%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%%
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      02/21/2025
%%%%
%%%% Purpose:   Gamma and Bessel Functions
%%%%

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\title{Gamma and Bessel Functions}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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\subject{Talks}
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% Since this a solution template for a generic talk, very little can
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will learn:
\begin{itemize}
\item the definition of the Gamma function,
\item the properties of the Gamma function,
\item how to solve Bessel's ordinary differential equation of order $p$,
\item notable Bessel's functions, and
\item properties of Bessel's functions.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Introducing the Gamma Function}

The \textbf{Gamma function} is denoted as $\Gamma(p)$ and is defined by the
integral,
\[
%\label{eq:spc:gammadefn}
\Gamma(p+1)
	= \int_{0}^{\infty} e^{-x} x^{p} \, dx	.
\]
\textbf{Remarks}:
\begin{itemize}
\item The improper integral converges for all $p \ge 0$.
\item For $p < 0$ the integrand becomes
unbounded as $x \to 0^{+}$.
\item The integral can be shown to converge for $p > -1$ and to diverge for
$p \le -1$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the case when $p = 0$,
\[
\Gamma(1)
	= \int_{0}^{\infty} e^{-x} \, dx
	= \lim_{M \to \infty} \int_{0}^{M} e^{-x} \, dx
	= \lim_{M \to \infty} \left( - e^{-M} + 1 \right)
	= 1	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Connection with Factorial}

\begin{lemma}
%\label{lma:spc:gammarecur}
For $p > 0$,
\[
%\label{eq:spc:gammarecur}
\Gamma(p+1) = p \, \Gamma(p)	.
\]
\end{lemma}

\vspace*{0.10in}

\uncover<2->{%
\begin{lemma}
%\label{lma:spc:gammafact}
If $p$ is a positive integer,
\[
%\label{eq:spc:gammafact}
\Gamma(p+1) = p!	.
\]
\end{lemma}

\vspace*{0.10in}

\textbf{Remark}: since the Gamma function is defined for non-integer
values of $p$, it generalizes the factorial.
}

\end{frame}
%%

\begin{frame}\frametitle{Example: $\Gamma(1/2)$}

By definition,
\[
\Gamma\left( \frac{1}{2} \right)
	= \int_{0}^{\infty} e^{-x} x^{-\frac{1}{2}} \, dx	.
\]
Making the substitution $u^{2} = x$ and $2 u \, du = dx$, the integral
can be re-written as
\[
\Gamma\left( \frac{1}{2} \right)
	= \int_{0}^{\infty} e^{-u^{2}} u^{-1} (2 u) \, du
	= 2 \int_{0}^{\infty} e^{-u^{2}} \, du
	= \int_{-\infty}^{\infty} e^{-u^{2}} \, du
	= \sqrt{\pi}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Pochhammer Symbol}

\begin{lemma}
%\label{lma:spc:gammaunit}
For $p > 0$,
\[
%\label{eq:spc:gammaunit}
\frac{\Gamma(p+n)}{\Gamma(p)}
	= p (p+1) (p+2) \cdots (p+n-1)
	= (p)_{n}	.
\]
\end{lemma}

\vspace*{0.10in}

\textbf{Remark}: the expression $(p)_{n}$ is called the
\textbf{Pochhammer symbol} or the \textbf{rising factorial}.

\vspace*{0.10in}

\uncover<2->{%
This lemma makes it possible to calculate $\Gamma(p)$ for all $p > 0$
provided $\Gamma(x)$ is known for $0 < x < 1$.
}

\end{frame}
%%

\begin{frame}\frametitle{Examples}

\[
\Gamma\left( \frac{3}{2} \right)
	= \Gamma\left( \frac{1}{2} + 1 \right)
	= \frac{1}{2} \Gamma\left( \frac{1}{2} \right)
	= \frac{\sqrt{\pi}}{2}
\]

\vspace*{0.10in}

\[
\Gamma\left( \frac{11}{2} \right)
	= \Gamma\left(5 + \frac{1}{2}\right)
	= \Gamma\left(\frac{1}{2}\right) \left(\frac{1}{2}\right) 
\left(\frac{3}{2}\right) \left(\frac{5}{2}\right)
\left(\frac{7}{2}\right) \left(\frac{9}{2}\right)
	= \frac{945 \sqrt{\pi}}{32}
\]

\end{frame}
%%

\begin{frame}\frametitle{Properties of the Gamma Function (1 of 2)}

\begin{lemma}
%\label{lma:spc:gammacont}
The Gamma function $\Gamma(x)$ is continuous for all $x > 0$.
\end{lemma}

\vspace*{0.10in}

\uncover<2->{%
The Gamma function is differentiable for $x > 0$.
\[
\Gamma'(x)
	= \frac{d}{d x} \int_{0}^{\infty} e^{-t} t^{x-1} \, dt
	= \int_{0}^{\infty} \frac{d}{d x} \left[ e^{-t} t^{x-1} \right] \, dt
	= \int_{0}^{\infty} e^{-t} (\ln t) t^{x-1} \, dt
\]

\uncover<3->{%
The Gamma function to be extended
to the real numbers in the set $\{ x \in \mathbb{R} \, | \, x \ne 0, -1, -2, \ldots
\}$.
If $n$ is a natural number and
if $-n < x < -n+1$ then $0 < x+n < 1$ and thus $\Gamma(x+n)$ is
already defined.
Recall that
\begin{align*}
\Gamma(x+n)
	&= \Gamma(x) (x)_{n}	\\
\Gamma(x)
	&= \frac{\Gamma(x+n)}{(x)_{n}}	.
\end{align*}
This extended form of the Gamma function is also continuous on the set
$\{ x \in \mathbb{R} \, | \, x \ne 0, -1, -2, \ldots \}$.
}
}

\end{frame}
%%

\begin{frame}\frametitle{Properties of the Gamma Function (2 of 2)}

\begin{theorem}
%\label{thm:spc:gammanegint}
If $n = 0, -1, -2, \ldots$ then $\lim_{x \to n} | \Gamma(x) | = \infty$.
\end{theorem}

\vspace*{0.10in}

\uncover<2->{%
\begin{itemize}
\item Define the reciprocal of the Gamma function at $x
= 0, -1, -2, \ldots$ to be $0$.
Thus $1/\Gamma(x)$ is continuous for all $x$ and $1/\Gamma(x) = 0$ at $x
= 0, -1, -2, \ldots$.
\item Define $f(x) = \ln | \Gamma(x) |$ then
\[
f'(x) = \frac{\Gamma'(x)}{\Gamma(x)}
\iff
\Gamma'(x) = \Gamma(x) f'(x)	.
\]
\item Function $f'(x)$ is called the
\textbf{digamma function} or \textbf{polygamma function} 
and it is denoted as
$\psi(x)$.
\item Adopting this notation allows the derivative of the Gamma function to
be written as $\Gamma'(x) = \psi(x) \Gamma(x)$.
\end{itemize}
}

\end{frame}
%%

\begin{frame}\frametitle{Illustration of the Gamma Function}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{gammafunction}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Bessel's Equation of Order $p \ge 0$}

\[
%\label{eq:spc:besseleqn}
x^{2} y'' + x y' + (x^{2} - p^{2}) y
	= 0
\mbox{ for $x > 0$,}
\]
The self-adjoint form is
\[
\left[ x \, y' \right]' + \left( x - \frac{p^{2}}{x} \right) y
	= 0
\mbox{ for $x > 0$.}
\]
The method of Frobenius will be used to solve this ODE.

\vspace*{0.10in}

The value $x_{0} = 0$ is a regular singular point with indicial
function $F(r) = r^{2} - p^{2}$, and exponents of
singularity $\pm p$.

\end{frame}
%%

\begin{frame}\frametitle{Solving Bessel's Equation (1 of 4)}

Assume $\displaystyle y(x) = \sum_{k=0}^{\infty}
a_{k} x^{k+r}$, differentiate formally with respect to $x$, substitute
the result into Bessel's equation
and re-index and combine the series to yield
\begin{align*}
0
	= {} & x^{2} \sum_{k=0}^{\infty} (k+r) (k+r-1) a_{k} x^{k+r-2} +
x \sum_{k=0}^{\infty} (k+r) a_{k} x^{k+r-1}	\\
	&+ (x^{2} - p^{2}) \sum_{k=0}^{\infty} a_{k} x^{k+r}	\\
	= {} & \sum_{k=0}^{\infty} [(k+r) (k+r-1) + (k+r) - p^{2}] a_{k} x^{k+r} +
\sum_{k=0}^{\infty} a_{k} x^{k+r+2}	\\
%0
	= {} & \sum_{k=0}^{\infty} [(k+r)^{2} - p^{2}] a_{k} x^{k+r} +
\sum_{k=0}^{\infty} a_{k} x^{k+r+2}	\\
	&= \sum_{k=0}^{\infty} F(r+k) a_{k} x^{k+r} +
\sum_{k=2}^{\infty} a_{k-2} x^{k+r}	\\
	&= F(r) a_{0} x^{r} + F(r+1) a_{1} x^{r+1} +
\sum_{k=2}^{\infty} [F(r+k) a_{k} + a_{k-2}] x^{k+r}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solving Bessel's Equation (2 of 4)}

\[
0
	= F(r) a_{0} x^{r} + F(r+1) a_{1} x^{r+1} +
\sum_{k=2}^{\infty} [F(r+k) a_{k} + a_{k-2}] x^{k+r}
\]
\begin{itemize}
\item Since $r = -p$ or $r = p$ then $F(r) = 0$ and for arbitrary $a_{0}$
the first term above is zero.
\item The second term has coefficient $a_{1} F(r+1) = a_{1} ((1+r)^{2} - p^{2})$
which must match the coefficient of $x^{r+1}$ on the left-hand side
of the equation.
This can be achieved by choosing $a_{1} = 0$ regardless of the values
of $p$ or $r$.
\item For $k \ge 2$ the following recurrence relation holds for any $r$ such
that $F(k+r) \ne 0$,
\[
a_{k} = -\frac{a_{k-2}}{F(k+r)}	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Solving Bessel's Equation (3 of 4)}

\[
a_{k} = -\frac{a_{k-2}}{F(k+r)}
\]
This equation is valid as long as $r$ is the larger of the two
exponents of singularity ($r = p$).
Since $a_{1} = 0$ then $a_{2k-1} = 0$ for $k \in \mathbb{N}$.
Since $F(k+p) \ne 0$ for $k \in \mathbb{N}$,
the recurrence relation can be simplified to
\[
a_{k}
	= -\frac{a_{k-2}}{k (k+2p)}	.
\]
Since $a_{0}$ is an arbitrary constant,
then
\begin{align*}
a_{2}
	&= -\frac{a_{0}}{2(2+2p)}
	= \frac{(-1)^{1} a_{0}}{2^{2} (1+p)}	\\
a_{4}
	&= -\frac{a_{2}}{4(4+2n)}
	= \frac{(-1)^{2} a_{0}}{2^{4} 2! (1+p)(2+p)}	\\
{}	&\vdotswithin{=} {}	\\
a_{2k}
	&= \frac{(-1)^{k} a_{0}}{2^{2k} k! (1+p)(2+p) \cdots (k+p)}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solving Bessel's Equation (4 of 4)}

One solution to Bessel's equation of order $p$
has the form,
\[
y_{1}(x)
	= a_{0} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2k} k!
          (1+p)(2+p) \cdots (k+p)} x^{2k+p}	.
\]
Traditionally the first solution to Bessel's equation of order $p$
has the arbitrary constant $a_{0}$ chosen as
\[
a_{0}
	= \frac{1}{2^{p} \Gamma(p+1)}	.
\]
This defines the
\textbf{Bessel function of the first kind of order $p$}:
\[
J_{p}(x)
%\label{eq:spc:besselgeneralsoln}
	= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k! \Gamma(k+p+1)} \left(
\frac{x}{2} \right)^{2k+p}	.
\]
\uncover<2->{%
When $p = n \in \{ 0, 1, \ldots \}$ then $J_{n}(x)$ is a
power series,
\[
J_{n}(x)
	= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k! (k+n)!} \left(
\frac{x}{2} \right)^{2k+n}	.
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Bessel Functions of the First Kind}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{besseljs}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Second Linearly Independent Solution (1 of 3)}

Since Bessel's equation is second order, there must be another
solution that is linearly independent of $J_{p}(x)$.

\vspace*{0.10in}

If $2 p$ is not an integer then a second solution is
\[
J_{-p}(x)
	= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k! \Gamma(k-p+1)} \left(
\frac{x}{2} \right)^{2k-p}	.
\]
By convention the second solution to Bessel's equation of order $p$,
linearly independent of $J_{p}(x)$ is chosen to be a linear
combination of $J_{p}(x)$ and $J_{-p}(x)$, namely
\[
%\label{eq:spc:defnYp}
Y_{p}(x)
	= \frac{J_{p}(x) \cos (p \pi) - J_{-p}(x)}{\sin (p \pi)}	.
\]
The function $Y_{p}(x)$ is called a
\textbf{Bessel function of the second kind of order $p$}.

\end{frame}
%%

\begin{frame}\frametitle{Second Linearly Independent Solution (2 of 3)}

If $p = n \in \mathbb{N}$ then $2 p$ is an integer.
Assume a second solution to Bessel's equation if order $n$, linearly
independent from $J_{n}(x)$ takes the form
\[
y_{2}(x)
	= a J_{n}(x) \ln x + x^{-n} \left( 1 + \sum_{k=1}^{\infty}
          c_{k} x^{k} \right)
\]
where $a$ and $c_{k}$ for $k \in \mathbb{N}$ are constants.

\vspace*{0.10in}

\uncover<2->{%
It can be shown that
\begin{align*}
y_{2}(x)
	= {} & \frac{-\ln x}{2^{n-1} (n-1)!} J_{n}(x) + x^{-n} \left[ 1 +
  \sum_{k=1}^{n-1} \frac{(n-k-1)!}{k! (n-1)!} \left(
    \frac{x}{2} \right)^{2k} \right]	\\
	&+ x^{-n} \left[ \sum_{k=0}^{\infty} \frac{(-1)^{k} (H_{k} +
    H_{n+k})}{k! (n+k)! (n-1)!} \left( \frac{x}{2} \right)^{2n+2k} \right]	,
\end{align*}
where $\displaystyle H_{k} = \sum_{j=1}^{k} \frac{1}{j}$ is the $k$th
Harmonic number.
}

\end{frame}
%%

\begin{frame}\frametitle{Second Linearly Independent Solution (3 of 3)}

By convention the second linearly independent solution to Bessel's
equation of order $n \in \mathbb{N}$ is expressed as
\begin{align*}
Y_{n}(x)
	= {} & \frac{2}{\pi} \left( (\gamma - \ln 2) J_{n}(x) - 2^{n-1} (n-1)! y_{2}(x)
              \right)	\\
	= {} & \frac{2}{\pi} \left( \gamma + \ln \frac{x}{2} \right) J_{n}(x) -
\frac{1}{\pi} \sum_{k=0}^{n-1} \frac{(n-k-1)!}{k!} \left(
    \frac{x}{2} \right)^{2k-n}	\\
	&- \frac{1}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^{k} (H_{k} +
    H_{n+k})}{k! (n+k)!} \left( \frac{x}{2} \right)^{2k+n}	.
\end{align*}
The constant $\gamma$ is called the
\textbf{Euler-M\'{a}scheroni constant}
or merely the
\textbf{Euler gamma}.
\[
\gamma \equiv \lim_{n \to \infty} \left( H_{n} - \ln n \right) \approx
0.577216	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Bessel's Equation of Order $p = 0$ (1 of 2)}

We have already found
\[
J_{0}(x)
	= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k!)^{2}} \left(
          \frac{x}{2} \right)^{2k}	.
\]
Assume a second solution linearly independent from $J_{0}(x)$ has the form
\[
y_{2}(x)
	= (\ln x) J_{0}(x) + \sum_{k=1}^{\infty} c_{k} x^{k}	,
\]
where $c_{k}$ for $k \in \mathbb{N}$ are constants.

\vspace*{0.10in}

It can be shown that
\[
y_{2}(x)
	= (\ln x) J_{0}(x) - \sum_{k=1}^{\infty} \frac{(-1)^{k}
  H_{k}}{(k!)^{2}} \left( \frac{x}{2} \right)^{2k}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Bessel's Equation of Order $p = 0$ (2 of 2)}

By convention the 
Bessel function of the second kind of order $0$ is expressed as
\[
Y_{0}(x)
%\label{eq:spc:bessel:besselY0}
	= \frac{2}{\pi} \left( \gamma + \ln \frac{x}{2} \right) J_{0}(x) -
\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}
  H_{k}}{(k!)^{2}} \left( \frac{x}{2} \right)^{2k}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Bessel Functions of the Second Kind}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{besselys}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Derivative Relationships}

\begin{theorem}
%\label{thm:spc:besselderiv}
The Bessel functions of the first kind of order $p$ satisfy the following
derivative relationships:
\begin{align*}
\frac{d}{d x} \left[ x^{-p} J_{p}(x) \right]
	&= -x^{-p}  J_{p+1}(x)	,	\\
\frac{d}{d x} \left[ x^{p} J_{p}(x) \right]
	&= x^{p}  J_{p-1}(x)	.
\end{align*}
\end{theorem}

\uncover<2->{%
\begin{corollary}
%\label{cor:spc:Besselrecurrence}
The Bessel functions of the first kind obey the following recurrence
relations.
\begin{align*}
%\label{eq:spc:Besselrecurrence1}
x J_{p}'(x) - p J_{p}(x)
	&= -x J_{p+1}(x)	\\
%\label{eq:spc:Besselrecurrence2}
x J_{p}'(x) + p J_{p}(x)
	&= x J_{p-1}(x)	\\
%\label{eq:spc:Besselrecurrence3}
J_{p+1}(x) + J_{p-1}(x)
	&= \frac{2 p}{x} J_{p}(x)
\end{align*}
\end{corollary}
}

\end{frame}
%%

\begin{frame}\frametitle{Zeros of Bessel Functions}

\begin{lemma}
%\label{lma:spc:bessel0zeros}
The positive zeros of $J_{0}(x)$ form a strictly increasing
sequence which tends to positive infinity.
\end{lemma}

\vspace*{0.10in}

\begin{theorem}
%\label{thm:spc:besselzeros}
Let $n$ be a nonnegative integer and suppose $\{ \lambda_{n, k}
\}_{k=1}^{\infty}$ are the positive zeros of $J_{n}(x)$ arranged in
increasing order.
Between each consecutive pair of zeros of $J_{n}(x)$ there exists a
unique zero of $J_{n+1}(x)$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Parametric Form of Bessel's Equation}

The \textbf{parametric form of Bessel's Equation} is
\[
%\label{eq:spc:besselparametric}
z^{2} \frac{d^{2} y}{d z^{2}} + z \frac{d y}{d z} + (\lambda^{2} z^{2} - p^{2}) y(z)
	= 0
\]
The function $J_{p}(x)$ solves Bessel's equation of order $p$ if and
only if the function $J_{p}(\lambda z)$ solves the parametric form of
Bessel's equation of order $p$.

\uncover<2->{%
\begin{theorem}
%\label{thm:spc:besselorthog}
Let $p \ge 0$ and let $\lambda_{p, k}$ be the $k$th zero of $J_{p}(x)$,
then
\[
%\label{eq:spc:besselorthog}
\int_{0}^{1} J_{p}(\lambda_{p, n} z) J_{p}(\lambda_{p, m} z) z \, dz
	= \frac{1}{2} \left( J_{p+1}(\lambda_{p, n}) \right)^{2} \delta_{m n}	,
\]
for $m, n \in \mathbb{N}$,
where $\delta_{m n}$ is the
\textbf{Kronecker delta function}.
\end{theorem}
}

\end{frame}
%%

\begin{frame}\frametitle{Bessel-Fourier Series}

If $f(x)$ is sufficiently smooth on $[0, 1]$,
$f(x)$ can be expressed as
\[
%\label{eq:spc:besselfourier}
f(x)
	= \sum_{n=1}^{\infty} a_{n} J_{p}(\lambda_{p, n} x)	,
\]
where
\[
%\label{eq:spc:besselcoeff}
a_{n}
	= \frac{2 \int_{0}^{1} f(x) J_{p}(\lambda_{p, n} x) x \, dx}%
{\left( J_{p+1}(\lambda_{p, n}) \right)^{2}}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Bessel Functions of Half-Integer Order (1 of 3)}

The \textbf{Bessel functions of half-integer order} are the functions
$J_{n+\frac{1}{2}}(x)$ and $Y_{n+\frac{1}{2}}(x)$ where $n$ is an integer.

\vspace*{0.10in}

Using the formula for $J_{p}(x)$
with $p = 1/2$,
\begin{align*}
J_{\frac{1}{2}}(x)
	&= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k! \Gamma(k+1+1/2)} \left(
\frac{x}{2} \right)^{2k+\frac{1}{2}}	\\
	&= \sqrt{\frac{2}{x}} \sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2k+1}}{
  2^{2k+1} k! \left( \frac{1}{2} \right) \left( \frac{3}{2} \right)
  \cdots \left( \frac{2k+1}{2} \right) \Gamma\left(\frac{1}{2}\right)}	\\
	&= \sqrt{\frac{2}{\pi x}}
\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2k+1}}{(2k+1)!}
	= \sqrt{\frac{2}{\pi x}} \sin x	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Bessel Functions of Half-Integer Order (2 of 3)}

Using the derivative realtionships between Bessel functions implies
\begin{align*}
J_{3/2}(x)
	&= \frac{1}{2x} J_{\frac{1}{2}}(x) - J_{\frac{1}{2}}'(x)	\\
	&= \frac{1}{2x} \sqrt{\frac{2}{\pi x}} \sin x - \sqrt{\frac{2}{\pi x}}
\cos x + \frac{1}{2 x} \sqrt{\frac{2}{\pi x}} \sin x	\\
	&= \sqrt{\frac{2}{\pi x}} \left( \frac{1}{x} \sin x - \cos x \right)	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Bessel Functions of Half-Integer Order (3 of 3)}

\begin{theorem}
%\label{thm:spc:besselnplushalf}
For $n = 0, 1, 2, \ldots$
\[
%\label{eq:spc:besselnplushalf}
J_{n+\frac{1}{2}}(x)
	= (-1)^{n} x^{n+1} \sqrt{\frac{2}{\pi x}} \left(
  \frac{d}{x \, dx} \right)^{n} \left[ \frac{\sin x}{x} \right]	.
\]
\end{theorem}

\vspace*{0.10in}

\begin{theorem}
%\label{thm:spc:besselnminushalf}
For $n = 0, 1, 2, \ldots$
\[
%\label{eq:spc:besselnminushalf}
J_{-n-\frac{1}{2}}(x)
	= x^{n+1} \sqrt{\frac{2}{\pi x}} \left(
  \frac{d}{x \, dx} \right)^{n} \left[ \frac{\cos x}{x} \right]	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Spherical Bessel's Equation}

The \textbf{spherical Bessel's equation} of order $n$ is
\[
%\label{eq:spc:sphericalbessel}
x^{2} y'' + 2 x \, y' + (x^{2} - n (n+1)) y
	= 0	.
\]
The solutions are the
\textbf{spherical Bessel function of the first kind}
\[
%\label{eq:spc:sphbesjn}
j_{n}(x) = \sqrt{\frac{\pi}{2 x}} J_{n+\frac{1}{2}}(x)
\]
and the \textbf{spherical Bessel function of the second kind}
\[
%\label{eq:spc:sphbesyn}
y_{n}(x) = \sqrt{\frac{\pi}{2 x}} Y_{n+\frac{1}{2}}(x)	.
\]
\uncover<2->{%
\begin{corollary}
%\label{cor:spc:sphericalbesselorthog}
For $n = 0, 1, 2, \ldots$,
\[
%\label{eq:spc:sphericalbesselorthog}
\int_{0}^{1} j_{n}(\lambda_{n+\frac{1}{2}, k} x)
j_{n}(\lambda_{n+\frac{1}{2}, l} x) x^{2} \, dx
	= \frac{1}{2} \left( j_{n+1}(\lambda_{n+\frac{1}{2}, k}) \right)^{2} \delta_{kl}
\]
where $\delta_{kl}$ is the Kronecker delta function.
\end{corollary}
}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

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