Decomposition of function in Chebyshev Polynomials Let's consider $f(x) = e^{ax}$. I know that there are variants how to decompose function in series. One of such is Fouries series. I found out that there is Chebyshev Polynomials. So, how to decompose this function  through Chebyshev Polynomials. I tried but a lot of calcutation had confused me. Could you descibe short solution.
Thanks.
 A: This is the general idea first: Roughly speaking, if you have a complete set of functions, say $g_n(x)$, which are orthogonal in some interval $I$, 
$$
\int_{I}{\rm d}\mu(x)~g_n^*(x)g_m(x) = \beta_n\delta_{mn} \tag{1}
$$
Then, you can expand a function $f(x)$ in the same interval as
$$
f(x) = \sum_n c_n g_n(x) \tag{2}
$$
where the expansion coefficients are calculated by multiplying both sides of Eq. (2) by $g_m^*(x)$ and integrating
\begin{eqnarray}
\int_I{\rm d}\mu(x)~g_m^*(x)f(x) &\stackrel{(2)}{=}& \int_I{\rm d}\mu(x)~g_m^*(x)\left(\sum_n c_n g_n(x)\right) \\ &=& \sum_n c_n \int_I{\rm d}\mu(x)~g_m^*(x)g_n(x) \\
&\stackrel{(1)}{=}& \sum_n c_n \beta_n\delta_{mn} = \beta_mc_m \tag{3}
\end{eqnarray}
In your case you have
$$
g_n(x) = T_n(x), ~~~ {\rm d}\mu(x) = \frac{{\rm d}x}{\sqrt{1 - x^2}}, ~~~ \beta_n = \frac{\pi}{2}(1 + \delta_{n0}), ~~~~ I = (-1,1)
$$
Putting everything together you can find
$$
c_n = \frac{2}{\pi(1 + \delta_{n0})}\int_{-1}^1{\rm d}x \frac{f(x)T_n(x)}{\sqrt{1 - x^2}} \tag{4}
$$
The challenge is now in calculating this last integral, but that obviously depends on the function $f(x)$
