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Special fourier series

  1. Dec 11, 2009 #1
    can anyone give me a hint on deriving this identity:
    sum(((p^n))/n)*sin(n*Q)= atan(2*p*sin(q)/(1-p^2)

    n = 1 to infinity
    p and q are polar coordinates
     
    Last edited: Dec 11, 2009
  2. jcsd
  3. Dec 11, 2009 #2
    [tex]\sum[/tex]p[tex]^{n}[/tex]sin(n*[tex]\o[/tex])=[tex]\\\\atan(\frac{2p*sin(\o)}{(1-p^{2})})[/tex]
     
  4. Dec 11, 2009 #3
    Having difficulty with this fourier series function. I know that it is an odd function....can anyone help me
     

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  5. Dec 11, 2009 #4
    see attached
     

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  6. Dec 11, 2009 #5

    LCKurtz

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    It is poor practice to hijack a thread about one question by posing another question. You should start a new thread with your question.
     
  7. Dec 11, 2009 #6

    LCKurtz

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    The function is periodic with period [itex]2\pi[/itex]. Why are you integrating over a range of [itex]3\pi[/itex]?
     
  8. Dec 12, 2009 #7
    I believe even though sin and cos are periodic over 2 pi a Fourier series is a representation over the given period and so the integrals must be taken over the period given. I forgot to multiply the integrals by [tex]\frac{1}{3\pi}[/tex]
     
  9. Dec 12, 2009 #8

    LCKurtz

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    No. You are given a function on [itex](-\pi,\pi)[/itex]. That function is extended periodically. The fact that you were asked to draw its graph over an interval of length [itex]3\pi[/itex] doesn't magically make the function periodic of that period.

    The problem asks you to determine whether the given function is even, odd, or neither. If it is one of the first two, that suggests using a half-range expansion.

    You are, as the saying goes, barking up the wrong tree.
     
  10. Dec 13, 2009 #9
    maybe this will work
     

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  11. Dec 13, 2009 #10
    any help on my problem?
     
  12. Dec 13, 2009 #11

    LCKurtz

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    What do you need help on? That last bmp looks like you have calculated the fourier series correctly. Do you see what the answer to the original question is?
     
  13. Dec 13, 2009 #12
    I was looking or hints on the proof of this identity
     
  14. Dec 15, 2009 #13
    [tex]\sum[/tex][tex]\frac{\rho}{n}[/tex][tex]\ast[/tex][tex]\rho[/tex][tex]^{n}[/tex]sin[tex]\sigma[/tex]=[tex]\frac{1}{2}[/tex][tex]\ast[/tex]tan[tex]^{-1}[/tex][tex]\left([/tex][tex]\frac{2*\rho*sin(\sigma)}{1-\rho^{2}}[/tex][tex]\right)[/tex][tex]\right)[/tex]

    this is a Fourier series listed in most references but I can't derive it. Any help?
     
  15. Dec 16, 2009 #14
    the actual identity is:
     

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  16. Dec 16, 2009 #15

    Mute

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    I would try the following: Write the left hand side as

    [tex]\mbox{Im}\left(\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\right),[/tex]

    where "Im" means take the imaginary part of the result, and I've written [itex] z = p \exp[i \phi][/itex].

    Take a derivative of this with respect to z, evaluate the series, then integrate the result with respect to z. Impose the initial condition z = 0 the sum = 0 to fix the arbitrary constant, then write z in terms of p and phi again and finally take the imaginary part of the result.
     
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