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Math Help - Fouriers series

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

    Expand the function f(x)=x\sin{x} into Fouriers series on an interval <-\pi,\pi>

    In my book, the solution of this example starts with.

    f(x)=\sin{x}\sum^{\infty}_{n=1}\frac{2}{n}(-1)^{n+1}\sin{nx}

    I know b_n i 0 because f(x) is an even function. But where did they get the (-1)^{n+1} from? Is that the solution for a_n?
    Last edited by Pinsky; January 31st 2008 at 01:53 PM.
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  2. #2
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    Quote Originally Posted by Pinsky View Post
    Expand the function f(x)=x\sin{x} into Fouriers series on an interval <-\pi,\pi>
    If \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx + b_n\sin nx

    Then a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos (nx) dx for n\geq 0 and b_n = 0 (because it is an even function).

    Thus, a_n = \frac{2}{\pi}\int_0^{\pi} x\sin x \cos (nx) dx for n\geq 0. Do this by integration by parts.
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