Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What are the Fourier series of $f(x)$ where $x \in [-\pi, \pi]$ defined by

$$f(n) = \begin{cases} 1, & \text{if $x \in$ [0,$\pi$)} \\ 0, & \text{if $x \in$ [$-\pi$,0)} \\ \end{cases} $$

And what are the complex Fourier series?

My result is

$$ \frac{\sqrt{2}}{2} + 2\sum_{n=0}^\infty \frac{\sin((2n+1)x)}{(2k+1)\pi}$$

and for the complex Fourier series

$$ \frac{\sqrt{2}}{2} + \sum_{n=-\infty}^\infty \frac{-ie^{inx}}{n\pi}$$

share|cite|improve this question
your answer is correct except for the constant term, which should be $1/2$. – Jonathan May 23 '13 at 15:08
thanks, forgot to multiply by $\frac{1}{\sqrt{2}}$ – Rayhunter May 23 '13 at 15:13
up vote 1 down vote accepted

As Jonathan said, your trigonometric series is correct except for the constant term and notational detail ($n$ versus $k$).

The exponential series is not correct, however. If you transform the sines to complex exponentials using $\sin t = \frac1{2i}(e^{it}-e^{-it})$, you will get only odd-indexed terms. Direct calculation confirms this: for $n\ne 0$ $$c_n = \frac{1}{2\pi}\int_0^\pi e^{-i nx}\,dx = -i \frac{1-(-1)^n}{2\pi n }$$ which is $-i/(\pi n )$ when $n$ is odd, and $0$ otherwise.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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