# Fourier series

1. Nov 28, 2006

### daniel_i_l

Lets say that you have a function f and a FS:
$$\sum_{n=1}^{\inf}a_{n}cos(xn) + b_{n}sin(xn)$$
what is the FS of
$$g(x) = (f(x) + f(-x))/2$$
I think that the answer is
$$\sum_{n=1}^{\inf}a_{n}cos(xn)$$
am i right?
thanks.

2. Nov 28, 2006

### dextercioby

Of course, the sine expansion from the Fourier series for an even function goes away, since sine is an odd function.

So,yes, the Fourier series will contain only cosines.

Daniel.

3. Nov 28, 2006

### HallsofIvy

Staff Emeritus
Yes, g(x) is the "even part" of f.

If h(x)= (f(x)- f(-x))/2, the "odd part" of f, then the Fourier series for h would be
$$\sum_{n=1}^{\inf}a_{n}sin(xn)$$

Since sine is an odd function and cosine is an even function they divide f into even and odd parts.

If you want a more "formal" demonstration of that just note that g(x)sin(nx) and h(x)cos(nx) are odd functions themselves. Their integrals from -a to a must be 0.