# Find the fourier series for $f(x) = x$ for $0<x<1$

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!

• looks to me that your period must be at least 2. and $$f(x) = x \qquad \text{for } -1 < x < 1$$ Commented Dec 7, 2017 at 18:24
• 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." Commented Dec 7, 2017 at 18:40
• 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?
– Paul
Commented Dec 7, 2017 at 18:40