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\begin{document}


{\bf Question}

Bessel functions (of order $v$) are denoted by $J_v(x)$. They
arise very frequently in the solution of wave problems with
cylindrical symmetry. It is often required to know where the zero
values of $J_v(x)$ are as a function o f$x$ (being related to
eigenvalues, energy levels or frequencies of vibration etc. or the
problem in question).

The large $x$ asymptotic expansion for $J_0(x)$ is given by,

$$J_0(x)\sim\sqrt{\ds\frac{2}{\pi}}\left[\ds\frac{\cos(x-\frac{\pi}{4})}
{x^{\frac{1}{2}}}+\ds\frac{\sin(x-\frac{\pi}{4})}{8x^{\frac{3}{2}}}\right]
+O\left(\ds\frac{1}{x^{\frac{5}{2}}}\right)$$

\begin{description}
\item[(i)]
Show that the roots as $x\to+\infty$ are given by $\cdots$

$$x\sim\left(n-\ds\frac{1}{4}\right)\pi+\ds\frac{1}{8(n+\frac{1}{4})\pi}+\cdots,\
n\to+\infty$$

\item[(ii)]
Compare these with the first five numerically evaluate roots, and
comment on this, given that $n$ is a measure of the size of $x$.

\end{description}

\begin{tabular}{|c|c|c|c|c|c|} Index & 1 & 2 & 3 & 4 & 5\\ Root
& 2.40482... & 5.52007... & 8.65372... & 11.79153... & 14.93091...
\end{tabular}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(i)]
Clearly $J_0(x)$ is approximately zero (to $O(x^{-2})$)

when

$\begin{array}{rcl}
\ds\frac{\cos(x-\frac{\pi}{4})}{x^{\frac{1}{2}}} & = &
-\ds\frac{\sin(x-\frac{\pi}{4})}{8x^{\frac{3}{2}}}\\ \Rightarrow
\cot\left(x-\ds\frac{\pi}{4}\right) & = & -\ds\frac{1}{8x}\ \
(\star)(\star) \end{array}$

so for $x\to\infty$

$\cot\left(x-\ds\frac{\pi}{4}\right)=0$ is a first approximation

$\begin{array}{rcl} -\ds\frac{\pi}{4}+x & = &
\left(m+\ds\frac{1}{2}\right)\pi\ \ \ m\ \rm{integer}\\ x & = &
\left(m+\ds\frac{3}{4}\right)\pi \end{array}$

Calling $m=n-1$ say we get ($n$ another integer)

$$x=\left(n-\ds\frac{1}{4}\right)\pi$$

Now to improve, let $n=\left(n-\ds\frac{1}{4}\right)\pi+\delta,\
\delta=o(1)$

Substitute into $(\star)(\star)$

$\cot\left(\left(n-\ds\frac{1}{2}\right)\pi+\delta\right)=-\ds\frac{1}{8[(n-\frac{1}{4})\pi+\delta]}$

$-\tan\delta=-\ds\frac{1}{8[(n-\frac{1}{4})\pi+\delta]}$

So expand to $O(\delta)$ on LHS

$$-\delta==-\ds\frac{1}{8[(n-\frac{1}{4})\pi+\delta]}$$

Hence roots are
$x=\left(n-\ds\frac{1}{4}\right)\pi+=-\ds\frac{1}{8[(n-\frac{1}{4})\pi+\delta]}+\cdots$

$\left(=o\left(\ds\frac{1}{n}\right)\right),\ \ n\to \infty$

\item[(ii)]

\end{description}
${}$

$\begin{array}{rccccc} n = & 1 & 2 & 3 & 4 & 5\\ \rm{exact} = &
2.40482.. & 5.52007.. & 8.65372.. & 11.79153.. & 14.93091..\\
\rm{approx.} = & 2.40925.. & 5.52052.. & 8.65385.. & 11.79158.. &
14.93094..\\ \% \rm{error} = & 0.18\% & 0.008\% & 0.001\% &
0.0004\% & 0.0002\% \end{array}$

$\left|\ds\frac{\rm{exact-approx}}{\rm{exact}}\right|\times 100$

\ \ \ \ \ \ \ so it seems even $n=1$ is a large parameter!!!



\end{document}