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% Ernest Yeung was supported by Mr. and Mrs. C.W. Yeung, Prof. Robert A. Rosenstone, Michael Drown, Arvid Kingl, Mr. and Mrs. Valerie Cheng, and the Foundation for Polish Sciences, Warsaw University.

\subsection*{Topological Manifolds }

$M$ topological manifold of $\text{dim}{n}$, or topological $n$-manifold
\begin{itemize}
\item locally Euclidean, $\text{dim}{n}$ - $\forall \, p \in M$, $\exists \, $ neighborhood $U \equiv U_p$ s.t. $U_p \approx^{\text{homeo}} \text{ open } V \subset \mathbb{R}^n$
\end{itemize}

\exercisehead{1.1} Recall, $M$ locally Euclidean $\dim{n}$ $\forall \, p \in M$, $\exists \,$ neighborhood homeomorphic to open subset. \\
open subset $\mathcal{O} \subseteq \mathbb{R}^n$ homeomorphic to open ball and $\mathcal{O}$ homeomorphic is $\mathbb{R}^n$ since $\mathbb{R}^n$ homeomorphic to open ball.

To see this explicitly, that open ball $B_{\epsilon}(x_0) \subseteq \mathbb{R}^n $ homeomorphic to $\mathbb{R}^n$

Consider $\begin{aligned} & \quad \\
  & T: B_{\epsilon}(x_0) \to \mathbb{R}^n \\
  & T(B_{\epsilon}(x_0)) = B_{\epsilon}(0) \\
  & T(x) = x- x_0 \end{aligned}$  \\
$T^{-1}(x) = x+x_0$.  Clearly $T$ homeomorphism.

Consider $\begin{aligned} & \quad \\
& \lambda : \mathbb{R}^n \to \mathbb{R}^n \\
  & \lambda(x) = \lambda x \\
  & \lambda^{-1}(x) = \frac{1}{ \lambda } x \end{aligned}$ for $\lambda >0$.  Clearly $\lambda$ homeomorphism.

Consider $B \equiv B_1(0)$.

Consider $\begin{aligned} & \quad \\
  & g: \mathbb{R}^n \to \mathbb{R}^n \\
  & g(x) = \frac{x}{ 1 + |x| } \end{aligned}$

$g$ cont.

Let $\begin{aligned} & \quad \\
  & f:B \to \mathbb{R}^n \\
  & f(x) = \frac{x}{ 1 - |x| } \end{aligned} $

How was $f$ guessed at?

$|g(x) | = \left| \frac{x}{1 + |x| } \right| = \frac{r}{ 1 + r}$ .  Note $0 \leq |g(x) | <1$ \\
So $g(\mathbb{R}^n) = B$

For $|g(x)| = |y|$, $y \in B$, $|y|(1+r) = r$, \, $r = \frac{ |y| }{ 1 - |y| }$

This is well-defined, since $0 \leq |y| < 1$ and $0 < 1 - |y| \leq 1$
\[
\begin{aligned}
  & gf(x) =\frac{  \frac{ x}{ 1 - |x| } }{ 1 + \frac{|x|}{ 1 - |x| } } = x \\
  & fg(x) = \frac{ \frac{x}{ 1 + |x| } }{ 1 - \frac{|x| }{ 1 + |x| } } = x
\end{aligned}
\]
$f$ homeomorphism between $B$ and $\mathbb{R}^n$.  $B$ and $\mathbb{R}^n$ homeomorphic.  So an open ball in $\mathbf{R}^n$ is homeomorphic to $\mathbb{R}^n$

\hrulefill

In practice, both the Hausdorff and second countability properties are usually easy to check, especially for spaces that are built out of other manifolds, because both properties are inherited by subspaces and products (Lemmas A.5 and A.8).  In particular, it follows easily that any open subset of a topological $n$-manifold is itself a topological $n$-manifold (with the subspace topology, of course).

\subsubsection*{Coordinate Charts}

chart on $M$, $(U, \varphi)$ where open $U \subset M$ and \\
homeomorphism $\varphi : U \to \mathbb{R}^n$, $\varphi(U)$ open.

\subsubsection*{Examples of Topological Manifolds}

Example 1.3. (Graphs of Continuous Functions) \\
Let open $U \subset \mathbb{R}^n$ \\
Let $F: U \to \mathbb{R}^k$ cont. \\
graph of $F$: $\Gamma(F) = \lbrace (x,y) \in \mathbb{R}^n \times \mathbb{R}^k | x \in U , y = F(x) \rbrace$ with subspace topology. \\
$\pi_1 : \mathbb{R}^n \times \mathbb{R}^k \to \mathbb{R}^n$ projection onto first factor. \\
$\varphi_k:\Gamma(F) \to U$ restriction of $\pi_1$ to $\Gamma(F)$ \\
$\varphi_F(x,y) = x$, $(x,y) \in \Gamma(F)$ \\

\textbf{Example 1.4 (Spheres)} $S^n = \lbrace x \in \mathbb{R}^{n+1} | |x| = 1 \rbrace$ \\
Hausdorff and second countable because it's topological subspace of $\mathbb{R}^n$   \\

\textbf{Example 1.5 (Projective Spaces)}

$U_i \subset \mathbb{R}^{n+1} - 0$ where $x^i \neq 0$  \\
$V_i = \pi(U_i)$

Let $a\in U_i$.
\[
|x-a|^2 = (x^1 - a^1)^2 + \dots + (x^i - a^i)^2 + \dots + (x^{n+1} - a^{n+1})^2 < \frac{ (n+1) \epsilon^2 }{ n + 1 } = \epsilon^2
\]
$\forall \, a^i \in \mathbb{R}$, $\exists \,  x^i$, s.t. $(x^i - a^i)^2 < \frac{ \epsilon^2}{ n +1}$, by choice of $0< x^i < a^i + \frac{ \epsilon}{ \sqrt{ n+ 1} }$ with $0<x^i$ for $i=i$ index.  \\
$U_i$ indeed open set, \emph{saturated open set}.

open $U_i \subset \mathbb{R}^{n+1} - 0$, $x^i \neq 0$

From Lemma A.10, recall
(d) restriction of $\pi$ to any saturated open or closed subset of $X$ is a quotient map.

natural map $\pi: \mathbb{R}^{n+1} - 0 \to \mathbb{R}P^n$ given quotient topology.

By Tu, Prop. 7.14, $\sim $ on $\mathbb{R}^{n+1} - 0  \to \mathbb{R}P^n$ open equivalence relation.  \\
$\Longrightarrow  \left. \pi \right|_{U_i}(U_i) = V_i$ open.

\[
\begin{aligned}
   & \varphi_i : V_i \to \mathbb{R}^n \\
& \varphi_i [ x^1 \dots x^{n+1} ] = \left( \frac{ x^1 }{ x^i } \dots \frac{ x^{i-1}}{ x^i } , \frac{ x^{i+1}}{ x^i} \dots \frac{x^{n+1}}{ x^i} \right)
\end{aligned}
\]

$\varphi_i$ well-defined since
\[
\varphi_i[tx^1 \dots tx^{n+1}] = \left( \frac{ tx^1 }{ tx^i } \dots \frac{ t\widehat{x}^i }{ tx^i } \dots \frac{ tx^{n+1}}{ tx^i } \right)= \left( \frac{x^1}{ x^i } \dots \frac{ \widehat{x}^i }{ x^i } \dots \frac{ x^{n+1}}{ x^i } \right) = \varphi_i[x^1 \dots x^{n+1} ]
\]

$\varphi_i$ cont. since $\varphi_i \pi$ cont.

\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=2em, column sep=3em, minimum width=1em]
  {
    U_i \subset \mathbb{R}^{n+1} - 0 &  \\
    V_i \subset \mathbb{R}P^n  & \mathbb{R}^n  \\ };
  \path[-stealth]
  (m-1-1) edge node [right] {$\varphi_i \pi$} (m-2-2)
  edge node [left] { $\pi$} (m-2-1)
  (m-2-1) edge node [below] {$\varphi$} (m-2-2);
\end{tikzpicture}

\[
\begin{gathered}
  \begin{aligned}
    & \varphi_i : U_i \subset \mathbb{R}P^n \to \mathbb{R}^n \\
    & \varphi_i[x^1 \dots x^{n+1} ] = \left( \frac{x^1}{x^i} \dots \frac{ \widehat{x}^i }{ x^i } \dots \frac{x^{n+1}}{ x^i } \right) \\
    & \varphi^{-1}_i(u^1 \dots u^n) = [ u^1 \dots u^{i-1}, 1 , u^i \dots u^n ]
\end{aligned} \\
  \varphi_i^{-1} \varphi_i[x^1 \dots x^{n+1} ] = \left[ \frac{x^1 }{ x^i } \dots \frac{x^{i-1}}{ x^i } , 1 , \frac{x^{i+1}}{ x^i} \dots \frac{x^{n+1}}{ x^i } \right] = [x^1 \dots x^{i-1}, x^i , x^{i+1} \dots x^{n+1} ] \\
  \varphi_i \varphi^{-1}_i(u^1 \dots u^n) = (u^1 \dots u^{i-1} , u^i  \dots u^n )
\end{gathered}
\]
cont. $\varphi_i$ bijective, $\varphi_i^{-1}$ cont. $\varphi_i$ homeomorphism.

From a previous edition:

\exercisehead{1.2}

Let $ \begin{aligned} & \quad \\
  & \phi_t : \mathbb{R}^{n+1}\backslash \lbrace  0 \rbrace \to \mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace \\
  & \phi_t(x) = tx \end{aligned}$ \\
$\phi_t$ invertible, $\phi_t^{-1} = \phi_{\frac{1}{t}}$ \\
$\phi_t$, $\phi_t^{-1}$ \, $C^1$ (and $C^{\infty}$), $\phi_t$ homeomorphism.  \\

Let $U$ open in $\mathbb{R}^{n+1}\backslash \lbrace 0 \rbrace$.  Then $\phi_t(U)$ open in $\mathbb{R}^{n+1}\backslash \lbrace 0 \rbrace$.  \\
Thus $\pi^{-1}([U]) = \bigcup_{t \in \mathbb{R}} \phi_t(U)$ open in $\mathbb{R}^{n+1}\backslash \lbrace 0 \rbrace$.  \\
Thus $[U]$ open in $\mathbb{R}P^n$.  $\sim$ open.  \\

Note $\begin{aligned} & \quad  \\
  & \pi : \mathbb{R}^{n+1} \backslash \lbrace 0 \rbrace \to \mathbb{R}P^n \\
  & \pi(x) = \frac{x}{ \| x\| } \end{aligned}$ \\
$\mathbb{R}^n$ 2nd. countable, $\mathbb{R}P^n$ 2nd. countable.

\exercisehead{1.3}

$S^n$ compact. \\
$\pi :\mathbb{R}^{n+1}\backslash \lbrace 0 \rbrace \to \mathbb{R}P^n$ \\
$\pi(x) = [ \frac{x}{ \| x \| } ]$ \\
Let $x \in \mathbb{R}P^n$ \\
\phantom{Let } $y = \frac{x}{ \| x \| } \in S^n$ and $ \left. \pi \right|_{S^n}(y) = [x]$ \\
$\left. \pi \right|_{S^n}$ surjective.

\exercisehead{1.6}

First, note that $\sim $ on $\mathbb{R}^{n+1} -0$ in the definition of $\mathbb{R}P^n$ is an open $\sim$ i.e. open equivalence relation.

This is because of the following:
$\forall \, U \subset \mathbb{R}^{n+1}-0$, \\
$\pi^{-1}(\pi(U)) = \bigcup_{t\in \mathbb{R}} tU$, set of all pts. equivalent to some pt. of $U$.  \\
multiplication by $t\in \mathbb{R}$ homeomorphism of $\mathbb{R}^{n+1} - 0$, so $tU$ open $\forall \, t \in \mathbb{R}$.  \\
$\pi^{-1}(\pi(U))$ open i.e. $\pi(U)$ open (for $\pi$ is cont.).

Let $X = \mathbb{R}^{n+1} -0$.  \\
Consider $R = \lbrace ( x,y) \in X \times X | x\sim y \text{ or } y = tx \text{ for some } t \in \mathbb{R} \rbrace$

$y = tx$ means $y_i = tx_i$ \, $\forall \, i = 0\dots n$.  Then $\frac{ x_i}{y_i} = \frac{ x_j}{ y_j}$ \, $\forall \, i,j = 0 \dots n$.  Hence $x_i y_j - y_i x_j = 0$ \, $\forall \, i , j$.

Let $\begin{aligned} & \quad \\
  & f : X \times X \to \mathbb{R} \\
  & f(x,y) = \sum_{ i \neq j} (x_i y_j - y_i x_j)^2 \end{aligned}$

\[
\begin{aligned}
  & \frac{ \partial f}{ \partial x_i} = \sum_{i \neq j } 2 (x_i y_j - y_i x_j) ( y_j - y_j) = 0 \\
  & \frac{ \partial f}{ \partial y_j} = \sum_{j \neq i } 2 (x_i y_j - y_i x_j) ( x_i - x_i) = 0
\end{aligned}
\]

Nevertheless, $f$ is $C^1$ so $f$ cont.

So $f^{-1}(0) = R$.

$0$ closed, so $f^{-1}(0) = R$ closed.  By theorem, since $\sim$ open, $\mathbb{R}P^n = \mathbb{R}^{n+1} - 0 / \sim$ Hausdorff.

cf. \url{http://math.stackexchange.com/questions/336272/the-real-projective-space-rpn-is-second-countable}

topological space is second countable if its topology has countable basis.  \\
\quad $\mathbb{R}^n$ second countable since $\mathcal{B} = \lbrace B_r{ ( q)} | r,q \in \mathbb{Q} \rbrace$ is a countable basis.  $\forall \, x \in \mathbb{R}^n$ \\

If $X$ is second countable, with countable basis $\mathcal{B}$,
\begin{enumerate}
\item If $Y \subseteq X$, $Y$ also second countable with countable basis $\lbrace B | B \in \mathcal{B}, \, Y \bigcap B \neq \emptyset \rbrace$
\item If $Z == X/\sim $, $\lbrace \lbrace [x] | x \in B \rbrace | B \in \mathcal{B} \rbrace$ is a countable basis for $Z$ since $\mathcal{B}$ countable.  \\

It is a basis since

\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=2em, column sep=3em, minimum width=1em]
  {
X = \bigcup_{B \in \mathcal{B}} B    \\
Z = X/ \sim = \pi{ \left( \bigcup_{B \in \mathcal{B}} B \right) } = \bigcup_{B \in \mathcal{B}} \pi{ (B)} = \bigcup_{ B \in \mathcal{B}} \lbrace [x] | x  \in B \rbrace \\
  };
%  \path[-stealth]
  \path[->]
  (m-1-1) edge node [right] {$\pi$} (m-2-1);
\end{tikzpicture}

\end{enumerate}

Now let $Y = \mathbb{R}^n-0$ and \\
\phantom{Now let } $Z = \mathbb{R}P^n = \mathbb{R}^n-0 / \sim$

\exercisehead{1.7}

$S^n$ compact so $S^n/\lbrace \pm \rbrace$ compact by Theorem, as $\pi_S(S^n) = S^n/ \lbrace \pm \rbrace$, as $\pi_S$ cont. surjective ($\forall \, [x] \in S^n/\lbrace \pm \rbrace$, $\exists \, x \in S^n$ s.t. $\pi_S(S^n) = [x]$) \\
$g$ cont. bijective as defined above so since $g(S^n/\lbrace \pm \rbrace) = \mathbb{R}P^n$, $\mathbb{R}P^n$ compact.

\textbf{Example 1.8 (Product Manifolds)}
\[
\begin{aligned}
  & M_1 = \bigcup_{\alpha \in \mathfrak{A}_1} U_{\alpha}^{(1)} \\
  & M_i = \bigcup_{\alpha \in \mathfrak{A}_i} U_{\alpha}^{(i)}
\end{aligned} \quad \quad \quad \,
M_1 \times \dots \times M_n = \bigcup_{ \begin{aligned} & \alpha_1 \in \mathfrak{A}_1 \\
& \vdots \\
    & \alpha_n \in \mathfrak{A}_n \end{aligned} }  U_{\alpha_1} \times \dots \times U_{\alpha_n} \quad \quad \, (\text{by def.})
\]

$\forall \, p = (p_1 \dots p_n) \in M_1 \times \dots M_n$, consider $p_i \in M_i$.  Choose coordinate chart $(U_{j_i}, \varphi_{j_i})$, $\varphi_i(U_i) \subset \mathbb{R}^{n_i}$.  Then,  \\

Consider $\varphi : U_1 \times \dots \times U_n \to \mathbb{R}^{m_1} \times \dots \times \mathbb{R}^{m_n} = \mathbb{R}^{ m_1 + \dots + m_n}$, $(U_1 \times \dots \times U_n, \varphi_1 \times \dots \times \varphi_n)$ \\
$\varphi = \varphi_1 \times \dots \times \varphi_n$\\
$\varphi$ also a homeomorphism.  $\begin{gathered} \varphi \psi^{-1} \\ (\varphi_1 \times \dots \times \varphi_n) \circ(\psi_1 \times \dots \times \psi_n)^{-1} \end{gathered}$ also a diffeomorphism, cont. bijective and $C^{\infty}$ \\
$\lbrace (U, \varphi)  = (U_1 \times \dots \times U_n, \varphi_1 \times \dots \times \varphi_n) | (U_i, \varphi_i) \in \lbrace (U_i, \varphi_i) | U_i \in M_i \rbrace \rbrace$ also an atlas.

\subsubsection*{Topological Properties of Manifolds}

\begin{lemma}[1.10]
$\forall \, $ topological $M$, $M$ has countable basis of precompact coordinate balls
\end{lemma}

\begin{proof}
  First consider $M$ can be covered by single chart.  \\
Suppose $\varphi : M \to \widehat{U} \subseteq \mathbb{R}^n$ global coordinate map. \\
Let $\mathcal{B} = \lbrace B_r(x) | \text{ open } B_r(x) \subseteq \mathbb{R}^n \text{ s.t. } r\in \mathbb{Q}, \, x \in \mathbb{Q}, \, \text{ i.e. $x$ rational coordinates }, B_{r'}(x) \subseteq \widehat{U}, \text{ for some } r' > r \rbrace$ \\
Clearly, $\forall \, B_r(x)$ precompact in $\widehat{U}$ \\
\quad \, $\mathcal{B}$ countable basis for topology of $\widehat{U}$

$\varphi$ homeomorphism, it follows $\lbrace \varphi^{-1}(B) | B \in \mathcal{B} \rbrace$ countable basis for $M$

Let $M$ arbitrary, \\
\quad By def., $\forall \, p \in M$, $p \in $ domain $U$ of a chart

Prop. A.16, $\forall \, $ open cover of second-countable space has countable subcover. \\
\quad $M$ covered by countably many charts $\lbrace (U_i, \varphi_i) \rbrace$ \\
$\forall \, U_i$, $U_i$ has countable basis of coordinate balls precompact in $U_i$ \\
union of all these coordinates bases is countable basis for $M$.   \\

If $V \subseteq U_i$ one of these balls,  \\
\quad then $\overline{V}$ compact in $U_i$.  $M$ Hausdorff, so $\overline{V}$ closed.  \\
\quad $\overline{V}$ in $M$ is same as $\overline{V}$ in $U_i$, so $V$ precompact in $M$.
\end{proof}

\subsubsection*{Connectivity}

\subsubsection*{Local Compactness and Paracompactness}

\subsubsection*{Fundamental Groups of Manifolds}

\subsection*{Smooth Structures}

If open $U \subset \mathbb{R}^n$, $V \subset \mathbb{R}^m$, \\
$F:U \to V$ smooth (or $C^{\infty}$) if $\forall \, $ cont. partial derivative of all orders exists.  \\

$F$ diffeomorphism if $F$ \emph{smooth}, bijective, and has a \emph{smooth} inverse.  \\
Diffeomorphism is a homeomorphism.  \\

$(U, \varphi), (V,\psi)$ smoothly compatible if $UV = \emptyset$ or
\[
\psi \varphi^{-1}: \varphi(UV) \to \psi(UV) \quad \quad \, \text{ diffeomorphism }
\]

atlas $\mathcal{A} \equiv \lbrace (U, \varphi ) \rbrace$ s.t. $\bigcup U = M$.  Smooth atlas if $\forall \, (U,\varphi), (V, \psi) \in \mathcal{A}$, $(U,\varphi), (V,\psi)$ smoothly compatible.  \\

Smooth structure on topological $n$-manifold $M$ is a maximal smooth atlas.   \\

Smooth manifold $(M, \mathcal{A})$ where $M$ topological manifold, $\mathcal{A}$ smooth structure on $M$.   \\

\begin{proposition}[1.17] Let $M$ topological manifold.
\begin{enumerate}
\item[(a)] $\forall \, $ smooth atlas for $M$ is contained in ! maximal smooth atlas.
\item[(b)] 2 smooth atlases for $M$ determine the same maximal smooth atlas iff union is smooth atlas.
\end{enumerate}
\end{proposition}

\begin{proof}
Let $\mathcal{A}$ smooth atlas for $M$   \\
$\overline{ \mathcal{A} } \equiv $ set of all charts that are smoothly compatible with every chart in $\mathcal{A}$

\textbf{Want}: $\overline{ \mathcal{A}}$ smooth atlas, i.e. $\forall \, (U, \varphi), (V \psi) \in \overline{\mathcal{A}}$, $\psi \varphi^{-1} : \varphi(UV) \to \psi(UV)$ smooth.

Let $x = \varphi(p) \in \varphi(UV)$ \\
$p\in M$, so $\exists \,$ some chart $(W, \theta) \in \mathcal{A}$ s.t. $p \in W$.  \\
By given, $\theta \varphi^{-1}, \psi \theta^{-1}$ smooth where they're defined.   \\
$p \in UVW$, so $\psi \varphi^{-1} = \psi \theta^{-1} \theta \varphi^{-1}$ smooth on $x$.  \\
Thus $\psi \varphi^{-1}$ smooth in a neighborhood of each pt. in $\varphi(UV)$.  Thus $\overline{\mathcal{A}}$ smooth atlas.

To check maximal,

\end{proof}

\subsection*{Local Coordinate Representations}

\begin{proposition}[1.19]
  $\forall \, $ smooth $M$ has countable basis of regular coordinate balls
\end{proposition}

\exercisehead{1.20}

smooth manifold $M$ has smooth structure \\
\quad Suppose single smooth chart $\varphi$ has entire $M$ as domain
\[
\varphi: M \to \widehat{U} \subseteq \mathbb{R}^n
\]

Let $\widehat{B} = \lbrace \widehat{B}_r(x) \subseteq \mathbb{R}^n | r \in \mathbb{Q}, \, x \in \mathbb{Q}, \, \widehat{B}_{r'}(x) \subseteq \widehat{U} \text{ for some } r' > r \rbrace$ \\

\quad $\forall \, \widehat{B}_r(x)$ precompact in $\widehat{U}$ \\
\quad $\widehat{\mathcal{B}}$ countable basis for topology of $\widehat{U}$

$\varphi$ homeomorphism,

\quad Let $\begin{aligned}
  & \quad \\
  & \varphi^{-1}( \widehat{B}_r{(0)}) = B \\
  & \varphi^{-1}{(\widehat{B}_{r'}{(0)} )} = B' \end{aligned}$

$\varphi$ homeomorphism and since $\widehat{B}$ countable basis, $\lbrace B \rbrace$ countable basis of regular coordinate basis. \\

Suppose arbitrary smooth structure.  \\
\quad By def., $\forall \, p \in M$, $p$ in some chart domain \\
\quad Prop. A.16., $\forall \, $ open cover of second countable space has countable subcover \\
\quad $M$ covered by countably many charts $\lbrace (U_i, \varphi_i) \rbrace$ \\

$\forall \, U_i , \, U_i$ has countable basis of coordinate balls precompact in $U_i$ \\
union of all these coordinate charts is countable basis for $M$.

If $V \subseteq U_i$, 1 of these balls,  \\
$\begin{aligned}
  & \quad \\
  & \varphi(V) = B_r(0) \\
  & \varphi{(\overline{V})} = \overline{B}_r(0)
\end{aligned}$

and $\varphi(B') = B_{r'}{(0)}$, $r' > r$ for countable basis for $U_i$ \\
So $V$ regular coordinate ball.

\hrulefill

\subsection*{Examples of Smooth Manifolds}

\subsubsection*{More Examples}

\textbf{Example 1.25 (Spaces of Matrices) } Let $M(m\times n,\mathbb{R}) \equiv $ set of $m\times n$ matrices with real entries.

\textbf{Example 1.26 (Open Submanifolds)}

$\forall \, $ open subset $U \subseteq M$ is itself a $\text{dim}{M}$ manifold. \\
EY : \emph{ $\forall \, $ open subset $U \subseteq M$ is itself a $\text{dim}{M}$ manifold. }

\textbf{Example 1.27 (The General Linear Group) }

general linear group $GL(n,\mathbb{R}) = \lbrace A | \text{det}{A} \neq 0 \rbrace$ \\
\quad $\text{det}:A \to \mathbb{R}$ is cont. (by def. of $\text{det}{A} = \epsilon^{ i_1 \dots i_n}a_{1 i_1} \dots a_{n i_n}$ \\
\quad $\text{det}^{-1}{ ( \mathbb{R} - 0 )}$ is open since $\mathbb{R}-0$ open so $GL(n,\mathbb{R})$ open \\
\quad $GL(n,\mathbb{R}) \subseteq M(n,\mathbb{R})$, $M(n,\mathbb{R})$ $n^2$-dim. vector space. \\
\quad \quad so $GL(n,\mathbb{R})$ smooth $n^2$-dim. manifold.

\textbf{Example 1.28 (Matrices of Full Rank)}

Suppose $m <n$ \\

Let $M_n(m\times n, \mathbb{R}) \subseteq M(m\times n, \mathbb{R})$ with matrices of rank $m$ \\
if $A \in M_m(m\times n, \mathbb{R})$, \\
\quad $\text{rank}{A}=m$

means that $A$ has some nonsingular $m \times m$ submatrix.  (EY 20140205 ???)

\textbf{Example 1.31 (Spheres)}

\[
\begin{aligned}
  & \varphi_i^{\pm} : S^n \to B_1^n(0) \subset \mathbb{R}^n \quad \, (B_1^n(0) \text{ disk of radius $1$}) \\
  & \varphi_i^{\pm}(x_1 \dots x_{n+1}) = (x_1 \dots \widehat{x}_i \dots x_{n+1} )= (y_1 \dots y_n)
\end{aligned}
\]
Note $x_1^2 + \dots + x_i^2 + \dots + x_{n+1}^2 = 1$.  \quad \, $x_i = \pm \sqrt{ 1 - (x_1^2 + \dots + \widehat{x}_i^2 + \dots + x_{n+1}^2 ) }$

\[
\begin{aligned}
  & (\varphi_i^{\pm})^{-1}(y_1 \dots y_n) = (y_1 \dots \pm \sqrt{ 1 - (y_1^2 + \dots + y_n^2) } \dots y_n) = (y_1 \dots y_{i-1}, \pm \sqrt{ 1-  |y|^2 }, y_i\dots y_n ) \\
 & \varphi_i^{\pm} (\varphi_j^{\pm})^{-1}(y_1 \dots y_n ) = (y_1 \dots \widehat{y}_i \dots y_{j-1} , \pm \sqrt{ 1 - |y|^2 }, y_j \dots y_n) \\
 & \varphi_i^{\pm} (\varphi_j^{\mp})^{-1}(y_1 \dots y_n ) = (y_1 \dots \widehat{y}_i \dots y_{j-1} , \mp \sqrt{ 1 - |y|^2 }, y_j \dots y_n) \\
\end{aligned}
\]
\[
\varphi_j^{\pm}(\varphi_i^{\mp})^{-1} \varphi_i^{\pm} (\varphi_j^{\pm})^{-1}(y_1 \dots y_n) = \varphi_j^{\pm}(y_1 \dots \pm y_i \dots y_{j-1}, \pm \sqrt{ 1 - |y|^2 } \dots y_n ) = (y_1 \dots \pm y_i \dots y_j \dots y_n)
\]
This is symmetrical in $i,j$ and so true if $i,j$ reverse.  \\

So $\varphi_i^{\pm}(\varphi_j^{\pm})^{-1}$ diff. and bijective.  Likewise for $\varphi_i^{\pm}(\varphi_j^{\mp})^{-1}$.  \\
So $\begin{aligned} & \varphi_i^{\pm} (\varphi_j^{\pm})^{-1} \\
  & \varphi_i^{\pm} (\varphi_j^{\mp})^{-1} \end{aligned}$ diffeomorphisms.

\begin{lemma}[1.35] \textbf{(Smooth Manifold Chart Lemma)}
Let $M$ be a set, suppose given $\lbrace U_{\alpha} | U_{\alpha} \subset M \rbrace$, given maps $\varphi_{\alpha}: U_{\alpha} \to \mathbb{R}^n$ s.t.
\begin{enumerate}
\item[(i)] $\forall \, \alpha$, $\varphi_{\alpha}$ bijection between $U_{\alpha}$ and open $\varphi_{\alpha}(U_{\alpha}) \subseteq \mathbb{R}^n$
\item[(ii)] $\forall \, \alpha, \beta$, $\varphi_{\alpha}(U_{\alpha} \bigcap U_{\beta})$, $\varphi_{\beta}(U_{\alpha} \bigcap U_{\beta})$ open in $\mathbb{R}^n$
\end{enumerate}
\end{lemma}

\textbf{Example 1.36 (Grassmann Manifolds)}

%$G_k(V) = $ set of all $k$-dim. linear subspaces of $V$, $\text{dim}{V} = n \geq k $

%Let $P,Q$ complementary subspaces of $V$, $\begin{aligned} & \quad \\ & \text{dim}{P} = k \\ & \text{dim}{Q} = n- k \end{aligned}$

%direct sum decomposition $V = P \oplus Q$

%graph of any linear $X: P \to Q$

%\[
%\Gamma(X) = \lbrace v + X v | v \in P \rbrace \subset V
%\]
%$k$-dim. subspace

%Now $\Gamma(X) Q = \emptyset$ since $\Gamma(A) \subset P \oplus Q$ and $P$ and $Q$ are complementary.

%If $KQ = \emptyset$, then $K \subset P \oplus Q$, $P\neq 0$.

$G_k(V) = \lbrace S | S \subseteq V \rbrace$ \quad \, $S$ $k$-dim. linear subspace of $V$ \\
$\text{dim}V = n$, $V$ vector space  \\

if $V = \mathbb{R}^n$, $G_k(\mathbb{R}^n) \equiv G_{k,n} \equiv G(k,n)$ (notation) \quad \, $G_1(\mathbb{R}^{n+1}) = \mathbb{R}P^n$

Let $V = P \oplus Q$, \, $\begin{aligned}
  & \text{dim}P =k \\
  & \text{dim}Q = n-k \end{aligned}$ \\

linear $X: P \to Q$

\[
\Gamma(X) = \lbrace v + Xv | v \in P \rbrace , \quad \, \Gamma(X) \subseteq V, \, \text{dim}\Gamma(X) = k
\]
$\Gamma(X) \bigcap Q = 0$ since $\forall \, w \in \Gamma(X)$, $w$ has a $P$ piece, and $Q$ complementary to $P$ \\

Converse: $\forall \, $ subspace $S \subseteq V$, s.t. $S\bigcap Q = 0$ \\
\quad \, let $\begin{aligned}
  & \quad \\
  & \pi_P : V \to P \\
  & \pi_Q: V \to Q
\end{aligned}$ \quad \, projections by direct sum decomposition $V = P\oplus Q$ \\

$\left. \pi_P \right|_S : S \to P$ isomorphism  \\
$\Longrightarrow X = ( \left. \pi_Q \right|_S ) \cdot ( \left. \pi_P \right|_S)^{-1}$, \, $X: P \to Q$ \\

Let $v\in P$.  $v+ Xv = v +  \left. \pi_Q \right|_S (  \left. \pi_P \right|_S)^{-1} v$.  Let $v  \in \left. \pi_P \right|_S (S)$ \quad \, $\Gamma(X) = S$ \\

Let $L(P;Q) = \lbrace f | \text{ linear } f : P \to Q \rbrace$, \, $L(P;Q)$ vector space \\
$U_Q \subseteq G_k(V)$, \, $U_Q = \lbrace S | \text{dim}S =k, \, S \text{ subspace }, \, S \bigcap Q = 0 \rbrace$ \\
$\begin{aligned}
  & \Gamma : L(P;Q) \to U_Q \\
  & X \mapsto \Gamma(X) \end{aligned}$ \\

$\Gamma$ bijection by above \\
$\varphi = \Gamma^{-1} : U_Q \to L(P;Q)$  \\

By choosing bases for $P,Q$, identify $L(P;Q)$ with $M((n-k)k; \mathbb{R})$ and hence with $\mathbb{R}^{k(n-k)}$ \\

think of $(U_Q, \varphi)$ as coordinate chart. \\
$\varphi(U_Q) = L(P;Q)$

\subsection*{Problems}

\problemhead{1.7} (This was Problem 1.5 in previous editions)

\[
S^n = \lbrace ( x_1 \dots x_{n+1} \rbrace \in \mathbb{R}^{n+1} | \sum_{i=1}^{n+1} x_i^2 = 1 \rbrace \subset \mathbb{R}^{n+1}
\]

Let $\begin{aligned} & \quad \\
  & N = (0 \dots 0, 1) \\
  & S = (0 \dots 0, -1) \end{aligned}$ \quad \, $x\in S^n$.
\begin{enumerate}
\item[(a)]
 Consider $t(x-N) + N = tx + (1-t)N$ when $x_{n+1} =0$
\[
\begin{gathered}
  tx_{n+1} + (1-t) = 0 \text{ or } tx_{n+1} + -1 + t = 0 \\
  \Longrightarrow \frac{ 1}{ 1- x_{n+1} } = t \quad \left( \text{ or } \frac{1}{ 1 + x_{n+1} } \right)
\end{gathered}
\]

\[
\begin{aligned}
  & \pi_1: S^n - N \to \mathbb{R}^n \\
  & \pi_1(x_1 \dots x_{n+1}) = \left( \frac{x_1}{ 1 - x_{n+1} } \dots \frac{x_n}{ 1 - x_{n+1} } , 0 \right) \\
  & \pi_2 : S^n -S \to \mathbb{R}^n \\
  & \pi_2(x_1 \dots x_{n+1}) = \left( \frac{x_1}{ 1 + x_{n+1}} \dots \frac{x_n}{ 1 + x_{n+1} }, 0 \right)
\end{aligned}
\]
Note that $-\pi_2(-x) = \pi_1$ and $\pi_1 \equiv \sigma$, $\pi_2 \equiv \widetilde{\sigma}$ in Massey's notation.
\item[(b)] Note, for $y_i = \frac{x_i}{ 1 - x_{n+1}}$
\[
\begin{gathered}
  y_1^2 + \dots +y_n^2 = |y|^2 = \frac{1-  x_{n+1}^2}{ (1-x_{n+1})^2 } = \frac{1+ x_{n+1}}{ 1- x_{n+1}} \text{ or } x_{n+1}  = \frac{ |y|^2 - 1 }{ |y|^2 + 1 }  \\
  x_i = y _i (1- x_{n+1}) = \frac{2y_i}{ 1 + |y|^2 }
\end{gathered}
\]

\[
\begin{aligned}
  & \pi_1^{-1}: \mathbb{R}^n \to S^n - N \\
  & \pi_1^{-1}(y_1 \dots y_n) = \left( \frac{2y_1}{ 1 + |y|^2 } \dots \frac{2y_n}{1+ |y|^2 } , \frac{ |y|^2 - 1 }{ |y|^2 + 1 } \right) \\
  & \pi_2^{-1}(y_1 \dots y_n) = \left( \frac{2y_1}{ 1 + |y|^2 } \dots \frac{2y_n}{1+ |y|^2 } , \frac{ 1 - |y|^2  }{ |y|^2 + 1 } \right)
\end{aligned}
\]

$\pi_1,\pi_2$ diff., bijective, and $(S^n - N ) \bigcup (S^n-S) = S^n$ \\

\item[(c)] Computing the transition maps for the stereographic projections.

Consider $(S^n - N)(S^n-S) = S^n - N \bigcup S$
\[
\begin{aligned}
  & \pi_1 \pi_2^{-1}(y_1 \dots y_n) = \left( \frac{y_1 }{ |y|^2} \dots \frac{y_n}{|y|^2} , 0 \right) \\
  & \pi_2 \pi_1^{-1}(y_1 \dots y_n) = \left( \frac{y_1 }{ |y|^2} \dots \frac{y_n}{|y|^2} , 0 \right)
\end{aligned}
\]
since, for example,
\[
\begin{gathered}
\frac{   \frac{2y_i }{ 1 + |y|^2} }{ 1 - \frac{ 1 - |y|^2}{ 1 + |y|^2 }} = \frac{y_i }{ |y|^2 }
\end{gathered}
\]

$\pi_1 \pi_2^{-1}$ bijective and $C^{\infty}$, $\pi_1 \pi_2^{-1}$ diffeomorphism.   \\

$\lbrace (S^n-N, \pi_1), (S^n - S, \pi_2) \rbrace$ \, $C^{\infty}$ atlas or differentiable structure.
\[
\begin{aligned}
  & \partial_j \frac{y_i}{ |y|^2} = \frac{ - 2y_i y_j }{ (y_1^2 + \dots + y_n^2 )^2 } \\
  & \partial_j \frac{ y_j}{ |y|^2} = \frac{ (y_1^2 + \dots + y_n^2) - 2y_j^2 }{ |y|^4} = \frac{ y_1^2 + \dots + \widehat{y}_j^2 + \dots + y_n^2 - y_j^2 }{ |y|^4}
\end{aligned}
\]
\[
\text{det}{ (\partial_j \pi_1 \pi_2^{-1}(y) ) } = \sum_{ \sigma \in S_n} \text{sgn}{ (\sigma)} \partial_{\sigma_1} \frac{y_1}{ |y|^2} \dots \partial_{\sigma_n} \frac{y_n}{|y|^2} = 0 \text{ only if $y=0$. But that's excluded }
\]
\item[(d)] Consider $\begin{gathered}
  \lbrace ( S^n\backslash, \pi_1), (S^n \backslash S, \pi_2) \rbrace \\
  \mathcal{A} = \lbrace (U_i^{\pm}, \varphi_i^{\pm}) \rbrace \end{gathered}$  \\

Now
\[
\begin{aligned}
  & S^n \backslash N \bigcap U_i^+ = \begin{cases} U_i^+ & \text{ if } i \neq n + 1 \\ U_{n+1}^+ \backslash N & \text{ if } i = n+1 \end{cases} \\
  & S^n \backslash N \bigcap U_i^- = U_i^-
\end{aligned}
\]

\[
\pi_1(\varphi_i^{\pm})^{-1}(y_1 \dots y_n) = \pi_1(y_1 \dots y_{i-1}, \pm \sqrt{ 1 - |y|^2 }, y_i \dots y_n) = \left( \frac{y_1 }{ 1 - y_n} \dots \frac{y_{i-1} }{ 1 - y_n}, \frac{ \pm \sqrt{ 1 - |y|^2 } }{ 1 - y_n } , \frac{y_i}{ 1 - y_n} \dots \frac{y_{n-1}}{ 1 - y_n }, 0 \right)
\]

Note $-1< y_n <1$ on $\varphi_i^{\pm}(S^n\backslash \bigcap U_i^{\pm})$
\[
\varphi_i^{\pm} \pi_1^{-1}(y_1 \dots y_n) = \varphi_i^{\pm}\left( \frac{ 2y_1}{ 1 + |y|^2} \dots \frac{2y_n}{ 1 + |y|^2} , \frac{|y|^2 - 1 }{ |y|^2 + 1 } \right) = \left( \frac{2y_1 }{ 1 + |y|^2} \dots \frac{ 2 \widehat{y}_i }{ 1 + |y|^2 } \dots \frac{ 2y_n }{ 1 + |y|^2 }, \frac{|y|^2 -1}{ |y|^2 + 1 } \right)
\]
$\pi_{1,2}(\varphi_i^{\pm})^{-1}, \varphi_i^{\pm}\pi_{1,2}^{-1}$ are diffeomorphisms (bijective and differentiable).

So $\lbrace(S^n\backslash N, \pi_1), (S^n\backslash S , \pi_2) \rbrace \bigcup \mathcal{A}$ also a $C^{\infty}$ atlas.  \\
So $\lbrace(S^n\backslash N, \pi_1), (S^n\backslash S , \pi_2) \rbrace$ ,  $\mathcal{A}$ equivalent.

\end{enumerate}


\chapter{Infinite Series}
We previously discussed some special types of series and formulas for their nth term.
Now, we'll look at sums of infinitely many terms and eventually see how summing variables rather than just numbers allows us to approximate functions.

\input{./infinite_series/power_series.tex}
\input{./infinite_series/taylor_series.tex}
\input{./infinite_series/error.tex}
\input{./infinite_series/convergence.tex}
\input{./infinite_series/radius_of_convergence.tex}


%%==========================================
%% Section 2.02: $\lambda$-DIFFERENTIABILITY
%%==========================================
\documentclass[../dissertation]{subfiles}

\begin{document}

\section{$\lambda$-Differentiability}\label{sec2:diff}

We begin this section by proving a useful variant of Young's inequality
(Technical Lemma \ref{tlma2:1}) that we use in a number of proofs in this
dissertation. We then prove that that the operators $T_{\star, \lambda, u}$ are
continuous in the spectral parameter $\lambda$ (Proposition
\ref{prop2:Tlamcont}). The remainder of this section the focuses on proving the
$\lambda$-differentiability of $G_{\star}^+$. We direclty use the results
from all four propositions in this section in proving that the Jost solution
boundary $M_1^+$ is differentiable in $\lambda$, and hence has a
linearization in $\lambda$. The $\lambda$ linearization of $M_1^+$ is ultimately
used in the proof that the direct scattering map $\mathscr D$ is Lipschitz
continuous.

Since the proof of the $\lambda$-continuity of $T_{\star, \lambda, u}$ calls
Technical Lemma \ref{tlma2:1}, we start with the proof of that lemma.

\begin{tlma}\label{tlma2:1}
	For $f \in \inn{\dotarg}^s L^1(\mathbb R)$ and
	$\inn{\dotarg}^s g \in L^\infty(\mathbb R)$ the inequality
	\begin{align} \label{eq3:tlma}
		\| f * g\|_{\inn{\dotarg}^s L^\infty}
			&\leq \|f\|_{\inn{\dotarg}^s L^1} \|\inn{\dotarg}^s g\|_{L^\infty}
	\end{align}
	holds for $s\geq0$. Alternatively, if $\inn{\dotarg}^s f \in L^1(\mathbb R)$
	and $g \in \inn{\dotarg}^s L^\infty(\mathbb R)$ the estimate
	\begin{align}\label{eq3:tlma2}
		\| f * g\|_{\inn{\dotarg}^s L^\infty}
			&\leq \nm{\inn{\dotarg}^s f}_{L^1}  \nm{g}_{\inn{\dotarg}^s L^\infty}
	\end{align}
	holds instead for $s \geq 0$.
\end{tlma}
\begin{proof}
	It is straightforward to show that
	\begin{align*}
		\frac{\inn{x'}}{\inn{x-x'}\inn{x}} \leq 1,
	\end{align*}
	for all $x, x' \in \mathbb R$, as
	\begin{align*}
		\left(\frac{\inn{x'}}{\inn{x-x'}\inn{x}}\right)^2
			= \frac{1+(x')^2}{1+(x-x')^2 + x^2 + x^2(x-x')^2}
			\leq \frac{1+(x')^2}{1+(x')^2}.
	\end{align*}
	As such, we find for $s> 0$
	\begin{align*}
		\| f * g\|_{\inn{\dotarg}^s L^\infty}
			&= \sup_{x\in\mathbb R} \inn{x}^{-s}
				\int_{\mathbb R}
					\big|
						f(x') \, g(x-x')
					\big|
				\, \mathrm{d}x'
				\\
			&= \sup_{x\in\mathbb R}
				\int_{\mathbb R}
					\Big|
						\big[\inn{x'}^{-s} f(x')\big]
						\big[\inn{x-x'} g(x-x') \big]
					\Big|
						\left(
							\frac{\inn{x'}}{\inn{x-x'}\inn{x}}
						\right)^s
				\, \mathrm{d}x'
				\\
			&\leq
				\left\|
						\big[\inn{\dotarg}^{-s} f \big]
						*
						\big[\inn{\dotarg}^{s} g \big]
				\right\|_{L^\infty} \\
			&\leq
				\|f\|_{\inn{\dotarg}^s L^1} \|\inn{\dotarg}^s g\|_{L^\infty}
	\end{align*}
	by Minkowski's integral inequality \cite[Theorem 1.2.10]{Grafakos}.
	If $s=0$, then \eqref{eq3:tlma} automatically holds by
	\cite[Theorem 1.2.10]{Grafakos}. An analogous argument also verifies
	\eqref{eq3:tlma2}.
\end{proof}

\begin{prop}\label{prop2:Tlamcont}
	For $u \in X \cap \inn{\dotarg}^{-2}L^\infty(\mathbb R)$, the operator
	$T_{\star, \lambda, u} : \inn{\dotarg}L^\infty(\mathbb R) \to
	\inn{\dotarg}L^\infty(\mathbb R)$ given by
	\begin{align*}
		T_{\star, \lambda, u} : f\mapsto \big[G_\star^+(\cdot; \lambda)\big]*(u\,f)
	\end{align*}
	is continuous in the parameter $\lambda \in \mathbb R$ in the sense the limit
	\[
		\lim_{h\to0} \| T_{\star, \lambda+h, u} - T_{\star, \lambda, u}
			 \|_{\inn{\dotarg}L^\infty\toitself}
			=0
	\]
	holds pointwise for each fixed $\lambda \in \mathbb R$.
\end{prop}
\begin{proof}
	To simplify notation, let $T_{\star, \lambda, u}$ be denoted by
	$T_\lambda$. Then, we see from Technical Lemma \ref{tlma2:1}
	that
	\begin{align}\label{eq2:Tcont}
		\|(T_{\lambda+h} - T_\lambda)f\|_{\inn{\dotarg}L^\infty}
			&=
				\left\|
					\big[
						G_\star^+(\dotarg, \lambda+h)
						- G_\star^+(\dotarg, \lambda)
					\big]
					*uf
				\right\|_{\inn{\dotarg}L^\infty}  \\
			&\leq
				\left\|
						G_\star^+(\dotarg, \lambda+h)
						- G_\star^+(\dotarg, \lambda)
				\right\|_{\inn{\dotarg}L^\infty}
				\left\|
					\inn{\dotarg} u f
				\right\|_{L^\infty}
				\nonumber
	\end{align}
	for all $f \in \inn{\dotarg}L^\infty(\mathbb R)$ as
	$u \in X \cap \inn{\dotarg}^{-2}L^\infty(\mathbb R)$ implies
	$\inn{\dotarg} u\, f\in \inn{\dotarg}L^\infty(\mathbb R)$.
	Noting that the argument in the proof of Proposition
	\ref{prop2:ldervG} (which does not depend on this Proposition) implies
	\begin{align}\label{eq2:Glamcont}
		\lim_{h\to\infty}
			\left\|
					G_\star^+(\dotarg, \lambda+h)
					- G_\star^+(\dotarg, \lambda)
			\right\|_{\inn{\dotarg}L^\infty}
			=0,
	\end{align}
	Proposition \ref{prop2:Tlamcont} follows from \eqref{eq2:Tcont}
	and \eqref{eq2:Glamcont}.
\end{proof}

\begin{rmk}\label{rmk2:notation}
	The notation in proofs of Propositions \ref{prop2:ldervG} through
	\ref{prop2:ghconv} can get unnecessarily complicated. To avoid this,
	in these proof we use the notation $G(x, \lambda)$ and $G(\lambda)$
	as stand-ins for $G_\star^+(x; \lambda)$.
\end{rmk}

\begin{prop}\label{prop2:ldervG}
	For each fixed $x\ne 0$ and $\lambda \in \mathbb R$, the Green's function
	boundary $G_\star^+$ ($\star = L \text{, or } R$) is differentiable in
	the spectral parameter $\lambda$, and
	\begin{align*}
		\frac{\partial}{\partial \lambda} G_\star^+(x; \lambda)
			= \frac{1}{2\pi}
				\int_{{\Gamma_\star}} e^{ix\xi}
					\left(
						\frac{\partial}{\partial \lambda} \frac{1}{p(\xi; \lambda)}
					\right)
				\, \mathrm{d}\xi.
	\end{align*}
\end{prop}
\begin{rmk}
	Since
	\[
		\frac{\partial}{\partial \lambda} \frac{1}{p(\xi; \lambda)}
			=
				\frac{\zeta'(\lambda)}
				{\Big(\xi - \zeta(\lambda)\big(1-e^{-2\xi}\big)\Big)^2},
	\]
	the function $\frac{\partial}{\partial \lambda} \frac{1}{p(\xi; \lambda)}$
	decays exponentially to zero as $\xi\to -\infty$, and, for $\xi > 0$, decays
	like $1/\xi^2$. As such, the integral
	\begin{align*}
		\int_{\Gamma_\star}
			\left(
				\frac{\partial}{\partial \lambda} \frac{1}{p(\xi; \lambda)}
			\right)
		\, \mathrm{d}\xi.
	\end{align*}
	converges absolutely.
\end{rmk}
\begin{proof}[Proof of Proposition \ref{prop2:ldervG}]
	In accordance with Remark \ref{rmk2:notation}, we write $G_\star^+(x; \lambda)$
	as either $G(x, \lambda)$ or as $G(\lambda)$. We further define $G_h$ to be the
	difference quotient
	\[
		G_h:= \frac{G(\lambda + h)- G(\lambda)}{h}.
	\]
	\label{sym2:Gh}
	and seek to prove that $\lim_{h\to0} G_h$ converges. For $|h|>0$ sufficiently
	small, we may assume by analyticity that $G(\lambda)$ and $G(\lambda+h)$
	share the same contour of integration $\Gamma$. If $\lambda = 0$, the
	contour $\Gamma$ runs along with a single small semi-circular detour below the
	real axis to avoid passing through $\xi = 0$ and $\xi=h$. In which case
	\begin{align*}
		G_h(\lambda) =
			\frac{1}{2\pi}\int_\Gamma e^{ix\xi}\,
				\frac{1}{h}\,
				\left[
					\frac{1}{p(\xi; \lambda +h)}
					-\frac{1}{p(\xi; \lambda)}
				\right]
			\, \mathrm{d}\xi
			=
				\int_\Gamma e^{ix\xi}
					\left(\frac{1}{p_\lambda(\xi)}\right)_{h}
				\, \mathrm{d}\xi,
	\end{align*}
	where we define
	\begin{align*}
		\left(\frac{1}{p_\lambda(\xi)}\right)_{h}
			:= \frac{1}{h}\,
				\left[
					\frac{1}{p(\xi; \lambda +h)}
					-\frac{1}{p(\xi; \lambda)}
				\right]
	\end{align*}
	\label{sym2:pdiffquot}
	as the difference quotient of $1/p$ with respect to $\lambda$.
	Since
	\[
		\left(\frac{1}{p_\lambda(\xi)}\right)_{h}
			= \frac{1}{h}
				\frac{
					\big[
						\zeta(\lambda + h) - \zeta(h)
					\big]
					\big(1-e^{-2\xi}\big)
				}
				{p(\xi; \lambda+h)\,p(\xi;\lambda)}
	\]
	decays exponentially as $\xi \to -\infty$ and decays as
	$1/\xi^2$ for positive $\xi$,
	\[
		\left(\frac{1}{p_\lambda(\xi)}\right)_{h} \in L^1(\Gamma)
	\]
	for each fixed $h\ne0$. Further, using the continuity of $\zeta$ and the reverse
	triangle inequality, we also have
	\[
		\left|\left(\frac{1}{p_\lambda(\xi)}\right)_{h}\right|
			\lesssim_\lambda
				\frac{
					\big|1-e^{-2\xi}\big|
				}{
					|\xi|\big| \xi - \zeta(\lambda) \big(1-e^{-2\xi}\big)\big|
				}
			=: \iota(\xi).
	\]
	Since $\iota$ is continuous in $\xi$ on $\Gamma$ and $\iota(\xi)
	= \mathcal O(1/\xi^2)$ for large $|\xi|$, one application of the
	Dominated Convergence Theorem completes this proof.
\end{proof}

\begin{prop}\label{prop2:lderivGspace}
	The partial derivative $\frac{\partial}{\partial \lambda} G_\star^+$
	($\star = L \text{, or } R$)
	of the Green's function boundary value $G_\star^+$ lies in $\inn{x}^s L_x^1(\mathbb R)$
	for $s> 3$ and all $\lambda \in \mathbb R$.
	If $\lambda \in \mathbb R$ and $\lambda \ne 0$, then
	$\frac{\partial}{\partial \lambda} G_\star^+
	\in \inn{x}^s L_x^1(\mathbb R)$ for $s > 2$.
\end{prop}
\begin{proof}
	Define
	\[
		g(\xi; \lambda)
			:= e^{ix\xi}\, \frac{\partial}{\partial \lambda} \frac{1}{p},
	\]
	Through direct computation, one can show
	\begin{align}\label{eq2:gzerores}
		\Res_{\xi=0} g
			=
				\frac{
					2 e^{2\lambda}\big(e^{2\lambda}-2\lambda-1\big)
				}{
					\big(
						1 - e^{2\lambda}+ 2\lambda e^{2\lambda}
					\big)^2
				},
	\end{align}
	\begin{align}\label{eq3:glamres}
		\Res_{\xi=\lambda} g
			=
				\frac{
					2 e^{2\lambda} - 2 - 4 \lambda e^{2\lambda}
					+ i
					\big[
						x -
						2 x e^{2\lambda} + x e^{4\lambda} + 2x\lambda
						- 2 x \lambda e^{2\lambda}
					\big]
				}
				{\big(e^{2\lambda} - 2 \lambda -1\big)^2}
				\, e^{i x \lambda},
	\end{align}
	and
	\begin{align}\label{eq2:gcollapsingpoles}
		\lim_{\lambda\to0}
				\big[
					\Res_{\xi=0} g + \Res_{\xi=\lambda} g
				\big]
			= -\frac{1}{2} \, x^2 + i \, \frac{1}{3} \, x.
	\end{align}
	In particular, equations \eqref{eq2:gzerores}, \eqref{eq3:glamres},
	and \eqref{eq2:gcollapsingpoles} imply
	\begin{align}\label{eq3:gsummary}
		\begin{aligned}
			\big|\Res_{\xi=0} g \big| \lesssim_\lambda 1, \qquad
		\big|\Res_{\xi=\lambda} g \big| \lesssim_\lambda |x|, \\
		\lim_{\lambda\to0}
				\big[
					\Res_{\xi=0} g + \Res_{\xi=\lambda} g
				\big]
			= \mathcal O(x^2)
		\end{aligned}
	\end{align}

	Further, on our work in the proof of Lemma \ref{prop2:ldervG}
	also implies that
	\begin{align}\label{eq3:pldervspace}
		\frac{\partial}{\partial \lambda} \frac{1}{p}
		 \in L_\xi^1(\mathbb R \pm i \pi).
	\end{align}
	As such, after applying the contour shift demonstrated in Figure
	\ref{fig1:GammaContour} to the contour of integration for
	$\frac{\partial}{\partial \lambda} G_L^+$,
	Lemma \ref{prop2:lderivGspace} is an immediate
	consequence of \eqref{eq3:gsummary} and \eqref{eq3:pldervspace}.
\end{proof}

\begin{prop}\label{prop2:ghconv}
	The difference quotient
	\[
		G_h(x; \lambda) := \frac{G_L^+(x; \lambda+h) - G_L^+(x; \lambda)}{h}
	\]
	converges to $\frac{\partial}{\partial \lambda} G_L^+$ in
	$\inn{x}^s L_x^1(\mathbb R)$ for $s > 3$ and all real $\lambda$.
	If $\lambda$ is real and non-zero, then
	this convergence happens in $\inn{x}^s L_x^1(\mathbb R)$ for
	$s > 2$.
\end{prop}
\begin{proof}
	Direct computation yields the following results
	\begin{align*}
		\Res_{\xi=0}
				\left[
					e^{ix\xi}
					\left(
						\frac{1}{p_\lambda(\xi)}
					\right)_h
				\right]
			&=
				\frac{
					2 e^{2\lambda}
					\big(
						h \, e^{2(\lambda+h)}
						+ \lambda
						- e^{2h}(\lambda + h)
					\big)
				}{
					h
					\big(
						1 + e^{2\lambda}(2\lambda - 1)
					\big)
					\big(
						1 + e^{2(\lambda+h)(2\lambda + 2h - 1)}
					\big)
				} \\
		\Res_{\xi=\lambda}
				\left[
					e^{ix\xi}
					\left(
						\frac{1}{p_\lambda(\xi)}
					\right)_h
				\right]
			&= - \frac{1}{h}\frac{e^{2\lambda}-1}{e^{2\lambda}-2\lambda-1}
				\, e^{ix\lambda} \\
		\Res_{\xi=\lambda+h}
				\left[
					e^{ix\xi}
					\left(
						\frac{1}{p_\lambda(\xi)}
					\right)_h
				\right]
			&=
				 \frac{1}{h}
				 \frac{e^{2(\lambda+h)}-1}{e^{2(\lambda+h)}-2(\lambda+h)-1}
				\, e^{ix(\lambda+h)}
	\end{align*}
	As such, through further direct computation, we find the
	limits
	\begin{align*}
		\lim_{h\to0}
			\Res_{\xi=0}
				\left\{
				\left(
					e^{ix\xi}
					\frac{\partial}{\partial \lambda}
						\frac{1}{p_\lambda(\xi)}
				\right)
				-
				\Res_{\xi=0}
				\left[
					e^{ix\xi}
					\left(
						\frac{1}{p_\lambda(\xi)}
					\right)_h
				\right]
				\right\}
			=0,
	\end{align*}
	and
	\begin{align*}
		\lim_{h\to0}
			\Res_{\xi=\lambda}
				\left\{
				\left(
					e^{ix\xi}
					\frac{\partial}{\partial \lambda}
						\frac{1}{p_\lambda(\xi)}
				\right)
				-
				\sum_{k \in \{\lambda, \lambda+h\}}
				\Res_{\xi=k}
				\left[
					e^{ix\xi}
					\left(
						\frac{1}{p_\lambda(\xi)}
					\right)_h
				\right]
				\right\}
			=0
	\end{align*}
	hold pointwise for each $x \in \mathbb R$. Thus, since
	$\left(\frac{1}{p_\lambda(\xi)}\right)_h \in L_\xi^1(\mathbb R \pm i\pi)$
	implies by Fourier theory that $\int_{\mathbb R\pm i\pi} e^{ix\xi}
	\left(\frac{1}{p_\lambda(\xi)}\right)_h \, \mathrm{d}\xi$ is continuous and bounded
	(in $x \in \mathbb R$), we may complete this proof by doing a contour shift
	and then applying Dominated Convergence to
	$\int_{\mathbb R} \inn{x}^{-s} \, G_h (x) \, \mathrm{d}x$.
	% \begin{align}
	% 	\Res_{\xi=\lambda}
	% 			\left[
	% 				e^{ix\xi}
	% 				\left(
	% 					\frac{1}{p_\lambda(\xi)}
	% 				\right)_h
	% 			\right]
	% 		= - \frac{1}{h}\frac{e^{2\lambda}-1}{e^{2\lambda}-2\lambda-1}
	% 			\, e^{ix\lambda}
	% \end{align}
	% \begin{align}
	% 	\Res_{\xi=\lambda+h}
	% 			\left[
	% 				e^{ix\xi}
	% 				\left(
	% 					\frac{1}{p_\lambda(\xi)}
	% 				\right)_h
	% 			\right]
	% 		=
	% 			 \frac{1}{h}
	% 			 \frac{e^{2(\lambda+h)}-1}{e^{2(\lambda+h)}-2(\lambda+h)-1}
	% 			\, e^{ix(\lambda+h)}
	% \end{align}
	% \begin{align}
	% 	&\lim_{\lambda\to0}
	% 		\left\{
	% 			\sum_{k \in \{0, \,\lambda,\, \lambda+h\}}
	% 			\Res_{\xi=k}
	% 			\left[
	% 				e^{ix\xi}
	% 				\left(
	% 					\frac{1}{p_\lambda(\xi)}
	% 				\right)_h
	% 			\right]
	% 			% +\Res_{\xi=\lambda}
	% 			% \left[
	% 			% 	e^{ix\xi}
	% 			% 	\left(
	% 			% 		\frac{1}{p_\lambda(\xi)}
	% 			% 	\right)_h
	% 			% \right]
	% 			% +\Res_{\xi=\lambda+h}
	% 			% \left[
	% 			% 	e^{ix\xi}
	% 			% 	\left(
	% 			% 		\frac{1}{p_\lambda(\xi)}
	% 			% 	\right)_h
	% 			% \right]
	% 		\right\} \\
	% 	&\qquad =
	% 		\frac{1}{h}
	% 		\left[
	% 			\frac{e^{2h}-1}{e^{2h}-2h-1}\,e^{ihx}
	% 			-
	% 			\frac{2\big(e^{2h}(h+1)-1\big)}
	% 				{e^{2h}(6h-3)+3}
	% 			- \frac{1}{3}
	% 			- ix
	% 		\right]
	% 		\nonumber
	% \end{align}
	% \begin{align}
	% 	&\lim_{h\to0}
	% 		\lim_{\lambda\to0}
	% 			\left\{
	% 				\sum_{k \in \{0, \,\lambda,\, \lambda+h\}}
	% 				\Res_{\xi=k}
	% 				\left[
	% 					e^{ix\xi}
	% 					\left(
	% 						\frac{1}{p_\lambda(\xi)}
	% 					\right)_h
	% 				\right]
	% 			\right\} \\
	% 	&\qquad=
	% 		- \frac{1}{2} x^2 + i \frac{1}{3} x \nonumber
	% \end{align}
\end{proof}

\end{document}


\newcommand{\definedas}{\stackrel{\triangle}{=}}
\newcommand{\spc}{\;\;\;\;\;}
\newcommand{\mspc}{\;\;\;\;\;\;\;\;\;\;}
\newcommand{\lspc}{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

\documentstyle[fleqn,11pt]{article}

\renewcommand{\baselinestretch}{1.1}
\newcommand{\eg}[1]{\begin{quote}{\tt #1} \end{quote}}
\setlength{\textwidth}{15cm}
\addtolength{\oddsidemargin}{-1cm}

\title{                    CHANGEVR,  \\
                         A REDUCE Facility   \\
                              to   \\
                    Perform Change of Independent Variable(s)\\
                               in \\
                    Differential Equations\\[2cm]
        }
\author{
          G. \"{U}\c{c}oluk
	     \thanks{Email address: UCOLUK@TRMETU.BITNET}
                     \\
                               Department of Physics \\
                               Middle East Technical University \\
                               Ankara, Turkey
       }
\date{October 1989}

\begin{document}
\maketitle
\newpage

\section{Introduction}
The mathematics behind the change of independent variable(s) in differential
equations is quite straightforward. It is basically the application of the
chain rule. If the dependent variable of the differential equation is $F$,
the independent variables are $x_{i}$ and the new independent variables are
$u_{i}$ (where ${\scriptstyle i=1\ldots n}$) then the first derivatives are:
\[
    \frac{\partial F}{\partial x_{i}} = \frac{\partial F}{\partial u_{j}}
                                        \frac{\partial u_{j}}{\partial x_{i}}
\]
We assumed Einstein's summation convention. Here the problem is to
calculate the $\partial u_{j}/\partial x_{i}$ terms if the change of variables
is given by
\[
    x_{i} = f_{i}(u_{1},\ldots,u_{n})
\]
The first thought might be solving the above given equations for $u_{j}$ and
then differentiating them with respect to $x_{i}$, then again making use of the
equations above, substituting new variables for the old ones in the calculated
derivatives. This is not always a  preferable way to proceed. Mainly because
the functions $f_{i}$ may not always be easily invertible. Another approach
that makes use of the Jacobian is better. Consider the above given equations
which relate the old variables to the new ones. Let us differentiate them:
\begin{eqnarray*}
  \frac{\partial x_{j}}{\partial x_{i}} & = &
        \frac{\partial f_{j}}{\partial x_{i}}   \\
  \delta_{ij} & = &
        \frac{\partial f_{j}}{\partial u_{k}}
        \frac{\partial u_{k}}{\partial x_{i}}
\end{eqnarray*}
The first derivative is nothing but the $(j,k)$ th entry of the Jacobian matrix.

So if we speak in matrix language
\[ {\bf 1 = J \cdot D} \]
where we defined the Jacobian
\[ {\bf J}_{ij} \definedas  \frac{\partial f_{i}}{\partial u_{j}} \]
and the matrix of the derivatives we wanted to obtain as
\[ {\bf D}_{ij} \definedas  \frac{\partial u_{i}}{\partial x_{j}}. \]
If the Jacobian has a non-vanishing determinant then it is invertible and
we are able to write from the matrix equation above:
\[ {\bf  D = J^{-1}} \]
so finally we have what we want
\[
   \frac{\partial u_{i}}{\partial x_{j}} = \left[{\bf J^{-1}}\right]_{ij}
\]

The higher derivatives are obtained by the successive application of the chain
rule and using the definitions of the old variables in terms of the new ones. It

can be easily verified that the only derivatives that are needed to be
calculated are the first order ones which are obtained above.

\section{How to Use CHANGEVR}
{\bf This facility requires the matrix package to be present in the session}.
So if it is not autoloaded in your REDUCE implementation, say
\eg{LOAD\_PACKAGE MATRIX;}
in the REDUCE environment. Then load {\tt CHANGEVR} by the statement:
\eg{LOAD\_PACKAGE CHANGEVR\$}
Now the REDUCE function {\tt CHANGEVAR} is ready to use. {\bf Note:  The
package is named CHANGEVR, but the function has the name CHANGEVAR}.  The
function {\tt CHANGEVAR} has (at least) four different arguments.  Here we
give a list them:
\begin{itemize}
\item {\bf FIRST ARGUMENT} \\
     Is a list of the dependent variables of the differential equation.
     They shall be enclosed in a pair of curly braces and separated by commas.
     If there is only one dependent variable there is no need for the curly
     braces.
\item {\bf SECOND ARGUMENT}  \\
     Is a list of the {\bf new} independent variables. Similar to what is said
     for the first argument, these shall also be separated by commas,
     enclosed in curly braces and the curly braces can be omitted if there is
     only one new variable.
\item {\bf THIRD ARGUMENT}  \\
     Is a list of equations separated by commas, where each of the equation
     is of the form
      \eg{{\em old variable} = {\em a function in new variables}}
     The left hand side cannot be a non-kernel structure. In this argument
     the functions which give the old variables in terms of the new ones are
     introduced. It is possible to omit totally the curly braces which enclose
     the list. {\bf Please note that only for this argument it is allowed to
     omit the curly braces even if the list has \underline{more than one}
     items}.
\item {\bf LAST ARGUMENT}  \\
     Is a list of algebraic expressions which evaluates to  differential
     equations, separated by commas, enclosed in curly braces.
     So, variables in which differential equations are already stored may be
     used freely. Again it is possible to omit the curly braces if there is
     only {\bf one} differential equation.
\end{itemize}

If the last argument is a list then the result of {\tt CHANGEVAR} is also a
list.

It is possible to display the entries of the inverse Jacobian, explained
in the introduction.  To do so, turn {\tt ON} the flag {DISPJACOBIAN} by a
statement: \eg{ON DISPJACOBIAN;}

\section{AN EXAMPLE\ldots\ldots The 2-dim. Laplace Equation}
The 2-dimensional Laplace equation in cartesian coordinates is:
\[
   \frac{\partial^{2} u}{\partial x^{2}} +
   \frac{\partial^{2} u}{\partial y^{2}} = 0
\]
Now assume we want to obtain the polar coordinate form of Laplace equation.
The change of variables is:
\[
   x = r \cos \theta, \mspc  y = r \sin \theta
\]
The solution using {\tt CHANGEVAR}  (of course after it is properly loaded)
is as follows
\eg{CHANGEVAR(\{u\},\{r,theta\},\{x=r*cos theta,y=r*sin theta\}, \\
    \hspace*{2cm}     \{df(u(x,y),x,2)+df(u(x,y),y,2)\} )}
Here we could omit the curly braces in the first and last arguments (because
those lists have only one member) and the curly braces in the third argument
(because they are optional), but you cannot leave off the curly braces in the
second argument. So one could equivalently write
\eg{CHANGEVAR(u,\{r,theta\},x=r*cos theta,y=r*sin theta,        \\
    \hspace*{2cm}     df(u(x,y),x,2)+df(u(x,y),y,2) )}

If you have tried out the above example, you will notice that the denominator
contains a $\cos^{2} \theta + \sin^{2} \theta$ which is actually equal to $1$.
This has of course nothing to do with the {\tt CHANGEVAR} facility introduced
here.  One has to be overcome these pattern matching problems by the
conventional methods REDUCE provides (a {\tt LET} statement, for example,
will fix it).

Secondly you will notice that your {\tt u(x,y)} operator has changed to
{\tt u(r,theta)} in the result. Nothing magical  about this. That is just what
we do with pencil and paper. {\tt u(r,theta)} represents the  the transformed
dependent variable.

\section{ANOTHER EXAMPLE\ldots\ldots An Euler Equation}
Consider a differential equation which is of Euler type, for instance:
\[
   x^{3}y''' - 3 x^{2}y'' + 6 x y' - 6 y = 0
\]
Where prime denotes differentiation with respect to $x$. As is well known,
Euler type of equations are solved by a change of variable:
\[
   x = e^{u}
\]
So our  {\tt CHANGEVAR} call reads as follows:
\eg{CHANGEVAR(y, u, x=e**u, x**3*df(y(x),x,3)-   \\
    \hspace*{2cm}   3*x**2*df(y(x),x,2)+6*x*df(y(x),x)-6*y(x))}

\end{document}


\input{docs/preamble}
\title{Photon Statistics}
\author{Max Bigras and David Frawley}

\begin{document}

\maketitle

\begin{abstract}
We explored the statistics for photons arriving from a constant intensity and a randomly scattered light source. We measured the number of photons arriving per a unit time. By constructing probability distributions and calculating $\chi^{2}$ values we found the distribution for constant intensity light to agree with a Poisson distribution and the distribution for randomly scattered light to agree with a Bose-Einstein distribution.
\end{abstract}

\section{Introduction}
We used a photomultiplier tube (PMT) and a photon counter to directly count the number of photons arriving per a unit time from two different light sources. By overlaying our experimental distribution over a theoretical distribution based on statistics we illustrated the probabilistic nature of the photodetection process itself \cite{koc}.

\section{Theory}
The probability for ejection of an electron from the PMT photocathode depends only on intensity light \cite{manual}. We are interested in the probability for obtaining $n$ counts in some large time interval $T$, for a constant intensity light source and a randomly scattered light source.

For a constant intensity light source and from statistics the probability of detecting $n$ photons in time $T$ is

\begin{equation}
P(n,T) = \frac{n_{\mathrm{av}}^{n}}{n!}e^{-n_{\mathrm{av}}}
\label{poisson}
\end{equation}
where $n_{\mathrm{av}}$ is the average number of photons counted in time $T$.

But this is the Poisson distribution. So we expect a constant intensity light source to cause electrons to be ejected from the cathod of the PMT with a Poisson distribution.

What if the intensity of light varies in space randomly? This is the case for randomly scattered light. If one can determine the intensity of the light at any spatial location,  and perform a weighted average then the probability for detecting $n$ photons in time $T$ is

\begin{equation}
P(n,T) = \frac{n_{\mathrm{av}}^{n}}{(n_{\mathrm{av}}+1)^{n+1}}
\label{bose}
\end{equation}

But this is the Bose-Einstein distribution. So we expect a randomly scattered light source to cause electrons to be ejected from the cathod of the PMT with a Bose-Einstein distribution.

\section{Apparatus}
Our apparatus is shown in Figure \ref{apparatus}. We used photons produced by an HeNe laser, which emits around $10^{18}$ photons per a second. We are interested in counting a few thousand photons per a second, so we are required to reduce the intensity of the laser. We attenuate the laser in three ways, first we polarize the laser beam, then we spread it out with a short focal length lens and finally we pass the light through two pinholes ($<$100 $\mu$m diameter). Any surviving photons are detected by the the PMT and finally counted by the photon counter. The photon counter detects pulses above a discriminator level and counts the number of pulses per a unit time.
\begin{figure}[H]
  \includegraphics[totalheight=0.5\textwidth]{figs/apparatus}
  \caption{Diagram for apparatus}
  \label{apparatus}
\end{figure}

It's important to count pulses coming from photons and not from noise in the PMT. We measured how the count rate varies with increasing discriminator level. Initially there is a fast change as the noise gets filtered out, shown in Figure \ref{dlevel}, then the curve flattens out and again dips down as we began to also filter out the pulses from the photons. Because we're interested in detecting photons away from the noise we chose our discriminator level to be in the flat region.
\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/dlevel}
  \caption{Finding the optimum discriminator level}
  \label{dlevel}
\end{figure}

\section{Analysis}
Figures \ref{ci1000}, \ref{ci3000} and \ref{ci10000} show our probability distributions for constant intensity light. As shown by the $\chi^{2}$ values and visually our experimental distributions agree with the theoretical Poisson distribution.
\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/ci1000}
  \caption{Theoretical and experimental Poisson distribution for constant intensity light with an average photon count = 1.09 photons/ms  and $\chi^{2}$ = 0.254}
  \label{ci1000}
\end{figure}

\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/ci3000}
  \caption{Theoretical and experimental Poisson distribution for constant intensity light with an average photon count = 3.004 photons/ms and $\chi^{2}$ = 0.933}
  \label{ci3000}
\end{figure}

\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/ci10000}
  \caption{Theoretical and experimental Poisson distribution for constant intensity light with average photon count = 9.987 photons/ms and $\chi^{2}$ = 3.487 and $\chi^{2}$ = 0.843}
  \label{ci10000}
\end{figure}

Figures \ref{pt1000}, \ref{pt3000} and \ref{pt10k} show our probability distributions for randomly scattered light. As shown by the $\chi^{2}$ values and visually our experimental distributions agree with the theoretical Bose-Einstein distribution. One observation that can be made is the probability where 0 events were observed is always lower than predicated. Further, for small counts, just away from 0, the probability is always higher than predicated. We suspect that our resolution for detecting dark may not be fine enough, in which case dark points would be counted as light. Also the HeNe laser that we used is known to emit some blue light in addition to the constant intensity light that we expect, this would also cause dark points to be counted as light. In either or both cases we'd expect to see a sagging at the origin and an uplifting just away from the origin, the exact behavior that our data is exhibiting.

\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/pt1000}
  \caption{Theoretical and experimental Bose-Einstein distribution for randomly scattered light with average photon count = 1.045 and $\chi^{2}$ = 7.864}
  \label{pt1000}
\end{figure}

\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/pt3000}
  \caption{Theoretical and experimental Bose-Einstein distribution for randomly scattered light with average photon count = 3.743 and $\chi^{2}$ = 8.078}
  \label{pt3000}
\end{figure}

\begin{figure}[H]
  \includegraphics[totalheight=0.6\textwidth]{figs/pt10k}
  \caption{Theoretical and experimental Bose-Einstein distribution for randomly scattered light with average photon count = 10.443 and $\chi^{2}$ = 3.487}
  \label{pt10k}
\end{figure}

\section{Conclusion}
We explored the statistics for photons arriving from a constant intensity and a randomly scattered light source. By constructing probability distributions and calculating $\chi^{2}$ values we found the distribution for constant intensity light to agree with a Poisson distribution and the distribution for randomly scattered light to agree with a Bose-Einstein distribution. We also provided explanation for the slight sagging and uplifting behavior for small counts.

\begin{thebibliography}{99}

\bibitem{manual} Physics Dept., ``Photon Counting and the Statistics of Light'', Quantum Lab, California Polytechnic University, (2013).

\bibitem{koc} P. Koczyk, P. Wiewior, and C. Radzewicz, ``Photon counting statistics-Undergraduate experiment'', Institute of Experimental Physics, Warsaw University, Poland, 1995.

\end{thebibliography}

\end{document}


\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{indentfirst}
\title{CS131 B1: HW 5}
\author{Duy Nguyen}
\date{4 November 2016}

\begin{document}

\maketitle

\subsection*{1. Prove that $1^2 - 2^2 + 3^2 - ... + (-1)^{n-1} n^2 = (-1)^{n-1} n(n+1)/2$, whenever $n$ is a positive integer.}

\textbf{Base case:}
$$1^2 = (-1)^0 1(1+1)/2$$
$$1^2 = \frac{2}{2} = 1$$

\textbf{Induction hypothesis:} Assume that
$$1^2 - 2^2 + 3^2 - ... + (-1)^{k-1} k^2 = (-1)^{k-1} \frac{k(k+1)}{2}$$

\textbf{Induction goal:}
$$1^2 - 2^2 + 3^2 - ... + (-1)^{k-1} k^2 + (-1)^{k} (k+1)^2 = (-1)^{k} \frac{(k+1)(k+2)}{2}$$

\textbf{Proof:}
$$1^2 - 2^2 + 3^2 - ... + (-1)^{k-1} k^2 + (-1)^{k} (k+1)^2 = (-1)^{k-1} \frac{k(k+1)}{2} + (-1)^{k} (k+1)^2$$
We look at the right side of the equation:
$$ = (-1)^{k-1} \frac{k(k+1)}{2} + (-1)^{k} (k+1)^2$$
$$ = (-1)^{k} (k+1) ((-1)^{-1}\frac{k}{2} + (k+1))$$
$$ = (-1)^{k} (k+1) (\frac{-k}{2} + \frac{2k}{2} + \frac{2}{2})$$
$$ = (-1)^{k} (k+1) \frac{-k + 2k +2}{2}$$
$$ = (-1)^{k} \frac{(k+1) (k+2)}{2}$$

We have proved the hypothesis through induction.

\subsection*{2. Prove that $n!<n^n$, where $n$ is an integer greater than 1.}

\textbf{Base case:}
Pick the next integer greater than 1. With $n=2$, we have $2<2^2 = 4$ so $2<4$.

\textbf{Induction hypothesis:} Assume that
$$k! < k^k$$

\textbf{Induction goal:}
$$(k+1)! < (k+1)^{k+1}$$

\textbf{Proof:}
$$(k+1)!< (k+1)^{k+1}$$
$$(k+1)*k!<(k+1)*(k+1)^k$$
Cancel $k+1$ from each side of the inequality, we have
$$k! < (k+1)^k$$
Compare this to what we have assumed ($k! < k^k$) we note that this inequality is true, since $k$ is always positive and thus $k<k+1$ so $k^k<(k+1)^k$ thus $k! < (k+1)^k$ is true.

\subsection*{3. Prove that 3 divides $n^3 +2n$ whenever n is a positive integer.}

\textbf{Base case:}
$$1^3 + 2\cdot 3 = \frac{6}{3} = 2$$

\textbf{Inductive hypothesis:} Assume that
$k^3 + 2n$ is divisible by $3$.

\textbf{Inductive goal:} $(k+1)^3 + 2(k+1)$ is divisible by $3$.

\textbf{Proof:}
We have
$$(k+1)^3 + 2(k+1)$$
$$k^3 + 3k^2 + 3k + 1 + 2k + 2$$
$$(k^3 + 2k) + 3(k^2+k +1)$$

Notice that $k^3 + 2k$ is divisible by $3$ as we have assumed and $3(k^2+k +1)$ is also divisible by $3$. Thus the hypothesis is proven through induction.

\subsection*{4. Show that $n$ cents payment for $n \geq 8$ can be formed using just 3-cents and 5-cents stamps.}

\textbf{Base case:} We notes the result of $P(n)$ for $n = 8, 9, 10$

$$8= 5+3$$
$$9 = 3 + 3+ 3$$
$$10=5+5$$

\textbf{Inductive hypothesis:} We note our result is true $\forall k: 8\leq k\leq n$ where $n = 10$.

\textbf{Inductive goal:} $\forall n >  10$ our result is true.

\textbf{Proof:}
We note that for $n>10$, $P(n)$ is just either $P(8), P(9),$ or $P(10)$ adding with the appropriate amount of 3-stamps. In another word, for $P(n)$ for $n>10$, we can write $n$ as $x + 3^k$, where $x = 8,9, $or$ 10$ and $k$ is an positive integer $>0$. In this way, since $P(8), P(9),$ and $P(10)$ are true, we can reduce any $P(n)$ back to any of these three cases.

Thus the hypothesis is true through induction.

\end{document}


\chapter{Conclusion and Future Work}
\label{chap:conclusion-future-work}

In this dissertation,
we contributed to the research needs highlighted in~\cref{chap:introduction}.

The applicability of SSAT to the analysis of VLSI systems was examined.
We formulated a framework for the property evaluation of probabilistic design.
The average-case and worst-case analyses are encoded
as random-exist and exist-random quantified SSAT formulas, respectively.

Motivated by the emerging VLSI applications,
we further devised novel algorithms for random-exist and exist-random quantified SSAT formulas.
The proposed algorithms leverage the success from SAT/QBF-solving and model-counting communities
and advance the state-of-the-art of SSAT solving beyond the conventional DPLL-based search.
For random-exist quantified SSAT formulas,
we used minterm generalization and weighted model counting as subroutines
and employed SAT solvers and weighted model counters as plug-in engines.
For exist-random quantified SSAT formulas,
we proposed clause-containment learning,
which was inspired by clause selection from QBF solving.
Under the framework of clause-containment learning,
we explored three heuristics to strengthen learnt clauses.
Moreover, unlike previous exact approaches,
the proposed algorithms can solve approximate SSAT by deriving upper and lower bounds of satisfying probabilities.
Our evaluation showed the benefits of the proposed solvers over a wide range of formula instances.
Furthermore, our implementations and the benchmark suite of SSAT instances are open-source
for other researchers to base their work on top of our results.

To generalize SSAT beyond the PSPACE-complete complexity class for more complex problems,
we extended DQBF to its stochastic variant DSSAT and proved its NEXPTIME-completeness.
Compared to the PSPACE-complete SSAT,
DSSAT is more powerful to succinctly model NEXPTIME-complete decision problems with uncertainty.
We demonstrated the DSSAT formulation of the analysis to probabilistic/approximate partial design
and gave a polynomial-time reduction from the NEXPTIME-complete Dec-POMDP to DSSAT.

We highlight several directions for future investigation.
First, in order to improve the scalability of probabilistic property evaluation,
we are interested in approximate approaches based on simulation techniques.
In addition to the conventional Monte Carlo method,
circuit simulation based on symbolic sampling~\cite{KravetsDAC19ECOSampling} may have much potential.
Another line of on-going work is to develop solvers for arbitrarily quantified SSAT and DSSAT formulas.
Recently, clause selection has been adapted to solve random-exist quantified SSAT formulas
and combined with the clause-containment learning in a recursive manner to solve general SSAT~\cite{Chen2021}.
It is also extended to DQBF~\cite{Tentrup2019},
which might provide a promising framework for DSSAT solving.
From the view of practical implementation,
SSAT solvers will benefit from a tight integration with model-counting components.
More advanced data structures, e.g., d-DNNF~\cite{Darwiche2001,Darwiche2002dDNNF}, could be also integrated.
In particular, \textit{incremental} model counting might be a key step to boost the performance of SSAT solvers
if the computational efforts among different counting queries can be effectively shared.
Motivated by approximate model counting,
we also hope to pursue a similar formulation for SSAT solving.
Specifically, we envisage a unified SSAT framework that allows users to control the solution precision,
in order to trade inexactness for better scalability.
Finally, we would like to bring SSAT to different research fields, especially to machine-learning applications.
The SSAT solvers developed in this dissertation have been applied to verify the fairness of supervised-learning algorithms~\cite{Ghosh2021}.
According to the reported data~\cite{Ghosh2021},
using the proposed SSAT solvers achieves several orders of magnitude improvement over the state-of-the-art tools.
This success shows the great potential and benefit of SSAT solving.

\iffalse
    We have proposed a formal framework to probabilistic property
    evaluation, under the worst-case and average-case scenarios.
    Connections between probabilistic property evaluation and existing
    solving techniques have been established. A novel BDD-based SSAT solver is proposed. A comparative experimental study has been performed to assess the capabilities
    of different methods. Among the considered solutions, the proposed BDD-based SSAT solver, which makes use of circuit structures to construct BDD, currently tends to be the most robust in our experiments. Nevertheless, there are cases solvable only by approximate weighted model counting, but not by other methods. As the BDD-based method has its memory explosion problem, SSAT and model
    counting approaches based on CNF formula might be more viable than the BDD one if their efficiency would be improved in the future. Our results may benefit the synthesis of probabilistic design, perhaps not only for silicon but also for genetic circuits, which are intrinsically stochastic. For future investigation, Monte-Carlo simulation may be incorporated to our proposed formal methods.

    In this paper, we focused on solving random-exist quantified SSAT formulas.
    In contrast to the previous DPLL-based algorithms, we proposed a novel algorithm using SAT solver and weighted model counter as underlying engines to improve computational efficiency.
    Leveraging the great success of modern SAT solving techniques, the proposed algorithm outperforms the state-of-the-art method in the experiment on random $k$-CNF and strategic companies formulas.
    Moreover, unlike previous exact SSAT methods, the proposed algorithm can be easily modified to solve approximate SSAT by deriving upper and lower bounds of satisfying probability.
    We demonstrated the applicability of our SSAT solver to VLSI circuit analysis.
    While the state-of-the-art solver fails to compute the exact satisfying probability, the proposed method succeeded in finding bounds of the formulas.
    In several cases, the derived bounds are very close to, or even match the exact satisfying probability.
    This approximation flexibility of our method can be helpful when SSAT is applied to real-world applications.
    %This work might shed light on the application of solving approximate SSAT to real-world problems.
    For future work, we intend to extend the proposed algorithm to arbitrary quantified SSAT formulas.

    We developed a new approach to solving E-MAJSAT formulas. In contrast to prior methods based on DPLL search or knowledge compilation, we proposed the clause containment learning technique, inspired by clause selection recently developed in QBF evaluation, and design a novel algorithm to solve E-MAJSAT efficiently. Under the framework of clause containment learning, three enhancement techniques were proposed to improve the computational efficiency.
    Experiment results show the benefit of our method.
    %that our method achieves significant performance gains and memory savings over prior SSAT methods, and also provides useful lower bound information for cases where no information can be given by prior methods.
    For future work, we intend to solve SSAT with general prefix structure.

    In this paper, we extended DQBF to its stochastic variant DSSAT and proved its NEXPTIME-completeness.
    Compared to the PSPACE-complete SSAT, DSSAT is more powerful to succinctly model NEXPTIME-complete decision problems with uncertainty.
    The new formalism can be useful in applications such as artificial intelligence and system design.
    Specifically, we demonstrated the DSSAT formulation of the analysis to probabilistic/approximate partial design, and gave a polynomial-time reduction from the NEXPTIME-complete Dec-POMDP to DSSAT.
    We envisage the potential broad applications of DSSAT and plan solver development for future work.
    %\textcolor{blue}{
    We note that recent developments of \textit{clausal abstraction} for QBF~\cite{JanotaM15,RabeT15} and DQBF~\cite{Tentrup19} might provide a promising framework for DSSAT solving.
    Clausal abstraction has been lifted to SSAT~\cite{ChenHJ21}, and we are investigating its feasibility for DSSAT.
    %}
\fi


\chapter{Assignment: Overfitting}
\label{hw:overfitting}

\newthought{Overfitting is something we try to avoid at all times.} But overfitting comes in many shapes and sizes. For this exercise we will use a \emph{blood-loneliness} data set with the File widget. This data set relates gene expressions in blood with a measure of loneliness obtained from a sample of elderly persons. Let's try to model loneliness with logistic regression and see how well the model performs.

\marginnote{To load the blood loneliness data set copy and paste the below URL to the URL field of the File widget.
\break\break
\url{http://file.biolab.si/datasets/blood-loneliness-GDS3898.tab}}

\begin{figure}[h]
  \centering
  \includegraphics[scale=0.7]{overfitting1.png}%
  \caption{$\;$}
  \label{fig:wf1}
\end{figure}

\begin{enumerate}
    \item Train the Logistic Regression model on the data and observe its performance. What is the result?
    \item We have many features in our data. What if we select only the most relevant ones, the genes that actually matter? Use Rank to select the top 20 best performing features.

    \begin{figure}[h]
        \centering
        \includegraphics[scale=0.7]{overfitting2.png}%
        \caption{$\;$}
        \label{fig:wf2}
      \end{figure}

    \item How are the results now? What happened? Is there a better way of performing feature selection?
\end{enumerate}