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\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\title[THE VOLUME OF $n$-BALLS]{THE VOLUME OF $n$-BALLS}
\author{Cerbi Ritchey}
\institute{}
\date{May 3rd, 2017}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
% Uncomment these lines for an automatically generated outline.
%\begin{frame}{Outline}
% \tableofcontents
%\end{frame}
\section{Introduction}
\begin{frame}{Introduction}
\begin{itemize}
\item Our goal is to derive a formula for the volume of n-dimensional balls in $\mathbb{R}^n$.
\item Let's begin with some familiar definitions, and we will rely on our intuition to start.
\end{itemize}
\begin{block}{Definition}
For a natural number $n\geq1$, an ($n-1$)-dimensional sphere of radius $r$ is the set of all points in $\mathbb{R}^n$ which are a fixed distance $r$ from a given center point.
\end{block}
\begin{itemize}
\item We take the center to be the origin and denote the ($n-1$)-sphere of radius $r$ in $\mathbb{R}^n$ by $\mathbb{S}^{n-1}(r)$. That is,
\begin{block}
\begin{center}
\item $\mathbb{S}^{n-1}(r)={\{(x_1,x_2,...,x_n)\in \mathbb{R}^n \hspace{.1cm} | \hspace{.1cm} x_1^2+x_2^2+...+x_n^2=r^2 \}}$
\end{center}
\end{block}
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item First, notice when $n=1$, the 0-sphere is just the two points on the real line at $r$ and $-r$.
\item When $n=2$, we have $\mathbb{S}^1(r)=\{(x_1,x_2)\in\mathbb{R}^2\ \hspace{.1cm} | \hspace{.1cm} x_1^2+x_2^2=r^2\} $
\item Taking $n=3$, $\mathbb{S}^2(r)$ is the sphere in $\mathbb{R}^3$ given by $\mathbb{S}^2(r)=\{(x_1,x_2,x_3)\in\mathbb{R}^3 \hspace{.1cm} | \hspace{.1cm} x_1^2+x_2^2+x_3^2=r^2\}$
\item Now, $\mathbb{S}^n(r)$ is harder to visualize in higher dimensions, but we can use our intuition of lower dimensional spheres to help us.
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item Let's try and visualize the 3-sphere. If we take a 0-sphere, which is just the endpoints of a line segment in $\mathbb{R}^1$, and rotate it about the origin, what do we have?
\end{itemize}
\end{frame}
\begin{frame}
\begin{block}
{Correct. We have a 1-sphere in $\mathbb{R}^2$, also known as a circle.}
\end{block}
\end{frame}
\begin{frame}
\begin{itemize}
\item Now, if we take this circle and rotate every point about any axis going through the center point and lying in the $\mathbb{R}^2$ plane, we will have the 2-sphere in $\mathbb{R}^3$.
\item We think of the 3-sphere in the same way. If we take the 2-sphere in $\mathbb{R}^3$, and rotate every point about any axis going through the center point, we will have the 3-sphere in $\mathbb{R}^4$.
\item This is difficult to visualize, but this inductive process we are doing in our minds is actually exactly what we will do mathematically.
\item So.....let the fun begin :)
\end{itemize}
\end{frame}
\begin{frame}
\begin{block}{Recall:}
Orthogonal matrices represent linear transformations that preserve the dot product of vectors. They represent isometries of Euclidean space (distance preserving) and denote rotations or reflections.
\end{block}
\begin{itemize}
\item By definition, orthogonal matrices have determinant $\pm1$. The matrices in the group of orthogonal matrices in $\mathbb{R}^n$ with determinant $+1$ represent the rotations. These are called special orthogonal matrices and are given by
\begin{block}
{SO($n)=\{A:A^TA=I;detA=1\}$}
\end{block}
\end{itemize}
\end{frame}
\begin{frame}
Consider the following rotation given as a square matrix in SO($n+1$).
$$
A_j=
\begin{bmatrix}
I_{j-1}&0&0\\
0&R&0\\
0&0&I_{n-j}
\end{bmatrix}
$$
for $1\le{j}\le{n}$, where
$$
R=
\begin{bmatrix}
cos\theta&-sin\theta\\
sin\theta&cos\theta
\end{bmatrix}
$$
is a $2\times2$(counter-clockwise) rotation matrix, $I_k$ is the $k\times{k}$ identity matrix, and $j$ specifies where the rotation matrix is placed.
\end{frame}
\begin{frame}
For example, $A_1$ is the $(n+1)\times{(n+1)}$ matrix
$$
A_1=
\begin{bmatrix}
R&0\\
0&I_{n-1}\\
\end{bmatrix}
=
\begin{bmatrix}
cos\theta&-sin\theta&0\\
sin\theta&cos\theta&0\\
0&0&1\\
\end{bmatrix}
$$
\begin{itemize}
\item Notice that in this case and in general, the determinant will always be 1.
\item Also, since $A_j^T=A_j^{-1}$, $A_j$ is in the special orthogonal group.
\end{itemize}
\end{frame}
\begin{frame}
Okay cool. So how does this help us?
\begin{itemize}
\item To help us see that these matrices generate spheres in $\mathbb{R}^{n+1}$, let's look at the case with $n=3$ to find a parametrization of a 3-sphere in $\mathbb{R}^4$. We start with the point $P=(1,0,0,0)$ in $\mathbb{R}^4$ and inductively apply our rotations. Applying the rotation $A_1$ to $P$ for $0\le{\theta}<{2\pi}$, we have
\end{itemize}
$$
\begin{bmatrix}
cos\theta&-sin\theta&0&0\\
sin\theta&cos\theta&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{bmatrix}
\begin{bmatrix}
1\\
0\\
0\\
0\\
\end{bmatrix}
=(cos\theta,sin\theta,0,0)
$$
\begin{itemize}
\item Notice that this is a parametrization of the circle $S^1\subset\mathbb{R}^4$ lying in the $x_1x_2$-plane.
\end{itemize}
\end{frame}
\begin{frame}
Now let's apply the rotation $A_2$ to our circle in the $x_1x_2$-plane. This gives
$$
\begin{bmatrix}
1&0&0&0\\
0&cos\phi&-sin\phi&0\\
0&sin\phi&cos\phi&0\\
0&0&0&1\\
\end{bmatrix}
\begin{bmatrix}
cos\theta\\
sin\theta\\
0\\
0\\
\end{bmatrix}
=
(cos\theta,cos\phi{sin\theta},sin\phi{sin\theta},0)
$$
where $0\le{\theta}\le{\pi}$ and $0\le{\phi}<2\pi$. This gives us a 2-sphere lying in $x_1x_2x_3$-space. Notice that the parametrization resembles spherical coordinates.
\end{frame}
\begin{frame}
Continuing in this way, let the new variable $\psi$ range from 0 to $2\pi$, and letting $\phi$ and $\theta$ range from 0 to $\pi$, we have our parametrization of the 3-sphere in $\mathbb{R}^4$:
$$
\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&cos\psi&-sin\psi\\
0&0&sin\psi&cos\psi\\
\end{bmatrix}
\begin{bmatrix}
cos\theta\\
cos\phi{sin\theta}\\
sin\phi{sin\theta}\\
0\\
\end{bmatrix}
=
(cos\theta,cos\phi{sin\theta},cos\psi{sin\phi}{sin\theta},sin\psi{sin\phi}sin{\theta})
$$
\begin{itemize}
\item "Thus we can see that rotations in higher dimensions can be realized as the action of a linear transformation in which there is one free parameter. This parameter does a rotation in two dimensions and leaves all other dimensions fixed."
\end{itemize}
\end{frame}
\begin{frame}
Continuing in this way, we now have a parametrization of the unit sphere $\mathbb{S}^{n-1}\subset{\mathbb{R}^n}$ given by
$$x_1=cos\theta{_1}$$
\\
$$x_2=sin\theta{_1}cos\theta{_2} $$
\\
$$x_3=sin\theta{_1}sin\theta{_2}cos\theta{_3}$$
\\
$$\vdots$$
\\
$$x_{n-1}=sin\theta{_1}\hdots{sin\theta_{n-2}}cos\theta{_{n-1}}$$
\\
$$x_n=sin\theta{_1}\hdots{sin\theta{_n-2}sin\theta_{n-1}},$$
\\
where $0\le{\theta{_{n-1}}}<2\pi$ and $0\le{\theta_i}\le{\pi}$, for $i=1,2,\hdots,n-2$. This will be useful, I promise.
\end{frame}
\begin{frame}
The following integral formula for $\int{sin^m{\theta}}d\theta$ will help us. For any integer $m\ge2$, we have
$$
\int_0^\pi{sin^m\theta}=-\frac{sin^{m-1}\theta{cos\theta}}{m}\Big|_{\theta=0}^{\theta=\pi}+\int_0^\pi{sin^{m-2}\theta{d\theta}}
=\int_0^\pi{sin^{m-2}\theta{d\theta}}
$$
\\
Note that when $m$ is even, say $m=2k$, then
$$
\int_0^\pi{sin^{2k}\theta{d\theta}}=\frac{2k-1}{2k}\cdot{\frac{2k-3}{2k-2}}\cdots{\frac{3}{4}}\cdot{\frac{1}{2}}\cdot{\pi}
$$
Similarly, when $m$ is odd, say $m=2k+1$, then
$$
\int_0^\pi{sin^{2k+1}\theta{d\theta}}=\frac{2k}{2k+1}\cdot{\frac{2k-2}{2k-1}}
\cdots{\frac{4}{5}}\cdot{\frac{2}{3}}\cdot{2}
$$
\end{frame}
\section{Some \LaTeX{} Examples}
\subsection{Tables and Figures}
\begin{frame}{Tables and Figures}
\begin{itemize}
\item Use \texttt{tabular} for basic tables --- see Table~\ref{tab:widgets}, for example.
\item You can upload a figure (JPEG, PNG or PDF) using the files menu.
\item To include it in your document, use the \texttt{includegraphics} command (see the comment below in the source code).
\end{itemize}
% Commands to include a figure:
%\begin{figure}
%\includegraphics[width=\textwidth]{your-figure's-file-name}
%\caption{\label{fig:your-figure}Caption goes here.}
%\end{figure}
\begin{table}
\centering
\begin{tabular}{l|r}
Item & Quantity \\\hline
Widgets & 42 \\
Gadgets & 13
\end{tabular}
\caption{\label{tab:widgets}An example table.}
\end{table}
\end{frame}
\subsection{Mathematics}
\begin{frame}{Readable Mathematics}
Let $X_1, X_2, \ldots, X_n$ be a sequence of independent and identically distributed random variables with $\text{E}[X_i] = \mu$ and $\text{Var}[X_i] = \sigma^2 < \infty$, and let
$$S_n = \frac{X_1 + X_2 + \cdots + X_n}{n}
= \frac{1}{n}\sum_{i}^{n} X_i$$
denote their mean. Then as $n$ approaches infinity, the random variables $\sqrt{n}(S_n - \mu)$ converge in distribution to a normal $\mathcal{N}(0, \sigma^2)$.
\end{frame}
\end{document}
