number of Rotor Types & \\bigstrut\\\\\n \\hline\n c1 & 0.5 \\bigstrut\\\\\n \\hline\n c2 & 116 \\bigstrut\\\\\n \\hline\n c3 & 0.4 \\bigstrut\\\\\n \\hline\n c4 & 0 \\bigstrut\\\\\n \\hline\n c5 & 5 \\bigstrut\\\\\n \\hline\n c6 & 21 \\bigstrut\\\\\n \\hline\n x & 0 \\bigstrut\\\\\n \\hline\n Theta & 0,5,10,15,20 \\bigstrut\\\\\n \\hline\n \\end{tabular}%\n \\label{CpCurveTab2}%\n\\end{table}%\n\nThe Fig (\\ref{WinResImg6}) shows the C_{P} curve graphs obtained from the Cp Curve Generator App for different values of the blade pitch angles.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{CpCurve_Eq2_1}\n\\caption{CpCurve Generator App-Equation 2 Simulation }\n\\label{WinResImg6} %% to refer use, \\ref{}\n\\end{figure}\n\n\n\\subsection{Hypothetical Wind Turbine Plant Information}\n\\\n\\\n\\\n\\\nThe Table (\\ref{DataTab1}) gives the information of the hypothetical wind power plant, which was fed into the Data Acquisition App and the Wind Energy Estimation App for testing purposes due lack of real WTPP data.\n\n\\begin{table}[H]\n \\centering\n \\captionHypothetical Wind Turbine Power Plant Information Table}\n \\begin{tabular}{|l|l|l|l|l|}\n \\hline\n \\multicolumn{5}{|c|}{\\textbf{HYPOTHETICAL WIND TURBINE POWER PLANT}} \\bigstrut\\\\\n \\hline\n \\multicolumn{5}{|c|}{\\textbf{SITE INFORMATION}} \\bigstrut\\\\\n \\hline\n \\textbf{LATITUDE} & \\multicolumn{4}{l|}{23.275} \\bigstrut\\\\\n \\hline\n \\textbf{LONGITUDE} & \\multicolumn{4}{l|}{72.682} \\bigstrut\\\\\n \\hline\n \\textbf{ALTITUDE (m)} & \\multicolumn{4}{l|}{100} \\bigstrut\\\\\n \\hline\n \\textbf{PLANT CAPACITY (MW)} & \\multicolumn{4}{l|}{138} \\bigstrut\\\\\n \\hline\n \\multicolumn{5}{|c|}{\\textbf{WIND TURBINE PLANT INFORMATION}} \\bigstrut\\\\\n \\hline\n \\textbf{PARAMETERS} & \\multicolumn{4}{c|}{\\textbf{WIND GENERATOR TYPES}} \\bigstrut\\\\\n \\hline\n & \\textbf{Type1} & \\textbf{Type2} & \\textbf{Type3} & \\textbf{Type4} \\bigstrut\\\\\n \\hline\n \\textbf{No. of Sub-Models} & 1 & 2 & 3 & 4 \\bigstrut\\\\\n \\hline\n \\textbf{No. of Turbines} & 5 & 5,10 & 5,10,15 & 5,10,15,20 \\bigstrut\\\\\n \\hline\n \\textbf{Cut-In Wind Speed} & 4 & 4,4 & 4,4,4 & 4,4,4,3.5 \\bigstrut\\\\\n \\hline\n \\textbf{Cut-Out Wind Speed} & 24 & 24,19 & 24,19,25 & 24,19,25,23 \\bigstrut\\\\\n \\hline\n \\textbf{Rotor Radius} & 38.5 & 38.5,19.5 & 38.5,19.5,35.5 & 38.5,19.5,35.5,38.5 \\bigstrut\\\\\n \\hline\n \\textbf{Hub-Height} & 80 & 80,55 & 80,55,73 & 80,50,73,80 \\bigstrut\\\\\n \\hline\n \\end{tabular}%\n \\label{DataTab1}%\n\\end{table}%\n\n\n\n\\subsection{Wind Turbine Power Curve GUI}\n\\\n\\\n\\\n\\\nThe Fig (\\ref{WinResImg7},\\ref{WinResImg8},\\ref{WinResImg9},\\ref{WinResImg10}) illustrates the Power (kW) vs Wind Speed (m/s) graphs of the four different wind turbines used in the simulation as observed in the Wind Power Curves sub-module.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{NegMicon}\n\\caption{Power vs Wind Speed Curve-Neg Micon 1.5MW}\n\\label{WinResImg7} %% to refer use, \\ref{}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{Vestas}\n\\caption{Power vs Wind Speed Curve-Vestas 0.6MW}\n\\label{WinResImg8} %% to refer use, \\ref{}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{Enercon}\n\\caption{Power vs Wind Speed Curve-Enercon 2MW }\n\\label{WinResImg9} %% to refer use, \\ref{}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{GE}\n\\caption{Power vs Wind Speed Curve-GE 1.5MW}\n\\label{WinResImg10} %% to refer use, \\ref{}\n\\end{figure}\n\n\n\\subsection{Wind Energy Estimation App Output}\n\\\n\\\n\\\n\\\nThe Fig (\\ref{WinResImg11}) shows the monthly energy output to the grid obtained from the four different types of wind turbines present in the type four category of the hypothetical wind power plant as computed in the Wind Energy Estimation App. The wind speed data was randomly generated and the temperature data was taken from Meteonorm, both at the temporal resolution of 60 minutes.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{WindGraph1}\n\\caption{Hypothetical Wind Plant Month Wise Energy Estimation For Different Wind Generator Types}\n\\label{WinResImg11} %% to refer use, \\ref{}\n\\end{figure}\n\nWe can clearly see that the energy output of the Enercon and GE compared to Neg Micon and Vestas is higher, as the number of these turbines is higher.\n\n\n\n\n", "meta": {"hexsha": "f1542d496725e9801834d9e1b9191bd7a351d450", "size": 30851, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "8_ProjectReport/2_ProjectProposal/Latex Support Files/Chapters/Ch5.tex", "max_stars_repo_name": "ninadkgaikwad/ARMA_TimeSeries_Forecasting_Project", "max_stars_repo_head_hexsha": "18329e436f823d55d2aad02b1d81d8cdda506ab2", 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YES\n2. YES", "lm_q1_score": 0.875787001374006, "lm_q2_score": 0.798186775339273, "lm_q1q2_score": 0.6990416025107693}} {"text": "%---------------------------Edge Ratio----------------------------------\n\\section{Edge Ratio}\n\nThe edge ratio of a tetrahedron is: \n\\[\n\\frac{L_{\\max}}{L_{\\min}}.\n\\]\n\n\\tetmetrictable{edge ratio}%\n{$1$}% Dimension\n{$[1,3]$}% Acceptable range\n{$[1,DBL\\_MAX]$}% Normal range\n{$[1,DBL\\_MAX]$}% Full range\n{$1$}% Equilateral tet\n{--}% Citation\n{v\\_tet\\_edge\\_ratio}% Verdict function name\n\n", "meta": {"hexsha": "c809884dc7404e91f33590d3650517baf0f0581a", "size": 462, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetEdgeRatio.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetEdgeRatio.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetEdgeRatio.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 25.6666666667, "max_line_length": 72, "alphanum_fraction": 0.4567099567, "num_tokens": 130, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.875787001374006, "lm_q2_score": 0.7981867729389246, "lm_q1q2_score": 0.6990416004085753}} {"text": "\\lab{Simplex Method}{Simplex Method}\n\\objective{Implement the Simplex Algorithm to solve linear constrained optimization problems.}\n\\label{lab:Simplex}\n\nThe Simplex Algorithm numbers among the most important algorithms invented within the last 100 years.\nIt provides a straightforward method for finding optimal solutions to linear constrained optimization problems.\nThe algorithm obtains the solution by traversing the edges of the feasible region defined by the constraints.\nThe theory of convex optimization guarantees that the optimal point will be found among the vertices of the feasible\nregion, and so a carefully implemented Simplex Algorithm will discover the exact solution in a finite number of steps.\n\n\\section*{Standard Form}\nThe Simplex Algorithm accepts a linear constrained optimization problem, also known as a \\emph{linear program},\nin the form given below:\n\n\\begin{align*}\n\\text{maximize}\\qquad &c^Tx \\\\\n\\text{subject to}\\qquad A&x \\leq b \\\\\n &x \\geq 0\n\\end{align*}\nNote that any linear program can be converted to standard form, so there is no loss of\ngenerality in restricting our attention to this particular formulation.\n\nSuch an optimization problem defines a region in space called the \\emph{feasible region}, the set of points\nsatisfying the constraints. Because the constraints are all linear, the feasible region forms a geometric object\ncalled a \\emph{polytope}, having flat faces and edges (see Figure \\ref{fig:polytope}).\nThe Simplex Algorithm jumps among the vertices of the feasible region searching for an optimal point.\nIt does this by moving along the edges of the feasible region in such a way that the objective function\nis always increased after each move.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{feasiblePolytope.pdf}\n\\caption{The feasible region for a linear program. The optimal point\nis one of the vertices of the polytope.}\n\\label{fig:polytope}\n\\end{figure}\n\nImplementing the Simplex Algorithm is straightforward, provided one carefully follows the procedure.\nWe will break the algorithm into several small steps, and write a function to perform each one.\nTo become familiar with the execution of the Simplex algorithm, it is helpful to work several examples by hand.\n\n\\section*{The Simplex Solver}\nOur program will be more lengthy than many other lab exercises and will consist of a collection of functions working\ntogether to produce a final result.\nIt is important to clearly define the task of each function and how all the functions will work together.\nIf this program is written haphazardly, it will be much longer and more difficult to read than it needs to be.\nWe will walk you through the steps of implementing the Simplex Algorithm as a Python class.\n%Since the Simplex Algorithm assumes that all the variables are non-negative, we do not need any special logic for it.\n%what is this previous statement saying? What 'special logic' would you need if the variables were negative?\n\nFor demonstration purposes, we will use the following linear program.\n\\begin{align*}\n\\text{maximize}\\qquad & 3x_0 + 2x_1 \\\\\n\\text{subject to}\\qquad\n& x_0 - x_1 \\leq 2 \\\\\n& 3x_0 + x_1 \\leq 5 \\\\\n& 4x_0 + 3x_1 \\leq 7 \\\\\n& x_0, x_1 \\geq 0.\n\\end{align*}\n\n\\subsection*{Accepting a Linear Program}\nOur first task is to determine if we can even use the Simplex algorithm.\nAssuming that the problem is presented to us in standard form, we need\nto check that the feasible region is nonempty.\n\n% Problem 1: Check Feasibility\n\\begin{problem}\nWrite a class that accepts the arrays \\li{c}, \\li{A}, and \\li{b} of a linear optimization problem in standard form.\nIn the constructor, check that the system is feasible at the origin\\footnote{For simplicity, we only check feasibility at the origin. A more robust solver sets up the auxiliary problem and solves it to find a starting point if the origin is infeasible.}.\nThat is, check that $Ax \\leq b$ componentwise when $x = 0$. % TODO: bell <=\nRaise a \\li{ValueError} if the problem is not feasible at the origin.\n\\label{prob:initsolver}\n\\end{problem}\n\n\\subsection*{Adding Slack Variables}\nOur next step is to convert the inequality constraints $Ax \\leq b$ into equality constraints\nby introducing one slack variable for each constraint.\nIf the constraint matrix $A$ is an $m \\times n$ matrix, then there are $m$ slack variables,\none for each row of $A$.\nGrouping all of the slack variables into a vector of length $m$, denoted $z$, our\nconstraints now take the form $Ax + z = b$.\nIn our example, this gives \n\\[\n\\z = \\begin{bmatrix}\n x_2 \\\\\n x_3 \\\\\n x_4\n\\end{bmatrix},\n\\]\n\nWhen adding slack variables, it is useful to represent all of your variables, both the original primal variables and\nthe additional slack variables, in a convenient manner.\nOne effective way is to refer to a variable by its subscript.\nFor example, we can use the integers $0$ through $n-1$ to refer to the original (non-slack) variables $x_0$ through\n$x_{n-1}$, and we can use the integers $n$ through $n+m-1$ to track the slack variables (where the slack variable\ncorresponding to the $i$-th row of the constraint matrix is represented by the index $n+i-1$).\n\nWe also need some way to track which variables are basic (non-zero) and which variables are nonbasic (those that have value $0$).\nA useful representation for the variables is a Python list (or NumPy array), where the elements of the list are integers.\nSince we know how many basic variables we have ($m$, to be precise), we can partition the list so that all the basic\nvariables are kept in the first $m$ locations, and all the non-basic variables are stored at the end of the list.\nThe ordering of this list is important. In particular, if $i \\leq m$, the $i$-th element of the list represents\nthe basic variable corresponding to the $i$-th row of $A$. Henceforth we will refer to this list as the \\emph{index list}.\n\nInitially, the basic variables are simply the slack variables, and their values correspond to the values of the matrix $b$.\nIn our example, we have 2 primal variables $x_0$ and $x_1$, and we must add 3 slack variables.\nThus, we instantiate the following index list:\n\\begin{lstlisting}\n>>> L = [2, 3, 4, 0, 1]\n\\end{lstlisting}\nNotice how the first $3$ entries of the index list are $2, 3, 4$, the indices representing the slack variables.\nThis reflects the fact that the basic variables at this point are exactly the slack variables.\n\nAs the Simplex Algorithm progresses, however, the basic variables change, and it will be necessary to swap\nelements in our index list. Suppose the variable represented by the index $4$ becomes nonbasic, while\nthe variable represented by index $0$ becomes basic. We must swap these two entries in the index list.\nThis can be done in a single, efficient line of Python code:\n\\begin{lstlisting}\n>>> L[2], L[3] = L[3], L[2]\n>>> L\n[2, 3, 0, 4, 1]\n\\end{lstlisting}\nNow our index list tells us that the current basic variables have indices $2, 3, 0$.\n\n% Problem 2: Slack variables.\n\\begin{problem}\nDesign and implement a way to store and track all of the basic and non-basic variables.\n\nHint: Using integers that represent the index of each variable is useful for Problem \\ref{prob:blands}.\n\\label{prob:slackvars}\n\\end{problem}\n\n\\subsection*{Creating a Tableau}\nAfter we have determined that our program is feasible, we need to create the \\emph{tableau} (sometimes called the \\emph{dictionary}, a construct that tracks the state of the algorithm.\nYou may structure the tableau to suit your specific implementation.\nRemember that your tableau will need to include in some way the slack variables that you created in Problem \\ref{prob:slackvars}.\n\nThere are many different ways to build your tableau.\nOne way is to mimic the tableau that is often used when performing the Simplex Algorithm by hand.\nDefine\n\\[\n\\bar{A} = \\begin{bmatrix}\n A & I_m\n\\end{bmatrix},\n\\]\nwhere $I_m$ is the $m \\times m$ identity matrix,\nand define\n\\[\n\\bar{c} = \\begin{bmatrix}\n c\\\\\n 0\\\\\n \\vdots\\\\\n 0\n\\end{bmatrix}.\n\\]\nThat is, $\\bar{c} \\in \\mathbb{R}^{n+m}$ such that the first $n$ entries are $c$ and the final $m$ entries are zeros.\nThen the initial tableau has the form\n\\begin{equation}\nT = \\begin{bmatrix}\n 0 & -\\bar{c}^T & 1 \\\\\n b & \\bar{A} & 0\n \\end{bmatrix}.\n\\label{eqn:hand_tab}\n\\end{equation}\n\nThe columns of the tableau correspond to each of the variables (both primal and slack), and the rows of the tableau\ncorrespond to the basic variables. Using the convention introduced above\nof representing the variables by indices in the index list, we have the following correspondence:\n\\[\n\\text{column } i \\Leftrightarrow \\text{index } i-2, \\qquad i = 2, 3, \\ldots, n+m+1,\n\\]\nand\n\\[\n\\text{row } j \\Leftrightarrow L_{j-1}, \\qquad j = 2, 3, \\ldots, m+1,\n\\]\nwhere $L_{j-1}$ refers to the $(j-1)$-th entry of the index list.\n\nFor our example problem, the initial index list is\n\\[\nL = (2, 3, 4, 0, 1),\n\\]\nand the initial tableau is\n\\begin{equation*}\nT = \\begin{bmatrix}\n 0 & -3 & -2 & 0 & 0 & 0 & 1\\\\\n 2 & 1 & -1 & 1 & 0 & 0 & 0\\\\\n 5 & 3 & 1 & 0 & 1 & 0 & 0\\\\\n 7 & 4 & 3 & 0 & 0 & 1 & 0\n \\end{bmatrix}.\n\\end{equation*}\nThe third column corresponds to index $1$, and the fourth row corresponds to index $4$, since this is the\nthird entry of the index list.\n\nThe advantage of using this kind of tableau is that it is easy to check the progress of your algorithm by hand.\nThe disadvantage is that pivot operations require careful bookkeeping to track the variables and constraints.\n\n% I (Jared) propose that we don't introduce the below tableau in the lab. It is less intuitive, and I think could lead to more confusion.\n\\begin{comment}\nWe can also use a tableau of the following format:\n\\begin{equation}\nT = \\begin{bmatrix}\n 0 & c^T & 0 \\\\\n 0 & I_n & 0\\\\\n b & -A & 0\n\\end{bmatrix}.\n\\label{eqn:matrix_tab}\n\\end{equation}\nHere, $T$ is a square matrix of size $(n+m+1) \\times (n+m+1)$.\nThe advantage of this form of the tableau is that all the pivot bookkeeping is built into the matrix.\nFor our example problem, the initial tableau of this form is\n\\begin{equation}\nT = \\begin{bmatrix}\n 0 & 3 & 2 & 0 & 0 & 0 \\\\\n 0 & 1 & 0 & 0 & 0 & 0 \\\\\n 0 & 0 & 1 & 0 & 0 & 0 \\\\\n 2 &-1 & 1 & 0 & 0 & 0 \\\\\n 5 &-3 &-1 & 0 & 0 & 0 \\\\\n 7 &-4 &-3 & 0 & 0 & 0\n\\end{bmatrix}.\n\\label{eqn:matrix_inittab}\n\\end{equation}\n\\end{comment}\n\n\\begin{problem}\nAdd a method to your Simplex solver that will create the initial tableau that you will use.\nUsing a NumPy array to represent your tableau will simplify several parts of the Simplex algorithm.\n\\label{prob:maketableau}\n\\end{problem}\n\n\\subsection*{Pivoting}\nPivoting is the mechanism that really makes Simplex useful.\nPivoting refers to the act of swapping basic and nonbasic variables, and transforming the tableau appropriately.\nThis has the effect of moving from one vertex of the feasible polytope to another vertex in a way that increases\nthe value of the objective function.\nDepending on how you store your variables, you may need to modify a few different parts of your solver to reflect this swapping.\n\nWhen initiating a pivot, you need to determine which variables will be swapped.\nIn the tableau representation, you first find a specific element on which to pivot, and the row and column that contain the pivot\nelement correspond to the variables that need to be swapped.\nRow operations are then performed on the tableau so that the pivot column becomes an elementary vector.\n\nLet's break it down, starting with the pivot selection. We need to use some care when choosing the pivot element.\nTo find the pivot column, search from left to right along the top row of the tableau\n(ignoring the first column), and stop once you encounter the first negative value. The index corresponding\nto this column will be designated the \\emph{entering index}, since after the full pivot operation, it will enter\nthe basis and become a basic variable.\n\nUsing our initial tableau $T$ in the example, we stop at the second column:\n\\[ T = \\left[ \\:\n\\begin{array}{*{7}{c}}\n\\cline{2-2}\n0 & \\multicolumn{1}{|c}{-3} & \\multicolumn{1}{|c}{-2} & 0 & 0 & 0 & 1\\\\\n2 & \\multicolumn{1}{|c}{1} & \\multicolumn{1}{|c}{-1} & 1 & 0 & 0 & 0\\\\\n5 & \\multicolumn{1}{|c}{3} & \\multicolumn{1}{|c}{1} & 0 & 1 & 0 & 0\\\\\n7 & \\multicolumn{1}{|c}{4} & \\multicolumn{1}{|c}{3} & 0 & 0 & 1 & 0\\\\\n\\cline{2-2}\n\\end{array}\n\\right] \\]\nWe now know that our pivot element will be found in the second column. Our entering index is thus $0$.\n\nNext, we select the pivot element from among the positive entries in the pivot column (ignoring the entry in the first row).\n\\emph{If all entries in the pivot column are non-positive, the problem is unbounded and has no solution.} In this case, the algorithm\nshould terminate.\nOtherwise, assuming our pivot column is the $j$-th column of the tableau and that the positive entries of this column are\n$T_{i_1, j}, T_{i_2, j}, \\ldots, T_{i_k, j}$, we calculate the ratios\n\\[\n\\frac{T_{i_1,1}}{T_{i_1,j}}, \\frac{T_{i_2,1}}{T_{i_2,j}}, \\ldots, \\frac{T_{i_k,1}}{T_{i_k,j}},\n\\]\nand we choose our pivot element to be one that minimizes this ratio. If multiple entries minimize the ratio, then we utilize\n\\emph{Bland's Rule}, which instructs us to choose the entry in the row corresponding to the smallest index.\n(Obeying this rule is important, as it prevents the possibility of the algorithm cycling back on itself infinitely.)\nThe index corresponding to the pivot row is designated as the \\emph{leaving index}, since after the full pivot operation,\nit will leave the basis and become a nonbasic variable.\n\nIn our example, we see that all entries in the pivot column (ignoring the entry in the first row, of course) are positive,\nand hence they are all potential choices for the pivot element. We then calculate the ratios, and obtain\n\\[\n\\frac{2}{1} = 2,\\quad \\frac{5}{3} = 1.66...,\\quad \\frac{7}{4} = 1.75.\n\\]\nWe see that the entry in the third row minimizes these ratios. Hence, the element in the second column, third row is our designated\npivot element, and our leaving index is $L_2 = 3$:\n\n\\[ T = \\left[ \\:\n\\begin{array}{*{7}{c}}\n\n0 & -3 & -2 & 0 & 0 & 0 & 1\\\\\n2 & 1 & -1 & 1 & 0 & 0 & 0\\\\\\cline{2-2}\n5 & \\multicolumn{1}{|c}{3} & \\multicolumn{1}{|c}{1} & 0 & 1 & 0 & 0\\\\\\cline{2-2}\n7 & 4 & 3 & 0 & 0 & 1 & 0\\\\\n\\end{array}\n\\right] \\]\n\n\\begin{comment}\nIf we are using the tableau representation in equation \\ref{eqn:matrix_tab}, pivot operations are reduced to a simple matrix equation:\n\\[T = T + T_m \\otimes T_n,\\]\nwhere $T_m$ is the column corresponding to the variable entering the basis and $T_n$ is a normalized vector corresponding to the variable leaving the basis. The result of the equation is the new tableau.\n\nFor example, for the initial tableau, \\ref{eqn:matrix_inittab}, we will demonstrate the first pivot operation.\nWe can do the entire pivot with a single outer product.\nthe first pivot should occur with $x_1$ leaving and $x_4$ entering.\nIn other words, we want to pivot at row $i = 4$ and column $j = 1$ in the tableau (the indices are offset by one because of the objective function and the row of constraints).\n\nThe row corresponding to $x_4$ is\n\\[\n\\begin{bmatrix} 5 &-3 &-1 & 0 & 0 & 0\\end{bmatrix}.\n\\]\nThis represents the equation\n\\[\nx_4 = 5 - 3x_1 - x_2.\n\\]\nOur eventual goal is to solve for $x_1$ and substitute into the remaining rows of the tableau.\nA simple method to accomplish this is to rewrite the equation so that we have zero on the left-hand side:\n\\[\n0 = 5 - 3x_1 - x_2 - x_4.\n\\]\nNow, we can normalize this equation so that the coefficient of $x_1$ is $-1$.\nThis is always accomplished by dividing the equation by the negative of the coefficient of $x_1$:\n\\begin{equation}\n0 = \\frac{5}{3} - x_1 - \\frac{1}{3}x_2 - \\frac{1}{3}x_4.\n\\label{eq:zero-equation}\n\\end{equation}\nThis is represented by the vector\n\\[\n\\begin{bmatrix} 5/3 & -1 & -1/3 & 0 & -1/3 & 0\\end{bmatrix}.\n\\]\nSince this left-hand side is zero, I can add any scalar multiple of this equation to any of the equations for $x_i$ and still have an equation for $x_i$.\nFor example, the equation for $x_5$ is\n\\[\nx_5 = 7 - 4x_1 - 3x_2.\n\\]\nThus, I can add $-4$ times \\eqref{eq:zero-equation} to this equation without changing the left-hand side:\n\\[ x_5 = 7 - 4x_1 - 3x_2 = 7 - 4x_1 - 3x_2 + -4\\left(\\frac{5}{3} - x_1 - \\frac{1}{3}x_2 - \\frac{1}{3}x_4\\right) = \\frac{1}{3} - \\frac{5}{3}x_2 + \\frac{4}{3} x_4.\n\\]\nNotice that we end up with an equation that does not include $x_1$ and now has $x_4$, just like we wanted.\nIn fact, this works in all of our equations, including those for the objective function and even for $x_1$!\nSince the coefficient of $x_1$ in \\eqref{eq:zero-equation} is $-1$, when we scale it by the coefficient of $x_1$ in any particular row, the $x_1$ cancels out.\n\\[\nT = T + \\begin{bmatrix}3 \\\\ 1 \\\\ 0 \\\\ -1 \\\\ -3 \\\\ -4\\end{bmatrix}\\begin{bmatrix} 5/3 & -1 & -1/3 & 0 & -1/3 & 0\\end{bmatrix}.\n\\]\nThe column vector is just the second column of $T$, which is the column containing the coefficients of $x_1$ in each row.\nWhen we compute this sum, we obtain the tableau.\n\\[\nT = \\begin{bmatrix}\n 5 & 0 & 1 & 0 & -1 & 0 \\\\\n 5/3 & 0 &-1/3 & 0 &-1/3 & 0 \\\\\n 0 & 0 & 1 & 0 & 0 & 0 \\\\\n 1/3 & 0 & 4/3 & 0 & 1/3 & 0 \\\\\n 0 & 0 & 0 & 0 & 1 & 0 \\\\\n 1/3 & 0 & -5/3 & 0 & 4/3 & 0\n\\end{bmatrix}.\n\\]\n\\end{comment}\n\\begin{problem}\nWrite a method that will determine the pivot row and pivot column according to Bland's Rule.\n\\begin{comment}\n\\begin{definition}[Bland's Rule]\nChoose the nonbasic variable with the smallest index that has a positive coefficient in the objective function\nas the leaving variable. Choose the basic variable with the smallest index among all the binding basic variables.\n\\end{definition}\n\nBland's Rule is important in avoiding cycles when performing pivots.\nThis rule guarantees that a feasible Simplex problem will terminate in a finite number of pivots.\n\\end{comment}\n\\label{prob:blands}\n\\end{problem}\n\nThe next step is to swap the entering and leaving indices in our index list.\nIn the example, we determined above that these indices are $0$ and $3$. We swap these two elements in our index list,\nand the updated index list is now\n\\[\nL = (2, 0, 4, 3, 1),\n\\]\nso the basic variables are now given by the indices $2, 0, 4$.\n\nFinally, we perform row operations on our tableau in the following way: divide the pivot row by the value of the pivot entry.\nThen use the pivot row to zero out all entries in the pivot column above and below the pivot entry. In our example, we first divide\nthe pivot row by 3, and then zero out the two entries above the pivot element and the single entry below it:\n\\begin{align*}\n\\begin{bmatrix}\n 0 & -3 & -2 & 0 & 0 & 0 & 1\\\\\n 2 & 1 & -1 & 1 & 0 & 0 & 0\\\\\n 5 & 3 & 1 & 0 & 1 & 0 & 0\\\\\n 7 & 4 & 3 & 0 & 0 & 1 & 0\n \\end{bmatrix} &\\rightarrow\n\\begin{bmatrix}\n 0 & -3 & -2 & 0 & 0 & 0 & 1\\\\\n 2 & 1 & -1 & 1 & 0 & 0 & 0\\\\\n 5/3 & 1 & 1/3 & 0 & 1/3 & 0 & 0\\\\\n 7 & 4 & 3 & 0 & 0 & 1 & 0\n \\end{bmatrix}\\rightarrow\\\\\n\\begin{bmatrix}\n 5 & 0 & -1 & 0 & 1 & 0 & 1\\\\\n 2 & 1 & -1 & 1 & 0 & 0 & 0\\\\\n 5/3 & 1 & 1/3 & 0 & 1/3 & 0 & 0\\\\\n 7 & 4 & 3 & 0 & 0 & 1 & 0\n \\end{bmatrix} &\\rightarrow\n\\begin{bmatrix}\n 5 & 0 & -1 & 0 & 1 & 0 & 1\\\\\n 1/3 & 0 & -4/3 & 1 & -1/3 & 0 & 0\\\\\n 5/3 & 1 & 1/3 & 0 & 1/3 & 0 & 0\\\\\n 7 & 4 & 3 & 0 & 0 & 1 & 0\n \\end{bmatrix}\\rightarrow\\\\\n\\begin{bmatrix}\n 5 & 0 & -1 & 0 & 1 & 0 & 1\\\\\n 1/3 & 0 & -4/3 & 1 & -1/3 & 0 & 0\\\\\n 5/3 & 1 & 1/3 & 0 & 1/3 & 0 & 0\\\\\n 1/3 & 0 & 5/3 & 0 & -4/3 & 1 & 0\n \\end{bmatrix}.\n\\end{align*}\nThe result of these row operations is our updated Tableau, and the pivot operation is complete.\n\n% Problem 5: Pivoting\n\\begin{problem}\nAdd a method to your solver that checks for unboundedness and performs a single pivot operation from start to completion.\nIf the problem is unbounded, raise a \\li{ValueError}.\n\\end{problem}\n\n\\subsection*{Termination and Reading the Tableau}\nUp to this point, our algorithm accepts a linear program, adds slack variables, and creates the initial tableau. After\ncarrying out these initial steps, it then performs the pivoting operation iteratively until the optimal point is found.\nBut how do we determine when the optimal point is found? The answer is to look at the top row of the tableau. More specifically,\nbefore each pivoting operation, check whether all of the entries in the top row of the tableau (ignoring the entry in the first\ncolumn) are nonnegative. If this is the case, then we have found an optimal solution, and so we terminate the algorithm.\n\nThe final step is to report the solution. The ending state of the tableau and index list tell us everything we need to know.\nThe maximum value attained by the objective function is found in the upper leftmost entry of the tableau. The nonbasic variables,\nwhose indices are located in the last $n$ entries of the index list, all have the value $0$. The basic variables, whose indices\nare located in the first $m$ entries of the index list, have values given by the first column of the tableau. Specifically, the basic\nvariable whose index is located at the $i$-th entry of the index list has the value $T_{i+1, 1}$.\n\nIn our example, suppose that our algorithm terminates with the tableau and index list in the following state:\n\\[\nT = \\begin{bmatrix}\n5.2 & 0 & 0 & 0 & .2 & .6 & 1\\\\\n.6 & 0 & 0 & 1 & -1.4 & .8 & 0\\\\\n1.6 & 1 & 0 & 0 & .6 & -.2 & 0\\\\\n.2 & 0 & 1 & 0 & -.8 & .6 & 0\\\\\n\\end{bmatrix}\n\\]\n\\[\nL = (2, 0, 1, 3, 4).\n\\]\nThen the maximum value of the objective function is $5.2$. The nonbasic variables have indices $3, 4$ and have the value $0$.\nThe basic variables have indices $2, 0,$ and $1$, and have values $.6, 1.6$, and $.2$, respectively.\nIn the notation of the original problem statement, the solution is given by\n\\begin{align*}\nx_0 &= 1.6\\\\\nx_1 &= .2.\n\\end{align*}\n\n% SimplexSolver.solve()\n\\begin{problem}\nWrite an additional method in your solver called \\li{solve()} that obtains the optimal solution, then returns the maximum value, the basic variables, and the nonbasic variables.\nThe basic and nonbasic variables should be represented as two dictionaries that map the index of the variable to its corresponding value.\n\nFor our example, we would return the tuple \\li{(5.2, \\{0: 1.6, 1: .2, 2: .6\\}, \\{3: 0, 4: 0\\})}.\n%The correct format of this tuple is critical, as this tuple of information will be used judge whether or not your solver works!\n\\end{problem}\n\nAt this point, you should have a Simplex solver that is simple to use. The following code demonstrates how your solver is\nexpected to behave:\n\n\\begin{lstlisting}\n>>> import SimplexSolver\n\n# Initialize objective function and constraints.\n>>> c = np.array([3., 2])\n>>> b = np.array([2., 5, 7])\n>>> A = np.array([[1., -1], [3, 1], [4, 3]])\n\n# Instantiate the simplex solver, then solve the problem.\n>>> solver = SimplexSolver(c, A, b)\n>>> sol = solver.solve()\n>>> print(sol)\n(5.200,\n {0: 1.600, 1: 0.200, 2: 0.600},\n {3: 0, 4: 0})\n\\end{lstlisting}\n\nIf the linear program were infeasible at the origin or unbounded, we would expect the solver to alert the user by raising an error.\n\nNote that this simplex solver is \\emph{not} fully operational.\nIt can't handle the case of infeasibility at the origin.\nThis can be fixed by adding methods to your class that solve the \\emph{auxiliary problem}, that of finding an initial feasible tableau when the problem is not feasible at the origin.\nSolving the auxiliary problem involves pivoting operations identical to those you have already implemented, so adding this functionality\nis not overly difficult.\n\n\\section*{The Product Mix Problem}\nWe now use our Simplex implementation to solve the \\emph{product mix problem}, which in its basic form can be expressed as a simple linear program.\nSuppose that a manufacturer makes $n$ products using $m$ different resources (labor, raw materials, machine time available, etc).\nThe $i$-th product is sold at a unit price $p_i$, and there are at most $m_j$ units\nof the $j$-th resource available. Additionally, each unit of the $i$-th product requires $a_{j,i}$ units of resource $j$.\nGiven that the demand for product $i$ is $d_i$ units per a certain time period, how do we choose the optimal amount\nof each product to manufacture in that time period so as to maximize revenue, while not exceeding the available resources?\n\nLet $x_1, x_2, \\ldots, x_n$ denote the amount of each product to be manufactured.\nThe sale of product $i$ brings revenue in the amount of $p_ix_i$.\nTherefore our objective function, the profit, is given by\n\\[\n\\sum_{i=1}^n p_ix_i.\n\\]\nAdditionally, the manufacture of product $i$ requires $a_{j,i}x_i$ units of resource $j$.\nThus we have the resource constraints\n\\[\n\\sum_{i=1}^n a_{j,i}x_i \\leq m_j \\text{ for } j = 1, 2, \\ldots, m.\n\\]\n\nFinally, we have the demand constraints which tell us not to exceed the demand for the products:\n\\[\nx_i \\leq d_i \\text{ for } i = 1, 2, \\ldots, n\n\\]\n\nThe variables $x_i$ are constrained to be nonnegative, of course. We therefore have a linear program in the appropriate form that is feasible at the origin.\nIt is a simple task to solve the problem using our Simplex solver.\n\n\\begin{problem}\nSolve the product mix problem for the data contained in the file \\li{productMix.npz}. In this problem, there are 4 products and 3 resources.\nThe archive file, which you can load using the function\n\\li{np.load}, contains a dictionary of arrays. The array with key \\li{'A'} gives the resource coefficients $a_{i,j}$ (i.e. the $(i,j)$-th entry\nof the array give $a_{i,j}$). The array with key \\li{'p'} gives the unit prices $p_i$. The array with key \\li{'m'} gives the available resource\nunits $m_j$. The array with key \\li{'d'} gives the demand constraints $d_i$.\n\nReport the number of units that should be produced for each product.\n\\end{problem}\n%---------------------------------------------------------------------------------------------------\n%\n%\n% \\section*{Auxiliary Problems}\n%\n% When one of the entries of $b$ is negative, we need to run an auxiliary problem to find a feasible point.\n% Try adding $x_0$ as the last variable.\n% That way you just need to add a single row and column to your current tableau $T$.\n% You can check that subtracting $x_0$ from each of the constraints is the same as adding $x_0$ to each of the expressions for the slack variables.\n% Since we also need write $x_0$ in terms of itself ($x_0 = x_0$), this is equivalent to setting the last column equal to 1 for all the rows corresponding to slack variables, plus one new row.\n% We also need to change the objective function to $-x_0$.\n% Let's let $N$ be the matrix for the auxiliary problem.\n% Then we can construct it from $T$ using the code\n% \\begin{lstlisting}\n% >>> N = zeros((s+1,s+1))\n% >>> N[0:s,0:s] = T\n% >>> N[n:s+1,s] = 1\n% >>> N[0,s] = -1\n% \\end{lstlisting}\n%\n% Before you run simplex on this auxiliary tableau, make sure you do a pivot on the last column and the row corresponding to the smallest entry of $b$.\n% Check that the objective function value is 0 when simplex finishes running.\n% If not, your initial problem is infeasible.\n% If the problem is feasible, get the system of equations from $N$ and put them back into $T$, leaving off the last row and column of $N$ (for $x_0$) and the first row (corresponding to the objective function).\n% \\begin{lstlisting}\n% T[1:s,0:s] = N[1:s,0:s]\n% \\end{lstlisting}\n% The last thing you need to do is insert your previous objective function.\n% However, you need to re-write it in terms of the current nonbasic variables.\n% Fortunately, $T$ currently contains all of your variables written in terms of the nonbasic variables.\n% You can check (mathematically, or however you want to satisfy yourself) that\n% \\begin{lstlisting}\n% T[0,0:s] = T[0,0:s]*T\n% \\end{lstlisting}\n% will put the objective function into the first row, now written in terms of the new nonbasic variables.\n\n\\section*{Beyond Simplex}\nThe \\emph{Computing in Science and Engineering} journal listed Simplex as one of the top ten algorithms of the twentieth century \\cite{Nash2000}.\nDespite its popularity, like any other algorithm, Simplex has its drawbacks.\n\nIn 1972, Victor Klee and George Minty Cube published a paper with several examples of worst-case polytopes for the Simplex algorithm \\cite{Klee1972}.\nIn their paper, they give several examples of polytopes that the Simplex algorithm struggles to solve.\n\nConsider the following linear program from Klee and Minty.\n\\begin{align*}\n\\text{max } & 2^{n-1}x_1 & + & 2^{n-2}x_2 & + & \\cdots & + & 2x_{n-1} & + & x_n\\\\\n\\text{subject to } & x_1 & & & & & & & &\\leq 5\\\\\n& 4x_1 & + & x_2 & & & & & &\\leq 25\\\\\n& 8x_1 & + & 4x_2 & + & x_3 & & & &\\leq 125\\\\\n& \\vdots & & & & & & & &\\vdots\\\\\n& 2^n x_1 & + & 2^{n-1} x_2 & + & \\cdots & + & 4x_{n-1} & + x_n &\\leq 5\\\\\n\\end{align*}\n\n% When $n = 3$, we have the initial tableau\n% \\begin*{equation*}\n% \\begin{bmatrix}\n% 0 & 4 & 2 & 1 & 0 & 0 & 0\\\\\n% 0 & 1 & 0 & 0 & 0 & 0 & 0\\\\\n% 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\\n% 0 & 0 & 0 & 1 & 0 & 0 & 0\\\\\n% 5 & -1 & 0 & 0 & 0 & 0 & 0\\\\\n% 25 & -4 & -1 & 0 & 0 & 0 & 0\\\\\n% 125 & -8 & -4 & -1 & 0 & 0 & 0\\\\\n% \\end{bmatrix}\n% \\end{equation*}\n%\n% After the first pivot with $x_1$ leaving and $s_1$ entering, we have the tableau\n%\n% \\begin{equation*}\n% \\begin{bmatrix}\n% 20 & 0 & 2 & 1 & -4 & 0 & 0\\\\\n% 5 & 0 & 0 & 0 & -1 & 0 & 0\\\\\n% 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\\n% 0 & 0 & 0 & 1 & 0 & 0 & 0\\\\\n% 0 & 0 & 0 & 0 & 1 & 0 & 0\\\\\n% 5 & 0 & -1 & 0 & 4 & 0 & 0\\\\\n% 85 & 0 & -4 & -1 & 8 & 0 & 0\\\\\n% \\end{bmatrix}\n% \\end{equation*}\n%\n% \\begin{problem}\n% What is the final tableau for this Klee-Minty example with $n=3$?\n% How many iterations does it take to arrive at the final tableau?\n% \\end{problem}\n%\n% \\begin{problem}\n% Using problem 1 as a guide, guess the optimum value of the Klee-Minty example with $n=20$.\n% Then find the maximum value for the Klee-Minty example with $n=20$.\n% How many iterations does it take to arrive at the final tableau?\n% How long does the program take to run for only $20$ variables?\n% \\end{problem}\n\nKlee and Minty show that for this example, the worst case scenario has exponential time complexity.\nWith only $n$ constraints and $n$ variables, the simplex algorithm goes through $2^n$ iterations.\nThis is because there are $2^n$ extreme points, and when starting at the point $x=0$, the simplex algorithm goes through all of the extreme points before reaching the optimal point $(0,0,\\dots, 0, 5^n)$.\nOther algorithms, such as interior point methods, solve this problem much faster because they are not constrained to follow the edges.\n\n\\printbibliography\n", "meta": {"hexsha": "959e015ebaaf7c38e75e83c47afbef558169476d", "size": 30561, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Vol2B/Simplex/Simplex.tex", "max_stars_repo_name": "joshualy/numerical_computing", "max_stars_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Vol2B/Simplex/Simplex.tex", "max_issues_repo_name": "joshualy/numerical_computing", "max_issues_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Vol2B/Simplex/Simplex.tex", "max_forks_repo_name": "joshualy/numerical_computing", "max_forks_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-11-05T14:45:03.000Z", "max_forks_repo_forks_event_max_datetime": "2019-11-05T14:45:03.000Z", "avg_line_length": 49.0545746388, "max_line_length": 254, "alphanum_fraction": 0.7063904977, "num_tokens": 9235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.798186768138228, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.6990415832655008}} {"text": "% !TEX root = Main.tex\n\\section{Word Embeddings}\n\\textbf{Distributional Model:}\\\\\n$p_\\theta(w|w')$ = Pr[$w$ occurs in context of $w'$]\\\\\n\\textbf{Log-likelihood:}\\\\\n$L(\\theta; \\mathbf{w}) = \\sum_{t=1}^T\\sum_{\\Delta \\in I}{\\log p_\\theta(w^{(t+\\Delta)}|w^{(t)})}$\\\\\n\\textbf{Latent Vector Model:} $w \\rightarrow (\\mathbf{x}_w, b_w) \\in \\mathbb{R}^{D+1} \\\\p_{\\theta}(w|w') = \\frac{\\exp[\\langle \\mathbf{x}_w,\\mathbf{x}_{w'}\\rangle + b_w]}{\\sum_{v\\in V}{\\exp[\\langle \\mathbf{x}_v,\\mathbf{x}_{w'}\\rangle + b_v ]}}$ (soft-max).\\\\\n\\textbf{Modifications:}\\\\\n$\\log p_{\\theta}(w|w') = \\langle y_{w} , x_{w'} \\rangle + b_w$, word $y_w$, c'txt $x_{w'}$\\\\\nuse GloVe objective\\\\\nnegative sampling (logistic classification)\n\n\\subsection*{GloVe (Weighted Square Loss)}\n\\textbf{Co-occurence Matrix:}\\\\\n$\\mathbf{N} = (n_{ij}) \\in \\mathbb{R}^{|V|\\times|C|} = \\# of word w_i$ in context $w_j$\\\\\n\\textbf{Objective:} $H(\\theta;\\mathbf{N})$\\\\\n$= \\sum_{n_{ij} > 0} f(n_{ij})(\\log n_{ij} - \\log \\exp[\\langle \\mathbf{x}_i, \\mathbf{y}_j \\rangle + b_i + d_j])^2$\\\\\nwith $f(n) = \\min\\{1, (\\frac{n}{n_{max}})^\\alpha\\}$, $\\alpha \\in (0;1]$.\\\\\nunnormalized distr. $\\rightarrow$ 2-sided loss function\\\\\n1. sample $(i,j) u.a.r, s.t. n_{ij}>0$\\\\\n2. $\\mathbf{x}_i^{new} \\leftarrow \\mathbf{x}_i + 2\\eta f(n_{ij})(\\log n_{ij} - \\langle \\mathbf{x}_i, \\mathbf{y}_j \\rangle)\\mathbf{y}_j$\\\\\n3. $\\mathbf{y}_j^{new} \\leftarrow \\mathbf{y}_j + 2\\eta f(n_{ij})(\\log n_{ij} - \\langle \\mathbf{x}_i, \\mathbf{y}_j \\rangle)\\mathbf{x}_i$\n\n\\subsection*{Discussion}\nWord embeddings can model analogies and relatedness, but antonyms are usually not well captured.\n", "meta": {"hexsha": "a9533732e9fca587cd6025f237ff2f68ca2ec7ac", "size": 1607, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "WordEmbedding.tex", "max_stars_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_stars_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-09-24T20:33:35.000Z", "max_stars_repo_stars_event_max_datetime": "2019-02-21T15:52:46.000Z", "max_issues_repo_path": "WordEmbedding.tex", "max_issues_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_issues_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "WordEmbedding.tex", "max_forks_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_forks_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2019-01-14T16:21:22.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-18T17:17:10.000Z", "avg_line_length": 61.8076923077, "max_line_length": 257, "alphanum_fraction": 0.6253889235, "num_tokens": 665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.948154530420204, "lm_q2_score": 0.7371581626286834, "lm_q1q2_score": 0.6989398515326197}} {"text": "% !TEX root = main.tex\n\n%-------------------------------------------------\n\\section{Appendix}\\label{appx:expe}\n\n%-----------------------------\n%\\subsection*{Bernoulli's law of large numbers}\\label{subsec:}\n\nIn the proof of Theorem~\\ref{thm:wlln}, we used Chebyshev's inequality to show that\n\\[\n\\prob(|\\bar{X}_n - \\mu| \\geq \\epsilon) \\leq \\frac{\\sigma^2}{n\\epsilon^2} \\qquad\\forall\\ \\epsilon>0.\n\\]\nWe say that the \\emph{rate} at which $\\bar{X}_n\\to\\mu$ is of order $O(1/n)$ as $n\\to\\infty$.\n%\nIn the proof of the following theorem, we use Bernstein's inequality to show that the sample mean of Bernoulli random variables satisfies\n\\[\n\\prob\\left(|\\bar{X}_n - \\mu| > \\epsilon\\right) \\leq e^{-\\frac{1}{2}n\\epsilon^2} \\qquad\\forall\\ \\epsilon>0.\n\\]\nIn this case, the rate at which $\\bar{X}_n\\to\\mu$ as $n\\to\\infty$ is said to be \\emph{exponentially fast}. \n\n% theorem\n\\begin{theorem}[Bernoulli's Law of Large Numbers]\\label{thm:bernoulli_lln}\nLet $X_1,X_2,\\ldots$ be independent, with each $X_i\\sim\\text{Bernoulli}(p)$, and let $\\bar{X}_n = \\frac{1}{n}\\sum_{i=1}^n X_i$ be the sample mean of the first $n$ variables in the sequence.\nThen for every $\\epsilon > 0$, \n\\[\n\\prob(|\\bar{X}_n - p| > \\epsilon) \\to 0 \\text{\\quad as\\quad} n\\to\\infty.\n\\]\n\\end{theorem}\n\n% proof\n\\begin{proof}\nLet $\\epsilon>0$ and define $S_n=\\sum_{i=1}^n X_i$. Then\n\\[\n\\prob\\left(\\bar{X}_n - p > \\epsilon\\right)\n\t= \\prob\\big[S_n > n(p+\\epsilon)\\big]\n\\]\n\n%\\bit\n%\\it Recall Bernstein's inequality: $\\prob(X>a) \\leq e^{-ta}\\expe(e^{tX})$ for all $t>0$.\n%\\eit\n\nApplying Bernstein's inequality (Theorem~\\ref{thm:bernstein}) to the random variable $S_n$ with $a = n(p+\\epsilon)$,\n\\begin{align*}\n\\prob\\big[S_n > n(p+\\epsilon)\\big]\n\t& \\leq e^{-tn(p+\\epsilon)}\\expe(e^{tS_n}) \\\\\n\t& = e^{-tn(p+\\epsilon)}\\big[1-p+pe^t\\big]^n \\\\\n\t& = e^{-tn\\epsilon}\\big[ e^{-tp}(1-p+pe^t) \\big]^n \\\\\n\t& = e^{-tn\\epsilon}\\big[ (1-p)e^{-tp} + pe^{t(1-p)} \\big]^n\n\\end{align*}\n\nUsing the inequality $e^x \\leq x + e^{x^2}$, which holds for all $x\\in\\R$, % (see exercises),\n\\begin{align*}\n(1-p)e^{-tp} + pe^{t(1-p)} \n\t& \\leq (1-p)\\big[-tp + e^{t^2p^2}\\big] + p\\big[t(1-p) + e^{t^2(1-p)^2}\\big] \\\\\n\t& = (1-p)e^{t^2p^2} + pe^{t^2(1-p)^2} \\\\\n\t& \\leq (1-p)e^{t^2} + pe^{t^2} \\\\\n\t& = e^{t^2}.\n\\end{align*}\nHence, for all $t>0$,\n\\[\n\\prob\\left(\\bar{X}_n -p > \\epsilon\\right) = \\prob\\big[S_n > n(p+\\epsilon)\\big] \\leq e^{-tn\\epsilon}e^{t^2n} = e^{tn(t-\\epsilon)}.\n\\]\n\\bit\n\\it This inequality is valid for all $t>0$. \n\\it We choose $t$ so that the right-hand side is made as small as possible. \n\\it Because $e^x$ is an increasing function, this corresponds to the minimum value of the exponent. \n\\it We differentiate the exponent $tn(t-\\epsilon)$ with respect to $t$ and set this equal to zero.\n\\eit\n\nThis yields the value $t = \\frac{1}{2}\\epsilon$, so\n\\[\n\\prob\\left(\\bar{X}_n -p > \\epsilon\\right) = \\prob\\big[S_n > n(p+\\epsilon)\\big] \\leq e^{-\\frac{1}{4}n\\epsilon^2}.\n\\]\nA similar argument shows that \n\\[\n\\prob\\left(\\bar{X}_n -p < -\\epsilon\\right) = \\prob\\big[S_n < n(p-\\epsilon)\\big] \\leq e^{-\\frac{1}{4}n\\epsilon^2},\n\\]\nThus we have\n\\[\n\\prob\\left(|\\bar{X}_n - p| > \\epsilon\\right) \\leq e^{-\\frac{1}{2}n\\epsilon^2} \\to 0 \\text{\\quad as $n\\to\\infty$,}\n\\]\nas required.\n\\end{proof}\n", "meta": {"hexsha": "c3c544c395c209a8d6c7e6efe87debaaa5aa8852", "size": 3208, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/06D_appendix.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/06D_appendix.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/06D_appendix.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 38.1904761905, "max_line_length": 189, "alphanum_fraction": 0.6119077307, "num_tokens": 1266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.884039278690883, "lm_q1q2_score": 0.6988598527000132}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n\n\n \n \n\n\\subsection*{error\\_ellipse.m} \n\n\\begin{par}\nERROR\\_ELLIPSE - plot an error ellipse, or ellipsoid, defining confidence region ERROR\\_ELLIPSE(C22) - Given a 2x2 covariance matrix, plot the associated error ellipse, at the origin. It returns a graphics handle of the ellipse that was drawn.\n\\end{par} \\vspace{1em}\n\n\\begin{verbatim} ERROR_ELLIPSE(C33) - Given a 3x3 covariance matrix, plot the\n associated error ellipsoid, at the origin, as well as its projections\n onto the three axes. Returns a vector of 4 graphics handles, for the\n three ellipses (in the X-Y, Y-Z, and Z-X planes, respectively) and for\n the ellipsoid.\\end{verbatim}\n \n\\begin{verbatim} ERROR_ELLIPSE(C,MU) - Plot the ellipse, or ellipsoid, centered at MU,\n a vector whose length should match that of C (which is 2x2 or 3x3).\\end{verbatim}\n \n\\begin{verbatim} ERROR_ELLIPSE(...,'Property1',Value1,'Name2',Value2,...) sets the\n values of specified properties, including:\n 'C' - Alternate method of specifying the covariance matrix\n 'mu' - Alternate method of specifying the ellipse (-oid) center\n 'conf' - A value betwen 0 and 1 specifying the confidence interval.\n the default is 0.5 which is the 50% error ellipse.\n 'scale' - Allow the plot the be scaled to difference units.\n 'style' - A plotting style used to format ellipses.\n 'clip' - specifies a clipping radius. Portions of the ellipse, -oid,\n outside the radius will not be shown.\\end{verbatim}\n \n\\begin{verbatim} NOTES: C must be positive definite for this function to work properly.\\end{verbatim}\n \\begin{par}\nby AJ Johnson \\begin{verbatim}http://www.mathworks.de/matlabcentral/fileexchange/4705\\end{verbatim}\n\\end{par} \\vspace{1em}\n\n\\begin{lstlisting}\nfunction h=error_ellipse(varargin)\n\\end{lstlisting}\n\n\n\\subsection*{Code} \n\n\n\\begin{lstlisting}\ndefault_properties = struct(...\n 'C', [], ... % The covaraince matrix (required)\n 'mu', [], ... % Center of ellipse (optional)\n 'conf', 0.95, ... % Percent confidence/100\n 'scale', 1, ... % Scale factor, e.g. 1e-3 to plot m as km\n 'style', '', ... % Plot style\n 'clip', inf); % Clipping radius\n\nif length(varargin) >= 1 & isnumeric(varargin{1})\n default_properties.C = varargin{1};\n varargin(1) = [];\nend\n\nif length(varargin) >= 1 & isnumeric(varargin{1})\n default_properties.mu = varargin{1};\n varargin(1) = [];\nend\n\nif length(varargin) >= 1 & isnumeric(varargin{1})\n default_properties.conf = varargin{1};\n varargin(1) = [];\nend\n\nif length(varargin) >= 1 & isnumeric(varargin{1})\n default_properties.scale = varargin{1};\n varargin(1) = [];\nend\n\nif length(varargin) >= 1 & ~ischar(varargin{1})\n error('Invalid parameter/value pair arguments.')\nend\n\nprop = getopt(default_properties, varargin{:});\nC = prop.C;\n\nif isempty(prop.mu)\n mu = zeros(length(C),1);\nelse\n mu = prop.mu;\nend\n\nconf = prop.conf;\nscale = prop.scale;\nstyle = prop.style;\n\nif conf <= 0 | conf >= 1\n error('conf parameter must be in range 0 to 1, exclusive')\nend\n\n[r,c] = size(C);\nif r ~= c | (r ~= 2 & r ~= 3)\n error(['Don''t know what to do with ',num2str(r),'x',num2str(c),' matrix'])\nend\n\nx0=mu(1);\ny0=mu(2);\n\n% Compute quantile for the desired percentile\nk = sqrt(qchisq(conf,r)); % r is the number of dimensions (degrees of freedom)\n\nhold_state = get(gca,'nextplot');\n\nif r==3 & c==3\n z0=mu(3);\n\n % Make the matrix has positive eigenvalues - else it's not a valid covariance matrix!\n if any(eig(C) <=0)\n error('The covariance matrix must be positive definite (it has non-positive eigenvalues)')\n end\n\n % C is 3x3; extract the 2x2 matricies, and plot the associated error\n % ellipses. They are drawn in space, around the ellipsoid; it may be\n % preferable to draw them on the axes.\n Cxy = C(1:2,1:2);\n Cyz = C(2:3,2:3);\n Czx = C([3 1],[3 1]);\n\n [x,y,z] = getpoints(Cxy,prop.clip);\n h1=plot3(x0+k*x,y0+k*y,z0+k*z,prop.style,'linewidth',2);hold on\n [y,z,x] = getpoints(Cyz,prop.clip);\n h2=plot3(x0+k*x,y0+k*y,z0+k*z,prop.style,'linewidth',2);hold on\n [z,x,y] = getpoints(Czx,prop.clip);\n h3=plot3(x0+k*x,y0+k*y,z0+k*z,prop.style,'linewidth',2);hold on\n\n\n [eigvec,eigval] = eig(C);\n\n [X,Y,Z] = ellipsoid(0,0,0,1,1,1);\n XYZ = [X(:),Y(:),Z(:)]*sqrt(eigval)*eigvec';\n\n X(:) = scale*(k*XYZ(:,1)+x0);\n Y(:) = scale*(k*XYZ(:,2)+y0);\n Z(:) = scale*(k*XYZ(:,3)+z0);\n h4=surf(X,Y,Z);\n colormap gray\n alpha(0.3)\n camlight\n if nargout\n h=[h1 h2 h3 h4];\n end\nelseif r==2 & c==2\n % Make the matrix has positive eigenvalues - else it's not a valid covariance matrix!\n if any(eig(C) <=0)\n error('The covariance matrix must be positive definite (it has non-positive eigenvalues)')\n end\n\n [x,y,z] = getpoints(C,prop.clip);\n h1=plot(scale*(x0+k*x),scale*(y0+k*y),prop.style,'linewidth',2);\n set(h1,'zdata',z+1)\n if nargout\n h=h1;\n end\nelse\n error('C (covaraince matrix) must be specified as a 2x2 or 3x3 matrix)')\nend\n%axis equal\n\nset(gca,'nextplot',hold_state);\n\n%---------------------------------------------------------------\n% getpoints - Generate x and y points that define an ellipse, given a 2x2\n% covariance matrix, C. z, if requested, is all zeros with same shape as\n% x and y.\n\\end{lstlisting}\n\n\\begin{lstlisting}\nfunction [x,y,z] = getpoints(C,clipping_radius)\n\nn=100; % Number of points around ellipse\np=0:pi/n:2*pi; % angles around a circle\n\n[eigvec,eigval] = eig(C); % Compute eigen-stuff\nxy = [cos(p'),sin(p')] * sqrt(eigval) * eigvec'; % Transformation\nx = xy(:,1);\ny = xy(:,2);\nz = zeros(size(x));\n\n% Clip data to a bounding radius\nif nargin >= 2\n r = sqrt(sum(xy.^2,2)); % Euclidian distance (distance from center)\n x(r > clipping_radius) = nan;\n y(r > clipping_radius) = nan;\n z(r > clipping_radius) = nan;\nend\n\n%---------------------------------------------------------------\nfunction x=qchisq(P,n)\n% QCHISQ(P,N) - quantile of the chi-square distribution.\nif nargin<2\n n=1;\nend\n\ns0 = P==0;\ns1 = P==1;\ns = P>0 & P<1;\nx = 0.5*ones(size(P));\nx(s0) = -inf;\nx(s1) = inf;\nx(~(s0|s1|s))=nan;\n\nfor ii=1:14\n dx = -(pchisq(x(s),n)-P(s))./dchisq(x(s),n);\n x(s) = x(s)+dx;\n if all(abs(dx) < 1e-6)\n break;\n end\nend\n\n%---------------------------------------------------------------\nfunction F=pchisq(x,n)\n% PCHISQ(X,N) - Probability function of the chi-square distribution.\nif nargin<2\n n=1;\nend\nF=zeros(size(x));\n\nif rem(n,2) == 0\n s = x>0;\n k = 0;\n for jj = 0:n/2-1;\n k = k + (x(s)/2).^jj/factorial(jj);\n end\n F(s) = 1-exp(-x(s)/2).*k;\nelse\n for ii=1:numel(x)\n if x(ii) > 0\n F(ii) = quadl(@dchisq,0,x(ii),1e-6,0,n);\n else\n F(ii) = 0;\n end\n end\nend\n\n%---------------------------------------------------------------\nfunction f=dchisq(x,n)\n% DCHISQ(X,N) - Density function of the chi-square distribution.\nif nargin<2\n n=1;\nend\nf=zeros(size(x));\ns = x>=0;\nf(s) = x(s).^(n/2-1).*exp(-x(s)/2)./(2^(n/2)*gamma(n/2));\n\n%---------------------------------------------------------------\nfunction properties = getopt(properties,varargin)\n%GETOPT - Process paired optional arguments as 'prop1',val1,'prop2',val2,...\n%\n% getopt(properties,varargin) returns a modified properties structure,\n% given an initial properties structure, and a list of paired arguments.\n% Each argumnet pair should be of the form property_name,val where\n% property_name is the name of one of the field in properties, and val is\n% the value to be assigned to that structure field.\n%\n% No validation of the values is performed.\n%\n% EXAMPLE:\n% properties = struct('zoom',1.0,'aspect',1.0,'gamma',1.0,'file',[],'bg',[]);\n% properties = getopt(properties,'aspect',0.76,'file','mydata.dat')\n% would return:\n% properties =\n% zoom: 1\n% aspect: 0.7600\n% gamma: 1\n% file: 'mydata.dat'\n% bg: []\n%\n% Typical usage in a function:\n% properties = getopt(properties,varargin{:})\n\n\n% Process the properties (optional input arguments)\nprop_names = fieldnames(properties);\nTargetField = [];\nfor ii=1:length(varargin)\n arg = varargin{ii};\n if isempty(TargetField)\n if ~ischar(arg)\n error('Propery names must be character strings');\n end\n f = find(strcmp(prop_names, arg));\n if length(f) == 0\n error('%s ',['invalid property ''',arg,'''; must be one of:'],prop_names{:});\n end\n TargetField = arg;\n else\n % properties.(TargetField) = arg; % Ver 6.5 and later only\n properties = setfield(properties, TargetField, arg); % Ver 6.1 friendly\n TargetField = '';\n end\nend\nif ~isempty(TargetField)\n error('Property names and values must be specified in pairs.');\nend\n\\end{lstlisting}\n", "meta": {"hexsha": "6e5640c159984a82651d956fe9a878fef4f865a4", "size": 8663, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/tex/error_ellipse.tex", "max_stars_repo_name": "SJTUGuofei/pilco-matlab", "max_stars_repo_head_hexsha": "a0b48b7831911837d060617903c76c22e4180d0b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 53, "max_stars_repo_stars_event_min_datetime": "2016-12-17T15:15:48.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-09T16:59:27.000Z", "max_issues_repo_path": "doc/tex/error_ellipse.tex", "max_issues_repo_name": "sahandrez/quad_pilco", "max_issues_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-04-24T11:02:23.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-24T11:09:45.000Z", "max_forks_repo_path": "doc/tex/error_ellipse.tex", "max_forks_repo_name": "sahandrez/quad_pilco", "max_forks_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 36, "max_forks_repo_forks_event_min_datetime": "2017-04-19T06:55:25.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-19T10:19:12.000Z", "avg_line_length": 28.4967105263, "max_line_length": 252, "alphanum_fraction": 0.6321135865, "num_tokens": 2685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8840392725805822, "lm_q2_score": 0.7905303186696747, "lm_q1q2_score": 0.6988598478696351}} {"text": "\\chapter{Generalized eigenvectors maximize signal-to-noise}\n\\label{apx:GEvecs-maximise-SNR}\n\nIn this appendix, we show that the weight vector that maximizes the ratio of signal to noise variance (i.e. the solution $\\what$ to \\cref{eq:argmax_R}) is equivalent to the first generalized eigenvector $\\w_1$ of the ordered pair of covariance matrices $(\\Rss, \\Rnn)$ (as defined in \\cref{sec:generalized-eigenproblem}).\n\nThis ratio of variances in \\cref{eq:argmax_R} is a quotient of quadratic forms, namely the so called ``generalized Rayleigh quotient'' of $(\\Rss, \\Rnn)$.\n\nFormally, the generalized Rayleigh quotient of a non-zero vector $\\w \\in \\reals^N$ and the ordered, symmetric matrix pair $(\\A,\\B)$ is the scalar $r(\\w)$ defined as:\n%\n\\begin{equation}\n\\label{eq:Rayleigh}\nr(\\w) = \\frac{\\w^T \\A \\w}\n {\\w^T \\B \\w}\n\\end{equation}\n\nRecall that a vector $\\w_i$ is a generalized eigenvector of $(\\A,\\B)$, with corresponding generalized eigenvalue $\\lambda_i$, if\n%\n\\begin{equation}\n\\label{eq:generalized-eigenproblem-AB}\n\\A \\w_i = \\lambda_i \\B \\w_i\n\\end{equation}\n\nWe must then prove the following:\n\n\n\n\\section{Theorem}\n\nThe generalized eigenvector $\\w_1$ corresponding to the largest generalized eigenvalue $\\lambda_1$ of $(\\A,\\B)$, is also the vector $\\what$ that maximises the generalized Rayleigh quotient $r(\\w)$ of $(\\A,\\B)$.\n\n\n\n\n\\section{Proof}\n\nAs a first step, we will show that if $\\what$ is the maximum of $r(\\w)$, that it is indeed an eigenvector of $(\\A,\\B)$. In the second step, we will show that the largest eigenvalue $\\lambda_1$ of $(\\A,\\B)$ corresponds to the maximum of $r(\\w)$.\n\nIf $\\what$ is a maximum of $r(\\w)$, then\n\\begin{equation}\n\\label{eq:critical}\n\\grad{r(\\what) = \\vb{0}}.\n\\end{equation}\n\nWorking out the partial derivatives that comprise the gradient of $r(\\w)$, we find:\n\\begin{align*}\n\\grad{r(\\w)} &= \\frac{2 \\A \\w \\qty(\\w^T \\B \\w) \n - 2 \\B \\w \\qty(\\w^T \\A \\w)}\n {\\qty(\\w^T \\B \\w)^2}\n\\end{align*}\n\nWith \\cref{eq:critical}, we then have the following condition for our maximising vector $\\what$:\n\\[\n2 \\A \\what \\qty(\\what^T \\B \\what) = 2 \\B \\what \\qty(\\what^T \\A \\what)\n\\]\nor\n\\begin{align*}\n\\A \\what &= \\frac{\\what^T \\A \\what}\n {\\what^T \\B \\what} \\; \\B \\what \\\\[1em]\n\\A \\what &= r(\\what) \\; \\B \\what\n\\end{align*}\nThis is the generalized eigenvalue/eigenvector definition (\\cref{eq:generalized-eigenproblem-AB}) for $\\w_i = \\what$ and $\\lambda_i = r(\\what)$.\n\nWe have thus shown that if $\\what$ is a maximum of $r(\\w)$, that it is an eigenvector of $(\\A,\\B)$, with $r(\\what)$ its corresponding eigenvalue.\n\nAs the second step, we now show that $r(\\what)$ is the \\emph{largest} eigenvalue of $(\\A,\\B)$. We follow the reasoning of \\citeauthor{Trefethen1997}, who prove a related result for the ordinary Rayleigh quotient \\cite[p. 204]{Trefethen1997}.\n\nWe will rewrite the generalized Rayleigh quotient $r(\\w)$ by writing the arbitrary vector $\\w$ as a linear combination of the generalized eigenvectors $\\w_i$ of $(\\A,\\B)$: $\\w = \\sum_i c_i \\w_i$. Then:\n\\begin{align*}\nr(\\w) &= \\frac{\\qty(\\sum_i c_i \\w_i)^T \\A \\qty(\\sum_i c_i \\w_i)}\n {\\qty(\\sum_i c_i \\w_i)^T \\B \\qty(\\sum_i c_i \\w_i)} \\\\[1em]\n % \n &= \\frac{\\sum_i c_i^2 \\w_i^T \\A \\w_i}\n {\\sum_i c_i^2 \\w_i^T \\B \\w_i} \\\\[1em]\n % \n &= \\frac{\\sum_i c_i^2 \\lambda_i \\w_i^T \\B \\w_i}\n {\\sum_i c_i^2 \\w_i^T \\B \\w_i}.\n\\end{align*}\n\nGeneralized eigenvectors are defined up to a scaling factor. We may therefore define our $\\w_i$ to be scaled such that $\\w_i^T \\B \\w_i = 1$. We then have:\n\\[\nr(\\w) = \\frac{\\sum_i c_i^2 \\lambda_i}{\\sum_i c_i^2}.\n\\]\n%\nEach generalized Rayleigh quotient is thus a convex combination of generalized eigenvalues $\\lambda_i$. The maximum of a convex combination of one-dimensional points is obtained in the largest of these points. If $\\lambda_1$ is thus the largest generalized eigenvalue of $(\\A,\\B)$, then $\\max{r(\\w)} = \\lambda_1$.\n\nWe have thus shown that $\\argmax r(\\w) = \\w_1$, where $\\w_1$ is an eigenvector of $(\\A,\\B)$, and that its corresponding eigenvalue $\\lambda_1 = \\max r(\\w)$ is the largest of the eigenvalues of $(\\A,\\B)$.\n\n\\qed\n", "meta": {"hexsha": "6782fc8c280bfb1d474bbb8574ccee766d7a23a4", "size": 4186, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "modules/Appendices/GEVecs-maximise-SNR.tex", "max_stars_repo_name": "tfiers/master-thesis", "max_stars_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-23T01:39:24.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-23T01:39:24.000Z", "max_issues_repo_path": "modules/Appendices/GEVecs-maximise-SNR.tex", "max_issues_repo_name": "tfiers/master-thesis", "max_issues_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 46, "max_issues_repo_issues_event_min_datetime": "2018-09-18T16:38:12.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-10T22:37:35.000Z", "max_forks_repo_path": "modules/Appendices/GEVecs-maximise-SNR.tex", "max_forks_repo_name": "tfiers/master-thesis", "max_forks_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.0337078652, "max_line_length": 320, "alphanum_fraction": 0.6602962255, "num_tokens": 1362, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711870587667, "lm_q2_score": 0.8221891370573388, "lm_q1q2_score": 0.6988370768114492}} {"text": "\\section{Duality in Optimization}\n\n\\subsection{The Primal Problem}\n\n\\begin{frame}\n \\frametitle{The Primal Problem}\n \n \\begin{itemize}\n \\item Consider the \\structure{\\emph{primal optimization problem}}: \\\\[.4cm]\n \\begin{center}\n \\tikz[baseline]{\n \\node[fill=bl1!100,anchor=base,rounded corners=3pt] (d1) {\n \\color{bl3}\n $\\begin{aligned}\n \\displaystyle \n \\mbox{minimize~} & \\qquad f_0(\\vec{x}) \\\\[.3cm]\n \\mbox{subject to} & \\qquad f_i(\\vec{x}) \\leq 0, \\quad i=1,2,\\dots, m \\\\\n & \\qquad h_i(\\vec{x}) = 0, \\quad i=1,2,\\dots, p\n \\end{aligned}$\n };\n }\\\\[.4cm]\n \\end{center}\n with variable $\\vec{x}\\in\\real^n$. \\\\[.5cm]\n \\item The function $f_0(\\vec{x})$ is \\structure{not} required to be \\structure{convex}.\n \\end{itemize}\n\\end{frame}\n\n\n\\subsection{The Lagrangian}\n\n\\begin{frame}\n \\frametitle{The Lagrangian}\n \n \\begin{citeblock}{Lagrangian}\n\n The \\structure{\\emph{Lagrangian} $L$} of the aforementioned problem is defined as\n \\begin{displaymath}\n L(\\vec{x},\\vec{\\lambda}, \\vec{\\nu}) = \n f_0(\\vec{x}) + \\sum_{i=1}^m \\lambda_i f_i(\\vec{x}) + \\sum_{i=1}^p \\nu_i h_i(\\vec{x})\n \\end{displaymath}\n \\pause\n \n \\begin{itemize}\n \\item $\\lambda_i$ is the Lagrange multipliers associated with the $i$-th \\structure{inequality} \\\\\n constraint $f_i(\\vec{x})\\leq 0$. \\\\[.15cm] \\pause\n \\item $\\nu_i$ is the Lagrange multiplier associated with the $i$-th \\structure{equality} constraint $h_i(\\vec{x})= 0$. \\\\[.15cm] \\pause\n \\item The vectors $\\vec{\\lambda}$ and $\\vec{\\nu}$ are called {\\em Lagrange multiplier vectors} \\\\\n or simply \\structure{\\emph{dual variables}}. \n \\end{itemize}\n \\end{citeblock}\n\\end{frame}\n\n\n\\subsection{Lagrange Dual Function}\n\n\\begin{frame}\n \\frametitle{Langrange Dual Function}\n \n \\begin{citeblock}{Lagrange dual function}\n\n The \\structure{\\emph{Lagrange dual function}} is defined as the infimum of the Lagrangian \\\\\n over $\\vec{x}$\n \\begin{eqnarray*}\n g(\\vec{\\lambda,\\nu}) \n & = & \\inf_{\\vec{x}}L(\\vec{x},\\vec{\\lambda}, \\vec{\\nu}) \\\\ \\pause \n & = & \\inf_{\\vec{x}} \\left( \n f_0(\\vec{x}) + \\sum_{i=1}^m\\lambda_i f_i(\\vec{x}) + \n \\sum_{i=1}^p\\nu_i h_i(\\vec{x})\n \\right)\n \\end{eqnarray*}\n \\end{citeblock}\n \\pspread\n \n \\structure{Note:} \n \n \\begin{itemize}\n \\item The Lagrange dual function is a \\structure{pointwise affine function} \\\\\n in the dual variables. \\pause \n \\item The \\structure{Lagrange dual function is concave} \\\\\n (even if the original problem is not convex).\n \\end{itemize}\n\n\\end{frame}\n\n\n\\begin{frame}{Optimal Value and Lower Bound}\n\n \\begin{lemma}\n Let $p^*$ be the optimal value of the optimization problem. \\\\\n For any $\\vec\\lambda \\succeq 0$ and any $\\vec \\nu$ the following bound is valid:\n \\begin{displaymath}\n g(\\vec{\\lambda,\\nu}) \\leq p^* \n \\end{displaymath} \n \\end{lemma}\n\\end{frame}\n\n\n\\begin{frame}{Optimal Value and Lower Bound \\cont}\n \n \\begin{center}\n \\resizebox{.8\\linewidth}{!}{\n \\alt<13->{\n \\input{\\texfigdir/lagrange13.pstex_t}\n }{\\alt<12>{\n \\input{\\texfigdir/lagrange12.pstex_t}\n }{\\alt<11>{\n \\input{\\texfigdir/lagrange11.pstex_t}\n }{\\alt<10>{\n \\input{\\texfigdir/lagrange10.pstex_t}\n }{\\alt<9>{\n \\input{\\texfigdir/lagrange09.pstex_t}\n }{\\alt<8>{\n \\input{\\texfigdir/lagrange08.pstex_t}\n }{\\alt<7>{\n \\input{\\texfigdir/lagrange07.pstex_t}\n }{\\alt<6>{\n \\input{\\texfigdir/lagrange06.pstex_t}\n }{\\alt<5>{\n \\input{\\texfigdir/lagrange05.pstex_t}\n }{\\alt<4>{\n \\input{\\texfigdir/lagrange04.pstex_t}\n }{\\alt<3>{\n \\input{\\texfigdir/lagrange03.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/lagrange02.pstex_t}\n }{\n \\input{\\texfigdir/lagrange01.pstex_t}\n }}}}}}}}}}}}\n }\n \\end{center}\n\\end{frame} \n\n\n\\begin{frame}{Optimal Value and Lower Bound \\cont}\n\n \\begin{columns}\n \\column{.6\\linewidth}\n \\begin{center}\n \\resizebox{\\linewidth}{!}{\n \\input{\\texfigdir/dual.pstex_t}\n }\n \\end{center}\n \\column{.4\\linewidth}\n \\vspace{.3cm}\n \\begin{itemize}\n \\item Neither $f_0(x)$ nor $f_1(x)$ is convex,\n \\item but the dual function $g(\\lambda)$ is concave!\n \\end{itemize}\n \\end{columns}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Optimal Value and Lower Bound \\cont}\n\n Let $\\tilde{\\vec{x}}$ be a \\structure{feasible point} of the optimization problem. \\\\[.3cm]\n If $\\vec{\\lambda}\\succeq 0$, we have due to the \\structure{$m$ inequality} and \\structure{$p$ equality} constraints:\n \\begin{displaymath}\n \\sum_{i=1}^m \\lambda_i f_i(\\tilde{\\vec{x}}) + \n \\sum_{i=1}^p\\nu_i h_i(\\tilde{\\vec{x}}) \n \\leq 0~,\n \\end{displaymath}\n \\pause\n\n Thus we have\n \\begin{displaymath}\n L(\\tilde{\\vec{x}},\\vec{\\lambda},\\vec{\\nu}) =\n f_0 ({\\tilde{\\vec{x}}}) + \n \\sum_{i=1}^m\\lambda_i f_i(\\tilde{\\vec{x}}) + \n \\sum_{i=1}^p\\nu_i h_i(\\tilde{\\vec{x}}) \n \\leq f_0 ({\\tilde{\\vec{x}}})~.\n \\end{displaymath}\n\\end{frame}\n \n \n\\begin{frame}\n \\frametitle{Optimal Value and Lower Bound \\cont}\n \n Using the \\structure{definition of the dual function} we get:\n \n \\begin{displaymath}\n g(\\vec{\\lambda,\\nu}) = \n \\inf_{{\\vec{x}}} L({\\vec{x}},\\vec{\\lambda}, \\vec{\\nu}) \\leq\n L(\\tilde{\\vec{x}}, \\vec{\\lambda}, \\vec{\\nu}) \\leq\n% f_0(\\tilde{\\vec{x}}) + \n% \\sum_{i=1}^m\\lambda_i f_i(\\tilde{\\vec{x}}) + \n% \\sum_{i=1}^p\\nu_i h_i(\\tilde{\\vec{x}}) \\leq \n f_0(\\tilde{\\vec{x}})\n \\end{displaymath}\n \\pause\n \n \\vspace{.25cm}\n The inequality $g(\\vec{\\lambda,\\nu}) \\leq f_0(\\tilde{\\vec{x}})$ holds for \\structure{every feasible point $\\tilde{\\vec{x}}$}.\\\\[.5cm]\n \\pause\n\n Consequently, the dual function $g(\\vec{\\lambda,\\nu})$ is also smaller or equal to the optimal value $p^*$:\n \n \\begin{displaymath}\n g(\\vec{\\lambda,\\nu}) \\leq p^*\n \\end{displaymath}\n \\hfill \\qed %\\structure{\\ensuremath{\\blacksquare}}\n\\end{frame}\n\n\n\\subsection{The Lagrange Dual Problem}\n\n\\begin{frame}\n \\frametitle{The Lagrange Dual Problem}\n \n \\structure{Problem:} how to find the best lower bound for the primal problem \\\\[1cm]\n \n The Lagrange dual problem is given by the optimization problem:\n \n \\begin{eqnarray*}\n \\mbox{maximize} & & g(\\vec{\\lambda}, \\vec{\\nu}) \\\\[.3cm]\n \\mbox{subject to} & & \\vec{\\lambda}\\succeq 0\n \\end{eqnarray*}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{The Lagrange Dual Problem \\cont}\n \n \\begin{citeblock}{Optimal duality gap}\n\n Let ${p}^*$ be the optimal value of the primal problem and \\\\\n ${d}^*$ the optimal value of the Lagrange dual problem. \\\\[.3cm] \\pause\n \n \\begin{itemize}\n \\item The difference $p^*-d^*$ is the \\structure{\\emph{optimal duality gap}}. \\\\[.3cm] \\pause\n \\item If $p^*=d^*$, the duality gap is zero. \\\\\n In this case we talk about \\structure{\\emph{strong duality}}. \\\\[.3cm] \\pause\n \\item If $p^*> d^*$, we have \\structure{\\emph{weak duality}}.\n \\end{itemize}\n \\end{citeblock}\n\\end{frame}\n\n\n\\subsection{Slater's condition}\n\n\\begin{frame}\n \\frametitle{Slater's Condition}\n\n \\begin{citeblock}{Theorem}\n\n Given a \\structure{\\emph{convex} primal optimization problem}:\n \\small\n \\begin{eqnarray*}\n \\mbox{minimize~} & & \\quad f_0(\\vec{x}) \\\\\n \\mbox{subject to} & & \\quad f_i(\\vec{x}) \\leq 0, \\quad i=1,2,\\dots, m \\\\\n & & \\quad \\mat{A} \\vec{x} = \\vec{b}\n \\end{eqnarray*}\n \\normalsize\n with $f_0, f_1, \\ldots, f_m$ being convex. \\\\[.3cm]\n\n If there exists an $\\vec{x} \\in \\text{relint}~\\big\\{\\mathcal{D} = \\cap_{i=0}^m \\text{dom}(f_i) \\big\\}$ with\n \\small\n \\begin{align*}\n & f_i(\\vec{x}) < 0, \\quad i=1,\\dots,m \\\\\n & \\mat{A}\\vec{x} = \\vec{b} \n \\end{align*}\n then \\structure{strong duality} holds.\n \\end{citeblock}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Refinement of Slater's Condition}\n\n \\begin{citeblock}{Theorem}\n\n Given a \\structure{\\emph{convex} primal optimization problem}. \\\\[.2cm]\n\n If the first $k$ constraint functions $f_1, \\ldots, f_k$ are \\structure{\\emph{affine}}, and \\\\\n if there exists an $\\vec{x} \\in \\text{relint}~\\mathcal{D}$ with\n \\small\n \\begin{displaymath}\n \\renewcommand\\arraystretch{1.4} \n \\begin{array}{l l}\n f_i(\\vec{x}) \\le 0, \\quad i=1,\\dots,k & \\text{\\footnotesize (affine constraints)}\\\\\n f_i(\\vec{x}) < 0, \\quad i=k+1, \\ldots, m \\qquad \\quad & \\text{\\footnotesize (convex constraints)}\\\\\n \\mat{A}\\vec{x} = \\vec{b} & \\\\\n \\end{array}\n \\end{displaymath}\n then \\structure{strong duality} holds.\n \\end{citeblock}\n \\pause\n\n \\vspace{.3cm} \n \\structure{Note:} the refined Slater's condition reduces to \\structure{feasibility} when the constraints are all linear equalities and inequalities, and $\\text{dom}(f_0)$ is open. \n\\end{frame}\n\n\n\\subsection{KKT Optimality Conditions}\n\n\\begin{frame}\n \\frametitle{Karush-Kuhn-Tucker Optimality Conditions}\n\n Let $\\vec x^*$ be a primal and $(\\vec \\lambda^*, \\vec \\nu^*)$ dual optimal points with zero duality gap. \\\\[.3cm]\n\n For the primal optimal point $\\vec x^*$, the gradient with respect to $\\vec x$ of $L(\\vec x, \\vec \\lambda^*, \\vec \\nu^*)$ is \\vec{0}:\n \n \\begin{displaymath}\n \\nabla L(\\vec x^*, \\vec \\lambda^*, \\vec \\nu^*) = \n \\nabla f_0(\\vec{x}^*) + \n \\sum_{i=1}^m \\lambda_i^* \\nabla f_i(\\vec{x}^*) + \n \\sum_{i=1}^p \\nu_i^* \\nabla h_i(\\vec{x}^*) = \n \\vec{0}\n \\end{displaymath}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Karush-Kuhn-Tucker Optimality Conditions}\n \n The following four conditions are called KKT conditions: \\\\[3mm] \\pause\n \n \\begin{enumerate}\n \\item \\structure{Primal constraints:} \n \\begin{itemize}\n \\item $f_i(\\vec{x}) \\leq 0, \\quad i = 1, 2, \\dots, m$\n \\item $h_i(\\vec{x}) = 0, \\quad i = 1, 2, \\dots, p$ \\\\[3mm] \\pause\n \\end{itemize}\n \\item \\structure{Dual constraints:} $\\vec \\lambda \\succeq 0$ \\\\[3mm] \\pause\n \\item \\structure{Complementary slackness:} $ \\lambda_i \\, f_i(\\vec{x})= 0$ \\\\[3mm] \\pause\n \\item \\structure{Gradient of the Lagrangian $L$ is zero:}\n \\begin{displaymath}\n \\nabla L(\\vec x, \\vec \\lambda, \\vec \\nu) = \n \\nabla f_0(\\vec{x}) + \n \\sum_{i=1}^m\\lambda_i \\nabla f_i(\\vec{x}) + \n \\sum_{i=1}^p\\nu_i \\nabla h_i(\\vec{x}) = \n \\vec{0}\n \\end{displaymath}\n \\end{enumerate}\n \\pause\n \n If strong duality holds and if $\\vec x^*$ and $(\\vec \\lambda^*, \\vec \\nu^*)$ are optimal points, \\\\\n then the KKT conditions hold.\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Karush-Kuhn-Tucker Optimality Conditions}\n\n \\alt<15->{\n \\structure{Complementary slackness:} {\\color{gr3} $\\lambda_i^* \\cdot f_i(\\vec{x}^*) = 0$}\n }{\n \\structure{Complementary slackness}\n }\n \n \\def\\phan{\\phantom{\\stackrel{\\color{red} =}{\\cancel{\\color{black} \\le}}}}\n\n \\begin{eqnarray*}\n \\alt<10->{ \n \\tikz[baseline]{ \n \\node[fill=bl1,anchor=base,rounded corners=2pt] (start) {\n \\color{bl3} $ f_0(\\vec{x}^*) $\n };\n }\n }{\n \\tikz[baseline]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (res) {\n \\color{black} $ f_0(\\vec{x}^*) $\n };\n }\n }\n \\pause\n & = & g(\\vec{\\lambda}^*, \\vec{\\nu}^*) \\\\ \\pause\n & = & \\inf_{\\vec{x}} \\Bigg( f_0(\\vec{x}) + \\sum_{i=1}^m \\lambda_i^* f_i(\\vec{x}) + \\sum_{i=1}^p \\nu_i^* h_i(\\vec{x}) \\Bigg) \\\\ \\pause\n \\alt<12->{\n & \\stackrel{\\color{red} =}{\\cancel{\\color{black} \\le}} & \n }{\n &\\le& \n }\n f_0(\\vec{x}^*) + \n \\alt<15>{\n \\tikz[baseline,remember picture]{ \n \\node[fill=gr1,anchor=base,rounded corners=2pt] (c1) {\n \\color{gr3} $ \\displaystyle \\sum_{i=1}^m $ \n \\tikz \\node[fill=white,anchor=base, rounded corners=2pt] (cs) {\\color{gr3} $ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.35cm}, decorate, color=gr3] (cs.west) -- node[pos=0.5,xshift=0mm,yshift=-5.5mm] {\n \\tiny $= 0$\n } (cs.east); \n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=black] (c1.west) -- node[pos=0.5,xshift=.4mm,yshift=-1.1cm] {\n \\scriptsize $\\stackrel{\\color{red}=}{\\cancel{\\color{black}\\le}} 0$\n } (c1.east); \n }\n }{\\alt<14>{\n \\tikz[baseline,remember picture]{ \n \\node[fill=gr1,anchor=base,rounded corners=2pt] (c1) {\n \\color{gr3} $ \\displaystyle \\sum_{i=1}^m $ \n \\tikz \\node[anchor=base, rounded corners=2pt] (cs) {\\color{gr3} $ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=black] (c1.west) -- node[pos=0.5,xshift=.4mm,yshift=-1.1cm] {\n \\scriptsize $\\stackrel{\\color{red}=}{\\cancel{\\color{black}\\le}} 0$\n } (c1.east); \n }\n }{\\alt<13>{\n \\tikz[baseline,remember picture]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (c1) {\n $ \\displaystyle \\sum_{i=1}^m $\n \\tikz \\node[anchor=base, rounded corners=2pt] (cs) {$ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=black] (c1.west) -- node[pos=0.5,xshift=.4mm,yshift=-1.1cm] {\n \\scriptsize $\\stackrel{\\color{red}=}{\\cancel{\\color{black}\\le}} 0$\n } (c1.east); \n }\n }{\\alt<9->{\n \\tikz[baseline,remember picture]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (c1) {\n $ \\displaystyle \\sum_{i=1}^m$\n \\tikz \\node[anchor=base, rounded corners=2pt] (cs) {$ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=black] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize $\\phan\\le 0\\phan$} (c1.east); \n }\n }{\\alt<8>{\n \\tikz[baseline,remember picture]{ \n \\node[fill=bl1,anchor=base,rounded corners=2pt] (c1) {\n \\color{bl3} $ \\displaystyle \\sum_{i=1}^m$\n \\tikz \\node[fill=bl1, anchor=base, rounded corners=2pt] (cs) {$ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=bl3] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize $\\phan\\le 0\\phan$} (c1.east); \n }\n }{\\alt<7>{\n \\tikz[baseline,remember picture]{ \n \\node[fill=bl1,anchor=base,rounded corners=2pt] (c1) {\n \\color{bl3} $ \\displaystyle \\sum_{i=1}^m$\n \\tikz \\node[fill=bl1, anchor=base, rounded corners=2pt] (cs) {$ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=white] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize \\phantom{$\\le 0$}} (c1.east); \n }\n }{\n \\tikz[baseline,remember picture]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (c1) {\n \\color{black} $ \\displaystyle \\sum_{i=1}^m $\n \\tikz \\node[anchor=base, rounded corners=2pt] (cs) {$ \\lambda_i^* f_i(\\vec{x}^*) $};\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=white] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize \\phantom{$\\le 0$}} (c1.east); \n }\n }}}}}}\n + \n \\alt<7->{\n \\tikz[baseline]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (c1) {\n $ \\displaystyle \\sum_{i=1}^p \\nu_i^* h_i(\\vec{x}^*) $\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=black] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize $= 0$} (c1.east); \n }\n }{\\alt<6->{\n \\tikz[baseline]{ \n \\node[fill=bl1!100,anchor=base,rounded corners=2pt] (c1) {\n \\color{bl3} $ \\displaystyle \\sum_{i=1}^p \\nu_i^* h_i(\\vec{x}^*) $\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=bl3] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize $= 0$} (c1.east); \n }\n }{\\alt<5>{\n \\tikz[baseline]{ \n \\node[fill=bl1!100,anchor=base,rounded corners=2pt] (c1) {\n \\color{bl3} $ \\displaystyle \\sum_{i=1}^p \\nu_i^* h_i(\\vec{x}^*) $\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=white] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize \\phantom{$= 0$}} (c1.east); \n }\n }{\n \\tikz[baseline]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (c1) {\n \\color{black} $ \\displaystyle \\sum_{i=1}^p \\nu_i^* h_i(\\vec{x}^*) $\n };\n \\draw [thick, decoration={brace, mirror, raise=0.75cm}, decorate, color=white] (c1.west) -- node[pos=0.5,yshift=-1.1cm] {\\scriptsize \\phantom{$= 0$}} (c1.east); \n }\n }}} \\\\\n \\alt<11->{\n & \\stackrel{\\color{red} =}{\\cancel{\\color{black} \\le}} & \n }{\\alt<9->{\n &\\le& \n }{}}\n \\alt<10->{ \n \\tikz[baseline]{ \n \\node[fill=bl1,anchor=base,rounded corners=2pt] (res) {\n \\color{bl3} $ f_0(\\vec{x}^*) $\n };\n }\n }{\\alt<9>{\n \\tikz[baseline]{ \n \\node[fill=white,anchor=base,rounded corners=2pt] (res) {\n \\color{black} $ f_0(\\vec{x}^*) $\n };\n }\n }{}} \n \\phan\n \\end{eqnarray*}\n\n \\onslide<15>\n \\hfill \\qed\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Karush-Kuhn-Tucker Optimality Conditions \\cont}\n\n \\structure{Conclusions \\scriptsize ~(Boyd 2004, Sec. 5.5.3)}\n\n \\begin{itemize}\n \\item For \\structure{\\emph{any} optimization problem} with differentiable objective and constraint functions for which strong duality obtains, any pair of primal and dual optimal points must satisfy the KKT conditions. \\\\[.2cm] \\pause\n \\item For \\structure{any \\emph{convex} optimization problem} with differentiable objective and and constraint functions, any points that satisfy the KKT conditions are primal and dual optimal, and have zero duality gap. \\\\[.2cm] \\pause\n \\item If a \\structure{\\emph{convex} optimization problem} with differentiable objective and constraint functions satisfies Slater's condition, then the KKT conditions provide necessary and sufficient conditions for optimality.\n \\end{itemize}\n\\end{frame}\n\n\n\\subsection{Lessons Learned}\n\n\\begin{frame}\n \\frametitle{Lessons Learned}\n \n \\begin{itemize}\n \\item Formalization of the primal problem using the Lagrangian \\\\[.5cm]\n \\item Lagrange dual function \\\\[.5cm]\n \\item Duality gap \\\\[.5cm]\n \\item Karush-Kuhn-Tucker optimality conditions\n \\end{itemize}\n\\end{frame}\n\n\\input{nextTime.tex}\n\n\\subsection{Further Readings}\n\n\\begin{frame}\n \\frametitle{Further Readings}\n\n \\begin{itemize}\n \\item S.~Boyd, L.~Vandenberghe: \\\\\n \\structure{Convex Optimization}, \\\\\n Cambridge University Press, 2004. \\\\\n \\point{\\small \\url{http://www.stanford.edu/~boyd/cvxbook/}} \\\\[.25cm]\n \\item Jorge Nocedal, Stephen Wright: \\\\\n \\structure{Numerical Optimization}, \\\\\n Springer, New York, 1999.\n \\end{itemize}\n\\end{frame}\n\n\n\\subsection{Comprehensive Questions}\n\n\\begin{frame}\n \\frametitle{Comprehensive Questions}\n\n \\begin{itemize}\n \\item What is the Lagrangian of a constrained objective function? \\\\[1cm]\n \\item What is the Lagrange dual function? \\\\[1cm]\n \\item What is the duality gap? \\\\[1cm]\n \\item What are the Karush-Kuhn-Tucker optimality conditions?\n \\end{itemize}\n\\end{frame}\n", "meta": {"hexsha": "cdd74dd0af53fe95c6ade32929266e7e93296353", "size": 19915, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "13_optimization_2.tex", "max_stars_repo_name": "akmaier/pr-slides", "max_stars_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2021-01-11T07:27:09.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-31T19:21:31.000Z", "max_issues_repo_path": "13_optimization_2.tex", "max_issues_repo_name": "akmaier/pr-slides", "max_issues_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": 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YES\n2. YES", "lm_q1_score": 0.8499711870587668, "lm_q2_score": 0.8221891305219504, "lm_q1q2_score": 0.6988370712565576}} {"text": "\\section{Co-Synthesis of Encodings} \\label{sec:co-synthesis}\n%\nThe functions described in the previous section only makes\nsense in the context of a fixed encoding, but in general, a\ndata type accepts more than one.\n%\nAs encodings are just expressions we would like to\nautomatically derive them as a result of synthesis.\n%\nTo do so we first need to find a way to give an abstract\nspecification of the meaning of a data type and we need to\ndo it in a way that is independent of an underlying\nencoding.\n%\nIn the following, we see how we can extend the ideas\npresented in \\cref{sec:synthesis} to express these abstract\nspecifications in a natural way.\n\nIn general, our approach consist of specifying the behavior\nof data types in relation with the functions that operates\non them.\n%\nWe call this a \\emph{co-specification} of a data type and\nits functions.\n%\nWe also refer to the process of synthesizing an encoding\ntogether with the functions that operate on it as\n\\emph{co-synthesis}.\n\nOur next key insight is that we can build natural and\nconcise co-specifications by looking at the introduction and\nelimination forms of a data type.\n%\nConsider booleans for example.\n%\nTheir introduction corresponds to the values \\true and \\false\nand their elimination corresponds to case analysis using the\nif-then-else (\\f{ite}) function.\n%\nUsing the ideas presented in \\cref{sec:io} we can co-specify\n\\true, \\false and \\f{ite} as:\n%\n\\begin{align*}\n \\f{?ite}\\;\\f{?true}\\;x\\;y &\\equiv x \\\\\n \\f{?ite}\\;\\f{?false}\\;x\\;y &\\equiv y\n\\end{align*}\n\nThe co-specification for natural numbers requires a little\nbit more work.\n%\nFirst, we identify $0$ and \\f{succ} as the introduction\nforms of natural numbers in its usual inductive definition.\n%\nSecond, we note that the elimination form of natural numbers\ncorrespond to the ability of defining recursive functions on\nthem.\n%\nInstead of spelling out a general recursive principle we\nspecify the recursive behavior by picking a particular\nrecursive definition.\n%\nIn our case we specify the elimination form in terms of the\nrecursive definition of the function \\f{plus}.\n%\nWe can now co-specify $0$, \\f{succ} and \\f{plus} as:\n\\begin{align*}\n \\f{?plus}\\;\\f{?}0\\;y &\\equiv y\\\\\n \\f{?plus}\\;(\\f{?succ}\\;x)\\;y &\\equiv \\f{?succ}\\;(\\f{?plus}\\;x\\;y)\\\\\n\\end{align*}\n\nWe conclude by showing how these ideas also apply to\nnon-recursive data types by showing a co-specification for pairs.\n%\nThe introduction form corresponds to the constructor\n\\f{pair} and the elimination forms to the first and\nsecond projections.\n%\n\\begin{align*}\n \\f{?first}\\;(\\f{?pair}\\;x\\;y) &\\equiv x\\\\\n \\f{?second}\\;(\\f{?pair}\\;x\\;y) &\\equiv y\\\\\n\\end{align*}\n", "meta": {"hexsha": "1cf689804181c5b02e23c223415975a58e153e61", "size": 2648, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/co-synthesis.tex", "max_stars_repo_name": "DavidThien/elsa", "max_stars_repo_head_hexsha": "2bf97839d1fc210d12dac919b34ec4e0143f80e6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/co-synthesis.tex", "max_issues_repo_name": "DavidThien/elsa", "max_issues_repo_head_hexsha": "2bf97839d1fc210d12dac919b34ec4e0143f80e6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/co-synthesis.tex", "max_forks_repo_name": "DavidThien/elsa", "max_forks_repo_head_hexsha": "2bf97839d1fc210d12dac919b34ec4e0143f80e6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.2926829268, "max_line_length": 69, "alphanum_fraction": 0.7454682779, "num_tokens": 734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711794579722, "lm_q2_score": 0.8221891239865619, "lm_q1q2_score": 0.698837059452375}} {"text": "\\documentclass[10pt]{article}\n\n\\usepackage{dsfont}\n\\usepackage{amssymb}\n\\usepackage{mathtools}\n\\usepackage{geometry}\n\\geometry{a4paper}\n\\usepackage[parfill]{parskip}\n\\usepackage{graphicx}\n\\usepackage{epstopdf}\n\\usepackage{listings}\n\\lstset{language = C++}\n\\usepackage{url}\n\n\\newcommand{\\re}[1]{\\ensuremath{\\operatorname{Re}(#1)}}\n\\newcommand{\\im}[1]{\\ensuremath{\\operatorname{Im}(#1)}}\n\n%\\newcommand{\\Vr}{\\re{V}}\n%\\newcommand{\\Vi}{\\im{V}}\n%\\newcommand{\\Ir}{\\re{I}}\n%\\newcommand{\\Ii}{\\im{I}}\n%\\newcommand{\\Sr}{\\ensuremath{P}}\n%\\newcommand{\\Si}{\\ensuremath{Q}}\n\n\\newcommand{\\Vr}{{V_R}}\n\\newcommand{\\Vi}{{V_I}}\n\\newcommand{\\Ir}{{I_R}}\n\\newcommand{\\Ii}{{I_I}}\n\\newcommand{\\Sr}{\\ensuremath{P}}\n\\newcommand{\\Si}{\\ensuremath{Q}}\n\n\\newcommand{\\Id}{\\mathds{1}}\n\n\\title{Power Flow Theory}\n\\author{Dan Gordon}\n\\date{}\n\n\\begin{document}\n\\maketitle\n\n\\section{Nomenclature}\n\\begin{align*}\nV &= \\Vr + \\Vi = \\text{complex voltage.} \\\\\nI &= \\Ir + j\\Ii = \\text{complex current.} \\\\\nS &= P + jQ = \\text{complex power.} \\\\\ny &= g + jb = \\text{complex admittance} \\\\\nZ &= R + jX = \\text{complex impedance} \\\\\ny_{ik} &= \\text{complex admittance between buses $i$ and $k$.} \\\\\nY_{ik} &= G_{ik} + jB_{ik} = \\text{nodal admittance matrix element $i, k$.} \\\\\n\\theta_{i} &= \\text{the voltage angle of bus $i$.} \\\\\n\\theta_{ik} &= \\theta_i - \\theta_k = \\text{the voltage angle difference between buses $i$ and $k$.} \\\\\nI_{\\text{ZIP},i} &= \\text{total complex current injection from load at bus $i$.} \\\\\nI_{\\text{gen},i} &= \\text{complex current injection due to the voltage controlled generation at bus $i$.} \\\\\nS_{\\text{gen},i} &= \\text{complex power injection due to the voltage controlled generation at bus $i$.} \\\\\nS_{ci} &= \\text{constant power injection ZIP component of load at bus $i$.} \\\\\nI_{ci} &= \\text{constant current injection ZIP component of load.} \\\\\ny_{ci} &= \\text{constant impedance ZIP component of load.} \\\\\n\\delta_{ik} &= \\text{the Kronecker delta, $\\delta_{ik} = 1$ if $i = k$, 0 otherwise.}\n\\end{align*}\n\n\\section{AC power}\nFor AC power, we can write $V(t) = V_\\text{max} \\cos(\\theta_V + \\omega t)$ and $I(t) = I_\\text{max} \\cos(\\theta_I + \\omega t)$, where $V_\\text{max}$ and $I_\\text{max}$ are the maximum power and voltage over a cycle. However, it is more usual to use RMS quantities $V_\\text{RMS} = V_\\text{max}/\\sqrt{2}$ and $I_\\text{RMS} = I_\\text{max}/\\sqrt{2}$, because the average power over a cycle then mimics the usual relationships such as $P = V^2/R$\\footnote{For a resistive load, the average power over a cycle works out to be $P_\\text{avg} = I_\\text{RMS}V_\\text{RMS}$, replicating the usual relationship for DC power. If we used the maximum rather than RMS values of $V$ and $I$, then this relationship would have an extra factor of 1/2. All the other usual DC relationships also hold for the RMS values, such as $V_\\text{RMS} = I_\\text{RMS}R$. We are not often concerned with knowing the instantaneous voltage and current, so the factors of $\\sqrt{2}$ in Eq. \\ref{EQ_V_I_RMS} are not too inconvenient.}, so,\n\\begin{align}\n\tV(t) &= \\sqrt{2}V_\\text{RMS} \\cos(\\theta_V + \\omega t), \\notag \\\\\n\tI(t) &= \\sqrt{2}I_\\text{RMS} \\cos(\\theta_I + \\omega t)\n\t\\label{EQ_V_I_RMS}\n\\end{align}\n\nIt is convenient to use the complex phasor notation $V := V_\\text{RMS} \\exp(j\\theta_V), \\, I := I_\\text{RMS} \\exp(j\\theta_I)$, which allows us to unambiguously code the voltage magnitude and phase shift in a single complex number. If we ever need to recover, for example, the instantaneous voltage, we can use\n\\begin{align}\n\tV(t) = \\sqrt{2}\\re{V \\exp(j\\omega t)}\n\\end{align}\n\nThe instantaneous power is\n\\begin{align}\n\tP(t) &= 2I_\\text{RMS}V_\\text{RMS}\\cos(\\theta_I + \\omega t)\\cos(\\theta_V + \\omega t) \\notag \\\\\n\t &= I_\\text{RMS}V_\\text{RMS}\\left[\\cos(\\theta_V - \\theta_I) + \\cos(\\theta_V + \\theta_I + 2 \\omega t)\\right] \\notag \\\\\n\t\t &= \\re{I^*V + IV\\exp(2j \\omega t)}\n\\end{align}\nThis inspires us to define the instantaneous complex power,\n\\begin{align}\n\tS(t) &= I^*V + IV\\exp(2j \\omega t)\n\\end{align}\nwhose average over a cycle is\n\\begin{align}\n\tS &= I^*V,\n\\end{align}\nWe define the real power and reactive power phasors to be \n\\begin{align}\n\tP := \\re{S} = I_\\text{RMS}V_\\text{RMS}\\cos(\\theta_V - \\theta_I), \\notag \\\\\n\tQ := \\im{S} = I_\\text{RMS}V_\\text{RMS}\\sin(\\theta_V - \\theta_I)\n\\end{align}\nwhich shows that $P$ and $Q$ are ``in phase'' and ``out of phase'' components of the product of $I$ and $V$. $P$ is just the average power dissipated by the resistive component of the load, while $Q$ is the magnitude of the instantaneous power absorbed by the reactive part of the load.\n\n\\subsection{Reactive power, voltage and line losses}\nThe loss of real power over a section of transmission line is given by\n\\begin{align}\n\tP_\\text{loss} &= \\left|I\\right|^2 R\n\\end{align}\nWe can express $I$ as a function of the voltage and power injection to the downstream (or upstream) bus, e.g.\n\\begin{align}\n\tI &= \\frac{S^*_\\text{out}}{V^*_\\text{out}}\n\\end{align}\nTherefore, we have\n\\begin{align}\n\tP_\\text{loss} &= \\frac{\\left|S_\\text{out}\\right|^2}{\\left|V_\\text{out}\\right|^2}R\n\\end{align}\n\nThis equation explains, firstly, why transmission lines are more efficient at high voltage (because, e.g. doubling the voltage means the line loss is reduced by a factor of four), and secondly, why utilities are keen to reduce the reactive power flowing through lines (because doubling the reactive power multiplies the line losses by four). Line losses are bad for three reasons: they are a waste of energy, they correspond to dissipated heat and therefore thermal limits, and they also correspond to voltage drops in the line.\n\nWe have\n\\begin{align}\n\t\\left|\\frac{V_\\text{out}}{V_\\text{in}}\\right|^2 &= \\left|\\frac{V_\\text{in} - IZ}{V_\\text{in}}\\right|^2 \\notag \\\\\n\t&= \\left|1 - \\frac{S^*_\\text{in}Z}{\\left|V_\\text{in}\\right|^2}\\right|^2 \\notag \\\\\n\t&= 1 - \\frac{2(PR + QX)}{\\left|V_\\text{in}\\right|^2} + \\frac{\\left|S\\right|^2\\left|Z\\right|^2}{\\left|V_\\text{in}\\right|^4}\n\\end{align}\nAt high voltages, the middle term in the sum dominates the voltage drop. High voltage lines normally also have a larger reactance than resistance, and in such cases we see that reactive power is a more significant contributor to the voltage drop than is real power.\n\\section{Three phase balanced power}\nIn balanced positive sequence three-phase systems, the phase to neutral voltages (or, simply, the phase voltages), at a bus, are \n\\begin{align}\n(V_A, V_B, V_C) = (V_A, V_A\\exp(-j2\\pi/3), V_A\\exp(j2\\pi/3)).\n\\end{align}\nThe line to line voltages, also just called the line voltages, are \n\\begin{align}\n(V_{AB}, V_{BC}, V_{CA}) &= (V_A - V_B, V_B - V_C, V_C - V_A) \\notag \\\\\n&= \\sqrt{3}\\exp(j\\pi/6)(V_A, V_B, V_C).\n\\end{align}\nThus, the line to line voltages have a magnitude $\\sqrt{3}$ times that of the phase voltages, and they also form a positive sequence set.\n\nLoads and sources can be connected in either $\\Delta$ or $Y$ configuration. Fig. \\ref{FIG_YY_DD} shows these connections.\n\\begin{figure}[!h]\n\t\\begin{center}\n\t\t\\includegraphics[width=(12cm)]{YSrcYLd.png}\n\t\t\\includegraphics[width=(12cm)]{DSrcDLd.png}\n\t\\end{center}\n\t\\caption{\n\t\tTop: A Y-connected source connected to a Y-connected load. Bottom: A $\\Delta$-connected source connected to a $\\Delta$-connected load.\n\t}\n\t\\label{FIG_YY_DD}\n\\end{figure}\n\nThe currents along the phase wires, $I_A, I_B, I_C$ are known as the ``line currents''. Note that this can be confusing when compared to the use of ``line voltage'' to mean line to line voltage.\n\nFor a $\\Delta$-connected load or source operating under balanced positive sequence conditions, we can easily show, using Ohm's law, that $I_{BC} = \\exp(-j2\\pi/3) I_{AB}$ and $I_{CA} = \\exp(j2\\pi/3) I_{AB}$, i.e. the delta currents also form a positive sequence set. Kirchoff's current law can then be used to show that,\n\\begin{align}\n(I_{AB}, I_{BC}, I_{CA}) &= \\frac{1}{\\sqrt{3}}\\exp\\left(j\\pi/6\\right)(I_A, I_B, I_C)\n\\end{align}\nmeaning the $\\Delta$ currents are a factor of $\\sqrt{3}$ less than the line currents. Note that this is opposite to the relationship that holds for phase voltages vs line to line voltages.\n\nNote that for every $\\Delta$ connected load, there is an equivalent $Y$-connected load. For balanced power flow, if the impedance on each arm of the $Y$ is $Z_Y$, then the equivalent impedance for each side of the $\\Delta$ is $Z_\\Delta = 3Z_Y$.\n\n\\section{Symmetric components}\nLet \n\\begin{align}\n\tA &= \\left[\n\t\t\\begin{array}{lll}\n\t\t\t1 & 1 & 1 \\\\\n\t\t\t1 & \\alpha^2 & \\alpha \\\\\n\t\t\t1 & \\alpha & \\alpha^2\n\t\t\\end{array}\n\t\\right],\\;\n\tA^{-1} = \\frac{1}{3}\\left[\n\t\t\\begin{array}{lll}\n\t\t\t1 & 1 & 1 \\\\\n\t\t\t1 & \\alpha & \\alpha^2 \\\\\n\t\t\t1 & \\alpha^2 & \\alpha\n\t\t\\end{array}\n\t\\right]\n\\end{align}\nwhere $\\alpha := \\exp(j2\\pi/3)$.\n\nWe define a transformation to symmetric components:\n\\begin{align}\n\tV_{012} = A^{-1}V_{ABC}\n\\end{align}\nwhich applies for all vectors e.g. to $I$ as well as $V$, and\n\\begin{align}\n\tZ_{012} = A^{-1}Z_{ABC}A\n\\end{align}\nwhich applies for matrices, for example, impedance matrices $Z$, as shown above.\n\nSuppose we have a balanced positive sequence voltage $V^+_{ABC} = V_A[1, \\alpha^2, \\alpha]^T$. Then, under the transformation to symmetric components, we find $V_{012} = V_A[0, 1, 0]^T$. Similarly, a negative sequence voltage $V^-_{ABC} = V_A[1, \\alpha, \\alpha^2]^T$ transforms to $V_{012} = V_A[0, 0, 1]^T$. Finally, the unbalanced \\emph{zero sequence} voltage $V^-_{ABC} = V_A[1, 1, 1]^T$ transforms to $V_{012} = V_A[1, 0, 0]^T$. Therefore, the transformation $A^{-1}$ decomposes $V_{ABC}$ into a positive sequence component $V_1$, a negative sequence component $V_2$, and a zero sequence component $V_0$. In summary:\n\\begin{align}\nV_{ABC}^+ = V_A\\left[\\begin{array}{l} \n\t1 \\\\ \\alpha^2 \\\\ \\alpha \n\\end{array}\\right] &\\Longleftrightarrow \nV_{012}^+ = V_A\\left[\\begin{array}{l} \n\t0 \\\\ 1 \\\\ 0\n\\end{array}\\right] & \\text{(positive sequence voltage)}\\notag \\\\\nV_{ABC}^- = V_A\\left[\\begin{array}{l} \n\t1 \\\\ \\alpha \\\\ \\alpha^2\n\\end{array}\\right] &\\Longleftrightarrow \nV_{012}^- = V_A\\left[\\begin{array}{l} \n\t0 \\\\ 0 \\\\ 1\n\\end{array}\\right] & \\text{(negative sequence voltage)}\\notag \\\\\nV_{ABC}^0 = V_A\\left[\\begin{array}{l} \n\t1 \\\\ 1 \\\\ 1\n\\end{array}\\right] &\\Longleftrightarrow \nV_{012}^0 = V_A\\left[\\begin{array}{l} \n\t1 \\\\ 0 \\\\ 0\n\\end{array}\\right] & \\text{(zero sequence voltage)}\\notag \\\\\n\\end{align}\n\n\\subsection{Reduction of a three-phase balanced problem to single phase}\n\nIn balanced power flow, there is only a positive sequence component, which suggests that symmetric components could be used to reduce a three-phase problem to an equivalent single-phase problem, since the negative and zero sequence components will always be zero. However, using the treatment above, it turns out that we then need to remember to multiply by an extra factor of three when calculating the total power. This is not ideal, because total power is an important physical quantity that corresponds to the amount of actual energy consumed or generated. Therefore, we shall develop a non-standard treatment of symmetric components.\n\nLet the scaled sequence transformation matrix $B$ be defined as\n\\begin{align}\n\tB := \\frac{1}{\\sqrt{3}}A &= \\frac{1}{\\sqrt{3}}\\left[\n\t\t\\begin{array}{lll}\n\t\t\t1 & 1 & 1 \\\\\n\t\t\t1 & \\alpha^2 & \\alpha \\\\\n\t\t\t1 & \\alpha & \\alpha^2\n\t\t\\end{array}\t\\right], \\notag \\\\\n\tB^{-1} := \\sqrt{3} A^{-1} &=\\frac{1}{\\sqrt{3}}\\left[\n\t\t\\begin{array}{lll}\n\t\t\t1 & 1 & 1 \\\\\n\t\t\t1 & \\alpha & \\alpha^2 \\\\\n\t\t\t1 & \\alpha^2 & \\alpha\n\t\t\\end{array} \\right]\n\\end{align}\nNote that $B^{-1} = B^\\dag$, so $B$ is a unitary matrix\\footnote{$B$ being unitary implies that the transformation that it describes will have nice properties like preserving scalar (dot) products. The total power, $S$, is a scalar product: $S = I^\\dag V$, so this transformation preserves the value obtained for the total power.}. The transformation to scaled symmetric components is:\n\\begin{align}\n\tV_{012'} &= B^{-1}V_{ABC} \\notag \\\\\n\tZ_{012'} &= B^{-1}Z_{ABC}B\n\\end{align}\n\nUnder this transformation, positive, negative and zero sequence voltages transform as follows\n\\begin{align}\nV_{ABC}^+ = V_A\\left[\\begin{array}{l} \n\t1 \\\\ \\alpha^2 \\\\ \\alpha \n\\end{array}\\right] &\\Longleftrightarrow \nV_{012'}^+ = \\sqrt{3}V_A\\left[\\begin{array}{l} \n\t0 \\\\ 1 \\\\ 0\n\\end{array}\\right] & \\text{(positive sequence voltage)}\\notag \\\\\nV_{ABC}^- = V_A\\left[\\begin{array}{l} \n\t1 \\\\ \\alpha \\\\ \\alpha^2\n\\end{array}\\right] &\\Longleftrightarrow \nV_{012'}^- = \\sqrt{3}V_A\\left[\\begin{array}{l} \n\t0 \\\\ 0 \\\\ 1\n\\end{array}\\right] & \\text{(negative sequence voltage)}\\notag \\\\\nV_{ABC}^0 = V_A\\left[\\begin{array}{l} \n\t1 \\\\ 1 \\\\ 1\n\\end{array}\\right] &\\Longleftrightarrow \nV_{012'}^0 = \\sqrt{3}V_A\\left[\\begin{array}{l} \n\t1 \\\\ 0 \\\\ 0\n\\end{array}\\right] & \\text{(zero sequence voltage)}\\notag \\\\\n\\end{align}\nSo for example, a positive sequence balanced three phase voltage will have a single nonzero sequence component, the positive sequence component $V_1 = \\sqrt{3}V_A$, with magnitude equal to the line voltage. A similar relationship holds for the line currents: a positive sequence balanced three phase current will have $I_1 = \\sqrt{3}I_A$, however, this time, there is no simple interpretation of $\\sqrt{3}I_A$ that is analogous to $\\sqrt{3}V_A$ being the line to line voltage.\n\nNow consider a three phase transmission line. The equation for the voltage drop in each wire is:\n\\begin{align}\n\t\\Delta V = ZI\n\\end{align}\nwhere $Z$ is the phase impedance matrix for the line. If the power flow is to be perfectly balanced, then there must be a symmetry under exchange of any two phases, meaning all wires must either be equidistant from each other, or else transposed\\footnote{Line transposition occurs in transmission systems, where the physical locations of the phases on the poles on are interchanged at regular intervals, so for example the lines are arranged as ABC--BCA--CAB.}, and must have equal impedances and symmetrical mutual impedances. $Z$ must therefore have the form:\n\\begin{align}\n\tZ_{ABC} = \\left[\n\t\t\t\\begin{array}{lll}\n\t\t\t\tz & m & m \\\\\n\t\t\t\tm & z & m \\\\\n\t\t\t\tm & m & z\n\t\t\t\\end{array}\n\t\t\\right]\n\t\t\\label{EQ_ZABC}\n\\end{align}\nwhere $z$ is the single line impedance, and $m$ is the mutual coupling impedance. Under the transformation to scaled symmetric components, this becomes\n\\begin{align}\n\tZ_{012} = \\left[\n\t\t\t\\begin{array}{lll}\n\t\t\t\tz+2m & 0 & 0 \\\\\n\t\t\t\t0 & z-m & 0 \\\\\n\t\t\t\t0 & 0 & z-m\n\t\t\t\\end{array}\n\t\t\\right]\n\t\t\\label{EQ_Z012}\n\\end{align}\nThis matrix is diagonal, which confirms that a perfectly symmetric network can be decomposed into three independent networks: a positive sequence network, a negative sequence network, and a zero sequence network. Note also that the positive and negative sequences are equal. In balanced power flow, a single positive sequence impedance only needs to be specified for normal operation, whereas the zero-sequence impedance needs to be additionally specified for unbalanced conditions such as non-symmetric faults. Shunt impedances or loads can be represented in a similar manner, but have $m = 0$. $\\Delta$ connected shunt impedances or loads can be treated using the equivalent $Y$ impedance: each impedance in the $\\Delta$ load must be divided by 3 to get the equivalent $Y$-impedance.\n\nIn summary, we can model a balanced three phase power problem as a single phase problem by making the following interpretations:\n\\begin{itemize}\n\t\\item The voltage in the equivalent single phase system is the line to line voltage in the three phase system (but, to be pedantic, with the same angle as A). To get the phase A voltage, divide the single phase voltage by $\\sqrt{3}$.\n\t\\item To get the current along each phase of a three phase line, divide the single phase current by $\\sqrt{3}$. This is most important when considering line limits.\n\t\\item Power in the equivalent single phase system is the \\emph{total} power of all phases in the three phase system. Thus, for example, the power at an equivalent single phase $PQ$ bus is the sum of the power for all phases in the three phase system. This is regardless of whether the load or source is $\\Delta$ or $Y$-connected.\n\t\\item Three-phase lines represented as in Eq. \\ref{EQ_ZABC} have an equivalent single phase impedance of $z - m$.\n\t\\item A Y-connected load/shunt impedance, with impedance $z$ on each phase, becomes a single impedance $z$ in the equivalent single phase system.\n\t\\item A $\\Delta$-connected load/shunt impedance, with impedance $z$ between each pair of phases, becomes a single load impedance $z/3$ in the equivalent single phase system.\n\t\n\\end{itemize}\n\n\\section{The powerflow problem}\n\nWe start with a collection \\emph{buses} -- terminal-like conductors that have a single associated voltage, $V_i$, where $i$ is the bus index. We can consider \\emph{ground} to be a special bus (to which we assign index 0), at which $V$ is always zero.\n\nCurrent can flow between buses, via the network. The current flowing from bus $i$ to $k$ is $I_{ik}$. \n\nCurrent that flows into a bus from ground is an \\emph{injection} into the bus. We use the notation $I_{\\text{inj},i}$ to denote a current injection from ground to bus $i$. The generated power is $S_{\\text{inj}, i} = I_{\\text{inj}, i}^*(V_i - V_i)$. The generated power -- that is, the power \\emph{injection} into the bus -- is the negative of this: $S_{\\text{inj},i} := I_{\\text{inj},i}^*V_i$.\n\n\\emph{Kirchoff's current law} states that the total current flowing into a bus (from other buses and ground) is zero -- in other words, current in equals current out; electrical charges don't accumulate at buses. Thus\n\\begin{align}\n\tI_{i} - \\sum_{k \\ne i}{I_{ik}} &= 0\n\t\\label{EQ_KCL}\n\\end{align}\n\nThe aim of the powerflow problem is, essentially, to find all bus voltages and currents\\footnote{This will also trivially allow us to calculate power flows in the system -- we will discuss this later.}. To do so, we need to define the properties of the network to provide a functional relationship between current and voltage at the buses. We start with a simple formulation, where the current flow between any two buses is ohmic: it is defined by a fixed impedance between them. This describes a network of passive \\emph{lines}. Later we will generalise this relationship, so that, for example, we can include transformers.\n\nSuppose that $y_{ik} = y_{ki}$ is an impedance between buses $i$ and $k$. Then Ohm's law states that\n\\begin{align}\n\tI_{ik} &= y_{ik}(V_i - V_k)\n\\end{align}\nSubstituting this into KCL, Eq. (\\ref{EQ_KCL}), gives\n\\begin{align}\n\tI_{\\text{inj},i} - \\sum_{k \\ne i}{y_{ik}(V_i - V_k)} &= 0\n\t\\label{EQ_PF_1}\n\\end{align}\nwhich can be written as the matrix equation\n\\begin{align}\n\tI_\\text{inj} - YV &= 0\n\\end{align}\nwhere\n\\begin{align}\n\tY_{ik} &=\n\t\t\\begin{cases}\n\t\t\t-y_{ik}&\\text{if $i \\ne k$} \\\\\n\t\t\t\\sum_{l \\ne i} y_{il}& \\text{if $i = k$}\n\t\t\\end{cases}\n\t\\label{EQ_YNODE_OHMIC}\n\\end{align}\nHowever, in what follows, we do not make further use of the particular form of $Y$ expressed in Eq. (\\ref{EQ_YNODE_OHMIC}) above. We simply require that $Y$ be a constant matrix, so that there is a linear relationship between $I$ and $V$. We shall see later that including components such as shunts and transformers allows us to retain the linear relationship while modifying the particular form of $Y$.\n\nDue to the basic relation $S = I^*V$, the power loss between bus $i$ and $k$ is $S_{ik} = I_{ik}^*(V_i - V_k)$. Note that power is a scalar: if we reverse the order of the buses, then the power loss is unchanged, in contrast to, say, a current, where reversing the order of the buses will negate the current.\n\nMore commonly, though, we deal with the power entering or leaving a bus rather than the power lost in a line. Consider a current leaving a bus. If the current is a draw to ground, or if the system is radial and the current is in the direction of the downstream loads, all of this current will eventually end up at ground. The voltage drop in this process will just be the voltage of the bus. Thus, the power eventually transferred to ground will be \n\\begin{align}\n\tS_\\text{draw} = I_\\text{draw}^*V_\\text{bus}\n\\end{align}\nor, equivalently, for injections\n\\begin{align}\n\tS_\\text{inj} = I_\\text{inj}^*V_\\text{bus}\n\\end{align}\nNote the way in which this concept of power is different to the concept of power loss in a line: the voltage in the equation is always the voltage of a bus relative to ground, and the current is always an injection or draw on a bus (rather than between two buses). There is an implied directionality: if the power draw from a bus is positive, then power is considered to be flowing from the bus to ground and we have a load.\n\nThe one piece of the puzzle we haven't yet dealt with is the current injections $I_\\text{inj}$ into the buses. If the current injections were all constant, then the powerflow equations could be immediately solved by standard linear algebra. However, typically, the current injections from ground will depend on the bus voltage, and it is here that the real complexity of powerflow modelling comes into play.\n\nCurrent injections take the form of \\emph{loads} or \\emph{generation}. Load currents can often be modeled using the \\emph{ZIP} model:\n\\begin{align}\n\tI_\\text{ZIP} = \\frac{I_cV}{|V|} + \\frac{S^*_c}{V^*} - y_cV\n\\end{align}\nWhere $I_\\text{ZIP}$ is the total current injection of the ZIP, $I_c$ and $S_c$ are constant current\\footnote{Note the expression $I_cV/|V|$ for the constant current component. The voltage dependence ensures that the complex power $S$ does not depend on the absolute phase of the voltage. Instead we have $S = I^*_c |V|$, which is much more reasonable for realistic devices.} and power injections and $y_c$ is a constant impedance. The ZIP model is a good model for many types of load, but could also be used for certain types of generation - e.g. fixed power generation. If the real power injection $P = \\re{I_\\text{ZIP}^*V}$ is negative, then we have a load; if it is positive, then we have a generator.\n\nIn addition to the ZIP loads at a bus, we may include generators that apply some kind of extra voltage control on the bus to maintain power stability in the system. A bus that only has a ZIP load is termed a PQ bus. The term PQ means constant P, constant Q.\\footnote{Confusingly, though, the presence of fixed current and fixed impedance components of the ZIP means that the total power injection may not be constant. This confusion arises because ZIP loads provide a generalisation to the original concept of PQ buses.}\n\nWe consider two other types of generator bus: SL (slack/swing/infinite/reference) buses, and PV buses. SL generators keep the total complex voltage of the bus fixed by injecting variable $P$ and $Q$. $PV$ buses keep the voltage magnitude $V$ and power $P$ fixed, while injecting a variable current\\footnote{Note that the voltage angle is relative to the slack bus, and so is a non-local property of the network, hence it usually does not make sense to control it anywhere except at the slack bus.}. We write the additional generator current as a power $S_{gen}$ to simplify the later analysis.\n\\begin{align}\nI_{gen} &= \\frac{S_{gen}^*}{V^*}\n\\end{align}\nNote that SL and PV buses may also have ZIP loads attached.\n\nWe now write the final form of the powerflow equations:\n\\begin{align}\nI_{\\text{inj},i} &= \\frac{S^*_{ci} + S^*_{\\text{gen},i}}{V^*_i} + \\frac{I_{ci}V_i}{|V_i|} - y_{ci}V_i - \\sum_{k=0}^NY_{ik}V_k = 0\n\\label{EQ_POWERFLOW_COMPLEX}\n\\end{align}\nThe term $-y_{ci}V_i$ may in practice be absorbed into $Y$ by adding $y_{ci}$ to $Y_{ii}$; we retain it here as an explicit bus shunt for bookkeeping purposes.\n\nReal and imaginary components are:\n\\begin{align}\n\t\\Ir_i &= \\frac{(P_{ci} + P_{\\text{gen},i})\\Vr_i + (Q_{ci} + Q_{\\text{gen},i})\\Vi_i}{|V|^2_i} \\notag \\\\\n\t &+ \\frac{\\Ir_{ci}\\Vr_{i} - \\Ii_{ci}\\Vi_i}{|V_i|} -g_{ci}\\Vr_i + b_{ci}\\Vi_i \\notag \\\\\n\t &+ \\sum_{k=0}^N\\left(-G_{ik}\\Vr_k + B_{ik}\\Vi_k\\right) = 0\\\\\n\t\\Ii_i &= \\frac{(P_{ci} + P_{\\text{gen},i})\\Vi_i - (Q_{ci} + Q_{\\text{gen},i})\\Vr_i}{|V|^2_i} \\notag \\\\\n\t &+ \\frac{\\Ir_{ci}\\Vi_{i} + \\Ii_{ci}\\Vr_{i}}{|V_i|} -g_{ci}\\Vi_i - b_{ci}\\Vr_i \\notag \\\\\n\t&+ \\sum_{k=0}^N\\left(-G_{ik}\\Vi_k - B_{ik}\\Vr_k\\right) = 0\n\\end{align}\n\nThe possible unknowns are the generator power $P_\\text{gen}, Q_\\text{gen}$ and the voltages $\\Vr, \\Vi$.\n\n\\subsection{PQ buses}\nFor PQ buses, there is no additional (non-ZIP) generation, and hence $S_{\\text{gen}}$ can be set to zero. The unknown variables are the components of $V$.\n\\subsection{Slack buses}\nFor SL buses, $V$ is constant and the power flow equations give an explicit expression for $S$. All quantities at slack buses can immediately be found and are therefore considered constants in the powerflow equations.\n\\subsection{PV buses}\nFor PV buses, the power flow equations also hold, with both components of $V$ being unknown as was the case for PQ buses. $Q_{\\text{gen}}$ is an additional variable, and an extra constraint also applies:\n\\begin{align}\n\\Delta |V|^2 = \\Vr^2 + \\Vi^2 - V_\\text{PV}^2 = 0\n\\label{EQ_POWERFLOW_PV_CONSTRAINT}\n\\end{align}\nwhere $V_{\\text{PV}}$ is the voltage magnitude setpoint for the PV bus.\n\n\\subsection{Newton-Raphson equations}\nConsidering PQ buses only for the moment, the unknowns are the real and imaginary parts of $V$, and these equations can be solved using the Newton-Raphson method. Letting the function to which we want to find the zero be $f = [\\Ir, \\Ii]$, the unknows be $x = [\\Vr, \\Vi]$, we wish to solve $f(x) = 0$. Using the Jacobian\n\\begin{align}\nJ_{ik}(x) = \\frac{\\partial f_i(x)}{\\partial x_k}\n\\end{align}\nthe NR method calculates the update to $x$ at each iteration as the solution to the linear equations\n\\begin{align}\n-f_{(n)} &= J(x_{(n)})(x_{(n+1)}-x_{(n)}) = J(x_{(n)})\\Delta x_{(n,n+1)}\n\\label{EQ_NR}\n\\end{align}\nThe Jacobian elements for variables $V$ are given by:\n\\begin{align}\n\t\\frac{\\partial \\Ir_i}{\\partial \\Vr_{k}} \n\t\t&= \\left[-\\frac{2\\Vr_k[(P_{ck} + P_{\\text{gen},k})\\Vr_k + (Q_{ck} + Q_{\\text{gen},k})\\Vi_k]}{|V|_k^4} + \\frac{(P_{ck} + P_{\\text{gen},k})}{|V|_k^2}\\right]\\delta_{ik} \\notag \\\\\n\t\t&+ \\frac{\\Vi_i(\\Ii_i\\Vr_i + \\Ir_i\\Vi_i)}{|V_i|^3}\\delta_{ik} \\notag \\\\\n\t\t&- (G_{ik} + g_{ci}\\delta_{ik}) \\\\\n\t\\frac{\\partial \\Ir_i}{\\partial \\Vi_{k}} \n\t\t&= \\left[-\\frac{2\\Vi_k[(P_{ck} + P_{\\text{gen},k})\\Vr_k + (Q_{ck} + Q_{\\text{gen},k})\\Vi_k]}{|V|_k^4} + \\frac{(Q_{ck} + Q_{\\text{gen},k})}{|V|_k^2} \\right]\\delta_{ik} \\notag \\\\\n\t\t&- \\frac{\\Vr_i(\\Ii_i\\Vr_i + \\Ir_i\\Vi_i)}{|V_i|^3}\\delta_{ik} \\notag \\\\\n\t\t&+ (B_{ik} + b_{ci}\\delta_{ik}) \\\\\n\t\\frac{\\partial \\Ii_i}{\\partial \\Vr_{k}}\n\t\t&= \\left[-\\frac{2\\Vr_k[(P_{ck} + P_{\\text{gen},k})\\Vi_k - (Q_{ck} + Q_{\\text{gen},k})\\Vr_k]}{|V|_k^4} - \\frac{(Q_{ck} + Q_{\\text{gen},k})}{|V|_k^2} \\right]\\delta_{ik} \\notag \\\\\n\t\t&+ \\frac{\\Vi_i(\\Ii_i\\Vi_i - \\Ir_i\\Vr_i)}{|V_i|^3}\\delta_{ik} \\notag \\\\\n\t\t&- (B_{ik} + b_{ci}\\delta_{ik}) \\\\\n\t\\frac{\\partial \\Ii_i}{\\partial \\Vi_{k}}\n\t\t&= \\left[-\\frac{2\\Vi_k[(P_{ck} + P_{\\text{gen},k})\\Vi_k - (Q_{ck} + Q_{\\text{gen},k})\\Vr_k]}{|V|_k^4} + \\frac{(P_{ck} + P_{\\text{gen},k})}{|V|_k^2} \\right] \\delta_{ik} \\notag \\\\\n\t\t&+ \\frac{\\Vr_i(\\Ir_i\\Vr_i - \\Ii_i\\Vi_i)}{|V_i|^3}\\delta_{ik} \\notag \\\\\n\t\t&- (G_{ik} + g_{ci}\\delta_{ik})\n\\end{align}\n\n\\subsubsection{Treatment of the slack bus}\nThere are no variables associated with the slack bus. The voltage is constant, and the power can be found after the rest of the buses are solved.\n\n\\subsubsection{Treatment of PV buses}\nAs explained earlier, $PV$ buses add an extra variable, $Q_g$. This adds the following non-zero elements to the Jacobian:\n\\begin{align}\n\\frac{\\partial \\Ir_k}{\\partial Q_{\\text{gen},k}} &= \\frac{\\Vi_k}{|V|_k^2} \\\\\n\\frac{\\partial \\Ii_k}{\\partial Q_{\\text{gen},k}} &= -\\frac{\\Vr_k}{|V|_k^2}\n\\end{align}\nThere is also an extra constraint, Eq. (\\ref{EQ_POWERFLOW_PV_CONSTRAINT}), reproduced here:\n\\begin{align}\n\\Delta |V|^2 = \\Vr^2 + \\Vi^2 - V_\\text{PV}^2 = 0\n\\label{EQ_POWERFLOW_PV_CONSTRAINT_AGAIN}\n\\end{align}\nwith corresponding elements in the Jacobian being:\n\\begin{align}\n\\frac{\\partial \\Delta|V|^2_k}{\\partial \\Vr_k} &= 2 \\Vr_k \\notag \\\\\n\\frac{\\partial \\Delta|V|^2_k}{\\partial \\Vi_k} &= 2 \\Vi_k\n\\label{EQ_PV_JAC_Q}\n\\end{align}\nThe NR update corresponding to this constraint row is:\n\\begin{align}\n\t-\\Vr_k^2 - \\Vi_k^2 + V_{\\text{PV},k}^2 &= 2\\Vr_k\\Delta\\Vr_k + 2\\Vi_k\\Delta\\Vi_k\n\\end{align}\nwhich gives\n\\begin{align}\n\\Delta \\Vr_k &= \\frac{V^2_{\\text{PV},k} - \\Vr_k^2 - \\Vi_k^2- 2\\Vi_k\\Delta \\Vi_k}{2\\Vr_k}\n\\end{align}\n\nUsing these expressions, $\\Delta \\Vr$ may be eliminated from the NR equations for the PV bus. First write the Jacobian as if all buses were PQ. Let $k$ be any PV bus. Take the column corresponding to $\\Delta \\Vr_k$, and add its product with $-\\Vi_k/\\Vr_k$ to the matching column for $\\Delta \\Vi_k$. Add its product with $(|V|^2_{\\text{PV},k} - \\Vr_k^2 - \\Vi_k^2)/(2\\Vr_k)$ to $f$ in Eq. (\\ref{EQ_NR}). The column and the corresponding element of $x$ for $\\Delta \\Vr_k$ will now be replaced with a column and elment for $\\Delta Q_{\\text{gen},k}$. Set the column to zero, and then set the block diagonal elements, using Eq. (\\ref{EQ_PV_JAC_Q}). Reinterpret the corresponding element of $\\Delta x$ in Eq. (\\ref{EQ_NR}) as now corresponding to $\\Delta Q_{\\text{gen},k}$ instead of $\\Delta \\Vr_k$.\n\n\\section{Power flow in polar coordinates}\nThe preceding section treated the power flow problem in rectangular coordinates, in which the mathematical expressions are simpler and the Newton-Raphson method works well in rectangular coordinates. In fact, it is more common to work in polar coordinates. Doing so can give a clearer physical insight, and is conducive to the development of approximations such as are used in the DC approximation or fast decoupled power flow. Taking the complex conjugate of Eq. \\ref{EQ_POWERFLOW_COMPLEX} and multiplying by $V_i$ gives\n\\newcommand{\\M}[1]{\\ensuremath{|V_{#1}|}}\n\\newcommand{\\cs}[1]{\\ensuremath{\\cos(\\theta_{#1})}}\n\\newcommand{\\sn}[1]{\\ensuremath{\\sin(\\theta_{#1})}}\n\\begin{align}\n\tS_i &= S_{ci} + S_{gi} + I^*_{ci}\\M{i} - y^*_{ci}\\M{i}^2 - V_i\\sum_{k = 0}^N{Y^*_{ik}V^*_k} \\notag \\\\\n\t&= S_{ci} + S_{gi} + I^*_{ci}M_i - y^*_{ci}\\M{i}^2 - \\M{i}\\sum_{k = 0}^N{Y^*_{ik}\\M{k}\\exp{(j\\theta_{ik})}}\n\\end{align}\nwhere we introduce the polar notation $V_i := \\M{i} \\exp{(j\\theta_i)}$ and $\\theta_{ik} := \\theta_i - \\theta_k$. $S_i$ is the \\emph{total} power injection into a bus, including all loads, branch flows and generation, and should be zero in a correct power flow solution. The real and reactive components of this are:\n\\begin{align}\n\tP_i &= P_{ci} + P_{gi} + \\Ir_{ci}\\M{i} - g_i\\M{i}^2 \\notag \\\\\n\t&- \\M{i}\\sum_{l = 0}^N{\\left[G_{il}\\cs{il} + B_{il}\\sn{il}\\right]\\M{l}} \\notag \\\\\n\tQ_i &= Q_{ci} + Q_{gi} - \\Ii_{ci}\\M{i} + b_i\\M{i}^2 \\notag \\\\\n\t&- \\M{i}\\sum_{l = 0}^N{\\left[G_{il}\\sn{il} - B_{il}\\cs{il}\\right]\\M{l}} \\notag \\\\\n\\end{align}\nHowever, for PV buses, $Q_{g}$ is a free variable, and can immediately be set to the value that makes the error $Q_i$ zero.\n\nWe first solve:\n\\begin{align}\n\tP_i &= P_{ci} + P_{gi} + \\Ir_{ci}\\M{i} - g_i\\M{i}^2 \\notag \\\\\n\t&- \\M{i}\\sum_{l = 0}^N{\\left[G_{il}\\cs{il} + B_{il}\\sn{il}\\right]\\M{l}} , & i \\in \\mathrm{PQ} \\cup \\mathrm{PV}\\notag \\\\\n\tQ_i &= Q_{ci} - \\Ii_{ci}\\M{i} + b_i\\M{i}^2 \\notag \\\\\n\t&- \\M{i}\\sum_{l = 0}^N{\\left[G_{il}\\sn{il} - B_{il}\\cs{il}\\right]\\M{l}}, & i \\in \\mathrm{PQ}\n\\end{align}\nfor variables $\\{\\M{\\text{PQ}}, \\theta_\\text{PQ, PV}\\}$. These equations are independent of $Q_g$, which can be therefore be calculated in a single final step as:\n\\begin{align}\n\tQ_{gi} &= -Q_{ci} + \\Ii_{ci}\\M{i} - b_i\\M{i}^2 \\notag \\\\\n\t&+ \\M{i}\\sum_{l = 0}^N{\\left[G_{il}\\sn{il} - B_{il}\\cs{il}\\right]\\M{l}}, & i \\in \\mathrm{PV}\n\\end{align}\n\nNote for computations (particularly in what follows), it may be more efficient to cache $\\sin(\\theta)$ and $\\cos(\\theta)$, and to use\n\\begin{align}\n\t\\sin(\\theta_{il}) &:= \\sin(\\theta_i)\\cos(\\theta_l) - \\cos(\\theta_i)\\sin(\\theta_l) \\notag \\\\\n\t\\cos(\\theta_{il}) &:= \\cos(\\theta_i)\\cos(\\theta_l) + \\sin(\\theta_i)\\sin(\\theta_l)\n\\end{align}\n\n\n\\subsection{The NR method in polar coordinates}\nFor PQ buses, the possible unknowns are $\\M{}$ and $\\theta$. For PV buses, the possible unknowns are $Q_g$ and $\\theta$. The Jacobian is therefore\n\\begin{align}\n\t\\frac{\\partial P_i}{\\partial \\M{k}} &= \n\t\t\\begin{cases}\n\t\t\t-\\M{i}\\left[G_{ik}\\cs{ik} + B_{ik}\\sn{ik}\\right] & k \\ne i \\\\\n\t\t\t\\Ir_{ci} - 2g_i\\M{i} - 2 \\M{i}G_{ii} - \\sum_{\\stackrel{l = 0}{l \\ne i}}^N{\\left[G_{il}\\cs{il} + B_{il}\\sn{il}\\right]\\M{l}} & k = i\n\t\t\\end{cases} & k \\in \\text{PQ} \\notag \\\\\n\t\\frac{\\partial P_i}{\\partial \\theta_k} &=\n\t\t\\begin{cases}\n\t\t\t\\M{i}\\left[-G_{ik}\\sn{ik} + B_{ik}\\cs{ik}\\right]\\M{k}, &k \\ne i \\\\\n\t\t\t\\M{i}\\sum_{\\stackrel{l = 0}{l \\ne i}}^N{\\left[G_{il}\\sn{il} - B_{il}\\cs{il}\\right]\\M{l}}, &k = i\n\t\t\\end{cases} \\notag \\\\\n\t\\frac{\\partial Q_i}{\\partial \\M{k}} &= \n\t\t\\begin{cases}\n\t\t\t-\\M{i}\\left[G_{ik}\\sn{ik} - B_{ik}\\cs{ik}\\right], & k \\ne i \\\\\n\t\t\t-\\Ii_{ci} + 2b_i\\M{i} + 2\\M{i}B_{ii} - \\sum_{l = 0}^N{\\left[G_{il}\\sn{il} - B_{il}\\cs{il}\\right]\\M{l}}, & k = i, \\\\\n\t\t\\end{cases} & i, k \\in \\text{PQ} \\notag \\\\\n\t\\frac{\\partial Q_i}{\\partial \\theta_k} &= \n\t\t\\begin{cases}\n\t\t\t\\M{i}\\left[G_{ik}\\cs{ik} + B_{ik}\\sn{ik}\\right]\\M{k}, & k \\ne i \\\\\n\t\t\t-\\M{i}\\sum_{\\stackrel{l = 0}{l \\ne i}}^N{\\left[G_{il}\\cs{il} + B_{il}\\sn{il}\\right]\\M{l}}, & k = i \\\\\n\t\t\\end{cases}, & i \\in PQ\n\\end{align}\n\nThis gives us enough information to carry out the NR method for polar coordinates. Note that the value of $Q_g$ is not needed for any of the iterations, and so can be calculated once-only after convergence is achieved.\n\n\\subsection{Fast-decoupled load flow}\nEspecially in transmission networks, we have that $G_{ik} \\ll B_{ik}$ and $\\theta_{ik} \\approx 0$. We also assume that shunts are mainly reactive (i.e. there are no large resistive loads), so that $g_i \\approx 0$, and that there are no large constant current loads, so that $I_c \\approx 0$. Under these conditions, the Jacobian above can be simplified:\n\\begin{align}\n\t\\frac{\\partial P_i}{\\partial \\M{k}} &= 0, & k \\in \\text{PQ} \\notag \\\\\n\t\\frac{\\partial P_i}{\\partial \\theta_k} &= \n\t\t\\begin{cases}\n\t\t\t\\M{i}B_{ik}\\M{k}, & i \\ne k \\\\\n\t\t\t-\\M{i}\\sum_{\\stackrel{l = 0}{l \\ne i}}^NB_{il}\\M{l}, &i = k\n\t\t\\end{cases} \\notag \\\\\n\t\\frac{\\partial Q_i}{\\partial \\M{k}} &= \n\t\t\\begin{cases}\n\t\t\t\\M{i}B_{ik}, & k \\ne i \\\\\n\t\t\t-\\Ii_{ci} + 2b_i\\M{i} + 2 \\M{i}B_{ii} + \\sum_{\\stackrel{l = 0}{l \\ne i}}^N{B_{il}\\M{l}}, & k = i,\n\t\t\\end{cases} & i, k \\in \\text{PQ} \\notag \\\\\n\t\\frac{\\partial Q_i}{\\partial \\theta_k} &= 0, & i \\in PQ\n\\end{align}\nNote that the problem then separates into two independent problems: one for $\\{P, \\theta\\}$ and one for $\\{Q, \\M{}\\}$. \n\nFor a flat-start, we have $\\M{k} \\approx 1$, and the Jacobian simplifies further. The fast decoupled load flow method normally uses this flat start Jacobian throughout the iteration process. The error function $f$ is assumed to be unchanged from the full NR method, so if it converges, it will converge to the correct solution to the full AC power flow problem. In practice it requires many more iterations than the full NR method, but each iteration is much faster, and the convergence properties can be much better.\n\n\\section{Generalised Power Flow Theory}\nLet the voltage at bus $i$ be $V_i$, and let the voltage difference between buses $i$ and $k$ be $V_{ik} := V_i - V_k$. We represent ground as a special bus, with index zero, which must always have a voltage of zero: $V_0 = 0$. Let $S^*_{g,i}$ be the power injected into bus $i$. Let $y_{c,ik}$ be a constant admittance between buses $i$ and $k$, $S_{c,ik}$ be a constant power flow between these buses and $I_{c,ik}$ be a constant magnitude current flow whose phase varies with $V$, so that the resulting actual current is\\footnote{This ensures that the controlled current loads can't have a power that depends on the local absolute phase.} $I_{c,ik}V_{ik}/|V_{ik}|$.\n\nThe generated current injection at bus $i$ is then equal to the sum of the current flowing from $i$ to all other connected buses:\n\\begin{align}\n\t\\frac{S^*_{g,i}}{V^*_i} - \\sum_{k \\ne i} \\left[y_{c,ik}V_{ik} + \\frac{S^*_{c,ik}}{V^*_{ik}} + \\frac{I_{c,ik}V_{ik}}{|V_{ik}|}\\right] = 0\n\t\\label{EQ_POWERFLOW_B}\n\\end{align}\n\nTo simplify the calculation for the terms proportional to $y$, we define the bus admittance matrix, $Y$:\n\\begin{align}\n\tY_{c,ik} &=\n\t\t\\begin{cases}\n\t\t\t-y_{c,ik}&\\text{if $i \\ne k$} \\\\\n\t\t\t\\sum_{l \\ne i} y_{c,il}& \\text{if $i = k$}\n\t\t\\end{cases}\n\t\\label{EQ_YNODE_OHMIC_B}\n\\end{align}\nThe sum over $k$ in this equation includes the special ground bus, $k = 0$. Such terms represent shunt admittances to ground. Equation \\ref{EQ_POWERFLOW_B} then becomes\n\\begin{align}\n\t\\frac{S^*_{g,i}}{V^*_i} - \\sum_{k \\ne 0}{Y_{c,ik}V_{k}} - \\sum_{k \\ne i}\\left[\\frac{S^*_{c,ik}}{V^*_{ik}} + \\frac{I_{c,ik}V_{ik}}{|V_{ik}|}\\right] = 0\n\t\\label{EQ_POWERFLOW_B1}\n\\end{align}\nSince ground is treated implicitly, there is no need to consider terms where $i = 0$, and, because $V_0$ = 0, we may safely omit terms in the first sum where $k = 0$, as shown in the equation. Note that the second sum \\emph{does} include terms with $k = 0$; these are usual constant current and constant power components of a ZIP load model. However, this equation allows more general loads than the usual ZIP model, as constant current and power components are also possible \\emph{between} buses, not just from a single bus to ground. This allows us to consider FACTS devices, as well as various types of delta loads, for example.\n\n\\subsection{NR method in rectangular coordinates}\nLet the mismatch, $\\Delta I$, equal the LHS of Eq. \\ref{EQ_POWERFLOW_B1}:\n\\begin{align}\n\\Delta I_i = \\frac{S^*_{g,i}}{V^*_i} - \\sum_{k \\ne 0}{Y_{c,ik}V_{k}} - \\sum_{k \\ne i}\\left[\\frac{S^*_{c,ik}}{V^*_{ik}} + \\frac{I_{c,ik}V_{ik}}{|V_{ik}|}\\right]\n\\label{EQ_MISMATCH}\n\\end{align}\n\nConsidering for the moment PQ buses only, the unknowns are the real and imaginary voltage components, $x = [\\Vr, \\Vi]$. We wish to find values for these variables such that the mismatch, $\\Delta I$, is zero. Using the Jacobian\n\\begin{align}\nJ_{ik}(x) = \\frac{\\partial \\Delta I_i(x)}{\\partial x_k},\n\\end{align}\nthe NR method calculates the update to $x$ at each iteration as the solution to the linear equations\n\\begin{align}\n-\\Delta I_{(n)} &= J(x_{(n)})(x_{(n+1)}-x_{(n)}) \\notag \\\\\n &= J(x_{(n)})\\Delta x_{(n,n+1)}\n\\label{EQ_NR}\n\\end{align}\n\nThe Jacobian is given below:\n\\begin{align}\n\t\\frac{\\partial \\Delta I_i}{\\partial \\Vr_l} \n\t\t&= \\delta_{il}\\frac{-S^*_{g,i}}{{V^*_i}^2} \\notag \\\\\n\t\t&- Y_{c,il} \\notag \\\\\n\t\t&+ \\delta_{il} \\sum_{k \\ne i}{\\frac{S^*_{c,ik}}{{V^*_{ik}}^2}} \\notag \\\\\n\t\t&+ (1-\\delta_{il})\\frac{-S^*_{c,il}}{{V^*_{il}}^2} \\notag \\\\\n\t\t&+ \\delta_{il} \\sum_{k \\ne i}{\\frac{I_{c,ik}(-|V_{ik}|^2 + \\Vr_{ik}V_{ik})}{|V_{ik}|^3}} \\notag \\\\\n\t\t&+ (1 - \\delta_{il})\\frac{I_{c,il}(|V_{il}|^2 - \\Vr_{il}V_{il})}{|V_{il}|^3} \\notag \\\\\n\t\\frac{\\partial \\Delta I_i}{\\partial \\Vi_l} \n\t\t&= \\delta_{il}\\frac{j S^*_{g,i}}{{V^*_i}^2} \\notag \\\\\n\t\t&- j Y_{c,il} \\notag \\\\\n\t\t&+ \\delta_{il} \\sum_{k \\ne i}{\\frac{-j S^*_{c,ik}}{{V^*_{ik}}^2}} \\notag \\\\\n\t\t&+ (1-\\delta_{il}) \\frac{j S^*_{c,il}}{{V^*_{il}}^2} \\notag \\\\\n\t\t&+ \\delta_{il} \\sum_{k \\ne i}{\\frac{I_{c,ik}(-j|V_{ik}|^2 + \\Vi_{ik}V_{ik})}{|V_{ik}|^3}} \\notag \\\\\n\t\t&+ (1 - \\delta_{il})\\frac{I_{c,il}(j|V_{il}|^2 - \\Vi_{il}V_{il})}{|V_{il}|^3}\t\n\\end{align}\nTo solve these equations, we transform each complex row of the mismatch vector and the Jacobian into separate real and imaginary rows, and then use a sparse solver on the resulting system of real equations.\n\n\\subsubsection{Treatment of the slack bus}\nThere are no variables associated with the slack bus. The voltage is constant, and the power can be found after the rest of the buses are solved.\n\n\\subsubsection{Treatment of the ground bus}\nSimilarly, there are no variables associated with the special ground bus, and indeed this bus may be treated completely implicitly. Its voltage is, of course, zero. We have already seen how constant admittance shunts to ground are absorbed into the definition of the bus admittance matrix, $Y$. Similarly, there are columns in $S_c$ and $I_c$ involving the ground bus that need to be taken into account in various sums.\n\n\\subsubsection{Treatment of PV buses}\nAs explained earlier, $PV$ buses add an extra variable, $Q_g$. This adds the following non-zero elements to the Jacobian:\n\\begin{align}\n\\frac{\\partial \\Delta I_i}{\\partial Q_{g,l}} &= \\delta_{il}\\frac{-i}{V^*_i}\n\\end{align}\nThere is also an extra constraint, Eq. (\\ref{EQ_POWERFLOW_PV_CONSTRAINT}), reproduced here:\n\\begin{align}\n\\Delta |V|^2 = |V|^2 - V_\\text{PV}^2 = 0\n\\label{EQ_POWERFLOW_PV_CONSTRAINT_AGAIN}\n\\end{align}\nwith corresponding elements in the Jacobian being:\n\\begin{align}\n\\frac{\\partial \\Delta|V|^2_k}{\\partial \\Vr_k} &= 2 \\Vr_k \\notag \\\\\n\\frac{\\partial \\Delta|V|^2_k}{\\partial \\Vi_k} &= 2 \\Vi_k\n\\label{EQ_PV_JAC_Q}\n\\end{align}\nThe NR update corresponding to this constraint row is:\n\\begin{align}\n\t-|V_k|^2 + V_{\\text{PV},k}^2 &= 2\\Vr_k\\Delta\\Vr_k + 2\\Vi_k\\Delta\\Vi_k\n\\end{align}\nwhich gives\n\\begin{align}\n\\Delta \\Vr_k &= \\frac{V^2_{\\text{PV},k} - |V_k|^2- 2\\Vi_k\\Delta \\Vi_k}{2\\Vr_k}\n\\end{align}\n\nUsing these expressions, $\\Delta \\Vr$ may be eliminated from the NR equations for the PV bus. First write the Jacobian as if all buses were PQ. Let $k$ be any PV bus. Take the column corresponding to $\\Delta \\Vr_k$, and add its product with $-\\Vi_k/\\Vr_k$ to the matching column for $\\Delta \\Vi_k$. Add its product with $(|V|^2_{\\text{PV},k} - \\Vr_k^2 - \\Vi_k^2)/(2\\Vr_k)$ to $f$ in Eq. (\\ref{EQ_NR}). The column and the corresponding element of $x$ for $\\Delta \\Vr_k$ will now be replaced with a column and elment for $\\Delta Q_{g,k}$. Set the column to zero, and then set the block diagonal elements, using Eq. (\\ref{EQ_PV_JAC_Q}). Reinterpret the corresponding element of $\\Delta x$ in Eq. (\\ref{EQ_NR}) as now corresponding to $\\Delta Q_{g,k}$ instead of $\\Delta \\Vr_k$.\n\n\\section{Formalism for modelling branches}\n\\subsection{Branch admittance ($Y$) parameters}\nConsider a network consisting of three buses, $a$, $b$ and $c$. Due to the linear nature of the nodal admittance relation, the nodal admittance matrix for the network $Y$, can be decomposed as follows:\n\\begin{align}\n\tY = \\begin{bmatrix}\n\t\tY^{ab}_{00} & Y^{ab}_{01} & 0 \\\\ Y^{ab}_{10} & Y^{ab}_{11} & 0 \\\\ 0 & 0 & 0\n\t\\end{bmatrix} + \\begin{bmatrix}\n\t\tY^{ac}_{00} & 0 & Y^{ac}_{01} \\\\ 0 & 0 & 0 \\\\ Y^{ac}_{10} & 0 & Y^{ac}_{11}\n\t\\end{bmatrix} + \\begin{bmatrix}\n\t\t0 & 0 & 0 \\\\ 0 & Y^{bc}_{00} & Y^{bc}_{01} \\\\ 0 & Y^{bc}_{10} & Y^{bc}_{11}\n\t\\end{bmatrix}\n\t\\label{EQ_BRANCH_DECOMP}\n\\end{align}\nwhere, e.g., $Y^{ab}$, is the nodal admittance matrix for buses $a$ and $b$ in isolation from the rest of the network. If nodes $a$ and $b$ were connected by a line with admittance $y_{ab}$, then we would have\n\\begin{align}\n\tY^{ab} &= \n\t\\begin{bmatrix}\n\t\ty_{ab} & -y_{ab} \\\\\n\t\t-y_{ab} & y_{ab}\n\t\\end{bmatrix}\n\\end{align}\nIn this kind of decomposition, the matrix $Y^{ab}$ represents the properties of the \\emph{branch} between nodes $a$ and $b$ and is termed the \\emph{branch admittance matrix}. The global properties of the full $Y$ matrix are reduced to the elements of the $2 \\times 2$ branch admittance matrices, which can be derived in isolation from each other.\n\nWhat about three phase systems? A three-phase bus can be represented as a set of three single phase buses. A three phase branch links two three phase buses, and thus involves six single phase buses; it is represented by a $6 \\times 6$ branch admittance matrix. Considering the network for these six buses in isolation from all other buses allows us to derive its elements, which could, for example, represent the properties of a three-wire overhead transmission line.\n\nThe modelling task is then to derive expressions for the branch admittance matrices of all the different kinds of lines and transformers. The full $Y$ matrix may then always be reconstructed from these components. \n\nSometimes, instead of using the branch admittance matrix representation, alternate expressions are used. Often, the inverse of the branch matrix, the $Z$ matrix, is used\n\\begin{align}\n\tZ = Y^{-1}\n\\end{align}\n\nAnother common representation is the ``ABCD'' representation, where\n\\begin{align}\n\t\\begin{bmatrix}\n\t\tV_1 \\\\ I_1\n\t\\end{bmatrix} &= \\begin{bmatrix}\n\t\tA & -B \\\\ C & -D\n\t\\end{bmatrix}\\begin{bmatrix}\n\t\tV_2 \\\\ I_2\n\t\\end{bmatrix}\n\\end{align}\n$V_1, I_1$ and $V_2, I_2$ are the current and voltage at buses $1$ and $2$ respectively.\\footnote{Take care: in most treatments, you will not see the negative signs on $B$ and $D$. This is because in these treatments $I_1$ is the current \\emph{into} bus 1 and $I_2$ is the current \\emph{out of} bus 2. To avoid choosing arbitrary directions of current flow, we have instead been working with the convention that all currents and powers are \\emph{injections into} a bus.} For multi-phase buses, these are vector quantities, and $A, B, C, D$ are matrices. We can reconstruct the $Y$ matrix as follows:\n\\begin{align}\n\tY = \\begin{bmatrix}\n\t\tDB^{-1} & C - DB^{-1}A \\\\\n\t\t-B^{-1} & B^{-1}A\n\t\\end{bmatrix}\n\\end{align}\nGiven a $Y$ matrix, we can also construct the $ABCD$ matrix:\n\\begin{align}\n\t\\begin{bmatrix} A & -B \\\\ C & -D \\end{bmatrix} &=\n\t\\begin{bmatrix}\n\t\t-Y_{21}^{-1}Y_{22} & -Y_{21}^{-1} \\\\\n\t\tY_{12} - Y_{11}Y_{21}^{-1}Y_{22} & -Y_{11}Y_{21}^{-1}\n\t\\end{bmatrix}\n\\end{align}\nwhere, again, all elements may be considered to be block submatrices.\n\nThe main point of the ABCD representation is that branches may be ``cascaded'' together. For example, consider two n-phase lines that are joined head to tail at a central n-phase bus. We can eliminate the central bus by multiplying the ABCD matrices:\n\\begin{align}\n\t\\begin{bmatrix}\n\t\tV_1 \\\\ I_1\n\t\\end{bmatrix} &=\n\t\\begin{bmatrix}\n\t\tA_{12} & -B_{12} \\\\ C_{12} & -D_{12}\n\t\\end{bmatrix}\n\t\\begin{bmatrix}\n\t\tV_2 \\\\ I_2\n\t\\end{bmatrix} \\notag \\\\\n\t&=\n\t\\begin{bmatrix}\n\t\tA_{12} & -B_{12} \\\\ C_{12} & -D_{12}\n\t\\end{bmatrix}\n\t\\begin{bmatrix}\n\t\tA_{23} & -B_{23} \\\\ C_{23} & -D_{23}\n\t\\end{bmatrix}\n\t\\begin{bmatrix}\n\t\tV_3 \\\\ I_3\n\t\\end{bmatrix}\n\\end{align}\nThis defines a composition law for $ABCD$ parameters.\n\\section{Transmission Lines}\n\\subsection{Single phase transmission lines}\nShort transmission lines (below 80 km in length) can be modelled as a single admittance $y$ (or equivalently an impedance $z = 1/y$) between two buses:\n\\begin{align}\n\tY = \\begin{bmatrix}\n\t\ty & -y \\\\ -y & y\n\t\\end{bmatrix}\n\\end{align}\nThe real part of $z$ is positive (a resistance), and the imaginary part is also positive (an inductive reactance). Both the resistance and the inductive reactance are proportional to the length of the line. The inductive reactance is usually larger than the resistance.\n\nLonger transmission lines ($> 80$ km) also develop a capacitance between the line and ground. If the length is less than about 240 km, then this is often modelled using the $\\pi$-model, see Fig. \\ref{FIG_PI_LINE}. The $Y$-matrix is:\n\\begin{align}\n\tY = \\begin{bmatrix}y_s + \\frac{j b_c}{2} & -y_s \\\\ -y_s & y_s + \\frac{j b_c}{2} \\end{bmatrix}\n\\end{align}\n\\begin{figure}[!h]\n\t\\begin{center}\n\t\t\\includegraphics[width=(5cm)]{pi_line}\n\t\\end{center}\n\t\\caption{\n\t\tThe $\\pi$ model of a transmission line.\n\t}\n\t\\label{FIG_PI_LINE}\n\\end{figure}\nVery long lines require a more complex \\emph{distributed model}.\n\n\\subsection{Multi-phase transmission lines}\nThe equations above apply also to multi-phase lines, except that the scalar elements become block submatrices whose size is the number of phases.\n\n\\subsection{Line parameters}\nWe have discussed above the structure of the $Y$ matrix for transmission lines -- but we still need to know how to find the values of the admittances in our equations. \\emph{Carson's Equations} allow us to do this, for N-phase lines which may also contain one or more multi-grounded neutral wires. They take into account the geometry of the transmission line and the conductivity of the ground. We do not, however, go into these equations in this document.\n\n\\section{Transformers}\n\\subsection{Single phase transformers}\nFor an ideal transformer with a single (possibly complex) turns ratio $n = n_1/n_0$, we have, using the ABCD parameters:\n\\begin{align}\n\\begin{bmatrix}V_1 \\\\ I_1\\end{bmatrix} &= \\begin{bmatrix}n & 0 \\\\ 0 & 1/n^*\\end{bmatrix}\\begin{bmatrix}V_0 \\\\ I_0 \\end{bmatrix}\n\\end{align}\nThis can't be modelled correctly in the formalism of nodal admittance matrices, in the same way that a line with zero impedance can't. Two equivalent circuits for an ideal transformer are shown in Fig. \\ref{FIG_IDEAL_TRANS_EQUIV}.\n\\begin{figure}[!h]\n\t\\begin{center}\n\t\t\\includegraphics[width=(\\textwidth-2cm)]{ideal_transformer_equiv.png}\n\t\\end{center}\n\t\\caption{\n\t\tTwo equivalent circuits for an ideal transformer.\n\t}\n\t\\label{FIG_IDEAL_TRANS_EQUIV}\n\\end{figure}\nThe relationships could be included in the power flow equations specially, but more commonly real transformers are used, as described below; these may be expressed in the nodal admittance formalism.\n\nA real transformer includes a leakage impedance (due to finite resistance of copper windings and core losses) and a shunt magnetising impedance. There are various ways of drawing an equivalent circuit. We use the general branch model described in Section \\ref{SEC_GEN_BRANCH_MODEL}, with the series admittance set to a leakage admittance $y_l$ (mainly inductive) and the shunt admittances forming the magnetising admittance $y_m$, see Fig. \\ref{FIG_REAL_TRANS}.\n\\begin{figure}[!h]\n\t\\begin{center}\n\t\t\\includegraphics[width=(\\textwidth-2cm)]{transformer.png}\n\t\\end{center}\n\t\\caption{\n\t\tAn equivalent circuit for a real transformer.\n\t}\n\t\\label{FIG_REAL_TRANS}\n\\end{figure}\n\nThe nodal admittance matrix is:\n\\begin{align}\n\\begin{bmatrix}I_0 \\\\ I_1 \\end{bmatrix} &= \n\\begin{bmatrix}\\frac{y_l + \\frac{1}{2}y_m}{|n|^2} & -\\frac{y_l}{n^*} \\\\ -\\frac{y_l}{n} & y_l + \\frac{1}{2}y_m\\end{bmatrix}\n\\begin{bmatrix}V_0 \\\\ V_1 \\end{bmatrix}\n\\label{EQ_REAL_TRANS}\n\\end{align}\nwhere $n = n_1/n_0$ is the ideal turns ratio, $y_l$ is the leakage impedance and $y_m$ is the magnetising admittance. Often, the magnetising admittance is very small and is neglected. $y_l$ can be found by short circuiting the secondary, while $y_m$ can be found using an open circuit test.\n\n\\subsection{Three-phase transformers}\nThe nodal admittance matrices of three-phase transformers may be derived from the single-phase expression, above, combined with information about the connections between phases.\n\\subsubsection{Wye-Wye}\nWe have\n\\begin{align}\nI_{A} &= \\frac{y_l}{|n|^2}V_A - \\frac{y_l}{n^*}V_a \\\\\nI_{B} &= \\frac{y_l}{|n|^2}V_B - \\frac{y_l}{n^*}V_b \\\\\nI_{C} &= \\frac{y_l}{|n|^2}V_C - \\frac{y_l}{n^*}V_c \\\\\nI_a &= -\\frac{y_l}{n}V_A + y_l V_a \\\\\nI_b &= -\\frac{y_l}{n}V_B + y_l V_b \\\\\nI_c &= -\\frac{y_l}{n}V_C + y_l V_c \\\\\n\\end{align}\nso we can immediately write down the nodal admittance relationship:\n\\begin{align}\n\\begin{bmatrix}I_A \\\\ I_B \\\\ I_C \\\\ I_a \\\\ I_b \\\\ I_c\\end{bmatrix} &=\ny_l \\begin{bmatrix}\n\t1/|n|^2 & 0 & 0 & -1/n^* & 0 & 0 \\\\\n\t0 & 1/|n|^2 & 0 & 0 & -1/n^* & 0 \\\\\n\t0 & 0 & 1/|n|^2 & 0 & 0 & -1/n^* \\\\\n\t-1/n & 0 & 0 & 1 & 0 & 0 \\\\\n\t0 & -1/n & 0 & 0 & 1 & 0 \\\\\n\t0 & 0 & -1/n & 0 & 0 & 1\n\\end{bmatrix}\n\\begin{bmatrix}V_A \\\\ V_B \\\\ V_C \\\\ V_a \\\\ V_b \\\\ V_c\\end{bmatrix}\n\\end{align}\nwith the nodal admittance matrix being specified by the matrix on the right, including the factor of $y_l$.\n\n\\subsubsection{Delta-Delta}\nWe have\n\\begin{align}\nI_{AB} &= \\frac{y_l}{|n|^2}(V_A - V_B) - \\frac{y_l}{n^*}(V_a - V_b) \\\\\nI_{BC} &= \\frac{y_l}{|n|^2}(V_B - V_C) - \\frac{y_l}{n^*}(V_b - V_c) \\\\\nI_{CA} &= \\frac{y_l}{|n|^2}(V_C - V_A) - \\frac{y_l}{n^*}(V_c - V_a) \\\\\nI_{ab} &= -\\frac{y_l}{n}(V_A - V_B) + y_l (V_a - V_b) \\\\\nI_{bc} &= -\\frac{y_l}{n}(V_B - V_C) + y_l (V_b - V_c) \\\\\nI_{ca} &= -\\frac{y_l}{n}(V_C - V_A) + y_l (V_c - V_a)\n\\end{align}\nAlso, by the KCL, we have\n\\begin{align}\nI_A &= I_{AB} - I_{CA} \\\\ &= \\frac{y_l}{|n|^2}(2V_A - V_B - V_C) - \\frac{y_l}{n^*}(2V_a - V_b - V_c) \\\\\nI_B &= I_{BC} - I_{AB} \\\\ &= \\frac{y_l}{|n|^2}(2V_B - V_C - V_A) - \\frac{y_l}{n^*}(2V_b - V_c - V_a) \\\\\nI_C &= I_{CA} - I_{BC} \\\\ &= \\frac{y_l}{|n|^2}(2V_C - V_A - V_B) - \\frac{y_l}{n^*}(2V_c - V_a - V_b) \\\\\nI_a &= I_{ab} - I_{ca} \\\\ &= -\\frac{y_l}{n}(2V_A - V_B - V_C) + y_l(2V_a - V_b - V_c) \\\\\nI_b &= I_{bc} - I_{ab} \\\\ &= -\\frac{y_l}{n}(2V_B - V_C - V_A) + y_l(2V_b - V_c - V_a) \\\\\nI_c &= I_{ca} - I_{bc} \\\\ &= -\\frac{y_l}{n}(2V_C - V_A - V_B) + y_l(2V_c - V_a - V_b)\n\\end{align}\nso we can immediately write down the nodal admittance relationship:\n\\begin{align}\n\\begin{bmatrix}I_A \\\\ I_B \\\\ I_C \\\\ I_a \\\\ I_b \\\\ I_c\\end{bmatrix} &=\ny_l \\begin{bmatrix}\n\t2/|n|^2 & -1/|n|^2 & -1/|n|^2 & -2/n^* & 1/n^* & 1/n^* \\\\\n\t-1/|n|^2 & 2/|n|^2 & -1/|n|^2 & 1/n^* & -2/n^* & 1/n^* \\\\\n\t-1/|n|^2 & -1/|n|^2 & 2/|n|^2 & 1/n^* & 1/n^* & -2/n^* \\\\\n\t-2/n & 1/n & 1/n & 2 & -1 & -1 \\\\\n\t1/n & -2/n & 1/n & -1 & 2 & -1 \\\\\n\t1/n & 1/n & -2/n & -1 & -1 & 2\n\\end{bmatrix}\n\\begin{bmatrix}V_A \\\\ V_B \\\\ V_C \\\\ V_a \\\\ V_b \\\\ V_c\\end{bmatrix}\n\\end{align}\nwith the nodal admittance matrix being specified by the matrix on the right, including the factor of $y_l$.\n\n\\subsubsection{Delta-GWye}\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=7cm]{DeltaGWye.pdf}\n\\caption{Schematic of a Delta-GWye transformer}\n\\label{FIG_DELTA_GWYE}\n\\end{center}\n\\end{figure}\nConsidering Fig. \\ref{FIG_DELTA_GWYE} and neglecting the magnetising impedance, we have,\n\\begin{align}\nI_{AB} &= \\frac{y_l}{|n|^2}(V_A - V_B) - \\frac{y_l}{n^*}V_a \\\\\nI_{BC} &= \\frac{y_l}{|n|^2}(V_B - V_C) - \\frac{y_l}{n^*}V_b \\\\\nI_{CA} &= \\frac{y_l}{|n|^2}(V_C - V_A) - \\frac{y_l}{n^*}V_c \\\\\nI_a &= -\\frac{y_l}{n}(V_A - V_B) + y_l V_a \\\\\nI_b &= -\\frac{y_l}{n}(V_B - V_C) + y_l V_b \\\\\nI_c &= -\\frac{y_l}{n}(V_C - V_A) + y_l V_c \\\\\n\\end{align}\nAlso, by the KCL, we have\n\\begin{align}\nI_A &= I_{AB} - I_{CA} \\\\\n&= \\frac{y_l}{|n|^2}(2V_A - V_B - V_C) + \\frac{y_l}{n^*}(V_c - V_a) \\\\\nI_B &= I_{BC} - I_{AB} \\\\\n&= \\frac{y_l}{|n|^2}(2V_B - V_C - V_A) + \\frac{y_l}{n^*}(V_a - V_b) \\\\\nI_C &= I_{CA} - I_{BC} \\\\\n&= \\frac{y_l}{|n|^2}(2V_C - V_A - V_B) + \\frac{y_l}{n^*}(V_b - V_c)\n\\end{align}\nso we can immediately write down the nodal admittance relationship:\n\\begin{align}\n\\begin{bmatrix}I_A \\\\ I_B \\\\ I_C \\\\ I_a \\\\ I_b \\\\ I_c\\end{bmatrix} &=\ny_l \\begin{bmatrix}\n\t2/|n|^2 & -1/|n|^2 & -1/|n|^2 & -1/n^* & 0 & 1/n^* \\\\\n\t-1/|n|^2 & 2/|n|^2 & -1/|n|^2 & 1/n^* & -1/n^* & 0 \\\\\n\t-1/|n|^2 & -1/|n|^2 & 2/|n|^2 & 0 & 1/n^* & -1/n^* \\\\\n\t-1/n & 1/n & 0 & 1 & 0 & 0 \\\\\n\t0 & -1/n & 1/n & 0 & 1 & 0 \\\\\n\t1/n & 0 & -1/n & 0 & 0 & 1\n\\end{bmatrix}\n\\begin{bmatrix}V_A \\\\ V_B \\\\ V_C \\\\ V_a \\\\ V_b \\\\ V_c\\end{bmatrix}\n\\end{align}\nwith the nodal admittance matrix being specified by the matrix on the right, including the factor of $y_l$.\n\n\\subsubsection{Open Delta-open Delta}\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=7cm]{OpenDelta.png}\n\\caption{Schematic of an open-delta transformer connection. The complex turns ratios (primary/secondary) are $n_A$ and $n_C$, and the leakage admittances are $y_{LA}$ and $y_{LC}$. There is a tie admittance $y_\\text{tie}$ between the middle terminals.}\n\\label{FIG_OPEN_DELTA}\n\\end{center}\n\\end{figure}\nWe use the open delta model shown in \\ref{FIG_OPEN_DELTA}. Note that the two middle phases are tied together using an admittance, to prevent a floating neutral. There are other grounding options available, for example, to ground the middle phase, and the model can be easily modified for these. Using Eq. (\\ref{EQ_REAL_TRANS}) (and neglecting magnetising impedance), we can derive the following equations:\n\\begin{align}\n\tI_A &= \\frac{y_{LA}}{|n_A|^2}(V_A - V_B) - \\frac{y_{LA}}{n_A^*}(V_a - V_b) \\notag \\\\\n\t&= V_A\\left(\\frac{y_{LA}}{|n_A|^2}\\right) + V_B\\left(-\\frac{y_{LA}}{|n_A|^2}\\right) + V_a\\left(-\\frac{y_{LA}}{n_A^*}\\right) + V_b\\left(\\frac{y_{LA}}{n_A^*}\\right) \\notag \\\\\n\t%\n\tI_B &= y_\\text{tie}(V_B - V_b) - I_A - I_C \\notag \\\\\n\t&= V_A\\left(-\\frac{y_{LA}}{|n_A|^2}\\right) + V_B \\left(y_\\text{tie} + \\frac{y_{LA}}{|n_A|^2} + \\frac{y_{LC}}{|n_C|^2}\\right) + V_C\\left(-\\frac{y_{C}}{|n_C|^2}\\right) \\notag \\\\\n\t&+ V_a\\left(\\frac{y_{LA}}{n_A^*}\\right) + V_b\\left(-y_\\text{tie} -\\frac{y_{LA}}{n_A^*} - \\frac{y_{LC}}{n_C^*}\\right) + V_c\\left(\\frac{y_{LC}}{n_C^*}\\right) \\notag \\\\\n\t%\n\tI_C &= \\frac{y_{LC}}{|n_C|^2}(V_C - V_B) - \\frac{y_{LC}}{n_C^*}(V_c - V_b) \\notag \\\\\n\t&= V_B\\left(-\\frac{y_{LC}}{|n_C|^2}\\right) + V_C\\left(\\frac{y_{LC}}{|n_C|^2}\\right) + V_b\\left(\\frac{y_{LC}}{n_C^*}\\right) + V_c\\left(-\\frac{y_{LC}}{n_C^*}\\right) \\notag \\\\\n\t%\n\tI_a &= -\\frac{y_{LA}}{n_A}(V_A - V_B) + y_{LA}(V_a - V_b) \\notag \\\\\n\t&= V_A\\left(-\\frac{y_{LA}}{n_A}\\right) + V_B\\left(\\frac{y_{LA}}{n_A}\\right) + V_a\\left(y_{LA}\\right) + V_b\\left(-y_{LA}\\right)\\notag \\\\\n\t%\n\tI_b &= -y_\\text{tie}(V_B - V_b) - I_a - I_c \\notag \\\\\n\t&= V_A\\left( \\frac{y_{LA}}{n_A}\\right) + V_B\\left(-y_\\text{tie} - \\frac{y_{LA}}{n_A} - \\frac{y_{LC}}{n_C}\\right) + V_C\\left( \\frac{y_{LC}}{n_C}\\right) \\notag \\\\\n\t&+ V_a\\left(-y_{LA}\\right) + V_b\\left(y_\\text{tie} + y_{LA} + y_{LC}\\right) + V_c\\left(-y_{LC}\\right) \\notag \\\\\n\t%\n\tI_c &= -\\frac{y_{LC}}{n_C}(V_C - V_B) + y_{LC}(V_c - V_b) \\notag \\\\\n\t&= V_B\\left(\\frac{y_{LC}}{n_C}\\right) + V_C\\left( -\\frac{y_{LC}}{n_C}\\right) + V_b\\left(-y_{LC}\\right) + V_c\\left(y_{LC}\\right)\n\\end{align}\nSince $I = YV$, we can deduce the bus admittance matrix $Y$ as:\n\\begin{align}\nY &= \n\\begin{bmatrix}\n\t\\frac{y_{LA}}{|n_A|^2} & -\\frac{y_{LA}}{|n_A|^2} & 0 & -\\frac{y_{LA}}{n_A^*} & \\frac{y_{LA}}{n_A^*} & 0 \\\\\n\t%\n\t-\\frac{y_{LA}}{|n_A|^2} & \\left(y_\\text{tie} + \\frac{y_{LA}}{|n_A|^2} + \\frac{y_{LC}}{|n_C|^2}\\right) & -\\frac{y_{C}}{|n_C|^2} & \\frac{y_{LA}}{n_A^*} & \\left(-y_\\text{tie} -\\frac{y_{LA}}{n_A^*} - \\frac{y_{LC}}{n_C^*}\\right) & \\frac{y_{LC}}{n_C^*} \\\\\n\t%\n\t0 & -\\frac{y_{LC}}{|n_C|^2} & \\frac{y_{LC}}{|n_C|^2} & 0 & \\frac{y_{LC}}{n_C^*} & -\\frac{y_{LC}}{n_C^*} \\\\\n\t%\n\t-\\frac{y_{LA}}{n_A} & \\frac{y_{LA}}{n_A} & 0 & y_{LA} & -y_{LA} & 0 \\\\\n\t%\n\t\\frac{y_{LA}}{n_A} & \\left(-y_\\text{tie} - \\frac{y_{LA}}{n_A} - \\frac{y_{LC}}{n_C}\\right) & \\frac{y_{LC}}{n_C} & -y_{LA} & \\left(y_\\text{tie} + y_{LA} + y_{LC}\\right) & -y_{LC} \\\\\n\t%\n\t0 & \\frac{y_{LC}}{n_C} & -\\frac{y_{LC}}{n_C} & 0 & -y_{LC}\\ & y_{LC}\n\\end{bmatrix}\n\\end{align}\n\n\\section{General single phase branch model}\n\\label{SEC_GEN_BRANCH_MODEL}\nSingle phase transformers and transmission lines can both be incorporated within a single general branch model, shown in Fig. \\ref{FIG_GEN_BRANCH_MODEL}.\n\\begin{figure}[!h]\n\t\\begin{center}\n\t\t\\includegraphics[width=(9cm)]{branch.png}\n\t\\end{center}\n\t\\caption{\n\t\tA general branch model, similar but more general than that used by {\\sc Matpower}.\n\t}\n\t\\label{FIG_GEN_BRANCH_MODEL}\n\\end{figure}\nThe $Y$ matrix for this model is defined by the following equation:\n\\begin{align}\n\t\\begin{bmatrix}I_0 \\\\ I_1\\end{bmatrix} &=\n\t\\begin{bmatrix}\\frac{y_a + y_b}{|n|^2} & -\\frac{y_a}{n^*} \\\\ -\\frac{y_a}{n} & y_a + y_c \\end{bmatrix}\n\t\\begin{bmatrix}V_0 \\\\ V_1\\end{bmatrix}\n\\end{align}\nNormally, $y_c = y_b$ for both transformers and transmission lines, as per the {\\sc Matpower} manual. For a transmission line, $y_b$ is capacitative, and for a transformer it has a resistive and an inductive component and is associated with the ``no load'' current needed in the primary windings. For both transformers and lines, $y_a$ has a resistive component and an inductive component.\n\n\\section{Units and the per-unit system}\nThe following self-consistent system of units is often used in modelling electricity networks:\n\\begin{description}\n\t\\item[Voltage:]kV (kilovolts)\n\t\\item[Power:]MW/MVAR/MVA (megawatts etc.)\n\t\\item[Current:]kA (kiloamps)\n\t\\item[Resistance:]$\\Omega$ (ohms)\n\t\\item[Admittance:]$S$ (siemens)\n\\end{description}\nUnder this system, relationships like Ohm's Law hold without the need to include any additional constants. The alternative is to use SI units ($V/W/A/\\Omega/S$\\ldots).\n\nThe \\emph{per-unit} system is also often used. The idea is to express electrical quantities as a \\emph{per-unit} dimensionless quantity times a \\emph{base unit}, e.g.\n\\begin{align}\nV &= V_\\text{pu}V_\\text{base} \\notag \\\\\nS &= S_\\text{pu}P_\\text{base} \\notag \\\\\nI &= I_\\text{pu}I_\\text{base} \\notag \\\\\n&\\text{etc.}\n\\end{align}\n\nBase units $V_\\text{base}$ and $P_\\text{base}$ are normally first defined for the network. $V_\\text{base}$ is set to the nominal operating voltage of a bus (e.g. 11~kV), which means that $V_\\text{base}$ may vary from bus to bus. On the other hand, a single network-wide value for $P_\\text{base}$ is usually chosen to be a convenient value such as 100~MVA. Other base quantities may then be derived:\n\\begin{align}\n\tI_\\text{base} &= P_\\text{base}/V_\\text{base} \\notag \\\\\n\tZ_\\text{base} &= V_\\text{base}^2 / P_\\text{base} \\notag \\\\\n\tY_\\text{base} &= P_\\text{base} / V_\\text{base}^2\n\\end{align}\nNote that there is some ambiguity about which bus $V_\\text{base}$ is referenced to in the expressions above. Consider for example a line that runs between two buses: logically there could be several ways to define the per-unit current: using the voltage at the first bus, the voltage at the second bus, or the voltage drop on the line. Depending on how things are done, different equations will need to be used for the network. The ambiguity is removed if we distinguish the power entering bus A along a line AB as being distinct from the power entering bus B along the same line, i.e. we only work with bus injections as opposed to line flows. A set of conventions consistent with \\textsc{Matpower} is given below:\n\\begin{itemize}\n\\item Per-unit quantities in lines are calculated using the base-voltage at the second (to) bus. This makes sense in Matpower, since a common branch model is used where all transformers occur at the first (from) bus, with the bulk of the line occurring between the transformer and the second (to) bus.\n\\item Per-unit quantities associated with buses (such as bus shunt admittance, current loads etc. are calculated, naturally enough, using the base voltage of the bus.\n\\end{itemize}\n\n\\section{The Swing Equations}\nA synchronous generator has a flywheel with moment of inertia $J$. There is a mechanical torque (e.g. from the turbine) $T_m$ pushing the rotor and an electrical torque $T_e$ that comes from the alternator and usually acts against the mechanical torque. We define the torque imbalance $\\Delta T = T_m - T_e$. When $\\Delta T$ is positive the rotor will speed up and when it is negative it will slow down. We define the rotor angle $\\theta$ which is assumed to be zero at $t = 0$. Its first time derivative is the angular velocity $\\omega = \\dot{\\theta}$.\n\nThe equation \\footnote{For those not familiar with the notation, a dot above a variable indicates a time derivative, e.g. $\\dot{\\theta} := \\partial \\theta / \\partial t$. Similarly, a double dot indicates a second time derivative, etc.} governing the motion of the rotor is\n\\begin{align}\n\tJ \\ddot{\\theta} &= \\Delta T\n\\end{align}\nwhich is analogous to Newton's $F = ma$.\n\nFor a rotor running at the constant mandated network angular frequency $\\omega_\\text{nom}$, we have $\\theta = \\omega_\\text{nom} t$. Because the rotor may deviate from this frequency, we define the angular error\n\\begin{align}\n\t\\Delta \\theta &= \\theta - \\omega_\\text{nom} t. \\label{EQ_DEFN_DELTA_THETA}\n\\end{align}\nThis quantity is basically equal to the voltage phase of the generator internal bus. In terms of this new variable, the generator equation is:\n\\begin{align}\n\tJ \\Delta \\ddot{\\theta} &= \\Delta T\n\\end{align}\n\nSince the power $P = \\omega T$, we multiply both sides of the second equation by $\\omega$ to obtain a relation for the power mismatch:\n\\begin{align}\n\tJ\\omega \\Delta \\ddot{\\theta} &= \\Delta P \\notag \\\\\n\tJ\\left(\\omega_\\text{nom} + \\Delta \\dot{\\theta}\\right) \\Delta\\ddot{\\theta} &= \\Delta P\n\\end{align}\nwhere $\\Delta P$ is the input mechanical power minus the electrical power supplied to the network. When the rotor is running far from the mandated frequency, then this equation would need to be applied as is. However, at frequencies near the network frequency, $\\omega_\\text{nom} \\gg \\dot{\\theta}$, and so we can write\n\\begin{align}\n\tJ\\omega_\\text{nom} \\Delta\\ddot{\\theta} &= L_\\text{nom} \\Delta\\ddot{\\theta} = \\Delta P \\label{EQ_SWING}\n\\end{align}\nwhere $L_\\text{nom} = J \\omega_\\text{nom}$ is the angular momentum of the rotor at the network frequency. This is the swing equation, and when solved will give the bus internal voltage angle as a function of time.\n\nA note about units: if, as is common, we measure power in $\\mathrm{MVA}$, then $L$ will need to be measured in commensurate units of $\\mathrm{Mkg\\cdot m^{-2}}$, i.e. we will need to divide the SI value by $10^6$.\n\n\\subsection{Multiple-pole generators}\nGenerators may have multiple pairs of north-south magnetic dipoles. In such cases, the rotor angular frequency will be scaled with respect to the network frequency. For example, a generator with two dipoles running on a 50~Hz network will turn at 25~Hz, because the north-south fields will pass the coils at twice the rate of the rotor's rotation.\n\nSuch cases are easily taken care of by absorbing the scaling factor into an effective moment of inertia $J_\\text{eff} = NJ$, where N is the number of dipoles.\n\n\\subsection{The swing equations and the power flow equations}\nWe derive the electrical power at all generators using the power flow equations. The phase angle $\\phi$ is fixed at the generator buses by the swing equations. The voltage magnitude $V$ could also be fixed (either to its fixed value, or scaled according to the current frequency), making all of the generator buses into effective slack buses. To ensure feasibility of the AC power flow equations, we could then convert all or some ZIP loads into constant impedance loads, or add a high constant impedance component to all loads to ensure feasibility.\n\n\\subsection{The swing equations at a three phase bus}\nFor three phase problems, the three terminals of the bus are connected to a single rotor, with a given angular momentum. The phase relationship between the bus terminals is enforced by the windings in the stator. The problem is what to do with this situation, in regards to the swing equations etc.\n\nThe answer is the obvious: that there is a single machine at each generator, which is input for the swing equations. \n\\end{document}", "meta": {"hexsha": "2e88d4781122cb7b9e32c6a7ce6f538a56f99339", "size": 66103, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/background/power_flow_theory.tex", "max_stars_repo_name": "dexterurbane/SmartGridToolbox", "max_stars_repo_head_hexsha": "ff2eb98e28b0c0ea9690ec6f522ccf1c306f79b7", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "docs/background/power_flow_theory.tex", "max_issues_repo_name": "dexterurbane/SmartGridToolbox", "max_issues_repo_head_hexsha": "ff2eb98e28b0c0ea9690ec6f522ccf1c306f79b7", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/background/power_flow_theory.tex", "max_forks_repo_name": "dexterurbane/SmartGridToolbox", "max_forks_repo_head_hexsha": "ff2eb98e28b0c0ea9690ec6f522ccf1c306f79b7", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.4278752437, "max_line_length": 1002, "alphanum_fraction": 0.6887887085, "num_tokens": 22081, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.84997116805678, "lm_q2_score": 0.8221891305219504, "lm_q1q2_score": 0.6988370556333305}} {"text": "%!TEX root = paper.tex\n% \\section{Synthesizing \\DIs}\n\n\\subsubsection{PCA-inspired \\View Derivation}\\label{candidatesec}\n\n\n\\setlength{\\textfloatsep}{0mm}\n\\begin{algorithm}[t]\n\t\\caption{Procedure to generate linear \\views.}\n\t\\label{fig:pca}\n\t\\LinesNumbered\n\t\\small{ \n \t\\SetKwInOut{Inputs}{Inputs}\n\t\t\\SetKwInOut{Output}{Output}\n\t\t\\Inputs{\n\t\t\tA dataset $D \\subset \\DDom^m$\n\t\t}\n\t\t\\Output{\n A set $\\{(F_1,\\gamma_1),\\ldots,(F_K,\\gamma_K)\\}$ of \\views and importance factors\n \t}\n $D_N \\gets D {\\mbox{ after dropping non-numerical attributes}}$ \\label{algo2:line1}\\\\ \n $D'_N \\gets [\\vec{1} ; D_N]$ \\label{algo2:line2} \\\\\n $\\{\\vec{w}_1,\\ldots,\\vec{w}_K\\} \\gets {\\mbox{ eigenvectors of }} {D_N'}^T D_N'$ \\label{algo2:line31} \\\\\n\t\t\t\t\\ForEach{$1 \\le k \\le K$}{\n\t\t\t\t\t$\\vec{w}_k' \\gets {\\mbox{ $\\vec{w}_k$ with first element removed}}$\\label{algo2:line32} \\\\\n $F_k \\gets \\lambda\\vec{A}: \\frac{\\vec{A}^T\\vec{w}_k'}{||\\vec{w}_k'||}$\\label{algo2:line33} \\\\\n\t\t\t\t\t%{\\mbox{ for $i=1,\\ldots,k$}}$ \\label{algo2:line3} \\\\\n\t\t\t\t\t$\\gamma_k \\gets \\frac{1}{\\log(2+\\stddev{F_k(D_N)})}$ \\label{algo2:line34}\n\t\t\t\t\t%\\af{I fixed it with $\\log$ since that is what we do. Any argument needed for that? Also, I fixed $\\stddev{F_i(D)}$ to $\\stddev{F_i(D_N)}$. Check if that's correct.}\n\t\t\t\t\t% {\\mbox{ for $i=1,\\ldots,k$}}$ \\label{algo2:line4} \\\\\n\t\t\t\t}\n \\Return $\\{(F_1,\\frac{\\gamma_1}{Z}),\\ldots,(F_K,\\frac{\\gamma_K}{Z})\\}$, where\n $Z = \\sum_k \\gamma_k$\\label{algo2:line35}\n\t}\n\\end{algorithm}\n\\setlength{\\textfloatsep}{5pt}\n \nTheorem~\\ref{THM:MAIN} sets the requirements for good \\views (see\nalso~\\cite{tveten2019principal, tveten2019tailored,\nDBLP:journals/tnn/KunchevaF14} that make similar observations in different\nways). It indicates that we can start with any arbitrary \\views and then\niteratively improve them. However, we can get the desired set of best \\views in\none shot using an algorithm inspired by principal component analysis (PCA).\n\\revisetwo{PCA relies on computing eigenvectors. \\label{pcadetails} There exist\ndifferent algorithms for computing eigenvectors (from the infinite space of\npossible vectors). The general mechanism involves applying numerical approaches\nto iteratively converge to the eigenvectors (up to a desired precision) as no\nanalytical solution exists in general.} Algorithm~\\ref{fig:pca} returns \\views that\ncorrespond to the principal components of a\nslightly modified version of the given dataset: % (Fig.~\\ref{fig:pcadetails}).\n% Algorithm~\\ref{fig:pca} details our approach for discovering \\views for\n% constructing \\dis:\n\n% The\n% only additional technical point is the addition of a dummy attribute (column)\n% to the dataset, which is set to the constant $1$ for each tuple. This is done\n% because we are not normalizing the data (e.g., to make the mean of each column\n% zero). It turns out that adding a constant dummy column\n% lets us avoid normalizing the data. % achieves the same effect.\n% We avoid normalizing just to keep the values unchanged so that the generated \\di is\n% interpretable on original values. In detail, we have:\n\\smallskip\n\n\\begin{description}\n\t%\n \\looseness-1\n \\item[Line~\\ref{algo2:line1}] Drop all non-numerical attributes from $D$ to\n get the numeric dataset $D_N$. This is necessary because PCA only applies\n to numerical values. Instead of dropping, one can also consider embedding\n techniques to convert non-numerical attributes to numerical ones.\n\n \\item[Line~\\ref{algo2:line2}] Add a new column to $D_N$ that consists of\n the constant $1$, to obtain the modified dataset $D_N' := [\\vec{1} ; D_N]$,\n where $\\vec{1}$ denotes the column vector with $1$ everywhere. We do this\n transformation to capture the additive constant within principal\n components, which ensures that the approach works even for unnormalized\n data.\n\n \\item[Line~\\ref{algo2:line31}] Compute $K$ eigenvectors of the square\n matrix ${D_N'}^{T}D_N'$, where $K$ denotes the number of columns in $D'_N$.\n\tThese eigenvectors provide coefficients to construct \\views.\n\n \\item[Lines~\\ref{algo2:line32}--\\ref{algo2:line33}] Remove the first\n element (coefficient for the newly added constant column) of all\n eigenvectors and normalize them to generate \\views. Note that we no longer\n need the constant element of the eigenvectors since we can appropriately\n adjust the bounds, $\\lb$ and $\\ub$, for each \\view by evaluating it on\n $D_N$.\n %% bounds of\n %% the \\dis by shifting the mean by the corresponding constant amount. \n % (The\n % 2-norm $||\\vec{w}'_k||$ is $\\sqrt{{\\vec{w}_k}^{'T} \\vec{w}'_k}$.)\n\n \\item[Line~\\ref{algo2:line34}] Compute importance factor for each \\view.\n Since \\views with smaller standard deviations are more discerning (i.e.,\n stronger), we assign each \\view an importance factor ($\\gamma$) that is\n inversely proportional to its standard deviation over $D_N$.\n\n \\item[Line~\\ref{algo2:line35}] Return the linear \\views with corresponding\n normalized importance factors.\n\n\\end{description}\n% \\\\\n% (a) Drop all non-numerical attributes from $D$ to get $D_N$ (line~\\ref{algo2:line1}).\n% %Let $\\vec{w}$ denote the unknown weights that define\n% %the linear \\view $F(\\vec{A}) = \\vec{A}^T\\vec{w}$ on $D_N$. We prefer $\\vec{w}$\n% %that minimizes $\\stddev{D_N\\vec{w}}$.\n% %\n% %\n% %\n% %We are interested in learning linear functionals $F(A_1,\\ldots,A_k)$ of the form\n% %$w_0 + w_1 A_1 + w_2 A_2 + \\cdots + w_k A_k$, where $x1,\\ldots,A_k$ are the $k$ numerical columns\n% %in $D_N$ and $w_0, w_1,\\ldots,w_k$ are $k+1$ unknown weights.\n% %%Since data can be noisy, we want to allow for the equality relation to be not necessarily an exact equality.\n% %\n% %Without loss of generality, we ignore the constant term $w_0$ in the \\invariant. This is because\n% %We find $\\vec{w}$ as follows:\n% \\\\\n% %\n% (b) Add a new column to $D_N$ that consists of the constant $1$, that is,\n% $D_N' := [\\vec{1} ; D_N]$, where $\\vec{1}$ denotes the column vector with $1$\n% everywhere (line~\\ref{algo2:line2}).\n% %If $D_N$ has $k-1$ columns, then $D_N'$ has $k$ columns.\n% %\n% \\\\\n% %\n% (c) Compute $k$ eigenvectors of the square matrix ${D_N'}^{T}D_N'$, where $k$\n% denotes the number of columns in $D'_N$\n% (line~\\ref{algo2:line31}). % $\\vec{v} := [v_1;v_2;\\ldots;v_k]$ corresponding to the lowest\n% %eigenvalue,\n% %\n% \\\\\n% %\n% (d) Remove the first element from each eigenvector (line~\\ref{algo2:line32})\n% and normalize them to generate \\views (line~\\ref{algo2:line33}). (The 2-norm\n% $||\\vec{v}||$ of $\\vec{v}$ is $\\sqrt{\\vec{v}^T\\vec{v}}$.)\n% \\\\ (e) Compute\n% importance factor for each \\view (line~\\ref{algo2:line34}).\n% \\\\ (f) Return the\n% linear \\views with corresponding normalized importance factors\n% (line~\\ref{algo2:line35}).\n\n\\smallskip\n\nWe now claim that the \\views returned by Algorithm~\\ref{fig:pca} include the\n\\view with minimum standard deviation and \n%. Furthermore, if $D$ is sufficiently large, then \nthe correlation between any two \\views \n%returned Algorithm~\\ref{fig:pca} \nis 0. This indicates that we cannot further improve\nthe \\views, and, thus they are optimal.\n%,and $[a;b]$ denotes $\\left[\\begin{matrix}a \\\\ b\\end{matrix}\\right]$.\n\n\\begin{theorem}[Correctness of Algorithm~\\ref{fig:pca}]\\label{THM:ALGO2-CORRECTNESS}\n\t%\n Given a numerical dataset $D$ over the schema $\\mathcal{R}$, let\n $\\mathcal{F} = \\{F_1,F_2,\\ldots, F_K\\}$ be the set of linear \\views\n returned by Algorithm~\\ref{fig:pca}. Let $\\sigma^* =\n \\min_k^{K}\\stddev{F_k(D)}$. If $\\mu(A_k(D)) = 0$ for all attribute $A_k$ in\n $\\mathcal{R}$, then,\\footnote{When the condition $\\forall A_k \\;\n \\mu(A_k(D)) = 0$ does not hold, slightly modified variants of the claim\n hold. However, by normalizing $D$ (i.e., by subtracting attribute mean\n $\\mu(A_k(D))$ from each $A_k(D)$), it is always possible to satisfy the\n condition.}\n%\n\t\\begin{enumerate}[label=(\\arabic*)]\n \\item $\\sigma^* \\leq \\stddev{F(D)}$ $\\forall F=\\vec{A}^T\\vec{w}$ where $||\\vec{w}||\\geq 1$, and\n \\item $\\forall F_j, F_k \\in \\mathcal{F}$ s.t.\\ $F_j\\neq F_k$, \n % the corresponding eigenvalues $\\varepsilon_j \\neq \\varepsilon_k$, \n $\\rho_{F_j, F_k} = 0$.\n\t\t% (assuming, wlog, Algorithm~\\ref{fig:pca} always picks orthogonal eigenvectors).\n % $\\lim_{|D| \\to \\infty} \\rho_{F_j, F_k} = 0$, \\textcolor{red}{under the assumption that\n % the columns of $D$ converge to some mean value and the\n % eigenvalues of $D^TD$ converge to some fixed values.}\n\t\\end{enumerate}\n % Furthermore, any linear \\view $F$ s.t.\\ $\\stddev{F(D)} = \\sigma^*$ \n % is equal to a linear\n % combination of $F_j$'s whose standard deviation is $\\sigma^*$ on $D$.\n\\end{theorem}\n% Let $\\vecw$ be the (column) vector consisting of the unknown weights $w_1,\\ldots,w_k$.\n% Since $d\\vecw$ may not identically be $0$ for all $d\\in D_N$, \n% we can cast the problem of finding $w_1,\\ldots,w_k$ as an optimization problem\n% that tries to minimize the sum $\\sum_{d\\in D_N} (d\\vecw)^2$, which we shall call {\\em{error}}. \n% In matrix notation, this error can be written\n% as $\\vecw^T D_N^TD_N \\vecw$.\n% We restrict $\\vecw$ to unit vectors to make the optimization problem well posed.\n\n\n\n\n\\ignore{\nIn Algorithm~\\ref{fig:pca}, we use the above procedure to compute linear \\views.\n\\af{I will pull this before, and will describe the procedure while referring\nto the corresponding lines in the algorithm. I can do it if you agree.}\nAlgorithm~\\ref{fig:pca} computes \\views using {\\em{all}} eigenvectors, rather\nthan just the eigenvector corresponding to the lowest eigenvalue. \n\\endignore}\n\nUsing \\views $F_1, \\ldots, F_K$, and importance factors\n$\\gamma_1,\\ldots,\\gamma_K$, returned by Algorithm~\\ref{fig:pca}, we generate\nthe simple (conjunctive) \\invariant with $K$ conjuncts:\n%\n$\n \\bigwedge_k \\lb_k \\leq F_k(\\vec{A}) \\leq \\ub_k\n$.\n%\nWe compute the bounds $\\lb_k$ and $\\ub_k$ following Section~\\ref{synth-bounds}\nand use the importance factor $\\gamma_k$ for the $k^{th}$ conjunct in the\nquantitative semantics.\n\n\\begin{example}\\label{ex:int} Algorithm~\\ref{fig:pca} finds the projection of the \\di of\nExample~\\ref{ex:tml}, but in a different form. The actual airlines dataset has\nan attribute $\\mathtt{distance}\\; (\\mathtt{DIS})$ that represents miles travelled by a flight. In\nour experiments, we found the following \\di\\footnote{For ease of exposition, we\nuse $F(\\vec{A}) \\approx 0$ to denote $\\epsilon_1 \\le F(\\vec{A}) \\le\n\\epsilon_2$, where $\\epsilon_i \\approx 0$.} over the dataset of daytime flights:\n%\n{\n\\begin{align}\\label{eq:one}\n0.7 \\times \\mathtt{AT} - 0.7 \\times \\mathtt{DT} - 0.14 \\times \\mathtt{DUR} - 0.07 \\times \\mathtt{DIS} \\approx 0\n\\end{align}\n}\n%\nThis \\invariant is not quite interpretable by itself, but it is in fact a\nlinear combination of two expected and interpretable\n\\invariants:~\\footnote{\\revisetwo{We developed a tool~\\cite{DBLP:conf/sigmod/FarihaTRG20} to explain causes of\nnon-conformance.~\\citeTechRep}}\n%\n{\n\\begin{align}\n\t\\mathtt{AT} - \\mathtt{DT} - \\mathtt{DUR} &\\approx 0 \\label{eq:two}\\\\\n\t\\mathtt{DUR} - 0.12 \\times \\mathtt{DIS} &\\approx 0 \\label{eq:three}\n\\end{align}\n}\n%\nHere, (\\ref{eq:two}) is the one mentioned in Example~\\ref{ex:tml} and\n(\\ref{eq:three}) follows from the fact that average aircraft speed is about\n$500$ mph implying that it requires $0.12$ minutes per mile. $0.7$ $\\times$\n(\\ref{eq:two}) + $0.56$ $\\times$ (\\ref{eq:three}) yields:\n%\n{\n\\begin{align*}\n&0.7 \\times (\\mathtt{AT} - \\mathtt{DT} - \\mathtt{DUR}) + 0.56 \\times \\mathtt{DUR} - 0.56 \\times 0.12 \\times \\mathtt{DIS} \\approx 0 \\\\\n\\implies & 0.7 \\times \\mathtt{AT} - 0.7 \\times \\mathtt{DT} - 0.14 \\times \\mathtt{DUR} - 0.07 \\times \\mathtt{DIS} \\approx 0\n\\end{align*}\n}\n%\nWhich is exactly the \\di~(\\ref{eq:one}). Algorithm~\\ref{fig:pca} found the\noptimal \\view of~(\\ref{eq:one}), which is a linear combination of the \\views\nof~(\\ref{eq:two}) and~(\\ref{eq:three}). The reason is: there is a correlation\nbetween the \\views of~(\\ref{eq:two}) and~(\\ref{eq:three}) over the dataset\n(Theorem~\\ref{THM:MAIN}). One possible explanation of this correlation is:\nwhenever there is an error in the reported duration of a tuple, it violates\nboth~(\\ref{eq:two}) and~(\\ref{eq:three}). Due to this natural correlation,\nAlgorithm~\\ref{fig:pca} returned the optimal \\view of~(\\ref{eq:one}), that\n``covers'' both \\views of~(\\ref{eq:two}) or~(\\ref{eq:three}).\n\n%\n\\end{example}\n\n\\ignore{ \n\nThe reason for including \\views corresponding to all eigenvalues in the\nsimple \\invariant---in addition to the one with smallest standard\ndeviation---is that it only strengthens the \\invariant. Note that we use importance factors $\\gamma_i$ to give\nmore importance to the \\views with smaller standard deviations. One obvious\nquestion is: can our procedure miss some linear \\views with small standard\ndeviations? The following result states that we do not miss any linear \\views\ncorresponding to the {\\em{smallest}} achievable standard deviation.\n\n% Consider the eigenvalues and eigenvectors of the positive semi-definite matrix $D_N^TD_N$.\n% Let $\\eig_1,\\ldots,\\eig_k$ be the $k$ eigenvalues, and \n% $\\vecv_1, \\ldots,\\vecv_k$ be the corresponding $k$ eigenvectors of $D_N^TD_N$. Assuming that the eigenvectors are normalized to have norm $1$, we have\n% $$\n % \\vecv_j^T D_N^TD_N \\vecv_j = \\vecv_j^T \\eig_j\\vecv_j = \\eig_j\n% $$\n% Since $D_N^TD_N$ is real-valued, symmetric, and positive semi-definite, we know that $\\eig_j$'s are all non-negative reals.\n% WLOG, assume $\\eig_1 \\leq \\eig_2 \\leq \\cdots \\leq \\eig_k$.\n% Define $k$ linear functionals $f_1,f_2,\\ldots,f_k$,\n% where $f_i$ is defined as $f_i(A_1,\\ldots,A_k) = \\vecv_{j1}A_1 + \\cdots + \\vecv_{jk}A_k$.\n% For these $k$ choices,\n% the spectrum, $\\eig_1, \\ldots, \\eig_k$, of the matrix $D_N^TD_N$, provides us with different choices for the error value. \n% %\n% Since we want to minimize error \\af{I don't think this argument is correct. As\n% mentioned in Section 2, we want neither overfit nor underfit \\invariants. If we\n% wanted to minimize error, you could always learn a NULL \\invariant, why even\n% just one with lowest eigenvalue?}, we should pick $\\vecv_1$, the eigenvector\n% corresponding to $\\eig_1$, to construct our \\invariant. In fact, we should\n% use all eigenvectors corresponding to small eigenvalues, since some prior\n% work~\\cite{tveten2019principal, tveten2019tailored,\n% DBLP:journals/tnn/KunchevaF14} showed that low eigenvalue principal components\n% are more sensitive to a general change. Rather than use some heuristic to\n% determine which eigenvalues are ``small'' and which are not, in our\n% implementation, we just use all eigenvectors \\af{I think we can get rid of\n% defer it to the Experiment section where I would keep a subsection for\n% implementation specifics.}, but {\\em{weigh them differently}} according to the\n% standard deviation of $f_i(D_N)$, as we stated before. \\todo{Talk about\n% $\\gamma$ being $\\frac{1}{\\log(2 + \\stddev{D})}$. And then talk about\n% normalizing them}. Note that the above derivation shows that $\\stddev{f_j}$ is\n% related to the value $\\eig_j$ of the corresponding eigenvalue.\n% Algorithm~\\ref{fig:pca} summarizes the procedure for computing the linear\n% functionals from a data set $D$, which are then used to define the actual\n% \\invariants as described in previous section.\\af{I am not sure if the algorithm\n% is very useful, as PCA is a very well known technique.}\\af{This is the heart of\n% this section, and one of our core novelty. I think we should revise this with\n% better justification and intuition. At present, it looks very weak.}\n\n\\begin{theorem}\n Let $D$ be a dataset and\n $F_1,F_2,\\ldots$ be the linear \\views returned by the procedure in Algorithm~\\ref{fig:pca} \n when on $D$.\n Let $\\sigma^* = \\min(\\{\\stddev{F_i(D)} | i=1,2,\\ldots\\})$.\n If there is a linear \\view $F$ s.t. $\\stddev{F(D)} = \\sigma^*$, then $F$ will be in the linear subspace spanned by all the linear\n $F_j$ with standard deviation $\\sigma^*$ on $D$.\n\\end{theorem}\n\\af{I now have high level idea of what this theorem is trying to say, but I still \nthink we can omit it in this paper.}\n% .. Dw.Dw = Dv.Dv and ||w||=||v||=1\n% ||av + bw||^2 = a^2 + b^2 + 2ab cos(theta)\n% D(av+bw).D(av+bw) = (aDv + bDw).(aDv + bDw) = a^2 S + b^2 S + 2ab (Dv.Dw) = aaS + bbS + 2abS v^Tw = S!!\n\n\nWe observe that the most common way in which PCA is used involves {\\em{throwing\naway}} the low variance components, whereas we give {\\em{most importance}} to\nthese components. This role reversal is one of our key insights in synthesizing good\n\\dis.\n\n\\endignore}\n\n\n\n\\subsection{Compound \\DIs}\\label{sec:disjunctive}\n% We now describe computation of \\emph{disjunctive} linear \\invariants. \nThe quality of our PCA-based simple linear \\invariants relies on how many low\nvariance linear \\views we are able to find on the given dataset. For many\ndatasets, it is possible we find very few, or even none, such linear \\views. In\nthese cases, it is fruitful to search for compound \\invariants; we first focus\non {\\em{disjunctive \\invariants}} (defined by $\\psi_A$ in our language grammar).\n\nThe PCA-based approach fails in cases where there exist different piecewise linear\ntrends within the data, as it will result into low-quality \\invariants, with\nvery high variances. In such cases, partitioning the dataset and then learning\n\\invariants separately on each partition will result in significant improvement\nof the learned \\invariants. A disjunctive \\invariant is a compound \\invariant\nof the form $\\bigvee_k((A = c_k) \\maxand \\phi_k)$, where each $\\phi_k$ is a\n\\invariant for a specific partition of $D$. Finding disjunctive \\invariants\ninvolves horizontally partitioning the dataset $D$ into smaller disjoint\ndatasets $D_1, D_2, \\ldots, D_L$. Our strategy for partitioning $D$ is to use\ncategorical attributes with a small domain in $D$; in our implementation, we\nuse those attributes $A_j$ for which $|\\{t.A_j | t\\in D\\}|\\le50$. If $A_j$ is\nsuch an attribute with values $v_1, v_2, \\ldots, v_L$, we partition $D$ into\n$L$ disjoint datasets $D_1, D_2,\\ldots, D_L$, where $D_l = \\{t \\in D | t.A_j =\nv_l\\}$. Let $\\phi_1, \\phi_2, \\ldots, \\phi_L$ be the $L$ simple \\dis we learn\nfor $D_1, D_2, \\ldots, D_L$ using Algorithm~\\ref{fig:pca}, respectively. We\ncompute the following disjunctive \\di for $D$:\\\\\n% (recall that the quantitative semantics\n% of \\dis does not satisfy the typical Boolean algebra laws for $\\vee$\n% and $\\wedge$)\n%\n\\indent $\n ((A_j = v_1) \\maxand \\phi_1) \\vee \n ((A_j = v_2) \\maxand \\phi_2) \\vee \n\\cdots \\vee\n ((A_j = v_L) \\maxand \\phi_L)\\\\[-0.7em]\n$\n\n\\looseness-1 We repeat this process and partition $D$ across multiple\nattributes and generate a compound disjunctive \\invariant for each attribute.\nThen we generate the final compound conjunctive \\di ($\\Psi$) for $D$, which is\nthe conjunction of all these disjunctive \\invariants. Intuitively, this final\n\\di forms a set of \\emph{overlapping} hyper-boxes around the data tuples.\n\n\n\\subsection{Theoretical Analysis}\\label{sec:complexity} \n\n\\subsubsection{Runtime Complexity}\\looseness-1 Computing simple \\invariants\ninvolves two computational steps: (1)~computing $X^TX$, where $X$ is an\n$n\\times m$ matrix with $n$ tuples and $m$ attributes, which~takes\n$\\mathcal{O}(nm^2)$ time, and (2)~computing the eigenvalues and eigenvectors of\nan $m\\times m$ positive definite matrix, which has complexity\n$\\mathcal{O}(m^3)$~\\cite{DBLP:conf/stoc/PanC99}. Once we obtain the linear\n\\views using the above two steps, we need to compute the mean and variance of\nthese \\views on the original dataset, which takes $\\mathcal{O}(nm^2)$ time. In\nsummary, the overall procedure is cubic in the number of attributes and linear\nin the number of tuples.\n%\nFor computing disjunctive \\invariants, we greedily pick attributes that take at\nmost $L$ (typically small) distinct values, and then run the above procedure\nfor simple \\invariants at most $L$ times. This adds just a constant factor\noverhead per attribute.\n\n\\subsubsection{Memory Complexity} The procedure can be implemented in\n$\\mathcal{O}(m^2)$ space. The key observation is that $X^T X$ can be computed\nas $\\sum_{i=1}^{n} t_i t_i^T$, where $t_i$ is the $i^{th}$ tuple in the\ndataset. Thus, $X^TX$ can be computed incrementally by loading only one tuple\nat a time into memory, computing $t_i t_i^T$, and then adding that to a running\nsum, which can be stored in $\\mathcal{O}(m^2)$ space. Note that instead of such\nan incremental computation, this can also be done in an embarrassingly parallel\nway where we horizontally partition the data (row-wise) and each partition is\ncomputed in parallel.\n\n% Due to such low time and memory complexity, our approach\n% scales gracefully to large datasets.\n\n\\revisetwo{\\subsubsection{Implication, Redundancy, and Minimality}\\label{sec:impl}\nDefinition~\\ref{def:stronger} gives us the notion of \\emph{implication} on\n\\dis: for a dataset $D$, satisfying $\\phi_1$ that is stronger than $\\phi_2$\nimplies that $D$ would satisfy $\\phi_2$ as well.\n%\nLemma~\\ref{LEMMA:MAIN} and Theorem~\\ref{THM:MAIN} associate \\emph{redundancy}\nwith correlation: correlated projections can be combined to construct a new\nprojection that makes the correlated projections redundant.\nTheorem~\\ref{THM:ALGO2-CORRECTNESS} shows that our PCA-based procedure finds\na non-redundant (orthogonal and uncorrelated) set of projections. For disjunctive\n\\invariants, it is possible to observe redundancy across partitions. However,\nour quantitative semantics ensures that redundancy does not affect the\nviolation score.\n%\nAnother notion relevant to data profiles (e.g., FDs) is \\emph{minimality}. In\nthis work, we do not focus on finding the minimal set of \\dis. Towards\nachieving minimality for \\dis, a future direction is to explore techniques for\noptimal data partitioning. However, our approach computes only $m$ \\dis for\neach partition. Further, for a single tuple, only $m_N \\cdot m_C$ \\dis are applicable,\nwhere $m_{N}$ and $m_C$ are the number of numerical and categorical attributes\nin $D$ (i.e., $m = m_{N} + m_C$). The quantity $m_N \\cdot m_C$ is upper-bounded by\n$\\frac{m^2}{4}$. }\n\n\n", "meta": {"hexsha": "6913cf966252b8d683295bfdcc2aa0c7f15f57d0", "size": 22212, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Paper/4_synthesizing-data-invariants_2.tex", "max_stars_repo_name": "afariha/ConformanceConstraintsReproducibility", "max_stars_repo_head_hexsha": "ee419285ab32464f063225fbdeba005a043d2033", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Paper/4_synthesizing-data-invariants_2.tex", "max_issues_repo_name": "afariha/ConformanceConstraintsReproducibility", "max_issues_repo_head_hexsha": "ee419285ab32464f063225fbdeba005a043d2033", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-12-09T09:30:49.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-09T09:30:49.000Z", "max_forks_repo_path": "Paper/4_synthesizing-data-invariants_2.tex", "max_forks_repo_name": "afariha/ConformanceConstraintsReproducibility", "max_forks_repo_head_hexsha": "ee419285ab32464f063225fbdeba005a043d2033", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-09T05:22:18.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-09T05:22:18.000Z", "avg_line_length": 51.7762237762, "max_line_length": 171, "alphanum_fraction": 0.7097965064, "num_tokens": 6871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9086179068309441, "lm_q2_score": 0.7690802423634963, "lm_q1q2_score": 0.6988000800013552}} {"text": "\\section{Asymptotic Theory}\n\nAsymptotic theory, large sample theory, or limit theory, deals with the question of the limiting behavior of sequences of random variables. \n\nThe notion of convergence for random variables is more involved than is usual for, for example, a sequence of numbers or functions in calculus. There are different types of convergence, and they do not necessarilty imply one another.\n\nThe chain of implication is:\n\n\\begin{equation}\n\\mathrm{quadratic\\ mean} \\rightarrow \\mathrm{probability} \\rightarrow \\mathrm{distribution}\n\\end{equation}\n\nFor point mass distributions only, convergence in distribution implies convergence in probability. Similarlty, implication rules exist for functions of random variables when convergence for the underlying random variables is given. See Theorems 5.4, 5.5 and 5.17 in \\citeasnoun{wasserman2013all} (pages 73-74, 81). \n\n\\subsection{Preasymptotics}\nThe behavior in the intermediate regime, for example where $n$ large but not yet asymptotic. Asymptotic behavior kicks in at very different rates for different underlying processes. The ``baseline\" in terms of speed is represented by normally distributed random variables.\n\n\n%convergence in probability\n\\subsection{Convergence in Probability}\n\n\\begin{equation}\n\\mathbb{P}(|X_n - X| > \\epsilon) \\rightarrow 0\n\\end{equation}\n\nas $n\\rightarrow \\infty$. Convergence in Probability means that the distribution of $X_n$ becomes sharper and sharper around $X$ as $n\\rightarrow \\infty$. At $n=\\infty$, it has point mass distribution concentrated at $X$.\n\n\n% convergence in distribution\n\\subsection{Convergence in Distribution}\nWhere $F$ is the CDF of $X_n$,\n\n\\begin{equation}\n\\lim_{n\\rightarrow \\infty} F_n(t) = F(t)\n\\end{equation}\n\nFor all $t$ for which $F$ is continuous. That means, convergence is satisfied even when the equality is violated at points of discontinuity.\n\n\n% convergence in quadratic mean\n\\subsection{Convergence in Quadratic Mean, Convergence in $L_2$}\n\n\\begin{equation}\n\\mathbb{E}(X_n - X)^2 \\rightarrow 0\n\\end{equation}\n\nas $n\\rightarrow \\infty$.\n\n\n% almost sure convergence \n\\subsection{Almost Sure Convergence}\n\n$X_n$ converges \\textit{almost surely} to $X$ if:\n\n\\begin{equation}\n\\mathbb{P}(\\{\\omega: X_n(\\omega) \\rightarrow X(\\omega) \\}) = 1\n\\end{equation}\n\nWhich I would read as the value of the measurable map $X_n$ converging to the value of the measurable map $X$ everywhere on the probability space except possibly on a set of probability measure $0$.\n\n\n% l1 convergence\n\\subsection{$L_1$ Convergence}\n\n$L_1$ convergence requires $\\mathbb{E}|X_n - X| \\rightarrow 0$ as $n\\rightarrow 0$.\n\n\n\n% WLLN\n\\subsection{Weak Law of Large Numbers}\n\nThe sample mean of i.i.d. variables $\\overline{X}_n$ converges in probability to the mean of the distribution $\\mu$i. It can be proven with Chebyshev's inequality that:\n\n\\begin{equation}\n\\mathbb{P}(|\\overline{X}_n-\\mu|>\\epsilon) \\leq \\frac{\\mathbb{V}(\\overline{X}_n)}{\\epsilon^2}=\\frac{\\sigma^2}{n \\epsilon^2}\n\\end{equation}\n\nIn words, the probability that the sample mean deviates from the population mean by more than $\\epsilon$ has an upper bound that decreases inversely proportional to $n\\epsilon^2$. The probability becomes more and more centered around the mean $\\mu$.\n\n% SLLN\n\\subsection{Strong Law of Large Numbers}\nThe strong law of large number gives almost surely convergence of the sample mean to the population mean.\n\nIf $\\mathbb{E}|X_1| < \\infty$, then $\\overline{X_n}\\xrightarrow{as}\\mu$.\n\n\n\n% CLT\n\\subsection{Central Limit Theorem}\nThe distribution of the sample mean converges in distribution to a normal distribution with variance $\\sigma^2/n$ and mean $\\mu$.\n\nIf $Z_n = \\frac{\\overline{X}_n - \\mu}{\\sqrt{\\mathbb{V}(\\overline{X}_n)}}$ then $\\lim_{n\\rightarrow \\infty} \\mathbb{P}(Z_n \\leq z) = \\Phi(z)$, where $\\Phi(z)$ is the CDF of a standard normal distribution. \n\nIt turns out that when $Z_n$ is obtained by normalizing not by the (most likely unknown) population variance $\\sigma$ but by the sample variance $S_n^2$, the CLT still holds. The accuracy of this is given by the Berry-Ess\\'een Inequality.\n\n\n% multivariate CLT\n\\subsection{Multivariate Central Limit Theorem}\nGiven $\\mathbf{X_1, ... ,X_n}$ i.i.d random vectors where each vector:\n\n\\begin{equation}\n\\mathbf{X_i} = \\left(\\begin{array}{c}X_{1i}\\\\ X_{2i} \\\\ \\vdots \\\\ X_{ki} \\end{array}\\right)\n\\end{equation}\n\nThen the population mean:\n\n\\begin{equation}\n\\mathbf{\\mu} = \\left(\\begin{array}{c} \\mu_1 \\\\ \\mu_2 \\\\ \\vdots \\\\ \\mu_k \\end{array} \\right) = \\left(\\begin{array}{c} \\mathbb{E}(X_{i1}) \\\\ \\mathbb{E}(X_{i2}) \\\\ \\vdots \\\\ \\mathbb{E}(X_{ik}) \\end{array} \\right)\n\\end{equation}\n\nThe variance is given by the matrix $\\Sigma$ as before. The sample mean:\n\n\\begin{equation}\n\\overline{\\mathbf{X}} = \\left(\\begin{array}{c}\\overline{X_{1}}\\\\ \\overline{X_{2}} \\\\ \\vdots \\\\ \\overline{X_{k}} \\end{array}\\right)\n\\end{equation}\n\nThen $\\sigma^{-\\frac{1}{2}} (\\overline{X}-\\mu)$ converges in distribution ot $\\mathscr{N}(0,1)$.\n\n\n% proof\n\\subsection{Proof of the Central Limit Theorem}\n\\citeasnoun{wasserman2013all}, page 81.\n\nGiven i.i.d random variables $X_i$, the transformation $Y_i = \\frac{X_i-\\mu}{\\sigma}$ gives i.i.d. random variables with zero mean and unit variance. Let $\\psi(t)$ be the MGF of $Y_i$. Since $Y_i$ are i.i.d., the sum $\\sum_{i=1}^n Y_i$ has MGF $\\psi(t)^n$. The normalized sample mean $Z_n = \\frac{1}{\\sqrt{n}}\\sum_{i=1}^n Y_i$ has MGF $\\Xi_n(t)=\\psi(t/\\sqrt{n})^n$. Two random variables that have the same MGF in an open interval about the point $t=0$ have the same distribution, probably because the Laplace transform is injective. Therefore, if $\\psi_n(t)\\rightarrow \\psi(t)$ in some open interval around $t=0$, then their underlying random variables $Z_n \\xrightarrow{dist}Z_n$ converge in distribution. Taking the Taylor expansion of $\\epsilon_n(t)$:\n\n\\begin{equation}\n\\epsilon_n(t) = \\left(1+0+\\frac{t^2}{2! n} + ... \\right)^n \\rightarrow e^{t^2/2}\n\\end{equation}\n\nWhich is the MGF of $\\mathscr{N}(0,1)$\n\n% Delta Method\n\\subsection{Delta Method}\n\nThe delta method allows statements regarding the convergence of functions of random variables, whenever the input random variable converges in distribution to a normal distribution. \n\nIf $Y_n$ has a limiting normal distribution, and $g(Y_n )$ is a smooth function so that $g'(\\mu) \\neq 0$, then if:\n\n\\begin{equation}\n\\frac{\\sqrt{n}(Y_n - \\mu)}{\\sigma} \\xrightarrow{dist} \\mathscr{N}(0,1)\n\\end{equation}\n\nThen:\n\n\\begin{equation}\n\\frac{\\sqrt{n}(g(Y_n)-g(\\mu))}{|g'(\\mu)|\\sigma}\\xrightarrow{dist}\\mathscr{N}(0,1)\n\\end{equation}\n\nRewriting, if $Y_n \\xrightarrow{dist}\\mathscr{N}(\\mu,\\frac{\\sigma^2}{n})$ then $g(Y_n)\\xrightarrow{dist}\\mathscr{N}(g(\\mu),(g'(\\mu))^2\\frac{\\sigma^2}{n})$.\n\n\\subsubsection{Multivariate Delta Method}\n\nIf $\\mathbf{Y_n} \\xrightarrow{dist}\\mathscr{\\mathbf{\\mu},\\mathbf{\\Sigma}}$ then the scalar-valued function $g(\\mathbf{Y_n}) \\xrightarrow{dist}\\mathscr{N}(g(\\mathbf{\\mu}),\\frac{1}{n} (\\nabla g(\\mu))^T \\mathbf{\\Sigma} (\\nabla g(\\mu)) )$.\n\nUse case would be functions of several sample means, where the underlying samples have non-trivial covariance (cf. \\cite{wasserman2013all}, p. 80).\n\n", "meta": {"hexsha": "bed36cf1366bae43822862375982e707f5c8b177", "size": 7126, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/chapters/sections/proba_asymptotictheory.tex", "max_stars_repo_name": "jpbm/probabilism", "max_stars_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "notes/chapters/sections/proba_asymptotictheory.tex", "max_issues_repo_name": "jpbm/probabilism", "max_issues_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/chapters/sections/proba_asymptotictheory.tex", "max_forks_repo_name": "jpbm/probabilism", "max_forks_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.3885350318, "max_line_length": 754, "alphanum_fraction": 0.7301431378, "num_tokens": 2139, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673178375734, "lm_q2_score": 0.8596637505099167, "lm_q1q2_score": 0.6987925671191849}} {"text": "\\section{Technical Cryptography}\n\\label{sec:tech_crypto}\n\nIn this section we present in detail the specifics of the\ncryptography we will be using for MadNetwork.\nAt times we will be verbose in our algorithmic details\nand design choices to allow for others to understand our decisions.\n\nWe will begin by comparing this work with the\nEthereum Distributed Key Generation whitepaper~\\cite{ethdkg}\nin Sec.~\\ref{ssec:ethdkg_comparison};\nour design is based this paper.\nSome of the implementation-specific details are covered in\nSec.~\\ref{ssec:pk_curve_specifics}.\nIn Sec.~\\ref{ssec:math_def}, we discuss the mathematics\nrelated to pairing-based cryptography; this is integral to\nour work as it is required for our group signatures.\nWe describe the distributed key generation protocol in\nSec.~\\ref{ssec:dkg};\nthe specific method of shared secret encryption is\ndescribed in Sec.~\\ref{ssec:secret_enc}.\nWe discuss how to construct group signatures in\nSec.~\\ref{ssec:crypto_group_sig}.\nGroup signatures from pairing-based cryptography require\na hash-to-curve function, and we talk about our construction\nbased on~\\cite{ft2012bnhashtocurve,boneh2019h2cBLS12}\nin Sec.~\\ref{ssec:hash-to-curve}.\n\nWe follow~\\cite{ethdkg} in our definition of\n$\\parens{t,n}$-thresholded system, where we need $t+1$ actors\nfor consensus.\nUnfortunately, this is different than what was used previously,\nwhere $\\parens{t,n}$-thresholded system meant $t$ actors were\nneeded to agree.\nWe keep this difference for ease of comparison with the referenced paper.\n\n\\input{tex/tcrypt_comparison.tex}\n\\input{tex/tcrypt_spec_imp.tex}\n\\input{tex/tcrypt_math_def.tex}\n\\input{tex/tcrypt_dkg.tex}\n\\input{tex/tcrypt_sse.tex}\n\\input{tex/tcrypt_grpsig.tex}\n\\input{tex/tcrypt_h2c.tex}\n", "meta": {"hexsha": "ec238e70b836edbd39aa4dfe83a5f6208e93ca72", "size": 1732, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/technical_crypto.tex", "max_stars_repo_name": "MadBase/MadNet-Whitepaper", "max_stars_repo_head_hexsha": "53e230263dac0c8d51a7c248e2ae5fca29ec249c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/technical_crypto.tex", "max_issues_repo_name": "MadBase/MadNet-Whitepaper", "max_issues_repo_head_hexsha": "53e230263dac0c8d51a7c248e2ae5fca29ec249c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/technical_crypto.tex", "max_forks_repo_name": "MadBase/MadNet-Whitepaper", "max_forks_repo_head_hexsha": "53e230263dac0c8d51a7c248e2ae5fca29ec249c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2021-01-25T15:44:15.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-13T21:19:44.000Z", "avg_line_length": 39.3636363636, "max_line_length": 73, "alphanum_fraction": 0.8013856813, "num_tokens": 453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9219218262741297, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6986271757486662}} {"text": "\\chapter{Exponential and Logarithmic Functions}\n\\section{Exponential Functions}\nAn exponential function is a defined as any function, $f(x)$ such that the\nindependent variable $x$ is bound up somewhere in the exponent. $f(x)=3^x$ is\nan example of such a function:\n\n\\graph{3^x}\n\nThis classification of functions have a few simple properties:\n\n\\begin{itemize}\n \\item{They are defined for all $x\\in(-\\infty,\\infty)$}\n \\item{Their range is: $(0, \\infty)$}\n \\item{They have intercepts at $(0,1)$}\n \\item{They are increasing on their entire domain}\n \\item{The $x$-axis serves as a horizontal asymptote}\n \\item{They are smooth and continuous}\n\\end{itemize}\n\n\\subsection{Interest Functions}\nThese types of functions have applications in a few different areas. For us, we\nwill be primarily concerned with manually and continually compouding interest:\n\n\\begin{itemize}\n \\item{For $n$ compoundings per year: $A=P(1+\\frac{r}{n})^{nt}$}\n \\item{For coninuous compounding: $A=Pe^{rt}$}\n\\end{itemize}\n\n\\section{Logarithmic Functions}\nThe parent function of the logarithmic family is defined as follows:\n\\begin{equation}\n f(x)=\\log_ax\n\\end{equation}\n\n\\graph{ln(x)}\n\n\\subsection{Properties of Logarithms}\n\\begin{itemize}\n \\item{$\\log_a1=0$}\n \\item{$\\log_a=1$}\n \\item{$\\log_aa^x=x$}\n \\item{If $\\log_ax=\\log_ay$ then $x=y$}\n\\end{itemize}\n\n\\section{Properties of Logarithms}\nA logarithm of any base $a$ can be represented as a quotient of two other\nlogarithms of base $b$, according to the following (Change of Base rule)\n\\begin{equation}\n \\log_ax=\\frac{\\log_bx}{\\log_ba}\n\\end{equation}\n\nThe product property states that you can expand a logarithm that is the product\nof two numbers $u$ and $v$ into two logarithms that are added together, as in\nthe following generalization:\n\\begin{equation}\n \\log_a(uv)=\\log_au+\\log_av\n\\end{equation}\n\nThe quotient property states what is already implied, that a logarithm\nrepresented as a quotient of two numbers $u$ and $v$ can be expanded equally as\na an expression of two logarithms which get subtracted, as in the following\ngeneralization:\n\\begin{equation}\n \\log_a(\\frac{u}{v})=\\log_au-\\log_av\n\\end{equation}\n\nThe power property is implied from the product rule. You can expand out a power\nwrapped up in a logarithm into a scalar that is multiplied to the logarithmic\nexpression, as in the following generalization:\n\\begin{equation}\n \\log_au^n=n\\log_au\n\\end{equation}\n\n\\section{Solving Logarithmic Functions}\nYou can apply the rules that have been previously explained in order to solve\nfor expressions bound up in logarithmic or exponential expressions. The genral\ncase for solving these types of problems is explained below:\n\n\\begin{enumerate}\n \\item{Reduce all variables that are bound in exponential expressions to\n loosely hanging variables that are included in logarithmic expressions}\n \\item{Apply the three rules that are stated above}\n \\item{Expand}\n \\item{Simplify}\n\\end{enumerate}\n", "meta": {"hexsha": "b569a39dbc566f91de0b6d652391f13030bfcdc2", "size": 2939, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pre_calculus_h/chapters/3_exponential_logarithmic_functions.tex", "max_stars_repo_name": "ttaylorr/finals", "max_stars_repo_head_hexsha": "41f6a03e4a082768c10b9a77a0e042cfce2a6d7f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "pre_calculus_h/chapters/3_exponential_logarithmic_functions.tex", "max_issues_repo_name": "ttaylorr/finals", "max_issues_repo_head_hexsha": "41f6a03e4a082768c10b9a77a0e042cfce2a6d7f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pre_calculus_h/chapters/3_exponential_logarithmic_functions.tex", "max_forks_repo_name": "ttaylorr/finals", "max_forks_repo_head_hexsha": "41f6a03e4a082768c10b9a77a0e042cfce2a6d7f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.1744186047, "max_line_length": 80, "alphanum_fraction": 0.7567199728, "num_tokens": 820, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9019206844384594, "lm_q2_score": 0.7745833841649233, "lm_q1q2_score": 0.6986127760006858}} {"text": "% !TEX root = main.tex\n\n%-------------------------------------------------\n\\section{Transformations}\\label{sec:transfs}\n\nRandom variables can be transformed into other random variables, and the distribution of a transformed variable can be deduced from the distribution of the original variable. Transformations of discrete distributions are relatively straightforward: here we focus on transformations of continuous distributions.\n\nApplying a transformation $g:\\R\\to\\R$ to a random variable $X:\\Omega\\to\\R$ involves the \\emph{composition} of these two functions,\n\\[\n\\begin{array}{rlcl}\ng(X) : \t& \\Omega & \\to \t\t& \\R \\\\\n\t\t& \\omega & \\mapsto\t& g\\big[X(\\omega)\\big],\n\\end{array}\n\\]\n\nThis can be interpreted in two ways:\n\\ben \n\\it $g(X)$ is a random variable on the probability space $(\\Omega,\\prob)$;\n\\it $g$ is a random variable on the probability space $(\\R,\\prob_X)$,\n\\een\nwhere $\\prob_X$ is the distribution of $X$. Here we focus on the first interpretation with $Y=g(X)$ denoting the transformed variable.\n\n%-----------------------------\n\\subsection{Support}\nMany PDFs are non-zero only over certain subsets of $\\R$. When we transform continuous distributions we need only consider these subsets, provided we ensure that those over which PDFs are zero are carried over correctly into the transformed space. For technical reasons, it is not quite enough to focus on the \\emph{range} of the random variable in question: instead we must consider the smallest closed set that contains the range (a closed set is one that contains all its limit points). This is called the \\emph{support} of the associated PDF.\n\n\\begin{definition}\nThe \\emph{support} of an arbitrary function $h:\\R\\to\\R$, denoted by $\\supp(h)$, is the smallest closed set for which $h(x)=0$ for all $x\\notin\\supp(h)$. \n\\end{definition}\n\nLet $Y = g(X)$ where $g:\\R\\to\\R$ is a transformation, and let $f_X$ and $f_Y$ denote the PMFs/PDFs of $X$ and $Y$ respectively. The support of the transformed variable $Y = g(X)$ is given by\n\\[\n\\supp(f_Y) = \\big\\{g(x) : x\\in\\supp(f_X)\\big\\}.\n\\]\nIn fact, $\\supp(f_Y)$ should be defined as the \\emph{closure} of this set, which is obtained by adding in its limit points where necessary. We will not pursue such matters here.\n\n%-----------------------------\n\\subsection{Linear transformations}\n\nLet $X$ be a random variable and let $Y = aX + b$ where $a\\neq 0$. \n\nThe CDF of of the transformed variable $Y$ can be expressed in terms of the CDF of $X$ as follows:\n\\[\nF_Y(y) \n\t= \\prob(Y\\leq y) \n\t= \\prob(aX+b\\leq y)\n\t= \\begin{cases}\n\t\t\\displaystyle\\prob\\left(X\\leq\\frac{y-b}{a}\\right) = \\phantom{1 -\\ } F_X\\left(\\frac{y-b}{a}\\right)\t\t& \\text{if $a>0$,} \\\\[3ex]\n\t\t\\displaystyle\\prob\\left(X>\\frac{y-b}{a}\\right)\t = 1 - F_X\\left(\\frac{y-b}{a}\\right)\t& \\text{if $a<0$.}\n\t\\end{cases}\n\\]\nUsing the chain rule, the PDF of $Y$ can then be expressed in terms of the PDF of $X$:\n\\smallskip\n\\[\nf_Y(y) = \\frac{d}{dy}F_Y(y) = \n\t\\left\\{\n\t\\begin{array}{ll}\n\t\t\\phantom{-}\\displaystyle\\frac{1}{a}f_X\\left(\\frac{y-b}{a}\\right)\t& \\text{if $a>0$,} \\\\[3ex]\n\t\t-\\displaystyle\\frac{1}{a}f_X\\left(\\frac{y-b}{a}\\right)\t\t\t\t& \\text{if $a<0$.}\n\t\\end{array}\n\t\\right\\}\n\t= \\frac{1}{|a|}f_X\\left(\\frac{y-b}{a}\\right).\n\\]\n% example: linear transf.\n\\begin{example}\nLet $X\\sim\\text{Uniform}[0,1]$. Find the distribution of the random variable $Y = 3X + 7$.\n\\begin{solution}\nLet $g(x) = 3x+7$ denote the transformation. The PDF of $X\\sim\\text{Uniform}[0,1]$ is\n\\[\nf_X(x) = \\begin{cases}\n\t1\t& 0\\leq x\\leq 1 \\\\\n\t0\t& \\text{otherwise.}\n\\end{cases}\n\\]\nFirst we see that $\\supp(f_X)$ is transformed as follows:\n\\[\n\\supp(f_Y) = \\{g(x) : x\\in\\supp(f_X)\\} = \\{3x+7 : x\\in [0,1]\\} = [7,10].\n\\]\nFrom the above discussion,\n\\[\nf_Y(y) = \\displaystyle\\frac{1}{3}f_X\\left(\\frac{y-7}{3}\\right) \n\t= \\begin{cases}\n\t\t1/3\t& \\text{if $7\\leq y\\leq 10$,} \\\\\n\t\t0\t& \\text{otherwise,}\n\t\\end{cases}\n\\]\nso $Y\\sim\\text{Uniform}[7,10]$. We see that the original distribution has been scaled by a factor of $3$ and shifted $7$ units to the right.\n\\end{solution}\n\\end{example}\n\nThese ideas can be extended to any one-to-one transformation of $X$.\n\n%-----------------------------\n\\subsection{Transformations of CDFs}\n\n\\begin{theorem}\\label{thm:transf_cdf}\nIf $g:\\R\\to\\R$ is one-to-one over $\\supp(f_X)$ the CDF of $Y=g(X)$ is\n\\[\nF_Y(y) = \\begin{cases}\nF_X\\big[g^{-1}(y)\\big]\t\t& \\text{if $g$ is increasing, and} \\\\[1ex]\n1 - F_X\\big[g^{-1}(y)\\big]\t& \\text{if $g$ is decreasing.} \n\\end{cases}\n\\]\n\\end{theorem}\n\n\\begin{proof}\n\\ben\n\\it If $g$ is increasing, $g(x)\\leq y$ implies that $x\\leq g^{-1}(y)$ so\n\\[\nF_Y(y) \n\t= \\prob(Y\\leq y) \n\t= \\prob\\big[g(X)\\leq y\\big] \n\t= \\prob\\big[X\\leq g^{-1}(y)\\big]\n\t= F_X\\big[g^{-1}(y)\\big].\n\\]\n\\it If $g$ is decreasing, $g(x)\\leq y$ implies that $x\\geq g^{-1}(y)$ so\n\\begin{align*}\nF_Y(y) \n\t= \\prob(Y\\leq y) \n\t= \\prob\\big[g(X)\\leq y\\big] \n\t& = \\prob\\big[X\\geq g^{-1}(y)\\big] \\\\\n\t& = 1 - \\prob\\big[X\\leq g^{-1}(y)\\big] \\quad\\text{(because $X$ is a continuous r.v.)}\\\\\n\t& = 1 - F_X\\big[g^{-1}(y)\\big].\n\\end{align*}\n\\een\n\\end{proof}\t\n\n\\begin{example}\nLet $X\\sim\\text{Uniform}[0,1]$ and let $Y = -\\displaystyle\\frac{1}{\\lambda}\\log X$ where $\\lambda>0$. Show that $Y\\sim\\text{Exponential}(\\lambda)$ where $\\lambda$ is a rate parameter.\n\\begin{solution}\nThe CDF and PDF of $X\\sim\\text{Uniform}[0,1]$ are, respectively,\n\\[\\begin{array}{lcl}\nF_X(x) = \\begin{cases}\n\t0\t& x < 0, \\\\ \n\tx\t& 0\\leq x\\leq 1 \\\\\n\t1\t& x > 1\n\\end{cases}\n& \\text{and} &\nf_X(x) = \\begin{cases}\n\t1\t& 0\\leq x\\leq 1 \\\\\n\t0\t& \\text{otherwise.}\n\\end{cases}\n\\end{array}\\]\nLet $g(x) = -\\log x/\\lambda$ denote the transformation. Because $\\lambda>0$ we see that $\\supp(f_X)$ is transformed to\n\\[\n\\supp(f_Y) = \\{g(x) : x\\in\\supp(f_X)\\} =\\{-\\log/\\lambda : x\\in[0,1]\\} = [0,\\infty).\n\\]\nThe transformation is strictly decreasing over $\\supp(f_X)$, and its inverse is $g^{-1}(y) = e^{-\\lambda y}$ over $\\supp(f_Y)$. Hence, by Theorem~\\ref{thm:transf_cdf},\n\\[\nF_Y(y) \n\t= 1 - F_X\\big[g^{-1}(y)\\big]\n\t= 1 - F_X(e^{-\\lambda y})\n\t= 1 - e^{-\\lambda y} \\quad\\text{for}\\quad y\\geq 0.\n\\]\nThis is the CDF of the $\\text{Exponential}(\\lambda)$ distribution (where $\\lambda$ is a rate parameter).\n\\end{solution}\n\\end{example}\n\n%-----------------------------\n\\subsection{Transformations of PDFs}\n\n\\begin{theorem}\\label{thm:transf_pdfs}\nIf $g:\\R\\to\\R$ is one-to-one over $\\supp(f_X)$ the PDF of $Y=g(X)$ is\n\\[\nf_Y(y) = f_X\\big[g^{-1}(y)\\big]\\left|\\frac{d}{dy}g^{-1}(y)\\right| %\\quad\\text{for all}\\quad y\\in \\supp(f_Y).\n\\]\n\\end{theorem}\n\n\\begin{proof}\nFor clarity of notation, let $h(y)$ denote the inverse function $g^{-1}(y)$.\n\\ben\n\\it If $g$ is increasing then $F_Y(y) = F_X\\big[g^{-1}(y)\\big]$, so by the chain rule (and using the fact that $h$ is also increasing),\n\\begin{align*}\nf_Y(y)\n\t= \\frac{d}{dy} F_Y(y)\n\t& = \\frac{d}{dy} F_X\\big[h(y)\\big] \\\\\n\t& = \\frac{d}{dh(y)} F_X\\big[h(y)\\big]\\cdot \\frac{dh(y)}{dy} \\\\\n\t& = f_X\\big[h(y)\\big]\\left|\\frac{dh(y)}{dy}\\right|, \\text{\\qquad\\qquad because } \\frac{dh(y)}{dy}>0\\text{ over $\\supp(f_Y)$.}\n\\end{align*}\n\n\\it If $g$ decreasing, $F_Y(y) = 1 - F_X\\big[g^{-1}(y)\\big]$, so by the chain rule (and using the fact that $h$ is also decreasing),\n\\begin{align*}\nf_Y(y)\n\t= \\frac{d}{dy} F_Y(y)\n\t& = \\frac{d}{dy} \\big[1 - F_X[h(y)]\\big] \\\\\n\t& = 0 - \\frac{d}{dh(y)} F_X\\big[h(y)\\big]\\cdot \\frac{dh(y)}{dy} \\\\\n\t& = - f_X\\big[h(y)\\big] \\frac{dh(y)}{dy} \\\\\n\t& = f_X\\big[h(y)\\big]\\left|\\frac{dh(y)}{dy}\\right|, \\text{\\qquad\\qquad because} \\frac{dh(y)}{dy}<0\\text{ over $\\supp(f_Y)$.}\n\\end{align*}\n\\een\n\\end{proof}\n\n% remark\n\\begin{remark}\nThe term $\\left|\\frac{d}{dy}g^{-1}(y)\\right|$ in Theorem~\\ref{thm:transf_pdfs} is a \\emph{scale factor}, which ensures that $f_Y$ integrates to one.\n\\end{remark}\n\n\\begin{example}\nLet $X$ be a continuous random variable with the following PDF,\n\\[\nf_X(x) = \\begin{cases}\n\t1/x^2\t& \\text{for $x > 1$,} \\\\\n\t0\t\t& \\text{otherwise.}\n\\end{cases}\n\\]\nFind the PDF of $Y=1/X$.\n\\end{example}\n\n\\begin{solution}\nLet $g(x) = 1/x$. Since $\\supp(f_X)=[x,\\infty)$, the support of $f_Y$ is \n\\[\n\\supp(f_Y) = \\{g(x): x\\in\\supp(f_X)\\} = \\{1/x :x\\in[0,\\infty)\\} = [0,1].\n\\]\nThe transformation is one-to-one over $\\supp(f_X)$, and its inverse is $g^{-1}(y) = 1/y$ over $\\supp(f_Y)$. Hence, by Theorem~\\ref{thm:transf_pdfs}, the PDF of $Y$ is given by\n\\begin{align*}\nf_Y(y)\n\t& = f_X\\big[g^{-1}(y)\\big]\\left|\\frac{d}{dy}g^{-1}(y)\\right| \\\\\n\t& = f_X\\left(\\frac{1}{y}\\right)\\left|\\frac{d}{dy}\\left(\\frac{1}{y}\\right)\\right| \\\\\n\t& = y^2 \\left|-\\frac{1}{y^2}\\right| \n\t= \\begin{cases}\n\t\t1\t& \\text{for } 01)=0$. For $u\\in[0,1]$, because the inverse $F^{-1}$ exists for all $x\\in\\R$ we have that\n\\begin{align*}\nF_U(u) = P(U\\leq u) \n\t& = P\\big(F(X)\\leq u\\big) \\\\\n\t& = P\\big(X\\leq F^{-1}(u)\\big) \\\\\n\t& = F\\big(F^{-1}(u)\\big) \\\\\n\t& = u,\n\\end{align*}\nwhich is the CDF of the continuous uniform distribution on $[0,1]$.\n\\end{proof}\n\n% corollary\n\\begin{corollary}\nLet $F$ be a CDF whose inverse exists for all $x\\in\\R$, and let $U\\sim\\text{Uniform}[0,1]$. Then $F$ is the CDF of the random variable $X = F^{-1}(U)$.\n\\end{corollary}\n\nAlthough it is difficult to generate truly random numbers, there are fast deterministic algorithms which generate numbers that are approximately random - such numbers are called \\emph{pseudo-random numbers}. Many such algorithms can generate numbers that are approximately uniformly distributed in $[0,1]$. Using the probability integral transform we can convert these into pseudo-random numbers from other continuous distributions.\n\n\\ben\n\\it First we obtain a uniformly distributed pseudo-random number $u\\in [0,1]$.\n\\it The number $x = F^{-1}(u)$ is then a pseudo-random number from the distribution $F$.\n\\een\n\n% example: \n\\begin{example}\nGiven an algorithm which generates uniformly distributed pseudo-random numbers in the range $[0,1]$, show how to obtain a pseudo-random number from the exponential distribution having rate parameter $2$.\n\\begin{solution}\nThe CDF of the exponential distribution with rate parameter $2$ is\n\\[\nF(x) = \\begin{cases}\n\t1 - e^{-2x}\t& x>0 \\\\\n\t0\t\t\t& \\text{otherwise.}\n\\end{cases}\t\n\\]\nFirst we invert $F$:\n\\begin{align*}\nu = 1 - e^{-2x} \n\t& \\iff e^{-2x} = 1-u \\\\\n\t& \\iff e^x = \\frac{1}{\\sqrt{1-u}} \\\\\n\t& \\iff x = \\log\\left(\\frac{1}{\\sqrt{1-u}}\\right)\n\\end{align*}\nThen we generate a pseudo-random number $u$ from the $\\text{Uniform}[0,1]$ distribution and\n\\[\nx = \\log\\left(\\frac{1}{\\sqrt{1-u}}\\right)\n\\]\nwhich is a pseudo-random number from the $\\text{Exponential}(\\lambda)$ distribution.\n\\end{solution}\n\\end{example}\n\n\n\\begin{exercise}\n\\begin{questions}\n\n\\question % transf\nLet $X\\sim\\text{Uniform}(-1,1)$. Find the CDF and PDF of $X^2$.\n\n\\begin{answer}\nThe PDF of $X$ is \n\\[\nf_X(x) = \\left\\{\\begin{array}{ll}\n\t1/2\t& -1\\leq x\\leq 1 \\\\\n\t0\t& \\text{otherwise}\n\\end{array}\\right.\t\n\\]\t\nFor $x\\in[-1,1]$, \n\\[\n\\prob(X\\leq x) \n\t= \\int_{-\\infty}^x f_X(t)\\,dt\n\t= \\int_{-1}^x \\frac{1}{2}\\,dt\n\t= \\left[\\frac{t}{2}\\right]_{-1}^x\n\t= \\frac{1}{2}(x+1).\n\\]\nThe CDF of $X$ is:\n\\[\nF(x) = \\left\\{\\begin{array}{ll}\n\t0\t\t\t\t \t& x < -1, \\\\\n\t\\frac{1}{2}(x+1) \t& -1\\leq x\\leq 1, \\\\\n\t1\t\t\t\t\t& x > 1.\n\\end{array}\\right.\t\n\\]\nLet $Y=X^2$. For $0\\leq y\\leq 1$ we have\n\\begin{align*}\n\\prob(Y\\leq y)\n\t= \\prob(X^2\\leq y)\n\t& = \\prob(-\\sqrt{y}\\leq X\\leq\\sqrt{y}) \\\\\n\t& = \\prob(X\\leq\\sqrt{y}) - \\prob(X\\leq-\\sqrt{y}) \\\\\n\t& = \\sqrt{y}.\n\\end{align*}\nHence the CDF of $Y$ is\n\\[\nF_Y(y) = \\left\\{\\begin{array}{ll}\n\t0\t\t \t& y < 0, \\\\\n\t\\sqrt{y} \t& 0\\leq y\\leq 1, \\\\\n\t1\t\t\t& y > 1.\n\\end{array}\\right.\t\n\\]\nand the PDF of $Y$ is\n\\[\nf_Y(y) = \\left\\{\\begin{array}{ll}\n\\frac{1}{2}y^{-1/2}\t& 0\\leq y\\leq 1, \\\\\n0\t\t\t\t\t& \\text{otherwise}.\n\\end{array}\\right.\t\n\\]\n\\end{answer}\n\n\\question % transf\nSuppose that $X$ has the exponential distribution with rate parameter $\\lambda>0$. (The PDF of $X$ is $f(x) = \\lambda\\exp(-\\lambda x)$ for $x \\geq 0$ and zero otherwise.)\n%\\[\n%f(x) = \\left\\{\\begin{array}{ll}\n%\t\\lambda\\exp(-\\lambda x)\t& \\text{for } x \\geq 0, \\\\\n%\t0\t\t\t\t\t\t& \\text{otherwise.}\n%\\end{array}\\right.\n%\\]\nFind the PDFs of $Y=X^2$ and $Z=e^X$.\n\n\\begin{answer}\n\\ben\n\\it % << (i)\nThe transformation $g(x)=x^2$ is one-to-one and increasing over $[0,\\infty)$; its inverse function is\n\\[\ng^{-1}(y) = \\sqrt{y},\\text{\\quad which has first derivative \\quad} \\frac{d}{dy}g^{-1}(y) = \\frac{1}{2\\sqrt{y}}.\n\\]\n\nSince $\\supp(f_X)=[0,\\infty)$ it follows immediately that $\\supp(f_Y)=[0,\\infty)$.\n\\par\nFor $y>0$, \n\\[\nf_Y(y) \n\t= f_X\\big[g^{-1}(y)\\big]\\left|\\frac{d}{dy}g^{-1}(y)\\right| \n\t= \\lambda\\exp(\\lambda\\sqrt{y}) \\left| \\frac{1}{2\\sqrt{y}}\\right|\n\t= \\frac{\\lambda}{2\\sqrt{y}}\\exp(-\\lambda\\sqrt{y}).\n\\]\nHence the PDF of $Y=X^2$ is given by\n\\[\nf_Y(y) = \\left\\{\\begin{array}{ll}\n\t\\displaystyle\\frac{\\lambda}{2\\sqrt{y}}\\exp(-\\lambda\\sqrt{y}) & y\\geq 0, \\\\[2ex]\n\t0 & \\text{otherwise}.\n\\end{array}\\right.\n\\]\n\n\\it % << (ii)\nThe transformation $g(x)=e^x$ is one-to-one and increasing over $[0,\\infty)$; its inverse function is\n\\[\ng^{-1}(z) = \\log y \\text{\\quad and\\quad} \\frac{d}{dy}g^{-1}(z) = \\frac{1}{z}.\n\\]\n\nSince $\\supp(f_X)=[0,\\infty)$ it follows immediately that $\\supp(f_Z)=[1,\\infty)$.\n\\par\nFor $z\\geq 1$,\n\\[\nf_Z(z) \n\t= f_X\\big[g^{-1}(z)\\big]\\left|\\frac{d}{dz}g^{-1}(z)\\right|\n\t= \\lambda\\exp(-\\lambda\\log z)\\left|\\frac{1}{z}\\right| \n\t= \\lambda z^{-(\\lambda+1)}.\n\\]\nHence the PDF of $Z=e^X$ is given by\n\\[\nf_Z(z) = \\left\\{\\begin{array}{ll}\n\t\\lambda z^{-(\\lambda+1)} & z\\geq 1, \\\\\n\t0 & \\text{otherwise}.\n\\end{array}\\right.\n\\]\n\\een\n\\end{answer}\n\n\\question % transf\nA continuous random variable $U$ has PDF\n$f(u) = 12u^{2}(1-u)$ for $0 < u < 1$ and zero otherwise.\n%\\[\n%f(u) = \\left\\{\\begin{array}{ll}\n%\t12u^{2}(1-u) \t& \\text{for}\\quad 0 < u < 1, \\\\\n%\t0\t\t\t\t& \\text{otherwise.}\n%\\end{array}\\right.\t\n%\\]\nFind the PDF of $V = (1 - U)^{2}$.\n\\begin{answer}\n\\bit\n\\it The transformation $g(u) = (1 - u)^{2}$ is one-to-one and decreasing over $[0,1]$. \n\\it The inverse transformation is $g^{-1}(v) = 1 - v^{1/2}$, for which $\\displaystyle \\frac{d}{dv}g^{-1}(v) = -\\frac{1}{2v^{1/2}}$.\n\\it Since $\\supp(f_U)=(0,1)$ it follows that $\\supp(f_V)=(0,1)$. \n\\eit\nHence for $0 1.\n\t\\end{array}\\right.\n\\]\nFor the second part we use the fact that $F_Y(Y)\\sim\\text{Uniform}(0,1)$. First we invert $F_Y$ by letting $u=F_Y(y)$ from which we obtain.\n\\[\ny = F_Y^{-1}(u) = u^{1/3}.\n\\]\nNext we obtain a pseudo-random number $u$ from the $\\text{Uniform}[0,1]$ distribution, then compute\n\\[\ny = u^{1/3},\n\\]\nwhich is a pseudo-random number from the distribution of $Y$.\n\\end{answer}\n\n\\end{questions}\n\\end{exercise}\n", "meta": {"hexsha": "4e4284f29cd9a549a82071c2160c38b5f035d8cb", "size": 16119, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/03C_transformations.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/03C_transformations.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/03C_transformations.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 34.2957446809, "max_line_length": 546, "alphanum_fraction": 0.6211923817, "num_tokens": 6169, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.9019206679615432, "lm_q1q2_score": 0.6986127632379403}} {"text": "\\documentclass[letterpaper]{article}\n\n\\usepackage[utf8x]{luainputenc}\n\\usepackage{aeguill}\n%\\usepackage{nopageno}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{mathrsfs}\n\\usepackage{fullpage}\n\\usepackage{fancyhdr}\n\\setlength{\\headheight}{12pt}\n\\pagestyle{fancy}\n\\chead{Linear Algebra}\n\\lhead{September 23, 2015}\n\\rhead{Jon Allen}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\n%\\renewcommand{\\labelenumii}{\\alph{enumii}.}\n%\\renewcommand{\\labelenumiii}{\\alph{enumiii}.}\n%\\renewcommand{\\labelenumi}{(\\arabic{enumi})}\n\\section*{Excercise Set 1}\n\\begin{enumerate}\n\\item\nProve the matrix\n\\[A=\\left[\\begin{array}{cc}a&b\\\\c&d\\end{array}\\right]\\]\nis invertible if and only if $ad-bc\\ne 0$. Find $A^{-1}$ in this case by solving a system of two equations with two unknowns.\n\nWe need some matrix $A^{-1}$ such that $AA^{-1}=I_2$.\nAssume $A^{-1}$ exists, then let $A^{-1}=\\left[\\begin{array}{cc}a'&b'\\\\c'&d'\\end{array}\\right]$ and we have\n\\begin{align*}\n AA^{-1}&= \\left[\\begin{array}{cc}a&b\\\\c&d\\end{array}\\right]\n \\left[\\begin{array}{cc}a'&b'\\\\c'&d'\\end{array}\\right]\\\\\n &=\\left[\\begin{array}{cc}aa'+bc'&ab'+bd'\\\\ca'+dc'&cb'+dd'\\end{array}\\right]\n =\\left[\\begin{array}{cc}1&0\\\\0&1\\end{array}\\right]\\\\\n\\end{align*}\nIn particular\n\\begin{align*}\n aa'+bc'&=1&\n ca'+dc'&=0\\\\\n\\end{align*}\nNow if $c=0$ then $d=0$ or $c'=0$. We know that $d\\ne 0$ because $cb'+dd'=1$. And so if $c=0$ then $c'=0$. But then $aa'+bc'=1$ so $a\\ne 0$ in this case. We will proceed in two cases then for $c\\ne 0$ and $a\\ne 0$.\n\nIf we assume that $c\\ne 0$ then we have\n\\begin{align*}\n ca'+dc'&=0&\n a'+\\frac{d}{c}c'&=0&\n -aa'-\\frac{ad}{c}c'&=0\\\\\n\\end{align*}\nNow if we add $aa'+bc'=1$ to the above result we get $\\left(b-\\frac{ad}{c}\\right)c'=1$. Solving for $c'$ gives us $c'=-\\frac{c}{ad-bc}$. Using we can solve $ca'+dc'=0$ to find $a'=\\frac{d}{ad-bc}$. Moving on we have\n\\begin{align*}\n cb'+dd'&=1&b'+\\frac{d}{c}d'&=\\frac{1}{c}&-ab'-\\frac{ad}{c}d'&=-\\frac{a}{c}\n\\end{align*}\nAdding in $ab'+bd'=0$ leads us to $\\frac{bc-ad}{c}d'=\\frac{-a}{c}$, or $d'=\\frac{a}{ad-bc}$. And finally substituting $d'$ in for $cb'+dd'=1$ leads to $cb'+d\\frac{a}{ad-bc}=cb'+1+\\frac{bc}{ad-bc}=1$ or $b'=-\\frac{b}{ad-bc}$.\n\nThe case when $a\\ne 0$ is similar.\n\\begin{align*}\n ab'+bd'&=0&cb'+dd'&=1\\\\\n -cb'-\\frac{cb}{a}d'&=0&cb'+dd'&=1\\\\\n \\left(d-\\frac{cb}{a}\\right)d'&=1&d'&=\\frac{a}{ad-bc}\\\\\n\\end{align*}\nIf we continue as we did when $c\\ne 0$ then we will arrive at $A^{-1}=\\frac{1}{ad-bc}\\left[\\begin{array}{cc}d&-b\\\\-c&a\\end{array}\\right]$ just as we did when $c\\ne 0$.\n\nIn this way we see that when we assume that $A^{-1}$ exists and that $ad-bc=0$ we arrive at a contradiction. Namely that $\\frac{1}{ad-bc}$ is undefined and therefore $A^{-1}$ can't exist. However, if we assume that $ad-bc\\ne 0$ we find our inverse, and so it must exist. Thus we have proven both directions of the hypothesis.\n$\\Box$\n\\item\nLet\n\\[A=\\left[\\begin{array}{ccc}0&0&1\\\\1&0&2\\\\0&1&-3\\end{array}\\right]\\]\n \\begin{enumerate}\n \\item\n Show that $A^3+3A^2-2A-I_3=\\mathbf{0}$\n \\begin{align*}\n A^3+3A^2-2A-I_3&=0\\\\\n (A^2+3A-2I_3)A&=I_3\\\\\n \\left( \\left[\\begin{array}{ccc}0&0&1\\\\1&0&2\\\\0&1&-3\\end{array}\\right]^2\n +3\\left[\\begin{array}{ccc}0&0&1\\\\1&0&2\\\\0&1&-3\\end{array}\\right]\n +\\left[\\begin{array}{ccc}-2&0&0\\\\0&-2&0\\\\0&0&-2\\end{array}\\right]\n \\right)A&=I_3\\\\\n \\left( \\left[\\begin{array}{ccc}0&1&-3\\\\0&2&-5\\\\1&-3&11\\end{array}\\right]\n +\\left[\\begin{array}{ccc}0&0&3\\\\3&0&6\\\\0&3&-9\\end{array}\\right]\n +\\left[\\begin{array}{ccc}-2&0&0\\\\0&-2&0\\\\0&0&-2\\end{array}\\right]\n \\right)A&=I_3\\\\\n \\left[\\begin{array}{ccc}-2&1&0\\\\3&0&1\\\\1&0&0\\end{array}\\right]\n \\left[\\begin{array}{ccc}0&0&1\\\\1&0&2\\\\0&1&-3\\end{array}\\right]&=I_3\\\\\n \\left[\\begin{array}{ccc}1&0&0\\\\0&1&0\\\\0&0&1\\end{array}\\right]&=I_3\n \\end{align*}\n \\item\n Use part (a) to see that $A$ is invertible and compute $A^{-1}$\n \n Obviously $A^{-1}=A^2+3A-2I_3=\\left[\\begin{array}{ccc}-2&1&0\\\\3&0&1\\\\1&0&0\\end{array}\\right]$\n \\end{enumerate}\n\\item\nLet $A\\in \\mathcal{M}_n$ be a diagonal matrix. Prove that $A$ is invertible if and only if $\\text{ent}_{ii}(A)\\ne0$ for all $i\\le n$. Find $A^{-1}$ in this case.\n\nLet us assume that there exists some $\\text{ent}_{ii}(A)=0$ and that $A^{-1}$ exists with $\\text{ent}_{jk}(A^{-1})=a_{jk}^{-1}$ and $\\text{ent}_{jk}(A)=a_{jk}$. Then $\\text{ent}_{ii}(AA^{-1})=\\sum\\limits_{k=1}^n{a_{ik}a_{ki}^{-1}}=\\sum\\limits_{k=1}^n{0a_{ki}^{-1}}=0$. But $AA^{-1}=I_n$ and $\\text{ent}_{ii}(I_n)=1$ and so we have a contradiction.\n\nNow lets assume that $\\text{ent}_{ii}(A)=c_i$ with $i\\le n$ and $c_i\\ne 0$. Lets choose another diagonal matrix and name it $A^{-1}$. We will assign $\\text{ent}_{ii}(A^{-1})=c_{i}^{-1}$. Now we see that if we choose any $i\\le n,j\\le n$ such that $j\\ne i$ then we see that $\\text{ent}_{ij}(AA^{-1})=\\sum\\limits_{k=1}^n{a_{ik}a_{kj}^{-1}}=0+\\dots+a_{ii}a_{ij}^{-1}+a_{ij}a_{jj}^{-1}+\\dots0=0+\\dots+a_{ii}0+0a_{jj}^{-1}+\\dots0=0$.\nBut $\\text{ent}_{ii}(AA^{-1})=\\sum\\limits_{k=1}^n{a_{ik}a_{ki}^{-1}}=0+\\dots+a_{ii}a_{ii}^{-1}+\\dots+0=cc^{-1}=1$. And so we see that $AA^{-1}$ is a diagonal matrix with all ones for its entries. But that is $I_n$ and so we have found the inverse of $A^{-1}$.\n$\\Box$\n\\item\nIf $A,B\\in \\mathcal{M_n}$ are invertible such that $A+B\\ne 0$, does it follow that $(A+B)^{-1}$ exists? Prove or find a counterexample.\n\n\\[\n\\left[\\begin{array}{cc}3&2\\\\8&4\\end{array}\\right]+\n\\left[\\begin{array}{cc}3&2\\\\4&4\\end{array}\\right]\n=\\left[\\begin{array}{cc}6&4\\\\12&16\\end{array}\\right]\n\\]\n\nNotice that $3\\cdot 4-8\\cdot 2=-4$ and $3\\cdot 4-2\\cdot 4=4$ but $6\\cdot8-4\\cdot 12=6\\cdot4\\cdot2-4\\cdot2\\cdot6=0$ and so by excercise 1 we have found a counterexample.\n\\end{enumerate}\n\\section*{Excercise Set 2}\n\n\\[A=\n\\left[\\begin{array}{ccccc}\n a_{11}&a_{12}&a_{13}&a_{14}&a_{15}\\\\\n a_{21}&a_{22}&a_{23}&a_{24}&a_{25}\\\\\n a_{31}&a_{32}&a_{33}&a_{34}&a_{35}\\\\\n a_{41}&a_{42}&a_{43}&a_{44}&a_{45}\n\\end{array}\\right]\\]\n\n\\begin{enumerate}\n\\item\nFor the matrix $A$ in the example above, determine $E_{3\\to c3}$ where $c$ is any nonzero real number.\n\n\\[E_{3\\to c3}=\\left[\\begin{array}{cccc}1&0&0&0\\\\0&1&0&0\\\\0&0&c&0\\\\0&0&0&1\\\\\\end{array}\\right]\\]\n\n\\item\nFor the matrix $A$ in the example above, determine $E_{4\\to4+c2}$ where $c$ is any nonzero real number.\n\n\\[E_{4\\to 4+c2}=\\left[\\begin{array}{cccc}1&0&0&0\\\\0&1&0&0\\\\0&0&1&0\\\\0&c&0&1\\\\\\end{array}\\right]\\]\n\n\\item\nProve that each of $E_{i\\leftrightarrow k},E_{i\\to ci}, E_{i\\to i+ck}$ is invertible by finding an inverse. Prove that the inverse of an elementary matrix is an elementary matrix. \n\nWe assume that $i\\ne k$.\nWe posit that $E_{i\\leftrightarrow k}$ is the identity matrix, save that $\\text{ent}_{ii}E_{i\\leftrightarrow k}=\\text{ent}_{kk}E_{i\\leftrightarrow k}=0$ and $\\text{ent}_{ki}E_{i\\leftrightarrow k}=\\text{ent}_{ik}E_{i\\leftrightarrow k}=0$\n\nWe shall now verify this claim.\nFirst we choose some $j\\not\\in\\{i,k\\}$ and any $l$.\nNow we see that for any $A\\in \\mathcal{M}_n$ we have $\\text{ent}_{jl}(E_{i\\leftrightarrow k}A)=\\sum\\limits_{m=1}^n{\\text{ent}_{jm}(E_{i\\leftrightarrow k})a_{ml}}=a_{jl}$ because all entries in the elementary matrix row are zero except the $j$ element which is one.\nAnd so as we wish $E_{i\\leftrightarrow k}$ leaves everything off of rows $i$ and $k$ undisturbed.\nNow lets see what happens for $\\text{ent}_{il}(E_{i\\leftrightarrow k}A)$ and $\\text{ent}_{kl}(E_{i\\leftrightarrow k}A)$.\n\\begin{align*}\n \\text{ent}_{il}(E_{i\\leftrightarrow k}A)\n &=\\sum\\limits_{m=1}^n{\\text{ent}_{im}(E_{i\\leftrightarrow k})a_{ml}}=a_{kl}\\\\\n \\text{ent}_{kl}(E_{i\\leftrightarrow k}A)\n &=\\sum\\limits_{m=1}^n{\\text{ent}_{km}(E_{i\\leftrightarrow k})a_{ml}}=a_{il}\n\\end{align*}\nThus we see that the $i$ and $k$ rows of $A$ and its product with the elementary matrices have been switched. This occurs because $\\text{ent}_{ki}(E_{i\\leftrightarrow k})=\\text{ent}_{ik}(E_{i\\leftrightarrow k})=1$ but all other entries on these two rows is zero.\n\nNow that we know what $E_{i\\leftrightarrow k}$ is, we see that for\n$i\\ne l$ we have $\\text{ent}_{il}(E_{i\\leftrightarrow k}E_{i\\leftrightarrow k})=\\text{ent}_{kl}(E_{i\\leftrightarrow k})=0$.\nAnd we have $\\text{ent}_{ii}(E_{i\\leftrightarrow k}E_{i\\leftrightarrow k})=\\text{ent}_{ki}(E_{i\\leftrightarrow k})=1$.\nSimilarly \nwe have $\\text{ent}_{kl}(E_{i\\leftrightarrow k}E_{i\\leftrightarrow k})=\\text{ent}_{il}(E_{i\\leftrightarrow k})=0$ when $k\\ne l$ and\n$\\text{ent}_{kk}(E_{i\\leftrightarrow k}E_{i\\leftrightarrow k})=\\text{ent}_{ik}(E_{i\\leftrightarrow k})=1$ when $k=l$. Thus $E_{i\\leftrightarrow k}^{2}=I_n$ and so this matrix is it's own inverse.\n\nFor the matrix $E_{i\\to ci}$ we must have $c\\ne 0$.\nNow if we take the identity matrix, but change $\\text{ent}_{ii}$ to be $c$ then I claim we have our elementary matrix.\nAs above, when $j\\ne i$ we have $\\text{ent}_{jk}(E_{i\\to ci}A)=a_{jk}$ for any $A\\in \\mathcal{M}_n$ and $k\\le n$.\nAnd so our matrix leaves all rows but $i$ untouched. Now if we examine row $i$ we see that $\\text{ent}_{ik}(E_{i\\to ci}A)=\\sum\\limits_{l=1}^n{\\text{ent}_{il}(E_{i\\to ci})a_{li}}=ca_{ik}$ as desired. So we have found our elementary matrix.\n\nNow let us examine $E_{i\\leftrightarrow (1/c)i}E_{i\\leftrightarrow ci}$. We have already seen that $\\text{ent}_{jk}(E_{i\\leftrightarrow (1/c)i}E_{i\\to ci})=\\text{ent}_{jk}(E_{i\\to ci})=\\text{ent}_{jk}(I_n)$. And $\\text{ent}_{ik}(E_{i\\leftrightarrow (1/c)i}E_{i\\to ci})=\\frac{1}{c}\\text{ent}_{ik}(E_{i\\to ci})$. Of course when $k\\ne i$ then this is zero and when $k=i$ then this entry will be $\\frac{1}{c}c=1$. Thus we have exactly the identity matrix. And so we have found our inverse.\n\nFinally we look at $E_{i\\to i+ck}$. As before we start with the identity matrix. This time the only entry on $I_n$ that we change is $\\text{ent}_{ik}(E_{i\\to i+ck})=c$.\n\nObviously if we have $j\\ne i, l\\le n$ and any $A\\in\\mathcal{M}_n$ then $\\text{ent}_{jl}(E_{i\\to i+ck}A)=\\sum\\limits_{m=1}^n{\\text{ent}_{jm}(E_{i\\to i+ck})a_{ml}}=a_{jl}$.\nBut $\\text{ent}_{il}(E_{i\\to i+ck}A)=\\sum\\limits_{m=1}^n{\\text{ent}_{im}(E_{i\\to i+ck})a_{ml}}=a_{il}+ca_{kl}$. Fantastic, we have found something that takes the entry from row k and multiplies it by c and adds it to the entry in row i, then places the result in row i. This then is the elementary matrix we require.\n\nNow if we take $E_{i\\to i-ck}$ for \n$j\\ne i, l\\le n$ we have $$\\text{ent}_{jl}(E_{i\\to i-ck}E_{i\\to i+ck})=\\sum\\limits_{m=1}^n{\\text{ent}_{jm}(E_{i\\to i-ck})\\text{ent}_{jm}(E_{i\\to i+ck})}=\\text{ent}_{jl}(I_n)$$\n\nAnd $\\text{ent}_{il}(E_{i\\to i-ck}E_{i\\to i+ck})=\\sum\\limits_{m=1}^n{\\text{ent}_{im}(E_{i\\to i-ck})\\text{ent}_{im}(E_{i\\to i+ck})}=\\text{ent}_{il}(E_{i\\to i+ck})-c\\cdot\\text{ent}_{kl}(E_{i\\to i+ck})$. Of course if $i=l$ then this will give us $1-0=1$ and if $k=l$ then we have $c-c=0$. This will leave us with our identity. And so we have our inverse.\n\nFor each of the three elementary matrices we have found an inverse, and thus one must exist. Further, these inverses are simply elementary matrices themselves.\n$\\Box$\n\\item\nDefine a relation $\\sim$ on $\\mathcal{M}_{m\\times n}$ given by $A\\sim B$ if and only if there exists a $P\\in \\mathcal{M}_{m\\times n}$ such that $A=PB$ where $P\\in \\mathcal{M}_{m\\times n}$ is a product of elementary matrices.\n\nWe know that $A=I_nA$ and $I_n$ is an elementary matrix and so this relation is reflexive.\n\nWe also know that every elementary matrix has an inverse and that inverse is an elementary matrix. And so if $P=E_1E_2\\dots E_n$ then $P^{-1}$ exists. In fact $P^{-1}=E_n^{-1}\\dots E_1^{-1}$ which is the product of elementary matrices also. Therefore if $A=PB$ then $P^{-1}A=B$ and so commutativity is preserved with this relation.\n\nNow of we have $P_1,P_2$ both products of elementary matrices, then it follows that $P_1P_2$ is also the product of elementary matrices. And so if $A=P_1B$ and $B=P_2C$ then $A=P_1(P_2C)=(P_1P_2)C$ and so the relation is transitive.\n\nThose are the three properties we need for the relation to fit the definition of an equivalence relation.\n$\\Box$\n\\end{enumerate}\n\\end{document}\n%hw 1.5:1,2b,3ce,10,11,13,14,15\n", "meta": {"hexsha": "e0742841b8b2024eb58d36de127ac65e4fe361c7", "size": 12075, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "linear/linear-hw-2015-09-23.tex", "max_stars_repo_name": "ylixir/school", "max_stars_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "linear/linear-hw-2015-09-23.tex", "max_issues_repo_name": "ylixir/school", "max_issues_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "linear/linear-hw-2015-09-23.tex", "max_forks_repo_name": "ylixir/school", "max_forks_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.5647668394, "max_line_length": 485, "alphanum_fraction": 0.6573913043, "num_tokens": 4815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059414036511, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.6985653994377389}} {"text": "\\section{Arc Length}{}{}\\label{sec:Arc Length}\nHere is another geometric application of the integral: Find the length\nof a portion of a curve. As usual, we need to think about how we might\napproximate the length, and turn the approximation into an integral.\n\nWe already know how to compute one simple arc length, that of a line\nsegment. If the endpoints are $\\ds P_0(x_0,y_0)$ and $\\ds P_1(x_1,y_1)$\nthen the length of the segment is the distance between the points,\n$\\ds \\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$, from the Pythagorean theorem, as\nillustrated in Figure~\\ref{fig:length of a line segment}.\n\n\\figure[H]\n%\\texonly\n\\centerline{\\vbox{\\beginpicture\n\\normalgraphs\n%\\sevenpoint\n\\setcoordinatesystem units <1.5truecm,1.5truecm>\n\\setplotarea x from 0 to 5, y from 0 to 3\n\\axis bottom /\n\\axis left /\n\\putrule from 2 1 to 4.5 1\n\\putrule from 4.5 1 to 4.5 2.5\n\\plot 2 1 4.5 2.5 /\n\\put {$(x_1,y_1)$} [bl] <3pt,3pt> at 4.5 2.5\n\\put {$(x_0,y_0)$} [tr] <-3pt,-3pt> at 2 1\n\\put {$x_1-x_0$} [t] <0pt,-3pt> at 3.25 1\n\\put {$y_1-y_0$} [l] <3pt,0pt> at 4.5 1.75\n\\put {$\\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$} [br] <-3pt,3pt> at 3.25 1.75\n\\endpicture}}\n%\\endtexonly\n%\\figrdef{fig:length of a line segment}\n%\\htmlfigure{Integration_applications-arc_length.html}\n\\caption{\\label{fig:length of a line segment}\nThe length of a line segment.}\n%\\endcaption\n\\endfigure\n\nNow if the graph of $f$ is ``nice'' (say, differentiable) it appears\nthat we can approximate the length of a portion of the curve with line\nsegments, and that as the number of segments increases, and their\nlengths decrease, the sum of the lengths of the line segments will\napproach the true arc length; see \nFigure~\\ref{fig:approximating arc length}.\n\n\\figure[H]\n%\\texonly\n\\centerline{\\vbox{\\beginpicture\n\\normalgraphs\n%\\sevenpoint\n\\setcoordinatesystem units <1.5truecm,0.8truecm>\n\\setplotarea x from 0 to 8, y from 0 to 5\n\\axis bottom /\n\\axis left /\n\\setquadratic\\plot \n1.000 1.000 1.150 1.953 1.300 2.694 1.450 3.246 1.600 3.636 \n1.750 3.884 1.900 4.012 2.050 4.041 2.200 3.990 2.350 3.874 \n2.500 3.711 2.650 3.515 2.800 3.299 2.950 3.075 3.100 2.853 \n3.250 2.644 3.400 2.454 3.550 2.290 3.700 2.157 3.850 2.060 \n4.000 2.000 4.150 1.979 4.300 1.996 4.450 2.050 4.600 2.138 \n4.750 2.255 4.900 2.396 5.050 2.554 5.200 2.721 5.350 2.886 \n5.500 3.039 5.650 3.168 5.800 3.258 5.950 3.294 6.100 3.261 \n6.250 3.140 6.400 2.912 6.550 2.556 6.700 2.051 6.850 1.374 \n7.000 0.500 /\n\\multiput {$\\bullet$} at 1 1 2.050 4.041 3.250 2.644\n4.300 1.996 5.500 3.039 6.250 3.140 7 0.5 /\n\\setlinear\\plot\n1 1 2.050 4.041 3.250 2.644\n4.300 1.996 5.500 3.039 6.250 3.140 7 0.5 /\n\\endpicture}}\n%\\endtexonly\n%\\figrdef{fig:approximating arc length}\n%\\htmlfigure{Integration_applications-arc_length_line_segments.html}\n\\caption{\\label{fig:approximating arc length}\nApproximating arc length with line segments.}\n%\\endcaption\n\\endfigure\n\nNow we need to write a formula for the sum of the lengths of the line\nsegments, in a form that we know becomes an integral in the limit. So\nwe suppose we have divided the interval $[a,b]$ into $n$ subintervals\nas usual, each with length $\\Delta x =(b-a)/n$, and endpoints $\\ds\na=x_0$, $\\ds x_1$, $\\ds x_2$, \\dots, $\\ds x_n=b$. The length of a\ntypical line segment, joining $\\ds (x_i,f(x_i))$ to $\\ds\n(x_{i+1},f(x_{i+1}))$, is $\\ds\\sqrt{(\\Delta x )^2\n +(f(x_{i+1})-f(x_i))^2}$. By the Mean Value Theorem, %(\\xrefn{thm:mvt}), \nthere is a number $\\ds t_i$ in $\\ds (x_i,x_{i+1})$\nsuch that $\\ds f'(t_i)\\Delta x=f(x_{i+1})-f(x_i)$, so the length of\nthe line segment can be written as\n$$\n \\sqrt{(\\Delta x)^2 + (f'(t_i))^2\\Delta x^2}=\n \\sqrt{1+(f'(t_i))^2}\\,\\Delta x.\n$$\nThen arc length is:\n$$\n \\lim_{n\\to\\infty}\\sum_{i=0}^{n-1} \\sqrt{1+(f'(t_i))^2}\\,\\Delta x=\n \\int_a^b \\sqrt{1+(f'(x))^2}\\,dx.\n$$\nNote that the sum looks a bit different than others we have\nencountered, because the approximation contains a $\\ds t_i$ instead of an\n$\\ds x_i$. In the past we have always used left endpoints (namely, $\\ds x_i$)\nto get a representative value of $f$ on $\\ds [x_i,x_{i+1}]$; now we are\nusing a different point, but the principle is the same.\n\nTo summarize, to compute the length of a curve on the interval\n$[a,b]$, we compute the integral\n$$\\int_a^b \\sqrt{1+(f'(x))^2}\\,dx.$$ \nUnfortunately, integrals of this form are typically difficult or\nimpossible to compute exactly, because usually none of our methods for\nfinding antiderivatives will work. In practice this means that the\nintegral will usually have to be approximated.\n\n\\begin{example}{Circumference of a Circle}{Circumference of a Circle}\\label{Circumference of a Circle} \nLet $\\ds f(x) = \\sqrt{r^2-x^2}$, the upper half circle of radius\n$r$. The length of this curve is half the circumference, namely $\\pi\nr$. Compute this with the arc length formula.\n\\end{example}\n\n\\begin{solution}\nThe derivative $f'$ is $\\ds \\ds -x/\\sqrt{r^2-x^2}$ so the integral is\n$$\n \\int_{-r}^r \\sqrt{1+{x^2\\over r^2-x^2}}\\,dx\n =\\int_{-r}^r \\sqrt{r^2\\over r^2-x^2}\\,dx\n =r\\int_{-r}^r \\sqrt{1\\over r^2-x^2}\\,dx.\n$$\nUsing a trigonometric substitution, we find the antiderivative, namely\n$\\ds \\arcsin(x/r)$. Notice that the integral is improper at both\nendpoints, as the function $\\ds \\sqrt{1/(r^2-x^2)}$ is undefined when\n$x=\\pm r$. So we need to compute\n$$\n \\lim_{D\\to-r^+}\\int_D^0 \\sqrt{1\\over r^2-x^2}\\,dx +\n \\lim_{D\\to r^-}\\int_0^D \\sqrt{1\\over r^2-x^2}\\,dx.\n$$\nThis is not difficult, and has value $\\pi$, so the original integral,\nwith the extra $r$ in front, has value $\\pi r$ as expected.\n\\end{solution}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\Opensolutionfile{solutions}[ex]\n\\section*{Exercises for \\ref{sec:Arc Length}}\n\n\\begin{enumialphparenastyle}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the arc length of $\\ds f(x)=x^{3/2}$ on $[0,2]$.\n\\begin{sol}\n $\\ds (22\\sqrt{22}-8)/27$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the arc length of $\\ds f(x) = x^2/8-\\ln x$\non $[1,2]$.\n\\begin{sol}\n $\\ln(2)+3/8$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n\nFind the arc length of $\\ds f(x) = (1/3)(x^2 +2)^{3/2}$\non the interval $[0,a]$.\n\\begin{sol}\n $\\ds a+a^3/3$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the arc length of $f(x)=\\ln(\\sin x)$ on the\ninterval $[\\pi/4,\\pi/3]$.\n\\begin{sol}\n $\\ds \\ln((\\sqrt2+1)/\\sqrt3)$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Let $a>0$. Show that the length of $y=\\cosh x$ on\n$[0,a]$ is equal to $\\ds \\int _0 ^a \\cosh x\\,dx$.\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the arc length of $f(x)=\\cosh x$ on $[0, \\ln 2]$.\n\\begin{sol}\n $3/4$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Set up the integral to find the arc length of $\\sin x$ \non the interval $[0,\\pi]$; do not evaluate the integral. If you have\naccess to appropriate software, approximate the value of the integral.\n\\begin{sol}\n $\\approx 3.82$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Set up the integral to find the arc length of $\\ds y=xe^{-x}$\non the interval $[2,3]$; do not evaluate the integral. If you have\naccess to appropriate software, approximate the value of the integral.\n\\begin{sol}\n $\\approx 1.01$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the arc length of $\\ds y=e^x$ on the interval $[0,1]$.\n(This can be done exactly; it is a bit tricky and a bit long.)\n\\begin{sol}\n $\\ds \\sqrt{1+e^2}-\\sqrt2+\n{1\\over2}\\ln\\left({\\sqrt{1+e^2}-1\\over\\sqrt{1+e^2}+1}\\right)+\n{1\\over2}\\ln(3+2\\sqrt2)$\n\\end{sol}\n\\end{ex}\n\n\\end{enumialphparenastyle}", "meta": {"hexsha": "c4251843795df3a0a50f31c5bf9c854ace1a3498", "size": 7337, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "8-applications-of-integration/8-7-arclength.old2.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "8-applications-of-integration/8-7-arclength.old2.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "8-applications-of-integration/8-7-arclength.old2.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.6088888889, "max_line_length": 103, "alphanum_fraction": 0.6673027123, "num_tokens": 2827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059511841119, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.6985653956439885}} {"text": "\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{amssymb,amsmath}\n\\usepackage{pdflscape}\n\\usepackage{subfigure}\n\\usepackage{dcolumn}\n%\\usepackage{rotating}\n\\usepackage{fancyhdr}\n\\pagestyle{fancyplain}\n\n\\newcommand\\T{\\rule{0pt}{2.6ex}}\n\\newcommand\\B{\\rule[-1.2ex]{0pt}{0pt}}\n\n\\begin{document}\n\\rhead{Prepared by Adam Beardsley on \\today}\n\n\\centerline{\\sc \\large Interpreting Discrete Fourier Transforms}\n\\vspace{.5pc}\n\n\\begin{abstract}\nWe present a systematic way of interpreting the discrete Fourier transform as an approximation to the continuous Fourier transform. The argument is made for general conventions, and several examples are given.\n\\end{abstract}\n\n\\section{Introduction}\n\nOften times it is necessary to approximate a continuous Fourier transform (CFT) with a discrete Fourier transform (DFT) when an analytic form of the data is not available. For example, a DFT can be performed on a sampling of time-ordered electric field measurements to approximate the frequency spectrum of the field. In order to perform these transforms, we utilize the numerical packages of programming languages such as IDL, MATLAB, or Mathematica. However, the definition of a Fourier transform is not completely constrained, and freedoms exist that give rise to multiple conventions in the normalization\\footnote{There are of course constraints limiting these freedoms, but for this argument we will leave them completely free.}, and scaling of the Fourier domain variables. In particular, the numerical package being used may not follow the desired convention. Below we outline a method for interpreting the results of a DFT, in terms of the CFT desired. Then we provide several examples of common conventions.\n\n\\section{Derivation}\\label{sec:derivation}\n\nWe start by assuming we have a continuous function, $f_c(x)$, of which we would like to compute a continuous Fourier transform of the form\n\\begin{equation}\n\\label{eq:continuous}\nF_c(u;C_c,p_c)=C_c\\int_{-\\infty}^{\\infty}dx \\ f_c(x) e^{i p_c u x}\n\\end{equation}\nwhere the constants $C_c$ and $p_c$ are determined by convention of the transform. However, in reality we often have a discrete sample of the continuous function, $f_d(j)$, where $j$ indexes the samples and runs from $0$ to $N-1$, $N$ being the number of samples. Then it is necessary to use a discrete Fourier transform. Numerical packages have various conventions for their DFTs, but in general they can be expressed in the form\n\\begin{equation}\n\\label{eq:discrete}\nF_d(k;C_d,p_d)=C_d\\sum\\limits_{j=0}^{N-1}f_d(j)e^{i p_d j k}\n\\end{equation}\nwhere $k$ indexes the Fourier domain samples, and also runs from $0$ to $N-1$. In some cases, the indices can be shifted. For example, MATLAB counts arrays from 1 rather than 0, so the sum would run from $1$ to $N$, and the values in the exponent are adjusted accordingly. For simplicity, we assume counting from 0. The translation is straighforward, and an example is given in Section \\ref{sec:examples}. We wish to approximate the CFT with the DFT.\n\nIn order to proceed, we assume the function $f_c(x)$ is only significantly non-zero within a domain $\\mathcal{D}$. Furthermore, we will assume $\\mathcal{D}=[0,x_{max}]$. This constraint is not actually necessary, and a translation of the domain results in a phase in the Fourier transform, as seen in Section \\ref{sec:shifting}. Then we can relate the indices in Eq. \\eqref{eq:discrete} to the axis values with the following expressions:\n\\begin{subequations}\n\\label{eq:axes}\n\\begin{align}\nx(j)=j\\Delta x \\label{eq:x_axis}\\\\\nu(k)=k\\Delta u \\label{eq:u_axis}\n\\end{align}\n\\end{subequations}\nwhere $\\Delta x = x_{max}/(N-1)$, and $\\Delta u$ will be derived below. We can now approximate the CFT with a finite sum.\n\n\\begin{eqnarray}\n\\label{eq:approx_cont}\nF_c(u;C_c,p_c) \t& \\approx\t&\tC_c\\int_0 ^{x_{max}} dx \\ f_c(x)e^{i p_c u x} \\nonumber \\\\\n\t\t\t\t\t\t\t\t& \\approx\t&\tC_c\\sum\\limits_{j=0}^{N-1}\\Delta x f_c(x(j))e^{i p_c u j \\Delta x} \\nonumber \\\\\n\t\t\t\t\t\t\t\t& =\t\t\t\t&\tC_c \\Delta x \\sum\\limits_{j=0}^{N-1} f_d(j) e^{i p_c u j \\Delta x}\n\\end{eqnarray}\n\nIn order to force the sum in Eq. \\eqref{eq:approx_cont} to the form of the DFT (Eq. \\eqref{eq:discrete}), we write $u(k)$ in the following way.\n\n\\begin{eqnarray}\n\\label{eq:delta_u}\nu(k) = k\\Delta u\t&\t= & \\frac{p_d k}{p_c \\Delta x} \\nonumber \\\\\n\t\t\\Rightarrow\t\\Delta u\t&\t=\t&\t\\frac{p_d}{p_c \\Delta x} \n\\end{eqnarray}\n\nThis expression for $\\Delta u$ defines our Fourier axis, which is an extremely important part of interpreting a Fourier transform and in particular comparing to other computational conventions.\n\nWe can now simplify Eq. \\ref{eq:approx_cont}. \n\n\\begin{eqnarray}\n\\label{eq:related}\nF_c(u;C_c,p_c)\t&\t\\approx\t&\tC_c \\Delta x \\sum\\limits_{j=0}^{N-1}f_d(j)e^{i p_d j k} \\nonumber \\\\\n\t\t\t\t\t\t\t\t&\t=\t\t\t\t&\t\\frac{C_c}{C_d} \\Delta x F_d(k;C_d,p_d) \\nonumber \\\\\n\t\t\t\t\t\t\t\t&\t=\t\t\t\t&\t\\frac{C_c}{C_d} \\Delta x F_d(u / \\Delta u;C_d,p_d)\n\\end{eqnarray}\n\nIn the first line of Eq. \\ref{eq:related} we introduced an intermediate index $k=u/\\Delta u$ for clarity in replacing the sum with the DFT. Of course any computer package will compute the DFT for integer values of $k$, and it should be noted that this approximation for the CFT is only used for $u$ equal to integer multiples of $\\Delta u$.\n\nTo summarize, we have related an arbitrary convention of the continuous Fourier transform to an arbitrary convention of the discrete Fourier transform (Eq. \\ref{eq:related}). In addition, we have shown the resulting Fourier axis is described by Eq. \\ref{eq:delta_u}. For reference, several examples are discussed in Section \\ref{sec:examples}.\n\n\\section{Shifting Domains}\\label{sec:shifting}\n\nThe above derivation applied to functions significantly non-zero only within a domain $\\mathcal{D}=[0,x_{max}]$. We can relax that condition slightly by allowing $\\mathcal{D}\\rightarrow[x_{min},x_{max}]$. The same process can be followed as in Section \\ref{sec:derivation}, with a simple translation of the variable of integration. The result is a mode-dependent phase, and we can generalize Eq. \\ref{eq:related} as\n\\begin{equation}\n\\label{eq:related_general}\nF_c(u;C_c,p_c) \\approx \\frac{C_c}{C_d} \\Delta x e^{i p_c u x_{min}}F_d(u / \\Delta u;C_d,p_d)\n\\end{equation}\nwhere now $\\Delta x = (x_{max}-x_{min})/(N-1)$.\n\nA particular case of interest is to shift the real and Fourier domains such that they are centered at zero. This can be achieved by a couple shifts of the arrays. In this case, $x_{min} = -\\Delta x (N-1)/2$. The phases in Eq. \\ref{eq:related_general} are now negative, but can be shifted properly to make them positive. This amounts to exchanging the positive and negative $x$ data in $f_d(j)$. This is illustrated in Step A in Fig. \\ref{fig:shifting}.\n\nOnce the DFT is performed, our approximation for $F_c(u)$ is still only valid for positive (or in some cases only negative) values of $u$. But we note that by the periodic nature of a complex phase, for $k>(N-1)/2$, and for $p_d = 2 \\pi / N$ (as is normally the case), we can make the substituion $k\\rightarrow-N+k$ and the sum in Eq. \\ref{eq:discrete} remains unchanged. In other words, we can relabel and shift our $u$ axis to center our Fourier axis on $u=0$, as shown in Steps C and D in Fig. \\ref{fig:shifting}. All these operations can be easily implemented using shifting functions built into many software packages. In MATLAB, for example, \\texttt{ifftshift} (Step A) and \\texttt{fftshift} (Step D) are provided for this very reason.\n\n\n\\begin{figure}\n\\begin{center}\n\\centerline{\\includegraphics[width=7in]{FFT_shifts.jpg}}\n\\end{center}\n\\caption{Centering axes on 0. \\emph{Step A}: Shift the real axis to be fed into DFT package. \\emph{Step B}: Perform discrete Fourier transform. \\emph{Step C}: Relabel Fourier modes to equivalent negative modes. \\emph{Step D}: Shift Fourier axis to recover correct order.}\n\\label{fig:shifting}\n\\end{figure}\n\n\\section{Examples}\\label{sec:examples}\n\nHere we apply the interpretation of Section \\ref{sec:derivation} specifically to the default MATLAB discrete Fourier transform. Further examples are listed in Table \\ref{table:examples}. For the forward DFT, MATLAB uses the convention\n\n\\begin{equation}\nF_d(k) = \\sum\\limits_{j=1}^N f_d(j)e^{-2 \\pi i (j-1)(k-1)/N}, \\nonumber\n\\end{equation}\nso that $C_d=1$ and $p_d = -2 \\pi / N$. Note the shift in indices due to the fact that MATLAB counts from 1, rather than zero. This can be accounted for by writing the axes as\n\\begin{eqnarray}\nx(j) = (j-1)\\Delta x \\nonumber \\\\\nu(k) = (k-1)\\Delta u \\nonumber\n\\end{eqnarray}\nwhere $\\Delta u = p_d/(p_c \\Delta x) = -2 \\pi / (p_c N \\Delta x)$.\n\nLet us now assume that we have sampled a function of position, $f_d(j)$, with resolution $\\Delta x = 3$m, and $N = 1001$ ($x_{range} = 3000$m). We are interested in the Fourier transform of the form\n\\begin{equation}\nF_c(u) = \\frac{1}{\\sqrt{2 \\pi}}\\int_{-\\infty}^{\\infty} dx \\ f_c(x) e^{i u x} \\nonumber\n\\end{equation}\nso that $C_c=1/\\sqrt{2\\pi}$ and $p_c=1$. Then we use Eq. \\ref{eq:related} to approximate the CFT as\n\\begin{equation}\nF_c(u) \\approx \\frac{3\\mbox{m}}{\\sqrt{2\\pi}} F_d(-u \\frac{3003\\mbox{m}}{2\\pi}+1;1,2\\pi/N) \\nonumber\n\\end{equation}\nwhere $F_d(k,1,2\\pi/N)$ is the array output from MATLAB's \\texttt{fft} function, given $f_d(j)$. Note that because the sign of $p_c$ is opposite the sign of $p_d$, the approximation for the CFT will only be valid for negative Fourier modes. In most applications this difference will not be significant, but in some cases it may be important to account for the axis reversal.\n\nSimilar steps can be taken for any DFT convention, and some common default parameters are listed in Table \\ref{table:examples}. We also list several common CFT conventions and their corresponding conversions from common numerical packages in Table \\ref{table:applied_examples}.\n\n\n%Figure format\n%\\begin{figure}\n%\\begin{center}\n%\\includegraphics[width=4in]{file.jpg}\n%\\end{center}\n%\\caption{Put a caption here, if you dare}\n%\\label{fig:figure_label}\n%\\end{figure}\n\n\n%appendix if needed\n%\\clearpage\n%\\appendix\n%\\section{appendix section}\\label{app_section}\n\n\n\\begin{table}\n\\begin{center}\n\\centerline{\\begin{tabular}{c c c c c c} \n\\hline\nPackage\t\\T \\B\t\t\t&\t$F_d(k)$\t&\t$C_d$\t&\t$p_d$\t&\t$x(j)$\t&\t$u(k)$\t\\\\ \\hline\nIDL\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{N}\\sum\\limits_{j=0}^{N-1} f_d(j)e^{-2\\pi i j k /N}$\t\t\t\t\t&\t$1/N$\t\t\t\t\t\t\t\t\t&\t$-2\\pi/N$\t&\t$j\\Delta x$\t\t\t&\t$-k\\frac{2\\pi}{p_c N \\Delta x}$\t\t\t \\\\\nMATLAB\t\\T \\B\t\t\t&\t$\\sum\\limits_{j=1}^N f_d(j)e^{2\\pi i (j-1)(k-1)/N}$\t\t\t\t\t\t\t\t\t\t&\t1\t\t\t\t\t\t\t\t\t\t\t&\t$-2\\pi/N$\t&\t$(j-1)\\Delta x$\t&\t$-(k-1)\\frac{2\\pi}{p_c N \\Delta x}$\t \\\\\nMathematica\t\\T \\B\t&\t$\\frac{1}{\\sqrt{N}}\\sum\\limits_{j=1}^N f_d(j)e^{2\\pi i (j-1)(k-1)/N}$\t&\t$\\frac{1}{\\sqrt{N}}$\t&\t$2\\pi/N$\t&\t$(j-1)\\Delta x$\t&\t$(k-1)\\frac{2\\pi}{p_c N \\Delta x}$\t \\\\\nPython (numpy) \\T \\B & $\\sum\\limits_{j=0}^{N-1} f_d(j)e^{-2\\pi i j k /N}$\t\t\t\t\t&\t1\t\t\t\t\t\t\t\t\t&\t$-2\\pi/N$\t&\t$j\\Delta x$\t\t\t&\t$-k\\frac{2\\pi}{p_c N \\Delta x}$\t\t\t \\\\\n\\hline\n\\end{tabular}}\n\\caption{Example default parameters for the forward discrete Fourier transform for various numerical packages}\n\\label{table:examples}\n\\end{center}\n\\end{table}\n\n\\begin{table}\n\\begin{center}\n\\begin{tabular}{c c c c}\n\\hline\nDFT Package\t\\T \\B\t&\t$F_c(u)$ Desired\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t&\tAmplitude Factor\t\t\t\t\t&\t$\\left|\\Delta u\\right|$\t\t\\\\ \\hline\nIDL\t\t\\T \\B\t\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{i u x}$\t&\t$\\Delta x N/\\sqrt{2\\pi}$\t&\t$2\\pi/(N\\Delta x)$ \t\t\t\t\\\\\n\t\t\t\\T \\B\t\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{i u x}$\t\t\t\t&\t$\\Delta x N/2\\pi$\t\t\t\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t\t\\\\\n\t\t\t\\T \\B\t\t\t\t&\t$\\int dx \\ f_c(x) e^{2\\pi i u x}$\t\t\t\t\t\t\t\t\t&\t$\\Delta x N$\t\t\t\t\t\t\t&\t$1/(N\\Delta x)$\t\t\t\t\t\t\\\\ \\hline\nMATLAB\t\\T \\B\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{i u x}$\t&\t$\\Delta x/\\sqrt{2\\pi}$\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t \t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{i u x}$\t\t\t\t&\t$\\Delta x/2\\pi$\t\t\t\t\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{2\\pi i u x}$\t\t\t\t\t\t\t\t\t&\t$\\Delta x$\t\t\t\t\t\t\t\t&\t$1/(N\\Delta x)$\t\t\t\t\t\t\\\\ \\hline\nMathematica\t\\T \\B\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{i u x}$\t&\t$\\Delta x\\sqrt{N/2\\pi}$\t\t&\t$2\\pi/(N\\Delta x)$ \t\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{i u x}$\t\t\t\t&\t$\\Delta x\\sqrt{N}/2\\pi$\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{2\\pi i u x}$\t\t\t\t\t\t\t\t\t&\t$\\Delta x\\sqrt{N}$\t\t\t\t&\t$1/(N\\Delta x)$\t\t\t\t\t\t\\\\ \\hline\nPython (numpy)\t\\T \\B\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{i u x}$\t&\t$\\Delta x/\\sqrt{2\\pi}$\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t \t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{i u x}$\t\t\t\t&\t$\\Delta x/2\\pi$\t\t\t\t\t\t&\t$2\\pi/(N\\Delta x)$\t\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{2\\pi i u x}$\t\t\t\t\t\t\t\t\t&\t$\\Delta x$\t\t\t\t\t\t\t\t&\t$1/(N\\Delta x)$\t\t\t\t\t\t\\\\ \\hline\n\\end{tabular}\n\\caption{Common CFT conventions listed with the normalization factor and axis scaling factors for sample numerical packages.}\n\\label{table:applied_examples}\n\\end{center}\n\\end{table}\n\n\\section{Inverse Fourier Transform}\n\nThe above argument holds for the inverse Fourier transform as well. Define the inverse CFT as\n\\begin{equation}\n\\label{eq:inverse_continuous}\nf_c(x;C'_c,p'_c)=C'_c\\int_{-\\infty}^{\\infty}du \\ F_c(k) e^{i p'_c u x}\n\\end{equation}\nand the inverse DFT as\n\\begin{equation}\n\\label{eq:inverse_discrete}\nf_d(j;C'_d,p'_d)=C'_d\\sum\\limits_{k=0}^{N-1}F_d(k)e^{i p'_d j k}.\n\\end{equation}\nThen the result for the approximation of the inverse CFT is\n\\begin{equation}\n\\label{eq:inverse_related}\nf_c(x;C'_c,p'_c) \\approx \\frac{C'_c}{C'_d} \\Delta u f_d(x / \\Delta x;C'_d,p'_d)\n\\end{equation}\nwhere $\\Delta x = p_d'/p_c' \\Delta u$.\n\nWe summarize the default parameters for the inverse DFT for various numerical packages in Table \\ref{table:inverse_examples}, and provide a few applied examples to common CFTs in Table \\ref{table:inverse_applied_examples}.\n\n\\begin{table}\n\\begin{center}\n\\centerline{\\begin{tabular}{c c c c c c}\n\\hline\nPackage\t\\T \\B\t\t\t&\t$f_d(j)$\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t&\t$C'_d$\t\t\t\t\t\t\t\t&\t$p'_d$\t\t&\t$u(k)$\t\t\t\t\t&\t$x(j)$\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\\\\ \\hline\nIDL\t\t\t\\T \\B\t\t\t&\t$\\sum\\limits_{k=0}^{N-1} F_d(k)e^{2\\pi i j k /N}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$1$\t\t\t\t\t\t\t\t\t\t&\t$2\\pi/N$\t&\t$k\\Delta u$\t\t\t&\t$j\\frac{2\\pi}{p'_c N \\Delta u}$\t\t\t\t\\\\\nMATLAB\t\\T \\B\t\t\t&\t$\\frac{1}{N}\\sum\\limits_{k=1}^N F_d(k)e^{-2\\pi i (j-1)(k-1)/N}$\t\t\t\t\t&\t$1/N$\t\t\t\t\t\t\t\t\t&\t$2\\pi/N$\t&\t$(k-1)\\Delta u$\t&\t$(j-1)\\frac{2\\pi}{p'_c N \\Delta u}$\t\t\\\\\nMathematica\t\\T \\B\t&\t$\\frac{1}{\\sqrt{N}}\\sum\\limits_{k=1}^N F_d(k)e^{-2\\pi i (j-1)(k-1)/N}$\t&\t$\\frac{1}{\\sqrt{N}}$\t&\t$-2\\pi/N$\t&\t$(k-1)\\Delta u$\t&\t$-(j-1)\\frac{2\\pi}{p'_c N \\Delta u}$ \t\\\\\nPython (numpy)\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{N}\\sum\\limits_{k=0}^{N-1} F_d(k)e^{2\\pi i j k /N}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$1/N$\t\t\t\t\t\t\t\t\t\t&\t$2\\pi/N$\t&\t$k\\Delta u$\t\t\t&\t$j\\frac{2\\pi}{p'_c N \\Delta u}$\t\t\t\t\\\\\n\\hline\n\\end{tabular}}\n\\caption{Example default parameters for the inverse discrete Fourier transform for various numerical packages}\n\\label{table:inverse_examples}\n\\end{center}\n\\end{table}\n\n\n\\begin{table}\n\\begin{center}\n\\centerline{\\begin{tabular}{c c c c}\n\\hline\nDFT Package\t\\T \\B\t&\t$f_c(x)$ Desired\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t&\tAmplitude Factor\t\t\t\t\t&\t$\\left|\\Delta x\\right|$\t\\\\ \\hline\nIDL\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{-i u x}$\t&\t$\\Delta u/\\sqrt{2\\pi}$\t\t&\t$2\\pi/(N \\Delta u)$\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{-i u x}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$\\Delta u$\t\t\t\t\t\t\t\t&\t$2\\pi/(N \\Delta u)$\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{-2\\pi i u x}$\t\t&\t$\\Delta u/2\\pi$\t\t\t\t\t\t&\t$1/(N \\Delta u)$\t\t\t\t\\\\ \\hline\nMATLAB\t\\T \\B\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{-i u x}$\t&\t$\\Delta u N/\\sqrt{2\\pi}$\t&\t$2\\pi/(N \\Delta u)$ \t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{-i u x}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$\\Delta u N$\t\t\t\t\t\t\t&\t$2\\pi/(N \\Delta u)$\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{-2\\pi i u x}$\t\t&\t$\\Delta u N/2\\pi$\t\t\t\t\t&\t$1/(N \\Delta u)$\t\t\t\t\\\\ \\hline\nMathematica\t\\T \\B\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{-i u x}$\t&\t$\\Delta u\\sqrt{N/2\\pi}$\t\t&\t$2\\pi/(N \\Delta u)$ \t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{-i u x}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$\\Delta u\\sqrt{N}$\t\t\t\t&\t$2\\pi/(N \\Delta u)$\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{-2\\pi i u x}$\t\t&\t$\\Delta u\\sqrt{N}/2\\pi$\t\t&\t$1/(N \\Delta u)$\t\t\t\t\\\\ \\hline\nPython (numpy)\t\\T \\B\t\t\t&\t$\\frac{1}{\\sqrt{2\\pi}}\\int dx \\ f_c(x) e^{-i u x}$\t&\t$\\Delta u N/\\sqrt{2\\pi}$\t&\t$2\\pi/(N \\Delta u)$ \t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\int dx \\ f_c(x) e^{-i u x}$\t\t\t\t\t\t\t\t\t\t\t\t&\t$\\Delta u N$\t\t\t\t\t\t\t&\t$2\\pi/(N \\Delta u)$\t\t\t\\\\\n\t\t\t\t\\T \\B\t\t\t&\t$\\frac{1}{2\\pi}\\int dx \\ f_c(x) e^{-2\\pi i u x}$\t\t&\t$\\Delta u N/2\\pi$\t\t\t\t\t&\t$1/(N \\Delta u)$\t\t\t\t\\\\ \\hline\n\\end{tabular}}\n\\caption{Common inverse CFT conventions listed with the normalization factor and axis scaling factors for sample numerical packages.}\n\\label{table:inverse_applied_examples}\n\\end{center}\n\\end{table}\n\n\n\n\\end{document}", "meta": {"hexsha": "af6a20a3f4398f4a62224381f1b94670d59f5f53", "size": 16644, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "001_DFT_Interpretation/DFT_Interpretation.tex", "max_stars_repo_name": "EoRImaging/Memos", "max_stars_repo_head_hexsha": "216dbda634c1686be25cda25bb258664067a3aad", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 1, 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1022, "alphanum_fraction": 0.6521869743, "num_tokens": 6042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519528019683106, "lm_q2_score": 0.8198933447152497, "lm_q1q2_score": 0.698510432345327}} {"text": "\\documentclass[10pt,a4paper]{article}\n\n% Packages\n\\usepackage[utf8]{inputenc}\n\\usepackage{amssymb}\n\\usepackage{mathtools}\n\\usepackage{gensymb}\n\\usepackage{amsmath}\n\\usepackage{MnSymbol}\n\\usepackage{pgfplots}\n\n\\title{\\textbf{Mathematical proofs}}\n\\author{Michal Špano}\n\\date{March 2022}\n\n\\begin{document}\n\n\\maketitle\n\n\\section*{The inner angle of an $n-sided$ convex regular polygon}\n\n% Introduction\nSuppose an $n-sided$ convex regular polygon. Its 4 consecutive vertices are shown in the figures, $N_1, N_2, N_3, ..., N_i$ respectively. \nThus $\\Delta_{N_1,N_2,S} \\cong \\Delta_{N_2,N_3,S}$, i.e. such $n-sided$ polygon consists of $n$ congruent isosceles triangles. \nThat is, $|\\measuredangle N_2 N_1 S| = |\\measuredangle N_1 N_2 S| \\Rightarrow |\\measuredangle N_1 N_2 S| = |\\measuredangle S N_2 N_3|$, denoted as $\\beta, \\beta'$ respectively. \n\n% Section I\nAn angle $\\alpha = |\\measuredangle N_1 S N_2| = \\cfrac{360}{n}$, since such an angle multiplied by $n$ makes for a perfect circle of $360\\degree$. Likewise $\\alpha = 180\\degree - 2 \\beta$.\n\n% Section I\nLet $\\Phi$ be an inner angle of the polygon, such that $\\phi = 2 \\beta$ (shown in the figure at the vertex $N_2$).\n\n% Steps of computations\nExpress in terms of $\\beta$: $\\alpha = 180 - 2 \\beta \\iff \\beta = \\cfrac{-\\alpha + 180}{2}$\n\nSubstitute $\\alpha$ for $\\alpha = \\cfrac{360}{n}$: $\\beta = \\cfrac{-\\cfrac{360}{n} + 180}{2} \\iff \\beta = \\cfrac{180n - 360}{2n}$\n\nExpress in terms of $\\Phi$: $\\Phi = 2 \\beta = 2 \\Big( \\cfrac{180n - 360}{2n} \\Big) = \\cfrac{180n - 360}{n}$ \\\\\n\nThe expression can be further simplified to the following form:\n\n% End of the proof\n$$\\Phi = \\cfrac{(n-2) \\pi}{n}$$.\n\n% Include a picture of a unit circle\n\\begin{figure}[htp]\n \\centering\n \\includegraphics[width=4.cm]{polygon_export.png}\n \\caption{$n-sided$ polygon with its 4 vertices in a plane}\n\\end{figure}\n\n\\newpage\n\n\\section*{The number of diagonals of an $n-sided$ convex regular polygon}\n\nSuppose a geometrical locus of $n$ points on the plane, \n% Steps of the proof\ni.e. a set of points $A_1, A_2, ..., A_n$ \nsuch that $A_1, A_2, ..., A_n$ create an $n-sided$ convex regular polygon.\nThe number of different abscissas in the geometrical locus is ${n \\choose 2}$ \nand denoted as $N_a$. It implies that no abscissa in the locus is given by more than two points, \ni.e. each point is unique. Likewise, $\\overlinesegment{A_1A_2} \\equiv \\overlinesegment{A_2A_1}$ \nholds for any 2 points in the locus, thus such abscissas are counted as one. It can be inferred \nthat the number of diagonals, denoted as $N_D$, is the same as \\textit{the difference of the number of \nabscissas and the number of sides}: \n\n% End of the proof\n$$N_D = N_s - n$$\n\n\\section*{The similarity coefficient}\n\n% Start of the proof\nSuppose a scalene triangle $\\Delta_{ABC}$. We assume that there exists a triangle $\\Delta_{A'B'C'}$,\nsuch that $\\Delta_{ABC} \\cong \\Delta_{A'B'C'}$. It implies that there exists some constant $c$,\nsuch that any abscissa created in the locus of points (from the original triangle) is equal to the\nproduct of $c$ and the corresponding abscissa of the similar triangle. Symbolically: \n\n% General expression\n$\\exists! c \\in \\mathbb{Q}: |V_1 V_2| = c \\times |V_1' V_2'| \\land c \\gneqq 1$, where $V_1, V_2$ \nare the vertices of the original triangle.\n\n% Section 1 - perimeters\nThen, it can be easily inferred, that for the \\textbf{perimeters} of the triangles \nholds the following: Let $P = \\overlinesegment{AB} + \\overlinesegment{BC} + \\overlinesegment{AC}$, similarly \n$P' = \\overlinesegment{A'B'} + \\overlinesegment{B'C'} + \\overlinesegment{A'C'}$. Likewise\n% Enumerating the similar sides with a constant\n$\n\\overlinesegment{AB} = c \\times \\overlinesegment{A'B'} \\land \n\\overlinesegment{BC} = c \\times \\overlinesegment{B'C'} \\land\n\\overlinesegment{AC} = c \\times \\overlinesegment{A'C'}\n$. \\\\\n\n% Steps of computation - perimeter\n$P = c \\times \\overlinesegment{A'B'} + c \\times \\overlinesegment{B'C'} + c \n\\times \\overlinesegment{A'C'} \\Rightarrow$\n$P = c \\times \\Big( \\overlinesegment{A'B'} + \\overlinesegment{B'C'} + \\overlinesegment{A'C'} \\Big) = c \\times P'$\n\n$$P = c \\times P'$$.\n\n% Conclusion of section 1\nIt implies that the ratio of the lengths of the sides of the similar triangles, \ni.e. the ratio of their perimeters, is equal to the similarity coefficient:\n$c = \\cfrac{P}{P'}$. \\\\\n\n% Section 2 - areas\nSimilar methodology is applied to the similarity coefficient of the areas of the triangles:\n\n% Steps of computation - area\nLet $A = \\dfrac{\\overlinesegment{AB} \\times \\overlinesegment{H_{AB}}}{2} \\land \nA' = \\dfrac{\\overlinesegment{A'B'} \\times \\overlinesegment{H'_{AB}}}{2}$,\nwhere $H_{AB}$ and $H'_{AB}$ are the heights of the triangles. Thus,\n$A = \\dfrac{c \\times \\overlinesegment{A'B'} c \\times \\overlinesegment{H'_{AB}}}{2} = \nc^2 \\times \\Bigg( \\dfrac{\\overlinesegment{A'B'} \n\\times \\overlinesegment{H'_{AB}}}{2} \\Bigg) = c^2 \\times A'$\n\n$$A = c^2 \\times A'$$\n\n% Conclusion of section 2\nIt implies, that the ratio of the areas equals to the similarity coefficient \ntaken to the second power: $c^2 = \\cfrac{A}{A'}$. \\\\\n\n% Conclusion\nTo sum up, such a methodology is also applicable to derive the similarity coefficient \nof the volume of the tetrahedron. Moreover, it is also applicable to derive the similarity coefficient\nof any planar polygon or solid figure (a scalene triangle per the given example is just an exemplary instance). \nStill, the following holds: $P = c \\times P' \\land A = c^2 \\times A' \\land V = c^3 \\times V'$ \nbased on the principles of \\textbf{congruence}.\n\n% End of the proof\n\n\\section*{Function bounded above and/or below}\n\n% Start of the proof - bounded above and/or below function\n\nSuppose a function $f$ which has a finite domain, say $D_f$. A function is bounded above if: \n$\\exists! a \\in \\mathbb{R}, \\forall x \\in D_f \\subset \\mathbb{R}: f(x) \\leq a$. That is to say,\nthat every function value is smaller or equal to some constant, say $a$. Similarly, a function\nis bounded below if: $\\exists! b \\in \\mathbb{R}, \\forall x \\in D_f \\subset \\mathbb{R}: f(x) \n\\geq b$. That implies, that every function value is greater or equal to some constant, say $b$.\nLastly, a function is bounded if both of the previous statements hold, i.e.:\n$\\exists! a \\in \\mathbb{R}, \\exists! b \\in \\mathbb{R}, \\forall x \\in D_f \\subset \\mathbb{R}:\nf(x) \\leq a \\land f(x) \\geq b$. Hence, some bounded function $f$ has the following range of \nvalues: $[b;a]$, where $\\{a,b\\} \\in \\mathbb{R}$. For example, the function $f(x) = sin(x)$ \nis bounded, such that $[-1;1]$ is its range of values. \\\\\n\n% Sample sin(x) plot\n\\begin{figure}[htp]\n \\begin{center} % Center the image\n \\begin{tikzpicture}\n \\begin{axis}[domain=0:2*pi,samples=100,smooth,xlabel={$x$},ylabel={$f(x)$}]\n \\addplot[color=blue] {sin(deg(x))}; % sine function\n \\addplot[color=black] {0}; % vertical line y = 0\n \\end{axis}\n \\end{tikzpicture}\n \\end{center}\n\n % Caption the figure\n \\caption{$sin(x)$, where $x \\in [0;2 \\pi]$}\n\\end{figure}\n\n% End of the proof\n\n\\newpage\n\n\\section*{The standard rules of logarithms}\n\n% Start of the proof - standard rules of logarithms\n\nThe following rules of logarithms hold for $x,y,a,s$, such that:\n\n% Variable enumeration\n\\[\\forall x,y \\in\\mathbb{R^+}, \\forall a > 0 \\land a \\neq 1, \\forall s \\in\\mathbb{R}\\]\n\n\\begin{enumerate}\n \\item{$\\boldsymbol{\\log_a 1 = 0}$} % Rule number 1\n\n For any $a$ holds the following $a^0 = 1$. Therefore $log_a 1 = 0$.\n \n \\item{$\\boldsymbol{\\log_a a = 1}$} % Rule number 2\n\n For any $a$ holds the following $a^1 = a$. Therefore $log_a a = 1$.\n\n \\item{$\\boldsymbol{a^{\\log_a x} = x}$} % Rule number 3\n\n Let $l = \\log_a x$. Therefore $a^{l} = x$. Combining the two rules, we obtain the following:\n $a^{\\log_a x} = x$.\n\n \\item{$\\boldsymbol{\\log_a (x \\times y) = \\log_a x + \\log_a y}$} % Rule number 4\n \n Suppose some $x,y$, where $x = a^{\\log_a x}$, $y = a^{\\log_a y}$. If we compute the product\n of them, we obtain that $x \\times y = a^{\\log_a x} \\times a^{\\log_a y}$. Furthermore, we \n have that $a^{\\log_a x + \\log_a y} = x \\times y$. According to the previous rule, we \n obtain that $\\log_a (x \\times y) = \\log_a x + \\log_a y$.\n\n \\item{$\\boldsymbol{\\log_a \\Big( \\frac{x}{y} \\Big) = \\log_a x - \\log_a y}$} % Rule number 5\n \n Similarly, suppose some $x,y$, where $x = a^{\\log_a x}$, $y = a^{\\log_a y}$. If we compute \n the quotient of them, we obtain that $\\cfrac{x}{y} = \\cfrac{a^{\\log_a x}}{a^{\\log_a y}}$. \n Furthermore, we obtain that $a^{\\log_a x - \\log_a y} = \\cfrac{x}{y}$. According to the third \n rule, we have that $\\log_a \\big(\\frac{x}{y}\\big) = \\log_a x - \\log_a y$.\n\n \\item{$\\boldsymbol{\\log_a x^{s} = s \\times \\log_a x}$} % Rule number 6\n\n Suppose some $x$, where $x = a^{\\log_a x}$. Power both sides of the equation by $s$, we \n obtain that $x^{s} = (a^{\\log_a x})^{s}$. Therefore $a^{s \\times \\log_a x} = x^{s}$.\n Having applied the third rule, it can be inferred that $\\log_a x^{s} = s \\times \\log_a x$.\n\n \\item{$\\boldsymbol{\\log_a x = \\dfrac{\\log_b x}{\\log_b a}}$} % Rule number 7\n\n Suppose some $x$, where $x = a^{\\log_a x}$. Then, for some $\\log$ with the base $b$ and the \n argument $x$ holds $\\log_b x = \\log_b a^{\\log_a x}$. Applying the sixth rule, we obtain that\n $\\log_b x = \\log_b a \\times \\log_a x$. Divide the equation by $\\log_b a$, thus obtaining\n $\\log_a x = \\cfrac{\\log_b x}{\\log_b a}$.\n\n\\end{enumerate}\n \n% End of the pr\n\n\\newpage\n\n% Start of the proof - sum of n terms of an arithmetic sequence\n\\section*{The sum of $n$ terms of an arithmetic sequence}\n\nSuppose an arithmetic sequence (also called a progression or a series) $\\{a_{n}\\}_{n=1}^{\\infty}$. That is, a sequence of $n$ terms, where:\n$$\\exists! d \\in \\mathbb{R} - \\{0\\}, \\forall i,j \\in \\mathbb{N}: d = a_{i} - a_{j} \\land i - j = 1$$ \n\nThat implies, that there exists exactly one constant, say $d$, such that the difference between every 2\nconsecutive terms is $d$. We label the sum of the first $n$ terms of the sequence as $S_{n}=\\sum_{i=1}^n a_{i}$.\nMore explicitly, we have $S_{n} = a_{1} + a_{2} + \\dots + a_{n-1} + a_{n}$. Similarly, if we reverse the \nsequence, we obtain $S_{n} = a_{n} + a_{n-1} + \\dots + a_{2} + a_{1}$. Add the two sums, we obtain:\n\n$$2 \\times S_{n} = (a_{1} + a_{n}) + (a_{2} + a_{n-1}) + \\dots + (a_{n-1} + a_{2}) + (a_{n} + a_{1})$$\n\nObserve the similarity between the sums. E. g.\n$(1 + 4) + (2 + 3) + \\dots + (n + k) = (k + n) + \\dots + (3 + 2) + (4 + 1)$\nThus, it can be infered that $2 \\times S_{n} = n \\times (a_{n} + a_{1})$.\nLastly, we have that: $$S_{n} = \\cfrac{n \\times (a_{n} + a_{1})}{2}$$\n\n\\textbf{Conclusion:} The sum of $n$ terms of an arithmetic sequence is $\\dfrac{n}{2} (a_{n} + a_{1})$.\n\n% End of the proof\n\n% Start of the proof - sum of n terms of an arithemtic sequence by mathematical induction\n\n\\section*{The sum of $n$ terms of an arithmetic sequence (proof)}\n\nSuppose you are given a simple arithmetic sequence, where the difference between every 2 consecutive terms \n$d$ is 1. We assume the following (using the notation from the previous section): \n$$S_{n} = 1 + 2 + \\dots + (n - 1) + n = \\sum_{i=1}^{n} i = \\frac{n}{2} (n + 1)$$\n\nFormulate a statement, say $S(n)$, which is to be proven.\n\n$$S(n): \\forall n \\in \\mathbb{N}: 1 + 2 + \\dots + n = \\frac{n}{2} (n + 1)$$\n\nFirst and foremost, we let $n=1$: $S(1): 1 = \\frac{1}{2} (1 + 1)$. That is true. \\linebreak % Step 1\nSecondly, assume that $S(k)$ holds for some arbitrary $k \\in \\mathbb{N}$. % Step 2\nThat is our \\textbf{induction hypothesis} (also called the induction assumption). We obtain:\n$$S(k): 1 + 2 + \\dots + k = \\frac{k}{2} (k + 1)$$\n\nThirdly, $S(k) \\implies S(k^{+})$, so we let $n=k+1$. Steps of the proof: % Step 3\n\n$$S(k + 1): 1 + 2 + \\dots + k + (k + 1) = \\cfrac{(k+1)(k+2)}{2}$$ % Steps of the proof\n$$S(k + 1): S(k) + (k + 1) = \\cfrac{(k+1)(k+2)}{2}$$\n$$S(k + 1): \\frac{k}{2} (k + 1) + (k + 1) = \\cfrac{(k+1)(k+2)}{2}$$\n$$S(k + 1): \\cfrac{k(k+1) + 2(k+1)}{2} = \\cfrac{(k+1)(k+2)}{2}$$\n$$S(k + 1): \\cfrac{(k+1)(k+2)}{2} = \\cfrac{(k+1)(k+2)}{2}$$\n\nWe obtain that $S(k+1)$ holds and so must $S(k)$ (our induction hypothesis). That suffices to prove $S(n)$.\nFinally, we have that $S(n)$ holds for all $n \\in \\mathbb{N}$, and thus \\textbf{the proof is complete}. % Step 4\n\n% End of the proof\n\n\\end{document}\n", "meta": {"hexsha": "002b3d4ca6f17bced64c56387e4cb5978038a2c3", "size": 12391, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "study-materials/Mathematics/proofs/src/main.tex", "max_stars_repo_name": "michalspano/study-materials", "max_stars_repo_head_hexsha": "a1d69bcf84ae654ba247587f717168225aefd588", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2022-02-10T07:33:18.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-30T15:29:13.000Z", "max_issues_repo_path": "study-materials/Mathematics/proofs/src/main.tex", "max_issues_repo_name": "michalspano/study-materials", "max_issues_repo_head_hexsha": "a1d69bcf84ae654ba247587f717168225aefd588", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "study-materials/Mathematics/proofs/src/main.tex", "max_forks_repo_name": "michalspano/study-materials", "max_forks_repo_head_hexsha": "a1d69bcf84ae654ba247587f717168225aefd588", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.2226277372, "max_line_length": 188, "alphanum_fraction": 0.6503914131, "num_tokens": 4314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.819893335913536, "lm_q2_score": 0.8519527963298947, "lm_q1q2_score": 0.6985104202237826}} {"text": "\\subsection{part a}\n\nWe used get\\_fog and opt\\_app to find first order time delay transfer function (FOTF).\n\\begin{itemize}\n \\item frequency\n $$\n G(s) = e^{-1.45s} \\dfrac{0.375}{0.9587s + 1}\n $$\n \\item transfer function\n $$\n G(s) = e^{-1.39s} \\dfrac{0.375}{0.9428s + 1}\n $$\n \\item optimum\n $$\n G(s) = e^{-1.38s} \\dfrac{0.383}{s + 1.021}\n $$\n\\end{itemize}\n\\begin{figure}[H]\n \\caption{system and FOTD step responde}\n \\centering\n \\includegraphics[width=12cm]{../Figure/Q1/a/FOTD.png}\n\\end{figure}\nI used below cost function to see witch one fits better.\n$$\n\\text{Cost} = \\int_{0}^{8} \\vert G(t) - G'(t)\\vert dt,\\qquad \\text{$G'$ is FOTD transfer function}\n$$\n\\begin{itemize}\n \\item frequency\n \n Cost = 1.5949\n \\item transfer function\n \n Cost = 1.3208\n \\item optimum\n \n Cost = 1.0345\n\\end{itemize}\nOptimum hase minimum cost so we choise FOTD that used optimum function.\n", "meta": {"hexsha": "2dd00f0d70f6f2077bc552c5418d6ecdf760bb26", "size": 941, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW/HW VI/Report/Q1/a.tex", "max_stars_repo_name": "alibaniasad1999/Principle-Of-Controller-Design", "max_stars_repo_head_hexsha": "2a6285f627377a5e5edfb32c92e054ab213d311a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "HW/HW VI/Report/Q1/a.tex", "max_issues_repo_name": "alibaniasad1999/Principle-Of-Controller-Design", "max_issues_repo_head_hexsha": "2a6285f627377a5e5edfb32c92e054ab213d311a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW/HW VI/Report/Q1/a.tex", "max_forks_repo_name": "alibaniasad1999/Principle-Of-Controller-Design", "max_forks_repo_head_hexsha": "2a6285f627377a5e5edfb32c92e054ab213d311a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.1282051282, "max_line_length": 98, "alphanum_fraction": 0.6110520723, "num_tokens": 343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8723473813156294, "lm_q2_score": 0.8006920068519376, "lm_q1q2_score": 0.6984815754176438}} {"text": "\\section{Elliptic systems: linear elasticity}\n\nIn this section we review the foundations of the analysis of elliptic\n(systems of) partial differential equations (\\putindex{PDE}) and apply\nthem, again with the purpose of reminding us of the analytical tools\nin mind, to the Lamé-Navier-equations of linear elasticity. While this\nis expected to be knowledge you already bring into this class, it will\nhelp us putting the analysis of mixed finite element methods into\nperspective. A short derivation of the Lamé-Navier equations is in\n\\cref{sec:lame-navier} in the appendix. For comparison,\nconsider~\\cite{Braess97,Braess13}.\n\n\\subsection{Example: weak form of the Lamé-Navier equations}\n\\begin{Definition}{weak-lame-navier}\n The weak formulation of the Lamé-Navier boundary value problem in\n linear elasticity with homogeneous displacement boundary conditions,\n namely\n \\begin{gather}\n \\label{eq:mixedintro:lame1}\n \\begin{aligned}\n -\\div \\sigma(\\vx) &= \\vf(\\vx) & \\vx&\\in\\domain,\\\\\n \\sigma(\\vx) &= 2\\mu \\strain \\vu(\\vx) + \\lambda \\operatorname{tr} \\strain \\vu(\\vx) \\id\\\\\n% \\vu(\\vx) &= 0 & x&\\in\\Gamma_D,\n \\end{aligned}\n \\end{gather}\n is: find $\\vu\\in \\vV = \\vH^1_{0}(\\domain;\\R^d)$ such that\n \\begin{gather}\n \\label{eq:mixedintro:lame2}\n a(\\vu,\\vv) \\equiv 2\\mu\\form(\\strain \\vu, \\strain \\vv)\n + \\lambda \\form(\\div \\vu, \\div \\vv)\n = \\form(\\vf,\\vv)\n \\quad\\forall \\vv\\in \\vV.\n \\end{gather}\n Here, $\\strain \\vu = \\nicefrac12\\bigl(\\nabla u + (\\nabla u)^\\transpose\\bigr)$ is the symmetric gradient.\n\\end{Definition}\n\n\\begin{Problem}{frobenius}\n Given the vector space of square matrices $X = \\R^{d\\times d}$ with the\n Frobenius inner product\n \\begin{gather}\n \\label{eq:mixedintro:frobenius}\n \\scal(A,B) = A:B = \\sum_{ij} a_{ij}b_{ij}.\n \\end{gather}\n Show that the subspaces of symmetric and skew-symmetric matrices,\n respectively, are orthogonal to each other and $X$ is the direct sum\n of those.\n\\begin{solution}\n\\begin{enumerate}\n\\item First, we note that for every matrix $A$, there holds\n \\begin{gather}\n A = A_S + A_A, \\quad A_S = \\frac{A+A^\\transpose}2, \\quad A_A = \\frac{A-A^\\transpose}2.\n \\end{gather}\n Furthermore, we see that $A_S$ is symmetric and $A_A$ is\n skew-symmetric. The sum of two symmetric matrices is symmetric,\n the same for skew-symmetric. Thus, the sets $X_S$ and $X_A$ of\n symmetric and skew-symmetric matrices, respectively, are vector\n spaces and $X=X_S+X_A$.\n \n\\item The only matrix which is symmetric and skew-symmetric is\n zero. Thus, $X=X_S\\oplus X_A$.\n \n\\item It remains to show orthogonality. To this end, we first note\n that for any skew-symmetric matrix $A$, there holds $a_{ii}=0$\n for all diagonal elements. Thus, the inner product of $A$ with a\n symmetric matrix $S$ is\n \\begin{align}\n \\scal(A,S) &= \\sum_{i\\neq j} a_{ij}s_{ij}\\\\\n &= \\sum_{i 0$ such that\n \\begin{gather}\n \\label{eq:infsup:elliptic}\n a(u,u) \\ge \\ellipa \\norm{u}_V^2 \\qquad \\forall u\\in V.\n \\end{gather}\n\\end{Assumption}\n\nNote that typically the space $V$ and its inner product are chosen\nsuch that the assumption holds, not the other way round.\n\nThe analytical results can be summarized by\n\n\\begin{Theorem*}{fem-toolbox}{The finite element toolbox (elliptic)}\n Let \\slideref{Assumption}{coercive} hold. Let furthermore\n $V_h\\subset V$. Then, there holds\n \\begin{enumerate}\n \\item (\\putindex{Lax-Milgram lemma}) There is a unique solution to\n the weak formulation in\n \\slideref{Definition}{elliptic-abstract-weak}. By the subspace\n property, this result is inherited by the Galerkin\n approximation. There holds\n \\begin{gather}\n \\norm{u}_V \\le \\tfrac1{\\ellipa} \\norm{f}_{V^*},\n \\qquad\n \\norm{u_h}_V \\le \\tfrac1{\\ellipa} \\norm{f}_{V^*}.\n \\end{gather}\n \\item (\\putindex{Céa's lemma}) The error of the Galerkin\n approximation admits the estimate (\\putindex{quasi-optimality})\n \\begin{gather}\n \\norm{u-u_h}_V \\le \\tfrac{M}{\\ellipa}\\inf_{v_h\\in V_h}\\norm{u-v_h}_V.\n \\end{gather}\n \\end{enumerate}\n\\end{Theorem*}\n\n\\begin{intro}\n The form $a(\\cdot,\\cdot)$ is symmetric and thus semi-definite on $V$. It can\n also be bounded easily by the $H^1$-norm. But, for well-posedness of\n the weak formulation, we also require ellipticity. This question is\n indeed not trivial and rests on the fact that for a function\n $u\\in V$, such that $\\nabla u$ is skew-symmetric everywhere, there\n holds $\\strain u\\equiv 0$. Thus, such functions must be excluded by\n the boundary conditions. Note, that in particular for rigid body\n translations and rotations $\\strain u = 0$. Therefore, the Dirichlet\n boundary conditions must exclude such solutions.\n \n The condition needed for well-posedness is called Korn inequality,\n and it will be posed as an assumption. We will give a proof for a\n simple case and refer the readers to a plethora of articles on more\n complicated cases.\n\\end{intro}\n\n% \\begin{Lemma*}{korn-inequality}{Korn inequality}\n% There is a constant $c_K>0$ such that\n% \\begin{gather}\n% \\label{eq:mixedintro:korn}\n% c_K \\norm{\\vv}_{\\vH^1(\\domain)}\n% \\le \\norm{\\strain \\vv}_{\\vL^2(\\Omega)}\n% \\qquad\\forall \\vv\\in \\vH^1_0(\\domain) \\cap \\vH^2(\\domain).\n% \\end{gather}\n% \\end{Lemma*}\n\n\\begin{Problem*}{korn-inequality}{Korn inequality}\n Prove that the inequality\n \\begin{gather}\n \\label{eq:mixedintro:korn}\n c_K \\norm{\\vv}_{\\vH^1(\\domain)}\n \\le \\norm{\\strain \\vv}_{\\vL^2(\\Omega)}\n \\qquad\\forall \\vv\\in \\vH^1_0(\\domain) \\cap \\vH^2(\\domain).\n \\end{gather}\n holds with a constant\n \\begin{gather}\n c_K = \\frac1{\\sqrt2(1+c_P)},\n \\end{gather}\n where $c_P$ is the constant of the \\putindex{Poincaré-Friedrichs inequality}.\n Use the intermediate result\n \\begin{gather}\n 2 \\norm{\\strain u}^2_{L^2(\\domain)} = \\norm{\\nabla u}_{L^2(\\domain)}^2\n + \\norm{\\div u}_{L^2(\\domain)}^2,\n \\end{gather}\n and prove it by integration by parts.\n\\end{Problem*}\n\nWhile the inequality is of high importance in the mathematical and\nnumerical analysis of problems in continuum mechanics, it a peripheral\ntopic to this class. We point out, that there is a whole family of\nKorn inequalities ensuring the definiteness of the strain bilinear\nform on certain spaces. We have only stated the simplest version\nhere. See~\\cite{DuvautLions76,DesvillettesVillani02} for some indepth\nanalysis.\n\n\\begin{Problem}{elasticity-standard-h1}\n Let the space $V=H^1_0(\\domain;\\R^d)$ be equipped with its standard inner\n product $\\scal(\\vu,\\vv)$ with the bilinear form of the\n Lamé-Navier equations and the corresponding norm $\\norm{.}_{\\vH^1}$.\n\n Show using the elliptic finite element toolbox in \\slideref{Theorem}{fem-toolbox}\n \\begin{enumerate}\n \\item The weak formulation has a unique solution for which there holds\n \\begin{gather}\n \\norm{\\vu}_{\\vH^1} \\le \\frac1{2c_K\\mu} \\norm{\\vf}_{\\vH^{-1}}.\n \\end{gather}\n \\item There holds the error estimate\n \\begin{gather}\n \\norm{\\vu-\\vu_h}_{\\vH^1}\n \\le \\frac{2\\mu+d\\lambda}{2c_K\\mu}\n \\inf_{\\vv_h\\in V_h} \\norm{\\vu-\\vv_h}_{\\vH^1}.\n \\end{gather}\n Hint: there are divergence-free functions ($\\div \\vv=0$) and others.\n \\end{enumerate}\n \\begin{solution}\n \\begin{enumerate}\n \\item For well-posedness, we want to apply the Lax-Milgram\n lemma. Thus, we have to show that the bilinear form is bounded\n and elliptic.\\marginpar{Todo}\n \\item For the $H^1$-error estimate we make use of Korn's inequality to note\\marginpar{Todo: this is outdated}\n \\begin{align}\n \\norm{u}_V^2=2\\mu\\norm{\\strain u}_0^2+ \\lambda \\norm{\\div u}_0^2\\geq\n 2\\mu(c_K^2\\norm{u}_1^2-\\norm{u}_0^2)\n \\end{align}\n and on the other hand\n \\begin{align}\n \\norm{u}_V^2=2\\mu\\norm{\\strain u}_0^2+ \\lambda \\norm{\\div u}_0^2\\leq\n (2\\mu+\\lambda d^2)\\norm{u}_1^2\n \\end{align}\n where we used\n \\begin{align}\n \\norm{\\nabla \\cdot u}_0^2 = \\norm{\\sum_i \\partial_i u_i}_0^2 &\\leq (\\sum_{i,j} \\norm{\\partial_j u_i}_0\\delta_{i,j})^2\\\\\n & \\leq \\sum_{i,j} \\left(\\norm{\\partial_j u_i}_0^2\\right) d^2 \\leq \\norm{u}_1^2 d^2.\n \\end{align}\n Using Poincaré's inequality\n \\begin{align}\n \\norm{u}_0\\leq C_P |u|_1\n \\end{align}\n we further notice\n \\begin{align}\n \\norm{u}_V^2+2\\mu\\norm{u}_0^2\\leq (1+C_P^2) \\norm{u}_V^2.\n \\end{align}\n Combining these estimates gives\n \\begin{align}\n 2\\mu c_K^2\\norm{u}_1^2\\leq\n \\norm{u}_V^2+2\\mu \\norm{u}_0^2\\leq\n (1+C_P^2)\\norm{u}_V^2\\leq\n (1+C_P^2)(2\\mu+\\lambda d^2)\\norm{u}_1^2.\n \\end{align}\n and therefore\n \\begin{align}\n \\norm{u-u_h}_1^2\\leq \\inf_{v\\in V_h} \\frac{(1+C_P^2)(2\\mu+\\lambda d^2)}{2\\mu c_K^2}\\norm{u-v_h}_1^2.\n \\end{align}\n If the solution is even $H_0^1(\\Omega)$-regular, we can improve the estimate to\n \\begin{align}\n \\norm{u}_1^2\\leq \\inf_{v\\in V_h}\\frac{(\\mu+(\\lambda+\\mu) d^2)}{\\mu}\\norm{u-v_h}_1^2.\n \\end{align}\n by using the identity\n \\begin{align}\n (\\strain u, \\strain v) = (\\nabla u, \\nabla v) + (\\nabla \\cdot u, \\nabla \\cdot v).\n \\end{align}\n\n \\item A more refined estimate would be\n \\begin{align}\n \\norm{u-u_h}_1^2\\leq \\inf_{v\\in V_h}\\frac{(1+C_P^2)}{2\\mu c_K^2}\\left(2\\mu\\norm{u-v_h}_1^2+\\lambda \\norm{\\nabla\\cdot (u-v_h)}_0^2\\right).\n \\end{align}\n Choosing a divergence-preserving interpolation operator, the second term vanishes.\n \\end{enumerate}\n\\end{solution}\n\\end{Problem}\n\nIf the bilinear form is symmetric, there is an even more elegant way of\ndoing the analysis: we can use the bilinear form as an inner product\nitself. This leads to our second, albeit somewhat more restricted toolbox.\n\n\\begin{Theorem*}{fem-toolbox-riesz}{The finite element toolbox (symmetric)}\n Let $a(\\cdot,\\cdot)$ be a symmetric, positive definite bilinear form\n on a space $V$. We equip $V$ with this bilinear form and the \\define{energy norm}:\n \\begin{gather}\n \\scal(u,v)_V = a(u,v)\n ,\\qquad \\norm{v}_V = \\sqrt{a(v,v)} \\qquad \\forall u,v\\in V.\n \\end{gather}\n Let furthermore $V_h\\subset V$. Then, there holds\n \\begin{enumerate}\n \\item (Riesz representation theorem) There is a unique solution to\n the continuous and discrete weak formulations in\n \\slideref{Definition}{elliptic-abstract-weak}, respectively. There holds\n \\begin{gather}\n \\norm{u}_V = \\norm{f}_{V^*},\n \\qquad\n \\norm{u_h}_V = \\norm{f}_{V^*}.\n \\end{gather}\n \\item (\\putindex{Céa's lemma}) The approximation error admits the\n estimate\n \\begin{gather}\n \\norm{u-u_h}_V \\le \\inf_{v_h\\in V_h}\\norm{u-v_h}_V.\n \\end{gather}\n \\end{enumerate}\n\\end{Theorem*}\n\n\\begin{Problem}{elasticity-standard}\n Let the space $V=H^1_0(\\domain;\\R^d)$ be equipped with the inner\n product $\\scal(\\vu,\\vv) = a(\\vu,\\vv)$, the bilinear form of the\n Lamé-Navier equations and the corresponding norm $\\norm{.}_\\vV$.\n\n Show using the symmetric finite element toolbox \\slideref{Theorem}{fem-toolbox-riesz}\n \\begin{enumerate}\n \\item The weak formulation has a unique solution for which there holds\n \\begin{gather}\n \\norm{\\vu}_\\vV \\le \\norm{\\vf}_{\\vV^*}.\n \\end{gather}\n \\item The ``energy estimate'' for conforming finite element\n approximation with a space $\\vV_h\\subset \\vV$\n \\begin{gather}\n \\norm{\\vu-\\vu_h}_\\vV = \\inf_{\\vv_h\\in \\vV_h} \\norm{\\vu-\\vv_h}_\\vV.\n \\end{gather}\n \\end{enumerate}\n\\begin{solution}\nWe consider the problem:\n Find $u\\in V = H^1_{\\Gamma_D}(\\domain;\\R^d)$ such that\n \\begin{gather}\n a(u,v) \\equiv 2\\mu\\form(\\strain u, \\strain v)\n + \\lambda \\form(\\div u, \\div v)\n = \\form(f,v)\n \\quad\\forall v\\in V.\n \\end{gather}\n where the norm on $V$ is given through $a$ as\n \\begin{align}\n \\norm{u}_V^2=a(u,u)=2\\mu\\norm{\\strain u}_0^2+ \\lambda \\norm{\\div u}_0^2\n \\end{align}\n \\begin{enumerate}\n \\item Testing symmetrically ($v=u$) gives\n \\begin{align}\n \\norm{u}_V=\\frac{a(u,u)}{\\norm{u}_V}= \\frac{\\form(f,u)}{\\norm{u}_V} \\leq \\norm{f}_{V^*}.\n \\end{align}\n \\item Applying Céa's lemma gives\n \\begin{align}\n \\norm{u-u_h}_V&=\\frac{a(u-u_h, u-u_h)}{\\norm{u-u_h}_V}\\\\\n &=\\frac{a(u-u_h, u-v_h)}{\\norm{u-u_h}_V}\\leq \\norm{u-v_h}_V\\quad \\forall v_h \\in V_h\n \\end{align} and hence $\\norm{u-u_h}_V = \\inf_{v_h\\in V_h} \\norm{u-v_h}_V$.\n \\item For the $H^1$-error estimate we make use of Korn's inequality to note\n \\begin{align}\n \\norm{u}_V^2=2\\mu\\norm{\\strain u}_0^2+ \\lambda \\norm{\\div u}_0^2\\geq\n 2\\mu(c_K^2\\norm{u}_1^2-\\norm{u}_0^2)\n \\end{align}\n and on the other hand\n \\begin{align}\n \\norm{u}_V^2=2\\mu\\norm{\\strain u}_0^2+ \\lambda \\norm{\\div u}_0^2\\leq\n (2\\mu+\\lambda d^2)\\norm{u}_1^2\n \\end{align}\n where we used\n \\begin{align}\n \\norm{\\nabla \\cdot u}_0^2 = \\norm{\\sum_i \\partial_i u_i}_0^2 &\\leq (\\sum_{i,j} \\norm{\\partial_j u_i}_0\\delta_{i,j})^2\\\\\n & \\leq \\sum_{i,j} \\left(\\norm{\\partial_j u_i}_0^2\\right) d^2 \\leq \\norm{u}_1^2 d^2.\n \\end{align}\n Using Poincaré's inequality\n \\begin{align}\n \\norm{u}_0\\leq C_P |u|_1\n \\end{align}\n we further notice\n \\begin{align}\n \\norm{u}_V^2+2\\mu\\norm{u}_0^2\\leq (1+C_P^2) \\norm{u}_V^2.\n \\end{align}\n Combining these estimates gives\n \\begin{align}\n 2\\mu c_K^2\\norm{u}_1^2\\leq\n \\norm{u}_V^2+2\\mu \\norm{u}_0^2\\leq\n (1+C_P^2)\\norm{u}_V^2\\leq\n (1+C_P^2)(2\\mu+\\lambda d^2)\\norm{u}_1^2.\n \\end{align}\n and therefore\n \\begin{align}\n \\norm{u-u_h}_1^2\\leq \\inf_{v\\in V_h} \\frac{(1+C_P^2)(2\\mu+\\lambda d^2)}{2\\mu c_K^2}\\norm{u-v_h}_1^2.\n \\end{align}\n If the solution is even $H_0^1(\\Omega)$-regular, we can improve the estimate to\n \\begin{align}\n \\norm{u}_1^2\\leq \\inf_{v\\in V_h}\\frac{(\\mu+(\\lambda+\\mu) d^2)}{\\mu}\\norm{u-v_h}_1^2.\n \\end{align}\n by using the identity\n \\begin{align}\n (\\strain u, \\strain v) = (\\nabla u, \\nabla v) + (\\nabla \\cdot u, \\nabla \\cdot v).\n \\end{align}\n\n \\item A more refined estimate would be\n \\begin{align}\n \\norm{u-u_h}_1^2\\leq \\inf_{v\\in V_h}\\frac{(1+C_P^2)}{2\\mu c_K^2}\\left(2\\mu\\norm{u-v_h}_1^2+\\lambda \\norm{\\nabla\\cdot (u-v_h)}_0^2\\right).\n \\end{align}\n Choosing a divergence-preserving interpolation operator, the second term vanishes.\n \\end{enumerate}\n\\end{solution}\n\\end{Problem}\n\n\\begin{Theorem*}{interpolation-elliptic}{Interpolation estimate}\n Let $\\mesh_h$ be a finite element mesh of size $h$, that is, the\n maximal diameter of a mesh cell is $h$. Let the finite element space\n be chosen piecewise polynomial such that\n \\begin{gather}\n V_h = \\bigl\\{ v \\in H^1_0(\\domain) \\;\\big|\\;\n v_{|\\cell} \\in \\mathcal P(\\cell) \\;\\forall \\cell\\in\\mesh_h \\bigr\\}.\n \\end{gather}\n For each cell $\\cell\\in\\mesh_h$, let $\\mathcal P(\\cell)$ be a\n polynomial space containing $\\P_k$. Then, there is an interpolation\n operator $I_h\\colon H^1_0(\\domain)\\cap H^{k+1}(\\domain)\\to V_h$,\n such that for $m\\le k$ there holds\n \\begin{gather}\n \\abs{u-I_h u}_{m} \\le c_I h^{k+1-m}\\abs{u}_{k+1},\n \\end{gather}\n with an \\define{interpolation constant} $c_I$ independent of $h$ and\n $u$, but depending on $k$, $m$, and \\putindex{shape regularity} of the mesh.\n\\end{Theorem*}\n\n\\begin{Theorem}{elliptic-convergence}\n On a sequence of meshes $\\mesh_h$ with $h\\to 0$ and uniform shape\n regularity, the error of the finite element solution $\\vu_h$ of the\n Lamé-Navier equations is bounded by\n \\begin{gather}\n \\norm{\\vu-\\vu_h}_{1} \\le c_I h^k \\frac{2\\mu+d\\lambda}{2c_K\\mu}\n \\abs{\\vu}_{k+1},\n \\end{gather}\n if the true solution $\\vu$ is sufficiently regular.\n\\end{Theorem}\n\n\\subsection{Where things go wrong}\n\\begin{example}\n We study the finite element approximation of the following problem:\n a square sheet of elastic material is hanging from the top, subject\n to gravity acting as body force pointing downward. We choose $\\mu=1$\n and vary $\\lambda$ from 1 to\n $10^5$. Figure~\\ref{fig:elasticity-compressibility} shows\n approximations with standard bilinear finite elements (red) and a\n ``good'' approximation (blue). In fact, the red solution for\n $\\lambda = 10^5$ is almost identical with the undeformed\n configuration, although the material is not as hard. This phenomenon\n was discovered early in finite element history and is called\n ``locking''.\n\\end{example}\n\n\\begin{figure}[tp]\n \\centering\n \\subfigure[{$\\lambda = 1$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-0}}\n \\subfigure[{$\\lambda = 10$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-1}}\n \\subfigure[{$\\lambda = 100$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-2}}\n \\subfigure[{$\\lambda = 1000$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-3}}\n \\subfigure[{$\\lambda = 10000$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-4}}\n \\subfigure[{$\\lambda = 100000$}]{\\includegraphics[width=.45\\textwidth]{./graph/elasticity/stalactite-5}}\n \\caption{Approximation with standard finite elements (red) and\n ``good'' elements (blue) for $\\mu=1$ and different values of\n $\\lambda$.}\n \\label{fig:elasticity-compressibility}\n\\end{figure}\n\n\\begin{example}\n In a second example, we cook up a right hand side function $\\vf$\n such that we know the solution to the Lamé-Navier equations. Indeed,\n we take a given solution $\\vu$ and apply the differential operator\n to compute $\\vf$.\n\n The \\cref{fig:elasticity-locking} shows the $H^1$-seminorm of the\n error over the mesh size (represented by its negative dyadic\n logarithm). The black triangle denotes the convergence order $h$,\n the expected convergence of the bilinear finite elements used in\n this example.\n\n We see that for $\\lambda=1$ and $\\lambda=10$, we get the expected\n curves with the correct convergence rate. For larger values of\n $\\lambda$ things seem to go wrong. There is no convergence in the\n beginning, but it seems that the expected rates are recovered on\n finer meshes.\n\\end{example}\n\n\\begin{figure}[tp]\n \\centering\n \\includegraphics[width=.9\\textwidth]{mixed/graph/elasticity/locking}\n \\caption{Errors for a manufactured solution with different values of $\\lambda$.}\n \\label{fig:elasticity-locking}\n\\end{figure}\n\n\\begin{intro}\n As we could see in the preceding examples, approximation\n of the solution to the Lamé-Navier equations becomes difficult, if\n $\\lambda \\gg \\mu$. In this case, the material is called almost\n incompressible, since the divergence measures compression or\n dilation and the dominating divergence term forces the divergence of\n the solution to be small. These cases are important in engineering\n and they initiated a lot of the research that resulted in the topics\n of this class. The phenomenon observed here, that the approximation\n of the solution to an elasticity problem deteriorates with\n increasing $\\lambda$ is called \\define{locking}.\n\\end{intro}\n\n\\begin{Problem}{divergence-q1}\n On $\\domain = (-1,1)^2$ use the finite element mesh\n \\begin{center}\n \\includegraphics[width=.25\\textwidth]{./fig/patch1m.tikz}, \n \\end{center}\n and on each cell the bilinear shape function space $\\Q_1$, that is\n \\begin{gather}\n \\vV_h = \\bigl\\{ \\vv\\in \\vH^1_0(\\domain) \\;\\big|\\;\n \\vv_{|\\cell_i} \\in \\Q_1\\quad i=1,2,3,4\\bigr\\}.\n \\end{gather}\n Show that for functions in $\\vV_h$ there holds\n \\begin{gather}\n \\norm{\\div \\vv_h}_{L^2(\\domain)} = 0\n \\qquad\\Rightarrow\\qquad\n \\vv_h = 0.\n \\end{gather}\n\\end{Problem}\n\n\\subsection{A mixed formulation}\n\n\\begin{Definition}{robust-discretization}\n We call a discretization of a parameterized (system of) PDE robust\n with respect to the parameters or simply \\define{robust}, if in an\n error estimate of the abstract form\n \\begin{gather}\n \\norm{u-u_h}_X \\le c h^\\alpha \\norm{u}_Y\n \\end{gather}\n the constant $c$ is independent of the parameters.\n\\end{Definition}\n\nA way to approach this problem is the introduction of an auxiliary variable\n\\begin{gather}\n p = -\\lambda \\div u.\n\\end{gather}\nEntering this definition into the Lamé-Navier equations, we obtain\nthe following weak formulation.\n\n\\begin{Definition}{displacement-pressure}\n The \\define{displacement-pressure formulation} of the Lamé-Navier\n equations reads: find a pair $(u,p) \\in V\\times Q$ such that\n \\begin{subequations}\n \\begin{gather}\n \\label{eq:lame-navier-mixed}\n \\begin{aligned}\n 2\\mu\\form(\\strain u, \\strain v) &- \\form(p,\\div v) &=&\\form(f,v)\n &\\forall v&\\in V = \\vH^1_0(\\domain)\\\\\n -\\form(q,\\div u) &-\\tfrac1\\lambda \\form(p,q) &=&0\n &\\forall q&\\in Q \\subset L^2(\\domain).\\\\\n \\end{aligned}\n \\end{gather}\n \\pause\n Equivalently, we write this in a single equation as\n \\begin{multline}\n 2\\mu\\form(\\strain u, \\strain v) - \\form(p,\\div v)\n - \\form(q,\\div u) -\\tfrac1\\lambda \\form(p,q)\n = \\form(f,v)\n \\\\\n \\qquad\\forall v\\in V, q\\in Q,\n \\end{multline}\n \\pause\n or in the nonsymmetric, (semi-)definite version\n \\begin{multline}\n 2\\mu\\form(\\strain u, \\strain v) + \\form(p,\\div v)\n - \\form(q,\\div u) +\\tfrac1\\lambda \\form(p,q)\n = \\form(f,v)\n \\\\\n \\qquad\\forall v\\in V, q\\in Q,\n \\end{multline} \n \\end{subequations}\n\\end{Definition}\n\n\\begin{remark}\n The three forms have different purposes and will be used\n accordingly. The first one highlights the fact that we now have a\n system of equations, each equation tested with its own test\n function. The second and third stress the fact that we now have a\n bilinear form on the product space $X=V\\times Q$.\n \n The second form is symmetric, but we will see later that is\n indefinite. Thus, non of our tools from functional analysis\n apply. In contrast, the third form is nonsymmetric, but we have\n \\begin{multline}\n 2\\mu\\form(\\strain u, \\strain u) + \\form(p,\\div u)\n - \\form(p,\\div u) +\\tfrac1\\lambda \\form(p,p)\n \\\\\n = 2\\mu\\form(\\strain u, \\strain u) + \\tfrac1\\lambda \\form(p,p)\n \\ge \\tfrac{2\\mu}{c_K} \\norm{u}_{H^1}^2 + \\tfrac1\\lambda \\form(p,p).\n \\end{multline}\n Thus, we have ellipticity with respect to the norm\n \\begin{gather}\n \\norm{(u,p)}_X^2 = \\norm{u}_{H^1}^2 + \\norm{p}_{L^2}^2.\n \\end{gather}\n Nevertheless, the ellipticity constant depends on $\\lambda$, and for\n large $\\lambda$, we loose sharpness of estimates again.\n\\end{remark}\n\n\\begin{Definition}{lame-navier-strong}\n Integrating the first equation by parts, we obtain the\n \\textbf{strong form} of the \\putindex{displacement-pressure\n formulation}\n \\begin{gather}\n \\arraycolsep.2ex\n \\begin{matrix}\n% Check signs\n - 2\\mu \\div \\strain u &+& \\nabla p &=& f \\\\\n \\div u &+& \\tfrac1\\lambda p &=& 0\n \\end{matrix}\n \\end{gather}\n\\end{Definition}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Stokes equations}\n\n\\begin{intro}\n When we write the Lamé-Navier equations in \\putindex{displacement-pressure formulation} according to\n \\slideref{Definition}{displacement-pressure}, there is no\n parameter $\\lambda$ tending to infinity when the material becomes\n less and less compressible. Instead, there is the parameter\n $1/\\lambda$ tending to zero. Thus, we can simply consider the case\n of incompressible material by setting $1/\\lambda=0$ or, in our\n abstract framework~\\eqref{eq:mixedintro:1} setting $c(\\cdot,\\cdot) = 0$. The\n resulting system is not only important for incompressible\n elasticity, but also models the slow flow of a very viscous liquid,\n so called \\putindex{creeping flow}.\n\\end{intro}\n\n\\begin{Definition}{stokes-eq1}\n The \\define{Stokes equations} in strong form are\n \\begin{gather}\n \\label{eq:mixedintro:stokes-strong1}\n \\arraycolsep.2ex\n \\begin{matrix}\n -2\\mu \\div \\strain \\vu &+& \\nabla p &=& \\vf \\\\\n \\div \\vu && &=& 0.\n \\end{matrix}\n \\end{gather}\n In weak form, they are: find $\\vu\\in \\vV \\subset \\vH^1(\\domain)$\n and $p\\in Q \\subset L^2(\\domain)$ such that\n \\begin{gather}\n \\label{eq:mixedintro:stokes-weak1}\n \\begin{aligned}\n 2\\mu\\form(\\strain \\vu, \\strain \\vv) &- \\form(\\div \\vv,p) &=&\\form(\\vf,\\vv)\n + \\text{bdry}\n &\\forall \\vv&\\in \\vV\\\\\n -\\form(\\div \\vu,q) & &=&0+ \\text{bdry}\n &\\forall q&\\in Q.\\\\ \n \\end{aligned}\n \\end{gather}\n The subspaces $\\vV$ and $Q$ are determined by boundary conditions.\n\\end{Definition}\n\n\\begin{Definition}{solenoidal}\n A vector-valued function $\\vu$ is called \\define{divergence-free} or\n \\define{solenoidal}, if there holds\n \\begin{gather}\n \\div \\vu = 0.\n \\end{gather}\n Flow described by a solenoidal function is called\n \\define{incompressible}.\n\\end{Definition}\n\n\\begin{Lemma}{stokes-equivalence}\n Let $\\vV=\\vH^1_0(\\domain)\\cap \\vH^2(\\domain)$. Then, for any\n solenoidal function $\\vu\\in vV$ there holds\n \\begin{gather}\n \\label{eq:mixedintro:6}\n 2\\mu\\form(\\strain \\vu, \\strain \\vv) = \\mu\\form(\\nabla \\vu, \\nabla \\vv)\n \\qquad\\forall \\vv\\in \\vV.\n \\end{gather}\n\\end{Lemma}\n\n\\begin{proof}\n We have ('$:$' denoting the \\putindex{double contraction} of\n \\putindex{Frobenius inner product})\n \\begin{align}\n \\strain \\vu: \\strain \\vv\n &= \\frac14\\sum_{i,j=1}^d\\left[ (\\d_i u_j+\\d_j u_i) (\\d_i v_j+\\d_j v_i)\n \\right]\n \\\\\n &= \\frac12\\sum_{i,j=1}^d\\left[ \\d_i u_j\\d_iv_j + \\d_i u_j\\d_j v_i\\right].\n \\end{align}\n The first term is the desired result, thus we have to eliminate the\n other one. We integrate by parts to obtain\n \\begin{align}\n \\int_\\domain \\d_i u_j\\d_j v_i \\dx\n &= - \\int_\\domain \\d_{ij}u_j v_i \\dx\n% \\\\ &\n = \\int_\\domain \\d_j u_j \\d_i v_i \\dx \n \\end{align}\n Entering in the\n previous equation and summing over $i$ and $j$, we obtain\n \\begin{gather}\n 2(\\strain \\vu,\\strain \\vv)\n = (\\nabla \\vu,\\nabla \\vv)+(\\div \\vu,\\div \\vv)\n = (\\nabla \\vu,\\nabla \\vv).\n \\end{gather}\n\\end{proof}\n\n\\begin{intro}\n In order to simplify subsequent discussion, it is customary to use\n the previous lemma to simplify the Stokes equations and to replace\n the strain tensor by the gradient. As a result, we can avoid the use\n of a Korn inequality and operate directly with the inner product of\n $\\vH^1$. We note though that this formulation, while mathematically\n simpler, is physically wrong if $\\vu\\neq \\vH^1_0(\\domain)$, that is,\n having nonzero boundary values.\n\\end{intro}\n\n\\begin{Definition}{stokes-eq2}\n The simplified \\define{Stokes equations} in strong form are\n \\begin{gather}\n \\label{eq:mixedintro:stokes-strong2}\n \\arraycolsep.2ex\n \\begin{matrix}\n - \\nu \\Delta \\vu &+& \\nabla p &=& \\vf \\\\\n \\div \\vu && &=& 0.\n \\end{matrix}\n \\end{gather}\n In weak form, they are: find $\\vu\\in \\vV \\subset \\vH^1(\\domain)$\n and $p\\in Q \\subset L^2(\\domain)$ such that\n \\begin{gather}\n \\label{eq:mixedintro:stokes-weak2}\n \\begin{aligned}\n \\nu\\form(\\nabla \\vu, \\nabla \\vv) &- \\form(\\div \\vv,p) &=&\\form(\\vf,\\vv)\n + \\text{bdry}\n &\\forall \\vv&\\in \\vV\\\\\n -\\form(\\div \\vu,q) & &=&0+ \\text{bdry}\n &\\forall q&\\in Q.\\\\ \n \\end{aligned}\n \\end{gather}\n The subspaces $V$ and $Q$ are determined by boundary conditions.\n\\end{Definition}\n\n\\begin{Definition}{stokes-boundary2}\n Typical boundary conditions for the Stokes problem are\n \\begin{xalignat}3\n \\text{no-slip:}&& \\vu &= 0,\\\\\n \\text{free:}&& \\d_n \\vu + p \\n &= 0 ,\\\\\n \\text{slip:}&& \\vu_n &= 0 & \\d_n \\vu_\\tau &= 0,\\\\\n \\text{friction:}&& \\vu_n &= 0 & \\d_n \\vu_\\tau &= \\alpha \\vu_\\tau.\n \\end{xalignat}\n Here, $\\vu_n$ and $\\vu_\\tau$ are the normal and tangential components of\n $\\vu$ at the boundary.\n\\end{Definition}\n\n\\begin{remark}\n Very much the same way as for elliptic equations, boundary\n conditions on the function $\\vu$ itself are essential boundary\n conditions which have to be incorporated into the space $V$, while\n those involving normal derivatives are the result of integration by\n parts and thus of the type of natural boundary conditions.\n\n All of the conditions above can also be imposed with nonzero data,\n where the physical meaning of such a condition might be debatable in\n some cases. Mathematically, inhomogeneous essential boundary\n conditions are achieved by lifting an arbitrary function with this\n boundary condition and modifying the right hand side of the\n equation, while inhomogeneous conditions on the normal derivative\n are implemented by boundary integrals on the right hand side.\n\\end{remark}\n\n\\begin{remark}\n Physically, the condition $\\vu_n=0$ models an impermeable wall. It\n says in particular, that no mass is lost through this boundary and\n is thus related to the first principle of mass conservation.\n\n The conditions on the tangential velocity model the fact that\n molecules very close to the wall stick to the wall. While this claim\n is not supported by a first principle, it has been verified by\n measurements to very high accuracy. Nevertheless, in the study of\n turbulent flow, a friction condition comes up quite naturally.\n\\end{remark}\n\n\\begin{Lemma}{divergence-compatibility}\n For any solenoidal $\\vu$ function there holds\n \\begin{gather}\n \\int_{\\d\\domain} \\vu\\cdot \\n\\ds = 0.\n \\end{gather}\n Furthermore, if the space $\\vV$ in the Stokes equations is chosen such that for any $\\vv\\in \\vV$ there holds $\\vv_n=0$ on the whole\n boundary $\\d\\domain$, then the pressure $p$ is determined by the\n Stokes equations only up to a constant.\n\\end{Lemma}\n\n\\begin{proof}\n The first statement is simple application of the Gauss theorem\n \\begin{gather}\n \\int_{\\domain} \\div \\vu\\dx = \\int_{\\d\\domain} \\vu\\cdot \\n\\ds.\n \\end{gather}\n For the second statement, we note that the only term in the\n equations which determines the pressure is $\\form(\\div \\vv,\n p)$. Integrating by parts and using the boundary condition, we\n obtain\n \\begin{gather}\n \\int_\\domain \\div \\vv\\,p\\dx\n = -\\int_\\domain \\vv\\cdot\\nabla p\\dx +\n \\int_{\\d\\domain} \\vv\\cdot \\n \\,p\\ds\n = -\\int_\\domain \\vv\\cdot \\nabla p\\dx.\n \\end{gather}\n Since the gradient of a constant is zero, we can add any constant\n function to a given solution $p$ without changing the term\n $\\form(\\div \\vv,p)$, thus leaving $p$ determined only up to a\n constant.\n\\end{proof}\n\n\\begin{Notation}{pressure-constant}\n If in Definition~\\ref{Definition:stokes-eq2} the space $\\vV$ is chosen\n such that for all $\\vv\\in \\vV$ there holds $\\vv\\cdot \\n =0$ on the whole\n boundary $\\d\\domain$, then the pressure solution $p$ cannot be\n determined uniquely in $L^2(\\domain)$. In such cases, we choose the\n pressure space\n \\begin{gather}\n L^2_0(\\domain) = L^2(\\domain)/\\R\n = \\biggl\\{q\\in L^2(\\domain)\\bigg| \\int_\\domain q\\dx = 0\\bigg\\}. \n \\end{gather}\n\\end{Notation}\n\n\\begin{remark}\n As we could see from the preceding lemma, solvability of\n equation~\\eqref{eq:mixedintro:stokes-weak2} depends on some\n compatibility of the spaces $\\vV$ and $Q$ we had not seen in the\n elliptic case. Indeed, we need a whole new tool from functional\n analysis to replace the Lax-Milgram lemma. This tool will be studied\n in Chapter~\\ref{sec:mixed-wellposedness}.\n\\end{remark}\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: \"main\"\n%%% End: \n", "meta": {"hexsha": "2faed5bb6fddaf716dc0e439e525aa7a931d891f", "size": 34515, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "mixed/mixedintro.tex", "max_stars_repo_name": "arimiftari/notes", "max_stars_repo_head_hexsha": "737b95ed6a4163bd1d395c0379410513dcb03ef1", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "mixed/mixedintro.tex", "max_issues_repo_name": "arimiftari/notes", "max_issues_repo_head_hexsha": "737b95ed6a4163bd1d395c0379410513dcb03ef1", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-05-24T07:31:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-08-31T12:58:14.000Z", "max_forks_repo_path": "mixed/mixedintro.tex", "max_forks_repo_name": "arimiftari/notes", "max_forks_repo_head_hexsha": "737b95ed6a4163bd1d395c0379410513dcb03ef1", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2018-05-15T19:28:53.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-05T19:07:29.000Z", "avg_line_length": 39.4006849315, "max_line_length": 141, "alphanum_fraction": 0.6813849051, "num_tokens": 11375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104788995148792, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.6982585427510372}} {"text": "\\documentclass{article}\n\n\\usepackage[stdmargin, noindent]{../../rajeev}\n\\usepackage{mathtools}\n\\usepackage{dirtytalk}\n\n\\pagestyle{fancy}\n\\rhead{\\today}\n\\lhead{Math 291H HW \\#2}\n\n\\begin{document}\n\n\\begin{center}\n \\Large \\textbf{Math 291H Homework \\#2}\n\\end{center}\n\\begin{center}\n \\Large Rajeev Atla\n\\end{center}\n\n\nHonors Pledge Statement: \\say{The writeup of this submission is my own work alone.}\n\n\\problem{1.19}\n\n\\subsection*{a}\n\n\\begin{align*}\n \\pars{1, 1, 0} + t \\pars{1, -1, 2} &= \\pars{2, 0, 2} + s \\pars{-1, 1, 0} \\\\\n \\pars{t, -t, 2t} - \\pars{-s, s, 0} &= \\pars{1, -1, 2} \\\\\n \\pars{t + s, -t-s, 2t} &= \\pars{1, -1, 2} \\\\\n\\end{align*}\n\nThis implies that $t=1$, which means that $s=0$.\nResubstituting, we find the point of intersection to be \\boxed{\\pars{2, 0, 2}}.\n\n\\subsection*{b}\n\nLet $\\bm{v}_1 = \\pars{1, -1, 2}$ and $\\bm{v}_2 = \\pars{-1, 1, 0}$.\nBoth these vectors are in the plane.\nWe next find the normal vector $\\bm{n}$.\n\n$$\n\\bm{n} = \\bm{v}_1 \\times \\bm{v}_2 = \\pars{-2, -2, 0}\n$$\n\nThe general form of a plane is $\\bm{n} \\cdot \\bm{x} = d$.\nSubstituting $\\bm{x} = \\pars{1, 1, 0}$, we see that $d=-4$.\nThe equation for the plane is therefore\n\n$$\n-2 x + -2y = -4\n$$\n\nWe can simplify this a little to get\n\n$$\n\\boxed{x + y = 2}\n$$\n\n\\problem{1.20}\nDefining $\\bm{n} := \\pars{2, -1, 3}$ and $\\bm{x}_0 := \\pars{2, 0, 0}$, we can write the given equation as\n\n$$\n\\bm{n} \\cdot \\pars{\\bm{x} - \\bm{x}_0} = 0\n$$\n\nWe normalize $\\bm{n}$, getting\n\n$$\n\\bm{u} = \\frac{1}{\\sqrt{14}} \\pars{2, -1, 3}\n$$\n\nThe minimal distance between this plane and $\\bm{p}$ is then\n\n$$\n\\abs{\\pars{\\bm{x}_0 - \\bm{p}} \\cdot \\bm{u}} = \\frac{1}{\\sqrt{14}} \\abs{\\pars{2, 3, 0} \\cdot \\pars{2, -1, 3}} = \\boxed{\\frac{1}{\\sqrt{14}}}\n$$\n\n\n\\problem{1.24}\n\nLet $k \\in \\RR$, so that\n\n$$\nh_{\\bm{u}} \\pars{\\bm{x}} = k \\bm{e}_1\n$$\n\nSince $\\norm{h_{\\bm{u}} \\pars{\\bm{x}}} = \\norm{\\bm{x}}$, \\boxed{k= \\pm 7}.\n\nBy construction of the Householder reflection,\n\n\\begin{align*}\n \\bm{u} &= \\pm \\frac{\\bm{x} - k \\bm{e}_1 }{\\norm{\\bm{x} - k \\bm{e}_1}} \\\\\n &= \\pm \\frac{\\pars{5 \\pm 7, 2, 4, 2}}{\\norm{\\pars{5 \\pm 7, 2, 4, 2}}} \\\\\n &= \\boxed{\\frac{\\pars{6, 1, 2, 1}}{\\sqrt{42}}, -\\frac{\\pars{6, 1, 2, 1}}{\\sqrt{42}}, \\frac{\\pars{-1, 1, 2, 1}}{\\sqrt{10}}, - \\frac{\\pars{-1, 1, 2, 1}}{\\sqrt{10}} }\\\\\n\\end{align*}\n\n(In the first two lines, the outer $\\pm$ signs are independent of the inner ones.)\n\n\\problem{1.27}\nWe define the orthonormal basis\n\n\\begin{align*}\n \\bm{u}_1 &= \\frac{\\bm{v}_1 \\times \\bm{v}_2}{\\norm{\\bm{v}_1 \\times \\bm{v}_2}} \\\\\n &= \\frac{1}{\\sqrt{14}}\\pars{-3, -2, 1} \\\\\n \\bm{u}_2 &= \\frac{1}{\\norm{\\bm{v}_1}} \\bm{v}_1 \\times \\bm{u}_1 \\\\\n &= \\frac{1}{ \\sqrt{10}} \\pars{-1, 0, -3} \\\\\n \\bm{u}_3 &= \\bm{u}_1 \\times \\bm{u}_2 \\\\\n &= \\frac{1}{\\sqrt{35}} \\pars{3, -5, -1} \\\\\n\\end{align*}\n\nWe can then define\n\n$$\n\\bm{b} = \\bm{x}_1 - \\bm{x}_2 = \\pars{0, 3, 1}\n$$\n\nWe then have\n\n$$\nt = \\frac{\\bm{b} \\cdot \\bm{u}_2}{\\bm{v}_2 \\cdot \\bm{u}_2} = \\frac{-5}{10} = - \\frac{1}{2}\n$$\n\nThe point on the first line that corresponds to this $t$-value is\n$$\n\\boxed{\\pars{\\frac{3}{2}, -\\frac{1}{2}, -\\frac{5}{2}}}\n$$\n\nSimilarly,\n\n$$\ns = \\frac{ t \\pars{\\bm{v}_2 - \\bm{b} } \\cdot \\bm{u}_3 }{\\bm{v}_1 \\cdot \\bm{u}_3} = \\frac{- \\frac{1}{2} \\pars{-1, 0, 2} \\cdot \\pars{3, -5, -1}}{\\pars{1, -4, -2} \\cdot \\pars{3, -5, -1}} = \\frac{\\frac{5}{2}}{25} = \\frac{1}{10}\n$$\n\n$$\n\\boxed{\\pars{\\frac{33}{10}, \\frac{3}{2}, \\frac{9}{10}}}\n$$\n\nThe distance between these points is\n\n$$\n\\abs{\\bm{b} \\cdot \\bm{u}_1} = \\boxed{\\frac{5}{\\sqrt{14}}}\n$$\n\n\n\\problem{1.29}\n\n\\subsection*{a}\n\nIt's easy to see that $\\norm{\\bm{u}_i} = 1$.\nIn addition, we can compute\n\n\\begin{align*}\n \\bm{u}_1 \\cdot \\bm{u}_2 &= \\frac{1}{9} \\pars{1 \\cdot 2 + 2 \\cdot 1 - 2 \\cdot 2} = 0 \\\\\n \\bm{u}_1 \\cdot \\bm{u}_3 &= \\frac{1}{9} \\pars{1 \\cdot 2 + 2 \\cdot \\pars{-2} -2 \\cdot \\pars{-1}} = 0 \\\\\n \\bm{u}_2 \\cdot \\bm{u}_3 &= \\frac{1}{9} \\pars{2 \\cdot 2 + 1 \\cdot \\pars{-2} + 2 \\cdot \\pars{-1}} = 0 \\\\\n\\end{align*}\n\nWe see that\n\n$$\n\\bm{u}_1 \\times \\bm{u}_2 = \\frac{1}{3} \\pars{2, -2, -1} = \\bm{u}_3\n$$\n\nTherefore, this is a right-handed orthonormal basis.\n\n\\subsection*{b}\n\n$$\n\\bm{u} = \\frac{\\bm{u}_1 - \\bm{e}_1}{\\norm{\\bm{u}_1 - \\bm{e}_1}} = \\frac{\\frac{1}{3} \\pars{-2, 2, -2}}{\\frac{1}{3} \\norm{\\pars{-2, 2, -2}}} = \\boxed{\\frac{1}{\\sqrt{3}} \\pars{-1, 1, -1}}\n$$\n\n\\subsection*{c}\n\n\\begin{align*}\n h_{\\bm{u}} \\pars{\\bm{u}_2} &= \\bm{u}_2 - 2 \\pars{\\bm{u}_2 \\cdot \\bm{u}} \\bm{u} \\\\\n &= \\frac{1}{3} \\pars{2, 1, 2} - \\frac{2}{9} \\pars{-3} \\pars{-1, 1, -1} \\\\\n &= \\pars{\\frac{2}{3}, \\frac{2}{3}, \\frac{1}{3}} + \\pars{-\\frac{2}{3}, \\frac{2}{3}, -\\frac{2}{3}} \\\\\n &= \\boxed{\\pars{0, 1, 0} = \\bm{e}_2} \\\\\n h_{\\bm{u}} \\pars{\\bm{u}_3} &= \\bm{u}_3 - 2 \\pars{\\bm{u}_3 \\cdot \\bm{u}} \\bm{u} \\\\\n &= \\frac{1}{3} \\pars{2, -2, -1} - \\frac{2}{9} \\pars{-3} \\pars{-1, 1, -1} \\\\\n &= \\pars{\\frac{2}{3}, - \\frac{2}{3}, - \\frac{1}{3}} + \\pars{- \\frac{2}{3}, \\frac{2}{3}, - \\frac{2}{3}} \\\\\n &= \\boxed{\\pars{0, 0, 1} = \\bm{e}_1} \\\\\n\\end{align*}\n\n\n\n\\problem{1.32}\n\n\\subsection*{a}\n\nDefine $V := V_1 \\cap V_2$.\nSuppose we have vectors $\\bm{r}_1, \\bm{r}_2 \\in V$ and scalars $a, b \\in \\RR$.\nBy definition, $\\bm{r}_1 \\in V_1, V_2$ and $\\bm{r}_2 \\in V_1, V_2$.\nBy definition of a subspace, $a \\bm{r}_1 \\in V_1, V_2$ and $b \\bm{r}_1 \\in V_1, V_2$.\nSince these vectors are in both subspaces, we can say that\n\n$$\na \\bm{r}_1 + b \\bm{r}_2 \\in V_1, V_2\n$$\n\nThis is what we wanted to show.\n\n\\subsection*{b}\n\nIf we scale up $\\bm{z}$ by a factor of $\\alpha$, we get\n\n$$\n\\alpha \\bm{z} = \\alpha \\bm{x} + \\alpha \\bm{y}\n$$\n\nSince $\\alpha \\bm{x} \\in V_1$ and $\\alpha \\bm{x} \\in V_2$ by the definition of a subspace, $\\alpha \\bm{z} \\in V_1 + V_2$ by the construction of $V_1 + V_2$.\nSimilarly, let\n\n$$\n\\bm{z'} = \\bm{z}_1 + \\bm{z}_2 = \\bm{x}_1 + \\bm{y}_1 + \\bm{x}_2 + \\bm{y}_2 = \\pars{\\bm{x}_1 + \\bm{x}_1} + \\pars{\\bm{y}_1 + \\bm{y}_2}\n$$\n\nBy the definition of a subspace, $\\bm{x}_1 + \\bm{x}_2 \\in V_1$.\nSimilarly, $\\bm{y}_1 + \\bm{y}_2 \\in V_2$.\nTherefore, by construction, $\\bm{z}_1 + \\bm{z}_2 \\in V_1 + V_2$.\n\n\\problem{1.33}\n\nDefine $V_3 := V_1 + V_2$ and $V := V_1 \\cap V_2$.\nWe then define an orthonormal basis for $V$.\n\n$$\n\\set{\\bm{u}_1, \\dots, \\bm{u}_p}\n$$\n\nTherefore, $\\dim \\pars{V} = p$.\nWe can extend $\\set{\\bm{u}_1, \\dots, \\bm{u}_p}$ to find orthonormal bases for $V_1$ and $V_2$.\nBy definition, $V \\subseteq V_1$ and $V \\subseteq V_2$.\nTherefore, $\\spn V \\subseteq \\spn V_1$ and $\\spn V \\subseteq \\spn V_2$.\nSimilarly, $\\spn V_1 \\subseteq \\spn V_3$ and $\\spn V_2 \\subseteq V_3$.\nWe can therefore write and define\n\n\\begin{align*}\n S &:= \\set{\\bm{u}_1, \\cdots, \\bm{u}_p, \\bm{v}_1, \\cdots, \\bm{v}_q} \\\\\n T &:= \\set{\\bm{u}_1, \\cdots, \\bm{u}_p, \\bm{w}_1, \\cdots, \\bm{w}_r} \\\\\n V_1 &= \\spn \\set{\\bm{u}_1, \\cdots, \\bm{u}_p, \\bm{v}_1, \\cdots, \\bm{v}_q} \\\\\n V_2 &= \\spn \\set{\\bm{u}_1, \\cdots, \\bm{u}_p, \\bm{w}_1, \\cdots, \\bm{w}_r} \\\\\n\\end{align*}\n\nIt's easy to see that $V_1 \\subseteq \\spn \\pars{S \\cup T}$ and $V_2 \\subseteq \\spn \\pars{S \\cup T}$.\nTherefore, $V_3 \\subseteq \\spn \\pars{S \\cup T} $.\nIn addition, $\\pars{S \\cup T} \\subseteq V_3$, so $\\spn \\pars{S \\cup T} \\subseteq V_3$.\nTherefore,\n\n\\begin{align*}\n V_3 &= \\spn \\pars{S \\cup T} \\\\\n &= \\spn \\set{\\bm{u}_1, \\cdots, \\bm{u}_p, \\bm{v}_1, \\cdots, \\bm{v}_q, \\bm{w}_1, \\cdots, \\bm{w}_r} \\\\\n \\dim V_3 &= p+q+r \\\\\n\\end{align*}\n\nWe can now verify that\n\n\\begin{align*}\n \\dim V_3 + \\dim V &= \\dim V_1 + \\dim V_2 \\\\\n \\pars{p + q + r} + \\pars{p} &= \\pars{p+q} + \\pars{p+r} \\\\\n\\end{align*}\n\n\n\n\n\n\n\n\n\n\n\\end{document}", "meta": {"hexsha": "d3abcdb60e27ac76ed02550160734153498c0e3d", "size": 7633, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homework/Assignment 2/hw2.tex", "max_stars_repo_name": "RajeevAtla/Math-291", "max_stars_repo_head_hexsha": "1aab6358d90b23ef62d57d8e67ae22124961a903", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homework/Assignment 2/hw2.tex", "max_issues_repo_name": "RajeevAtla/Math-291", "max_issues_repo_head_hexsha": "1aab6358d90b23ef62d57d8e67ae22124961a903", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homework/Assignment 2/hw2.tex", "max_forks_repo_name": "RajeevAtla/Math-291", "max_forks_repo_head_hexsha": "1aab6358d90b23ef62d57d8e67ae22124961a903", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.9597069597, "max_line_length": 224, "alphanum_fraction": 0.5367483296, "num_tokens": 3475, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.6982585419947454}} {"text": "\\section*{Problem 1: Gaussian Random Projections and Inner Products}\n\nLet $\\phi(x) = \\frac{1}{\\sqrt{m}}Ax$ represent our random projection of\n$x \\in \\mathbb{R}^d$, with $A$ an $m \\times d$ projection matrix with each entry\nsampled i.i.d from $N\\left(0,1\\right)$. Note that each row of $A$ is a random\nprojection vector, $v^{(i)}$.\n\nThe \\emph{norm preservation theorem} states that for all $x \\in \\mathbb{R}^d,$\nthe norm of the random projection $\\phi\\left(x\\right)$ approximately maintains\nthe norm of the original $x$ with high probability:\n\\begin{equation}\n \\mathbb{P}\\left(\n (1 - \\epsilon) \\lVert x \\rVert^2\n \\leq\n \\lVert \\phi\\left(x\\right) \\rVert^2\n \\leq\n \\left(1 + \\epsilon\\right) \\lVert x \\rVert^2\n \\right) \n \\geq 1 - 2\\exp\\left(-\\left(\\epsilon^2 - \\epsilon^3\\right)m/4\\right),\n \\label{eqn:1thm}\n\\end{equation}\nwhere $\\epsilon \\in \\left(0,1/2\\right).$\n\nUsing the norm preservation theorem, prove that for any $u, v \\in \\mathbb{R}^d$\nsuch that $\\lVert u \\rVert \\leq 1$ and $\\lVert v \\rVert \\leq 1$,\n\\begin{equation}\n \\mathbb{P}\\left(\\lvert u \\cdot v - \\phi\\left(u\\right)\\cdot\\phi\\left(v\\right)\\rvert \\geq \\epsilon\\right) \\leq 4\\exp\\left(\n -\\left(\\epsilon^2 - \\epsilon^3\\right)m/4\n \\right)\n \\label{eqn:1}\n\\end{equation}\n\n\\subsection*{Solution}\n\\begin{proof}\n First note that\n \\begin{equation*}\n (1 - \\epsilon) \\lVert u + v \\rVert^2\n \\leq\n \\lVert \\phi\\left(u + v\\right) \\rVert^2\n \\leq\n \\left(1 + \\epsilon\\right) \\lVert u + v \\rVert^2\n \\end{equation*}\n implies that\n \\begin{equation}\n \\lVert u + v \\rVert^2 - 2\\epsilon\n \\leq\n \\lVert \\phi\\left(u + v\\right) \\rVert^2\n \\leq\n \\lVert u + v \\rVert^2 + 2\\epsilon\n \\label{eqn:1sup}\n \\end{equation}\n by triangle inequality and the assumption of the norms of $u$ and $v$.\n\n Thus, the probability of the event in Equation \\ref{eqn:1sup} than that of\n Equation \\ref{eqn:1thm}.\n \n Using this and taking the additive inverse, we have that\n \\begin{align}\n \\mathbb{P}\\left(\n \\left\\lVert \\phi(u + v) \\right\\rVert^2\n \\not\\in\n \\left[\n \\lVert u + v \\rVert^2 - 2\\epsilon,\n \\lVert u + v \\rVert^2 + 2\\epsilon\n \\right]\n \\right)\n &\\leq 2\\exp\\left(-\\left(\\epsilon^2 - \\epsilon^3\\right)m/4\\right)\n \\label{eqn:1conda}\\\\\n \\mathbb{P}\\left(\n \\left\\lVert \\phi(u - v) \\right\\rVert^2\n \\not\\in\n \\left[\n \\lVert u - v \\rVert^2 - 2\\epsilon,\n \\lVert u - v \\rVert^2 + 2\\epsilon\n \\right]\n \\right)\n &\\leq 2\\exp\\left(-\\left(\\epsilon^2 - \\epsilon^3\\right)m/4\\right).\n \\label{eqn:1condb}\n \\end{align}\n\n By the countable sub-additivity property of probability distributions, we have\n that the probability of both these events occurring is at most\n $4\\exp\\left(-\\left(\\epsilon^2 - \\epsilon^3\\right)m/4\\right)$. Thus, we are\n done if we can show\n $\\left\\{\\left\\lvert u \\cdot v - \\phi(u) \\cdot \\phi(v) \\right\\rvert \\geq\n \\epsilon\\right\\}$ subsets these two conditions.\n\n If we have the pair\n \\begin{align}\n \\left\\lVert \\phi(u + v) \\right\\rVert^2\n &\\leq \\left\\lVert\n u + v\n \\right\\rVert^2 - 2\\epsilon \\Rightarrow\n \\left\\lVert\n u + v\n \\right\\rVert^2 -\n \\left\\lVert \\phi(u + v) \\right\\rVert^2\n \\geq\n 2\\epsilon\n \\\\\n \\left\\lVert \\phi(u - v) \\right\\rVert^2\n &\\geq \\left\\lVert\n u - v\n \\right\\rVert^2 + 2\\epsilon \\Rightarrow\n \\left\\lVert\n u - v\n \\right\\rVert^2 -\n \\left\\lVert \\phi(u - v) \\right\\rVert^2\n \\geq\n 2\\epsilon,\n \\end{align}\n we can use the linearity of $\\phi$ and the expansion\n $\\lVert u \\pm v \\rVert^2 = \\lVert u \\rVert^2 + \\lVert v \\rVert^2 \\pm 2u \\cdot\n v$, we can add the two inequalities to obtain\n \\begin{align}\n 4\\left(\n u \\cdot v\n -\n \\phi\\left(u\\right)\\cdot\\phi\\left(v\\right)\n \\right)\n \\geq 4\\epsilon.\n \\nonumber\n \\end{align}\n\n Thus, we have that the conditions in Equations \\ref{eqn:1conda} and\n \\ref{eqn:1condb} imply\n $u \\cdot v - \\phi\\left(u\\right)\\cdot\\phi\\left(v\\right) \\geq \\epsilon$.\n\n Similarly, we show that the pair\n \\begin{align}\n \\left\\lVert \\phi(u + v) \\right\\rVert^2\n &\\geq \\left\\lVert\n u + v\n \\right\\rVert^2 + 2\\epsilon \n \\\\\n \\left\\lVert \\phi(u - v) \\right\\rVert^2\n &\\leq \\left\\lVert\n u - v\n \\right\\rVert^2 - 2\\epsilon\n \\end{align}\n implies\n $u \\cdot v - \\phi\\left(u\\right)\\cdot\\phi\\left(v\\right) \\leq -\\epsilon$, which\n gives us\n $\\left\\lvert u \\cdot v - \\phi\\left(u\\right)\\cdot\\phi\\left(v\\right)\n \\right\\rvert \\geq \\epsilon.$\n\\end{proof}\n", "meta": {"hexsha": "89d8c042cba08342e9a7c48b60cb133be54019a7", "size": 4526, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw3/problem1/problem1.tex", "max_stars_repo_name": "kspathak/cse547", "max_stars_repo_head_hexsha": "2379c6435c871720aa7da53d3c8066a628e81830", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw3/problem1/problem1.tex", "max_issues_repo_name": "kspathak/cse547", "max_issues_repo_head_hexsha": "2379c6435c871720aa7da53d3c8066a628e81830", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw3/problem1/problem1.tex", "max_forks_repo_name": "kspathak/cse547", "max_forks_repo_head_hexsha": "2379c6435c871720aa7da53d3c8066a628e81830", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-02-18T01:39:20.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-18T01:39:20.000Z", "avg_line_length": 30.7891156463, "max_line_length": 122, "alphanum_fraction": 0.6129032258, "num_tokens": 1731, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581097540519, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.6981715271795728}} {"text": "\\chapter{Shortest Path}\n\nGiven a weighted graph, one interesting problem is to find the\npath with minimal weight.\n\n\\section{Single Source}\n\nThis instance of the shortest path problem begins at a specific vertex\nin the graph, hence ``Single Source.''\n\n\\subsection{Dijkstra's Algorithm}\n\nGiven an acyclic graph $G=(V,E)$ in which every edge has a positive\nweight and a single vertex $s$ from which to start the path, we begin\nby labeling all vertices $u \\in E | weight(u) = \\infty$, then label\n$weight(s) = 0$. We also create a min-heap of the vertices using\nweight as key.\n\nWe pop-best a vertex $v$ from our heap. For each edge $e$ incident to\n$v$ and some other vertex $u$, we check if $weight(v) + weight(e)$ is\nless than $weight(u)$, in which case we set $weight(u) = weight(v) +\nweight(e)$ and decrease the key of $u$ in our heap. We continue this\nprocess until no vertices remain in the heap.\n\nIt should be easy to see that if we have a particular target vertex,\nwe can halt the algorithm as soon as our target is the root of the\nheap, because we have found the shortest path to it.\n\n\\subsection{Analysis of Dijkstra's Algorithm}\n\nIntuitively, this algorithm has an upper bound of $O(|V|)$ times\nhowever long it takes us to extract the minimum vertex from our heap,\nplus $O(|E|)$ times however long it takes us to decrease the key,\nsince we have to check every vertex and we may have to check every\nedge. Using a standard heap, this gives us $O(|V|log|V| +\n|E|log|V|)$.\n\n\\section{All Sources}\n\nThis instance of the shortest path problem is concerned with several\nsources, in which case Dijkstra's Algorithm does not suffice.\n\n\\subsection{Floyd-Warshall Algorithm}\n\nThe Floyd-Warshall Algorithm is\n\\hyperlink{sec:floyd_warshall}{discussed in detail in the chapter on\n Dynamic Programming}.\n\n\\section{Arbitrary Weights}\n\nA keen reader will have noticed that both Dijkstra's and Floyd\nWarshall Algorithms assume positive edge weights. Neither of which\nare generally useful when there may be edges of negative weight.\n\n\\subsection{Bellman-Ford Algorithm}\n\nThe Bellman-Ford Algorithm can be applied with edges of arbitrary\nweight. Note that even Bellman-Ford doesn't deal with cycles of\nnegative weight, since these can be used to make any path have an\narbitrarily small total weight.\n\nThe Bellman-Ford Algorithm is beyond the scope of these notes, for\nmore information please see\n\\href{https://en.wikipedia.org/wiki/Bellman-Ford}{Wikipedia's entry on\n the Bellman-Ford Algorithm}, or CLRS.\n", "meta": {"hexsha": "f8424a4c2769b197c6e14e18f7e49a4e9af7da3b", "size": 2498, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "shortest_path.tex", "max_stars_repo_name": "SteamedPears/AllTheAlgorithms", "max_stars_repo_head_hexsha": "13a04cc4a6bd8dec5e35c1a42b96680d47a98962", "max_stars_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2016-10-12T19:16:53.000Z", "max_stars_repo_stars_event_max_datetime": "2017-05-01T03:13:05.000Z", "max_issues_repo_path": "shortest_path.tex", "max_issues_repo_name": "SteamedPears/AllTheAlgorithms", "max_issues_repo_head_hexsha": "13a04cc4a6bd8dec5e35c1a42b96680d47a98962", "max_issues_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "shortest_path.tex", "max_forks_repo_name": "SteamedPears/AllTheAlgorithms", "max_forks_repo_head_hexsha": "13a04cc4a6bd8dec5e35c1a42b96680d47a98962", "max_forks_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.8484848485, "max_line_length": 70, "alphanum_fraction": 0.7646116894, "num_tokens": 624, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8774767746654976, "lm_q2_score": 0.7956581000631542, "lm_q1q2_score": 0.6981715033798943}} {"text": "\n%These notes summarize the lecture notes from the Linear Modelling course at Sheffield's School of Mathematics and Statistics, MSc degree programme. The original notes were written by Dr.\\ Kevin Walters and Dr.\\ Jeremy Oakley. This summary is completely derived from these notes and from other MSc sources. Any errors are most probably mine.\n\n%Everything is in matrix form unless a lower case letter with a subscript (such as $x_i$) is used (even there, I might deviate from this convention if I need to index sub-matrices; it's best to look at the context to decide what is meant).\n\n\n\\section{Background}\n\n\\subsection{Derivations of combinations of functions}\n\n\\begin{equation}\n (uv)' = uv' + vu'\n\\end{equation}\n\n\\begin{equation}\n(u/v)' = \t\\frac{vu' - uv'}{v^2}\n\\end{equation}\n\n\n\\subsection{Some key distributional results}\n\n\n\\subsection{Some very basic matrix algebra facts}\n\n\\textbf{Inverse (2x2)}: \n\n$\\begin{pmatrix}\na & b \\\\\nc & d\\\\\n\\end{pmatrix}^{-1}\n= \\frac{1}{ad-bc} \n\\begin{pmatrix}\nd & -b \\\\\n-c & a\\\\\n\\end{pmatrix}$\n\n\n\\textbf{Inverse of non-singular matrices}. \nIf A and B are non-singular matrices then $(AB)^{?1} = B^{-1}A^{-1}$.\n\n\\textbf{Symmetric square matrix}: $A=A^T$. \n\nIf a symmetric matrix A is non-singular then $A^{-1}$ is also symmetric.\n\n$AA^{-1}=A^{-1}A=I$ given A is square and invertible.\n \n If the symmetric matrix $A$ is non-singular then $A^{-1}$ is also symmetric.\n \n \\textbf{Multiplication is distributive}: Multiplication is distributive over addition and subtraction, so $(A-B)(C-D)=AC -BC-AD+BD$.\n\n\\textbf{Transpose}: $(A+B)^T =A^T+B^T$ and $(AB)^T =B^TA^T$\n\n\\textbf{Sum of squares}: $\\sum x_i^2 = \\mathbf{x}^T \\mathbf{x}$.\n\n\\textbf{Symmetry under multiplication}: If A is $n\\times p$, then $AA^T$ and $A^TA$ are symmetric.\n\n\n\n\n\\textbf{Trace of a square matrix}: \n\n\\begin{enumerate}\n\\item\n$tr(A)=\\sum a_{ii}$\n\\item\n$tr(A + B) = tr(A) + tr(B)$\n\\item \n$tr(cA) = ctr(A)$\n\\item\n$tr(AB) = tr(BA)$\n\\end{enumerate}\n\n\\textbf{Idempotent}: $A^2 = AA = A$. Example: $A = I_n$; this is the only non-singular idempotent matrix. \n\nIf A is idempotent and if $A\\neq I_n$, then A is singular and its trace is equal to its rank $n-p$, for some $p>0$.\n\n\\textbf{Inverse of a matrix product}:\nIf A and B are non-singular matrices then $(AB)^{-1} = B^{-1}A^{-1}$.\n\n\\textbf{Rank}: the number of linearly independent columns or rows of A.\n\nHow to determine linear independence: \n\n\\section{Basic facts}\n\n\\begin{equation}\ny=X\\beta+\\epsilon\n\\end{equation}\n\n\\begin{tabular}{@{}ll@{}ll@{}}\n$E(y) = X\\beta = \\mu$ & & $E(\\epsilon)=0$ \\\\\n$Var(y) = \\sigma^2 I_n $ & & $Var(\\epsilon) = \\sigma^2 I_n$ \\\\\n%y+X\\hat{\\beta} + e$ \\\\\n%%$E(y) = X\\beta = \\mu$ & Var\\\\\n% $\\epsilon \\sim N_p (0,\\sigma^2I_n)$ & No \\verb!\\part! divisions. \\\\\n%\\verb!article! & No \\verb!\\part! or \\verb!\\chapter! divisions. \\\\\n%\\verb!letter! & Letter (?). \\\\\n%\\verb!slides! & Large sans-serif font.\n\\end{tabular}\n\n\\begin{equation}\ny = X\\hat{\\beta} + e\n\\end{equation}\n\n\\begin{fmpage}{\\linewidth}\nNote that $S_{xx}=\\frac{1}{(X'X)^{-1}}$, and \n$G=(X'X)^{-1}$, so that $S_{xx}=\\frac{1}{G}$.\n\\end{fmpage}\n\n\\begin{tabular}{@{}ll@{}ll@{}}\nResults for $\\hat{\\beta}$ & Results for $e$\\\\\n$E(\\hat{\\beta}) = \\beta$ & $E(e) = 0$\\\\\n$Var(\\hat{\\beta}) = \\sigma^2 (X^T X)^{-1} = \\frac{\\sigma^2}{S_{xx}}=\\sigma^2G$ & $Var(e)=\\sigma^2 M$ \\\\\n$\\hat{\\beta} \\sim N_p(\\beta,\\sigma^2 (X^T X)^{-1})$ & $Var(e_i)=\\sigma^2 m_{ii} $\\\\ \n& $E(e_i^2)= \\sigma^2 m_{ii}$\\\\\n$\\hat{\\beta} = (X^T X)^{-1} X^T y$, $X$ has full rank & $E(\\sum e_i^2) = \\sigma^2 (n-p)$\\\\\n\\end{tabular}\n\n\\medskip\n\\textbf{Sum of Squares}:\n\n\\begin{equation}\nS(\\hat{\\beta}) = \\sum e_i^2 = e^T e = (y-X\\hat{\\beta})^T (y-X\\hat{\\beta}) = y^T y - y^T X \\hat{\\beta} = S_r\n\\end{equation}\n\nAlternatively: $S_r= y^Ty - \\hat{\\beta}^T X^T X\\hat{\\beta}=y^Ty - \\hat{\\beta}^T X^T y$ (see review exercises 2).\n\n\\textbf{Estimation of error variance: $e=My$}\n\n\\begin{equation}\ne = y - X\\hat{\\beta} = y - X (X^T X)^{-1} X^T y = My\n\\end{equation}\n\n\\noindent\nwhere\n\n\\begin{equation}\nM = I_n - X (X^T X)^{-1} X^T \n\\end{equation}\n\nM is symmetric, idempotent $n\\times n$.\n\nNote that $MX=0$, which means that \n\n\\begin{equation}\nE(e)=E(My) = ME(y)= MX\\beta = 0\n\\end{equation}\n\nAlso, $Var(e) = Var(My) = M Var(y) M^T = \\sigma^2 I_n M$.\n\n\\medskip\n\\textbf{Important properties of M}:\n\n\\begin{itemize}\n\\item $M$ is singular because every idempotent matrix except $I_n$ is singular.\n\\item $trace(M)=rank(M)=n-p$.\n\\end{itemize}\n\n\\medskip\n\n\n\\textbf{Residual mean square}:\n\\begin{equation}\n\\hat{\\sigma}^2 = \\frac{\\sum e_i^ 2}{n-p} \\quad E(\\hat{\\sigma}^2)=\\sigma^2\n\\end{equation}\n\nThe square root of $\\hat{\\sigma}^2$, $\\hat{\\sigma}$ is the \\textbf{residual standard error}.\nNote: The phrase ``standard error'' here should not be misinterpreted to mean standard error in the sense of ``SE''. \n\n\\textbf{Variance-covariance matrix}:\n\nIn a model like \\begin{verbatim}fm<-lm(Maint ~ Age, data = data)\\end{verbatim}, the variance-covariance matrix is:\n\n\\begin{equation}\n\\begin{pmatrix}\nVar(\\hat{\\beta}_0) & Cov(\\hat{\\beta}_0,\\hat{\\beta}_1) \\\\\nCov(\\hat{\\beta}_0,\\hat{\\beta}_1) & Var(\\hat{\\beta}_1)\\\\\n\\end{pmatrix}\n\\end{equation}\n\nThe correlation between the two parameter estimates is therefore:\n\n\\begin{equation}\nCorr(\\hat{\\beta}_0,\\hat{\\beta}_1) = \\frac{Cov(\\hat{\\beta}_0,\\hat{\\beta}_1)}{SE(\\hat{\\beta}_0) SE(\\hat{\\beta}_1)}\n\\end{equation}\n\n\nExample (tractor data):\n\n\\begin{verbatim}\n> vcov(fm)\n (Intercept) Age\n(Intercept) 21591 -4624.0\nAge -4624 1267.9\n\\end{verbatim}\n\nWe can check the correlation calculation using\n\n\\begin{verbatim}\n> cov2cor(vcov(fm))\n (Intercept) Age\n(Intercept) 1.00000 -0.88378\nAge -0.88378 1.00000\n\\end{verbatim}\n\n\n\n\\subsection{Some short-cuts for hand-calculations}\n\n\\begin{tabular}{@{}ll@{}}\n$S_{xx} = \\sum (x_i - \\bar{x})^2$ & $= \\sum x_i^2 - n\\bar{x}^2$\\\\\n$S_{yy} = \\sum (y_i - \\bar{y})^2$ & $= \\sum y_i^2 - n\\bar{y}^2$\\\\ \n$S_{xy} = \\sum (x_i -\\bar{x})(y_i -\\bar{y})$ & = $\\sum x_i y_i - n\\bar{x}\\bar{y}$\\\\\n\\end{tabular}\n\n\\begin{equation}\n\\hat{\\beta} = (X^T X)^{-1} X^T y = \n\\begin{pmatrix} \n\\bar{y} - \\bar{x} \\frac{S_{xy}}{S_{xx}}\\\\\n\\frac{S_{xy}}{S_{xx}}\n\\end{pmatrix}\n\\end{equation}\n\n\\begin{equation}\nX^T X = \\begin{pmatrix}\nn & \\sum_{i=1}^n x_i\\\\\n\\sum_{i=1}^n x_i & \\sum_{i=1}^n x_i^2\\\\\n\\end{pmatrix}\n\\end{equation}\n\n\\begin{equation}\n(X^T X)^{-1} = \\frac{1}{nS_{xx}} \n\\begin{pmatrix}\n S_{xx}+n\\bar{x}^2 & -n\\bar{x} \\\\\n -n\\bar{x} & n\\\\\n\\end{pmatrix}\n\\end{equation}\n\n\\noindent\nNote that $\\sum_{i=1}^n x_i = n\\bar{x}$.\n\n\\begin{equation}\nX^T y = \\begin{pmatrix}\nn\\bar{y} \\\\\nS_{xy} + n\\bar{x}\\bar{y}\n\\end{pmatrix}\n\\end{equation}\n\n\nSee \\cite[25]{DraperSmith} for a full exposition.\n\n\\subsection{Gauss-Markov conditions}\n\nThis imposes distributional assumptions on $\\epsilon = y - X \\beta$.\n\n$E(\\epsilon)=0$ and $Var(\\epsilon)=\\sigma^2 I_n$,\n\n\\subsection{Gauss-Markov theorem}\n\nLet $a$ be any $p \\times 1$ vector and suppose that $X$ has rank $p$. Of all estimators of $\\theta = a^T \\beta$ that are unbiased and linear functions of $y$, the estimator $\\hat{\\theta} = a^T \\hat{\\beta}$ has minimum variance. Note that $\\theta$ is a scalar.\n\nNote: no normality assumption required! But if $\\epsilon \\sim N(0,\\sigma^2)$, $\\hat{\\beta}$ have smaller variances than any other estimators.\n\n\\textbf{Minimum variance unbiased linear estimators}:\nto-do\n\n\\subsection{$R^2$ or Coefficient of determination}\n\n\\begin{tabular}{@{}ll@{}}\n$S_{TOTAL} = (y-\\bar{y})^T(y-\\bar{y})$ $= y^T y - n\\bar{y}^2$ & \\\\\n$S_{REG} = (X\\hat{\\beta}-\\bar{y})^T (X\\hat{\\beta}-\\bar{y})$ & \\\\\n$S_r = \\sum e_i^2 = (y-X\\hat{\\beta})^T (y-X\\hat{\\beta})$ & \\\\\n\\end{tabular}\n\n\\begin{equation}\nS_{TOTAL} = S_{REG}+ S_r\n\\end{equation}\n\n\\begin{equation}\nR^2 = \\frac{S_{TOTAL}-S_r}{S_{TOTAL}} = \\frac{S_{REG}}{S_{TOTAL}}\n\\end{equation}\n\nFor $y = 1_n \\beta_0 + \\epsilon$, then $R^2 = \\frac{S_{REG}}{S_{TOTAL}} = 0$ because $X\\hat{\\beta} = \\bar{y}$. So $S_{REG} = (X\\hat{\\beta} - \\bar{y})^T (X\\hat{\\beta} - \\bar{y}) = 0$.\n\nIn simple linear regression, $R^2 = r^2$. $R^2$ is a generalization of $r^2$.\n\nAdjusted $R^2= R_{Adj}^2$. $R_{Adj}^2= 1-\\frac{S_r/(n-p)}{S_{TOTAL}/(n-1)}$. \n\n$R^2$ increases with increasing numbers of explanatory variables, therefore $R_{Adj}^2$ is better. \n\n\n\\section{Hypothesis testing}\n\n\\subsection{Some theoretical background}\n\n\\textbf{Multivariate normal}:\n\nLet $X^T = < X_1,\\dots,X_p>$, where $X_i$ are univariate random variables.\n\nX has a multivariate normal distribution if and only if every component of $X$ has a univariate normal distribution.\n\n\n\\textbf{Linear transformations}:\n\nLet $A, b$ be constants. Then, $Ax + b\\sim N_q (A\\mu + b, A\\Sigma A^T)$.\n\n\\textbf{Standardization}:\n\nNote that $\\Sigma$ is positive definite (it's a variance covariance matrix), so $\\Sigma = CC^T$. \n$C$ is like a square root (not necessarily unique).\n \nIt follows ``immediately'' that \n\n\\begin{equation}\nC^{-1} (X-\\mu) \\sim N_p (0_p, I_p)\n\\end{equation}\n\nIf $\\Sigma$ is a diagonal matrix, then $X_1,\\dots,X_n$ are independent and uncorrelated.\n\n\\textbf{Quadratic forms}:\n\nRecall distributional result: If we have $n$ independent standard normal random variables, their sum of squares is $\\chi_n^2$.\n\nLt $z = C^{-1} (X-\\mu)$, and $\\Sigma=CC^T$. The sum of squares $z^T z$ is:\n\n\\begin{equation}\n\\begin{split}\nz^T z & = [C^{-1} (X-\\mu)]^T [C^{-1} (X-\\mu)]\\\\\n& = (X-\\mu)^T [C^{-1}]^T [C^{-1}](X-\\mu) \\quad \\dots (AB)^T=B^T A^T\\\\\n\\end{split}\n\\end{equation} \n\nNote that $ [C^{-1}]^T = [C^{T}]^{-1}$. Therefore, \n\n\\begin{equation}\n\\begin{split}\n[C^{-1}]^T [C^{-1}] & = [C^T]^{-1} [C^{-1}]\\\\\n& = (C^T C)^{-1}\\\\\n& = (C C^T)^{-1}\\\\\n& = \\Sigma^{-1}\\\\\n\\end{split}\n\\end{equation} \n\nTherefore: $z^T z = (X-\\mu)^T \\Sigma^{-1} (X-\\mu)\\sim \\chi_p^2$, where $p$ is the number of parameters.\n\n\\textbf{Quadratic expressions involving idempotent matrices}\n\nGiven a matrix $K$ that is idempotent, symmetric. Then:\n\n\\begin{equation}\nx^T K x = x^T K^2 x = x^T K^T K x\n\\end{equation}\n\nLet $x\\sim N_n(\\mu,\\sigma^2 I_n)$, and let $K$ be a symmetric, idempotent $n \\times n$ matrix such that $K\\mu=0$. Let $r$ be the rank or trace of $K$. Then we have the \n\n\\begin{fmpage}{\\linewidth}\n\\textbf{sum of squares property}:\n\n\\begin{equation}\nx^T K x \\sim \\sigma^2 \\chi_r^2\n\\end{equation}\n\\end{fmpage}\n\nThe above generalizes the fact that if we have $n$ independent standard normal random variables, their sum of squares is $\\chi_n^2$.\n\nTwo points about the sum of squares property:\n\\begin{itemize}\n\\item\nRecall that the expectation of a chi-squared random variable is its degrees of freedom. It follows that:\n\n\\begin{equation}\nE(x^T K x) = \\sigma^2 r \n\\end{equation}\n\nIf $K\\mu\\neq 0$, $E(x^T K x) = \\sigma^2 r+\\mu^T K\\mu$. \n\n\\item If $K$ is idempotent, so is $I-K$. This allows us to split $x^T x$ into two components sums of squares:\n\n\\begin{equation}\nx^T x = x^T K x+x^T (I-K) x\n\\end{equation}\n\\end{itemize}\n\n\\textbf{Partition sum of squares}:\n\n[helps prove independence of SSs] \n\n\\begin{enumerate}\n\\item Let $K_1, K_2,\\dots, K_q$ be \\textbf{symmetric idempotent $n \\times n$ matrices} such that\n\\item $\\sum K_i= I_n$ and \n\\item $K_iK_j =0$, for all $i\\neq j $. \n\\item Let $x\\sim N_n(\\mu, \\sigma^2I_n)$.\n \\end{enumerate}\n \n Then we have the following partitioning into \\textbf{independent} sums of squares:\n \n \\begin{equation}\nx^T x = \\sum x^T K_i x\n\\end{equation}\n\nIf $K_i \\mu = 0$, then $ x^T K_i x\\sim \\sigma^2 \\chi_{r_i}^2$, where $r_i$ is the rank of $K_i$.\n\n\\textbf{Example}:\n\n\\begin{equation}\ny^T y = y^T M y + y^T (I-M) y \n\\end{equation}\n\nLet $K_1 = M$ and $K_2 = (I-M)$. It is easy to check that all four conditions above are satisfied; therefore the sums of squares are independent.\n\nNote that \n\n\\begin{equation}\n y^T M y = e^T e \\sim \\chi^2_{n-p}\n\\end{equation}\n\nand\n\n\\begin{equation}\n y^T (I-M y) = \\hat{\\beta}^T (X^T X) \\hat{\\beta} \\sim \\chi^2_{p}\n\\end{equation}\n\nRecall distributional result: if $X\\sim \\chi_v^2, Y\\sim \\chi_w^2$ and $X,Y$ independent then $\\frac{X/v}{Y/w}\\sim F_{v,w}$. \n\nTherefore, $ \\frac{\\frac{y^T (I-M y)}{n-p}}{\\frac{y^T M y}{p}}\\sim F_{p,n-p}$.\n\n\\subsection{Confidence intervals for $\\hat{\\beta}$}\n\nNote that $\\hat{\\beta} \\sim N_p (\\beta,\\sigma^2 (X^T X)^{-1})$, and that \n$\\frac{\\hat{\\sigma}^2}{\\sigma^2} \\sim \\frac{\\chi^2_{n-p}}{n-p}$.\n\nFrom distributional theory we know that $T=\\frac{X}{\\sqrt{Y/v}}$, when $X\\sim N(0,1)$ and $Y\\sim \\chi^2_{v}$. \n\nLet \n $x_i$ be a column vector containing the values of the explanatory/regressor variables for a new observation $i$. Then if we define:\n\n\\begin{equation}\nX=\\frac{x_i^T \\hat{\\beta} - x_i^T \\beta}{\\sqrt{\\sigma^2 x_i^T (X^T X)^{-1}x_i}} \\sim N(0,1)\n\\end{equation}\n\n\\noindent\nand \n\n\\begin{equation}\nY=\\frac{\\hat{\\sigma}^2}{\\sigma^2} \\sim \\frac{\\chi^2_{n-p}}{n-p}\n\\end{equation}\n\n\nIt follows that $T=\\frac{X}{\\sqrt{Y/v}}$:\n\n\\begin{equation}\nT= \\frac{x_i^T \\hat{\\beta} - x_i^T \\beta}{\\sqrt{\\hat{\\sigma}^2 x_i^T (X^T X)^{-1}x_i}} = \n\\frac{ \\frac{x_i^T \\hat{\\beta} - x_i^T \\beta}{\\sqrt{\\sigma^2 x_i^T (X^T X)^{-1}x_i}}}{\\sqrt{\\frac{\\hat{\\sigma}^2}{\\sigma^2}}}\n \\sim t_{n-p}\n\\end{equation}\n\nI.e., a 95\\% CI:\n\n\\begin{equation}\nx_i^T \\hat{\\beta} \\pm t_{n-p,1-\\alpha/2}\\sqrt{\\hat{\\sigma}^2 x_i^T(X^T X)^{-1}x_i}\n\\end{equation}\n\nCf.\\ a prediction interval:\n\n\\begin{equation}\nx_i^T \\hat{\\beta} \\pm t_{n-p,1-\\alpha/2}\\sqrt{\\hat{\\sigma}^2 (1+x_i^T(X^T X)^{-1}x_i)}\n\\end{equation}\n\nNote that \n\n\\begin{enumerate}\n\\item\nA prediction interval will be wider about the edges; this is because the term $\\hat{\\sigma}^2 (1+x_i^T(X^T X)^{-1}x_i)$ in the prediction interval formula is minimized at the mean value of the \npredictor variable. When $x_i = \\bar{x}$ we have the smallest value for the term, and so the further away the $x_i$ value from $\\bar{x}$, the larger the interval. \n\\item \nThe width of the prediction interval stays much more constant around the range of observed values.\nThis is because 1 is much larger than $x_i^T(X^T X)^{-1}x_i)$; so if $x_i$ is near the mean value for $x$ then this term will not change much. \n\n\\end{enumerate}\n\n\\subsection{Distributions of estimators and residuals}\n\nCovar$(\\hat{\\beta},e)=0$: \n\n Var$\\begin{pmatrix}\n\t \\hat{\\beta} \\\\\n\te \\\\\n\t\\end{pmatrix}\n\t= \n\t\\begin{pmatrix}\n\t Var(\\hat{\\beta}) & 0 \\\\\n\t 0 & Var(e) \\\\\n\t\\end{pmatrix}\n\t= \n\t\\begin{pmatrix}\n\t \\sigma^2 (X^T X)^{-1} & 0 \\\\\n\t 0 & \\sigma^2 M \\\\\n\t\\end{pmatrix}\n\t$.\n\t\n\\textbf{Confidence intervals for components of $\\beta$}\t\n\n\nLet $G=(X^T X)^{-1}$, and $g_{ii}$ the $i$-th diagonal element. \n\n\\begin{equation}\n\\hat{\\beta}_i \\sim N(\\beta_i, \\sigma^2 g_{ii})\n\\end{equation}\n\nSince $\\hat{\\beta}$ and $S_r$ are independent, we have:\n\n\\begin{equation}\n\\frac{\\hat{\\beta}_i - \\beta_i}{\\hat{\\sigma}\\sqrt{g_{ii}}} \\sim t_{n-p}\n\\end{equation}\n\nThe 95\\% CI:\n\n\n\\begin{equation}\n\\hat{\\beta}_i \\pm t_{n-p,(1-\\alpha)/2} \\hat{\\sigma} \\sqrt{g_{ii}} \n\\end{equation}\n\n\\subsection{Maximum likelihood estimators}\n\nFor $\\sigma^2$:\n\n\nLet $X_i$, $i=1,\\dots,n$ be a random variable with PDF $f(x; \\sigma) = \\frac{1}{2\\sigma} exp (-\\frac{\\mid x \\mid}{\\sigma})$. Find $\\hat \\sigma$, the MLE of $\\sigma$.\n\n\n\\begin{equation}\n\tL(\\sigma) = \\prod f(x_i; \\sigma) = \\frac{1}{(2\\sigma)^n} exp (-\\sum \\frac{(x_i - \\mu)^2}{\\sigma^2})\n\\end{equation}\n\nLet $\\ell$ be log likelihood. Then:\n\n\\begin{equation}\n\t\\ell (x; \\sigma) = - n\\log 2 - n\\log \\sigma - \\sum (x_i - \\mu)^2 /\\sigma^2\n\\end{equation}\n\nDifferentiating and equating to zero to find maximum:\n\n\\begin{equation}\n\t\\ell ' (\\sigma) = - \\frac{n}{\\sigma} + \\sum (x_i - \\mu)^2 /\\sigma^3 =\n\t 0\n\\end{equation}\n\nRearranging the above, the MLE for $\\sigma$ is:\n\n\\begin{equation}\n\t\\hat \\sigma^2 = \\sum (x_i - \\mu)^2 /n\n\\end{equation}\n\nSince $S_r\\sim \\chi^2_{n-p}$, $E(S_r)=\\sigma^2 (n-p)$. So we need to correct $S_r$ as $S_r/n-p$ to get $E(S_r)=\\sigma^2$. \n\n\\subsection{Hypothesis testing}\n\nA general format for specifying null hypotheses: $H_0: C\\beta = c$, where $C$ is a $q\\times p$ matrix and $c$ is a $q\\times 1$ vector of known constants. The matrix $C$ effectively asserts specific values for $q$ linear functions of $\\beta$. In other words, it asserts $q$ null hypotheses stated in terms of (components of) the parameter vector $\\beta$.\n\n E.g., given:\n\n\\begin{equation}\ny_i = \\beta_0 + \\beta_1 x_i + \\beta_2 x_i^2+\\epsilon_i\n\\end{equation}\n\n\\noindent\nwe can test $H_0: \\beta_1=1, \\beta_2=2$ by setting \n\n$C=\\begin{pmatrix} \n0 & 1 & 0\\\\\n0 & 0 & 1\\\\\n\\end{pmatrix}$\nand $c=\\begin{pmatrix} \n1\\\\\n2\\\\\n\\end{pmatrix}$.\n\nThe alternative is usually the negation of the null, i.e., $H_1: C\\beta\\neq c$, which means that at least one of the $q$ linear functions does not take its hypothesized value. \n\n\\textbf{Constructing a test}:\n\n\\begin{equation}\nC\\hat{\\beta} \\sim N_q (c,\\sigma^2 C (X^T X)^{-1} C^T)\n\\end{equation}\n\nSo, if $H_0$ is true, by sum of squares property:\n\n\\begin{equation}\n(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c) \\sim \\sigma^2 \\chi_q^2\n\\end{equation}\n\nIn other words:\n\n\\begin{equation}\n\\frac{(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c)}{ \\sigma^2} \\sim \\chi_q^2\n\\end{equation}\n\nNote that $\\hat{\\beta}$ is independent of $\\hat{\\sigma}^2$, and recall that\n\n\\begin{equation}\n\\frac{\\hat{\\sigma}^2 }{\\sigma^2} \\sim \\frac{\\chi_{n-p}^2}{n-p} \\Leftrightarrow \n\\frac{\\hat{\\sigma}^2 (n-p)}{\\sigma^2} \\sim \\chi_{n-p}^2\n\\end{equation} \n\nRecall distributional result: if $X\\sim \\chi_v^2, Y\\sim \\chi_w^2$ and $X,Y$ independent then $\\frac{X/v}{Y/w}\\sim F_{v,w}$. \n\n\nIt follows that if $H_0$ is true, and setting\n$X=\\frac{(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c)}{ \\sigma^2}$,\n$Y=\\frac{\\hat{\\sigma}^2 (n-p)}{\\sigma^2}$, and setting the degrees of freedom to $v=q$ and $w=n-p$:\n\n\\begin{equation}\n\\frac{X/v}{Y/w}=\n\\frac{\\frac{(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c)}{ \\sigma^2}/q}{\\frac{\\hat{\\sigma}^2 (n-p)}{\\sigma^2}/(n-p)}\n\\end{equation}\n\nSimplifying:\n\n\\begin{equation}\n\\frac{(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c)}{q\\hat{\\sigma}^2} \\sim F_{q,n-p}\n\\end{equation}\n\n%The above test is the \\textbf{generalized likelihood ratio test}.\n\nThis is a \\textbf{one-sided test} even though the original alternative was two-sided.\n\n\\textbf{Special cases of hypothesis tests}:\n\nWhen $q$ is 1, we have only one hypothesis to test, the $i$-th element of $\\beta$. Given:\n\n\\begin{equation}\ny_i = \\beta_0 + \\beta_1 x_i + \\beta_2 x_i^2+\\epsilon_i\n\\end{equation}\n\n\\noindent\nwe can test $H_0: \\beta_1=0$ by setting \n\n$C=\\begin{pmatrix} \n0 & 1 & 0\\\\\n\\end{pmatrix}$\nand $c=0$.\n\n\nUsing the fact that $X\\sim t(v)\\Leftrightarrow X^2 \\sim F(1,v)$, we have\n\n\n\\begin{equation}\n\\frac{\\hat{\\beta}_i - c_i}{\\hat{\\sigma}\\sqrt{g_{ii}}} \\sim t_{n-p}\n\\end{equation}\n \n\\subsection{Sum of squares}\n\nThis is a very important section!\n\n\\begin{fmpage}{\\linewidth}\n\nRecall:\nIf $K$ is idempotent, so is $I-K$. This allows us to split $x^T x$ into two components sums of squares:\n\n\\begin{equation}\nx^T x = x^T K x+x^T (I-K) x\n\\end{equation}\n\n Let $K_1, K_2,\\dots, K_q$ be symmetric idempotent $n \\times n$ matrices such that\n $\\sum K_i= I_n$ and $K_iK_j =0$, for all $i\\neq j $. Let $x\\sim N_n(\\mu, \\sigma^2)$.\n Then we have the following partitioning into independent sums of squares:\n \n \\begin{equation}\nx^T x = \\sum x^T K_i x\n\\end{equation}\n\nIf $K_i \\mu = 0$, then $ x^T K_i x\\sim \\sigma^2 \\chi_{r_i}^2$, where $r_i$ is the rank of $K_i$.\n\\end{fmpage}\n\nWe can use the sum of squares property just in case $K$ is idempotent, and $K\\mu =0$ . Below, $K=M$ and $\\mu=E(y)=X\\beta$.\n\nConsider the sum of squares partition:\n\n\\begin{equation}\ny^T y = \\explain{\\underline{y^T M y}}{S_r= e^T e} + \\explain{\\underline{y^T (I-M) y}}{\\hat{\\beta}^T (X^T X)\\hat{\\beta}}\n\\end{equation}\n\nNote that the preconditions for sums of squares partitioning are satisfied:\n\\begin{enumerate}\n\\item $M$ is idempotent (and symmetric), rank=trace=$n-p$.\n\\item $I-M$ is idempotent (and symmetric), rank=trace=$p$.\n\\item $ME(y) = 0$ because $ME(y)=MX\\beta$ and $MX=0$.\n\\end{enumerate}\n\nWe can therefore partition the sum of squares into two independent sums of squares:\n\n\\begin{equation}\ny^T y = \\explain{\\underline{y^T M y}}{e^T e \\sim \\sigma^2 \\chi_{n-p}^2} \\hbox{~~~~~~~~}+\\hbox{~~~~~~~~} \n\\explain{\\underline{y^T (I-M) y}}{ \\sim \\sigma^2 \\chi_p^2 \\newline \\hbox{ iff } X\\beta=0, i.e., \\beta=0}\n\\end{equation}\n\nSo, iff we have $H_0: \\beta=0$, we can partition sum of squares as above. Saying that $\\beta=0$ is equivalent to saying that $X$ has rank $p$ and $X\\beta=0$.\n\n\\subsection{Testing the effect of a subset of regressor variables}\n\nLet:\n\n\\begin{equation}\nC= (0_{p-q} I_q) \\quad c=0, \\hbox{ and } \\beta=\\begin{pmatrix} \\beta_1\\\\ \\beta_2 \\end{pmatrix}\n\\end{equation}\n\n\nHere, $\\beta_{1,2}$ are vectors (sub-vectors?), not components of the $\\beta$ vector.\nThen, $C\\times \\beta = \\beta_2$ and $H_0: \\beta_2=0$. Note that order of elements in $\\beta$ is arbitrary; i.e., any subset of $\\beta$ can be tested.\n\nSince $C\\times \\beta = \\beta_2$ and $c=0$, we can construct a sum of squares:\n\n\\begin{equation}\n(C\\hat{\\beta} - c)^T [C (X^T X)^{-1} C^T]^{-1} (C\\hat{\\beta} - c) \\sim \\sigma^2 \\chi_q^2\n\\end{equation}\n\nThis becomes (since $C\\beta=\\hat{\\beta}_2$):\n\n\\begin{equation}\n\\hat{\\beta}_2^T [C (X^T X)^{-1} C^T]^{-1} \\hat{\\beta}_2 \\sim \\sigma^2 \\chi_q^2\n\\end{equation}\n\nWe can rewrite this as: $\\hat{\\beta}_2^T G_{qq}^{-1} \\hat{\\beta}_2$, where $G_{qq}= C (X^T X)^{-1} C^T$ ($G_{qq}$ should not be confused with $g_{ii}$) is a $q\\times q$ submatrix of $G=(X^T X)^{-1}$. \n\nNote that $\\hat{\\beta}$ is independent of $\\hat{\\sigma}^2$, and \nrecall that $\\frac{\\hat{\\sigma}^2 (n-p)}{\\sigma^2} \\sim \\chi_{n-p}^2$. We can now construct the F-test as before:\n\n\\begin{equation}\n\\frac{\\hat{\\beta}_2^T C (X^T X)^{-1} C^T \\hat{\\beta}_2}{q\\hat{\\sigma}^2} = \n\\frac{\\hat{\\beta}_2^T G \\hat{\\beta}_2}{q\\hat{\\sigma}^2}\n \\sim F_{q,n-p}\n\\end{equation}\n\n\n\\textbf{Sums of squares}:\n\nWe can construct three idempotent matrices:\n\n\\begin{itemize}\n\\item\n$M = I_n - X(X^T X)^{-1} X^T$\n\\item\n$M_1 = X(X^T X)^{-1} X^T - [X(X^T X)^{-1} C^T] [\\explain{\\underline{C (X^T X)^{-1} C^T}}{G}]^{-1} \n[C(X^T X)^{-1} X^T]$\n\n(that is: $M_1 = X(X^T X)^{-1} X^T - M_2$)\n\n\\item $M_2 = [X(X^T X)^{-1} C^T] [\\explain{\\underline{C (X^T X)^{-1} C^T}}{G}]^{-1} \n[C(X^T X)^{-1} X^T]$\n\\end{itemize}\n\nNote that $M+M_1+M_2=I_n$ and $MM_1=MM_2=M_1M_2=0$. I.e., sum of squares partition property applies. We have three independent sums of squares:\n\n\\begin{enumerate}\n\\item $S_r = y^T M y$\n\\item $S_1 = y^T M_1 y = \\hat{\\beta}^T X^T X \\hat{\\beta}- \\hat{\\beta}_2^T G_{qq}^{-1} \\hat{\\beta}_2$\n\\item $S_2 = y^T M_2 y = \\hat{\\beta}_2^T G_{qq}^{-1} \\hat{\\beta}_2$\n\\end{enumerate}\n\nSo: $y^T y = S_r + S_1 + S_2$. Then:\n\n\\begin{itemize}\n\\item It is unconditionally true that $S_r \\sim \\sigma^2 \\chi^2_{n-p}$.\n\\item If $H_0: \\beta=0$ is true, then $E(\\hat{\\beta}_2) = \\beta_2 = 0$. It follows from the sum of squares property that $S_2 \\sim \\sigma^2 \\chi_q^2$. \n\\item Regarding $S_1$: \nWe can prove that $M_1 = X_1 (X_1^T X_1)^{-1}X_1^T$, where $X_1$ contains the first $p-q$ columns of $X$. It follows that:\n\n$S_1 = y^T M_1 y =y^T X_1 (X_1^T X_1)^{-1}X_1^T y$\n\nNote that $X_1 (X_1^T X_1)^{-1}X_1^T$ is idempotent. If $\\beta=0$, i.e., if $E(y) =X\\beta = 0$, we can use the sum of squares property and conclude that\n\n$S_1 \\sim \\sigma^2 \\chi_{p-q}^2$\n\nThe degrees of freedom are $p-q$ because the rank=trace of $X_1 (X_1^T X_1)^{-1}X_1^T$ is $n-p$.\n\n\\textbf{Thus, $S_1$ is testing $\\beta_1=0$ but under the assumption that $\\beta_2=0$}.\n\n\\end{itemize}\n\n\\textbf{Analysis of variance}\n\n%\\begin{table}[htdp]\n%\\caption{default}\n%\\begin{center}\n\\begin{tabular}{|l|c|c|c|c|}\n\\hline\nSources & SS & df & MS & MS ratio\\\\\n of variation & & & & \\\\\n\\hline\nDue to $X_1$ & $S_1$ & $p-q$ & $S_1/(p-q)$ & $F_1$ \\\\\nif $\\beta_2=0$ d& & & & $F_{p-q,n-p}$\\\\\n\\hline\nDue to $X_2$ & $S_2$ & $q$ & $S_2/q$ & $F_2$\\\\\n& & & & $F_{q,n-p}$\\\\\n\\hline\nResiduals & $S_r$ & $n-p$ & $\\hat{\\sigma}^2$ & \\\\\n\\hline\nTotal & $y^T y$ & n & &\\\\\n\\hline\n\\end{tabular}\n%\\end{center}\n%\\label{default}\n%\\end{table}%\n\nNote:\n\n\\begin{enumerate}\n\\item \nThe ANOVA tests are \\textbf{performed in order}: First we test $H_0: \\beta_2=0$. Then, if this test does not reject the null, we test $H_0: \\beta_1 = 0$ \\textbf{on the assumption (which may or may not be true)} that $\\beta_2=0$. \n\\item What happens if we reject the first hypothesis?\n\\end{enumerate}\n\n\\textbf{The null or minimal model (constant term)}\n\nWe can set $C=I_p$ and $c=0$. This tests whether all coefficients are zero. But this states that $E(y)=0$, whereas it should have a non-zero value (e.g., reading times). We include the constant term to accommodate this desire to have $E(y)=\\mu=\\neq 0$. In matrix format: let $\\beta$ be the parameter vector; then, $\\beta_1=\\mu$ is the first, constant, term, and the rest of the parameters are the vector $\\beta_2$ ($p-1\\times 1$).\nThe first column of $X$ will be $X_1=1_n$.\n\n\\begin{enumerate}\n\\item\n$S_1=y^T (X_1^T X_1)^{-1} X_1^T y = (\\sum y)^2/n = n\\bar{y}^2$\n\\item\n$S_r = y^Ty - \\hat{\\beta}^T X^T X\\hat{\\beta}$\n\\item\n$S_2 = y^T y -S_1 - S_r = \\hat{\\beta}^T X^T X\\hat{\\beta}-n\\bar{y}^2$\n\\end{enumerate}\n\nIt is normal to omit the row in the ANOVA table corresponding to the constant term.\n\n\\medskip\n\\textbf{Testing whether all predictors (besides the constant term) are zero}\n\nTo test whether $p$ predictor variables have any effect on $y$,we set $q=p-1$, and our anova table looks like this:\n\n\\begin{tabular}{|l|l|l|l|l|}\n\\hline\nSources & SS & df & MS & MS \\\\\nof variation & & & & ratio \\\\\n\\hline\n%Due to $X_1$ if $\\beta_2=0$ & $S_1$ & $p-q$ & $S_1/(p-q)$ & ($F_1$) $ F_{p-q,n-p}$\\\\\n%\\hline\nDue & $S_2$ & $p-1$ & $\\frac{S_2}{(p-1)}$ & $F_2$\\\\\n to regressors & & & & $F_{p-1,n-p}$\\\\\n\\hline\nResiduals & $S_r$ & $n-p$ & $\\hat{\\sigma}^2$ & \\\\\n\\hline\nTotal & $S_{yy}=$ & n-1 & &\\\\\n (\\textbf{adjusted}) & $(y-\\bar{y})^T(y-\\bar{y})$ & & & \\\\\n& $=y^T y - n\\bar{y}^2$ & & & \\\\\n\\hline\n\\end{tabular}\n\nNote that $S_{yy}=\\sum (y_i - \\bar{y})^2$ is the residual sum of squares that we get after fitting the constant $\\hat{\\mu}=\\bar{y}$.\n\n\\medskip\n\\textbf{Testing a subset of predictors $\\beta_2$}\n\n\\begin{tabular}{|l|l|l|l|l|}\n\\hline\nSources & SS & df & MS & MS \\\\\nof variation & & & & ratio \\\\\n\\hline\nDue to $X_1$ & $S_1$ & $p-q-1$ & $\\frac{S_1}{(p-q-1)}$ & ($F_1$) \\\\\nif $\\beta_2=0$ & & & & $F_{p-q-1,n-p}$\\\\\n(test of $\\beta_1$) & & & & \\\\\n\\hline\nDue & $S_2$ & $q$ & $\\frac{S_2}{q}$ & $F_2$\\\\\n to $X_2$ & & & & $F_{q,n-p}$\\\\\n(test of $\\beta_2$) & & & & \\\\\n\\hline\nResiduals & $S_r$ & $n-p$ & $\\hat{\\sigma}^2$ & \\\\\n\\hline\nTotal & $S_{yy}$ & n-1 & &\\\\\n\\hline\n\\end{tabular}\n%Note: the lecture notes have total SS as $y^T y$ but I think that's a typo.\n\n%Used at the very beginning of a document:\n%\\verb!\\documentclass{!\\textit{class}\\verb!}!. Use\n%\\verb!\\begin{document}! to start contents and \\verb!\\end{document}! to\n%end the document.\n\n\n\\section{Checking model assumptions}\n\n\\subsection{Standardized residuals (\\texttt{stdres} in R)}\n\nRecall that $Var(e)=\\sigma^2 M$, where $M = I_n - X (X^T X)^{-1} X^T \\quad \\hbox{M is symmetric, idempotent } n\\times n$. The diagonals of $M$ are all less than 1, and are not all equal (i.e., not equal variance), and off-diagonals are not 0 (i.e., the residuals are correlated). Correcting for unequal variance is done by the \\textbf{scaled residual}:\n\n\\begin{equation}\ne_i* = \\frac{e_i}{\\sqrt{m_{ii}}}\n\\end{equation}\n\nNote: $Var(e_i*) = \\sigma^2$ because $e_i \\sim N(0,\\sigma^2 m_{ii})$, therefore $e_i=\\frac{e_i}{\\sqrt{m_{ii}}}*\\sim N(0,\\sigma^2)$. \n\n\nThe \\textbf{standardized residuals} are \n\n\\begin{equation}\ns_i = \\frac{e_i*}{\\hat{\\sigma}}\n\\end{equation}\n\nThis is approximately $t_{n-p}$ (approximately because $e_i*$ and $\\hat{\\sigma}$ are not independent).\nSince $s_i\\sim t_{n-p}$, we can designate a residual as an outlier if $\\mid s_i \\mid > t_{crit}$ where $t_{crit}$ is the critical t-value.\n\n\\subsection{Standardized deletion residuals (\\texttt{studres} in R)}\n\nThis is a more exact way to test for outliers than the above discussion. Define:\n\n\\begin{equation}\n\\hat{\\beta}_{-i} = (X_{-i}^T X_{-i})^{-1} X_{-i}^T y_{-i}\n\\end{equation}\n\n\\noindent\nwhere the $-i$ refers to removing data point $i$.\nStandardized deletion residuals are\n\n\\begin{equation}\ns_{-i} = \\frac{e_i}{\\hat{\\sigma}_{-i}\\sqrt{m_{ii}}}\n\\end{equation}\n\nWe can compute $s_{-i}$ from $s_{i}$:\n\n\\begin{equation}\ns_{-i} = \\frac{s_i \\sqrt{n-p-1}}{\\sqrt{n-p-s_{i}^2}} \\sim t_{n-p-1}\n\\end{equation}\n\nIf $n$ is large, $s_{-i}\\approx s_i$. \n\n\\subsection{Correcting for multiple testing}\n\n\\v{S}id\\'ak correction: \n``suppose we are performing $n$ tests and in each test we specify the probability of making a type I error to be $\\beta$ (note: don't confuse this as type II error). Then, if the tests are independent, the probability of at least one false positive claim in the $n$ tests is given by \n\n\\begin{equation}\n1-(1-\\beta)^n = \\alpha \\Leftrightarrow \\beta = 1-(1-\\alpha)^{1/n}\n\\end{equation}\n\nThis correction ``has a stronger bound [than the Bonferroni] and so has greater statistical power.''\n\n\\subsection{Checks}\n\n\\begin{enumerate}\n\\item Normality: qqnorm etc. Hist is a useful addition to qqplot in large samples. For small samples, use scaled or standardized residuals if sample size is small (not sure why).\n\\item Independence: index-plots: residuals against observation number. Not useful for small samples. Or: compute correlation between $e_i, e_{i+1}$ pairs of residuals.\n\\item Homoscedasticity: residuals against fitted. Fan out suggests violation. A quadratic trend in a plot of residuals against predictor x could suggest that a quadratic predictor term is needed; note that $X^T e = 0$. (review exercises 3), so we will never have a perfect straight line in such a plot. Alternative: Bartlett's test.\n\\end{enumerate}\n\n\\subsection{Formal tests of normality}\n\nKomogorov-Smirnov and Shapiro-Wilk. Only useful for large samples ; not very powerful and not much better than diagnostic plots. Tests may be useful as follow-ups if non-normality is suspected.\n\n\\subsection{Influence and leverage (\\texttt{lm.influence\\$hat} in R)}\n\nA point can influence the parameter estimates without being an exceptional outlier. Influence does not depend on ``outlyingness''. Potential to influence (e.g., by being an extreme x value) is called leverage; once the y value is also extreme, we have influence. I.e., it takes an extreme x and y value to be influential, and it takes only an extreme x value to have leverage.\n\nLeverage more formally defined: recall that $M = I_n - X (X^T X)^{-1} X^T$. Define a hat matrix $H=I-M=X (X^T X)^{-1} X^T$. It's called a hat matrix because it puts a hat on y: $\\hat{y} = X \\hat{\\beta} = Hy$.\nSince $x_i^T$ is the $i$-th row of $X$, we have $h_{ii} = x_i^T (X^T X)^{-1}x_i$. The measure for leverage is:\n\n\\begin{equation}\nh_{ii} = 1 - m_{ii}\n\\end{equation}\n\nNotice that $h_{ii}$ is a scalar, so $\\hbox{trace}(h_{ii}=h_{ii}$.\nSo (because for a square matrix A,B, tr(AB)= tr(BA)):\n\n\\begin{equation}\nh_{ii} = tr(x_i^T (X^T X)^{-1}x_i)=tr(x_i^T x_i (X^T X)^{-1})\n\\end{equation}\n\nSince $X^T X = \\sum_{i=1}^n x_i x_i^T$, $h_{ii}$ represents the magnitude of $ x_i x_i^T$ relative to the sum of the values for all observations. Note that $h_{ii}$ only depends on X. \n\nAlso note that\n\n\\begin{equation}\n\\sum_{i=1}^n h_{ii} = tr(X^T X (X^T X)^{-1}) = tr(I_p)=p \\quad mean(h_{ii})=p/n\n\\end{equation}\n\n$h_{ii}$ measures leverage because $Var(e_i)=\\sigma^2 m_{ii} = \\sigma^2(1-h_{ii})$ and $Var(\\hat{y}_i) = \\sigma^2 h_{ii}$. Therefore $h_{ii}$ has to lie between 0 and 1. When it is close to one, the fitted value will be close to the actual value of $y_i$---signalling potential for leverage (aside by SV: the explanation sounds circular to me---this statement says it has leverage by definition. Also, I don't know why I should care that a data point has \\textit{potential} to influence the estimates). \n\nA cutoff one can use to identify high leverage points is $h_{ii} > 2p/n$ or $h_{ii} > 3p/n$.\n\nThe leverage of a data point is directly related to how far away it is from the mean:\n\n\\begin{equation}\nh_{ii} = n^{-1} + \\frac{(x_i - \\bar{x})^2}{S_{xx}}\n\\end{equation}\n\nIn \\texttt{lm.influence}, ``coefficients is the matrix whose i-th row contains the change in the estimated coefficients which results when the i-th case is dropped from the regression. sigma is a vector whose i-th element contains the estimate of the residual standard error obtained when the i-th case is dropped from the regression'' (p.\\ 71 of lecture notes).\n\n\\subsection{Cook's distance D: A measure of influence}\n\nLet $s_i$ be the i-th standardized residual, $\\hat{\\beta}_{-i}$ the estimate of the vector of parameters with the i-th row removed.\n\n\\begin{equation}\nD_i = \\frac{(\\hat{\\beta}-\\hat{\\beta}_{-i})^T(X^T X)^{-1}(\\hat{\\beta}-\\hat{\\beta}_{-i})}{p\\hat{\\sigma}^2} = \\frac{s_i^2 h_{ii}}{p(1-h_{ii})}\n\\end{equation}\n\nA data point is influential if it is outlying as well as high leverage. Cutoff for Cook's distance is $\\frac{4}{n}$.\n\n\\textbf{Procedure for checking model fit}:\nto-do, see p 73\n\n\\subsection{Transformations}\n\nSuppose $Y$ is a random variable whose variance depends on its mean. I.e., $E(y)=\\mu, Var(y)=g(\\mu)$.\nThe function $g(\\cdot)$ is known.\n\nWe seek a transformation from $y$ to $z = f(y)$ such that $Var(z)$ is (approximately) constant. \n\nExpand $f(\\cdot)$ in a Taylor series expansion, keeping only the first-order term:\n\n\\begin{equation}\nz= f(y)\\approx f(\\mu) + (y-\\mu)f'(\\mu)\n\\end{equation}\n\nThen: $E(z)=f(\\mu)$ and $Var(z)=g(\\mu) f'(\\mu)^2$. The variance needs to be constant at, say, $k^2$:\n\n\\begin{equation}\nk^2 = g(\\mu) f'(\\mu)^2 \\Rightarrow f'(\\mu)=\\frac{k}{\\sqrt{g(\\mu)}}\n\\end{equation}\n\nSo, \n\n\\begin{equation}\nf(\\mu) = \\int f'(\\mu) = k\\int \\frac{1}{\\sqrt{g(\\mu)}}\n\\end{equation}\n\n\\begin{fmpage}{\\linewidth}\n\\begin{equation}\nf(\\mu) = k\\int [\\sqrt{g(\\mu)}]^{-1/2}\\, d\\mu\n\\end{equation}\n\\end{fmpage}\n\n\\textbf{Example 1}: Let $g(\\mu)=a\\mu$; then $f(\\mu)=2k\\sqrt{\\frac{\\mu}{a}}$. So, $z=\\sqrt{\\mu}$.\n\n\\textbf{Example 2}: Let $g(\\mu)=a\\mu^2$; then $f(\\mu)=k\\sqrt{\\frac{1}{a}}\\log \\mu$. So, $z=\\log \\mu$.\n\n\\textbf{Estimating a transformation}: $\\exists \\lambda$ such that \n\n\\begin{equation}\nf_\\lambda (y_i) = x_i^T \\beta + \\epsilon_i \\quad \\epsilon_i \\sim N(0, \\sigma^2)\n\\end{equation}\n\nWe use maximum likelihood estimation to estimate $\\lambda$. Note that\n\n$L(\\beta_\\lambda, \\sigma^2_\\lambda, \\lambda; y) \\propto$\n\n\\begin{equation}\n(\\frac{1}{\\sigma})^n \\exp [-\\frac{1}{2\\sigma^2} \\sum [f_\\lambda(y_i)- x_i^T \\beta ]^2] [\\prod \\explain{f'_\\lambda(y_i)}{\\hbox{Jacobian}}] \n\\end{equation}\n\nFor fixed $\\lambda$, we estimate $\\hat{\\beta}$ and $\\hat{\\sigma}^2$ in the usual MLE way, and then we turn our attention to $\\lambda$:\n\n\\begin{equation}\nL(\\hat{\\beta}_\\lambda, \\hat{\\sigma}^2_\\lambda, \\lambda; y) = S_\\lambda^{-n/2}\\prod f'_\\lambda(y_i) \n\\end{equation}\n\nTaking logs:\n\n\\begin{equation}\n\\ell = c-\\frac{n}{2} \\log S_\\lambda + \\sum \\log f'_\\lambda(y_i)\n\\end{equation}\n\n\\textbf{Box-Cox family}:\n\n\\begin{equation}\nf_\\lambda (y) = \\left\\{ \n\\begin{array}{l l}\n \\frac{y^\\lambda - 1}{\\lambda} & \\lambda \\neq 0\\\\\n \\log y & \\quad \\lambda=0\\\\\n\\end{array}\n\\right.\n\\end{equation}\n\nWe assume that $f_\\lambda (y) \\sim N(x_i^T \\beta,\\sigma^2)$. So we have to just estimate $\\lambda$ by MLE, along with $\\beta$.\n\n\\textbf{Box-Cox by hand}:\n\nSince $f_\\lambda=\\frac{y^\\lambda-1}{\\lambda}$, it follows that $f'_\\lambda(y)= y^{\\lambda-1}$.\n\nNow, for different $\\lambda$ you can figure out the log likelihoods by hand by solving this equation:\n\n\\begin{equation}\n\\ell = c-\\frac{n}{2} \\log \\explain{S_\\lambda}{\\hbox{Residual sum of squares}} + (\\lambda-1)\\sum \\log (y_i)\n\\end{equation}\n\n\\section{Factors}\n\n\\subsection{Overcoming multicollinearity through parameterization}\n\nto-do: multicollinearity explanation\n\nIf the model matrix $X$ is not full rank (this is true when we include a column for the intercept), then we can put constraints on the predictors (through parameterization). E.g., treatment contrasts (corner-point constraints), sum contrasts, etc.\n\nto-do constraints on p.\\ 94-95\n\n%\\subsection{Merging factor levels}\n\n%This is done to increase group size. \n\n\\subsection{Model selection}\n\n\\textbf{$S_r$ and $R^2$ can't be used for model selection}:\n``$S_r$ will always decrease when we add more regressor variables, so the best fitting model is always the full model which contains all the possible regressor variables - so $S_r$ is not a good model selection tool. For the same reason, the coefficient of determination $R^2$ is not a useful measure in model selection. $R^2$ will not decrease as the number of parameters increases (i.e. it is a non-decreasing function of the number of parameters in the model).''\n\n\\textbf{Penalized likelihood methods for model comparison}:\n\nWe can compare models using log likelihood: \n\n\\begin{equation}\n\\ell = n \\log \\hat{\\sigma}^2 + z(p)\n\\end{equation}\n\n\\noindent\nwhere $z$ is some penalty function. ``Then we declare that the optimal model is that which minimizes $\\ell$. We can think of $z$ as an ad hoc adjustment that tries to give simpler models credit for having fewer regressor variables.''\n\nAIC etc cannot be used to compare across datasets, but can be used to compare non-nested models (cf.\\ ANOVA, which allows only nested models to be compared).\n\n\\textbf{AIC}: here, $z(p)= 2p$. To calculate AIC:\n\n\\begin{equation}\nAIC = 2p + n\\log \\frac{S_r}{n}\n\\end{equation}\n\n[Note: does not match up with the AIC function output in R.]\n\n\\textbf{Where does AIC come from?} From the fact that maximum likelihood of $\\hat{\\sigma}^2$ is:\n\n\\begin{equation}\n L(\\sigma) \\propto (\\hat{\\sigma}^2)^{-n/2}\n\\end{equation}\n\n$-2\\times loglik=n\\log \\hat{\\sigma}^2=n\\log \\frac{S_r}{n}$ is a good model selection tool: smaller values (smaller $S_r$) will mean better fit.\n\n\\textbf{BIC}: $z(p)= p\\log n$. This penalty will be large for large $n$, compared to AIC. \n\n\\textbf{Mallow's $C_p$}: \n\n\\begin{equation}\nC_p = \\frac{S_r}{\\hat{\\sigma}_f^2} - n + 2p_r\n\\end{equation}\n\n$\\hat{\\sigma}_f^2$ is the residual mean square of the full model, $S_r$ is residual \nsums of squares of reduced model, $p_r$ is the number of regressors in the reduced model. \n\nWe want a small $C_p$ and $C_p \\approx p$. \n\n\\textbf{Best subsets method}: in leaps library, regsubsets command.\n\n\\begin{verbatim}\nb<-regsubsets(model specification)\nsummary(b)$rsq, $cp, $bic\n\\end{verbatim}\n\n\\textbf{Backward elimination}: fit full model, and remove the t-value that's smallest, and so on.\nForward elimination goes in the other direction.\n\n\\textbf{Stepwise selection}: step function in MASS. Incrementally add a predictor as above, but once one is added, try removing other predictors with smallest t-value. Repeat until nothing can be added or deleted.\n\n\\begin{verbatim}\nstep(fm,scope=list(upper/lower=formula),\n direction=\"forward/backward/both\")\n\\end{verbatim}\n\n\\section{Generalized least squares}\n\nLet $Var(\\epsilon)=\\sigma^2 \\Sigma$, where $\\Sigma$ is known and non-singular. If $\\Sigma\\neq I_n$ then we either have correlation or non-equal variance or both. If $\\Sigma$ is known, we only need to estimate $\\sigma$ and we are back in least squares theory, with some modification:\n\n\\begin{equation}\ny \\sim N(X\\beta,\\sigma^2 \\Sigma)\n\\end{equation}\n\nLikelihood $L(\\beta,\\sigma^2; y)$ is now:\n\n\\begin{equation}\n (2\\pi)^{-n/2}\\mid \\sigma^2 \\Sigma\\mid^{-1/2} \\exp[-\\frac{1}{2\\sigma^2}(y-X\\beta)^T \\Sigma^{-1} (y-X\\beta)]\n\\end{equation}\n\nThe MLE of $\\beta$ minimizes:\n\n\\begin{equation}\nS=(y-X\\beta)^T \\Sigma^{-1} (y-X\\beta)\n\\end{equation}\n\ninstead of $(y-X\\beta)^T (y-X\\beta)$.\n\n\\textbf{Least squares estimators}:\n\n$\\hat{\\beta}: (X^T \\Sigma^{-1} X)^{-1}X^T \\Sigma^{-1} y$\n\n$E(\\hat{\\beta})=\\beta$\n\n$Var(\\hat{\\beta})=\\sigma^2 (X^T \\Sigma^{-1}X)^{-1}$\n\n\nEstimator of $\\sigma^2$ is $\\frac{S_v}{n-p}$, where\n\n$S_v=y^T \\Sigma^{-1} y - \\hat{\\beta}^T X^T \\Sigma^{-1} X\\hat{\\beta} = y^T \\Sigma^{-1} y - \\hat{\\beta}^T X^T \\Sigma^{-1} y$.\n\n\\subsection{Weighted least squares}\n\nSuppose $\\Sigma=diag(c1,\\dots,cn)$ (uncorrelated, but not homoscedastic) and so $\\Sigma^{-1}=diag(1/\nc1,\\dots,1/cn)$. Let $w_i = 1/c_i$. \n\nThe sum of squares will be \n$S=\\sum_{i=1}^n w_i (y_i - x_i^T \\beta)^2$. So each squared residual is weighted by that observation's variance; observations with large variance are less reliable, and are down-weighted.\n\nCompared to $X^T X = \\sum x_i x_i^T$ in WLS we have $X^T \\Sigma^{-1} X = \\sum w_i x_i x_i^T$. \nAnd instead of $X^T y = \\sum x_i y_i$ in WLS we have $X^T \\Sigma^{-1}y = \\sum w_i x_i y_i$.\n\nSo $\\hat{\\beta}=(X^T \\Sigma^{-1} X)^{-1}X^T \\Sigma^{-1}y= ( \\sum w_i x_i x_i^T)^{-1} (\\sum w_i x_i y_i)$\n\n``Weighted LS is appealing because it allows us to adjust for different (known) variances in the observations in an intuitive way that is easy to implement. The variances of the different observations might be known from pilot or previous studies'' (or are estimated from data).\n\n[SV: I don't get this ``known variance'' business. How can we ever \\textbf{know} what the variance is? Pilot or previous data will only yield estimates.]\n\nThe main disadvantage of WLS is that weights have to specified in advance.\n\n\\subsection{OLS vs WLS}\n\nWith dataset wls1.txt, if we fit:\n\n\\begin{verbatim}\nsummary(m0<-lm(y~X.x,data))\nsummary(m0.wls<-lm(y~X.x,data,\n weights=I(1/X.x^2)))\n\\end{verbatim}\n\nThe coefs will be the same in each, but SEs of coefs will be smaller in WLS fit (because $S_r$ will be down-weighted.\n\n\\textbf{Effect of scaling weights}:\n\nMultiplying the weights by some constant will change residual standard error, but leaves SEs and coefs unchanged. This is because $Var(\\hat{\\beta})=\\sigma^2 (X^T \\Sigma^{-1} X)^{-1}$, so whatever factor $\\sigma^2$ gets multiplied by, it will be cancelled out because it also appears in $ \\Sigma^{-1}$.\n\nDifferently put:\n\n``Let $w_i'$ be the scaled weights, $\\hat{\\sigma}'$ be the residual standard error in the analysis with the scaled weights, $S_r'$ be the residual sum of squares for the scaled analysis and $(\\Sigma^{-1})'$ be the weight matrix for the scaled analysis. Then if $w_i' = 16w_i$, we see that $\\hat{\\sigma}' = 4\\hat{\\sigma}$ since $S_r' = 16S_r$.''\n\nThe SEs will not change because: ``If \n$(\\Sigma^{-1})'=16\\Sigma^{-1}$ and \n$\\hat{\\sigma}' = 4\\hat{\\sigma}$ \nthen $Var(\\hat{\\beta}') = \n(\\hat{\\sigma}^2)' (X^T (\\Sigma^{-1})'X)^{-1} =\n(\\hat{\\sigma}^2) (X^T (\\Sigma^{-1})X)^{-1}$.\nSo the standard errors don’t change.''\n\n\\begin{verbatim}\nsummary(m0.wls<-lm(y~X.x,data,\n weights=I(16*1/X.x^2)))\n\\end{verbatim}\n\n\\begin{enumerate}\n\\item\nIf you get the weights wrong, SEs will increase. So, if unsure about weights use OLS.\n\\item\n``If looking at the standardized residuals shows that there are observations that may be outliers and they reside in regions that will be given high weight then it may be safer to use OLS rather than WLS.''\n\\item One can estimate the weights from the data, but one needs lots of replicates for this, with fewer replicates the SEs will increase.\n\\item If you don't have enough replicates, you can group x values close to each other.\n\\item Outliers can dramatically influence estimates in WLS.\n\\end{enumerate}\n\nConclusion: WLS is a powerful tool, if weights are known, but outliers must be studied carefully.\n\n\\subsection{Using group means in WLS with replicated data}\n\nWithout replicates in the data, we have:\n\n\\begin{equation}\nX^T y = \\sum_i^n x_i y_i\n\\end{equation}\n\nIf we have $k$ replicates:\n\n\\begin{equation}\nX^T y = \\sum_{i=1}^k \\sum_{j=1}^{n_i} x_i y_{ij} = \n\\sum_{i=1}^k x_i \\sum_{j=1}^{n_i} y_{ij} \n= \\sum_{i=1}^k n_i x_i \\bar{y}_i\n\\end{equation}\n\nThis is equivalent to having $\\bar{y}_i$ as observations for the replicate sets, and $n_i$ as weights in WLS.\nIf we had $\\bar{y}_i$ as observations, then their variances would be $\\sigma^2/n_i$, so we would have unequal variances and would use WLS with $w_i = n_i$. \\textbf{Example}: tractor data.\n\n\\subsection{Replication}\n\nDefine replicates here as repeated measurements that are mutually independent (cf.\\ replicates which are not independent, as in linear mixed model theory).\nSince all $x_i^T$ within a replicate set come from the same distribution, their variance should be an estimate of $\\sigma^2$.\n\nLet $y_{ij}$ be the $j$th observation in the $i$th replicate set, where $j=1,\\dots,n_i$ and $n_i$ is the size of the $i$th replicate set ($i=1,\\dots,k$). When $n_i = 1$ we have no replication, and for higher values we have replication. \\textit{Within} each replicate, we can produce an estimator:\n\n\\begin{equation}\n\\hat{\\sigma}^2 = (n_i - 1)^{-1} \\sum_{j=1}^{n_{i}} (y_{ij} - \\bar{y}_i)^2 = (n_i - 1)^{-1}S_i\n\\end{equation}\n\n$\\bar{y}_i=n_i^{-1} \\sum_{j=1}^{n_i} y_{ij}$ is the mean of the replicate set.\n$S_i \\sim \\sigma^2 \\chi_{n_i - 1}^2$.\n\nIn a one-factor model, $S_r = \\sum_{i=1}^k S_i$ and dfs are $n-k = \\sum_{i=1}^k (n_i -1)$. \nSo, in the general case, $df_r=n-k$, $S_r = \\sum_{i=1}^k S_i\\sim \\chi_{df_R}^2$. The ratio $\\hat{\\sigma}^2/df_R$ is an unbiased estimator of $\\sigma^2$.\n\nThe distributional fact being used here is that the sum of independent chi-squared distributions has a chi-squared distribution, and the degrees of freedom is the sum of dfs of the RVs being summed.\n\n\\textbf{The replication estimator}: $\\hat{\\sigma}_R^2=\\frac{S_R}{df_R}$ is the replication estimator of $\\sigma^2$.\n\n``The $\\hat{\\sigma}_R^2$ is independent of the particular form that we have used for the model. We obtain the same replication sum of squares with the same degrees of freedom whether we postulate a linear, quadratic, or some other relationship between Maintenance cost and Age.'' (to-do: don't get this). This is the strength of the replication sum of squares.\n\nThe $S_R$ is in general not equal to $S_r$.\n\n\\subsection{Partitioning replication sum of squares}\n\nto-do\n\n\\bibliographystyle{plain}\n\\bibliography{/Users/shravanvasishth/Dropbox/Bibliography/bibcleaned}\n", "meta": {"hexsha": "894d4ae963d7d5151556dc65b7b51236f3c4a89c", "size": 45728, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "LinearModels.tex", "max_stars_repo_name": "vasishth/MScStatisticsNotes", "max_stars_repo_head_hexsha": "bd4f4076fa785785b419c13580fe214120e4087e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 37, "max_stars_repo_stars_event_min_datetime": "2015-02-14T09:10:46.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-07T14:38:57.000Z", "max_issues_repo_path": "LinearModels.tex", "max_issues_repo_name": "ma0511/MScStatisticsNotes", "max_issues_repo_head_hexsha": "33acde1f79d3ec803f6491b7fee0897e126ddbe6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "LinearModels.tex", "max_forks_repo_name": "ma0511/MScStatisticsNotes", "max_forks_repo_head_hexsha": "33acde1f79d3ec803f6491b7fee0897e126ddbe6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 25, "max_forks_repo_forks_event_min_datetime": "2015-02-14T12:07:34.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-21T08:30:19.000Z", "avg_line_length": 34.800608828, "max_line_length": 504, "alphanum_fraction": 0.6599457663, "num_tokens": 16465, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127492339909, "lm_q2_score": 0.817574478416099, "lm_q1q2_score": 0.6981372705678371}} {"text": "\\section{Options, Part 2}\n\n\\subsection*{Binomial model: risk-neutral pricing}\n\n\nSolve by replication $\\delta$ shares of the stock, $b$ dollars of riskless bond\n\n$\\delta u S_0 + b (1+r) = C_u$ \\\\\n$\\delta d S_0 + b (1+r) = C_d$\n\nSolution: $\\delta = \\frac{C_u-C_d}{(u-d)S_0}$ , \n$b=\\frac{1}{1+r}\\frac{uC_d-dC_u}{u-d}$ \\\\\nThen: $C_0 = \\delta S_0 + b$ \\\\\n$ C_0 = \\frac{C_u-C_d}{(u-d)} + \\frac{1}{1+r} \\frac{uC_d-dC_u}{u-d}$ \\\\\n\n\\subsection*{Risk neutral probability}\n\n$q_u=\\frac{(1+r)-d}{u-d}$, $q_d = \\frac{u-(1+r)}{u-d}$ Then\n\n$C_0=\\frac{q_u C_u + q_d C_d}{1+r} = \\frac{E^Q[C_T]}{1+r}$ \nwhere $E^Q[\\cdot]$ is the expectation under probability $Q=(1,1-q)$, which is\ncalled the risk-neutral probability\n\n\\subsection*{State prices and risk-neutral probabilities}\n$\\Phi_u = \\frac{q}{1+r}$ , $\\Phi_d = \\frac{1-q}{1+r}$\n\n$\\Phi_{uu} = \\frac{q^2}{(1+r)^2}$ , $\\Phi_{ud} = \\Phi_{du} = \\frac{q(1-q)}{(1+r)^2}$ , $\\Phi_{dd} = \\frac{(1-q)^2}{(1+r)^2}$\nWith state prices, can price any state-contingent payoff as a portfolio of\nstate-contingent claims: mathematically equivalent to the risk-neutral\nvaluation formula.\n\n\n\\subsection*{Implementing binominal model}\n\n\\begin{itemize}\n\t\\item As we reduce the length of the time step, holding the maturity fixed, the\n\tbinomial distribution of log returns converges to Normal distribution.\n\t\\item Key model parameters $u$, and $d$ need to be chosen to reflect the\n\tdistribution of the stock return\n\\end{itemize}\n\nOnce choice is: $u=exp(\\sigma \\frac{T}{n})$, $d=\\frac{1}{u}$, $p=\\frac{1}{2} + \\frac{1}{2} \\frac{u}{\\sigma} \\sqrt{\\frac{T}{n}}$\n\n\\subsection*{Black-Scholes-Merton formula}\n\n$C_0 = S_0 N(x) - K e^{-rT} N (x-\\sigma \\sqrt(T))$ \\\\\n$x = \\frac{ln(\\frac{S_0}{K e^{-rT}})}{\\sigma \\sqrt{T}} + \\frac{1}{2} \\sigma \\sqrt{T}$ \\\\\n\nIn Excel \\texttt{$N(x)$=NORM.S.DIST(x,TRUE)}\nThe call is equivalent to a levered long position in the stock;\n $S_0 N(x) $ is the amount invested in the stock;\n$K e^{-rT} N (x-\\sigma \\sqrt(T))$ is the dollar amount borrowed.\n\nEquivalent formulation:\n$ C(S_t, t) = N(d_1)S_t - N(d_2)Ke^{-r(T - t)} $ \\\\\n$ d_1 = \\frac{1}{\\sigma\\sqrt{T - t}}\\left[\\ln\\left(\\frac{S_t}{K}\\right) + \\left(r + \\frac{\\sigma^2}{2}\\right)(T - t)\\right] $\\\\\n$ d_2 = d_1 - \\sigma\\sqrt{T - t} $ \n\nThe price of a corresponding put option based on put-call parity is:\n$ P(S_t, t) = Ke^{-r(T - t)} - S_t + C(S_t, t) = N(-d_2) Ke^{-r(T - t)} - N(-d_1) S_t $\n\n\\subsection*{Option Greeks}\n\nDelta: $\\delta = \\frac{\\partial C}{\\partial S}$\nOmega: $\\Omega = \\frac{\\partial C}{\\partial S} \\frac{S}{C}$\nGamma: $\\Gamma = \\frac{\\partial \\delta}{\\partial S} = \\frac{\\partial^2 C}{\\partial S^2} $\nTheta: $\\Theta = \\frac{\\partial C}{\\partial S}$\nVega: $\\mathcal{V} = \\frac{\\partial C}{\\partial S}$\n", "meta": {"hexsha": "20458728cbe20c398c63c87f5fa4a5970f2400c8", "size": 2712, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "15.415.2x/assets/week_13.tex", "max_stars_repo_name": "j053g/cheatsheets", "max_stars_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-12-14T08:49:07.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-07T17:26:15.000Z", "max_issues_repo_path": "15.415.2x/assets/week_13.tex", "max_issues_repo_name": "j053g/cheatsheets", "max_issues_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "15.415.2x/assets/week_13.tex", "max_forks_repo_name": "j053g/cheatsheets", "max_forks_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.3043478261, "max_line_length": 128, "alphanum_fraction": 0.6286873156, "num_tokens": 1015, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127492339909, "lm_q2_score": 0.817574478416099, "lm_q1q2_score": 0.6981372705678371}} {"text": "\\section{Multi-class Case}\nWe now consider the situation where there are $k > 2$ classes. \n\nWe now define the analogous notion of `margin,' which will be important when we introduce the SVM model later.\n\\begin{definition}\\label{margin_k_classes}\n Suppose that $k$ subsets of $\\mathbb{R}^n$, $A_1,...,A_k$ are linearly separable by the matrix $W$ and vector $b$,\n and let $d$ be a metric on $\\mathbb{R}^n$ as before. The margin of separation with respect to\n $d$ is given by\n \\begin{equation}\n m(W,b; A_1,..., A_k) = \\sup \\{\\epsilon:\\text{$A^{\\epsilon}_1,..., A^\\epsilon_k$ are still linearly separated by $W$ and $b$}\\}\n \\end{equation}\nwhere, as before, the $\\epsilon$-enlargement of $A_i$ is given by\n\\begin{equation}\n A^\\epsilon_i = \\{y:\\text{there is an $x\\in A_i$ such that $\\|x-y\\| < \\epsilon$}\\}\n\\end{equation}\n\\end{definition}\nThis notion of margin measures how much we can perturb each data point without triggering a misclassification. This is \nquite a useful notion especially given the freedom to choose the metric $d$.\n\nWe close this section with a technical lemma relating compactness and the margin of separation, which can safely be skipped.\n\\begin{lemma}{\\label{kclass_margin}}\n Suppose that the sets $A_1,...,A_k$ are separated by the matrix $W$ and vector $b$ and that $A_1,...,A_k$ are compact\n with respect to the topology induced by a metric $d$. Then\n \\begin{equation}\n m(W,b; A_1,...,A_k) > 0\n\\end{equation}\n\n\\end{lemma}\n\n\\begin{proof}\n\tAccording to Lemma \\ref{Interplation}, we only need to prove that for each $A_i$, there exists a positive number $\\epsilon_i$ such that $A_i^{\\epsilon_i}$ are still in $\\Gamma_i(W,b)$. In fact, $\\Gamma_i(W,b)$ is an open set, so $(\\Gamma_i(W,b))^C$ and $A_i$ are two disjoint closed set. Thus we have\n\t\\begin{equation}\n\td_i = d(A_i,(\\Gamma_i(W,b))^C) > 0\n\t\\end{equation}\n\tWe can take $\\epsilon_i = d_i$. Actually, $d_i$ is the largest value that $\\epsilon_i $ can take. And we have\n\t\\begin{equation}\n\tm(W,b; A_1,...,A_k) = \\min_{i}~d_i > 0\n\t\\end{equation}\n\\end{proof}\n\n\n\\subsection{k-class hard-margin Case}\nIn our examination of the $k$-class case, we first consider the situation where our data is linearly separable.\n\nAssuming that our dataset is finite, and thus compact, we see that there exists an $\\epsilon > 0$ such that\n\\begin{equation}\n w_i x+b_i \\geq w_jx+b_j + \\epsilon\n\\end{equation}\nfor all $x\\in A_i$ and $j\\neq i$, and a rescaling of $W$ and $b$ by $2\\epsilon^{-1}$ means that we can assume without loss of \ngenerality that\n\\begin{equation}\n w_i x+b_i \\geq w_jx+b_j + 2\n\\end{equation}\nfor all $x\\in A_i$ and $j\\neq i$. Or we can formulate it as\n\\begin{equation}\\label{separation_constraint_k}\nw_i x+b_i \\geq \\max_{j\\neq i} (w_jx + b_j) + 2,~~\\forall x\\in A_i.\n\\end{equation}\n\n%In order to simplify notation, we index our data points $x_1,...,x_N$ and introduce labels $y_1,...,y_N$, where\n%\\begin{equation}\n% y_i = e_j~\\text{if $x_i\\in A_j$}\n%\\end{equation}\n%i.e. $y_i$ is the standard basis vector corresponding to the class to which $x_i$ belongs. A clever and compact way of writing\n%the linear separation constraint is now\n%\\begin{equation}\\label{separation_constraint_k}\n %(y_i - e_j)\\cdot (Wx_i+b) \\geq \\|y_i - e_j\\|_1\n%\\end{equation}\n%for all $i = 1,...,N$ and $j = 1,...,k$.\n\nWe note that the condition (\\ref{separation_constraint_k}) is a collection of $Nk$ linear constraints on the matrix $W$ and\nvector $b$. This means that we can determine whether our data is linearly separable by checking whether a given linear\nprogram is feasible.\n\nOf course, if our data is linearly separable, then we would like to find a matrix $W$ and vector $b$ which maximizes the\nmargin of separation as defined in Definition \\ref{margin_k_classes}. To this end, we have the following lemma.\n\\begin{lemma}\n Assume that our data is separated by the matrix $W$ and vector $b$, and that (\\ref{separation_constraint_k}) holds.\n Then the margin of separation is bounded below by\n \\begin{equation}\n m(W,b;A_1,...,A_k)\\geq \\frac{1}{\\max_{1\\leq i\\leq k}\\|w_i\\|}\n \\end{equation}\n where $\\|w_i\\|$ is the dual norm of the $i$-th row of $w$.\n\\end{lemma}\n\n\\begin{proof}\n\tUse the notations in the proof of lemma \\ref{kclass_margin}. Because $m(W,b, A_1,...,A_k) = \\min_{i}~d_i$, we only need to prove for each $i$, $d_i = d(A_i,\\partial\\Gamma_i(W,b)) \\geq 1/\\max_{1\\leq i\\leq k}\\|w_i\\|$.\\\\\n\tTake a point $x$ from $A_i$ and a point $y$ from $\\partial\\Gamma_i(W,b)$ arbitrarily. There exists $j\\neq i$ such that\n\t$y \\in H_{ij}$. So we have\n\t\\begin{equation}\n\t\\|x-y\\| \\geq d(x^i,H_{ij}) = \\frac{|(w_i-w_j)x+(b_i-b_j)|}{\\|w_i-w_j\\|}\\geq \\frac{2}{\\|w_i-w_j\\|}\\geq\n\t\\frac{1}{\\max_{1\\leq i\\leq k}\\|w_i\\|}.\n\t\\end{equation}\n So we have\n\t\\begin{equation}\n\td_i = \\inf_{y\\in \\partial\\Gamma_i(W,b)} \\|x-y\\| \\geq \\frac{1}{\\max_{1\\leq i\\leq k}\\|w_i\\|}.\n\t\\end{equation}\n\tAbove all, we can obtain\n\t\\begin{equation}\n\tm_d(W,b;A_1,...,A_k) = \\min_i~d_i \\geq \\frac{1}{\\max_{1\\leq i\\leq k}\\|w_i\\|}.\n\t\\end{equation}\n\t\n\\end{proof}\n\nThis leads to the following convex optimization problem\n\\begin{align}{\\label{kclass_hard_op}}\n \\min_{W,~b~}~~&\\max_{1\\leq i\\leq k}\\|w_i\\|\\\\\n s.t.~~&w_i x+b_i \\geq \\max_{j\\neq i} (w_jx + b_j) + 2,~~\\forall x\\in A_i.\n\\end{align}\nNaturally, we are interested in the relation between (\\ref{kclass_hard_op}) for $k=2$ and $(\\ref{2class_hard_op})$. It seems that we have done some relaxation in (\\ref{kclass_hard_op}), but actually it is equivalent to $(\\ref{2class_hard_op})$ in 2-class situations.\n\\begin{lemma}\nThe optimization problem (\\ref{kclass_hard_op}) is equivalent to the problem $(\\ref{2class_hard_op})$ when $k=2$.\n\\end{lemma} \n\\begin{proof}\nWe'll set $k=2$ in the following statement. When $k=2$, the constraint (\\ref{separation_constraint_k}) is equivalent to the constraint (\\ref{separation_condition_2}), So we use (\\ref{separation_condition_2}) to denote the constraint of both two problems. \\\\\nWe define three sets $\\Omega,\\Omega_1,\\Omega_2$ as\n\\begin{gather}\n\\Omega = \\{w~|~\\exists b\\in \\mathbb{R},~\\text{s.t.}~wx+b\\geq 2, \\forall x \\in A_1~\\text{and}~wx+b\\leq -2, \\forall x\\in A_2\\}.\\\\\n\\Omega_1 = \\{(w_1,w_2)~|~w_1-w_2\\in\\Omega\\}\\\\\n\\Omega_2 = \\{(w_1,w_2)~|~w_1-w_2\\in\\Omega~\\text{and}~w_1 = -w_2\\}\n\\end{gather}\nThe constraint (\\ref{separation_condition_2}) can be written as\n\\begin{equation}\n(w_1,w_2)\\in \\Omega_1.\n\\end{equation}\nObserving that $\\Omega_2 \\subset\\Omega_1$, we'll show that for problem (\\ref{kclass_hard_op}), nothing will change if we use the subset $\\Omega_2$ to replace $\\Omega_1$ in the constraint.\\\\\nTake $(w_1,w_2)\\in \\Omega_1$ arbitrarily, and denote \n\\[\n\\tilde{w}_1 = \\frac{w_1-w_2}{2},~\\tilde{w}_2 = -\\frac{w_1-w_2}{2},\n\\]\nthen we have $(\\tilde{w}_1,\\tilde{w}_2)\\in \\Omega_2$ and\n\\begin{equation}\n\\max_{1\\leq i\\leq 2}\\|\\tilde{w}_i\\| = \\frac{1}{2} \\|\\tilde{w}_1-\\tilde{w}_2\\| = \\frac{1}{2}\\|w_1-w_2\\| \\leq \\max_{1\\leq i\\leq 2}\\|w_i\\|.\n\\end{equation}\nThe inequalities above show that the element in $\\Omega_1\\backslash\\Omega_2$ won't affect the solution of (\\ref{kclass_hard_op}) at all. So we can use $\\Omega_2$ to replace $\\Omega_1$ in problem (\\ref{kclass_hard_op}), which means (\\ref{kclass_hard_op}) is equivalent to \n\\begin{equation}\\label{proof_op1}\n\t\\min_{(w_1,w_2)\\in\\Omega_2} \\max_{1\\leq i\\leq 2}\\|w_i\\|.\n\\end{equation}\nAlso, easy to observe that the problem (\\ref{separation_condition_2}) is equivalent to \n\\begin{equation}\\label{proof_op2}\n\\min_{(w_1,w_2)\\in \\Omega_2} \\| w_1-w_2\\|.\n\\end{equation}\nFor all $(w_1,w_2)\\in \\Omega_2$, we have\n\\begin{equation}\n\t\\max_{1\\leq i\\leq 2}\\|w_i\\| = \\frac{1}{2} \\| w_1-w_2\\|,\n\\end{equation}\nwhich implies that the problem (\\ref{proof_op1}) is equivalent to the problem (\\ref{proof_op2}). \\\\\nConsequently, the problem (\\ref{2class_hard_op}) is equivalent to the problem (\\ref{kclass_hard_op}) when $k=2$.\n\\end{proof}\n\n\\subsection{k-class soft-margin Case}\nIf the data is not max-mat linearly separable, then we replace the hard constraint (\\ref{separation_constraint_k})\nby a penalty term. A common choice of penalty is the hinge or ReLU loss, which only penalizes those constraints which are\nnot satisfied and leads to an optimization problem of the form\n\\begin{equation}{\\label{kclass_soft_op}}\n \\min_{W,~b}\\max_{1\\leq i\\leq k}\\|w_i\\| + C\\displaystyle\\sum_{i=1}^k\\displaystyle\\sum_{x\\in A_i}\\text{ReLU}\\Big( \\max_{j\\neq i} (w_jx + b_j) + 2 - (w_ix+b_i)\\Big)\n\\end{equation}\nwhere $C$ is a parameter controlling how much the penalty term is weighted.\n\nWhen $k =2$, denoting $w_1-w_2$ as $w$, $b_1-b_2$ as $b$, we can write the 2-class soft margin optimization as \n\\begin{equation}{\\label{2-class soft-margin optimization}}\n\\min_{w,b} \\|w\\| + C\\sum_{i=1}^2 \\sum_{x\\in A_i} ReLU((-1)^i (wx+b)+2)\n\\end{equation}\n\n\\section{Optimizaiton Problem in SVM}\nFrom the discussion above, we transformed the classification problem into some optimization problems. In this section, we'll use Lagrange duality to transform the primal problem into dual problem. That will provide convinience for introducing the kernel method in SVM.\\\\\nWe first take a look at the hard-margin case as a transition to the soft-margin case. Trivially, the problem (\\ref{kclass_hard_op}) is equivalent to the following problem:\n\\begin{equation}\n\\begin{aligned}\n\\min_{W,~b} ~~& \\max_{1\\leq i\\leq k}\\|w_i\\|,\\\\\ns.t. ~~& \\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\leq 0,~\\forall i,j.\\\\\n\\end{aligned}\n\\end{equation}\nFor the introduction of inner product, here we take $\\|\\cdot\\|=\\|\\cdot\\|_2$, and use $\\displaystyle\\sum_{l=1}^k \\|w_i\\|_2^2$ to replace $\\max_{1\\leq i\\leq k}\\|w_i\\|$. We can obtain a convex quadratic optimization problem for hard-margin case:\n\\begin{equation}{\\label{kclass_hard_CQP}}\n\\begin{aligned}\n\\min_{W,~b} ~~& \\frac{1}{2}\\displaystyle\\sum_{l=1}^k \\|w_l\\|_2^2,\\\\\ns.t. ~~& \\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\leq 0,~\\forall i,j.\\\\\n\\end{aligned}\n\\end{equation}\n\nNow let's turn to the soft-margin case. Similarly, we take $\\|\\cdot\\|=\\|\\cdot\\|_2$, and use $\\displaystyle\\sum_{l=1}^k \\|w_i\\|_2^2$ to replace $\\max_{1\\leq i\\leq k}\\|w_i\\|$.\n\n\\begin{equation}{\\label{kclass_soft_OP}}\n\\min_{W,~b} \\frac{1}{2}\\displaystyle\\sum_{i=1}^k \\|w_i\\|_2^2 + C\\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k\\text{ReLU}\\Big(\\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\Big)\n\\end{equation}\n\n\\begin{theorem}[Representer Theorem]\nThe solution of (\\ref{kclass_soft_OP}) must belong to $span\\{x_1,\\cdots,x_N\\}$, which means for each $i = 1,2,\\cdots,k$, if $w_i^*$ is the optimal $w_i$ of (\\ref{kclass_soft_OP}), then there must exist $\\alpha_1^i,\\cdots,\\alpha_N^i\\in \\mathbb{R}$, such that \n\\[\nw_i^* = \\sum_{j = 1}^N \\alpha_j^i x_j.\n\\]\n\\end{theorem}\n\n\\begin{proof}\nDenote $span\\{x_1,\\cdots,x_N\\}$ as $X$ and the orthogonal projection operater on $X$ in $\\mathbb{R}^n$ as $P$. Denote the objective function as $L(w_1,\\cdots,w_k,b)$.\\\\\nIn fact, for each $i = 1,\\cdots,k$, we have\n\\begin{align}\n\\|w_i\\|_2^2 = ~&\\|Pw_i + (w_i-Pw_i)\\|_2^2 = \\|Pw_i\\|_2^2+\\|w_i-Pw_i\\|_2^2\\geq \\|Pw_i\\|_2^2,\\\\\nw_ix+b_i &= (Pw_i) x + b_i,~~\\forall x \\in\\{x_1,\\cdots,x_N\\}.\n\\end{align}\nSo easy to observe that\n\\begin{align}\n\\sum_{i=1}^k &\\|Pw_i\\|_2^2\\leq \\sum_{i=1}^k \\|w_i\\|_2^2,\\\\\n\\max_{j\\neq i} &((Pw_j)x + b_j) + 2 - ((Pw_i)x+b_i) = \\max_{j\\neq i} (w_jx + b_j) + 2 - (w_ix+b_i).\n\\end{align}\nSo we have\n\\begin{equation}\nL(Pw_1,\\cdots,Pw_k,b)\\leq L(w_1,\\cdots,w_k,b),\n\\end{equation}\nwhich means the optimal $w_i$ for each $i\\in\\{1,\\cdots,k\\}$ must belong to $X$.\n\\end{proof}\n\nWe can obtain a similar convex quadratic optimization problem to (\\ref{kclass_hard_CQP}) by introducing some slack variables $\\xi$.\n\n\\begin{equation}{\\label{kclass_soft_CQP}}\n\\begin{aligned}\n\\min_{W,b,\\xi} ~~& \\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2+C\\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\xi_{ij},\\\\\ns.t. ~~& \\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)-\\xi_{ij} \\leq 0,~\\forall i,j.\\\\\n & \\xi_{ij} \\geq 0,~\\forall i,j.\n\\end{aligned}\n\\end{equation}\n\n\\begin{lemma}\nThe problem (\\ref{kclass_soft_CQP}) is equivalent to the probelm (\\ref{kclass_soft_OP}).\n\\end{lemma}\n\\begin{proof}\nDenote the objetive functions as $f_1,f_2$:\n\\begin{equation}\n\\begin{split}\n&f_1(W,b) = \\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2 + C\\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k\\text{ReLU}\\Big(\\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\Big)\\\\\n&f_2(W,b,\\xi) = \\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2+C\\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\xi_{ij}.\n\\end{split}\n\\end{equation}\nEasy to observe that \n\\begin{equation}\n\tf_2(W,b,\\xi)\\leq f_1(W,b),\n\\end{equation}\nequality holds iff \n\\begin{equation}\n \\xi_{ij} = \\text{ReLU}\\Big(\\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\Big).\n\\end{equation}\n\\end{proof}\n\n\nIt's easy to observe that the Slater's condition holds in the problem (\\ref{kclass_soft_CQP}). That means we can obtain the optimal points of problem (\\ref{kclass_soft_CQP}) by solving its dual problem.\\\\\nThe Lagrangian function of (\\ref{kclass_soft_CQP}) is\n\\begin{equation}\n\\begin{aligned}\n\t&L(W,b,\\xi,\\lambda,\\mu) \\\\\n\t= &\\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2+C\\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\xi_{ij} + \\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\lambda_{ij}\\Big(\\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)-\\xi_{ij}\\Big) - \\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\mu_{ij} \\xi_{ij}\\\\\n\t= &\\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2 + \\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k (C-\\lambda_{ij}-\\mu_{ij})\\xi_{ij} + \\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k \\lambda_{ij}\\Big(\\|y_i - e_j\\|_1 - (y_i - e_j)\\cdot (Wx_i+b)\\Big)\\\\\n\t= &\\frac{1}{2}\\displaystyle\\sum_{j=1}^k \\|w_j\\|_2^2 + \\displaystyle\\sum_{i=1}^N\\displaystyle\\sum_{j=1}^k (C-\\lambda_{ij}-\\mu_{ij})\\xi_{ij} + \\displaystyle\\sum_{i=1}^N\\Big(\\delta_i-t_i\\cdot (Wx_i)-t_i\\cdot b\\Big)\n\\end{aligned}\n\\end{equation}\nwhere $\\delta_i = \\sum_{j=1}^k\\lambda_{ij}\\|y_i - e_j\\|_1$, $t_i = \\sum_{j=1}^k\\lambda_{ij}(y_i - e_j)$, and $\\lambda_{ij},\\mu_{ij}\\geq 0$.\\\\\nThe dual problem of (\\ref{kclass_soft_CQP}) is \n\\begin{equation}{\\label{dual_problem}}\n\\max_{\\lambda,\\mu\\geq 0} \\min_{W,b,\\xi} L(W,b,\\xi,\\lambda,\\mu). \n\\end{equation}\nAccording to KKT conditions, we have \n\\begin{equation}\n\\begin{aligned}\n\\nabla_{W}L(W^*,b^*,\\xi^*,\\lambda,\\mu) &= W_* - \\displaystyle\\sum_{i=1}^N t_ix_i^{T} = 0,\\\\\n\\nabla_{b}L(W^*,b^*,\\xi^*,\\lambda,\\mu) &= -\\displaystyle\\sum_{i=1}^N t_i = 0,\\\\\n\\nabla_{\\xi_{ij}}L(W^*,b^*,\\xi^*,\\lambda,\\mu) &= C-\\lambda_{ij}-\\mu_{ij} = 0,~\\forall i,j..\n\\end{aligned}\n\\end{equation}\nTake the equations above into dual problem (\\ref{dual_problem}), we have\n\\begin{equation}\n\\begin{aligned}\n&\\min_{W,b,\\xi} L(W,b,\\xi,\\lambda,\\mu) \\\\\n= &L(W^*,b^*,\\xi^*,\\lambda,\\mu)\\\\ \n= &\\frac{1}{2} Tr\\Big((\\displaystyle\\sum_{i=1}^N t_i x_i^T)(\\displaystyle\\sum_{l=1}^N x_l t_l^T)\\Big) -\\displaystyle\\sum_{i=1}^N t_i\\cdot\\Big((\\displaystyle\\sum_{l=1}^N t_l x_l^T)x_i\\Big)+ \\displaystyle\\sum_{i=1}^N \\delta_i\\\\\n= &-\\frac{1}{2}\\displaystyle\\sum_{i=1}^N\\sum_{l=1}^N (t_i\\cdot t_l)(x_i\\cdot x_l)+ \\displaystyle\\sum_{i=1}^N \\delta_i\\\\\n= &-\\frac{1}{2}\\displaystyle\\sum_{i=1}^N\\sum_{l=1}^N\\Big(\\displaystyle\\sum_{j=1}^k\\sum_{m=1}^k \\lambda_{ij}\\lambda_{lm}(y_i-e_j)\\cdot(y_l-e_m)\\Big)(x_i\\cdot x_l)+ \\displaystyle\\sum_{i=1}^N\\sum_{j=1}^k\\lambda_{ij}\\|y_i - e_j\\|_1\n\\end{aligned}\n\\end{equation}\nSo our dual problem can be transformed into the following form:\n\\begin{equation}{\\label{SVM_dual}}\n\\begin{aligned}\n\\min_{\\lambda}&~~ -\\frac{1}{2}\\displaystyle\\sum_{i=1}^N\\sum_{l=1}^N\\Big(\\displaystyle\\sum_{j=1}^k\\sum_{m=1}^k \\lambda_{ij}\\lambda_{lm}(y_i-e_j)\\cdot(y_l-e_m)\\Big)(x_i\\cdot x_l)+ \\displaystyle\\sum_{i=1}^N\\sum_{j=1}^k\\lambda_{ij}\\|y_i - e_j\\|_1,\\\\\ns.t.&~~\\displaystyle \\sum_{i=1}^N\\sum_{j=1}^k\\lambda_{ij}(y_i - e_j) = 0,\\\\\n &~~0\\leq \\lambda_{ij}\\leq C,~\\forall i,j.\n\\end{aligned}\n\\end{equation}\n\n\\section{Kernel Methods}\nWe denote the input space as $X$. Recalling that in the beginning of this chapter, we mentioned that a common way of obtaining a nonlinear classification model is to set $\\mathscr{H} = \\{\\bm{f}: \\bm{f}(x) = W\\bm{\\phi}(x)+b\\}$, where $\\bm{\\phi}$ is a feature mapping from $X$ to a feature space $\\mathcal{H}$. In other words, we use two steps to obtain a nonlinear classification model. First, we map input space $X$ to a feature space $\\mathcal{H}$. Second, use linear classifier to do classification on $\\bm\\phi(X)\\subset \\mathcal{H}$.\\\\\nNotice that if we use SVM on feature space, the optimization problem (\\ref{SVM_dual}) turns to be\n\\begin{equation}{\\label{SVM_dual}}\n\\begin{aligned}\n\\min_{W,~b}&~~ -\\frac{1}{2}\\displaystyle\\sum_{i=1}^N\\sum_{l=1}^N\\Big(\\displaystyle\\sum_{j=1}^k\\sum_{m=1}^k \\lambda_{ij}\\lambda_{lm}(y_i-e_j)\\cdot(y_l-e_m)\\Big)\\langle\\bm\\phi(x_i),\\bm\\phi(x_l)\\rangle+ \\displaystyle\\sum_{i=1}^N\\sum_{j=1}^k\\lambda_{ij}\\|y_i - e_j\\|_1,\\\\\ns.t.&~~\\displaystyle \\sum_{i=1}^N\\sum_{j=1}^k\\lambda_{ij}(y_i - e_j) = 0,\\\\\n&~~0\\leq \\lambda_{ij}\\leq C,~\\forall i,j.\n\\end{aligned}\n\\end{equation}\nAn interesting thing is that we actually only need to use the inner product of feature space in computation. That inspires us to focus on the inner product in feature space instead of the feature mapping itself. We regard the inner product in feature space as a binary function on $X\\times X$ and define it as a kernel function. In other words, a kernel function is $k: X\\times X \\rightarrow \\mathbb{R}$ which\n\\begin{equation}\nk(x,x) = \\langle\\bm\\phi(x),\\bm\\phi(x')\\rangle.\n\\end{equation}\n\n\\begin{definition}\nLet $X$ be a non-empty set. Then a function $k:X\\times X \\rightarrow\\mathbb{K}$ is\ncalled a \\textbf{kernel} on $X$ if there exists a $\\mathbb{K}$-Hilbert space $\\mathcal{H}$ and a map $\\bm{\\phi}:X\\rightarrow \\mathcal{H}$ such that for all $x \\in X$ we have\n\\begin{equation}\nk(x,x) = \\langle\\bm\\phi(x),\\bm\\phi(x')\\rangle.\n\\end{equation}\nWe call $\\bm{\\phi}$ a feature map and $\\mathcal{H}$ a feature space of k.\n\\end{definition}\n\n\\begin{lemma}\nLet $X$ be a non-empty set and $f_n:X\\times X \\rightarrow \\mathbb{K}, n\\in \\mathbb{N}$, be functions such that $(f_n(x))\\in l_2$ for all $x\\in X$. Then \n\\begin{equation}\nk(x,x'):= \\sum_{n=1}^{\\infty} f_n(x) \\overline{f_n(x')},~x,x'\\in X,\n\\end{equation}\ndefines a kernel on $X$.\n\\end{lemma}\n\n\\begin{definition}\nA function $k:X\\times X \\rightarrow \\mathbb{R}$ is called \\textbf{positive definite} if for all $n\\in \\mathbb{N}, ~\\alpha_1,\\cdots,\\alpha_n\\in \\mathbb{R}$ and all $x_1,\\cdots,x_n \\in X$, we have\n\\begin{equation}{\\label{positive_definite_kernel}}\n\\displaystyle\\sum_{i=1}^n\\sum_{j=1}^n\\alpha_i \\alpha_j k(x_j,x_i)\\geq 0.\n\\end{equation}\nFurthermore, $k$ is said to be \\textbf{strictly positive definite} if, for mutually distinct $x_1,\\cdots,x_n \\in X$, equality in (\\ref{positive_definite_kernel}) only holds for $\\alpha_1=\\cdots=\\alpha_n=0$. Finnaly, $k$ is called \\textbf{symmetric} if $k(x,x')=k(x',x)$ for all $x,x' \\in X$.\n\\end{definition}\n\n\\begin{theorem}\nA function $k:X\\times X \\rightarrow \\mathbb{R}$ is a kernel if and only if it is symmetric and postive definite.\n\\end{theorem}\n\\begin{proof}\nIt is trivial that a kernel is a symmetric and postive definite function. We will mainly show the reverse implication.\\\\\nAssuming $k$ is a positive definite kernel defined on $X$, we'll proceed to construct a feature mapping $\\bm\\phi$ into a Hilbert space for which $k$ is the kernel. We'll suppose that $X$ is an infinite set in the following discussion.\n\\begin{enumerate}\n\t\\item Define the feature mapping $\\bm\\phi$ and construct a vector space $H_{pre}$.\\\\\n\tWe define\n\t$$\\bm\\phi:x\\longmapsto k(\\cdot,x)$$\n\tAccording to this mapping, we span the image set to a vector space\n\t$$H_{pre}=\\Big\\{\\sum_{i=1}^{n}\\alpha_{i}k(\\cdot,x_{i}):n\\in\\mathbb{N},\\ \\bm x_{i}\\in X,~\\alpha_{i}\\in\\mathbf{R},~i=1,\\cdots,N \\Big\\}$$\n\t\\item Make $H_{pre}$ an inner product space by defining inner product on $H_{pre}$.\\\\\n\tWe define a binary function \"$<\\cdot,\\cdot>$\" on $H_{pre}$: for any $f,g\\in \\bm F$\n\t$$ f(\\cdot)=\\sum_{i=1}^{n}\\alpha_{i}k(\\cdot,x_{i}) $$\n\t$$ g(\\cdot)=\\sum_{j=1}^{m}\\beta_{i}k(\\cdot,x_{j}) $$\n\t$$ =\\sum_{i=1}^{n}\\sum_{j=1}^{m}\\alpha_{i}\\beta_{j}k(x'_{j},x_{i}) $$\n\tThe bilinearity and symmetry are easy to prove. Because the function$k$ is positive definite, we have\n\t$$ =\\sum_{i,j=1}^{m}\\alpha_{i}\\alpha_{j}k(x_{i},x_{j}),\\forall\\ f\\in H_{pre} $$\\\\\n\tUse the symetric and positive definite property of Gram matrix, we can prove the Cauchy-Schwarz inequality\n\t\\begin{equation}\n\t^2\\leqslant\n\t\\end{equation}\n\tNotice that\n\t\\begin{equation}\\label{reproducing property}\n\t=\\sum_{i=1}^{m}\\alpha_{i}k(x_{i},x)=f(x)\n\t\\end{equation}\n\tTake $g(\\cdot)=k(\\cdot,x)$ into (45), we have\n\t$$ ^2=|f(x)|^2\\leq k(x,x) $$\n\tSo $=0$ implies $f(x)\\equiv 0$.\n\tAbove all, binary function \"$\\bm\\cdot$\" satisfies bilinearity, symmetry and positive definiteness which suffices to show that \"$<\\cdot,\\cdot>$\" is an inner product on $H_{pre}$. \\\\\n\tNow vector space $H_{pre}$ equipped with inner product \"$<\\cdot,\\cdot>$\" is an inner product space.\n\t\\item Complete the inner product space $H_{pre}$, we can get a Hilbert space $H$.\\\\\n\tLet $H$ be a completion of $H_{pre}$ and $I:H_{pre} \\rightarrow H$ be the corresponding isometric embedding. Then $H$ is a Hilbert space and we have\n\t$$\n\t_H=_{H_{pre}}=k(x,x')\n\t$$\n\tfor all $x,x'\\in X, i.e., x\\mapsto Ik(\\cdot,x), x\\in X$, defines a feature map of $k$.\n\tThis Hilbert space is called $\\textbf{reproducing\\ kernel\\ Hilbert\\ space\\ (RKHS)}$ because it satisfies equation (\\ref{reproducing property}) which is called \\textbf{reproducing property}.\n\\end{enumerate}\n\\end{proof}\n\n\\section{Relationship between SVM and Logistic Regression}\n\\includepdf[pages=-]{6DL/SVM-LR.pdf}\n", "meta": {"hexsha": "32ad9efa958708d47f986cb2c26f775a9ad7e493", "size": 21683, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/DL-SVM-k.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/DL-SVM-k.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/DL-SVM-k.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.4447439353, "max_line_length": 538, "alphanum_fraction": 0.6830696859, "num_tokens": 8193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744673038222, "lm_q2_score": 0.8539127585282744, "lm_q1q2_score": 0.6981372686776913}} {"text": "\\section{Principles of Data Reduction}\n\nIn this chapter, we explore how we can use functions of a sample $\\vec{X}$ to make inferences about an unknown parameter (of the distribution of the sample) $\\theta$.\n\n\\begin{definition}[Statistic]\n A statistic is any function of the data.\n\\end{definition}\n\nA statistic $T$ forms a partition of the sample space $\\X$ according to its image, $\\mathcal{T} = \\{t: \\exists \\vec{x} \\in \\X{} \\text{ s.t. } t = T(\\vec{x})\\}$. In this way a statistic provides a method of data reduction. An experimenter who observes only $T$ will treat as equal two samples $\\vec{x}, \\vec{y}$ for which $T(\\vec{x}) = T(\\vec{y})$.\n\n\\subsection{The Sufficiency Principle}\n\n\\begin{definition}[Sufficient Statistic]\n A statistic $T(\\vec{X})$ is a \\emph{sufficient statistic} for $\\theta$ if the conditional distribution of the sample $\\vec{X}$ given $T(\\vec{X})$ does not depend on $\\theta$.\n\\end{definition}\n\n\\begin{remark}\n We ignore the fact that all points have 0 probability for continuous distributions.\n\\end{remark}\n\n\\begin{definition}[The Sufficiency Principle]\n If $T(\\vec{X})$ is a sufficient statistic for $\\theta$, then any inference about $\\theta$ should depend on the sample $\\vec{X}$ only through $T(\\vec{X})$.\n\\end{definition}\n\n\\begin{theorem}\n If $p(\\vec{x}\\vert{}\\theta)$ is the pmf/pdf of the sample $\\vec{X}$ and $q(t\\vert{} \\theta)$ is the pmf/pdf of $T(\\vec{X})$, then $T(\\vec{X})$ is a sufficient statistic for $\\theta$ if $\\forall \\vec{x} \\in \\X{}$, $p(\\vec{x}\\vert{}\\theta) / q(t\\vert{}\\theta)$ is constant as a function of $\\theta$.\n\\end{theorem}\n\n\\begin{remark}[Niceness of the exponential family]\n It turns out that outside of the exponential family, it is rare to have a sufficient statistic that is of smaller dimension than the size of the sample.\n\\end{remark}\n\n\\begin{theorem}[Factorisation Theorem]\n Let $f(\\vec{x} \\vert{} \\theta)$ denote the pmf/pdf of a sample $\\vec{X}$. A statistic $T(\\vec{X})$ is a sufficient statistic for $\\theta$ if and only if there exists functions $g(t \\vert{} \\theta)$ and $h(\\vec{x})$ such that $\\forall \\vec{x} \\in \\X{}$, $\\forall \\theta \\in \\Theta$\n \\[\n f(\\vec{x} \\vert{} \\theta) = g(T(\\vec{x}) \\vert{} \\theta)h(\\vec{x}).\n \\]\n\\end{theorem}\n\n\\begin{remark}\n This theorem shows that the identity is a sufficient statistic. It is straightforward to show from this that any bijection of a sufficient statistic is a sufficient statistic.\n\\end{remark}\n\n\\begin{theorem}\n Let $X_1, \\dots, X_n$ be iid observations from a pmf/pdf $f(x \\vert{} \\theta)$ from an exponential family \n \\[\n f(x|\\vec{\\theta}) = h(x)c(\\vec{\\theta})\\exp\\left( \\sum_{i=1}^k w_i(\\vec{\\theta}) t_i(x) \\right),\n \\]\n where $\\vec{\\theta} = (\\theta_1, \\dots, \\theta_d)$, $d \\leq k$. Then $\\vec{T}(\\vec{X})$ defined by\n \\[\n T_i(\\vec{X}) = \\sum_{j=1}^k t_i(\\vec{X}_j)\n \\]\n is a sufficient statistic for $\\vec{\\theta}$.\n\\end{theorem}\n\n\\begin{definition}[Minimal sufficient statistic]\n A sufficient statistic $T(\\vec{X})$ is called a \\emph{minimal sufficient statistic} if, for any other sufficient statistic $T'(\\vec{X})$, $T$ is a function of $T'$.\n\\end{definition}\n\n\\begin{remark}\n By `function of' we mean that if $T'(\\vec{x}) = T'(\\vec{y})$ then $T(\\vec{x}) = T(\\vec{y})$ -- $T$ varies with respect to $X$ only insofar as it varies with $T'$. This means that each tile of the partition of the sample space according to the image of $T'$ is a subset of some tile in the partition according to $T$. This means that minimal sufficient statistics provide the coarsest possible tiling of the sample space and thus are the sufficient statistics that provide the greatest data reduction.\n\\end{remark}\n\n\\begin{theorem}\n Let $f(\\vec{x} \\vert{} \\theta)$ be the pmf/pdf of a sample $\\vec{X}$. Suppose there exists a function $T(\\vec{X})$ such that $\\forall \\vec{x}, \\vec{y} \\in \\X{}$ the ratio $f(\\vec{x}\\vert{}\\theta)/f(\\vec{y}\\vert{}\\theta)$ is constant as a function of $\\theta$ if an only if $T(\\vec{x}) = T(\\vec{y})$. Then $T(\\vec{X})$ is a minimal sufficient statistic for $\\theta$.\n\\end{theorem}\n\n\\begin{definition}[Necessary Statistic]\n A statistic is \\emph{necessary} if it can be written as a function of every sufficient statistic.\n\\end{definition}\n\n\\begin{theorem}\n A statistic is minimal sufficient if and only if it is a necessary and sufficient statistic.\n\\end{theorem}\n\n\\begin{definition}[Ancillary Statistic]\n A statistic $S(\\vec{X})$ whose distribution does not depend on the parameter $\\theta$ is called an \\emph{ancillary statistic}.\n\\end{definition}\n\n\\begin{definition}[First Order Ancillary]\n A statistic $V(\\vec{X})$ is \\emph{first order ancillary} if $\\E{}[V(\\vec{X})]$ is independent of $\\theta$.\n\\end{definition}\n\n\\begin{definition}[Complete Statistic]\n Let $f(t \\vert{} \\theta)$ be a family of pdfs or pmfs for a statistic $T(\\vec{X})$. The family of probability distributions is called \\emph{complete} if for every (measurable) function $g$\n \\[\n \\E{}[g(T)] = 0 \\,\\, \\forall \\theta \\implies \\P{}(g(T) = 0) = 1 \\,\\, \\forall \\theta.\n \\]\n Equivalently, $T(\\vec{X})$ is called a $\\emph{complete statistic}$.\n\\end{definition}\n\n(It is left unsaid that the function $g$ must be independent of $\\theta$.)\n\n\\begin{theorem}[Basu's Theorem]\n If $T(\\vec{X})$ is a complete and minimal sufficient statistic, then $T(\\vec{X})$ is independent of every ancillary statistic.\n\\end{theorem}\n\n\\begin{theorem}[Complete statistics in the exponential family]\n Let $X_1, \\dots, X_n$ be iid observations from a pmf/pdf $f(x \\vert{} \\theta)$ from an exponential family \n \\[\n f(x|\\vec{\\theta}) = h(x)c(\\vec{\\theta})\\exp\\left( \\sum_{i=1}^k w_i(\\vec{\\theta}) t_i(x) \\right),\n \\]\n where $\\vec{\\theta} = (\\theta_1, \\dots, \\theta_d)$, $d \\leq k$. Then $\\vec{T}(\\vec{X})$ defined by\n \\[\n T_i(\\vec{X}) = \\sum_{j=1}^k t_i(\\vec{X}_j)\n \\]\n is complete if $\\{(w_1(\\vec{\\theta}), \\dots, w_n(\\vec{\\theta}))\\}$ contains an open set in $\\R{}^k$.\n\\end{theorem}\n\n\\begin{remark}\n The open set criteria excludes curved exponential families.\n\\end{remark}\n\n\\begin{theorem}\n If a minimal sufficient statistic exists, then every complete statistic is minimal sufficient.\n\\end{theorem}\n\n\n \n\\subsection{The Likelihood Principle}\n\n\\begin{definition}[Likelihood]\n Let $f(\\vec{x}\\vert{} \\theta)$ denote the pmf/pdf of the sample $\\vec{X}$. Then the \\emph{likelihood function}, given an observation $\\vec{X} = \\vec{x}$, is\n \\[\n L(\\theta \\vert{} \\vec{x}) = f(\\vec{x} \\vert{} \\theta)\n \\]\n as a function of $\\theta$.\n\\end{definition}\n\n\\begin{definition}[Likelihood Principle]\n If $\\vec{x}$ and $\\vec{y}$ are two sample points such that $L(\\theta \\vert{} \\vec{x})$ is proportional to $L(\\theta \\vert{} \\vec{y})$, that is, there exists a constant $C(\\vec{x}, \\vec{y})$ such that \n \\[\n L(\\theta \\vert{} \\vec{x}) = C(\\vec{x}, \\vec{y})L(\\theta \\vert{} \\vec{y}) \\quad \\forall \\theta,\n \\]\n then the conclusions drawn from $\\vec{x}$ and $\\vec{y}$ should be identical.\n\\end{definition}\n\n\\subsection{A Slightly More Formal Construction}\n\\begin{definition}[Experiment]\n We define an \\emph{experiment} $E$ to be a triple $(\\vec{X}, \\theta, f(\\vec{x}\\vert{}\\theta))$, where $\\vec{X}$ is a random vector with pmf/pdf $f$.\\\\\n \n Knowing what experiment $E$ was performed, an experimenter will observer $\\vec{X} = \\vec{x}$. The conclusions they draw about $\\theta$ will be denoted $\\text{Ev}(E, \\vec{x})$, which stands for \\emph{the evidence about $\\theta$ arising from $E$ and $\\vec{x}$}.\n\\end{definition}\n\n\\begin{definition}[Formal Sufficiency Principle]\n Consider an experiment $E = (\\vec{X}, \\theta, f(\\vec{x}\\vert{}\\theta))$ and suppose that $T(\\vec{X})$ is a sufficient statistic for $\\theta$. If $\\vec{x}$ and $\\vec{y}$ are sample points satisfying $T(\\vec{x}) = T(\\vec{y})$, then $\\text{Ev}(E, \\vec{x}) = \\text{Ev}(E, \\vec{y})$.\n\\end{definition}\n\n\\begin{definition}[Conditionality Principle]\n Suppose that $E_1 = \\{X_1, \\theta, f_1(x_1\\vert{} \\theta)\\}$ and $E_2 = \\{X_2, \\theta, f_2(x_2\\vert{} \\theta)\\}$ are two experiments, where only the unknown parameter $\\theta$ need be common between the two experiments. Consider the mixed experiment in which the random variable $J$ is observed, where $\\P{}(J = 1) = \\P{}(J = 2) = \\frac12$ (independent of $\\vec{X}_1, \\vec{X}_2, \\theta$), and then the experiment $E_J$ is performed. Formally, the experiment performed is $E^* = (\\vec{X}^*, \\theta, f(\\vec{x}^* \\vert{} \\theta))$, where $\\vec{X}^* = (j, \\vec{X})j$ and $f^*(\\vec{x}^* \\vert{} \\theta) = f^*((j, \\vec{x}_j)\\vert{} \\theta) = \\frac12 f_j(\\vec{x}_j \\vert{}\\theta)$. Then\n \\[\n \\text{Ev}(E^*, (j, \\vec{x}_j)) = \\text{Ev}(E_j, \\vec{x_j}).\n \\]\n That is, information about $\\theta$ depends only on the experiment run (not on the fact that the particular experiment was chosen).\n\\end{definition}\n\n\\begin{definition}{Formal Likelihood Principle}\n Suppose that we have two experiments, $E_1 = (\\vec{X}_1, \\theta, f_1(\\vec{x}_1 \\vert{} \\theta))$ and $E_2 = (\\vec{X}_2, \\theta, f_2(\\vec{x}_2 \\vert{} \\theta))$ where the unknown parameter $\\theta$ is the same in both experiments. Suppose that $\\vec{x}_1^*$ and $\\vec{x}_2^*$ are sample points from $E_1$ and $E_2$ respectively, such that \n \\[\n L(\\theta \\vert{} \\vec{x}_2^*) = CL(\\theta \\vert{} \\vec{x}_1^*)\n \\]\n for all $\\theta$ and for some constant $C(\\vec{x}_1^*, \\vec{x}_2^*)$ that is independent of $\\theta$. Then\n \\[\n \\text{Ev}(E_1, \\vec{x}_1^*) = \\text{Ev}(E_2, \\vec{x}_2^*).\n \\]\n\\end{definition}\n\n\\begin{remark}\n Note that this is more general from the other likelihood principle since it concerns two experiments (that we can of course set to be equal).\n\\end{remark}\n\n\\begin{corollary}[Likelihood Principle Corollary]\n If $E = (\\vec{X}, \\theta, f(\\vec{x} \\vert{} \\theta))$ is an experiment then $\\text{Ev}(E, \\vec{x})$ should depend on $E$ and $\\vec{x}$ only through $L(\\theta \\vert{} \\vec{x})$.\n\\end{corollary}\n\n\\begin{theorem}[Birnbaum's Theorem]\n The Formal Likelihood Principle follows from the Formal Sufficiency Principle and the Conditionality Principle. The converse is also true.\n\\end{theorem}\n\n\\begin{remark}\n Many common statistical procedures violate the Formal Likelihood Principle -- the topic of the applicability of these principles is not settled. For instance, checking the residuals of a model (to grade the model) violates the Sufficiency Principle. These notions are model dependent, so may not be applicable until \\emph{after} we have decided on a model.\n\\end{remark}\n \n\\subsection{The Equivariance Principle}\n\n\\begin{definition}[Measurement Equivariance]\n Inferences should not depend on the measurement scale used.\n\\end{definition}\n\n\\begin{definition}[Formal Invariance]\n If two inference problems have the same formal structure, in terms of the mathematical model used, then the same inference procedure should be used, regardless of the physical realisation.\n\\end{definition}\n\n\\begin{definition}[Equivariance Principle]\n If $\\vec{Y} = g(\\vec{X})$ is a change of measurement scale such that the model $\\vec{Y}$ has the same formal structure as the model $\\vec{X}$, then an inference procedure should be both measurement equivariant and formally invariant.\n\\end{definition}\n\n\\begin{definition}\n Let $\\mathcal{F} = \\{f(\\vec{x} \\vert{} \\theta): \\theta \\in \\Theta\\}$ be a set of pdfs or pmfs for $\\vec{X}$, and let $\\mathcal{G}$ be group of transformations on the sample space $\\X{}$. Then $\\mathcal{F}$ is \\emph{invariant under the group} $\\mathcal{G}$ if for every $\\theta \\in \\Theta$ and $g \\in \\mathcal{G}$ there exists a unique $\\theta' \\in \\Theta$ such that $\\vec{Y} = g(\\vec{X})$ has the distribution $f(\\vec{y} \\vert{} \\theta')$ if $\\vec{X}$ has the distribution $f(\\vec{x} \\vert{} \\theta')$.\n\\end{definition}\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n\n \n \n \n \n ", "meta": {"hexsha": "61f27b47e62f44f2579b9b5720cd17771a15f42d", "size": 11963, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/chapters/chapter6/content.tex", "max_stars_repo_name": "brynhayder/statistical_inference", "max_stars_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-01-25T05:52:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-10-12T07:20:16.000Z", "max_issues_repo_path": "notes/chapters/chapter6/content.tex", "max_issues_repo_name": "brynhayder/statistical_inference", "max_issues_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-10-17T15:44:01.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-24T09:31:29.000Z", "max_forks_repo_path": "notes/chapters/chapter6/content.tex", "max_forks_repo_name": "brynhayder/statistical_inference", "max_forks_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-04-29T11:11:32.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-17T09:47:07.000Z", "avg_line_length": 54.1312217195, "max_line_length": 683, "alphanum_fraction": 0.6660536655, "num_tokens": 3588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835289107309, "lm_q2_score": 0.8354835452961425, "lm_q1q2_score": 0.6980327407708696}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{exercises}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 2.1 Using Cadabra's own product rule}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n \\nabla{#}::Derivative.\n \\partial{#}::PartialDerivative.\n\n # templates for covariant derivatives\n\n deriv1 := \\nabla_{a}{A?^{b}} -> \\partial_{a}{A?^{b}}\n + \\Gamma^{b}_{c a} A?^{c}.\n\n deriv2 := \\nabla_{a}{A?_{b}} -> \\partial_{a}{A?_{b}}\n - \\Gamma^{c}_{b a} A?_{c}.\n\n # create an object\n\n uv := \\nabla_{a}{v_{b} u^{b}}\n - \\partial_{a}{v_{b} u^{b}}. # cdb (ex-0201.101,uv)\n\n # apply the rules, then simplify\n\n product_rule (uv) # cdb (ex-0201.102,uv)\n substitute (uv,deriv1) # cdb (ex-0201.103,uv)\n substitute (uv,deriv2) # cdb (ex-0201.104,uv)\n distribute (uv) # cdb (ex-0201.105,uv)\n sort_product (uv) # cdb (ex-0201.106,uv)\n rename_dummies (uv) # cdb (ex-0201.107,uv)\n\\end{cadabra}\n\n\\begin{align}\n \\cdb{ex-0201.101} &= \\Cdb{ex-0201.102}\\\\\n &= \\Cdb{ex-0201.103}\\\\\n &= \\Cdb{ex-0201.104}\\\\\n &= \\Cdb{ex-0201.105}\\\\\n &= \\Cdb{ex-0201.106}\\\\\n &= \\Cdb{ex-0201.107}\n\\end{align}\n\n\\clearpage\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 2.1 Using hand crafted product rules}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n \\nabla{#}::Derivative.\n \\partial{#}::PartialDerivative.\n\n # templates for covariant derivatives\n\n deriv1 := \\nabla_{a}{A?^{b}} -> \\partial_{a}{A?^{b}}\n + \\Gamma^{b}_{c a} A?^{c}.\n\n deriv2 := \\nabla_{a}{A?_{b}} -> \\partial_{a}{A?_{b}}\n - \\Gamma^{c}_{b a} A?_{c}.\n\n # tempaltes for product rules\n\n deriv3 := \\nabla_{a}{A?_{b} B?^{c}} -> B?^{c} \\nabla_{a}{A?_{b}}\n + A?_{b} \\nabla_{a}{B?^{c}}.\n\n deriv4 := \\partial_{a}{A?_{b} B?^{c}} -> B?^{c} \\partial_{a}{A?_{b}}\n + A?_{b} \\partial_{a}{B?^{c}}.\n\n # create an object\n\n uv := \\nabla_{a}{v_{b} u^{b}}\n - \\partial_{a}{v_{b} u^{b}}. # cdb (ex-0201.201,uv)\n\n # apply the rules, then simplify\n\n substitute (uv,deriv3) # cdb (ex-0201.202,uv)\n substitute (uv,deriv4) # cdb (ex-0201.203,uv)\n substitute (uv,deriv1) # cdb (ex-0201.204,uv)\n substitute (uv,deriv2) # cdb (ex-0201.205,uv)\n distribute (uv) # cdb (ex-0201.206,uv)\n sort_product (uv) # cdb (ex-0201.207,uv)\n rename_dummies (uv) # cdb (ex-0201.208,uv)\n\\end{cadabra}\n\n\\begin{align}\n \\cdb{ex-0201.201} &= \\Cdb{ex-0201.202}\\\\\n &= \\Cdb{ex-0201.203}\\\\\n &= \\Cdb{ex-0201.204}\\\\\n &= \\Cdb{ex-0201.205}\\\\\n &= \\Cdb{ex-0201.206}\\\\\n &= \\Cdb{ex-0201.207}\\\\\n &= \\Cdb{ex-0201.208}\n\\end{align}\n\n\\end{document}\n", "meta": {"hexsha": "d62ce3e0cf4166f33ce8df5421cecd81676b9457", "size": 3436, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/exercises/ex-0201.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/exercises/ex-0201.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/exercises/ex-0201.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 33.359223301, "max_line_length": 94, "alphanum_fraction": 0.440628638, "num_tokens": 1155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835289107309, "lm_q2_score": 0.8354835391516133, "lm_q1q2_score": 0.6980327356372167}} {"text": "\\section{Abstract}\n\nA \\emph{Support Vector Machine} is a learning model used both for \\emph{classification} and \\emph{regression} tasks whose goal is to construct a \\emph{maximum margin separator}, i.e., a decision boundary with the largest distance from the nearest training data points.\n\nThe aim of this report is to compare the \\emph{primal}, the \\emph{Wolfe dual}~\\cite{fletcher2009support} and the \\emph{Lagrangian dual} formulations of this model in terms of \\emph{complexity}.\n\nFirstly, a detailed mathematical derivation of the model for all these formulations is given, then three algorithms are described to solve the optimization problem in case of \\emph{primal}, \\emph{Wolfe dual} or \\emph{Lagrangian dual} formulation of the problem, explaining their theoretical properties, i.e., \\emph{convergence rate} and \\emph{complexity}.\n\nFinally, some experiments are shown for \\emph{linearly} and \\emph{nonlinearly} separable generated datasets to compare the performance with different \\emph{hyperparameters} and different \\emph{kernels}, also comparing the \\emph{custom} results with \\emph{liblinear}~\\cite{fan2008liblinear} for the \\emph{primal} formulations, \\emph{libsvm}~\\cite{chang2011libsvm} and \\emph{cvxopt}~\\cite{vandenberghe2010cvxopt} for the \\emph{dual} ones.", "meta": {"hexsha": "0606e7bf74763c0a8438028e3948b209ebb30c49", "size": 1278, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notebooks/optimization/tex/abstract.tex", "max_stars_repo_name": "DonatoMeoli/NumericalOptimization", "max_stars_repo_head_hexsha": "e60144458026a6ddbe1612f92b838c342db572eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2020-05-22T09:17:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-15T18:23:56.000Z", "max_issues_repo_path": "notebooks/optimization/tex/abstract.tex", "max_issues_repo_name": "DonatoMeoli/NumericalOptimization", "max_issues_repo_head_hexsha": "e60144458026a6ddbe1612f92b838c342db572eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-25T08:29:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-25T09:03:13.000Z", "max_forks_repo_path": "notebooks/optimization/tex/abstract.tex", "max_forks_repo_name": "DonatoMeoli/NumericalOptimization", "max_forks_repo_head_hexsha": "e60144458026a6ddbe1612f92b838c342db572eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2020-10-10T13:38:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-13T20:23:37.000Z", "avg_line_length": 142.0, "max_line_length": 436, "alphanum_fraction": 0.7957746479, "num_tokens": 355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.837619947119304, "lm_q2_score": 0.8333246035907932, "lm_q1q2_score": 0.6980093103929351}} {"text": "\\lab{Algorithm}{The Knapsack Problem}{The Knapsack Problem}\n\\label{Ch:Knapsack}\n\n\\objective{This section teaches about NP-hard problems using the knapsack problem as an example.}\n\n\\section*{The Knapsack Problem}\n\n%Lab \\ref{Knapsack}\n\nThe two dimensional Knapsack problem is you are given a knapsack that can only hold a certain weight. You have a plethera of items that have value and weight. The goal is to find the combination of items that have the most value with out exceeding the weight.\n\nA three dimensional version the items would have volume and the knapsack can only hold an certain volume as well as weight. \n\nThis problem is has many different applications for disecion making in a wide variety of fields. Two examples are finding the least wasteful way to cut raw materials and the selection of capital investments and financial portfolios.\n\n\\begin{problem}\nWrite a function that solves the Knapsack problem by testing all the combination and choosing the one with the most value that meets the weight constraint. Only test it up to 15 objects.\n\\end{problem}\n\nThis way finds the optimal solution, but there is a problem. Given $n$ objects the number of combinations is $2^n$. The complecity of the problem grows fast. For example, if you had 50 items, it would take more than 10 years to compute the optimal solution. For all the following timing problems let your items have vaules between 1 and 100 and the weights between 1 and $\\frac{Capacity}{10}$.\n\n\\begin{problem}\nTime the Knapsack problem for $11-20$ items with a carrying capacity of $10,000$. Plot the times. What is the complexity of the algorithm for increasing the number of items?\n\\end{problem}\n\n\\begin{problem}\nTime that algorithm for the carrying capacities $10,000-90,000$ every mutiple of $10,000$ with $15$ items. Plot the times. What is the complexity of the algorithm for increasing the weight?\n\\end{problem}\n\n\\section*{NP-Hard}\n\nAny problem that is not polynomial time is know as NP-hard. The Knapsack Problem is a NP-hard problem. Using the branch and bound approach we can find the optimal solution in psuedo-polynomial time.\n\n\\section*{Branch and Bound method}\nWe calculate the optimal solution using only the first i items for weights 0 through W (where W is the max weight the knapsack can hold)\n\nWe do this by intializing 0th row to being empty.\nFor i$>$0 row we go through j (being the weight) from 1 thorugh W. If the ith element's weight is more than j then the optimal combination is the same as it was for i-1. If ith element's weight is less than j, then we compare the combination at i-1 at wieght j to the combination i-1 at the weight j-the ith element weight plus the ith element. Whichever one has a higher value is the optimal combination using the first ith elements less than or equal to the wieght of j. You continue this until you have done all n elements. The combination using n items with a weight of W is guaranteed to be the optimal combination.\n\nThis only does $W*n$ checks, so it is a lot faster. This works by eliminating combinations that could not be the optimal solution\n\n\\begin{problem}\nWrite a function that solves the Knapsack problem using the branch and bound method.\n\\end{problem}\n\n\\begin{problem}\nTime the Knapsack problem for $11-20$ items with a carrying capacity of $10,000$. Plot the times. What is the complexity of the algorithm for increasing the number of items? For the items use the same specifications as in the previous timing problems.\n\\end{problem}\n\n\\begin{problem}\nTime that algorithm for the carrying capacities $10,000-90,000$ every mutiple of $10,000$ with $15$ items. Plot the times. What is the complexity of the algorithm for increasing the weight?\n\\end{problem}\n", "meta": {"hexsha": "52133ecfce228cfb7e9f57814e01d4cadbe6570a", "size": 3709, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Applications/Knapsack/Knapsack.tex", "max_stars_repo_name": "abefrandsen/numerical_computing", "max_stars_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Applications/Knapsack/Knapsack.tex", "max_issues_repo_name": "abefrandsen/numerical_computing", "max_issues_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Applications/Knapsack/Knapsack.tex", "max_forks_repo_name": "abefrandsen/numerical_computing", "max_forks_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.9811320755, "max_line_length": 620, "alphanum_fraction": 0.7851172823, "num_tokens": 870, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8918110375304408, "lm_q1q2_score": 0.697987037391166}} {"text": "\\section*{Exercises}\n\n\\label{LU}\n\n\\begin{ex} \\label{p7.1}Find an $LU$ factorization of $\\begin{mymatrix}{rrr}\n1 & 2 & 0 \\\\\n2 & 1 & 3 \\\\\n1 & 2 & 3\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{ccc}\n1 & 2 & 0 \\\\\n2 & 1 & 3 \\\\\n1 & 2 & 3\n\\end{mymatrix} = \\begin{mymatrix}{ccc}\n1 & 0 & 0 \\\\\n2 & 1 & 0 \\\\\n1 & 0 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrr}\n1 & 2 & 0 \\\\\n0 & -3 & 3 \\\\\n0 & 0 & 3\n\\end{mymatrix}\n\\]\n\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of $\\begin{mymatrix}{rrrr}\n1 & 2 & 3 & 2 \\\\\n1 & 3 & 2 & 1 \\\\\n5 & 0 & 1 & 3\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{cccc}\n1 & 2 & 3 & 2 \\\\\n1 & 3 & 2 & 1 \\\\\n5 & 0 & 1 & 3\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n1 & 1 & 0 \\\\\n5 & -10 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrrr}\n1 & 2 & 3 & 2 \\\\\n0 & 1 & -1 & -1 \\\\\n0 & 0 & -24 & -17\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\begin{mymatrix}{rrrr}\n1 & -2 & -5 & 0 \\\\\n-2 & 5 & 11 & 3 \\\\\n3 & -6 & -15 & 1\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{rrrr}\n1 & -2 & -5 & 0 \\\\\n-2 & 5 & 11 & 3 \\\\\n3 & -6 & -15 & 1\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n-2 & 1 & 0 \\\\\n3 & 0 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrrr}\n1 & -2 & -5 & 0 \\\\\n0 & 1 & 1 & 3 \\\\\n0 & 0 & 0 & 1\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\begin{mymatrix}{rrrr}\n1 & -1 & -3 & -1 \\\\\n-1 & 2 & 4 & 3 \\\\\n2 & -3 & -7 & -3\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{rrrr}\n1 & -1 & -3 & -1 \\\\\n-1 & 2 & 4 & 3 \\\\\n2 & -3 & -7 & -3\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n-1 & 1 & 0 \\\\\n2 & -1 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrrr}\n1 & -1 & -3 & -1 \\\\\n0 & 1 & 1 & 2 \\\\\n0 & 0 & 0 & 1\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\allowbreak \\allowbreak\n\\begin{mymatrix}{rrrr}\n1 & -3 & -4 & -3 \\\\\n-3 & 10 & 10 & 10 \\\\\n1 & -6 & 2 & -5\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n \\begin{mymatrix}{rrrr}\n1 & -3 & -4 & -3 \\\\\n-3 & 10 & 10 & 10 \\\\\n1 & -6 & 2 & -5\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n-3 & 1 & 0 \\\\\n1 & -3 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrrr}\n1 & -3 & -4 & -3 \\\\\n0 & 1 & -2 & 1 \\\\\n0 & 0 & 0 & 1\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix\n$\\begin{mymatrix}{rrrr}\n1 & 3 & 1 & -1 \\\\\n3 & 10 & 8 & -1 \\\\\n2 & 5 & -3 & -3\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{rrrr}\n1 & 3 & 1 & -1 \\\\\n3 & 10 & 8 & -1 \\\\\n2 & 5 & -3 & -3\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n3 & 1 & 0 \\\\\n2 & -1 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrrr}\n1 & 3 & 1 & -1 \\\\\n0 & 1 & 5 & 2 \\\\\n0 & 0 & 0 & 1\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\begin{mymatrix}{rrr}\n3 & -2 & 1 \\\\\n9 & -8 & 6 \\\\\n-6 & 2 & 2 \\\\\n3 & 2 & -7\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{rrr}\n3 & -2 & 1 \\\\\n9 & -8 & 6 \\\\\n-6 & 2 & 2 \\\\\n3 & 2 & -7\n\\end{mymatrix} = \\begin{mymatrix}{rrrr}\n1 & 0 & 0 & 0 \\\\\n3 & 1 & 0 & 0 \\\\\n-2 & 1 & 1 & 0 \\\\\n1 & -2 & -2 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrr}\n3 & -2 & 1 \\\\\n0 & -2 & 3 \\\\\n0 & 0 & 1 \\\\\n0 & 0 & 0\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\begin{mymatrix}{rrr}\n-3 & -1 & 3 \\\\\n9 & 9 & -12 \\\\\n3 & 19 & -16 \\\\\n12 & 40 & -26\n\\end{mymatrix}$.\n%\\begin{sol}\n%\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find an $LU$ factorization of the matrix $\\begin{mymatrix}{rrr}\n-1 & -3 & -1 \\\\\n1 & 3 & 0 \\\\\n3 & 9 & 0 \\\\\n4 & 12 & 16\n\\end{mymatrix}$.\n\\begin{sol}\n\\[\n\\begin{mymatrix}{rrr}\n-1 & -3 & -1 \\\\\n1 & 3 & 0 \\\\\n3 & 9 & 0 \\\\\n4 & 12 & 16\n\\end{mymatrix} = \\begin{mymatrix}{rrrr}\n1 & 0 & 0 & 0 \\\\\n-1 & 1 & 0 & 0 \\\\\n-3 & 0 & 1 & 0 \\\\\n-4 & 0 & -4 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrr}\n-1 & -3 & -1 \\\\\n0 & 0 & -1 \\\\\n0 & 0 & -3 \\\\\n0 & 0 & 0\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find the $LU$ factorization of the coefficient matrix using Dolittle's\nmethod and use it to solve the system of equations.\n\\begin{equation*}\n\\begin{array}{c}\nx+2y=5 \\\\\n2x+3y=6\n\\end{array}\n\\end{equation*}\n\\begin{sol}\nAn $LU$ factorization of the coefficient matrix is\n\\[\n\\begin{mymatrix}{cc}\n1 & 2 \\\\\n2 & 3\n\\end{mymatrix} = \\begin{mymatrix}{cc}\n1 & 0 \\\\\n2 & 1\n\\end{mymatrix} \\begin{mymatrix}{cc}\n1 & 2 \\\\\n0 & -1\n\\end{mymatrix}\n\\]\nFirst solve\n\\[\n\\begin{mymatrix}{cc}\n1 & 0 \\\\\n2 & 1\n\\end{mymatrix} \\begin{mymatrix}{c}\nu \\\\\nv\n\\end{mymatrix} =\\begin{mymatrix}{c}\n5 \\\\\n6\n\\end{mymatrix}\n\\]\nwhich gives $\\begin{mymatrix}{c}\nu \\\\\nv\n\\end{mymatrix} =$ $\\begin{mymatrix}{r}\n5 \\\\\n-4\n\\end{mymatrix}$. Then solve\n\\[\n\\begin{mymatrix}{rr}\n1 & 2 \\\\\n0 & -1\n\\end{mymatrix} \\begin{mymatrix}{c}\nx \\\\\ny\n\\end{mymatrix} =\\begin{mymatrix}{r}\n5 \\\\\n-4\n\\end{mymatrix}\n\\]\nwhich says that $y=4$ and $x=-3$.\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find the $LU$ factorization of the coefficient matrix using Dolittle's\nmethod and use it to solve the system of equations.\n\\begin{equation*}\n\\begin{array}{c}\nx+2y+z=1 \\\\\ny+3z=2 \\\\\n2x+3y=6\n\\end{array}\n\\end{equation*}\n\\begin{sol}\nAn $LU$ factorization of the coefficient matrix is\n\\[\n\\begin{mymatrix}{rrr}\n1 & 2 & 1 \\\\\n0 & 1 & 3 \\\\\n2 & 3 & 0\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n2 & -1 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrr}\n1 & 2 & 1 \\\\\n0 & 1 & 3 \\\\\n0 & 0 & 1\n\\end{mymatrix}\n\\]\nFirst solve\n\\[\n \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n2 & -1 & 1\n\\end{mymatrix} \\begin{mymatrix}{c}\nu \\\\\nv \\\\\nw\n\\end{mymatrix} =\\begin{mymatrix}{c}\n1 \\\\\n2 \\\\\n6\n\\end{mymatrix}\n\\]\nwhich yields $u=1,v=2,w=6$. Next solve\n\\[\n\\begin{mymatrix}{rrr}\n1 & 2 & 1 \\\\\n0 & 1 & 3 \\\\\n0 & 0 & 1\n\\end{mymatrix} \\begin{mymatrix}{c}\nx \\\\\ny \\\\\nz\n\\end{mymatrix} =\\begin{mymatrix}{c}\n1 \\\\\n2 \\\\\n6\n\\end{mymatrix}\n\\]\nThis yields $z=6,y=-16,x=27$.\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find the $LU$ factorization of the coefficient matrix using Dolittle's\nmethod and use it to solve the system of equations.\n\\begin{equation*}\n\\begin{array}{c}\nx+2y+3z=5 \\\\\n2x+3y+z=6 \\\\\nx-y+z=2\n\\end{array}\n\\end{equation*}\n%\\begin{sol}\n%\\end{sol}\n\\end{ex}\n\n\\begin{ex} Find the $LU$ factorization of the coefficient matrix using Dolittle's\nmethod and use it to solve the system of equations.\n\\begin{equation*}\n\\begin{array}{c}\nx+2y+3z=5 \\\\\n2x+3y+z=6 \\\\\n3x+5y+4z=11\n\\end{array}\n\\end{equation*}\n\\begin{sol}\nAn $LU$ factorization of the coefficient matrix is\n\\[\n\\begin{mymatrix}{rrr}\n1 & 2 & 3 \\\\\n2 & 3 & 1 \\\\\n3 & 5 & 4\n\\end{mymatrix} = \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n2 & 1 & 0 \\\\\n3 & 1 & 1\n\\end{mymatrix} \\begin{mymatrix}{rrr}\n1 & 2 & 3 \\\\\n0 & -1 & -5 \\\\\n0 & 0 & 0\n\\end{mymatrix}\n\\]\nFirst solve\n\\[\n \\begin{mymatrix}{rrr}\n1 & 0 & 0 \\\\\n2 & 1 & 0 \\\\\n3 & 1 & 1\n\\end{mymatrix} \\begin{mymatrix}{c}\nu \\\\\nv \\\\\nw\n\\end{mymatrix} =\\begin{mymatrix}{c}\n5 \\\\\n6 \\\\\n11\n\\end{mymatrix}\n\\]\nSolution is: $\\begin{mymatrix}{c}\nu \\\\\nv \\\\\nw\n\\end{mymatrix} =$ $\\begin{mymatrix}{c}\n5 \\\\\n-4 \\\\\n0\n\\end{mymatrix}$. Next solve\n\\[\n\\begin{mymatrix}{rrr}\n1 & 2 & 3 \\\\\n0 & -1 & -5 \\\\\n0 & 0 & 0\n\\end{mymatrix} \\begin{mymatrix}{c}\nx \\\\\ny \\\\\nz\n\\end{mymatrix} =\\begin{mymatrix}{c}\n5 \\\\\n-4 \\\\\n0\n\\end{mymatrix}\n\\]\nSolution is: $\\begin{mymatrix}{c}\nx \\\\\ny \\\\\nz\n\\end{mymatrix} =\\begin{mymatrix}{c}\n7t-3 \\\\\n4-5t \\\\\nt\n\\end{mymatrix} ,t\\in \\R$.\n\\end{sol}\n\\end{ex}\n\n\\begin{ex} Is there only one $LU$ factorization for a given matrix? \\textbf{\nHint:\\ }Consider the equation\n\\begin{equation*}\n\\begin{mymatrix}{rr}\n0 & 1 \\\\\n0 & 1\n\\end{mymatrix} =\\begin{mymatrix}{rr}\n1 & 0 \\\\\n1 & 1\n\\end{mymatrix} \\begin{mymatrix}{rr}\n0 & 1 \\\\\n0 & 0\n\\end{mymatrix} .\n\\end{equation*}\nLook for all possible $LU$ factorizations.\n\\begin{sol}\nSometimes there is more than one $LU$ factorization as is the case in this\nexample. The given equation clearly gives an $LU$ factorization. However, it\nappears that the following equation gives another $LU$ factorization.\n\\[\n\\begin{mymatrix}{cc}\n0 & 1 \\\\\n0 & 1\n\\end{mymatrix} =\\begin{mymatrix}{cc}\n1 & 0 \\\\\n0 & 1\n\\end{mymatrix} \\begin{mymatrix}{cc}\n0 & 1 \\\\\n0 & 1\n\\end{mymatrix}\n\\]\n\\end{sol}\n\\end{ex}\n\n", "meta": {"hexsha": "5e89fe95ebea1d1daae9cf568d9bccdad2528fc2", "size": 7839, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/exercises/Matrices-LUFactorizationMultiplier.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "old/exercises/Matrices-LUFactorizationMultiplier.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/exercises/Matrices-LUFactorizationMultiplier.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 16.9308855292, "max_line_length": 81, "alphanum_fraction": 0.5604031126, "num_tokens": 3708, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8887588023318196, "lm_q2_score": 0.785308583400079, "lm_q1q2_score": 0.697949916043552}} {"text": "\\section{Computing Limits: Graphically}\\label{sec:LimitsGraphically}\r\n\r\nIn this section we look at an example to illustrate the concept of a limit \\ifont{graphically}. \r\n\r\nThe graph of a function $f(x)$ is shown below. We will\r\nanalyze the behaviour of $f(x)$ around $x=-5$, $x=-2$, $x=-1$ and $x=0$, and $x=4$. % and also as $x\\to\\pm\\infty$.\r\n\r\n$$\\includegraphics[width=5.5in]{images/limits-graphically}$$\r\n\r\nObserve that $f(x)$ is indeed a function (it passes the vertical line test). We now analyze the function at each point separately.\r\n\r\n$\\ifont{x=-5:}$ Observe that at $x=-5$ there is no closed circle, thus $f(-5)$ is undefined.\r\nFrom the graph we see that as $x$ gets closer and closer to $-5$ from the left, then $f(x)$ approaches $2$, so\r\n$$\\lim_{x\\to -5^-}f(x)=2.$$\r\nSimilarly, as $x$ gets closer and closer $-5$ from the right, then $f(x)$ approaches $-3$, so\r\n$$\\lim_{x\\to -5^+}f(x)=-3.$$\r\nAs the right-hand limit and left-hand limit are not equal at $-5$, we know that\r\n$$\\lim_{x\\to -5}f(x)\\quad\\mbox{does not exist.}$$\r\n\r\n$\\ifont{x=-2:}$ Observe that at $x=-2$ there is a closed circle at $0$, thus $f(-2)=0$.\r\nFrom the graph we see that as $x$ gets closer and closer to $-2$ from the left, then $f(x)$ approaches $3.5$, so\r\n$$\\lim_{x\\to -2^-}f(x)=3.5.$$\r\nSimilarly, as $x$ gets closer and closer $-2$ from the right, then $f(x)$ again approaches $3.5$, so\r\n$$\\lim_{x\\to -2^+}f(x)=3.5.$$\r\nAs the right-hand limit and left-hand limit are both equal to $3.5$, we know that\r\n$$\\lim_{x\\to -2}f(x)=3.5.$$\r\n\r\nDo not be concerned that the limit does not equal 0. This is a discontinuity, which is completely valid, and will be discussed in a later section.\r\n\r\nWe leave it to the reader to analyze the behaviour of $f(x)$ for $x$ close to $-1$ and $0$.\r\n \r\nSummarizing, we have:\r\n\\hspace{-2.3cm}{\r\n$$\\begin{array}{|c|c|c|c|}\r\n\\hline&~&~&~\\\\\r\nf(-5)~\\mbox{is undefined} & f(-2)=0 & f(-1)=-2 & f(0)=-2 \\\\\r\n~&~&~&~\\\\\r\n\\ds{\\lim_{x\\to -5^-}f(x)=2} & \\ds{\\lim_{x\\to -2^-}f(x)=3.5} & \\ds{\\lim_{x\\to -1^-}f(x)=0} & \\ds{\\lim_{x\\to 0^-}f(x)=-2} \\\\\r\n~&~&~&~\\\\\r\n\\ds{\\lim_{x\\to -5^+}f(x)=-3} & \\ds{\\lim_{x\\to -2^+}f(x)=3.5} & \\ds{\\lim_{x\\to -1^+}f(x)=-2} & \\ds{\\lim_{x\\to 0^+}f(x)=-2}\\\\\r\n~&~&~&~\\\\\r\n\\ds{\\lim_{x\\to -5}f(x)=\\mbox{DNE}} & \\ds{\\lim_{x\\to -2}f(x)=3.5} & \\ds{\\lim_{x\\to -1}f(x)=\\mbox{DNE}} & \\ds{\\lim_{x\\to 0}f(x)=-2}\\\\\r\n~&~&~&~\\\\\r\n\\hline\r\n\\end{array}$$\r\n}\r\n%\\end{solution}\r\n\r\n\r\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\n\\Opensolutionfile{solutions}[ex]\r\n\\section*{Exercises for Section \\ref{sec:LimitsGraphically}}\r\n\r\n\\begin{enumialphparenastyle}\r\n\r\n%%%%%%%%%%\r\n\\begin{ex}\r\nEvaluate the expressions by reference to this graph:\r\n$$\\includegraphics[width=3.5in]{images/limit-exercise-graph}$$\r\n\\begin{multicols}{3}\r\n\\begin{enumerate}\r\n\t\\item\t$\\ds \\lim_{x\\to 4} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to -3} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to 0} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to 0^-} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to 0^+} f(x)$\r\n\t\\item\t$\\ds f(-2)$\r\n\t\\item\t$\\ds \\lim_{x\\to 2^-} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to -2^-} f(x)$\r\n\t\\item\t$\\ds \\lim_{x\\to 0} f(x+1)$\r\n\t\\item\t$\\ds f(0)$\r\n\t\\item\t$\\ds \\lim_{x\\to 1^-} f(x-4)$\r\n\t\\item\t$\\ds \\lim_{x\\to 0^+} f(x-2)$\r\n\\end{enumerate}\r\n\\end{multicols}\r\n\\begin{sol}\r\n\\begin{multicols}{3}\r\n\\begin{enumerate}\r\n\t\\item\t$8$\r\n\t\\item\t$6$\r\n\t\\item\tdne\r\n\t\\item\t$-2$\r\n\t\\item\t$-1$\r\n\t\\item\t$8$\r\n\t\\item\t$7$\r\n\t\\item\t$6$\r\n\t\\item\t$3$\r\n\t\\item\t$-3/2$\r\n\t\\item\t$6$\r\n\t\\item\t$2$\r\n\\end{enumerate}\r\n\\end{multicols}\r\n\\end{sol}\r\n\\end{ex}\r\n\r\n\\end{enumialphparenastyle}", "meta": {"hexsha": "fe6763c4da82a8e0bbcb65e7abdf401a097dbc93", "size": 3455, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "3-limits/3-3-limits-graphically.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "3-limits/3-3-limits-graphically.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "3-limits/3-3-limits-graphically.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.9895833333, "max_line_length": 147, "alphanum_fraction": 0.5858176556, "num_tokens": 1423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339676722393, "lm_q2_score": 0.831143054132195, "lm_q1q2_score": 0.6979390545496509}} {"text": "\\documentclass[cyan,normal,en]{elegantnote}\n\n\\title{Computer Graphic Note}\n\\author{Jiechang Shi}\n\\version{1.1}\n\\date{\\today}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Class 1: Overview}\n\\begin{enumerate}\n\t\\item Frame Buffer\n\t\\begin{itemize}\n\t\t\\item Pixel: One element of a frame buffer\n\t\t\\item Pixel depth: Number of bytes per-pixel in the buffer\n\t\t\\item Resolution: Width $\\times$ Height\n\t\t\\item Buffer size: Total memory allocated for frame buffer\n\t\t\\item \\textbf{Exam Question}: Given Resolution and Pixel depth, Asked Buffer size\n\t\t\\item Z Value: For solving the hidden-surface removal(HSR) problem\n\t\\end{itemize}\n\\end{enumerate}\n\\newpage\n\\section{Class 2,3: Rasterization}\n\\begin{enumerate}\n\t\\item For each pixel on screen, the sample point of that pixel is the center point of that pixel, which have \\textbf{integer} coordinates. For examples, $(3,2)$.\n\t\\item A simple problem: Rasterizing Lines \\\\\n\t\\textbf{Program Description}: Given two endpoints, $P = (x_0,y_0)$ , $R=(x_1,y_1)$ find the pixels that make up the line. \\\\\n\t\\textbf{Note that}: Lines are infinitely thin so they rarely fall on pixel sample point.\\\\\n\t\\textbf{A Feasible Description}: Rasterize lines as \\textbf{closest} pixels to actual lines, with 2 requirement\n\t\\begin{itemize}\n\t\t\\item No Gap\n\t\t\\item Minimize error(distance to true line)\n\t\\end{itemize}\n\t\\textbf{To make this question simplify}: Only consider situation that $|x_1 - x_0| \\geq |y_1 - y_0| \\geq 0 \\land |x_1-x_0| \\neq 0$, which means the $-1 \\leq slope \\leq 1$. Otherwise we just exchange $x$ and $y$. \\\\\n\t\\textbf{A basic Algorithm}: $k=\\frac{y_1-y_0}{x_1-x_0}$, $d=y_0-kx_0$, for each $x_0 \\leq x \\leq x_1$, $y=ROUND(kx+d)$. This method by brute force is inefficient because of the multiplication and the function $ROUND()$.\\\\\n\t\\textbf{Basic Incremental Algorithm}: for each $x_0 \\leq x_i < x_{i+1} \\leq x_1$, $y_{i+1} = y_i +k$, However, the successsive addition of a real number can lead to a \\textbf{cumulative error buildup}!\\\\\n\t\\textbf{Midpoint Line Algorithm}: \n\t\\begin{itemize}\n\t\t\\item For $0 \\leq k \\leq 1$\n\t\t\\item For one approximate point $P=(x,y)$, we only have 2 choices for the next point $E=(x+1,y)$ and $NE=(x+1,y+1)$, we should choose the one which is closer to $k(x+1)+d$\n\t\t\\item Calculate the middle point $M=(x+1,y+\\frac{1}{2})$\n\t\t\\item If the \\textbf{Intersection point} $Q$ is below $M$, take $E$ as next, otherwise take $NE$ as next.\n\t\t\\item Note that: we consider this equation:\n\t\t$$ f(x,y)=ax+by+c=(y_1-y_0)x - (x_1-x_0)y + (x_1y_0-y_1x_0)$$\n\t\tWe assume $a>0$\n\t\t\\item For a point $(x,y)$\\\\\n\t\tif $f(x,y)=0$, $(x,y)$ lies on the line.\\\\\n\t\tif $f(x,y)<0$, $(x,y)$ lies upon the line.\\\\\n\t\tif $f(x,y)>0$, $(x,y)$ lies below the line.\n\t\t\\item So we have to test $f(M)=a(x+1)+b(y+\\frac{1}{2})+c=f(Former)+a+\\frac{b}{2}$\n\t\t\\item Assum $a>0$, if $f(M) > 0$ choose $NE$ otherwise choose $E$\n\t\t\\item Update $f(Former)$:\\\\\n\t\tIf we choose $E$, $f(Former)=f(Former)+a$\\\\\n\t\tIf we choose $NE$, $f(Former)=f(Former)+a+b$\\\\\n\t\tNote that:$a$ and $b$ are \\textbf{constant integer}, so here is no \\textbf{cumulative error issuse}\n\t\\end{itemize}\n\t\\item A harder problem: Triangles Rasterization\\\\\n\t\\textbf{Why Triangle}:\n\t\\begin{itemize}\n\t\t\\item Triangles (\\textit{tris}) are a simple explicit 3D surface representation.\n\t\t\\item Convex and concave polygons (\\textit{polys}) can be decomposed into triangles.\n\t\t\\item Tris are planar and unambiguously defined by three vertex(\\textit{verts}) coordinates (\\textit{coords}).\n\t\\end{itemize}\n\t\\textbf{Definition}: Find and draw \\textbf{pixel} samples \\textbf{inside} \\textit{tri} edges and interpolate parameters defined at \\textit{verts}\n\t\\item \\textbf{Rasterizxation and Hidden Surface Removal(HSR) Algorithm Classes}:\n\t\\begin{itemize}\n\t\t\\item Image order rasterization: ray tracing/ ray casting \\\\\n\t\t\\underline{traverse} pixel, process each in world-space\\\\\n\t\t\\underline{transform} rays from image-space to world-space\n\t\t\\item Object order rasterization: scan-line / LEE \\\\\n\t\t\\underline{traverse} triangles, process each in image-space\\\\\n\t\t\\underline{transform} objects from model-space to image-space\n\t\\end{itemize}\n\t\\item \\textbf{LEE Linear Expression Evaluation Algorithm}:\n\t\\begin{itemize}\n\t\t\\item We already discussed in \\textit{Midpoint Line Algorithm} that how to determine a point is on the left(up) or right(below) the line, just \\textit{a quick review here}:\\\\\n\t\tAssume the lien have a positive $slope$\\\\\n\t\tFor an Edge Equation $E$, for point $(x,y)$: $E(x,y)=dY(x-X)-dX(y-Y)$\\\\\n\t\tif $E(x,y)=0$, $(x,y)$ lies on the line.\\\\\n\t\tif $E(x,y)<0$, $(x,y)$ lies right(below) the line.\\\\\n\t\tif $E(x,y)>0$, $(x,y)$ lies left(up) the line.\n\t\t\\item For \\textbf{Rasterization}:\\\\\n\t\tCompute LEE result for all three edges.\\\\\n\t\tPixels with \\textbf{consistent sign} for all three edges are inside the \\textit{tri}.\\\\\n\t\tInclude \\textbf{edge pixels} on left or right edges.\n\t\t\\item LEE need to check every pixel in the bounding box.\n\t\t\\item LEE is very good in parallel(SIMD) system.\n\t\t\\item Furthermore: Given 3 random \\textit{verts} how to find CW edge cycle\\\\\n\t\tDetermine L/R and Top/Bot edges for edge-pixel ownership\n\t\\end{itemize}\n\t\\item \\textbf{Scan Line Rasterizer}:\n\t\\begin{itemize}\n\t\t\\item Sort vets by Y\n\t\t\\item Setup edge DDAs for edges\n\t\t\\item Sort edges by L or R(The long edge on left or right)\n\t\t\\item Start from Top Vertice, and switch DDA when hit the middle vertice.\t\t\n\t\\end{itemize}\n\t\\item \\textbf{Interpolate Z}:\\\\\n\tA general 3D plane equation has 4 terms: $Ax+By+Cz+D=0$\\\\\n\t$(A,B,C)$ is the normal of that plane, so $(X,Y,Z)_0 \\times (X,Y,Z)_1 = (A,B,C)$\\\\\n\tThen plug any vertex coord into equation and solve for $D$.\\\\\n\tGiven $(A,B,C,D)$ and any point $(x,y)$ can solve $z$\n\t\\item \\textbf{Used $Z$-buffer to remove hidden surfaces}\\\\\n\tInitial $Z$-buffer to MAXINT at the begining of every frame\\\\\n\tInterpolate vertex $Z$ values to get $Z_{pix}$\\\\\n\tOnly write new pixel to the buffer if $Z_{pix}