战巡

令$f_n(t)=(1+t^2)g_n(t)$，易证有
\[f_{n+1}(t)=(1+t^2)f'_n(t)\]
此时换元$t=\tan(x)$，有
\[f_{n+1}(\tan(x))=\sec^2(x)f'_n(\tan(x))=\frac{d}{dx}f_n(\tan(x))\]
易证$f_1(t)=1+t^2=\sec^2(x)$，即有
\[f_n(\tan(x))=\frac{d^n}{dx^n}\sec^2(x)\]\[f_n(t)=\frac{d^n}{dx^n}\sec^2(x)|_{x=\arctan(t)}\]\[g_n(t)=\frac{1}{1+t^2}·\frac{d^n}{dx^n}\sec^2(x)|_{x=\arctan(t)}\]

战巡

回复 7# 其妙

换元$x=2t$，有
\[f_{n+1}(2t)=2tf_n(2t)-f_{n-1}(2t)\]   
加上$f_1,f_2$，显然$f_n(2t)$就是第二类切比雪夫多项式，有
\[f_n(2t)=U_n(t)\]\[f_n(2\cos(\theta))=U_n(\cos(\theta))=\frac{\sin((n+1)\theta)}{\sin(\theta)}\]