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\begin{document}

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\section*{Sandbox}

:category: meta-syntax

\hypertarget{sandbox}{}\subsection*{{Markdown+itex2MML Sandbox}}\label{sandbox}

Play around \hyperlink{more}{below}. Your changes will, periodically, be rolled back.

\hypertarget{examples}{}\subsubsection*{{Some examples}}\label{examples}

\begin{equation}
\operatorname{min} w_h p_h + w_r p_r + w_l p_l
\label{objftn}\end{equation}
\begin{displaymath}
\left\{
\begin{aligned}
\nabla \times \vec{\mathbf{B}} - \frac{1}{c}\frac{\partial\vec{\mathbf{E}}}{\partial t} &= \frac{4\pi}{c}\vec{\mathbf{j}} \\
\nabla \cdot \vec{\mathbf{E}} &= 4 \pi \rho \\
\nabla \times \vec{\mathbf{E}}+\frac{1}{c}\frac{\partial\vec{\mathbf{B}}}{\partial t} &= \vec{\mathbf{0}} \\
\nabla \cdot \vec{\mathbf{B}} &= 0
\end{aligned}
\right.
\end{displaymath}
Here's an equation

\begin{equation}
{\int_{-\infty}^\infty e^{-a x^2/2} \mathrm{d}x} = \sqrt{\frac{2\pi}{a}}
\label{gaussian}\end{equation}
which we can later refer\footnote{You can also refer to it as \eqref{gaussian}. \emph{Chacun à  son goût!}.}  back to as \eqref{gaussian}.

Aligned equations:

\begin{displaymath}
\begin{aligned}
  a+b &= b+a \\
  a+(b+c) &= (a+b)+c
\end{aligned}
\end{displaymath}
The Dirac equation (boxed):

\begin{displaymath}
\boxed{(i\slash{D}+m)\psi = 0}
\end{displaymath}
Here's the table of Clifford\footnote{For more information, see \href{http://en.wikipedia.org/wiki/Clifford_algebra}{Wikipedia}.}  algebras over $\mathbb{R}$:

\begin{tabular}{r|c|c|c|c|c|c|c|c|c}
$j$&$0$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$\\
\hline 
$\mathcal{C}\ell_{j}^-$&$\mathbb{R}$&$\mathbb{C}$&$\mathbb{H}$&$\mathbb{H}\oplus\mathbb{H}$&$\mathbb{H}(2)$&$\mathbb{C}(4)$&$\mathbb{R}(8)$&$\mathbb{R}(8)\oplus\mathbb{R}(8)$&$\mathbb{R}(16)$\\
$\mathcal{C}\ell_{j}^+$&$\mathbb{R}$&$\mathbb{R}\oplus\mathbb{R}$&$\mathbb{R}(2)$&$\mathbb{C}(2)$&$\mathbb{H}(2)$&$\mathbb{H}(2)\oplus\mathbb{H}(2)$&$\mathbb{H}(4)$&$\mathbb{C}(8)$&$\mathbb{R}(16)$\\
\end{tabular}

where the generators of $\mathcal{C}\ell_{j}^\pm$ satisfy

\begin{displaymath}
\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}
\end{displaymath}
and $\mathcal{C}\ell_{n+8}^\pm = \mathcal{C}\ell_n^\pm \otimes \mathbb{R}(16)$.

\begin{displaymath}
\lim_{n \to \infty}
\sum_{k=1}^n \frac{1}{k^2}
= \frac{\pi^2}{6}
\end{displaymath}
\begin{displaymath}
\mathbf{V}_{1} \times \mathbf{V}_{2} =
    \begin{vmatrix}
        \mathbf{i} & \mathbf{j} & \mathbf{k} \\\\
        \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
        \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \\
    \end{vmatrix}
\end{displaymath}
\hypertarget{second_page}{}\subsection*{{Second Page}}\label{second_page}

\hypertarget{more}{}\subsubsection*{{More Examples}}\label{more}

\begin{displaymath}
\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}
\end{displaymath}
\begin{equation}
\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \href{https://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law}{\mu_0 I_\text{enc}}
\label{Ampere}\end{equation}
$H^1(\mathcal{Z}, \mathcal{O}(-k))$ Let $G=(V,E)$ be a graph, with $w:V\to [0,1]$ a weight function.

\begin{displaymath}
\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.
\end{displaymath}
\hypertarget{third_page}{}\subsection*{{Third Page}}\label{third_page}

\hypertarget{theorems}{}\subsubsection*{{Theorems}}\label{theorems}

\begin{udefn}
Let $H$ be a subgroup of a group $G$. A \emph{left coset} of $H$ in $G$ is a subset of $G$ that is of the form $x H$, where $x \in G$ and $x H = \{ x h : h \in H \}$.

Similarly a \emph{right coset} of $H$ in $G$ is a subset of $G$ that is of the form $H x$, where $H x = \{ h x : h \in H \}$.

\end{udefn}
\begin{lemma}
\label{LeftCosetsDisjoint}\hypertarget{LeftCosetsDisjoint}{}
Let $H$ be a subgroup of a group $G$, and let $x$ and $y$ be elements of $G$. Suppose that $x H \cap y H$ is non-empty. Then $x H = y H$.

\end{lemma}
\begin{proof}
Let $z$ be some element of $x H \cap y H$. Then $z = x a$ for some $a \in H$, and $z = y b$ for some $b \in H$. If $h$ is any element of $H$ then $a h \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $z h = x(a h)$ and $xh = z(a^{-1}h)$ for all $h \in H$. Therefore $z H \subset x H$ and $x H \subset z H$, and thus $x H = z H$. Similarly $y H = z H$, and thus $x H = y H$, as required.

\end{proof}
\begin{lemma}
\label{SizeOfLeftCoset}\hypertarget{SizeOfLeftCoset}{}
Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$.

\end{lemma}
\begin{proof}
Let $H = \{ h_1, h_2,\ldots, h_m\}$, where $h_1, h_2,\ldots, h_m$ are distinct, and let $x$ be an element of $G$. Then the left coset $x H$ consists of the elements $x h_j$ for $j = 1,2,\ldots,m$. Suppose that $j$ and $k$ are integers between $1$ and $m$ for which $x h_j = x h_k$. Then $h_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k$, and thus $j = k$, since $h_1, h_2,\ldots, h_m$ are distinct. It follows that the elements $x h_1, x h_2,\ldots, x h_m$ are distinct. We conclude that the subgroup $H$ and the left coset $x H$ both have $m$ elements, as required.

\end{proof}
\begin{theorem}
\label{Lagrange}\hypertarget{Lagrange}{}
\textbf{(Lagrange's Theorem).} Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.

\end{theorem}
\begin{proof}
Each element $x$ of $G$ belongs to at least one left coset of $H$ in $G$ (namely the coset $x H$), and no element can belong to two distinct left cosets of $H$ in $G$ (see Lemma \ref{LeftCosetsDisjoint}). Therefore every element of $G$ belongs to exactly one left coset of $H$. Moreover each left coset of $H$ contains $|H|$ elements (Lemma \ref{SizeOfLeftCoset}). Therefore $|G| = n |H|$, where $n$ is the number of left cosets of $H$ in $G$. The result follows.

\end{proof}
\begin{cor}
\label{OrderDivides}\hypertarget{OrderDivides}{}
Let $x$ be an element of a finite group $G$. Then the order of $x$ divides the order of $G$.

\end{cor}
\begin{theorem}
\label{}\hypertarget{}{}
Let $f : \Delta \longrightarrow \Delta,$ where $\Delta=\{z\in\mathbb{C}: \vert z \vert \lt 1\}$, be analytic with $a \in \Delta$. Then

\begin{displaymath}
\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert
\end{displaymath}
for all $\vert z \vert \le 1$ and

\begin{displaymath}
\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.
\end{displaymath}
Furthermore, equality holds iff $f$ realizes a conformal mapping of $\Delta$ onto itself.

\end{theorem}
\begin{proof}
Let $w=\frac{z-a}{1-\overline{a}z}$ and put $\phi(w)=\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}$. Define for $\abs{b}\lt 1$ $C_b(z)=\frac{z-b}{1-\overline{b}z}.$ All conformal maps from $\Delta$ to itself, sending $b$ to $0$, are of the form $C_b(z)e^{i\gamma}$ for $\gamma\in[0,2\pi].$ In this notation, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w),$ where $C_a^{-1}$ is the inverse of $C_a$ as a function. Note that $C_a(z)$ is conformal, so it has an inverse. It is clear that $\phi(0)=C_{f(a)}\circ f \circ C_a^{-1}(0)=C_{f(a)}(f(a))=0$. Since $C_a^{-1}: \Delta \longrightarrow \Delta$ and $f : \Delta \longrightarrow \Delta$ and $C_{f(a)}: \Delta \longrightarrow \Delta,$ then $\vert\phi(w)\vert\lt 1$ for $\vert w\vert \lt 1 .$ Applying Schwarz's lemma, we obtain $\vert\phi(w)\vert\le \vert w \vert$ for $\vert w \vert \le 1$. Furthermore, if equality holds, then $f(z)=e^{i\gamma'} z$ for $\gamma'\in [0,2\pi]$. Therefore,

\begin{equation}
\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert
\label{eqn1}\end{equation}
for all $\vert z \vert\le 1.$ Rearranging, we obtain

\begin{displaymath}
\left\vert\frac{f(z)-f(a)}{z-a}\right\vert\le\left\vert\frac{1-\overline{f(a)}f(z)}{1-\overline{a}z}\right\vert.
\end{displaymath}
If we take the limit as $z$ tends to $a$, we obtain

\begin{displaymath}
\left\vert f'(a)\right\vert \le \left\vert\frac{1-\vert f(a)\vert ^2}{1-\vert a \vert^2}\right\vert=\frac{1-\vert f(a)\vert^2}{1-\vert a \vert^2},
\end{displaymath}
or

\begin{displaymath}
\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.
\end{displaymath}
As said above, if equality holds in \eqref{eqn1}, then Schwarz's lemma tells us that $\phi(w)=e^{i\gamma'}w$. Thus, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma'}w,$ so $f(z)=C_{f(a)}^{-1}(e^{i\gamma'}C_a(z))$. Since $e^{i\gamma'}C_a(z)$ is conformal, $C_{f(a)}^{-1},$ the inverse function of $C_{f(a)}$, is conformal, and a composition of conformal maps is conformal, then $f$ is a conformal map of $\Delta$ onto itself. Conversely, if $f$ is a conformal map of $\Delta$ onto itself, then $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma}C_b(w),$ since a composition of conformal maps is conformal and because all conformal maps from $\Delta$ onto itself are of the form $e^{i\gamma}C_b(w).$ We also know that $\phi(0)=0,$ so $b=0$. Therefore,

\begin{displaymath}
\phi(w)=e^{i\gamma}C_0(w)=e^{i\gamma}w \Leftrightarrow \vert\phi(w)\vert=\vert w \vert \Leftrightarrow \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert=\left\vert\frac{z-a}{1-\overline{a}z}\right\vert
\end{displaymath}
for all $\vert z \vert\le 1$. In sum, equality holds in \eqref{eqn1} iff $f$ is a conformal map from $\Delta$ to itself.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
Someone needs to code $\backslash$abs, i.e. $\backslash$left$\backslash$vert \# $\backslash$right$\backslash$vert. This is terribly annoying. Also, when using \$\$, why must one enter a line before and after.

\end{remark}
\hypertarget{svg_graphics_created_in_svgedit}{}\paragraph*{{SVG graphics, created in SVG-Edit:}}\label{svg_graphics_created_in_svgedit}

\begin{displaymath}
 
\end{displaymath}
\begin{displaymath}
 
\end{displaymath}


$K^0\overline{K}^0$ Mixing

\hypertarget{yet_more_examples}{}\paragraph*{{Yet More examples}}\label{yet_more_examples}

\begin{itemize}%
\item SVG:

\end{itemize}
\begin{itemize}%
\item Animated SVG

\end{itemize}
\begin{itemize}%
\item Complicated commutative diagrams (equations in SVG)

\end{itemize}
\begin{itemize}%
\item SVG in equations.

\end{itemize}
In $SU(3)$, $ \includegraphics[width=2em]{young1}
 \otimes
 \includegraphics[width=1em]{young2}
 =
 \includegraphics[width=2em]{young3}
 \oplus
 \includegraphics[width=3em]{young4}$.

\begin{itemize}%
\item Cases:

\end{itemize}
\begin{displaymath}
r_{a+1} = \begin{cases}
    0 & \text{with prob.}\quad \exp(-\theta r_a) \\
    \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a)
  \end{cases}
\end{displaymath}
\begin{itemize}%
\item Stretchy Brackets:

\end{itemize}
\begin{displaymath}
q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}
\end{displaymath}
\begin{itemize}%
\item Linearity of Quadrature Rules

\end{itemize}
\begin{displaymath}
\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i }  = \alpha \sum_{i = 1}^N {f(x_i )w_i }  + \beta \sum_{i = 1}^N {g(x_i )w_i }
\end{displaymath}
\begin{itemize}%
\item Linearity of Integrals

\end{itemize}
\begin{displaymath}
{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx}  + \beta \int_a^b {g(x)\,dx} }
\end{displaymath}
\begin{itemize}%
\item Can we talk about $x_i^2$ inline? What about $\int_a^b x^2\,dx$? Inline fractions $\frac{x-x_2}{x_1-x_2}$?


\item Big fractions



\end{itemize}
\begin{displaymath}
p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}
\end{displaymath}
\begin{itemize}%
\item Diagram

\end{itemize}
\begin{displaymath}
\begin{matrix}
     P_1(Y) &\to& P_1(X) \\
     \downarrow &\Downarrow\mathrlap{\sim}& \downarrow \\
     T' &\to& T
\end{matrix}
\end{displaymath}
\begin{itemize}%
\item $\backslash$mathcal\{\} versus $\backslash$mathscr\{\}\begin{displaymath}
\begin{gathered}
\mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ}\\
\text{versus}\\
\mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ}
\end{gathered}
\end{displaymath}


\end{itemize}

\vspace{.5em} \hrule \vspace{.5em}
\begin{equation}
\itexarray{\arrayopts{\align{center}}
 
} \equiv \left({U(k)}^{n_1},\{v_i\}\right)
\label{quiver}\end{equation}
``This is \textbf{my text}'', says Anymouse.

Ruby code example:

\begin{verbatim}    class Person
      attr_reader :name, :age
      def initialize(name, age)
        @name, @age = name, age
      end
      def <=>(person) # Comparison operator for sorting
        @age <=> person.age
      end
      def to_s
        "#@name (#@age)"
      end
    end
 
    group = [
      Person.new("Bob", 33), 
      Person.new("Chris", 16), 
      Person.new("Ash", 23) 
    ]
 
    puts group.sort.reverse\end{verbatim}
A Python example:

\begin{verbatim}ListOfStrings(
              title = _("A List of some strings"),
              help  = _("A List of strings"),
              orientation = "horizontal"
              )\end{verbatim}


\end{document}