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Math Help - fourier series

  1. #1
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    fourier series

    fin the cosine series fo r the function defined by
    f(x) = 1/4 - x, 0<= x < 1/2
    f(x) = x - 3/4, 1/2<=x<=1

    for this, I chose L = 1/2, so t = 2pie x
    so the integral limits are 0 to pie and pie to 2 pie...
    How do I get it to go to -pie to pie? help!
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  2. #2
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    Extend you function (represented by dotted line) in the way shown below.
    Since f(-1) = f(1) we can extend f periodically by f(x) = f(x+L) where L=1.

    The Fourier series would be given as, \frac{a_0}{2} + \sum_{n=0}^{\infty} a_n \cos (\pi n x) + b_n \sin (\pi n x).

    Since f on [-1,1] is an even function it forces b_n = 0.

    Thus, a_n = 2\int_0^1 f(x)\cos (\pi n x) dx for n\geq 0.
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  3. #3
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    Hi,

    I still don't get it.
    f(-1) = f(1) ? how?

    i see that f(-1/2) = f (1/2)
    so L = 1/2
    so f(x) = f(x-1/2)
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  4. #4
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    Quote Originally Posted by tennis14321 View Post
    I still don't get it.
    f(-1) = f(1) ? how?
    f(-1)=1/4 = f(1)

    i see that f(-1/2) = f (1/2)
    so L = 1/2
    so f(x) = f(x-1/2)
    It seems you can do your problem two ways. One way is the approach I was trying to do above with L=1 but you can also do it with L=1/2. Because if you take the function f on [-1/2,1/2] and extend it periodically you would get the same function. Thus, in that case you can pick L=1/2 too.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    f(-1)=1/4 = f(1)


    It seems you can do your problem two ways. One way is the approach I was trying to do above with L=1 but you can also do it with L=1/2. Because if you take the function f on [-1/2,1/2] and extend it periodically you would get the same function. Thus, in that case you can pick L=1/2 too.

    Thanks so much.
    How do I go about extend it to 2Pie?
    subtitute x by t/2pie
    so the two integral are 0 to pie (1/4 - 2pie x) and pie to 2pie (2pie x - 3/4) ??? solving this, I should get a0 and an, is this correct?
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