Find the fourier series for the given function

$$f(x)=-x \quad \text{for } -L\le x < L, f(x+2L)=f(x)$$

this is a question from my book, and im just wondering about one thing and that is what does $f(x+2L)=f(x)$ mean? I think it got something with the period or something? im not sure what it means when it says $2L$ tho? but if it was just $f(x+2)=f(x)$ wouldnt that just mean period $2$ so $L=1$?


What's happening here is that the problem is only specifying the function $f$ on the set $[-L,L)$, and then adding the extra stipulation that $f(x+2L) = f(x)$ for all $x$. You are correct in thinking that it forces the function to be $2L$-periodic.

For example, if you want to know what $f(2L)$ is, say, then we cannot go to the formula $f(x) = -x$ because this formula is only valid for $x\in [-L,L)$. But we may use periodicity: $f(2L) = f(0)$ (by taking $x=0$ in $f(x+2L) = f(x)$), in which case $f(2L)= 0$.

The point is that we can make the definition of $f$ on the interval $[-L,L)$ and force $f$ to be periodic on the rest of $\mathbb{R}$ with the added assumption that $f(x+2L) = f(x)$.

  • $\begingroup$ Well, if we just asked for the Fourier series of the function $f$ without stating it is periodic, then the series will only work on $[-L,L)$. The condition that $f$ is periodic will ensure the Fourier series works everywhere. $\endgroup$ – Andrew May 20 '15 at 5:01

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