Operator for Laguerre polynomial

Question

Is there any operator that could truncate Laguerre polynomial so that the polynomial is only left with the highest order term?

Answer 1

With $L_{n}(x)=\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}\frac{x^{j}}{j!}$, umbral substitution gives

$$L_{n}((.)!L_{.}(-x))=\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}\frac{j!L_{j}(-x)}{j!}=\frac{(-x)^{n}}{n!},$$

i.e., the highest order term of the Laguerre polynomial, as noted by Gian-Carlo Rota in Finite Operator Calculus. (A reflection of the finite difference operator being an involution, i.e., being its own inverse.)

Using the ordinary generating function for the Laguerre polynomials, this operation could also be represented as the differential operator

$$R=\sum_{j=0}^{\infty }L_{j}(-x)D^j_{x=0}=\frac{exp(\frac{xD_y}{1-D_y})}{1-D_y}|_{y=0}$$

acting on $L_n(x)$ or $L_n(y)$ respectively, where $D_x=\frac{\mathrm{d} }{\mathrm{d} x}$; i.e., $$R L_{n}(x)=\frac{(-x)^{n}}{n!}.$$