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

J. M. ain't a mathematician
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Rob
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    Well, you could reverse the polynomial ($x^n L_n(1/x)$), differentiate the appropriate number of times, and then multiply the constant you get with $x^n$... why do you want to do this, anyway? – J. M. ain't a mathematician Jul 10 '12 at 14:39
  • The variable of Laguerre polynomial should not be changed. The operator should act on the polynomial and eliminate any term other than the highest order term. – Rob Jul 10 '12 at 14:47

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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., $$R L_{n}(x)=\frac{(-x)^{n}}{n!}.$$

Tom Copeland
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