Finding the FS of g(x): Is it Right?

[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.

 

[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.

 

[HallsofIvy]

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.