1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier coefficient help

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi there.

    The question is:

    if f(x) is a periodic function with period 2pi and the derivative is f'(x) continuous on [-pi, pi]

    show that the fourier coefficients of f' are kbk and -kak if the fourier coefficients of f are ak and bk respectively.

    2. Relevant equations



    3. The attempt at a solution

    This is my attempt. Doesn't quite get there:

    so ak' = 1/2pi int from -pi to pi of f'(x) cos kx dx

    use parts:
    f'(x) = du
    f(x) = u
    cos kx = v
    -1/k sin kx = dv

    i get

    1/2pi[f(x) cos kx/k + 1/k int from -pi to pi f(x) sin kx]

    of course the integral in the parts equation is just bk, and the f(x) cos kx/k evaluated from pi to -pi just becomes 2f(pi)cos kpi.

    So then I'm left with.

    1/pi[f(pi)cos k pi + (1/k)bk]

    I guess I can write it as

    [(-1)^k f(pi) + (1/k) bk]/pi

    What did i do wrong??
     
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    E.g. f(x) evaluated from -pi to pi is f(pi)-f(-pi). That's not 2f(pi).
     
  4. Jan 24, 2012 #3
    isn't it f(pi) cos kpi - f(-pi) cos -kpi

    I wasn't sure how to evaluate that, with k odd i think -2f(pi) and even its 2f(pi) isn't it?
     
  5. Jan 24, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No. f is periodic. So f(-pi)=f(pi) and cos(kpi)=cos(-kpi), right?
     
  6. Jan 24, 2012 #5
    Oh right. So then the first term is 0?
    But then I'm still left with (1/k pi)(bk).
     
  7. Jan 24, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    In your integration by parts you've got v=cos(kx). Then dv=(-k*sin(kx))dx. Not 1/k. And the pi should have been included into definition of bk.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier coefficient help
  1. Fourier coefficients (Replies: 8)

  2. Fourier coefficients (Replies: 1)

  3. Fourier coefficients (Replies: 2)

Loading...