Fourier series with half range 
Question What are $a_0,a_n,b_n$ equal to with range $-L\leq x \lt0$, rather than the standard $-L\leq x \leq L$?

For example:
$$f(x)=2x^2,\quad-1\leq x\leq0$$
Instead of $f(x)=2x^2,\quad-1\leq x\leq1$
$a_0 = \frac{1}{2L} \int_{-L}^L f(x) \, dx$
$a_n=\frac{1}{2L}\int_{-L}^L f(x) \cos nx \,dx$
$b_n=\frac{1}{2L} \int_{-L}^L f(x) \sin nx \,dx$

I assumed I would just take the Fourier Cosine series, e.g.
$$a_0 = \frac{1}{L}\int_0^L f(x) \, dx=\int_0^1 2x^2 \, dx=\frac23$$
$$a_n = \frac{2}{L}\int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx = 2\int_0^1 2x^2 \cos(n\pi x) \, dx=(-1)^n\frac{8}{n^2\pi^2}$$
But this doesn't appear to be right when subbed in to give:
$$f(x) = a_0+\sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{L}\right)=\frac23 + \sum_{n=1}^\infty (-1)^n\frac{8}{n^2\pi^2} \cos(n\pi x)$$
 A: Hey you have a few things that you are doing wrong. First of all the fact that the function is defined on the interval [-1,0] doesn't mean that you have to try and expand it. You can really just use the formula's for the coefficients on the range [-1,0]. The formula's bounds are between -L and L but the  bounds can be used between any two points with difference 2L which is the period. In your case the period is 1 so the integral of the function on [-L,L] is the same as the integral on the ranges [-1,0]. Another thing that you don't really seem to get is that you can't just arbitrarily calculate the cosine and sine series and expect it to equal the function. You can only say that when the function you are finding the Fourier series for is even for the cosine series and odd for the sine series. In your case the function is neither which means you have to calculate all of the coefficients. Hope that helps
A: Here is what happens to the Fourier Series of a function when the function is restricted from $[-1,1]$ to $[-1,0]$.
Multiplication of functions is equivalent to convolution of their Fourier series. The Fourier series for $[x\lt0]$ is
$$
\begin{align}
\frac12\int_{-1}^0e^{-i\pi nx}\,\mathrm{d}x
&=\left[\frac{i}{2\pi n}e^{-i\pi nx}\right]_{-1}^0\\
&=\frac{i}{2\pi n}\left(1-e^{i\pi n}\right)\\[2pt]
&=\frac{i}{2\pi n}\left(1-(-1)^n\right)\tag{1}
\end{align}
$$
for $n\ne0$ and $\frac12$ for $n=0$.
Using complex exponentials instead of trigonometric functions, we can write
$$
f(x)=\sum_{n=-\infty}^\infty c_ne^{i\pi n}\tag{2}
$$
where $c_0=a_0$ and
$$
c_n=\frac{a_n-ib_n}2\quad\text{and}\quad c_{-n}=\frac{a_n+ib_n}2\tag{3}
$$
If we extend $a_{-n}=a_n$ and $b_{-n}=-b_n$, then both sides of $(3)$ hold for all $n\ne0$.
The convolution of the coefficients gives
$$
\begin{align}
\tilde{a}_0
&=\frac12a_0+\frac{i}{\pi}\sum_{\substack{k=-\infty\\k\ne0}}^{\infty}\frac{1-(-1)^k}{2k}\frac{a_k-ib_k}2\\
&=\frac12a_0+\frac1\pi\sum_{k=1}^\infty\frac{b_{2k-1}}{2k-1}\tag{4}
\end{align}
$$
and for $n\ne0$,
$$
\frac{\tilde{a}_n-i\tilde{b}_n}2
=\frac{i}{\pi}\frac{1-(-1)^n}{2n}a_0+\frac12\frac{a_n-ib_n}2+\frac{i}{\pi}\sum_{\substack{k=-\infty\\k\ne0,n}}^{\infty}\frac{1-(-1)^{n-k}}{2(n-k)}\frac{a_k-ib_k}2\tag{5}
$$
which can be split into
$$
\begin{align}
\tilde{a}_n
&=\frac12a_n+\frac2\pi\sum_{k=1}^\infty\frac{[n-k\text{ is odd}]}{n^2-k^2}kb_k\tag{6}
\end{align}
$$
and
$$
\begin{align}
\tilde{b}_n
&=\frac12b_n-\frac{2[n\text{ is odd}]}{n\pi}a_0-\frac2\pi\sum_{k=1}^\infty\frac{[n-k\text{ is odd}]}{n^2-k^2}na_k\tag{7}
\end{align}
$$
