# Fourier Series

1. Mar 5, 2007

### Mindscrape

I am expanding the function $$f(t) = e^{i \omega t}$$ from (-π,π) as a complex fourier series where w is not an integer. I am stuck figuring out how the series expands with n.

$$c_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i \omega t} e^{-int} dt$$

Join exponentials

$$c_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i t(\omega - n)$$

Then integrate the function and evaluate the limits

$$c_n = \frac{1}{2 \pi} \frac{1}{i(\omega - n)} \frac{dw}{dt} (e^{i \pi (\omega - n)} - e^{ -i \pi(\omega - n)})$$

Use the identity for sinx

$$c_n = \frac{1}{\pi} \frac{1}{(\omega - n)} \frac{dw}{dt} sin(\pi \omega - \pi n)$$

This is the point when I am not quite sure if I am done, or I need to do something more. Should I just put this into back into the expression for the complex fourier series?

$$f(x) = \sum_{n=-\infty}^{n=+\infty} c_n e^{int}$$

Last edited: Mar 5, 2007
2. Mar 5, 2007

### Dr Transport

So far so good.....

3. Mar 5, 2007

### Mindscrape

So then the function as a fourier series would just be:

$$f(x) = \sum_{n=-\infty}^{n=+\infty} \frac{1}{\pi} \frac{1}{(\omega - n)} \frac{dw}{dt} sin(\pi \omega - \pi n) e^{int}$$