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Obtaining One Fourier Series from Another

  1. Jul 3, 2013 #1

    FeDeX_LaTeX

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    1. The problem statement, all variables and given/known data
    (a) On (-π,π), find the Fourier series of f(x) = x.
    (b) Hence, or otherwise, find the Fourier series of g(x) = x2
    (c) Hence, show that [tex]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/tex]

    2. Relevant equations
    [tex]f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L} \right)[/tex]

    where

    [tex]a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx[/tex]

    [tex]a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \left( \frac{n \pi x}{L} \right) dx[/tex]

    [tex]b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \left( \frac{n \pi x}{L} \right) dx[/tex]

    3. The attempt at a solution
    I found (a) as

    [tex]x = \sum_{n=1}^{\infty} \frac{2}{n} (-1)^{n+1} \sin(nx)[/tex]

    but for (b), how do I obtain the Fourier series for x2, without starting from scratch? I can derive that Fourier series fine and I use Parseval's identity for the last part and the result follows -- but I'm just puzzled about the 'hence' part of question (b). Are you supposed to integrate both sides from 0 to x? Doing that doesn't seem to give me the Fourier series for x2...
     
  2. jcsd
  3. Jul 3, 2013 #2

    micromass

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    Yes.
     
  4. Jul 3, 2013 #3

    FeDeX_LaTeX

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    But when I do that, I get:

    [tex]x^2 = \sum_{n=1}^{\infty} \frac{4}{n^2}(-1)^{n} (\cos(nx) - 1)[/tex]

    and it has also not included the first term, a0. What am I doing wrong?

    The actual result is supposed to be:

    [tex]x^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4}{n^2}(-1)^{n}(\cos(nx))[/tex]
     
  5. Jul 3, 2013 #4

    micromass

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    [tex] \sum_{n=1}^{+\infty} \frac{4}{n^2} (-1)^n (\cos(nx) - 1) = - \sum_{n=1}^{+\infty} \frac{4(-1)^n}{n^2} + \sum_{n=1}^{+\infty} \frac{4}{n^2} (-1)^n\cos(nx)[/tex]

    So there's your ##a_0##.
     
  6. Jul 3, 2013 #5

    FeDeX_LaTeX

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    Oh, I see... but is that term ##a_0## or ##a_0 / 2##?
     
  7. Jul 3, 2013 #6

    micromass

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    It's ##a_0/2##.
     
  8. Jul 3, 2013 #7

    FeDeX_LaTeX

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    OK, but then with Parseval's identity I don't seem to be able to get the result... what am I doing wrong below:

    [tex]\int_{-\pi}^{\pi} x^4 dx = \pi \sum_{n=1}^{\infty} \frac{32}{n^4} + \pi \sum_{n=1}^{\infty} \frac{16}{n^4} = \pi \sum_{n=1}^{\infty} \frac{48}{n^4}[/tex]

    which leads to

    [tex]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{120}[/tex]

    ?
     
  9. Jul 3, 2013 #8

    micromass

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    Please don't tell me you just did

    [tex]\left(\sum x_n\right)^2 = \sum x^2_n[/tex]

    Cause it looks like you did.
     
  10. Jul 3, 2013 #9

    FeDeX_LaTeX

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    Oops... applied Parseval's identity in the wrong sense there.
     
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