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Fourier series proof

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose, in turn, that the periodic function is symmetric or antisymmetric about the point x=a. Show that the Fourier series contains, respectively, only [tex]cos(k_{n}(x-a))[/tex] (including the [tex]a_0[/tex]) or [tex]sin(k_{n}(x-a))[/tex] terms.

    2. Relevant equations
    The Fourier expansion for the periodic function, f(x):
    [tex]f(x)=a_0 + \sum_{n=1}^{\infty}(a_{n}cos(k_{n}x) + b_{n}sin(k_{n}x))[/tex]

    [tex]a_0=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}f(x) dx[/tex]

    [tex]a_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)cos(k_{n}x) dx, n\neq0[/tex]

    [tex]b_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)sin(k_{n}x) dx, (b_0=0)[/tex]

    [tex]k= \frac{2n\pi}{\lambda}\qquad \lambda=period[/tex]

    Hint: An integral centered on a point where the integrand is antisymmetric will vanish.
    3. The attempt at a solution

    I believe I understand what they are saying, but I do not know how to show/prove it. I know that cosine is symmetric about a point, while sine is not. Initially I was thinking that since sine is antisymmetric, for f(x) to be symmetric, the sine terms cannot exist in the expansion (which is exactly what the problem states). I was thinking the same for the opposite case regarding cosines.

    The hint leads me to believe I am supposed to set up an integral. However, I do not understand how.

    If I am to set up an integral, the best I can figure is that the lower limit should be [tex]a-\lambda[/tex] and the upper limit should be [tex]a+\lambda[/tex]

    I think this because for the integral to be centered at point "a", we would need to go 1 full period in both directions.
  2. jcsd
  3. Mar 23, 2010 #2
    Something just dawned on me. Does anyone think the following is correct?:

    If f(x) is symmetric, then b_n is 0 since sine is antisymmetric. Therefore, the sin term in the Fourier series drops out.

    If f(x) is antisymmetric, then a_n is 0 since f(x) is antisymmetric and the cosine term in the Fourier series drops out.
  4. Mar 23, 2010 #3


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    Yup, that's exactly what the hint was getting at.
  5. Mar 23, 2010 #4
    Thanks! I can't believe I burned almost 3 hours on that, only to have the answer as soon as I posted it.
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