I've been studying about some properties of Laguerre polynomials and my professor came up with this problem. Prove that: $$f_m(t,x)=(-1)^m x^me^{-\frac{xt}{1-t}}=(1-x)^{m+1}\sum_{n=m}^{\infty}L_n^m(x)\frac{t^n}{n!}$$ In class we defined the Laguerre polynomials as $$L_n(x)=\sum_{k=0}^{n}(-1)^k \frac{(n!)^2}{(k!)^2(n-k)!}x^k$$ Such that the generating function is $$f(t,x)=\frac{1}{1-t}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{\infty}L_n(x)\frac{t^n}{n!}$$ And the generalized Laguerre polynomials are $$L_n^m(x)=\frac{d^m}{dx^m}L_n(x)$$ I've tried expanding the exponencial function in a series and after that expanding the new $\frac{1}{(1-t)^k}$ term in a Taylor series (just the same way as we did it with the Laguerre polynomials) but I can't find how to introduce the derivative to obtain the generalized Laguerre polynomials as well as the $(1-x)^{m+1}$ term. Could anyone with me some hint or any recommendation for other way to solve the problem? Thank you for your time!


This is wrong (e.g., it's clearly wrong for $m=0$). The $x$ outside the sum should be a $t$:

$$ \frac1{1-t}\left(-\frac t{1-t}\right)^m\mathrm e^{-\frac{xt}{1-t}}=\sum_{n=m}^\infty L_n^m(x)\frac{t^n}{n!}\;, $$

which follows directly by differentiating $f(t,x)$ $m$ times with respect to $x$.

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  • $\begingroup$ Oh thank you. It was my fault for not verifying particular cases and going directly to the general case. Now I think I caught the idea, thanks for the answer! $\endgroup$ – Luis A. Jan 12 at 9:19

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