Duality and Difference Operators for Matrix Valued Discrete Polynomials on the Nonnegative Integers

Abstract

In this paper we introduce a notion of duality for matrix valued orthogonal polynomials with respect to a measure supported on the nonnegative integers. We show that the dual families are closely related to certain difference operators acting on the matrix orthogonal polynomials. These operators belong to the so called Fourier algebras, which play a key role in the construction of the families. In order to illustrate duality, we describe a family of Charlier type matrix orthogonal polynomials with explicit shift operators which allow us to find explicit formulas for three term recurrences, difference operators and squared norms. These are the essential ingredients for the construction of different dual families.

Appendix A: Miscellaneous Proofs

1.1 Block Vandermonde Determinant

This subsection will mostly involve upper triangular matrices and we will focus our attention on their diagonal parts. So when two upper triangular matrices $\mathcal {M}_1, \mathcal {M}_2$ have the same diagonal, we will write

$$\begin{aligned} \mathcal {M}_1 = \mathcal {M}_2 + \text {s.u.t.} \end{aligned}$$

where $\text {s.u.t.}$ stands for strictly upper triangular. All the matrices $\rho _i^{(\lambda )}(n)$ in Sect. 8 satisfy the conditions in Corollary A.2, but we will collect a preliminary result for a slightly simpler case first.

Lemma A.1

Consider the n-dependent upper triangular matrix of the following form

$$\begin{aligned} \mathcal {M}(n) = n \mathcal {T} + \mathcal {T}_0 +\mathrm {s.u.t.} \end{aligned}$$

(A.1)

where $\mathcal {T}$ and $\mathcal {T}_0$ are constant diagonal matrices, and the $\text {s.u.t.}$ part is allowed to depend on n. The determinant of its block Vandermonde matrix does not depend on the $\text {s.u.t.}$ part or $\mathcal {T}_0$, and is given by

$$\begin{aligned} \det \begin{pmatrix} I &{} \mathcal {M}(n_0) &{} \dots &{} \mathcal {M}(n_0)^x \\ I &{} \mathcal {M}(n_1) &{} \dots &{} \mathcal {M}(n_1)^x \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} \mathcal {M}(n_x) &{} \dots &{} \mathcal {M}(n_x)^x \end{pmatrix} = \det (\mathcal {T})^{\frac{1}{2} x(x+1)} \left( \prod _{0\le s < t \le x} (n_t-n_s) \right) ^N, \quad x \in \mathbb {N}_{>0}, \end{aligned}$$

where $(n_j)_{j=0}^x$ is a list of complex values for which $\mathcal {M}(n_j)$ is defined.

Proof

This proof will be by induction on x. On occasion we will denote the block Vandermonde as $\det \left( \mathcal {M}(n_j)^k \right) _{j,k=0}^x$ even though this a slight abuse of notation in case $\mathcal {M}(n_j)$ is singular. We apply a block analogue of an elementary row operation to get

$$\begin{aligned}&\det \begin{pmatrix} I &{} \mathcal {M}(n_0) &{} \dots &{} \mathcal {M}(n_0)^x \\ I &{} \mathcal {M}(n_1) &{} \dots &{} \mathcal {M}(n_1)^x \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} \mathcal {M}(n_x) &{} \dots &{} \mathcal {M}(n_x)^x \end{pmatrix} \\&= \det \begin{pmatrix} I &{} \mathcal {M}(n_0) &{} \dots &{} \mathcal {M}(n_0)^x \\ I &{} \mathcal {M}(n_1) &{} \dots &{} \mathcal {M}(n_1)^x \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} \mathcal {M}(n_x) &{} \dots &{} \mathcal {M}(n_x)^x \end{pmatrix} \det \begin{pmatrix} I &{} -\mathcal {M}(n_0) &{} 0 &{} \dots &{} 0 \\ 0 &{} I &{} -\mathcal {M}(n_0) &{} \dots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \dots &{} 0 &{} I &{} -\mathcal {M}(n_0) \\ 0 &{} \dots &{} \dots &{} 0 &{} I \end{pmatrix} \\&= \det \begin{pmatrix} I &{} 0 &{}\dots &{} 0 \\ I &{} (n_1 - n_0)\mathcal {T} + \text {s.u.t.} &{}\dots &{} \mathcal {M}(n_1)^{x-1}(n_1 - n_0)\mathcal {T} + \text {s.u.t.} \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} (n_x - n_0)\mathcal {T} + \text {s.u.t.} &{} \dots &{} \mathcal {M}(n_x)^{x-1}(n_x - n_0)\mathcal {T} + \text {s.u.t.} \end{pmatrix}. \end{aligned}$$

In the last step we have used that $\mathcal {M}(n_j) - \mathcal {M}(n_0) = (n_j-n_0)\mathcal {T} + \text {s.u.t}$. To proceed we notice that $\mathcal {M}(n_j)^k - \mathcal {M}(n_j)^{k-1}\mathcal {M}(n_0) = (n_j-n_0)\mathcal {T}\left( n_j\mathcal {T}+\mathcal {T}_0\right) ^{k-1} +\text {s.u.t.}$ and reduce the size of the determinent using the Schur complement

$$\begin{aligned}&= \det \begin{pmatrix} (n_1 - n_0)\mathcal {T} + \text {s.u.t.} &{}\dots &{} (n_1 - n_0)\mathcal {T}(n_1 \mathcal {T}+\mathcal {T}_0)^{x-1} + \text {s.u.t.} \\ \vdots &{} \, &{} \vdots \\ (n_x - n_0)\mathcal {T} + \text {s.u.t.} &{} \dots &{} (n_x - n_0)\mathcal {T}(n_x \mathcal {T}+\mathcal {T}_0)^{x-1} + \text {s.u.t.} \end{pmatrix} \\&=\det \left( \text {diag}((n_j-n_0)\mathcal {T} + \text {s.u.t.})_{j=1}^x \right) \det \begin{pmatrix} I &{} n_1\mathcal {T} + \text {s.u.t.} &{}\dots &{} (n_1\mathcal {T}+\mathcal {T}_0)^{x-1} + \text {s.u.t.} \\ \vdots &{} \vdots &{} \, &{}\vdots \\ I &{} n_x\mathcal {T} + \text {s.u.t.} &{}\dots &{} (n_x\mathcal {T}+\mathcal {T}_0)^{x-1} + \text {s.u.t.} \end{pmatrix} \end{aligned}$$

The block diagonal matrix in the second line is meant to have exactly the entries of the first column of the matrix in the previous line; in particular it has the same s.u.t. parts. We need this so that the first column of the second determinant is exactly just identity matrices without any $\text {s.u.t.}$ part.

So we have a reduction of the block Vandermonde to one of a smaller size

$$\begin{aligned} \det \left( \mathcal {M}(n_j)^k \right) _{j,k=0}^x = \det (\mathcal {T})^x \left( \prod _{j=1}^x (n_j-n_0) \right) ^N \det \left( \mathcal {N}(n_{j+1})^k \right) _{j,k=0}^{x-1},\qquad \end{aligned}$$

(A.2)

with $\mathcal {N}(n) = n\mathcal {T}+\mathcal {T}_0 +\text {s.u.t.}$ So $\mathcal {N}(n)$ has the same diagonal entries as $\mathcal {M}(n)$. Then since the factor in front of the block Vandermonde determinant with $\mathcal {N}(n)$ in (A.2), does not depend on the specifics of the $\text {s.u.t.}$ we can continue reducing the size of the block Vandermonde until we get the desired result. $\square$

Corollary A.2

The determinant formula in Lemma A.1 still holds if we conjugate $\mathcal {M}(n)$ in (A.1) by an invertible matrix $\mathcal {Q}$ that does not depend on n.

Proof

Let $\mathcal {R}(n)=\mathcal {Q}\mathcal {M}(n)\mathcal {Q}^{-1}$. Then it is easy to see that

$$\begin{aligned} \det \begin{pmatrix} I &{} \mathcal {R}(n_0) &{} \dots &{} \mathcal {R}(n_0)^x \\ I &{} \mathcal {R}(n_1) &{} \dots &{} \mathcal {R}(n_1)^x \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} \mathcal {R}(n_x) &{} \dots &{} \mathcal {R}(n_x)^x \end{pmatrix} = \frac{\det \left( \text {diag}(\mathcal {Q})_{j=0}^x \right) }{\det \left( \text {diag}(\mathcal {Q})_{j=0}^x \right) } \det \begin{pmatrix} I &{} \mathcal {M}(n_0) &{} \dots &{} \mathcal {M}(n_0)^x \\ I &{} \mathcal {M}(n_1) &{} \dots &{} \mathcal {M}(n_1)^x \\ \vdots &{} \vdots &{} \, &{} \vdots \\ I &{} \mathcal {M}(n_x) &{} \dots &{} \mathcal {M}(n_x)^x \end{pmatrix} \end{aligned}$$

$\square$

1.2 Proof of Proposition 8.2

In this section we prove the first part of Proposition 8.2 which states that

$$\begin{aligned} R_n^{(\lambda )}(0) = -a(I+\mathcal {A}_{n+\lambda })R_{n-1}^{(\lambda +1)}(0), \end{aligned}$$

(A.3)

using the matrix introduced in (6.23) to express $\mathcal {A}_{n+\lambda } = (N+\lambda +n-J)^{-1} J A^*.$

Proof

Recall that the entries $R_n^{(\lambda )}(0)_{jk}=\xi _{j,k,n}^{(\lambda )}$ are given explicitly in Theorem 7.9 and that $A_{j+1,j} =\frac{\sqrt{N-j}}{\sqrt{a}}$. So we start off by writing (A.3) in terms of its entries

$$\begin{aligned} \xi _{j,k,n}^{(\lambda )} = -a \xi _{j,k,n-1}^{(\lambda +1)} - \frac{j \sqrt{a}\sqrt{N-j}}{(N+\lambda +n-j)} \xi _{j+1,k,n-1}^{(\lambda +1)}. \end{aligned}$$

Note that since the $\xi _{j,k,n}^{(\lambda )}$ are given by two different expressions (for $n+j\ge N$ and for $n+j<N$), we should prove the desired entrywise recursion for both cases. But it is not necessary to prove a mixed case because the two cases actually coincide for $n+j=N$, so for any values of the parameters the three terms can always be considered as belonging to the same case.

Before specifying to either case, we compute the following expressions which will come in handy later on in the proof

$$\begin{aligned} \frac{\mathfrak {X}(j,k,n-1,\lambda +1)}{\mathfrak {X}(j,k,n,\lambda )}&= -\frac{N+\lambda +1-j}{a(\lambda +1)}, \\ \frac{\mathfrak {X}(j+1,k,n-1,\lambda +1)}{\mathfrak {X}(j,k,n,\lambda )}&= \frac{\sqrt{N-j}}{\sqrt{a}}\frac{N+\lambda +n-j}{j(\lambda +1)}. \end{aligned}$$

To derive the $n+j>N$ case we will need a fairly standard hypergeometric identity that is easy to check in general (for parameters with which the series either converge or truncate)

$$\begin{aligned}{} & {} \mathfrak {d} \left( \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b},\mathfrak {c}}{\mathfrak {d}, \mathfrak {e}};1 \right) - \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b},\mathfrak {c}}{\mathfrak {d}+1, \mathfrak {e}};1 \right) \right) \\{} & {} \quad = \mathfrak {b} \left( \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b}+1,\mathfrak {c}}{\mathfrak {d}+1, \mathfrak {e}};1 \right) - \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b},\mathfrak {c}}{\mathfrak {d}+1, \mathfrak {e}};1 \right) \right) , \end{aligned}$$

It follows from both sides being equal to

$$\begin{aligned} \sum _{s=1} \frac{(\mathfrak {a})_s(\mathfrak {b})_s (\mathfrak {c})_s s}{s! (\mathfrak {d}+1)_s (\mathfrak {e})_s}. \end{aligned}$$

When we fill in the values $\mathfrak {a}=1-k$, $\mathfrak {b}=j-N$, $\mathfrak {c}=n+\lambda +1$, $\mathfrak {d}=\lambda +1$ and $\mathfrak {e} = 1-N$ and collect some similar terms we get

$$\begin{aligned}&\,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,j-N,n+\lambda +1}{\lambda +1, 1-N};1 \right) \\&\quad = \frac{N+\lambda +1-j}{\lambda +1} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,j-N,n+\lambda +1}{\lambda +2, 1-N};1 \right) \\&\qquad - \frac{N-j}{\lambda +1} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,j-N+1,n+\lambda +1}{\lambda +2, 1-N};1 \right) . \end{aligned}$$

After multiplying this equation by $\mathfrak {X}(j,k,n,\lambda )(1-N)_{k-1}$ we get the desired result for $n+j>N$ by using the ratios of different $\mathfrak {X}$ expressed earlier in this proof.

The $n+j \le N$ case is slightly different. Note that now instead of the factor of $(1-N)_{k-1}$ we have $(1-n-j)_{k-1}$ so it is no longer a common factor in the three terms of our desired result. The necessary hypergeometric identity is somewhat less standard,

$$\begin{aligned}{} & {} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b},\mathfrak {c}}{\mathfrak {d}, \mathfrak {e}};1 \right) \\{} & {} \quad = \frac{\mathfrak {c}}{\mathfrak {d}}\frac{\mathfrak {e}-\mathfrak {a}}{\mathfrak {e}} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b}+1,\mathfrak {c}+1}{\mathfrak {d}+1, \mathfrak {e}+1};1 \right) - \frac{\mathfrak {c}-\mathfrak {d}}{\mathfrak {d}} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{\mathfrak {a},\mathfrak {b}+1,\mathfrak {c}}{\mathfrak {d}+1, \mathfrak {e}};1 \right) , \end{aligned}$$

but it can be shown to hold using the methods described in [23, Recipe 5.4] for the cases where the series either converge or truncate. We then put in the values $\mathfrak {a}=1-k$, $\mathfrak {b}=-n$, $\mathfrak {c}=N-j+\lambda +1$, $\mathfrak {d}=\lambda +1$ and $\mathfrak {e}=1-n-j$, to get

$$\begin{aligned}&\,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,-n,N-j+\lambda +1}{\lambda +1, 1-n-j};1 \right) \\&\quad = \frac{N+\lambda +1-j}{\lambda +1} \frac{n+j-k}{n+j-1} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,-n+1,N-j+\lambda +2}{\lambda +2, 2-n-j};1 \right) \\&\qquad - \frac{N-j}{\lambda +1} \,_{3}F_{2} \left( \genfrac{}{}{0.0pt}{}{1-k,-n+1,N-j+\lambda +1}{\lambda +2, 1-n-j};1 \right) . \end{aligned}$$

After multiplying this equation by $\mathfrak {X}(j,k,n,\lambda )(1-n-j)_{k-1}$ we get the desired result for $n+j \le N$ by using the ratios of different $\mathfrak {X}$ expressed earlier in this proof. $\square$

1.3 Proof of Lemma 8.13

We want to find $\mathscr {W}_R^{(\lambda )}(0)$ so that we will have the zero moment courtesy of (8.12). For this we need $(\mathcal {D}_n^{(\lambda )})_{jj}$ and $R_n^{(\lambda )}(0)_{j,k} = \xi _{j,k,n}^{(\lambda )}$. The entrywise expression is

$$\begin{aligned} \left( \mathscr {W}_R^{(\lambda )}(0)\right) _{j,k} = \sum _{n=0}^\infty \sum _{r=1}^N \frac{ \xi _{r,j,n}^{(\lambda )} \xi _{r,k,n}^{(\lambda )} }{\left( \mathcal {D}_n^{(\lambda )}\right) _{rr}}. \end{aligned}$$

Since this expression is clearly symmetric, we will take $j\ge k$ from now on. It will prove useful to have the standard weight of the scalar dual Hahn polynomials nearby c.f. [44],

$$\begin{aligned} w_{dH}(x;\gamma ,\delta ,\mathcal {N}) = \frac{(2x+\gamma +\delta +1)(\gamma +x)!(x+\gamma +\delta )!\delta !(\mathcal {N}!)^2}{x!\gamma !(\delta +x)!(\mathcal {N}-x)!(x+\gamma +\delta +\mathcal {N}+1)!}. \end{aligned}$$

Let us introduce the following shorthand to distinguish between two different expressions for $\xi$ that hold for different parameter values,

$$\begin{aligned} \xi _{r,k,n}^{(\lambda )} = \mathfrak {X}(r,k,n,\lambda )\times {\left\{ \begin{array}{ll} \alpha _{r,k,n}^{(\lambda )} &{} n+r \ge N, \\ \beta _{r,k,n}^{(\lambda )} &{} k \le n+r< N, \\ 0 &{} n+r < k. \end{array}\right. } \end{aligned}$$

The explicit expressions for $\alpha$ and $\beta$ can be found in Theorem 7.9. We then write out the double sum, which we note will not have any cross terms

$$\begin{aligned} \begin{aligned} \left( \mathscr {W}_R^{(\lambda )}(0)\right) _{j,k}&= \sum _{n=0}^\infty \sum _{r=r_0}^N \frac{\mathfrak {X}(r,j,n,\lambda ) \mathfrak {X}(r,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{rr}} \alpha _{r,j,n}^{(\lambda )} \alpha _{r,k,n}^{(\lambda )} \\&\quad + \sum _{n=0}^{N-1} \sum _{r=r_1}^{N-n-1} \frac{\mathfrak {X}(r,j,n,\lambda ) \mathfrak {X}(r,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{rr}} \beta _{r,j,n}^{(\lambda )} \beta _{r,k,n}^{(\lambda )}. \end{aligned} \end{aligned}$$

We denote $r_0 = \max (1,N-n)$ and $r_1 = \max (1,j-n)$ and whenever a sum is over the empty set, we take it to be zero. It will come in handy to introduce another summation variable $s=n+r$ and rearrange the summations

$$\begin{aligned} \begin{aligned} \left( \mathscr {W}_R^{(\lambda )}(0)\right) _{j,k}&= \sum _{s=N}^\infty \sum _{r=1}^N \frac{\mathfrak {X}(r,j,s-r,\lambda ) \mathfrak {X}(r,k,s-r,\lambda )}{\left( \mathcal {D}_{s-r}^{(\lambda )}\right) _{rr}} \alpha _{r,j,s-r}^{(\lambda )} \alpha _{r,k,s-r}^{(\lambda )} \\&\qquad + \sum _{s=j}^{N-1} \sum _{n=0}^{s-1} \frac{\mathfrak {X}(s-n,j,n,\lambda ) \mathfrak {X}(s-n,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{s-n,s-n}} \beta _{s-n,j,n}^{(\lambda )} \beta _{s-n,k,n}^{(\lambda )}. \end{aligned} \end{aligned}$$

(A.4)

The common part of the summands is (note that the rightmost factor is not a factorial)

$$\begin{aligned}{} & {} \frac{\mathfrak {X}(r,j,n,\lambda ) \mathfrak {X}(r,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{rr}}\\{} & {} \quad = \frac{2^\lambda \sqrt{(N-j)!(N-k)!}(\lambda +n)!}{e^{a} a^{\lambda +\frac{1}{2}(j+k)}n!(\lambda !)^2} \frac{a^{r+n} (N+\lambda -r)!(N+1-r+n+\lambda )}{(r-1)!(N-r)!(N+n+\lambda )!}, \end{aligned}$$

where we have used the expression of the inverse of the diagonal part of the squared norm in terms of factorials

$$\begin{aligned} \left( \mathcal {D}_n^{(\lambda )}\right) _{rr}^{-1} = e^{-a} \left( \frac{2}{a}\right) ^\lambda \frac{(r-1)!}{n! a^n} \frac{(N-r+n+\lambda )!(N+1-r+n+\lambda )!}{(\lambda +n)!(N-r+\lambda )!(N+n+\lambda )!}. \end{aligned}$$

The distinct parts of the summands are

$$\begin{aligned} \alpha _{r,j,s-r}^{(\lambda )} \alpha _{r,k,s-r}^{(\lambda )}&= (-1)^{j+k}\frac{(N-1)!^2}{(N-k)!(N-j)!} \\&\qquad \times \mathcal {R}_{j-1}\bigl (\ell (N-r); \, \lambda , s-N, N-1\bigr ) \mathcal {R}_{k-1}\bigl (\ell (N-r); \, \lambda , s-N, N-1\bigr ) \\ \beta _{s-n,j,n}^{(\lambda )} \beta _{s-n,k,n}^{(\lambda )}&= (-1)^{j+k}\frac{(s-1)!^2}{(s-k)!(s-j)!} \\&\qquad \times \mathcal {R}_{j-1}\bigl (\ell (n); \, \lambda , N-s, s-1\bigr ) \mathcal {R}_{k-1}\bigl (\ell (n); \, \lambda , N-s, s-1\bigr ). \end{aligned}$$

Fortunately we can find a corresponding dual Hahn weight in the common parts as follows

$$\begin{aligned}&\frac{\mathfrak {X}(r,j,s-r,\lambda ) \mathfrak {X}(r,k,s-r,\lambda )}{\left( \mathcal {D}_{s-r}^{(\lambda )}\right) _{rr}}\\&\quad = \frac{2^\lambda \sqrt{(N-j)!(N-k)!}}{e^a a^{\lambda -s+\frac{1}{2}(j+k)}} \frac{w_{dH}(N-r;\lambda ,s-N,N-1)}{\lambda !(s-N)!((N-1)!)^2}, \end{aligned}$$

and similarly in

$$\begin{aligned}&\frac{\mathfrak {X}(s-n,j,n,\lambda ) \mathfrak {X}(s-n,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{s-n,s-n}}\\&\quad = \frac{ 2^\lambda \sqrt{(N-j)!(N-k)!} }{e^a a^{\lambda -s+\frac{1}{2}(j+k)} } \frac{ w_{dH}(n;\lambda ,N-s,s-1) }{\lambda !(N-s)!((s-1)!)^2} . \end{aligned}$$

The first sum in (A.4) is then

$$\begin{aligned}&\sum _{s=N}^\infty \sum _{r=1}^N \frac{\mathfrak {X}(r,j,s-r,\lambda ) \mathfrak {X}(r,k,s-r,\lambda )}{\left( \mathcal {D}_{s-r}^{(\lambda )}\right) _{rr}} \alpha _{r,j,s-r}^{(\lambda )} \alpha _{r,k,s-r}^{(\lambda )} \\&\qquad = \frac{e^{-a}2^\lambda (-1)^{j+k} a^{-\lambda -\frac{1}{2}(j+k)}}{\lambda !\sqrt{(N-j)!(N-k)!}} \sum _{s=N}^\infty \frac{a^s}{(s-N)!} \sum _{r=1}^N w_{dH}(N-r;\lambda ,s-N,N-1) \\&\qquad \qquad \times \mathcal {R}_{j-1}\bigl (\ell (N-r); \, \lambda , s-N, N-1\bigr ) \mathcal {R}_{k-1}\bigl (\ell (N-r); \, \lambda , s-N, N-1\bigr ), \end{aligned}$$

which vanishes when $j\ne k$ due to the orthogonality of the dual Hahn polynomials. When we do have $j=k$ we get

$$\begin{aligned}&\frac{e^{-a}2^\lambda a^{-\lambda -j}}{\lambda !(N-j)!} \sum _{s=N}^\infty \frac{a^s}{(s-N)!} \sum _{r=1}^N w_{dH}(N-r;\lambda ,s-N,N-1) \mathcal {R}_{j-1}\bigl (\ell (N-r); \, \lambda , s-N, N-1\bigr )^2 \\&\quad = e^{-a}\left( \frac{2}{a}\right) ^\lambda \frac{(j-1)!}{(\lambda +j-1)!} \sum _{s=N}^\infty \frac{a^{s-j}}{(s-j)!}. \end{aligned}$$

The second sum in (A.4) is then

$$\begin{aligned}&\sum _{s=j}^{N} \sum _{n=0}^{s-1} \frac{\mathfrak {X}(s-n,j,n,\lambda ) \mathfrak {X}(s-n,k,n,\lambda )}{\left( \mathcal {D}_n^{(\lambda )}\right) _{s-n,s-n}} \beta _{s-n,j,n}^{(\lambda )} \beta _{s-n,k,n}^{(\lambda )} \\&\qquad = (-1)^{j+k} \sum _{s={j}}^{N} \frac{ e^{-a} a^{s-\frac{1}{2}(j+k)} }{(s-k)!(s-j)!}\left( \frac{2}{a}\right) ^\lambda \frac{ \sqrt{(N-j)!(N-k)!} }{ (N-s)!\lambda !} \sum _{n=0}^{s-1} w_{dH}(n;\lambda ,N-s,s-1) \\&\qquad \times \mathcal {R}_{j-1}\bigl (\ell (n); \, \lambda , N-s, s-1\bigr ) \mathcal {R}_{k-1}\bigl (\ell (n); \, \lambda , N-s, s-1\bigr ). \end{aligned}$$

which also vanishes for $j\ne k$ and when we do have $j=k$ we get

$$\begin{aligned}&\sum _{s={j}}^{N-1} \frac{ e^{-a} a^{s-j} }{(s-j)!^2}\left( \frac{2}{a}\right) ^\lambda \frac{ (N-j)!}{ (N-s)!\lambda !} \sum _{n=0}^{s-1} w_{dH}(n;\lambda ,N-s,s-1) \mathcal {R}_{j-1}\bigl (\ell (n); \, \lambda , N-s, s-1\bigr )^2 \\&\quad = e^{-a}\left( \frac{2}{a}\right) ^\lambda \frac{(j-1)!}{(\lambda +j-1)!} \sum _{s={j}}^{N-1} \frac{a^{s-j}}{(s-j)!}. \end{aligned}$$

The only thing that is left to do is combine the sums and perform the summation over s

$$\begin{aligned} \left( \mathscr {W}_R^{(\lambda )}(0)\right) _{jj} = \frac{e^{-a}(j-1)!}{(\lambda +j-1)!} \left( \frac{2}{a}\right) ^\lambda \sum _{s=j}^\infty \frac{a^{s-j}}{(s-j)!} = \frac{(j-1)!}{(\lambda +j-1)!} \left( \frac{2}{a}\right) ^\lambda = \left( T^{(\lambda )}\right) _{jj}^{-1}. \end{aligned}$$

The entries of $T^{(\lambda )}$ were given in (6.16). Now all that is left is to use (8.12) to arrive at the zero moment

$$\begin{aligned} \mathscr {W}_0^{(\lambda )} = (I+A^*)^{-\lambda }(T^{(\lambda )})^{-1}(I+A)^{-\lambda } = \left( W^{(\lambda )}(0)\right) ^{-1}. \end{aligned}$$