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Fourier series expansion of Sin(x)

  1. Jun 27, 2008 #1
    May be simple, but I'm getting problem with doing Fourier series expansion of Sin(x) for -pi[tex]\leq[/tex]x[tex]\leq[/tex]pi
     
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  3. Jun 27, 2008 #2

    CompuChip

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    Re: Fourier Series question

    What does the formula for the Fourier expansion of a general function f(x) look like?
     
  4. Jun 29, 2008 #3
    Re: Fourier Series question

    f(x)=ao+[tex]\sum[/tex](ancos(nx)+bnsin(nx))

    where,
    ao=[tex]\frac{1}{\pi}[/tex][tex]\int[/tex]f(x) dx
    an=[tex]\frac{1}{2\pi}[/tex][tex]\int[/tex]f(x)cos(nx) dx
    bn=[tex]\frac{1}{2\pi}[/tex][tex]\int[/tex]f(x)sin(nx) dx
     
    Last edited by a moderator: Jun 29, 2008
  5. Jun 29, 2008 #4

    Defennder

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    Re: Fourier Series question

    Isn't it supposed to be 1/pi for both a_n and b_n and 1/2pi for a_0? And in the formulae for a_n, b_n, these are supposed to be definite integrals not indefinite ones.
     
    Last edited: Jun 29, 2008
  6. Jun 29, 2008 #5

    CompuChip

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    Re: Fourier Series question

    Anyway, for the sine you can either calculate all these coefficients, but you can also just read off the values from the general expression.

    HallsOfIvy, you may be right, didn't notice that. Asked the mods to look into it and maybe split the thread.
     
  7. Jun 29, 2008 #6
    Re: Fourier Series question

    Yes there is definite intergral from -infinity to +infinity. I missed that.
    Sorry

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  8. Jun 29, 2008 #7
    Re: Fourier Series question

    I've tried that, May be I might have done some mistake, The integral term while calculating a_o, a_n and b_n vanishes and results zero. That is why I kept this in this forum. Any way I'll try that.
    Thank You
     
  9. Jun 29, 2008 #8

    CompuChip

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    You mean they all vanished? Check again the ones for n = -1 and n = 1.
    Also, check the integration boundaries. They are not [itex](-\infty, \infty)[/itex].
     
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