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Fourier Series

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    The following function is periodic between -π and π:

    f(x) = |x|
    Find the Coefficients of the Fourier series and, by examining the Fourier series at x=π or otherwise, determine:
    1 + 1/32 + 1/52 + 1/72 ... = Σj=1 1/(2j - 1)2

    2. Relevant equations
    f(x) = a0/2 + ∑n=1 ancos(nx) + bn sin(nx)

    a0 = 1/π ∫π f(x) dx
    an = 1/π ∫π f(x) cos(nx) dx
    bn = 1/π ∫π f(x) sin(nx) dx
    3. The attempt at a solution
    So I've found the coefficients:
    a0 = π
    an = -2(1 - (-1)n)/πn2
    bn = 0 (as even function)

    and so f(x) = π/2 + ∑-2(1 - (-1)n)/πn2 cos(nx)

    I don't know how to do the last bit however, I don't really understand if I'm meant to come out with a number or something...
    Any help would be greatly appreciated thank you.
     
  2. jcsd
  3. May 12, 2016 #2

    stevendaryl

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    Yes. Just plug [itex]x=\pi[/itex] into your expression. (You can see that when [itex]n[/itex] is even, you get 0, so you only have to consider the case for [itex]n[/itex] is odd.)
     
  4. May 12, 2016 #3

    Samy_A

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    Homework Helper

    Yes, they expect a value for ##\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)²}##.

    You have computed the Fourier series, and found ##\displaystyle |x|=\frac{\pi}{2}- \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(1-(-1)^n)}{n²}\cos(nx)##.

    What you have to do next is actually stated in the question: "examining the Fourier series at x=π". In other words, set x=π in your Fourier series.
     
    Last edited: May 12, 2016
  5. May 12, 2016 #4
    Ok, Yeah I've done that and get

    π/2 +4/π∑1/n^2 (From n=1 to ∞ for n odd only)

    And can't see how to get rid of the π/2 or 4/π

    Thanks
     
  6. May 12, 2016 #5

    Charles Link

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    Homework Helper

    Take another look at it: You have ## \pi=\pi/2+(4/\pi)(...) ## The rest is algebra.
     
  7. May 13, 2016 #6
    Thank you for your help, after doing a few more questions I finally understood the point of the question and I got an answer of π2/8.
     
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