1. ## Fourier series

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.

2. 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.

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

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