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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:

    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


    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..
    the I realize the problem is for example :
    f(x)= sinx
    it's a problem I want to understand like how is it really solved?
  5. Nov 6, 2008 #4


    User Avatar
    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..
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