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I'm trying to understand a few properties of $T_n$, Chebychev polynomials of the first kind. That is, those of the form: $\cos(kz)=T_k(\cos(z))$. For example, $T_{2}(z)=2z^2-1$.

I don't understand why this isn't a trivial fact: for all $m,n$, show that $T_n$ commutes with $T_m$. Thinking about a concrete example, $T_2$ and $T_3$. It's clear that $(4x^3-3)(2x^2-1)=(2x^2-1)(4x^3-3)$ since multiplication of polynomial is commutative. Is there something deeper here where the commutative property is in doubt?

emka
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1 Answers1

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Commutativity here means $T_m(T_n(x))=T_n(T_m(x))$. The reason for this is that $T_m(T_n(x))=T_{mn}(x)$. The reason for that is that $$T_m(T_n(\cos y))=T_m(\cos ny)=\cos mny.$$

Angina Seng
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  • Either way, that seems like a simple result. I thought there might have been something deeper. – emka Aug 30 '17 at 16:56
  • @emka A simple result because the proof is simple? – snulty Aug 30 '17 at 17:02
  • @snulty I don't want to go that far. I just figured it would be something that involved far more computation. – emka Aug 30 '17 at 17:04
  • @Lord Shark, although I would think that just by looking at the proof that $x=\cos y$ restricts $-1 \leq x \leq 1$ for the equation $T_m(T_n(x))=T_{mn}(x)$? – snulty Aug 30 '17 at 17:08
  • @snulty Two polynomials agreeing on an infinite set are equal everywhere. – Angina Seng Aug 30 '17 at 17:09
  • Yeah and that accounts for all possible values of $m$ and $n$ then. Thats pretty handy – snulty Aug 30 '17 at 17:11