Question

Derive series expansion for pi/sinh(pi) with the help of fourier series representation of e^-ax , x belongs to [pi,-pi]

Answer 1

Given:

Given,  $\[\frac{\pi }{\sinh (\pi )}\]$ with the help of Fourier series representation of $\[{{e}^{-}}^{ax}\]$, $x$ belongs to [$\[\pi ,-\pi$ ].

To find:

We need to derive the series expansion.

Solution:

An infinite series in which the terms are constants multiplied by sine or cosine functions of integer multiples of the variable and which is used in the analysis of periodic functions.

A Fourier series is a periodic function composed of harmonically related sinusoids, combined by a weighted summation. With appropriate weights, one cycle (or period) of the summation can be made to approximate an arbitrary function in that interval (or the entire function if it too is periodic).

Applying the formula,

$\[f(x)=\frac{1}{2}{{a}_{0}}+\sum\nolimits_{n=1}^{\infty }{{{a}_{n}}\cos }\text{(nx)+}\sum\nolimits_{n=1}^{\infty }{{{b}_{n}}\sin }(nx)\]$

$\[{{a}_{0}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{f(x)}dx\]$

$& {{a}_{n}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{f(x)\cos (nx)\text{ }dx} \\ & \\ & {{b}_{n}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{f(x)\sin (nx)\text{ }dx}$

Here,

$\[{{a}_{0}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{f(x)}dx=\frac{1}{\pi a}\mathop{[-{{e}^{-ax}}]}_{-\pi }^{\pi }=\frac{{{e}^{\pi a}}-{{e}^{-\pi a}}}{\pi a}$

$\[{{a}_{n}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}\cos (nx)\text{ }dx}\]$                

$=\frac{1}{n\pi }\left[ {{e}^{-ax}}{Sin}(nx)+\frac{a}{n\pi } \right]+\frac{a}{n\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}{Sin}(nx)\text{ }dx}$

$& =0-\left[ \frac{a}{{{n}^{2}}\pi }{{e}^{-ax}}\cos (nx)\text{ } \right]_{-\pi }^{\pi }-\frac{{{a}^{2}}}{{{n}^{2}}\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}\cos (nx)\text{ }dx}$

$& =-\left[ \frac{a}{{{n}^{2}}\pi }{{e}^{-ax}}\cos (nx) \right]_{-\pi }^{\pi }-\frac{{{a}^{2}}}{{{n}^{2}}\pi }\int\limits_{-\pi }^{\pi }{{{e}^{-ax}}\cos (nx)\text{ }dx}$

$& \\ & =-\frac{a}{{{n}^{2}}\pi }\left[ {{e}^{-a\pi }}-{{e}^{a\pi }} \right]\cos (n\pi )-\frac{{{a}^{2}}}{{{n}^{2}}}{{a}_{n}} \\ & \\ & {{a}_{n}}=\frac{{{\left( -1 \right)}^{n}}\left[ {{e}^{-a\pi }}-{{e}^{a\pi }} \right]}{\left( {{n}^{2}}+{{a}^{2}} \right)\pi }$

                             

$\[{{b}_{n}}=\frac{1}{\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}\sin (nx)\text{ }dx}$                      

$& \text{ }=\frac{1}{n\pi }\left[ {{e}^{-ax}}\cos (nx) \right]_{-\pi }^{\pi }-\frac{a}{n\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}\cos (nx)\text{ }dx}$

$& \text{ }=\frac{{{\left( -1 \right)}^{n}}\left( {{e}^{\pi a}}-{{e}^{-\pi a}} \right)}{n\pi }-\frac{a}{{{n}^{2}}\pi }\left[ {{e}^{-ax}}\sin (nx) \right]_{-\pi }^{\pi }-\frac{{{a}^{2}}}{{{n}^{2}}\pi }\int_{-\pi }^{\pi }{{{e}^{-ax}}\sin (nx)\text{ }dx}$

$& {{b}_{n}}\text{ }=\frac{{{\left( -1 \right)}^{n}}\left( {{e}^{\pi a}}-{{e}^{-\pi a}} \right)}{n\pi }-0-\frac{{{a}^{2}}}{{{n}^{2}}}{{b}_{n}}$

$& {{b}_{n}}\text{ }=\frac{{{\left( -1 \right)}^{n}}\left( {{e}^{\pi a}}-{{e}^{-\pi a}} \right)n}{\left( {{n}^{2}}+{{a}^{2}} \right)\pi }$

   

${{e}^{-ax}}\text{ over }\!\![\!\!\text{ }\pi \text{,-}\pi \text{ }\!\!]\!\!\text{ }$; the Fourier series is,

${{e}^{-ax}}\text{ =}\frac{\left( {{e}^{\pi a}}-{{e}^{-\pi a}} \right)}{n}\left[ \frac{a}{2}+\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}a}{{{a}^{2}}+{{n}^{2}}}\cos }\text{(nx)+}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}n}{{{a}^{2}}+{{n}^{2}}}\sin }(nx) \right]$                

        $\text{ = }\frac{\left( {{e}^{\pi a}}-{{e}^{-\pi a}} \right)}{n}\left[ \frac{a}{2}+\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{a}^{2}}+{{n}^{2}}}\left\{ a\text{ Cos }nx+n\text{ Sin }nx \right\}} \right]$