The Largest Root of Associated Laguerre Polynomial

Question

The Laguerre polynomial $L_n(x)$ is the solution to the Laguerre differential equation \begin{equation*} x\,y'' + (1 - x)\,y' + n\,y = 0. \end{equation*}

The associated Laguerre polynomial $L_n^\alpha(x)$ is the solution to the more general Laguerre differential equation \begin{equation*} x\,y'' + (\alpha + 1 - x)\,y' + n\,y = 0. \end{equation*}

One can easily see that $L_n(x) = L_n^0(x)$.

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My question is:

1. Let $D$ be the derivative operator with respect to $x$. Some papers mention that iterating the operator $(I − \alpha D)$ for any $\alpha > 0$ for $x^n$ generates an associated Laguerre polynomial; that is, \begin{equation*} p_k(x) = (I − \alpha D)^k x^n, \end{equation*} But I'm not sure how the Laguerre differential equation is related to this differential operator notataion.
2. I'm wondering whether we can get the upper-bound of the largest root of $L_n^\alpha(x)$. Maybe it's hard to compute it. Is there any known result on this topic?

Answer 1

I may be misunderstanding, but I don't think $p_k(x)$ is always a Laguerre polynomial. For example, consider, $$f(x)=(1-2D)x^4=x^4-8x^3,$$ then for $T_\alpha=xD^2+(1+\alpha-x)D+n$, we have, $$T_\alpha[f(x)]=4(\alpha+2)x^3-6(\alpha+3)x^2\not=0,$$ for every $\alpha\in\mathbb{R}$. I do know that if we go to Rainville - Special Functions - Page 204, then we have the Rodrigues formula, $$L_n^{(\alpha)}(x)=\frac{1}{n!}x^{-\alpha}e^xD^n e^{-x} x^\alpha x^n=\frac{1}{n!}x^{-\alpha}(1-D)^n x^\alpha x^n,$$ hence, if $\alpha=0$ then, $$L_n(x)=\frac{1}{n!}(1-D)^nx^n.$$ Maybe a manipulation of this formula will give you what you want. If you are looking for a simple estimate on the roots, then the Enestrom-Kakeya Theorem, might be of use, Rahman & Schmeisser - Analytic Theory of Polynomials - Page 255. It states polynomials with positive coefficients, $\{c_k\}$, have all their roots in the annulus, $$\left\{z\in\mathbb{C}:\min\left(\frac{c_{k-1}}{c_k}\right)\le |z| \le \max\left(\frac{c_{k-1}}{c_k}\right) \right\}.$$ Using this crude estimate on $L_n(-x)$ and just staring at a table of Laguerre polynomials, it seems, $$\{r\in\mathbb{R}:L_n(r)=0\}\subseteq (0,n^2).$$ On that same wikipedia site is another estimate (currently without reference) which is a little better than mine, $$\{r\in\mathbb{R}:L_n^{(\alpha)}(r)=0\}\subseteq \left(0,n+\alpha+(n-1)\sqrt{n+\alpha}\right].$$