Question

Prove that Pn(1) = 1 and Pn(-1) = (-1)

Answer 1

$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$

$P_n(1)=P_n(-1)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(1-1)^n=1$