I found the Christoffel–Darboux formula of the Laguerre polynomials on wikipedia, https://en.wikipedia.org/wiki/Laguerre_polynomials

\begin{align} K_n^{(\alpha)}(x,y) &:= \frac{1}{\Gamma(\alpha+1)} \sum_{i=0}^n \frac{L_i^{(\alpha)}(x) L_i^{(\alpha)}(y)}{{\alpha+i \choose i}}\\ &{=}\frac{1}{\Gamma(\alpha+1)} \frac{L_n^{(\alpha)}(x) L_{n+1}^{(\alpha)}(y) - L_{n+1}^{(\alpha)}(x) L_n^{(\alpha)}(y)}{\frac{x-y}{n+1} {n+\alpha \choose n}} \\ &{=}\frac{1}{\Gamma(\alpha+1)}\sum_{i=0}^n \frac{x^i}{i!} \frac{L_{n-i}^{(\alpha+i)}(x) L_{n-i}^{(\alpha+i+1)}(y)}{{\alpha+n \choose n}{n \choose i}}; \end{align}

but there is no reference for the proof of the last equation. I have tried for several days, but failed.

I used the definition of Laguerre polynomials

\begin{equation} L_n^{(\alpha)} (x) = \sum_{i=0}^n (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!} \end{equation}

and the expansions

\begin{equation}\frac{x^n}{n!}= \sum_{i=0}^n (-1)^i {n+ \alpha \choose n-i} L_i^{(\alpha)}(x),\end{equation} \begin{equation}L_n^{(\alpha)}(x)= \sum_{i=0}^n L_{n-i}^{(\alpha+i)}(y)\frac{(y-x)^i}{i!},\end{equation} \begin{equation}L_n^{(\alpha+1)}(x)= \sum_{i=0}^n L_i^{(\alpha)}(x)\end{equation}

The generating function for Laguerre polynomials is

\begin{equation}\sum_n^\infty t^n L^{(\alpha)}_n(x)= \frac{1}{(1-t)^{\alpha+1}} e^{-\frac{tx}{1-t}}.\end{equation}

Some other references are listed as follows. I am not sure if they would help.

[1] https://academic.oup.com/qjmath/article-abstract/os-11/1/18/1574925/EXPANSIONS-AND-INTEGRAL-TRANSFORMS-FOR-PRODUCTS-OF?redirectedFrom=fulltext

[2] doi.org/10.1016/j.jat.2009.07.006

Can anyone help me on the proof of it?

After some work on the expansions to compare the coefficients of $x^jy^k$, I came to the formulas as follows

(a) \begin{equation}\sum_{i=\max\{j,k\}}^{n}{i\choose j}{i+\alpha \choose i-k}\end{equation}

(b) \begin{equation} \sum_{i_1=\max\{n-j,k\}}^n (-1)^{-n+i_1} {i_1 \choose n-j}{n+\alpha+1 \choose i_1-k} \end{equation} where $j=0,1,2...,n;k=0,1,2,...,n$ and I need now to proof (a)=(b)

  • $\begingroup$ Have you checked en.m.wikipedia.org/wiki/Gradshteyn_and_Ryzhik ? Also this rep can sometimes be useful: mathoverflow.net/questions/107159/… along with the Euler transform and generalized Vandermonde identities. $\endgroup$ – Tom Copeland Jun 23 '17 at 19:10
  • $\begingroup$ @TomCopeland, I have checked the book, but found no results. I also read the pdf in the second link you mentioned, and found the generating function of L_n^(alpha)(x)L_n^(alpha)(y). It includes a hypergeometric function and I don't know how to do with it. Besides, I tried to expand the series to compare the coefficients of the x^j y^k, but stuck at the proof of the equation of two products of combinomials. $\endgroup$ – ZHONG Shane Jun 26 '17 at 14:50

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