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 $L_n(x) = L_n^0(x)$.

My question is:

- 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.
- 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?