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Math Help - [SOLVED] Fourier series - Trigonometric

  1. #1
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    [SOLVED] Fourier series - Trigonometric

    Hello everyone!

    I have the following problem:

    I have a function g(t) that is 2*pi perodic as follow:

    g(t)=e^(-t)
    0<t<2*pi

    Calculate the trigonometric Fourier series for g(t)

    I need som guidance to solve it completely. I have started my solution but got stuck.

    Here how I started:
    (Page 1) http://img249.imageshack.us/img249/3...oblempage1.jpg
    (Page 2) http://img526.imageshack.us/img526/8...oblempage2.jpg

    I do appriciate any help to comlpete my solution.

    Thanks
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by 4Math View Post
    Hello everyone!

    I have the following problem:

    I have a function g(t) that is 2*pi perodic as follow:

    g(t)=e^(-t)
    0<t<2*pi

    Calculate the trigonometric Fourier series for g(t)

    I need som guidance to solve it completely. I have started my solution but got stuck.

    Here how I started:
    (Page 1) http://img249.imageshack.us/img249/3...oblempage1.jpg
    (Page 2) http://img526.imageshack.us/img526/8...oblempage2.jpg

    I do appriciate any help to comlpete my solution.

    Thanks
    When you integrate e^{-t}\cos(nt) by parts, twice, you get back to the function that you started with, but a different multiple of it. in other words, \int_0^{2\pi}\!\!\!e^{-t}\cos(nt)\,dt = \text{some stuff } + K\int_0^{2\pi}\!\!\!e^{-t}\cos(nt)\,dt, for some constant K different from 1. If you take that term over to the other side then you get (1-K)\int_0^{2\pi}\!\!\!e^{-t}\cos(nt)\,dt = \text{some stuff }, and then when you divide through by 1ľK you get the value of the integral.
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