Chebyshev polynomials semigroup property $T_n \circ T_m = T_{nm}$

real-analysis abstract-algebra semigroups chebyshev-polynomials orthogonal-polynomials

Here is an easy proof. $T_n(x)=\cos(n\arccos(x))$. Therefore $$T_n(T_m(x))=\cos(n\arccos(\cos(m\arccos(x))))=\cos(mn(\arccos(x)))=T_{mn}(x).$$


I prefer $$(T_n\circ T_m)(\cos x)=T_n(T_m(\cos x))=T_n(\cos mx)=\cos(nmx)=T_{nm}(\cos x).$$

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