Suppose I have some function which is represented as converging series of Chebyshev polynomials of first kind in $[-1;1]$: $$ f(x)=\sum\limits_{n=1}^\infty a_n T_{2n}(x) $$
I need to transform this series into power series, $$ f(x)= \sum\limits_{m=1}^\infty b_m x^{2m}. $$
I have to separate terms with $x^2$ from other values in my analytical assumptions. Is it possible?
I tried to do this by substitution of explicit form of $T_{2n}(x)$ into above series $$ f(x)=\sum\limits_{n=1}^\infty a_n n\sum\limits_{k=0}^n\frac{(-1)^k(2n-k-1)!}{k!(2n-2k)!}(2x)^{2(n-k)} $$
When I "collect" all of the coefficients for $x^2$: $$ b_m = \frac{2^{2m}}{(2m)!}\sum\limits_{n=m}^\infty a_n \frac{(-1)^{n}n(n+m-1)!}{(n-m)!} $$ but this series diverge rapidly for all of my $a_n$.
I tried to do a zeta function regularization but I'm not good in this field and I don't know if regularization is formally correct or not; is it correct to chose the first Bernoulli number $B_1=-1/2$ to get $\zeta(0)=\sum n^0 = 1+1+...=-B_1=1/2$?
Typically, in a Chebychev series, the coefficients converge pretty rapidly, so you only consider a finite number of them.
So you start with
$f(x) =\sum\limits_{n=1}^N a_n T_{2n}(x) $
and your result becomes
$b_m = \frac{2^{2m}}{(2m)!}\sum\limits_{n=m}^N a_n \frac{(-1)^{n}n(n+m-1)!}{(n-m)!} $.
Note that $b_m = 0$ for $m > N$.
Also note that, if we write $b_{m, N} = \frac{2^{2m}}{(2m)!}\sum\limits_{n=m}^N a_n \frac{(-1)^{n}n(n+m-1)!}{(n-m)!} $, then $b_{m, N}-b_{m, N-1} = \frac{2^{2m}}{(2m)!} a_{N} \frac{(-1)^{N}n(N+m-1)!}{(N-m)!} = a_{N} (-1)^{N}2^{2m}n\frac{(N+m-1)!}{(2m)!(N-m)!} = a_{N} (-1)^{N}2^{2m}N\binom{N+m}{2m}/(N+m) $, so this gives an idea of how quickly the $a_n$ have to convergs to restore the original polynomial.
It also shows why the Chebychev economization is a good thing.