Asymptotic equality of $\frac{1}{n!}\frac{d^n}{dx^n}(x^n(1-x)^n) $

Question

Consider the Shifted Legendre Polynomial $$\tilde P_n(x)=\frac{1}{n!}\frac{d^n}{dx^n}(x^n(1-x)^n) $$ where $n\in\mathbb{N}\cup\{0\}$

Question: What is the asymptotic equality of $\tilde P_n(x)$ as $n\to\infty$ where $0<x<1$?

By Rodrigues' formula for Legendre Polynomial $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}((x^2-1)^n)$$ then we have $\tilde P_n(x)=P_n(1-2x)$.

We have in this MSE post the following integral representation of $P_n(x)$ $$P_n(x)=\frac{1}{\pi}\int_0^\infty(x+\sqrt{x^2-1}\cos(\theta))^n d\theta $$

and as $n\to \infty$,

$$P_n(x) \sim \frac{1}{\sqrt{2\pi n}}\frac{(x+(x^2-1)^{1/2})^{n+1/2}}{(x^2-1)^{1/4}}$$ Any help will be highly appreciated. Thank you!

Answer 1

For any $0<x<1$ there is a unique $0<\alpha<\frac{\pi}{2}$ such that $x=\sin^2 \alpha$. Then $$ \widetilde P_n ( \sin ^2 \alpha ) = P_n (1 - 2\sin ^2 \alpha ) = P_n (\cos (2\alpha )). $$ Hence by the Wikipedia results, $$ \widetilde P_n (\sin ^2 \alpha ) = \sqrt {\frac{2}{{\pi n\sin (2\alpha )}}} \cos \left( {(2n + 1)\alpha - \tfrac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^{3/2} }}} \right) $$ and $$ \widetilde P_n (\sin ^2 \alpha ) = \sqrt {\frac{{2\alpha }}{{\sin (2\alpha )}}} J_0 ((2n + 1)\alpha ) + \mathcal{O}\!\left( {\frac{1}{n}} \right) $$ as $n\to+\infty$. The latter one has a removable singularity at $\alpha=0$, so allows $x=0$ as well.