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

Fourier coefficients

  1. Jan 20, 2014 #1

    joshmccraney

    User Avatar
    Gold Member

    hey pf!

    okay, so if you've studied PDEs you know the value of a fourier series, and the difficulty of determining a fourier coefficient. my question relates to finding this coefficient. briefly, i'll define a fourier series as [tex]f(x)=\sum_{n=0}^{\infty} A_n\cos\frac{n\pi x}{L}+B_n\sin\frac{n\pi x}{L}[/tex] when solving for [itex]A_n[/itex] one way (the only way i know) is to multiply both sides by [itex]\cos\frac{m\pi x}{L}[/itex]. now, when we integrate over [itex][-L,L][/itex] the sine term vanishes by its orthogonality with cosine and we are left with [tex]\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx=\sum_{n=0}^{\infty} A_n \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx[/tex], where the r.h.s=0 if [itex]m \neq n[/itex] thus we know [itex]m = n[/itex] (for now let's assume [itex]m=n \neq 0[/itex]). [itex]m = n \implies \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx=L[/itex] thus [tex]\sum_{n=1}^{\infty}A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx[/tex] but every book i see states simply [tex]A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx[/tex] where did the sum go???
     
  2. jcsd
  3. Jan 20, 2014 #2

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    Exactly. Which means only one term of the sum is left, the [itex]n=m[/itex] term, so this equation is identical to

    [tex]\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx[/tex]

    Does that make sense?

    jason
     
  4. Jan 20, 2014 #3

    joshmccraney

    User Avatar
    Gold Member

    wait, does this mean [tex]\sum\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx[/tex] is equivalent to [tex]\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx= \sum A_m \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{m\pi x}{L}dx[/tex]
    could you explain?
     
  5. Jan 20, 2014 #4

    joshmccraney

    User Avatar
    Gold Member

    never mind, i see what you mean now! thanks!
     
  6. Jan 20, 2014 #5

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    "Where did the sum go?" is the wrong question.

    The right question is "where do the parentheses go?"

    Answer:
    $$\int_{-L}^{L}f (x) \cos\frac{m\pi x}{L} dx = \sum_{n=0}^{\infty} \left( A_n \int_{-L}^{L} \cos \frac{m\pi x}{L} \cos \frac{n\pi x}{L} dx \right)$$

    and all the integrals on the right hand side are zero, except when ##m = n##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




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

  2. Fourier Coefficients (Replies: 3)

  3. Determining coefficients (Replies: 12)

Loading...