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Math Help - kth power of fourier series

  1. #1
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    kth power of fourier series

    I'm kinda stuck on this problem (attached).. I know it works (from a previous class), but I forgot how to actually prove it.

    Thank you!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: kth power of fourier series

    If you have:
    S_n^{f}=a_0+\sum_{n=1}^{n} a_n \cos(nx) + \sum_{n=1}^{n} b_n \sin(nx)

    You know the first derivative of:
    a_n \cos(nx) = - a_n \cdot n \cdot \sin(nx)
    and so the second derivative will be
     -a_n \cdot n^2 \cdot \cos(nx)
    ...
    So the kth derivative will be:
    \pm a_n \cdot n^k \cdot \mbox{trig}(nx)
    We don't know if we get a sinus or a cosinus, that's the reason we wright trig(nx).

    Do this also for the other sum.

    Does this help you? ...
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  3. #3
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    Re: kth power of fourier series

    thank you! i totally misread the question, and thought it was the series to the kth power, rather than the kth derivative of the series. looks simple enough, problem solved!
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