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Math Help - Fourier series

  1. #1
    Super Member Deadstar's Avatar
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    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.
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  2. #2
    Senior Member Peritus's Avatar
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    this is going to be a long one:


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

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

    <br />
 = \frac{1}<br />
{{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 }}<br />
{{\pi \left( {1 - jk} \right)}}<br /> <br />

    so:

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

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


    now we'll play a little with the sum:

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

    thus:

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

    Q.E.D.
    Last edited by Peritus; February 18th 2008 at 02:02 PM.
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  3. #3
    Super Member Deadstar's Avatar
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    Thanks! That must have taken a while to type out...
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Deadstar View Post
    Thanks! That must have taken a while to type out...
    Yes indeed, my biggest work in this forum so far.
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  5. #5
    Super Member Deadstar's Avatar
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    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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