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I was wondering whether there is a simple alogorithm of taking a power series, $\sum\limits_{n=0}^\infty a_n x^n$, and rewriting it in the form of Chebyshev expansion, i.e:

$$ \sum\limits_{n=0}^\infty b_n T_n(x) \quad \text{where} \quad T_n \; \text{is the } n\text{-th Chebyshev polynomial} $$

A quick search has not yielded anything, and I was wondering whether it is a known fact for people familiar with the subject?

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I think I found an answer to this question, so for the sake of closure I'll post it here:

For all $n\in \mathbb{N}$ one can write:

$$ x^n=\sum\limits_{m=0}^n b_m^{(n)} T_m(x) $$

An explicit formula can be found in the wikipedia page Chebyshev Polynomials. It says that:

$$ x^{2n+1}= 2^{-2n}\sum\limits_{m=0}^n \binom{2n+1}{n-m}T_{2m+1}(x) \quad \text{and} \quad x^{2n}= 2^{-2n}\sum\limits_{m=1}^n 2\binom{2n+1}{n-m}T_{2m}(x)+ 2^{-2n}\binom{2n}{n} $$

Assuming the power series converges, we can write:

$$\sum\limits_{n=0}^\infty a_nx^n= \sum\limits_{n=0}^\infty a_n \Bigg( \sum\limits_{m=0}^n b_m^{(n)} T_m(x) \Bigg)= \sum\limits_{m=0}^\infty \Bigg( \sum\limits_{n=m}^\infty a_n b_m^{(n)} \Bigg) T_m(x) =$$

$$ = \Bigg( \sum\limits_{n=0}^\infty \frac{a_{2n}}{2^{2n}}\binom{2n}{n} \Bigg) T_0(x) + \sum\limits_{m=0}^\infty \Bigg( \sum\limits_{n=m}^\infty \frac{a_{2n+1}}{2^{2n}} \binom{2n+1}{n-m} \Bigg) T_{2m+1}(x)+ \sum\limits_{m=1}^\infty 2\Bigg( \sum\limits_{n=m}^\infty \frac{a_{2n}}{2^{2n}} \binom{2n}{n-m} \Bigg) T_{2m}(x) $$

If someone does go through the time to read and finds an error, I would appreciate if someone could point it out.

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  • $\begingroup$ Don't you actually need absolute convergence of the power series as you are changing the order of summation? $\endgroup$ – emacs drives me nuts May 7 '20 at 15:54

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