Prove that $(-1)^n \text{Laguerre}_n(2) \leq 1$.

laguerre-polynomials

Theorem 8.22. on Szego's Orthogonal polynomials the asymptotics of the Laguerre polynomials (called there Fejer's formula) looks like $$L_n(x)=\frac{e^{x/2}}{2^{1/4}\sqrt{\pi}\cdot n^{1/4}}cos \left ( 2\sqrt{nx}-\pi /4\right )+O(n^{-3/4}),\, \text{for }x>0.$$ Take $x=2,$ we have then that $$|L_n(2)|\leq \frac{e}{2^{1/4}\sqrt{\pi}}\left |\frac{cos(2^{3/2}\sqrt n-\pi /4)}{n^{1/4}} \right |+\left |O(n^{-3/4})\right |\leq 1.2896\cdot \frac{1}{n^{1/4}}+\frac{C}{n^{3/4}},$$ for some constant $C.$

Even if the constant is something big, you can check the result for small values of $n.$ Here is a graph using $C=2$ which I would say is big according to simulation: enter image description here

Also, this is a very good survey about the whole problem


Plot of the function

This is a graph of $(-1)^n L_{n}^{(0)}(2)$ where $0 \leq n \leq 2500$. In this range $|(-1)^n L_{n}^{(0)}(2)| \leq 1$. This may be of interest. The Digital Library of Mathematical Functions, NIST gives $| L_{n}^{(0)}(2)| \leq e$ in equation 18.14.8 as above. Proving a better bound is probably difficult.

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