I am a bit confused by the complex Fourier Series in general, but I am working with complex Fourier Series on the form: $$\sum_{n=-\infty}^{\infty} c_ne^{inx}$$

Where the Fourier coefficient $c_n$ is defined as: $$ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(y)e^{-iny}dy$$

I am then find the $c_0$ and the $c_n$ coefficient of the function on the interval $]-\pi, \pi]$ defined by:

$$f(x) := \begin{cases} 0, & \text{for $-\pi<x<0$} \\ sin(x), & \text{for $0\le x\le \pi $} \end{cases}$$

This is what I have done so far:

Note that $sin(x) = \frac{e^{ix} - e^-{ix}}{2i}$, so we have that

$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^{0}0(x)e^{-inx}dx+\frac{1}{2\pi}\int_{0}^{\pi}sin(x)e^{-inx}dx = \frac{1}{2\pi}\int_{0}^{\pi}(\frac{e^{ix} - e^{-ix}}{2i})e^{-inx}dx$$

What am I supposed to do here in order to find $c_0$ and $c_n$ ?


For $c_0$, you can simply do $$c_0=\frac{1}{2\pi}\int_{0}^{\pi}\sin(x)dx.$$And $$c_n=\frac{1}{2\pi}\int_{0}^{\pi}(\frac{e^{ix} - e^{-ix}}{2i})e^{-inx}dx=\frac{1}{4\pi i}\int_{0}^{\pi}e^{ix(1-n)} - e^{-ix(1+n)}dx\\=\frac{1}{4\pi i}\left[\int_{0}^{\pi}e^{ix(1-n)}dx -\int_{0}^{\pi} e^{-ix(1+n)}dx\right]$$ and integrating by parts.

The above evaluates to $$c_n=\frac{1+e^{-i\pi n}}{2\pi(1-n^2)}.$$

EDIT: Something important missing so I add some more lines.

Bare in mind that the function is periodic on $\Large (-\pi,\pi]$, $$f(x)=\sum_{n=-\infty}^{\infty} c_ne^{inx}$$where $$c_n=\begin{cases} \frac{i}{4}&n=-1\\ -\frac{i}{4}&n=1\\ \frac{1}{\pi(1-n^2)}&n=2k\,\,\text{for some integer}\,k;k\ne1,-1\\ 0&\text{otherwise} \end{cases}$$ Since $$\frac{1}{4\pi i}\left[\int_{0}^{\pi}e^{ix(1-1)}dx -\int_{0}^{\pi} e^{-ix(1+1)}dx\right]=\frac{1}{4\pi i}\left[\pi -0\right]=\frac{-i}{4}.$$

  • $\begingroup$ So for $c_0$ I have that $\frac{1}{2 \pi} \left[\int_0^\pi sin(x)dx \right] = \frac{1}{2\pi} \left[-cos(x) \right]_{0}^{\pi} = \frac{1}{2\pi} \cdot 2= \frac{1}{\pi}$ To find the $c_n$ coefficient you have $$\frac{1}{4\pi i}\left[\int_{0}^{\pi}e^{ix(1-n)}dx -\int_{0}^{\pi} e^{-ix(1+n)}dx\right]$$ you simply integrate by parts and then we have the $c_n$ coefficient? $\endgroup$ – Martin Winther Jun 6 '18 at 14:54
  • $\begingroup$ Correct. Of course you'll have to simplify the expression at the end. Also you have to separate odd cases of $n$, including $n=1,-1$ in your final answer. They're all zero. $\endgroup$ – poyea Jun 6 '18 at 14:59
  • $\begingroup$ Okay thank you for you answer! One last question, if you were to calculate $c_1$ would you calculate it by just changing $n=1$ and calculate it the same way as for $c_0$ and $c_n$? $\endgroup$ – Martin Winther Jun 6 '18 at 15:10
  • $\begingroup$ @MartinWinther Yes in general. But in this question $c_i=0\,\forall \,\text{odd}\,i$. $\endgroup$ – poyea Jun 6 '18 at 15:16
  • $\begingroup$ Hmmm I tried to calculate $c_1$ and I get that: $c_1 = -\frac{i}{4}$ and for $c_{-1} = \frac{i}{4}$ $\endgroup$ – Martin Winther Jun 6 '18 at 18:06

I suggest that you use the fact that a primitive of $e^{i(1-n)x}$ is $\frac{e^{i(n-1)x}}{i(1-n)}$ (unless $n=1$) and that a primitive of $e^{-i(n+1)x}$ is $-\frac{e^{-i(n+1)x}}{i(n+1)}$ (unless $n=-1$).

  • $\begingroup$ I am not so familiar with that the that use, however, can I use the fact that $e^{i\pi n} = (-1)^n$? $\endgroup$ – Martin Winther Jun 6 '18 at 14:09
  • $\begingroup$ Since it is true, of course you can use it. $\endgroup$ – José Carlos Santos Jun 6 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.