[SOLVED] Fourier Series

**Ian1779** · August 19th 2008, 02:11 PM

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.

Please help!!

Thanks

**Chris L T521** · August 19th 2008, 06:23 PM

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.

Please help!!

Thanks

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

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

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

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

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

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

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

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

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

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

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

I hope this answers your question.

--Chris

**Ian1779** · August 20th 2008, 02:03 AM

Thanks very much - I can see clearly now the 'missing bit' I couldn't work out.

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