\chapter{Greek alphabet}

\section{Conventional letters}
\begin{longtable}{l| l| l| l}
    Greek & Latin & spelled greek & typical usage \\
    $\alpha$ & A & alpha & constants \\
    $\beta$ & B & beta & \\
    $\gamma$ & $\Gamma$  & gamma & \\
    $\delta$ & D  & delta & given functions  \\
    $\epsilon$, $\varepsilon$ & E & epsilon & \\
    $\zeta$ & Z & zeta & \\
    $\eta$ & H & eta & \\
    $\theta, \vartheta$ & $\Theta$ & theta & \\
    $\iota$ & I & iota & \\
    $\kappa, \varkappa$ & K & kappa & \\
    $\lambda$ & $\Lambda$ & lambda & \\
    $\mu$ & M & mu & \\
    $\nu$ & N & nu & \\
    $\xi$ & $\Xi$ & xi & \\
    $o$ & O & omicron & bounds \\
    $\pi, \varpi$ & $\Pi$ & pi & \\
    $\rho, \varrho $ & P & rho & \\
    $\sigma, \varsigma$ & $\Sigma$ & sigma & sum \\
    $\tau$ & T & tau &  time shift\\
    $\upsilon$ & $\Upsilon$ & upsilon & \\
    $\phi, \varphi$ & $\Phi$ & phi & wavelet \\
    $\chi$ & X & chi & \\
    $\psi$ & $\Psi$ & psi & \\
    $\omega$ & $\Omega$ & omega & \\
\end{longtable}

\section{Other symbols}
\begin{longtable}{l}
    $\partial$ \\
    % $\eth$ \\
    $\hbar$ \\
    $\imath$ \\
    $\ell$ \\
    $\Re$ \\
    $\Im$ \\
    $\wp$ \\
    $\nabla$ \\
    $\Box$ \\
    $\infty$ \\
    $\aleph$ \\
    $\beth$ \\
    $\gimel$ \\

\end{longtable}


\documentclass{subfile}

\begin{document}
	\section{Engel Form of Cauchy-Schwarz}\label{sec:engel}
	We mentioned earlier that there is a particular case of Cauchy-Schwarz inequality which is very useful for solving problems. It is known as \index{Engel form of Cauchy-Schwarz}\emph{Engel form of Cauchy-Schwarz}. Some people also call it \index{Titu's lemma}\emph{Titu's lemma} or \emph{T2's lemma}. This name became popular among the USA students who attended the IMO training camp after a lecture given by Titu Andreescu at Math Olympiad Summer Program (MOSP) at Georgetown University in June, $2001$. Even though it is a direct consequence of Cauchy-Schwarz inequality, it can be proven independently as well. Conversely, Cauchy-Schwarz inequality can be proven using this result.
		\begin{theorem}[\itshape Engel form of Cauchy-Schwarz]\label{thm:engel}
			Let $a_1,\ldots,a_n,b_1,\ldots,b_n$ be real numbers. Then
				\begin{align}
					\dfrac{a_1^2}{x_1}+\ldots+\dfrac{a_n^2}{x_1}
						& \geq\dfrac{(a_1+\ldots+a_n)^2}{x_1+\ldots+x_n}\label{ineq:engel}
				\end{align}
		\end{theorem}
	Some writers such as \textcite{mitrinovitch_1959, bellman_1955} also call this \index{Bergstr\"{o}m's inequality}\emph{Bergstr\"{o}m's inequality} due to \textcite{bergstrom_1949}.
		\begin{proof}[Proof using Cauchy-Schwarz]
			Use Cauchy-Schwarz inequality on
				\begin{align*}
					\frac{a_1}{\sqrt{x_1}},\ldots,\frac{a_{n}}{\sqrt{x_{n}}};\sqrt{x_1},\ldots,\sqrt{x_n}
				\end{align*}
			We get
				\begin{align*}
					\left(\dfrac{a_1^2}{\sqrt{x_1}^2}+\ldots+\dfrac{a_n^2}{\sqrt{x_{n}}^2}\right)\left(\sqrt{x_1}^2+\ldots+\sqrt{x_n}^2\right)
						& \geq\left(\dfrac{a_1}{\sqrt{x_1}}\sqrt{x_1}+\ldots+\dfrac{a_n}{\sqrt{x_n}}\sqrt{x_n}\right)^2\\
						& = (a_1+\ldots+a_n)^2
				\end{align*}
			This proves the theorem.
		\end{proof}

		\begin{proof}[Proof by induction]
			We will use induction to prove \ref{ineq:engel}. The inequality is trivial for $n=1$. For $n=2$,
				\begin{align*}
					\dfrac{a^2}{x}+\dfrac{b^2}{y}
						& \geq\dfrac{(a+b)^2}{x+y}\\
					\iff\dfrac{a^2y+b^2x}{xy}
						& \geq\dfrac{a^2+2ab+b^2}{x+y}\\
					\iff a^2xy+b^2x^2+a^2y^2+b^2xy
						& \geq a^2xy+2abxy+b^2xy\\
					\iff (ay-bx)^2
						& \geq0
				\end{align*}
			This is obviously true. Now, assume that the claim is true for
		\end{proof}
	We get a very useful result as a corollary of this result.
		\begin{theorem}
			Let $a_1,\ldots,a_n,b_1,\ldots,b_n$ be real numbers. Then
				\begin{align*}
					\dfrac{a_1^2}{b_1^2}+\ldots+\dfrac{a_n^2}{b_n^2}
						& \geq\left(\dfrac{a_1+\ldots+a_n}{b_1+\ldots+b_n}\right)^2
				\end{align*}
		\end{theorem}

		\begin{proof}
			By \nameref{thm:engel},
				\begin{align*}
					\dfrac{a_1^2}{b_1^2}+\ldots+\dfrac{a_n^2}{b_n^2}
						& \geq\dfrac{(a_1+\ldots+a_n)^2}{b_1^2+\ldots+b_n^2}\\
						& \geq\dfrac{(a_1+\ldots+a_n)^2}{(b_1+\ldots+b_n)^2}
				\end{align*}
			The last inequality follows from the fact that $b_1^2+\ldots+b_n^2\leq(b_1+\ldots+b_n)^2$.
		\end{proof}
	Another result that is similar to \nameref{thm:engel} is a special case of the \textit{Beckenbach inequality}.
		\begin{theorem}
			Let $x_{1},\ldots,x_{n}$ and $y_{1},\ldots,y_{n}$ be positive real numbers. Then
				\begin{align*}
					\dfrac{\sum\limits_{i=1}^{n}x_{i}^{2}}{\sum\limits_{i=1}^{n}x_{i}}+\dfrac{\sum\limits_{i=1}^{n}y_{i}^{2}}{\sum\limits_{i=1}^{n}y_{i}}
						& \geq \dfrac{\sum\limits_{i=1}^{n}(x_{i}+y_{i})^{2}}{\sum\limits_{i=1}^{n}(x_{i}+y_{i})}
				\end{align*}
		\end{theorem}
	We will start with another proof of \nameref{thm:nesbitt} with the \nameref{thm:engel}.
		\begin{proof}[Nesbitt inequality using Engel form]
			We will use the fact that $a^2+b^2+c^2\geq ab+bc+ca$.
				\begin{align*}
					S
						& = \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\\
						& = \dfrac{a^2}{ab+ca}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\\
						& \geq\dfrac{(a+b+c)^2}{ab+ca+bc+ab+ca+bc}\\
						& = \dfrac{(a+b+c)^2}{2(ab+bc+ca)}\\
						& = \dfrac{a^2+b^2+c^2+2(ab+bc+ca)}{ab+bc+ca}\\
						& = \dfrac{a^2+b^2+c^2}{2(ab+bc+ca)}+1\\
						& \geq\dfrac{1}{2}+1=\dfrac{3}{2}
				\end{align*}
		\end{proof}
	Let us use this inequality to solve some problems.
		\begin{problem}
			Let $n$ be a positive integer and $a_{1},\ldots,a_{n};b_{1},\ldots,b_{n}$ be positive real numbers. Show that
				\begin{align*}
					(a_{1}b_{1}+\ldots+a_{n}b_{n})\left(\dfrac{a_{1}}{b_{1}}+\ldots+\dfrac{a_{n}}{b_{n}}\right)
						& \geq (a_{1}+\ldots+a_{n})^{2}
				\end{align*}

				\begin{solution}
					By \nameref{thm:engel},
						\begin{align*}
							\dfrac{a_{1}}{b_{1}}+\ldots+\dfrac{a_{n}}{b_{n}}
								& = \dfrac{a_{1}^{2}}{a_{1}b_{1}}+\ldots+\dfrac{a_{n}^{2}}{a_{n}b_{n}}\\
								& \geq \dfrac{(a_{1}+\ldots+a_{n})^{2}}{a_{1}b_{1}+\ldots+a_{n}b_{n}}\\
							\iff (a_{1}b_{1}+\ldots+a_{n}b_{n})\left(\dfrac{a_{1}}{b_{1}}+\ldots+\dfrac{a_{n}}{b_{n}}\right)
								& \geq (a_{1}+\ldots+a_{n})^{2}
						\end{align*}
				\end{solution}
		\end{problem}

		\begin{remark}
			In order to apply the Engel form, we often need transformations like $a/b=a^{2}/b^{2}$. Sometimes we will divide the odd power into an even one and send the remaining one in the denominator, for example,
				\begin{align*}
					\dfrac{a^{2n+1}}{b}
						& = \dfrac{\left(a^{n}\right)^{2}}{\dfrac{b}{a}}
				\end{align*}
			so that we can then apply it. We will demonstrate these ideas in the following problems.
		\end{remark}

		\begin{problem}
			For positive real numbers $a,b,c$, prove that
				\begin{align*}
					\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}
						& \geq 1
				\end{align*}

				\begin{solution}
					We again resort to the same trick.
						\begin{align*}
							\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}
								& = \dfrac{a^{2}}{ab+2ca}+\dfrac{b^{2}}{bc+2ab}+\dfrac{c^{2}}{ca+2bc}\\
								& \geq \dfrac{(a+b+c)^{2}}{3(ab+bc+ca)}\\
								& = \dfrac{a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{3(ab+bc+ca)}\\
								& \geq \dfrac{ab+bc+ca+2(ab+bc+ca)}{3(ab+bc+ca)}
						\end{align*}
					In the last line, we used $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ which is well known by now.
				\end{solution}
		\end{problem}

		\begin{problem}
			Let $x,y,z$ be positive real numbers. Prove that
				\begin{align*}
					\dfrac{2}{x+y}+\dfrac{2}{y+z}+\dfrac{2}{z+x}
						& \geq \dfrac{9}{x+y+z}
				\end{align*}

				\begin{solution}
					This is again a similar problem.
						\begin{align*}
							\dfrac{2}{x+y}+\dfrac{2}{y+z}+\dfrac{2}{z+x}
								& = \dfrac{4}{2(x+y)}+\dfrac{4}{2(y+z)}+\dfrac{4}{2(z+x)}\\
								& \geq \dfrac{(2+2+2)^{2}}{2(x+y+y+z+z+x)}\\
								& = \dfrac{36}{4(x+y+z)}\\
								& = \dfrac{9}{x+y+z}
						\end{align*}
				\end{solution}
		\end{problem}

		\begin{problem}
			Let $x,y,z$ be positive real numbers. Prove that
				\begin{align*}
					\dfrac{x^{2}}{(x+z)(x+y)}+\dfrac{y^{2}}{(y+x)(y+z)}+\dfrac{z^{2}}{(z+y)(z+x)}
						& \geq \dfrac{3}{4}
				\end{align*}

				\begin{solution}
					First, see that $(x+z)(x+y)=x^{2}+xy+zx+yz$. Next, we apply \nameref{thm:engel}.
						\begin{align*}
								& \dfrac{x^{2}}{(x+z)(x+y)}+\dfrac{y^{2}}{(y+x)(y+z)}+\dfrac{z^{2}}{(z+y)(z+x)}\\
								& = \dfrac{x^{2}}{x^{2}+xy+zx+yz}+\dfrac{y^{2}}{y^{2}+yz+xy+zx}+\dfrac{z^{2}}{z^{2}+zx+yz+xy}\\
								& \geq \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3(xy+yz+zx)}\\
								& = \dfrac{3}{4}\left(\dfrac{4(x+y+z)^{2}}{3(x^{2}+y^{2}+z^{2}+2(xy+yz+zx)+xy+yz+zx)}\right)\\
								& = \dfrac{3}{4}\left(\dfrac{3(x+y+z)^{2}+(x+y+z)^{2}}{3((x+y+z)^{2}+xy+yz+zx)}\right)\\
								& = \dfrac{3}{4}\left(\dfrac{3(x+y+z)^{2}+x^{2}+y^{2}+z^{2}+2(xy+yz+zx)}{3(x+y+z)^{2}+3(xy+yz+zx)}\right)\\
								& \geq \dfrac{3}{4}\left(\dfrac{3(x+y+z)^{2}+3(xy+yz+zx)}{3(x+y+z)^{2}+3(xy+yz+zx)}\right)\\
								& = \dfrac{3}{4}\cdot1=\dfrac{3}{4}
						\end{align*}
					Thus, the inequality is proved.
				\end{solution}
		\end{problem}

		\begin{problem}[IMO Shortlist $1993$]
			For positive real numbers $a,b,c$ and $d$, prove that
				\begin{align*}
					\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}
						& \geq \dfrac{2}{3}
				\end{align*}

				\begin{solution}
					Writing
						\begin{align*}
							\frac{a}{b+2c+3d}
								& =\frac{a^{2}}{ab+2ca+3ad}
						\end{align*}
					and applying \nameref{thm:engel}, we get
						\begin{align*}
							S
								& = \dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\\
								& = \dfrac{a^{2}}{ab+2ca+3ad}+\dfrac{b^{2}}{bc+2bd+3ab}+\dfrac{c^{2}}{cd+2ca+3bc}+\dfrac{d^{2}}{ad+2bd+3cd}\\
								& \geq \dfrac{(a+b+c+d)^{2}}{4ab+4bc+4ca+4ad+4bd+4cd}\\
								& = \dfrac{2}{3}\left(\dfrac{3(a+b+c+d)^{2}}{8(ab+bc+ca+ad+bd+cd)}\right)\\
								& = \dfrac{2}{3}\left(\dfrac{3(a+b+c+d)^{2}}{4\left((a+b+c+d)^{2}-(a^{2}+b^{2}+c^{2}+d^{2})\right)}\right)\\
								& = \dfrac{2}{3}\left(\dfrac{4(a+b+c+d)^{2}-(a+b+c+d)^{2}}{4(a+b+c+d)^{2}-4(a^{2}+b^{2}+c^{2})}\right)
						\end{align*}
					By \nameref{thm:cs},
						\begin{align*}
							(a^{2}+b^{2}+c^{2}+d^{2})(1^{2}+1^{2}+1^{2}+1^{2})
								& \geq (a+b+c+d)^{2}\\
							\iff -4(a^{2}+b^{2}+c^{2}+d^{2})
								& \leq -(a+b+c+d)^{2}\\
							\iff 4(a+b+c+d)^{2}-4(a^{2}+b^{2}+c^{2}+d^{2})
								& \leq 4(a+b+c+d)^{2}-(a+b+c+d)^{2}
						\end{align*}
					Thus, we have $S\geq \frac{2}{3}\cdot1=\frac{2}{3}$.
				\end{solution}
		\end{problem}

		\begin{problem}[IMO $1995$, problem $2$]
			Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
				\begin{align*}
					\dfrac{1}{a^{3}(b+c)}+\dfrac{1}{b^{3}(c+a)}+\dfrac{1}{c^{3}(a+b)}
						& \geq \dfrac{3}{2}
				\end{align*}

				\begin{solution}
					Write
						\begin{align*}
							\frac{1}{a^{3}(b+c)}
								& =\frac{\frac{1}{a^{2}}}{a(b+c)}
						\end{align*}
					and apply \nameref{thm:engel}.
						\begin{align*}
							\dfrac{1}{a^{3}(b+c)}+\dfrac{1}{b^{3}(c+a)}+\dfrac{1}{c^{3}(a+b)}
								& = \dfrac{\dfrac{1}{a^{2}}}{a(b+c)}+\dfrac{\dfrac{1}{b^{2}}}{b(c+a)}+\dfrac{\dfrac{1}{c^{2}}}{c(a+b)}\\
								& \geq \dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2}}{ab+ca+bc+ab+ca+bc}\\
								& = \dfrac{\left(\dfrac{ab+bc+ca}{abc}\right)^{2}}{2(ab+bc+ca)}\\
								& = \dfrac{ab+bc+ca}{2}\\
								& \geq \dfrac{3\sqrt[3]{ab\cdot bc\cdot ca}}{2}\\
								& = \dfrac{3\sqrt[3]{(abc)^{2}}}{2}\\
								& = \dfrac{3}{2}
						\end{align*}
					Here, we get $ab+bc+ca\geq3\sqrt[3]{ab\cdot bc\cdot ca}$ by \nameref{thm:amgm}.
				\end{solution}
		\end{problem}

		\begin{problem}[Tournament of the Towns, $1998$]
			Let $a,b,c$ be positive real numbers. Prove that
				\begin{align*}
					\dfrac{a^{3}}{a^{2}+ab+b^{2}}+\dfrac{b^{3}}{b^{2}+bc+c^{2}}+\dfrac{c^{3}}{c^{2}+ca+a^{2}}
						& \geq \dfrac{a+b+c}{3}
				\end{align*}

				\begin{solution}
					We again make the numerator a square using

						\begin{align*}
							\frac{a^{3}}{a^{2}+ab+b^{2}}
								& =\frac{a^{4}}{a^{3}+a^{2}b+ab^{2}}
						\end{align*}
					Then
						\begin{align*}
							\dfrac{a^{3}}{a^{2}+ab+b^{2}}+\dfrac{b^{3}}{b^{2}+bc+c^{2}}+\dfrac{c^{3}}{c^{2}+ca+a^{2}}
								& = \dfrac{a^{4}}{a^{3}+a^{2}b+ab^{2}}+\dfrac{b^{4}}{b^{3}+b^{2}c+bc^{2}}+\dfrac{c^{4}}{c^{3}+c^{2}a+ca^{2}}\\
								& \geq \dfrac{(a^{2}+b^{2}+c^{2})^{2}}{a^{3}+b^{3}+c^{3}+ab(a+b)+bc(b+c)+ca(c+a)}\\
								& = \dfrac{(a^{2}+b^{2}+c^{2})^{2}}{(a+b+c)(ab+bc+ca)}\\
								& = \dfrac{a^{2}+b^{2}+c^{2}}{a+b+c}
						\end{align*}
					The last line follows since $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$. Then the conclusion follows because by \nameref{thm:cs},
						\begin{align*}
							(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2})
								& \geq (a+b+c)^{2}\\
							\iff \dfrac{a^{2}+b^{2}+c^{2}}{a+b+c}
								& \geq \dfrac{a+b+c}{3}
						\end{align*}
				\end{solution}
		\end{problem}
\end{document}


\clearpage
\section{Appendix}\label{sec-appendix}

\subsection{Compute transformers}\label{sec-appendix-transformers}

In this section we clarify some of the compute transformers used in this paper.

\hs{execute} uses the transformation based on the \hs{Identity} monad, feeding
\hs{fetch k = pure (store k)} to the compute:

\begin{minted}[xleftmargin=10pt]{haskell}
execute :: Compute Monad k v -> (k -> v) -> k -> Maybe v
execute compute store = fmap runIdentity . compute (pure . store)
\end{minted}
\vspace{1mm}
\begin{minted}[xleftmargin=10pt]{haskell}
newtype Identity a = Identity { runIdentity :: a }
\end{minted}
\vspace{1mm}
\begin{minted}[xleftmargin=10pt]{haskell}
instance Functor Identity where
    fmap f (Identity a) = Identity (f a)
\end{minted}
\vspace{1mm}
\begin{minted}[xleftmargin=10pt]{haskell}
instance Applicative Identity where
    pure a = Identity a
    Identity f <*> Identity a = Identity (f a)
\end{minted}
\vspace{1mm}
\begin{minted}[xleftmargin=10pt]{haskell}
instance Monad Identity where
    Identity a >>= f = f a
\end{minted}
\vspace{1mm}

\todo{AM}{Explain \hs{track}.}

Here is a draft implementation of \hs{inputs} used in the definition of
build system correctness in \S\ref{sec-build-correctness}:

\begin{minted}[xleftmargin=10pt]{haskell}
inputs :: Eq k => Task Monad k v -> Store i k v -> k -> [k]
inputs task store key = filter (isInput task) (closure deps key)
  where
    deps k = maybe [] snd (track task (\k -> getValue k store) k)

closure :: Eq a => (a -> [a]) -> a -> [a] -- Standard graph transitive closure

data Proxy a = Proxy

isInput :: Task Monad k v -> k -> Bool
isInput task = isNothing . task (const Proxy)
\end{minted}

\subsection{Compute examples}\label{sec-appendix-compute-examples}

\todo{AM}{Add some explanatory text.}

The \emph{Collatz sequence} $C_i$ is defined as follows:

\[
C_{i} = {\begin{cases}~n&{\text{for }}i=0\\~f(C_{i-1})&{\text{otherwise}},\end{cases}}\hspace{12pt}\text{where}\hspace{12pt}f(k)={\begin{cases}~k/2&{\text{if }}k\text{ is even}\\~3k+1&{\text{otherwise}}\end{cases}}
\vspace{2mm}
\]
\noindent
and $n$ is a positive integer parameter. The famous \emph{Collatz conjecture}
states that the Collatz sequence eventually reaches 1 for all possible values of
$n$. For example, if $n=6$, we reach 1 in eight steps:
$(6, 3, 10, 5, 16, 8, 4, 2, 1, \dots)$, after which the sequence loops forever:
$(4, 2, 1, 4, 2, 1, \dots)$.

We can express the computation of values in the Collatz sequence as a functorial
compute:

\begin{minted}[xleftmargin=10pt]{haskell}
data Collatz = Collatz Int

collatz :: Compute Functor Collatz Int
collatz get (Collatz k) | k <= 0    = Nothing
                        | otherwise = Just $ f <$> get (Collatz (k - 1))
  where
    f n | even n    = n `div` 2
        | otherwise = 3 * n + 1
\end{minted}

...

The \emph{generalised Fibonacci sequence} $F_i$ is defined as follows:

\[
F_{i} = {\begin{cases}~n&{\text{for }}i=0\\~m&{\text{for }}i=1\\~F_{i-1}+F_{i-2}&{\text{otherwise}}\end{cases}}
\vspace{2mm}
\]
\noindent
where $n$ and $m$ are integer parameters. By setting $n=0$ and $m=1$ we obtain
the famous \emph{Fibonacci sequence}: $(0, 1, 1, 2, 3, 5, 8, 13, \dots$), and if
$n=2$ and $m=1$, the result is the \emph{Lucas sequence}:
$(2, 1, 3, 4, 7, 11, 18, 29, \dots)$.

We can express the computation of values in the generalised Fibonacci sequence
as an applicative compute:

\begin{minted}[xleftmargin=10pt]{haskell}
data Fibonacci = Fibonacci Int

fibonacci :: Compute Applicative Fibonacci Int
fibonacci get (Fibonacci k) | k <= 1    = Nothing
                            | otherwise = Just $ (+) <$> get (Fibonacci (k - 1))
                                                     <*> get (Fibonacci (k - 2))
\end{minted}

...

The \emph{Ackermann function} $A(m, n)$ is defined as follows:

\[
A(m, n) = {\begin{cases}~n+1&{\text{for }}m=0\\~A(m-1, 1)&{\text{for }}n=0\\~A(m-1,A(m,n-1))&{\text{otherwise}}\end{cases}}
\vspace{2mm}
\]
\noindent
We can express the computation of the Ackermann function as a monadic compute:

\begin{minted}[xleftmargin=10pt]{haskell}
data Ackermann = Ackermann Int Int

ackermann :: Compute Monad Ackermann Int
ackermann get (Ackermann m n)
    | m < 0 || n < 0 = Nothing
    | m == 0    = Just $ return (n + 1)
    | n == 0    = Just $ get (Ackermann (m - 1) 1)
    | otherwise = Just $ do
        index <- get (Ackermann m (n - 1))
        get (Ackermann (m - 1) index)
\end{minted}


\section{Introduction} \label{sec:introduction}

Machine learning has been a heated research topic these days. With this amazing tool, we are now capable of predicting the price of the stock price based on history, doing the classifying by just inputing the pixels of images.

Supervised learning is one of the most important sections for machine learning. And Regression is probably the core of supervised learning. By firstly inputing in the computer some of the training data, the machine would be able to learn the characteristics of the dataset and make predictions.

Gaussian Process Regression(GPR) is a supervised learning regression method which are getting increasingly welcome in both the research field and the industry.
In a GPR, we take advantage of the flexibility and simplicity of a Gaussian Process and implement it into a regression problem. \\

There are still countless unsolved problems in the field of GPR.
In this paper, we would comprehensively introduce the concept of GPR, including most of the notable works.
Specifically, we focus on some marvellous kernel choosing works.
We also implemented several experiments to quantitatively analyze the performance of different \emph{Mean Functions}, \emph{Likelihood Functions} and the \emph{Inference Methods}
We seek to provide a practical overview in the field of GPR.

The structure of this paper is as follows:
In section \ref{sec:intro}, we provide a precise overview of a Gaussian Process Regression.
In section \ref{sec:related}, we briefly listed some of the notable progress on GPR.
Section \ref{sec:kernel} describes some great recent work on the methods for auto-construction of the kernels, as well as an interesting additive kernel which can have a better performance when solving some specific cases.
In section \ref{sec:experiment}, we conduct two experiments and use them to compare the performance of different methods and algorithms.
Conclusions are drawn in section \ref{sec:conclusion}.


\par
\section{Driver programs for the {\tt DFrontMtx} object}
\label{section:DFrontMtx:drivers}
\par
%=======================================================================
\begin{enumerate}
%-----------------------------------------------------------------------
\item
\begin{verbatim}
testGrid msglvl msgFile n1 n2 n3 maxzeros maxsize seed type
         symmetryflag sparsityflag pivotingflag tau droptol
         lockflag nrhs
\end{verbatim}
This driver program tests the serial {\tt FrontMtx\_factor()}
and {\tt FrontMtx\_solve()} methods for the linear system $AX = B$.
Use the script file {\tt do\_grid} for testing.
\par
\begin{itemize}
\item
The {\tt msglvl} parameter determines the amount of output.
Use {\tt msglvl = 1} for just timing output.
\item
The {\tt msgFile} parameter determines the message file --- if {\tt
msgFile} is {\tt stdout}, then the message file is {\it stdout},
otherwise a file is opened with {\it append} status to receive any
output data.
\item
{\tt n1} is the number of points in the first grid direction.
\item
{\tt n2} is the number of points in the second grid direction.
\item
{\tt n3} is the number of points in the third grid direction.
\item
{\tt maxzeros} is used to merge small fronts together into larger
fronts.
Look at the {\tt ETree} object for
the {\tt ETree\_mergeFronts\{One,All,Any\}()} methods.
\item
{\tt maxsize} is used to split large fronts into smaller
fronts.
See the {\tt ETree\_splitFronts()} method.
\item
The {\tt seed} parameter is a random number seed.
\item
The {\tt type} parameter specifies a real or complex linear system.
\begin{itemize}
\item
{\tt type = 1 (SPOOLES\_REAL)} for real,
\item
{\tt type = 2 (SPOOLES\_COMPLEX)} for complex.
\end{itemize}
\item
The {\tt symmetryflag} parameter specifies the symmetry of the matrix.
\begin{itemize}
\item
{\tt type = 0 (SPOOLES\_SYMMETRIC)} for $A$ real or complex symmetric,
\item
{\tt type = 1 (SPOOLES\_HERMITIAN)} for $A$ complex Hermitian,
\item
{\tt type = 2 (SPOOLES\_NONSYMMETRIC)}
\end{itemize}
for $A$ real or complex nonsymmetric.
\item
The {\tt sparsityflag} parameter signals a direct or approximate
factorization.
\begin{itemize}
\item
{\tt sparsityflag = 0 (FRONTMTX\_DENSE\_FRONTS)} implies a direct
factorization, the fronts will be stored as dense submatrices.
\item
{\tt sparsityflag = 1 (FRONTMTX\_SPARSE\_FRONTS)} implies an
approximate factorization.
The fronts will be stored as sparse submatrices, where
the entries in the triangular factors will be
subjected to a drop tolerance test --- if the magnitude of an entry
is {\tt droptol} or larger, it will be stored, otherwise it will be
dropped.
\end{itemize}
\item
The {\tt pivotingflag} parameter signals whether pivoting for
stability will be enabled or not.
\begin{itemize}
\item
If {\tt pivotingflag = 0 (SPOOLES\_NO\_PIVOTING)},
no pivoting will be done.
\item
If {\tt pivotingflag = 1 (SPOOLES\_PIVOTING)},
pivoting will be done to ensure that all
entries in $U$ and $L$ have magnitude less than {\tt tau}.
\end{itemize}
\item
The {\tt tau} parameter is an upper bound on the magnitude of the
entries in $L$ and $U$ when pivoting is enabled.
\item
The {\tt droptol} parameter is a lower bound on the magnitude of the
entries in $L$ and $U$ when the approximate factorization is enabled.
\item
When {\tt lockflag} is zero, the mutual exclusion lock for the
factor matrix is not enabled.
When {\tt lockflag} is not zero, the mutual exclusion lock is set.
This capability is here to test the overhead for the locks for a
serial factorization.
\item
The {\tt nrhs} parameter is the number of right hand sides to solve
as one block.
\end{itemize}
%-----------------------------------------------------------------------
\item
\begin{verbatim}
testQRgrid msglvl msgFile n1 n2 n3 seed nrhs type
\end{verbatim}
This driver program tests the serial {\tt FrontMtx\_QR\_factor()}
and {\tt FrontMtx\_QR\_solve()} methods for the least squares problem
$ \min_X \| F - A X \|_F$.
\par
\begin{itemize}
\item
The {\tt msglvl} parameter determines the amount of output.
Use {\tt msglvl = 1} for just timing output.
\item
The {\tt msgFile} parameter determines the message file --- if {\tt
msgFile} is {\tt stdout}, then the message file is {\it stdout},
otherwise a file is opened with {\it append} status to receive any
output data.
\item
{\tt n1} is the number of points in the first grid direction.
\item
{\tt n2} is the number of points in the second grid direction.
\item
{\tt n3} is the number of points in the third grid direction.
\item
The {\tt seed} parameter is a random number seed.
\item
The {\tt nrhs} parameter is the number of right hand sides to solve
as one block.
\item
The {\tt type} parameter specifies a real or complex linear system.
\begin{itemize}
\item
{\tt type = 1 (SPOOLES\_REAL)} for real,
\item
{\tt type = 2 (SPOOLES\_COMPLEX)} for complex.
\end{itemize}
\end{itemize}
%-----------------------------------------------------------------------
\end{enumerate}


\chapter{Quadratic Fields}
\label{chap:chap-four}
In this chapter, we are going to discover the most simplest number field that are different from
$\mathbb{Q}$, i.e. quadratic fields. Let denote it as $K$. A quadratic field is of degree 2 over $\mathbb{Q}$, it can be given by $K=\mathbb{\theta}$, where $\theta$ is a root of minimal polynomial $f(x)=x^2+a x+b$ of $\mathbb{Z}[x]$. Let $d=a^2-b$, then $K=\mathbb{Q}(\sqrt{d})$. Clearly, $d$ is not a square, otherwise $f(x)$ won't be a irreducible quadratic polynomial. Since $\mathbb{Q}(\sqrt{m^2d})=\mathbb{Q}(\sqrt{d})$, hence we may assume that $d$ is squarefree.

Since $n=2=r_1+2r_2$, the signature of quadratic field $K$ is either $(2,0)$ or $(0,1)$, which is called real quadratic field or complex (or imaginary) quadratic field respectively.
It's easy to show that the real quadratic field has positive discriminant and the complex quadratic field has negative discriminant.

\section{Discriminant, Integral Basis}
The integral basis and discriminant of quadratic field is easy as following theorem:
\begin{theorem}\label{thm:quad_disc}
Let $K=\mathbb{Q}(\sqrt{d})$ be a quadratic field with $d$ squarefree. Then if $d\equiv1(\operatorname{mod} 4)$, we have integral basis $(1,(1+\sqrt{d})/2)$ with discriminant $d(K)=d$; if $d\equiv2\text{ or }3(\operatorname{mod} 4)$, we have integral basis $(1,\sqrt{d})$ with discriminant $d(K)=4d$.
\end{theorem}

Let us denote $D=d(K)$ (namely fundamental discriminant) for these two cases, then we can see that it satisfies Stickelberger's criterion \ref{thm:stickelberger}. What's more, $K=\mathbb{Q} (\sqrt{D})$ has integral basis $(1,\omega)$, where $$\omega=\frac{D+\sqrt{D}}{2},$$ and therefore $O_K=\mathbb{Z}[\omega]$.

\section{Decomposition of Primes}
Note that Proposition \ref{thm:decomposition} and Proposition \ref{thm:kummer} immediately shows how prime numbers decompose in a quadratic field \citep{Xianke2006ANT}.

\begin{theorem}\label{thm:quad_decompose}
Let $K=\mathbb{Q}(\sqrt{D})$, where $D$ is the fundamental discriminant, i.e. $D=d(K)$, $O_K=\mathbb{Z}[\omega]$ where $\omega=(D+\sqrt{D})/2$ its ring of integers, and $p$ be a prime number. Then\footnote{$\left(\frac{D}{p}\right)$ is Kronecker symbol which can be seen in Appendix \ref{chap:appA}.}
\begin{enumerate}
\item If $\left(\frac{D}{p}\right)=0$, then $p$ is ramified, i.e. $pO_K=\wp^2$. More precisely, $$\wp=pO_K+\omega O_K,$$ except when $p=2$ and $D\equiv12(\operatorname{mod} 16)$.
\item If $\left(\frac{D}{p}\right)=-1$, then $p$ is inert, hence $pO_K=\wp$ is a prime ideal.
\item If $\left(\frac{D}{p}\right)=1$, then $p$ is split, and we have $pO_K=\wp_1\wp_2$, where
$$\wp_i=pO_K+(\omega-\frac{D\pm b}{2})O_K,$$ and $b$ is any solution to $b^2\equiv D(\operatorname{mod} 4p)$
\end{enumerate}
\end{theorem}

\section{Unit Group of Imaginary Quadratic Fields}
First we consider the unit group of quadratic field, by the Dirichlet's Unit theorem \ref{thm:Dirichlet}, we can divide the quadratic field into two cases. The imaginary quadratic field has simple unit group since $r=r_1+r_2-1=0$, i.e. the unit group is finite. In fact, the imaginary quadratic field is the only number field apart from $\mathbb{Q}$ who has finite units. A theorem for the unit group of Imaginary Quadratic fields could be found in \citep{Xianke2006ANT,cohen1993course} etc.:

\begin{theorem}\label{thm:quad_unit}
Let $K=\sqrt{D}$, where the fundamental discriminant $D<0$, then the group $U(K)=\mu(K)$ of units  is equal to the group of $\omega(D)-th$ roots of unity, where
$$\omega(D)=\left\{\begin{array}{ll}
2,& \text{if }D<-4\\
4,& \text{if }D=-4\\
6,& \text{if }D=-3
\end{array}\right.$$
\end{theorem}

\section{Class number of Imaginary Quadratic Fields}
Let us now consider the problem of computing the class number of imaginary quadratic field. Here we give a beautiful result from L-function. Since it's to far to enter into the details of the analytic theory of \textbf{L-functions}, so we just recall the results. Our main result is a corollary of Dirichlet's Theorem (We have reorganized this theorem to fit all imaginary quadratic number field. In the original theorem, it just states $D<-4$ cases.) \citep{cohen1993course}.

\begin{theorem}
If $D$ is a negative fundamental discriminant, then $$h(K)=\frac{\omega(D)}{4-2\left(\frac{D}{2}\right)}\sum_{1\leq r<|D|/2}\left(\frac{D}{r}\right),$$ where $\omega(D)$ is defined in Theorem \ref{thm:quad_unit}.
\end{theorem}

\begin{remark}
For $D<-4$, we have $$h(K)=\frac{1}{2-\left(\frac{D}{2}\right)}\sum_{1\leq r<|D|/2}\left(\frac{D}{r}\right).$$ Of course, we can get $h(-3)=h(-4)=1$ by calculation.
\end{remark}
More precise list for class number can be seen in the book \citep{Xianke2006ANT}.

\section{Unit group of Real Quadratic Fields}
For real quadratic fields, ZHANG \citep{Xianke2006ANT} has given a precise algorithm to compute it through the Pell's equation. Since the unit circle of a real quadratic field only hits two time real axis, then we have $\mu{K}=\{\pm1\}$. By Dirichlet Unit Theorem \ref{thm:Dirichlet}, we have $$U(K)=\{\pm1\}\times \epsilon^{\mathbb{Z}},$$ where $\epsilon$ generates an infinite cyclic group $\langle\epsilon\rangle=\epsilon^{\mathbb{Z}}$. There exists only one unit in the set (all of them can generate whole infinite cyclic group) $\{\pm\epsilon,\pm\epsilon^{-1}\}$ which is larger than 1, and we denote it as fundamental unit of $K$.
Now we just need to compute the fundamental unit. Let $K=\mathbb{Q}(\sqrt{d})$, where $d$ is squarefree. Let $\alpha=a+b\sqrt{d}(a,b\in\mathbb{Q})$ be unit of $K$, then $N(\alpha)=\pm1$.
If $d\equiv 2 \text{ or }3 \mod 4$, then $O_K=\mathbb{Z}[\sqrt{d}]$. A integer $\alpha=x+y\sqrt{d}(x,y\in\mathbb{Z})$ is unit if and only if $N(\alpha)=\pm1$, i.e. we have following Pell's equation:
\begin{equation}\label{eqn:pell1}
x^2-d y^2=\pm1
\end{equation}
From the Unit Theorem, we note that if $\epsilon=a_1+b_1\sqrt{d}$ is a fundamental unit of $K$($a_1,b_1>0$), then $$\epsilon^n=(a_1+b_1\sqrt{d})^n=a_n+b_n\sqrt{d}$$ is also a unit who larger than one. What's more, $(a_n,b_n)$ are natural number solutions of the Pell's equation \ref{eqn:pell1}. Note that $b_{n+1}=a_1b_n+a_nb_1$, hence $b_n$ is an increasing sequence.

Let us consider $b=1,2,3,\cdots$, if $d b^2\pm1$ is a square number (i.e. $a^2$), then we stop the process, i.e. we find the "smallest" solution of the Pell's equation \ref{eqn:pell1} and this solution is also the fundamental unit.

Let's consider $d=6$, $6b^2\pm1$ is a square number firstly when $b=2$, i.e. we have the fundamental unit of $\mathbb{Q}(\sqrt{6})$ is $\epsilon=5+2\sqrt{6}$.

Similarly, if $d\equiv1 \mod 4$, then $O_K=\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$, an integer of $K$ has form $\alpha=(a+b\sqrt{d})/2$, where $a\equiv b \mod 2$, $a,b\in\mathbb{Z}$. $\alpha$ be unit of $K$ if and only if $N(\alpha)=(a^2-b^2d)/4=\pm1$, hence, $(a,b)$ is a solution of the following Pell's equation:
\begin{equation}\label{eqn:pell2}
x^2-d y^2=\pm4
\end{equation}
if $\epsilon=(a_1+b_1\sqrt{d})/2$ is a fundamental unit of $K$($a_1,b_1>0$), then $$\epsilon^n=(\frac{a_1+b_1\sqrt{d}}{2})^n=\frac{a_n+b_n\sqrt{d}}{2}$$ is also a unit who larger than one. Similarly, consider $b=1,2,3,\cdots$, if $d b^2\pm4$ is a square number (i.e. $a^2$), then we stop the process and get the fundamental unit $(a+b\sqrt{d})/2$.

\section{Class Numbers of Real Quadratic Fields}
Also, as in the imaginary case, using L-function, we can also get a beautiful results\citep{cohen1993course}. We have modified the results through discarding regulators.

\begin{theorem}\label{thm:rquad_cl}
If $D$ is a positive fundamental discriminant, then $$h(K)=-\frac{1}{\operatorname{ln}(\epsilon)}\sum_{r=1}^{\lfloor(D-1)/2\rfloor}\left(\frac{D}{r}\right)\operatorname{ln}\sin\left(\frac{r\pi}{D}\right).$$
\end{theorem}

Now we see an example, let $D=4d=8$, then $\epsilon=1+\sqrt{2}$ by the result of unit group of real quadratic field. From the Theorem \ref{thm:rquad_cl}, we have $$h(K)=-\frac{1}{\operatorname{ln}(1+\sqrt{2})}\left(\operatorname{ln}\sin(\frac{\pi}{8})-\operatorname{ln}\sin(\frac{3\pi}{8})\right)=-\frac{\operatorname{ln}(\tan(\frac{\pi}{8}))}{\operatorname{ln}(1+\sqrt{2})}=1.$$


\documentclass{memoir}
\usepackage{notestemplate}

%\logo{~/School-Work/Auxiliary-Files/resources/png/logo.png}
%\institute{Rice University}
%\faculty{Faculty of Whatever Sciences}
%\department{Department of Mathematics}
%\title{Class Notes}
%\subtitle{Based on MATH xxx}
%\author{\textit{Author}\\Gabriel \textsc{Gress}}
%\supervisor{Linus \textsc{Torvalds}}
%\context{Well, I was bored...}
%\date{\today}

%\makeindex

\begin{document}

% \maketitle

% Notes taken on 07/09/21

Recall that a holomorphic isomorphism is a holomorphic map
\begin{align*}
	f:U\to V
\end{align*}
with a holomorphic inverse
\begin{align*}
	g:V\to U
\end{align*}
so that
\begin{align*}
	f\circ g &= \textrm{Id}_V\\
	g\circ f &= \textrm{Id}_U
\end{align*}
We say that \(f\) is an \textbf{automorphism} if \(U=V\), and refer to the set of automorphisms of \(U\) by \(\textrm{Aut}(U)\).

\subsection{Automorphisms of the Unit Disc}
\label{sub:automorphisms_of_the_unit_disc}

\begin{thm}[Schwarz' Lemma]
	Let \(f:D_1 \to D_1\) be a holomorphic automorphism of the unit disc with \(f(0) = 0\). Then
	\begin{align*}
		\left| f(z) \right| \leq \left| z \right|
	\end{align*}
	for all \(z \in D_1\). If for some \(z_0\neq 0\) we have \(\left| f(z_0) \right| = \left| z_0 \right| \), then
	\begin{align*}
		f(z) = e^{i \theta } z
	\end{align*}
	for some \(\theta \in \R\).

	In particular, \(\left| f'(0) \right| \leq 1\). If equality holds, then \(f(z) = e^{i\theta } z\) for some \(\theta \in \R\).
\end{thm}
In other words, an automorphism of the unit disc that fixes the origin has an implicit "contraction" requisite. If it isn't a contraction, then it must be a rotation.

\begin{proof}

\end{proof}%Lang p.210, via maximum modulus principle

\begin{anki}
TARGET DECK
Complex Qual::Complex Analysis
START
MathJaxCloze
Text: **Schwarz' Lemma**
Let \(f:D_1 \to D_1\) be a holomorphic automorphism of the unit disc with \(f(0) = 0\). Then
 {{c1::\(\begin{align*}
        	\left| f(z) \right| \leq \left| z \right|
        \end{align*}\)::modulus}}
for all \(z \in D_1\). If for some \(z_0\neq 0\) we have \(\left| f(z_0) \right| = \left| z_0 \right| \), then
 {{c1::\(\begin{align*}
        	f(z) = e^{i \theta } z
        \end{align*}\)::function is of form}}
for some \(\theta \in \R\).

In particular, {{c1::\(\left| f'(0) \right| \leq 1\)::modulus}}. If equality holds, then {{c1::\(f(z) = e^{i\theta } z\)::function is of form}} for some \(\theta \in \R\).
\end{thm}
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841377-->
END
\end{anki}

We can apply Schwarz Lemma to classify all the automorphisms of the unit disc. Because every automorphism must send exactly one point to zero, we can transform the general case to the origin.

\begin{thm}[Structure of Automorphisms of the Unit Disc]
\label{thm:classification_of_automorphisms_of_the_unit_disc}
	Let \(\sigma :D_1\to D_1\) be a holomorphic automorphism of the unit disc, and let \(z_0\in D_1\) be the point for which \(\sigma (z_0 )=0\). Then for some \(\theta \in \R\)
	\begin{align*}
		\sigma (z) = e^{i\theta } \frac{z-z_0}{1-\overline{z_0}z}
	\end{align*}
\end{thm}

Observe that when \(\theta =2\pi k\) for \(k \in \Z\), the automorphism of the disc satisfies \(\sigma (z_0) = 0\), \(\sigma (0) = z_0\), and \(\sigma \circ \sigma  = \textrm{Id}_{D_1}\). For this reason, we will denote by \(\sigma_{z_0}\) the \textbf{canonical automorphism of the unit disc centered at \(z_0\)} when \(\theta =2\pi k\) and \(\sigma (z_0)=0\).

\begin{anki}
START
MathJaxCloze
Text: \textbf{Structure of Automorphisms of the Unit Disc}
Let \(\sigma :D_1\to D_1\) be a holomorphic automorphism of the unit disc, and let \(z_0\in D_1\) be the point for which \(\sigma (z_0 )=0\). Then for some \(\theta \in \R\)
 {{c1::\(\begin{align*}
        	\sigma (z) = e^{i\theta } \frac{z-z_0}{1-\overline{z_0}z}
        \end{align*}\)}}
Extra: Observe that when \(\theta =2\pi k\) for \(k \in \Z\), the automorphism of the disc satisfies \(\sigma (z_0) = 0\), \(\sigma (0) = z_0\), and \(\sigma \circ \sigma  = \textrm{Id}_{D_1}\). For this reason, we will denote by \(\sigma_{z_0}\) the **canonical automorphism of the unit disc centered at \(z_0\)** when \(\theta =2\pi k\) and \(\sigma (z_0)=0\).
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841400-->
END
\end{anki}

\subsection{Isomorphism from the Upper Half Plane}
\label{sub:isomorphism_from_the_upper_half_plane}

We began this chapter with automorphisms of the unit disc because it turns out that we can obtain an isomorphism from larger parts of the complex plane to the disc. Then we can use the automorphisms of the disc to characterize more general sections of the complex plane.\\

We will now show that the entire upper half complex plane is holomorphically isomorphic to the unit disc.

\begin{thm}[Isomorphism between \(\mathbb{H}\) and \(D_1\)]
	Let \(\mathbb{H}\subset \C\) be the upper half plane. The map
	\begin{align*}
		f&:\mathbb{H} \to D_1\\
		f&:z \mapsto \frac{z-i}{z+i}
	\end{align*}
	is a holomorphic isomorphism onto the unit disc.
\end{thm}
The inverse map for \(f\) is given by
\begin{align*}
	g&:D_1 \to \mathbb{H}\\
	g&:z\mapsto -i\frac{z+1}{z-1}
\end{align*}
We encourage the reader to verify that \(g\) is indeed an inverse for \(f\).

\begin{anki}
START
MathJaxCloze
Text:
Let \(\mathbb{H}\subset \C\) be the upper half plane. The map
 {{c1::\(\begin{align*}
        	f&:\mathbb{H} \to D_1\\
        	f&:z \mapsto \frac{z-i}{z+i}
        \end{align*}\)}}
is a holomorphic isomorphism onto the unit disc.

The inverse map for \(f\) is given by
 {{c1::\(\begin{align*}
        	g&:D_1 \to \mathbb{H}\\
        	g&:z\mapsto -i\frac{z+1}{z-1}
        \end{align*}\)}}
Extra: It follows that
\(\begin{align*}
  	\textrm{Aut}(\mathbb{H}) = f^{-1} \textrm{Aut}(D) f
  \end{align*}\)
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841426-->
END
\end{anki}

\begin{cor}[Automorphism Group of \(\mathbb{H}\)]
	\begin{align*}
		\textrm{Aut}(\mathbb{H}) = f^{-1} \textrm{Aut}(D) f
	\end{align*}
	where
	\begin{align*}
		f&:\mathbb{H} \to D_1\\
		f&:z \mapsto \frac{z-i}{z+i}
	\end{align*}
\end{cor}
Of course, it remains to actually calculate what automorphisms of the upper half plane look like.

\begin{thm}[Structure of Automorphisms of \(\mathbb{H}\)]
	Every automorphism of \(\mathbb{H}\) is a linear fractional translation of the form
	\begin{align*}
		A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\
		S_A(z) =\frac{az+b}{c z + d}
	\end{align*}
	where \(a,b,c,d \in \R\) with \(\textrm{det}(A)\neq 0\).\\

	Two automorphisms \(S_A\) and \(S_{A'}\) are equal if and only if \(A' = \pm A\).
\end{thm}
Hence, the automorphism group \(\textrm{Aut}(\mathbb{H})\) is isomorphic to the projective special linear group-- that is,
 \begin{align*}
	 \textrm{Aut}(\mathbb{H}) \cong PSL_2(\R).
\end{align*}

\begin{anki}
START
MathJaxCloze
Text: **Structure of Automorphisms of \(\mathbb{H}\)**
Every automorphism of \(\mathbb{H}\) is a {{c1::linear fractional translation}} of the form
{{c1::\(\begin{align*}
        	A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\
        	S_A(z) =\frac{az+b}{c z + d}
        \end{align*}\)}}
where {{c1::\(a,b,c,d \in \R\)}} with {{c1::\(\textrm{det}(A)\neq 0\)}}.\\

Two automorphisms \(S_A\) and \(S_{A'}\) are equal if and only if {{c2::\(A' = \pm A\)}} .
Extra: Hence, the automorphism group \(\textrm{Aut}(\mathbb{H})\) is isomorphic to the projective special linear group-- that is,
\( \begin{align*}
  	 \textrm{Aut}(\mathbb{H}) \cong PSL_2(\R).
  \end{align*}\)
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841448-->
END
\end{anki}

\subsection{Isomorphisms Between Boundaries of Closed Curves}
\label{sub:isomorphisms_between_boundaries_of_closed_curves}

We note that so far, many results have been shown by utilizing the behavior of the function on the boundary of an open set. This is no coincidence-- we formalize this notion via the following theorems.

\begin{thm}[Surjectivity via Boundary Mapping]
	Let \(\Omega \) be a bounded connected open set. Suppose \(f\) is a non-constant holomorphic function on \(\Omega \) that is continuous on \(\partial\Omega \), which maps \(\partial\Omega \) into \(\partial D_1\), that is,
	\begin{align*}
		\left| f(z) \right| =1
	\end{align*}
	for all \(z \in \partial\Omega \). Then
	\begin{align*}
		f(\Omega ) = D_1.
	\end{align*}
\end{thm}
If we further assume that \(f\) is injective, then it follows that \(f\) is an isomorphism.

\begin{anki}
START
MathJaxCloze
Text: Let \(\Omega \) be a bounded connected open set. Suppose \(f\) is a non-constant holomorphic function on \(\Omega \) that is continuous on \(\partial\Omega \), which maps \(\partial\Omega \) into \(\partial D_1\), that is,
\(\begin{align*}
  	\left| f(z) \right| =1
  \end{align*}\)
for all \(z \in \partial\Omega \). Then
 {{c1::\(\begin{align*}
        	f(\Omega ) = D_1.
        \end{align*}\)}}
Extra: If we further assume that \(f\) is injective, then it follows that \(f\) is an isomorphism.
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841465-->
END
\end{anki}

\begin{proof}%Lang VII 4.2, pg. 227

\end{proof}

We can further generalize this via interiors of curves, but first we prove a lemma that will be essential:

\begin{lemma}
	Let \(\gamma \) be a piecewise smooth closed curve in an open set \(\Omega \subset \C\). Suppose that \(\gamma \) has an interior denoted by \(\textrm{Int}(\gamma )\). Then
	\begin{align*}
		\textrm{Int}(\gamma ) \cup \gamma
	\end{align*}
	is compact.
\end{lemma}
This follows from the compactness of \(\gamma \) combined with the continuity of the winding number within the interior.

\begin{anki}
START
MathJaxCloze
Text: Let \(\gamma \) be a piecewise smooth closed curve in an open set \(\Omega \subset \C\). Suppose that \(\gamma \) has an interior denoted by \(\textrm{Int}(\gamma )\). Then
\(\begin{align*}
  	\textrm{Int}(\gamma ) \cup \gamma
  \end{align*}\)
is {{c1::compact::topological property}}.
Extra: This follows from the compactness of \(\gamma \) combined with the continuity of the winding number within the interior.
Tags: analysis complex_analysis complex_topology
<!--ID: 1626293841487-->
END
\end{anki}

\begin{thm}[Condition for Conformality on Open Connected Sets]
	Let \(f\) be a non-constant holomorphic function on some open connected set \(\Omega \subset \C\), and let \(\gamma \) be a piecewise-smooth closed curve in \(\Omega \) homologous to \(0\).\\

	If \(\gamma \) and \(f\circ \gamma \) have interiors so that
	\begin{align*}
		f\circ \gamma \cap f(\textrm{Int}\gamma ) = \emptyset
	\end{align*}
	then \(f\) is injective on \(\textrm{Int}(\gamma )\) and hence \(f:\textrm{Int}(\gamma )\to f(\textrm{Int}\gamma )\) is an isomorphism. If in addition \(\textrm{Int}(f\circ \gamma )\) is connected, then
	\begin{align*}
		f(\textrm{Int}\gamma ) = \textrm{Int}(f\circ \gamma ).
	\end{align*}
\end{thm}

\begin{proof}

\end{proof}

\begin{anki}
START
MathJaxCloze
Text: Let \(f\) be a non-constant holomorphic function on some open connected set \(\Omega \subset \C\), and let \(\gamma \) be a piecewise-smooth closed curve in \(\Omega \) homologous to \(0\).\\

If \(\gamma \) and \(f\circ \gamma \) have interiors so that
{{c1::\(\begin{align*}
        	f\circ \gamma \cap f(\textrm{Int}\gamma ) = \emptyset
        \end{align*}\)::necessary condition}}
then \(f\) is injective on \(\textrm{Int}(\gamma )\) and hence \(f:\textrm{Int}(\gamma )\to f(\textrm{Int}\gamma )\) is an isomorphism. If in addition \(\textrm{Int}(f\circ \gamma )\) is connected, then
{{c1::\(\begin{align*}
        	f(\textrm{Int}\gamma ) = \textrm{Int}(f\circ \gamma ).
        \end{align*}\)::mapping of interior}}
Tags: analysis complex_analysis conformal_mappings
<!--ID: 1626293841503-->
END
\end{anki}

\begin{exmp}

\end{exmp}

\begin{exmp}%Unbounded example

\end{exmp}

% \printindex
\end{document}