Using a family of functions to find fourier series I'm given a family of functions $$T= \left \{\frac{1}{\sqrt{2\pi}},\frac{1}{\sqrt \pi} \cos n\pi, \frac{1}{\sqrt \pi} \sin n \pi: n=1,2,3,\ldots \right \} ,  $$ on the interval $[-\pi, \pi]$ orthonormal with respect to the inner product $$\langle f,g\rangle = \int_{- \pi}^{\pi}f(x) \bar{g}(x) \, dx$$
I'm asked to find a fourier series for the function $f(x)=x$ on $[- \pi, \pi]$ using the family $T$. That's my confusion the "using portion." How do I use this in the computation of fourier series? I know how to compute a fourier series given any $f(x)$ that is trivial but what role does $T$ play in my computations? Thanks for the help.
 A: In general, when we have a Hilbert space, and an orthonormal topological basis for that space, we can compute what the coefficient of a vector $v$ with any unit basis vector $e$ has to be. That is, precisely $\langle v, e \rangle$. These are the "Fourier" coeffiecients in a general sense. The entire point of Fourier analysis is that the above trigonometric functions form a basis for an appropriate Hilbert space of functions on $[-\pi, \pi]$. The usual expressions for computing Fourier coefficients are just the inner products. 
A: If the $\pi$ in the sinusoidal functions are supposed to be $x$, this is almost the same as the normal fourier series case, but it's shifted to the range $[-\pi,\pi]$ and then properly normalized for this range. The important parts are that they are orthonormal on some range with respect to some inner product. Then you can just use $g(x)\in T$ in the inner product to find the coefficients.
So finding the coefficients is very similar to the normal case, $a_n=\frac{1}{\sqrt{\pi}}\int_{-\pi}^{\pi}f(x)\cos(nx)dx$, $b_n=\frac{1}{\sqrt{\pi}}\int_{-\pi}^{\pi}f(x)\sin(nx)dx$, and $a_0=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)dx$ where $$f(x)=\frac{a_0}{\sqrt{2\pi}}+\frac{1}{\sqrt{\pi}}\sum_{n=1}^{\infty}a_n\cos(nx)+\frac{1}{\sqrt{\pi}}\sum_{n=1}^{\infty}b_n\sin(nx)$$
