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I'm working on a problem concerning fourier, but got stuck at two points.
Here is the question:

3)
Find the fourier series of the function $$f(x) = \begin{cases} 1, & \text{if $|x|<1$ } \\ 0, & \text{if $1\leq|x|<2$ } \end{cases}$$

Added is the solution:

The solution


In the first step I dont get why they use $f(x)=0$ if $-2\leq x\leq -1$ and $f(x)=0$ if $1\leq x\leq 2$
Why smaller/bigger or $\mathbf {equal}$ to $2$ and $-2$? Since in the problem it says $|x|<2$

Then in the final step after the coefficients are calculated I don't understand how they get to this fourier series, normally you use this rule of $a_0+ \sum a_ncos(\frac{n\pi x}{L})+ b_n sin(\frac{n\pi x}{L})$ but here in the final answer there are no cosines.

Thanks in advance :)

Amaluena
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  • $[-2,2]$ is the same as $(-2,2)$ when integrating a piecewise continuous function – reuns May 28 '17 at 11:05
  • A sine is just a cosine shifted by $\frac{\pi}{2}$. Your function is even so it a sum of cosines, but you can write it as a sum of sines with suitable phase shifts if you like. This has only been done to create a clean looking infinite sum, but there really is no need for it. – Paul May 28 '17 at 11:25

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