0
$\begingroup$

I was doing Fourier series problem sets but encountered a rather surprising "problem".

The first problem stated: find the FS for f(x)=|sin(x)| for -pi < x < pi. Thus, the implied period of the function that I used for calculations later on, was 2*pi. The second didn't specify an interval. So, I used P=pi.

The actual period of |sin(x)| is just pi.

Is it because in the problem it said "-pi < x < pi" the reason one must use 2pi as a period? This changes the angular frequency (1 for Period=pi and 2 for Period=2*pi). Or there's another reason?

Thanks!

1 Answers 1

0

If I understood your question, the reason the period of the function is $\;\pi\;$ is because (using trigonometric identity)

$$\sin(x+\pi)=\sin x\cos \pi+\sin \pi\cos x=-\sin x\implies$$

$$|\sin(x+\pi)|=|-\sin x|=|\sin x|$$

  • 0
    It was more along the lines why we use 2pi the first time and 1pi the second time? Because in the solution the author uses 2pi as a period (for when x is between -pi and pi). – 2017-02-17
  • 0
    @tummath if $\;k\;$ is **the** period for some function, then *any* integer multiple of $\;k\;$ is **a** period, so the author *perhaps* wanted to work, for some reason, with $\;2\pi\;$ instead of $\;\pi\;$ – 2017-02-17