4
$\begingroup$

I have the following expression: $$ \sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L_n^k(x))^2, $$ where $$ L_n^k(x)=\sum_{j=0}^n(-1)^j\binom{n+k}{n-j}\frac{x^j}{j!} $$ is the usual associated Laguerre polynomial and $k\in\mathbb N$. In particular $\int_0^{\infty}e^{-x}x^k(L^k_n(x))^2=\frac{(k+n)!}{n!}$.

I am trying to figure out a way to simplify this sum. I am not an expert on special functions, and I would appreciate some references or hints. I have been trying quite hard using a few recurrence relations and other formulas found in the literature.

What I would like to prove is the following: $$ \sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L_n^k(x))^2=e^x+P_{2n-1}(x), $$ where $P_{2n-1}(x)$ is a polynomial of degree $2n-1$ (if $n=0$, we set $P_{-1}=0)$. This is clearly true when $n=0$, and one can easily prove for small values of $n$ ($n=1,2,...$). An alternative way is to differentiate $2n$ times and prove that the resulting sum gives $e^x$, but this approach remains complicated (at least, for me).

I don't know if this is something known, I would appreciate a reference in that case.

$\endgroup$
0

3 Answers 3

8
$\begingroup$

Everything becomes simpler if add some parameters and start the sum at $k=-n$ instead of $k=0$. Note that if $k$ is a negative integer with $-n\le k \le -1$ then $L_n^k(x)$ is a polynomial in $x$ divisible by $x^{-k}$.

Then we have the formula $$\sum_{k=-n}^\infty \frac{n!}{(k+n)!}z^k L_n^k(x)L_n^k(y)=e^z\sum_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i,$$ which can be proved by expanding the right side and simplifying, using Vandermonde's theorem. Thus \begin{multline*} \sum_{k=0}^\infty \frac{n!}{(k+n)!}z^k L_n^k(x)L_n^k(y)\\ = e^z\sum_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i - \sum_{k=-n}^{-1} \frac{n!}{(k+n)!}z^k L_n^k(x)L_n^k(y). \end{multline*} If we set $z=x$ the terms in the first sum on the right with $i>0$ vanish and we get $$ \sum_{k=0}^\infty \frac{n!}{(k+n)!}x^k L_n^k(x)L_n^k(y) =e^x - \sum_{k=-n}^{-1} \frac{n!}{(k+n)!}x^k L_n^k(x)L_n^k(y), $$ so this gives an explicit formula for the OP's polynomials.

$\endgroup$
1
  • $\begingroup$ Thank you very much, very clear $\endgroup$
    – L. Proz
    Mar 14, 2022 at 7:05
4
$\begingroup$

$\newcommand{\bi}{\binom}$Let us prove the more general conjecture suggested by T. Amdeberhan.

Let \begin{equation*} \begin{aligned} &s_n(x,y) \\ &:=\sum_{k=0}^\infty\frac{n!}{(k+n)!}x^kL_n^k(x)L_n^k(y) \\ &=\sum_{k=0}^\infty\frac{n!}{(k+n)!} \sum_{j=0}^n \frac{(-1)^j}{j!}\bi{n+k}{n-j}y^j \, \sum_{i=0}^n \frac{(-1)^i}{i!}\bi{n+k}{n-i}x^{k+i} \\ &=n! \sum_{p=0}^\infty \sum_{j=0}^n y^j \, x^{p-j} \sum_{i=0}^n 1(i+j\le p)\frac{(-1)^{i+j}}{i!j!} \frac1{(p-i-j+n)!}\\ &\qquad\qquad\qquad\qquad\quad\times\bi{n+p-i-j}{n-i}\bi{n+p-i-j}{n-j}. \end{aligned} \tag{1}\label{1} \end{equation*} So, for natural $n$, after some algebra we get \begin{equation*} \frac{s_n(x,y)-s_{n-1}(x,y)}{(n-1)!} =\sum_{p=0}^\infty \sum_{j=0}^n c_{n,p,j}\, y^j x^{p-j}, \tag{2}\label{2} \end{equation*} where \begin{equation*} c_{n,p,j}:=\sum_{i=0}^n 1(i+j\le p)F(j,i) \tag{3}\label{3} \end{equation*} and \begin{equation*} F(j,i):=F_{n,p}(j,i):=\frac{(-1)^{i+j} (n p-i j) (n+p-i-j-1)!}{i! j! (n-i)! (n-j)! (p-i)! (p-j)!}; \end{equation*} note that $c_{n,p,j}=0$ if $j>p$.

Following Peter Taylor's answer, let \begin{equation*} R(j,i):=\frac{i \left(n+p+n p-i-j-i j\right)}{(j+1) \left(n p-i j\right)} \end{equation*} and then \begin{equation*} G(j,i):=R(j,i)F(j,i). \end{equation*} Then we get the Wilf–Zeilberger identity \begin{equation*} F(j+1,i)-F(j,i)=G(j,i+1)-G(j,i). \tag{4}\label{4} \end{equation*} Take now any $p\ge2n$. Then the factor $1(i+j\le p)$ in \eqref{3} is $1$ for $i,j\le n$, and hence, in view of the telescoping property provided by \eqref{4}, $c_{n,p,j}$ does not depend on $j\in\{0,\dots,n\}$: \begin{equation*} \begin{aligned} c_{n,p,j}&=\sum_{i=0}^n F(j,i) \\ &=\sum_{i=0}^n F(n,i) =\frac{(-1)^n n}{n!n!(n+q)!}\sum_{i=0}^n (-1)^i \bi ni=0. \end{aligned} \end{equation*} So, $c_{n,p,j}=0$ for $p\ge2n$.

That is, by \eqref{2}, $s_n(x,y)-s_{n-1}(x,y)$ is a polynomial (in $x,y$) of degree $\le2n-1$. Also, $s_0(x,y)=e^x$. Thus, $s_n(x,y)-e^x$ is a polynomial of degree $\le2n-1$.

Substituting here $x$ for $y$, we get the conjecture in the OP.

$\endgroup$
3
  • $\begingroup$ I thought that the sum in your question looked related to this one, and almost asked whether that was the case. $\endgroup$ Mar 11, 2022 at 21:21
  • $\begingroup$ @PeterTaylor : Yes, indeed. :-) $\endgroup$ Mar 11, 2022 at 22:05
  • $\begingroup$ Let me point out anothwe answer here: math.stackexchange.com/questions/4400266/…. Maybe it could be helpful also in order to prove the identity for $s_n(x,y)$. $\endgroup$
    – L. Proz
    Mar 14, 2022 at 7:10
3
$\begingroup$

Conjecture. It appears that this generalization holds: $$ \sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^kL_n^k(x)L_n^k(y)=e^x+P_{2n-1}(x,y), $$ where $P_{2n-1}(x,y)$ is a polynomial of degree $2n-1$.

$\endgroup$
2
  • $\begingroup$ Good observation! Indeed, the previous proof works almost literally in this most general setting. $\endgroup$ Mar 11, 2022 at 18:28
  • $\begingroup$ Glad you could concur. $\endgroup$ Mar 11, 2022 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.