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I've been asked to calculate the Fourier series for $f$, however, I've been given two $f(x)$'s (see below). I know how to calculate Fourier coefficients and the Fourier series but in this case, I'm not sure what to do since there are multiple $f(x)$'s. Do I just calculate the first one? Or do I need to do both? \begin{align*} & f(x) = \frac{x}{\pi },\text{ if } - \pi \le x < \pi , \\ & f(x) = f(x + 2\pi ). \end{align*}

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That's not multiple $f(x)$s, there is just one. $f$ is a periodic function with a period of $2\pi$.

In particular:

  • on the interval $[-\pi, \pi)$, the value of $f(x)$ is $\frac{x}{\pi}$.
  • On the interval $[-3\pi, -\pi)$, the value of $f(x)$ is equal to $f(x)=f(x+2\pi)=\frac{x+2\pi}{\pi}$.
  • On the interval $[\pi, 3\pi)$, the value of $f(x)$ is (because you can show that $f(x)=f(x-2\pi))$ $f(x)=\frac{x-2\pi}{\pi}$.

In general, on the interval $[(2k-1)\pi, (2k+1)\pi)$, the value of $f(x)$ is equal to $f(x)=\frac{x-2k\pi}{\pi}$.

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  • $\begingroup$ Ok, that clears a lot up, but then what is the function who's fourier series I need to find? Is it the last f(x) you put? Can I still find its fourier series if it has 'x' and 'k' in it? $\endgroup$
    – Maximus
    Apr 19, 2021 at 13:22
  • $\begingroup$ It is the linear function equal to $-1$ at $-\pi$, and $1$ at $\pi$, reproduced periodically outside the interval $(-\pi,\pi)$ with the global formula $\;f(x)=f(x\bmod 2\pi)$. $\endgroup$
    – Bernard
    Apr 19, 2021 at 14:08
  • $\begingroup$ @Maximus I will say it again: THERE IS ONLY ONE FUNCTION. $\endgroup$
    – 5xum
    Apr 20, 2021 at 6:18

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