The classical Christoffel-Darboux identity for Hermite polynomials reads $$\sum_{k=0}^n\frac{H_k(x)H_k(y)}{2^k k!}=\frac{1}{2^{n+1} n!}\frac{H_{n+1}(x)H_n(y)-H_n(x)H_{n+1}(y)}{x-y}.$$ I am interested in a similar identity, where one of the indices is shifted by one, explicitly $$F_n(x,y):=\sum_{k=0}^n\frac{H_{k+1}(x)H_k(y)}{2^{k+1} (k+1)!}=x\sum_{k=0}^n\frac{1}{k+1}\frac{H_{k}(x)H_k(y)}{2^{k} k!}-\sum_{k=0}^n\frac{k}{k+1}\frac{H_{k-1}(x)H_k(y)}{2^{k} k!},$$ where the recursion relation $H_{k+1}(x)=2xH_k(x)-2kH_{k-1}(x)$ was used. In the first term Christoffel-Darboux cannot be applied due to the $1/(k+1)$ prefactor and the second term is not quite $F_{n-1}(y,x)$ due to the $k/(k+1)$ prefactor.

Is there any hope of finding some closed expression for $F_n$?

Edit: By differentiating the classical identity in $y$, one finds \begin{align*}\sum _{k=0}^n \frac{ H_{k+1}(x) H_k(y)}{2^k k!}=\frac{1}{2^{n+2}(n+1)!}&\left(2\frac{ (n+2) H_{n+1}(x) H_{n+1}(y)- (n+1) H_{n+2}(x) H_n(y)}{y-x}\\+\frac{H_{n+2}(x) H_{n+1}(y)-H_{n+1}(x) H_{n+2}(y)}{(y-x)^2}\right)\end{align*} This is quite close to what I need, but the terms in the sum are off by a factor of $k/(k+1)$. Any ideas?

Edit 2: Taking the antiderivative of the classical identity in $x$ would exactly give what I am interested in. But I am not sure there is much hope in finding an explicit formula for an antiderivative of $$\frac{H_{n+1}(x)H_n(y)-H_n(x)H_{n+1}(y)}{x-y}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.