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Fourier Series Problem

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the Fourier series of
    f(x) = { 1 |x|<a
    { 0 a<|x|<[itex]\pi[/itex]
    for 0<a<[itex]\pi[/itex]
    extended as a 2-Pi periodic function for x [itex]\in[/itex]R
    to find
    [itex]\sum[/itex] Sin2(na)/n2
    2. Relevant equations[/b

    I got that the fourier series of f(x) was
    a/[itex]\pi[/itex]+[itex]\sum[/itex] (2/(m[itex]\pi[/itex]) sin(ma) sin(mx)


    3. The attempt at a solution
    I'm not sure what to do. I'm guessing that since its meant to use the Fourier series of f; it should be related to f(a), but f(a) is undefined, and the denominator is wrong...
    Could anyone please point me in the right direction?
     
  2. jcsd
  3. Aug 26, 2012 #2

    vela

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    Your f(x) is an even function, so you should have cosines instead of sines:
    $$f(x) = \frac{a}{\pi} + \sum_{m=1}^\infty \frac{2}{m\pi} \sin ma \cos mx.$$ Now if you look at the sum you're trying to evaluate, you need another power of m on the bottom and you want to turn the cosine into a sine. Any idea of how to get that?
     
  4. Aug 26, 2012 #3

    gabbagabbahey

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    The formulas you are likely using for the Fourier Series of a function [itex]f(x)[/itex] require the function to be piecewise smooth and periodic over the Reals, as well as being defined at each discontinuity as equal to the mean value of the one-sided limits, so you require

    [tex]f(x=\pm a) = \frac{1}{2} \left( \lim_{x \to \pm a^+}f(x) + \lim_{x \to \pm a^-}f(x) \right) = \frac{1}{2}[/tex]
     
  5. Aug 28, 2012 #4
    I integrate it from zero to a to get the extra sine term?
     
  6. Aug 28, 2012 #5

    vela

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    Sounds good. What do you get?
     
  7. Aug 28, 2012 #6
    a Pi/2. Thanks for the help :)
     
  8. Aug 28, 2012 #7

    uart

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    Ok, you're on the right track but not quite there yet. I think there's another term.
     
  9. Aug 28, 2012 #8
    I don't think there's another term - the 1/2 A0 term goes to zero when you integrate it from
    -a to a.
     
  10. Aug 28, 2012 #9

    uart

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    No actually it doesn't.

    BTW. Previously you said you would integrate from zero to pi? Either way will work, but zero to pi is easier.
     
  11. Aug 28, 2012 #10
    Integrating from zero to a got that it was Pi/2 (a+a^2/Pi); is that right?
    Thanks for pointing it out
     
  12. Aug 28, 2012 #11

    uart

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    Almost. Check your "signs".
     
  13. Aug 28, 2012 #12
    Minus, sorry, thanks - I need sleep...
     
  14. Aug 28, 2012 #13

    uart

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    Yep, minus.

    And as a nice little "sanity" check, notice that the sum is now zero when [itex]a = \pi[/itex]. :smile:
     
  15. Aug 28, 2012 #14
    Thank you :)
     
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