Fourier series

**Deadstar** · February 18th 2008, 11:31 AM

Define $f(t) = e^t$ on $[- \pi, \pi )$ and extend f to be $2\pi$ -periodic.

By evaluating the Fourier series for t = 0, show that;

$\sum_{k=0}^{\infty}\frac{(-1)^k}{1 + k^2} = \frac{1}{2}(1 + \frac{\pi}{sinh(\pi)})$

So for the first part, extending f to be $2\pi$ -periodic gives $f(t + 2\pi) = f(t)$ , for $(-\infty , +\infty)$ i think.

Then evaluating t at 0... well i couldnt make the last two fourier lectures so im a bit lost here.

**Peritus** · February 18th 2008, 12:44 PM

this is going to be a long one:

$\begin{gathered} e^t = \sum\limits_{k = - \infty }^\infty {c_k e^{jkt} } \hfill \\ \hfill \\ c_k = \frac{1} {{2\pi }}\int\limits_{ - \pi }^\pi {e^t e^{ - jkt} dt} \hfill \\ \end{gathered} $

$ c_k = \frac{1} {{2\pi }}\int\limits_{ - \pi }^\pi {e^t e^{ - jkt} dt} = \left. {\frac{1} {{2\pi \left( {1 - jk} \right)}}e^{\left( {1 - jk} \right)t} } \right|_{ - \pi }^\pi = \frac{1} {{2\pi \left( {1 - jk} \right)}}\left[ {e^{\left( {1 - jk} \right)\pi } - e^{ - \left( {1 - jk} \right)\pi } } \right] $

$ = \frac{1} {{2\pi \left( {1 - jk} \right)}}\left[ {e^\pi \left( { - 1} \right)^k - e^{ - \pi } \left( { - 1} \right)^k } \right] = \frac{{\left( { - 1} \right)^k \sinh \pi }} {{\pi \left( {1 - jk} \right)}} $

so:

$ e^t = \sum\limits_{k = - \infty }^\infty {\frac{{\left( { - 1} \right)^k \sinh \pi }} {{\pi \left( {1 - jk} \right)}}e^{jkt} } $

now we have to evaluate the series at t=0:
$ 1 = \sum\limits_{k = - \infty }^\infty {\frac{{\left( { - 1} \right)^k \sinh \pi }} {{\pi \left( {1 - jk} \right)}}} $

now we'll play a little with the sum:

$ \sum\limits_{k = - \infty }^\infty {\frac{{\left( { - 1} \right)^k \sinh \pi }} {{\pi \left( {1 - jk} \right)}} = \frac{{\sinh \pi }} {\pi }} \left[ {1 + \sum\limits_{k = 1}^\infty {\frac{{\left( { - 1} \right)^k }} {{\left( {1 - jk} \right)}} + \sum\limits_{k = 1}^\infty {\frac{{\left( { - 1} \right)^k }} {{\left( {1 + jk} \right)}}} } } \right] = $ $ \frac{{\sinh \pi }} {\pi }\left[ {1 + 2\sum\limits_{k = 1}^\infty {\frac{{\left( { - 1} \right)^k }} {{1 + k^2 }}} } \right] = \frac{{\sinh \pi }} {\pi }\left[ {2\sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }} {{1 + k^2 }} - 1} } \right] $

thus:

$ \begin{gathered} 1 = \frac{{\sinh \pi }} {\pi }\left[ {2\sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }} {{1 + k^2 }} - 1} } \right] \hfill \\ \hfill \\ \Leftrightarrow \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }} {{1 + k^2 }} = } \frac{1} {2}\left( {\frac{\pi } {{\sinh \pi }} + 1} \right) \hfill \\ \end{gathered} $

Q.E.D.

**Deadstar** · February 18th 2008, 01:05 PM

Thanks! That must have taken a while to type out...

**Peritus** · February 18th 2008, 01:07 PM

Originally Posted by Deadstar

Thanks! That must have taken a while to type out...

Yes indeed, my biggest work in this forum so far.

**Deadstar** · February 18th 2008, 01:48 PM

Now say if i wanted to find a similar expression for;

$\sum_{k=0}^{\infty}\frac{1}{1 + k^2}$ by considering the series $t = \pi$ .

Would i just stick absolute value signs on the equation and eventually get to the expression;

$e^{\pi} = \frac{sinh \pi}{\pi}(2 \sum_{k=0}^{\infty}\frac{1}{1 + k^2} -1)$ ? Somehow that just doesnt seem right...

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