1. Not finding help here? Sign up for a free 30min 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 series how can it be solved?

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    find fourier series for:
    sinnπ=0
    (n=0,+1,-1,+2,-2...)

    I really can't understand how fourier series works.. Like I tried solving problems..but till now, it didn't sink in my brain..
    I just want 2 know the basic way of solving a problem regarding fourier series..then it will be much easier to understand I guess..
    Thanx in advance..
     
  2. jcsd
  3. Nov 2, 2008 #2

    Mark44

    Staff: Mentor

    What's the function that you want to find the Fourier series for? sin (n*pi) = 0 for all integers, as you point out, but it's otherwise not very interesting. Is the function f(x) = 0?
    If so, all of the coefficients in the Fourier series would be 0.

    Is there more to this problem that you haven't shown us?
     
  4. Nov 6, 2008 #3
    hi mark
    sorry 4 being late..plus giving a completely wrong question..
    ok..
    the I realize the problem is for example :
    f(x)= sinx
    0<x<pi
    it's a problem I want to understand like how is it really solved?
     
  5. Nov 6, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    A Fourier series for a function f(x) is an infinite sum such that
    [tex]f(x)= \sum_{m=0}^\infty A_n cos(nx)+ B_n sin(nx)[/tex] (or other (x/L) or whatever inside the trig functions).

    Of course, the right side of that is periodic with period [itex]2\pi[/itex] so if f is not itself periodic, that can only be true on some interval (which is one reason why you might need that "/L" to alter the interval).

    But if f(x)= sin(x) is not only periodic with period [itex]2\pi[/itex], it is already of that form and it is obvious that [itex]A_n= 0[/itex] for all n, while [itex]B_1= 1[/itex] and [itex]B_n= 0[/itex] for all n greater than 1.

    More generally, if f(x) is integrable on the interval [0, L], then on that interval [itex]f(x)= A_n cos(2n x\pi/L)+ B_n sin(2n x\pi/L)[/itex] where
    [itex]A_0= \frac{1}{L}\int_0^L f(x) dx[/itex]
    [itex]A_n= \frac{1}{2L}\int_0^L f(x)cos(2n x/L)[/itex]
    for n> 0 and
    [itex]B_n= \frac{1}{2L}\int_0^L f(x)sin(2n x/L)[/itex].
     
  6. Nov 7, 2008 #5
    THANX a lot Hallsofivy..
    u were a lot of help..as usual lol..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?