Let $M_n(x) = x^n$ be the standard monomials. The binomial formula allows one to expand $M_n(ax+b)$ as a linear combination of $M_k(x)$, for $k \leq n$, giving

$$ M_n(ax+b) = (ax+b)^n = \sum_{k=0}^n \binom{n}{k} a^k x^k b^{n-k} = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} M_k(x). $$

Question: Do equivalent formulas exist for other polynomial sequences? I am mostly interested in the Chebyshev polynomials of the first kind, $T_n(x) = \cos(n \arccos(x))$, seeking a formula of the form

$$ T_n(ax+b) = \sum_{k=0}^{n} C(n, k, a, b) T_k(x). $$

I believe such a formula ought to exist, because one could always transform $T_n(x)$ into monomial basis, expand terms using the binomial formula, then transform back into a linear combination of $T_n(x)$'s. However, I would like to use it as a part of an applied algorithm, and thus am concerned about the computational complexity of the formula; going through a monomial basis would be an overkill.

Is anyone aware of any such formula?

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    $\begingroup$ Does $\langle T_n(ax+b), T_k(x) \rangle$ lead to anything interesting? $\endgroup$ – Vít Tuček Nov 13 '14 at 9:25
  • $\begingroup$ Sadly, I never got far with it. $\endgroup$ – ANSI C Mastah Nov 13 '14 at 20:30
  • $\begingroup$ Another shot in the dark: generating function $\endgroup$ – Vít Tuček Nov 13 '14 at 22:27

Why not use the recurrence $T_{n+1}(x) = 2 x T_n(x) - T_{n-1}(x)?$ When applying it to the transformed argument, you get

$$T_{n+1}(a x + b) = 2( a x + b) T_n(a x + b) - T_{n-1}(a x + b).$$ Now, write $$T_l(a x + b) = \sum_k C_{l, k, a, b} T_k(x).$$ This gives you a linear recurrence for the $C$s, since you can get rid of the $x,$ by writing $2 a x T_k(x) = a (T_{k+1}(x) + T_{k-1} (x)).$

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Check the paper

  • Roman, Steven M.; Rota, Gian-Carlo The umbral calculus. Advances in Math. 27 (1978), no. 2, 95–188.

for many such polynomial sequences. I do not remember a result about Chebyshev polynomials.

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  • $\begingroup$ I have looked at it and some related web pages and it seems to be talking about polynomial sequences of binomial type, for which the formula is indeed very simple: just the binomial formula with changed basis functions. Sadly, Chebyshev polynomials are not of that type: $$T_2(x+y) \neq T_2(x)T_0(y) + 2T_1(x)T_1(y)+T_0(x)T_2(y).$$ $\endgroup$ – ANSI C Mastah Nov 13 '14 at 20:25

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