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.
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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 $f(x)$
caverac
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