Legendre polynomial for $p_k(1)$

Question

Let be

$$p_k(x)=\frac{k!}{(2k)!}\frac{d^k}{dx^k}(x^2-1)^k,\quad k=0,\ldots,n\in\mathbb{N}$$

a Legendre polynomial. How can I prove that

$$p_k(1)=\frac{k!^2}{(2k)!}2^k\ ?$$

Answer 1

let $u=x-1$ so that $x^2-1=u(u+2)$ $$p_k(u+1)=\frac{k!}{(2k)!}\frac{d^k}{du^k}(u^k(u+2)^k)$$

$\frac{d^k}{du^k}(u^k(u+2)^k)$ evaluated at $u=0$ is just $k!$ times the coefficient of $u^k$ in the binomial expansion of $u^k(u+2)^k$ which is $2^k$

so $$p_k(1)= \frac{k!}{(2k)!}\cdot k!\cdot 2^k $$

Answer 2

Hint: use induction on $k$ and the recursive definition of $p_k$:

$$ p_0(x) = 1,\quad p_1(x) = x,\\ (k+1)p_{k+1}(x) = (2k+1)xp_k(x)-kp_{k-1}(x) $$