Kravchuk polynomials identity

Question

I am trying to solve a combinatorics problem making use of Kravchuk polynomials and I am stuck with a specific sum I cannot compute. It goes like this : \begin{equation} \sum_{m=0}^{n-1}(m+1)\begin{pmatrix}n-1\\m \end{pmatrix}K_m(x,n-1) \end{equation} This would be very easy to compute using the orthogonality relation if it were not for the $m+1$. Does anyone have a clue on how to proceed?

Answer 1

It seems I posted this question too early since two days later I found an answer! We will make use of the Kravchuk polynomials orthogonality relationship : \begin{equation} \sum_{x=0}^N \begin{pmatrix} N \\ x \end{pmatrix} K_m(x,N) K_n (x,N) = \frac{1}{\begin{pmatrix} N \\ n \end{pmatrix}} \delta_{mn} \end{equation} As well as the self duality of Kravchuk polynomials ($K_m(x,N)=K_x(n,N)$ for $n,x=0,1,...,N$) and their mirror symmetry ($K_m(x,N) = K_m(N-x,N)$)

We also need to distinguish the $n$ odd and even cases.

$n$ odd :

The $m+1$ factor in front of each term can be seen as having $m+1$ copies of the $\begin{pmatrix} n-1 \\ x \end{pmatrix} K_m(x,n-1)$ term. However from the minor symmetry and self duality we know that this is equal to the $\begin{pmatrix} n-1 \\ n-1-x \end{pmatrix} K_{n-1-m}(x,n-1)$ terms of which there are $n-m$ copies. Combining these identical terms we get $n+1$ copies total for each term, which does not depend on $m$. Therefore we can split the terms equally in two parts such that the sum becomes : \begin{equation} \sum_{m=0}^{n-1} (m+1) \begin{pmatrix} n-1 \\ m \end{pmatrix} K_m(x,n-1) = \frac{n+1}{2} \sum_{m=0}^{n-1} \begin{pmatrix} n-1 \\ m \end{pmatrix} K_m(x,n-1) \end{equation} in which we can introduce $K_m(0,n-1) = 1$ to get the orthogonality relationship \begin{equation} \frac{n+1}{2} \sum_{m=0}^{n-1} \begin{pmatrix} N \\ m \end{pmatrix} K_m(x,n-1)K_m(0,n-1)=\frac{n+1}{2}\delta_{x,0} \end{equation} (We can use this orthogonality relationship by summing over $m$ instead of $x$ due to the self-duality of Kravchuk polynomials)

$n$ even :

The beginning of the calculation is the same as what we did previously The problem here is that the term with $K_{n/2}(x,N)$ has no "dual" copies ; there are only $\frac{n+2}{2}$ copies of it. Thus we split our sum as follows : \begin{gather*} \sum_{m=0}^{n-1} (m+1) \begin{pmatrix} n-1 \\ m \end{pmatrix} K_m(x,n-1) = \frac{n+1}{2} \sum_{m=0}^{n-1} \begin{pmatrix} n-1 \\ m \end{pmatrix} K_m(x,n-1) + \frac{1}{2} K_{n/2}(x,n-1) \\ = \frac{n+1}{2} \delta_{x,0} + \frac{1}{2} K_{n/2}(x,n-1) \end{gather*}