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

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

  • 2
    $\begingroup$ your answer is correct except for the constant term, which should be $1/2$. $\endgroup$ – Jonathan May 23 '13 at 15:08
  • $\begingroup$ thanks, forgot to multiply by $\frac{1}{\sqrt{2}}$ $\endgroup$ – Rayhunter May 23 '13 at 15:13

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.


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