Calculate the Fourier series of the function

  • Thread starter Dustinsfl
  • Start date
  • #1
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Calculate the Fourier series of the function $f$ defined on the interval [itex][\pi, -\pi][/itex] by
$$
f(\theta) =
\begin{cases} 1 & \text{if} \ 0\leq\theta\leq\pi\\
-1 & \text{if} \ -\pi < \theta < 0
\end{cases}.
$$
[itex]f[/itex] is periodic with period [itex]2\pi[/itex] and odd since [itex]f[/itex] is symmetric about the origin.
So [itex]f(-\theta) = -f(\theta)[/itex].
Let [itex]f(\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{in\theta}[/itex].
Then [itex]f(-\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{-in\theta} = \cdots + a_{-2}e^{2i\theta} + a_{-1}e^{i\theta} + a_0 + a_{1}e^{-i\theta} + a_{2}e^{-2i\theta}+\cdots[/itex]
[itex]-f(\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{-in\theta} = \cdots - a_{-2}e^{-2i\theta} - a_{-1}e^{-i\theta} - a_0 - a_{1}e^{i\theta} - a_{2}e^{2i\theta}-\cdots[/itex]

[itex]a_0 = -a_0 = 0[/itex]

I have solved many Fourier coefficients but I can't think today.

What do I need to do next?
 

Answers and Replies

  • #2
22,089
3,296
Why don't you just find the Fourier coefficients by using the usual formula?? That is

[tex]a_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx[/tex]
 
  • #3
LCKurtz
Science Advisor
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Or use the half-range sine expansion since it is an odd function.
 

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