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A periodic function f with period $2\pi$ is defined by

$f(t) =$

\begin{array}{ll} t^2 & 0 \leq x\leq \pi \\ 0 & \pi \leq x\leq 2 \pi \\ \end{array}

(a) What is the value of the function t = $\pi /2 $ and $t = - \pi/2$

(b) Derive a Fourier series expansion for f in its most simplified form

(c) By considering the value of the series at $t = \pi$ deduce that

$\pi = \sqrt{\sum_{n=1}^\infty (\frac{6}{n^2})} $

I have found my Fourier expansion as:

$FS = \frac{\pi^2}{6} + \sum_{n=1}^\infty ({\frac{\pi (-1)^n}{n} + \frac{2(-1)^n}{n^3 \pi} - \frac{2}{n^3 \pi}) sin(nt)} + \sum_{n=1}^{\infty}{(\frac{2(-1)^ncos(nt)}{n^2})}$

But I don't know how to continue from there, anyone have any ideas?

Thank you so much!!

user2250537
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1 Answers1

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Hint: In general one want to say that $f(t) =$ FS. In case of discontinuous functions of f this relation doesn't hold on the place of discontinuity. What must FS equal at $t = \pi$.

Once you have figured that out, plug in pi into sines and cosines and do some moving around of the terms.

Piotr Benedysiuk
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  • I dont know :( what must FS equal to? – user2250537 Dec 05 '16 at 13:31
  • I get: $\pi^2 = \sum_{n=1}^\infty (\frac{2(-1)^n}{n^2})$ – user2250537 Dec 05 '16 at 13:33
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    Not quite. On the discontinuity you'd get the average of the value just left and just right of it. So you'd get $\frac{\pi^2}{2}$ there. Please note that cosine won't simply go away for example $\cos{\pi} \neq 1$. Also you forgot another term! Redo your calculations. – Piotr Benedysiuk Dec 05 '16 at 21:53