多项式$P_n$的恒等式

hbghlyj

对$n = 0, 1, 2, 3...,$ 设$P_n(x)=\dfrac{1}{2^{n} n !} \dfrac{d^{n}}{d x^{n}}\left(x^{2}-1\right)^{n}$
求证$$\sum_{k=0}^n(-1)^k e^{(n-2k)i\theta}=\sum _{\ell =0}^{n}P_{\ell }(\cos \theta )P_{n-\ell }(\cos \theta ).$$

hbghlyj

令$x=\cos\theta$, 左边变成:
$$\sum_{k=0}^n(-1)^k e^{(n-2k)i\theta}=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k2\cos((n-2k)\theta)\quad\big\{-(-1)^k\text{ if }2|n\big\}$$
$P_n$代入右边,$$\sum _{\ell =0}^{n}P_{\ell }(x)P_{n-\ell }(x)=\frac1{2^{2n}n!}\sum _{\ell =0}^{n}\binom n\ell \frac{d^\ell}{d x^\ell}\left(x^{2}-1\right)^\ell\frac{d^{n-\ell}}{d x^{n-\ell}}\left(x^{2}-1\right)^{n-\ell}$$
然后如何处理