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Homework Help: Fourier coefficient help

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi there.

    The question is:

    if f(x) is a periodic function with period 2pi and the derivative is f'(x) continuous on [-pi, pi]

    show that the fourier coefficients of f' are kbk and -kak if the fourier coefficients of f are ak and bk respectively.

    2. Relevant equations

    3. The attempt at a solution

    This is my attempt. Doesn't quite get there:

    so ak' = 1/2pi int from -pi to pi of f'(x) cos kx dx

    use parts:
    f'(x) = du
    f(x) = u
    cos kx = v
    -1/k sin kx = dv

    i get

    1/2pi[f(x) cos kx/k + 1/k int from -pi to pi f(x) sin kx]

    of course the integral in the parts equation is just bk, and the f(x) cos kx/k evaluated from pi to -pi just becomes 2f(pi)cos kpi.

    So then I'm left with.

    1/pi[f(pi)cos k pi + (1/k)bk]

    I guess I can write it as

    [(-1)^k f(pi) + (1/k) bk]/pi

    What did i do wrong??
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2


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    E.g. f(x) evaluated from -pi to pi is f(pi)-f(-pi). That's not 2f(pi).
  4. Jan 24, 2012 #3
    isn't it f(pi) cos kpi - f(-pi) cos -kpi

    I wasn't sure how to evaluate that, with k odd i think -2f(pi) and even its 2f(pi) isn't it?
  5. Jan 24, 2012 #4


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    No. f is periodic. So f(-pi)=f(pi) and cos(kpi)=cos(-kpi), right?
  6. Jan 24, 2012 #5
    Oh right. So then the first term is 0?
    But then I'm still left with (1/k pi)(bk).
  7. Jan 24, 2012 #6


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    In your integration by parts you've got v=cos(kx). Then dv=(-k*sin(kx))dx. Not 1/k. And the pi should have been included into definition of bk.
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