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I am getting confused on how to determine the bounds for integration for this problem. My reasoning is as follows:

$ f(x) = x $ for $ 0 < x < 1 $ and $f(-x) = -x = -f(x)$ therefore $f(x)$ is odd and the cosine term of the series will cancel out. Will the $ a_0 $ term cancel out as well? I was expecting it would since my integration is from $-1/2$ to $1/2$.

I say that because, $ f(x+P) = f(x) $ taking P: period and $ P = 2L $ therefore if my interval is [0,1] then: $$ P = 2L = 1 $$ and $$ L = 1/2 $$

That way the bounds of my integral are $[-1/2,1/2]$. But in the solutions it is indicated that the bounds of integration are $[0,1]$, which I think is the right answer cause we always find the Fourier series over the entire period and in this case it would be [0,1]. What am I understanding incorrectly? You can find the problem here: http://exampleproblems.com/wiki/index.php/FS8 Thanks!

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  • $\begingroup$ looks to me that your period must be at least 2. and $$ f(x) = x \qquad \text{for } -1 < x < 1 $$ $\endgroup$ Commented Dec 7, 2017 at 18:24
  • $\begingroup$ Your assumption that $f(-x)=-x$ is not a given. What is required is $f(x)=x$ for $0<x<1$ and its periodic extension elsewhere, thus period is $1$. Bristow-Johnson always hated me, for what, I don't know. Every time I do the DFT, he said "Your program is way to slow." $\endgroup$ Commented Dec 7, 2017 at 18:40
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    $\begingroup$ A fourier series describes a periodic function. You must first extend f(x) to a periodic function and then find its fourier series, which will also then represent f(x) in [0, 1]. One way is to say $f(x) = x$ in [0, 1] and $f(x) = f(x+1)$ outside this range, which has period 1 and is neither odd nor even. You could also say $f(x) = x$ in [-1, 1] and $f(x) = f(x+2)$, the so called odd extension of f with period 2. What is the even extension of f? $\endgroup$
    – Paul
    Commented Dec 7, 2017 at 18:40

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