Fourier series with respect to orthonormal sequence Let $H$ be the space of piecewise continuous $2 \pi$-periodic functions on the real line. For $f$ and $g$ in $H$, consider the inner product $<f,g>=\frac{1}{2\pi}\int_{- \pi}^{\pi}f(x)\overline {g(x)}dx$ and the functions $e_n(x)=e^{inx}$ for integer $n$ with $-\infty<n<\infty$.
Let $f(x)=x$ for $-\pi \le x < \pi$ and assume $f$ is extended to be a $2 \pi$ periodic function. Compute the Fourier series of $f$ with respect to the orthonormal sequence $\{e_n| -\infty < n< \infty \}.$
I know how to find the fourier series but when it's with respect to  sequence, I've never done that. How does it differ? Thank you.
 A: Suppose that $(x_n),n\in\mathbb N$ is an orthonormal sequence in a Hilbert space $V$ over a field $F$. Then for every vector $x \in V$, the number $\langle x,x_n\rangle\in F$ is called the $n$-th Fourier coefficient of $x$ with respect to the orthonormal sequence $(x_n)$, and the series
$$x\sim\sum_{n\in\mathbb N}\langle x,x_n\rangle x_n$$
is called the Fourier series of $x$ with respect to the orthonormal sequence.
A: The coefficients of the series are given by $a_n = \langle f,e_n(x) \rangle$. Does this sound familiar?
EDIT:
Thats the thing. There is no $b_n$. 
If your orthonormal sequence consisted of $\sin(nx)$ and $\cos(nx)$, you have both $a_n$ and $b_n$ where your $n$  is non-negative. But here we are using $e^{inx}$ for all integers $n$. 
The big picture is that to one choice of an orthonormal basis for the space of functions is using the sines and cosines. Another choice is to use the modes of the exponential function. In the first case, the $a_n$s correspond to $<\cos,f>$ and the $b_n$ correspond to $<\sin,f>$. In our case, there is only one "type" of function, i.e. $e^{inx}$ so only one set of coefficients which im calling $a_n$ corresponds to $<e^{inx},f>$ . but these $n$s run over the integers
