[SOLVED] Fourier series - Trigonometric

• June 3rd 2010, 03:03 AM
4Math
[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
• June 3rd 2010, 03:39 AM
Opalg
Quote:

Originally Posted by 4Math
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 1–K you get the value of the integral.