# Finding the value of pi/4 using Fourier series

1. Jun 17, 2013

### phosgene

1. The problem statement, all variables and given/known data

The function f(x) is defined by:

$f(x) = -1$ when $\pi < x <0$ and $0$ when $0<x<\pi$

Show that $\sum^{∞}_{0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$

2. Relevant equations

Fourier series for a function of period $2\pi$ $= a_{0} + \sum^{∞}_{1}a_{n}cos(nx) + b_{n}sin(nx)$

$a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}cos(nx)dx$

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

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

3. The attempt at a solution

f(x)=0 between 0 and pi, so I can ignore that interval in all of the integrals and integrate from -pi to pi.

Doing this, I get $a_{0}=-1/2$ and $a_{n}=0$.

Then I do $b_{n}=\frac{1}{\pi}\int_{-\pi}^{0}-sin(nx)dx=\frac{1}{\pi}[\frac{1-cos(n\pi)}{n}]$ since f(x)=-1 in the relevant interval.

$\frac{1}{\pi}[\frac{1-cos(n\pi)}{n}]=\frac{1}{\pi}[\frac{1-(-1)^{n}}{n}]=\frac{2}{\pi(2n+1)}$

I plug this value of $b_{n}$ into the Fourier series, I get $-1/2 + \sum_{0}^{∞}\frac{2}{\pi(2n+1)}sin(nx)$.

I can turn this into a series similar to the alternative harmonic series by letting $x=3\pi/2$, giving me $-1/2 + \sum_{0}^{∞}\frac{2(-1)^{n}}{\pi(2n+1)}$. Now the missing piece is showing that this is equal to 0, which I don't know how to do.

2. Jun 17, 2013

### clamtrox

But your original function is not defined when $x=3\pi/2$.

3. Jun 17, 2013

### phosgene

Oh, sorry, I forgot to add that $f(x+2\pi)=f(x)$

4. Jun 17, 2013

### clamtrox

Well then, what is f(3π/2)?

5. Jun 17, 2013

### phosgene

Ah, right. 0. I got caught up thinking about what sin(nx) was doing instead of f(x). Embarrassing! Thanks though.