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I have to compute the fourier series of these 2 functions:

1 . $$f(t)=e^{ita}\text{ for }-\pi < t < \pi ;\qquad a \in \mathbb{R}\backslash \mathbb{Z}$$


$$f(t)=|t| \text{ for }-\pi < t < \pi$$

I start with computing the complex fourier series of $f(t)=e^{ita}$ by calculating of $c_n $ the coefficient for $f(t) \sim \sum\limits_{k=-\infty}^{\infty}c_ne^{ikt}$:

$c_n = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi}e^{iat}e^{ikt}\mathrm dt = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi}e^{it(k+a)}\mathrm dt$

putting $u:=k+a$

gives: $\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}e^{itu}\mathrm dt = \frac{1}{2\pi}\left(\frac{2 \sin(\pi u)}{u}\right) = \frac{\sin(\pi (k+a))}{\pi(k+a)}$

setup final series: $f(t)\sim \sum\limits_{k=-\infty}^{\infty} \frac{ \sin(\pi(k+a))}{\pi(k+a)}e^{ikt}$

for $t=0$ : $\displaystyle {\sum\limits_{k=-\infty}^{\infty}}\frac{\sin(\pi(k+a))}{\pi(k+a)}=1$

and for $t=\pi$ : $\displaystyle{e^{\pi i a} = \frac{\sin(\pi(k+a))}{\pi(k+a)}e^{\pi i k} }$ ??

For the second function I take the real series of the form

$f(t)\sim \frac{a_0}{2}+\sum\limits_{k=0}^{\infty}b_k \sin(kt)+a_k \cos(kt)$,

we have $f(-t)=f(t)$ which mean $b_k=0$,

so we only have to computate $ a_0= \frac{1}{\pi}\left(\int\limits_{-\pi}^{0}-t\mathrm dt + \int\limits_{0}^{\pi} t \mathrm dt \right) = \pi $;

$ a_k= \frac{1}{\pi}\left(\int\limits_{-\pi}^{0}-t\cos(kt)\mathrm dt + \int\limits_{0}^{\pi}t\cos(kt)\mathrm dt\right) = \frac{2(\pi k \sin (\pi k)+ \cos(\pi k)-1}{k^2} = \frac{2(-1)^k-2}{\pi k^2}$

which gives for final series: $f(t)\sim \pi + \sum\limits_{k=0}^{\infty}\frac{2(-1)^k -2}{\pi k^2}\cos(kt)$.

In this series, we can put $t=0$ to get : $\pi = \sum\limits_{k=0}^{\infty}\frac{2(-1)^k -2}{\pi k^2} $ ??

Thanks for every help and input!

share|cite|improve this question
Same comment as for your previous question. If you want to plug a specific value $t_0$ and assert that the Fourier series converges to $f(t_0)$ there, you need to verify that, for instance, Dirichlet theorem applies. See pointwise convergence here:… – 1015 Feb 2 '13 at 18:25
You have my thank julien!!! – booth Feb 2 '13 at 18:51

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