# Fourier Series Expansions

1. Apr 16, 2007

### kreil

1. The problem statement, all variables and given/known data
Find the Fourier Series Expansion for:

(a) f(x) = [pi-2x, 0 < x < pi | pi+2x, -pi < x < 0]

(b) f(x) = [0, -pi < x < 0 | sin(x), 0 < x < pi]

2. Relevant equations

$$F(x)=\frac{a_0}{2}+\Sigma_{n=1}^{\infty}[a_ncos(nx)+b_nsin(nx)]$$

$$a_0=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx$$

$$a_n= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx$$

$$b_n= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)sin(nx)dx$$

3. The attempt at a solution

For (a) my final answer was:

$$f(x)=\frac{8}{\pi}(cos(x)+\frac{cos(3x)}{9}+...+\frac{cos(nx)}{n^2})$$

and i think this is correct, but for (b) i got kind of a funny answer imo;

$$f(x)=\frac{1}{\pi}+\frac{2}{\pi}(\frac{cos(3x)}{8}+\frac{cos(5x)}{24}+...+\frac{cos(nx)}{n^2-1})$$

if someone could work out b and see if they get the same answer i would appreciate it.

Josh

2. Apr 16, 2007

### Mindscrape

Those are supposed to be piecewise functions, right?

Hmm, in my denominator (for the second one) I got something different. Here's what I got
$$f(x) = \sum_{n=0}^\infty \frac{cosnx}{\pi(1+1/n^2)}$$

Anybody else get something similar?

3. Apr 18, 2007

### kreil

well heres my work for (b) so it can be checked:

f(x)=[0, -pi < x < 0 | sin(x), 0 < x < pi]

$$a_0= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx= \frac{1}{\pi} \int_{0}^{\pi}sin(x)dx= \frac{1}{\pi}(-cos(x))|_{0}^{\pi}=\frac{2}{\pi} \implies \frac{a_0}{2}=\frac{1}{\pi}$$

$$a_n=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx= \frac{1}{\pi} \int_0^{\pi}sin(x)cos(nx)dx=\frac{1}{\pi} \frac{cos(x)cos(nx)+nsin(x)sin(nx)}{n^2-1}|_0^{\pi}=\frac{1}{\pi}(\frac{1}{n^2-1}-\frac{cos(n \pi)}{n^2-1})=\frac{1}{\pi (n^2-1)}(1-(-1)^n)$$
$$\implies a_n=\frac{1}{\pi (n^2-1)}((-1)^{n+1}+1)$$

$$b_n=0$$

$$f(x)=\frac{1}{\pi}+\frac{2}{\pi}(\frac{cos(3x)}{8} +\frac{cos(5x)}{24}+...+\frac{cos(nx)}{n^2-1})$$

Last edited: Apr 18, 2007
4. Apr 18, 2007

### Mindscrape

How did you go from the integral of sin(x) cos(nx) to what you have? A reduction formula?

The b_n's are definitely zero. I forgot to put a_0 in my solution too.

5. Apr 23, 2007

### kreil

I used the mathematica integrator at http://integrals.wolfram.com since from my experience it is pretty accurate.