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Math Help - Fourier series

  1. #1
    Junior Member Pinsky's Avatar
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    Fourier series

    Expand the function f(x)=10-x on the interval <5,15>.

    How do i expand it when the interval isn't <-l,l> ?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Pinsky View Post
    Expand the function f(x)=10-x on the interval <5,15>.

    How do i expand it when the interval isn't <-l,l> ?
    Hmmmmm...

    Hint: Can you find the F transform of f(x + 10) = 10 - (x + 10) = -x on (-5, 5)?

    -Dan
    Last edited by topsquark; February 1st 2008 at 10:06 AM.
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  3. #3
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    Could you not add ten to your function? then it would be the same as taking the integral from -5 to 5?

    I'm not sure about that but that seems reasonable to me.
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  4. #4
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    Quote Originally Posted by Pinsky View Post
    Expand the function f(x)=10-x on the interval <5,15>.

    How do i expand it when the interval isn't <-l,l> ?
    Do what topsquark said: f(x+10) = g(x) where g(x) = x on (-5,5). This means, f(x) = g(x-10). Now we expand g = x into a Fourier series on (-5,5). This function is odd so we will just have sine terms. Meaning  \sum_{n=1}^{\infty} a_n \sin \frac{\pi n x}{5} where a_n = \frac{1}{5} \int_{-5}^5 x\sin \frac{\pi n x}{5} dx = \frac{2}{5} \int_0^5 x\sin \frac{\pi n x}{5} dx. Once you find that you will have Fourier for g(x) this means g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5} .
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  5. #5
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    g(x) would be -x not x.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TrevorP View Post
    g(x) would be -x not x.
    That would be my mistake, not TPH's, since he was mostly copying from my post, I think. Thanks for the catch and I have fixed it in my original post.

    -Dan
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  7. #7
    Junior Member Pinsky's Avatar
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    Isn't the formula fo a_n=\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx?

    U used sinus
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  8. #8
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    Quote Originally Posted by Pinsky View Post
    Isn't the formula fo a_n=\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx?

    U used sinus
    \frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx gives the coefficent for the cos terms in a Fourier Series ......
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  9. #9
    Junior Member Pinsky's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Do what topsquark said: f(x+10) = g(x) where g(x) = x on (-5,5). This means, f(x) = g(x-10). Now we expand g = x into a Fourier series on (-5,5). This function is odd so we will just have sine terms. Meaning  \sum_{n=1}^{\infty} a_n \sin \frac{\pi n x}{5} where a_n = \frac{1}{5} \int_{-5}^5 x\sin \frac{\pi n x}{5} dx = \frac{2}{5} \int_0^5 x\sin \frac{\pi n x}{5} dx *. Once you find that you will have Fourier for g(x) this means g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5} .
    *Why sinus?
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Pinsky View Post
    *Why sinus?
    The function g(x) = -x is an odd function. Therefore the F series expansion for it has to be odd as well. So we can have no cosine terms in the expansion since cosine is an even function.

    -Dan
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  11. #11
    Junior Member Pinsky's Avatar
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    Oh, ok! I was a bit to radical on what a_n and b_n represent.
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