Let a be a positive real number and f $2\pi$ periodic function defined by:
$f(x)= \begin{cases}0 & \text{if }-\pi\lt x\lt 0 \\ 1 &\text{if }0\le x\le a \\ 0 &\text{if }a\lt x\le \pi\end{cases}$
Find the complex form of Fourier series of $f(x)$
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Math1995
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What have you already tried to solve this problem? – J W May 29 '15 at 16:07
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Trying to find cn – Math1995 May 29 '15 at 16:10
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I get cn= (sin(na/2)/pi n )*e^(-ina/2) – Math1995 May 29 '15 at 16:33
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The complex form of the Fourier series associated to $f$ is of the form:
$$\sum_{n \in \mathbb Z} c_n e^{inx}$$
Where:
$$c_n = \begin{cases} \frac{a_n - ib_n}2 &\text{if $n \ge 1$} \\ \frac{a_0}2 &\text{if $n = 0$} \\ \frac{a_{-n} + ib_{-n}}2 &\text{if $n \le -1$} \end{cases}$$
We just have to find the $a_n$'s and $b_n$'s.
As $f = 0$ on $(-\pi, 0)$ and $(a, \pi]$, our integrals are just over $[0,a]$.
$$a_0 = \frac1{\pi} \int_0^a (1) dx = \frac{a}{\pi}$$
For $n > 0$, we have:
$$a_n = \frac{1}{\pi} \int_0^a \cos(nx)dx = \frac{\sin(na)}{n \pi}$$
$$b_n = \frac{1}{\pi} \int_0^a \sin(nx) dx = \frac{1 - \cos(na)}{n \pi}$$
Therefore:
$$c_n = \begin{cases} i\frac{e^{-ina} - 1}{2\pi n} &\text{if $n \ge 1$} \\ \frac{a}{2\pi} &\text{if $n = 0$} \\ i\frac{e^{-ina} - 1}{2\pi n} &\text{if $n \le -1$} \end{cases}$$
Therefore:
$$ f(x) \sim \frac{a}{2 \pi} +\frac{i}{2\pi} \sum_{n \in \mathbb Z^*} \left( \frac{e^{-ina} - 1}{n} \right) e^{inx} $$
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Writing $q=e^{i\tau}$ and $f(\tau)=\sum c_n q^n$, one can compute the $c_n$ directly by $c_n=\frac{1}{2n}\int_{-\pi}^\pi e^{-in\tau}f(\tau)d\tau$. There is no reason to first compute $a_n$ and $b_n$ in order to find $c_n$. – anon May 30 '15 at 02:36
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Thanks, I have found the limit of the partial sum of this series and I get 1 – Math1995 May 31 '15 at 11:45
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