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.
Best Answer
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.