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On variation of zeros of classical discrete orthogonal

polynomials

K. Castillo

CMUC, Department of Mathematics, University of Coimbra, 3001-501 Coimbra, Portugal

F. R. Rafaeli

FAMAT-UFU, Department of Mathematics, University of Uberlandia, 38408-100

Uberlandia, Minas Gerais, Brazil

A. Suzuki

FAMAT-UFU, Department of Mathematics, University of Uberlandia, 38408-100

Uberlandia, Minas Gerais, Brazil

Abstract

The purpose of this note is to establish, from the hypergeometric-type differenceequation introduced by Nikiforov and Uvarov, new tractable sufficient condi-tions for the monotonicity with respect to a real parameter of zeros of classicaldiscrete orthogonal polynomials. This result allows one to carry out a sys-tematic study of the monotonicity of zeros of classical orthogonal polynomialson linear, quadratic, q-linear, and q-quadratic grids. In particular, we analyzein 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, lit-tle q-Jacobi, little q-Laguerre/Wall, q-Bessel, q-Racah and dual q-Hahn poly-nomials.

Keywords: Classical discrete orthogonal polynomials, linear grid, quadraticgrid, q-linear grid, q-quadratic grid, discrete Stieltjes theorem, monotonicity,zeros.2010 MSC: 30C15, 05A30, 33C45, 33D15

1. Introduction

The properties of Jacobi polynomials, P(α,β)n (n = 1, 2, . . . ;α > −1, β > −1),

may well consult in [23, Chapter IV]; they are orthogonal on [−1, 1] with the

Email addresses: kenier@mat.uc.pt (K. Castillo), rafaeli@ufu.br (F. R. Rafaeli),alexandresuzuki@ufu.br (A. Suzuki)

June 23, 2020

weight function (1−X)α(1 +X)β, and

y = P (α,β)n (X) = 2F1

(−n, n+ α+ β + 1

1 + α

∣∣∣∣1−X

2

)

satisfy the homogeneous differential equation of second order

a y′′ + b y′ + c y = 0, (1.1)

where a = a(X) = −X2 + 1 and b = b(X ;α, β) = −(α + β + 2)X − α + β.

(Having the explicit expression of P(α,β)n , the parameter c does not play any

interesting role and it can be easily calculated.) Recall that the hypergeometricfunction iFj is formally defined by the series (see [1, (2.1.2)])

iFj

(α1, . . . , αi

β1 . . . , βj

∣∣∣∣ X)=

∞∑

k=0

(α1, . . . , αi)k(β1 . . . , βj)k

Xk

k!,

where

(α1, . . . , αi)k = (α1)k · · · (αi)k, (α1)k =

k∏

j=1

(α1 + j − 1);

by convention, the empty product is 1. From (1.1), Stieltjes proves an importantstatement concerning the dependence of the zeros of Jacobi polynomials on theparameters α and β (see [21]). Soon after his work has been accepted forpublication, Stieltjes implicitly acknowledges, in a note added at the end ofthe manuscript itself and also in a letter of February 3, 1887 to Hermite (see[22, Lettre 106]), that the monotonicity of the zeros of Jacobi polynomials waspreviously proved by A. Markov in [8]. However, in Stieltjes’ work, ut in multisaliis rebus, the “How” is more important than the “What” and, as he wroteto Hermite, “la demonstration que j’ai developpee pour les Acta Mathematicaest differente de celle de M. Markoff”. Historically, Stieltjes proves that given apositive definite real symmetric matrix with non-positive off-diagonal elements1,its inverse is also positive definite. As a consequence, putting aside a clevermanipulation of the differential equation (1.1), he shows that, since

b

a=

−(α+ β + 2)X − α+ β

−X2 + 1(1.2)

is a strictly decreasing function of X ∈ (−1, 1), and

∂

∂α

(b

a

)=

1

X − 1< 0,

∂

∂β

(b

a

)=

1

X + 1> 0, (1.3)

for each X ∈ (−1, 1), the zeros of Jacobi polynomials are strictly decreasingfunctions of α on (−1,∞) and strictly increasing functions of β on (−1,∞) (see

1These matrices are now known as Stieltjes’ matrices (see [24, Definition 3.4]).

2

[21, (6)-(7)]). The proof of Markov is entirely different from that of Stieltjesand is based on the weight function. While it is true that Stieltjes worked di-rectly with Jacobi polynomials —and Markov proves his result through a generaltheorem—, his argument furnishes similar results for a more general differentialequation (see [23, Section 6.22]). Stieltjes himself considered the ultrasphericalcase α = β and Szego noted that “the same method applies to Laguerre polyno-mials” (see [23, p. 123]). In particular, by definition, the old classical orthogonalpolynomials on the real line (Hermite, Jacobi, and Laguerre)2 are solutions ofa differential equation of the same type of (1.1) (see [20, Section 4.2]). How-ever, 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 wastraditionally the birthplace of some of the most beautiful families (see [25])—, itrarely will require more than the classical orthogonal polynomials. For classical(discrete) orthogonal polynomials on a uniform grid 3 (Charlier, Krawtchouk,Hahn, and Meixner polynomials), Markov’s theorem can be used (see [4, Chap-ter 7]) and, of course, for other discrete families 4. Here we purpose an alter-native approach, establishing a bridge between two works separated in time byalmost one century, on one hand the work of Stieltjes and, on the other hand,the work [14] by Nikiforov and Uvarov. This allows one to carry out a systematicstudy of the monotonicity of zeros of classical orthogonal polynomials on linear,quadratic, q-linear, and q-quadratic grids. Indeed, the purpose of this note isto prove that under suitable regularity conditions the Stieltjes result for Jacobipolynomials remains valid if we replace the differential equation (1.1), with aand b arbitrary polynomials of degree at most 2 and 1, respectively, and c anarbitrary constant (from now on, whenever we refer to (1.1), we are assumingthese conditions), by the following difference equation introduced in [14, (5)](see also [15, p. 127] and [11, p. 71]):

a(X)∆

∆x (s− 1/2)

(∇y(X)

∇X

)+

b(X)

2

(∆y(X)

∆X+

∇y(X)

∇X

)+ c y(X) = 0,

(1.4)

2We write “classical orthogonal polynomials on the real line" rather than simply “classicalorthogonal polynomials" because, for instance, from the algebraic point of view of Maroni (see[9]), the Jacobi polynomials exist and are “classical” even when −α,−β,−α− β + 1 6∈ N (see[17, Chapters 8 and 9] for a recent survey on the subject). Moreover, the Bessel polynomialsare classical in the same sense as the other three systems. As Maroni says, “comme dans leroman d’Alexandre Dumas, les trois mousquetaires etaient quatre en realite”.

3A non-empty (totally) ordered set of equidistant (respectively, non-equidistant) points iscalled uniform (respectively, non-uniform) grid, which in turn is an elementary example oflattice.

4For a “continuous” case as the Askey-Wilson polynomials, Askey and Wilson used a con-sequence of Markov’s theorem, which goes back to Szego (see [23, Theorem 6.12.2]), to studythe monotonicity of zeros of these polynomials (see [2, Section 7]).

3

or, equivalently,

a(s)∆

∆x (s− 1/2)

(∇y(X)

∇X

)+ b(X)

∆y(X)

∆X+ c y(X) = 0, (1.5)

where

a(s) = a(X)−1

2b(X)∆x

(s−

1

2

),

X = x(s)5 defines a class of grids with, generally nonuniform, step-size ∆X =∆x(s) = x(s+1)− x(s) and ∇X = ∇x(s) = ∆x(s− 1). (By abuse of notation,we use the same letter a for the function a(s) and the polynomial a(X).) Inwhat follows, we assume that x is a real-valued function defined on an intervalof the real line. Any solution of (1.4) can be brought in correspondence with thesolution of (1.1) by replacing s by s/h and then taking limit h → 0, wheneverit exists. It is important to highlight that (1.4) has polynomial solutions, in X ,whose difference-derivatives satisfy equations of the same kind if and only if, forq 6= 1 fixed, x is a linear, quadratic, q-linear, or q-quadratic grid of the form

x(s) =

{C1s

2 + C2s,

C3q−s + C4q

s,

where (C1, C2) 6= (0, 0) and (C3, C4) 6= (0, 0) (see [3, (1.68)]). The grids thatdepend on “q” are called q-linear if C3 or C4 is zero; otherwise it is q-quadratic.By using linear transformations (see [12, (3.4.1)]) we can reduce the expressionsfor the grids to simpler forms. In what follows, we assume that the grid x takeson the following canonical forms (see [12, p. 74]):

x(s) =

s (I)

s(s+ 1) (II)

qs (q > 1) (III)

1

2(qs − q−s) (q > 1) (IV)

1

2(qs + q−s) (q > 1) (V)

1

2(qs + q−s) (q = e2iθ, 0 < θ < π/2). (VI)

(1.6)

Under the above notation, we adopt the following definition of classical discreteorthogonal polynomials on the real line (COPRL), which is enough for ourpurpose:

5In first edition of “Special Functions of Mathematical Physics” (see [13]) the author onlyconsider the case x(s) = s. The second edition (see [15]) was significantly enriched with theequation (1.4); although A. A. Samarskii’s preface is the same in both editions.

4

Definition 1.1. Fix a ∈ R∪{−∞} and N ∈ N∪{∞} and define b = a+N . Fixq and let X = x(s) be a real-valued function given by (1.6), where the variables ranges over the finite interval [a, b] or the infinity interval [a,∞). A sequenceof polynomials, (Pn(X))N−1

n=0 , is said to be sequence of classical discrete orthogo-nal polynomials on the set {x(a), x(a + 1), . . . , x(b− 1)} or, simply, COPRL if :

i) Pn satisfy (1.4), x being a strictly monotone function on [a, b] or [a,∞)given, up to a linear transformation, by (1.6);

ii) a positive weight function ω satisfying the boundary conditions 6

ω(s)a(s)xk

(s−

1

2

)∣∣∣∣a,b

= 0 (k = 0, 1, . . . ) (1.7)

exists;

iii) the difference equation

∆

∆x

(s−

1

2

)(ω(s)a(s)

)= ω(s)b(X) (1.8)

holds.

Subsequently, when we say that a certain polynomial is a COPRL, we areassuming the definition and notation given in Definition 1.1. From (1.5), (1.7),and (1.8), we conclude that the COPRL satisfy the orthogonality condition (see[12, (3.3.4)])

b−1∑

s=a

Pm(X)Pn(X)ω(s)∆x

(s−

1

2

)= 0 (m 6= n). (1.9)

The polynomials (Pn(X))N−1n=0 given in Definition 1.1 are called simply discrete

orthogonal polynomials, in X , on the set {x(a), x(a + 1), . . . , x(b − 1)} with re-spect to a positive weight function ω if they satisfy the relation (1.9) insteadof the requirements i) − iii). Of course, COPRL are a special case of discreteorthogonal polynomials. From (1.9), we can see that the zeros of discrete orthog-onal polynomials on {x(a), x(a + 1), . . . , x(b− 1)} are real and distinct and arelocated in (min{x(a), x(b − 1)},max{x(a), x(b − 1)}) (see [23, Theorem 3.3.1]).In concluding this section we remark that it is possible to obtain a series repre-

6If a and b are finite, (1.7) can be written in the form ω(a)a(a) = 0 and ω(b)a(b) = 0. Ifa = −∞ and/or b = ∞, then (1.7) must hold for each k in the limiting sense.

5

sentation of COPRL (see [3, (4.19)] and [16, Section 3]):

Pn(X) = (−1)nγn

n∑

j=0

(n

j

)

q

(− q−n/2

)j ∆x(s− (n− 1)/2 + j

)n∏

k=0

∆x(s+ (j − k + 1)/2

) (1.10)

×

n∏

k=0

a(s− n+ j + k)

j∏

l=0

a(x(s+ l − 1)) + 1/2 b(x(s+ l − 1))∆x(s+ l − 1/2)

a(x(s+ l))− 1/2 b(x(s+ l))∆x(s+ l + 1/2),

where γn is a constant and(n

j

)

q

=(q; q)n

(q; q)j(q; q)n−j.

Furthermore (see [16]), COPRL represent special cases of hypergeometric seriesor q-hypergeometric series, the latter formally defined by (see [1, (10.9.4)])

iφj

(α1, . . . , αi

β1 . . . , βj

∣∣∣∣ q, X

)=

∞∑

k=0

(α1, . . . , αi; q)k(β1 . . . , βj ; q)k

(−1)(1−i+j)k

(k

2

)Xk

(q; q)k.

The outline of this note is as follows. In Section 2 we present an extensionof the Stieltjes work in the framework of COPRL. In Section 3 we study thevariation of zeros of some families of COPRL according to the type of underlyinggrid.

2. Main results

Unless otherwise stated we assume that a, b, and c appearing in (1.4) dependon a parameter t varying in a non-degenerate open interval of the real line.Rewrite (1.4) in the more suggestive form

Ay(s− 1) +B y(s+ 1) + C y(s) = 0, (2.11)

where y(s) = y(X ; t) and

A = A(s; t) =a(s; t)

∇X∆x(s− 1/2), B = B(s; t) =

a(s; t) + b(X ; t)∆x(s− 1/2)

∆X∆x(s− 1/2),

(2.12)

C = C(s; t) = c(t)−A(s; t)−B(s; t).

The following example will help motivate our main result.

Example 2.1. In 1960, Karlin and McGregor proved (see [5, (1.3)] and [7,(5.1)]) that the COPRL known as Hahn polynomials (see [6, Section 9.5])

y(s) = H(α,β)n (X) = 3F2

(−n, −X, α+ β + n+ 1

β + 1, 1−N

∣∣∣∣ 1)

6

(n = 1, . . . , N − 1; a = 0, b = N ;α > −1, β > −1), —which constitute the finitediscrete analogue of Jacobi polynomials considered by Stieltjes in [21]— satisfy(2.11) with X = x(s) = s, and A and B given by

A = A(X ;α) = X(−X + α+N), B = B(X ;β) = (X + β + 1)(−X +N − 1).

The function

B

A=

(X + β + 1)(−X +N − 1)

X(−X + α+N)(2.13)

is a positive and strictly decreasing function of X ∈ (a, b− 1), and

∂

∂α

(B

A

)=

(X + β + 1)(X −N + 1)

X(−X + α+N)2< 0, (2.14)

∂

∂β

(B

A

)=

−X +N − 1

X(−X + α+N)> 0. (2.15)

for each X ∈ (a, b − 1). Since this example corresponds to the grid (I), from

Markov’s theorem, Ismail proves that the zeros of H(α,β)n are decreasing functions

of α on (−1,∞) and increasing functions of β on (−1,∞) (see [4, Theorem7.1.2]). Comparison of (1.2) and (2.13), and (1.3) and (2.14)-(2.15) suggeststhat, as for Jacobi polynomials, the information on the monotonicity of the zerosof Hahn polynomials is stored in the rational function B/A. Therefore, it is notunreasonable to conjecture that the same happens with any COPRL.

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

Definition 2.1. Let A and B be given by (2.12). The function f , from now oncalled monotonicity function, is defined by

f(s; t) =B(s; t)

A(s; t)=

a(s; t) + b(X ; t)∆x(s− 1/2)

a(s; t)

∇X

∆X. (2.16)

Definition 2.2. Let X = x(s) be given by (1.6) and let x(yj(t)) (j = 1, . . . , n)be the zeros of a COPRL of degree n, say Pn(X ; t), depending on a parameter

t taking values on J ⊆ R. For any set I ⊆ R, the (nonempty) subset S(t)I (P ) of

(a, b− 1) is defined by

S(t)I (Pn) =

{y ∈ (a, b− 1)

∣∣ t ∈ I ∩ J ∧(∀j ∈ {1, . . . , n}

)[y = yj(t)

]}.

We will write it simply SI(Pn), S(t)I or SI when no confusion can arise7.

To prove our main result we need two lemmas. The first one was proved forthe Hahn polynomials by Levit (see [7, Theorem 3]). Here we reproduce in amore general framework, mutatis mutandis, his arguments.

7Note that SJ = SR and SI j SJ whenever I j J.

7

Lemma 2.1. Let X = x(s) be given by (1.6). Let x(y(t)) and x(z(t)) beconsecutive zeros of a COPRL depending on a real parameter t. Let f be themonotonicity function given by (2.16). Then |z(t) − y(t)| > 1 for those valuesof t such that f(·; t) > 0 on SR.

Proof. We give the proof only for the case in which x is a strictly increasingfunction. The same arguments apply to the case in which x is a strictly decreas-ing 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). (Recallthat the zeros of P are distinct and are located in (x(a), x(b − 1)). More-over, under our assumptions, y(t) < z(t).) Suppose that there exists t0 suchthat z(t0) = y(t0) + 1. Replacing s by z(t0) and t by t0 in (2.11) we haveP (x(z(t0) + 1); t0) = 0, because f(z(t0); t0) 6= 0, which contradicts the assump-tion that x(z(t0)) is the greatest zero of P (·; t0). (Here we have also used thefact that x is, in particular, a strictly increasing function on [a, b), and not onlyon [a, b− 1], because the possibility that z(t0) + 1 ∈ [b− 1, b) is not excluded.)Suppose now that z(t0) < y(t0) + 1. Replacing again s by z(t0) and t by t0 in(2.11) we have

−P (x(z(t0)− 1); t0)

P (x(z(t0) + 1); t0)= f(z(t0); t0) > 0.

Since z(t0) 6= y(t0) + 1, there is an odd number of zero of P (·; t0) in [x(z(t0)−1), x(z(t0) + 1)]. We next claim that there exists at least one integer m (1 ≤m ≤ N − 1) such that y(t0) ≤ a +m ≤ z(t0) or, what is the same, x(y(t0)) ≤x(a +m) ≤ x(z(t0)). Suppose the assertion were false. Hence

b−1∑

s=a

P 2(X ; t0)(X − x(y(t0))

)(X − x(z(t0))

)ω(s; t0)∆x

(s−

1

2

)> 0,

a contradiction with (1.9). Since z(t0) < y(t0) + 1, m is unique. Thus theonly zeros of P (·; t0) in [x(z(t0) − 1), x(z(t0) + 1)] are x(y(t0)) and x(z(t0)).This contradicts the fact that there is an odd number of zeros of P (·; t0) in[x(z(t0)− 1), x(z(t0) + 1)], and the lemma follows.

Lemma 2.2. Let X = x(s) be given by (1.6). Let P be a non-constant CO-PRL, in X, satisfying (2.11) and depending on a parameter t varying in anon-degenerate open interval of the real line containing t0. Suppose that A(·; t)and B(·; t) given by (2.12) admit partial derivatives with respect to t on a neigh-bourhood of t0. Assume that P (Y0; t0) = 0. Then there exist ǫ > 0 and δ > 0such that (Y0 − δ, Y0 + δ) × (t0 − ǫ, t0 + ǫ) is on the neighbourhood where P isdefined, and there exists Y : (t0 − ǫ, t0 + ǫ) → (Y0 − δ, Y0 + δ), such that

P (Y (t); t) = 0 (2.17)

and, for each t ∈ (t0 − ǫ, t0 + ǫ), Y is the unique solution of (2.17) with Y (t) ∈(Y0 − δ, Y0 + δ). Moreover, Y possess a continuous derivative on (t0 − ǫ, t0 + ǫ).

8

Proof. From (1.10) we see that the coefficients of P (·; t) are differentiable func-tions of t. Moreover, P (Y0; t0) = 0; from this it follows that

∂P

∂X(X ; t)

∣∣∣∣X=Y0,t=t0

6= 0,

because the zeros of P (·; t0) are distinct. Thus, the result is a direct consequenceof the implicit function theorem (see [19, Theorem 3.4.2]).

We shall refer to Theorem 2.1 below as discrete Stieltjes theorem.

Theorem 2.1. Assume the hypotheses and notation of Lemma 2.2. Let f bethe monotonicity function given by (2.16). Denote f1(s; ·) = (∂f/∂s)(s; ·) andf2(·; t) = (∂f/∂t)(·; t). Suppose that f(s; t) > 0 and f1(s; t) < 0 on S(t0−ǫ,t0+ǫ).In the case of the grid (IV), assume also that nf(s; ·) + f1(s; ·) ≤ 0. Supposefurthermore that f2(s; t) > 0 (respectively, f2(s; t) < 0) on S(t0−ǫ,t0+ǫ). Then Yis a strictly increasing (respectively, decreasing) function on (t0 − ǫ, t0 + ǫ) if xis a strictly increasing function, or else Y is a strictly decreasing (respectively,increasing) function on (t0 − ǫ, t0 + ǫ).

Proof. We give the proof only for the case in which x is a strictly increasingfunction. The same arguments apply to the case in which x is a strictly de-creasing function. Assume that P is monic and has fixed degree n. Let Yj

(j = 1, . . . , n) denote the zeros of P . Since the indeterminate of P takes valueson the real (open) interval x((a, b − 1)), there exist functions yj defined on aneighbourhood of t0 and taking values on (a, b− 1) such that Yj(t) = x(yj(t)).Moreover, since x is strictly increasing on (a, b − 1), x−1 is differentiable onx((a, b − 1)). Hence, by Lemma 2.2, there exists a neighbourhood of t0 whereyj is differentiable.

Let

P (X ; t) =

n∏

j=1

(X − Yj(t))

defined in a neighbourhood of t0. Replacing s by yj(t) in (2.11) we have

f(yj(t); t) = −P (x(yj(t)− 1); t)

P (x(yj(t) + 1); t). (2.18)

Taking the partial derivative of (2.18) with respect to t on the neighbourhoodof t0 leads to

y′j(t)f1(yj(t); t) + f2(yj(t); t)

=P (x(yj(t)− 1); t)

∂P

∂t(x(yj(t) + 1); t)− P (x(yj(t) + 1); t)

∂P

∂t(x(yj(t)− 1); t)

P 2(x(yj(t) + 1); t),

9

where

∂P

∂t(x(yj(t)± 1); t)

= P (x(yj(t)± 1); t)

n∑

k=1

dx

ds(s± 1)

∣∣∣∣s=yj(t)

y′j(t)−dx

ds(s)

∣∣∣∣s=yk(t)

y′k(t)

x(yj(t)± 1)− Yk(t).

Hence

f2(yj(t); t) =

n∑

k=1

ajk(t)y′k(t), (2.19)

where

ajj(t) = −f1(yj(t); t) + f(yj(t); t)

n∑

k=1

bjk(t)−

n∑

k=1k 6=j

ajk, (2.20)

with

bjk(t) =

dx

ds(s− 1)

∣∣∣∣s=yj(t)

−dx

ds(s)

∣∣∣∣s=yk(t)

x(yj(t)− 1)− Yk(t)−

dx

ds(s+ 1)

∣∣∣∣s=yj(t)

−dx

ds(s)

∣∣∣∣s=yk(t)

x(yj(t) + 1)− Yk(t),

and ajk(t) = f(yj(t); t) cjk(t) (j 6= k), where

cjk(t) =

(1

x(yj(t) + 1)− Yk(t)−

1

x(yj(t)− 1)− Yk(t)

)dx

ds(s)

∣∣∣∣s=yk(t)

.

We claim that ajk(t) < 0 (j 6= k). From now on, without loss of generality,we assume Yj(t) < Yk(t). Then, by Lemma 2.1, x(yj(t) + 1) < Yk(t). Since(dx/ds)(s)|s=yk(t)

> 0, ajk(t) < 0 as required. Consider now the canonical

forms (1.6) to conclude that ajj(t) > 0. We next claim that, for the grids(I) − (III) and (V) − (VI), bjk(t) > 0. Indeed, for the grids (I) and (III), theproof is straightforward. For the grid (II), we have

bjk(t) =4(

yj(t) + yk(t))(yj(t) + yk(t) + 2

)

and yj(t) ∈ S(t0−ǫ,t0+ǫ) ⊂ (−1/2,∞), the latter because x is strictly increasingon (−1/2,∞). Hence, by Lemma 2.1, bjk(t) > 0 as claimed. Since f(yj(t); t) >0, f1(yj(t); t) < 0, ajk(t) < 0, and bjk(t) ≥ 0, (2.20) implies ajj(t) > 0. Thecorresponding result for the grid (V) follows in the same way after noting that

bjk(t) = 2θ sinh(2θ) csch((yk(t) + yj(t)− 1)θ

)csch

((yk(t) + yj(t) + 1)θ

),

10

for q = e2θ (θ > 0), and yj(t) ∈ S(t0−ǫ,t0+ǫ) ⊂ (0,∞). And the same goes forthe grid (VI) using hyperbolic identities and noting that yj(t) ∈ S(t0−ǫ,t0+ǫ) ⊂(−π/(2θ), 0). Define

f(t) = (f2(y1(t); t), . . . , f2(yn(t); t))T,

A(t) = (ajk(t)), y(t) =(y′1(t), . . . , y

′n(t)

)T,

and rewrite (2.19) as f(t) = A(t)y(t). Observe that A(t) has positive diagonalentries and negative of off-diagonal entries. Moreover, from (2.20) we get

|ajj(t)| >

n∑

k=1k 6=j

|ajk(t)|.

Hence A(t) is a real irreducibly diagonally dominant matrix, and so, by [24,Corollary 1, p. 85] all the entries of A−1(t) are positive 8. Thus all the entriesof y(t) = A−1(t)f(t) are positive, and the theorem is proved for the grids(I)− (III) and (V)− (VI). The above argument does not work for the grid (IV),because, for q = e2θ (θ > 0),

bjk(t) = −2θ sinh(2θ) sech((yj(t) + yk(t) + 1)θ

)sech

((yj(t) + yk(t)− 1)θ

)< 0.

Indeed, −1 < bjk(t) < 0 and, therefore, under the additional hypothesis for thisgrid, the theorem follows from (2.20).

Remark 2.1. In some cases, the hypotheses of the discrete Stieltjes theorem arefulfilled in (a, b− 1) ⊃ SR. However, there are several, and important, exampleswhere this is only true on a subset of (a, b − 1) containing all the elements ofSR; see, for instance, the Racah polynomials in Section 3.4 below. Of course,since the precise location of the zeros of P is not known, these cases require amore careful analysis.

In Example 2.1, we can write H(α,β,N)n instead of H

(α,β)n . The interlacing

property between the zeros of H(α,β,N)n and H

(α,β,N+1)n was observed by Levit

(see [7, Theorem 6]). It was proved independently, and almost simultaneously,by Mesztenyi that [7, Theorem 6] is always true for discrete orthogonal polyno-mials on the grid (I) when the variable s ranges over a finite interval (see [10,Lemma 3]). In this way, the following general result might be of interest to thereader.

Theorem 2.2. Fix a ∈ R and N ∈ N and define b = a + N . Let X = x(s)be a real-valued function, where the variable s ranges over the finite interval[a, b+1]. Let (Pn(X ;N))N−1

n=1 be the sequence of discrete orthogonal polynomials,in X, on the set {x(a), x(a + 1), . . . , x(b − 1)}. Let (Pn(X ;N + 1))Nn=1 be

8Indeed, A(t) is a Stieltjes matrix.

11

the sequence of discrete orthogonal polynomials, in X, on the set {x(a), x(a +1), . . . , x(b)}. Assume that both sequence are orthogonal with respect to the samepositive weight function. Let Y1 < Y2 < · · · < Yn be the zeros of Pn(·;N). Thenone of the following situations holds:

i) The zeros of Pn(·;N + 1) are those of Pn(·;N), if Pn(x(b);N) = 0;

ii) Pn(·;N + 1) has a zero on (Yk, Yk+1), if Pn(x(b);N) 6= 0 and x(b) 6∈(Yk, Yk+1) for fixed k ∈ {1, . . . , n− 1};

iii) Pn(·;N+1) has a zero on each of the intervals (Yk, x(b)) and (x(b), Yk+1),if Pn(x(b);N) 6= 0 and x(b) ∈ (Yk, Yk+1) for fixed k ∈ {1, . . . , n− 1}.

Proof. To shorten notation, write P(N)n and P

(N+1)n instead of Pn(·;N) and

Pn(·;N + 1), respectively. The reader may check for himself that by expressing

P(N+1)n as a linear combination of the elements of the set

{1, P

(N)1 , . . . , P

(N)n

}

and using the orthogonality property, we obtain P(N+1)n (x(b)) = ζn P

(N)n (x(b))

and

P (N+1)n (X) =P (N)

n (X) (2.21)

− ηnP(N+1)n (x(b))

P(N)n−1(x(b))P

(N)n (X)− P

(N)n (x(b))P

(N)n−1(X)

X − x(b),

where ηn = ω(b)∆x(b − 1/2)∥∥P (N)

n−1

∥∥−2and

ζn = 1 + ηn

P

(N)n−1(x(b))

dP(N)n

dX(X)

∣∣∣∣∣X=x(b)

−dP

(N)n−1

dX(X)

∣∣∣∣∣X=x(b)

P (N)n (x(b))

.

Following a standard procedure, the rest of the proof follows as [10, Lemma3].

We end this section with the following consequence of Theorem 2.2, whichis valid, in particular, for COPRL.

Corollary 2.1. Assume the hypotheses and notation of Theorem 2.2. Supposefurthermore that x is a strictly monotone function on [a, b + 1]. Set Y0 =Yn+1 = x(b). Then Pn(·;N + 1) has a zero on each interval (Yk, Yk+1) for allk ∈ {1, . . . , n} if x is a strictly increasing function, or else Pn(·;N + 1) has azero on each interval (Yk, Yk+1) for all k ∈ {0, . . . , n− 1}.

Proof. We give the proof only for the case in which x is a strictly increasing func-tion. The same arguments apply to the case in which x is a strictly decreasingfunction. From (2.21), it follows that

sgn (Pn(Yn;N + 1)) = −sgn (Pn(Yn+1;N)) = −1.

Hence Pn(·;N + 1) has a zero on (Yn, Yn+1). The rest of the proof is a directconsequence of Theorem 2.2 ii).

12

3. Applications

In this section we apply the discrete Stieltjes theorem to specific familiesof COPRL. (Of course, Corollary 2.1 is applicable to any family of COPRLon 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 similarcases, only of some illustrative examples; the other cases are stated and theproofs are left as exercises for the reader. The reader also should satisfy himselfthat the hypotheses of Lemma 2.2 are fulfilled. As far as we know, only themonotonicity of zeros of COPRL on the grid (I) has been studied (see [4, Chapter7]) 9. We include this case for the sake of completeness.

3.1. The grid (I)

Examples of COPRL on X = x(s) = s (Hahn, Charlier, Krawtchouk, andMeixner polynomials).

The Hahn polynomials, H(α,β)n , are defined in Example 2.1.

Proposition 3.1. The zeros of H(α,β)n are strictly decreasing functions of α on

(−1,∞) and strictly increasing functions of β on (−1,∞).

Proof. It suffices to use (2.13) and (2.14)-(2.15) together with the discrete Stielt-jes theorem.

The Charlier polynomials (see [6, Section 9.14]),

y(s) = C(α)n (X) = 2F0

(−n, −X

−

∣∣∣∣ −1

α

)

(n = 1, 2, . . . ; a = 0, b = ∞;α > 0), satisfy (2.11) with A and B given by

A = A(X) = X, B = B(α) = α.

Proposition 3.2. The zeros of C(α)n are strictly increasing functions of α on

(0,∞).

The Krawtchouk polynomials (see [6, Section 9.11]),

y(s) = K(α)n (X) = 2F1

(−n, −X

1−N

∣∣∣∣1

α

)

(n = 1, . . . , N − 1; a = 0, b = N ; 0 < α < 1), satisfy (2.11) with A and B givenby

A = A(X ;α) = (1− α)X, B = B(X ;α) = α(−X +N − 1).

9Recall that for a continuous case on the grid (VI) as the Askey-Wilson polynomials, themonotonicity of their zeros was studied in [2, Section 7].

13

Proposition 3.3. The zeros of K(α)n are strictly increasing functions of α on

(0, 1).

The Meixner polynomials (see [6, Section 9.10]),

y(s) = M (α,β)n (X) = 2F1

(−n, −X

β

∣∣∣∣ 1−1

α

)

(n = 1, 2, . . . ; a = 0, b = ∞; 0 < α < 1, β > 0), satisfy (2.11) with A and Bgiven by

A = A(X) = X, B = B(X ;α, β) = α(X + β).

Proposition 3.4. The zeros of M(α,β)n are strictly increasing functions of α on

(0, 1) and strictly increasing functions of β on (0,∞).

3.2. The grid (II)

Examples of COPRL on X = x(s) = s(s+ 1) (Racah and dual Hahn polynomi-als).

The Racah polynomials (see [16, p. 236]),

y(s) = R(α,β)n (X) = 4F3

(−n, α+ β + n+ 1, a− s, s+ a + 1

2a + α+N + 1, β + 1, 1−N

∣∣∣∣ 1)

(3.22)

(n = 1, . . . , N − 1; a > −1/2, b = a + N ;α > −1,−1 < β < 2a + 1), satisfy(2.11) with A and B given by

A = A(s;α, β) =(s− a)(s+ a +N)(s− a− α−N)(s+ a− β)

2s(2s+ 1),

B = B(s;α, β) =(s+ a + 1)(s− a−N + 1)(s+ a + α+N + 1)(s− a + β + 1)

2(s+ 1)(2s+ 1).

Proposition 3.5. The zeros of R(α,β)n are strictly decreasing functions of α on

(−1,∞) and strictly increasing function of β on (−1, 2a + 1) if a ≥ 0, or else

the zeros of R(α,β)n are strictly decreasing functions of α on (−1,∞) for each

β ∈ (a, 2a + 1) and strictly increasing function of β on (a, 2a + 1).

Proof. We give the proof only for the case in which a ≥ 0. The proof for−1/2 < a < 0 is similar. Define the interval K =

(max

{a, β − a

}, a +N − 1

).

The monotonicity function

f =B

A=

s(s+ a + 1)(s− a−N + 1)(s+ a + α+N + 1)(s− a + β + 1)

(s+ 1)(s− a)(s+ a +N)(s− a− α−N)(s+ a− β)(3.23)

14

is a positive and strictly decreasing function of s ∈ K, and

∂f

∂α=

s(2s+ 1)(s+ a + 1)(s− a−N + 1)(s− a + β + 1)

(s+ 1)(s− a)(s+ a +N)(s− a− α−N)2(s+ a− β)< 0, (3.24)

∂f

∂β=

s(2s+ 1)(s+ a + 1)(s− a−N + 1)(s+ a + α+N + 1)

(s+ 1)(s− a)(s+ a +N)(s− a− α−N)(s+ a− β)2> 0, (3.25)

for each s ∈ K. It is immediate that S(β)(−1,2a] ⊂ K. We next claim that

S(β)(2a,2a+1) ⊂ K. Indeed, this is equivalent to prove that R

(α,β)n has no zeros

on(a(a + 1), (β − a)(β − a + 1)

)for β ∈ (2a, 2a + 1). For β ∈ (2a, 2a + 1),

R(α,β)n (a(a + 1)) = 1 and

R(α,β)n

((β − a)(β − a + 1)

)= 3F2

(−n, α+ β + n+ 1, 2a− β

2a + α+N + 1, 1−N

∣∣∣∣ 1)

=(2a− β +N − n)n(α+ β +N + 1)n

(2a + α+N + 1)n(N − n)n> 0,

the last equality being a consequence of Sheppard’s identity (see [1, Corollary

3.3.4]). Hence R(α,β)n has no zeros on

(a(a + 1), (β − a)(β − a + 1)

)or has at

least two zeros there. Suppose the second of these possibilities is true. Fromthe proof of Lemma 2.1, we have x(a+1) < x(β− a), which is impossible. Thus

S(β)(2a,2a+1) ⊂ K as claimed. The same proof actually shows that S

(α)(−1,∞) ⊂ K.

The result follows from the discrete Stieltjes theorem.

Remark 3.1. In [6, Section 9.2], the Racah polynomials (interchanging α andβ) are defined by

R(α,β,δ)n (x(s)) = 4F3

(−n, α+ β + n+ 1, −s, s+ δ −N + 1

α+ δ + 1, β + 1, 1−N

∣∣∣∣ 1)

(n = 1, . . . , N − 1; a = 0, b = N ;α > −1,−1 < β < δ−N +1, δ > N − 1), where

x(s) = s(s + δ − N + 1). In (3.22), we can write R(α,β,a)n instead of R

(α,β)n ;

although, implicitly, we have the agreement to omit those parameters that areassumed fixed. Clearly, x is not one of the canonical forms (1.6). However, xand x are related by a linear transformation. Therefore, for a = (δ−N)/2 fixed,

R(α,β,(δ−N)/2)n (x(s)) = R(α,β,δ)

n

(x

(s−

δ −N

2

)),

where

x

(s−

δ −N

2

)= x(s)−

(δ −N + 1)2 − 1

4.

Consequently, Proposition 3.5 remains valid if we replace R(α,β)n by R

(α,β,δ)n .

15

The dual Hahn polynomials (see [16, p. 236]),

y(s) = W (α)n (X) = 3F2

(−n, a− s, s+ a + 1

α+ 1, 1−N

∣∣∣∣ 1)

(3.26)

(n = 1, . . . , N − 1; a > −1/2, b = a +N ;−1 < α < 2a + 1), satisfy (2.11) withA and B given by

A = A(s;α) =(s− a)(s+ a +N)(s+ a− α)

2s(2s+ 1),

B = B(s;α) =(s+ a + 1)(−s+ a +N − 1)(s− a + α+ 1)

2(s+ 1)(2s+ 1).

Proposition 3.6. The zeros of W(α)n are strictly increasing functions of α on

(−1, 2a + 1) if a ≥ 0, or else the zeros of W(α)n are strictly increasing functions

of α on (a, 2a + 1).

Proof. We sketch the proof only for the case in which a ≥ 0. The proof for−1/2 < a < 0 is similar. Define the interval K =

(max

{a, α− a

}, a +N − 1

).

Note that the hypotheses of the discrete Stieltjes theorem are fulfilled in K. Thus

we only need to prove that S(2a,2a+1) ⊂ K10. Note that W(α)n (a(a + 1)) = 1 and

W (α)n

((α− a)(α− a + 1)

)= 2F1

(−n, 2a− α

1−N

∣∣∣∣ 1)=

(1−N + α− 2a)n(1−N)n

> 0,

the last equality being a consequence of Chu-Vandermonde’s identity ([1, Corol-lary 2.2.3]). The rest of the proof runs as in Proposition 3.5.

Remark 3.2. In [6, Section 9.6], the dual Hahn polynomials are defined by

W (α,β)n (x(s)) = 3F2

(−n, −s, s+ α+ β + 1

α+ 1, 1−N

∣∣∣∣ 1)

(n = 1, . . . , N − 1; a = 0, b = N ;α > −1, β > −1 or α < 1 − N, β < 1 − N),

where x(s) = s(s+ α+ β +1). In (3.26), we can write W(α,a)n instead of W

(α)n .

Hence, for a = (α+ β)/2 fixed,

W (α,(α+β)/2)n (x(s)) = W (α,β)

n

(x

(s−

α+ β

2

)),

where

x

(s−

α+ β

2

)= x(s)−

(α+ β + 1)2 − 1

4.

Consequently, Proposition 3.6 remains valid if we replace W(α)n by W

(α,β)n and

assume that α+ β is constant.

10It is immediate that S(−1,2a] ⊂ K.

16

3.3. The grid (III)3.3.1. 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 bigq-Laguerre).

The q-Meixner polynomials (see [6, Section 14.13]),

y(s) = M (α,β)n (X ; q) = 2φ1

(q−n, X

βq

∣∣∣∣ q, −qn+1

α

)

(n = 1, 2, . . . ; a = 0, b = ∞;α > 0, 0 ≤ β < q−1), satisfy (2.11) with A and Bgiven by

A = A(s;α, β) = (1 − qs)(1 + αβqs), B = B(s;α, β) = αqs(1− βqs+1).

Proposition 3.7. The zeros of M(α,β)n (·; q) are strictly increasing functions of

α on (0,∞) and strictly decreasing functions of β on [0, q−1).

Proof. The monotonicity function

f =B

A=

αqs(1− βqs+1)

(1 − qs)(1 + αβqs)

is a positive and strictly decreasing function of s ∈ (0,∞), and

∂f

∂α=

qs(1− βqs+1)

(1− qs)(1 + αβqs)2> 0,

∂f

∂β= −

αq2s(α+ q)

(1− qs)(1 + αβqs)2< 0,

for each s ∈ (0,∞). The result follows from the discrete Stieltjes theorem.

Remark 3.3. The q-Charlier polynomials (see [6, Section 14.23]) are given by

C(α)n (·; q) = M

(α,0)n (·; q).

The second family of Al-Salam-Carlitz polynomials (see [6, Section 14.25]),

y(s) = V (α)n (X ; q) = (−α)nq−(

n

2)2φ0

(q−n, X

−

∣∣∣∣ q,qn

α

)

(n = 1, 2, . . . ; a = 0, b = ∞; 0 < α < q−1), satisfy (2.11) with A and B given by

A = A(s;α) = (1− q−s)(α− q−s), B = B(s;α) = αq.

Proposition 3.8. The zeros of V(α)n (·; q) are strictly increasing functions of α

on (0, q−1).

Proof. Define the interval K =(max{− logq α, 0},∞

). Note that the hypothe-

ses of the discrete Stieltjes theorem are fulfilled in K. Thus we only need toprove that S(1,q−1) ⊂ K11. The rest of the proof runs as in Proposition 3.5; al-

11It is immediate that S(0,1] ⊂ K.

17

though given the simplicity of this case, we do not need to use q-hypergeometricidentities.

Remark 3.4. The first family of Al-Salam-Carlitz polynomials is given by U(α)n

(·; q−1) = V(α)n (·; q).

The q-Hahn polynomials (see [6, Section 14.6]),

y(s) = H(α,β)n (X ; q) = 3φ2

(q−n, αβqn+1, X

αq, q1−N

∣∣∣∣ q, q

)

(n = 1, . . .N − 1; a = 0, b = N ; 0 < α < q−1, 0 < β < q−1), satisfy (2.11) withA and B given by

A = A(s;α, β) = αq(1 − qs)(β − qs−N ),

B = B(s;α) = (1− qs−N+1)(1 − αqs+1).

Proposition 3.9. The zeros of H(α,β)n (·; q) are strictly decreasing functions of

α on (0, q−1) and strictly increasing functions of β on (0, q−1).

The q-Krawtchouk polynomials (see [6, Section 14.15]),

y(s) = K(α)n (X ; q) = 3φ2

(q−n, −αqn, X

0, q1−N

∣∣∣∣ q, q

)

(n = 1, . . . , N − 1; a = 0, b = N ;α > 0), satisfy (2.11) with A and B given by

A = A(s;α) = α(qs − 1), B = B(s) = 1− qs−N+1.

Proposition 3.10. The zeros of K(α)n (·; q) are strictly decreasing functions of

α on (0,∞).

The affine q-Krawtchouk polynomials (see [6, Section 14.16]),

y(s) = K(α)n (X ; q) = 3φ2

(q−n, 0, X

αq, q1−N

∣∣∣∣ q, q

)

(n = 1, . . . , N − 1; a = 0, b = N ; 0 < α < q−1), satisfy (2.11) with A and Bgiven by

A = A(s;α) = αqs−N+1(qs − 1), B = B(s;α) = (1− qs−N+1)(1− αqs+1).

Proposition 3.11. The zeros of K(α)n (·; q) are strictly decreasing functions of

α on (0, q−1).

Remark 3.5. The big q-Laguerre polynomials (see [6, Section 14.16]) are givenby

L(α,β)n (X ; q) = 3φ2

(q−n, 0, X

αq, βq

∣∣∣∣ q, q

)

(n = 1, 2, . . . ; 0 < α < q−1, β < 0). In particular, K(α)n (·; q) = L

(α,q−N )n (X ; q)

(see also Remark 3.6).

18

The quantum q-Krawtchouk polynomials (see [6, Section 14.14]),

y(s) = K(α)n (X ; q) =

(q−N , q)nαnqn2 2φ1

(q−n, X

q1−N

∣∣∣∣ q, αqn+1

)

(n = 1, . . . , N − 1; a = 0, b = N ;α > q1−N ), satisfy (2.11) with A and B givenby

A = A(s;α) = (1 − qs)(α − qs−N ), B = B(s) = −qs(1− qs−N+1).

Proposition 3.12. The zeros of K(α)n (·; q) are strictly decreasing functions of

α on (q1−N ,∞).

Proof. Define the interval K =(max{0, logq α + N}, N − 1

). Note that the

hypotheses of the discrete Stieltjes theorem are fulfilled in K. Thus we onlyneed to prove that S(q1−N ,q−N ) ⊂ K12. The rest of the proof runs as in Propo-sition 3.5; although given the simplicity of this case, we do not need to useq-hypergeometric identities.

3.3.2. Examples of COPRL on X = x(s) = qs (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).

The q-Bessel (see [6, Section 14.22]),

y(s) = B(α)n (X ; q) = 2φ1

(q−n, −αqn

0

∣∣∣∣ q, qX

)

(n = 1, 2, . . . ; a = 0, b = ∞;α > 0), satisfy (2.11) with A and B given by

A = A(s) = qs − 1, B = B(s;α) = α.

Proposition 3.13. The zeros of B(α)n (·; q) are strictly decreasing functions of

α on (0,∞).

The little q-Jacobi polynomials (see [6, Section 14.12]),

y(s) = P (α,β)n (X ; q) = 2φ1

(q−n, αβqn+1

αq

∣∣∣∣ q, qX

)

(n = 1, 2, . . . ; a = 0, b = ∞; 0 < α < q−1, β < q−1), satisfy (2.11) with A and Bgiven by

A = A(s) = q−s(qs − 1), B = B(s;α, β) = αq−s(βqs+1 − 1).

Proposition 3.14. The zeros of P(α,β)n (·; q) are strictly decreasing functions of

α on (0, q−1) and strictly increasing functions of β on (−∞, q−1).

12It is immediate that S[q−N ,∞) ⊂ K.

19

Corollary 3.1. The zeros of the special case of big q-Jacobi polynomials (see[6, Section 14.5]),

P (α,β,0)n (X ; q) =

(βq; q)n(αq; q)n

(−1)nαnqn+(n

2)P (β,α)n (α−1q−1X ; q)

(n = 1, 2, . . . ; 0 < α < q−1, 0 < β < q−1), are strictly increasing functions of αon (0, q−1) and strictly decreasing functions of β on (0, q−1).

Remark 3.6. P(α,0,0)n (·; q) = L

(α,0)n (·; q) (see also Remark 3.5).

The little q-Laguerre/Wall polynomials (see [6, Section 14.20]),

y(s) = L(α)n (X ; q) = 2φ1

(q−n, 0

αq

∣∣∣∣ q, qX

)

(n = 1, 2, . . . ; a = 0, b = ∞; 0 < α < q−1), satisfy (2.11) with A and B given by

A = A(s) = qs − 1, B = B(s;α) = αq−s.

Proposition 3.15. The zeros of L(α)n (·; q) are strictly decreasing functions of

α on (0, q−1).

Corollary 3.2. The zeros of the q-Laguerre polynomials (see [6, Section 14.21]),

L(α)n (·; q) = q−αn (q

α+1; q)n(q; q)n

L(qα)n (·; q)

(n = 1, 2, . . . ;α > −1), are strictly decreasing functions of α on (−1,∞).

3.4. The grid (VI)

Examples of COPRL on X = x(s) = (qs + q−s)/2 (0 < q < 1) (q-Racah anddual q-Hahn polynomials).

The q-Racah polynomials (see [16, p. 239]),

y(s) = R(α,β)n (X ; q) = 4φ3

(q−n, qα+β+n+1, qa−s, qs+a

q2a+α+N , qβ+1, q1−N

∣∣∣∣ q, q

)(3.27)

(n = 1, . . . , N − 1; a > 0, b = a +N ;α > −1,−1 < β < 2a), satisfy (2.11) withA and B given by

A = A(s;α, β)

= −4 qα+β+5/2(qs−a − 1)(qs+a+N−1 − 1)(qs−a−α−N − 1)(qs+a−β−1 − 1)

(q − 1)2(q2s − 1)(q2s−1 − 1),

B = B(s;α, β)

= −4 q3/2(qs+a − 1)(qs−a−N+1 − 1)(qs+a+α+N − 1)(qs−a+β+1 − 1)

(q − 1)2(q2s − 1)(q2s+1 − 1).

20

Proposition 3.16. The zeros of R(α,β)n (·; q) are strictly decreasing functions of

α on (−1,∞) and strictly increasing function of β on (−1, 2a) if a ≥ 1/2, or

else the zeros of R(α,β)n (·; q) are strictly decreasing functions of α on (−1,∞)

for each β ∈ (a− 1/2, 2a) and strictly increasing function of β on (a− 1/2, 2a).

Proof. We give the proof only for the case in which a ≥ 1/2. The proof for0 < a < 1/2 is similar. Define the interval K =

(max

{a, β− a+ 1

}, a+N − 1

).

The monotonicity function

f =B

A

=(qs+a − 1)(qs−a−N+1 − 1)(qs+a+α+N − 1)(qs−a+β+1 − 1)

qα+β+1(qs−a − 1)(qs+a+N−1 − 1)(qs−a−α−N − 1)(qs+a−β−1 − 1)

q2s−1 − 1

q2s+1 − 1

is a positive and strictly decreasing function of s ∈ K, and

0 >∂f

∂α=

log q (q2s − 1)(q2s−1 − 1)(qs+a − 1)(qs−a−N+1 − 1)(qs−a+β+1)

qα+β+1/(q2s+1 − 1)(qs−a − 1)(qs+a+N−1 − 1)(qs−a−α−N − 1)2(qs+a−β−1 − 1),

0 <∂f

∂β=

log q (q2s − 1)(q2s−1 − 1)(qs+a − 1)(qs−a−N+1 − 1)(qs+a+α+N )

qα+β+1(q2s+1 − 1)(qs−a − 1)(qs+a+N−1 − 1)(qs−a−α−N − 1)(qs+a−β−1 − 1)2,

for each s ∈ K. Thus we only need to prove that S(β)(2a−1,2a) ⊂ K13. Indeed, this

is equivalent to prove that R(α,β)n (·; q) has no zeros on

((qa + q−a)/2, (qβ−a+1 +

qa−β−1)/2)

for β ∈ (2a−1, 2a). For β ∈ (2a−1, 2a), R(α,β)n

((qa+q−a)/2; q

)= 1

and

R(α,β)n

((qβ−a+1 + qa−β−1)/2; q

)= 3φ2

(q−n, qα+β+n+1, q2a−β−1

q2a+α+N , q1−N

∣∣∣∣ q, q

)

=(q2a−β+N−n−1; q)n(q

α+β+N+1; q)n(q2a+α+N ; q)n(qN−n; q)n

> 0,

the last equality being a consequence of q-Pfaff-Saalschutz’s identity (see [1,

(10.10.3)]). We thus get, as in Proposition 3.5, S(β)(−1,2a) ⊂ K. The same proof

actually shows that S(α)(−1,∞) ⊂ K. The result follows from the discrete Stieltjes

theorem.

13It is immediate that S(β)(−1,2a−1]

⊂ K.

21

Remark 3.7. In [6, Section 14.2], the q-Racah polynomials (replacing α by qβ

and β by qα) are defined by

R(α,β,δ)n (x(s); q) = 4φ3

(q−n, qα+β+n+1, q−s, qs−N+1

δqα+1, qβ+1, q1−N

∣∣∣∣ q, q

)

(n = 1, . . . , N−1; a = 0, b = N ;α > −1,−1 < β < 1−N+logq δ, 0 < δ < qN−1),

where x(s) = δqs−N+1 + q−s. In (3.27), we can write R(α,β,a)n (·; q) instead of

R(α,β)n (·; q). Hence, for a = (1−N + logq δ)/2 fixed,

R(α,β,(1−N+logq δ)/2)n (x(s); q) = R(α,β,δ)

n

(x(s− (1−N + logq δ)/2

); q),

wherex(s− (1−N + logq δ)/2

)= 2q(1−N+logq δ)/2x(s).

Consequently, Proposition 3.16 remains valid if we replace R(α,β)n (·; q) by R

(α,β,δ)n

(·; q).

The dual q-Hahn polynomials (see [16, p. 239]),

y(s) = W (α)n (X ; q) = 3φ2

(q−n, qa−s, qs+a

qα+1, q1−N

∣∣∣∣ q, q

)(3.28)

(n = 1, . . . , N − 1; a > 0, b = a +N ;−1 < α < 2a), satisfy (2.11) with A and Bgiven by

A = A(s;α) = −4q−a+α−N+5/2(qs−a − 1)(qs+a+N−1 − 1)(qs+a−α−1 − 1)

(q − 1)2(q2s − 1)(q2s−1 − 1),

B = B(s;α) = −4qs+3/2(qs+a − 1)(qs−a−N+1 − 1)(qs−a+α+1 − 1)

(q − 1)2(q2s − 1)(q2s+1 − 1).

Proposition 3.17. The zeros of W(α)n (·; q) are strictly increasing functions of

α on (−1, 2a) if a ≥ 1/2, or else the zeros of W(α)n (·; q) are strictly increasing

functions of α on (a− 1/2, 2a).

Proof. We sketch the proof only for the case in which a ≥ 1/2. The proof for 0 <a < 1/2 is similar. Define the interval K =

(max

{a, α−a+1

}, a+N−1

). Note

that the hypotheses of the discrete Stieltjes theorem are fulfilled in K. Thus we

only need to prove that S(2a−1,2a) ⊂ K14. Note that W(α)n

((qa + q−a)/2; q

)= 1

and

W (α)n

((qα−a+1 + qa−α−1)/2; q

)= 2φ1

(q−n, q2a−α−1

q1−N

∣∣∣∣ q, q

)

= qn(2a−α−1) (q−2a+α−N−2; q)n(q1−N ; q)n

> 0,

14It is immediate that S(−1,2a−1] ⊂ K.

22

the last equality being a consequence of q-Chu-Vandermonde’s identity ([6,(1.11.5)]). The rest of the proof runs as in Proposition 3.16.

Remark 3.8. In [6, Section 14.7], the dual q-Hahn polynomials are defined by

W (α,β)n (x(s); q) = 3φ2

(q−n, qa−s, qs+a

qα+1, q1−N

∣∣∣∣ q, q

)

(n = 1, . . . , N − 1; a = 0, b = N ;α > −1, β > −1 or α < −N, β < −N),

where x(s) = δqs+α+β+1 + q−s. In (3.28), we can write W(α,a)n (·; q) instead of

W(α)n (·; q). Hence, for a = (α+ β + 1)/2 fixed,

W (α,(α+β+1)/2)n (x(s); q) = W (α,β)

n (x (s− (α+ β + 1)/2) ; q) ,

wherex (s− (α+ β + 1)/2) = 2q(α+β+1)/2x(s).

Consequently, Proposition 3.17 remains valid if we replace W(α)n (·; q) by W

(α,β)n

(·; q) and assume that α+ β is constant.

Acknowledgements

The authors thank the Warsaw University Library for kindly sending themthe book [13] and the Keldysh Institute of Applied Mathematics for making[14] available at https://keldysh.ru/papers/1983/prep1983_17.pdf after theyrequest. KC is supported by the Centre for Mathematics of the University ofCoimbra - UIDB/00324/2020, funded by the Portuguese Government throughFCT/ MCTES. FRR and AS are supported by the Fundacao de Amparo aPesquisa do Estado de Minas Gerais (FAPEMIG) Demanda Universal underthe grant APQ-03782-18, Conselho Nacional de Desenvolvimento Cientıfico eTecnologico (CNPq), and Coordenacao de Aperfeicoamento de Pessoal de NıvelSuperior (CAPES).

References

[1] G. E. Andrews, R. Askey, and R. Roy. Special Functions, volume 71 ofEncyclopedia of Mathematics and its Applications. Cambridge UniversityPress, Cambridge, 1999.

[2] R. Askey and J. Wilson. Some basic hypergeometric orthogonal polynomi-als that generalize jacobi polynomials. Mem. Amer. Math. Soc., 54(319),1985.

[3] N. M. Atakishiyev, M. Rahman, and S. K. Suslov. On classical orthogonalpolynomials. Constr. Approx., 11:181–226, 1995.

23

[4] M. E. H. Ismail. Classical and quantum orthogonal polynomials in onevariable, volume 98 of Encyclopedia of Mathematics and Its Applications.Cambridge University Press, Cambridge, 2005.

[5] S. Karlin and J. L. McGregor. The Hahn polynomials, formulas and anapplication. Technical Report 2, Applied Mathematics and Statistics Lab-oratorires, Stanford University, Stanford, CA, April 1960.

[6] R. Koekoek, P. A. Lesky, and R. F. Swarttouw. Hypergeometric OrthogonalPolynomials and Their q-Analogues. Springer Monographs in Mathematics,2010.

[7] R. J. Levit. The zeros of the Hahn polynomials. SIAM Rev., 9:191–203,1967.

[8] A. Markoff. Sur les racines de certaines equations (second note). Math.Ann., 27:177–182, 1886.

[9] P. Maroni. Une theorie algebrique des polynomes orthogonaux. applica-tion aux polynomes orthogonaux semi-classiques. In C. Brezinski, L. Gori,and A. Ronveaux, editors, Orthogonal Polynomials and Their Applications,volume 9 of IMACS Annals Comput Appl Math., pages 95–130, 1991.

[10] Ch. K. Mesztenyi. Orthogonal polynomials on a finite set. Technical ReportTR-66-30, University of Maryland, 1966.

[11] A. F. Nikiforov, S. K. Suslov, and V. B. Uvarov. Классические

ортогональные полиномы дискретной переменной (Russian)[Classical orthogonal polynomials of a discrete variable]. “Nauka”, Moscow,1985.

[12] A. F. Nikiforov, S. K. Suslov, and V. B. Uvarov. Classical orthogonalpolynomials of a discrete variable. Translated from the Russian. SpringerSeries in Computational Physics. Springer-Verlag, 1991.

[13] A. F. Nikiforov and V. B. Uvarov. Специальные функции

математической физики. (Russian) [Special Functions of Mathemati-cal Physics] With a preface by A. A. Samarskii. “Nauka”, Moscow, 1978.

[14] A. F. Nikiforov and V. B. Uvarov. Классические ортогональные

полиномы дискретной переменной на неравномерных сетках

(russian) [classical orthogonal polynomials of a discrete variable].Preprint 17, Keldysh Inst. Appl. Math., 1983.

[15] A. F. Nikiforov and V. B. Uvarov. Специальные функции

математической физики. (Russian) [Special Functions of Mathemati-cal Physics] With a preface by A. A. Samarskii. “Nauka”, Moscow, secondedition, 1984.

24

[16] A. F. Nikiforov and V. B. Uvarov. Polynomial solutions of hypergeometrictype difference equations and their classification. Integral Transforms Spec.Funct., 1:223–249, 1993.

[17] J. Petronilho. Orthogonal polynomials and special functions. Class notesfor a course given in the UC|UP Joint PhD Program in Mathematics, Uni-versity of Coimbra, 2018.

[18] G.-C. Rota and D. Sharp. Mathematics, Philosophy, and Artificial Intel-ligence... a dialogue with Gian-Carlo Rota and David Sharp. Los AlamosScience, Spring/Summer(12):92–104, 1985.

[19] B. Simon. Basic Complex Analysis. A Comprehensive Course in Analysis,Part 2A. American Mathematical Society, Providence, RI, 2015.

[20] B. Simon. Opetator Theory. A Comprehensive Course in Analysis, Part 4.American Mathematical Society, Providence, RI, 2015.

[21] T. J. Stieltjes. Sur les racines de l’equation Xn = 0. Acta Math., 9:385–400,1887.

[22] T. J. Stieltjes and Ch. Hermite. Correspondance d’Hermite et de Stieltjes.Vol. I. Gauthier-Villars, Paris, 1905.

[23] G. Szego. Orthogonal polynomials, volume 23. Amer. Math. Soc. Coll.Publ., Amer. Math. Soc., Providence, R. I., revised edition, 1959.

[24] R. S. Varga. Matrix iterative analysis. Prentice-Hall, Inc., Englewood Cliffs,N.J., 1962.

[25] N. Ja. Vilenkin and U. A. Klimyk. Representation of Lie groups and spe-cial functions. Vol. 1. Simplest Lie groups, special functions and integraltransforms. Translated from the Russian by V. A. Groza and A. A. Groza.,volume 72 of Mathematics and its Applications (Soviet Series). KluwerAcademic Publishers Group, Dordrecht, 1991.

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