\chapter{Mixed Potential Green's Functions for Abutted Half-Spaces}
\label{chap:mpgf}
In this chapter, we derive an efficient method of evaluating the
potential Green's functions for a geometry consisting of the abutment
of two dissimilar half-spaces.
The formulas derived herein will be used as the asymptotic forms for those of
Chapter~\ref{chap:gfstratified}, which deals with multiply stratified media.

 The structure for which the potential Green's
functions are desired consists of two layers, as shown in
Figure~\ref{fig:geom2}.
%
\begin{figure}[bp]
  \begin{center}
    \setlength{\unitlength}{0.2in}
      \small
    \begin{picture}(18,8.5)(-9,-4)
      \put(0,-3){\line(0,1){1.7}}  % z boundary
      \put(0,-1){\line(0,1){1.7}}   % z boundary
      \put(0,1){\line(0,1){1.7}}   % z boundary
      \put(0,3){\line(0,1){1.7}}   % z boundary
      \put(0,-3.5){\makebox(0,0){$z = 0$}}
      %
      %
      \put(-4,0){\makebox(0,0){%
        \begin{tabular}{@{}c@{}}
          Region 1: $z<0$ \\
          $\epsilon_1$, $\mu_1$, $k_1$, $\eta_1$
        \end{tabular}}}
      %
      \put(4,0){\makebox(0,0){%
        \begin{tabular}{@{}c@{}}
          Region 2: $z>0$ \\
          $\epsilon_2$, $\mu_2$, $k_2$, $\eta_2$
        \end{tabular}}}
      %
    \end{picture}
    \caption{The two-layer structure for which the Green's functions
          are desired.}
    \label{fig:geom2}
  \end{center}
\end{figure}
The coordinate system origin is located in the interface between
layers~1 and~2 and the $z$ axis points into layer~2. Both layers
are semi-infinite in the $z$ direction and translationally invariant in the $x$ and $y$
directions.  Each layer $i=1,2$ is characterized by a complex
permittivity $\epsilon_i$ and permeability $\mu_i$, each of which lies
either in
the fourth quadrant of the complex plane, or on the real axis.  The
medium intrinsic wavenumbers are $k_i = \omega\sqrt{\mu_i\epsilon_i}$.

We will use the shorthand notation $\displaystyle \summn$ to denote
the double sum $\displaystyle \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}.$

\section{Derivation of Modal Series}
\subsection{Magnetic Vector Potential}

The assumed source is the electric current density
\begin{equation}
  \label{eq:J}
  \J(\r) = \x  \summn
  \delta(\r-\vecrho'-m\s_1 -n\s_2) e^{-j(m\psi_1 + n\psi_2)}
\end{equation}
where, as usual
\begin{subequations}
  \begin{align}
    \r &= \x x + \y y + \z z \\
    \vecrho &= \x x + \y y \\
    \r' &= \x x' + \y y' + \z z' \\
    \vecrho' &= \x x' + \y y'.
  \end{align}
\end{subequations}
The electric current density $\J$
can also be expanded into a series of Floquet modes as follows:
\begin{equation}
  \label{eq:Jfloquet}
  \J(\r) =
  \x \frac{\delta(z)}{A} \summn e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
\end{equation}
The surface current density is
\begin{equation}
  \Js(\vecrho) = \int_{-\infty}^{\infty} \J(\r) \, \d z
  = \x \frac{1}{A} \summn  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
\end{equation}

The region~$i$ magnetic vector potential
 $\A^{(i)}$ satisfies the Helmholtz equation
 \begin{equation}
   \label{eq:Ahelmholtz}
   (\laplace + k_i^2) \A\eye = 0,
 \end{equation}
the radiation condition as $\abs{z} \rightarrow \infty,$
and the following interface conditions at $z=0$ obtained from the fundamental
boundary conditions on $\E$ and $\H$:
\begin{subequations}
  \begin{gather}
    A\one_x = A\two_x, \label{eq:Abc:a} \\
    %
    \frac{1}{\mu_1} A\one_z = \frac{1}{\mu_2} A\two_z, \label{eq:Abc:b} \\
    %
    \frac{1}{\mu_1}  \pdz{A\one_x} - \frac{1}{\mu_2} \pdz{A\two_x}
    = \x \bdot \Js, \label{eq:Abc:c} \\
    %
    \frac{1}{k_1^2}
    \left(
      \pdx{A\one_x} + \pdz{A\one_z}
    \right)
    =
    \frac{1}{k_2^2}
    \left(
      \pdx{A\two_x} + \pdz{A\two_z}
    \right). \label{eq:Abc:d}
  \end{gather}
\end{subequations}

It is well known that the interface conditions require two components
of $\A$.  We follow the standard procedure and choose $\A\eye = \x
A\eye_x + \z A\eye_z.$  Writing $\A$ as a series of Floquet modes we
have
\begin{equation}
  \label{eq:Afloquet1}
  \A\eye = \summn
  \left(
    \x A\eye_{mn} + \z B\eye_{mn}
  \right)
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho') - \gamma\eye_{mn} \abs{z}},
\end{equation}
where
\begin{equation}
  \label{eq:gamma}
  \gamma\eye_{mn} = \sqrt{\vecbeta_{mn} \bdot \vecbeta_{mn} - k_i^2}.
\end{equation}

Because the modal expansions consist of a series of orthogonal
functions, each term
(mode) must independently obey the stated boundary conditions.
To satisfy \eqref{eq:Abc:a}, let $A\one_{mn} = A\two_{mn} \equiv
a_{mn}.$  To satisfy \eqref{eq:Abc:b}, let $B\eye_{mn} = \mu_i
b_{mn}.$  The series for $\A$ can then be written as
\begin{equation}
  \label{eq:Afloquet2}
  \A\eye = \summn
  \left(
    \x a_{mn} + \z \mu_i b_{mn}
  \right)
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho') - \gamma\eye_{mn} \abs{z}}.
\end{equation}

From \eqref{eq:Abc:c} we obtain
\begin{equation*}
  \frac{1}{\mu_1} \gamma\one_{mn} a_{mn} + \frac{1}{\mu_2} \gamma\two_{mn}
  a_{mn} = \frac{1}{A}
\end{equation*}
so that
\begin{equation}
  \label{eq:amn}
  a_{mn} = \frac{\tilde{\mu}}{2A} \; \frac{\mu_1+\mu_2}{\mu_1
  \gamma\two_{mn} + \mu_2 \gamma\one_{mn}}
\end{equation}
where
\begin{equation}
  \tilde{\mu} = \frac{2\mu_1 \mu_2}{\mu_1 + \mu_2}.
\end{equation}

From \eqref{eq:Abc:d} we obtain
\begin{equation*}
  \frac{1}{\mu_1\epsilon_1}
  \left( -j\beta_{mnx} a_{mn} + \gamma\one_{mn} \, \mu_1 b_{mn} \right)
  =
  \frac{1}{\mu_2\epsilon_2}
  \left( -j\beta_{mnx} a_{mn} - \gamma\two_{mn} \, \mu_2 b_{mn} \right)
\end{equation*}
where $\beta_{mnx} = \x \bdot \vecbeta_{mn}.$
After some algebraic manipulations the expression for $b_{mn}$ is obtained.
\begin{equation}
  \label{eq:bmn}
  b_{mn} = \frac{\mu_1\epsilon_1 - \mu_2\epsilon_2}{A} \,
  \frac{-j\beta_{mnx}}{(\epsilon_1\gamma\two_{mn} + \epsilon_2\gamma\one_{mn})
     (\mu_1\gamma\two_{mn} + \mu_2\gamma\one_{mn})}
\end{equation}

\subsection{Scalar Electric Potential}
Using the Lorentz gauge the electric scalar potential $\Phi$ is
\begin{align}
  \Phi\eye(\r) &= \frac{j}{\omega \mu_i \epsilon_i} \divergence \A\eye
  \nonumber \\
  %
  &= \frac{j}{\omega\mu_i\epsilon_i} \summn
  \left(
    -j\beta_{mnx} a_{mn} - \gamma\eye_{mn} \sgn(z) \mu_i b_{mn}
  \right)
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho') - \gamma\eye_{mn} \abs{z}}
\end{align}
To aid in simplifying this summand, let us define $F\eye =
-j\beta_{x} a - \gamma\eye \sgn(z) \mu_i b$, where for simplicity we
have temporarily omitted the subscript $mn$. Proceeding with the
algebra:
\begin{align}
  F\eye &=
  \frac{-j\beta_x}{A\mu_i\epsilon_i}
  \left[
  \frac{\mu_1\mu_2}{\mu_1\gamma\two+\mu_2\gamma\one}
  -
  \frac{\gamma\eye \sgn(z) \mu_i(\mu_1\epsilon_1 - \mu_2\epsilon_2)}%
  {\left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) }
  \right]  \nonumber \\
  %
  &= -j\beta_x
  \frac{\mu_1\mu_2\epsilon_1\gamma\two+\mu_1\mu_2\epsilon_2\gamma\one
        -\gamma\eye \sgn(z) \mu_i(\mu_1\epsilon_1 - \mu_2\epsilon_2)}%
  {A\mu_i\epsilon_i
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) }
  %
\end{align}
Evaluated in Region~2 ($z>0$) this expression becomes
\begin{align}
  F\two &=-j\beta_x
  \frac{\mu_1\mu_2\epsilon_1\gamma\two+\mu_1\mu_2\epsilon_2\gamma\one
        -\gamma\two \mu_2(\mu_1\epsilon_1 - \mu_2\epsilon_2)}%
  {A\mu_2\epsilon_2
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) } \nonumber \\
  %
  &=-j\beta_x
  \frac{\mu_1\epsilon_1\gamma\two+\mu_1\epsilon_2\gamma\one
        -\gamma\two (\mu_1\epsilon_1 - \mu_2\epsilon_2)}%
  {A\epsilon_2
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) } \nonumber \\
  %
  &=-j\beta_x
  \frac{\mu_1\gamma\one
         + \mu_2\gamma\two}%
  {A
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) }
\end{align}
A similar derivation for Region~1 yields
\begin{align}
  F\one
  %
  &= -j\beta_x
  \frac{\mu_2\epsilon_1\gamma\two+\mu_2\epsilon_2\gamma\one
        + \gamma\one (\mu_1\epsilon_1 - \mu_2\epsilon_2)}%
  {A\epsilon_1
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) }  \nonumber \\
  %
  &= -j\beta_x
  \frac{\mu_1\gamma\one + \mu_2\gamma\two}%
  {A
    \left(\epsilon_1\gamma\two+\epsilon_2\gamma\one\right)
    \left(\mu_1\gamma\two+\mu_2\gamma\one\right) }  \nonumber \\
  %
  &= F\two.
\end{align}
The expression for $\Phi$ is therefore
\begin{align}
  \label{eq:Phi}
  \Phi\eye(\r)
  &=
  \frac{j}{\omega A} \summn
    -j\beta_{mnx}
  \frac{(\mu_1\gamma\one_{mn} + \mu_2\gamma\two_{mn})
     e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho') - \gamma\eye_{mn} \abs{z}} }%
  {    (\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn})
    (\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}) }
\end{align}

From the equation of continuity we find that the electric charge
density $q_e$ that gives rise to $\Phi$ is
\begin{equation}
  q_e = \frac{j}{\omega} \divergence \J =
  \frac{j}{\omega} \divergence [\x
  \delta(x-x')\delta(y-y')\delta(z)]
  = \frac{j}{\omega} \delta'(x-x') \delta(y-y') \delta(z)
\end{equation}
To determine an expression for the Green's function $G^\Phi$ for the
electric scalar potential, note that
\begin{equation}
  \Phi(\r) = \iiint G^\Phi(\r-\r_0) \, q_e(\r_0) \, \d V_0
  = \frac{-j}{\omega} \frac{\partial G^\Phi(\r-\vecrho')}{\partial x'}
  = \frac{j}{\omega}  \pdx{G^\Phi(\r-\vecrho')}
\end{equation}
so that
\begin{equation}
  \label{eq:pdxGphi}
  \pdx{G^\Phi} = \frac{\omega}{j} \Phi.
\end{equation}
Comparing \eqref{eq:Phi} and \eqref{eq:pdxGphi} we conclude that the
expression for the Green's function is
\begin{equation}
  G^\Phi(\r-\vecrho') =
  \frac{1}{2\bar{\epsilon}A} \summn
  \frac{(\epsilon_1+\epsilon_2)(\mu_1\gamma\one_{mn} + \mu_2\gamma\two_{mn})}%
  {(\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn})
    (\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}) }
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho') - \gamma\eye_{mn} \abs{z}}
\end{equation}
where
$\displaystyle  \bar{\epsilon}= \frac{\epsilon_1+\epsilon_2}{2}$.

\subsection{Summary of Source Plane Potentials}
For convenience we list here the four potential Green's functions that will
be required for the moment method analysis.  Those for the
 electric vector
potential $\F$ and magnetic scalar potential $\Psi$ are obtained from
those of $\A$ and $\Phi$, respectively, via duality.
\begin{subequations}
  \label{eq:allpots}
\begin{align}
  G^A_{xx}(\vecrho-\vecrho') & =
  \frac{\tilde{\mu}}{2A} \summn
  \frac{\mu_1 + \mu_2}%
  {\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}} \,
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}  \label{eq:allpots:A}\\
                                %
  G^\Phi(\vecrho-\vecrho') &=
  \frac{1}{2\bar{\epsilon}A} \summn
  \frac{(\epsilon_1+\epsilon_2)(\mu_1\gamma\one_{mn} + \mu_2\gamma\two_{mn})}%
  {(\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn})
    (\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}) }
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')} \label{eq:allpots:Phi} \\
  %
  G^F_{xx}(\vecrho-\vecrho') & = -
  \frac{\tilde{\epsilon}}{2A} \summn
  \frac{\epsilon_1 + \epsilon_2}%
  {\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn}} \,
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')} \\
                                %
  G^\Psi(\vecrho-\vecrho') &=
  \frac{1}{2\bar{\mu}A} \summn
  \frac{(\mu_1+\mu_2)(\epsilon_1\gamma\one_{mn} + \epsilon_2\gamma\two_{mn})}%
  {(\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn})
    (\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn}) }
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
  \end{align}
\end{subequations}
where
\begin{equation}
  \tilde{\mu} =
  \frac{2\mu_1\mu_2}{\mu_1+\mu_2},
  \quad
  \bar{\epsilon} = \frac{\epsilon_1 + \epsilon_2}{2},
  \quad
  \tilde{\epsilon} =
  \frac{2\epsilon_1\epsilon_2}{\epsilon_1+\epsilon_2},
  \quad
  \bar{\mu} = \frac{\mu_1 + \mu_2}{2}.
\end{equation}
It should be noted that these series represent only a formal solution
of Maxwell's equations.  They are at best only conditionally
convergent and are completely unsuitable for direct numerical
evaluation.  In fact, it is easy to see that these series can not be
absolutely convergent, for as $\vecrho\rightarrow\vecrho'$ the
Green's functions increase in magnitude without bound, a singularity
that is proportional to $1/\norm{\vecrho-\vecrho'}$.  It is desirable
to extract this singularity as an explicit term, so that it can be
integrated in closed form, as in \cite{wrgs:84}.
 The next section describes the technique used to perform this
 extraction and
accelerate the series for efficient evaluation.

\section{Series Acceleration}
We now manipulate the series in Equations~\eqref{eq:allpots} to
arrive at an efficient,
wide-band formulation.  In the process, the Green's functions will be
expressed as the sum of several absolutely, uniformly convergent
series.  Our strategy is to find the first few terms of an asymptotic
expansion of the summands in a suitably chosen variable.  Applying a Kummer's
transformation, these terms are
subtracted from the original summand, leaving a spectral sum with greatly
accelerated convergence.  The sum of the asymptotic terms is then
accelerated by means of the Poisson transformation (which converts it to a
spatial sum) after which it is recombined
with the spectral sum.  A careful choice of the expansion variable ensures
that all sums are rapidly convergent and  provides
the wide-band capability described below.

The basic idea used here is based on the approach presented in \cite{srzw:90}, where a first-order Kummer's
transformation is applied to the free-space periodic Green's function.
The technique was extended to a second-order transformation primarily by Kim McInturff of Raytheon Electromagnetic
Systems Division and was presented in \cite{smjj:91} and \cite{smjs:92}.
The extension of the technique to periodic sources
in stratified media is due to the present author and was first presented in \cite{simo:01b}.

\subsection{Kummer's Transformation}
Let $u>0$
be an appropriately chosen positive {\em smoothing factor}; define
\begin{equation}
  \kappa_{mn}  =  \sqrt{\vecbeta_{mn} \bdot \vecbeta_{mn} +u^2}
         \label{eq:kappa}
\end{equation}
and
\begin{equation}
  w_i  \equiv w(\mu_i,\epsilon_i) =
  \sqrt{\omega^2\mu_i\epsilon_i + u^2} =
  \sqrt{k_i^2+u^2}, \quad i=1,2 \label{eq:w}
\end{equation}
so that
\begin{equation}
  \gamma\eye_{mn}
  \equiv \gamma_{mn} (\mu_i,\epsilon_i)
  = \sqrt{\kappa_{mn}^2 - u^2 - \omega^2\mu_i\epsilon_i}
  = \kappa_{mn} \sqrt{1 - \left(\frac{w_i}{\kappa_{mn}}\right)^2}
\end{equation}
The original summands in \eqref{eq:allpots} will be expanded in series of reciprocal
powers
of $\kappa_{mn}$.  The smoothing factor $u$ should be chosen large enough to ensure that the
spatial series of asymptotic terms converges rapidly, yet small
enough so that after only a few terms, the differences between the original summands
of \eqref{eq:allpots} and their asymptotic expansions are negligible.

Consider first the summand for  magnetic vector potential given in
\eqref{eq:allpots:A}:
\begin{align}
  A_{mn}(\mu_1,\epsilon_1,\mu_2,\epsilon_2) &\equiv
  \frac{\mu_1 + \mu_2}%
  {\mu_1\gamma_{mn}(\mu_2,\epsilon_2) +
    \mu_2\gamma_{mn}(\mu_1,\epsilon_1)}
  =
  \frac{\mu_1 + \mu_2}%
  {\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}} \nonumber \\
  %
  &= \frac{1}{\kappa_{mn}} \,
  \frac{\mu_1 + \mu_2}%
  {\mu_1 \sqrt{1-w_2^2/\kappa_{mn}^2} +
    \mu_2 \sqrt{1-w_1^2/\kappa_{mn}^2}} \nonumber \\
    %
\end{align}
An asymptotic expansion of the summand $A_{mn}$ for large
$\kappa_{mn}$ can be found by determining the first few coefficients
in the MacLaurin series of the function
\begin{equation}
  \label{eq:AMac}
  f(x) = x \,
  \frac{\mu_1 + \mu_2}%
  {\mu_1 \sqrt{1-w_2^2 x^2} +
    \mu_2 \sqrt{1-w_1^2 x^2}}.
\end{equation}
This task was accomplished with the aid of a computer algebra
system with the result
\begin{equation}
  \label{eq:Aasymp}
  A_{mn} = \frac{c_1}{\kappa_{mn}} +
  \frac{c_3}{\kappa_{mn}^{3}} + O(\kappa_{mn}^{-5}),
\end{equation}
where
\begin{equation}
  \label{eq:cdef}
  c_1 = 1, \quad c_3(\mu_1,\epsilon_1,\mu_2,\epsilon_2)
  = \frac{\mu_1 w^2(\mu_2,\epsilon_2) + \mu_2 w^2(\mu_1,\epsilon_1)}{2(\mu_1 + \mu_2)}
  =\frac{\mu_1 w_2^2 + \mu_2 w_1^2}{2(\mu_1 + \mu_2)}.
\end{equation}

Now consider the summand for the electric scalar potential Green's
function, Equation~\eqref{eq:allpots:Phi}.
  \newcommand{\fac}[1]{\sqrt{1-\frac{w_{#1}^2}{\kappa_{mn}^2}}}
\begin{align}
  \Phi_{mn} &\equiv
  \frac{(\epsilon_1+\epsilon_2)
    (\mu_1\gamma\one_{mn} + \mu_2\gamma\two_{mn})}%
  {(\epsilon_1\gamma\two_{mn}+\epsilon_2\gamma\one_{mn})
    (\mu_1\gamma\two_{mn}+\mu_2\gamma\one_{mn}) }  \nonumber \\
  %
  &=
  \frac{1}{\kappa_{mn}} \,
  \frac{(\epsilon_1+\epsilon_2)\left(\mu_1\fac1 + \mu_2\fac2\right)}%
  {\left(\epsilon_1\fac2 + \epsilon_2\fac1\right) \left(\mu_1\fac2+\mu_2\fac1\right) }
\end{align}
A MacLaurin expansion of this summand in the variable $1/\kappa_{mn}$
yields the result
\begin{equation}
  \label{eq:Phiasymp}
  \Phi_{mn} = \frac{d_1}{\kappa_{mn}} +
  \frac{d_3}{\kappa_{mn}^{3}} + O(\kappa_{mn}^{-5}),
\end{equation}
where
\begin{subequations}
  \label{eq:ddef}
  \begin{align}
    &d_1 = 1, \\
    &d_3(\mu_1,\epsilon_1,\mu_2,\epsilon_2) =
    \frac{\mu_1[w_2^2(2\epsilon_1+\epsilon_2)-w_1^2\epsilon_1]
      + \mu_2[w_1^2(\epsilon_1+2\epsilon_2)-w_2^2\epsilon_2]}%
    {2(\mu_1+\mu_2)(\epsilon_1+\epsilon_2)}.
  \end{align}
\end{subequations}

The potential Green's functions can now be written as (note that we
are showing the explicit dependence of the Green's functions on the
material parameters of Regions~1 and~2):
\begin{subequations}
  \label{eq:gf}
  \begin{align}
    G^A_{xx}(\vecrho-\vecrho';{\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) & =
    \tilde{\mu}
    \left\{
      \Sigma_{M1}({\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) +
      \frac{u}{4\pi}
      \left[
        \Sigma_{S1} + \frac{c_3({\scriptstyle
            \mu_1,\epsilon_1,\mu_2,\epsilon_2})}{u^2} \Sigma_{S2}
        \right]
    \right\}
  \\
  G^\Phi(\vecrho-\vecrho';{\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) &=
  \frac{1}{\bar{\epsilon}}
  \left\{
    \Sigma_{M2}({\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) +
    \frac{u}{4\pi}
    \left[
        \Sigma_{S1} + \frac{d_3({\scriptstyle
            \mu_1,\epsilon_1,\mu_2,\epsilon_2})}{u^2} \Sigma_{S2}
        \right]
    \right\}
  \\
  G^F_{xx}(\vecrho-\vecrho';{\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) & =
  -\tilde{\epsilon}
  \left\{
    \Sigma_{M1}({\scriptstyle \epsilon_1,\mu_1,\epsilon_2,\mu_2}) +
      \frac{u}{4\pi}
      \left[
        \Sigma_{S1} + \frac{c_3({\scriptstyle
            \epsilon_1,\mu_1,\epsilon_2,\mu_2})}{u^2} \Sigma_{S2}
        \right]
    \right\}
  \\
  G^\Psi(\vecrho-\vecrho';{\scriptstyle \mu_1,\epsilon_1,\mu_2,\epsilon_2}) &=
  \frac{1}{\bar{\mu}}
  \left\{
    \Sigma_{M2}({\scriptstyle \epsilon_1,\mu_1,\epsilon_2,\mu_2}) +
    \frac{u}{4\pi}
    \left[
        \Sigma_{S1} + \frac{d_3({\scriptstyle
            \epsilon_1,\mu_1,\epsilon_2,\mu_2})}{u^2} \Sigma_{S2}
        \right]
    \right\}
  \end{align}
\end{subequations}
where
\begin{multline}
  \Sigma_{M1}(x_1,y_1,x_2,y_2)  =
  \frac{1}{2A}
  \summn \\
  \left[
    \frac{x_1+x_2}{x_1\gamma_{mn}(x_2,y_2) + x_2\gamma_{mn}(x_1,y_1)}
    -\frac{1}{\kappa_{mn}} - \frac{c_3(x_1,y_1,x_2,y_2)}{\kappa_{mn}^3}
  \right]
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
  \label{eq:sigmam1}
\end{multline}
\begin{multline}
  \Sigma_{M2}(x_1,y_1,x_2,y_2)  =
  \frac{1}{2A}
  \summn \\
  \biggl[
    \frac{(y_1+y_2)
      (x_1\gamma_{mn}(x_1,y_1) + x_2\gamma_{mn}(x_2,y_2))}%
    {(y_1\gamma_{mn}(x_2,y_2)+y_2\gamma_{mn}(x_1,y_1))
      (x_1\gamma_{mn}(x_2,y_2)+x_2\gamma_{mn}(x_1,y_1)) } \\
    -\frac{1}{\kappa_{mn}} - \frac{d_3(x_1,x_2,y_1,y_2)}{\kappa_{mn}^3}
  \biggr]
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
  \label{eq:sigmam2}
\end{multline}
\begin{align}
  \Sigma_{S1}(x_1,y_1,x_2,y_2)  &=
  \frac{2\pi}{uA}
  \summn
    \frac{1}{\kappa_{mn}}
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}
  \\
  \Sigma_{S2}(x_1,y_1,x_2,y_2)  &=
  \frac{2\pi u}{A}
  \summn
  \frac{1}{\kappa_{mn}^3}
  e^{-j\vecbeta_{mn} \bdot (\vecrho-\vecrho')}.
\end{align}
The modal series $\Sigma_{M1}$ and $\Sigma_{M2}$ are in a form
suitable for direct numerical evaluation.  They are absolutely
convergent since their summands decay as
$(\gamma\eye_{mn})^{-5}$.  Considered as a function of
$\vecrho-\vecrho'$, they are uniformly convergent and
extremely smooth, since all of their
singularities have been subtracted off.  This suggests that an
efficient method of evaluation is to tabulate the series on a grid of
points in the unit cell using the fast Fourier transform (FFT).  Then,
a low-order bivariate interpolation scheme can be used to evaluate the
functions at arbitrary points in the unit cell.  More will said on
this topic later, in Section~\ref{sec:modalFFT}.

The (soon to be) spatial sums $\Sigma_{S1}$ and $\Sigma_{S2}$ are still slowly
convergent, especially $\Sigma_{S1}$ which is at best only
conditionally convergent.  However, they are in a form suitable for
application of the Poisson summation formula, as discussed in the next section.

\subsection{Application of the Poisson Transformation}
We now wish to accelerate the convergence of the double sums
\begin{equation}
S_l =  \frac{1}{2A}
  \summn
  \frac{e^{-j\left[\left(\frac{\psi_1}
          {2\pi}+m\right)\vecbeta_1+\left(\frac{\psi_2}{2\pi}+n\right)
        \vecbeta_2\right]\bdot(\vecrho-\vecrho')}}
  {\kappa_{mn}^{l}}
\end{equation}
for $l=1$ and $3$.
For $m$ and $n$ defined as {\em real\/} variables, let
\begin{equation}
f_l(m,n)  =  \frac{\displaystyle
  e^{-j\left[\left(\frac{\psi_1}{2\pi}+m\right)\vecbeta_1+\left(\frac{\psi_2}
        {2\pi}+n\right)\vecbeta_2\right]\bdot(\vecrho-\vecrho')}}
{2A\left[\norm{\left(\frac{\psi_1}{2\pi}+m\right)\vecbeta_1
      +\left(\frac{\psi_2}{2\pi}+n\right)\vecbeta_2}^2+u^2\right]^{l/2}}
.
\end{equation}
Then by the Poisson summation formula \cite[p.~139]{stak:79}, \cite{papo:62}
\begin{equation}
S_l =   \summn f_l(m,n) = \summn F_l(2\pi m,2\pi n)
\end{equation}
where
\begin{equation}
  F_l(\zeta,\eta)  =  \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
    f_l(m,n)e^{j(\zeta m+\eta n)} \d m\, \d n.
\end{equation}
Let
\begin{equation}
  \mu=\frac{\psi_1}{2\pi}+m \quad \text{and} \quad \nu=\frac{\psi_2}{2\pi}+n,
\end{equation}
so
 \begin{equation}
   F_l(\zeta,\eta)  =
   \frac{e^{-j\left(\frac{\psi_1}{2\pi}\zeta+\frac{\psi_2}
         {2\pi}\eta\right)}}{2A}
   \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
   \frac{e^{-j(\mu\vecbeta_1
       +\nu\vecbeta_2)\bdot(\vecrho-\vecrho')}e^{j(\mu\zeta+\nu\eta)}}
   {(\norm{\mu\vecbeta_1+\nu\vecbeta_2}^2+u^2)^{l/2}}
   \d\mu\,\d\nu
   \label{eq:Fdef}
 \end{equation}
 In order to proceed
 we now introduce the change of variables
 \begin{equation}
   \mu\vecbeta_1 + \nu\vecbeta_2 = \x k_x  + \y k_y = \k .
 \end{equation}
 The differential elements of area in the $k_x$-$k_y$ and
 $\mu$-$\nu$ planes are related as follows:
 \begin{equation}
   \d k_x \, \d k_y =
   \d \mu \, \d \nu \norm{\vecbeta_1 \cross \vecbeta_2} =
   \d \mu \, \d \nu \frac{4\pi^2}{\norm{\s_1 \cross \s_2}} =
   \d \mu \, \d \nu \frac{4\pi^2}{A}.
 \end{equation}
 Recalling the properties of the direct and reciprocal lattice vectors
 \begin{subequations}
   \begin{align}
     \vecbeta_1\bdot\s_1 &= 2\pi, \quad \vecbeta_1\bdot\s_2 = 0 \\
     \vecbeta_2\bdot\s_2 &= 2\pi, \quad \vecbeta_2\bdot\s_1 = 0
   \end{align}
 \end{subequations}
 we see that
 \begin{equation}
   \mu = \k \bdot \s_1 /(2\pi), \quad \nu = \k \bdot \s_2/(2\pi),
   \nonumber
 \end{equation}
so that Equation~\eqref{eq:Fdef} can be written as
 \begin{equation}
   F_l(\zeta,\eta)  =
   \frac{
     e^{-j\left(\frac{\psi_1}{2\pi}\zeta+\frac{\psi_2}{2\pi}\eta\right)}}
   {4\pi}
   B_l\left(\vecrho-\vecrho'-\frac{\zeta\s_1 + \eta\s_2}{2\pi}\right)
   \end{equation}
where
 \begin{equation}
   B_l(\vecrho)  \equiv  \frac{1}{2\pi}
   \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
   \frac{e^{-j\k \bdot \vecrho}}{(\norm{\k}^2 +u^2)^{l/2}}
     \, \d k_x\,\d k_y
     \label{eq:Bl1}
   \end{equation}
The form of the integrand in \eqref{eq:Bl1}
immediately suggests a transformation to polar coordinates:
\begin{align*}
  k_x &= k_{\rho} \cos\theta,  \quad k_y =  k_{\rho} \sin\theta \\
  x &= \rho \cos\phi,  \quad y = \rho \sin\phi
\end{align*}
yielding
\begin{align}
  B_l(\vecrho)
  & =
  \frac{1}{2\pi} \int_0^\infty \int_0^{2\pi}
  \frac{e^{-j k_{\rho} {\rho} \cos(\theta-{\phi})}}
  {(k_{\rho}^2 + u^2)^{l/2}}
  \, k_{\rho} \, \d \theta \, \d k_{\rho}  \nonumber \\
  &=
  \int_0^{\infty} \!\!\!
  \frac{J_0(k_{\rho} {\rho})}
  {(k_{\rho}^2 + u^2)^{l/2}}
  \, k_{\rho} \, \d k_{\rho},
\end{align}
which is recognized as a Hankel (Fourier-Bessel) transform that
depends only on $\rho = \norm{\vecrho}$.
The case $l=1$ is given in \cite[Eqn.~(5.15.5)]{lebe:72}:
\begin{equation}
  B_1(\vecrho)
  =
   \int_0^{\infty} \!\!\!
  \frac{J_0(k_{\rho} {\rho})}
  {(k_{\rho}^2 + u^2)^{1/2}}
  \, k_{\rho} \, \d k_{\rho}
  = \frac{e^{-u{\rho}}}{{\rho}}.
\end{equation}
$B_3$ is easily evaluated by noting that
\begin{align}
  \pdu{B_1(\vecrho)}
  &=
  - e^{-u{\rho}} \nonumber \\
  &=
   \pdu{} \int_0^{\infty}
  \frac{J_0(k_{\rho} {\rho})}
  {(k_{\rho}^2 + u^2)^{1/2}} \, k_{\rho} \, \d k_{\rho}
  \nonumber \\
  &=
  -u \int_0^{\infty}
  \frac{J_0(k_{\rho} {\rho})}
  {(k_{\rho}^2 + u^2)^{3/2}} \, k_{\rho} \, \d k_{\rho}
  \nonumber \\
  &=
  -u B_3(\vecrho)
\end{align}
so that
\begin{equation}
  B_3(\vecrho) = \frac{1}{u} e^{-u {\rho}}.
\end{equation}
We then have
\begin{equation}
  F_l(2\pi m, 2\pi n) = \frac{1}{4\pi} e^{-j(m\psi_1 + n\psi_2)}
  B_l(\vecrho_{mn})
\end{equation}
where
\begin{equation}
  \vecrho_{mn} = \vecrho - \vecrho' - m\s_1 - n\s_2
\end{equation}
is the vector from the $(m,n)$th
translational image of the source point to the observation point.
The spatial series are thus
given explicitly by the formulas:
\begin{subequations}
  \begin{align}
    \Sigma_{S1}
    &=
    \summn
    \frac{e^{-u\rho_{mn}}}{u\rho_{mn}} e^{-j(m\psi_1 + n\psi_2)}
    \\
                                %
    \Sigma_{S2}
    &=
    \summn
    e^{-u\rho_{mn}} e^{-j(m\psi_1 + n\psi_2)}.
  \end{align}
\end{subequations}
Let us take note of the following facts regarding these series:
\begin{itemize}
\item The free-space singularity is explicitly recovered in the $m=n=0$
term of the first series in $\Sigma_{S1}$ and $\Sigma_{S2}$.
\item The summands exhibit exponential decay with $\rho_{mn}$.  An
  astute choice of $u$ will assure convergence using only a few rings
  in the summation lattice.
\item These series consist of frequency-independent terms, which are
  then multiplied by the frequency-dependent constants $c_3$ and $d_3$
  before being combined.  Therefore, it is possible to compute the
  contribution of the frequency-independent sums to the interaction
  matrix that occurs in the moment method procedure once, prior to
  sweeping the analysis frequency.  These contributions can be
  weighted appropriately and combined at each desired analysis
  frequency, avoiding redundant calculations and saving large amounts
  of execution time.  In this sense the formulation of the periodic
  Green's functions can be said to be ``wide-band.''  If this approach
  is pursued, it makes sense to choose $u$ relatively small, so that
  the convergence of the modal series is enhanced, since these will
  have to be evaluated at each frequency while the spatial series will
  be evaluated only once.
\end{itemize}


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\begin{document}
\large

MCB 135 Problem Set 4 \hfill Due Monday, March 2, 2015 at 2:30 PM

\section*{Problem 1: Positive Feedback (adapted from Alon 3.4, 15 points)}

In last week's problem set, you showed that a gene $Z$ with simple regulation decribed by
\begin{eqnarray*}
\frac{dZ}{dt} = \beta_0 - \alpha Z, \hspace{2 cm} Z(0)=0
\end{eqnarray*}
has a response time (i.e., time required to reach half of its steady-state expression level) given by $\tau=\frac{\ln 2}{\alpha}$. Now consider a protein $X$ which positively regulates its own expression:
\[ \frac{dX}{dt} = \beta_0 + \beta_1 X  - \alpha X, \hspace{1 cm} X(0)=0\]

\begin{enumerate}[a)]
\setlength{\itemsep}{0pt}
\item Identify any fixed points of this system and determine their stability. Consider the three cases (i) $\alpha < \beta_1$, (ii) $\alpha = \beta_1$, and (iii) $\alpha > \beta_1$.\\

{\color{red}
\begin{enumerate}[i)]
\item $\alpha < \beta_1$:\\

$dX/dt > 0$ for all non-negative values of $X$: this system has no physically-meaningful fixed points and grows exponentially without bound. (No deduction for mentioning that there is a fixed point at $X* = - \frac{\beta_0}{\beta_1 - \alpha} $ or indicating that it is unstable, though this value of $X$ is physically unrealizable.)

\item $\alpha = \beta_1$:\\

Now $dX/dt = \beta_0$: the system grows linearly without bound; no fixed points.

\item $\alpha > \beta_1$

Setting $dX/dt = 0$, we get:

\[ X^* = \frac{\beta_0}{\alpha - \beta_1} \]

This fixed point is stable since $dX/dt > 0$ for $X< X^*$ and $dX/dt < 0$ for $X > X^*$.

\end{enumerate}

}
\item When the response time of $X$ is well-defined, how does it compare to the response time of $Z$?\\

{
\color{red}
From the above we conclude that the response time is only well-defined when $\alpha > \beta_1$, in which case the response time is the time $\tau$ such that $X(\tau)=  \frac{ \beta_0}{2(\alpha - \beta_1)}$. Following a method analogous to that used on last week's problem set, we find an expression for $X(t)$:

\begin{eqnarray*}
\int \frac{dX}{\beta_0 + \left(\beta_1 - \alpha \right) X} & = & \int dt\\
\frac{1}{\beta_1 - \alpha} \ln \left[ \beta_0 + \left( \beta_1 - \alpha \right) X \right] & = & t + c_0\\
 \beta_0 + \left( \beta_1 - \alpha \right) X & = & c_1e^{-\left(\alpha - \beta_1 \right)t} \hspace{1 cm} X(0)=0 \implies c_1 = \beta_0\\
 X & = & \frac{\beta_0 \left[ 1 - e^{-\left(\alpha - \beta_1 \right)t}  \right]}{\alpha - \beta_1}
\end{eqnarray*}

We can then solve for $\tau$:

\begin{eqnarray*}
\frac{ \beta_0}{2(\alpha - \beta_1)} & = & \frac{\beta_0 \left[ 1 - e^{-\left(\alpha - \beta_1 \right)\tau}  \right]}{\alpha - \beta_1}\\
e^{-\left(\alpha - \beta_1 \right)\tau}  & = & \frac{1}{2}\\
\tau & = & \frac{\ln 2}{\alpha - \beta_1}
\end{eqnarray*}

Because the denominator is smaller than in the simple regulation case, the response time is longer with positive feedback.

}

\item Describe a biological scenario where positive regulation of this type would be preferable to simple regulation. Explain your reasoning.\\

{ \color{red}

Slowed responses can be used to institute a time delay, which may be useful for coordinating the timing of developmental processes.

}

\end{enumerate}

\section*{Problem 2: Global dynamics from local stability analysis (Ingalls 4.8.6, 35 points)}

\begin{enumerate}[a)]
\setlength{\itemsep}{0pt}
\item Consider the chemical reaction network with mass-action kinetics:
\begin{eqnarray*}
\ce{A + X ->[k_1] 2 X} \hspace{2 cm} \ce{X + X ->[k_2] Y}  \hspace{2 cm} \ce{Y ->[k_3] B}
\end{eqnarray*}
Assume that [A] and [B] are held constant.
\begin{enumerate}[i)]
\setlength{\itemsep}{0pt}
\item Write a differential equation model describing the concentrations of $X$ and $Y$.\\

{\color{red}

\begin{eqnarray*}
\frac{dX}{dt} = f(X,Y) & = & k_1A X - 2 k_2 X^2\\
\frac{dY}{dt} = g(X,Y) & = & k_2 X^2 - k_3 Y
\end{eqnarray*}

}

\item Verify that the system has two steady states.\\

{\color{red}

\begin{eqnarray*}
\frac{dX}{dt} & = & X \left( k_1A - 2 k_2 X \right) = 0 \implies X = 0 \textrm{ or } \frac{k_1A}{2k_2}\\
X = 0 \textrm{ and } \frac{dY}{dt}=0 & \implies & Y=0\\
X =  \frac{k_1A}{2k_2} \textrm{ and } \frac{dY}{dt}=0 & \implies & Y = \frac{k_1^2 k_2 A}{4 k_2^2 k_3}
\end{eqnarray*}

The fixed points are $(0,0)$ and $(\frac{k_1A}{2k_2}, \frac{k_1^2 k_2 A}{4 k_2^2 k_3})$.

}

\item Determine the system Jacobian at the steady states and characterize the local behavior of the system near these points.\\

{\color{red}

\begin{eqnarray*}
J & = & \begin{pmatrix} \frac{df}{dX} & \frac{df}{dY}\\ \frac{dg}{dX} & \frac{dg}{dY} \end{pmatrix}_{(X^*, Y^*)} = \begin{pmatrix} k_1 A -4 k_2 X & 0\\ 2k_2 X & - k_3 \end{pmatrix}_{(X^*, Y^*)}
\end{eqnarray*}

Linearizing around the fixed point $(0,0)$ therefore gives

\[ \frac{d}{dt} \begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} k_1 A & 0\\ 0 & - k_3 \end{pmatrix} \begin{pmatrix} X\\Y\end{pmatrix} \]

Since all constants in this system are non-negative, the Jacobian is guaranteed to have one positive and one negative eigenvalue: $\lambda_1 = k_1A > 0$ and $\lambda_2 = -k_3 < 0$. This means the origin is a \textit{saddle}. (For comparison to part iv, we also mention that the eigenvector corresponding to $\lambda_2$ is $\mathbf{v}_2 = (0, 1)^T$, so all trajectories where $X=0$ decay to the origin with time.)\\

Linearizing around the fixed point $(\frac{k_1A}{2k_2}, \frac{k_1^2 k_2 A}{4 k_2^2 k_3})$ gives

\[ \frac{d}{dt} \begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} - k_1 A& 0\\ k_1 A & - k_3 \end{pmatrix} \begin{pmatrix} X\\Y\end{pmatrix} \]

For this fixed point, both eigenvalues $\lambda_1=-k_1A$ and $\lambda_2 = -k_3$ are negative, so this is a stable fixed point.
}

\item By referring to the network, provide an intuitive description of the system behavior from any initial condition for which [X] = 0.\\

{\color{red}

If [X] is initially zero, then it will stay zero forever, because the only reaction that produces X requires some X to already be present. This also means that no more Y will be produced, since Y is produced from X. However,  Y is decaying into B: therefore, [Y] will approach zero as $t \to \infty$.\\

This matches our intuition from the formal stability analysis, since the origin is a saddle and trajectories along the y-axis approach the origin.
}

\item Sketch a phase portrait for the system that is consistent with your expectations from (iii) and (iv).\\

{\color{red}
Drawing not included in this file.
}
\end{enumerate}
\item Repeat for the related system
\begin{eqnarray*}
\ce{A + X ->[k_1] 2 X} \hspace{2 cm} \ce{X + Y ->[k_2] 2 Y}  \hspace{2 cm} \ce{Y ->[k_3] B}
\end{eqnarray*}
In this case, you'll find that the nonzero steady state is a center: it is surrounded by concentric periodic trajectories.\\

{\color{red}

\begin{eqnarray*}
\frac{dX}{dt} = f(X,Y) & = & k_1A X -  k_2 XY\\
\frac{dY}{dt} = g(X,Y) & = & k_2 XY - k_3 Y
\end{eqnarray*}

Find the steady states:

\begin{eqnarray*}
\frac{dX}{dt} =\left( k_1A -  k_2Y \right) = 0 & \implies & X = 0 \textrm{ or } Y = \frac{k_1A}{k_2}\\
\frac{dY}{dt} = Y \left( k_2 X - k_3 \right) = 0 & \implies & Y=0 \textrm{ or } X = \frac{k_3}{k_2}
\end{eqnarray*}

This gives two possible fixed points: $(0, 0)$ and $(\frac{k_3}{k_2}, \frac{k_1A}{k_2})$.

\begin{eqnarray*}
J & = & \begin{pmatrix} \frac{df}{dX} & \frac{df}{dY}\\ \frac{dg}{dX} & \frac{dg}{dY} \end{pmatrix}_{(X^*, Y^*)} = \begin{pmatrix} k_1 A - k_2 Y & -k_2 X\\ k_2 Y & k_2 X - k_3 \end{pmatrix}_{(X^*, Y^*)}
\end{eqnarray*}

Linearizing around the fixed point $(0,0)$ therefore gives

\[ \frac{d}{dt} \begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} k_1 A & 0\\ 0 & - k_3 \end{pmatrix} \begin{pmatrix} X\\Y\end{pmatrix} \]

$\ldots$and exactly as before, the origin has one positive and one negative eigenvalue, and is therefore a saddle.\\

Linearizing around the fixed point $(\frac{k_3}{k_2}, \frac{k_1A}{k_2}))$ gives

\[ \frac{d}{dt} \begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} 0 & -k_3 \\ k_1 A & 0 \end{pmatrix} \begin{pmatrix} X\\Y\end{pmatrix} \]

This matrix has two purely imaginary eigenvalues, so solutions form closed loops around this fixed point.\\

Drawing not included in this file.

}

\end{enumerate}

\section*{Problem 3: Practice with Laplace transforms (adapted from Meister 2009, 50 points)}

Absorption of a single photon by a photoreceptor neuron triggers a biochemical cascade that results in a change in current across the neuron's membrane called the ``single photon response," which lasts for a few hundred milliseconds before decaying away. The neuron's photon detection system is approximately linear and time-invariant: when several photons get absorbed, the resulting current is simply the sum of all their single photon responses.

\begin{enumerate}[a)]
\setlength{\itemsep}{0pt}
\item  We will represent the single photon response for a photon absorbed at time $t=0$ by $h(t)$. What is the photoreceptor's response $O(t)$ to an arbitrary time-varying light stimulus, in which the rate of photon absorptions varies with time as $I(t)$, where
\[ I(t) \, dt = \textrm{number of photons absorbed in short interval } dt \textrm{ ?} \]
Write the answer as a convolution integral, then apply a Laplace transform to find $\tilde{O}(s)$ in terms of $\tilde{I}(s)$ and $\tilde{h}(s)$.

{\color{red}
\begin{eqnarray*}
O(t) & = & I(t) \otimes h(t) = \int_{-\infty}^{\infty} I(t - \tau) h(\tau) \, d\tau = \int_{0}^{t} I(t - \tau) h(\tau) \, d\tau
 \end{eqnarray*}

 It is not necessary to prove that $\tilde{O}(s) = \tilde{h}(s)  \tilde{I}(s)$ as this is a property of the Laplace transform of the convolution integral which we stated in class. However, the derivation is given below for reference.\\

 To narrow the limits on the convolution integral, we have used the fact that $h(\tau)$ is nonzero only for positive $\tau$, and $I(t-\tau)$ is nonzero only for $\tau \leq t$.

\begin{eqnarray*}
 \tilde{O}(s) & = & \int_{0}^{\infty} e^{-st} \left[  \int_{0}^{t} I(t - \tau) h(\tau) \, d\tau \right] \, dt\\
 & = & \int_{0}^{\infty} \left[  \int_{0}^{t} e^{-st} I(t - \tau) h(\tau) \, d\tau \right] \, dt\\
  & = & \int_{0}^{\infty} \left[  \int_{\tau}^{\infty} e^{-st} I(t - \tau) h(\tau) \, dt \right] \, d\tau\\
    & = & \int_{0}^{\infty} \left[  \int_{\tau}^{\infty} e^{-st} I(t - \tau) \, dt \right] h(\tau)  \, d\tau\\
     \end{eqnarray*}

Now, making the substitution $\sigma = t - \tau$, $d\sigma = dt$:

\begin{eqnarray*}
 \tilde{O}(s) & = & \int_{0}^{\infty} \left[  \int_{0}^{\infty} e^{-s(\sigma + \tau)} I(\sigma) \, d\sigma \right] h(\tau)  \, d\tau\\
 & = & \int_{0}^{\infty} \left[  \int_{0}^{\infty} e^{-s\sigma} I(\sigma) \, d\sigma \right] e^{-s\tau} h(\tau)  \, d\tau\\
  & = & \int_{0}^{\infty} \tilde{I}(s) e^{-s\tau} h(\tau)  \, d\tau\\
 &= & \tilde{h}(s)  \tilde{I}(s)
 \end{eqnarray*}
}

\item One way to investigate a system's behavior is to apply a well-defined ``test" input and study the resulting output timecourse. Determine the Laplace transforms $\tilde{I}(s)$ for the two common types of test inputs given below, showing your work:
\begin{enumerate}[i)]
\setlength{\itemsep}{0pt}
\item  $I(t) = \delta(t)= \left\{
     \begin{array}{lr}
       \infty, & t = 0 \\
       0, & t \neq 0
     \end{array}
   \right.$ (Dirac delta/unit impulse)\\

{\color{red}
\begin{eqnarray*}
 \tilde{I}(s) & = & \int_{0}^{\infty} e^{-st} \delta(t) \, dt\\
 & = & \left. e{-st} \right|{t=0} = 1
 \end{eqnarray*}

We have used the fact that the integrand is non-zero at only one point ($t=0$) and that the Dirac delta distribution is defined such that

\[ \int_{-\infty}^{\infty} \delta(t) \, dt = 1 \]
}

\item  $I(t) = \theta(t)= \left\{
     \begin{array}{lr}
       0, & t < 0 \\
       1, & t \geq 0
     \end{array}
   \right.$ (Heaviside step function)\\

{\color{red}
\begin{eqnarray*}
 \tilde{I}(s) & = & \int_{0}^{\infty} e^{-st} \theta(t) \, dt\\
 & = & \int_{0}^{\infty} e^{-st} \, dt = \left. \frac{-e^{-st}}{s} \right|_0^{\infty}\\
 & = & \int_{-\infty}^{\infty} e^{-s\tau} I(t-\tau) h(\tau)\\
 & = & 0 - \left(-\frac{1}{s} \right) = \frac{1}{s}
 \end{eqnarray*}
}

\end{enumerate}
\end{enumerate}
When a very brief flash of light consisting of $N$ photons is delivered to a photoreceptor, the photon detection system's output is:
\[ O(t) = \left\{
     \begin{array}{lr}
       0, & t < 0 \\
       A \sin \left( \omega t \right) e^{-\alpha t}, & t \geq 0
     \end{array}
   \right. \]
   \begin{enumerate}[a)]
\setlength{\itemsep}{0pt}
\setcounter{enumi}{2}
\item Write an expression for $I(t)$ in this case, and then apply a Laplace transform to find an expression for $\tilde{I}(s)$.

{\color{red}
\begin{eqnarray*}
 I(t) & = & N \delta(t)\\
 \tilde{I}(s) & = & N
 \end{eqnarray*}
}

\item Find an expression for $\tilde{O}(s)$ (for the rest of the problem, you may consult a table of Laplace transforms\footnote{A thorough table is available at \url{http://www.dartmouth.edu/\~sullivan/22files/New\%20Laplace\%20Transform\%20Table.pdf}.}).

{\color{red}
\begin{eqnarray*}
 \tilde{O}(s) & = & \frac{A \omega}{\left(s + \alpha \right)^2 + \omega^2} \\
 \end{eqnarray*}
}

\item Find the impulse response $h(t)$ by determining $\tilde{h}(s)$ and finding the inverse Laplace transform.

{\color{red}
\begin{eqnarray*}
 \tilde{h}(s) & = & \frac{\tilde{O}(s)}{\tilde{I}(s)} = \frac{A \omega}{N \left[ \left(s + \alpha \right)^2 + \omega^2 \right]} \\
 \end{eqnarray*}
}

\end{enumerate}
An input of the form
\[ I(t) = \left\{
     \begin{array}{lr}
       0, & t < 0 \\
      M \textrm{ photons per second, } & t \geq 0
     \end{array}
   \right. \]
is then applied. Below you will determine what output $O(t)$ was expected under the assumption that the photon detection system is linear and time-invariant.
\begin{enumerate}[a)]
\setlength{\itemsep}{0pt}
\setcounter{enumi}{4}
\item What is $\tilde{I}(s)$?\\

{\color{red}
\[ \tilde{I}(s) = \frac{M}{s} \]
}

\item Use $\tilde{h}(s)$, calculated in part (e), to express $\tilde{O}(s)$ as a rational function in $s$.\\

{\color{red}
\begin{eqnarray*}
 \tilde{O}(s) & = & \tilde{h}(s)  \tilde{I}(s)\\
 & = & \frac{A M}{N} \frac{\omega}{s \left[ \left(s + \alpha \right)^2 + \omega^2 \right]}
 \end{eqnarray*}
}

\item Determine $O(t)$. What value does $O(t)$ approach as $t \to \infty$?

{\color{red}
\begin{eqnarray*}
 \tilde{O}(s) & = & \frac{A M}{N} \frac{\omega}{\alpha^2 + \omega^2} \left( \frac{\alpha^2 + \omega^2}{s \left[ \left(s + \alpha \right)^2 + \omega^2 \right]} \right)\\
 & = & \frac{A M}{N} \frac{\omega}{\alpha^2 + \omega^2} \left(   \frac{\left( s + \alpha \right)^2 + \omega^2}{s \left[ \left(s + \alpha \right)^2 + \omega^2 \right]}  - \frac{s \left( s + 2 \alpha \right)}{s \left[ \left(s + \alpha \right)^2 + \omega^2 \right]} \right)\\
  & = & \frac{A M}{N} \frac{\omega}{\alpha^2 + \omega^2} \left(   \frac{1}{s}  - \frac{s + \alpha}{\left(s + \alpha \right)^2 + \omega^2} - \frac{\alpha}{\omega} \frac{\omega}{\left(s + \alpha \right)^2 + \omega^2} \right)\\
  O(t) & = & \frac{A M}{N} \frac{\omega}{\alpha^2 + \omega^2} \left( 1 - e^{-\alpha t} \cos \omega t - \frac{\alpha}{\omega} e^{-\alpha t} \sin \omega t \right)\\
 & \to & \frac{A M}{N} \frac{\omega}{\alpha^2 + \omega^2} \textrm{ as } t \to \infty
 \end{eqnarray*}
}

\end{enumerate}

\end{document}


%!TEX root = forallx-ubc.tex
\chapter{Identity}
\label{ch.identity}

This chapter extends QL by adding logical resources for discussion of \emph{identity}. In the logical sense, identity is about which object something is. Two things aren't `identical' in our sense just because they look exactly alike. (Identical twins aren't \emph{logically} identical.) Neither does identity, in our logical sense, have to do with the way an individual conceives of oneself. (We're not getting into whether one \emph{identifies as} a woman, etc.) Instead, we will develop logical vocabulary to talk about whether, e.g., the person who robbed the bank is \emph{the same person as} the person who shot Uncle Ben.

It is already possible to talk about identity in QL, by assigning identity and related concepts to predicates in interpretation keys. (In this sense, it's possible to talk about pretty much anything in QL.) However, there are a few reasons why we may wish to have a more robust logical notion of identity than that. We begin by examining some of these motivations. Identity will strengthen QL, increasing its expressive power in many ways.

\section{Motivating identity as a logical category}
\label{sec.identitymotivation}

Consider this sentence:

\begin{earg}
\item[\ex{onlyyou}] Only you can prevent forest fires.
\end{earg}

How should we translate it into QL? Here is one simple option. We could use this symbolization key:

\begin{ekey}
\item[UD:] people
\item[Px:] only $x$ can prevent forest fires
\item[u:] you
\end{ekey}

And then we could translate sentence \ref{onlyyou} straightforwardly as $Pu$. This might be adequate for some purposes, but this translation misses out on a key fact about sentence \ref{onlyyou}. Notice that sentence \ref{onlyyou} is inconsistent with sentence \ref{onlyher}:

\begin{earg}
\item[\ex{onlyher}] Only your mother can prevent forest fires.
\end{earg}

If we add a name for your mother to the symbolization key above,

\begin{ekey}
\item[m:] your mother
\end{ekey}

we can translate sentence \ref{onlyher} as $Pm$. The problem is that $Pu$ and $Pm$ are obviously consistent in QL, as this simple model demonstrates:

\begin{partialmodel}
	UD & \{$u$, $m$\}\\
	\extension{P} & \{$u$, $m$\}
\end{partialmodel}

The `only' in each sentence has just been made a part of the predicate, in a way that ignores its logical force. (Compare the analogous discussion in the Introduction to Chapter \ref{ch.QL} that motivated extending SL with predicates and quantifiers.) If \emph{only} you can prevent forest fires, that means that no one \emph{else} --- i.e., no one who \emph{is not} you --- can prevent forest fires. We don't yet have the logical vocabulary to express these thoughts in a strong enough way.

These sentences give rise to a related issue:

\begin{earg}
\item[\ex{ppq}] Peter Parker lives in Queens.
\item[\ex{pps}] Peter Parker is Spider-Man.
\item[\ex{sq}] Spider-Man lives in Queens.
\end{earg}

Here is a natural interpretation key for these sentences, given the resources we've seen so far.

\begin{ekey}
\item[UD:] people and places
\item[Lxy:] $x$ lives in $y$
\item[Ixy:] $x$ is $y$
\item[p:] Peter Parker
\item[s:] Spider-Man
\item[q:] Queens
\end{ekey}

Then sentences \ref{ppq}--\ref{sq} can be translated thus:

\begin{earg}
\item[\ref{ppq}.] $Lpq$
\item[\ref{pps}.] $Ips$
\item[\ref{sq}.] $Lsq$
\end{earg}

But the argument from the English \ref{ppq} and \ref{pps} to \ref{sq} seems valid, whereas the corresponding argument form is obviously invalid in QL. This model satisfies the premises and falsifies the conclusion:

\begin{partialmodel}
	UD & \{p, s, q\}\\
	\extension{L} &\begin{tabular}{l|lll}
		$Lxy$   & p & s & q \\ \hline
		p & -     & -  & 1 \\
		s & -     & -  & 0 \\
		q & -     & -  & -
		\end{tabular}\\
	\extension{I} &\begin{tabular}{l|lll}
		$Ixy$   & p & s & q \\ \hline
		p & -     & 1  & - \\
		s & -     & -  & - \\
		q & -     & -  & -
		\end{tabular}
\end{partialmodel}

Given the interpretation key above, this is a model where Peter is Spider-Man, but where Peter and Spider-Man do not live in the same place. This does not correspond to a genuine possibility; if Peter and Spider-Man are the same person, then they must live in the same place. This is an important logical fact about identity. But in the formalism given, $I$ is just another predicate like $L$, susceptible to many different interpretations and extensions.

Here is a third example.

\begin{earg}
\item[\ex{m1}] Mozart composed \emph{Le Nozze di Figaro}.
\item[\ex{m2}] Mozart composed \emph{Don Giovanni}.
\item[\ex{m3}] Mozart composed at least two things.
\end{earg}

We introduce the obvious interpretation key:

\begin{ekey}
\item[UD:] people and operas
\item[Cxy:] $x$ composed $y$
\item[m:] Mozart
\item[n:] \emph{Le Nozze di Figaro}
\item[g:] \emph{Don Giovanni}
\end{ekey}

Then we might translate the argument thus:

\begin{earg}
\item[] $Cmn$
\item[] $Cmg$
\item[\therefore] $\exists x Cmx \eand \exists y Cmy$
\end{earg}

This is a valid argument form. But the problem is that $\exists x Cmx \eand \exists y Cmy$ is a very bad translation of `Mozart composed at least two things'. Notice that it is the conjunction of two existentials, each of which says that Mozart composed at least one thing. Suppose he composed only one thing; then each conjunct is true, so the conjunction is true. To say that he composed at least \emph{two} things, it's not enough to say that he composed at least one thing, and then to say again that he composed at least one thing --- we need some way to ensure that the $x$ and $y$ are \emph{not the same thing}.

What all three of these cases --- the forest fire case, the Spider-Man case, and the Mozart case --- show us is that if we want to reflect the logical relationships having to do with identity, we need special logical vocabulary to do so. Just as we introduced the `$\forall$' and `$\exists$' formalisms to reflect logical facts about quantifiers, we also need a special symbol for identity. That symbol will be `$=$'. Like our other logical symbols --- `\eif', `\enot', `$\forall$', etc. --- its meaning will be fixed as part of our language. Interpretation keys tell us what predicates like $Fx$ mean; they don't get to specify the meaning of the logical terms.

When we introduced quantifiers and predicates, we described the shift as switching to a new language, from SL to QL. In fact, however, QL isn't so much a change away from SL as it is an addition to it. (All SL sentences are QL sentences, and all SL entailments are also QL entailments.) Here we will make one more addition, corresponding to identity. We won't introduce a whole new name for our extended version of QL; it can be specified descriptively as `quantified logic with identity'. (The choice about whether to introduce a new language name is a purely conventional one; we're here just following ordinary practice.)

\section{=}

We revise the definitions of grammaticality and truth in QL to add the following stipulation:

For any two names \script{a} and \script{b}, `$\script{a}{=}\script{b}$' is an atomic sentence in QL. It is true in a model just in case, in that model, the referent of \script{a} is the same object as the referent of \script{b}; i.e., the model assigns \script{a} and \script{b} to the same object. For example, here is a model:

\begin{partialmodel}
	UD & \{Peter Parker, Aunt May\}\\
	\referent{p} & Peter Parker\\
	\referent{s} & Peter Parker\\
	\referent{m} & Aunt May
\end{partialmodel}

In this model, $p{=}s$ is true, and $p{=}m$ and $s{=}m$ are false. We'll introduce a familiar shorthand for negating identity claims; $s {\neq} m$ is our preferred abbreviation for \enot $s{=}m$. (Remember that `$s{=}m$' is an atomic sentence, so `\enot $s {=} m$' is the negation of that sentence. It is tempting, when one looks at that sentence, to attach the negation only to $s$, so that the sentence says that the `negation of $s$' is identical to $m$. But this way of reading it ultimately makes no sense; names do not have negations in QL. Only sentences do. This is part of the reason we will usually prefer to write $s {\neq} m$, rather than the potentially confusing \enot $s{=}m$. But if you do see or write the latter, remember that the negation applies to the whole identity claim.)

A logical notion of identity is just what is needed to overcome the shortcomings of the translations in the introduction to this chapter.

\section{Identity and `no one else'}

Consider again sentence \ref{onlyyou}:

\begin{ekey}
\item[\ref{onlyyou}.] Only you can prevent forest fires.
\end{ekey}

In the introduction we tried translating this into QL with a predicate `only  \blank can prevent forest fires'; now we can reflect its logical structure more fully. So we'll let our predicate just indicate the ability to prevent forest fires:

\begin{ekey}
\item[UD:] people
\item[Px:] $x$ can prevent forest fires
\item[u:] you
\end{ekey}

If only you can prevent forest fires, that means that you can prevent forest fires, and, moreover, no one \emph{else} --- no one who is not you --- can prevent forest fires. The first part is simple. You can prevent forest fires: $Pu$. Now we must say that no one else can prevent forest fires. Another way to say that is this: it is \emph{not} the case that there is someone \emph{who is not you} who can prevent forest fires. So you can think of that as a negated existential: $\enot \exists x (x {\neq} u \eand Px)$. Putting these two pieces together as a conjunction, we can translate sentence \ref{onlyyou} as:

\begin{earg}
\item[\ref{onlyyou}.] $Pu \eand \enot \exists x (x {\neq} u \eand Px)$
\end{earg}

A second way of characterizing the `no one else' clause would translate it as a universal claim, saying, of everything, that if it is not you, then it cannot prevent forest fires: $\forall x (x {\neq} u \eif \enot Px)$. This is equivalent to $\enot \exists x (x {\neq} u \eand Px)$; either is an acceptable translation. (By the end of this chapter we'll be able to prove this equivalence.)

Using similar reasoning, one can translate other kinds of sentences that talk about objects \emph{other than} ones just mentioned. For example, suppose we want to say that Rebecca only thinks about herself. She thinks about herself, and there is no one else that she thinks about. Using $r$ for Rebecca and $Txy$ for `$x$ thinks about $y$', we could say: $$Trr \eand \enot \exists x (x {\neq} r \eand Trx)$$

There are more examples of similar kinds of translation exercises in the practice problems at the end of this chapter.

\section{Identical objects satisfy identical predicates}

In \S\ref{sec.identitymotivation}, we saw that treating identity as a simple predicate prevented the validity of intuitively valid arguments like  this one:

\begin{earg}
\item[\ref{ppq}] Peter Parker lives in Queens.
\item[\ref{pps}] Peter Parker is Spider-Man.
\item[\ref{sq}] Spider-Man lives in Queens.
\end{earg}

The problem was that if the extension of the identity predicate can simply be presented in a model, without regard for the other features of the model, there was no guarantee that $p$ and $s$ --- the names for Peter Parker and Spider-Man --- were related in any particular way, just because \ntuple{Peter Parker, Spider-Man} happened to be assigned in the extension of the $I$ predicate.

But if we translate the identity predicate with the QL identity claim $p{=}s$, this requires that the names $s$ and $p$ be assigned to the \emph{very same object} in the model. Consequently, for any sentence \metaA{} involving the name $p$, \metaA{}\substitute{p}{s} --- the sentence that results from replacing each $p$ in \metaA{} with an $s$ --- will have the same truth value as \metaA{}. If $Fp$ is true, then the object named by $p$ is in the extension of $F$. And if $p{=}s$, then the object named by $p$ is \emph{the very same object as} the object named by $s$. So of course the object named by $s$ (that same object) must also be in the extension of $F$. In general, identical objects have identical properties.

(The converse is not true in general. There is no rule in QL that says that if all the same wffs are true of two different names, they must be two names for the same object.)

If Peter is Spider-Man, then everything true of Peter is also true of Spider-Man. So this argument form, unlike the first translation we attempted, is valid in QL:

\begin{earg}
\item[] $Lpq$
\item[] $p{=}s$
\item[\therefore] $Lsq$
\end{earg}

In \S\ref{sec.identity.trees} below we'll extend our tree method to claims with identity, which will give us a formal means for proving the validity of arguments like this one. (The natural deduction system for QL given in Chapter \ref{ch.QLND} will give us another.)

\section{Quantity}
\label{sec.quantity}

Identity also lets us talk about quantity in a way that wasn't available before. In \S\ref{sec.identitymotivation} we observed some challenges with translating claims about the number of objects that satisfy a particular description. For example, we struggled with sentences like `Mozart composed more than one opera'. But identity makes such claims straightforward.

\begin{ekey}
\item[UD:] people and operas
\item[Cxy:] $x$ composed $y$
\item[Ox:] $x$ is an opera
\item[m:] Mozart
\item[n:] \emph{Le Nozze di Figaro}
\item[g:] \emph{Don Giovanni}
\end{ekey}

We want to say that there are two things, that have the following features: they are different from each other, they are both operas, and they are both composed by Mozart. A double-existential sentence, listing those conjuncts, will suffice:
$$\exists x \exists y (x {\neq} y \eand Ox \eand Oy \eand Cmx \eand Cmy)$$
Strictly speaking, QL conjunctions only have two conjuncts, so to write this sentence scrupulously, we'd need to include brackets to signify that we're talking about conjunctions of conjunctions:
$$\exists x \exists y (x {\neq} y \eand [(Ox \eand Oy) \eand (Cmx \eand Cmy)])$$
In sentences like these --- and the longer ones listing more conditions that will feature soon --- it is usually preferable for readability to leave these internal brackets out. We will write as if our system allows arbitrary numbers of conjuncts, with the understanding that, should the need arise, we could always put them back in explicitly. (We'll rarely have occasion to do so.)

This sentence says that Mozart wrote at least two operas. To say that he wrote at least \emph{three} operas would require another quantifier with a new variable $z$, and the specification that $z$ be an opera, written by Mozart, and non-identical to both of the ones previously mentioned:
\begin{multline*}
\exists x \exists y \exists z (x {\neq} y \eand x {\neq} z \eand y {\neq} z \eand
Ox \eand Oy \eand Oz \eand Cmx \eand Cmy \eand Cmz)
\end{multline*}

As you can see, these sentences get long and complicated rather quickly.

Suppose we wanted to say that Mozart composed \emph{exactly} two operas --- no more, and no fewer. There are two reasonably natural choices for how to say that. One is to conjoin the first QL sentence of this section --- Mozart composed at least two operas --- with the negation of the second --- Mozart did not compose at least three operas. The result is:
\begin{multline*}
\exists x \exists y (x {\neq} y \eand Ox \eand Oy \eand Cmx \eand Cmy) \eand \\ \enot \exists x \exists y \exists z (x {\neq} y \eand x {\neq} z \eand y {\neq} z \eand Ox \eand Oy \eand Oz \eand Cmx \eand Cmy \eand Cmz)
\end{multline*}

A slightly more efficient way of saying the same thing, rather than conjoining a new negated triple existential, would be to add, within the scope of the first two quantifiers, a conjunct to the effect that every opera composed by Mozart is one of the two mentioned in that first conjunction. To say that any opera composed by Mozart is either $x$ or $y$ is to say that, for any $z$, if $z$ is an opera composed by Mozart, then it is either $x$ or $y$. In other words: $\forall z [(Oz \eand Cmz) \eif (z{=}x \eor z{=}y)]$. So another way to say that Mozart composed exactly two operas is:
\begin{multline*}
\exists x \exists y ((x {\neq} y  \eand  Ox \eand Oy \eand  Cmx \eand Cmy) \eand \forall z [(Oz \eand Cmz) \eif (z{=}x \eor z{=}y)])
\end{multline*}

So identity gives us a way to talk about quantities, albeit not a particularly efficient one.

\section{Definite descriptions}
\label{sec.defdesc}
In 1905, Bertrand Russell famously characterized \emph{definite descriptions} in terms of identity. A definite description is a description that implies that only one object satisfies it. Paradigmatically, definite descriptions are ones that use the definite article `the'. If I say `the baby is hungry', I'm saying that the \emph{one and only baby we might be talking about} is hungry. Russell was motivated in part by the apparent fact that one can use this sort of language in a meaningful way even if one is wrong about whether there's any baby around. If there is no baby --- the crying I'm hearing is a recording --- my statement is false, but it's still meaningful. For this reason, Russell was reluctant to suppose that we should understand `the baby' as a name. Remember, in QL, all names have to refer to objects in the UD. Instead, my sentence can be understood as an existentially quantified claim about a unique baby. If I say `the baby is hungry', according to Russell, I'm in effect saying three things: there is a baby, there's no other baby than that one, and that baby is hungry. That second element, the uniqueness claim, can be expressed in QL with identity.

\begin{ekey}
\item[UD:]People in this house
\item[Bx:]$x$ is a baby.
\item[Hx:]$x$ is hungry.
\item[j:]Jonathan
\end{ekey}

A sentence like `Jonathan is hungry' is straightforwardly translated as $Hj$. According to Russell's theory of definite descriptions, `the baby is hungry' has a much more complex logical form: there is a baby $x$, there is no baby other than $x$, and $x$ is hungry.

\begin{equation*}
\exists x [Bx \eand \enot \exists y (By \eand y {\neq} x) \eand Hx]
\end{equation*}

One of the interesting features of Russell's theory is that `the baby is not hungry' is not the negation of `the baby is hungry'. Instead, the negation applies only to the hunger predication in the last conjunct:

\begin{equation*}
\exists x [Bx \eand \enot \exists y (By \eand y {\neq} x) \eand \enot Hx]
\end{equation*}

The reason Russell designed his theory this way was that he thought that both of these sentences equally implied that there is a baby. If there is no baby, then you'd be mistaken in saying either `the baby is hungry' or `the baby is not hungry'. Consequently, one can't be the negation of the other. As a treatment of the truth conditions of English sentences, Russell's theory is controversial. Some philosophers of language think that sentences that seem to presuppose the existence of something that isn't there aren't straightforwardly false, but are rather defective in some other way --- perhaps they fail to be meaningful at all, or perhaps they take on some truth value other than true or false. These matters are beyond our present scope. We will remain neutral on whether Russell's theory is an accurate treatment of English; it is relevant for us because it provides an interesting and useful case for translation of sentences into QL with identity.

The general statement of Russell's theory of definite descriptions is, if you have a sentence applying a predicate \script{P} to the referent of a definite description \script{D}, translate it thus: there is something \script{x} that is \script{D}, there is nothing other than \script{x} that is \script{D}, and \script{x} is \script{P}.

\section{Identity and trees}
\label{sec.identity.trees}

Because we have extended QL to include claims of identity, we also need to modify our tree system. Notice that if we do not do so, the system will not be complete. For example, the rules offered in Chapter \ref{ch.QLsoundcomplete} do not provide a way to demonstrate that $Fa, \enot Fb, a{=}b \models \bot$, even though those sentences are obviously inconsistent. Since each of those sentences are atoms or negated atoms in QL, there is nothing to resolve, and there is no contradiction to close the tree. We need to add new rules to deal with identity.

Here is an obvious one:

\factoidbox{If a branch contains an identity claim \script{a}{=}\script{b} and some sentence \metaA{}, then you may add \metaA{}\substitute{\script{a}}{\script{b}} or \metaA{}\substitute{\script{b}}{\script{a}} to the branch.}

Recall that `\metaA{}\substitute{\script{a}}{\script{b}}' just means the sentence you get by replacing every `\script{a}' in \metaA{} with `\script{b}'. With this rule, we can easily demonstrate that the set of sentences mentioned above closes:

\begin{prooftree}
{
to prove={\{Fa, \enot Fb, a{=}b\} \vdash{}\bot}
}
	[Fa
	[\enot Fb, grouped
	[a{=}b, grouped
		[Fb, just={1, 3 =}, close={2, 4}]
	]
	]
	]
\end{prooftree}

We will add this rule to our tree system. Notice that it is not a \emph{resolution} rule --- we do not put a check mark next to either sentence referenced. They may be used again.

Since our tree system is supposed to be \emph{complete}, we need to \emph{ensure} that the tree will close; it's not enough that we have a rule that allows that it might. So, as in the case of the general rules discussed in Chapter \ref{ch.QLTrees}, we will also add an additional requirement governing branch completion. What we need is to make sure that we've made enough identity substitutions to demonstrate everything that needs demonstrating. The way we'll do this is by requiring, for every identity claim in the branch, that, for at least one of its directions (left-to-right or right-to-left) the substitution has been performed in that direction on every atomic sentence or negated atomic sentence in the branch.

More precisely stated, an open branch is complete only if, for any identity claim $\script{a}{=}\script{b}$ in the branch, either, for any atomic sentence or negated atomic sentence \metaA{} containing \script{a} in the branch, \metaA{}\substitute{\script{a}}{\script{b}} is in the branch, or, for any atomic sentence or negated atomic sentence \metaA{} containing \script{b} in the branch, \metaA{}\substitute{\script{b}}{\script{a}} is in the branch.

So, adding this requirement to the definition of branch completion given on page \pageref{branchcompletion.defined}, we get:

\factoidbox{
A branch is \define{complete} if and only if either (i) it is closed, or (ii) all of the following conditions are met:

\begin{enumerate}
\item Every resolvable sentence in the branch has been resolved;
\item For every general quantified sentence \metaA{} and every name \script{a} in the branch, the \script{a} instance of \metaA{} is in the branch; and
\item For every identity claim $\script{a}{=}\script{b}$ in the branch, either:
	\begin{enumerate}
	\item for every atomic sentence or negated atomic sentence \metaA{} containing \script{a} in the branch, \metaA{}\substitute{\script{a}}{\script{b}} is in the branch, or
	\item for every atomic sentence or negated atomic sentence \metaA{} containing \script{b} in the branch, \metaA{}\substitute{\script{b}}{\script{a}} is in the branch.
	\end{enumerate}
\end{enumerate}}

We must also add one more rule to our tree system. Previously we said that branches only close when they contain some sentence along with its negation. We add one more condition that suffices for branch closure. If any branch contains a sentence $\script{a}{\neq}\script{a}$, then that branch closes. In no model can something be non-identical with itself.

Adding this to the rule given on page \pageref{branchclosure.defined}:

\factoidbox{
A branch closes if and only if either (i) it contains some sentence \metaA{} along with its negation, \enot\metaA{}, or (ii) it contains, for some name \script{a}, the sentence $\script{a}{\neq}\script{a}$.
}

We need this rule, for example, to prove with a tree that $\forall x \:  a{=}x$ is inconsistent with $\exists x \:  x{\neq}a$:

\begin{prooftree}
{
to prove={\{\forall x \: a{=}x, \exists x \: x{\neq}a\} \vdash \bot}
}
	[\forall x \:  a{=}x, subs={b}
	[\exists x \:  x{\neq}a, grouped, checked=b
		[b {\neq} a, just=2 $\exists$
			[a{=}b, just=1 $\forall$
				[b{\neq}b, just={3, 4 {=}}, close=5
				]
			]
		]
	]
	]
\end{prooftree}

We must also take care, when constructing models from completed open branches, to respect identity claims. If for example the sentence $b{=}c$ appears in an open branch, we will not construct a model with two different objects, b and c --- instead, we'll just pick one of them as the object indicated, and assign both names to it. For example, consider this tree:

\begin{prooftree}
{}
	[\forall x (Rax \eif x{=}b), subs={a, b, c}
	[\exists x Rax, grouped, checked=c
		[Rac, just=2 $\exists$
			[Raa \eif a{=}b, just=1 $\forall$, checked
			[Rab \eif b{=}b, grouped, just=1 $\forall$, checked
			[Rac \eif c{=}b, grouped, just=1 $\forall$, checked
				[\enot Rac, just=6 \eif, close={3, 7}
				]
				[c{=}b
					[Rab, just={3, 7 {=}}
						[\enot Rab, just=5 \eif, close={8, 9}]
						[b{=}b
							[\enot Raa, just=4 \eif, open
							]
							[a{=}b
								[Rbb, just={8, 10 =}
									[Rbc, just={3, 10 =}, open
									]
								]
							]
						]
					]
				]
			]
			]
			]
		]
	]
	]
\end{prooftree}

We end up with two open branches. We only need one to demonstrate that the root is satisfiable, but it is useful practice to consider both. Let's begin with the shorter open branch to the left, that completes at line 10. Examining that branch, we see three names, $a$, $b$, and $c$, and three atomic sentences concerning the extension of $R$: $Rac$, $Rab$, and $\enot Raa$. Using our previous method for constructing models, we would have posited a three-object UD and related those objects via $R$ in the appropriate way. Now, however, we also have identity claims to consider. There are two identity claims in the branch. $b{=}b$ doesn't tell us anything interesting --- of \emph{course} that object is identical with itself --- but the presence of $c{=}b$ indicates that our model will not include separate objects b and c. We'll only posit one object in the UD for those two names, and another object for $a$. We can pick either letter for the object; let's call it $b$. (Remember, it doesn't matter what kinds of objects we pick for the UD; it could be Reed Richards for all it matters. We typically pick letters because they are easy to keep track of.)

So the model suggested by the first open branch here is:

\begin{partialmodel}
	UD & \{a, b\}\\
	\referent{a} & a\\
	\referent{b} & b\\
	\referent{c} & b\\
	\extension{R} &
	\begin{tabular}{l|lllll}
	$Rxy$   & $a$ & $b$ \\ \hline
	$a$   & 0 & 1   \\
	$b$   & - & - \\
	\end{tabular}
\end{partialmodel}

The longer open branch includes additional identity claims. It includes $a{=}b$ and $c{=}b$, which of course implies that all three of these names are names for the same object. So the UD for the model corresponding to the right branch will have only one object, which all three names denote.

\begin{partialmodel}
	UD & \{a\}\\
	\referent{a} & a\\
	\referent{b} & a\\
	\referent{c} & a\\
	\extension{R} &
	\begin{tabular}{l|lll}
	$Rxy$   & $a$  \\ \hline
	$a$   & 1    \\
	\end{tabular}
\end{partialmodel}

It it sometimes helpful, for an intuitive grip on what is going on in these trees, to think about an English translation alongside the formalisms. For example, suppose that $Rxy$ is interpreted as `$x$ loves $y$'. Line 1 of the tree says that person $a$ loves at most one person; line 2 says that $a$ loves at least one person. The tree works out two ways that might be: the one and only person $a$ loves could be $a$, or it could be someone else.

\practiceproblems

\solutions
\problempart
\label{pr.QL-ID-spies}
Using the symbolization key given, translate each English-language sentence into QL with identity. For sentences containing definite descriptions, assume Russell's theory.
\begin{ekey}
\item[UD:] people
\item[Kx:] $x$ knows the combination to the safe.
\item[Sx:] $x$ is a spy.
\item[Vx:] $x$ is a vegetarian.
\item[Txy:] $x$ trusts $y$.
\item[h:] Hofthor
\item[i:] Ingmar
\end{ekey}
\begin{earg}
\item Hofthor is a spy, but no vegetarian is a spy.
\item No one knows the combination to the safe unless Ingmar does.
\item No spy knows the combination to the safe.
\item Neither Hofthor nor Ingmar is a vegetarian.
\item Hofthor trusts a vegetarian.
\item Everyone who trusts Ingmar trusts a vegetarian.
\item Everyone who trusts Ingmar trusts someone who trusts a vegetarian.
\item Only Ingmar knows the combination to the safe.
\item Ingmar trusts Hofthor, but no one else.
\item The person who knows the combination to the safe is a vegetarian.
\item The person who knows the combination to the safe is not a spy.
\end{earg}

\solutions
\problempart
\label{pr.QL-ID-cards}
Using the symbolization key given, translate each English-language sentence into QL with identity. For sentences containing definite descriptions, assume Russell's theory.
\begin{ekey}
\item[UD:] cards in a standard deck
\item[Bx:] $x$ is black.
\item[Cx:] $x$ is a club.
\item[Dx:] $x$ is a deuce.
\item[Jx:] $x$ is a jack.
\item[Mx:] $x$ is a man with an axe.
\item[Ox:] $x$ is one-eyed.
\item[Wx:] $x$ is wild.
\end{ekey}
\begin{earg}
\item All clubs are black cards.
\item There are no wild cards.
\item There are at least two clubs.
\item There is more than one one-eyed jack.
\item There are at most two one-eyed jacks.
\item There are exactly two black jacks.
\item There are exactly four deuces.
\item The deuce of clubs is a black card.
\item One-eyed jacks and the man with the axe are wild.
\item If the deuce of clubs is wild, then there is exactly one wild card.
\item The man with the axe is not a jack.
\item The deuce of clubs is not the man with the axe.
\end{earg}

\solutions
\problempart
\label{pr.QLbuffy}
Using the symbolization key given, translate each English-language sentence into QL with identity.
\begin{ekey}
\item[UD:] people, generations, and monsters
\item[Gx:] $x$ is a generation.
\item[Hx:] $x$ is human.
\item[Sx:] $x$ is a slayer.
\item[Vx:] $x$ is a vampire.
\item[Dx:] $x$ is a demon.
\item[Wx:] $x$ is a werewolf.
\item[Fx:] $x$ is a force of darkness.
\item[Axy:] $x$ will stand against $y$.
\item[Bxy:] $x$ is born in generation $y$.
\item[Kxy:] $x$ will kick $y$.
\item[b:] Buffy
\item[f:] Faith
\item[w:] Willow
\end{ekey}
\begin{earg}
\item Buffy and Willow were born unto the same generation.
\item There is no more than one slayer born in each generation.
\item A slayer other than Buffy is one of the forces of darkness.
\item Willow will stand against any force of darkness other than a werewolf.
\item Faith will kick everyone except herself.
\item Buffy will kick anyone who stands against a slayer, unless they are also kicking vampires or demons.
\item In every generation a slayer is born.
\item In every generation a slayer is born. She will stand against the vampires, demons, and forces of darkness.
\item In every generation a slayer is born. She alone will stand against the vampires, demons, and forces of darkness.
\end{earg}

\problempart Using the symbolization key given, translate each English-language sentence into QL with identity. For sentences containing definite descriptions, assume Russell's theory.
\begin{ekey}
\item[UD:] animals in the world
\item[Bx:] $x$ is in Farmer Brown's field.
\item[Hx:] $x$ is a horse.
\item[Px:] $x$ is a Pegasus.
\item[Wx:] $x$ has wings.
\end{ekey}
\begin{earg}
\item There are at least three horses in the world.
\item There are at least three animals in the world.
\item There is more than one horse in Farmer Brown's field.
\item There are three horses in Farmer Brown's field.
\item There is a single winged creature in Farmer Brown's field; any other creatures in the field must be wingless.
\item The Pegasus is a winged horse.
\item The animal in Farmer Brown's field is not a horse.
\item The horse in Farmer Brown's field does not have wings.
\end{earg}

\solutions
\problempart Demonstrate each of the following, either by constructing a model, or by explaining why it's impossible to do so. If you wish, you can draw a tree to help you answer these questions; however, it is good conceptual practice to tackle some of these questions directly by thinking about just what you'd need to put in your model.
\label{pr.IdentityModels}
\begin{earg}
\item Show that $\{{\enot}Raa, \forall x (x{=}a \eor Rxa)\}$
is consistent.
%There are many possible answers. Here is one:
%\begin{partialmodel}
%UD & \{Harry, Sally\}\\
%\extension{R} &\{\ntuple{Sally, Harry}\}\\
%\referent{a} & Harry
%\end{partialmodel}
\item\ Show that $\{\forall x\forall y\forall z(x{=}y \eor y{=}z \eor x{=}z),
\exists x\exists y\ x{\neq} y\}$ is consistent.
%There are no predicates or constants, so we only need to give a UD.
%Any UD with 2 members will do.
\item\ Show that $\{\forall x\forall y\ x{=}y, \exists x\ x {\neq} a\}$ is inconsistent.
%We need to show that it is impossible to construct a model in which these are both true. Suppose $\exists x\ x {\neq} a\$ is true in a model. There is something in the universe of discourse that is \emph{not} the referent of $a$. So there are at least two things in the universe of discourse: \referent{a} and this other thing. Call this other thing \metaB{}--- we know $a {\neq} \metaB{}$. But if $a {\neq} \metaB{}$, then $\forall x\forall y\ x=y$ is false. So the first sentence must be false if the second sentence is true is true. As such, there is no model in which they are both true. Therefore, they are inconsistent.
\item Show that $\exists x (x {=} h \eand x {=} i)$ is contingent.
\item Show that \{$\exists x\exists y(Zx \eand Zy \eand x{=}y)$, $\enot Zd$, $d{=}s$\} is consistent.
\item Show that `$\forall x(Dx \eif \exists y Tyx)$ \therefore\ $\exists y \exists z\ y{\neq} z$' is invalid.
\end{earg}

\solutions
\problempart Construct a tree to test the following entailment claims. If they are false, provide a model that demonstrates this.
\label{pr.IdentityTrees}
\begin{earg}
\item\  $\models \forall x \forall y (x{=}y \eif y{=}x)$
\item\ $\models \forall x \exists y \: x{=}y$
\item\  $\models \exists x \forall y \: x{=}y$
\item   $\exists x \forall y \: x{=}y \models \forall x \forall y (Rxy \eiff Ryx)$
\item   $\models \enot \forall x \forall y \forall z [(Axy \eand Azx \eand y{=}z) \eif Axx] $
\item  $\forall x \forall y \: x{=}y \models \exists x Fx \eiff \forall x Fx$
\item $\forall x (x{=}a \eor x{=}b), Ga \eiff \enot Gb \models \enot \exists x \exists y \exists z (Gx \eand Gy \eand \enot Gz)$
\item $\forall x (Fx \eif x{=}f), \exists x (Fx \eor \forall y \: y{=}x) \models Ff$
\item $\exists x \exists y Dxy \models \forall x_{1} \forall x_{2} \forall x_{3} \forall x_{4} [(Dx_{1}x_{2} \eand Dx_{3}x_{4}) \eif (x_{2}{\neq}x_{3} \eor Dx_{1}x_{4})]$
\end{earg}

\problempart In \S \ref{sec.quantity} we looked at two different translations of `Mozart wrote exactly two operas'. Use trees to prove that they are equivalent.

\problempart Translate these arguments into QL with identity, and evaluate them for validity with a tree. (Don't be surprised or discouraged if some of these trees end up very complex.)
\label{pr.IdentityArguments}
\begin{earg}
\item Dudley will threaten anyone who threatens anyone. Therefore, Dudley will threaten himself.
\item The exam is easy. Therefore every exam Sheila took was easy. (Use Russell's theory of definite descriptions.)
\item Three wise men visited Jesus. Every wise man who visited Jesus gave Jesus a gift. Therefore, Jesus received more than one gift.
\item Worf is the only Klingon in Starfleet. Everyone in Starfleet is brave. All brave Klingons are warriors. Therefore, there is at least one brave warrior in Starfleet.
\item Worf is the only Klingon in Starfleet. Everyone in Starfleet is brave. All brave Klingons are warriors. Therefore, there is exactly one brave warrior in Starfleet.
\item Every person likes every kind of sandwich that is tasty. Jack is a person. Jack likes exactly one kind of sandwich. Therefore, no more than one kind of sandwich is tasty.
\end{earg}


\documentclass[letterpaper]{article}

\usepackage[utf8]{luainputenc}
\usepackage{fullpage}
\usepackage{nopageno}
\usepackage{amsmath}
\usepackage{amssymb}
\allowdisplaybreaks

\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}

\begin{document}
\title{Notes}
\date{11 février, 2015}
\maketitle
\section*{quiz}
\begin{enumerate}
\item
boundedness, continuity, integrable
\item
0
\end{enumerate}

\section*{theorem}
if $f_n\to f$ uniformly on $[a,b]$ and $f_n$ is integrable (riemann, although it's true for lebesgue also) then so is $f$ and $F_n=\int_x^x{f_n(x)\;\mathrm{d}x}\to\int_a^x{f(x)\;\mathrm{d}x}=F(x)$

\subsubsection*{proof}
$|F_n(x)-F(x)|=
\int_a^x{(f_n-f)\;\mathrm{d}t}
\le \int_a^x{|f_n-f|\;\mathrm{d}t}
\le \int_a^x{||f_n-f||_\infty\;\mathrm{d}t}
=||f_n-f||_\infty\int_a^x{\;\mathrm{d}t}
=||f_n-f||_{\infty}(x-a)\le||f_n-f||_{\infty}(b-a)$.
given $\epsilon>0$ find $N$ such that
$||f_n-f||_{\infty}<\epsilon/(b-a)\forall n\ge N$ then
$|F_n(x)-F(x)|\le \epsilon/(b-a)\cdot (b-a)=\epsilon$
so $||F_n-F||_{\infty}\le\epsilon\forall n\ge N$

\subsection*{corollary}
if $f_n\to f$ uniformaly on $[a,b]$ then $\lim\limits_{n\to\infty}\int_a^b{f_n(t)\mathrm{d}t}=\int_a^b{f(t)\mathrm{d}t}$
\subsection*{corollary}
if $\{f_n\}$ are continuously differentiable on $[a,b]$, $f_n'\to g$ uniformly  on $[a,b]$ ant there is $c\in [a,b]$ such that $f_n(c)=\gamma$ for some $\gamma$ then $f_n\to f$ uniformly with $f$ diff func s.t. $f'=g$ and $f(c)=\gamma$

\subsubsection*{proof}
notice $f_n(x)=f_n(0)+\int_c^x{f_n'(t)\mathrm{d}t}\to\gamma+\int_c^x{g(t)\mathrm{d}t}=f(x)$. $f$ is differentiable by the fundamental theorem of calculus on $\int_c^x{g(t)\;\mathrm{d}t}$. $f_n\to f$ uniformly because $\int_c^x{f_n'}\to\int_c^x{g}$ uniformally. $f'=g(f'(x)=0+g(x))$ by FTC. and $f(c)=\gamma+\int_c^c{g(t)\mathrm{d}t}=\gamma$

\subsection*{example}
$f_n(x)=\frac{e^{-n^2x^2}}{n}$
\begin{enumerate}
\item
$f_n\to0$ uniformly
\begin{align*}
  |\frac{e^{-n^2x^2}}{n}-0|\\
  &=|\frac{e^{-n^2x^2}}{n}|\\
  &=|\frac{1}{e^{n^2x^2}}\frac{1}{n}|\\
  &\le \frac{1}{n}\to 0\\
\end{align*}
\item
\begin{align*}
  f_n'&=\frac{-2n^2xe^{-n^2x^2}}{n}=-2nxe^{-n^2x^2}
\end{align*}
which is continuously differentiable
\item
$\lim\limits_{n\to\infty}-2nxe^{-n^2x^2}
=\lim\limits_{n\to\infty}\frac{-2nx}{e^{n^2x^2}}$
and with  l'hopitals rule

$\lim_{n\to\infty}\frac{-2n}{2xn^2e^{n^2x^2}}=\frac{-1}{xne^{n^2x^2}}=0$
\item
$f_n'\not\to 0$ uniformly on any neighborhood of $0$.
\begin{align*}
  ||-2nxe^{-n^2x^2-0}
  &=||-2nxe^{-n^2x^2}\\
  &=2nx||e^{-n^2x^2}\\
  &<\epsilon\\
  ||x||||e^{-n^2x^2}||<\frac{\epsilon}{2n}\\
  \intertext{x close enough to zero}
  ||e^{-n^2x^2}||&\ge \frac{1}{2}\\
  |||x||\frac{1}{2}\le ||x||||e^{-n^2x^2}||<\frac{\epsilon}{2n}\\
  ||x||<\frac{\epsilon}{n}
\end{align*}
when $x$ is close to zero.

point is that it does depend on $x$ and so not uniform.
\end{enumerate}
does this break our theorem?
$f_n(0)\to0=f(0)$, $f'=0=\lim_{n\to\infty}f_n'$. so we don't need all the hypotheses all the time.

\section*{commentary}
the whole point is:

derivatives and integrals are limits. integral is a limit of closer and closer approximations. uniform convergence means you can interchange limits. that is pull limits into integrals and derivates. this is what uinformity gives us. switching limits in various contexts.
\subsection*{lebesgue integral}
lebesgue integral and limits of functions:
\section*{thrm}
if $f_n\to f$ uniformally on $E$ then $\int_E{f_n\;\mathrm{d}m}\to \int_E{f\;\mathrm{d}m}$

\section*{thrm}
dominated convergence thm

if $f_n\to f$ pointwise and there is a $g$ with $\int_E{g\;\mathrm{d}m}>\infty$ with $|f_n(x)|<g(x)$ for all $x$ then $\lim_{n\to\infty}\int_E{f_n\;\mathrm{d}m}=\int_E{f\;\mathrm{d}m}$

if we have one function that bounds $f_n$ then you can get away with pointwise limits
\end{document}


As the background is estimated from data (and not simulation), the uncertainties associated with the background yield are either statistical in nature or associated with the details of the fitting procedure.
The latter uncertainties, which are systematic in nature, are those associated with estimating the background through a fit to events in the \mgg sidebands and are treated through the discrete profiling method mentioned in Sec.~\ref{sec:tth_bkg_models}.

In contrast to the relatively simple uncertainties associated with the background model, there are many sources of uncertainty may affect either the yield or \mgg shape of the Standard Model Higgs boson production modes (often both).

Those affecting only the yield are modeled with log-normal pdfs, which approximate the behavior of a Gaussian pdf in most cases but have one distinct advantage.
A Gaussian pdf is unbounded, meaning the quantity modeled by the pdf could, in principle, take on any value with non-zero probability.
When modeling the uncertainty of positive definite quantities like cross sections, luminosities, and efficiencies, the possibility of a negative value is unphysical.
A log-normal pdf is defined for a parameter $\theta$ as
\begin{equation} \label{eqn:tth_log_normal}
    p(\theta) = \frac{1}{\sqrt{2\pi} \ln(\kappa)} \exp \bigg[ -\frac{1}{2} \bigg(\frac{\ln(\theta/\hat{\theta})}{ \ln\kappa} \bigg)^2 \bigg] \frac{1}{\theta},
\end{equation}
where $\hat{\theta}$ is the most probable value and $\kappa$ is analogous to the width $\sigma$ of a Gaussian.

%On the other hand, those affecting the shape of the \mgg distribution (and likely also the yield) are incorporated directly into the signal model, manifesting as uncertainties in the fitted parameters of the Gaussian and Double Crystal Ball functions used to model each Higgs boson production mode.

The sources are divided into theoretical uncertainties, those relating to imperfect knowledge of Higgs boson production in the Standard Model, and experimental uncertainties, those related specifically to the CMS detector and the methods used to reconstuct each event.

\subsection{Theoretical Uncertainties}
Each theoretical uncertainty described below is calculated separately for each Higgs boson production mode.
Generally, the uncertainty for \ttH has the largest effect on the measurement of $\mu_{\ttH}$, though the uncertainties for other production modes can be significant as well.
Uncertainties are either calculated as an overall uncertainty on the normalization, a ``flat uncertainty'', or as individual variations on a per-event basis.
The flat uncertainties affect only normalization, while the individual uncertainties may also modify kinematics and result in event migration between signal regions.
%The following theoretical uncertainties are taken into consideration in the measurement of $\mu_{\ttH}$:
\begin{itemize}
    \item \emph{Strong coupling constant $(\alpha_s)$}: flat uncertainty in the value of the coupling constant of the strong force, $\alpha_s$. The magnitude is taken following the PDF4LHC prescription~\cite{pdf4lhc}, and is about 2\% for \ttH.
    \item \emph{PDF (parton density function)}: uncertainty due to imperfect knowledge of the structure of the proton. Two distinct PDF uncertainties are considered: a flat uncertainty, computed following the PDF4LHC prescription~\cite{pdf4lhc,higgs_handbook}, and per-event PDF weight variations, taken from the NNPDF3.0 PDF set~\cite{ball_pdf} with the \texttt{MC2Hessian} procedure~\cite{mc2hessian}. The flat uncertainty is about 3\% for \ttH and the per-event PDF weight uncertainties are typically $\leq1\%$.
    \item \emph{QCD scale}: the uncertainty in the renormalization and factorization scales. Values are taken following the recommendations of~\cite{florian2016handbook} and is nearly 10\% for \ttH, making it the single largest systematic uncertainty.
    \item \emph{\Hgg branching fraction}: estimated to be around 2\%~\cite{florian2016handbook}.
    \item \emph{\ggH contamination}: the standard model predictions of Higgs boson production via gluon fusion (\ggH) are not reliable in the \ttH regime with a high number of jets. Three distinct sources contribute to this uncertainty:
    \begin{enumerate}
        \item \emph{Parton shower}: the uncertainty in the gluon fusion yield at a high number of jets (i.e. the uncertainty in the parton shower modeling) is taken from the difference between the jet multiplicity in simulation and that observed in data for fully leptonic \ttbar events, where the dominant production mode is via gluon fusion.
        \item \emph{Gluon splitting modeling}: the uncertainty in gluon splitting to b quarks is taken from the difference between data and simulation in the ratio $\sigma (\ttb\bb) / \sigma (\ttb\jj)$.
        \item \emph{Statistical}: uncertainty in the \ggH estimate due to limited number of simulated events in the high-jet regime.
    \end{enumerate}
\end{itemize}

\subsection{Experimental Uncertainties}
Like the theoretical uncertainties, experimental uncertainties may either be described globally as a flat uncertainty or on a per-event basis.
Additionally, the per-event uncertainties may either modify the central weight of the event (i.e. normalization) or the shape of the \mgg distribution.
The uncertainties which affect the shape of the \mgg distribution are accounted for by performing separate fits of the signal models for the up and down variations of each uncertainty source.
These then manifest themselves as uncertainties in the fitted parameters of the Gaussian and Double Crystal Ball functions used to model each Higgs boson production mode.

The uncertainty sources affecting the shape of the \mgg distribution are:
\begin{itemize}
    \item \emph{Photon energy scale \& resolution}: the uncertainty associated with the corrections derived for the photon energy scales and resolution, described in Sec.~\ref{sec:evt_photon_energy}, is estimated by varying the shower shape variable $R_9$, the electron ID criteria, and the preselection $E_{\text{T}}$ requirement. The variations from each of these sources are added in quadrature with the statistical uncertainty to give the total uncertainty. Separate uncertainties are considered for both the scale and the resolution and each source is additionally split into contributions from {low $R_9$, high $R_9$} $\odot$ {barrel, endcap}.
    \item \emph{Residual \pT dependence of scale corrections}: photon energy scale corrections are derived in \Zee events with $\pT \sim 45$ GeV but applied in \Hgg events with $\pT \sim 60$ GeV, which may introduce error. This uncertainty is conservatively estimated as the magnitude of the correction itself, translating to a 0.1\% uncertainty in the overall photon energy scale.
    \item \emph{Differences between electrons and photons}: nearly all corrections, smearings, scale factors, etc. are derived on electrons in \Zee events, but applied on photons. Several differences between electrons and photons (and their reconstruction in the CMS detector) are used to estimate the uncertainty:
    \begin{enumerate}
        \item \emph{Modeling of the material budget}: in general, electrons shower earlier than photons when passing through the CMS detector. The uncertainty in the material between the interaction point and the ECAL then translates to an additional source of uncertainty.
        \item \emph{Non-uniformity of light collection}: differences in the light collection efficiency (LCE) along the length of ECAL crystals result in a different response to electrons and photons (again due to the fact that electrons shower earlier than photons). This uncertainty is estimated using the LCE model described in~\cite{Adams:2016viv}, derived from optical simulation~\cite{Gentit:2001ky}.
    \end{enumerate}
    \item \emph{Shower shape corrections}: the shower shape corrections described in Sec~\ref{sec:evt_photon_ss} may effect the photon energy scale. The uncertainty is estimated by comparing the energy scale before and after the application of corrections.
\end{itemize}

The remaining sources of uncertainty affect only the overall normalization of a given process, and include:
\begin{itemize}
    \item \emph{Shape of the b-tagging discriminant}: the b-tagging discriminant is corrected in simulation by a continuous reshaping factor (derived as a function of \pT, $\eta$, and jet flavor) such that the distributions between data and simulation agree. The uncertainty in the reshape factor for a given event is calculated as described in ~\cite{Sirunyan_2018_deepcsv} and has an impact of about 4\% on the \ttH signal strength measurement.
    \item \emph{Integrated luminosity}: the total uncertainty in the integrated luminosity is estimated to be about 2\%~\cite{CMS-PAS-LUM-17-001,CMS-PAS-LUM-17-004,CMS-PAS-LUM-18-002}.
    \item \emph{Trigger Scale Factor}: the efficiency of the HLT triggers used for this analysis are calculated using the tag-and-probe method. Simulation is then corrected for this trigger efficiency, with the uncertainty in the efficiency taken as a systematic uncertainty.
    \item \emph{Diphoton Preselection Scale Factor}: the efficiency of the diphoton preselection is calculated for both data and simulation with the tag-and-probe method and a scale factor is derived from the ratio between data and simulation. The uncertainty in this scale factor (binned by barrel/endcap and low/high $R_9$) is then used to calculate the associated systematic uncertainty.
    \item \emph{Photon Identification BDT Score}: the uncertainty in the photon ID BDT score is assumed to stem from the limited size of the training sample used to derive the corrections for its inputs, the shower shape and isolation variables. The magnitude of this uncertainty is estimated by splitting the original training sample in half and deriving two sets of corrections, taking the uncertainty as the difference between the two trainings.
    \item \emph{Jet Energy Scale \& Resolution}: the scaling and smearing factors derived for individual jets each have associated uncertainties, which are propagated to the final result by varying all factors up/down by their uncertainty.
    \item \emph{\met }: there are four individual uncertainty sources associated with the calculation of the \met:
    \begin{itemize}
        \item Jet energy scale: as described above.
        \item Jet energy resolution: as described above.
        \item Photon energy scale: the energy scales of pf photons used in the calculation of \met are varied within their uncertainties, with the result propagated through the \met calculation.
        \item Unclustered pf candidate energy scale: the energy scales of pf candidates not clustered within a jet are varied in the same way.
    \end{itemize}
    \item \emph{Lepton ID and Isolation}: both electrons and muons have scale factors derived with the tag-and-probe method to account for differences in efficiency between data and simulation, with the uncertainty in these scale factors dictating the resulting systematic uncertainty.
\end{itemize}

\subsection{Impact of Systematic Uncertainties}
By far, the largest systematic uncertainty is the uncertainty in the QCD renormalization and factorization scales, with an impact of about 10\% on $\mu_{\ttH}$.
Other large uncertainties include those associated with the shape of the b-tagging discriminant, the integrated luminosity, the parton density function, and the \Hgg branching ratio.
The impacts of the dominant systematic uncertainties on $\mu_{\ttH}$ are shown in Fig.~\ref{fig:tth_impacts}.
\begin{figure} [htbp!]
    \centering
    \includegraphics[width=0.9\linewidth]{figures/tth/impacts_mutth_obs_1.pdf}
    \caption{Impacts of the dominant systematic uncertainties on the measurement of $\mu_{\ttH}$.}
    \label{fig:tth_impacts}
\end{figure}


\input{man_init}
\setcounter{secnumdepth}{0}

\subsubsection{Railway traffic lights}
On a railway line a new station is built.
To guarantee a good throughput of express trains,
an extra track is made near the platforms.
This way the main track remains free for passing
express trains.

%figuur trein.eps
\begin{figure}[bth]
\centerline{\callpsfig{train.eps}{width=1\textwidth}}
\caption{Situation}
\end{figure}

To prevent accidents, a system with traffic lights is constructed.
There are four light ($S_{1a}, S_{1b}, S_{2a}, S_{2b}$)
that are controlled by two signals
($S_{1}\ en\ S_{2}$).
When a signal is at a logical '1', the
corresponding lights $S_{na}\ en\ S_{nb}$
are yellow and red respectively.
In the other case, they are both green.
To control the lights, there are three sensors
($d_{1},d_{2},d_{3}$).
They give a logical '1' when the train is in that segment.
It is your task to create a circuit that takes care that the lights are
controlled in such a way that, when the train drivers obey the signals,
there will not occur any accidents.
Further, the design should be such that when two trains are arriving
at the same time on the main track and side track, the train at the main
track gets priority.
Further it holds that:
\begin{itemize}
\item
The distance to stop a train is less than the distance
between $S_{na}\ and\ S_{nb}$.
\item
It is not possible that a train activitates two sensors at the same time.
\item
The trains are only going from left to right.
\end{itemize}
\end{document}


% $Id$

%\subsection{Design and Implementation Notes}

Internally all route-based communication calls are implemented as sparse matrix multiplications. The precompute step for all of the supported communication methods can be broken up into three steps:
\begin{enumerate}
\item Construction of the sparse matrix for the specific communication method.
\item Generation of the communication pattern according to the sparse matrix.
\item Encoding of the communication pattern for each participating PET in form of an XXE stream.
\end{enumerate}