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Fourier series of x/3

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the fourier series of x/3.


    2. Relevant equations



    3. The attempt at a solution

    So i believe that it is an odd function so a0, ak = 0. To find bk:

    bk = 1/3pi int (x sin kx dx) from -pi to pi. From integration by parts
    = 1/3pi ([x sin kx from -pi to pi]+ 1/k int (cos kx dx) from -pi to pi)
    = 1/3pi [(pi sin k pi + pi sin -k pi) +1/k^2 (sin k pi - sin -k pi)]
    I think that works out to be 0? Is that right?
     
  2. jcsd
  3. Jan 22, 2012 #2
    Well, by inspection [itex]b_k[/itex] cannot all be zero, because then the whole function would be zero.
     
  4. Jan 22, 2012 #3
    Sorry I made a big mistake. Forgot to mention the period which is from 0 to 2pi. Also my integration by parts was wrong which resulted in bk being 0 when it shouldn't be.

    First question, how do i do it with a period from 0 to 2pi. Is my integrand just from 0 to 2pi?
    The formula for bk is 1/L*integral from L to -L f(x) sin k pi x/L dx

    What is L in this case? 2pi??
     
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