# The values of $P_n(x)$ at the zeros of $P’_n(x)$

calculuslegendre polynomialslinear algebraorthogonal-polynomialspolynomials

I recently come across a problem with respect to Legendre polynomial as follow.

For any $$n \in \mathbb{N}$$, $$P_n(x) := \frac{1}{2^n n!}\frac{{\rm d}^n (x^2-1)^n}{ {\rm d} x^n}$$ is the classical Legendre polynomial of degree $$n$$, orthogonal on $$[-1,1]$$ with respect to Lebesgue measure.

$$\alpha_1,\cdots,\alpha_{n-1}$$ are $$n-1$$ distinct roots of the polynomial $$P'_n(x)$$, where $$P'_n(x)$$ is the derivate function of $$P_n(x)$$.

I need to estimate the quantity $$\left| \prod_{k=1}^{n-1} P_n(\alpha_k) \right|$$, but I don't know how to estimate it.
I guess that it may be exponentially decreasing as $$n$$ increases.

The graphs of these polynomials (up to $$n = 5$$) are shown below (the image is from Wikipedia):

According to the graph, it is easy to see that when $$n \geq 2$$, $$\left| \prod_{k=1}^{n-1} P_n(\alpha_k) \right| \leq \left| \prod_{k=1}^{n-1} \frac{1}{2} \right| = \frac{1}{2^{n-1}}$$, but I don't know how to prove it.

In fact, there's an exact formula for $$D_n:=\prod_{k=1}^{n-1}P_n(\alpha_k)$$: $$\bbox[5pt,border:2pt solid black]{|D_n|=2^{1-n}\prod_{k=1}^{n-1}k^{3k-n-1}(n+k)^{1-k}}$$ and $$D_n=(-1)^{n(n-1)/2}|D_n|$$. This can be deduced from the case $$\lambda=\mu=0$$ of the formula $$\operatorname{disc}P_n^{(\lambda,\mu)}=2^{-n(n-1)}\prod_{k=1}^n k^{k-2n+2}(k+\lambda)^{k-1}(k+\mu)^{k-1}(n+k+\lambda+\mu)^{n-k}$$ for the discriminant of Jacobi polynomials (see G. Szegő Orthogonal polynomials, theorem $$6.71$$); indeed, using the properties of resultants and discriminants, we have $$D_n=(nA_n)^{-n}\operatorname{res}(P_n,{P_n}')=(nA_n)^{-n}(-1)^{n(n-1)/2}A_n\operatorname{disc}P_n,$$ where $$A_n=2^{-n}\binom{2n}{n}$$ is the leading coefficient of $$P_n$$ (and $$nA_n$$ is the leading coefficient of $${P_n}'$$).
Expressing $$|D_n|$$ in terms of factorials and hyperfactorials, one gets the $$n\to\infty$$ asymptotics $$|D_n|\asymp\frac{C}{n^{1/4}}\left(\frac4{n\pi}\right)^{n/2},\quad C=\frac{A^3}{ \pi^{1/2}e^{1/8}2^{1/12}}\approx 0.9911753905908846\dots$$ where $$A$$ is the Glaisher–Kinkelin constant.