text:
\begin{document}
\title{On variation of zeros of classical discrete orthogonal polynomials}
\author{K. Castillo}
\ead{kenier@mat.uc.pt}
\address{CMUC, Department of Mathematics, University of Coimbra, 3001-501 Coimbra, Portugal}

\author{F. R. Rafaeli}
\ead{rafaeli@ufu.br}
\address{FAMAT-UFU, Department of Mathematics, University of Uberl\^andia, 38408-100 Uber{l\^a}ndia, Minas Gerais, Brazil}

\author{A. Suzuki}
\ead{alexandresuzuki@ufu.br}
\address{FAMAT-UFU, Department of Mathematics, University of Uberl\^andia, 38408-100 Uber{l\^a}ndia, Minas Gerais, Brazil}

\date{\today}

\begin{keyword} Classical discrete orthogonal polynomials \sep linear grid \sep quadratic grid \sep q-linear grid \sep q-quadratic grid \sep discrete Stieltjes theorem \sep monotonicity \sep zeros.
\MSC[2010]{30C15, 05A30, 33C45, 33D15}
\end{keyword}

\begin{abstract}
The purpose of this note is to establish, from the hypergeometric-type difference equation introduced by Nikiforov and Uvarov, new tractable sufficient conditions for the monotonicity with respect to a real parameter of zeros of classical discrete orthogonal polynomials. This result allows one to carry out a systematic study of the monotonicity of zeros of classical orthogonal polynomials on linear, quadratic, q-linear, and q-quadratic grids. In particular, we analyze in a simple and unified way the monotonicity of the zeros of Hahn, Charlier, Krawtchouk, Meixner, Racah, dual Hahn, q-Meixner, quantum q-Krawtchouk, q-Krawtchouk, affine q-Krawtchouk, q-Charlier, Al-Salam-Carlitz, q-Hahn, little q-Jacobi, little q-Laguerre/Wall, q-Bessel, q-Racah and dual q-Hahn polynomials.
 \end{abstract}
\maketitle
\section{Introduction}\label{intro}
The properties of {\em Jacobi polynomials}, $P_n^{(\alpha, \beta)}$ $(n=1,2,\dots; \alpha>-1, \beta>-1)$, may well consult in \cite[Chapter IV]{S75}; they are orthogonal on $[-1,1]$ with the weight function $(1-X)^\alpha(1+X)^\beta$, and
$$
y=P_n^{(\alpha, \beta)}(X)=\pFq{2}{1}{-n, n+\alpha+\beta+1}{1+\alpha}{\frac{1-X}{2}}
$$
satisfy the homogeneous differential equation of second order
\begin{align}\label{edo}
a \,y''+ b \, y'+c\,y=0,
\end{align}
where $a=a(X)=-X^2+1$ and $b=b(X; \alpha, \beta)= -(\alpha+\beta+2)X-\alpha+\beta$. (Having the explicit expression of $P_n^{(\alpha, \beta)}$, the parameter $c$ does not play any interesting role and it can be easily calculated.) Recall that the hypergeometric function ${}_{i}F_{j}$ is formally defined by the series (see \cite[(2.1.2)]{AAR99})
$$
\pFq{i}{j}{\alpha_1,\dots, \alpha_i}{\beta_1\dots, \beta_j}{X}=\sum_{k=0}^\infty \frac{(\alpha_1,\dots, \alpha_i)_k}{(\beta_1\dots, \beta_j)_k}\frac{X^k}{k!},
$$
where
$$
(\alpha_1,\dots, \alpha_i)_k=(\alpha_1)_k\cdots(\alpha_i)_k, \quad (\alpha_1)_k=\prod_{j=1}^k (\alpha_1+j-1);
$$
by convention, the empty product is $1$. From \eqref{edo}, Stieltjes proves an important statement concerning the dependence of the zeros of Jacobi polynomials on the parameters $\alpha$ and $\beta$  (see \cite{S87}). Soon after his work has been accepted for publication, Stieltjes implicitly acknowledges, in a note added at the end of the manuscript itself and also in a letter of February 3, 1887 to Hermite (see \cite[Lettre 106]{SH}), that the monotonicity of the zeros of Jacobi polynomials was previously proved by A. Markov in \cite{M86}. However, in Stieltjes' work, {\em ut in multis aliis rebus}, the ``How” is more important than the ``What” and, as he wrote to Hermite, {\em ``la d$\acute{e}$monstration que j'ai d$\acute{e}$velopp$\acute{e}$e pour les Acta Mathematica est diff$\acute{e}$rente de celle de M. Markoff''}. Historically, Stieltjes proves that given a positive definite real symmetric matrix with non-positive off-diagonal elements\footnote{These matrices are now known as Stieltjes' matrices (see \cite[Definition 3.4]{V62}).}, its inverse is also positive definite.  As a consequence, putting aside a clever manipulation of the differential equation \eqref{edo}, he shows that, since
\begin{align}\label{s1}
\frac ba=\frac{-(\alpha+\beta+2)X-\alpha+\beta}{-X^2+1}
\end{align}
 is a strictly decreasing function of $X\in (-1,1)$, and
\begin{align}\label{s2}
\frac{\partial}{\partial \alpha}\left(\frac ba\right)=\frac{1}{X-1}<0, \quad \frac{\partial}{\partial \beta}\left(\frac ba\right)=\frac{1}{X+1}>0,
\end{align}
for each $X\in (-1,1)$, the zeros of Jacobi polynomials are strictly decreasing functions of $\alpha$ on $(-1, \infty)$ and strictly increasing functions of $\beta$ on $(-1, \infty)$ (see \cite[(6)-(7)]{S87}). The proof of Markov is entirely different from that of Stieltjes and is based on the weight function. While it is true that Stieltjes worked directly with Jacobi polynomials ---and Markov proves his result through a general theorem---, his argument furnishes similar results for a more general differential equation (see \cite[Section $6.22$]{S75}). Stieltjes himself considered the ultraspherical case $\alpha=\beta$ and Szeg\H{o} noted that {\em ``the same method applies to Laguerre polynomials''} (see \cite[p. 123]{S75}). In particular, by definition, the {\em old classical orthogonal polynomials on the real line} (Hermite, Jacobi, and Laguerre)\footnote{We write ``classical orthogonal polynomials on the real line" rather than simply  ``classical orthogonal polynomials" because, for instance, from the algebraic point of view of Maroni (see \cite{M91}), the Jacobi polynomials exist and are ``classical'' even when $-\alpha, -\beta, -\alpha-\beta+1\not\in \mathbb{N}$  (see \cite[Chapters $8$ and $9$]{P18} for a recent survey on the subject). Moreover, the Bessel polynomials are classical in the same sense as the other three systems. As Maroni says, ``comme dans le roman d'Alexandre Dumas, les trois mousquetaires \'etaient quatre en r\'ealité''.} are solutions of a differential equation of the same type of \eqref{edo} (see \cite[Section 4.2]{Si15O}). However, only Jacobi and Laguerre polynomials depend on a real parameter and, therefore, the Stieltjes result is no longer applicable in this framework.

From a truly practical point of view ---for example in Physics, which was traditionally the birthplace of some of the most beautiful families (see \cite{VK91})---, it rarely will require more than the classical orthogonal polynomials. For classical (discrete) orthogonal polynomials on a uniform grid \footnote{A non-empty (totally) ordered set of equidistant (respectively, non-equidistant) points is called uniform (respectively, non-uniform)  grid, which in turn is an elementary example of lattice.} (Charlier, Krawtchouk, Hahn, and Meixner polynomials), Markov's theorem can be used (see \cite[Chapter $7$]{I05}) and, of course, for other discrete families \footnote{For a ``continuous'' case as the Askey-Wilson polynomials, Askey and Wilson used a consequence of Markov's theorem, which goes back to Szeg\H{o} (see \cite[Theorem 6.12.2]{S75}), to study the monotonicity of zeros of these polynomials (see \cite[Section $7$]{AW85}).}. Here we purpose an alternative approach, establishing a bridge between two works separated in time by almost one century, on one hand the work of Stieltjes and, on the other hand, the work \cite{NU83} by Nikiforov and Uvarov. This allows one to carry out a systematic study of the monotonicity of zeros of classical orthogonal polynomials on linear, quadratic, q-linear, and q-quadratic grids. Indeed, the purpose of this note is to prove that under suitable regularity conditions the Stieltjes result for Jacobi polynomials remains valid if we replace the differential equation \eqref{edo}, with $a$ and $b$ arbitrary polynomials of degree at most $2$ and $1$, respectively, and $c$ an arbitrary constant (from now on, whenever we refer to \eqref{edo}, we are assuming these conditions), by the following difference equation introduced in \cite[(5)]{NU83} (see also \cite[p. $127$]{NU84} and \cite[p. $71$]{NSU85}):
\begin{align}\label{edd}
a(X)\frac{\Delta}{\Delta x\left(s-1/2\right)}
\left(\frac{\nabla y(X)}{\nabla X}\right)
+\frac{b(X)}{2}\left(\frac{\Delta y(X)}{\Delta X}+\frac{\nabla y(X)}{\nabla X}\right)+c\,y(X)=0,
\end{align}
or, equivalently,
\begin{align}\label{edd2}
a(s)\frac{\Delta}{\Delta x\left(s-1/2\right)}
\left(\frac{\nabla y(X)}{\nabla X}\right)
+b(X)
\frac{\Delta y(X)}{\Delta X}+c\, y(X)=0,
\end{align}
where
\begin{align*}
a(s)=a(X)-\frac{1}{2}\,b(X)\Delta x\left(s-\frac12\right),
\end{align*}
$X=x(s)$\footnote{In first edition of ``{S}pecial {F}unctions of {M}athematical {P}hysics'' (see \cite{NU78}) the author only consider the case $x(s)=s$. The second edition (see \cite{NU84}) was significantly enriched with the equation \eqref{edd}; although A. A. Samarskii's preface is the same in both editions.}  defines a class of grids with, generally nonuniform, step-size $\Delta X=\Delta x(s)=x(s+1)-x(s)$ and $\nabla X=\nabla x(s)=\Delta x(s-1)$. (By abuse of notation, we use the same letter $a$ for the function $a(s)$ and the polynomial $a(X)$.) In what follows, we assume that $x$ is a real-valued function defined on an interval of the real line. Any solution of \eqref{edd} can be brought in correspondence with the solution of \eqref{edo} by replacing $s$ by $s/h$ and then taking limit $h \to 0$, whenever it exists. It is important to highlight that \eqref{edd} has polynomial solutions, in $X$, whose difference-derivatives satisfy equations of the same kind if and only if, for $q\neq1$ fixed, $x$ is a linear, quadratic, q-linear, or q-quadratic grid of the form
 \begin{align*}
x(s)=
\left\{
\begin{array}{ll}
C_1s^2+C_2s,\\[7pt]
C_3q^{-s}+C_4q^s,
\end{array}
\right.
\end{align*}
where $(C_1,C_2)\neq(0,0)$ and $(C_3,C_4)\neq(0,0)$ (see \cite[(1.68)]{ARS95}). The grids that depend on ``q'' are called q-linear if $C_3$ or $C_4$ is zero; otherwise it is q-quadratic. By using linear transformations (see \cite[(3.4.1)]{NSU}) we can reduce the expressions for the grids to simpler forms. In what follows, we assume that the grid $x$ takes on the following canonical forms (see \cite[p. 74]{NSU}):
 \begin{align}\label{grid}
x(s)=
\left\{
\begin{array}{lll}
s & \qquad \qquad (\mathrm{I})\\[7pt]
s(s+1) & \qquad \qquad (\mathrm{II})\\[7pt]
q^{s} \qquad \quad \quad \ \ \ \ \ \ \ (q>1) & \qquad \qquad (\mathrm{III})\\[7pt]
\dps\frac12(q^s-q^{-s}) \qquad (q>1) & \qquad \qquad (\mathrm{IV}) \\[7pt]
\dps\frac12(q^s+q^{-s}) \qquad (q>1) & \qquad \qquad (\mathrm{V})\\[7pt]
\dps\frac12(q^s+q^{-s}) \qquad \, (q=e^{2i\theta},\, 0<\theta<\pi/2). & \qquad \qquad (\mathrm{VI})
\end{array}
\right.
\end{align}
Under the above notation, we adopt the following definition of classical discrete orthogonal polynomials on the real line (COPRL), which is enough for our purpose:

\begin{definition}\label{def1}
Fix $\mathrm{a}\in\mathbb{R}\cup\{-\infty\}$ and $N\in\mathbb{N}\cup \{\infty\}$ and define $\mathrm{b}=\mathrm{a}+N$. Fix $q$ and let $X=x(s)$ be a real-valued function given by \eqref{grid}, where the variable $s$ ranges over the finite interval $[\mathrm{a}, \mathrm{b}]$ or the infinity interval $[\mathrm{a}, \infty)$. A sequence of polynomials, $(P_n(X))_{n=0}^{N-1}$,  is said to be sequence of classical discrete orthogonal polynomials on the set $\{x(\mathrm{a}), x(\mathrm{a+1}), \dots, x(\mathrm{b-1})\}$ or, simply, COPRL if $:$\\
\begin{enumerate}
\item  $P_n$ satisfy \eqref{edd}, $x$ being a strictly  monotone function on $[\mathrm{a}, \mathrm{b}]$ or $[\mathrm{a}, \infty)$ given, up to a linear transformation, by \eqref{grid};\\[1pt]
\item   a positive weight function $\omega$ satisfying the boundary conditions \footnote{If $\mathrm{a}$ and $\mathrm{b}$ are finite, \eqref{bc} can be written in the form $\omega(\mathrm{a})a(\mathrm{a})=0$ and $\omega(\mathrm{b})a(\mathrm{b})=0$. If $\mathrm{a}=-\infty$ and/or $\mathrm{b}=\infty$, then \eqref{bc} must hold for each $k$ in the limiting sense.}
\begin{align}\label{bc}
\omega(s) a(s) x^k\left.\left(s-\frac12\right)\right|_{\mathrm{a}, \mathrm{b}}=0 \quad (k=0,1,\dots)
\end{align}
exists;
\item the difference equation
\begin{align}\label{pearson}
\frac{\Delta}{\Delta x \left(s-\dps\frac12 \right)}\big(\omega(s)a(s)\big)=\omega(s)b(X)
\end{align}
holds.
\end{enumerate}
\end{definition}

Subsequently, when we say that a certain polynomial is a COPRL, we are assuming the definition and notation given in Definition \ref{def1}. From \eqref{edd2}, \eqref{bc}, and \eqref{pearson}, we conclude that the COPRL satisfy the orthogonality condition  (see \cite[(3.3.4)]{NSU})
\begin{align}\label{orth}
\sum_{s=\mathrm{a}}^{\mathrm{b}-1} P_m(X)P_n(X)\omega(s)\Delta x\left(s-\frac12\right)=0 \quad (m\not=n).
\end{align}
The polynomials $(P_n(X))_{n=0}^{N-1}$ given in Definition \ref{def1} are called simply discrete orthogonal polynomials, in $X$, on the set $\{x(\mathrm{a}), x(\mathrm{a+1}), \dots, x(\mathrm{b-1})\}$ with respect to a positive weight function $\omega$ if they satisfy the relation \eqref{orth} instead of the requirements $\mathrm{i})-\mathrm{iii})$. Of course, COPRL are a special case of discrete orthogonal polynomials.
From \eqref{orth}, we can see that the zeros of discrete orthogonal polynomials on $\{x(\mathrm{a}), x(\mathrm{a+1}), \dots, x(\mathrm{b-1})\}$ are real and distinct and are located in $(\min\{x(\mathrm{a}), x(\mathrm{b-1})\}, \max\{x(\mathrm{a}), x(\mathrm{b-1})\})$ (see \cite[Theorem 3.3.1]{S75}). In concluding this section we remark that it is possible to obtain a series representation of COPRL (see \cite[(4.19)]{ARS95} and \cite[Section 3]{NU93}):
\begin{align}\label{explicit}
&P_n(X)=(-1)^n \gamma_n \sum_{j=0}^n \dps\binom{n}{j}_q \big(-q^{-n/2}\big)^j\frac{\Delta x\big(s-(n-1)/2+j\big)}{\dps \prod_{k=0}^n \Delta x\big(s+(j-k+1)/2\big)}\\[7pt]
\nonumber&\times  \prod_{k=0}^n a(s-n+j+k)\prod_{l=0}^j \frac{a(x(s+l-1))+1/2\, b(x(s+l-1))\Delta x(s+l-1/2)}{a(x(s+l))-1/2\, b(x(s+l)) \Delta x(s+l+1/2)},
\end{align}
where $\gamma_n$ is a constant and
$$
\binom{n}{j}_q=\frac{(q; q)_n}{(q; q)_j (q; q)_{n-j}}.
$$
Furthermore (see \cite{NU93}), COPRL represent special cases of hypergeometric series or q-hypergeometric series, the latter formally defined by (see \cite[(10.9.4)]{AAR99})
$$
\pPq{i}{j}{\alpha_1,\dots, \alpha_i}{\beta_1\dots, \beta_j}{q, X}=\sum_{k=0}^\infty \frac{(\alpha_1,\dots, \alpha_i; q)_k}{(\beta_1\dots, \beta_j; q)_k}(-1)^{(1-i+j)k\dps\binom{k}{2}}\frac{X^k}{(q; q)_k}.
$$

The outline of this note is as follows. In Section \ref{ext} we present an extension of the Stieltjes work in the framework of COPRL. In Section \ref{app} we study the variation of zeros of some families of COPRL according to the type of underlying grid.
\section{Main results}\label{ext}

Unless otherwise stated we assume that $a$, $b$, and $c$ appearing in \eqref{edd} depend on a parameter $t$ varying in a non-degenerate open interval of the real line. Rewrite  \eqref{edd} in the more suggestive form
\begin{align}\label{DE2}
A\, y(s-1)+B\, y(s+1)+C\, y(s)=0,
\end{align}
where $y(s)=y(X; t)$ and
\begin{align}
\label{AB} A=A(s; t)&=\frac{a(s; t)}{\nabla X\Delta x(s-1/2)}, \quad B=B(s; t)=\frac{a(s; t)+b(X; t)\Delta x(s-1/2)}{\Delta X\Delta x(s-1/2)},\\[7pt]
\nonumber C=C(s; t)&=c(t)-A(s; t)-B(s; t).
\end{align}
The following example will help motivate our main result.

\begin{eje}\label{eje1}
In 1960, Karlin and McGregor proved  (see \cite[(1.3)]{KM60} and \cite[(5.1)]{L67}) that the COPRL known as Hahn polynomials (see \cite[Section 9.5]{KLS10})
$$
y(s)=H^{(\alpha, \beta)}_n(X)=\pFq{3}{2}{-n, -X, \alpha+\beta+n+1}{\beta+1, 1-N}{1}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>-1, \beta>-1)$, ---which constitute the finite discrete analogue of Jacobi polynomials considered by Stieltjes in \cite{S87}--- satisfy \eqref{DE2} with $X=x(s)=s$, and $A$ and $B$ given by
\begin{align*}
A=A(X; \alpha)=X(-X+\alpha+N),\quad B=B(X; \beta)=(X+\beta+1)(-X+N-1).
\end{align*}
The function
\begin{align}\label{S1}
\frac BA=\frac{(X+\beta+1)(-X+N-1)}{X(-X+\alpha+N)}
\end{align}
 is a positive and strictly decreasing function of $X \in(\mathrm{a}, \mathrm{b}-1)$, and
\begin{align}
\label{S21}\frac{\partial}{\partial \alpha}\left(\frac BA\right)&=\frac{(X+\beta+1)(X-N+1)}{X(-X+\alpha+N)^2}<0,\\[7pt]
\label{S22} \frac{\partial}{\partial \beta}\left(\frac BA\right)&=\frac{-X+N-1}{X(-X+\alpha+N)}>0.
\end{align}
for each $X \in(\mathrm{a}, \mathrm{b}-1)$. Since this example corresponds to the  grid $(\mathrm{I})$, from Markov's theorem, Ismail proves that the zeros of $H^{(\alpha, \beta)}_n$ are decreasing functions of $\alpha$ on $(-1, \infty)$ and increasing functions of $\beta$ on $(-1, \infty)$ (see \cite[Theorem $7.1.2$]{I05}).  Comparison of \eqref{s1} and \eqref{S1}, and \eqref{s2} and \eqref{S21}-\eqref{S22} suggests that, as for Jacobi polynomials, the information on the monotonicity of the zeros of Hahn polynomials is stored in the rational function $B/A$. Therefore, it is not unreasonable to conjecture that the same happens with any COPRL.
\end{eje}

The following two mathematical objects play a central role in our exposition.

\begin{definition}
Let $A$ and $B$ be given by \eqref{AB}. The function $f$, from now on called monotonicity function, is defined by
\begin{align}\label{f}
f(s; t)=\frac{B(s; t)}{A(s; t)}=\frac{a(s;t)+b(X; t)\Delta x(s-1/2)}{a(s; t)}\frac{\nabla X}{\Delta X}.
\end{align}
\end{definition}

\begin{definition}
Let $X=x(s)$ be given by \eqref{grid} and let $x(y_j(t))$ $(j=1,\dots,n)$ be the zeros of a COPRL of degree $n$, say $P_n(X; t)$, depending on a parameter $t$ taking values on $\mathrm{J}\subseteq \mathbb{R}$. For any set $\mathrm{I}\subseteq \mathbb{R}$, the (nonempty) subset $\mathrm{S}_\mathrm{I}^{(t)}(P)$ of $(\mathrm{a}, \mathrm{b}-1)$ is defined by
$$
\mathrm{S}_\mathrm{I}^{(t)}(P_n)=\Big\{y\in (\mathrm{a}, \mathrm{b}-1)\,\big|\, t\in \mathrm{I}\cap \mathrm{J}\,\wedge\, \big(\forall j\in \{1,\dots,n \}\big)\big[y=y_j(t)\big]\Big\}.
$$
We will write it simply $\mathrm{S}_\mathrm{I}(P_n)$, $\mathrm{S}_\mathrm{I}^{(t)}$ or $\mathrm{S}_\mathrm{I}$ when no confusion can arise\footnote{Note that $\mathrm{S}_\mathrm{J}=\mathrm{S}_\mathbb{R}$ and  $\mathrm{S}_\mathrm{I}\subseteqq\mathrm{S}_\mathrm{J}$ whenever $\mathrm{I}\subseteqq \mathrm{J}$.}.
\end{definition}

To prove our main result we need two lemmas. The first one was proved for the Hahn polynomials by Levit  (see \cite[Theorem $3$]{L67}). Here we reproduce in a more general framework, {\em mutatis mutandis}, his arguments.

\begin{lemma}\label{lemma}
Let $X=x(s)$ be given by \eqref{grid}. Let $x(y(t))$ and $x(z(t))$ be consecutive zeros of a COPRL depending on a real parameter $t$. Let $f$ be the monotonicity function given by \eqref{f}. Then $|z(t)-y(t)|>1$ for those values of $t$ such that $f(\cdot; t)>0$ on $\mathrm{S}_{\mathbb{R}}$.
\end{lemma}
\begin{proof}
We give the proof only for the case in which $x$ is a strictly increasing function. The same arguments apply to the case in which $x$ is a strictly decreasing function. There is no loss of generality in assuming that $x(y(t))<x(z(t))$ are the greatest pair of consecutive zeros of a COPRL, say $P(X; t)$. (Recall that the zeros of $P$ are distinct and are located in $(x(\mathrm{a}), x(\mathrm{b-1}))$. Moreover, under our assumptions, $y(t)<z(t)$.) Suppose that there exists $t_0$ such that $z(t_0)=y(t_0)+1$. Replacing $s$ by $z(t_0)$ and $t$ by $t_0$ in (\ref{DE2}) we have $P(x(z(t_0)+1);t_0)=0$, because $f(z(t_0); t_0)\neq0$, which contradicts the assumption that $x(z(t_0))$ is the greatest zero of $P(\cdot; t_0)$. (Here we have also used the fact that $x$ is, in particular, a strictly increasing function on $[\mathrm{a}, \mathrm{b})$, and not only on $[\mathrm{a}, \mathrm{b}-1]$, because the possibility that $z(t_0)+1\in [\mathrm{b}-1,\mathrm{b})$ is not excluded.) Suppose now that $z(t_0)<y(t_0)+1$. Replacing again $s$ by $z(t_0)$ and $t$ by $t_0$ in (\ref{DE2}) we have
$$
-\frac{P(x(z(t_0)-1); t_0)}{P(x(z(t_0)+1); t_0)}=f(z(t_0); t_0)>0.
$$
Since $z(t_0)\not=y(t_0)+1$, there is an odd number of zero of $P(\cdot; t_0)$ in $[x(z(t_0)-1),x(z(t_0)+1)]$. We next claim that there exists at least one integer $m$ $(1\leq m\leq N-1)$ such that $y(t_0)\leq \mathrm{a}+m \leq z(t_0)$ or, what is the same, $x(y(t_0)) \leq x(\mathrm{a}+m) \leq x(z(t_0))$. Suppose the assertion were false. Hence
$$
\sum_{s=\mathrm{a}}^{\mathrm{b}-1} \frac{\dps P^2(X; t_0)}{\big(X-x(y(t_0))\big)\big(X-x(z(t_0))\big)}\omega(s; t_0)\Delta x\left(s-\frac12\right)>0,
$$
a contradiction with \eqref{orth}. Since $z(t_0)<y(t_0)+1$, $m$ is unique. Thus the only zeros of $P(\cdot; t_0)$ in $[x(z(t_0)-1),x(z(t_0)+1)]$ are $x(y(t_0))$ and $x(z(t_0))$. This contradicts the fact that there is an odd number of zeros of $P(\cdot; t_0)$ in $[x(z(t_0)-1),x(z(t_0)+1)]$, and the lemma follows.
\end{proof}

\begin{lemma}\label{Lemma}
Let $X=x(s)$ be given by \eqref{grid}. Let $P$ be a non-constant COPRL, in $X$, satisfying \eqref{DE2} and depending on a parameter $t$ varying in a non-degenerate open interval of the real line containing $t_0$. Suppose that $A(\cdot; t)$ and $B(\cdot; t)$ given by \eqref{AB} admit partial derivatives with respect to $t$ on a neighbourhood of $t_0$. Assume that $P(Y_0; t_0)=0$. Then there exist $\epsilon>0$ and $\delta>0$ such that  $(Y_0-\delta, Y_0+\delta) \times (t_0-\epsilon, t_0+\epsilon)$  is on the neighbourhood where $P$ is defined, and there exists $Y: (t_0-\epsilon, t_0+\epsilon) \to (Y_0-\delta, Y_0+\delta)$, such that
\begin{align}\label{zero}
P(Y(t); t)=0
\end{align}
and, for each $t \in (t_0-\epsilon, t_0+\epsilon)$, $Y$ is the unique solution of \eqref{zero} with $Y(t)\in (Y_0-\delta, Y_0+\delta)$. Moreover, $Y$  possess a continuous derivative on $(t_0-\epsilon, t_0+\epsilon)$.
\end{lemma}
\begin{proof}
From \eqref{explicit} we see that the coefficients of $P(\cdot; t)$ are differentiable functions of $t$. Moreover, $P(Y_0; t_0)=0$; from this it follows that
$$
\left. \frac{\partial P}{\partial X}(X; t)\right|_{X=Y_0, t=t_0}\not=0,
$$
because the zeros of $P(\cdot; t_0)$ are distinct. Thus, the result is a direct consequence of the implicit function theorem (see \cite[Theorem 3.4.2]{Si15}).
\end{proof}

We shall refer to Theorem \ref{main} below as discrete Stieltjes theorem.

\begin{theorem}\label{main}
Assume the hypotheses and notation of Lemma \ref{Lemma}. Let $f$ be the monotonicity function given by \eqref{f}. Denote $f_1(s; \cdot)=(\partial f/\partial s) (s;\cdot)$ and $f_2(\cdot;t)=(\partial f/\partial t)(\cdot;t)$. Suppose that $f(s;t)>0$ and $f_1(s;t)<0$ on $\mathrm{S}_{(t_0-\epsilon, t_0+\epsilon)}$. In the case of the grid $(\mathrm{IV})$, assume also that $n f(s; \cdot)+f_1(s; \cdot)\leq 0$. Suppose furthermore that $f_2(s;t)>0$ (respectively, $f_2(s;t)<0$) on $\mathrm{S}_{(t_0-\epsilon, t_0+\epsilon)}$. Then $Y$ is a strictly increasing (respectively, decreasing) function on $(t_0-\epsilon, t_0+\epsilon)$ if $x$ is a strictly increasing function, or else $Y$ is a strictly decreasing (respectively, increasing) function on $(t_0-\epsilon, t_0+\epsilon)$.
\end{theorem}
\begin{proof}
We give the proof only for the case in which $x$ is a strictly increasing function. The same arguments apply to the case in which $x$ is a strictly decreasing function. Assume that $P$ is monic and has fixed degree $n$. Let $Y_j$ $(j=1,\dots,n)$ denote the zeros of $P$. Since the indeterminate of $P$ takes values on the real (open) interval $x((\mathrm{a}, \mathrm{b}-1))$, there exist functions $y_j$ defined on a neighbourhood of $t_0$ and taking values on $(\mathrm{a}, \mathrm{b}-1)$ such that $Y_j(t)=x(y_j(t))$. Moreover, since $x$ is strictly  increasing on $(\mathrm{a}, \mathrm{b}-1)$, $x^{-1}$ is differentiable on $x((\mathrm{a}, \mathrm{b}-1))$. Hence, by Lemma \ref{Lemma}, there exists a neighbourhood of $t_0$ where $y_j$ is differentiable.

Let
\begin{align*}
P(X; t)=\prod_{j=1}^{n}(X-Y_j(t))
\end{align*}
defined in a neighbourhood of $t_0$. Replacing $s$ by $y_j(t)$ in (\ref{DE2}) we have
\begin{align}\label{eq1}
f(y_j(t); t)=-\frac{P(x(y_j(t)-1); t)}{P(x(y_j(t)+1); t)}.
\end{align}
Taking the partial derivative of \eqref{eq1} with respect to $t$ on the neighbourhood of $t_0$ leads to
\begin{align*}
&y'_j(t)f_1(y_j(t); t)+f_2(y_j(t); t)\\[7pt]
\nonumber &\quad=\frac{P(x(y_j(t)-1);t)\dps \frac{\partial P}{\partial t}(x(y_j(t)+1); t)-P(x(y_j(t)+1);t)\dps \frac{\partial P}{\partial t}(x(y_j(t)-1); t) }{P^2(x(y_j(t)+1);t)},
\end{align*}
where
\begin{align*}
&\frac{\partial P}{\partial t}(x(y_j(t)\pm1); t)\\[7pt]
\nonumber &\quad=P(x(y_j(t)\pm1); t)\dps \sum_{k=1}^n \frac{\left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s\pm1)\right|_{s=y_j(t)} y'_j(t)-  \left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s)\right|_{s=y_k(t)} y'_k(t)}{x(y_j(t)\pm1)-Y_k(t)}.
\end{align*}
Hence
\begin{align}\label{f2}
f_{2}(y_j(t); t)=\sum\limits_{k=1}^{n} a_{j k}(t)y'_k(t),
\end{align}
where
\begin{align}
\label{aii}a_{jj}(t)=-f_{1}(y_j(t); t)+f(y_j(t); t)\sum_{k=1}^{n}b_{j k}(t)-\sum\limits_{\substack{k=1 \\ k\neq j}}^{n}a_{jk},
\end{align}
with
\begin{align*}
b_{jk}(t)= \frac{\left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s-1)\right|_{s=y_j(t)}-\left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s)\right|_{s=y_k(t)}}{x(y_j(t)-1)-Y_k(t)}-\frac{\left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s+1)\right|_{s=y_j(t)}-  \left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s)\right|_{s=y_k(t)}}{x(y_j(t)+1)-Y_k(t)},
\end{align*}
and $a_{jk}(t)=f(y_j(t); t)\,c_{jk}(t)$ $(j\not=k)$, where
\begin{align*}
c_{jk}(t)=\left(\frac{1}{x(y_j(t)+1)-Y_k(t)}-\frac{1}{x(y_j(t)-1)-Y_k(t)}  \right)\left.\dps\frac{\mathrm{d}x}{\mathrm{d}s}(s)\right|_{s=y_k(t)}.
\end{align*}
We claim that $a_{jk}(t)<0$ $(j\not=k)$. From now on, without loss of generality, we assume $Y_j(t)<Y_k(t)$. Then, by Lemma \ref{lemma}, $x(y_j(t)+1)<Y_k(t)$. Since $\left.(\mathrm{d}x/\mathrm{d}s)(s)\right|_{s=y_k(t)}$ $>0$, $a_{jk}(t)<0$ as required. Consider now the canonical forms \eqref{grid} to conclude that $a_{jj}(t)>0$. We next claim that, for the grids $(\mathrm{I})-(\mathrm{III})$ and $(\mathrm{V})-(\mathrm{VI})$, $b_{jk}(t)>0$. Indeed, for the grids $(\mathrm{I})$ and $(\mathrm{III})$, the proof is straightforward. For the grid $(\mathrm{II})$, we have
$$
b_{jk}(t)=\frac{4}{\big(y_j(t)+y_k(t)\big)\big(y_j(t)+y_k(t)+2\big)}
$$
and $y_j(t)\in \mathrm{S}_{(t_0-\epsilon, t_0+\epsilon)}\subset (-1/2, \infty)$, the latter because $x$ is strictly  increasing on $(-1/2, \infty)$. Hence, by Lemma \ref{lemma}, $b_{jk}(t)>0$ as claimed. Since $f(y_j(t); t)>0$, $f_{1}(y_j(t); t)<0$, $a_{jk}(t)<0$, and $b_{jk}(t)\geq 0$, \eqref{aii} implies $a_{jj}(t)>0$. The corresponding result for the grid $(\mathrm{V})$ follows in the same way after noting that
$$
b_{jk}(t)=2\theta \sinh (2 \theta) \csch\big((y_k(t)+y_j(t)-1)\theta\big)\csch\big((y_k(t)+y_j(t)+1)\theta \big),
$$
for $q=e^{2\theta}$ ($\theta>0$), and $y_j(t)\in \mathrm{S}_{(t_0-\epsilon, t_0+\epsilon)}\subset (0, \infty)$. And the same goes for  the grid $(\mathrm{VI})$ using hyperbolic identities and noting that $y_j(t)\in \mathrm{S}_{(t_0-\epsilon, t_0+\epsilon)}\subset (-\pi/(2\theta), 0)$. Define
\begin{align*}
&\quad\mathbf{f}(t)=(f_2(y_1(t); t), \dots, f_2(y_n(t); t))^\mathsf{T},\\[7pt]
\mathbf{A}&(t)=(a_{jk}(t)),\quad \mathbf{y}(t)=\big(y'_1(t),\dots, y'_n(t) \big)^\mathsf{T},
\end{align*}
and rewrite \eqref{f2} as $\mathbf{f}(t)=\mathbf{A}(t)\mathbf{y}(t)$. Observe that $\mathbf{A}(t)$ has positive diagonal entries and negative of off-diagonal entries. Moreover, from \eqref{aii} we get
$$
 |a_{jj}(t)|>\sum\limits_{\substack{k=1 \\ k\neq j}}^{n}|a_{jk}(t)|.
$$
Hence $\mathbf{A}(t)$ is a real irreducibly diagonally dominant matrix, and so, by \cite[Corollary 1, p. 85]{V62} all the entries of $\mathbf{A}^{-1}(t)$ are positive \footnote{Indeed, $\mathbf{A}(t)$ is a Stieltjes matrix.}. Thus all the entries of $\mathbf{y}(t)=\mathbf{A}^{-1}(t) \mathbf{f}(t)$ are positive, and the theorem is proved for the grids $(\mathrm{I})-(\mathrm{III})$ and $(\mathrm{V})-(\mathrm{VI})$. The above argument does not work for the grid $(\mathrm{IV})$, because, for $q=e^{2\theta}$ ($\theta>0$),
$$
b_{jk}(t)=-2\theta \sinh(2\theta)\sech\big((y_j(t)+y_k(t)+1)\theta\big)\sech\big((y_j(t)+y_k(t)-1)\theta\big)<0.
$$
Indeed, $-1<b_{jk}(t)<0$ and, therefore, under the additional hypothesis for this grid, the theorem follows from \eqref{aii}.
\end{proof}

\begin{obs}
In some cases, the hypotheses of the discrete Stieltjes theorem are fulfilled in $(\mathrm{a}, \mathrm{b}-1)\supset  \mathrm{S}_\mathbb{R}$. However, there are several, and important, examples where this is only true on a subset of $(\mathrm{a}, \mathrm{b}-1)$ containing all the elements of $\mathrm{S}_\mathbb{R}$; see, for instance, the Racah polynomials in Section \ref{latticeII} below. Of course, since the precise location of the zeros of $P$ is not known, these cases require a more careful analysis.
\end{obs}

In Example \ref{eje1}, we can write $H^{(\alpha, \beta, N)}_n$ instead of $H^{(\alpha, \beta)}_n$. The interlacing property between the zeros of $H^{(\alpha, \beta, N)}_n$ and $H^{(\alpha, \beta, N+1)}_n$ was observed by Levit (see \cite[Theorem 6]{L67}). It was proved independently, and almost simultaneously, by Mesztenyi that  \cite[Theorem 6]{L67} is always true for discrete orthogonal polynomials on the grid $(\mathrm{I})$ when the variable $s$ ranges over a finite interval (see \cite[Lemma 3]{Ma66}). In this way, the following general result might be of interest to the reader.

\begin{theorem}\label{N}
Fix $\mathrm{a}\in\mathbb{R}$ and $N\in\mathbb{N}$ and define $\mathrm{b}=\mathrm{a}+N$. Let $X=x(s)$ be a real-valued function, where the variable $s$ ranges over the finite interval $[\mathrm{a}, \mathrm{b}+1]$.  Let $(P_n(X; N))_{n=1}^{N-1}$ be the sequence of discrete orthogonal polynomials, in $X$, on the set $\{x(\mathrm{a}), x(\mathrm{a}+1), \dots,$ $ x(\mathrm{b}-1)\}$. Let $(P_n(X; N+1))_{n=1}^{N}$ be the sequence of discrete orthogonal polynomials, in $X$, on the set $\{x(\mathrm{a}), x(\mathrm{a}+1), \dots, x(\mathrm{b})\}$. Assume that both sequence are orthogonal with respect to the same positive weight function. Let $Y_1<Y_2<\cdots <Y_n$ be the zeros of $P_n(\cdot; N)$. Then one of the following situations holds:
\begin{enumerate}
\item The zeros of $P_n(\cdot; N+1)$ are those of $P_n(\cdot; N)$, if $P_n(x(\mathrm{b}); N)=0$;
\item  $P_n(\cdot; N+1)$ has a zero on $(Y_k, Y_{k+1})$, if $P_n(x(\mathrm{b}); N)\not=0$  and $x(\mathrm{b})\not\in (Y_k, Y_{k+1})$ for fixed $k\in \{1,\dots, n-1\}$;
\item  $P_n(\cdot; N+1)$ has a zero on each of the intervals $(Y_k, x(\mathrm{b}))$ and $(x(\mathrm{b}), Y_{k+1})$, if $P_n(x(\mathrm{b}); N)\not=0$ and $x(\mathrm{b})\in (Y_k, Y_{k+1})$ for fixed $k\in \{1,\dots, n-1\}$.
\end{enumerate}
\end{theorem}
\begin{proof}
To shorten notation, write $P_n^{(N)}$ and $P_n^{(N+1)}$ instead of $P_n(\cdot; N)$ and $P_n(\cdot; N+1)$, respectively. The reader may check for himself that by expressing $P_n^{(N+1)}$ as a linear combination of the elements of the set $\big\{1, P_1^{(N)}, \dots, P_{n}^{(N)}\big\}$ and using the orthogonality property, we obtain $P_n^{(N+1)}(x(\mathrm{b}))=\zeta_n\, P_n^{(N)}(x(\mathrm{b}))$ and
\begin{align}\label{linearcomb}
P_n^{(N+1)}(X)=& P_n^{(N)}(X)\\[7pt]
\nonumber&-\eta_n P_n^{(N+1)}(x(\mathrm{b}))\frac{P_{n-1}^{(N)}(x(\mathrm{b})) P_n^{(N)}(X)-P_n^{(N)}(x(\mathrm{b}))P_{n-1}^{(N)}(X)}{X-x(\mathrm{b})},
\end{align}
where $\eta_n=\omega(\mathrm{b})\Delta x(\mathrm{b}-1/2)\big\|P_{n-1}^{(N)}\big\|^{-2}$ and
\begin{align*}
\zeta_n&=1+\eta_n\left( P_{n-1}^{(N)}(x(\mathrm{b}))  \dps \left.\dps\frac{\mathrm{d}P_n^{(N)}}{\mathrm{d}X}(X)\right|_{X=x(\mathrm{b})}-\dps \left.\dps\frac{\mathrm{d}P_{n-1}^{(N)}}{\mathrm{d}X}(X)\right|_{X=x(\mathrm{b})} P_n^{(N)}(x(\mathrm{b}))\right).
\end{align*}
Following a standard procedure, the rest of the proof follows as \cite[Lemma 3]{Ma66}.
\end{proof}

We end this section with the following consequence of Theorem \ref{N}, which is valid, in particular, for COPRL.

\begin{coro}\label{CN}
Assume the hypotheses and notation of Theorem \ref{N}. Suppose furthermore that $x$ is a strictly monotone function on $[\mathrm{a}, \mathrm{b}+1]$. Set $Y_0=Y_{n+1}=x(\mathrm{b})$. Then $P_n(\cdot; N+1)$ has a zero on each interval $(Y_k, Y_{k+1})$ for all $k\in \{1,\dots, n\}$  if $x$ is a strictly increasing function, or else $P_n(\cdot; N+1)$ has a zero on each interval $(Y_k, Y_{k+1})$ for all $k\in \{0,\dots, n-1\}$.
\end{coro}
\begin{proof}
We give the proof only for the case in which $x$ is a strictly increasing function. The same arguments apply to the case in which $x$ is a strictly decreasing function. From \eqref{linearcomb}, it follows that
$$
\mathrm{sgn}\left(P_n(Y_n; N+1)\right)=-\mathrm{sgn}\left(P_n(Y_{n+1}; N)\right)=-1.
$$
Hence $P_n(\cdot; N+1)$ has a zero on $(Y_n, Y_{n+1})$. The rest of the proof is a direct consequence of Theorem \ref{N} $\mathrm{ii})$.
\end{proof}

\section{Applications}\label{app}
In this section we apply the discrete Stieltjes theorem to specific families of COPRL. (Of course, Corollary \ref{CN} is applicable to any family of COPRL on a finite grid, see for instance the Hahn, Krawtchouk, Racah, dual Hahn, q-Hahn, q-Krawtchouk, affine q-Krawtchouk, quantum q-Krawtchouk, q-Racah, and dual q-Hanh polynomials below.) We prove, or sketch the proof in similar cases, only of some illustrative examples; the other cases are stated and the proofs are left as exercises for the reader. The reader also should satisfy himself that the hypotheses of Lemma \ref{Lemma} are fulfilled. As far as we know, only the monotonicity of zeros of COPRL on the grid $(\mathrm{I})$ has been studied (see \cite[Chapter $7$]{I05}) \footnote{Recall that  for a continuous case on the grid $(\mathrm{VI})$  as the Askey-Wilson polynomials, the monotonicity of their zeros was studied in \cite[Section 7]{AW85}.}. We include this case for the sake of completeness.

\subsection{The grid $(\mathrm{I})$}\label{latticeI}
\subsubsection*{Examples of COPRL on $X=x(s)=s$ (Hahn, Charlier, Krawtchouk, and Meixner polynomials).}

The {\em Hahn polynomials}, $H^{(\alpha, \beta)}_n$, are defined in Example \ref{eje1}.
\begin{proposition}
The zeros of $H^{(\alpha, \beta)}_n$ are strictly decreasing functions of $\alpha$ on $(-1,\infty)$ and strictly increasing functions of $\beta$ on $(-1,\infty)$.
\end{proposition}
\begin{proof}
It suffices to use \eqref{S1} and \eqref{S21}-\eqref{S22} together with the discrete Stieltjes theorem.
\end{proof}

The {\em Charlier polynomials} (see \cite[Section 9.14]{KLS10}),
$$
y(s)=C_n^{(\alpha)}(X)=\pFq{2}{0}{-n, -X}{-}{-\frac{1}{\alpha}}
$$
$(n=1,2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; \alpha>0)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(X)=X, \quad B=B(\alpha)=\alpha.
\end{align*}

\begin{proposition}
The zeros of $C_n^{(\alpha)}$ are strictly increasing functions of $\alpha$ on $(0, \infty)$.
\end{proposition}

The {\em Krawtchouk polynomials} (see \cite[Section 9.11]{KLS10}),
$$
y(s)=K_n^{(\alpha)}(X)=\pFq{2}{1}{-n, -X}{1-N}{\frac{1}{\alpha}}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; 0<\alpha<1)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(X; \alpha)=(1-\alpha)X,\quad B=B(X; \alpha)=\alpha(-X+N-1).
\end{align*}

\begin{proposition}
The zeros of $K_n^{(\alpha)}$ are strictly increasing functions of $\alpha$ on $(0, 1)$.
\end{proposition}

The {\em Meixner polynomials} (see \cite[Section 9.10]{KLS10}),
$$
y(s)=M_n^{(\alpha, \beta)}(X)=\pFq{2}{1}{-n, -X}{\beta}{1-\frac{1}{\alpha}}
$$
$(n=1,2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; 0<\alpha<1, \beta>0)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(X)=X, \quad B=B(X; \alpha, \beta)=\alpha(X+\beta).
\end{align*}

\begin{proposition}
The zeros of $M_n^{(\alpha, \beta)}$ are strictly increasing functions of $\alpha$ on $(0, 1)$ and strictly increasing functions of $\beta$ on $(0,\infty)$.
\end{proposition}

\subsection{The grid $(\mathrm{II})$}\label{latticeII}
\subsubsection*{Examples of COPRL on $X=x(s)=s(s+1)$ (Racah and dual Hahn polynomials).}

The {\em Racah polynomials} (see \cite[p. 236]{NU93}),
\begin{align}\label{FRacah}
y(s)=R_n^{(\alpha, \beta)}(X)&=\pFq{4}{3}{-n, \alpha+\beta+n+1, \mathrm{a}-s, s+\mathrm{a}+1}{2\mathrm{a}+\alpha+N+1, \beta+1, 1-N}{1}
\end{align}
$(n=1,\dots, N-1; \mathrm{a}>-1/2, \mathrm{b}=\mathrm{a}+N; \alpha>-1, -1<\beta<2\mathrm{a}+1)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A&=A(s; \alpha, \beta)=\frac{(s-\mathrm{a})(s+\mathrm{a}+N)(s-\mathrm{a}-\alpha-N)(s+\mathrm{a}-\beta)}{2s(2s+1)},\\[7pt]
B&=B(s; \alpha, \beta)=\frac{(s+\mathrm{a}+1)(s-\mathrm{a}-N+1)(s+\mathrm{a}+\alpha+N+1)(s-\mathrm{a}+\beta+1)}{2(s+1)(2s+1)}.
\end{align*}

\begin{proposition}\label{Racah}
The zeros of $R_n^{(\alpha,\beta)}$ are strictly decreasing functions of $\alpha$ on $(-1, \infty)$ and strictly increasing function of $\beta$ on $(-1, 2\mathrm{a}+1)$ if $\mathrm{a}\geq 0$, or else the zeros of $R_n^{(\alpha,\beta)}$ are strictly decreasing functions of $\alpha$ on $(-1, \infty)$ for each $\beta\in (\mathrm{a}, 2\mathrm{a}+1)$ and strictly increasing function of $\beta$ on $(\mathrm{a}, 2\mathrm{a}+1)$.
\end{proposition}
\begin{proof}
We give the proof only for the case in which $\mathrm{a}\geq 0$. The proof for $-1/2<\mathrm{a}<0$ is similar. Define the interval $\mathrm{K}=\big(\max\big\{\mathrm{a}, \beta-\mathrm{a}\big\}, \mathrm{a}+N-1\big)$.
The monotonicity function
\begin{align}\label{fR}
f=\frac{B}{A}=\frac{s(s+\mathrm{a}+1)(s-\mathrm{a}-N+1)(s+\mathrm{a}+\alpha+N+1)(s-\mathrm{a}+\beta+1)}{(s+1)(s-\mathrm{a})(s+\mathrm{a}+N)(s-\mathrm{a}-\alpha-N)(s+\mathrm{a}-\beta)}
\end{align}
is a positive and strictly decreasing function of $s \in\mathrm{K}$, and
\begin{align}
\label{DfR1}\frac{\partial f}{\partial \alpha}&=\frac{s(2s+1)(s+\mathrm{a}+1)(s-\mathrm{a}-N+1)(s-\mathrm{a}+\beta+1)}{(s+1)(s-\mathrm{a})(s+\mathrm{a}+N)(s-\mathrm{a}-\alpha-N)^2(s+\mathrm{a}-\beta)}<0,\\[7pt]
\label{DfR2}\frac{\partial f}{\partial \beta}&=\frac{s(2s+1)(s+\mathrm{a}+1)(s-\mathrm{a}-N+1)(s+\mathrm{a}+\alpha+N+1)}{(s+1)(s-\mathrm{a})(s+\mathrm{a}+N)(s-\mathrm{a}-\alpha-N)(s+\mathrm{a}-\beta)^2}>0,
\end{align}
for each $s \in\mathrm{K}$. It is immediate that $\mathrm{S}_{(-1,2\mathrm{a}]}^{(\beta)}\subset \mathrm{K}$. We next claim that $\mathrm{S}_{(2\mathrm{a}, 2\mathrm{a}+1)}^{(\beta)}\subset \mathrm{K}$. Indeed, this is equivalent to prove that $R_n^{(\alpha,\beta)}$ has no zeros on $\big(\mathrm{a}(\mathrm{a}+1), (\beta-\mathrm{a})(\beta-\mathrm{a}+1)\big)$ for $\beta \in (2\mathrm{a}, 2\mathrm{a}+1)$. For $\beta \in (2\mathrm{a}, 2\mathrm{a}+1)$, $R_n^{(\alpha, \beta)}(\mathrm{a}(\mathrm{a}+1))=1$ and
\begin{align*}
 R_n^{(\alpha, \beta)}\big((\beta-\mathrm{a})(\beta-\mathrm{a}+1)\big)&=\pFq{3}{2}{-n, \alpha+\beta+n+1, 2\mathrm{a}-\beta}{2\mathrm{a}+\alpha+N+1, 1-N}{1}\\[7pt]
 &=\frac{(2\mathrm{a}-\beta+N-n)_n(\alpha+\beta+N+1)_n}{(2\mathrm{a}+\alpha+N+1)_n(N-n)_n}>0,
\end{align*}
the last equality being a consequence of Sheppard's identity (see \cite[Corollary 3.3.4]{AAR99}). Hence $R_n^{(\alpha, \beta)}$ has no zeros on $\big(\mathrm{a}(\mathrm{a}+1), (\beta-\mathrm{a})(\beta-\mathrm{a}+1)\big)$ or has at least two zeros there. Suppose the second of these possibilities is true. From the proof of Lemma \ref{lemma}, we have $x(\mathrm{a}+1)<x(\beta-\mathrm{a})$, which is impossible. Thus $\mathrm{S}_{(2\mathrm{a}, 2\mathrm{a}+1)}^{(\beta)}\subset \mathrm{K}$ as claimed. The same proof actually shows that $\mathrm{S}_{(-1,\infty)}^{(\alpha)}\subset \mathrm{K}$. The result follows from the discrete Stieltjes theorem.
\end{proof}

\begin{obs}
In \cite[Section 9.2]{KLS10}, the Racah polynomials (interchanging $\alpha$ and $\beta$) are defined by
$$
\widetilde{R}_n^{(\alpha, \beta, \delta)}(\widetilde{x}(s))=\pFq{4}{3}{-n, \alpha+\beta+n+1, -s, s+\delta-N+1}{\alpha+\delta+1, \beta+1, 1-N}{1}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>-1, -1<\beta<\delta-N+1, \delta>N-1)$, where $\widetilde{x}(s)=s(s+\delta-N+1)$.  In \eqref{FRacah}, we can write $R^{(\alpha, \beta, \mathrm{a})}_n$ instead of $R^{(\alpha, \beta)}_n$; although, implicitly, we have the agreement to omit those parameters that are assumed fixed. Clearly, $\widetilde{x}$ is not one of the canonical forms \eqref{grid}. However, $\widetilde{x}$ and $x$ are related by a linear transformation. Therefore, for  $\mathrm{a}=(\delta-N)/2$ fixed,
$$
R^{(\alpha, \beta, (\delta-N)/2)}_n(x(s))=\widetilde{R}_n^{(\alpha, \beta, \delta)}\left(\widetilde{x}\left(s-\frac{\delta-N}{2}\right)\right),
$$
where
$$
\widetilde{x}\left(s-\frac{\delta-N}{2}\right)=x(s)-\frac{(\delta-N+1)^2-1}{4}.
$$
Consequently, Proposition \ref{Racah} remains valid if we replace $R^{(\alpha, \beta)}_n$ by $\widetilde{R}_n^{(\alpha, \beta, \delta)}$.
\end{obs}

The {\em dual Hahn polynomials} (see \cite[p. 236]{NU93}),
\begin{align}\label{DualHahneq}
y(s)=W_n^{(\alpha)}(X)&=\pFq{3}{2}{-n, \mathrm{a}-s, s+\mathrm{a}+1}{\alpha+1, 1-N}{1}
\end{align}
$(n=1,\dots, N-1; \mathrm{a}>-1/2, \mathrm{b}=\mathrm{a}+N; -1<\alpha<2\mathrm{a}+1)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A&=A(s; \alpha)=\frac{(s-\mathrm{a})(s+\mathrm{a}+N)(s+\mathrm{a}-\alpha)}{2s(2s+1)},\\[7pt]
B&=B(s; \alpha)=\frac{(s+\mathrm{a}+1)(-s+\mathrm{a}+N-1)(s-\mathrm{a}+\alpha+1)}{2(s+1)(2s+1)}.
\end{align*}

\begin{proposition}\label{DualHahn}
The zeros of $W_n^{(\alpha)}$ are strictly increasing functions of $\alpha$ on $(-1, 2\mathrm{a}+1)$ if $\mathrm{a}\geq 0$, or else the zeros of $W_n^{(\alpha)}$ are strictly increasing functions of $\alpha$ on $(\mathrm{a}, 2\mathrm{a}+1)$.
\end{proposition}
\begin{proof}
We sketch the proof only for the case in which $\mathrm{a}\geq 0$. The proof for $-1/2<\mathrm{a}<0$ is similar. Define the interval $\mathrm{K}=\big(\max\big\{\mathrm{a}, \alpha-\mathrm{a}\big\}, \mathrm{a}+N-1\big)$. Note that the hypotheses of the discrete Stieltjes theorem are fulfilled in $\mathrm{K}$. Thus we only need to prove that $\mathrm{S}_{(2\mathrm{a}, 2\mathrm{a}+1)}\subset \mathrm{K}$\footnote{It is immediate that $\mathrm{S}_{(-1, 2\mathrm{a}]} \subset \mathrm{K}$.}.  Note that $W_n^{(\alpha)}(\mathrm{a}(\mathrm{a}+1))=1$ and
\begin{align*}
 W_n^{(\alpha)}\big((\alpha-\mathrm{a})(\alpha-\mathrm{a}+1)\big)=\pFq{2}{1}{-n, 2\mathrm{a}-\alpha}{1-N}{1}=\frac{(1-N+\alpha-2\mathrm{a})_n}{(1-N)_n}>0,
 \end{align*}
the last equality being a consequence of Chu-Vandermonde's identity (\cite[Corollary 2.2.3]{AAR99}). The rest of the proof runs as in Proposition \ref{Racah}.
\end{proof}

\begin{obs}
In \cite[Section 9.6]{KLS10}, the dual Hahn polynomials are defined by
$$
\widetilde{W}_n^{(\alpha, \beta)}(\widetilde{x}(s))=\pFq{3}{2}{-n, -s, s+\alpha+\beta+1}{\alpha+1, 1-N}{1}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>-1, \beta>-1\; \text{or}\; \alpha<1-N, \beta<1-N)$, where $\widetilde{x}(s)=s(s+\alpha+\beta+1)$.  In \eqref{DualHahneq}, we can write $W^{(\alpha, \mathrm{a})}_n$ instead of $W^{(\alpha)}_n$. Hence, for $\mathrm{a}=(\alpha+\beta)/2$ fixed,
$$
W^{(\alpha, (\alpha+\beta)/2)}_n(x(s))=\widetilde{W}_n^{(\alpha, \beta)}\left(\widetilde{x}\left(s-\frac{\alpha+\beta}{2}\right)\right),
$$
where
$$
\widetilde{x}\left(s-\frac{\alpha+\beta}{2}\right)=x(s)-\frac{(\alpha+\beta+1)^2-1}{4}.
$$
Consequently, Proposition \ref{DualHahn} remains valid if we replace $W^{(\alpha)}_n$ by $\widetilde{W}_n^{(\alpha, \beta)}$ and assume that $\alpha+\beta$ is constant.
\end{obs}

\subsection{The grid $(\mathrm{III})$}\label{qlinear}
\subsubsection{Examples of COPRL on $X=x(s)=q^{-s}$ $(0<q<1)$ (q-Meixner, Al-Salam-Carlitz, q-Charlier, q-Hahn, q-Krawtchouk, affine q-Krawtchouk, and quantum q-Krawtchouk) and some related cases (q-Charlier and big q-Laguerre).}
\vspace{.2cm}
The {\em q-Meixner polynomials} (see \cite[Section 14.13]{KLS10}),
$$
y(s)=M_n^{(\alpha, \beta)}(X; q)=\pPq{2}{1}{q^{-n}, X}{\beta q}{q, -\frac{q^{n+1}}{\alpha}}
$$
$(n=1,2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; \alpha>0, 0\leq\beta<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s; \alpha, \beta)=(1-q^s)(1+\alpha \beta q^s), \quad B=B(s; \alpha, \beta)=\alpha q^s(1-\beta q^{s+1}).
\end{align*}

\begin{proposition}
The zeros of $M_n^{(\alpha, \beta)}(\cdot; q)$ are strictly increasing functions of $\alpha$ on $(0, \infty)$ and strictly decreasing functions of $\beta$ on $[0,q^{-1})$.
\end{proposition}
\begin{proof}
The monotonicity function
$$
f=\frac{B}{A}=\frac{\alpha q^s(1-\beta q^{s+1})}{(1-q^s)(1+\alpha \beta q^s)}
$$
is a positive and strictly decreasing function of $s \in (0, \infty)$, and
\begin{align*}
\frac{\partial f}{\partial \alpha}&=\frac{q^s(1-\beta q^{s+1})}{(1-q^s)(1+\alpha \beta q^s)^2}>0,\\[7pt]
\frac{\partial f}{\partial \beta}&=-\frac{\alpha q^{2s} (\alpha+q)}{(1-q^s)(1+\alpha \beta q^s)^2}<0,
\end{align*}
for each $s \in(0, \infty)$. The result follows from the discrete Stieltjes theorem.
\end{proof}
\begin{obs}
The q-Charlier polynomials (see \cite[Section 14.23]{KLS10}) are given by $C_n^{(\alpha)}(\cdot; q)=M_n^{(\alpha, 0)}(\cdot; q)$.
\end{obs}

The {\em second family of Al-Salam-Carlitz polynomials} (see \cite[Section 14.25]{KLS10}),
$$
y(s)=V_n^{(\alpha)}(X; q)=(-\alpha)^n q^{-\binom{n}{2}} \pPq{2}{0}{q^{-n}, X}{-}{q, \frac{q^n}{\alpha}}
$$
$(n=1,2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; 0<\alpha<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s; \alpha)=(1-q^{-s})(\alpha-q^{-s}),\quad B=B(s; \alpha)=\alpha q.
\end{align*}

\begin{proposition}\label{aSC}
The zeros of $V_n^{(\alpha)}(\cdot; q)$ are strictly increasing functions of $\alpha$ on $(0, q^{-1})$.
\end{proposition}
\begin{proof}
Define the interval $\mathrm{K}=\big(\max\{-\log_q \alpha, 0\}, \infty\big)$. Note that the hypotheses of the discrete Stieltjes theorem are fulfilled in $\mathrm{K}$. Thus we only need to prove that $\mathrm{S}_{(1, q^{-1})}\subset \mathrm{K}$\footnote{It is immediate that $\mathrm{S}_{(0, 1]} \subset \mathrm{K}$.}.  The rest of the proof runs as in Proposition \ref{Racah}; although given the simplicity of this case, we do not need to use q-hypergeometric identities.
\end{proof}

\begin{obs}
The first family of Al-Salam-Carlitz polynomials is given by $U_n^{(\alpha)}$ $(\cdot; q^{-1})=V_n^{(\alpha)}(\cdot; q)$.
\end{obs}

The {\em q-Hahn polynomials} (see \cite[Section 14.6]{KLS10}),
$$
y(s)=H_n^{(\alpha, \beta)}(X; q)=\pPq{3}{2}{q^{-n}, \alpha \beta q^{n+1}, X}{\alpha q, q^{1-N}}{q, q}
$$
$(n=1,\dots N-1; \mathrm{a}=0, \mathrm{b}=N; 0<\alpha<q^{-1}, 0<\beta<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A&=A(s; \alpha, \beta)=\alpha q (1-q^s)(\beta-q^{s-N}),\\[7pt]
B&=B(s; \alpha)=(1-q^{s-N+1})(1-\alpha q^{s+1}).
\end{align*}
\begin{proposition}
The zeros of $H_n^{(\alpha, \beta)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, q^{-1})$ and strictly increasing functions of $\beta$ on $(0,q^{-1})$.
\end{proposition}

The {\em q-Krawtchouk polynomials} (see \cite[Section 14.15]{KLS10}),
$$
y(s)=K_n^{(\alpha)}(X; q)=\pPq{3}{2}{q^{-n}, -\alpha q^{n}, X}{0, q^{1-N}}{q, q}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>0)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s; \alpha)=\alpha (q^s-1),\quad B=B(s)=1-q^{s-N+1}.
\end{align*}
\begin{proposition}
The zeros of $K_n^{(\alpha)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, \infty)$.
\end{proposition}

The {\em affine q-Krawtchouk polynomials} (see \cite[Section 14.16]{KLS10}),
$$
y(s)=\widehat{K}_n^{(\alpha)}(X; q)=\pPq{3}{2}{q^{-n},0, X}{\alpha q, q^{1-N}}{q, q}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; 0<\alpha<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s; \alpha)=\alpha q^{s-N+1} (q^s-1),\quad B=B(s; \alpha)=(1-q^{s-N+1})(1-\alpha q^{s+1}).
\end{align*}
\begin{proposition}
The zeros of $\widehat{K}_n^{(\alpha)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, q^{-1})$.
\end{proposition}
\begin{obs}\label{bqL1}
The big q-Laguerre polynomials (see \cite[Section 14.16]{KLS10}) are given by
$$
L_n^{(\alpha, \beta)}(X; q)=\pPq{3}{2}{q^{-n}, 0, X}{\alpha q, \beta q}{q, q}
$$
$(n=1,2,\dots ; 0<\alpha<q^{-1}, \beta<0)$. In particular, $\widehat{K}_n^{(\alpha)}(\cdot; q)=L_n^{(\alpha, q^{-N})}(X; q)$ (see also Remark \ref{bqL2}).
\end{obs}

The {\em quantum q-Krawtchouk polynomials} (see \cite[Section 14.14]{KLS10}),
$$
y(s)=\widetilde{K}_n^{(\alpha)}(X; q)=\frac{(q^{-N},q)_n}{\alpha^n q^{n^2}} \pPq{2}{1}{q^{-n}, X}{q^{1-N}}{q, \alpha q^{n+1}}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>q^{1-N})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s; \alpha)=(1-q^s)(\alpha-q^{s-N}),\quad B=B(s)=-q^s(1-q^{s-N+1}).
\end{align*}
\begin{proposition}
The zeros of $\widetilde{K}_n^{(\alpha)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(q^{1-N}, \infty)$.
\end{proposition}
\begin{proof}
Define the interval $\mathrm{K}=\big(\max\{0, \log_q \alpha+N\}, N-1\big)$. Note that the hypotheses of the discrete Stieltjes theorem are fulfilled in $\mathrm{K}$. Thus we only need to prove that $\mathrm{S}_{(q^{1-N}, q^{-N})}\subset \mathrm{K}$\footnote{It is immediate that $\mathrm{S}_{[q^{-N}, \infty)}\subset \mathrm{K}$.}.  The rest of the proof runs as in Proposition \ref{Racah}; although given the simplicity of this case, we do not need to use q-hypergeometric identities.
\end{proof}

\subsubsection{Examples of COPRL on $X=x(s)=q^{s}$ $(0<q<1)$ (q-Bessel, little q-Jacobi, and little q-Laguerre/Wall) and some related cases (big q-Jacobi, big q-Laguerre, and q-Laguerre).}
\vspace{.2cm}
The {\em q-Bessel} (see \cite[Section 14.22]{KLS10}),
$$
y(s)=B_n^{(\alpha)}(X; q)=\pPq{2}{1}{q^{-n}, -\alpha q^{n}}{0}{q, q X}
$$
$(n=1, 2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; \alpha>0)$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s)=q^{s}-1,\quad B=B(s; \alpha)=\alpha.
\end{align*}
\begin{proposition}
The zeros of $B_n^{(\alpha)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, \infty)$.
\end{proposition}

The {\em little q-Jacobi polynomials} (see \cite[Section 14.12]{KLS10}),
$$
y(s)=P_n^{(\alpha, \beta)}(X; q)=\pPq{2}{1}{q^{-n}, \alpha \beta q^{n+1}}{\alpha q}{q, q X}
$$
$(n=1, 2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; 0<\alpha<q^{-1}, \beta<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s)=q^{-s} (q^{s}-1),\quad B=B(s; \alpha, \beta)=\alpha q^{-s} (\beta q^{s+1}-1).
\end{align*}
\begin{proposition}
The zeros of $P_n^{(\alpha, \beta)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, q^{-1})$ and strictly increasing functions of $\beta$ on $(-\infty,q^{-1})$.
\end{proposition}

\begin{coro}
The zeros of the special case of big q-Jacobi polynomials (see \cite[Section 14.5]{KLS10}),
$$
P_n^{(\alpha, \beta, 0)}(X; q)=\frac{(\beta q; q)_n}{(\alpha q; q)_n} (-1)^n \alpha^n q^{n+\binom{n}{2}} P_n^{(\beta, \alpha)}(\alpha^{-1} q^{-1} X; q)
$$
 $(n=1, 2,\dots; 0<\alpha<q^{-1}, 0<\beta<q^{-1})$, are strictly increasing functions of $\alpha$ on $(0, q^{-1})$ and strictly decreasing functions of $\beta$ on $(0,q^{-1})$.
\end{coro}

\begin{obs}\label{bqL2}
$P_n^{(\alpha, 0, 0)}(\cdot; q)=L_n^{(\alpha, 0)}(\cdot; q)$ (see also Remark \ref{bqL1}).
\end{obs}

The {\em little q-Laguerre/Wall polynomials} (see \cite[Section 14.20]{KLS10}),
$$
y(s)=L_n^{(\alpha)}(X; q)=\pPq{2}{1}{q^{-n}, 0}{\alpha q}{q, q X}
$$
$(n=1, 2,\dots; \mathrm{a}=0, \mathrm{b}=\infty; 0<\alpha<q^{-1})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A=A(s)=q^{s}-1,\quad B=B(s; \alpha)=\alpha q^{-s}.
\end{align*}
\begin{proposition}
The zeros of $L_n^{(\alpha)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(0, q^{-1})$.
\end{proposition}

\begin{coro}
The zeros of the q-Laguerre polynomials (see \cite[Section 14.21]{KLS10}),
$$
\widehat{L}_n^{(\alpha)}(\cdot; q)=q^{-\alpha n}\frac{(q^{\alpha+1}; q)_n}{(q; q)_n}L_n^{(q^\alpha)}(\cdot; q)
$$
 $(n=1, 2,\dots; \alpha>-1)$, are strictly decreasing functions of $\alpha$ on $(-1, \infty)$.
\end{coro}

\subsection{The grid $(\mathrm{VI})$}\label{latticeII}
\subsubsection*{Examples of COPRL on $X=x(s)=(q^{s}+q^{-s})/2$ $(0<q<1)$ (q-Racah and dual q-Hahn polynomials).}

The {\em q-Racah polynomials} (see \cite[p. 239]{NU93}),
\begin{align}\label{qFRacah}
y(s)=R_n^{(\alpha, \beta)}(X; q)&=\pPq{4}{3}{q^{-n}, q^{\alpha+\beta+n+1}, q^{\mathrm{a}-s},q^{s+\mathrm{a}}}{q^{2\mathrm{a}+\alpha+N}, q^{\beta+1}, q^{1-N}}{q,q}
\end{align}
$(n=1,\dots, N-1; \mathrm{a}>0, \mathrm{b}=\mathrm{a}+N; \alpha>-1, -1<\beta<2\mathrm{a})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A&=A(s; \alpha, \beta)\\[7pt]
&=-\frac{4\,q^{\alpha+\beta+5/2}(q^{s-\mathrm{a}}-1)(q^{s+\mathrm{a}+N-1}-1)(q^{s-\mathrm{a}-\alpha-N}-1)(q^{s+\mathrm{a}-\beta-1}-1)}{(q-1)^2(q^{2s}-1)(q^{2s-1}-1)},\\[7pt]
B&=B(s; \alpha, \beta)\\[7pt]
&=-\frac{4\, q^{3/2}(q^{s+\mathrm{a}}-1)(q^{s-\mathrm{a}-N+1}-1)(q^{s+\mathrm{a}+\alpha+N}-1)(q^{s-\mathrm{a}+\beta+1}-1)}{(q-1)^2(q^{2s}-1)(q^{2s+1}-1)}.
\end{align*}

\begin{proposition}\label{qRacah}
The zeros of $R_n^{(\alpha,\beta)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(-1, \infty)$ and strictly increasing function of $\beta$ on $(-1, 2\mathrm{a})$ if $\mathrm{a}\geq 1/2$, or else the zeros of $R_n^{(\alpha,\beta)}(\cdot; q)$ are strictly decreasing functions of $\alpha$ on $(-1, \infty)$ for each $\beta\in (\mathrm{a}-1/2, 2\mathrm{a})$ and strictly increasing function of $\beta$ on $(\mathrm{a}-1/2, 2\mathrm{a})$.
\end{proposition}
\begin{proof}
We give the proof only for the case in which $\mathrm{a}\geq 1/2$. The proof for $0<\mathrm{a}<1/2$ is similar. Define the interval $\mathrm{K}=\big(\max\big\{\mathrm{a}, \beta-\mathrm{a}+1\big\}, \mathrm{a}+N-1\big)$.
The monotonicity function
\begin{align*}
f&=\frac{B}{A}\\[7pt]
&=\frac{(q^{s+\mathrm{a}}-1)(q^{s-\mathrm{a}-N+1}-1)(q^{s+\mathrm{a}+\alpha+N}-1)(q^{s-\mathrm{a}+\beta+1}-1)}{q^{\alpha+\beta+1}(q^{s-\mathrm{a}}-1)(q^{s+\mathrm{a}+N-1}-1)(q^{s-\mathrm{a}-\alpha-N}-1)(q^{s+\mathrm{a}-\beta-1}-1)}\frac{q^{2s-1}-1}{q^{2s+1}-1}
\end{align*}
is a positive and strictly decreasing function of $s \in\mathrm{K}$, and
\begin{align*}
&0>\frac{\partial f}{\partial \alpha}=\\[7pt]
&\frac{\log q\,(q^{2s}-1)(q^{2s-1}-1)(q^{s+\mathrm{a}}-1)(q^{s-\mathrm{a}-N+1}-1)(q^{s-\mathrm{a}+\beta+1})}{q^{\alpha+\beta+1/}(q^{2s+1}-1)(q^{s-\mathrm{a}}-1)(q^{s+\mathrm{a}+N-1}-1)(q^{s-\mathrm{a}-\alpha-N}-1)^2(q^{s+\mathrm{a}-\beta-1}-1)},\\[7pt]
&0<\frac{\partial f}{\partial \beta}=\\[7pt]
&\frac{\log q\,(q^{2s}-1)(q^{2s-1}-1)(q^{s+\mathrm{a}}-1)(q^{s-\mathrm{a}-N+1}-1)(q^{s+\mathrm{a}+\alpha+N})}{q^{\alpha+\beta+1}(q^{2s+1}-1)(q^{s-\mathrm{a}}-1)(q^{s+\mathrm{a}+N-1}-1)(q^{s-\mathrm{a}-\alpha-N}-1)(q^{s+\mathrm{a}-\beta-1}-1)^2},
\end{align*}
for each $s \in\mathrm{K}$. Thus we only need to prove that $\mathrm{S}^{(\beta)}_{(2\mathrm{a}-1, 2\mathrm{a})}\subset \mathrm{K}$\footnote{It is immediate that $\mathrm{S}^{(\beta)}_{(-1, 2\mathrm{a}-1]}\subset \mathrm{K}$.}. Indeed, this is equivalent to prove that $R_n^{(\alpha,\beta)}(\cdot; q)$ has no zeros on $\big((q^{\mathrm{a}}+q^{-\mathrm{a}})/2, (q^{\beta-\mathrm{a}+1}+q^{\mathrm{a}-\beta-1})/2\big)$ for $\beta \in (2\mathrm{a}-1, 2\mathrm{a})$. For $\beta \in (2\mathrm{a}-1, 2\mathrm{a})$, $R_n^{(\alpha, \beta)}\big((q^{\mathrm{a}}+q^{-\mathrm{a}})/2; q\big)=1$ and
\begin{align*}
 R_n^{(\alpha, \beta)}\big((q^{\beta-\mathrm{a}+1}+q^{\mathrm{a}-\beta-1})/2; q\big)&=\pPq{3}{2}{q^{-n}, q^{\alpha+\beta+n+1}, q^{2\mathrm{a}-\beta-1}}{q^{2\mathrm{a}+\alpha+N}, q^{1-N}}{q,q}\\[7pt]
 &=\frac{(q^{2\mathrm{a}-\beta+N-n-1}; q)_n (q^{\alpha+\beta+N+1}; q)_n}{(q^{2\mathrm{a}+\alpha+N} ;q)_n (q^{N-n}; q)_n}>0,
\end{align*}
the last equality being a consequence of q-Pfaff-Saalsch\"utz's identity (see \cite[(10.10.3)]{AAR99}).  We thus get, as in Proposition \ref{Racah}, $\mathrm{S}^{(\beta)}_{(-1, 2\mathrm{a})}\subset \mathrm{K}$. The same proof actually shows that $\mathrm{S}_{(-1,\infty)}^{(\alpha)}\subset \mathrm{K}$. The result follows from the discrete Stieltjes theorem.
\end{proof}

\begin{obs}
In \cite[Section 14.2]{KLS10}, the q-Racah polynomials (replacing $\alpha$ by $q^{\beta}$ and $\beta$ by $q^{\alpha}$) are defined by
$$
\widetilde{R}_n^{(\alpha, \beta, \delta)}(\widetilde{x}(s); q)=\pPq{4}{3}{q^{-n}, q^{\alpha+\beta+n+1}, q^{-s},q^{s-N+1}}{\delta q^{\alpha+1}, q^{\beta+1}, q^{1-N}}{q,q}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>-1, -1<\beta<1-N+\log_q \delta, 0<\delta<q^{N-1})$, where $\widetilde{x}(s)=\delta q^{s-N+1}+q^{-s}$.  In \eqref{qFRacah}, we can write $R^{(\alpha, \beta, \mathrm{a})}_n(\cdot; q)$ instead of $R^{(\alpha, \beta)}_n(\cdot; q)$. Hence, for  $\mathrm{a}=(1-N+\log_q \delta)/2$ fixed,
$$
R^{(\alpha, \beta, (1-N+\log_q \delta)/2)}_n(x(s); q)=\widetilde{R}_n^{(\alpha, \beta, \delta)}\left(\widetilde{x}\left(s-(1-N+\log_q \delta)/2\right); q\right),
$$
where
$$
\widetilde{x}\left(s-(1-N+\log_q \delta)/2\right)=2 q^{(1-N+\log_q \delta)/2} x(s).
$$
Consequently, Proposition \ref{qRacah} remains valid if we replace $R^{(\alpha, \beta)}_n(\cdot; q)$ by $\widetilde{R}_n^{(\alpha, \beta, \delta)}$ $(\cdot; q)$.\end{obs}

The {\em dual q-Hahn polynomials} (see \cite[p. 239]{NU93}),
\begin{align}\label{DqH}
y(s)=W_n^{(\alpha)}(X; q)&=\pPq{3}{2}{q^{-n}, q^{\mathrm{a}-s},q^{s+\mathrm{a}}}{q^{\alpha+1}, q^{1-N}}{q,q}
\end{align}
$(n=1,\dots, N-1; \mathrm{a}>0, \mathrm{b}=\mathrm{a}+N; -1<\alpha<2\mathrm{a})$, satisfy \eqref{DE2} with $A$ and $B$ given by
\begin{align*}
A&=A(s; \alpha)=-\frac{4 q^{-\mathrm{a}+\alpha-N+5/2}(q^{s-\mathrm{a}}-1)(q^{s+\mathrm{a}+N-1}-1)(q^{s+\mathrm{a}-\alpha-1}-1)}{(q-1)^2(q^{2s}-1)(q^{2s-1}-1)},\\[7pt]
B&=B(s; \alpha)=-\frac{4 q^{s+3/2}(q^{s+\mathrm{a}}-1)(q^{s-\mathrm{a}-N+1}-1)(q^{s-\mathrm{a}+\alpha+1}-1)}{(q-1)^2(q^{2s}-1)(q^{2s+1}-1)}.
\end{align*}

\begin{proposition}\label{DqHahn}
The zeros of $W_n^{(\alpha)}(\cdot; q)$ are strictly increasing functions of $\alpha$ on $(-1, 2\mathrm{a})$ if $\mathrm{a}\geq 1/2$, or else the zeros of $W_n^{(\alpha)}(\cdot; q)$ are strictly increasing functions of $\alpha$ on $(\mathrm{a}-1/2, 2\mathrm{a})$.
\end{proposition}
\begin{proof}
We sketch the proof only for the case in which $\mathrm{a}\geq 1/2$. The proof for $0<\mathrm{a}<1/2$ is similar. Define the interval $\mathrm{K}=\big(\max\big\{\mathrm{a}, \alpha-\mathrm{a}+1\big\}, \mathrm{a}+N-1\big)$. Note that the hypotheses of the discrete Stieltjes theorem are fulfilled in $\mathrm{K}$. Thus we only need to prove that $\mathrm{S}_{(2\mathrm{a}-1, 2\mathrm{a})}\subset \mathrm{K}$\footnote{It is immediate that $\mathrm{S}_{(-1, 2\mathrm{a}-1]} \subset \mathrm{K}$.}.  Note that $W_n^{(\alpha)}\big((q^{\mathrm{a}}+q^{-\mathrm{a}})/2; q\big)=1$ and
\begin{align*}
 W_n^{(\alpha)}\big((q^{\alpha-\mathrm{a}+1}+q^{\mathrm{a}-\alpha-1})/2; q\big)&=\pPq{2}{1}{q^{-n}, q^{2\mathrm{a}-\alpha-1}}{q^{1-N}}{q, q}\\[7pt]
 &=q^{n(2\mathrm{a}-\alpha-1)}\frac{(q^{-2\mathrm{a}+\alpha-N-2}; q)_n}{(q^{1-N}; q)_n}>0,
 \end{align*}
the last equality being a consequence of q-Chu-Vandermonde's identity (\cite[(1.11.5)]{KLS10}). The rest of the proof runs as in Proposition \ref{qRacah}.
\end{proof}

\begin{obs}
In \cite[Section 14.7]{KLS10}, the dual q-Hahn polynomials are defined by
$$
\widetilde{W}_n^{(\alpha, \beta)}(\widetilde{x}(s); q)=\pPq{3}{2}{q^{-n}, q^{\mathrm{a}-s},q^{s+\mathrm{a}}}{q^{\alpha+1}, q^{1-N}}{q,q}
$$
$(n=1,\dots, N-1; \mathrm{a}=0, \mathrm{b}=N; \alpha>-1, \beta>-1\; \text{or}\; \alpha<-N, \beta<-N)$, where $\widetilde{x}(s)=\delta q^{s+\alpha+\beta+1}+q^{-s}$.  In \eqref{DqH}, we can write $W^{(\alpha, \mathrm{a})}_n(\cdot; q)$ instead of $W^{(\alpha)}_n(\cdot; q)$. Hence, for  $\mathrm{a}=(\alpha+\beta+1)/2$ fixed,
$$
W^{(\alpha, (\alpha+\beta+1)/2)}_n(x(s); q)=\widetilde{W}_n^{(\alpha, \beta)}\left(\widetilde{x}\left(s-(\alpha+\beta+1)/2\right); q\right),
$$
where
$$
\widetilde{x}\left(s-(\alpha+\beta+1)/2\right)=2 q^{(\alpha+\beta+1)/2} x(s).
$$
Consequently, Proposition \ref{DqHahn} remains valid if we replace $W^{(\alpha)}_n(\cdot; q)$ by $\widetilde{W}_n^{(\alpha, \beta)}$ $(\cdot; q)$ and assume that $\alpha+\beta$ is constant.
\end{obs}

\section*{Acknowledgements}
The authors thank the Warsaw University Library for kindly sending them the book \cite{NU78} and the Keldysh Institute of Applied Mathematics for making \cite{NU83} available at \url{https://keldysh.ru/papers/1983/prep1983_17.pdf} after they request. KC is supported by the Centre for Mathematics of the University of Coimbra - UIDB/00324/2020, funded by the Portuguese Government through FCT/ MCTES. FRR and AS are supported by the Funda\c{c}\~ao de Amparo \`a Pesquisa do Estado de Minas Gerais (FAPEMIG) Demanda Universal under the grant APQ-03782-18, Conselho Nacional de Desenvolvimento Cient\'\i fico e Tecnol\'ogico (CNPq), and Coordena\c{c}\~ao de Aperfei\c{c}oamento de Pessoal de N\'\i vel Superior (CAPES).
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TITLE: Smolin on Cosmological selection and neutron stars
QUESTION [3 upvotes]: Regarding the cosmological selection hypothesis and testable predictions, Lee Smolin asserted the following:

"Smolin: I did make two predictions which were eminently checkable by
  astrophysical and cosmological observations, and both of them could
  easily have been falsified by observations over the last 20 years, and
  both have been confirmed by observations so far.
One of them concerns the masses of neutron stars and the prediction is
  there can't be a neutron star heavier than about twice the mass of the
  sun. This continues to be confirmed by the best measurements of the
  masses of neutron stars."

What is he referring to? As far as I know, the neutron star mass limit is a prediction of GR, and doesn't suggest that any fine-tuning is involved in it. Is this correct?
http://www.space.com/21335-black-holes-time-universe-creation.html

REPLY [3 votes]: The question is dealt with on pages 302 to 304 of Smolin's book The Life of the Cosmos.
The reasoning is based on a proposal by Hans Bethe and Gerald Brown (I don't have the reference to hand) that in a neutron star kaons can become light (by a mechanism analogous to superconductivity), and indeed can become light enough for an electron to decay into a kaon and a neutrino. Smolin argues that if Bethe and Brown are correct this decreases the mass at which a neutron star will collapse to somewhere around 1.5 solar masses (the book doesn't explain the calculation).
Smolin's idea is based on the tendancy of the universe to maximise black hole formation. Because the Bethe-Brown mechanism would increase the number of black holes formed, Smolin predicts the Bethe-Brown mechanism does occur and therefore than no neutron stars will be found with a mass of greater than two solar masses.
I should caution that Smolin's idea of cosmological natural selection is not widely accepted, and relies upon the unproven idea that a black hole nucleates a new universe with slightly different physical laws. Until such time as evidence emerges to support it I should treat the idea with some caution. Having said this, Smolin's book is a fascinating read. I thoroughly enjoyed reading it and I recommend it to everyone (who isn't too credulous!).
Later:
Smolin's paper The status of cosmological natural selection (last updated in 2008) places the mass limit for kaon condensate stars at 1.6 solar masses. That means that if the J1614-2230 mass measurement is accurate it does indeed rule out Smolin's idea. Having said this, we don't know the equation of state for such dense objects and I'm sure there is wiggle room.


TITLE: Is the Moyal-Liouville equation $\frac{\partial \rho}{\partial t}= \frac{1}{i\hbar} [H\stackrel{\star}{,}\rho]$ used in applications?
QUESTION [18 upvotes]: This answer by Qmechanic shows that the classical Liouville equation can be extended to quantum mechanics by the use of Moyal star products, where it takes the form
$$
\frac{\partial \rho}{\partial t}~=~ \frac{1}{i\hbar} [H\stackrel{\star}{,}\rho].
$$
I've seen this mentioned as a possible classical interpretation of quantum mechanics, but I've never seen this used in anger, so I want to ask: is this representation actually useful in understanding dynamics? In what contexts does it provide insights about physical processes that are harder to obtain through alternative routes?
I'm not really looking for an exhaustive list of places where this representation turns up, but I would like to see a few representative examples to get a better feel of how it looks like in applications.

REPLY [2 votes]: The Moyal-Liouville equation can be regarded as an application of the Wigner transformation to the von Neumann equation. This means you are thinking in the context of density matrices.
There is of course also an equivalent approach based on the single particle Green's function. Its Dyson equation in the phase space representation reads
$$
( \epsilon - H(x,p) + i0^+)~\star~G^\mathrm{R}(x,p) = \mathrm{id},
$$
with the $\star$-product in 4-vector notation defined by
$$
\star \equiv \exp \left\lbrace
\frac{i \hbar}{2} \left(
\overset{\leftarrow}{\partial}_{x^\mu} \overset{\rightarrow}{\partial}_{p_\mu} -
\overset{\leftarrow}{\partial}_{p_\mu} \overset{\rightarrow}{\partial}_{x^\mu}
\right)
\right\rbrace
$$
Let's say you are trying to describe the effect of a homogeneous magnetic field $B$ which acts on your orbital degree of freedom in a system of crystal electrons. If you were about to use standard perturbation theory you are facing a problem: the associated vector potential $A$ is space dependent, and this breaks the translational symmetry of the crystal. In general, you can not make use of the Bloch theorem and the situation seems hopeless.
The canonical trick would then be to introduce a periodic modulation of the magnetic field on a length scale $\lambda$. Perform your calculations and then take the limit $\lambda\to\infty$ at the end of the day. This way it is possible to keep your Bloch waves intact at all stages of the derivation.
There is however a much easier way based on the Dyson equation above. If you perform the shift $p \to p + e A(x) \equiv \pi$, you can make a change of variables in the Dyson equation, altering the definition of the $\star$-product to
$$
\star \rightarrow \exp \left\lbrace
\frac{i \hbar}{2} \left(
\overset{\leftarrow}{\partial}_{x^\mu} \overset{\rightarrow}{\partial}_{\pi_\mu} -
\overset{\leftarrow}{\partial}_{\pi_\mu} \overset{\rightarrow}{\partial}_{x^\mu}
- e F^{\mu\nu} \overset{\leftarrow}{\partial}_{\pi^\mu} \overset{\rightarrow}{\partial}_{\pi^\nu}
\right)
\right\rbrace .
$$
Simply by expanding the Dyson equation in powers of the field tensor $F$, you have now a way to treat magnetic fields in a way which is much simpler then based on standard quantum mechanics.
References
[1] Onoda, Sugimoto, Nagaosa (2006) https://arxiv.org/abs/cond-mat/0605363
[2] Zhu et al. (2012) https://arxiv.org/abs/1201.3446


TITLE: Difference between unpolarized beam and half H+V polarized beam
QUESTION [0 upvotes]: Let's say that we have got two light beams. Beam A is not polarized at all. Beam B is composed of half H polarized photons + half V polarized photons.
Is there any way to measure those two beams in order to distinguish one from the other ?
The same question applies for a single photon : Is it possible to distinguish a unpolarized single photon from a H or V polarized photon ?
Thanks a lot for your answers !

REPLY [0 votes]: This

beam B is composed of half H polarized photons + half V polarized photons

is an insufficient specification of the polarization state of beam B, because you do not specify whether the two sources are combined in a coherent or incoherent fashion.

If the two components are incoherent, then the combination is unpolarized, and it is indistinguishable from beam A.
If the two components are coherent with each other, then their superposition will be fully polarized, but the polarization state of the combination will depend on the relative phase of the superposition.
If the two beams are in phase (in the usual convention) then the combination will be linearly polarized along an axis at 45° from both, and rotating a linear polarizer will produce a Malus-law pattern that will allow you to distinguish it from beam A.
If the beams are 90° out of phase, on the other hand, then you will have a circular polarization, and you will need a circular analyzer to provide a suitable benchmark.

Either way, if the combination is coherent (or even partially coherent) then the thing to do is to perform a full polarization tomography measurement on the combination to measure its Stokes parameters, which encode the full information about the polarization of the beam.

As for the second half,

The same question applies for a single photon: Is it possible to distinguish a unpolarized single photon from a H or V polarized photon?

it's important to note that unpolarized single photons are only possible if you allow your system to occupy mixed quantum states, which can be undesirable in several circumstances.
The core of your question, however, is one of quantum state discrimination, and it's not completely a trivial question.

If you have a single copy of the photon, then it cannot be done.
If you have multiple copies then you can pass the photon through a polarizing beam splitter and measure the output. If you get hits on two channels, then you know it's not a single such polarization, whereas if you only get hits on a single output port, then the more hits you get the more confident you can be about the purity of the state along that polarization.


TITLE: Universal Linearizer of Alternating Multi-$F[x]$-Linear Maps is Same as that of Multi-$F$-Linear Maps.
QUESTION [1 upvotes]: Let $V$ be a an $n$-dimensional vector space over a field $F$. Let $M=F[x]\otimes_F V$. We can consider $M$ as an $F[x]$-module by extending scalars using the inclusion $F\to F[x]$.
Fact 1. There is an $F$-module $\Lambda^k_FM$ and a multi-$F$-linear map $\pi:M^k\to \Lambda^k_F M$ such that all alternating multi-$F$-linear maps defined on $M^k$ factor through $\pi$ to give a unique $F$-linear map from $\Lambda^k_F M$.
Fact 2. There is an $F[x]$-module $\Lambda^k_{F[x]}M$ and a multi-$F[x]$-linear map $\Pi:M^k\to \Lambda^k_{F[x]} M$ such that all alternating multi-$F[x]$-linear maps defined on $M^k$ factor through $\Pi$ to give a unique $F[x]$-linear map from $\Lambda^k_{F[x]}M$.
Note that we are treating $\Lambda^k_F M$ and $\Lambda^k_{F[x]} M$ merely as notations. They may have nothing to do with each other.

Question. Is it true that the gadget $\Pi:M^k\to \Lambda^k_{F[x]}M$ fills the bill for the gadget $\pi:M^k\to \Lambda^k_F M$?

In other words, if suppose we have a multi-$F$-linear map $f:M^k\to N$, then does there exists a unique $F$-linear map $\tilde f:\Lambda^k_{F[x]} M\to N$ whose composition with $\Pi$ is same as $f$?
The motivation for this question is from my discussion with @MooS in this thread.

REPLY [1 votes]: No, this is not true.  For instance, take $n=k=2$, so $M=F[x]^2$, $\Lambda^k_{F[x]}M=F[x]$, and $\Pi:M^2\to F[x]$ is the map that sends $((a,b),(c,d))$ to $ad-bc$ (for $a,b,c,d\in F[x]$).  Write $a_n$ for the degree $n$ coefficient of a polynomial $a\in F[x]$.  Let $N=F$ and define $f:M^2\to N$ by $f((a,b),(c,d))=a_0d_1-b_1c_0$.  Then $f$ is $F$-bilinear and alternating.  However, $f$ cannot factor though $\Pi$: for instance, $\Pi((1,1),(x,x))=0$ but $f((1,1),(x,x))=1$.


TITLE: A list of Multiple Zeta values of depth three
QUESTION [8 upvotes]: The multiple zeta function of depth three has a following integral representation:
\begin{eqnarray}
\zeta(t^{-1},p,q,r) &:=& \sum\limits_{m_1 > m_2 > m_3 > 0} \frac{t^{m_1}}{m_1^p} \frac{1}{m_2^q} \frac{1}{m_3^r} \\
&=& \int\limits_0^t \frac{[\log(t/\xi)]^{p-1}}{(p-1)!} \int\limits_0^{\xi} \frac{[\log(\xi/\xi_1)]^{q-1}}{(q-1)!} \cdot \frac{Li_r(\xi_1)}{1-\xi_1} d\xi_1 \frac{1}{1-\xi} d\xi
\end{eqnarray}
By using the formula above we calculated the following list of values at plus unity for weights from four to six. We have:
\begin{eqnarray}
\zeta(2,1,1) &=& \zeta(4)\\
\hline\\
\zeta(3,1,1) &=& -  \zeta(2) \zeta(3) + 2 \zeta(5)\\
\zeta(2,2,1) &=& + 3\zeta(2) \zeta(3) -\frac{11}{2} \zeta(5)\\
\zeta(2,1,2) &=& - 2\zeta(2) \zeta(3) + \frac{9}{2} \zeta(5)\\
\hline \\
\zeta(4,1,1) &=& -  \zeta(3)^2 + \frac{23}{16} \zeta(6)\\
\zeta(3,2,1) &=& +3  \zeta(3)^2 - \frac{203}{48} \zeta(6)\\
\zeta(3,1,2) &=& -\frac{3}{2}  \zeta(3)^2 + \frac{53}{24} \zeta(6)\\
\zeta(2,3,1) &=& -\frac{3}{2}  \zeta(3)^2 + \frac{53}{24} \zeta(6)\\
\zeta(2,2,2) &=& \frac{3}{16} \zeta(6) \\
\zeta(2,1,3) &=& + \zeta(3)^2 - \frac{13}{16} \zeta(6)\\
\hline
\zeta(5,1,1)&=&-\frac{5}{4} \zeta(3) \zeta(4)-2 \zeta(2)\zeta(5) +5 \zeta(7)\\
\zeta(4,2,1)&=&+\frac{-236 \zeta(3) \zeta(2)^2+28776 \zeta(5) \zeta(2)+18902 \zeta(3) \zeta(4)-72267 \zeta(7)}{5232}\\
\zeta(4,1,2)&=&-\frac{451}{327} \zeta(3) \zeta(2)^2+\frac{5}{2} \zeta(5) \zeta(2)-\frac{395 \zeta(3) \zeta(4)}{1308}+\frac{5 \zeta(7)}{8}\\
\zeta(3,3,1)&=&-\frac{77}{109} \zeta(3)\zeta(2)^2-\frac{9}{2} \zeta(5) \zeta(2)+\frac{385}{218} \zeta(3) \zeta(4)+\frac{61 \zeta(7)}{8}\\
\zeta(3,2,2)&=&+\frac{15908 \zeta(3) \zeta(2)^2-39240 \zeta(5) \zeta(2)-27998 \zeta(3) \zeta(4)+51339 \zeta(7)}{5232}\\
\zeta(3,1,3)&=&+\frac{1600 \zeta(3) \zeta(2)^2-3673 \zeta(3) \zeta(4)-327\zeta(7)}{1308}\\
\zeta(2,4,1)&=&+\frac{59 \zeta(3) \zeta(2)^2}{1308}+5 \zeta(5) \zeta(2)-\frac{3565 \zeta(3) \zeta(4)}{2616}-\frac{109 \zeta(7)}{16}\\
\zeta(2,3,2)&=&-\frac{1519}{654} \zeta(3) \zeta(2)^2-\frac{11}{2} \zeta(5) \zeta(2)+\frac{7595 \zeta(3) \zeta(4)}{1308}+\frac{75 \zeta(7)}{8}\\
\zeta(2,2,3)&=&+\frac{844 \zeta(3) \zeta(2)^2+62784 \zeta(5) \zeta(2)-9958 \zeta(3) \zeta(4)-95157 \zeta(7)}{5232}\\
\zeta(2,1,4)&=&+\frac{1}{8} (14 \zeta(3) \zeta(4)-44 \zeta(2) \zeta(5)+61 \zeta(7))\\
\hline\\
\zeta(6,1,1)&=&+\frac{1}{2} \zeta (3)^2 \zeta(2)-3 \zeta (3) \zeta (5)+\frac{61 \zeta(8)}{24}\\
\zeta(5,2,1)&=&+\frac{7}{4} \zeta(6,2)-\zeta (3)^2 \zeta(2)+\frac{7 \zeta (3) \zeta (5)}{2}-\frac{289\zeta(8)}{144}\\
\zeta(5,1,2)&=&-\zeta(6,2)-\frac{3}{2} \zeta (3)^2 \zeta(2)+\frac{9 \zeta (3) \zeta (5)}{2}-\frac{145 \zeta(8)}{72}\\
\zeta(4,3,1)&=&-\frac{25}{4} \zeta(6,2)+\frac{1}{2} \zeta (3)^2\zeta(2)+\frac{21 \zeta (3) \zeta (5)}{2}-\frac{677 \zeta(8)}{48}\\
\zeta(4,2,2)&=&+\frac{9}{2} \zeta(6,2)+3 \zeta (3)^2 \zeta(2)-20 \zeta (3) \zeta (5)+\frac{1271\zeta(8)}{72}\\
\zeta(4,1,3)&=&+\frac{5}{2} \zeta(6,2)+\frac{1}{2} \zeta (3)^2 \zeta(2)-\frac{15 \zeta (3) \zeta (5)}{2}+\frac{583 \zeta(8)}{72}\\
\zeta(3,4,1)&=&+\frac{15}{4} \zeta(6,2)-2 \zeta (3)^2\zeta(2)+\frac{673 \zeta(8)}{144}\\
\zeta(3,3,2)&=&+\frac{13}{4} \zeta(6,2)+\frac{9}{2} \zeta (3)^2 \zeta(2)-23 \zeta (3) \zeta (5)+\frac{857 \zeta(8)}{48}\\
\zeta(3,2,3)&=&-10 \zeta(6,2)-6\zeta (3)^2 \zeta(2)+\frac{89 \zeta (3) \zeta (5)}{2}-\frac{245 \zeta(8)}{6}\\
\zeta(3,1,4)&=&+\frac{3}{2} \zeta (3)^2 \zeta(2)-\frac{11 \zeta (3) \zeta (5)}{2}+\frac{241\zeta(8)}{72}\\
\zeta(2,5,1)&=&+\frac{7}{4} \zeta(6,2)+2 \zeta (3)^2 \zeta(2)-12 \zeta (3) \zeta (5)+\frac{487 \zeta(8)}{48}\\
\zeta(2,4,2)&=&-10 \zeta(6,2)-5 \zeta (3)^2 \zeta(2)+40 \zeta (3)\zeta (5)-\frac{677 \zeta(8)}{18}\\
\zeta(2,3,3)&=&+\frac{27}{4} \zeta(6,2)+2 \zeta (3)^2 \zeta(2)-\frac{45 \zeta (3) \zeta (5)}{2}+\frac{1111 \zeta(8)}{48}\\
\zeta(2,2,4)&=&+\frac{11}{2} \zeta(6,2)+2\zeta (3)^2 \zeta(2)-20 \zeta (3) \zeta (5)+\frac{121 \zeta(8)}{6}\\
\zeta(2,1,5)&=&-\frac{5}{2} \zeta(6,2)-\zeta (3)^2 \zeta(2)+\frac{21 \zeta (3) \zeta (5)}{2}-\frac{181 \zeta(8)}{18}\\
\hline\\
\zeta(7,1,1)&=&-3 \zeta (7) \zeta(2)-\frac{9}{4} \zeta (5) \zeta(4)-\frac{7}{4} \zeta (3) \zeta(6)+\frac{28 \zeta (9)}{3}+\frac{\zeta (3)^3}{6}\\
\zeta(6,2,1)&=&+11 \zeta (7) \zeta(2)+\frac{13}{2} \zeta(5) \zeta(4)+\frac{9}{2} \zeta (3) \zeta(6)-\frac{2189 \zeta (9)}{72}-\frac{\zeta (3)^3}{3}\\
\zeta(6,1,2)&=&+7 \zeta (7) \zeta(2)-\frac{1}{4} \zeta (5) \zeta(4)-\frac{5}{3} \zeta (3)\zeta(6)-\frac{313 \zeta (9)}{36}-\frac{\zeta (3)^3}{3}\\
\zeta(5,3,1)&=&-17 \zeta (7) \zeta(2)-\frac{23}{4} \zeta (5) \zeta(4)-\frac{3}{4} \zeta (3) \zeta(6)+\frac{845 \zeta (9)}{24}+\frac{\zeta(3)^3}{6}\\
\zeta(5,2,2)&=&-21 \zeta (7) \zeta(2)+\frac{7}{4} \zeta (5) \zeta(4)-\frac{8}{3} \zeta (3) \zeta(6)+\frac{2513 \zeta (9)}{72}+\frac{2 \zeta (3)^3}{3}\\
\zeta(5,1,3)&=&-7 \zeta (7)\zeta(2)+\frac{1}{2} \zeta (5) \zeta(4)+\frac{5}{4} \zeta (3) \zeta(6)+\frac{121 \zeta (9)}{12}-\frac{\zeta (3)^3}{3}\\
\zeta(4,4,1)&=&+18 \zeta (7) \zeta(2)+5 \zeta (5) \zeta(4)+\frac{4}{3}\zeta (3) \zeta(6)-\frac{328 \zeta (9)}{9}-\frac{\zeta (3)^3}{3}\\
\zeta(4,3,2)&=&+14 \zeta (7) \zeta(2)-\frac{35}{2} \zeta (5) \zeta(4)-\frac{8}{3} \zeta (3) \zeta(6)-\frac{53 \zeta (9)}{36}+\frac{2\zeta (3)^3}{3}\\
\zeta(4,2,3)&=&+28 \zeta (7) \zeta(2)+10 \zeta (5) \zeta(4)+\frac{5}{3} \zeta (3) \zeta(6)-59 \zeta (9)-\frac{\zeta (3)^3}{3}\\
\zeta(4,1,4)&=&-3 \zeta (5) \zeta(4)-\frac{41}{12} \zeta (3)\zeta(6)+\frac{115 \zeta (9)}{18}+\frac{2 \zeta (3)^3}{3}\\
\zeta(3,5,1)&=&-10 \zeta (7) \zeta(2)+\frac{5}{4} \zeta (5) \zeta(4)+\frac{5}{4} \zeta (3) \zeta(6)+\frac{341 \zeta (9)}{24}-\frac{\zeta(3)^3}{3}\\
\zeta(3,4,2)&=&-14 \zeta (7) \zeta(2)+\frac{15}{2} \zeta (5) \zeta(4)-\frac{7}{3} \zeta (3) \zeta(6)+\frac{593 \zeta (9)}{36}+\frac{2 \zeta (3)^3}{3}\\
\zeta(3,3,3)&=&-\frac{1}{2} \zeta (3)\zeta(6)+\frac{\zeta (9)}{3}+\frac{\zeta (3)^3}{6}\\
\zeta(3,2,4)&=&-28 \zeta (7) \zeta(2)-\frac{11}{2} \zeta (5) \zeta(4)+\frac{29}{4} \zeta (3) \zeta(6)+46 \zeta (9)-\frac{4 \zeta(3)^3}{3}\\
\zeta(3,1,5)&=&+7 \zeta (7) \zeta(2)-\frac{1}{4} \zeta (5) \zeta(4)-\frac{1}{2} \zeta (3) \zeta(6)-\frac{131 \zeta (9)}{12}+\frac{\zeta (3)^3}{6}\\
\zeta(2,6,1)&=&+7 \zeta (7) \zeta(2)-\frac{7}{4}\zeta (5) \zeta(4)-\frac{43}{12} \zeta (3) \zeta(6)-\frac{461 \zeta (9)}{72}+\frac{2 \zeta (3)^3}{3}\\
\zeta(2,5,2)&=&-11 \zeta (7) \zeta(2)+\frac{41}{6} \zeta (3) \zeta(6)+\frac{439 \zeta(9)}{36}-\frac{4 \zeta (3)^3}{3}\\
\zeta(2,4,3)&=&+31 \zeta (7) \zeta(2)-\frac{15}{2} \zeta (5) \zeta(4)+\zeta (3) \zeta(6)-\frac{1567 \zeta (9)}{36}-\frac{\zeta (3)^3}{3}\\
\zeta(2,3,4)&=&-32 \zeta (7)\zeta(2)+12 \zeta (5) \zeta(4)-\frac{25}{6} \zeta (3) \zeta(6)+\frac{1567 \zeta (9)}{36}+\frac{2 \zeta (3)^3}{3}\\
\zeta(2,2,5)&=&+31 \zeta (7) \zeta(2)-\zeta (5) \zeta(4)-\frac{25}{6} \zeta (3)\zeta(6)-\frac{3319 \zeta (9)}{72}+\frac{2 \zeta (3)^3}{3}\\
\zeta(2,1,6)&=&-11 \zeta (7) \zeta(2)+\frac{1}{4} \zeta (5) \zeta(4)+\frac{37}{12} \zeta (3) \zeta(6)+\frac{551 \zeta (9)}{36}-\frac{\zeta(3)^3}{3}\\
\hline\\
\zeta(8,1,1)&=&+\zeta (3) \zeta (5) \zeta(2)+\frac{1}{2} \zeta (3)^2 \zeta(4)+\frac{333 \zeta(10)}{80}-4 \zeta (3) \zeta (7)-2 \zeta (5)^2\\
\zeta(7,2,1)&=&-\zeta (3)^2 \zeta(4)-\frac{377\zeta(10)}{60}-\zeta(2) \zeta(6,2)+\frac{9}{4} \zeta(8,2)+\frac{9 \zeta (3) \zeta (7)}{2}+\frac{9 \zeta (5)^2}{4}\\
\zeta(7,1,2)&=&-7 \zeta (3) \zeta (5) \zeta(2)-\zeta (3)^2\zeta(4)-\frac{77 \zeta(10)}{48}+\zeta(2) \zeta(6,2)-\zeta(8,2)+10 \zeta (3) \zeta (7)+5 \zeta (5)^2\\
\zeta(6,3,1)&=&-4 \zeta (3) \zeta (5) \zeta(2)+\frac{1}{2} \zeta (3)^2\zeta(4)-\frac{3219 \zeta(10)}{160}+\frac{5}{2} \zeta(2) \zeta(6,2)-\frac{35}{4} \zeta(8,2)+\frac{29 \zeta (3) \zeta (7)}{2}+\frac{37 \zeta (5)^2}{4}\\
\zeta(6,2,2)&=&+8 \zeta (3) \zeta (5)\zeta(2)+2 \zeta (3)^2 \zeta(4)+\frac{539 \zeta(10)}{16}+\zeta(2) \zeta(6,2)+3 \zeta(8,2)-28 \zeta (3) \zeta (7)-18 \zeta (5)^2\\
\zeta(6,1,3)&=&+16 \zeta (3) \zeta (5)\zeta(2)-\frac{1}{2} \zeta (3)^2 \zeta(4)+\frac{511 \zeta(10)}{160}-\frac{7}{2} \zeta(2) \zeta(6,2)+\frac{7}{2} \zeta(8,2)-21 \zeta (3) \zeta (7)-9 \zeta (5)^2\\
\zeta(5,4,1)&=&-5 \zeta(3) \zeta (5) \zeta(2)-\zeta (3)^2 \zeta(4)+\frac{9209 \zeta(10)}{240}+\frac{21}{2} \zeta(8,2)-10 \zeta (3) \zeta (7)-\frac{27 \zeta (5)^2}{2}\\
\zeta(5,3,2)&=&+35 \zeta (3) \zeta (5) \zeta(2)+2\zeta (3)^2 \zeta(4)-\frac{5683 \zeta(10)}{160}-\frac{25}{2} \zeta(2) \zeta(6,2)+\frac{19}{4} \zeta(8,2)-\frac{63 \zeta (3) \zeta (7)}{2}-\frac{3 \zeta (5)^2}{4}\\
\zeta(5,2,3)&=&-50 \zeta (3)\zeta (5) \zeta(2)-2 \zeta (3)^2 \zeta(4)-\frac{719 \zeta(10)}{20}+10 \zeta(2) \zeta(6,2)-\frac{49}{4} \zeta(8,2)+\frac{161 \zeta (3) \zeta (7)}{2}+\frac{163 \zeta(5)^2}{4}\\
\zeta(5,1,4)&=&-10 \zeta (3) \zeta (5) \zeta(2)+\frac{1}{2} \zeta (3)^2 \zeta(4)-\frac{1027 \zeta(10)}{480}+\frac{5}{2} \zeta(2) \zeta(6,2)-\frac{7}{2} \zeta(8,2)+14 \zeta (3)\zeta (7)+\frac{9 \zeta (5)^2}{2}\\
\zeta(4,5,1)&=&+15 \zeta (3) \zeta (5) \zeta(2)-\frac{1173 \zeta(10)}{32}-\frac{5}{2} \zeta(2) \zeta(6,2)-7 \zeta(8,2)-3 \zeta (3) \zeta (7)+9 \zeta(5)^2\\
\zeta(4,4,2)&=&-40 \zeta (3) \zeta (5) \zeta(2)+\frac{249 \zeta(10)}{40}+10 \zeta(2) \zeta(6,2)-\frac{15}{2} \zeta(8,2)+49 \zeta (3) \zeta (7)+15 \zeta (5)^2\\
\zeta(4,3,3)&=&-20 \zeta (3) \zeta(5) \zeta(2)+\frac{3233 \zeta(10)}{40}+10 \zeta(2) \zeta(6,2)+\frac{7}{2} \zeta(8,2)-7 \zeta (3) \zeta (7)-\frac{59 \zeta (5)^2}{2}\\
\zeta(4,2,4)&=&+80 \zeta (3) \zeta (5) \zeta(2)+\zeta(3)^2 \zeta(4)-\frac{\zeta(10)}{6}-20 \zeta(2) \zeta(6,2)+\frac{35}{2} \zeta(8,2)-105 \zeta (3) \zeta (7)-35 \zeta (5)^2\\
\zeta(4,1,5)&=&-11 \zeta (3) \zeta (5) \zeta(2)+\zeta (3)^2\zeta(4)+\frac{4937 \zeta(10)}{480}+\frac{5}{2} \zeta(2) \zeta(6,2)+7 \zeta (3) \zeta (7)+2 \zeta (5)^2\\
\zeta(3,6,1)&=&-13 \zeta (3) \zeta (5) \zeta(2)-\zeta (3)^2 \zeta(4)+\frac{53\zeta(10)}{40}+\zeta(2) \zeta(6,2)+\frac{7}{4} \zeta(8,2)+\frac{35 \zeta (3) \zeta (7)}{2}+\frac{21 \zeta (5)^2}{4}\\
\zeta(3,5,2)&=&+20 \zeta (3) \zeta (5) \zeta(2)+2 \zeta (3)^2\zeta(4)+\frac{2303 \zeta(10)}{32}+\frac{5}{2} \zeta(2) \zeta(6,2)+\frac{13}{2} \zeta(8,2)-60 \zeta (3) \zeta (7)-\frac{81 \zeta (5)^2}{2}\\
\zeta(3,4,3)&=&+30 \zeta (3) \zeta (5)\zeta(2)-\frac{2703 \zeta(10)}{20}-20 \zeta(2) \zeta(6,2)+17 \zeta (3) \zeta (7)+50 \zeta (5)^2\\
\zeta(3,3,4)&=&-10 \zeta (3) \zeta (5) \zeta(2)+\frac{1}{2} \zeta (3)^2\zeta(4)+\frac{2191 \zeta(10)}{40}+10 \zeta(2) \zeta(6,2)-\frac{7}{2} \zeta(8,2)-11 \zeta (3) \zeta (7)-\frac{41 \zeta (5)^2}{2}\\
\zeta(3,2,5)&=&-52 \zeta (3) \zeta (5) \zeta(2)-4 \zeta(3)^2 \zeta(4)-\frac{241 \zeta(10)}{5}+10 \zeta(2) \zeta(6,2)-\frac{63}{4} \zeta(8,2)+\frac{195 \zeta (3) \zeta (7)}{2}+\frac{159 \zeta (5)^2}{4}\\
\zeta(3,1,6)&=&+17 \zeta (3) \zeta (5)\zeta(2)+\frac{1}{2} \zeta (3)^2 \zeta(4)+\frac{891 \zeta(10)}{160}-\frac{7}{2} \zeta(2) \zeta(6,2)+\frac{7}{2} \zeta(8,2)-25 \zeta (3) \zeta (7)-10 \zeta (5)^2\\
\zeta(2,7,1)&=&+6 \zeta(3) \zeta (5) \zeta(2)+2 \zeta (3)^2 \zeta(4)+\frac{419 \zeta(10)}{20}+\frac{9}{4} \zeta(8,2)-\frac{41 \zeta (3) \zeta (7)}{2}-\frac{43 \zeta (5)^2}{4}\\
\zeta(2,6,2)&=&-16 \zeta (3) \zeta (5)\zeta(2)-4 \zeta (3)^2 \zeta(4)-\frac{3249 \zeta(10)}{40}-\zeta(2) \zeta(6,2)-\frac{21}{2} \zeta(8,2)+63 \zeta (3) \zeta (7)+41 \zeta (5)^2\\
\zeta(2,5,3)&=&+20 \zeta (3) \zeta (5)\zeta(2)+\frac{5969 \zeta(10)}{80}+11 \zeta(8,2)-56 \zeta (3) \zeta (7)-\frac{89 \zeta (5)^2}{2}\\
\zeta(2,4,4)&=&-40 \zeta (3) \zeta (5) \zeta(2)-\zeta (3)^2 \zeta(4)-\frac{289\zeta(10)}{48}+10 \zeta(2) \zeta(6,2)-10 \zeta(8,2)+56 \zeta (3) \zeta (7)+20 \zeta (5)^2\\
\zeta(2,3,5)&=&+28 \zeta (3) \zeta (5) \zeta(2)+2 \zeta (3)^2 \zeta(4)-\frac{425\zeta(10)}{16}-10 \zeta(2) \zeta(6,2)+\frac{23}{4} \zeta(8,2)-\frac{63 \zeta (3) \zeta (7)}{2}+\frac{17 \zeta (5)^2}{4}\\
\zeta(2,2,6)&=&+8 \zeta (3) \zeta (5) \zeta(2)+2 \zeta (3)^2\zeta(4)+\frac{3817 \zeta(10)}{80}+\frac{15}{2} \zeta(8,2)-35 \zeta (3) \zeta (7)-23 \zeta (5)^2\\
\zeta(2,1,7)&=&-6 \zeta (3) \zeta (5) \zeta(2)-\zeta (3)^2 \zeta(4)-\frac{256\zeta(10)}{15}+\zeta(2) \zeta(6,2)-\frac{7}{2} \zeta(8,2)+18 \zeta (3) \zeta (7)+9 \zeta (5)^2\\
\hline\\
\end{eqnarray}
For each weight it turns out that it is pretty easy to calculate the quantity on the very top and that the complexity increases as we move from the top to the bottom. Now my question is are all multiple zeta values defined above reduce-able to single zeta values and if not what is the lowest weight when this is not the case.  Another question is can we establish recurrence relations (just as we did in Calculating alternating Euler sums of odd powers for the respective quantities of depth two ) between those values and solve them for any given weight.

REPLY [3 votes]: This is not going to be a full answer but a starting point  that will eventually lead to establishing recurrence relations for the quantities in question. Let us define:
\begin{eqnarray}
{\mathfrak F}^{(p,q,r)}_\xi &:=&
(-1)^q \sum\limits_{l=p+1}^{q+p+1} \binom{q}{l-p-1} (l-1)! Li_{r+l}(1) [\log(\xi)]^{q+p+1-l}+\\
&&(-1)^p \sum\limits_{l=q+1}^{q+p+1} \binom{p}{l-q-1} (-1)^l (l-1)! Li_{r+l}(\xi) [\log(\xi)]^{q+p+1-l}
\end{eqnarray}
and $\theta_j := (j+r) \% 2$.
Now, by using An integral involving product of poly-logarithms and a power of a logarithm. we expressed the Multiple Zeta Values (MZVs) as follows.
\begin{eqnarray}
\zeta(p,q,r)&=&
\sum\limits_{j=q}^{\lfloor \frac{q-1+r}{2} \rfloor}
(-1)^{j-q} \binom{j-1}{q-1} \int\limits_0^1 \frac{[\log(1/\xi)]^{p-1}}{(p-1)!} \cdot \frac{Li_j(\xi) Li_{r+q-j}(\xi) Li_0(\xi)}{\xi} d\xi+\\
&& \sum\limits_{j=0}^{q-1}(-1)^{\lfloor\frac{j+r}{2}\rfloor-j} \binom{\lfloor\frac{j+r-1}{2}\rfloor}{j}\frac{1}{2^{\theta_j}} \frac{1}{(p-1)!(q-1-j-\theta_j)!} \int\limits_0^1 {\mathfrak F}^{(p-1,q-1-j-\theta_j,0)}_\xi \cdot \frac{[Li_{\theta_j+\lfloor \frac{j+r}{2} \rfloor}(\xi)]^2}{\xi} d\xi
\end{eqnarray}
Note 1: Let $q=r=1$ and $p\ge 2$. Then the only term that survives in the right hand side above is the $(j=0)$ term in the second sum. Then we have:
\begin{eqnarray}
&&\zeta(p,1,1) = \frac{(-1)^{p-1}}{2(p-1)!} \int\limits_0^1 \frac{Li_0(\xi)}{\xi} [\log(\xi)]^{p-1} [Li_1(\xi)]^2 d\xi\\
&&= \frac{(-1)^{p-1}}{2 (p-1)!} \int\limits_0^1 \frac{[\log(1-\xi)]^{p-1} [\log(\xi)]^2}{\xi} d\xi\\
&&=\frac{(-1)^{p-1}}{2 (p-1)!}\left[ -\frac{1}{3} \Psi^{(p+1)}(1) + \frac{1}{2} \sum\limits_{j=1}^{p-2} \binom{p-1}{j} \left\{ \Psi^{(j+1)}(1) \Psi^{(p-1-j)}(1) + \Psi^{(j)}(1) \Psi^{(p-j)}(1)\right\}
-\frac{1}{3} \sum\limits_{1 \le j < j_1 \le p-2} \binom{p-1}{j,j_1-j,p-1-j_1} \Psi^{(j)}(1) \Psi^{(j_1-j)}(1) \Psi^{(p-1-j_1)}(1)\right]\\
&&=\left\{\zeta(4),2 \zeta (5)-\zeta (3) \zeta(2),\frac{23 \zeta(6)}{16}-\zeta (3)^2,-2 \zeta (5) \zeta(2)-\frac{5}{4} \zeta (3) \zeta(4)+5 \zeta (7),\cdots\right\}
\end{eqnarray}
Now let us take $p \leftarrow p-1$, $q=2$ and $r=1$. The again the only term which survives in the right hand side is the $(j=0)$ term in the second sum. Then we have:
\begin{eqnarray}
&&\zeta(p-1,2,1) =\\
&& \frac{1}{2(p-1)!} \int\limits_0^1 {\mathfrak F}^{(p-2,0,0)}_\xi \cdot \frac{[Li_1(\xi)]^2}{\xi} d\xi\\
&&=
\zeta(3) \zeta(p-1) - \frac{1}{2} \sum\limits_{l=1}^{p-1} \int\limits_0^1 \frac{[Li_1(\xi)]^2}{\xi} Li_l(\xi) \cdot \frac{[\log(1/\xi)]^{p-1-l}}{(p-1-l)!}d\xi\\
&&=
\zeta(3) \zeta(p-1) - \frac{1}{2} \sum\limits_{l=1}^{p-1} \\
&&
\left(
\sum\limits_{l_1=2}^l \binom{p-1-l_1}{p-1-l} (-1)^{l-l_1} \zeta(p-l_1+1,1) \zeta(l_1) + \sum\limits_{l_1=2}^{p-l} \binom{p-1-l_1}{l-1} (-1)^l \zeta(p-l_1+1,1) \zeta(l_1) + \sum\limits_{l_1=1}^{p-l} \left\{ \zeta(p-l_1+1,l_1,1)+\zeta(p-l_1+1,1,l_1)+\zeta(p-l_1+1,l_1+1)+\zeta(p+1,1)\right\}
\right)
\end{eqnarray}
for $p\ge 3$. By simplifying we get the following identity:
\begin{eqnarray}
\zeta(2,1,p) + \zeta(2,p,1)+\zeta(p,2,1) = \zeta(3) \zeta(p) - \zeta(2,p+1)-\zeta(p+2,1)
\end{eqnarray}
for $p \ge 2$.
Now let us take $p\leftarrow p-1$, $q=1$ and $r=2$. We encounter an integral with three poly-logarithms in it. That integrals has been evaluated in my answer to Generating function for cubes of Harmonic numbers I . Taking this into account and simplifying the result we get the following identity:
\begin{eqnarray}
2 \zeta(2,1,p)  + \zeta(2,p,1) - \sum\limits_{j=2}^p \zeta(j,p-j+1,2) \cdot(1_{j<p} + 2 \cdot 1_{j=p})=\\
\zeta(p+3) + 2 \zeta(3) \zeta(p)-\zeta(2) \zeta(p+1) - \zeta(2) \zeta(p,1)-\zeta(2,p+1) - \zeta(p+2,1)
\end{eqnarray}
valid for $p\ge 2$.
Finally let us take $p \leftarrow p-2$, $q=3$ and $r=1$. Then only the $(j=0)$ term in the second sum survives. The only non-trivial integral we come across is the one that contains a square of a polylog of order one, another polylog and a power of a log. That integral has been calculated in my answer to Generating function for cubes of Harmonic numbers I .By simplifying everything we end up with the following identity:
\begin{eqnarray}
&&2 \zeta(p-2,3,1)+\sum\limits_{l=2}^{p-1} \sum\limits_{j=1}^l \sum\limits_{i=1}^{p-1-l}(l-1) A_{l,j} \zeta(p+1-l-i,i+j,l+1-j)=\\
&&\sum\limits_{l=1}^{p-1} \sum\limits_{j=1}^l(l-1)A_{l,j} \cdot \\
&&\left( \zeta(p-l+j+1,l+1-j)+\zeta(p-l+j,l+2-j)+\zeta(p-l+j,l+1-j,1)+\zeta(p-l+j,1,l+1-j)\right) +\\
&&-2(p-2) \zeta(3) \zeta(p-1)-(\zeta(2)^2-3 \zeta(4))\zeta(p-2)
\end{eqnarray}
where $A_{l,j} :=(1_{l<j}+2 \cdot 1_{l=j})$. Here $p\ge 4$.
Note 2: In the generic case of arbitrary $p$, $q$ and $r$ there are only two types of integrals that we are dealing with firstly an integral that involves a square of a polylogarithm and a power of a log and secondly an integral that involves products of three polylogarithms and a power of a log. From An integral involving product of poly-logarithms and a power of a logarithm. we know that the first type of integral always reduces to single zeta values and to bi-variate zeta values (where the later in many cases reduce to single zeta values as well). The second type of integral is more complicated however we conjecture that it also always reduces to single, bi-variate and to triple-variate zeta values all of which have the same weight. As such we will obtain a system of linear equations for all triple-zeta values of a given weight. We will derive those equations and solve them asap.


TITLE: Property of "provability predicate"
QUESTION [1 upvotes]: In A Simple Proof of Gödel’s Incompleteness Theorems, the provability predicate $P$ is defined:

Let $g$ denote the Gödel number function.
Let $\operatorname{Proof}(x,y)$ be a binary predicate that translates "$x$ is the Gödel number of a proof of the formula whose Gödel number is $y$".
Define $\operatorname{Pr}(y) = \exists x \operatorname{Proof}(x,y)$.
Define $P(X) = Pr(g(X))$.

The properties of $P$ are stated but not proved:

If $\vdash X$, then $\vdash P(X)$.
$\vdash P(X \implies Y) \implies (P(X) \implies P(Y))$.
$\vdash P(X) \implies P(P(X))$.

Each seems to be intuitively obvious, but how does one go about proving those properties?

REPLY [5 votes]: These three properties are famous under the name Hilbert-Bernays provability conditions.

The middle one is the easiest. You need to have done a lot of fiddlywork proving things about how Gödel numbers for symbol sequences work in general,  of course, but once you have an appropriate box of tools, it's pretty simple: Your proof system contains an explicit rule (number 5 in the paper you link to) that says a proof of $Y$ can consist of a proof of $X$ together with a proof of $X\Rightarrow Y$. All you need to to is prove that the proof you construct has a Gödel number -- and that the precise definition of $Pr$ (which the author helpfully leaves it to you to work out in detail) works as advertised.

The first one is already pretty tedious -- though not particularly difficult, once you get in the rhythm. You work by induction on the structure of the proof of $X$ -- or in other words, by long induction on the length of the proof of $X$. The rules 4, 5, 6 in your system say that a proof of something can be constructed as a combination of (necessarily shorter) proofs of various other things. When the proof of $X$ concludes with one of these rules, apply the induction hypothesis to those shorter proofs, giving you $P(\cdots)$ for those, and then combine them in the same way as the middle one.
If your proof of $X$ is by one of the rules 1, 2, 3, there are no shorter proof to appeal to -- so what you do there is just verify that your $Pr$ considers each instance of these axioms to be a proof of itself.
Once you cross the t's and dot the i's you have a proof of the first property.

The third property is where it really gets hairy.
To begin with it doesn't actually hold in the system of the paper you link to -- because the proof system that was presented cannot even represent $P$. By definition, $P(X)$ means $\exists x\, \mathit{Proof}(x, g(X))$, but the system of the paper only covers propositional logic. It contains neither a Gödel number for the $\exists$ symbol, nor proof rules to allow you to prove things with $\exists$ in them, nor a way to consider $\exists$ an abbreviation.  And without that, the notation $P(P(X))$ doesn't even make sense!
Also, there don't seem to be any axioms that tell you how arithmetic works, so proving anything about numbers is going to be tough.
But all of that is fixable. Let's suppose we have fixed it. Then, of course, we need to define a new better $P$ for our extended system, and prove the two the Hilbert-Bernays properties once again for the extended system. Much work, not particularly enlightening, but it can be done.
Then you hope that you have not only extended your proof system enough to express $P$ in it, but also to formalize the proofs of ordinary mathematics. Because the strategy is now to prove that everything you did to prove the first property can be formalized in the system you are considering.  If you manage to convince yourself of that, the third property is nothing more than the system proving the first property about itself.


TITLE: Theorem dual to trivial kernel $\iff$ injective in $\boldsymbol{Grp}$
QUESTION [0 upvotes]: Cokernels are "dual" to kernels. I've been told that if X dual to Y and there's a theorem about X then there's a "dual" theorem about Y.
Theorem: In $\boldsymbol{Grp}$ a homomorphism has trivial kernel iff it is injective.
However it is not true that a group homomorphism has trivial cokernel iff it's surjective. So what would instead be the dual theorem to the above theorem? (or have I been lied to and those dual theorems don't always exist?)

REPLY [3 votes]: The dual theorem holds in the opposite category $\boldsymbol{Grp}^{op}$, which is not equivalent to $\boldsymbol{Grp}$.
This sort of duality holds in general when a theorem is true for all categories, or at least some subset of categories that are self dual. If for all categories $\mathcal C$, some theorem holds for $\mathcal C$, then the dual theorem holds for $\mathcal C^{op}$. Since the original theorem also holds for the category $\mathcal C^{op}$, the dual theorem holds for $\mathcal (C^{op})^{op} = \mathcal C$.
This particular theorem doesn't hold for all categories (where kernels, cokernels, injectivity and surjectivity can be defined). Thus, the dual theorem is not guaranteed to hold.


TITLE: Proof of series inequality using triangle inequality
QUESTION [2 upvotes]: Prove the following inequality:
$$|x+x_1+x_2+...+x_n|\ge|x|-(|x_1|+|x_2|+...+|x_n|)$$
I've spent a long time on this one. I've come up with something but I don't think it's correct.
According to the triangle inequality $|x+y|\le|x|+|y|$:
$$|x_1+x_2+...+x_n|\le|x_1|+|x_2|+...+|x_n|$$
So then
$$|x|+|x_1|+|x_2|+...+|x_n|\ge|x|-(|x_1|+|x_2|+...+|x_n|)$$
 $$\Downarrow$$
$$|x_1|+|x_2|+...+|x_n|\ge-|x_1|-|x_2|-...-|x_n|$$
Which is supposed to be proof of the first statement according to some people here but I think we only proved that a bigger number than the one in the L.H.S. is bigger than the one in the R.H.S., which doesn't mean that the L.H.S. is bigger than the R.H.S.
Thanks for your help.

REPLY [2 votes]: Using the triangle inequality, $$|x|=|x+x_1+\cdots +x_n-(x_1+\cdots + x_n)|\le $$
$$\le|x+x_1+\cdots+x_n|+|-x_1|+\cdots +|-x_n|=$$
$$=|x+x_1+\cdots+x_n|+|x_1|+\cdots +|x_n|$$