# [SOLVED] Fourier Series

• Aug 19th 2008, 02:11 PM
Ian1779
[SOLVED] Fourier Series
Hi

I am having some problems trying to answer part of this question

I have attached file as I don't quite know how to upload the image... yet

I have done the first part and calculated the F(t)

I understand what happens to the series when you input t=0 as it that means cos 0 = 1 and sin 0 = 0.

Same again with the third part - the best value i think to use is t=pi as cos pi = -1 and sin pi = 0.

What i'm stuck on is how I apply this to evaluate the series.

Thanks
• Aug 19th 2008, 06:23 PM
Chris L T521
Quote:

Originally Posted by Ian1779
Hi

I am having some problems trying to answer part of this question

I have attached file as I don't quite know how to upload the image... yet

I have done the first part and calculated the F(t)

I understand what happens to the series when you input t=0 as it that means cos 0 = 1 and sin 0 = 0.

Same again with the third part - the best value i think to use is t=pi as cos pi = -1 and sin pi = 0.

What i'm stuck on is how I apply this to evaluate the series.

Thanks

Applying $\displaystyle t=0$ to the series $\displaystyle F(t)=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{\cos(2n+1)t} {(2n+1)^2}+\sum_{n=0}^{\infty}(-1)^{n}\frac{\sin(n+1)t}{n+1}$, we get

$\displaystyle F(0)=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

Since $\displaystyle F(0)=0$, we see that $\displaystyle 0=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$.

This implies that $\displaystyle \frac{\pi}{4}=\frac{2}{\pi}\sum_{n=0}^{\infty}\fra c{1}{(2n+1)^2}\implies\color{red}\boxed{\sum_{n=0} ^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$

So we have shown that $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2} {8}$ evaluating the series about $\displaystyle t=0$.

$\displaystyle \mathbb{Q.E.D.}$

Now we need to find the value of $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$.

The appropriate t value here would be $\displaystyle t=\frac{\pi}{2}$.

The nice thing about the value $\displaystyle t=\frac{\pi}{2}$ is that the cosine series disappears and we are left with $\displaystyle F\left(\frac{\pi}{2}\right)=\frac{\pi}{4}+\sum_{n= 0}^{\infty}\frac{(-1)^n}{2n+1}$

I'm not 100% sure why this is the case [maybe someone else can explain why], but if we let $\displaystyle F\left(\frac{\pi}{2}\right)=\frac{\pi}{2}$, then $\displaystyle \frac{\pi}{2}=\frac{\pi}{4}+\sum_{n=0}^{\infty}\fr ac{(-1)^n}{2n+1}\implies\color{red}\boxed{\sum_{n=0}^{\ infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}}$

$\displaystyle \mathbb{Q.E.D.}$