Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series

  1. Apr 1, 2005 #1
    I have tried to find the Fourier series for a function [itex]u(x)[/itex]:

    [tex]
    u(x)=\sin((1+3\cos(t))t)
    [/tex]

    The function is odd, hence the Fourier coefficients [itex]a_n[/itex] equal zero and the [itex]b_n[/itex]s are given as

    [tex]
    b_n=\frac{4}{T}\int_0^{T/2}u(t)\sin(n\omega t)\,\text{d}t
    [/tex]

    where [itex]T=2\pi[/itex] and [itex]\omega=2\pi/T=1[/itex].

    Then, my problem is that u(x)*sin(n*t) is not easily integrated. I would then like to ask if there could be any way getting around this integration-problem, perhaps if I wrote u(x) as either the real og imaginary part of a complex function? I would appreciate any help.
     
  2. jcsd
  3. Apr 1, 2005 #2
    Well, it is sort of close to a Bessel integral. But not really. It doesn't look promising at all. Neither Maple or Mathematica will do it, but that doesn't mean anything. I've had precisely one hour of sleep in the last day and a half, so I may be missing something~
     
    Last edited: Apr 1, 2005
  4. Apr 1, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    So your function is

    [tex] u(x)= \sin \left[\left(1+3\cos t\right)t\right] [/tex]...Hmm.

    U could you leave it like that,namely the coeff. "b_{n}",because you can't evaluate that integral.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier series
  1. Fourier Series (Replies: 3)

  2. Fourier Series (Replies: 1)

  3. Fourier Series (Replies: 1)

  4. Fourier series (Replies: 1)

  5. Fourier Series (Replies: 1)

Loading...