Fourier Series

1. Oct 23, 2006

Brewer

Can someone explain this to me simply. I just plain do not get Fourier Series.

I've got a question that says:
show that
f(x)=exp(-cx) for 0<x<pi
=exp(cx) for -pi<x<0

can be written as a cosine series thats way too complicated for me to work out how to write here.

I have no idea where to start. My lecture notes are of very little help, and all the websites and textbooks seem to be of no help at all.

My only thought of where to start is to write e as functions of cos and sin, but there are no sin's in the answer, so I'm a little dubious about this. If anyone has a way of explaining this to dumbasses, then I'd love to hear it!

2. Oct 23, 2006

mathman

Any function can be considered the sum of an even function and an odd function. The Fourier series for even functions have only cosine terms, while for odd functions have only sine terms. The function you have is even, so the sine terms have zero coefficients.

3. Oct 24, 2006

Brewer

I still struggle to understand what you mean though. How are the coefficients worked out?

Thanks though.

4. Oct 24, 2006

HallsofIvy

Almost all "Fourier coefficient" problems can be calculated using "integration by parts".

In this case, where your interval is from $-\pi$ to $\pi$ the coefficient of sin(nx) in the Fourier series for f(x) would be
$$\frac{1}{\pi}\int_{-\pi}^\pi f(x)sin(nx)dx$$
the "constant term" (corresponding to cos(0x)) would be
$$\frac{1}{2\pi}\int_{-\pi}^\pi f(x) dx[/itex] and the coefficient of cos(nx) for n> 0 [tex]\frac{1}{\pi}\int_{-\pi}^\pi f(x)cos(nx)dx$$

In the case of your function, the coefficient for sin(nx) is
$$\frac{1}{\pi}\int_{-\pi}^0 e^{cx}sin(nx)dx+\frac{1}{\pi}\int_0^\pi e^{-cx}sin(nx)dx$$
Now, in the first let $u= e^{cx}$ and $dv= sin(nx)dx$. That will give an integral in $e^{cx}cos(nx)$. Do the same thing and you will get back to $e^{cx}sin(nx)$. Solve that algebraically for the integral.

However, as mathman pointed out, since this function is even and sine is odd, f(x)sin(nx) is an odd function. It's anti-derivative will be even so evaluating at $-\pi$ and $\pi$ and subtracting gives 0.

Essentially, the Fourier sine terms give the "odd part" of a function and the cosine terms give the "even part". Since this function is even, it has no sine terms. Getting coefficient 0 is exactly right.