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Thread: Fourier series

  1. #1
    Super Member Deadstar's Avatar
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    Fourier series

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

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

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

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


    $\displaystyle \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}
    $

    $\displaystyle
    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]

    $

    $\displaystyle
    = \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:

    $\displaystyle
    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:
    $\displaystyle
    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:

    $\displaystyle
    \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] =
    $$\displaystyle
    \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:

    $\displaystyle
    \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.
    Last edited by Peritus; Feb 18th 2008 at 01:02 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;

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

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

    $\displaystyle 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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