\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

Now consider the Bessel function as both its order, $n$ and
argument $x$ get large. By setting $x=nr$, show that

$$J_n(nr) \sim \sqrt{\ds\frac{2}{\pi
n}}(r^201)^{-\frac{1}{4}}\cos\left\{n\left[\sqrt{r^2-1}
-\rm{arccos}\left(\ds\frac{1}{r}\right)\right]-\ds\frac{\pi}{4}\right\}.$$





\vspace{.5in}

{\bf Answer}

Start with the integral representation:

\begin{eqnarray*} J_n(x) & = & \df{1}{\pi}\ds\int_0^\pi
\cos(nt-x\sin t)\,dt\\ & = & \sum_{\pm}\df{1}{2\pi}\ds\int_0^\pi
e^{\pm int}e^{\mp ix\sin t}\,dt \end{eqnarray*}

\un{NB} and \un{now} $x \to + \infty,\ n \to +\infty$

We set $x=nr$ by hint of question.

$J_n(x)=\sum_{\pm}\df{1}{2\pi}\ds\int_0^\pi e^{\pm int \mp inr\sin
t}\, dt$

so that $h^{(\pm)}(t)=\pm r\sin t \mp t$

and we consider just $n\to+\infty,\ r$ fixed.

Focus on $\begin{array}{rcl}h^{(+)}(t) & = & +r\sin t-t\\
{h^{(+)}}'(t) & = & r\cos t-1\\ {h^{(+)}}''(t) & = & -r\sin t
\end{array}$

So stationary point at $t=\arccos\left(\df{1}{r}\right)
\Rightarrow r \geq 1$ for real contribution

$\Rightarrow \begin{array}{rcl}
h^{(+)}\left(\arccos\left(\df{1}{r}\right)\right) & = &
r\sin\left[\arccos\left(\df{1}{r}\right)\right]-\arccos\left(\df{1}{r}\right)\\
{h^{(+)}}''\left(\arccos\left(\df{1}{r}\right)\right) & = & -r\sin
\left[\arccos\left(\df{1}{r}\right)\right]\end{array}$

This can be simplified by considering the right-angled triangle:

\setlength{\unitlength}{.5in}
\begin{picture}(2,2)
\put(0,0){\line(1,0){2}}

\put(2,0){\line(0,1){1}}

\put(0,0){\line(2,1){2}}

\put(0,0){\makebox(0,0)[l]{$t$}}

\put(1,0){\makebox(0,0)[t]{1}}

\put(1,1.2){\makebox(0,0){$r$}}

\put(2,.5){\makebox(0,0){$\sqrt{r^2-1}$}}

\put(1.8,0){\line(0,1){.2}}

\put(1.8,.2){\line(1,0){.2}}
\end{picture}

Therefore
$h^{(+)}\left[\arccos\left(\df{1}{r}\right)\right]=\sqrt{r^2-1}-\arccos\left(\df{1}{r}\right)$

${h^{(+)}}''\left[\arccos\left(\df{1}{r}\right)\right]=-\sqrt{r^2-1}$

NB We assume \un{$r>1$}

Locally at stationary point we have

$\begin{array}{rcl}& &
-in\left[h^{(+)}(t)-h^{(+)}\left(\arccos\left(\df{1}{r}\right)
\right)\right]\\ & = &
-in\df{(-\sqrt{r^2-1})}{2}\left(t-\arccos\left(\df{1}{r}\right)\right)^2\\
& = &
\df{\in}{2}\sqrt{r^2-1}\left(t-\arccos\left(\df{1}{r}\right)\right)^2\
\ r \geq 1\end{array}$

\un{Twists}:

Locally require

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\put(2.5,2.5){\line(1,1){1.5}}

\put(2,2){\circle*{.125}}

\put(0.4,3.6){\makebox(0,0){H}}

\put(3.6,0.4){\makebox(0,0){H}}

\put(0,4){\makebox(0,0){(t)}}

\put(2.5,2.2){\makebox(0,0){$+\frac{\pi}{4}$}}

\put(1.6,1.2){\makebox(0,0){V}}

\put(3,3.2){\makebox(0,0){V}}
\end{picture} $\left\{\begin{array}{l}Re[-in[\left(h^{(+)}(t)-h^{(+)}\left(\arccos\left(\df{1}{r}\right)\right)\right]<0\\ Re\left[\df{i\sqrt{r^2-1}}{2}\left(t-\arccos\left(\df{1}{r}\right)\right)^2\right]<0\\
\left[t-\arccos\left(\df{1}{r}\right)\right]^2=Re^{i\theta_+}\\
\Rightarrow
Re\left[\df{i\sqrt{r^2-1}}{2}R^2e^{2i\theta_+}\right]<0\\
\rm{since}\ r \geq 1\\ \Rightarrow
Re\left[ie^{2i\theta_+}\right]<0\\ \Rightarrow -\sin 2\theta_+
<0\\ \sin 2\theta_+ >0\end{array}\right.$

Angle of twist from:

\begin{eqnarray*}
Im\left\{in\left[h^{(+)}(t)-h^{(+)}\left(\arccos\df{1}{r}\right)\right]\right\}
& = & 0\\ \Rightarrow \cos 2\theta_+ & = & 0\\ \Rightarrow
\theta_+ & = & \un{+\df{\pi}{4}}
\end{eqnarray*}

PICTURE \vspace{2in}

Can use equation (3.99) with $n$ playing the role of $x$.

\begin{eqnarray*} J_+ & = & \df{1}{2\pi}\ds\int_0^\pi
e^{+in(t-r\sin t)}\, dt\\ & \sim &
\df{1}{2\pi}\sqrt{\df{2\pi}{|n{h^(+)}''(t_0)|}}e^{-ing^(+)(t_0)+i\theta_+}f^(+)(t_0)\\
& & t_0=\arccos\left(\df{1}{r}\right)\\ & & f^(+)(t)=1,\ \mbox{now
since both exponentials contribute}\\ & \sim & \df{1}{2\pi}
\sqrt{\df{2\pi}{|-n\sqrt{r^2-1}|}}e^{-in(\sqrt{r^2-1}-\arccos(\frac{1}{r}))+i\frac{\pi}{4}}\\
& \sim &
\sqrt{\df{1}{2n\pi\sqrt{r^2-1}}}e^{-in(\sqrt{r^2-1}-\arccos(\frac{1}{r}))+i\frac{\pi}{4}}
\end{eqnarray*}

$J_{-}$ is similar due to $+/-$ difference in $h^{(-)}(t)$.

$h^(-)\left(\arccos\left(\df{1}{r}\right)\right)=-\sqrt{r^2-1}+\arccos\left(\df{1}{r}\right)$

${h^(-)}''\left(\arccos\left(\df{1}{r}\right)\right)=+\sqrt{r^2-1}\
\ (r>1)$

$\theta_{-}=-\df{\pi}{4}$

so using (3.99) with $n$ playing the role of $x$:

\begin{eqnarray*} J_{-} & = & \df{1}{2\pi}\ds\int_0^\pi
e^{-in(t-r\sin t)}\,dt\\ & \sim &
\df{1}{2\pi}\sqrt{\df{2\pi}{|n{h^(-)}''(t_0)|}}e^{-inh^{(-)}(t_0)+i\theta-f^(-)(t_0)}\\
& & t_0=\arccos\left(\df{1}{r}\right),\ f^(-)(t)=1\\ & \sim &
\df{1}{2\pi}\sqrt{\df{2\pi}{|n\sqrt{r^2-1}|}}e^{-in(-\sqrt{r^2-1}+\arccos(\frac{1}{r}))
-i\frac{\pi}{4}}\\ & \sim & \df{1}{\sqrt{2\pi
n\sqrt{r^2-1}}}e^{+in(\sqrt{r^2-1}-\arccos(\frac{1}{r}))-\frac{i\pi}{4}}\end{eqnarray*}

The total integral approximation is obtained by adding $J_+$ and
$J_{-}$.

\begin{eqnarray*} J & = & J_+ + J_{-}\\ & \sim & \df{1}{\sqrt{2\pi
n\sqrt{r^2-1}}}\left\{\begin{array}{l}e^{+in(\sqrt{r^2-1}-\arccos(\frac{1}{r}))
-\frac{\pi}{4}}\\
+e^{-in(\sqrt{r^2-1}-\arccos(\frac{1}{r}))+\frac{\pi}{4}}\end{array}\right\}\\
& \sim & \sqrt{\ds\frac{2}{\sqrt{\pi
n\sqrt{r^2-1}}}}\cos\left(n\left[\sqrt{r^2-1}-\arccos\left(\df{1}{r}\right)\right]
-\df{\pi}{4}\right)\\ & \sim & \sqrt{\df{2}{\pi
n}}(r^2-1)^{-\frac{1}{4}}\cos\left(n\left[\sqrt{r^2-1}
-\arccos\left(\df{1}{r}\right)\right]-\df{\pi}{4}\right)\end{eqnarray*}

as required.\ \ $n\to +\infty$


\end{document}