% -*- root: 00-main.tex -*-

\renewcommand{\theequation}{A.\arabic{equation}}
\renewcommand{\thesubsection}{Appendix \arabic{subsection}}

\section*{Appendix}

\subsection{Simplifying the regularization term}\label{app:reg_term}
The exponentials of the Thikonov regularization prior \eqref{eq:regseg-thikonov} have the general form
  $\vec{v}^T \mathbf{M} \vec{v}$.
If $\mathbf{M}$ is a $n \times n$ diagonal matrix such that $\mathbf{M} = \vec{m} \, \mathbf{I}_n$,
  then:

\begin{equation*}
\vec{v}^T \mathbf{M} \vec{v} = \vec{m} \cdot (\vec{v}^T \mathbf{I}_n \vec{v}) = \vec{m} \cdot \vec{v}^{\circ2},
\end{equation*}
  where we have introduced the Hadamard power notation\footnote{The Hadamard power of a matrix or a vector
  is the power of its elements $\mathbf{M}^{\circ p} = ({m_{ij}}^{p})$}.

In general, the anisotropy of the distortion field is aligned with the
  voxel coordinate system, so
  $\mathbf{A}$ and $\mathbf{B}$ of \eqref{eq:regseg-energy} can be simplified to diagonal matrices
  to regularize the registration process, such that
  $\mathbf{A}= \, \boldsymbol{\alpha}\,\vec{I}_n$ and
  $\mathbf{B}= \, \boldsymbol{\beta}\,\vec{I}_n$.
By substituting into equation \eqref{eq:regseg-energy}, we obtain:

  \begin{align}
  E(\vec{u}) &= \const + \, \underset{l}{\sum} \int_{\Omega_l} \mdist{f'}{l} \,d\vec{r} \,
  \ifthenelse{\boolean{review}}{+}{+ \notag\\ &+}
  \int_{\Omega} \frac12 \left[ \boldsymbol{\alpha} \cdot \vec{u}^{\circ2} + \boldsymbol{\beta} \cdot (\nabla \vec{u})^{\circ2} \right] \,d\vec{r}.
  \label{eq:regseg-app_energy}
  \end{align}

\subsection{Application of the shape-gradients}\label{app:shape_gradients}
The computation of gradients at the locations of the active contours in the
  instant $t$ is based on the work of \cite{herbulot_segmentation_2006}.
Let $F(\vec{r})$ be an ``arbitrary'' function over the image domain
  $\Omega = \Omega_l \cup \Omega_m$ split in two regions $l$ and
  $m$, and $\Gamma_{l,m}$ a closed boundary between them.
We now derive the domain integral w.r.t. $t$:

  \begin{equation}
  \frac{\partial}{\partial t} \int_\Omega F(\vec{r}) d\vec{r} =
  \int_\Omega \frac{\partial}{\partial t}F(\vec{r}) d\vec{r}
  - \int_{\Gamma_{l,m}} F(\vec{r}) \left\langle \frac{\partial \Gamma_{l,m} }{\partial t},
  N_{\Gamma_{l,m}}\right\rangle d\vec{r},
  \end{equation}
%
  where $\left\langle\frac{\partial\Gamma_{l,m}}{\partial t}, N_{\Gamma_{l,m}}\right\rangle$ is
  the projection of the boundary movement on the unit inward normal $N_{\Gamma_{l,m}}$.
Assuming that the region descriptors $\{\boldsymbol{\mu}_l, \boldsymbol{\Sigma}_l\}$ vary slowly enough, we can consider
  that $\frac{\partial}{\partial t} F(\vec{r}) = 0$ and thus:

  \begin{equation}
  \frac{\partial}{\partial t} \int_\Omega F(\vec{r}) d\vec{r} =
  - \int_{\Gamma_{l,m}} F(\vec{r}) \left\langle \frac{\partial \Gamma_{l,m} }{\partial t},
  N_{\Gamma_{l,m}}\right\rangle d\vec{r}.
  \label{eq:regseg-shape_gradients}
  \end{equation}

The equation \eqref{eq:regseg-shape_gradients} is discretized as follows.
First, the surface between limiting regions $l$ and $m$ ($\Gamma_{l,m}$) is explicitly represented by
  a discrete set of vertices $\vec{v}_i$, with $i \in \{0, \ldots, N_p -1 \}$.
Consequently, the inwards normal of the surface $N_{\Gamma_{l,m}}$ is represented by the discrete
  set of normals $\hat{\vec{n}}_i$ at each vertex of the mesh.
The resulting summation is, therefore, discrete and the integral operator is replaced by the sum:

  \begin{align}
  \frac{\partial}{\partial t} \int_\Omega F(\vec{r}) d\vec{r} &=
  \underbracket{\cancel{\int_\Omega \frac{\partial}{\partial t}F(\vec{r}) d\vec{r} }}_{\text{Functional's evolution}}
  - \underbracket{\int_{\Gamma_{l,m}} F(\vec{r}) \left\langle \frac{\partial \Gamma_{l,m}}{\partial t},
  N_{\Gamma_{l,m}}\right\rangle d\vec{r}}_{\text{Shape's evolution}} \notag \\
  & = - \underset{p}{\sum} \frac{1}{A_p} \underset{i}{\sum} \, a_i \, F(\vec{v}_i) \left\langle \underbracket{\frac{\partial \vec{v}_i}{\partial t}}_{\text{speed of }\vec{v}_i},
  \hat{\vec{n}}_{i}\right\rangle.
  \label{eq:regseg-shape_gradient_orig}
  \end{align}
where $a_i$ is the area corresponding to vertex $\vec{v}_i$, and $A_p = \sum_i a_i$ is the total area of the surface $p$.
In the following, we will refer as $w_{p,i} = a_i / A_p $ to the area contribution of $\vec{v}_i$ to the
  total area of the surface it belongs to.
For simplicity, the sum over $p$ can be also removed, as the vertices belong to only one of the total $P$ contours.

Then, the speed of $\vec{v}_i$ is discretized using the artificial time-step parameter $\delta$, as the displacement
  $\frac{\partial \vec{v}_i}{\partial t} = \vec{v}_i(\delta = t+1) - \vec{v}_i(\delta = t)$:

  \begin{equation}
  \frac{\partial}{\partial t} \int_\Omega F(\vec{r}) \, d\vec{r} =
  - \underset{i}{\sum} w_{p,i} F(\vec{v}_i) \frac{\partial \vec{v}_i}{\partial t} \cdot \hat{\vec{n}}_i.
  \label{eq:regseg-shape_gradient_disc1}
  \end{equation}

Since the energy functional is defined over competing regions, the displacement of $\vec{v}_i$ will cause
  an energy exchange between the limiting regions, and therefore $F(\vec{r})$ must be split in
  two terms, $F_{in}(\vec{r})$ corresponding to the interior region and $F_{out}(\vec{r})$ to the exterior:

  \begin{equation}
  \frac{\partial}{\partial t} \int_\Omega F(\vec{r}) \, d\vec{r} =
  - \underset{i}{\sum} \, \frac{\partial \vec{v}_i}{\partial t} \cdot
  \underbracket{w_{p,i} \, \Big[ F_{out}(\vec{v}_i) - F_{in}(\vec{v}_i) \Big] \hat{\vec{n}}_i}_{\bar{s}_i \text{ in Figure 1}}.
  \label{eq:regseg-shape_gradient_disc2}
  \end{equation}

Then, we identify the shape gradient contribution $\vec{g}_k$ on the coefficients $\vec{u}_k$ of the B-spline grid
\revcomment[R\#1-C4]{to obtain the definition of $\vec{g}_k$ given in \eqref{eq:regseg-final_gradient}}:

  \begin{equation}
  \label{eq:regseg-gradient_wshape}
  \begin{split}
  \vec{g}_k &= - \underset{i}{\sum} \left\langle \frac{\partial \vec{v}_i'}{\partial \vec{u}_k}, \bar{s}_i'\right\rangle \\
  \text{with }
  \bar{s}_i' &= w_i \left[ \mdist{f_i'}{out} - \mdist{f_i'}{in} \right] \, \hat{\vec{n}}_i, \\
  \text{and }
  \frac{\partial \vec{v}_i'}{\partial \vec{u}_k} &=
  \frac{\partial}{\partial \vec{u}_k} \left\{ \vec{v}_i + \sum_k \psi_k(\vec{v}_i) \vec{u}_k \right\} = \psi_k(\vec{v}_i)\, \hat{\vec{e}},
  \end{split}
  \end{equation}%
\revcomment[R\#1-C4]{where $\psi_k$ and $\vec{u}_k$ define our B-spline deformation model (\autoref{eq:regseg-nodes_tfm})} and
  $\hat{\vec{e}}$ is the coordinates system's unit vector.


\section{Features}
\label{sec:Features}
Each sample is split into windowed and overlapping blocks. Different features are calculated on each block. The blocks have a length of 20~ms, 50~\% overlap and are Hann windowed.\\
Features used are the RMS, peak, crest factor, spectral centroid, logarithmic spectral centroid, spectral variance, spectral skewness, spectral flatness, spectral brightness and mean spectral absolute slope. They are described in detail as follows.\\
For every feature, the block is given as $x$, $x[k]$ is a frame within the block, $K$ is the number of frames within one block. The FFT of the block is $X$, $X[n]$ is an FFT bin, $2 N$ is the FFT-Length. The features are calculated as follows:
\begin{description}
    \item[RMS:]
        The RMS is a measure for the energy of the block.
        \[
            \mathrm{RMS} = \sqrt{\frac{1}{K}\sum_{k=0}^{K-1}{x[k]^2}}
        \]
    \item[Peak:]
        The peak is simply the maximum of all absolute values within the block.
        \[
            \mathrm{PK} = \max{|x[k]|}
        \]
    \item[Crest Factor:]
        The ratio of the peak to the RMS value is the crest factor.
        \[
            \mathrm{CF} = \frac{\mathrm{PK}}{\mathrm{RMS}}
        \]
    \item[Spectral Centroid:]
        The spectral centroid is the normalized frequency weighted mean of the absolute spectrum. The frequency is normalized to the sampling frequency.
        \[
            \mathrm{SC} = \sum_{n=0}^{N}{|X[n]| \cdot \frac{n}{2N}}
        \]

    \item[Logarithmic Spectral Centroid:]
        To account for human frequency perception, the spectral centroid is calculated relative to a logarithmic frequency axis as well.
        \[
            \mathrm{LSC} = \sum_{n=0}^{N}{|X[n]| \cdot \log{\frac{n}{2 N}+1}}
        \]
    \item[Spectral Variance:]
        The spectral variance gives a value of the change of the absolute spectrum over all frequencies.
        \[
            \mathrm{SV} = \sum_{n=0}^{\frac{K}{2}}{(|X[n]|-\overline{|X[n]|})^2}
        \]
        with the spectral mean
        \[
            \overline{|X[n]|} = \frac{1}{N+1}\sum_{n=0}^{N}{|X[n]|}.
        \]
    \item[Spectral Skewness:]
    		The skewness of the absolute spectrum shows how much its distribution leans towards high or low frequencies.
        \[
            \mathrm{SS} = \sum_{n=0}^{\frac{K}{2}}{(|X[n]|-\overline{|X[n]|})^3}.
        \]
    \item[Spectral Flatness:]
        Given the spectral mean $\overline{|X[n]|}$ and the geometric spectral mean
        \[
            \overline{|X[n]|}_\mathrm{geom} = \sqrt[N+1]{\prod_{n=0}^{N}{|X[n]|}},
        \]
        the spectral flatness is calculated as the ratio of them.
        \[
        		\frac{\overline{|X[n]|}_\mathrm{geom}}{\overline{|X[n]|}}
        \]
    \item[Spectral Brightness:]
        The spectral brightness is the ratio of high to the low frequency energy. To get the energy of low and high frequency, the weighted sum of the absolute spectrum is taken. The weighting of the high frequency bins is given as
        \begin{eqnarray*}
            w_\mathrm{high} &=& 0.5 - \frac{\cos(\omega)}{2}\\
            w_\mathrm{low} &=& \frac{\cos(\omega)}{2} + 0.5
        \end{eqnarray*}
        with $\omega$ as $N$ logarithmic ordered values in the range $[0;~\pi]$. The point of intersection between the low and high frequency weighting was set to approximately 2~kHz.
        \[
            \mathrm{SB} = \frac{\sum_{n=0}^{\frac{K}{2}}{|X[n]|\cdot w_\mathrm{high}}}{\sum_{n=0}^{\frac{K}{2}}{|X[n]|\cdot w_\mathrm{low} }}
        \]
    \item[Mean Absolute Spectral Slope:]
        The mean of the absolute spectral slope gives the amount of change in the spectrum over the frequency. It is calculated by
        \[
            \mathrm{MASS} = \frac{1}{N}\sum_{n=1}^{N}{||X[n]|-|X[n-1]||}
        \]
\end{description}
The features of all blocks of all samples are saved within a Pandas DataFrame\footnote{http://pandas.pydata.org/}. This makes them easily searchable and saveable.


%-------------------------------------------------------------------------------
\section{Working Example}
%-------------------------------------------------------------------------------

\begin{figure}
  \centering
  \begin{subfigure}[b]{0.48\columnwidth}
    \includegraphics[width=\linewidth]{figs/sosp-working_ex3}
    \caption{\label{fig:working_ex1}Proximal cost function at $\bar{x} = 7$.}
  \end{subfigure}
  \hspace{0.1cm}
  \begin{subfigure}[b]{0.48\columnwidth}
    \includegraphics[width=\linewidth]{figs/sosp-working_ex2}
    \caption{\label{fig:working_ex2}Evaluated Proximal Gradient for $\bar{x} = 7$.}
  \end{subfigure}
  \vspace{-15pt}
  \caption{\label{fig:working_ex}Example of Proximal Gradient Evaluation.}
  \vspace{-15pt}
\end{figure}

To give a better intuition for how proximal gradients work in practice, we provide a working example. Consider the function $f(x) =   x \texttt{\&} 4$ shown in figure \ref{fig:ex_funcs}, and suppose we are evaluating it with $\bar{x} = 7$. This type of bitwise operation is common in programs that parse a file for input, since they frequently need to check if flag bits are set in a header or section header of the file. However, discrete functions like bitwise operations are nondifferentiable so we instead evaluate the proximal gradient.

To evaluate the proximal gradient on $f\left(x\right)$, the sampling region must first be defined. \tc{x \& 4} has a Lipschitz Constant of $K=4$ (any bitwise operation with a constant has a Lipschitz Constant with the value of the most significant bit in the constant), so assuming a proximal scaling factor of $\lambda = 1$, the sampling distance bound is $2K = 8$. Therefore, a total of 16 samples must be taken, for $x = [-1, 15]$, and the cost function of the proximal operator, $f(x) + ||x - \bar{x}||_2^2$ evaluated for each possible $x$. Figure \ref{fig:working_ex1} shows the value of the cost function near $x = 7$, there is a clear minimum at $x=8$, where $f\left(8\right) + ||8-7||_2^2 = 1$, so the proximal operator evaluates to $x=8$. The proximal gradient is then $prox_{\nabla f}\left(7\right) = \frac{f\left(7\right) - f\left(8\right)}{7 - 8} = -4$.


\section{Euclidean Structure}

\subsection{Exercise 1}
Applying Cauchy-Schwarz gives that $(x, y) \leq \norm{x}\norm{y} = \norm{x}$, which yields
the desired result.

\subsection{Exercise 2}
Both of these just follow from the fact that a linear space with Euclidean structure is
isomorphic to $R^k$, which has the desired properties. The following feel like cop-out
solutions, but I feel they're fair given this isn't supposed to be a real analysis text.

(i) Let $x_k = \sum_i a_i^{(k)} x^{(i)}$ and $x_j = \sum_i a_i^{(j)} x^{(i)}$. Then we have that
$\norm{x_k - x_j} \to 0 \implies \abs{a_i^{(k)} - a_i^{(j)}} \to 0$. Since $\mathbb{R}$ is
complete, $a_i^{(n)} \to a_i$ for some $a_i \in \mathbb{R}$. Thus $x_k \to x = \sum a_i x^{(i)}$.

(ii) Same logic as (i): the individual $a_i^{(n)}$ have convergent subsequences.

\subsection{Exercise 3}
We also need to assume $X$ is finite dimensional (I think).

(i) Since $X$ is finite dimensional, we have that $x = \sum a_i e^{(i)}$ for some basis $e^{(i)}$. Let
$v$ be the vector whose components are $v_i = \norm{Ae^{(i)}}$. Then
\begin{align*}
        \norm{Ax} &= \norm{\sum a_i Ae^{(i)}} \\
                  &= \norm{(x, v)} \\
                  &\leq \norm{x} \norm{v}
\end{align*}
Since $\norm{v}$ is a constant, $\norm{Ax}$ is bounded on the unit sphere.

(ii) We have that $\norm{A} = \max_x \frac{\norm{Ax}}{\norm{x}} = \max_x \norm{Ax}$.
 The result then follows from $(Ax, y) \leq \norm{Ax}$.

(iii) Let $v$ be as in (i). Then $\norm{Ax_i - Ax_j} \leq \norm{A} \norm{x_i - x_j}$. Since
$\norm{A} \leq \norm{v}$ from (i), we are done (we can make $ \norm{x_i - x_j}$ as small
as we'd like).

\subsection{Exercise 4}
Follows immediately from $(Ax, y) = (x, A^{*}y)$ and Exercise 3 (ii).

\subsection{Exercise 5}
Let $x = y_1 + y_1^{\perp}$ and $z = y_2 + y_2^{\perp}$. Then
\begin{align*}
        (P_Y x, z) &= (y_1, y_2 + y_2^{\perp}) \\
                   &= (y_1, y_2) + 0 \\
                   &= (y_1, y_2) + (y_1^{\perp}, y_2) \\
                   &= (x, P_Y z)
\end{align*}

\subsection{Exercise 6}
Reflection across $x_3 = 0$ sends $(x_1, x_2, x_3) \to (x_1, x_2, -x_3)$. Hence it can be represented as
\begin{align*}
       A =
       \begin{pmatrix}
               1 & 0 & 0 \\
               0 & 1 & 0 \\
               0 & 0 & -1
       \end{pmatrix}
\end{align*}
Which has determinant -1 (product of diagonal terms).

\subsection{Exercise 7}
(a) If $A$ is orthogonal, then $A^{*}A = I$. Since $A^{*} = A^{\top}$ in this case, we immediately get
that the column vectors of $A$ are pairwise orthogonal unit vectors. Similarly, a matrix $A$ with
pairwise orthogonal unit vectors satisfies $A^{*} A = I$, implying that it is orthogonal.

(b) $A^{*} A = I \implies A^{*} = A^{-1} \implies A A^{*} = I$ so $A$ orthogonal implies $A^{*}$ orthogonal.
The result then follows from plugging $A^{*}$ into (a).

\subsection{Exercise 8}
The proof is almost identical to the non-complex case.
\begin{align*}
        (x + ty, x + ty) &= (x, x) + (ty, x) + (x, ty) + \norm{t}^2 (y, y) \\
                         &= (x, x) + t(y, x) + \bar{t}(x, y) + \norm{t}^2 (y, y)
\end{align*}
Plugging in $t = \frac{(x, y)}{(y, y)}$ and using the fact that $(x + ty, x + ty) \geq 0$ for all
complex $t$ gives the desired inequality.

\subsection{Exercise 9}
\subsubsection{Theorem 4}
The proof of Theorem 4 is the same, except we use $y = \sum \bar{b_k} x^{(k)}$ instead.

\subsubsection{Theorem 5}
No changes need to be made to the proof of Theorem 5.

\subsubsection{Theorem 6}
Again, no changes need to be made.

\subsubsection{Theorem 7}
No changes need to be made here either, since $(y, y^{\perp}) = 0 \implies \overline{(y, y^{\perp})} = 0$.

\subsection{Exercise 10}
\subsubsection{Theorem 8}
Parts (i)-(iii) remain the same. For part (iv), we see that
\begin{align*}
        (Ax, y) = (x, A^{*}y) = \overline{(A^{*}y, x)} = \overline{(y, A^{**}x)} = (A^{**}x, y)
\end{align*}

\subsubsection{Theorem 9}
(i) $\norm{kA} = \max_x \frac{\sqrt{(kAx, kAx)}}{\norm{x}} = \sqrt{k\bar{k}} \norm{A} = \norm{k} \norm{A}$.

(ii) Applying triangle inequality to the definition of norm gives the result.

(iii) Comes immediately from $\norm{A(Bx)} \leq \norm{A} \norm{Bx}$.

\subsection{Exercise 11}
Follows from $\norm{M(x) - M(y)} = (\bar{x} - \bar{y}) (x - y) = \norm{x - y}$.

\subsection{Exercise 12}
Same idea as proof of Theorem 10.

\subsection{Exercise 13}
As in Exercise 7, $M^{*} M = I \implies M^{*} = M^{-1} \implies M M^{*} = I$. Similarly,
$M^{*} = M^{-1} \implies (M^{*})^{-1} M^{-1} = I$.

\subsection{Exercise 14}
Associativity follows from associativity of composition. From Exercise 13, if $M$ is unitary, so is $M^{-1}$.
Finally, $I$ is also unitary, so the unitary maps are a group with unit $I$.

\subsection{Exercise 15}
Again, same idea as in Theorem 10: $\det M^{*} \det M = 1$.

\subsection{Exercise 16}
\begin{align*}
        (Mf, Mg) = \int_{-1}^{1} m^2(s) f(s) \bar{g}(s) = \int_{-1}^{1} f(s) \bar{g}(s)
\end{align*}


\subsection{Looking from a \emph{modular} perspective}

Given a Riordan array $\mathcal{M}$, we are interested to look at it
doing a transformation of the matrix $\lbrace m_{nk}\rbrace_{n,k\in\mathbb{N}}$,
building a new one defined by $\lbrace m_{nk}\mod\,p\rbrace_{n,k\in\mathbb{N}}$,
where $p\in\mathbb{N}$ is a module, usually a \emph{prime} number.

This transformation allows us to represent any array $\mathcal{M}$ pictorially,
associating a colour to each \emph{remainder class} of $\equiv_{p}$ relation:
inspecting the produced coloured triangle it is possible to recognize some
patterns in the colouring and we will see that for same classes of Riordan
arrays, these colourings are pretty interesting, while for other classes they
seems to be quite random.

We will attempt to formally prove what can be see in the pictures, in
particular we will discuss the Pascal array $\mathcal{P}$ and the Catalan array
$\mathcal{C}$.  We believe that this kind of study is important in order to
understand and to make conjectures about the combinatorial structure underlying
those arrays and, why not, any array $\mathcal{M}$ in general.

\input{Chapters/back-to-the-basics/main-idea-modular-characterization/assembling-include-figures}

In \autoref{fig:main:idea:modular:characterization} is shown this kind of study,
and it will be addressed in \autoref{ch:modular:characterization}.

\subsection{Looking for characterizations, again}

Altough pretty interesting and fairly general characterizations exists for
a coefficient $m_{nk}$ in a Riordan array $\mathcal{M}$, we would like to
rework some of them, looking from different points of view.

In particular we provide a new characterization, called $h$-\emph{characterization},
which allows us to rewrite a Riordan array $\mathcal{M}=(d(t),h(t))$ only
using function $h$, namely $\mathcal{M}_{h(t)}(\alpha(h(t)),h(t))$, for some
function $\alpha$ in the ``variable'' $h(t)$. If the given array $\mathcal{M}$
is in the \emph{Renewal} subgroup, then the $h$-characterization shows the $A$-sequence's
\ac{gf} in the first component of array $\mathcal{M}_{h(t)}$, automagically.

Moreover, we provide another derivation of a concept similar to $A$-matrix characterization,
found by reasoning from a \emph{column}'s \ac{gf} point of view, instead of
focusing on the combination for a generic element $m_{n+1,k+1}$. This study will be
addressed in \autoref{ch:h:characterization}.

\subsection{Python implementation}

In order to support the present work, we have coded a bunch of classes in the
\emph{Python} programming language, that implement a subset of \emph{Riordan
group} theory. We rest on \emph{Sage} \cite{sage} mathematical framework in
order to do hard math computation and our implementation aims to be truly
object oriented and very minimal, in order to be extended easily.

Nonetheless, it is possible to play with Riordan arrays, inverting them, doing
raw matrix expansions and building a modular representation producing
\LaTeX\,ready code.  All pictures in this document are generated by compilation
of \emph{TikZ} statements, produced by our implementation.


\chapter{Game Theory}
\label{chap:game-theory}

As a warm-up, we include the following result.

\begin{lemma}
	For every game $G$, there exists at least one individually rational strategy profile.
	In other words, $\mathcal{S}_1(G) \ne \emptyset$.
\end{lemma}

\begin{proof}
	For every player $i \in P$ we denote $m_i \in S_i$ the strategy that lower-bounds utility of player $i$:
	\[
		m_i = \text{arg} \max_{t_i \in S_i} \min_{\vect_{-i} \in S_{-i}} u_i(t_i, \vect_{-i}).
	\]
	We claim that the strategy profile $\mm = (m_1, \dots, m_n) \in S$ is individually rational.
	If $\mm$ was not individually rational, there would have to be some player $j \in P$ and a strategy $s_j^* \in S_j$ that would ensure a better worst-case payoff: $\min_{\vecs_{-j} \in S_{-j}}u_j(s_j^*, \vecs_{-j}) > u_j(\mm)$ which would contradict the choice of~$m_j$.
\end{proof}

The main goal of this work is to create a gaming platform of human vs. computer, which trains the human to play towards a PTE.
This \enquote{training} should be conducted by the human playing many games and trying to figure out how to maximize their payoffs.
The primary means of affecting this process is through choosing an appropriate strategy chosen by the computer in each play.
How precisely this response should be chosen is not clear and needs to be determined by experiments.

This chapter introduces three new equilibria which might be useful for the cause described in the previous paragraph: the perfectly transparent best response equilibrium and the perfectly transparent optimal profile equilibrium.
Then, we discuss some relations between these two equilibria and between the other equilibria introduced in \autoref{chap:background}.

We start by defining the most natural way to translate the preemption process for determining a PTE to a notion of best response.

\begin{definition}[Perfectly transparent best response]
	In a game (possibly with ties) $G = (P, S, \uu)$ we say a strategy $s_i^* \in S_i$ is a perfectly transparent best response to a strategy profile $\vecs_{-i} \in S_{-i}$ of the opponent players if it is the Nashian best response to $\vecs_{-i}$ across all profiles that survive the maximum possible number of preemption rounds.
	The weak/strict notion is inherited from the Nashian best response.
	Formally $s_i^*$ is a (\textit{weak}) perfectly transparent best response if either $(s_i^*, \vecs_{-i})$ is a PTE or $\exists k \in \N: (s_i^*, \vecs_{-i}) \in \mathcal{S}_k(G), (s_i, \vecs_{-i}) \notin \mathcal{S}_{k+1}(G) \forall s_i \in S_i$, and
	\[
		\forall (s_i', \vecs_{-i}) \in \mathcal{S}_k(G): u_i(s_i^*, \vecs_{-i}) \ge u_i(s_i', \vecs_{-i}).
	\]
	It is a \textit{strict} perfectly transparent best response if either $(s_i^*, \vecs_{-i})$ is a PTE or $\exists k \in \N: (s_i^*, \vecs_{-i}) \in \mathcal{S}_k(G), (s_i, \vecs_{-i}) \notin \mathcal{S}_{k+1}(G) \forall s_i \in S_i$, and
	\[
		\forall (s_i', \vecs_{-i}) \in \mathcal{S}_k(G): s_i' \ne s_i^* \implies u_i(s_i^*, \vecs_{-i}) > u_i(s_i', \vecs_{-i}).
	\]
\end{definition}

A best response naturally induces an equilibrium.

\begin{definition}[Perfectly transparent best response equilibrium]
	We say a strategy profile $\vecs = (s_1, \dots, s_n) \in S$ is a (\textit{weak}) perfectly transparent best response equilibrium (PTBRE) if $s_i$ is a (weak) perfectly transparent best response to $\vecs_{-i}$ for all players $i \in P$.

	Similarly, $\vecs$ is a \textit{strict} PTBRE if $s_i$ is a strict perfectly transparent best response to $\vecs_{-i}$ for all players $i \in P$.
\end{definition}

It is easy to see that for any game, if a strategy profile is a PTE, then it is also a PTBRE.
Our initial hope was that the other direction also holds, so PTE and PTBRE are equivalent.
Unfortunately, this is not true, as we can see in the following sections.

Next, we can consider the following situation: we are playing a game of 2 players, and we know that our opponent's strategy will always be a perfectly transparent best response to our strategy.
In this case, what strategy should we choose to maximize our payoff?
This is captured in the following definition.

\begin{definition}[Perfectly transparent $i$-best profile]
	For a game $G = (P, S, \uu)$ of two players with $P = \{i, j\}$
	we say a strategy profile $\vecs = (s_1, s_2) \in S$ is a (\textit{weak}) perfectly transparent $i$-best profile if
	$j$'s strategy $s_j$ is a (weak) perfectly transparent best response to $s_i$, and
	\[
		\forall s_i' \in S_i: u_i(\vecs) \ge u_i\left(s_i', b(s_i')\right),
	\]
	where $b(s_i')$ is some weak perfectly transparent best response of player $j$ to $s_i'$ (note that there may be multiple best responses to $s_i'$ but they must all have the same utility for player $i$).

	It is a \textit{strict} perfectly transparent $i$-best profile if $s_j$ is a strict perfectly transparent best response to $s_i$ and
	\[
		\forall s_i' \in S_i: s_i' \ne s_i \implies u_i(\vecs) > u_i\left(s_i', b(s_i')\right).
	\]
	Note that we still consider $b(s_i')$ to be a weak best response even in the strict definition.
	This is to ensure that every strict best profile is also a weak best profile.
\end{definition}

Again, this notion naturally induces an equilibrium.

\begin{definition}[Perfectly transparent best profile equilibrium]
	For a game of two players, we say that a strategy profile $\vecs \in S$ is a (\textit{weak}) perfectly transparent best profile equilibrium (PTBPE) if $\vecs$ is a (weak) perfectly transparent $i$-best profile for both players $i \in P$.

	Similarly, $\vecs$ is a \textit{strict} PTBPE if it is a strict perfectly transparent $i$-best profile for both $i \in P$.
\end{definition}

Another situation that we consider is the following: we are playing a game of two players, and the strategy that we are going to choose is a perfectly transparent best response to the opponent's strategy.
What is the best payoff that we can get---that is, what payoff would we get if the opponent knew we would choose a perfectly transparent best response and were optimizing for \textit{our} payoff instead of theirs?
This is captured by the following definition.

\begin{definition}[Perfectly transparent $i$-optimal profile]
	For a given game $G = (P, S, \uu)$ and player $i \in P$ we say a strategy profile $\vecs = (s_1, \dots, s_n) \in S$ is a (\textit{weak}) perfectly transparent $i$-optimal profile if
	$s_i$ is a (weak) perfectly transparent best response to $\vecs_{-i}$ and
	\[
		\forall \vecs'_{-i} \in S_{-i}: u_i(\vecs) \ge u_i\left(b(\vecs'_{-i}), \vecs'_{-i}\right),
	\]
	where $b(\vecs'_{-i})$ is some weak perfectly transparent best response to $\vecs'_{-i}$ (note that there may be multiple best responses to $\vecs'_{-i}$ but they must all have the same utility for player $i$).

	It is a \textit{strict} perfectly transparent $i$-optimal profile if $s_i$ is a strict perfectly transparent best response to $\vecs_{-i}$ and
	\[
		\forall \vecs_{-i}' \in S_{-i}: \vecs_{-i}' \ne s_{-i} \implies u_i(\vecs) > u_i\left(b(\vecs_{-i}'), \vecs_{-i}'\right).
	\]
	Note that even in the strict definition, we still consider $b(\vecs_{-1})$ to be a weak best response.
	This is to ensure that every strict optimal profile is also a weak optimal profile.
\end{definition}

This definition also induces an equilibrium.

\begin{definition}[Perfectly transparent optimal profile equilibrium]
	We say that a strategy profile $\vecs \in S$ is a (\textit{weak}) perfectly transparent optimal profile equilibrium (PTOPE) if $\vecs$ is a (weak) perfectly transparent $i$-optimal profile for all $i \in P$.

	Similarly, $\vecs$ is a \textit{strict} PTOPE if it is a strict perfectly transparent $i$-optimal profile for all $i \in P$.
\end{definition}

While the other equilibria model selfish behavior, PTOPE models altruistic behavior.
It can be seen as the non-Nashian counterpart to the Berge equilibrium \cite{berge1957theorie}.

Recall that for any game $G = (P, S, \uu)$, a strategy profile $\vecs \in S$ is individually rational if and only if it survives the first preemption round: $\vecs \in \mathcal{S}_1(G)$.
This means that every PTE must be individually rational.
If we think of these equilibria as sets of strategy profiles in any possible game, we can express this as PTE $\subset$ IR.
It is not hard to see that the inclusion is indeed strict as there are games in which $\mathcal{S}_2 \subsetneq \mathcal{S}_1$, so not all individually rational profiles are PTEs.

In the rest of this chapter, we present several such inclusion results that help to understand the relations between individual rationality, perfectly transparent best response equilibria, perfectly transparent equilibria, perfectly transparent optimal profile equilibria, and minimax rationalizability.

These results were obtained by first computing equilibria for all games in large datasets \cite{bigDataset}, and analyzing the results using Apache Spark.
Most of the counterexamples found in the Appendix come from these datasets.
Moreover, not finding counterexamples for certain inclusions initiated many hypotheses that we proved at a later point.`'
An overview of all inclusions that we prove or disprove can be found in \autoref{sec:inclusions-overview} of the Appendix.

\section{Games with Ties}
\label{sec:games-with-ties}

We start by presenting results about the most general setup: games with ties.
All the proofs presented in this section naturally hold also for the more specific cases (symmetric games, games without ties).
The results are summarized by a Venn diagram in \autoref{fig:venn-with-ties}.

We consider the most interesting result of this section to be that both PTBPE and PTOPE are subsets of PTE.
This means that both selfishness and altruism can achieve a perfectly transparent equilibrium under the non-Nashian assumptions.

\begin{figure}[b]
	\centering
	\begin{tikzpicture}[line width=0.25pt]
		\draw (0,0) ellipse (1 and 0.5);
		\draw (-0.2,0.2) ellipse (1.5 and 1.1);
		\draw (1.75,0) ellipse (1 and 0.5);
		\draw (2,0.2) ellipse (1.5 and 1.1);
		\draw (0.9,0.3) ellipse (2 and 2.5);
		\draw (0.3,0.4) ellipse (3.3 and 3.3);
		\draw (1.7,-0.2) ellipse (3.5 and 2.5);
		\draw[rounded corners=2ex] (-3.2,4) rectangle (5.5,-3.2);
		\draw (4.1,-0.3) ellipse (3.7 and 3);
		\node at (-0.25,0) {sPTOPE};
		\node at (-0.3,0.8) {PTOPE};
		\node at (1.95,0) {sPTBPE};
		\node at (1.85,0.8) {PTBPE};
		\node at (0.4,1.6) {sPTBRE};
		\node at (-1.5,2.0) {PTBRE};
		\node at (4.3,0) {PTE};
		\node at (-2.5,3) {IR};
		\node at (6.5,0) {MR};
	\end{tikzpicture}
	\caption{A Venn diagram depicting the inclusion of different equilibria in games with ties.}
	\label{fig:venn-with-ties}
\end{figure}

\begin{observation}
	\label{th:strict-sub-weak}
	We can see from our definition of PTOPE, PTBPE, and PTBRE that any strategy profile satisfying the strict definition of each equilibrium also satisfies the weak definition of the respective equilibrium.
	Thus, we have sPTOPE $\subset$ PTOPE, sPTBPE $\subset$ PTBPE, and sPTBRE $\subset$ PTBRE.
\end{observation}

\begin{observation}
	\label{th:ptope-subset-ptbre}
	For any game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTOPE, then it is a PTBRE.
	Similarly, if $\vecs$ is a strict PTOPE, then it is a strict PTBRE.
	Thus, we have PTOPE $\subset$ PTBRE, and sPTOPE $\subset$ sPTBRE.
\end{observation}

\begin{proof}
	If $\vecs$ is a PTOPE, it is by definition a perfectly transparent $i$-optimal profile for every player $i \in P$.
	If $\vecs$ is a perfectly transparent $i$-optimal profile, then, again by definition, $s_i$ is a perfectly transparent best response to $\vecs_{-i}$.
	This being true for all $i \in P$ is precisely the definition of a PTBRE.
	The same argument holds for strict equilibria.
\end{proof}

\begin{lemma}
	\label{th:ptope-subset-pte}
	If a game $G = (P, S, \uu)$ has a PTOPE $\vecs \in S$, then $\vecs$ is also a PTE, i.e., we have the inclusion PTOPE $\subset$ PTE.
\end{lemma}

\begin{proof}
	For the sake of deriving a contradiction, suppose that there is a game $G = (P, S, \uu)$ with a strategy profile $\vecs = (s_1, \dots, s_n) \in S$ such that $\vecs$ is a PTOPE but not a PTE and let $k$ be the last preemption round in which $\vecs$ is not eliminated (i.e., $\vecs \in \mathcal{S}_k(G)$ but $\vecs \notin \mathcal{S}_{k+1}(G)$).
	Because $\vecs$ is eliminated in the $(k+1)$\textsuperscript{st} round, there must be some player $i \in P$ with a strategy $s_i^* \ne s_i$ which causes the elimination:
	\[
		\forall \vecs_{-i}' \in S_{-i}: (s_i^*, \vecs_{-i}') \in \mathcal{S}_k(G) \implies u_i(s_i^*, \vecs_{-i}') > u_i(\vecs).
	\]
	Let $\vecs_{-i}^* \in S_{-i}$ be the opponents' profile that minimizes $u_i(s_i^*, \vecs_{-i}^*)$, subject to $(s_i^*, \vecs_{-i}^*)\in \mathcal{S}_k(G)$ (there must be at least one such profile to cause $\vecs \notin \mathcal{S}_{k+1}(G)$).
	Let $b \in S_i$ be the perfectly transparent best response to $\vecs_{-i}^*$.
	By the choice of $\vecs_{-i}^*$, we must have $u_i(b, \vecs_{-i}^*) \ge u_i(s_i^*, \vecs_{-i}^*) > u_i(\vecs)$.
	This means that $\vecs$ is not a perfectly transparent $i$-optimal profile, contradicting $\vecs$ being a PTOPE.
\end{proof}

\begin{lemma}
	\label{th:ptbpe-subset-pte}
	If a game $G = (P, S, \uu)$ of two players has a PTBPE $\vecs \in S$, then $\vecs$ is a PTE; we have PTBPE $\subset$ PTE.
\end{lemma}

\begin{proof}
	The proof is very similar to the proof of \autoref{th:ptope-subset-pte}.
	We suppose for the sake of contradiction that there is a game $G = (P, S, \uu)$ with $P = \{1, 2\}$, and a strategy profile $\vecs = (s_1, s_2) \in S$ such that $\vecs$ is a PTBPE but not a PTE and let $k$ be the last preemption round in which $\vecs$ is not eliminated (i.e., $\vecs \in \mathcal{S}_k(G)$ but $\vecs \notin \mathcal{S}_{k+1}(G)$).
	Because $\vecs$ is eliminated in the $(k+1)$\textsuperscript{st} round, there must be some player $i \in P$ with a strategy $s_i^* \ne s_i$ which causes the elimination:
	\[
		\forall s_j' \in S_j: (s_i^*, s_j') \in \mathcal{S}_k(G) \implies u_i(s_i^*, s_j') > u_i(\vecs),
	\]
	where $j = 3 - i$ is the opponent of $i$ and there is at least one $s_j' \in S_j$ which satisfies $(s_i^*, s_j') \in \mathcal{S}_k(G)$.
	Let $b \in S_j$ be the perfectly transparent best response of player $j$ to $s_i^*$.
	We must have $(s_i^*, b) \in \mathcal{S}_k$ and so $u_i(s_i^*, b) > u_i(\vecs)$.
	This means that $\vecs$ is not a perfectly transparent $i$-best profile, contradicting $\vecs$ being a PTBPE.
\end{proof}

\begin{lemma}
	\label{th:ptbpe-subset-ptbre}
	If a game $G = (P, S, \uu)$ of two players has a PTBPE $\vecs \in S$, then $\vecs$ is a PTBRE.
	Similarly, if $\vecs$ is a strict PTBPE, then it is a strict PTBRE.
	We have PTBPE $\subset$ PTBRE, and sPTBPE $\subset$ sPTBRE.
\end{lemma}

\begin{proof}
	Suppose that a game $G = (P, S, \uu)$ with $P = \{i, j\}$ has a PTBPE $\vecs = (s_i, s_j) \in S$.
	By definition, this means that $\vecs$ is a perfectly transparent $i$-best profile, hence $s_j$ is a perfectly transparent best response to $s_i$.
	Moreover, $\vecs$ is a perfectly transparent $j$-best profile, hence $s_i$ is a perfectly transparent best response to $s_j$.
	This implies that $\vecs$ is a PTBRE.
	The same argument holds for strict equilibria.
\end{proof}

\begin{observation}
	\label{th:pte-subset-ir}
	In any game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTE, then it is individually rational.
	We have PTE $\subset$ IR.
\end{observation}

\begin{proof}
	By definition, a profile is IR if it survives the first preemption round, i.e., $\vecs \in \mathcal{S}_1$.
	If $\vecs$ is a PTE, it survives all preemption rounds.
	Thus, it must satisfy $\vecs \in \mathcal{S}_1$.
\end{proof}

\begin{lemma}
	\label{th:ptbre-subset-ir}
	For any game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTBRE, then it is individually rational.
	Thus, we have PTBRE $\subset$ IR.
\end{lemma}

\begin{proof}
	Let us suppose that there is a game $G = (P, S, \uu)$ and a strategy profile $\vecs = (s_1, \dots, s_n) \in S$ such that $\vecs$ is a PTBRE but not individually rational.
	This is equivalent to $s_i$ being a perfectly transparent best response to $\vecs_{-i}$ for all $i \in P$, and $\vecs \notin \mathcal{S}_1(G)$.
	Note that we must also have $(s_i', \vecs_{-i}) \notin \mathcal{S}_1(G)\ \forall s_i' \in S_i,\ \forall i \in P$ because otherwise $s_i$ would not be a perfectly transparent best response to $\vecs_{-i}$ for some player $i$.
	This means the perfectly transparent best response of every player $i \in P$ to $\vecs_{-i}$ is simply a Nashian best response (since all considered strategy profiles are eliminated in the same preemption round).
	Thus, $\vecs$ is a Nash equilibrium, which is always individually rational.
\end{proof}

\begin{lemma}
	\label{th:ptbpe-subset-mr}
	In any game $G = (P, S, \uu)$ of two players with $P = \{i, j\}$, if a strategy profile $\vecs \in S$ is a PTBPE, then it is minimax rationalizable.
	We have PTBPE $\subset$ MR.
\end{lemma}

\begin{proof}
	Suppose for the sake of contradiction that there is a strategy profile $\vecs \in S$, which is a PTBPE but not MR.
	This means there is some $k \in \N$ such that $\vecs \in \mathcal{R}_k$ and $\vecs \notin \mathcal{R}_{k+1}$.
	Suppose WLOG that player $i$ caused the elimination of $\vecs$ in the $k+1$\textsuperscript{st} round.
	This means, player $i$ has some strategy $s_i' \in S_i'$ such that
	\[
		\forall (s_i', s_j') \in \mathcal{R}_k: u_i(s_i', s_j') > u_i(\vecs),
	\]
	and there is at least one $s_j' \in S_j$ such that $(s_i', s_j') \in \mathcal{R}_k$.
	Now, we can observe that $(s_i', s_j) \in \mathcal{R}_k$; otherwise either the whole row or column corresponding to $(s_i', s_j)$ would have to be eliminated in some previous round but we know that $\vecs \in \mathcal{R}_k$ and $(s_i', s_j') \in \mathcal{R}_k$ for some $s_j' \in S_j$.
	Since $u_i(s_i', s_j) > u_i(\vecs)$, it cannot be that $s_i$ is a perfectly transparent best response to $s_j$. Hence, $\vecs$ cannot be a perfectly transparent $j$-best profile, a contradiction.
\end{proof}

All the described inclusions are strict.
In \autoref{tab:overview-nosym-dup} an overview of all inclusions can be found.
For each pair of equilibria $A, B$, it refers to either a proof or a counterexample of the claim \enquote{$A \subseteq B$} (note that $A \subseteq B \land B \not\subseteq A$ implies strict inclusion: $A \subset B$).

Minimax rationalizability intersects with all the other equilibria, as shown in \autoref{tab:PTOPE-eq-minimax}.

\section{Symmetric Games with Ties}
Every proof from the previous section holds even for symmetric games with ties.
However, some counterexamples that we gave are non-symmetric.
In particular, for the following claims we do not have any counterexamples that would prove them:
\begin{enumerate}
	\item sPTOPE $\not\subseteq$ MR. For non-symmetric games proved by \autoref{tab:PTOPE-ne-minimax}.
	\item PTOPE $\not\subseteq$ MR. For non-symmetric games holds transitively because sPTOPE~$\subset$~PTOPE (\autoref{th:strict-sub-weak}), and sPTOPE~$\not\subseteq$~MR (point 1).
	\item PTE $\not\subseteq$ MR. For non-symmetric games holds because PTOPE $\subset$ PTE (\autoref{th:ptope-subset-pte}), and PTOPE $\not\subset$ MR (point 2).
\end{enumerate}
Since we have not found any counterexamples in our datasets, we believe the opposite statements to be true.

\begin{conjecture}
	For any game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTOPE, then it is minimax rationalizable.
	We have PTOPE $\subset$ MR.
\end{conjecture}

\begin{conjecture}
	For any game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTE, then it is minimax rationalizable.
	We have PTE $\subset$ MR.
\end{conjecture}

Due to limited time constraints, we have not been able to prove these conjectures.

All other inclusions are the same as in \autoref{sec:games-with-ties}.
An overview of all inclusion proofs and counterexamples can be found in \autoref{tab:overview-sym-dup}.

\section{Games without Ties}
We can observe that in games without ties, similar to Nash equilibria, weak and strict equilibria are also equivalent in the case of PTBRE and PTOPE.
Thus, in this section, we do not distinguish between them.
Moreover, all the \enquote{is subset of} relations are inherited from \autoref{sec:games-with-ties}.
The only difference is that PTE is not intersecting with PTBRE anymore (the example in \autoref{tab:ties-pte-not-sub-ptbre} indeed has ties).
All inclusions for games without ties are depicted by a Venn diagram in \autoref{fig:venn-general-without-ties}.

\begin{observation}
	\label{th:pte-subset-ptbre}
	If a strategy profile is a PTE, then it must be a PTBRE because there is only one strategy profile to choose from in the last preemption round.
	We have PTE $\subset$ PTBRE.
\end{observation}

\begin{figure}[t]
	\centering
	\begin{tikzpicture}[line width=0.25pt]
		\draw (-0.9,0) ellipse (1.1 and 1.1);
		\draw (0.9,0) ellipse (1.1 and 1.1);
		\draw (0,0) circle (2.2);
		\draw (0,0) circle (3.3);
		\draw[rounded corners=2ex] (-3.5,3.5) rectangle (3.5,-3.5);
		\draw (2.8,0) ellipse (3 and 2.8);
		\node at (-1,0) {PTOPE};
		\node at (1,0) {PTBPE};
		\node at (-0.5,1.45) {PTE};
		\node at (-1.3,2.45) {PTBRE};
		\node at (-2.5,3.0) {IR};
		\node at (4.3,0) {MR};
	\end{tikzpicture}
	\caption{A Venn diagram depicting the inclusion of different equilibria in games without ties.}
	\label{fig:venn-general-without-ties}
\end{figure}

An overview of all inclusion proofs and counterexamples can be found in \autoref{tab:overview-nosym-nodup}.

\section{Symmetric Games without Ties}
In this section, we restrict ourselves to symmetric games without ties.
All the \enquote{subset} relations from the previous section still hold since symmetric games are a special case of general games.
Moreover, none of the equilibria collapses (i.e., they are still strict subsets, even when restricted to symmetric games); this because all the provided counterexamples are symmetric games.
The only new relation that emerges is between minimax rationalizability and~PTE.
All the inclusions are depicted in \autoref{fig:venn-sym-without-ties}.
Furthermore, we present some other interesting observations about symmetric games.

\begin{figure}[t]
	\centering
	\begin{tikzpicture}[line width=0.25pt]
		\draw (-0.9,0) ellipse (1.1 and 1.1);
		\draw (0.9,0) ellipse (1.1 and 1.1);
		\draw (0,0) circle (2.2);
		\draw (0,0) circle (3.3);
		\draw[rounded corners=2ex] (-3.5,3.5) rectangle (3.5,-3.5);
		\draw (2.8,0) ellipse (5 and 3.2);
		\node at (-1,0) {PTOPE};
		\node at (1,0) {PTBPE};
		\node at (-0,1.35) {PTE};
		\node at (-1.3,2.45) {PTBRE};
		\node at (-2.5,3.0) {IR};
		\node at (5,0) {MR};
	\end{tikzpicture}
	\caption{A Venn diagram depicting the inclusion of equilibria in symmetric games without ties.}
	\label{fig:venn-sym-without-ties}
\end{figure}

\begin{observation}
	 In a symmetric game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTE, then it lies on the diagonal, i.e., we have $\vecs = (a, a, \dots, a)$ for some strategy $a$.
\end{observation}

\begin{proof}
	If there was a PTE somewhere else than on the diagonal, we would necessarily have multiple PTEs by symmetry, contradicting the uniqueness of PTE.
\end{proof}

\begin{corollary}
	If a symmetric game $G = (P, S, \uu)$ has a PTE $\vecs \in S$ and $\vecs$ is the perfectly transparent $i$-optimal profile for some player $i \in P$, then $\vecs$ is a PTOPE.
\end{corollary}

\begin{proof}
	Since $\vecs$ is on the diagonal and it is the perfectly transparent $i$-optimal profile for some $i \in P$, by symmetry, it must actually be the perfectly transparent $j$-optimal profile for all $j \in P$, which is the definition of PTOPE.
\end{proof}

\begin{lemma}[Fourny \cite{Fourny20}]
	\label{th:pte-sym-subset-mr}
	In a symmetric game $G = (P, S, \uu)$, if a strategy profile $\vecs \in S$ is a PTE, then it is minimax rationalizable.
	Thus, we have PTE\textsuperscript{sym} $\subset$ MR\textsuperscript{sym}.
\end{lemma}

\begin{proof}
	Suppose, for the sake of deriving a contradiction, that there is a game $G = (P, S, \uu)$ and a strategy profile $\vecs \in S$ that is a PTE but not minimax rationalizable.
	By definition, this means that there is some player $i \in P$ such that $(s_i, \vecs_{-i}')$ is not minimax rationalizable for any $\vecs' \in S$.
	By symmetry, if this statement holds for one player, then it must hold for all players.
	Let $s_j$ be the strategy that minimax-dominates $s_i$ (i.e. causes that $s_i$ is not minimax rationalizable).
	We denote $\vecs^* = (s_j, s_j, \dots, s_j)$.
	In the round where strategy $s_i$ is eliminated, strategy profile $\vecs^*$ must survive---otherwise, strategy $s_j$ would be eliminated as well, but we supposed that $s_j$ is the strategy that caused the elimination of $s_i$.
	Moreover, we must have $u_i(\vecs^*) > u_i(\vecs)\ \forall i \in P$ because $s_j$ minimax-dominates $s_i$.
	This means $\vecs$ is not Pareto-optimal which contradicts $\vecs$ being a PTE.
\end{proof}

An overview of all inclusion proofs and counterexamples can be found in \autoref{tab:overview-sym-nodup}.


\documentclass[12pt]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{color}
\usepackage{geometry}
 \geometry{
 a4paper,
 total={210mm,297mm},
 left=20mm,
 right=20mm,
 top=20mm,
 bottom=20mm,
}
\usepackage[parfill]{parskip}
\usepackage[colorlinks=true]{hyperref}
\hypersetup{urlcolor=blue, citecolor=red}

\newtheorem{theorem}{Theorem}
\newtheorem{definition}{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{rem}[theorem]{Remark}
\newtheorem{pro}[theorem]{Proposition}
\newtheorem{con}[theorem]{Conjecture}

\renewcommand{\vec}[1]{\ensuremath{\overrightarrow{#1}}}

\begin{document}
\begin{center}
{\huge \bf Idealized 2 links} \\
\end{center}
\vspace{1ex}
\begin{flushright}
Ferenc A. Bartha\\
fab2@rice.edu
\end{flushright}

\subsection*{Generalized Coordinates}
We use the following coordinates and notation:
\begin{itemize}
 \item The middle mass point is given in polar coordinates w.r. to the origin and the fixed $x$-, $y$-axes.
 \begin{itemize}
  \item the connection to the origin (fixed) is a weightless rod,
  \item the weight of the mass point is $m_1$,
  \item radius$\, = d_1$, constant,
  \item angle from $x$-axis to the direction $\vec{e_{r_1}} = \theta_1$.
 \end{itemize}
 \item The upper mass point is given in polar coordinates w.r. to middle point
 \begin{itemize}
  \item the connection to the middle point is a weightless rod,
  \item the weight of the mass point is $m_2$,
  \item radius$\, = d_2$, constant,
  \item angle from the direction $\vec{e_{r_1}}$ to the direction $\vec{e_{r_2}} = \theta_2$.
 \end{itemize}
 \item Consequently, we describe our system by $\theta_1$ and $\theta_2$ and their derivatives.
\end{itemize}

\begin{figure}[!ht]
	\centering
	\includegraphics[scale=0.8]{01_Idealized_2links.png}
    \caption{The coordinates}
\end{figure}

Note that none of these coordinate systems (describing the middle or the upper point) is an inertial coordinate system. Both are rotating (thus accelerating) compared to the one fixed in the origin with axis $x$ and $y$.

\newpage

\subsection*{The Lagrangian}
We assume that no external forces act on the system, so we have conservation of energy, that is in our case conservation of kinetic energy.
The Lagrangian is
$$\mathcal{L}(\theta_1, \dot \theta_1, \theta_2, \dot \theta_2) = \frac{1}{2} m_1 |\vec{v_1}|^2 + \frac{1}{2} m_2 |\vec{v_2}|^2.$$
\vspace{1ex}
\noindent
We have
\begin{itemize}\itemsep 0.7ex
 \item conservation of energy: $\mathcal{L} = $ constant,
 \item the position vector of the middle mass from the origin is $d_1 \vec{e_{r_1}}$,
 \item the velocity vector is the derivative $\vec{v_1} = \dot{d_1} \vec{e_{r_1}} + d_1 \dot{\vec{e_{r_1}}} = d_1 \dot{\theta_1} \vec{e_{\theta_1}}$\\
 as $\dot d_1 = 0$ and $\dot{\vec{e_{r_1}}} = \dot{\theta_1} \vec{e_{\theta_1}}$ (see {\bf Appendix}),
 \item thus, the speed$^2$ is $|\vec{v_1}|^2 = (d_1 \dot{\theta_1})^2$,
 \item the position vector of the upper mass from the origin is $d_1 \vec{e_{r_1}} + d_2 \vec{e_{r_2}}$,
 \item the velocity vector is $\vec{v_2} = \vec{v_1} + \dot{d_2} \vec{e_{r_2}} + d_2 \dot{\vec{e_{r_2}}} = d_1 \dot{\theta_1} \vec{e_{\theta_1}} + d_2 (\dot{\theta_1} + \dot{\theta_2}) \vec{e_{\theta_2}}$\\
 as $\dot d_2 = 0$ and $\dot{\vec{e_{r_2}}} = (\dot{\theta_1} + \dot{\theta_2}) \vec{e_{\theta_2}}$  (see {\bf Appendix}),
 \item thus, the speed$^2$ is $|\vec{v_2}|^2 = (d_1 \dot{\theta_1})^2 + (d_2 (\dot{\theta_1} + \dot{\theta_2}))^2 + 2 d_1 d_2 \dot{\theta_1} (\dot{\theta_1} + \dot{\theta_2}) \mathrm{cos}(\mathrm{angle}(\vec{e_{\theta_1}}, \vec{e_{\theta_2}}))$\\
 that is $|\vec{v_2}|^2 = (d_1 \dot{\theta_1})^2 + (d_2 (\dot{\theta_1} + \dot{\theta_2}))^2 + 2 d_1 d_2 \dot{\theta_1} (\dot{\theta_1} + \dot{\theta_2}) \mathrm{cos}\theta_2$.
\end{itemize}

\vspace{1ex}
Thus,
\begin{equation*}
\begin{split}
\mathcal{L}(\theta_1, \dot \theta_1, \theta_2, \dot \theta_2) = &\,\frac{1}{2} (m_1 + m_2) d_1^2 \dot{\theta_1}^2
             + \frac{1}{2} m_2 d_2^2 (\dot{\theta_1} + \dot{\theta_2})^2
             + m_2 d_1 d_2 \dot{\theta_1} (\dot{\theta_1} + \dot{\theta_2}) \mathrm{cos}\theta_2.
\end{split}
\end{equation*}

\subsection*{Equations of Motion}
The equations of motion are derived by taking\\ {\it(Lagrange equations of the $2nd$ kind a.k.a Euler--Lagrange equations)}
$$d/dt [ \partial\mathcal{L}/\partial \dot q_j ] = \partial\mathcal{L}/\partial q_j$$
for all generalized coordinates $q_j \in \{\theta_1, \theta_2 \}$.

This gives us
$$\frac{d}{dt}\left[(m_1 + m_2) d_1^2 \dot{\theta_1}
             + m_2 d_2^2 (\dot{\theta_1} + \dot{\theta_2})
             + m_2 d_1 d_2 (2 \dot{\theta_1} + \dot{\theta_2}) \mathrm{cos}\theta_2\right] = 0 \quad \text{ for } q_j = \theta_1$$
and
\begin{equation*}
 \frac{d}{dt}\left[m_2 d_2^2 (\dot{\theta_1} + \dot{\theta_2})
             + m_2 d_1 d_2 \dot{\theta_1} \mathrm{cos}\theta_2 \right] =
  - m_2 d_1 d_2 \dot{\theta_1} (\dot{\theta_1} + \dot{\theta_2}) \mathrm{sin}(\theta_2) \quad \text{ for } q_j = \theta_2.
\end{equation*}

Writing in the time derivatives we get
\begin{equation*}
\begin{split}
  (m_1 + m_2) d_1^2 \ddot{\theta_1}
 + m_2 d_2^2 (\ddot{\theta_1} + \ddot{\theta_2})
 + m_2 d_1 d_2 [(2 \ddot{\theta_1} + \ddot{\theta_2}) \mathrm{cos}\theta_2 - (2 \dot{\theta_1} + \dot{\theta_2})\dot{\theta_2} \mathrm{sin}\theta_2] = 0
\end{split}
\end{equation*}
that results in
\begin{equation}\label{eq-theta1}
\begin{split}
 \left( m_2 d_2^2 + m_2 d_1 d_2 \mathrm{cos}\theta_2 \right) \ddot{\theta_2} =
 &- \left( (m_1 + m_2) d_1^2 + m_2 d_2^2 + 2 m_2 d_1 d_2 \mathrm{cos}\theta_2 \right) \ddot{\theta_1}\\
 &+ m_2 d_1 d_2 (2 \dot{\theta_1} + \dot{\theta_2}) \dot{\theta_2} \mathrm{sin}\theta_2
 \qquad \qquad \qquad \text{ for } q_j = \theta_1
\end{split}
\end{equation}
and
\begin{equation*}
 m_2 d_2^2 (\ddot{\theta_1} + \ddot{\theta_2})
 + m_2 d_1 d_2 [\ddot{\theta_1} \mathrm{cos}\theta_2 - \dot{\theta_1} \dot{\theta_2} \mathrm{sin}\theta_2] =
- m_2 d_1 d_2 \dot{\theta_1} (\dot{\theta_1} + \dot{\theta_2}) \mathrm{sin}(\theta_2)
\end{equation*}
that is transformed to
$$m_2 d_2^2 \ddot{\theta_2} =
- m_2 d_2^2 \ddot{\theta_1} - m_2 d_1 d_2 \ddot{\theta_1} \mathrm{cos}\theta_2
- m_2 d_1 d_2 \dot{\theta_1}^2 \mathrm{sin}(\theta_2)
$$
and simplified to
\begin{equation}\label{eq-theta2}
 \ddot{\theta_2} = - \left( \left[ 1 + \frac{d_1}{d_2} \mathrm{cos}\theta_2 \right] \ddot{\theta_1} +
 \frac{d_1}{d_2} \dot{\theta_1}^2 \mathrm{sin}(\theta_2) \right)
  \qquad \qquad \qquad \qquad \text{ for } q_j = \theta_2
\end{equation}

\subsection*{Further Transformation of the Equations of Motion}
We want to solve the equations \eqref{eq-theta1} and \eqref{eq-theta2} for
$\ddot{\theta_1}$ and $\ddot{\theta_1}$
(some tools can model them as they readily are, but we need now a more standard form).

Substituting \eqref{eq-theta2} into \eqref{eq-theta1} yields
\begin{equation*}
\begin{split}
 - \left( m_2 d_2^2 + m_2 d_1 d_2 \mathrm{cos}\theta_2 \right) &\left( \left[ 1 + \frac{d_1}{d_2} \mathrm{cos}\theta_2 \right] \ddot{\theta_1} + \frac{d_1}{d_2} \dot{\theta_1}^2 \mathrm{sin}(\theta_2) \right) = \\
  &- \left( (m_1 + m_2) d_1^2 + m_2 d_2^2 + 2 m_2 d_1 d_2 \mathrm{cos}\theta_2 \right) \ddot{\theta_1}\\
 &+ m_2 d_1 d_2 (2 \dot{\theta_1} + \dot{\theta_2}) \dot{\theta_2} \mathrm{sin}\theta_2
\end{split}
\end{equation*}
that is transformed to
\begin{equation*}
\begin{split}
 \big( m_1 d_1^2 + m_2 d_1^2 -& m_2 d_1^2 \mathrm{cos}^2\theta_2 \big) \ddot{\theta_1} =\\
 & m_2 d_1 d_2 (2 \dot{\theta_1} + \dot{\theta_2}) \dot{\theta_2} \mathrm{sin}\theta_2
 + \left( m_2 d_2^2 + m_2 d_1 d_2 \mathrm{cos}\theta_2 \right) \frac{d_1}{d_2} \dot{\theta_1}^2 \mathrm{sin}(\theta_2)
\end{split}
\end{equation*}
and simplified to
\begin{equation}\label{eq-theta1b}
 \left( \tfrac{m_1}{m_2} + \mathrm{sin}^2\theta_2 \right) \ddot{\theta_1} =
  \tfrac{d_2}{d_1} (2 \dot{\theta_1} + \dot{\theta_2}) \dot{\theta_2} \mathrm{sin}\theta_2
 + \left( \tfrac{d_2}{d_1} + \mathrm{cos}\theta_2 \right) \dot{\theta_1}^2 \mathrm{sin}(\theta_2)
\end{equation}

Substitute $\ddot{\theta_1}$ from \eqref{eq-theta1b} into \eqref{eq-theta2} and we reach the desired form.

\subsection*{Deriving the Initial Conditions}
Our experiment is the following. The initial alignment is $\theta_1(0) = \frac{\pi}{2}$ and $\theta_2(0) = 0$. A momentary impulse gives velocity to the middle point and does not affect the upper point.
\begin{itemize}
 \item $\vec{v_1}_{, init} = \vec{v_1}(0) = d_1 \dot{\theta_1}(0) \vec{e_{\theta_1}}(0)$ \\
 \hspace*{0.5cm} $\Rightarrow$ we obtain the initial value $\dot{\theta_1}(0)$.

 \item The upper point is not affected that is $\vec{0} = \vec{v_2}(0) = d_1 \dot{\theta_1}(0) \vec{e_{\theta_1}}(0) + d_2 (\dot{\theta_1}(0) + \dot{\theta_2}(0)) \vec{e_{\theta_2}}(0)$\\
 \hspace*{0.5cm} $\Rightarrow$ we obtain the initial value $\dot{\theta_2}(0) = - \frac{d_1}{d_2} \dot{\theta_1}(0) - \dot{\theta_1}(0)$\\
 \hspace*{1.1cm} as $\vec{e_{\theta_1}}(0) = \vec{e_{\theta_2}}(0)$ due to the initial alignment.
\end{itemize}

Recall that $\dot\theta_2$ is a relative coordinate with respect to the middle point that is in motion at time $0$.
Thus, as the upper point is not in motion viewing from the fixed coordinate system centered at the origin, it is moving
when viewed from the middle point, therefore, it has nonzero {\bf relative} velocity $d_2 (\dot{\theta_1}(0) + \dot{\theta_2}(0)) \vec{e_{\theta_2}}(0) = - d_1 \dot{\theta_1}(0) \vec{e_{\theta_2}}(0) = - d_1 \dot{\theta_1}(0) \vec{e_{\theta_1}}(0) = - \vec{v_1}(0)$.

\newpage
\subsection*{Appendix}
The derivation for $\dot{\vec{e_{r_1}}}(t) = \dot{\theta_1}(t) \vec{e_{\theta_1}}(t)$ is pretty straightforward. Centered at the origin,
$\vec{e_{r_1}}(t)$ and $\vec{e_{\theta_1}}(t)$ form a coordinate system that rotates with speed $\dot{\theta_1}(t)$ around the origin
(thus, rotates with respect to the inertial coordinate system). This rotation is actually around the $z$-axis (lifting it into 3D).

Given a position vector $\vec{p}_{relative}(t)$ of a moving point in this rotating coordinate system, its velocity (within this rotating system) is $\dot{\vec{p}}_{relative}(t)$. Looking from the origin in the inertial coordinate system, the effect of the rotation needs to be taken into account:
$$\dot{\vec{p}}_{global}(t) = \dot{\vec{p}}_{relative}(t) + \dot\theta_1(t) \vec{k} \times \vec{p}_{global}(t).$$
Here $\vec{k}$ is the axis of rotation, the direction of the $z$-axis.

Applying this for $\vec{p}_{global}(t) = \vec{e_{r_1}}(t)$, we get
$$\dot{\vec{e_{r_1}}}(t) = \dot{(1,0)} + \dot\theta_1(t) \vec{k} \times \vec{e_{r_1}}(t) = \dot\theta_1(t) \vec{k} \times \vec{e_{r_1}}(t).$$
Now as $\vec{e_{\theta_1}}(t) = \vec{k} \times \vec{e_{r_1}}(t)$ by definition, we have shown the sought relation.

\vspace{1cm}
Similar argument works for $\dot{\vec{e_{r_2}}}(t) = (\dot{\theta_1}(t) + \dot{\theta_2}(t)) \vec{e_{\theta_2}}(t)$ as well.
Translate $\dot{\vec{e_{r_2}}}(t)$ to the origin and observe that it actually rotates with speed
$(\dot{\theta_1}(t) + \dot{\theta_2}(t))$ around the $z$-axis.

\subsection*{Further reading}
\begin{itemize}
 \item \href{http://en.wikipedia.org/wiki/Lagrangian_mechanics#Euler.E2.80.93Lagrange_equations}{Wikipedia - Lagrangian Mechanics}
 \item \href{http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec08.pdf}{MIT OpenCourseware - Dynamics - Relative Motion using Rotating Axis}
 \item \href{http://en.wikibooks.org/wiki/Kinematics/2D_Coordinate_Systems}{Wikibooks - 2D Coordinate Systems}
 \item \href{http://en.wikipedia.org/wiki/Double_pendulum}{Wikipedia - Double Pendulum}
\end{itemize}
\end{document}


% LaTeX file for a 1 page document
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{physymb}
\usepackage{braket}
\title{EP 307 Assignment 2}

\author{Manish Goregaokar\\120260006}

\begin{document}
\maketitle

\section*{Problem 1}

\begin{align*}
\Braket{e_1|e_1}=\Braket{e_2|e_2} &= 2\\
 \Braket{e_3|e_3} &=4\\
 \Braket{e_1|e_2} &=i\sqrt{2}\\
 \Braket{e_1|e_3} &=1+i\\
 \Braket{e_2|e_3} &=2
\end{align*}

First, we normalize the states to get:

\begin{align*}
\Braket{a_1|a_1}=\Braket{a_2|a_2} &= 1\\
 \Braket{e_3|a_3} &=1\\
 \Braket{a_1|a_2} &=i\frac1{\sqrt{2}}\\
 \Braket{a_1|a_3} &=\frac{1+i}{2\sqrt{2}}\\
 \Braket{a_2|a_3} &=\frac1{\sqrt{2}}
\end{align*}

\newcommand{\proj}[1]{\hat{p}_{a_{#1}}}
Taking $\proj i = \Ket{a_i}\Bra{a_i}$ to be the projection operator with respect to $a_i$, by Gram-Schmidt orthogonalization procedure,

\begin{align*}
&\Ket{\phi_1}=\Ket{a_1}\\
&\Ket{\phi_2}=\Ket{a_2}-\proj1\Ket{a_2}\\
&\Ket{\phi_3}=\Ket{a_3}-\proj1\Ket{a_3}-\proj2\Ket{a_3}\\
\end{align*}
\renewcommand{\proj}[2]{\Ket{a_{#1}}\Braket{a_{#1}|#2}}
Thus
\begin{align*}
&\Ket{\phi_1}=\Ket{a_1}\\
&\Ket{\phi_2}=\Ket{a_2}-\proj1{a_2}\\
&\Ket{\phi_3}=\Ket{a_3}-\proj1{a_3}-\proj2{a_3}\\
\end{align*}
and
\begin{align*}
&\Ket{\phi_1}=\Ket{a_1}\\
&\Ket{\phi_2}=\Ket{a_2}+\frac{1}{\sqrt{2}i}\Ket{a_1}\\
&\Ket{\phi_3}=\Ket{a_3}-\frac{1+i}{2\sqrt{2}}\Ket{a_1}-\frac{1}{\sqrt{2}}\Ket{a_2}\\
\end{align*}

We can calculate \begin{align*}&\Braket{\phi_1|\phi_1}=1\\
&\Braket{\phi_2|\phi_2}=1-\frac{1}{2}=\frac12\\ &\Braket{\phi_3|\phi_3}=1+\frac{(1-i)(1+i)}{8}+\frac12=\frac74
\end{align*}
and from here we get an orthonormal basis

\begin{align*}
&\Ket{\psi_1}=\Ket{a_1}\\
&\Ket{\psi_2}=\sqrt{2}\left(\Ket{a_2}+\frac{1}{\sqrt{2}i}\Ket{a_1}\right)\\
&\Ket{\psi_3}=\frac{2}{\sqrt{7}}\left(\Ket{a_3}-\frac{1+i}{2\sqrt{2}}\Ket{a_1}-\frac{1}{\sqrt{2}}\Ket{a_2}\right)\\
\end{align*}

Rewriting in terms of the original vectors

\begin{align*}
&\Ket{\psi_1}=\frac{1}{\sqrt{2}}\Ket{e_1}\\
&\Ket{\psi_2}=\Ket{e_2}+\frac{1}{\sqrt{2}i}\Ket{e_1}\\
&\Ket{\psi_3}=\frac{4}{\sqrt{7}}\left(\Ket{e_3}-\frac{1+i}{2\sqrt{2}}\Ket{e_1}-\frac{1}{\sqrt{2}}\Ket{e_2}\right)\\
\end{align*}

$\Ket{\psi_1},\Ket{\psi_2},\Ket{\psi_3}$ are our orthonormal basis vectors.

\section*{Problem 2}
First, we normalize them by taking integrals in $[-1,1]$, and get $c_0=\frac1{\sqrt{2}},c_1=\sqrt{\frac32},c_2=\sqrt{\frac52}$

Applying the orthogonalization process,

\begin{align*}
\Ket{\phi_1}&=\Ket{\psi_1}\\
\Ket{\phi_2}&=\Ket{\psi_2}-\Ket{\psi_1}\Braket{\psi_1|\psi_2}\\
\Ket{\phi_3}&=\Ket{\psi_3}-\Ket{\psi_1}\Braket{\psi_1|\psi_3}-\Ket{\psi_2}\Braket{\psi_2|\psi_3}
\end{align*}

calculating the inner products, we get

\begin{align*}
\Ket{\phi_1}&=\Ket{\psi_1}\\
\Ket{\phi_2}&=\Ket{\psi_2}-0\Ket{\psi_1}\\
\Ket{\phi_3}&=\Ket{\psi_3}-\frac{\sqrt5}{2}\frac23\Ket{\psi_1}-0\Ket{\psi_2}
\end{align*}

which evaluates to

\begin{align*}
\phi_1&=\frac1{\sqrt{2}}\\
\phi_2&=\sqrt{\frac32}x\\
\phi_3&=\sqrt{\frac52}\left(x^2-\frac13\right)
\end{align*}
\section*{Problem 3}

$$\hat Bg(x)=g(-x)$$
The operator is linear, since $\hat B(\alpha f(x)+\beta g(x))=\alpha f(-x)+\beta g(-x)=\alpha \hat Bf(x)+\beta\hat B g(x)$

It also is Hermitian, as \begin{align*}\Bra{f}(\hat B\Ket{g})&=\int_{-\infty}^\infty \bar{f}(x)\hat B g(x) dx\\
&= \int_{-\infty}^\infty \bar{f}(x) g(-x) dx\\
&= \int_{-\infty}^\infty \bar{f}(-x)g(x)dx\\
&= \int_{-\infty}^\infty \bar{f}(-x)g(x)dx\\
&=\int_{-\infty}^\infty \overline{\hat B f(x)}g(x)\\
&=\Braket{\overline{\hat B} f|g}\\
\therefore \Braket{f|\hat B g}&=\Braket{\overline{\hat B} f|g}
\end{align*}

To find eigenvalues, $\hat B g(x)=b g(x)=g(-x)$

Since $b g(x)=g(-x)$ and $b g(-x)=g(x)$, $b=\pm 1$ (neglecting the trivial $b=0$ solution).

Thus, we have eigenvectors:
\begin{align*}
&\sum_n a_n x^{2n} & \text{(even function)}\qquad &\text{for eigenvalue $b=1$}\\
&\sum_n b_n x^{2n+1}& \text{(odd function)}\qquad &\text{for eigenvalue $b=-1$}
\end{align*}
To calculate the commutator,
\begin{align*}
 [B,\hat{x}^n]f&=\hat B \hat x^n f(x)-\hat xn \hat B f(x)\\
 &=\hat B x^n f(x) - \hat x^n f(-x)\\
 &= (-x)^{n}f(-x)-x^n f(-x)\\
 &= f(-x)((-x)^{n}-x^n)
\end{align*}

For this to be zero for all states $f$, $n$ must be even. Thus the commutator is zero only for even $n$
\section*{Problem 4}

$$\hat \Omega = \Ket{\psi}\Bra{\phi}$$

For it to be Hermitian, $\hat \Omega^\dagger =\Omega$. Applying it to the state $\Ket{\Psi}=\Ket{\psi}+\Ket{\phi}$,

\begin{align*}
\Ket{\psi}\Braket{\phi|\Psi}&=\Ket{\phi}\Braket{\psi|\Psi}\\
\Ket{\psi}\Braket{\phi|\psi}+\Ket{\phi}\Braket{\psi|\psi} &= \Ket{\psi}\Braket{\phi|\phi}+\Ket{\phi}\Braket{\psi|\phi}\\
\left(\Braket{\phi|\psi}-\Braket{\phi|\phi}\right)\Ket{\psi} &= \left(\Braket{\psi|\phi}-\Braket{\psi|\psi}\right)\Ket{\phi}
\end{align*}

For this to hold, either $\Ket{\psi}=c\ket{\phi}$ or  $\left(\Braket{\phi|\psi}-\Braket{\phi|\phi}\right)=0=\left(\Braket{\psi|\phi}-\Braket{\psi|\psi}\right)$

In the latter case, we get $\Braket{\psi|\psi}=\Braket{\phi|\psi}=\Braket{\phi|\phi}$, which can only happen if the two states are equivalent.

Thus condition for the operator to be hermitian is that $\Ket{\psi}=c\Ket{\phi}$.

For it to be a projection operator, $c=\frac1{\Braket{\phi|\phi}}$

\section{Problem 5}

$$\hat B \psi(x)=\int_{-\infty}^x x' \psi(x')dx'$$

The eigenvalue problem $\hat B \psi =\lambda\psi$ thus becomes $\int_{-\infty}^x x' \psi(x')dx' = \lambda \psi(x)$.

Differentiating with respect to $x$, this is \begin{align*}
x\psi(x)&=\lambda\psi'(x)\\
\implies \frac{\psi'(x)}{\psi(x)}&=\frac{x}{\lambda}\\
\implies \int_{\psi(0)}^\psi\frac{\psi'(x)}{\psi(x)}&=\int_0^x\frac{x}{\lambda}\\
\implies \log\frac{\psi(x)}{\psi(0)} &= \frac1{\lambda} \frac{x^2}2\\
\implies \psi(x)=\psi(0)e^{\frac{x^2}{2\lambda}}
\end{align*}

This is an acceptable (square-integrable) wavefunction only when $\lambda$ is negative.
\end{document}

\documentclass[12pt]{}

Course material/EP 307 - Quantum Mechanics/Assignment 2/assign2.tex