第1类Chebyshev多项式定义为\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x).\end{aligned}证明
$${x^n+\frac{1}{x^n}\over2}=T_n\left(x+\frac{1}{x}\over2\right)$$
证明对$n$归纳\begin{aligned}
&\left(x+\frac{1}{x}\right)\left(x^{n+1}+\frac{1}{x^{n+1}}\right)=x^{n+2}+\frac{1}{x^{n+2}}+x^n+\frac{1}{x^n}\\t&={x+\frac1x\over2}\\
x^{n+2}+\frac{1}{x^{n+2}}\over2 & =2\left(x+\frac{1}{x}\over2\right)\left(x^{n+1}+\frac{1}{x^{n+1}}\over2\right)-\left(x^n+\frac{1}{x^n}\over2\right) \\
& =2t T_{n+1}(t)-T_n(t)=T_{n+2}(t)
\end{aligned}
注意到$\cosh x=\frac{e^x+e^{-x}}2$ 把$x$换成$e^x$ 这等价于$T_n(x)=\cosh(n\operatorname{arcosh}x)\quad{\text{ if }}~x\geq 1$ | 
