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

The first problem stated:

Find the Fourier series 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 $\pi$, but 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 $2\pi $ as a period? This changes the angular frequency ($1$ for period $\pi$ and $2$ for period $2\pi$). Or is there another reason?

Thanks!

Ѕᴀᴀᴅ
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tummath
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1 Answers1

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

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

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

DonAntonio
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  • 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). – tummath Feb 17 '17 at 09:53
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    @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 $;\pi;$ – DonAntonio Feb 17 '17 at 10:04