$T_n$ 可交换

hbghlyj

\begin{aligned}
T_1(x) & =x \\
T_2(x) & =2 x^2-1 \\
T_3(x) & =4 x^3-3 x \\
& \vdots \\
T_n(x) & =2 x T_{n-1}(x)-T_{n-2}(x) .
\end{aligned}Using the defining equation above, it is easy to verify that $T_n ◦ T_m = T_m ◦ T_n$ for all $m, n ≥ 1$.

如何证明$T_n ◦ T_m = T_m ◦ T_n$呢？对n、m归纳？

hbghlyj

用$T_n(x)=\cos n\left(\cos ^{-1}(x)\right),\forall x\in[-1,1]$容易能推出$$T_n\circ T_m(x)=\cos mn\left(\cos ^{-1}(x)\right)=T_m\circ T_n(x),\forall x\in[-1,1]$$
所以$T_n\circ T_m-T_m\circ T_n$有无穷多个根，所以=0.

但是文章让使用上面的递推式来证，

设$n\le m$，对$n$归纳，当$n=1$时$T_n\circ T_m=T_m=T_m\circ T_n$.
假设对$k<n$都成立$T_k\circ T_m=T_m\circ T_k$，要证明$T_n ◦ T_m=T_m ◦ T_n$
$T_n ◦ T_m(x) =2xT_{n-1}(T_m(x))-T_{n-2}(T_m(x))=2xT_m\circ T_{n-1}(x)-T_m\circ T_{n-2}(x)=???$

然后怎么做呢