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Reversing Fourier series

  1. Jul 4, 2006 #1
    Anyone have any clues what function f(x) satisfies the following integral (for an integer n > 0):

    [itex]\int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}[/itex]
     
    Last edited: Jul 4, 2006
  2. jcsd
  3. Jul 4, 2006 #2

    J77

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    So an alternating function at +1 and -1?
     
  4. Jul 4, 2006 #3
    Yes. In other words, the function whose Fourier series would be:

    sin(x)-sin(2x)+sin(3x)...
     
  5. Jul 4, 2006 #4
    Nevermind, figured it out.
     
  6. Jul 6, 2006 #5
    If you define:

    [tex] \int_{0}^{\infty}dxW(x)f(x)sin(nx)=(-1)^{n+1} [/tex]

    where x is W(x)=0 iff x>2pi, and W8x)=1 iff x<2pi, the integral above is just a fourier sine transform with inverse:

    [tex] W(x)f(x)=-\frac{2}{\pi}\int_{0}^{\infty}dne^{n\pi i}sin(nx) [/tex]

    which is equal to [tex] f(x)W(x)= 2i(\delta (x+i \pi)-\delta (x-i \pi )) [/tex]

    which is real...:bigrin:
    : :bigrin:
     
    Last edited: Jul 6, 2006
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