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Fourier Series Question

  1. Oct 22, 2008 #1

    TFM

    User Avatar

    1. The problem statement, all variables and given/known data

    Find the Fourier series corresponding to the following functions that are periodic over the
    interval [tex][-\pi,\pi][/tex]

    [tex] f(x) = 1, -\pi/2 < x< \pi/2; f(x) [/tex] otherwise.

    2. Relevant equations

    Fourier Series:

    [tex] f(x) = \frac{1}{2}a_0 + \sum^\infty_{n=1}a_n cos\frac{2*\pi*n*x}{l} + \sum^\infty_{n=1} b_n sin\frac{2*\pi*n*x}{l}[/tex]

    [tex] \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx [/tex]

    [tex] a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos frac{2*\pi*n*x}{l}dx [/tex]

    [tex] a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin frac{2*\pi*n*x}{l}dx [/tex]

    3. The attempt at a solution

    So far I have:

    [tex] a_0 = 1 [/tex]

    [tex] a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2} [/tex]

    [tex] b_n = -\frac{1}{\pi n}[cos(nx)]^{\pi}_{-\pi} [/tex]

    But I am not sure what to do now. I seem to be mainly confused about the n's

    TFM
     
  2. jcsd
  3. Oct 22, 2008 #2

    TFM

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    My equations didn't come out quite right...Sorry. Should be:


    [tex]
    a_0 = \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx
    [/tex]

    [tex]
    a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos \frac{2*\pi*n*x}{l} dx
    [/tex]

    [tex]
    b_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin \frac{2*\pi*n*x}{l}dx
    [/tex]

    TFM
     
  4. Oct 22, 2008 #3
    Note that:

    [tex] \sin( n \cdot \pi /2)= 1 [/tex] if n = 1,3,5...
    [tex] \sin( n \cdot \pi /2) = 0 [/tex] if n = 2,4,6,...


    and

    [tex] \cos( n \cdot \pi ) = (-1)^{n} [/tex]
     
  5. Oct 22, 2008 #4

    TFM

    User Avatar

    So for [tex] a_n [/tex]

    [tex]
    a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2}
    [/tex]

    n = 1,3,5...

    so

    [tex] a_n = \frac{1}{\pi 1}[sin(1x)]^{\pi/2}_{-\pi/2}, [/tex]

    [tex] \frac{1}{\pi * 3}[sin(3x)]^{\pi/2}_{-\pi/2}, [/tex]

    [tex] \frac{1}{\pi * 5}[sin(5x)]^{\pi/2}_{-\pi/2} ... [/tex]


    and

    [tex]
    \cos( n \cdot \pi ) = (-1)^{n}
    [/tex]

    cos of n pi always = -1? (as long as pi is a whole number)

    TFM
     
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