Calculating Fourier Series Expansions for Piecewise Functions

[kreil]

Homework Statement

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]

Homework 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

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

 

[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?

 

[kreil]

well here's 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})

 

[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.

 

[kreil]

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