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{\bf Question}

Sketch the graphs of $\cot x$ and $\ds\frac{1}{x}$. Show that the
solutions of $x\cot x=1$ behave like

$$x=(1+\frac{1}{2})\pi-\ds\frac{1}{(n+\frac{1}{2})\pi}+\cdots,\ n\
\rm{integer}.$$

\vspace{.5in}

{\bf Answer}

PICTURE  \vspace{3in}

Look for $x\cot x=1 \Rightarrow \cot x=\ds\frac{1}{x}$.

As $x\to \infty\ \ds\frac{1}{x}\to 0$

Therefore if it crosses $\cot x$ it does so near to $\cot x=0$.

$\cot x=0$ at $x=\left(n+\ds\frac{1}{2}\right)\pi$ where $n$ is an
integer.

(since $\tan\left(n+\ds\frac{1}{2}\right)\pi \rightarrow \pm\
\infty$)

Therefore first guess is
$x=\left(n+\ds\frac{1}{2}\right)\pi+\delta$ where
$\delta\ll\left(n+\ds\frac{1}{2}\right)\pi$

$\begin{array}{rcl}
\cot\left[\left(n+\ds\frac{1}{2}\right)\pi+\delta\right] & = &
\ds\frac{1}{\tan((n+\frac{1}{2}\pi)+\delta)}\\ & & \\& = &
\ds\frac{1-\tan(n+\frac{1}{2})\pi\tan\delta}{\tan(n+\frac{1}{2})\pi+\tan\delta}\\&
& \\ & = & -\tan\delta \end{array}$

Therefore $-\tan\delta=\ds\frac{1}{(n+\frac{1}{2})\pi+\delta}$ is
the equation to solve for $\delta$.

$\left.\begin{array}{l}
\tan\delta=\delta+\ds\frac{\delta^3}{3}+O(\delta^5)\\ \rm{
Therefore}\
-\delta-\ds\frac{\delta^3}{3}+O(\delta^5)=\ds\frac{1}{(n+\frac{1}{2})\pi+\delta}\\
\Rightarrow
-\left(n+\ds\frac{1}{2}\right)\pi\delta+O(\delta^2)=1\\
\Rightarrow \delta \approx
-\ds\frac{1}{(n+\frac{1}{2})\pi}\end{array}\right\}$

so
$x=\left(n+\ds\frac{1}{2}\right)\pi-\ds\frac{1}{(n+\frac{1}{2})\pi}+\cdots\
\ n\to\infty$ (hence $x\to\infty$)




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