Relation between the local maxima and the local minima for approximating the generalized Laguerre polynomial I have already asked my question in the link below:
Minima approximation for Laguerre polynomials
I have suggested to anyone to give me the approximations of the minima for the Laguerre polynomial, but someone have suggested the zero Laguerre polynomials. But i want the general formula for the approximation of the local minima for the generalized Laguerre polynomial. Since, we have the formula given in the Book "Orthogonal polynomials-Szego 1939" in page 240:
$$Max_{a > 0}e^{-x/2} x^{\alpha/2+1/4}|L_{n}^{(\alpha)}(x)| \sim n^{\alpha/2-1/12} \mbox{ for n is very large and } x \in R^{+} \mbox{ and } \alpha \mbox{ is a real number } $$
My question is: Can the local minima of the Laguerre polynomial be written as the following
$$Min_{a > 0}e^{-x/2} x^{\alpha/2+1/4}|L_{n}^{(\alpha)}(x)| \sim n^{k} \mbox{ for n is very large and } x \in R^{+} \mbox{ and } \alpha \mbox{ is a real number } $$
where $-1 \leq k \leq \alpha/2-1/12$? If so why? I have seen that because for that case $n^{k} \leq n^{\alpha/2}$. But I want someone to suggest to me a demonstration if that approximation is true? If not can we deduce another approximation?
 A: $\newcommand{\al}{\alpha}$You are citing Szegő's book incorrectly. The correct citation is this: for every real $\al$ and every real $a>0$,
\begin{equation*}
    \max_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| \sim n^{\al/2-1/12} \tag{1}\label{1}
\end{equation*}
as $n\to\infty$.
There can be no corresponding result with $\min$ in place of $\max$ -- because, for large enough $n$, the polynomial $L_n^{(\al)}$ oscillates, so that for every real $\al$, every real $a>0$, and all large enough $n$
\begin{equation*}
    \min_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| =0. \tag{2}\label{2}
\end{equation*}
Indeed, by Theorem 6.31.3 (p. 129) of G. Szegő, Orthogonal polynomials, 1939, for $\al>-1$ and the $\nu$th smallest root $x_{\nu n}$ of $L_n^{(\al)}(x)$ we have
\begin{equation*}
    x_{\nu n}>\frac{(j_\nu/2)^2}{n+(\al+1)/2}
\end{equation*}
where $\nu=1,\dots,n$ and, according to the first sentence after that theorem,
$(j_\nu/2)^2\sim\pi^2\nu^2/4$ as $\nu\to\infty$. So, for any fixed natural $k$, uniformly over  all $\nu\in\{n-k+1,\dots,n\}$ we have $x_{\nu n}\gtrsim\frac{\pi^2}4\,n$.
Therefore and because $\frac{\pi^2}4>1$, for any real $\al>-1$, any fixed natural $k$, and all large enough $n$, there are at least $k$ roots of $L_n^{(\al)}(x)$ in the interval $[n,\infty)$, and hence \eqref{2} holds.
By Rolle's theorem and the formula
\begin{equation*}
    \frac d{dx}\,[x^\al e^{-x}\,L_n^{(\al)}(x)]=(n+1)x^{\al-1} e^{-x}\,L_{n+1}^{(\al-1)}(x) 
\end{equation*}
at the top of page 287, it follows that for any fixed natural $k$, any real $\al>-1$, and all large enough $n$, there are at least $k-1$ roots of $L_{n+1}^{(\al-1)}(x)$ in the interval $[n,\infty)$. That is, for any fixed natural $k$, any real $\al>-1-1$, and all large enough $n$, there are at least $k-1$ roots of $L_n^{(\al)}(x)$ in the interval $[n-1,\infty)$. Continuing thus, we conclude that, for any fixed natural $k$, any real $\al>-1-(k-1)$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$.
That is, for any real $\al$, any fixed natural $k>-\al$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$.
Thus, we have \eqref{2} for every real $\al$, every real $a>0$, and all large enough $n$.
