Springer Monographs in Mathematics
For further volumes: www.springer.com/series/3733
Roelof Koekoek r Peter A. Lesky Ren´e F. Swarttouw
r
Hypergeometric Orthogonal Polynomials and Their q-Analogues
With a Foreword by Tom H. Koornwinder
Roelof Koekoek Delft Institute of Applied Mathematics Delft University of Technology P.O. Box 5031 2600 GA Delft The Netherlands
[email protected] Peter A. Lesky (1927–2008) University of Stuttgart, Germany
Ren´e F. Swarttouw Department of Mathematics Faculty of Sciences VU University Amsterdam De Boelelaan 1081a 1081 HV Amsterdam The Netherlands
[email protected] Foreword by Tom H. Koornwinder Korteweg-de Vries Institute of Mathematics Faculty of Science University of Amsterdam P.O. Box 94248 1090 GE Amsterdam The Netherlands
[email protected]
ISSN 1439-7382 ISBN 978-3-642-05013-8 e-ISBN 978-3-642-05014-5 DOI 10.1007/978-3-642-05014-5 Springer Heidelberg Dordrecht London New York Library of Congress Control Number: 2010923797 Mathematics Subject Classification (2000): 33C45; 33D45 © Springer-Verlag Berlin Heidelberg 2010 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover illustration: The Authors Cover design: deblik Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)
Foreword
The present book is about the Askey scheme and the q-Askey scheme, which are graphically displayed right before chapter 9 and chapter 14, respectively. The families of orthogonal polynomials in these two schemes generalize the classical orthogonal polynomials (Jacobi, Laguerre and Hermite polynomials) and they have properties similar to them. In fact, they have properties so similar that I am inclined (following Andrews & Askey [34]) to call all families in the (q-)Askey scheme classical orthogonal polynomials, and to call the Jacobi, Laguerre and Hermite polynomials very classical orthogonal polynomials. These very classical orthogonal polynomials are good friends of mine since almost the beginning of my mathematical career. When I was a fresh PhD student at the Mathematical Centre (now CWI) in Amsterdam, Dick Askey spent a sabbatical there during the academic year 1969–1970. He lectured to us in a very stimulating way about hypergeometric functions and classical orthogonal polynomials. Even better, he gave us problems to solve which might be worth a PhD. He also pointed out to us that there was more than just Jacobi, Laguerre and Hermite polynomials, for instance Hahn polynomials, and that it was one of the merits of the Higher Transcendental Functions (Bateman project) that it included some newer stuff like the Hahn polynomials (see [198, §10.23]). Note that the emphasis in this section of the Bateman project is on Chebyshev’s (or Tchebichef’s) polynomials of a discrete variable, the special case of Hahn polynomials where we have equal weights on equidistant points. This special case is very important for applications, in particular in numerical analysis, see for instance Savitzky & Golay [468] (this paper from 1964 has now 2778 citations in Google Scholar) and Meer & Weiss [404]. Ironically, as Askey later wrote in his comments on [494], Chebyshev already published in 1875 on what we now call the Hahn polynomials of general parameters. Of course, Askey told us during 1969–1970 also about the limit transitions Jacobi → Laguerre, Jacobi → Hermite and Laguerre → Hermite (formulas (9.8.16), (9.8.18) and (9.12.13) in this volume). During the seventies there grew a greater awareness that these three limit relations were part of a larger system of such limits, for instance also involving some discrete orthogonal polynomials like Hahn and Meixner polynomials. The idea to present these limits graphically was born at an v
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Oberwalfach meeting in 1977 on “Combinatorics and Special Functions” organized by George Andrews and Dominique Foata, also attended by me. In Dick Askey’s words (personal communication): “I gave a talk about many of the classical type orthogonal polynomials and it fell flat. Few there appreciated it. Later in the week Michael Hoare, a physicist then at Bedford College, talked about some very nice work [155] he had done with R.D. Cooper and Mizan Rahman. In this talk he had an overhead of the polynomials they had dealt with, starting with Hahn polynomials at the top and moving down to limiting cases with arrows illustrating the limits which they had used. The audience did not seem to care much about the probability problem, but they were very excited about the chart1 he had shown and wanted copies. If there was that much interest in his chart, I thought that it should be extended to include all of the classical type polynomials which had been found. This was included in the Memoir [72, Appendix] Jim Wilson and I wrote. We missed one case, since we had found the symmetric continuous Hahn polynomials, but had not realized that the symmetry was not needed. That was done by Atakishiyev and Suslov [81].” During the conference Polynˆomes Orthogonaux et Applications in Bar-le-Duc, France in 1984, Jacques Labelle presented a poster of size 89 × 122 cm containing what he called Tableau d’Askey [362]. For some years I had it hanging on the wall of my office, but it gradually faded away. At this 1984 Bar-le-Duc conference Andrews & Askey [34] talked about the qanalogues of the polynomials in the Askey scheme, which had already been around for some seven years, starting with the work of Askey together with his PhD student Jim Wilson. This culminated into their joint memoir [72] in 1985. As Andrews & Askey wrote in [34]: “A set of orthogonal polynomials is classical if it is a special case or a limiting case of the Askey-Wilson polynomials or q-Racah polynomials.” It took some time before also the q-Askey scheme was graphically displayed. As Labelle wrote in [362], one would need a 3-dimensional chart, because there are both arrows within the q-Askey scheme and from the q-Askey scheme to the Askey scheme. But if one is satisfied with just the arrows of the first type, then one can find the q-Askey scheme just before chapter 14 in the present volume. The present book is a merger of the report The Askey-scheme of hypergeometric orthogonal polynomials and its q-analogue by Koekoek and Swarttouw (1994, and thoroughly revised and updated in 1998) and of a series of papers by Lesky on the classification of these polynomials. The report of Koekoek and Swarttouw had its roots in a regular seminar on orthogonal polynomials at Delft University of Technology in 1990 or so, when Koekoek and Swarttouw were PhD students there, and where I also participated. The Koekoek-Swarttouw report, in its various versions, has become very well-known, a real standard reference although it was lacking during all those years the status of a publication at a recognized publisher. At present, Google Scholar gives 686 citations for the version at arXiv (arXiv:math/9602214v1 1
An extension of [155, p.285, Figure 2], called The seven-fold way of orthogonal polynomials and The seven-fold way of probability distributions
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[math.CA]), 106 citations for the 1998 report version and 69 citations for the 1994 report version (all these numbers are crude lower bounds). Many citations are from applied fields. The pleasant systematics of the report, giving for each family a limited but important list of formulas, always in the same order, will have greatly contributed to the success of the report. It is used not only to look up formulas of a family of polynomials one has already in mind, but maybe even more in the “inverse way”: one has a formula and one wonders if this is a formula for one of the polynomials in the (q-)Askey scheme. Given the classification results of Bochner and Hahn which are mentioned in the Preface of this book, it is natural to look for similar results in the case of a quadratic lattice z(s) = as2 + bs + c or a q-quadratic lattice z(s) = aqs + bq−s + c. The task is to classify for such z(s) all systems of polynomials pn of degree n (n = 0, 1, 2, . . . , N or n = 0, 1, 2, . . .) such that the functions s → pn (z(s)) are eigenfunctions of a secondorder difference operator in s. This was done by Gr¨unbaum & Haine [256] under the additional assumption that the pn are orthogonal polynomials and next by Ismail [275] without this assumption. The result was just what we already knew for the (q-) quadratic lattice cases in the (q-)Askey scheme. In an earlier elegant paper Leonard [371] had shown that a finite system of orthogonal polynomials of which the dual is again a finite system of orthogonal polynomials, must be (q-)Racah or one of its limit cases which are finite systems. Finally, the condition that z(s) is a (q-)linear or (q-) quadratic lattice was dropped for finite systems in Terwilliger [499] and for infinite systems in Vinet & Zhedanov [507]. Again nothing more came out than was already in the (q-)Askey scheme. In particular in [256] and [371] the notion of bispectrality is important: on the one hand orthogonal polynomials pn (x) are eigenfunctions of a second-order difference operator in the n-variable with eigenvalue x, on the other hand, if they are in the (q-)Askey scheme then they are eigenfunctions of a secondorder operator in the x-variable with eigenvalue depending on n. While the classification results mentioned in the previous paragraph could be written up in relatively short papers, one might also try to structure the families in the (q-)Askey scheme in a more refined way and to list precisely for which values of the parameters there will exist a positive orthogonality measure, and to give such an orthogonality measure possibly in an explicit way. Such tasks can be quite laborious and very technical. On the one hand, as mentioned in the Preface, such work was done in the books by Nikiforov and Uvarov (Suslov being also an author for the second book), and continued subsequently by many authors, in particular belonging to the Spanish school of orthogonal polynomials. On the other hand, rather independently, this was done in a long series of papers by Lesky, which material now constitutes a large part of the present book. An interesting aspect of Lesky’s approach, also reflected in the present book, is the special consideration for finite systems of orthogonal polynomials. In particular, Romanovski introduced in 1929 two finite systems of Jacobi polynomials and one finite system of Bessel polynomials which are here included in sections 9.8 (Jacobi), 9.9 (Pseudo Jacobi) and 9.13 (Bessel). As already proposed by Lesky [384] in 1998, these families, in particular the Bessel polynomials, should be embedded
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in the Askey scheme, together with suitable limit arrows. This is indeed done in the Askey scheme in this book, right before chapter 9. Why are the three finite orthogonal systems just mentioned much less known than for instance the Krawtchouk and Hahn polynomials, which are also finite systems? Are they possibly of different nature? From the point of view of Favard’s theorem (see also the discussion in §3.1 of the book), whenever one has a three-term recurrence relation x pn (x) = pn+1 (x) + Bn pn (x) +Cn pn−1 (x)
(Bn ∈ R, Cn > 0)
for n = 0, 1, . . . , N − 1, yielding monic polynomials pn of degree n for n = 0, 1, . . . , N, then these polynomials will form a finite system of orthogonal polynomials. This finite system can be extended to an infinite system of orthogonal polynomials by extending the above recurrence relation for n = N, N +1, . . . with arbitrary choices for Bn ∈ R and Cn > 0. On the other hand, any infinite system of orthogonal polynomials pn yields a finite system by only considering pn for n ≤ N. In this way the notion of a finite system seems quite arbitrary, but the finite systems will come up in an essential way if one requires that the pn are eigenfunctions of some second-order operator. Still one may wonder, for a given finite system p0 , p1 , . . . , pN , if there are natural choices for pN+1 , BN ∈ R and CN > 0 such that x pN (x) = pN+1 (x) + BN pN (x) +CN pN−1 (x). If so, then pN+1 will have N + 1 distinct real zeros interlacing the N zeros of pN and there is a natural way of realizing the orthogonality of the finite system p0 , p1 , . . . , pN by suitable weights on the zeros of pN+1 . Let us consider this for the example of Krawtchouk polynomials given in §9.11. For N a generic complex number the normalized recurrence relation (9.11.4) is satisfied for all n = 0, 1, 2, . . . by n −n, −x −1 n n−k ;p p (−N + k)n−k (x − k + 1)k . =∑ pn (x) = (−N)n pn 2 F1 −N k=0 k This is continuous in N ∈ C, and similarly are the coefficients in the normalized recurrence relation (9.11.4) continuous in N. If we let N a positive integer and if n > N then only the terms with N < k ≤ n in the expression for pn (x) survive. In particular, we then find: pN+1 (x) = x(x − 1)(x − 2) . . . (x − N), pN+2 (x) = x(x − 1)(x − 2) . . . (x − N) x − N − 1 + p(N + 2) = x − N − 1 + p(N + 2) pN+1 (x), which is compatible with CN+1 = 0. Since CN = N p(1 − p) > 0, we can realize the orthogonality of the Krawtchouk polynomials p0 , p1 , . . . , pN by weights on the zeros {0, 1, . . . , N} of pN+1 (x), and this gives the usual orthogonality for Krawtchouk
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polynomials. The Hahn and Racah polynomials and their q-analogues can be treated in a similar way. The situation is very different with the finite system of Jacobi polynomials (α ,β ) given by (9.8.3). There α + β + 1 < 0 and β > −1. Take the integer N Pn (α ,β ) such that N < −(α + β + 1)/2 ≤ N + 1. Then the Jacobi polynomials Pn for n = 0, 1, . . . , N form a finite orthogonal system. Then the coefficient of pn−1 (x) in the normalized recurrence relation (9.8.5) is positive for n = 0, 1, . . . , N − 1. If moreover 2n + α + β + 2 = 0 then (9.8.5) for n = N has no singularities and the coefficient of pN−1 (x) there is still positive. If also 2n + α + β + 4 = 0 then (9.8.5) for n = N + 1 has no singularities and the coefficient of pN (x) there is negative. So one might real(α ,β ) ize the orthogonality of the finite system by weights on the zeros of PN+1 (x) (still assuming 2n + α + β + 2 = 0). But there is no explicit expression for these zeros and there is no evident second-order operator adapted to these zeros which has the Jacobi polynomials as eigenfunctions. Therefore the orthogonality (9.8.3) is a much better choice. The finite orthogonal system (9.8.3) of Jacobi polynomials discussed above can (α ,β ) in the case that α + β + 1 < 0. also be obtained in terms of Jacobi functions φμ The finitely many values of μ for which the Jacobi functions are L2 with respect to the suitable weight function, then yield the finite system of Jacobi polynomials. See Flensted-Jensen [206, Appendix 1] for details. This example is an illustration of an important principle: If one has a finite system of orthogonal polynomials which “ends” in a somewhat unnatural way, then look for a natural completion to some infinite dimensional L2 -space, possibly using a continuous (generalized) orthogonal system like the Jacobi functions. Similarly, Koelink & Stokman [321, §7] observed finite orthogonal systems of Askey-Wilson polynomials as specializations to discrete parts of the spectrum for the Askey-Wilson function transform. The so-called Macdonald-Koornwinder polynomials are analogues of AskeyWilson polynomials in several variables associated with root system BCn . Apart from q they have five parameters, one more than in the one-variable case. There are strong indications that all families and arrows in the (q-)Askey scheme have their analogues in this multi-variable case. See for instance Stokman [491]. The families in the (q-)Askey scheme have found interpretations in various settings, for instance in the representation theory of specific Lie or finite or quantum groups, in combinatorics and in probability theory. Some of the limits in the scheme could also be brought over to limit relations of the structures where the polynomials were interpreted. In principle, the limit formulas in chapters 9 and 14 can be applied to all suitable identities there, and the resulting identities will be again present in these chapters on the suitable places. It is quite straightforward, although a little technical, to do this for the identities in the following categories: (normalized) recurrence relation, (q-) difference or differential equation, shift operators and Rodrigues-type formula. For the orthogonality relation taking the limit will not always be possible or evident. It is fun to walk downwards from one of the generating functions (14.1.13)–(14.1.15) for Askey-Wilson polynomials. An immediate question then is if it will be possible
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to walk upwards from all the generating functions for Jacobi polynomials listed in (9.8.11)–(9.8.15). Certainly there are formulas for Jacobi polynomials which have not yet found all their way up to the Askey-Wilson polynomials, for instance the addition formula for Jacobi polynomials, see [332]. Potentially, there are many further formulas which could be filled in for the various families in the (q-)Askey scheme. One category is formed by the structure relations and raising and lowering operators, see [351] and references given there. Related with this are the generators and relations for the Zhedanov algebra which one can associate with each family in the scheme, see [254]. Something which in future should also be written down for all families in the scheme are the so-called nonsymmetric versions of the various orthogonal polynomials, to be understood in the framework of Cherednik’s double affine Hecke algebra or some degenerate version of it. See this in Noumi & Stokman [423] for the case of Askey-Wilson polynomials. Finally, Bessel functions, although not being polynomials, are very natural companions of Jacobi polynomials, from which they can be obtained as limit cases. Remarkably enough, several kinds of q-Bessel functions occur as limits of families in the q-Askey scheme, see for instance Koelink & Stokman [322]. Tom Koornwinder, August 2009 University of Amsterdam
Preface
In 1929 S. Bochner (see [106]) found all families of polynomials satisfying a secondorder differential equation with polynomial coefficients. This led to the continuous classical orthogonal polynomials named after Jacobi, Laguerre and Hermite. The Bessel polynomials also appear in this study by S. Bochner. However, these polynomials can only be seen as continuous classical orthogonal polynomials in the case of a finite system (in case of positive-definite orthogonality). The continuous classical orthogonal polynomials are treated in chapter 4. In 1949 W. Hahn (see [261]) found orthogonal polynomial solutions of secondorder q-difference equations. This class of families of orthogonal polynomials is known as the Hahn class of orthogonal polynomials. Many other families of orthogonal polynomials such as the discrete classical orthogonal polynomials have been very well known for a long time, but a classification of all of these families did not exist. A first attempt to combine both the continuous and discrete classical orthogonal polynomials was made in 1985 by R. Askey and J.A. Wilson (see [72]) by introducing the so-called Askey scheme of hypergeometric orthogonal polynomials. In [72] the continuous classical orthogonal polynomials were introduced as limiting cases of the Wilson polynomials and the discrete classical orthogonal polynomials as limiting cases of the Racah polynomials. These polynomials are treated in chapter 7. In [72] R. Askey and J.A. Wilson also introduced q-analogues of the Wilson polynomials, which are known as the Askey-Wilson polynomials. We also mention the books [417] by A.F. Nikiforov and V.B. Uvarov (1988) and [416] by A.F. Nikiforov, S.K. Suslov and V.B. Uvarov (1991) in this perspective. Furthermore we refer to [256], [275] and [507] for characterizations of the AskeyWilson polynomials. In 1994 the first and third author (see [318]) published a preliminary version of their report on the Askey scheme of hypergeometric orthogonal polynomials and its q-analogue. All known q-analogues of the families of orthogonal polynomials belonging to the Askey scheme were arranged into a q-analogue of this Askey scheme. In 1998 (see [319]) a completely revised and updated version of this report appeared,
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containing more formulas for these families of orthogonal polynomials. However, a classification of the orthogonal polynomials was still missing. In the meanwhile, the second author studied the classification of several kinds of both continuous and discrete orthogonal polynomials. This led to several publications on orthogonal polynomial solutions of several kinds of eigenvalue problems. In [385] four types of q-orthogonal polynomials in x and three types of q-orthogonal polynomials in q−x were introduced. These seven types form a comprehensive basis for the classical q-orthogonal polynomials in x and in q−x . These polynomials are treated in chapter 10 and chapter 11. The intention of this book is to give a classification of all families of classical orthogonal polynomials and their q-analogues, the classical q-orthogonal polynomials. In order to do this we make the following observations: • We only consider positive-definite orthogonality in terms of an inner product. • We consider both infinite and finite systems of orthogonal polynomials. The characterization of classical orthogonal polynomials can be written in terms of eigenvalue problems as follows: the polynomial solutions of all degrees n = 0, 1, 2, . . . satisfy a three-term recurrence relation (with respect to the degree n) from which we can deduce necessary and sufficient conditions by use of the theorem of Favard. These conditions lead to all possible weight functions for orthogonality. It is also possible to have finite systems of orthogonal polynomials. We consider the following cases: 1. Second-order differential equations (chapter 4). The polynomial solutions lead to the continuous classical orthogonal polynomials; three infinite systems of Hermite, Laguerre and Jacobi polynomials and three finite systems of Jacobi, Bessel and pseudo Jacobi polynomials. 2. Second-order difference equations with real coefficients (chapter 5). The polynomial solutions lead to the first part of the discrete classical orthogonal polynomials; two infinite systems of Charlier and Meixner polynomials and two finite systems of Krawtchouk and Hahn polynomials. 3. Second-order difference equations with complex coefficients (chapter 6). The polynomial solutions lead to the second part of the discrete classical orthogonal polynomials; two infinite and finite systems of Meixner-Pollaczek and continuous Hahn polynomials. Second-order difference equations can also be seen as three-term recurrence relations in terms of the argument x. The connection between the three-term recurrence relation with respect to the degree n and the one with respect to the argument x leads to the concept of duality.
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4. The concept of duality leads to polynomial solutions in x(x+u), with x ∈ R and u ∈ R a constant, of second-order difference equations with real coefficients (chapter 7). The polynomial solutions lead to the third part of the discrete classical orthogonal polynomials; two infinite and finite systems of dual Hahn and Racah polynomials. 5. The concept of duality also leads to polynomial solutions in z(z + u), with z ∈ C and u ∈ R a constant, of second-order difference equations with complex coefficients (chapter 8). The polynomial solutions lead to the fourth part of the discrete classical orthogonal polynomials; two infinite and finite systems of continuous dual Hahn and Wilson polynomials. In chapter 9 we list all families of hypergeometric orthogonal polynomials belonging to the Askey scheme. In each case we use the most common notation and we list the most important properties of the polynomials such as a representation as hypergeometric function, orthogonality relation(s), the three-term recurrence relation, the second-order differential or difference equation, the forward shift (or degree lowering) and backward shift (or degree raising) operator, a Rodrigues-type formula and some generating functions. Moreover, in each case we mention the connection between various families by given the appropriate limit relations. 6. Hahn’s q-operator leads to eigenvalue problems in terms of second-order qdifference equations (chapter 10). The polynomial solutions lead to the first part of the classical q-orthogonal polynomials, containing the Stieltjes-Wigert, the q-Laguerre, the little q-Jacobi, the little q-Laguerre, the q-Bessel, the big q-Jacobi, the big q-Laguerre, the Al-Salam-Carlitz and the discrete q-Hermite polynomials. 7. By changing x into q−x , we obtain even more second-order q-difference equations (chapter 11). The polynomial solutions lead to the second part of the classical q-orthogonal polynomials, containing the q-Meixner, the q-Krawtchouk, the quantum q-Krawtchouk, the affine q-Krawtchouk, the q-Hahn and the q-Charlier polynomials. 8. The concept of duality can also be applied to the case of q-orthogonal polynomials, which leads to polynomial solutions in q−x + uqx , with x ∈ R and u ∈ R a constant, of second-order q-difference equations with real coefficients (chapter 12). The polynomial solutions lead to the third part of the classical q-orthogonal polynomials, containing the q-Racah, the dual q-Racah, the dual q-Hahn, the dual qKrawtchouk and the dual q-Charlier polynomials. 9. By changing qx into az , we also obtain polynomial solutions in az + uz a , with z ∈ C and u ∈ R and a ∈ C constants, of second-order q-difference equations with complex coefficients (chapter 13).
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Real polynomial solutions can only exist for |z| = 1. These lead to the fourth part of the classical q-orthogonal polynomials, containing the Askey-Wilson, the continuous q-Hahn, the continuous q-Jacobi, the continuous dual q-Hahn, the q-MeixnerPollaczek, the Al-Salam-Chihara, the continuous q-Laguerre and the continuous (big) q-Hermite polynomials. In chapter 14 we list all families of basic hypergeometric orthogonal polynomials belonging to the q-analogue of the Askey scheme. Again, in each case we use the most common notation and we list the most important properties of the polynomials such as a representation as basic hypergeometric function, orthogonality relation(s), the three-term recurrence relation, the second-order q-difference equation, the forward shift (or degree lowering) and backward shift (or degree raising) operator, a Rodrigues-type formula and some generating functions. Moreover, in each case we also indicate the limit relations between various families of q-orthogonal polynomials and the limit relations (q → 1) to the classical hypergeometric orthogonal polynomials belonging to the Askey scheme. Roelof Koekoek, Peter A. Lesky† and Ren´e F. Swarttouw
†
Unfortunately, Peter Lesky died in Innsbruck (Austria) on February 12, 2008 at the age of 81.
Contents
Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Definitions and Miscellaneous Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 The Gamma and Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 The Shifted Factorial and Binomial Coefficients . . . . . . . . . . . . . . . . . 1.4 Hypergeometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 The Binomial Theorem and Other Summation Formulas . . . . . . . . . . 1.6 Some Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Transformation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 The q-Shifted Factorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 The q-Gamma Function and q-Binomial Coefficients . . . . . . . . . . . . . 1.10 Basic Hypergeometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 The q-Binomial Theorem and Other Summation Formulas . . . . . . . . 1.12 More Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 Transformation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.14 Some q-Analogues of Special Functions . . . . . . . . . . . . . . . . . . . . . . . 1.15 The q-Derivative and q-Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.16 Shift Operators and Rodrigues-Type Formulas . . . . . . . . . . . . . . . . . .
1 1 3 4 5 7 8 10 11 13 15 16 18 19 22 24 26
2
Polynomial Solutions of Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . 2.1 Hahn’s q-Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Regularity Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Determination of the Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . 2.4.1 First Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Second Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Existence of a Three-Term Recurrence Relation . . . . . . . . . . . . . . . . . 2.6 Explicit Form of the Three-Term Recurrence Relation . . . . . . . . . . . .
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Contents
Orthogonality of the Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Favard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Orthogonality and the Self-Adjoint Operator Equation . . . . . . . . . . . . 3.3 The Jackson-Thomae q-Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Rodrigues Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 53 55 59 62 71
Part I Classical Orthogonal Polynomials 4
Orthogonal Polynomial Solutions of Differential Equations . . . . . . . . . . Continuous Classical Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . 4.1 Polynomial Solutions of Differential Equations . . . . . . . . . . . . . . . . . 4.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
79 79 79 80 83
5
Orthogonal Polynomial Solutions of Real Difference Equations . . . . . . . 95 Discrete Classical Orthogonal Polynomials I . . . . . . . . . . . . . . . . . . . . . . . . 95 5.1 Polynomial Solutions of Real Difference Equations . . . . . . . . . . . . . . 95 5.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions 101
6
Orthogonal Polynomial Solutions of Complex Difference Equations . . . Discrete Classical Orthogonal Polynomials II . . . . . . . . . . . . . . . . . . . . . . . 6.1 Real Polynomial Solutions of Complex Difference Equations . . . . . . 6.2 Classification of the Real Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
7
Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Discrete Classical Orthogonal Polynomials III . . . . . . . . . . . . . . . . . . . . . . 7.1 Motivation for Polynomials in x(x + u) Through Duality . . . . . . . . . . 7.2 Difference Equations Having Real Polynomial Solutions with Argument x(x + u) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Hypergeometric Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Three-Term Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 The Self-Adjoint Difference Equation . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Orthogonality Relations for Dual Hahn Polynomials . . . . . . . . . . . . . 7.8 Orthogonality Relations for Racah Polynomials . . . . . . . . . . . . . . . . .
123 123 123 130 131 141 141 141 142 144 148 150 156 158 162
Contents
8
Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Discrete Classical Orthogonal Polynomials IV . . . . . . . . . . . . . . . . . . . . . . 8.1 Real Polynomial Solutions of Complex Difference Equations . . . . . . 8.2 Orthogonality Relations for Continuous Dual Hahn Polynomials . . . 8.3 Orthogonality Relations for Wilson Polynomials . . . . . . . . . . . . . . . .
xvii
171 171 171 173 177
Askey Scheme of Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . 183 9
Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Wilson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Racah . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Continuous Dual Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Continuous Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Dual Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Meixner-Pollaczek . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8.1 Gegenbauer / Ultraspherical . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8.2 Chebyshev . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8.3 Legendre / Spherical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9 Pseudo Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.10 Meixner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.11 Krawtchouk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.12 Laguerre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.13 Bessel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.14 Charlier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.15 Hermite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
185 185 190 196 200 204 208 213 216 222 225 229 231 234 237 241 244 247 250
Part II Classical q-Orthogonal Polynomials 10 Orthogonal Polynomial Solutions of q-Difference Equations . . . . . . . . . . Classical q-Orthogonal Polynomials I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Polynomial Solutions of q-Difference Equations . . . . . . . . . . . . . . . . . 10.2 The Basic Hypergeometric Representation . . . . . . . . . . . . . . . . . . . . . 10.3 The Three-Term Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Solutions of the q-Pearson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
257 257 257 258 266 267 293 307
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations . . . . Classical q-Orthogonal Polynomials II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Polynomial Solutions in q−x of q-Difference Equations . . . . . . . . . . . 11.2 The Basic Hypergeometric Representation . . . . . . . . . . . . . . . . . . . . . 11.3 The Three-Term Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . .
323 323 323 324 328
xviii
11.4 11.5 11.6 11.7 11.8
Contents
Orthogonality and the Self-Adjoint Operator Equation . . . . . . . . . . . . Rodrigues Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions of the q−1 -Pearson Equation . . . . . . . . . . . . . . . . . . . . . . . . . Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classical q-Orthogonal Polynomials III . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Motivation for Polynomials in q−x + uqx Through Duality . . . . . . . . 12.2 Difference Equations Having Real Polynomial Solutions with Argument q−x + uqx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 The Basic Hypergeometric Representation . . . . . . . . . . . . . . . . . . . . . 12.4 The Three-Term Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Solutions of the q-Pearson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Orthogonal Polynomial Solutions in az + uz a of Complex q-Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classical q-Orthogonal Polynomials IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Real Polynomial Solutions in az + uz a with u ∈ R \ {0} and a, z ∈ C \ {0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Solutions of the q-Pearson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
329 333 334 344 354 369 369 369 370 373 377 379 383 389 395 395 395 398 401 407
Scheme of Basic Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . 413 14 Basic Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . 14.1 Askey-Wilson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 q-Racah . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Continuous Dual q-Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Continuous q-Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Big q-Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.1 Big q-Legendre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 q-Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.7 Dual q-Hahn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.8 Al-Salam-Chihara . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.9 q-Meixner-Pollaczek . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.10 Continuous q-Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.10.1 Continuous q-Ultraspherical / Rogers . . . . . . . . . . . . . . . . . . 14.10.2 Continuous q-Legendre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
415 415 422 429 433 438 443 445 450 455 460 463 469 475
Contents
14.11 Big q-Laguerre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.12 Little q-Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.12.1 Little q-Legendre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.13 q-Meixner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.14 Quantum q-Krawtchouk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.15 q-Krawtchouk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.16 Affine q-Krawtchouk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.17 Dual q-Krawtchouk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.18 Continuous Big q-Hermite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.19 Continuous q-Laguerre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.20 Little q-Laguerre / Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.21 q-Laguerre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.22 q-Bessel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.23 q-Charlier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.24 Al-Salam-Carlitz I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.25 Al-Salam-Carlitz II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.26 Continuous q-Hermite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.27 Stieltjes-Wigert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.28 Discrete q-Hermite I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.29 Discrete q-Hermite II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xix
478 482 486 488 493 496 501 505 509 514 518 522 526 530 534 537 540 544 547 550
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
Chapter 1
Definitions and Miscellaneous Formulas
1.1 Orthogonal Polynomials A system of polynomials {pn (x)}∞ n=0 with degree[pn (x)] = n for all n ∈ {0, 1, 2, . . .} is called orthogonal on an interval (a, b) with respect to the weight function w(x) ≥ 0 if b
a
pm (x)pn (x)w(x) dx = 0,
m = n,
m, n ∈ {0, 1, 2, . . .}.
(1.1.1)
Here w(x) is continuous or piecewise continuous or integrable, and such that b
0< a
x2n w(x) dx < ∞ for all
n ∈ {0, 1, 2, . . .}.
More generally, w(x) dx may be replaced in this definition by a positive measure d α (x), where α (x) is a bounded nondecreasing function on [a, b] ∩ R with an infinite number of points of increase, and such that b
0< a
x2n d α (x) < ∞
for all
n ∈ {0, 1, 2, . . .}.
If this function α (x) is constant between its (countably many) jump points then we have the situation of positive weights wx on a countable subset X of R. Then the system {pn (x)}∞ n=0 is orthogonal on X with respect to these weights as follows:
∑ pm (x)pn (x)wx = 0,
m = n,
m, n ∈ {0, 1, 2, . . .}.
(1.1.2)
x∈X
The case of weights wx (x ∈ X) on a finite set X of N + 1 points yields orthogonality for a finite system of polynomials {pn (x)}Nn=0 :
∑ pm (x)pn (x)wx = 0,
m = n,
m, n ∈ {0, 1, 2, . . . , N}.
(1.1.3)
x∈X
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 1, © Springer-Verlag Berlin Heidelberg 2010
1
2
1 Definitions and Miscellaneous Formulas
The orthogonality relations (1.1.1), (1.1.2) or (1.1.3) determine the polynomials {pn (x)}∞ n=0 up to constant factors, which may be fixed by a suitable normalization. We set:
σn =
b
a
σn =
{pn (x)}2 w(x) dx,
∑ {pn (x)}2 wx ,
n = 0, 1, 2, . . . ,
(1.1.4)
n = 0, 1, 2, . . .
(1.1.5)
n = 0, 1, 2, . . . , N,
(1.1.6)
x∈X
or
σn =
∑ {pn (x)}2 wx ,
x∈X
respectively and pn (x) = kn xn + lower order terms,
n = 0, 1, 2, . . . .
(1.1.7)
Then we have the following normalizations: 1. σn = 1 for all n = 0, 1, 2, . . .. In that case the system of polynomials is called orthonormal. If moreover kn > 0, for instance, then the polynomials are uniquely determined. 2. kn = 1 for all n = 0, 1, 2, . . .. In that case the system of polynomials is called monic. Also in this case the polynomials are uniquely determined. Polynomials {pn (x)}∞ n=0 with degree[pn (x)] = n for all n ∈ {0, 1, 2, . . .} which are orthogonal with respect to a (piecewise) continuous or integrable weight function w(x) as in (1.1.1) are called continuous orthogonal polynomials. Polynomials {pn (x)}Nn=0 with degree[pn (x)] = n for all n ∈ {0, 1, 2, . . . , N} and possibly N → ∞ which are orthogonal with respect to a countable subset X of R as in (1.1.2) or (1.1.3) are called discrete orthogonal polynomials. By using the Kronecker delta, defined by ⎧ ⎨ 0, m = n, δmn := m, n ∈ {0, 1, 2, . . .}, (1.1.8) ⎩ 1, m = n, orthogonality relations can be written in the form b a
pm (x)pn (x)w(x) dx = σn δmn ,
m, n ∈ {0, 1, 2, . . .}
(1.1.9)
or (with possibly N → ∞)
∑ pm (x)pn (x)wx = σn δmn ,
x∈X
respectively.
m, n ∈ {0, 1, 2, . . . , N},
(1.1.10)
1.2 The Gamma and Beta Function
3
1.2 The Gamma and Beta Function For Re z > 0 the gamma function can be defined by the gamma integral Γ(z) :=
∞
t z−1 e−t dt,
Re z > 0.
(1.2.1)
0
This gamma function satisfies the well-known functional equation Γ(z + 1) = zΓ(z) with
Γ(1) = 1,
(1.2.2)
which shows that Γ(n + 1) = n! for n ∈ {0, 1, 2, . . .}. For non-integral values of z, the gamma function also satisfies the reflection formula
π , sin π z
Γ(z)Γ(1 − z) = Hence we have Γ(1/2) = ∞ −∞
e−x dx = 2 2
z∈ / Z.
(1.2.3)
√ π , which implies that
∞
e−x dx = 2
0
∞ 0
√ π.
(1.2.4)
α = 0.
(1.2.5)
2z = 0, −1, −2, . . .
(1.2.6)
t −1/2 e−t dt = Γ(1/2) =
More general we have ∞ −∞
−α 2 x2 −2β x
e
dx =
π β 2 /α 2 e , α2
α , β ∈ R,
Further we have Legendre’s duplication formula √ Γ(z)Γ(z + 1/2) = 21−2z π Γ(2z), z ∈ C, and Stirling’s asymptotic formula √ Γ(z) ∼ 2π zz−1/2 e−z ,
Re z → ∞.
For z = x + iy with x, y ∈ R we also have √ Γ(x + iy) ∼ 2π |y|x−1/2 e−|y|π /2 ,
(1.2.7)
|y| → ∞
(1.2.8)
and for the ratio of two gamma functions we have the asymptotic formula Γ(z + a) ∼ za−b , Γ(z + b)
a, b ∈ C,
|z| → ∞.
(1.2.9)
For Re x > 0 and Re y > 0 the beta function can be defined by the integral 1
B(x, y) := 0
t x−1 (1 − t)y−1 dt,
Re x > 0,
Re y > 0.
(1.2.10)
4
1 Definitions and Miscellaneous Formulas
The connection between the beta function and the gamma function is given by the relation Γ(x)Γ(y) , Re x > 0, Re y > 0. (1.2.11) B(x, y) = Γ(x + y) There is another beta integral due to Cauchy, which can be written as 1 2π
∞ −∞
dt (r + it)ρ (s − it)σ
=
(r + s)1−ρ −σ Γ(ρ + σ − 1) Γ(ρ )Γ(σ )
(1.2.12)
for Re r > 0, Re s > 0 and Re(ρ + σ ) > 1.
1.3 The Shifted Factorial and Binomial Coefficients The shifted factorial – or Pochhammer symbol – is defined by k
(a)0 := 1 and (a)k := ∏(a + i − 1),
k = 1, 2, 3, . . . .
(1.3.1)
i=1
This can be seen as a generalization of the factorial since (1)n = n!,
n = 0, 1, 2, . . . .
The binomial coefficient can be defined by α Γ(α + 1) . := β Γ(β + 1)Γ(α − β + 1) For integer values of the parameter β we have (−α )k α := (−1)k , k = 0, 1, 2, . . . k k! and when the parameter α is an integer too, we have n n! := , k = 0, 1, 2, . . . , n, k k! (n − k)!
n = 0, 1, 2, . . . .
The latter formula can be used to show that 1
2n = 2 n 4n , n = 0, 1, 2, . . . . n n!
(1.3.2)
1.4 Hypergeometric Functions
5
1.4 Hypergeometric Functions The hypergeometric function r Fs is defined by the series ∞ a1 , . . . , ar (a1 , . . . , ar )k zk , ; z := ∑ r Fs b1 , . . . , b s k=0 (b1 , . . . , bs )k k!
(1.4.1)
where (a1 , . . . , ar )k := (a1 )k · · · (ar )k . Of course, the parameters must be such that the denominator factors in the terms of the series are never zero. When one of the numerator parameters ai equals −n, where n is a nonnegative integer, this hypergeometric function is a polynomial in z. Otherwise the radius of convergence ρ of the hypergeometric series is given by ⎧ ∞ ⎪ ⎪ ⎪ ⎪ ⎨ ρ= 1 ⎪ ⎪ ⎪ ⎪ ⎩ 0
if r < s + 1 if r = s + 1 if r > s + 1.
A hypergeometric series of the form (1.4.1) is called balanced or Saalsch¨utzian if r = s + 1, z = 1 and a1 + a2 + . . . + as+1 + 1 = b1 + b2 + . . . + bs . Many limit relations between hypergeometric orthogonal polynomials are based on the observations that a1 , . . . , ar−1 , μ a1 , . . . , ar−1 ; z = r−1 Fs−1 ;z , (1.4.2) r Fs b1 , . . . , bs−1 , μ b1 , . . . , bs−1 a1 , . . . , ar−1 , λ ar z a1 , . . . , ar−1 = r−1 Fs ; ; ar z , (1.4.3) lim r Fs b1 , . . . , b s λ b1 , . . . , b s λ →∞ a1 , . . . , ar a1 , . . . , a r z (1.4.4) ; λ z = r Fs−1 ; lim r Fs b1 , . . . , bs−1 , λ bs b1 , . . . , bs−1 bs λ →∞ and
lim r Fs
λ →∞
a1 , . . . , ar−1 , λ ar a1 , . . . , ar−1 ar z . ; z = r−1 Fs−1 ; b1 , . . . , bs−1 , λ bs b1 , . . . , bs−1 bs
(1.4.5)
Mostly, the left-hand side of (1.4.2) occurs as a limit case where some numerator parameter and some denominator parameter tend to the same value. All families of discrete orthogonal polynomials {Pn (x)}Nn=0 are defined for n = 0, 1, 2, . . . , N, where N is a positive integer. In these cases something like (1.4.2) occurs in the hypergeometric representation when n = N. In these cases we have to be aware of the fact that we still have a polynomial (in that case of degree N). For instance, if we take n = N in the hypergeometric representation (9.5.1) of the Hahn
6
1 Definitions and Miscellaneous Formulas
polynomials, we have QN (x; α , β , N) =
(N + α + β + 1)k (−x)k . (α + 1)k k! k=0 N
∑
So these cases must be understood by continuity. In cases of discrete orthogonal polynomials, we need a special notation for some of the generating functions. We define N
[ f (t)]N :=
∑
k=0
f (k) (0) k t , k!
for every function f for which f (k) (0), k = 0, 1, 2, . . . , N exists. As an example of the use of this Nth partial sum of a power series in t we remark that the generating function (9.11.12) for the Krawtchouk polynomials must be understood as follows: the Nth partial sum of ∞ k x t t m −x t (−x)m t ;− =∑ − e 1 F1 ∑ −N p p k=0 k! m=0 (−N)m m! equals N
Kn (x; p, N) n t n! n=0
∑
for x = 0, 1, 2, . . . , N. The classical exponential function ez and the trigonometric functions cos z and sin z can be expressed in terms of hypergeometric functions as − ;z , (1.4.6) ez = 0 F0 − − z2 cos z = 0 F1 1 ; − (1.4.7) 4 2
and sin z = z 0 F1
− 3 2
;−
z2 4
.
(1.4.8)
Further we have the well-known Bessel function of the first kind Jν (z), which can be defined by 1 ν z z2 − ; − . (1.4.9) F Jν (z) := 2 0 1 Γ(ν + 1) ν +1 4
1.5 The Binomial Theorem and Other Summation Formulas
7
1.5 The Binomial Theorem and Other Summation Formulas One of the most important summation formulas for hypergeometric series is given by the binomial theorem: ∞ a (a)n n ;z = ∑ z = (1 − z)−a , |z| < 1, (1.5.1) 1 F0 − n=0 n! which is a generalization of Newton’s binomium n n −n n (−n)k k ;z = ∑ (−z)k = (1 − z)n , z =∑ 1 F0 − k=0 k! k=0 k We also have Gauss’s summation formula a, b Γ(c)Γ(c − a − b) ;1 = , 2 F1 c Γ(c − a)Γ(c − b)
n = 0, 1, 2, . . . . (1.5.2)
Re(c − a − b) > 0
and the Vandermonde or Chu-Vandermonde summation formula (c − b)n −n, b ;1 = , n = 0, 1, 2, . . . . 2 F1 c (c)n On the next level we have the summation formula −n, a, b (c − a)n (c − b)n ;1 = , 3 F2 c, 1 + a + b − c − n (c)n (c − a − b)n
n = 0, 1, 2, . . . ,
which is called the Saalsch¨utz or Pfaff-Saalsch¨utz summation formula. For a very-well-poised 5 F4 we have the summation formula 1 + a/2, a, b, c, d ;1 5 F4 a/2, 1 + a − b, 1 + a − c, 1 + a − d Γ(1 + a − b)Γ(1 + a − c)Γ(1 + a − d)Γ(1 + a − b − c − d) = . Γ(1 + a)Γ(1 + a − b − c)Γ(1 + a − b − d)Γ(1 + a − c − d) The limit case d → −∞ leads to 1 + a/2, a, b, c Γ(1 + a − b)Γ(1 + a − c) ; −1 = . 4 F3 a/2, 1 + a − b, 1 + a − c Γ(1 + a)Γ(1 + a − b − c)
(1.5.3)
(1.5.4)
(1.5.5)
(1.5.6)
(1.5.7)
Finally, we mention Dougall’s bilateral sum ∞
Γ(n + a)Γ(n + b) Γ(a)Γ(1 − a)Γ(b)Γ(1 − b)Γ(c + d − a − b − 1) = , (1.5.8) Γ(c − a)Γ(d − a)Γ(c − b)Γ(d − b) n=−∞ Γ(n + c)Γ(n + d)
∑
which holds for Re(a + b) + 1 < Re(c + d). By using
8
1 Definitions and Miscellaneous Formulas
Γ(n + a)Γ(n + b) =
Γ(a)Γ(1 − a)Γ(b)Γ(1 − b) , Γ(1 − a − n)Γ(1 − b − n)
n ∈ Z,
Dougall’s bilateral sum can also be written in the form ∞
1 Γ(n + c)Γ(n + d)Γ(1 − a − n)Γ(1 − b − n) n=−∞
∑
=
Γ(c + d − a − b − 1) Γ(c − a)Γ(d − a)Γ(c − b)Γ(d − b)
(1.5.9)
for Re(a + b) + 1 < Re(c + d).
1.6 Some Integrals For the 2 F1 hypergeometric function we have Euler’s integral representation 1 Γ(c) a, b ;z = t b−1 (1 − t)c−b−1 (1 − zt)−a dt, (1.6.1) 2 F1 c Γ(b)Γ(c − b) 0 where Re c > Re b > 0, which holds for z ∈ C \ (1, ∞). Here it is understood that argt = arg(1 − t) = 0 and that (1 − zt)−a has its principal value. We also have Barnes’ integral representation i∞ a, b 1 Γ(a + s)Γ(b + s)Γ(−s) Γ(a)Γ(b) ; z = (−z)s ds F (1.6.2) 2 1 c Γ(c) 2π i −i∞ Γ(c + s) for |z| < 1 with arg(−z) < π , where the path of integration is deformed if necessary, to separate the decreasing poles s = −a − n and s = −b − n from the increasing poles s = n for n ∈ {0, 1, 2, . . .}. Such a path always exists if a, b ∈ / {. . . , −3, −2, −1}. Secondly, we have the Mellin-Barnes integral or Barnes’ first lemma
i∞ 1 Γ(a + s)Γ(b + s)Γ(c − s)Γ(d − s) ds 2π i −i∞ Γ(a + c)Γ(a + d)Γ(b + c)Γ(b + d) = Γ(a + b + c + d)
(1.6.3)
for Re(a + b + c + d) < 1, where the contour must also be taken in a such a way that the increasing poles and the decreasing poles remain separate. By using analytic continuation, one can avoid the condition Re(a + b + c + d) < 1. The Barnes integral (1.6.3) can be seen as a continuous analogue of Gauss’ summation formula (1.5.3). We also have a continuous analogue of the Pfaff-Saalsch¨utz summation formula (1.5.5) given by the integral, which is also called Barnes’ second lemma,
1.6 Some Integrals
9
i∞ Γ(a + s)Γ(b + s)Γ(c + s)Γ(1 − d − s)Γ(−s) 1 ds 2π i −i∞ Γ(e + s) Γ(a)Γ(b)Γ(c)Γ(1 + a − d)Γ(1 + b − d)Γ(1 + c − d) = Γ(e − a)Γ(e − b)Γ(e − c)
(1.6.4)
with d + e = a + b + c + 1, where the contour must be taken in a such a way that the increasing poles and the decreasing poles stay separated. A continuous analogue of the summation formula (1.5.6) for a very-well-poised F 5 4 is given by Bailey’s integral (see also [90])
i∞ Γ(1 + a/2 + s)Γ(a + s)Γ(b + s)Γ(c + s)Γ(d + s)Γ(b − a − s)Γ(−s) 1 ds 2π i −i∞ Γ(a/2 + s)Γ(1 + a − c + s)Γ(1 + a − d + s) Γ(b)Γ(c)Γ(d)Γ(b + c − a)Γ(b + d − a) = (1.6.5) 2 Γ(1 + a − c − d)Γ(b + c + d − a)
which can also be written in a more symmetric form 1 2π i
i∞ Γ(a + s)Γ(b + s)Γ(c + s)Γ(d + s)
Γ(a + 2s) Γ(−s)Γ(b − a − s)Γ(c − a − s)Γ(d − a − s) ds × Γ(−a − 2s) 2 Γ(b)Γ(c)Γ(d)Γ(b + c − a)Γ(b + d − a)Γ(c + d − a) = Γ(b + c + d − a) −i∞
(1.6.6)
and also in the following form due to Wilson (see also [512])
i∞ Γ(a + s)Γ(b + s)Γ(c + s)Γ(d + s)Γ(a − s)Γ(b − s)Γ(c − s)Γ(d − s) 1 ds 2π i −i∞ Γ(2s)Γ(−2s) 2 Γ(a + b)Γ(a + c)Γ(a + d)Γ(b + c)Γ(b + d)Γ(c + d) , (1.6.7) = Γ(a + b + c + d)
where the contours must be taken in a such a way that the increasing poles and the decreasing poles remain separate. We also mention the integral 1 2π i
i∞ −i∞
Γ(a + s)Γ(b − s)z−s ds = Γ(a + b)
za . (1 + z)a+b
(1.6.8)
Again the contour must be taken in such a way that the increasing poles of Γ(b − s) and the decreasing poles of Γ(a + s) remain separate. The integrals (1.6.3) through (1.6.7) are all special cases of a more general formula (id est formula (4.5.1.2) in [471]), which has more interesting and useful special cases such as 1 2π i
i∞ Γ(a + s)Γ(b − s) −i∞
Γ(c + s)Γ(d − s)
ds =
Γ(a + b)Γ(c + d − a − b − 1) , Γ(c + d − 1)Γ(c − a)Γ(d − b)
(1.6.9)
10
1 Definitions and Miscellaneous Formulas
where the contour must be taken in a such a way that the increasing poles and the decreasing poles remain separate and
i∞ Γ(a + s)Γ(b + s)Γ(c + s)Γ(−s)Γ(b − a − s)Γ(c − a − s) 1 ds 2π i −i∞ Γ(a + 2s)Γ(−a − 2s) 1 = Γ(b)Γ(c)Γ(b + c − a), (1.6.10) 2
where the contour must be taken in a such a way that the increasing poles and the decreasing poles remain separate. If a, b, c, d are positive or b = a and/or d = c and the real parts are positive, Wilson’s integral (1.6.7) can be written in the form 1 2π
∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2
dx Γ(2ix) Γ(a + b)Γ(a + c)Γ(a + d)Γ(b + c)Γ(b + d)Γ(c + d) . = Γ(a + b + c + d) 0
The limit case d → ∞ leads to 1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix) 2 dx = Γ(a + b)Γ(a + c)Γ(b + c). 2π 0 Γ(2ix)
(1.6.11)
(1.6.12)
1.7 Transformation Formulas In this section we list a number of transformation formulas which can be used to transform hypergeometric representations and other formulas into equivalent but different forms. First of all we have Euler’s transformation formula: a, b c − a, c − b c−a−b ; z = (1 − z) ;z . (1.7.1) 2 F1 2 F1 c c Another transformation formula for the 2 F1 series, which is also due to Euler, is a, b a, c − b z −a ; z = (1 − z) ; . (1.7.2) F F 2 1 2 1 c c z−1 This transformation formula is also known as the Pfaff or Pfaff-Kummer transformation formula. As a limit case of the Pfaff-Kummer transformation formula we have Kummer’s transformation formula for the confluent hypergeometric series:
a c−a ; −z . (1.7.3) ; z = ez 1 F1 1 F1 c c
1.8 The q-Shifted Factorial
11
If we reverse the order of summation in a terminating 1 F1 series, we obtain a 2 F0 series; in fact we have (−x)n 1 −n −n, −a − n + 1 ; x = ; − , n = 0, 1, 2, . . . . (1.7.4) F F 1 1 2 0 a − (a)n x If we apply this technique to a terminating 2 F1 series, we find −n, b (b)n −n, −c − n + 1 1 ;x = ; , n = 0, 1, 2, . . . . (1.7.5) (−x)n 2 F1 2 F1 c −b − n + 1 (c)n x On the next level we have Whipple’s transformation formula for a terminating balanced 4 F3 series: −n, a, b, c ;1 4 F3 d, e, f (e − a)n ( f − a)n −n, a, d − b, d − c ;1 (1.7.6) = 4 F3 d, a − e − n + 1, a − f − n + 1 (e)n ( f )n provided that a + b + c + 1 = d + e + f + n.
1.8 The q-Shifted Factorial The theory of q-analogues or q-extensions of classical formulas and functions is based on the observation that 1 − qα = α. q→1 1 − q lim
Therefore the number (1 − qα )/(1 − q) is sometimes called the basic number [α ]. For q = 0 and q = 1 we define [α ] :=
1 − qα , 1−q
(1.8.1)
which implies that [0] = 0,
[n] =
1 − qn n−1 k = ∑q , 1−q k=0
n = 1, 2, 3, . . . .
(1.8.2)
Now we can give a q-analogue of the Pochhammer symbol (a)k defined by (1.3.1): k
(a; q)0 := 1 and (a; q)k := ∏(1 − aqi−1 ), i=1
k = 1, 2, 3, . . . .
(1.8.3)
12
1 Definitions and Miscellaneous Formulas
It is clear that
(qα ; q)k = (α )k . q→1 (1 − q)k lim
(1.8.4)
The symbols (a; q)k are called q-shifted factorials. For negative subscripts we define (a; q)−k :=
1
a = q, q2 , q3 , . . . , qk ,
,
k
∏(1 − aq
−i
k = 1, 2, 3, . . . .
(1.8.5)
)
i=1
Now we have (a; q)−n =
1 (aq−n ; q)n
(−qa−1 )n (n) q2 , (qa−1 ; q)n
=
n = 0, 1, 2, . . . .
(1.8.6)
If we replace q by q−1 , we obtain n
(a; q−1 )n = (a−1 ; q)n (−a)n q−(2) ,
a = 0.
(1.8.7)
We can also define ∞
(a; q)∞ = ∏ (1 − aqk ),
0 < |q| < 1.
k=0
This implies that (a; q)n =
(a; q)∞ , (aqn ; q)∞
0 < |q| < 1,
(1.8.8)
(a; q)∞ , (aqλ ; q)∞
0 < |q| < 1,
(1.8.9)
and, for any complex number λ , (a; q)λ =
where the principal value of qλ is taken. Finally, we list a number of transformation formulas for the q-shifted factorials, where k and n are nonnegative integers: (a; q)n+k = (a; q)n (aqn ; q)k ,
(1.8.10)
(aqn ; q)k (a; q)k = , (a; q)n (aqk ; q)n
(1.8.11)
(aqk ; q)n−k =
(a; q)n , (a; q)k
k = 0, 1, 2, . . . , n, n
(a; q)n = (a−1 q1−n ; q)n (−a)n q(2) , n 2
(aq−n ; q)n = (a−1 q; q)n (−a)n q−n−( ) ,
a = 0, a = 0,
(1.8.12) (1.8.13) (1.8.14)
1.9 The q-Gamma Function and q-Binomial Coefficients
13
(aq−n ; q)n (a−1 q; q)n a n = , a = 0, b = 0, (bq−n ; q)n (b−1 q; q)n b
q k k (a; q)n (a; q)n−k = −1 1−n − q(2)−nk , a = 0, k = 0, 1, 2, . . . , n, (a q ; q)k a (a; q)n−k (a; q)n (b−1 q1−n ; q)k b k = , (b; q)n−k (b; q)n (a−1 q1−n ; q)k a a = 0, b = 0, k = 0, 1, 2, . . . , n, (q−n ; q)k =
k (q; q)n (−1)k q(2)−nk , (q; q)n−k
(aq−n ; q)k =
k = 0, 1, 2, . . . , n,
(a−1 q; q)n (a; q)k q−nk , (a−1 q1−k ; q)n
(1.8.15) (1.8.16)
(1.8.17) (1.8.18)
a = 0,
(1.8.19)
a n−k (k)−(n) (a−1 q; q)n − q2 2 , (aq ; q)n−k = −1 (a q; q)k q a = 0, k = 0, 1, 2, . . . , n,
(1.8.20)
(a; q)2n = (a; q2 )n (aq; q2 )n ,
(1.8.21)
(a2 ; q2 )n = (a; q)n (−a; q)n ,
(1.8.22)
−n
(a; q)∞ = (a; q )∞ (aq; q )∞ ,
0 < |q| < 1,
(1.8.23)
(a2 ; q2 )∞ = (a; q)∞ (−a; q)∞ ,
0 < |q| < 1.
(1.8.24)
2
2
We remark that by using (1.8.22) we have 1 − a2 q2n (a2 q2 ; q2 )n (aq; q)n (−aq; q)n = = . 2 1−a (a2 ; q2 )n (a; q)n (−a; q)n
(1.8.25)
1.9 The q-Gamma Function and q-Binomial Coefficients The q-gamma function is defined by Γq (x) :=
(q; q)∞ (1 − q)1−x , (qx ; q)∞
0 < q < 1.
This is a q-analogue of the gamma function given by (1.2.1). In fact we have lim Γq (x) = Γ(x).
q→1
Note that the q-gamma function satisfies the functional equation
(1.9.1)
14
1 Definitions and Miscellaneous Formulas
Γq (z + 1) =
1 − qz Γq (z) with 1−q
Γq (1) = 1,
(1.9.2)
which is a q-extension of the functional equation (1.2.2) for the ordinary gamma function. If we take the principal values of qx and (1 − q)1−x definition (1.9.1) holds for 0 < |q| < 1. For q > 1 the q-gamma function can be defined by Γq (x) =
(q−1 ; q−1 )∞ (x) q 2 (q − 1)1−x , (q−x ; q−1 )∞
The q-binomial coefficient is defined by n (q; q)n n := = , k q n−k q (q; q)k (q; q)n−k
q > 1.
(1.9.3)
k = 0, 1, 2, . . . , n,
(1.9.4)
where n denotes a nonnegative integer. This definition can be generalized in the following way. For arbitrary complex α we have k α (q−α ; q)k := (−1)k qkα −(2) . (1.9.5) k q (q; q)k Or more general, for all complex α and β and 0 < |q| < 1 we have Γq (α + 1) α (qβ +1 ; q)∞ (qα −β +1 ; q)∞ = := . β q Γq (β + 1)Γq (α − β + 1) (q; q)∞ (qα +1 ; q)∞
(1.9.6)
For instance this implies that n+α (qα +1 ; q)n = . (q; q)n n q Note that
α α Γ(α + 1) = lim . = q→1 β q β Γ(β + 1)Γ(α − β + 1)
Finally, we remark that n 1 qk =∑ , (q; q)n k=0 (q; q)k
n = 0, 1, 2, . . . ,
which can easily be shown by induction, and that n n (k ) q 2 (−a)k , n = 0, 1, 2, . . . , (a; q)n = ∑ k q k=0 which is a special case of (1.11.2).
(1.9.7)
(1.9.8)
1.10 Basic Hypergeometric Functions
15
1.10 Basic Hypergeometric Functions The basic hypergeometric or q-hypergeometric function r φs is defined by the series a1 , . . . , ar ; q, z r φs b1 , . . . , b s ∞
:=
(a1 , . . . , ar ; q)k
zk
∑ (b1 , . . . , bs ; q)k (−1)(1+s−r)k q(1+s−r)(2) (q; q)k , k
(1.10.1)
k=0
where (a1 , . . . , ar ; q)k := (a1 ; q)k · · · (ar ; q)k . Again we assume that the parameters are such that the denominator factors in the terms of the series are never zero. If one of the numerator parameters ai equals q−n , where n is a nonnegative integer, this basic hypergeometric function is a polynomial in z. Otherwise the radius of convergence ρ of the basic hypergeometric series is given by ⎧ ∞ if r < s + 1 ⎪ ⎪ ⎪ ⎪ ⎨ ρ = 1 if r = s + 1 ⎪ ⎪ ⎪ ⎪ ⎩ 0 if r > s + 1. The special case r = s + 1 reads ∞ a1 , . . . , as+1 (a1 , . . . , as+1 ; q)k zk ; q, z = ∑ . s+1 φs b1 , . . . , b s k=0 (b1 , . . . , bs ; q)k (q; q)k This basic hypergeometric series was first introduced by Heine in 1846; therefore it is sometimes called Heine’s series. A basic hypergeometric series of this form is called balanced or Saalsch¨utzian if z = q and a1 a2 · · · as+1 q = b1 b2 · · · bs . The q-hypergeometric function is a q-analogue of the hypergeometric function defined by (1.4.1) since a q 1 , . . . , q ar a1 , . . . , a r 1+s−r ; q, (q − 1) z = F ; z . lim r φs r s q→1 qb1 , . . . , q bs b1 , . . . , b s This limit will be used frequently in chapter 14. In all cases the hypergeometric series involved is in fact a polynomial so that convergence is guaranteed. In the sequel of this paragraph we also assume that each (basic) hypergeometric function is in fact a polynomial. We remark that a1 , . . . , a r a1 , . . . , ar−1 z = r−1 φs ; q, ; q, z . lim r φs ar →∞ b1 , . . . , b s ar b1 , . . . , b s
16
1 Definitions and Miscellaneous Formulas
k In fact this is the reason for the factors (−1)(1+s−r)k q(1+s−r)(2) in the definition (1.10.1) of the basic hypergeometric function. Many limit relations between basic hypergeometric orthogonal polynomials are based on the observations that a1 , . . . , ar−1 , μ a1 , . . . , ar−1 φ ; q, z = φ ; q, z , (1.10.2) r s r−1 s−1 b1 , . . . , bs−1 , μ b1 , . . . , bs−1 a1 , . . . , ar−1 , λ ar a1 , . . . , ar−1 z = r−1 φs ; q, ; q, ar z , (1.10.3) lim r φs b1 , . . . , b s λ b1 , . . . , b s λ →∞ a1 , . . . , ar a1 , . . . , a r z , (1.10.4) ; q, λ z = r φs−1 ; q, lim r φs b1 , . . . , bs−1 , λ bs b1 , . . . , bs−1 bs λ →∞
and
lim r φs
λ →∞
a1 , . . . , ar−1 , λ ar a1 , . . . , ar−1 ar z . ; q, z = r−1 φs−1 ; q, b1 , . . . , bs−1 , λ bs b1 , . . . , bs−1 bs
(1.10.5)
Mostly, the left-hand side of (1.10.2) occurs as a limit case when some numerator parameter and some denominator parameter tend to the same value. All families of discrete orthogonal polynomials {Pn (x)}Nn=0 are defined for n = 0, 1, 2, . . . , N, where N is a positive integer. In these cases something like (1.10.2) occurs in the basic hypergeometric representation when n = N. In these cases we have to be aware of the fact that we still have a polynomial (in that case of degree N). For instance, if we take n = N in the basic hypergeometric representation (14.6.1) of the q-Hahn polynomials, we have QN (q−x ; α , β , N|q) =
(αβ qN+1 ; q)k (q−x ; q)k k q. (α q; q)k (q; q)k k=0 N
∑
So these cases must be understood by continuity.
1.11 The q-Binomial Theorem and Other Summation Formulas A q-analogue of the binomial theorem (1.5.1) is called the q-binomial theorem: ∞ a (a; q)n n (az; q)∞ ; q, z =∑ φ z = , 0 < |q| < 1, |z| < 1. (1.11.1) 1 0 − (z; q)∞ n=0 (q; q)n For a = q−n with n a nonnegative integer we find −n q ; q, z = (zq−n ; q)n , n = 0, 1, 2, . . . . 1 φ0 −
(1.11.2)
1.11 The q-Binomial Theorem and Other Summation Formulas
17
In fact this is a q-analogue of Newton’s binomium (1.5.2). Note that the case a = 0 of (1.11.1) is the limit case (n → ∞) of (1.9.7). Gauss’s summation formula (1.5.3) and the Vandermonde or Chu-Vandermonde summation formula (1.5.4) have the following q-analogues: c a, b c (a−1 c, b−1 c; q)∞ ; q, = (1.11.3) , 0 < |q| < 1, < 1, 2 φ1 c ab (c, a−1 b−1 c; q)∞ ab −n q ,b cqn (b−1 c; q)n ; q, = , n = 0, 1, 2, . . . (1.11.4) 2 φ1 c b (c; q)n
and 2 φ1
(b−1 c; q)n n q−n , b ; q, q = b , c (c; q)n
n = 0, 1, 2, . . . .
(1.11.5)
By taking the limit b → ∞ in (1.11.3) we also obtain a summation formula for the 1 φ1 series:
a c (a−1 c; q)∞ = ; q, , 0 < |q| < 1. (1.11.6) 1 φ1 a (c; q)∞ c By taking the limit c → ∞ in (1.11.4) we obtain a summation formula for a terminating 2 φ0 series: −n q ,b qn 1 ; q, = n , n = 0, 1, 2, . . . . φ (1.11.7) 2 0 − b b By taking the limit a → ∞ in (1.11.6) we obtain 1 − ; q, c = , 0 φ1 c (c; q)∞
0 < |q| < 1.
On the 3 φ2 level we have the summation formula q−n , a, b (a−1 c, b−1 c; q)n ; q, q = , 3 φ2 c, abc−1 q1−n (c, a−1 b−1 c; q)n
n = 0, 1, 2, . . . ,
(1.11.8)
(1.11.9)
which is a q-analogue of the Saalsch¨utz or Pfaff-Saalsch¨utz summation formula (1.5.5). On the 6 φ5 level we have the Jackson summation formula √ √ q a, −q a, a, b, c, d aq √ ; q, 6 φ5 √ bcd a, − a, ab−1 q, ac−1 q, ad −1 q aq −1 −1 −1 −1 −1 −1 (aq, ab c q, ab d q, ac d q; q)∞ = , < 1 (1.11.10) −1 −1 −1 −1 −1 −1 (ab q, ac q, ad q, ab c d q; q)∞ bcd for a non-terminating very-well-poised 6 φ5 . We also have the summation formula
18
1 Definitions and Miscellaneous Formulas
6 φ5
=
√
√ √ q a, −q a, a, b, c, q−n aqn+1 √ ; q, a, − a, ab−1 q, ac−1 q, aqn+1 bc
(aq, ab−1 c−1 q; q)n , (ab−1 q, ac−1 q; q)n
n = 0, 1, 2, . . .
(1.11.11)
for a terminating very-well-poised 6 φ5 , which is also due to Jackson. By taking the limit c → 0 we obtain √ √ q a, −q a, a, b, 0, q−n qn √ √ φ ; q, 6 4 a, − a, ab−1 q, aqn+1 b (aq; q)n −n = b , n = 0, 1, 2, . . . . (1.11.12) (ab−1 q; q)n If we take the limit b → 0 we obtain √ √ q a, −q a, a, 0, 0, q−n qn−1 √ √ ; q, 6 φ3 a, − a, aqn+1 a n+1 = (−1)n a−n q−( 2 ) (aq; q) , n = 0, 1, 2, . . . . n
(1.11.13)
1.12 More Integrals In this section we list some q-extensions of the beta integral. For instance, we will need the following integral: ∞ 0
xc−1
(−ax, −bq/x; q)∞ π (ab, qc , q1−c ; q)∞ dx = , (−x, −q/x; q)∞ sin π c (bqc , aq−c , q; q)∞
(1.12.1)
which holds for 0 < q < 1, Re c > 0 and |aq−c | < 1. If we set b = 1 in (1.12.1), we find that ∞ (−ax; q)∞ π (a, q1−c ; q)∞ xc−1 dx = (1.12.2) (−x; q)∞ sin π c (aq−c , q; q)∞ 0 for 0 < q < 1, Re c > 0 and |aq−c | < 1, and by taking the limit c → 1 in (1.12.1), we obtain ∞ (−ax, −bq/x; q)∞ (ab, q; q)∞ dx = − ln q (1.12.3) (−x, −q/x; q)∞ (bq, a/q; q)∞ 0 for 0 < q < 1 and aq−1 < 1. Finally, we mention the Askey-Wilson q-beta integral 1 2π
1
where
w(x) 1 √ dx = 2 2π −1 1−x
π 0
w(cos θ ) d θ =
(abcd; q)∞ , (1.12.4) (ab, ac, ad, bc, bd, cd, q; q)∞
1.13 Transformation Formulas
19
2 (e2iθ ; q)∞ h(x, 1)h(x, −1)h(x, q1/2 )h(x, −q1/2 ) = iθ iθ iθ w(x) = h(x, a)h(x, b)h(x, c)h(x, d) (ae , be , ce , deiθ ; q)∞ =
(e2iθ , e−2iθ ; q)∞ , (aeiθ , ae−iθ , beiθ , be−iθ , ceiθ , ce−iθ , deiθ , de−iθ ; q)∞
x = cos θ
with h(x, α ) =
k 2 2k 1 − 2 α xq + α q ∏ ∞
k=0
2 = (α eiθ ; q)∞ = α eiθ , α e−iθ ; q , ∞
x = cos θ .
The Askey-Wilson integral (1.12.4) is a q-analogue of Wilson’s integral (1.6.11) and holds for max(|a|, |b|, |c|, |d|) < 1 and a, b, c, d positive or b = a and/or d = c with positive real parts.
1.13 Transformation Formulas In this section we list a number of transformation formulas which can be used to transform basic hypergeometric representations and other formulas into equivalent but different forms. First of all we have Heine’s transformation formulas for the 2 φ1 series: −1 (az, b; q)∞ a, b b c, z ; q, z = ; q, b (1.13.1) φ φ 2 1 2 1 c az (c, z; q)∞ (b−1 c, bz; q)∞ abc−1 z, b c = ; q, (1.13.2) 2 φ1 bz (c, z; q)∞ b a−1 c, b−1 c (abc−1 z; q)∞ abz = ; q, . (1.13.3) 2 φ1 c (z; q)∞ c The latter formula is a q-analogue of Euler’s transformation formula: a, b c − a, c − b c−a−b ; z = (1 − z) ; z . F F 2 1 2 1 c c Another transformation formula for the 2 φ1 series is (az; q)∞ a, b a, b−1 c ; q, z = ; q, bz , φ φ 2 1 2 2 c c, az (z; q)∞ which is a q-analogue of the Pfaff-Kummer transformation formula (1.7.2). Limit cases of Heine’s transformation formulas are
(1.13.4)
(1.13.5)
20
1 Definitions and Miscellaneous Formulas
2 φ1
a, 0 (az; q)∞ z ; q, z = ; q, c 1 φ1 c az (c, z; q)∞ −1 1 a c ; q, az , = 1 φ1 c (z; q)∞
2 φ1
1 φ1
(1.13.6) (1.13.7)
z 1 0, 0 ; q, z = ; q, c φ 1 1 c (c, z; q)∞ 0 − 1 ; q, cz , φ = 0 1 c (z; q)∞
(1.13.8) (1.13.9)
−1 (a, z; q) a c, 0 ∞ ; q, a ; q, z = 2 φ1 z (c; q)∞ c −1 a c, 0 az −1 = (ac z; q)∞ · 2 φ1 ; q, c c
a
and
2 φ1
(1.13.10) (1.13.11)
a, b (az, b; q)∞ z, 0 ; q, z = ; q, b 2 φ1 0 az (z; q)∞ b (bz; q)∞ ; q, az . = 1 φ1 bz (z; q)∞
(1.13.12) (1.13.13)
The q-analogues of (1.7.4) and (1.7.5) are −n −n −1 1−n q q ,a q (q−1 z)n aqn+1 ; q, z = ; q, 1 φ1 2 φ1 a (a; q)n 0 z for n = 0, 1, 2, . . ., and −n q ,b ; q, z 2 φ1 c (b; q)n −n−(n) 2 (−z)n φ = q 2 1 (c; q)n
q−n , c−1 q1−n cqn+1 ; q, b−1 q1−n bz
for n = 0, 1, 2, . . .. A limit case of the latter formula is −n −n q ,b q ,0 q n n ; q, zq = (b; q)n z 2 φ1 ; q, 2 φ0 − b−1 q1−n bz for n = 0, 1, 2, . . .. The next transformation formula is due to Jackson:
(1.13.14)
(1.13.15)
(1.13.16)
1.13 Transformation Formulas
2 φ1
21
−n −1 q−n , b q , b c, 0 (bc−1 q−n z; q)∞ ; q, z = φ ; q, q 2 3 c c, b−1 cqz−1 (bc−1 z; q)∞
for n = 0, 1, 2, . . .. Equivalently, we have −n −n −1 (b−1 q; q)∞ aq q , a, 0 q ,a c ; q, q = −1 1−n ; q, 3 φ2 2 φ1 b, c c (b q ; q)∞ b for n = 0, 1, 2, . . .. Other transformation formulas of this kind are given by: −n q ,b ; q, z 2 φ1 c −n −1 −1 1−n q , qz , c q (b−1 c; q)n bz n ; q, q = 3 φ2 (c; q)n q bc−1 q1−n , 0 −n q , b, bc−1 q−n z (b−1 c; q)n ; q, q = 3 φ2 bc−1 q1−n , 0 (c; q)n
(1.13.17)
(1.13.18)
(1.13.19) (1.13.20)
for n = 0, 1, 2, . . ., or equivalently −n −n −1 (b; q)n n q , a, b q ,b c q ; q, q = (1.13.21) a 2 φ1 ; q, 3 φ2 c, 0 b−1 q1−n (c; q)n a −n q ,a (a−1 c; q)n n bq = (1.13.22) a 2 φ1 ; q, ac−1 q1−n (c; q)n c for n = 0, 1, 2, . . .. Limit cases of these formulas are −n −n q ,b q , b, bzq−n −n ; q, z = b ; q, q , φ φ 2 0 3 2 − 0, 0
n = 0, 1, 2, . . . , (1.13.23)
or equivalently 3 φ2
q−n , a, b ; q, q 0, 0
−n q ,0 q = (b; q)n an 2 φ1 ; q, b−1 q1−n a −n n q , a bq = an 2 φ0 ; q, − a
(1.13.24) (1.13.25)
for n = 0, 1, 2, . . .. On the next level we have Sears’ transformation formula for a terminating balanced 4 φ3 series:
22
1 Definitions and Miscellaneous Formulas
4 φ3
= =
q−n , a, b, c ; q, q d, e, f
(a−1 e, a−1 f ; q)n n a 4 φ3 (e, f ; q)n
q−n , a, b−1 d, c−1 d ; q, q d, ae−1 q1−n , a f −1 q1−n
(a, a−1 b−1 e f , a−1 c−1 e f ; q)n (e, f , a−1 b−1 c−1 e f ; q)n −n −1 −1 q , a e, a f , a−1 b−1 c−1 e f × 4 φ3 ; q, q , a−1 b−1 e f , a−1 c−1 e f , a−1 q1−n
(1.13.26)
(1.13.27)
provided that de f = abcq1−n . Sears’ transformation formula is a q-analogue of Whipple’s transformation (1.7.6). Finally, we have a quadratic transformation formula which is due to Singh: a2 , b2 , c, d a2 , b2 , c2 , d 2 2 2 ; q, q = 4 φ3 ; q , q , (1.13.28) 4 φ3 abq1/2 , −abq1/2 , −cd a2 b2 q, −cd, −cdq which is valid when both sides terminate.
1.14 Some q-Analogues of Special Functions For the exponential function we have two different natural q-extensions, denoted by eq (z) and Eq (z), which can be defined by ∞ 0 zn 1 ; q, z = ∑ = , 0 < |q| < 1, |z| < 1 (1.14.1) eq (z) := 1 φ0 − (q; q) (z; q)∞ n n=0 and Eq (z) := 0 φ0
n ∞ − q(2) n ; q, −z = ∑ z = (−z; q)∞ , − n=0 (q; q)n
0 < |q| < 1.
These q-analogues of the exponential function are related by eq (z)Eq (−z) = 1. They are q-extensions of the exponential function since lim eq ((1 − q)z) = lim Eq ((1 − q)z) = ez .
q→1
q→1
Note that (1.14.1) can be seen as the special case a = 0 of (1.11.1). If we assume that 0 < |q| < 1 and |z| < 1, we may define
(1.14.2)
1.14 Some q-Analogues of Special Functions ∞ eq (iz) + eq (−iz) (−1)n z2n =∑ 2 n=0 (q; q)2n
(1.14.3)
∞ eq (iz) − eq (−iz) (−1)n z2n+1 =∑ . 2i n=0 (q; q)2n+1
(1.14.4)
cosq (z) := and sinq (z) :=
23
These are q-analogues of the trigonometric functions cos z and sin z. On the other hand, we may define 2n ∞ Eq (iz) + Eq (−iz) (−1)n q( 2 ) z2n =∑ Cosq (z) := 2 (q; q)2n n=0
and
2n+1 ∞ Eq (iz) − Eq (−iz) (−1)n q( 2 ) z2n+1 Sinq (z) := =∑ . 2i (q; q)2n+1 n=0
(1.14.5)
(1.14.6)
Then we have eq (iz) = cosq (z) + i sinq (z)
and Eq (iz) = Cosq (z) + i Sinq (z).
⎧ ⎨ cosq (z)Cosq (z) + sinq (z)Sinq (z) = 1
Further we have
⎩
sinq (z)Cosq (z) − cosq (z)Sinq (z) = 0.
The q-analogues of the trigonometric functions can be used to find different forms of formulas appearing in this book, although we will not use them. Some q-analogues of the Bessel functions are given by 0, 0 (qν +1 ; q)∞ z ν z2 (1) , |z| < 2 (1.14.7) φ ; q, − Jν (z; q) := 2 1 qν +1 (q; q)∞ 2 4 and (2)
Jν (z; q) :=
(qν +1 ; q)∞ z ν 0 φ1 (q; q)∞ 2
−
q
; q, − ν +1
qν +1 z2 4
.
(1.14.8)
These q-Bessel functions are connected by (2)
Jν (z; q) = (−
z2 (1) ; q)∞ · Jν (z; q), 4
|z| < 2.
They are q-extensions of the Bessel function of the first kind since (k)
lim Jν ((1 − q)z; q) = Jν (z),
q→1
k = 1, 2.
These q-Bessel functions were introduced by F.H. Jackson in 1905 and are therefore referred to as Jackson’s q-Bessel functions. A third q-analogue of the Bessel
24
1 Definitions and Miscellaneous Formulas
function is given by (3) Jν (z; q) :=
(qν +1 ; q)∞ ν z 1 φ1 (q; q)∞
0 qν +1
; q, qz
2
.
(1.14.9)
This third q-Bessel function is also known as the Hahn-Exton q-Bessel function. This is also a q-extension of the Bessel function of the first kind since (3)
lim Jν ((1 − q)z; q) = Jν (2z).
q→1
1.15 The q-Derivative and q-Integral For q = 1 the q-derivative operator Dq is defined by ⎧ f (z) − f (qz) ⎪ ⎨ , z = 0 (1 − q)z Dq f (z) := ⎪ ⎩ z = 0. f (0),
(1.15.1)
Further we define
and Dqn f := Dq Dqn−1 f ,
Dq0 f := f
n = 1, 2, 3, . . . .
(1.15.2)
It is not very difficult to see that lim Dq f (z) = f (z)
q→1
if the function f is differentiable at z. The q-derivative operator is a special case of Hahn’s q-operator, which will be defined in section 2.1. This operator also generalizes the differentiation operator D = d/dx. An easy consequence of the definition (1.15.1) is Dq [ f (γ x)] = γ (Dq f ) (γ x)
(1.15.3)
for all γ ∈ C, or more general
Dqn [ f (γ x)] = γ n Dqn f (γ x),
n = 0, 1, 2, . . . .
(1.15.4)
Further we have Dq [ f (x)g(x)] = f (qx)Dq g(x) + g(x)Dq f (x)
(1.15.5)
which is often referred to as the q-product rule. This can be generalized to a qanalogue of Leibniz’ rule:
1.15 The q-Derivative and q-Integral
25
n n n Dqn−k f (qk x) Dqk g (x), Dq [ f (x)g(x)] = ∑ k=0 k q
n = 0, 1, 2, . . . .
(1.15.6)
The q-integral is defined by z 0
∞
f (t) dqt := z(1 − q) ∑ f (qn z)qn ,
0 < q < 1.
(1.15.7)
n=0
This definition is due to J. Thomae (1869) and F.H. Jackson (1910). Jackson also defined a q-integral on (0, ∞) by ∞ 0
f (t) dqt := (1 − q)
∞
∑
f (qn )qn ,
0 < q < 1.
(1.15.8)
n=−∞
If the function f is continuous on [0, z], we have z
lim
z
f (t) dqt =
q→1 0
f (t) dt. 0
Of course, definition (1.15.7) implies that b a
∞
∞
n=0
n=0
f (t) dqt = b(1 − q) ∑ f (bqn )qn − a(1 − q) ∑ f (aqn )qn ,
0
and definition (1.15.8) implies that ∞ −∞
f (t) dqt = (1 − q)
∞
∑
{ f (qn ) + f (−qn )} qn ,
0 < q < 1.
n=−∞
A function f is called q-integrable if the appropriate sum converges absolutely. We remark that we have z
F(z) = 0
and
b a
f (t) dqt
Dq F(z) = f (z)
=⇒
Dq f (t) dqt = f (b) − f (a).
The q-productrule (1.15.5) implies the q-integration by parts formula b a
b g(x)Dq f (x) dq x = f (x)g(x) − a
a
b
f (qx)Dq g(x) dq x
(1.15.9)
for suitable functions f and g. Finally, we mention the following q-integral formula (see [238], formula (2.10.18))
26
1 Definitions and Miscellaneous Formulas
b −1 a qt, b−1 qt, ct; q ∞ dq t (dt, et, f t; q)∞ a −1
q, a bq, ab−1 q, cd −1 , ce−1 , c f −1 ; q ∞ = (b − a)(1 − q) , (ad, ae, a f , bd, be, b f ; q)∞
(1.15.10)
provided that c = abde f . By taking the limit f → 0, we obtain −1
b −1 a qt, b−1 qt; q ∞ q, a bq, ab−1 q, abde; q ∞ dqt = (b − a)(1 − q) . (1.15.11) (dt, et; q)∞ (ad, ae, bd, be; q)∞ a Limit cases of (1.15.11) are
∞ −1 a qt; q ∞ q, a, a−1 q, ade, a−1 d −1 e−1 q; q ∞ dqt = (1 − q) (ad, ae, d, d −1 q, e, e−1 q; q)∞ a (dt, et; q)∞
(1.15.12)
and ∞
1 dq t (dt, et; q)∞ −∞
q, −q, −1, −de, −d −1 e−1 q; q ∞ = (1 − q) . (d, −d, d −1 q, −d −1 q, e, −e, e−1 q, −e−1 q; q)∞
(1.15.13)
1.16 Shift Operators and Rodrigues-Type Formulas We need some more differential and difference operators to formulate the Rodriguestype formulas. These operators can also be used to formulate the second-order differential or difference equations but this is mostly avoided. As usual, we will use the notation df d f (x) d f (x) = (x) = . f (x) = dx dx dx Further we define Δ f (x) = f (x + 1) − f (x), (1.16.1)
and
∇ f (x) = f (x) − f (x − 1)
(1.16.2)
δ f (x) = f (x + 12 i) − f (x − 12 i).
(1.16.3)
The ordinary difference operator Δ given by (1.16.1) is also a special case of Hahn’s q-operator defined in section 2.1. Note that (1.16.3) implies that
δ 2 f (x) = f (x + i) − f (x) − f (x) + f (x − i) = f (x + i) − 2 f (x) + f (x − i),
1.16 Shift Operators and Rodrigues-Type Formulas
27
δ x = x + 12 i − (x − 12 i) = i and δ x2 = (x + 12 i)2 − (x − 12 i)2 = 2ix. Further we have for λ (x) := x(x + γ + δ + 1)
Δ λ (x) = 2x + γ + δ + 2 and ∇λ (x) = 2x + γ + δ . In a similar way we have for μ (x) := q−x + γδ qx+1
Δ μ (x) = q−x−1 (1 − q)(1 − γδ q2x+2 ) and ∇μ (x) = q−x (1 − q)(1 − γδ q2x ) and hence for λ (x) := q−x + cqx−N
Δ λ (x) = q−x−1 (1 − q)(1 − cq2x−N+1 ) and ∇λ (x) = q−x (1 − q)(1 − cq2x−N−1 ). Also note that
Δ q−x = q−x−1 (1 − q) and ∇q−x = q−x (1 − q). Finally, we define Dq f (x) := with Here we have
δq f (x) , δq x
x = cos θ
δq f (eiθ ) = f (q 2 eiθ ) − f (q− 2 eiθ ). 1
δq x = − 12 q− 2 (1 − q)(eiθ − e−iθ ), 1
1
x = cos θ .
(1.16.4)
Chapter 2
Polynomial Solutions of Eigenvalue Problems
2.1 Hahn’s q-Operator Let P denote the space of polynomials over C. In [261] W. Hahn introduced the linear operator Aq,ω defined by p(qx + ω ) − p(x) ω , p ∈ P, x ∈ R \ (Aq,ω p) (x) := qx + ω − x 1−q
(2.1.1)
for all q ∈ R \ {−1, 0}1 , ω ∈ R and (q, ω ) = (1, 0). This class of operators includes the q-derivative operator Dq (ω = 0), the difference operator Δ (q = 1 and ω = 1) and also the differentiation operator D as a limit case (q → 1 and ω = 0). In order to avoid the latter limiting process, we introduce the operator Aq,ω in a second way. From (2.1.1) we obtain that Aq,ω (1) = 0, Aq,ω (x) = 1 and the product rule (Aq,ω (p1 p2 )) (x) = (Aq,ω p1 ) (x)p2 (x) + p1 (qx + ω ) (Aq,ω p2 ) (x)
(2.1.2)
for p1 , p2 ∈ P. Now we have Theorem 2.1. For all q ∈ R \ {−1, 0} and ω ∈ R there exists a unique linear operator Aq,ω on P satisfying Aq,ω (x) = 1 and the product rule (2.1.2). Proof. The product rule (2.1.2) implies that Aq,ω (1 · 1) = Aq,ω (1) · 1 + 1 · Aq,ω (1). Hence we have Aq,ω (1) = 0. So Aq,ω is uniquely defined on the basis {1, x, x2 , . . .} of the space P by Aq,ω (xn+1 ) = Aq,ω (x)xn + (qx + ω )Aq,ω (xn ),
n = 1, 2, 3, . . .
and the initial values Aq,ω (1) = 0 and Aq,ω (x) = 1.
1 An essential property of the operator A q,ω is that its action on a polynomial of degree n leads to a polynomial of degree n − 1 for all n = 1, 2, 3, . . .. For q = −1 this property does not hold (for instance A−1,ω (x2 ) = ω ). That is why the case q = −1 is excluded.
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 2, © Springer-Verlag Berlin Heidelberg 2010
29
30
2 Polynomial Solutions of Eigenvalue Problems
As a consequence of this theorem, (2.1.2) together with Aq,ω (x) = 1 can be seen as a definition of Aq,ω . This definition holds for all q ∈ R \ {−1, 0} and ω ∈ R, and therefore includes the differentiation operator D = A1,0 .
2.2 Eigenvalue Problems We consider the eigenvalue problem2 ϕ (x) Aq,2ω yn (x) + ψ (x) (Aq,ω yn ) (x) = λn yn (qx + ω )
(2.2.1)
for polynomials yn of degree n, where Aq,2ω yn = Aq,ω (Aq,ω yn ) with λn ∈ C and n ∈ {0, 1, 2, . . .}. In [106], [372] and [379], for instance, it is shown that polynomial solutions of any degree n, where n ∈ {0, 1, 2, . . .}, can only exist if ϕ is a polynomial of degree at most 2 and ψ is a polynomial of degree 1, say
ϕ (x) = ex2 + 2 f x + g,
ψ (x) = 2ε x + γ ,
e, f , g, ε , γ ∈ C,
ε = 0.
(2.2.2)
In order to calculate the eigenvalues λn , we define (cf. (1.8.2)) 1 [−1] = − , q
n−1
[0] = 0,
[n] =
∑ qk ,
n = 1, 2, 3, . . . .
k=0
Note that this definition holds for all q ∈ R \ {−1, 0}. Now we obtain [n] − qn−k [k] = [n − k],
n ≥ k,
n, k ∈ {0, 1, 2, . . .}
(2.2.3)
and [n][n − 1] − qn−k [k][k − 1] = [n − k][n + k − 1],
n ≥ k,
n, k ∈ {0, 1, 2, . . .}. (2.2.4)
Further we have Aq,ω (xn ) =
(qx + ω )n − xn n−1 = ∑ (qx + ω )n−1−k xk qx + ω − x k=0
= xn−1
n−1
∑ qk + r(x) = [n]xn−1 + r(x),
n = 2, 3, 4, . . . ,
(2.2.5)
k=0
where r is a polynomial of degree at most n − 2. The eigenvalues λn can now be obtained by comparing the coefficients of xn in (2.2.1):
λn = 2
[n] (e[n − 1] + 2ε ), qn
n = 0, 1, 2, . . . .
It will turn out to be convenient to take yn (qx + ω ) on the right-hand side.
(2.2.6)
2.2 Eigenvalue Problems
31
So we have: if (2.2.1) has a polynomial solution of degree n, then λn given by (2.2.6) is the corresponding eigenvalue. Note that this result is also valid for q = 1 and ω = 0. So the eigenvalue problem (2.2.1) can be written in the form (ex2 + 2 f x + g) Aq,2ω yn (x) + (2ε x + γ ) (Aq,ω yn ) (x) =
[n] (e[n − 1] + 2ε )yn (qx + ω ), qn
n = 0, 1, 2, . . . .
(2.2.7)
In the case of the difference operator Δ (i.e. q = 1 and ω = 1) this reads (ex2 +2 f x+g) Δ 2 yn (x)+(2ε x+ γ ) (Δ yn ) (x) = n(e(n−1)+2ε )yn (x+1) (2.2.8) for n = 0, 1, 2, . . . and in the case of the differentiation operator D (i.e. q = 1 and ω = 0) (ex2 + 2 f x + g)yn (x) + (2ε x + γ )yn (x) = n(e(n − 1) + 2ε )yn (x)
(2.2.9)
for n = 0, 1, 2, . . .. We remark that the operator Aq,ω is not invariant under translations for q = 1, unlike the operators Δ and D. This can be seen as follows. If we apply the operator Aq,ω to the polynomial p( · + c), where c ∈ R is a constant, we obtain (Aq,ω p( · + c)) (x) =
p(qx + ω + c) − p(x + c) qx + ω − x
(2.2.10)
and, if we apply the operator Aq,ω to the polynomial p first and then replace the argument x by x + c, we have p(q(x + c) + ω ) − p(x + c) q(x + c) + ω − (x + c) p(qx + ω + c) − p(x + c) , = qx + ω − x
(Aq,ω p( · )) (x + c) =
(2.2.11)
where ω = ω + c(q − 1). Both results coincide only if q = 1. Now we will give another version of the operator equation (2.2.7) in the case that (q, ω ) = (1, 0). In order to do this, we write (Aq,ω yn ) (x) =
yn (qx + ω ) − yn (x) =: pn (x) qx + ω − x
and 2 pn (qx + ω ) − pn (x) Aq,ω yn (x) = (Aq,ω pn ) (x) = qx + ω − x 2 yn (q x + [2]ω ) − (1 + q)yn (qx + ω ) + qyn (x) . = q(qx + ω − x)2
32
2 Polynomial Solutions of Eigenvalue Problems
Then the operator equation (2.2.7) can be written in the form C(qx + ω )yn (q2 x + [2]ω ) − {C(qx + ω ) + D(qx + ω )} yn (qx + ω ) + D(qx + ω )yn (x) = λn yn (qx + ω ), where C(qx + ω ) =
ex2 + 2 f x + g q(qx + ω − x)2
and
D(qx + ω ) = qC(qx + ω ) −
2ε x + γ . qx + ω − x
If we now replace x by (x − ω )/q, we obtain the so-called symmetric form C(x)yn (qx + ω ) − {C(x) + D(x)} yn (x) + D(x)yn ((x − ω )/q) = λn yn (x) (2.2.12) for n = 0, 1, 2, . . . with C(x) =
e(x − ω )2 + 2 f q(x − ω ) + gq2 q(qx + ω − x)2 and
D(x) = qC(x) −
2ε (x − ω ) + γ q . qx + ω − x
(2.2.13)
Finally, we will derive a third version of the operator equation (2.2.7), which involves the operators Aq,ω and A1/q,−ω /q . First of all we have yn (q((x − ω )/q) + ω ) − yn ((x − ω )/q) q((x − ω )/q) + ω − (x − ω )/q yn ((x − ω )/q) − yn (x) = A1/q,−ω /q yn (x). = (x − ω )/q − x
(Aq,ω yn ) ((x − ω )/q) =
In the case of the q-derivative operator Dq (i.e. ω = 0), this reads (Dq yn ) (x/q) = Aq,0 yn (x/q) = A1/q,0 yn (x) = D1/q yn (x). In the case of the difference operator Δ (i.e. q = 1 and ω = 1), this reads (Δ yn ) (x − 1) = (A1,1 yn ) (x − 1) = (A1,−1 yn ) (x) =: ∇yn (x). Now we have
A1/q,−ω /q yn (qx + ω ) − A1/q,−ω /q yn (x) Aq,ω A1/q,−ω /q yn (x) = qx + ω − x yn (qx + ω ) − (1 + q)yn (x) + qyn ((x − ω )/q) = (qx + ω − x)2 = q−1 Aq,2ω yn ((x − ω )/q).
Hence we obtain for n = 0, 1, 2, . . .
2.3 The Regularity Condition
33
qϕ ((x − ω )/q) Aq,ω A1/q,−ω /q yn (x) + ψ ((x − ω )/q) A1/q,−ω /q yn (x) = λn yn (x).
(2.2.14)
In the case of the q-derivative operator Dq (i.e. ω = 0), this reads qϕ (x/q) Dq D1/q yn (x) + ψ (x/q) D1/q yn (x) = λn yn (x)
(2.2.15)
for n = 0, 1, 2, . . .. In the case of the difference operator Δ (i.e. q = 1 and ω = 1), this reads
ϕ (x − 1) (Δ (∇yn )) (x) + ψ (x − 1) (∇yn ) (x) = λn yn (x),
n = 0, 1, 2, . . . . (2.2.16)
2.3 The Regularity Condition In this section we will point out in which cases the eigenvalue problem (2.2.1) has essentially unique polynomial solutions yn (x) of degrees n = 0, 1, 2, . . . , N for some positive integer N with possibly N → ∞. Solutions are called essentially unique if they are determined up to a factor independent of x. We have Theorem 2.2. Let N denote a positive integer (possibly N → ∞). Then the following statements are equivalent: 1. For each n = 0, 1, 2, . . . , N there exists a solution yn of the eigenvalue problem (2.2.1) and all eigenspaces are one dimensional. 2. For m, n ∈ {0, 1, 2, . . . , N} with m = n we have λm = λn . Proof. Assume that λm = λn for m = n. Then there is either no polynomial solution for one of the degrees m and n or the solutions ym and yn belong to the same eigenspace. This shows that the first statement implies the second. Now we use induction to show that the second statement implies the first. For n = 0 we have λ0 = 0 and the one-dimensional eigenspace generated by y0 (x) = 1. Now we assume that n ∈ {1, 2, 3, . . .}. Suppose that the polynomials yν (x) are solutions of degree ν for ν = 0, 1, 2, . . . , n − 1. Then the (monic) polynomial yn (x) of degree n given by yn (x) = xn + is a solution of (2.2.1) if
n−1
∑ αν yν (x)
ν =0
with αν ∈ C
(2.3.1)
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2 Polynomial Solutions of Eigenvalue Problems
ϕ (x)Aq,2ω (xn ) + ψ (x)Aq,ω (xn ) + ϕ (x)
Aq,2ω
= λn (qx + ω ) + n
n−1
∑ αν yν
ν =0
(x) + ψ (x) Aq,ω
n−1
n−1
∑ αν yν
(x)
ν =0
∑ αν yν (qx + ω )
ν =0
holds. The polynomial ϕ (x)Aq,2ω (xn ) + ψ (x)Aq,ω (xn ) has degree at most n. Hence we may write
ϕ (x)Aq,2ω (xn ) + ψ (x)Aq,ω (xn ) = βn (qx + ω )n +
n−1
∑ βν yν (qx + ω )
ν =0
with βn , βν ∈ C. Combining the last two equations, we get
βn (qx + ω ) + n
n−1
∑ (βν + λν αν )yν (qx + ω ) = λn
ν =0
(qx + ω ) + n
n−1
∑ αν yν (qx + ω )
ν =0
and therefore (βn − λn )(qx + ω )n +
n−1
∑ (αν (λν − λn ) + βν ) yν (qx + ω ) = 0.
ν =0
Since λν = λn , this implies that the numbers αν are uniquely determined by this equation. So in fact this means that the (monic) polynomial solution yn given by (2.3.1) is uniquely determined, which implies that the corresponding eigenspace is one dimensional. Now we use (2.2.3) and (2.2.4) to find from (2.2.6): qn (λn − λm ) = e([n][n − 1] − qn−m [m][m − 1]) + 2ε ([n] − qn−m [m]) = e[n − m][n + m − 1] + 2ε [n − m] = [n − m](e[n + m − 1] + 2ε ), n ≥ m, m, n ∈ {0, 1, 2, . . .}. Hence we have
λn − λm =
[n − m] (e[n + m − 1] + 2ε ), qn
n ≥ m,
m, n ∈ {0, 1, 2, . . .}.
(2.3.2)
Since q = −1, it follows that [n − m] = 0 for m = n, so λm = λn is equivalent to e[n + m − 1] + 2ε = 0. Therefore, theorem 2.2 leads to: Corollary 2.3. Let N denote a positive integer (possibly N → ∞). Then the eigenvalue problem (2.2.1) has polynomial solutions yn of degree n for all n = 0, 1, 2, . . . , N with one-dimensional eigenspaces if and only if the regularity condition
2.4 Determination of the Polynomial Solutions
e[n] + 2ε = 0,
35
n = 0, 1, 2, . . . , 2N − 2
(2.3.3)
holds.
2.4 Determination of the Polynomial Solutions We want to obtain a two-term recurrence relation for the coefficients of the polynomial solutions yn of (2.2.1). In order to achieve this, we introduce so-called generalized binomial coefficients xn; c with x ∈ R, c ∈ C and n ∈ {0, 1, 2, . . .}, such that
x; c ·; c ·; c (2.4.1) (x) = 0 and Aq,ω (x) = Aq,ω n−1 0 n for n = 1, 2, 3, . . .. This can be done by n x; c x; c x + cqi−1 − [i − 1]ω := ∏ := 1 and , n 0 [i] i=1
n = 1, 2, 3, . . . .
(2.4.2)
Note that these generalized binomial coefficients depend on q and ω . However, for ease of expression we omit these in the notation of the symbol . In the case of the difference operator (i.e. q = 1 and ω = 1) these generalized binomial coefficients reduce to the ordinary binomial coefficients and
x+c x+c x+c , n = 1, 2, 3, . . . . = 0 and Δ = Δ n−1 0 n In the case of the differentiation operator D (i.e. q = 1 and ω = 0), we have D(1) = 0
and D
(x + c)n−1 (x + c)n = , n! (n − 1)!
n = 1, 2, 3, . . . .
2.4.1 First Approach To simplify the expressions, we write v = ω + c(1 − q). Then we have n x; c x + c − [i − 1]v =∏ , n = 1, 2, 3, . . . . n [i] i=1 Now we write n
yn (x) =
∑ an,k
k=0
x; c , k
an,n = 0,
n = 0, 1, 2, . . . .
(2.4.3)
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2 Polynomial Solutions of Eigenvalue Problems
For a given an,n = 0 we want to determine the other coefficients an,k in such a way that yn (x) given by (2.4.3) satisfies (2.2.7). By using (2.2.3), (2.2.4), (2.4.1), x; c x; c x; c , k = 1, 2, 3, . . . , + ([k − 1]v − c) = [k] x k−1 k k−1 x; c x; c x; c x + [k − 1] {([k − 1] + [k − 2]) v − 2c} = [k][k − 1] k−1 k k−2 x ; c + ([k − 2]v − c)2 , k = 2, 3, 4, . . . , k−2 qx + ω ; c k x; c k−1 x ; c + vq , k = 1, 2, 3, . . . , =q k k−1 k
2
by substituting (2.4.3) into (2.2.7) and by comparing the coefficients, we find [n − k] (e[n + k − 1] + 2ε ) an,k + e[n][n − 1]v + 2ε [n − k]v − e[k] (([k] + [k − 1])v − 2c) qn−k + (2ε c − 2 f [k] − γ )qn−k an,k+1 − e([k]v − c)2 + 2 f ([k]v − c) + g qn−k an,k+2 = 0 (2.4.4) for k = n − 1, n − 2, n − 3, . . . , 0 with the convention that an,n+1 := 0. Hence, if the regularity condition (2.3.3) holds, then (2.4.4) gives us the coefficients an,k for k = n − 1, n − 2, n − 3, . . . , 0 in terms of an,n = 0. The three-term recurrence relation (2.4.4) can be rewritten in a such a way that the possibility of a two-term recurrence relation becomes apparent: [n − k] (e[n + k − 1] + 2ε ) an,k − (e[k − 1]v − 2ec + 2 f ) [k − 1]qn−k+1 an,k+1 + [n − k − 1] (e[n + k] + 2ε ) van,k+1 − (e[k]v − 2ec + 2 f ) [k]vqn−k an,k+2 + ((e + 2ε )v + {2c(e + ε ) − γ − 2 f } q) qn−k−1 an,k+1 − (ec2 − 2 f c + g)qn−k an,k+2 = 0. This implies that if (e + 2ε )v2 + {2c(e + ε ) − γ − 2 f } vq = −(ec2 − 2 f c + g)q2 holds, then the recurrence relation (2.4.4) can be written in the form s(k)an,k + t(k)an,k+1 + v s(k + 1)an,k+1 + t(k + 1)an,k+2 = 0 with and
s(k) := [n − k] (e[n + k − 1] + 2ε )
(2.4.5)
2.4 Determination of the Polynomial Solutions
37
t(k) := (e + 2ε )v + 2ε c − γ + 2(ec − f )[k] − e[k − 1]2 vq q qn−k−1 . Now we consider the two-term recurrence relation s(k)an,k + t(k)an,k+1 = 0,
k = n − 1, n − 2, n − 3, . . . , 0.
(2.4.6)
If there exists a number c satisfying (2.4.5) and if the coefficients an,k satisfy (2.4.6), then they also satisfy (2.4.4). Therefore, the coefficients an,k can be determined by (2.4.6) in terms of an,n = 0 provided that there exists a number c such that (2.4.5) holds. In the case that v = 0, and c satisfies (2.4.5), we can write the two-term recurrence relation for the coefficients an,k in a different form. If we multiply (2.4.6) by v and use (2.4.5), we obtain [n − k] (e[n + k − 1] + 2ε ) van,k − e([k − 1]v − c)2 + 2 f ([k − 1]v − c) + g qn−k+1 an,k+1 = 0 (2.4.7) for k = n − 1, n − 2, n − 3, . . . , 0. With this first approach the case of the differentiation operator D (i.e. q = 1 and ω = v = 0) can be treated completely. If there exists a c satisfying (2.4.5), id est ec2 − 2 f c + g = 0, then the coefficients of the polynomial solutions n
yn (x) =
∑ an,k
k=0
(x + c)k , k!
an,n = 0,
n = 0, 1, 2, . . .
(2.4.8)
of the second-order differential equation (2.2.9) satisfy the two-term recurrence relation (2.4.6), id est (n − k) (e(n + k − 1) + 2ε ) an,k + (2(ec − f )k + 2ε c − γ ) an,k+1 = 0
(2.4.9)
for k = n − 1, n − 2, n − 3, . . . , 0. If (2.4.5) has no solution for c, id est e = f = 0 and g = 0, then we find from (2.4.4) with c = γ /2ε the two-term recurrence relation 2ε (n − k)an,k − gan,k+2 = 0,
an,n+1 = 0,
k = n − 1, n − 2, n − 3, . . . , 0, (2.4.10)
which leads to the (symmetric) Hermite polynomials.
2.4.2 Second Approach In order to deal with the cases where c cannot be determined by (2.4.5), we use a second approach. For this purpose we will modify (2.4.3) in such a way that in the numerator of the generalized binomial coefficient x + c occurs as the penultimate factor. By using v = ω + c(1 − q) and the definition (2.4.2), we obtain
38
2 Polynomial Solutions of Eigenvalue Problems
k x + cqi−1 + [k − 2]vqi+1−k − [i − 1]ω x ; c + [k − 2]vq2−k =∏ k [i] i=1 x + c + [k − i − 1]vqi+1−k [i] i=1 k
=∏
for k = 1, 2, 3, . . .. Note that
x ; c + [k − 2]vq2−k · ; c + [k − 2]vq2−k Aq,ω (x) = , k k−1
(2.4.11)
k = 1, 2, 3, . . . .
These alternative generalized binomial coefficients can be used to build a second form for the solutions: n x ; c + [k − 2]vq2−k , bn,n = 0, n = 0, 1, 2, . . . . yn (x) = ∑ bn,k (2.4.12) k k=0 For a given bn,n = 0 we want to determine the other coefficients bn,k in such a way that yn (x) given by (2.4.12) satisfies (2.2.7). By using x ; c + [k − 2]vq2−k x ; c + [k − 2]vq2−k x ; c + [k − 2]vq2−k = [k] + (v − c) x k−1 k k−1 for k = 1, 2, 3, . . ., 2−k x ; c + [k − 2]vq2−k 2 x ; c + [k − 2]vq = [k][k − 1] x k−2 k 2−k 2−k x ; c + [k − 2]vq 2 x ; c + [k − 2]vq + [k − 1](v − 2c) +c k−1 k−2 for k = 2, 3, 4, . . ., x ; c + [k − 2]vq2−k x ; c + [k − 2]vq2−k x ; c + [k − 2]vq2−k x = [k − 1] −c k−2 k−1 k−2 for k = 2, 3, 4, . . . and 1−k qx + ω ; c + [k − 2]vq2−k k x ; c + [k − 1]vq =q k k 2−k x ; c + [k − 2]vq2−k k x ; c + [k − 2]vq =q + vq k k−1 for k = 1, 2, 3, . . ., we find by substituting (2.4.12) into (2.2.7)
2.4 Determination of the Polynomial Solutions n
x ; c + [k − 1]vq1−k bn,k k n + ∑ (e[k − 1] + 2ε ) [k]q1−k − 1 vqn + (2[k − 1](ec − f ) + 2ε c − γ ) qn
∑ [n − k] (e[n + k − 1] + 2ε ) qk
k=0
39
k=1
x ; c + [k − 2]vq2−k bn,k k−1 n 2−k 2 n x ; c + [k − 2]vq bn,k = 0. − ∑ (ec − 2 f c + g)q k−2 k=2 ×
Hence, if ec2 − 2 f c + g = 0 can be solved for c, only two sums remain. These lead to a two-term recurrence relation for the coefficients bn,k : [n − k] (e[n + k − 1] + 2ε ) bn,k + (e[k] + 2ε )([k + 1]q−k − 1)v + (2[k](ec − f ) + 2ε c − γ )} qn−k bn,k+1 = 0
(2.4.13)
for k = n − 1, n − 2, n − 3, . . . , 0. The case of the difference operator Δ (i.e. q = ω = v = 1) can be treated by using both approaches. The first approach leads to solutions of the form
n x+c , an,n = 0, n = 0, 1, 2, . . . , yn (x) = ∑ an,k (2.4.14) k k=0 where c is a solution of (2.4.5), id est e(c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ and the coefficients satisfy (2.4.6), which reads (n − k) (e(n + k − 1) + 2ε ) an,k − e(k − 1 − c)2 + 2 f (k − 1 − c) + g an,k+1 = 0
(2.4.15)
for k = n − 1, n − 2, n − 3, . . . , 0. The second approach leads to solutions of the form
n x+c+k−2 , bn,n = 0, n = 0, 1, 2, . . . . (2.4.16) yn (x) = ∑ bn,k k k=0 If c satisfies ec2 − 2 f c + g = 0, then the coefficients satisfy the two-term recurrence relation (2.4.13), which reads (n − k) (e(n + k − 1) + 2ε ) bn,k + ek2 + 2(ec − f + ε )k + 2ε c − γ bn,k+1 = 0 for k = n − 1, n − 2, n − 3, . . . , 0.
(2.4.17)
40
2 Polynomial Solutions of Eigenvalue Problems
2.5 Existence of a Three-Term Recurrence Relation In this section we will show that the monic polynomial solutions yn of the operator equation (2.2.1) satisfy a three-term recurrence relation of the form yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
cn , dn ∈ C,
n = 1, 2, 3, . . . .
(2.5.1)
For simplification we introduce the operator Sq,ω on the space P of polynomials defined by (Sq,ω p) (x) := p(qx + ω ), p ∈ P, x ∈ R. (2.5.2) Then the following commutation relations hold: Aq,ω Sq,ω = qSq,ω Aq,ω
−1 and Sq,−1 ω Aq,ω = qAq,ω Sq,ω .
Further we will use the notation p(x) := Sq,−1 ω p (x) = p((x − ω )/q),
p ∈ P,
x∈R
(2.5.3)
(2.5.4)
for convenience. Theorem 2.4. Let Λ be a linear functional on P defined by
Λ [1] = 1 and Λ [yn ] = 0,
n = 1, 2, 3, . . . ,
(2.5.5)
where yn denotes a polynomial solution of the operator equation (2.2.1). Then the following distributional equation holds for every polynomial p ∈ P: p] = 0. Λ [qϕ(Aq,ω p) + ψ
(2.5.6)
Furthermore, for all m, n ∈ {0, 1, 2, . . .} we have
Λ [qϕ (Aq,ω ym ) (Aq,ω yn )] = −λnΛ [ym yn ]. Proof. By using (2.5.2), we obtain from (2.2.1) for every polynomial p∗ (x) =
n
∑ αk yk (x),
αk ∈ C
k=0
that ϕ (x) Aq,2ω p∗ (x) + ψ (x) (Aq,ω p∗ ) (x) n = ∑ αk ϕ (x) Aq,2ω yk (x) + ψ (x) (Aq,ω yk ) (x) k=0 n
=
∑ αk λk (Sq,ω yk ) (x).
k=0
(2.5.7)
2.5 Existence of a Three-Term Recurrence Relation
41
Applying Sq,−1 ω to both sides and using the commutation relation (2.5.3), we obtain by using the notation (2.5.4) ∗ ∗ (x) Sq,−1 qϕ(x) Aq,ω Sq,−1 ω Aq,ω p (x) + ψ ω Aq,ω p (x) =
n
∑ αk λk yk (x).
k=0
From (2.2.5) it can be deduced that for each polynomial p ∈ P there exists a polynomial p∗ such that ∗ p(x) = Sq,−1 ω Aq,ω p (x). Now we use the fact that λ0 = 0 and Λ [yk ] = 0 for k = 1, 2, 3, . . . to conclude that p] = Λ [qϕ (Aq,ω p) + ψ
n
∑ αk λkΛ [yk ] = 0,
k=0
which proves (2.5.6). To prove (2.5.7), we apply Sq,−1 ω to the operator equation (2.2.1), use the commutation relation (2.5.3) and multiply the result by ym : (x) Sq,−1 qϕ(x) Aq,ω Sq,−1 ω Aq,ω yn (x)ym (x) + ψ ω Aq,ω yn (x)ym (x) = λn ym (x)yn (x). If we apply the product rule (2.1.2) to p1 (x) = Sq,−1 ω Aq,ω yn (x) and p2 (x) = ym (x), then we find Aq,ω Sq,−1 ω Aq,ω yn ym (x) = Aq,ω Sq,−1 ω Aq,ω yn (x)ym (x) + (Aq,ω yn ) (x) (Aq,ω ym ) (x). Combining the last two results, we find that qϕ(x) Aq,ω Sq,−1 ω Aq,ω yn ym (x) − (Aq,ω yn ) (x) (Aq,ω ym ) (x) (x) Sq,−1 +ψ ω Aq,ω yn (x)ym (x) = λn ym (x)yn (x). Finally, we apply Λ on both sides of this equation to obtain Sq,−1 Λ qϕ Aq,ω Sq,−1 ω Aq,ω yn ym + ψ ω Aq,ω yn ym − Λ [qϕ (Aq,ω yn ) (Aq,ω ym )] = λnΛ [ym yn ]. The first term equals zero in view of (2.5.6) with p(x) = Sq,−1 ω Aq,ω yn (x)ym (x). This completes the proof of (2.5.7). The following theorem states the ”weak” orthogonality of the polynomial solutions of the eigenvalue problem (2.2.1). Theorem 2.5. Let the regularity condition (2.3.3) hold for the operator equation (2.2.1) with polynomial solutions yn with n = 0, 1, 2, . . .. Then the linear functional Λ given by (2.5.5) satisfies
42
2 Polynomial Solutions of Eigenvalue Problems
Λ [ym yn ] = 0 for m = n,
m, n ∈ {0, 1, 2, . . .}.
(2.5.8)
Proof. From (2.5.7) we get for m, n ∈ {0, 1, 2, . . .} ⎧ ⎨ Λ [qϕ (Aq,ω yn ) (Aq,ω ym )] = −λnΛ [ym yn ] ⎩
Λ [qϕ (Aq,ω ym ) (Aq,ω yn )] = −λmΛ [ym yn ].
Subtracting these two equations, we find (λn − λm )Λ [ym yn ] = 0,
m, n ∈ {0, 1, 2, . . .}.
The regularity condition (2.3.3) implies that λm = λn for m = n, which implies (2.5.8). If the linear functional Λ in the preceding theorem is quasi-definite, it is well known that a recurrence relation of the form (2.5.1) exists for all n = 0, 1, 2, . . .. See for instance [146]. Now we will prove: Theorem 2.6. With the assumptions of the preceding theorem, assume that there exists a number N ∈ {1, 2, 3 . . .} such that
Λ [y2n ] = 0,
for n = 0, 1, 2, . . . , N − 1 and Λ [y2N ] = 0.
(2.5.9)
Then there exists a three-term recurrence relation of the form yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
n = 1, 2, 3, . . . , N + 1
(2.5.10)
with dN = 0. Proof. The monic polynomial yn+1 can be written as n
yn+1 (x) = xyn (x) + ∑ αk yk (x), (n)
(n)
αk ∈ C,
n = 0, 1, 2, . . . .
(2.5.11)
k=0
(n)
Now we want to show that αk = 0 for k = 0, 1, 2, . . . , n − 2. To do this we multiply this equation by yν for ν ∈ {0, 1, 2, . . . , n − 2} and apply the linear functional Λ to both sides, which leads to (n)
αk Λ [y2k ] = 0,
k = 0, 1, 2, . . . , n − 2. (n)
Now we use (2.5.9) to conclude that αk
= 0 for k = 0, 1, 2, . . . , n − 2 for all n = (n)
(n)
2, 3, 4, . . . , N + 1, which proves (2.5.10) with cn = −αn and dn = −αn−1 . In order to show that dN = 0, we start with (2.5.10) for n = N, multiply by yN−1 and apply the linear functional Λ to find
Λ [yN+1 yN−1 ] = Λ [xyN yN−1 ] − cN Λ [yN yN−1 ] − dN Λ [y2N−1 ].
2.5 Existence of a Three-Term Recurrence Relation
43
Since Λ [xyN yN−1 ] = Λ [y2N ] = 0 and Λ [y2N−1 ] = 0, we get dN =
Λ [y2N ] = 0. Λ [y2N−1 ]
This completes the proof.
Now we will show that a finite system {yn }Nn=0 of polynomial solutions of the eigenvalue problem (2.2.1) which satisfies a three-term recurrence relation of the form (2.5.10) can be extended to an infinite system {yn }∞ n=0 of polynomials which satisfies a three-term recurrence relation of the form (2.5.1). First we will prove: Theorem 2.7. Let the monic solutions {yn }Nn=0 of the eigenvalue problem (2.2.1) satisfy the three-term recurrence relation (2.5.10) and let the regularity condition (2.3.3) with N → ∞ hold. Then we may write yN+k = yk yN ,
k = 0, 1, 2, . . .
(2.5.12)
with monic polynomials yk of degree k which are solutions of the eigenvalue problem (Aq,ω yk ) = ϕ Aq,2ω yk + ψ λk (Sq,ω yk ) ,
k = 0, 1, 2, . . . ,
(2.5.13)
is a λk = λN+k − λN , ϕ is a polynomial of degree at most 2 with ϕ = 0 and ψ where polynomial of degree 1 exactly. Proof. Since dN = 0, we have from (2.5.10) yN+1 (x) = (x − cN )yN (x) = y1 (x)yN (x) with
y1 (x) = x − cN
and yN+2 (x) = (x − cN+1 )yN+1 (x) − dN+1 yN (x) = ((x − cN+1 )(x − cN ) − dN+1 ) yN (x) = y2 (x)yN (x) with y2 (x) = (x − cN+1 )(x − cN ) − dN+1 . Together with y0 (x) = 1 this proves (2.5.12) for k = 0, 1, 2. Substitution of yN+k = yk yN in the eigenvalue problem (2.2.1) gives ϕ Aq,2ω ( yk yN ) + ψ (Aq,ω ( yk yN )) = λN+k (Sq,ω ( yk yN )) . By using the product rule (2.1.2), we find yk yN ) = (Aq,ω yN ) (Sq,ω yk ) + yN (Aq,ω yk ) , Aq,ω ( yk yN ) = (Aq,ω yN ) yk + (Sq,ω yN ) (Aq,ω yk ) Aq,ω ( and
44
2 Polynomial Solutions of Eigenvalue Problems
Aq,2ω ( yk yN ) = Aq,2ω yN (Sq,ω yk ) + (Aq,ω yN ) (Aq,ω yk ) + Sq,2ω yN Aq,2ω yk + (Aq,ω Sq,ω yN ) (Aq,ω yk ) . Hence ϕ Sq,2ω yN Aq,2ω yk + (ϕ (Aq,ω Sq,ω yN ) + ϕ (Aq,ω yN ) + ψ yN ) (Aq,ω yk ) + ϕ Aq,2ω yN + ψ (Aq,ω yN ) (Sq,ω yk ) = λN+k (Sq,ω yN ) (Sq,ω yk ) . Together with
ϕ Aq,2ω yN + ψ (Aq,ω yN ) = λN (Sq,ω yN ) ,
we obtain ϕ Sq,2ω yN Aq,2ω yk + (ϕ (Aq,ω Sq,ω yN ) + ϕ (Aq,ω yN ) + ψ yN ) (Aq,ω yk ) = (λN+k − λN ) (Sq,ω yN ) (Sq,ω yk ) . This equation holds for k = 0, 1, 2. For k = 1 this reads (ϕ (Aq,ω Sq,ω yN ) + ϕ (Aq,ω yN ) + ψ yN )C1 = (λN+1 − λN ) (Sq,ω yN ) (Sq,ω y1 ) with C1 = Aq,ω y1 (= 0). Since λN+1 − λN = 0, the polynomial on the left-hand side can be defined of this equation contains Sq,ω yN as a factor. Hence the polynomial ψ by ϕ (Aq,ω Sq,ω yN ) + ϕ (Aq,ω yN ) + ψ yN := ψ . Sq,ω yN For k = 2 we find (Sq,ω yN ) = (λN+2 − λN ) (Sq,ω yN ) (Sq,ω y2 ) ϕ C2 Sq,2ω yN + (Aq,ω y2 ) ψ with C2 = Aq,2ω y2 (= 0). Since λN+2 − λN = 0, this implies that Sq,ω yN divides ϕ Sq,2ω yN . Hence the polynomial ϕ can be defined by
ϕ :=
ϕ Sq,2ω yN . Sq,ω yN
So we have found that (Aq,ω yk ) = ϕ Aq,2ω yk + ψ λk (Sq,ω yk ) ,
λk = λN+k − λN .
If (2.3.2) is used, the regularity condition (2.3.3) implies that [k] λk = λN+k − λN = N+k (e[2N + k − 1] + 2ε ) (= 0), q
k = 1, 2, 3, . . . .
This proves that for (2.5.13) polynomial solutions of all degrees exist.
2.6 Explicit Form of the Three-Term Recurrence Relation
45
The preceding theorem allows us to prove the existence of a recurrence relation of the form (2.5.1). Let dn = 0 for n = 1, 2, 3, . . . , N − 1 and dN = 0. Then we have the recurrence relation (2.5.10). The preceding theorem shows that this recurrence relation can be continued as long as it is possible to continue the recurrence relation yk+1 (x) = (x − ck ) yk (x) − dk yk−1 (x) for yk . This recurrence relation either holds for all k = 0, 1, 2, . . . or there is a number K ∈ {1, 2, 3, . . .} such that dk = 0 for k = 0, 1, 2, . . . , K − 1 and dK = 0. Then the preceding theorem can be applied again. This process can be continued until we arrive at: Theorem 2.8. If the regularity condition (2.3.3) with N → ∞ holds, then there exist numbers cn , dn ∈ C such that the polynomial solutions {yn }∞ n=0 of the eigenvalue problem (2.2.1) satisfy a three-term recurrence relation of the form (2.5.1).
2.6 Explicit Form of the Three-Term Recurrence Relation To determine the polynomial solutions in section 2.4, we needed a two-term recurrence relation for the coefficients. In this section we will not need such a two-term recurrence relation. Therefore we only need the representation (2.4.3) and the threeterm recurrence relation (2.4.4). We will not need (2.4.5) here. In the case of the differentiation operator D (i.e. q = 1 and ω = v = 0), we simply use the form n xk yn (x) = ∑ an,k , an,n = 0, n = 0, 1, 2, . . . . k! k=0 Since we do not need a two-term recurrence relation for the coefficients, it suffices to take c = 0. In that case (2.4.4) reduces to the three-term recurrence relation (n − k) (e(n + k − 1) + 2ε ) an,k − (2 f k + γ )an,k+1 − gan,k+2 = 0
(2.6.1)
for k = n − 1, n − 2, n − 3, . . . , 0 with an,n+1 := 0. In the case of the q-derivative operator Dq (i.e. q = 1 and ω = 0), we need an extra observation in order to deal with the lack of translation invariance. Recall that (2.2.10) and (2.2.11) imply that (Aq,ω p( · )) (x + c) = Aq,ω p( · + c) (x) with ω = ω + c(q − 1) and c ∈ R. For q = 1 we can put c = ω /(1 − q), which yields ω = 0. Hence, since Dq := Aq,0 , the operator equation (2.2.1) can be written in terms of the q-derivative operator as ϕ (x + c) Dq2 yn ( · + c) (x) + ψ (x + c) (Dq yn ( · + c)) (x) = λn yn (qx + c), n = 0, 1, 2, . . . (2.6.2)
46
2 Polynomial Solutions of Eigenvalue Problems
with c = ω /(1 − q). This translation does not affect the possible orthogonality. In fact the regularity condition (2.3.3) is preserved since the leading coefficients of ϕ (x) and ψ (x) are equal to the leading coefficients of ϕ (x + c) and ψ (x + c), respectively. Similarly, the recurrence relation (2.5.1) for the polynomials yn is transformed into the recurrence relation yn+1 (x + c) = (x − cn + c)yn (x + c) − dn yn−1 (x + c),
n = 0, 1, 2, . . .
(2.6.3)
for the polynomials yn (x + c). While dn remains unchanged, cn is replaced by cn − c. However, cn is real if and only if cn − c is real. Together with the theorem by Favard (see the next chapter) this implies that the polynomials yn are orthogonal in the positive-definite or quasi-definite sense if and only if this is the case for the polynomials yn (x + c). In the case of the q-derivative operator Dq , we have ω = 0 and therefore c = ω /(1 − q) = 0 and v = ω + c(1 − q) = 0. If we set ω = c = v = 0 into (2.4.4), we obtain the three-term recurrence relation [n − k] (e[n + k − 1] + 2ε ) an,k − (2 f [k] + γ ) qn−k an,k+1 − gqn−k an,k+2 = 0 (2.6.4) for k = n − 1, n − 2, n − 3, . . . , 0 with an,n+1 := 0. Note that (2.6.4) for q = 1 equals (2.6.1). This implies that the case of the differentiation operator D needs no special treatment. As a generalization of the difference operator for q = 1 and ω = 0, we use another observation. From the definition (2.1.1) it follows that p(qρ x + ω ) − p(ρ x) qρ x + ω − ρ x 1 p(ρ (qx + ω /ρ )) − p(ρ x) = Aq,ω /ρ p(ρ · ) (x) = ρ (qx + ω /ρ − x) ρ
(Aq,ω p( · )) (ρ x) =
for ρ = 0 and p ∈ P. For q = 1 and ρ = ω (= 0) this yields Aq,ω /ρ = A1,1 = Δ and the operator equation (2.2.1) reads 1 1 ϕ (ω x) Δ 2 yn (ω · ) (x) + ψ (ω x) (Δ yn (ω · )) (x) = λn yn (ω x + ω ). ω2 ω
(2.6.5)
Similar to the case q = 1 and ω = 0 above, it can easily be seen by means of the three-term recurrence relation that the possible orthogonality is not affected by the dilatation x → ω x. In the case of the difference operator Δ (i.e. q = 1 and ω = v = 1), we may use the form
n x , an,n = 0, n = 0, 1, 2, . . . . (2.6.6) yn (x) = ∑ an,k k k=0 Since we do not need a two-term recurrence relation for the coefficients, we may simply take c = 0. In that case (2.4.4) reads the three-term recurrence relation
2.6 Explicit Form of the Three-Term Recurrence Relation
47
(n − k) (e(n + k − 1) + 2ε ) an,k + {e (n(n − 1) − k(2k − 1)) + 2ε (n − k) − 2 f k − γ } an,k+1 − (ek2 + 2 f k + g)an,k+2 = 0
(2.6.7)
for k = n − 1, n − 2, n − 3, . . . , 0 with an,n+1 := 0. In order to obtain the explicit form of the three-term recurrence relation (2.5.1), we substitute the monic polynomials n
yn (x) =
∑ αk
(n) k
x,
(n)
αn = 1,
n = 0, 1, 2, . . .
(2.6.8)
k=0
in the recurrence relation (2.5.1) and y1 (x) = x − c0 . Comparison of the coefficients of xn and xn−1 yields (1)
(n)
c0 = −α0 and
(2)
(n+1)
and cn = αn−1 − αn (1)
d1 = −α0 − c1 α0
(n)
,
n = 1, 2, 3, . . .
(n+1)
(n)
and dn = αn−2 − αn−1 − cn αn−1
(2.6.9) (2.6.10)
(n)
(n)
for n = 2, 3, 4, . . .. Hence it suffices to compute αn−1 for n = 1, 2, 3, . . . and αn−2 for n = 2, 3, 4, . . .. In the case of the q-derivative operator Dq , we find from (2.6.4) for k = n − 1 (e[2n − 2] + 2ε ) an,n−1 = q (2 f [n − 1] + γ ) an,n ,
n = 1, 2, 3, . . .
and for k = n − 2 and n = 2, 3, 4, . . . [2] (e[2n − 3] + 2ε ) an,n−2 − q2 (2 f [n − 2] + γ ) an,n−1 − q2 gan,n = 0. Hence, if the regularity condition (2.3.3) holds for n = 1, 2, 3, . . ., we find an,n−1 =
q (2 f [n − 1] + γ ) an,n , e[2n − 2] + 2ε
n = 1, 2, 3, . . .
and an,n−2 =
q3 (2 f [n − 1] + γ ) (2 f [n − 2] + γ ) + q2 g (e[2n − 2] + 2ε ) an,n [2] (e[2n − 3] + 2ε ) (e[2n − 2] + 2ε )
for n = 2, 3, 4, . . .. Comparing (2.4.3) and (2.6.8), we find by using (2.4.2) that an,n = [n]!,
(n)
αn−1 =
with
an,n−1 an,n−1 = [n] , [n − 1]! an,n
n = 1, 2, 3 . . .
n
[0]! := 1,
[n]! := ∏[i], i=1
n = 1, 2, 3, . . .
48
2 Polynomial Solutions of Eigenvalue Problems
and
(n)
αn−2 =
an,n−2 an,n−2 = [n][n − 1] , [n − 2]! an,n
n = 2, 3, 4, . . . .
Hence we have (n)
αn−1 =
q[n] (2 f [n − 1] + γ ) , e[2n − 2] + 2ε
n = 1, 2, 3, . . .
and (n)
αn−2 =
q2 [n][n − 1] {q (2 f [n − 1] + γ ) (2 f [n − 2] + γ ) + g (e[2n − 2] + 2ε )} [2] (e[2n − 3] + 2ε ) (e[2n − 2] + 2ε )
for n = 2, 3, 4, . . .. By using (2.6.9), we conclude that c0 = −γ q/2ε and q[n] (2 f [n − 1] + γ ) q[n + 1] (2 f [n] + γ ) − e[2n − 2] + 2ε e[2n] + 2ε q {[n] (2 f [n − 1] + γ ) (e[2n] + 2ε ) − [n + 1] (2 f [n] + γ ) (e[2n − 2] + 2ε )} = (e[2n − 2] + 2ε ) (e[2n] + 2ε ) qn (e[n − 1] + 2ε ) (2 f [n](1 + q) + γ q) − eγ [n + 1]qn−1 (2.6.11) =− (e[2n − 2] + 2ε ) (e[2n] + 2ε )
cn =
for n = 1, 2, 3, . . .. By using (2.6.10), we obtain q2 [2] {γ q(2 f + γ ) + g (e[2] + 2ε )} q {2ε (2 f (1 + q) + γ q) − eγ [2]} γ q + · [2] (e + 2ε ) (e[2] + 2ε ) 2ε (e[2] + 2ε ) 2ε 2 2 2 2 q 4 ε ( f γ − g ε ) − eγ q (e[2] + 2ε ) 4ε ( f γ − gε ) − eγ = = (2ε )2 (e + 2ε ) (e[2] + 2ε ) 4ε 2 (e + 2ε )
d1 = −
and for n = 2, 3, 4, . . . dn =
q2 [n][n − 1] {q (2 f [n − 1] + γ ) (2 f [n − 2] + γ ) + g (e[2n − 2] + 2ε )} [2] (e[2n − 3] + 2ε ) (e[2n − 2] + 2ε ) q2 [n][n + 1] {q (2 f [n] + γ ) (2 f [n − 1] + γ ) + g (e[2n] + 2ε )} [2] (e[2n − 1] + 2ε ) (e[2n] + 2ε ) q[n] (2 f [n − 1] + γ ) + e[2n − 2] + 2ε qn (e[n − 1] + 2ε ) (2 f [n](1 + q) + γ q) − eγ [n + 1]qn−1 × (e[2n − 2] + 2ε ) (e[2n] + 2ε ) −
=
qn+1 [n] (e[n − 2] + 2ε ) (e[2n − 3] + 2ε ) (e[2n − 2] + 2ε )2 (e[2n − 1] + 2ε ) × qn−1 (2 f [n − 1] + γ ) 2 f {e[n − 1] + 2ε } − qn−1 eγ − g (e[2n − 2] + 2ε )2 .
(2.6.12)
2.6 Explicit Form of the Three-Term Recurrence Relation
49
Note that the latter formula also holds for n = 1. In the case of the differentiation operator D, we have q = 1. In that case we get cn = −
2 f n (e(n − 1) + 2ε ) − γ (e − ε ) , 2 (e(n − 1) + ε ) (en + ε )
n = 0, 1, 2, . . .
(2.6.13)
and dn =
n (e(n − 2) + 2ε ) 4 (e(2n − 3) + 2ε ) (e(n − 1) + ε )2 (e(2n − 1) + 2ε ) × {2 f (n − 1) + γ } {2 f (e(n − 1) + 2ε ) − eγ } − 4g (e(n − 1) + ε )2 , n = 1, 2, 3, . . . .
(2.6.14)
In the case of the difference operator Δ , we find from (2.6.7) with k = n − 1 (e(2n − 2) + 2ε ) an,n−1 = e(n2 − 4n + 3) − 2ε + 2 f (n − 1) + γ an,n for n = 1, 2, 3, . . ., and with k = n − 2 2 (e(2n − 3) + 2ε ) an,n−2 − e(n2 − 8n + 10) − 4ε + 2 f (n − 2) + γ an,n−1 − e(n − 2)2 + 2 f (n − 2) + g an,n = 0 for n = 2, 3, 4, . . .. Hence, if the regularity condition (2.3.3) holds for n = 1, 2, 3, . . ., we find an,n−1 =
e(n − 1)(n − 3) − 2ε + 2 f (n − 1) + γ an,n , 2 (e(n − 1) + ε )
n = 1, 2, 3, . . .
and for n = 2, 3, 4, . . . an,n−2 =
e(n − 2)2 + 2 f (n − 2) + g
2 (e(2n − 3) + 2ε ) 2 + e(n − 8n + 10) − 4ε + 2 f (n − 2) + γ ×
(e(n − 1)(n − 3) − 2ε + 2 f (n − 1) + γ ) an,n . 4 (e(n − 1) + ε ) (e(2n − 3) + 2ε )
Comparing (2.6.8) and (2.6.6), we find that an,n−1 n − 1 n (2(n − 1)( f − e) − (n + 1)ε + γ ) (n) an,n = n!, αn−1 = n = − an,n 2 2 (e(n − 1) + ε ) for n = 1, 2, 3, . . . and
50
2 Polynomial Solutions of Eigenvalue Problems (n)
αn−2
an,n−2 n 3n − 1 n − 2 an,n−1 − + = 3 4 2 (n − 2)! (n − 2)! (n − 2)(3n − 1) n − 2 an,n−1 an,n−2 − + = n(n − 1) 24 2 an,n an,n n(n − 1) = 4 (e(n − 1) + ε ) (e(2n − 3) + 2ε ) 1 (n + 1)(3n + 2) (e(n − 1) + ε ) (e(2n − 3) + 2ε ) × 6 − en(n − 1)2 (e(2n − 3) + 2ε ) + e2 (n − 1)2 (n − 2)2 − 2 (2 f (n − 1) + γ ) (e(2n − 3) + nε ) − γ (e(n − 1) + 2ε ) + (2 f (n − 1) + γ ) (2 f (n − 2) + γ ) + 2g (e(n − 1) + ε )
for n = 2, 3, 4, . . .. By using (2.6.9), we conclude that c0 = 1 − γ /2ε and n (2(n − 1)( f − e) − (n + 1)ε + γ ) 2 (e(n − 1) + ε ) (n + 1) (2n( f − e) − (n + 2)ε + γ ) − 2(en + ε ) n (e(n − 1) + 2ε ) (2(e − f ) + ε ) + (e − ε )(γ − 2ε ) = 2 (e(n − 1) + ε ) (en + ε )
cn =
(2.6.15)
for n = 1, 2, 3, . . .. By using (2.6.10), we find d1 = −
1 4(e + ε )(e + 2ε ) − 2e(e + 2ε ) 2(e + ε )(e + 2ε )
− 2(2 f + γ )(e + 2ε ) − γ (e + 2ε ) + (2 f + γ )γ + 2g(e + ε )
+ =
2ε (2(e − f ) + ε ) + (e − ε )(γ − 2ε ) 2ε − γ · 2ε (e + ε ) 2ε
4 ε ( f γ − g ε ) − eγ 2 . 4ε 2 (e + 2ε )
and for n = 2, 3, 4, . . .
2.6 Explicit Form of the Three-Term Recurrence Relation
dn =
51
n(n − 1) 4 (e(n − 1) + ε ) (e(2n − 3) + 2ε ) 1 (n + 1)(3n + 2) (e(n − 1) + ε ) (e(2n − 3) + 2ε ) × 6 − en(n − 1)2 (e(2n − 3) + 2ε ) + e2 (n − 1)2 (n − 2)2 − 2 (2 f (n − 1) + γ ) (e(2n − 3) + nε ) − γ (e(n − 1) + 2ε ) + (2 f (n − 1) + γ ) (2 f (n − 2) + γ ) + 2g (e(n − 1) + ε ) −
n(n + 1) 4(en + ε ) (e(2n − 1) + 2ε ) 1 (n + 2)(3n + 5)(en + ε ) (e(2n − 1) + 2ε ) × 6 − en2 (n + 1) (e(2n − 1) + 2ε ) + e2 n2 (n − 1)2 − 2(2 f n + γ ) (e(2n − 1) + (n + 1)ε ) − γ (en + 2ε ) + (2 f n + γ ) (2 f (n − 1) + γ ) + 2g(en + ε )
−
=−
n (2(e − f ) + ε ) (e(n − 1) + 2ε ) + (e − ε )(γ − 2ε ) 2 (e(n − 1) + ε ) (en + ε ) n (2(n − 1)( f − e) − (n + 1)ε + γ ) × 2 (e(n − 1) + ε ) n (e(n − 2) + 2ε )
4 (e(2n − 3) + 2ε ) (e(n − 1) + ε )2 (e(2n − 1) + 2ε ) × e(n − 1)2 (e(n − 1) + 2ε )2 + 2(n − 1) (e(n − 1) + 2ε ) (2eg + 2 f (ε − f ) − eγ ) + 4ε (gε − f γ ) + eγ 2 .
Note that the latter formula also holds for n = 1.
(2.6.16)
Chapter 3
Orthogonality of the Polynomial Solutions
3.1 Favard’s Theorem In this section we consider the possible orthogonality of polynomials satisfying a three-term recurrence relation of the form (2.5.1). Hereby we use Favard’s theorem (see for instance [146]): Theorem 3.1. Let yn denote the monic polynomial of degree n ∈ {0, 1, 2, . . .} satisfying the three-term recurrence relation (2.5.1) yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
cn , dn ∈ C,
n = 1, 2, 3, . . . .
Then there exists a unique linear functional Λ with
Λ [1] = 1 and Λ [ym yn ] = 0 for m = n,
m, n ∈ {0, 1, 2, . . .}.
(3.1.1)
This linear functional Λ is quasi-definite if and only if dn = 0 for all n = 1, 2, 3, . . .. This linear functional Λ is positive-definite if and only if cn ∈ R for all n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. Proof. For the fulfillment of (3.1.1), we must have that Λ [y0 ]=1 and Λ [yn ] = 0 for n = 1, 2, 3, . . .. Then the linear functional Λ is uniquely determined on P. Furthermore, (2.5.1) can be written as xyn (x) = yn+1 (x) + cn yn (x) + dn yn−1 (x),
n = 1, 2, 3, . . .
(3.1.2)
and applying the linear functional Λ , we obtain
Λ [xyn (x)] = 0,
n = 2, 3, 4, . . . .
Next we use x2 yn (x) = xyn+1 (x) + cn xyn (x) + dn xyn−1 (x),
n = 1, 2, 3, . . .
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 3, © Springer-Verlag Berlin Heidelberg 2010
53
54
3 Orthogonality of the Polynomial Solutions
to conclude that Λ [x2 yn (x)] = 0 for n = 3, 4, 5, . . .. Hence, by using induction, we have proved (3.1.1). Now we multiply (3.1.2) by yn−1 (x) to obtain xyn (x)yn−1 (x) = yn+1 (x)yn−1 (x) + cn yn (x)yn−1 (x) + dn y2n−1 (x) for n = 1, 2, 3, . . .. Now we use (2.5.11) to write n−1
xyn−1 (x) = yn (x) − ∑ αk
(n−1)
yk (x),
(n−1)
αk
∈ C,
n = 1, 2, 3, . . . ,
k=0
apply the linear functional Λ and find by using (3.1.1)
Λ [y2n ] = dnΛ [y2n−1 ],
n = 1, 2, 3, . . . .
(3.1.3)
A quasi-definite linear functional Λ for a monic polynomial system {yn }∞ n=0 is characterized by Λ [y2n ] = 0 for all n = 0, 1, 2, . . .. In view of (3.1.3), this is equivalent to dn = 0 for all n = 1, 2, 3, . . .. A positive-definite linear functional Λ for a monic polynomial system {yn }∞ n=0 is characterized by Λ [y2n ] > 0 and each yn has real coefficients for all n = 0, 1, 2, . . .. In view of (2.5.1) and (3.1.3), this is equivalent to cn ∈ R for all n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. Remark. From (3.1.3) together with Λ [y20 ] = 1 it follows that n
Λ [y2n ] = ∏ dk ,
n = 1, 2, 3, . . . .
(3.1.4)
k=1
Finite orthogonal polynomial systems {yn }Nn=0 with N + 1 polynomials occur if Λ [y2n ] = 0 for n = 0, 1, 2, . . . , N. These polynomials satisfy a three-term recurrence relation of the form yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
n = 0, 1, 2, . . . , N − 1
with y−1 (x) := 0. Further we define y∗N+1 by y∗N+1 (x) = (x − c∗N )yN (x) − dN yN−1 (x) with c∗N arbitrary. Then (3.1.3) also holds for n = N. Now we define {y∗n }∞ n=0 such that y∗n (x) = yn (x), n = 0, 1, 2, . . . , N. The polynomials of degree higher than N + 1 can be obtained by
3.2 Orthogonality and the Self-Adjoint Operator Equation
55
∗ y∗N+2 (x) = (x − c∗N+1 )y∗N+1 (x) − dN+1 yN (x), ∗ ∗ ∗ ∗ y∗N+1 (x), yN+3 (x) = (x − cN+2 )yN+2 (x) − dN+2 .. . ∗ ∞ where the coefficients {c∗n }∞ n=N and {dn }n=N+1 can be chosen freely according to the following rules:
• In the quasi-definite case: dn∗ = 0 for n = N + 1, N + 2, N + 3, . . .. • In the positive-definite case: c∗n ∈ R for n = N, N + 1, N + 2, . . . and dn∗ > 0 for n = N + 1, N + 2, N + 3, . . .. Let Λ ∗ be the linear functional given by
Λ ∗ [y∗0 ] = 1,
Λ ∗ [y∗n ] = 0,
n = 1, 2, 3, . . . ,
where only the first N + 1 polynomials {y∗n }Nn=0 satisfy the eigenvalue problem (2.2.1). Then Favard’s theorem implies that {y∗n }∞ n=0 is an orthogonal system. Then we call {yn }Nn=0 a finite orthogonal system with N + 1 polynomials. See also [373].
3.2 Orthogonality and the Self-Adjoint Operator Equation Let q ∈ R \ {−1, 0}, ω ∈ R and (q, ω ) = (1, 0). Then the definition (2.1.1) of Hahn’s q-operator Aq,ω can be extended to f (qx + ω ) − f (x) ω , x ∈ C\ , (3.2.1) (Aq,ω f ) (x) := qx + ω − x 1−q for arbitrary complex-valued functions f whose domain contains the number qx + ω ∈ C as well as x ∈ C. For two such functions, the product rule (2.1.2) is still valid: (Aq,ω ( f1 f2 )) (x) = (Aq,ω f1 ) (x) f2 (x) + f1 (qx + ω ) (Aq,ω f2 ) (x).
(3.2.2)
Likewise, the definition (2.5.2) of the operator Sq,ω can be extended: (Sq,ω f ) (x) := f (qx + ω ), and
x∈C
f(x) := Sq,−1 ω f (x) = f ((x − ω )/q),
x ∈ C.
(3.2.3) (3.2.4)
Assume that w is a complex-valued function for which w(x) and (Sq,ω w) (x) are defined for infinitely many x ∈ C. If we multiply the operator equation (2.2.1) by (Sq,ω w) (x), we find by using (3.2.3)
56
3 Orthogonality of the Polynomial Solutions
(Sq,ω w) (x)ϕ (x) Aq,2ω yn (x) + (Sq,ω w) (x)ψ (x) (Aq,ω yn ) (x) = λn (Sq,ω w) (x) (Sq,ω yn ) (x). By using (3.2.4), we note that this equation coincides with the self-adjoint operator equation (3.2.5) (Aq,ω (wϕAq,ω yn )) (x) = λn (Sq,ω w) (x) (Sq,ω yn ) (x) if the so-called Pearson operator equation1 (Aq,ω (wϕ)) (x) = (Sq,ω w) (x)ψ (x)
(3.2.6)
holds. From this Pearson operator equation the function w can be determined. Let us state the Pearson operator equation in another form. By using the definition (3.2.1), the product rule (3.2.2) and the definition (3.2.4), we find for the left-hand side of (3.2.6) w(qx + ω ) − w(x) ϕ(qx + ω ) − ϕ(x) ϕ(x) + w(qx + ω ) qx + ω − x qx + ω − x w(qx + ω )ϕ (x) − w(x)ϕ(x) . = qx + ω − x
(Aq,ω (wϕ)) (x) =
By using (3.2.4), we can write the right-hand side of (3.2.6) as (Sq,ω w) (x)ψ (x) = w(qx + ω )ψ (x). Hence we have w(x)ϕ(x) = w(qx + ω ) (ϕ (x) − (qx + ω − x)ψ (x)) .
(3.2.7)
If (2.2.2) and (2.2.13) are used, this can be written as w(x)C(x) = qw(qx + ω )D(qx + ω ).
(3.2.8)
Now we examine the relationship between the function w and the orthogonality functionals for the polynomials {yn }∞ n=0 . For this purpose we multiply (3.2.5) by (Sq,ω ym ) (x) and subtract from the resulting equation the same equation with m and n exchanged. Then we apply Sq,−1 ω to the result and use the commutation rules (2.5.3) to find (λn − λm )w(x)ym (x)yn (x) Aq,ω yn ) (x)ym (x) = q Aq,ω Sq,−1 ω (wϕ Aq,ω ym ) (x)yn (x). − q Aq,ω Sq,−1 ω (wϕ
1
(3.2.9)
The Pearson operator equation also plays an important role in stochastics, where it is used to derive stochastic distribution functions. This shows the connection between orthogonal polynomials and stochastic distribution functions. See for instance the book [469] by W. Schoutens.
3.2 Orthogonality and the Self-Adjoint Operator Equation
57
This leads to two kinds of orthogonality. A. For ω = 0, id est Aq,0 = Dq , we have the q-integration by parts formula (1.15.9) b a
Aq,0 f1 (x) f2 (x) dq x
b b Sq,0 f1 (x) Aq,0 f2 (x) dq x = f1 (x) f2 (x) − a
(3.2.10)
a
for arbitrary complex-valued functions f1 and f2 which are q-integrable on the interval (a, b). Then we have for (3.2.9) (λn − λm )
b
b
a
w(x)ym (x)yn (x) dq x
−1 wϕAq,0 yn (x)ym (x) Aq,0 Sq,0 a
−1 wϕAq,0 ym (x)yn (x) dq x − Aq,0 Sq,0 −1 −1 Aq,0 yn (x)ym (x) (wϕ) (x) Sq,0 = q Sq,0
b −1 Aq,0 ym (x)yn (x) . − Sq,0
=
q
a
In view of the regularity condition (2.3.3), we have λm = λn for m = n. So we have Theorem 3.2. Let {yn }∞ n=0 denote the polynomial solutions of the eigenvalue problem (2.2.1), let the regularity condition (2.3.3) hold for n = 0, 1, 2, . . . and let w denote a complex valued function which is q-integrable on the interval (a, b) on the real line and which satisfies the Pearson operator equation (3.2.6). Then we have the orthogonality relation b
m, n ∈ {0, 1, 2, . . .}
(3.2.11)
with weight function w if the boundary conditions −1 −1 Sq,0 (wϕ) (a) = 0 and Sq,0 (wϕ) (b) = 0
(3.2.12)
a
w(x)ym (x)yn (x) dq x = 0,
m = n,
hold. Here a continuous extension of wϕ might be necessary. If the necessary convergence conditions hold, the integral in (3.2.11) can also be taken over (a, ∞), (−∞, b) or (−∞, ∞) with appropriate boundary conditions. Since we used the definition (3.2.1), which is not valid for (q, ω ) = (1, 0), the differentiation operator D has not been covered. This case needs a separate treatment. The differential equation (2.2.9) has the self-adjoint form
58
3 Orthogonality of the Polynomial Solutions
wϕ yn (x) = λn w(x)yn (x),
n = 0, 1, 2, . . .
if w satisfies the Pearson differential equation (wϕ ) (x) = w(x)ψ (x),
(3.2.13)
where ϕ (x) = ex2 + 2 f x + g, ψ (x) = 2ε x + γ and λn = n (e(n − 1) + 2ε ). Partial integration over an interval (a, b) with a, b ∈ R leads to the orthogonality relation b
w(x)ym (x)yn (x) dx (λn − λm ) a b = w(x)ϕ (x) ym (x)yn (x) − yn (x)ym (x) =0 a
(3.2.14)
for m = n and m, n ∈ {0, 1, 2, . . .} if the boundary conditions w(a)ϕ (a) = 0 and w(b)ϕ (b) = 0
(3.2.15)
hold. If the necessary convergence conditions hold, the integral in (3.2.14) can also be taken over (a, ∞), (−∞, b) and (−∞, ∞) with appropriate boundary conditions. B. For N ∈ {1, 2, 3, . . .} we consider the set of points xν := Aqν + [ν ]ω ,
A ∈ C,
ν = 0, 1, 2, . . . , N + 1,
(3.2.16)
which implies that xν +1 = qxν + ω . Then we have the summation by parts formula N
∑ (Aq,ω f1 ) (xν ) f2 (xν ) ((q − 1)xν + ω )
ν =0
N+1 = f1 (xν ) f2 (xν ) ν =0
−
N
∑ (Sq,ω f1 ) (xν ) (Aq,ω f2 ) (xν ) ((q − 1)xν + ω )
(3.2.17)
ν =0
for arbitrary complex-valued functions f1 and f2 whose domain contains the set of points {xν }N+1 ν =0 . Hence we have for (3.2.9)
3.3 The Jackson-Thomae q-Integral
59
N
(λn − λm ) ∑ w(xν )ym (xν )yn (xν ) ((q − 1)xν + ω ) ν =0
Aq,ω yn ) (xν )ym (xν ) = ∑ q Aq,ω Sq,−1 ω (wϕ N
ν =0
Aq,ω ym ) (xν )yn (xν ) ((q − 1)xν + ω ) − Aq,ω Sq,−1 ω (wϕ ) (xν ) Sq,−1 = q Sq,−1 ω (wϕ ω (Aq,ω yn ) (xν )ym (xν ) N+1 − Sq,−1 (A y ) (x )y (x ) . q,ω m ν n ν ω ν =0
In view of the regularity condition (2.3.3), we have λm = λn for m = n. So we have Theorem 3.3. Let {yn }∞ n=0 denote the polynomial solutions of the eigenvalue problem (2.2.1), let the regularity condition (2.3.3) hold for n = 0, 1, 2, . . . and let w denote a complex-valued function whose domain contains the set of points {xν }N+1 ν =0 and which satisfies the Pearson operator equation (3.2.6). Then we have the orthogonality relation N
∑ w(xν )ym (xν )yn (xν ) ((q − 1)xν + ω ) = 0,
ν =0
m = n,
m, n ∈ {0, 1, 2, . . . , N}
with weight function w if the boundary conditions −1 −1 Sq,ω (wϕ) (x0 ) = 0 and Sq,ω (wϕ) (xN+1 ) = 0
(3.2.18)
(3.2.19)
hold. Here a continuous extension of wϕ might be necessary. Note that the boundary conditions (3.2.19) have to be satisfied for two points of ) . the set (3.2.16). This means that two of these points must be zeros of Sq,−1 ω (wϕ If this cannot be achieved, the above sums can be generalized by Jackson-Thomae integrals (see the next section), which also include the case where infinite sums are used instead of the finite sums (3.2.18).
3.3 The Jackson-Thomae q-Integral In the previous section we developed a basis for the representation of orthogonality functionals for the polynomial solutions of the eigenvalue problem (2.2.1). It remains to deal with the difficulty of satisfying the boundary conditions (3.2.19) by elements of the set (3.2.16). This problem can be solved by decomposing the functional into two (infinite) sums, which can be seen as a Stieltjes integral. It will be necessary to distinguish the cases 0 < |q| < 1 and |q| > 1. Therefore we introduce the set of points
60
3 Orthogonality of the Polynomial Solutions
D(x, q, ω ) :=
∞ ⎧ k ⎨ xq + [k]ω k=0 , ⎩
0 < |q| < 1
∞ q−k−1 (x − [k + 1]ω ) k=0 , |q| > 1.
Let A, B ∈ C and consider the two sets of points D(A, q, ω ) and D(B, q, ω ). For 0 < ∗ ∞ |q| < 1 these sets of points can be described by the sequences {xν }∞ ν =0 and {xν }ν =0 , given by xν = Aqν + [ν ]ω
xν∗ = Bqν + [ν ]ω ,
and
ν = 0, 1, 2, . . . ,
respectively. For |q| > 1 we have xν =
1 (A − [ν + 1]ω ) qν +1
and xν∗ =
1 (B − [ν + 1]ω ) , qν +1
ν = 0, 1, 2, . . . ,
respectively. Note that for 0 < |q| < 1 we have ω 1 − qν ω = lim (Aqν + [ν ]ω ) = lim Aqν + ν →∞ ν →∞ 1−q 1−q and for |q| > 1 we obtain A − [ν + 1]ω A(1 − q) − (1 − qν +1 )ω ω . = lim = ν +1 ν →∞ ν →∞ q (1 − q)qν +1 1−q lim
For a complex-valued function f with domain containing D(A, q, ω ) the JacksonThomae integral is defined by A ω 1−q
f (x) dq,ω x :=
⎧ ∞ ⎪ ⎪ f (xν ) ((1 − q)xν − ω ) , 0 < |q| < 1 ⎪ ⎨ ν∑ =0 ⎪ ⎪ ⎪ ⎩
∞
∑
ν =0
f (xν ) ((q − 1)xν + ω ) , |q| > 1,
provided that the sums converge. If the domain of f also contains D(B, q, ω ) and all corresponding sums converge, we define B A
B
f (x) dq,ω x :=
ω 1−q
f (x) dq,ω x −
A ω 1−q
f (x) dq,ω x.
(3.3.1)
Then we have the following analogue of the fundamental theorem of calculus: Theorem 3.4. Let the domain of the complex-valued function f contain D(A, q, ω ) ∪ D(B, q, ω ). If all corresponding sums converge and if there exists a continuous extension of f for the point ω /(1 − q), then we have B A
(Aq,ω f ) (x) dq,ω x = f (B) − f (A).
(3.3.2)
3.3 The Jackson-Thomae q-Integral
61
Proof. Consider the case that 0 < |q| < 1. First we prove the assertion for A = ω /(1− q). With xν∗ = Bqν + [ν ]ω we get B ω 1−q
∞
(Aq,ω f ) (x) dq,ω x = = =
∑ (Aq,ω f ) (xν∗ ) ((1 − q)xν∗ − ω )
ν =0 ∞
∑
f (qxν∗ + ω ) − f (xν∗ ) ((1 − q)xν∗ − ω ) qxν∗ + ω − xν∗
ν =0 ∞
∑
ν =0
f (xν∗ ) − f (xν∗ +1 )
= f (x0∗ ) − lim f (xν∗ +1 ) = f (B) − f (ω /(1 − q)). ν →∞
In fact since 0 < |q| < 1, we have ω . lim Bqν +1 + [ν + 1]ω = ν →∞ 1−q So the continuity of f at the point ω /(1 − q) implies that lim f (xν∗ +1 ) = f (ω /(1 − q)).
ν →∞
Similarly we get A ω 1−q
(Aq,ω f ) (x) dq,ω x = f (A) − f (ω /(1 − q)).
By using (3.3.1), we have proved (3.3.2) in the case that 0 < |q| < 1. In the case that |q| > 1, we first set A = ω /(1 − q) again. With xν∗ = qν1+1 (B − [ν + 1]ω ) we get B ω 1−q
(Aq,ω f ) (x) dq,ω x
∞
=
∑ (Aq,ω f ) (xν∗ ) ((q − 1)xν∗ + ω )
ν =0 ∞
=
∑
ν =0
f (qxν∗ + ω ) − f (xν∗ ) ((q − 1)xν∗ + ω ) qxν∗ + ω − xν∗
= f (qx0∗ + ω ) − f (x0∗ ) +
∞
∑
ν =1
f (xν∗ −1 ) − f (xν∗ )
= f (B) − f (x0∗ ) + f (x0∗ ) − lim f (xν∗ ) = f (B) − f (ω /(1 − q)). ν →∞
In fact since |q| > 1, we have lim
1
ν →∞ qν +1
(B − [ν + 1]ω ) =
ω . 1−q
62
3 Orthogonality of the Polynomial Solutions
So the continuity of f at the point ω /(1 − q) implies that lim f (xν∗ ) = f (ω /(1 − q)).
ν →∞
Similarly we get A ω 1−q
(Aq,ω f ) (x) dq,ω x = f (A) − f (ω /(1 − q)).
By using (3.3.1), we have proved (3.3.2) in the case that |q| > 1.
Now the summation by parts formula (3.2.17) and the orthogonality relation (3.2.18) with the boundary conditions (3.2.19) can easily be extended to Jackson-Thomae integrals: Theorem 3.5. Let the function w : D(A, q, ω ) ∪ D(B, q, ω ) → C satisfy the selfadjoint operator equation (3.2.5) and the boundary conditions −1 −1 Sq,ω (wϕ) (A) = 0 and Sq,ω (wϕ) (B) = 0. If the regularity condition (2.3.3) holds for n = 0, 1, 2, . . ., then the polynomial solutions {yn }∞ n=0 of the eigenvalue problem (2.2.1) satisfy the orthogonality relation B A
w(x)ym (x)yn (x) dq,ω x = 0,
m = n,
m, n ∈ {0, 1, 2, . . .},
provided that convergence holds.
3.4 Rodrigues Formulas In this section we will show that polynomial solutions {yn }∞ n=0 of the eigenvalue problem (2.2.7) satisfy a so-called Rodrigues formula. In order to do this, we start with the differential equation (2.2.9). The procedure in this special case will give a motivation for the procedure starting from the q-operator equation (2.2.7), for which the difference equation (2.2.8) is a special case. The differential equation (2.2.9) can be written as
ϕ (x)yn (x) + ψ (x)yn (x) = λn yn (x),
n = 0, 1, 2, . . .
(3.4.1)
with
ϕ (x) = ex2 + 2 f x + g,
ψ (x) = 2ε x + γ
and λn = n (e(n − 1) + 2ε ) .
In self-adjoint form this reads w(x)ϕ (x)yn (x) = λn w(x)yn (x)
(3.4.2)
3.4 Rodrigues Formulas
63
where w satisfies the Pearson differential equation (w(x)ϕ (x)) = w(x)ψ (x).
(3.4.3)
Since λ0 = 0, (3.4.2) can be written as w(x)ϕ (x)yn (x) = (λn − λ0 )w(x)yn (x).
(3.4.4)
This can be generalized to (k) (k−1) w(x) {ϕ (x)}k yn (x) = (λn − λk−1 )w(x) {ϕ (x)}k−1 yn (x)
(3.4.5)
for k = 1, 2, 3, . . .. To prove this, we use induction on k. For k = 1 (3.4.5) equals (3.4.4). For the left-hand side of (3.4.5) we obtain by using the Pearson differential equation (3.4.3) (k) w(x) {ϕ (x)}k yn (x) (k+1) (k) = w(x) {ϕ (x)}k yn (x) + w(x)ϕ (x) {ϕ (x)}k−1 yn (x) (k+1)
(x) = w(x) {ϕ (x)}k yn (k) + (w(x)ϕ (x)) {ϕ (x)}k−1 + (k − 1)w(x) {ϕ (x)}k−1 ϕ (x) yn (x) (k+1)
(x) + = w(x) {ϕ (x)}k yn (k) ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k−1 yn (x).
(3.4.6)
Hence (3.4.5) is equivalent to (k+1)
w(x) {ϕ (x)}k yn
(k) (x) + ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k−1 yn (x) (k−1)
= (λn − λk−1 )w(x) {ϕ (x)}k−1 yn
(x).
Multiplying by ϕ (x) and differentiating, we obtain (k+1) (k) (x) + ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k yn (x) w(x) {ϕ (x)}k+1 yn (k) + ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k yn (x) (k−1) = (λn − λk−1 ) w(x) {ϕ (x)}k yn (x) (k)
+ (λn − λk−1 )w(x) {ϕ (x)}k yn (x). Now we use the fact that ψ (x) = 2ε and ϕ (x) = 2e, so that
λn − λk−1 = λn − λk + ψ (x) + (k − 1)ϕ (x)
(3.4.7)
64
3 Orthogonality of the Polynomial Solutions
to conclude that (k+1) (k) w(x) {ϕ (x)}k+1 yn (x) + ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k yn (x) (k−1) (k) = (λn − λk−1 ) w(x) {ϕ (x)}k yn (x) + (λn − λk )w(x) {ϕ (x)}k yn (x). Since we have w(x)ϕ (x) {ϕ (x)}k−1 = ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k−1 , we obtain (k+1) (k) w(x) {ϕ (x)}k+1 yn (x) + ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k yn (x) (k−1) = (λn − λk−1 ) ψ (x) + (k − 1)ϕ (x) w(x) {ϕ (x)}k−1 yn (x) (k)
+ (λn − λk )w(x) {ϕ (x)}k yn (x). Hence, if (3.4.5) holds for k, we have (k+1) (k) (x) = (λn − λk )w(x) {ϕ (x)}k yn (x), w(x) {ϕ (x)}k+1 yn which equals (3.4.5) with k replaced by k + 1. This proves that (3.4.5) holds for all k = 1, 2, 3, . . .. (n) If we assume that the polynomials {yn }∞ n=0 are monic, id est yn (x) = n!, and if the regularity condition (2.3.3) holds, then we have by using (3.4.5) (w(x) {ϕ (x)}n )
(n)
(n) 1 (n) w(x) {ϕ (x)}n yn (x) n! (n−1) 1 (n−1) = (λn − λn−1 ) w(x) {ϕ (x)}n−1 yn (x) n! (n−2) 1 (n−2) = (λn − λn−1 )(λn − λn−2 ) w(x) {ϕ (x)}n−2 yn (x) n! .. . n 1 = ∏ (λn − λn−k ) w(x)yn (x), n = 1, 2, 3, . . . . n! k=1 =
Hence, together with y0 (x) = 1, we have the Rodrigues formula yn (x) =
Kn n Kn d n D (w(x) {ϕ (x)}n ) = (w(x) {ϕ (x)}n ) w(x) w(x) dxn
for n = 1, 2, 3, . . . with
(3.4.8)
3.4 Rodrigues Formulas
Kn =
65
n! n
∏ (λn − λn−k )
=
k=1
1 n
∏ (e(2n − k − 1) + 2ε )
,
n = 1, 2, 3, . . . .
(3.4.9)
k=1
Now we use a similar procedure for the q-operator equation (2.2.7), which can be written as ϕ (x) Aq,2ω yn (x) + ψ (x) (Aq,ω yn ) (x) = λn yn (qx + ω ), n = 0, 1, 2, . . . (3.4.10) where
ϕ (x) = ex2 + 2 f x + g,
ψ (x) = 2ε x + γ ,
λn =
[n] (e[n − 1] + 2ε ) . qn
In self-adjoint form this reads Aq,ω w Sq,−1 ω ϕ Aq,ω yn (x) = λn (Sq,ω w) (x) (Sq,ω yn ) (x)
(3.4.11)
where w satisfies the Pearson operator equation Aq,ω w Sq,−1 (x) = (Sq,ω w) (x)ψ (x). ωϕ
(3.4.12)
Since λ0 = 0, (3.4.11) can be written as Aq,ω w Sq,−1 ω ϕ Aq,ω yn (x) = (λn − λ0 ) (Sq,ω w) (x) (Sq,ω yn ) (x).
(3.4.13)
Applying the operator Sq,−1 ω on both sides of (3.4.13), we obtain by using (2.5.3)
Aq,ω
−2 −1 λn − λ0 w(x)yn (x). Sq,ω ϕ Sq,ω (Aq,ω yn ) (x) = q
Sq,−1 ωw
(3.4.14)
This formula can be generalized to k −k −i−1 −k k (x) Sq,ω w Aq,ω ∏ Sq,ω ϕ Sq,ω Aq,ω yn i=1
λn − λk−1 −k+1 k−1 −i−1 = Sq,ω w Aq,k−1 (x) (3.4.15) ∏ Sq,ω ϕ Sq,−k+1 ω ω yn q i=1 for k = 1, 2, 3, . . .. To prove this we use induction on k. For k = 1 (3.4.15) equals 0
(3.4.14) since the empty operator product ∏ at the right-hand side must be interi=1
preted as the identity operator. For simplicity we leave out the argument x in the sequel. Then we obtain, by using the product rule (3.2.2) twice, for the left-hand side of (3.4.15)
66
3 Orthogonality of the Polynomial Solutions
k k Sq,−k Aq,ω Sq,−k ∏ Sq,−i−1 ωw ω ϕ ω Aq,ω yn =
i=1
−i−1 −k k S A S ϕ A y q, ω n ∏ q,ω q,ω q,ω
Sq,−k ωw
k
i=1
+ Sq,−k+1 ω
Aq,kω yn
k −k −i−1 Aq,ω Sq,ω w ∏ Sq,ω ϕ
i=1
−i−1 ∏ Sq,ω ϕ Aq,ω Sq,−kω Aq,kω yn
= Sq,−k ωw
k
i=1
+ Sq,−k+1 ω
Aq,kω yn
k−1
∏
Sq,−i−1 ω ϕ
Aq,ω
Sq,−k ωw
Sq,−k−1 ω ϕ
i=1
Sq,−k+1 Sq,−k ω w ωϕ k−1 −i−1 × Aq,ω ∏ Sq,ω ϕ .
+ Sq,−k+1 ω
Aq,kω yn
(3.4.16)
i=1
Now we use (2.5.3) and the Pearson operator equation (3.4.12) to obtain −1 −k−1 −k Aq,ω Sq,−k = A S w S ϕ q, ω ω q,ω q,ω w Sq,ω ϕ −1 = q−k Sq,−k ω Aq,ω w Sq,ω ϕ = q−k Sq,−k ω ((Sq,ω w) ψ ) −k = q−k Sq,−k+1 . w S ψ q,ω ω
(3.4.17)
Further we use the following extension of the product rule (3.2.2) Aq,ω
N
∏ fi i=1
N−1
=
∏ fi
N−2
(Aq,ω fN ) +
i=1
∏ fi i=1
N
+ . . . + (Aq,ω f1 ) ∏ (Sq,ω fi ) i=2
to find that
(Aq,ω fN−1 ) (Sq,ω fN )
3.4 Rodrigues Formulas
67
Aq,ω
k−1
∏
Sq,−i−1 ω ϕ
k−2
∏
=
i=1
Sq,−i−1 ω ϕ
i=1
Aq,ω Sq,−k ωϕ
k−2
∏
+
i=1
Sq,−i−1 ω ϕ
Aq,ω Sq,−k+1 ϕ ω
k−2
∏
+...+
Sq,−i−1 ω ϕ
Aq,ω Sq,−2 ωϕ
i=1
k−2
∏
=
Sq,−i−1 ω ϕ
k−1
∑ Aq,ω
i=1
j−1 Sq,−k+ ϕ . (3.4.18) ω
j=1
Now we use (3.4.16), (3.4.17) and (3.4.18) to obtain k −k −i−1 −k k Aq,ω Sq,ω w ∏ Sq,ω ϕ Sq,ω Aq,ω yn
= Sq,−k ωw
i=1
−i−1 ∏ Sq,ω ϕ Aq,ω Sq,−kω Aq,kω yn
k
i=1
+
Sq,−k+1 ω w
∏
Sq,−i−1 ω ϕ
k Sq,−k+1 A y n q,ω ω
i=1
× q
k−1
−k
Sq,−k ωψ
k−1
+ ∑ Aq,ω
j−1 Sq,−k+ ϕ ω
.
(3.4.19)
j=1
Hence, combining (3.4.15) and (3.4.19), we have k −k −i−1 Sq,ω w ∏ Sq,ω ϕ Aq,ω Sq,−kω Aq,kω yn i=1
+ Sq,−k+1 ω w
k−1
∏
Sq,−i−1 ω ϕ
Sq,−k+1 Aq,kω yn ω
i=1
−k+ j−1 ϕ × q−k Sq,−k ω ψ + ∑ Aq,ω Sq,ω
λn − λk−1 −k+1 Sq,ω w = q
k−1
j=1
k−1
∏
Sq,−i−1 ω ϕ
Sq,−k+1 Aq,k−1 ω ω yn . (3.4.20)
i=1
−2 Now we apply Sq,−1 ω on both sides of (3.4.20), multiply by Sq,ω ϕ and apply Aq,ω on both sides, then we obtain by using (2.5.3)
68
3 Orthogonality of the Polynomial Solutions
Aq,ω
k+1
∏
Sq,−k−1 ω w
+ Aq,ω
Sq,−i−1 ω ϕ
i=1
q
q−k Sq,−k−1 Aq,k+1 ω ω yn
Sq,−i−1 ω ϕ
−k
Sq,−k−1 ω ψ
k Sq,−k A y n q,ω ω
i=1
×
k
∏
Sq,−k ωw
k−1
+∑q
−k+ j−1
j−2 Sq,−k+ (Aq,ω ϕ ) ω
j=1
k λn − λk−1 −k −i−1 −k k−1 Aq,ω Sq,ω w = ∏ Sq,ω ϕ Sq,ω Aq,ω yn . (3.4.21) q i=1 Hence q−k Aq,ω
k+1
∏
Sq,−k−1 ω w
+ Sq,−k w ω
Sq,−i−1 ω ϕ
i=1
k
∏
Sq,−i−1 ω ϕ
Sq,−k−1 Aq,k+1 ω ω yn
k Sq,−k A y q,ω n ω
i=1
× Aq,ω
k−1
q−k Sq,−k−1 ω ψ+∑
j−2 q−k+ j−1 Sq,−k+ (Aq,ω ϕ ) ω
j=1
+ q
−k
Sq,−k ωψ +
k−1
∑q
−k+ j−1
j−1 Sq,−k+ (Aq,ω ϕ ) ω
j=1
k k Sq,−k Sq,−k × Aq,ω ∏ Sq,−i−1 ωw ω ϕ ω Aq,ω yn
λn − λk−1 −k = q Sq,−k ωw q
i=1
k
∏
Sq,−i−1 ω ϕ
k Sq,−k A y ω q,ω n
i=1
λn − λk−1 −k+1 k−1 Sq,ω Aq,ω yn + q k −k −i−1 × Aq,ω Sq,ω w ∏ Sq,ω ϕ . i=1
Now we use the fact that
λn − λk−1 = λn − λk + q−k e[k − 1](1 + qk−1 ) + 2ε
to conclude that
(3.4.22)
3.4 Rodrigues Formulas
69
λn − λk−1 −k −k q Sq,ω w q =q
−k−1
(λn − λk )
k
∏
−i−1 Sq,ω ϕ
i=1
+ q−2k−1 Sq,−k ωw
k
i=1
k
∏
k Sq,−k ω Aq,ω yn
−i−1 −k k S S A ϕ y n ∏ q,ω q,ω q,ω
Sq,−k ωw
Sq,−i−1 ω ϕ
i=1
× Sq,−k ω
Aq,kω yn e[k − 1](1 + qk−1 ) + 2ε .
(3.4.23)
Now we use (3.4.17) and (3.4.18) to find that k −k −i−1 Aq,ω Sq,ω w ∏ Sq,ω ϕ
= Aq,ω
i=1
Sq,−k ωw
k−1
∏
=
Sq,−i−1 ω ϕ
Sq,−k−1 ω ϕ
k−1
∏ i=1
Sq,−k + Sq,−k+1 ω w ω ϕ Aq,ω =
k−1
∏
Sq,−i−1 ω ϕ
i=1
−k+1 −k −k −i−1 Sq,ω ϕ Sq,ω w q Sq,ω ψ
k−1
∏ i=1
+ Sq,−k+1 ω w
k−1
∏
Sq,−i−1 ω ϕ
k−1
i=1
=
−k−1 Aq,ω Sq,−k w S ϕ q,ω ω
i=1
Sq,−i−1 ω ϕ
∑ Aq,ω
j−1 Sq,−k+ ϕ ω
j=1
−k+1 −i−1 Sq,ω ϕ Sq,ω w
k−1
∏ i=1
× q
−k
Sq,−k ωψ +
k−1
∑q
−k+ j−1
j−1 Sq,−k+ (Aq,ω ϕ ) ω
.
j=1
Since ψ (x) = 2ε x + γ , we have −2k−1 −k−1 Sq,ω (Aq,ω ψ ) = q−2k−1 2ε Aq,ω q−k Sq,−k−1 ω ψ =q and since ϕ (x) = ex2 + 2 f x + g, we have (Aq,ω ϕ ) (x) = e(1 + q)x + eω + 2 f and therefore
and
Aq,2ω ϕ (x) = e(1 + q)
(3.4.24)
70
3 Orthogonality of the Polynomial Solutions
k−1
∑
Aq,ω
j−2 q−k+ j−1 Sq,−k+ (Aq,ω ϕ ) ω
j=1
=q
k−1
∑q
−2k−1
2 j−2
j−2 2 Aq,ω ϕ Sq,−k+ ω
j=1
k−1
= q−2k−1 e(1 + q) ∑ q2 j−2 = q−2k−1 e j=1
=q
−2k−1
1 − q2k−2 1−q
e[k − 1](1 + qk−1 ).
(3.4.25)
If (3.4.15) holds for k, by combining (3.4.22), (3.4.23), (3.4.24) and (3.4.25), we obtain (3.4.15) with k replaced by k + 1. This proves that (3.4.15) holds for all k = 1, 2, 3, . . .. n If we assume that the polynomials {yn }∞ n=0 are monic, id est Aq,ω yn (x) = [n]!, and the regularity condition (2.3.3) holds, then we have by using (3.4.15) −n n −i−1 n Aq,ω Sq,ω w ∏ Sq,ω ϕ 1 = An [n]! q,ω
i=1
Sq,−n ωw
−i−1 −n n ∏ Sq,ω ϕ Sq,ω Aq,ω yn
n
i=1
λn − λn−1 n−1 −n+1 n−1 −i−1 −n+1 n−1 = Aq,ω Sq,ω Aq,ω yn Sq,ω w ∏ Sq,ω ϕ q [n]! i=1
=
(λn − λn−1 )(λn − λn−2 ) q2 [n]! −n+2 n−2 −i−1 −n+2 n−2 n−2 × Aq,ω Sq,ω Aq,ω yn Sq,ω w ∏ Sq,ω ϕ i=1
.. . n
∏ (λn − λn−k )
=
k=1
qn [n]!
w yn = q
−n
λn − λn−k ∏ [k] k=1 n
w yn ,
n = 1, 2, 3, . . . .
Hence, together with y0 (x) = 1, we have the Rodrigues formula −n n −k−1 Kn n Aq,ω Sq,ω w ∏ Sq,ω ϕ (x) yn (x) = w(x) k=1 for (q, ω ) = (1, 0) and n = 1, 2, 3, . . ., where by using (2.3.2)
(3.4.26)
3.5 Duality
71 n
[k] = k=1 λn − λn−k
Kn = qn ∏
qn(n+1) n
∏ (e[2n − k − 1] + 2ε )
,
n = 1, 2, 3, . . . .
(3.4.27)
k=1
In the case of the difference operator Δ (i.e. q = 1 and ω = 1), this reads n Kn n yn (x) = Δ w(x − n) ∏ ϕ (x − k − 1) , n = 1, 2, 3, . . . , (3.4.28) w(x) k=1 where Kn is given by (3.4.9).
3.5 Duality In this section we consider the concept of dual polynomial systems. The following definition of dual polynomial systems is due to D.A. Leonard (see [371]): Definition 3.1. Let N ∈ {1, 2, 3, . . .} or N → ∞ and let {κn }Nn=0 and {λn }Nn=0 denote two (finite or infinite) sequences of complex numbers with
κm = κn
and λm = λn ,
m = n.
Then two (finite or infinite) polynomial systems {yn }Nn=0 and {zn }Nn=0 in P with degree[yn ] = n and degree[zn ] = n for n = 0, 1, 2, . . . , N are called dual polynomial systems with respect to the sequences {κn }Nn=0 and {λn }Nn=0 if yn (κm ) = zm (λn ),
m, n = 0, 1, 2, . . . , N.
(3.5.1)
The numbers κn ∈ C and λn ∈ C are called eigenvalues. Remark. If yn = zn for n = 0, 1, 2, . . . , N in the above definition, then the (finite or infinite) polynomial system {yn }Nn=0 is called self-dual. Now we want to construct a dual polynomial system {zn }Nn=0 to a system of polynomial solutions {yn }Nn=0 of the eigenvalue problem (2.2.1). We will prove: Theorem 3.6. Let N ∈ {1, 2, 3, . . .} or N → ∞, q ∈ R \ {−1, 0} and ω ∈ R with (q, ω ) = (1, 0). Define the (finite or infinite) sequence {κn }Nn=0 by ω n , n = 0, 1, 2, . . . , N. κn := q x0 + [n]ω , x0 ∈ C \ 1−q Let {yn }Nn=0 denote a (finite or infinite) system of polynomial solutions of the eigenvalue problem (2.2.12) and let C and D be functions defined by (2.2.13), (D(x0 ) = ) qC(x0 ) −
2ε (x0 − ω ) + γ q =0 qx0 + ω − x0
72
and
3 Orthogonality of the Polynomial Solutions
C(κn ) = 0,
n = 0, 1, 2, . . . , N − 1.
If the regularity condition (2.3.3) holds, then there exists a (finite or infinite) polynomial system {zn }Nn=0 in P such that {yn }Nn=0 and {zn }Nn=0 are dual polynomial systems with respect to the sequence {λn }Nn=0 of eigenvalues of (2.2.12) and the sequence {κn }Nn=0 , id est we have (3.5.1). The polynomials {zn }Nn=0 satisfy the threeterm recurrence relation C(κn )zn+1 (x) − {C(κn ) + D(κn )} zn (x) + D(κn )zn−1 (x) = xzn (x)
(3.5.2)
for n = 0, 1, 2, . . . , N − 1 with the convention that z−1 (x) := 0. Proof. Since degree[y0 ] = 0, we have y0 (x) = c ∈ C \ {0}. Hence from (3.5.1) we have z0 (λ0 ) = y0 (κ0 ) = c. Since degree[z0 ] = 0, we conclude that z0 (x) = c = 0. Since C(κn ) = 0 for n = 0, 1, 2, . . . , N − 1, the polynomial system {zn }Nn=0 is uniquely determined by the three-term recurrence relation (3.5.2). If we substitute x = κm in the symmetric form (2.2.12) of the operator equation, we find by using (3.5.1) C(κm )yn (κm+1 ) − {C(κm ) + D(κm )} yn (κm ) + D(κm )yn (κm−1 ) = λn yn (κm ). This equation can be seen as a three-term recurrence relation for {yn (κm )}Nm=0 . The coefficients of this recurrence relation coincide with those of (3.5.2). So we may conclude that zm (λn ) = yn (κm ) if the initial conditions z0 (λn ) = yn (κ0 ) coincide for n = 0, 1, 2, . . . , N. This can be achieved if and only if yn (κ0 ) = 0 for n = 0, 1, 2, . . . , N. To prove the last assumption, we first remark that y0 (κ0 ) = c = 0. Next we assume that y1 (κ0 ) = 0. Then we have C(κ0 )y1 (κ1 ) −C(κ0 )y1 (κ0 ) = λ1 y1 (κ0 ) = 0. Hence y1 (κ0 ) = y1 (κ1 ) = 0, which implies that y1 (x) = 0 since C(κ0 ) = 0, κ0 = κ1 and degree[y1 ] = 1. Hence y1 (κ0 ) = 0. Similarly for each n ∈ {2, 3, 4, . . .} the assumption that yn (κ0 ) = 0 would imply that yn (κ1 ) = yn (κ2 ) = yn (κ3 ) = · · · = yn (κn ) = 0 and therefore yn (x) = 0 since κm = κn if m = n and degree[yn ] = n. Hence yn (κ0 ) = 0 for n = 0, 1, 2, . . . , N. Sometimes it is useful to have a bounded sequence {κn }Nn=0 as we will see later on. In the preceding theorem this is not the case for |q| > 1. Therefore we will restate this theorem for a different sequence {κn }Nn=0 : Theorem 3.7. Let N ∈ {1, 2, 3, . . .} or N → ∞ and let {κn }Nn=0 be a finite or infinite sequence defined by 1 ω , n = 0, 1, 2, . . . , N. κn = n (x0 − [n]ω ), x0 ∈ C \ q 1−q
3.5 Duality
73
Assume that the hypotheses of the preceding theorem hold with (C(x0 ) = ) and
e(x0 − ω )2 + 2 f q(x0 − ω ) + gq2 =0 q(qx0 + ω − x0 )2
D(κn ) = 0,
n = 0, 1, 2, . . . , N − 1.
Then there exists a (finite or infinite) sequence of dual polynomials {zn }Nn=0 satisfying the three-term recurrence relation D(κn )zn+1 (x) − {D(κn ) +C(κn )} zn (x) +C(κn )zn−1 (x) = xzn (x) for n = 0, 1, 2, . . . , N − 1 with the convention that z−1 (x) = 0. In some cases a theorem by G.K. Eagleson (see [189] in a generalized form) establishes a correlation between the orthogonality of two dual polynomial systems. ∞ For a given sequence {xn }∞ n=0 of real numbers and a sequence {wn }n=0 of weights with wn > 0 for n = 0, 1, 2, . . . we define the Hilbert space ∞ 2 2 L ({xn }, {wn }) = f : R → C ∑ wn | f (xn )| < ∞ n=0 with the scalar product
∞
( f , g) =
∑ wn f (xn )g(xn ),
n=0
where g denotes the complex conjugate of g. Now we will prove Theorem 3.8. Let N ∈ {1, 2, 3, . . .} or N → ∞ and let {xν }Nν =0 be a finite or infinite sequence of real numbers. Let {yn }Nn=0 be a finite or infinite sequence of polynomials with degree[yn ] = n satisfying the orthogonality relation N
∑ wν ym (xν )yn (xν ) = σn δmn ,
ν =0
wν > 0,
σn > 0,
m, n = 0, 1, 2, . . . , N.
∞
be a complete orthonormal set in L2 ({xn }, {wn }). Then For N → ∞ let √yσn n n=0 the dual orthogonality relation N
1
∑ σn yn (xμ )yn (xν ) =
n=0
δμν , wν
μ , ν = 0, 1, 2, . . . , N
holds. Proof. We prove the theorem for N → ∞. Consider the function
(3.5.3)
74
3 Orthogonality of the Polynomial Solutions
⎧ ⎨ 1 for x = xν , uν (x) =
⎩
x ∈ R.
(3.5.4)
0 for x = xν ,
∞ Because of the completeness of the orthonormal set √yσn n we have the repren=0 sentation ∞ yn (x) yn uν (x) = ∑ uν , √ √ , σn σn n=0 where the Fourier coefficients can be computed as ∞ yn yn (xi ) wν yn (xν ) uν , √ = ∑ wi uν (xi ) √ = √ . σn σn σn i=0 By using Parseval’s identity, we obtain ∞ ∞ w w y (x )y (x ) yn yn ν μ n ν n μ (uν , uμ ) = ∑ uν , √ uμ , √ =∑ . σ σ σn n n n=0 n=0 On the other hand we have ∞
(uν , uμ ) = ∑ wi uν (xi )uμ (xi ) = wμ δμν . i=0
Comparing both results for (uν , uμ ) leads to (3.5.3).
In terms of duality, the last result can be restated as follows. Corollary 3.9. Let N ∈ {1, 2, 3, . . .} or N → ∞ and let the finite or infinite polynomial systems {yn }Nn=0 and {zn }Nn=0 be dual with respect to the finite or infinite sequences {κn }Nn=0 and {λn }Nn=0 and let the orthogonality relation (3.1.1) hold for xν = κν . Then we also have the dual orthogonality relation zμ (λn )zν (λn ) δμν = , σn wν n=0 N
∑
μ , ν = 0, 1, 2, . . . , N.
∞
in L2 ({xn }, {wn }) is necesIn the case that N → ∞, the completeness of √yσn n n=0 sary.
∞
in the infinite case (i.e. N → ∞) The completeness of the orthonormal set √yσn n n=0
can easily be established for a bounded sequence {xn }∞ n=0 : Theorem 3.10. Let {xn }∞ n=0
be a bounded sequence. Then the orthonormal set
√yn σn
∞
n=0
is complete in L2 ({xn }, {wn }).
Proof. It suffices to show that the functions uν defined by (3.5.4) can be approximated in the sense of the norm ||·|| of L2 ({xn }, {wn }) by polynomials. Since {xn }∞ n=0
3.5 Duality
75
is a bounded sequence, it is contained in some compact interval [a, b] on the real line. First we show that the function uν can be approximated by continuous functions on [a, b]. For a given number ε > 0 we choose n0 ∈ {1, 2, 3, . . .} such that ∞
∑
μ =n0
wμ < ε
and we choose a function f ∈ C[a, b] with | f (xμ )| ≤ 1 for all μ ∈ {0, 1, 2, . . .} and ⎧ ⎨ 1 for μ = ν , f (xμ ) =
⎩
0 for μ = ν , μ < n0 .
Then we have || f − uν || =
∞
∞
μ =0
μ =n0
∑ wμ | f (xμ ) − uν (xμ )|2 = ∑
wμ | f (xμ ) − uν (xμ )|2 ≤
∞
∑
μ =n0
wμ < ε .
Now it remains to prove that functions in C[a, b] can be approximated by polynomials. According to the Weierstrass approximation theorem, for a given number ε > 0 and a function f ∈ C[a, b], there exists a polynomial p ∈ P such that sup | f (x) − p(x)| < ε . x∈[a,b]
Hence || f − p|| =
∞
∞
μ =0
μ =0
∑ wμ | f (xμ ) − p(xμ )|2 < ε ∑ wμ .
Part I
Classical Orthogonal Polynomials
Chapter 4
Orthogonal Polynomial Solutions of Differential Equations Continuous Classical Orthogonal Polynomials
4.1 Polynomial Solutions of Differential Equations In the case of the ordinary differentiation operator D = A1,0 , we have to deal with (cf. (2.2.9)): (ex2 + 2 f x + g)yn (x) + (2ε x + γ )yn (x) = n(e(n − 1) + 2ε )yn (x)
(4.1.1)
with e, f , g, ε , γ ∈ C and n = 0, 1, 2, . . .. We look for monic polynomial solutions of the form n (x + c)k , an,n = n!, n = 0, 1, 2, . . . yn (x) = ∑ an,k (4.1.2) k! k=0 where the coefficients satisfy the three-term recurrence relation (n − k) (e(n + k − 1) + 2ε ) an,k + (2(ec − f )k + 2ε c − γ ) an,k+1 − (ec2 − 2 f c + g)an,k+2 = 0,
an,n+1 = 0,
k = n − 1, n − 2, n − 3, . . . , 0.
If c can be chosen such that ec2 − 2 f c + g = 0, this leads to the two-term recurrence relation (cf. (2.4.9)) (n − k) (e(n + k − 1) + 2ε ) an,k + (2(ec − f )k + 2ε c − γ ) an,k+1 = 0
(4.1.3)
for k = n − 1, n − 2, n − 3, . . . , 0. If ec2 − 2 f c + g = 0 has no solution for c, id est e = f = 0 and g = 0, then we find from (2.4.4) with c = γ /2ε the two-term recurrence relation (cf. (2.4.10)) 2ε (n − k)an,k − gan,k+2 = 0,
an,n+1 = 0,
k = n − 1, n − 2, n − 3, . . . , 0. (4.1.4)
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 4, © Springer-Verlag Berlin Heidelberg 2010
79
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4 Orthogonal Polynomial Solutions of Differential Equations
In section 2.6 we found that the monic polynomial solutions {yn }∞ n=0 satisfy the three-term recurrence relation yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
n = 1, 2, 3, . . . ,
(4.1.5)
n = 0, 1, 2, . . .
(4.1.6)
with initial values y0 (x) = 1 and y1 (x) = x − c0 , where cn = −
2 f n (e(n − 1) + 2ε ) − γ (e − ε ) , 2 (e(n − 1) + ε ) (en + ε )
and dn =
n (e(n − 2) + 2ε ) 4 (e(2n − 3) + 2ε ) (e(n − 1) + ε )2 (e(2n − 1) + 2ε ) × {2 f (n − 1) + γ } {2 f (e(n − 1) + 2ε ) − eγ } − 4g (e(n − 1) + ε )2 , n = 1, 2, 3, . . . .
(4.1.7)
4.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions In this chapter we classify all positive-definite orthogonal polynomial solutions of the second-order differential equation (4.1.1). These solutions are called continuous classical orthogonal polynomials. Such a classification was first given in 1929 by S. Bochner in [106]. By Favard’s theorem (theorem 3.1), a polynomial yn (x) of degree n satisfying both the differential equation (4.1.1) and the three-term recurrence relation (4.1.5) is orthogonal with respect to a positive-definite linear functional only if cn ∈ R for all n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. We will consider three different cases depending on the form of ϕ (x) = ex2 + 2 f x + g: Case I. Degree[ϕ ] = 0: e = f = 0 and we may choose g = 1. Then we have cn = −
γ , 2ε
n = 0, 1, 2, . . .
and dn = −
n , 2ε
n = 1, 2, 3, . . . .
Hence positive-definite orthogonality occurs for ε < 0 and γ ∈ R. Case II. Degree[ϕ ] = 1: e = 0 and we may choose 2 f = 1. Then we have cn = −
2n + γ , 2ε
n = 0, 1, 2, . . .
and dn =
n(n − 1 + γ − 2gε ) , 4ε 2
n = 1, 2, 3, . . . .
4.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions
81
Hence we have positive-definite orthogonality for 2gε < γ and g, ε , γ ∈ R. Case III. Degree[ϕ ] = 2: we may choose e = 1. Then we have cn = −
2 f n(n − 1 + 2ε ) − γ (1 − ε ) , 2(n − 1 + ε )(n + ε )
n = 0, 1, 2, . . .
and dn =
n(n − 2 + 2ε ) Dn , 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . ,
where Dn = {2 f (n − 1) + γ } {2 f (n − 1 + 2ε ) − γ } − 4g(n − 1 + ε )2 ,
n = 1, 2, 3, . . . .
Positive-definite orthogonality requires that f , g, ε , γ ∈ R. For ε > 0 we have n(n − 2 + 2ε ) > 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . ,
where the case 2ε = 1 must be understood by continuity. Now we have Dn = {2 f (n − 1) + γ } {2 f (n − 1 + 2ε ) − γ } − 4g(n − 1 + ε )2 = {2 f (n − 1) + γ } {2 f (n − 1) − γ } + 4 f ε {2 f (n − 1) + γ } − 4g(n − 1 + ε )2 = 4( f 2 − g)(n − 1 + ε )2 − (γ − 2 f ε )2 , n = 1, 2, 3, . . . . Hence for ε > 0 the positivity of dn > 0 for all n = 1, 2, 3, . . . requires that f 2 − g > 0, which implies that the polynomial ϕ (x) = x2 + 2 f x + g has two different real zeros. For ε < 0 we write 2ε = −2N − t
with N ∈ {1, 2, 3, . . .}
− 1 < t ≤ 1.
and
Note that this notation does not cover the (trivial) range −1 ≤ 2ε < 0. However, we will be able to find finite orthogonal polynomial systems consisting of N + 1 polynomials. Now we have n(n − 2 + 2ε ) < 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . , N
which does not longer hold for n = N + 1. So in this case we have dn > 0 for
n = 1, 2, 3, . . . , N
⇐⇒
Dn < 0 for
n = 1, 2, 3, . . . , N.
In this case the polynomial ϕ (x) = x2 + 2 f x + g might have two different real zeros if f 2 − g > 0, two equal real zeros if f 2 − g = 0 or two nonreal (complex conjugate) zeros if f 2 − g < 0.
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4 Orthogonal Polynomial Solutions of Differential Equations
Concluding, positive-definite orthogonality occurs for Case IIIa. ε > 0 and 4( f 2 − g)ε 2 > (γ − 2 f ε )2 . This implies that f 2 > g, which means that the polynomial ϕ has two different real zeros. Moreover, we also find three finite orthogonal polynomial systems in the following three different cases: Case IIIb. The polynomial ϕ has two different real zeros: f 2 > g and 4( f 2 − g)ε 2 < (γ − 2 f ε )2 . Case IIIc. The polynomial ϕ has two equal real zeros: f 2 = g and γ = 2 f ε . Case IIId. The polynomial ϕ has two non-real (complex conjugate) zeros: f 2 < g. As indicated in section 3.2, the orthogonality relations can be obtained in each case as follows. The differential equation (4.1.1) can be written in the self-adjoint form wϕ yn (x) = λn w(x)yn (x), n = 0, 1, 2, . . . (4.2.1) if w satisfies the Pearson differential equation (cf. (3.2.13)) (wϕ ) (x) = w(x)ψ (x),
(4.2.2)
where ϕ (x) = ex2 + 2 f x + g, ψ (x) = 2ε x + γ and λn = n (e(n − 1) + 2ε ). If we multiply (4.2.1) by ym (x) and subtract from the resulting equation the same equation with m and n exchanged, then integration by parts over an interval (a, b) with a, b ∈ R and a < b leads to (cf. (3.2.14)) b
w(x)ym (x)yn (x) dx (λn − λm ) a b = w(x)ϕ (x) ym (x)yn (x) − yn (x)ym (x) a
(4.2.3)
for m, n ∈ {0, 1, 2, . . .}. If the regularity condition (2.3.3) holds, we have λm = λn for m = n. This leads to an orthogonality relation on the interval (a, b) with respect to the weight function w(x) if the boundary conditions w(a)ϕ (a) = 0 and w(b)ϕ (b) = 0
(4.2.4)
hold. Here we have a < b with a, b ∈ R or possibly a → −∞ and/or b → ∞. In the latter cases we have orthogonality on (a, ∞), (−∞, b) or (−∞, ∞) and the corresponding improper integrals of the first kind exist if the moments b
Mn :=
a
w(x)xn dx,
n = 0, 1, 2, . . . ,
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
83
where a = −∞ and/or b = ∞, are all finite. In the case of finite polynomial systems {yn (x)}Nn=0 consisting of N + 1 polynomials with N ∈ {1, 2, 3, . . .} we must have that b
Mn :=
w(x)xn dx,
n = 0, 1, 2, . . . , 2N
a
are finite. For the computation of the squared norm (cf. (3.1.4)) we must have Λ [y20 ] = Λ [1] = 1. Therefore we define b
d0 :=
w(x) dx
(4.2.5)
a
which leads to the squared norm
σn :=
b a
n
w(x){yn (x)}2 dx = ∏ dk ,
n = 0, 1, 2, . . . , N
(4.2.6)
k=0
with N ∈ {1, 2, 3, . . .} or N → ∞. Finally, we remark that the possible positive-definite orthogonal polynomials satisfying both the differential equation (4.1.1) and the three-term recurrence relation (4.1.5) are uniquely determined by Favard’s theorem. However, the orthogonality relation can be obtained in several different ways. In the next section we will derive a weight function, a second-order differential equation, a three-term recurrence relation, a hypergeometric representation, a Rodrigues formula and an orthogonality relation for the positive-definite orthogonal polynomials in all six cases. More details and properties will be given in chapter 9.
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions Case I. We have e = f = 0, g = 1 and ε < 0. In this case we have ϕ (x) = 1 and ψ (x) = 2ε x + γ . Hence by using the Pearson differential equation (4.2.2), we obtain w (x) = 2ε x + γ w(x) which leads to a positive-definite weight function of the form w(x) = eε x
2 +γ x
for the Hermite polynomials. The boundary conditions (4.2.4) lead to the interval of orthogonality (−∞, ∞). The weight function for the Hermite polynomials is connected with the normal distribution in stochastics.
84
4 Orthogonal Polynomial Solutions of Differential Equations
The differential equation (4.1.1) reads yn (x) + (2ε x + γ )yn (x) = 2ε nyn (x),
n = 0, 1, 2, . . .
and the three-term recurrence relation (4.1.5) can be written as n γ yn (x) + yn−1 (x), n = 1, 2, 3, . . . yn+1 (x) = x + 2ε 2ε with y0 (x) = 1 and y1 (x) = x + γ /2ε . For the coefficients of the representation (4.1.2) we use (4.1.4) with c = γ /2ε to obtain 2ε (n − k)an,k = an,k+2 ,
k = n − 1, n − 2, n − 3, . . . , 0,
an,n+1 = 0,
an,n = n!.
Hence we have an,n−2k+1 = 0
and an,n−2k =
n! , (4ε )k k!
k = 0, 1, 2, . . . , n/2
which implies by using (4.1.2) n/2
an,n−2k γ n−2k 1 γ n−2k x + x+ = n! ∑ k 2ε 2ε k=0 (n − 2k)! k=0 (4ε ) k!(n − 2k)! n n (−n)2k γ n−2k (−n/2)k (−(n − 1)/2)k γ n−2k x+ x+ = ∑ =∑ k k 2ε 2ε ε k! k=0 (4ε ) k! k=0
1 −n/2, −(n − 1)/2 γ n ; = x+ 2 , n = 0, 1, 2, . . . , 2 F0 − 2ε ε x+ γ n/2
yn (x) =
∑
2ε
where n/2 denotes the largest integer smaller than or equal to n/2. The Rodrigues formula (3.4.8) equals e− ε x − γ x n ε x 2 + γ x , D e (2ε )n 2
yn (x) =
n = 0, 1, 2, . . . .
By using (1.2.5) we find that ∞
d0 :=
−∞
∞
w(x) dx =
−∞
e
ε x 2 +γ x
dx =
π −γ 2 /4ε e > 0. −ε
Together with (3.1.4), this leads to the orthogonality relation ∞ 2 π n! ε x 2 +γ x e ym (x)yn (x) dx = e−γ /4ε δmn , m, n = 0, 1, 2, . . . . n − ε (−2 ε ) −∞
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
85
Case II. We have e = 0, 2 f = 1 and 2gε < γ . By writing g = −a, we have ϕ (x) = x − a and ψ (x) = 2ε x + γ . Hence by using the Pearson differential equation (4.2.2), we obtain α w (x) 2ε x + γ − 1 = = 2ε + , x = a w(x) x−a x−a with γ = α + 1 − 2aε , which leads to a positive-definite weight function of the form w(x) = (x − a)α e2ε x ,
a<x
for the Laguerre polynomials. The boundary conditions (4.2.4) lead to the interval of orthogonality (a, ∞) and the conditions α + 1 > 0 and ε < 0. Note that we have
α +1 > 0
⇐⇒
γ + 2aε > 0
⇐⇒
γ − 2gε > 0,
which is equivalent to 2gε < γ . The weight function for the Laguerre polynomials is connected to the gamma distribution in stochastics. The differential equation (4.1.1) reads (x − a)yn (x) + {2ε (x − a) + α + 1} yn (x) = 2ε nyn (x),
n = 0, 1, 2, . . .
and the three-term recurrence relation (4.1.5) can be written as
2n − 2aε + α + 1 n(n + α ) yn (x) − yn+1 (x) = x + yn−1 (x), 2ε 4ε 2
n = 1, 2, 3, . . .
with y0 (x) = 1 and y1 (x) = x − a + (α + 1)/2ε . For the coefficients of the representation (4.1.2) we use (4.1.3) with c = −a to obtain 2ε (n − k)an,k = (k + α + 1)an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0,
an,n = n!.
Hence we have an,k =
(k + α + 1)n−k (α + 1)n (−n)k (−2ε )k , n! = (2ε )n (α + 1)k (2ε )n−k (n − k)!
k = 0, 1, 2, . . . , n
which leads to n
yn (x) =
∑ an,k
k=0
(α + 1)n n (−n)k (2ε )k (a − x)k (x − a)k = ∑ (α + 1)k k! (2ε )n k=0 k!
−n (α + 1)n ; 2 F ε (a − x) , n = 0, 1, 2, . . . . = 1 1 (2ε )n α +1
The Rodrigues formula (3.4.8) equals yn (x) =
(x − a)−α e−2ε x n D (x − a)n+α e2ε x , n (2ε )
n = 0, 1, 2, . . . .
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4 Orthogonal Polynomial Solutions of Differential Equations
By using (1.2.1) we find that ∞
d0 :=
∞
w(x) dx = a
a
(x − a)α e2ε x dx =
Γ(α + 1)e2aε > 0. (−2ε )α +1
Together with (3.1.4), this leads to the orthogonality relation ∞ a
(x − a)α e2ε x ym (x)yn (x) dx =
Γ(n + α + 1)e2aε n! δmn , (−2ε )2n+α +1
m, n = 0, 1, 2, . . . .
Case IIIa. We have e = 1, f 2 > g, ε > 0 and 4( f 2 − g)ε 2 > (γ − 2 f ε )2 . In this case we may write ϕ (x) = x2 + 2 f x + g = (x − a)(x − b) with 2 f = −a − b and g = ab. Then we obtain by using the Pearson differential equation (4.2.2)
α β w (x) 2(ε − 1)x + γ + a + b = = − , w(x) (x − a)(x − b) x−a b−x
x = a,
x = b,
a
with 2ε = α + β +2 and γ = −a(β +1)−b(α +1), which leads to a positive-definite weight function of the form w(x) = (x − a)α (b − x)β ,
a<x
for the Jacobi polynomials. Note that we have 4( f 2 − g)ε 2 − (γ − 2 f ε )2 >0 (b − a)2
⇐⇒ ⇐⇒
−4gε 2 − γ 2 + 4 f εγ >0 (b − a)2 (α + 1)(β + 1) > 0.
The boundary conditions (4.2.4) lead to the interval of orthogonality (a, b) and the conditions α + 1 > 0 and β + 1 > 0. The weight function for the Jacobi polynomials is connected to the beta distribution in stochastics. The differential equation (4.1.1) reads (x − a)(x − b)yn (x) + {(α + 1)(x − b) + (β + 1)(x − a)} yn (x) = n(n + α + β + 1)yn (x) for n = 0, 1, 2, . . . and the three-term recurrence relation (4.1.5) can be written as yn+1 (x)
2n(n + α + β + 1)(a + b) + {a(β + 1) + b(α + 1)}(α + β ) = x− yn (x) (2n + α + β )(2n + α + β + 2) −
n(n + α )(n + β )(n + α + β )(b − a)2 yn−1 (x) (2n + α + β − 1)(2n + α + β )2 (2n + α + β + 1)
for n = 1, 2, 3, . . . with y0 (x) = 1 and y1 (x) = x − (a(β + 1) + b(α + 1))/(α + β + 2).
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
87
For the coefficients of the representation (4.1.2) we use (4.1.3) to find (n − k)(n + k + α + β + 1)an,k = − {(2k + α + β + 2)c + (a + b)(k + 1) + aβ + bα } an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0, where (c + a)(c + b) = 0. So if we choose c = −a, we have (n − k)(n + k + α + β + 1)an,k = (a − b)(k + α + 1)an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0 with an,n = n!, and if we choose c = −b, we obtain (n − k)(n + k + α + β + 1)an,k = (b − a)(k + β + 1)an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0 with an,n = n!. Hence for c = −a we obtain an,k = =
(a − b)n−k (k + α + 1)n−k n! (n − k)! (n + k + α + β + 1)n−k (a − b)n (α + 1)n (−n)k (n + α + β + 1)k (n + α + β + 1)n (α + 1)k
1 b−a
k ,
which leads to (a − b)n (α + 1)n (x − a)k ∑ an,k k! = (n + α + β + 1)n 2 F1 k=0 n
−n, n + α + β + 1 x − a ; α +1 b−a
for n = 0, 1, 2, . . ., and for c = −b we obtain an,k = =
(b − a)n−k (k + β + 1)n−k n! (n − k)! (n + k + α + β + 1)n−k (b − a)n (β + 1)n (−n)k (n + α + β + 1)k (n + α + β + 1)n (β + 1)k
1 a−b
k ,
which leads to (b − a)n (β + 1)n (x − b)k ∑ an,k k! = (n + α + β + 1)n 2 F1 k=0 n
−n, n + α + β + 1 x − b ; β +1 a−b
for n = 0, 1, 2, . . .. The Rodrigues formula (3.4.8) equals yn (x) = (−1)n
(x − a)−α (b − x)−β n D (x − a)n+α (b − x)n+β , (n + α + β + 1)n
By using (1.2.10) we find that
n = 0, 1, 2, . . . .
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4 Orthogonal Polynomial Solutions of Differential Equations
b
d0 :=
b
w(x) dx = a
a
(x − a)α (b − x)β dx =
Γ(α + 1)Γ(β + 1) (b − a)α +β +1 > 0. Γ(α + β + 2)
Together with (3.1.4), this leads to the orthogonality relation b a
=
(x − a)α (b − x)β ym (x)yn (x) dx Γ(n + α + β + 1)Γ(n + α + 1)Γ(n + β + 1)n! (b − a)2n+α +β +1 δmn Γ(2n + α + β + 1)Γ(2n + α + β + 2)
for m, n = 0, 1, 2, . . .. Case IIIb. We have e = 1, f 2 > g, 4( f 2 − g)ε 2 < (γ − 2 f ε )2 and 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1. Here we also have ϕ (x) = x2 + 2 f x + g = (x − a)(x − b) with 2 f = −a − b and g = ab. Then we find by using the Pearson differential equation (4.2.2)
α β w (x) 2(ε − 1)x + γ + a + b = = + , w(x) (x − a)(x − b) x−a x−b
x = a,
x = b,
a
with 2ε = α + β +2 and γ = −a(β +1)−b(α +1), which leads to a positive-definite weight function of the form w(x) = (x − a)α (x − b)β ,
a
for the Jacobi polynomials. Now we have 4( f 2 − g)ε 2 − (γ − 2 f ε )2 <0 (b − a)2
⇐⇒ ⇐⇒
−4gε 2 − γ 2 + 4 f εγ <0 (b − a)2 (α + 1)(β + 1) < 0.
The boundary conditions (4.2.4) lead to the interval of orthogonality (b, ∞) and the conditions β + 1 > 0 and α + β + 1 < −2N. The differential equation, the three-term recurrence relation and the hypergeometric representation are the same as in the previous case. The Rodrigues formula (3.4.8) can now be written as yn (x) =
(x − a)−α (x − b)−β n D (x − a)n+α (x − b)n+β , (n + α + β + 1)n
n = 0, 1, 2, . . . , N.
By using (1.2.10) we find that ∞
d0 :=
∞
w(x) dx = b
b
=
(x − a)α (x − b)β dx
Γ(β + 1)Γ(−α − β − 1) (b − a)α +β +1 > 0. Γ(−α )
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
89
Together with (3.1.4), this leads to the orthogonality relation ∞ b
=
(x − a)α (x − b)β ym (x)yn (x) dx
Γ(n + β + 1)Γ(−2n − α − β − 1)Γ(−2n − α − β )n! (b − a)2n+α +β +1 δmn Γ(−n − α − β )Γ(−n − α )
for m, n = 0, 1, 2, . . . , N. Case IIIc. We have e = 1, f 2 = g, γ = 2 f ε and 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1. In this case we have ϕ (x) = x2 + 2 f x + g = (x − a)2 with f = −a. Then we obtain by using the Pearson differential equation (4.2.2)
α β w (x) 2(ε − 1)x + γ + 2a = + = , 2 w(x) (x − a) x − a (x − a)2
x = a
with 2ε = α + 2 and γ = β − a(α + 2), which leads to a positive-definite weight function of the form β w(x) = (x − a)α e− x−a , x > a for the Bessel polynomials. The boundary conditions (4.2.4) lead to the interval of orthogonality (a, ∞) and the conditions α + 1 < −2N and β > 0. The differential equation (4.1.1) reads (x − a)2 yn (x) + ((α + 2)(x − a) + β ) yn (x) = n(n + α + 1)yn (x),
n = 0, 1, 2, . . . , N
and the three-term recurrence relation (4.1.5) can be written as
4n(n + α + 1)a + {a(α + 2) − β }α yn (x) yn+1 (x) = x − (2n + α )(2n + α + 2) +
n(n + α )β 2 yn−1 (x) (2n + α − 1)(2n + α )2 (2n + α + 1)
for n = 1, 2, 3, . . . , N − 1 with y0 (x) = 1 and y1 (x) = x − (a(α + 2) − β )/(α + 2). For the coefficients of the representation (4.1.2) we use (4.1.3) to find (n − k)(n + k + α + 1)an,k = β an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0,
an,n = n!
where c = −a. Hence we obtain an,k =
(−1)k (−n)k (n + α + 1)k β n−k β n−k n! = , (n − k)! (n + k + α + 1)n−k (n + α + 1)n
which leads to (x − a)k βn ∑ an,k k! = (n + α + 1)n 2 F0 k=0 n
x−a −n, n + α + 1 ;− − β
90
4 Orthogonal Polynomial Solutions of Differential Equations
for n = 0, 1, 2, . . . , N. The Rodrigues formula (3.4.8) equals β
β (x − a)−α e x−a n D (x − a)2n+α e− x−a , yn (x) = (n + α + 1)n
n = 0, 1, 2, . . . , N.
By using (1.2.1) we find that ∞
d0 :=
∞
w(x) dx = a
a
β
(x − a)α e− x−a dx = β α +1 Γ(−α − 1) > 0.
Together with (3.1.4), this leads to the orthogonality relation ∞ a
β
(x − a)α e− x−a ym (x)yn (x) dx =
β 2n+α +1 Γ(−2n − α − 1)Γ(−2n − α )n! δmn Γ(−n − α )
for m, n = 0, 1, 2, . . . , N. Case IIId. We have e = 1, f 2 < g and 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and 2 2 2 −1 < t ≤ 1. In this case we have ϕ (x) = x + 2 f x + g = (x + f ) + g − f = (x + f + 2 iζ )(x + f − iζ ) with ζ = g − f . Then we obtain by using the Pearson differential equation (4.2.2) for x + f = ±iζ 2(ε − 1)(x + f ) + γ − 2 f ε w (x) = w(x) (x + f )2 + ζ 2 2(ε − 1)(x + f ) γ −2fε γ −2fε + , = + (x + f )2 + ζ 2 2ζ {ζ + i(x + f )} 2ζ {ζ − i(x + f )} which leads to a positive-definite weight function of the form ε −1 {(γ −2 f ε )/ζ } arctan x+ f ζ w(x) = ζ 2 + (x + f )2 e = {ζ + i(x + f )}ε −1−iν {ζ − i(x + f )}ε −1+iν with ν = (γ −2 f ε )/2ζ for the pseudo Jacobi polynomials. The boundary conditions (4.2.4) lead to the interval of orthogonality (−∞, ∞). The weight function for the pseudo Jacobi polynomials is connected to the student’s t-distribution in stochastics. The differential equation (4.1.1) reads (x + f )2 + ζ 2 yn (x) + (2ε x + γ )yn (x) = n(n − 1 + 2ε )yn (x), n = 0, 1, 2, . . . , N and the three-term recurrence relation (4.1.5) can be written as
2 f n(n − 1 + 2ε ) − γ (1 − ε ) yn (x) yn+1 (x) = x + 2(n − 1 + ε )(n + ε ) n(n − 2 + 2ε ) 4(n − 1 + ε )2 ζ 2 + (γ − 2 f ε )2 yn−1 (x) + 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
91
for n = 1, 2, 3, . . . , N − 1 with y0 (x) = 1 and y1 (x) = x + γ /2ε . For the coefficients of the representation (4.1.2) we use (4.1.3) to find (n − k)(n + k − 1 + 2ε )an,k = − (2(c − f )k + 2ε c − γ ) an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0, where c2 − 2 f c + g = 0 If we choose c = f + iζ , we have
⇐⇒
(c − f )2 + ζ 2 = 0.
(n − k)(n + k − 1 + 2ε )an,k = (γ − 2 f ε − 2iζ (k + ε )) an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0 and if we choose c = f − iζ , we have (n − k)(n + k − 1 + 2ε )an,k = (γ − 2 f ε + 2iζ (k + ε )) an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. Hence for c = f + iζ we obtain (−2iζ )n−k (k + ε + iν )n−k n! (n − k)!(n + k − 1 + 2ε )n−k (−2iζ )n (ε + iν )n (−n)k (n − 1 + 2ε )k 1 = , (n − 1 + 2ε )n (ε + iν )k (2iζ )k
an,k =
which leads to n
∑ an,k
k=0
(−2iζ )n (ε + iν )n (x + f + iζ )k = 2 F1 k! (n − 1 + 2ε )n
−n, n − 1 + 2ε x + f + iζ ; ε + iν 2iζ
for n = 0, 1, 2, . . . , N and for c = f − iζ we obtain (2iζ )n−k (k + ε − iν )n−k n! (n − k)!(n + k − 1 + 2ε )n−k (2iζ )n (ε − iν )n (−n)k (n − 1 + 2ε )k 1 = , (n − 1 + 2ε )n (ε − iν )k (−2iζ )k
an,k =
which leads to n
∑ an,k
k=0
(2iζ )n (ε − iν )n (x + f − iζ )k = 2 F1 k! (n − 1 + 2ε )n
−n, n − 1 + 2ε x + f − iζ ; ε − iν −2iζ
for n = 0, 1, 2, . . . , N. The Rodrigues formula (3.4.8) equals 1−ε −2ν arctan x+ f ζ x+ f (x + f )2 + ζ 2 e n 2 2 n+ε −1 2ν arctan ζ (x + f ) + ζ D e yn (x) = (n − 1 + 2ε )n
for n = 0, 1, 2, . . . , N. By using the Cauchy integral (1.2.12), we obtain
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4 Orthogonal Polynomial Solutions of Differential Equations
∞
d0 :=
−∞
∞
w(x) dx = =
−∞
{ζ + i(x + f )}ε −1−iν {ζ − i(x + f )}ε −1+iν dx
2π Γ(1 − 2ε )(2ζ )2ε −1 |Γ(1 − ε + iν )|2
> 0.
Together with (3.1.4), this leads to the orthogonality relation ∞ −∞
=
(x + f )2 + ζ 2
ε −1
f 2ν arctan x+ ζ
e
ym (x)yn (x) dx
2π Γ(1 − 2n − 2ε )Γ(2 − 2n − 2ε )(2ζ )2n+2ε −1 n! Γ(2 − n − 2ε ) |Γ(1 − n − ε + iν )|2
δmn ,
m, n = 0, 1, 2, . . . , N.
4.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
93
In this chapter we have proved:
Theorem 4.1. The positive-definite orthogonal polynomial solutions yn (x) of the differential equation (4.1.1) (ex2 + 2 f x + g)yn (x) + (2ε x + γ )yn (x) = n(e(n − 1) + 2ε )yn (x) for n = 0, 1, 2, . . . consist of three infinite systems Case I. Hermite polynomials with orthogonality on (−∞, ∞) with respect to 2 w(x) = eε x +γ x ; e = f = 0, g = 1 and ε < 0 Case II. Laguerre polynomials with orthogonality on (a, ∞) with respect to w(x) = (x − a)α e2ε x ; e = 0, 2 f = 1, ε < 0 and α > −1 Case IIIa. Jacobi polynomials with orthogonality on (a, b) with a < b with respect to w(x) = (x − a)α (b − x)β ; e = 1, f 2 > g, ε > 0, α > −1 and β > −1 and three finite systems of N + 1 polynomials Case IIIb. Jacobi polynomials with orthogonality on (b, ∞) with respect to w(x) = (x − a)α (x − b)β ; a < b, e = 1, f 2 > g, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, α < −2N and β > −1 Case IIIc. Bessel polynomials with orthogonality on (a, ∞) with respect to w(x) = (x − a)α e−β /(x−a) ; e = 1, f 2 = g, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, α < −2N − 1 and β > 0 Case IIId. Pseudo Jacobi polynomials with orthogonality on (−∞, ∞) with ε −1 {(γ −2 f ε )/ζ } arctan x+ f ζ respect to w(x) = ζ 2 + (x + f )2 e ; e = 1, f 2 < g, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, ζ = g − f 2 > 0.
The finite cases of the Jacobi, Bessel and pseudo Jacobi polynomials were discovered in 1929 by V. Romanovski in [463]. These six systems constitute the class of continuous classical orthogonal polynomials.
Chapter 5
Orthogonal Polynomial Solutions of Real Difference Equations Discrete Classical Orthogonal Polynomials I
5.1 Polynomial Solutions of Real Difference Equations In the case of the difference operator Δ = A1,1 , we have to deal with (cf. (2.2.8)): (ex2 +2 f x+g) Δ 2 yn (x)+(2ε x+ γ ) (Δ yn ) (x) = n(e(n−1)+2ε )yn (x+1) (5.1.1) for n = 0, 1, 2, . . . with e, f , g, ε , γ ∈ R. This difference equation can also be written in the form (cf. (2.2.16)) e(x − 1)2 + 2 f (x − 1) + g (Δ (∇yn )) (x) + (2ε (x − 1) + γ ) (∇yn ) (x) = n(e(n − 1) + 2ε )yn (x) or in the form (cf. (2.2.12)) C(x)yn (x + 1) − {C(x) + D(x)} yn (x) + D(x)yn (x − 1) = n(e(n − 1) + 2ε )yn (x) for n = 0, 1, 2, . . ., where (cf. (2.2.13)) C(x) = e(x − 1)2 + 2 f (x − 1) + g and D(x) = C(x) − 2ε (x − 1) − γ . We look for monic polynomial solutions of the form (cf. (2.4.14)) n x+c , an,n = n!, n = 0, 1, 2, . . . , yn (x) = ∑ an,k k k=0
(5.1.2)
(5.1.3)
where the coefficients satisfy the two-term recurrence relation (cf. (2.4.15))
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 5, © Springer-Verlag Berlin Heidelberg 2010
95
96
5 Orthogonal Polynomial Solutions of Real Difference Equations
(n − k) (e(n + k − 1) + 2ε ) an,k − e(k − 1 − c)2 + 2 f (k − 1 − c) + g an,k+1 = 0
(5.1.4)
for k = n − 1, n − 2, n − 3, . . . , 0 provided that c satisfies e(c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ .
(5.1.5)
In cases where c cannot be obtained from this equation, we use the representation (cf. (2.4.16)) n x+c+k−2 , bn,n = n!, n = 0, 1, 2, . . . , (5.1.6) yn (x) = ∑ bn,k k k=0 where the coefficients satisfy the two-term recurrence relation (cf. (2.4.17)) (n − k) (e(n + k − 1) + 2ε ) bn,k + ek2 + 2(ec − f + ε )k + 2ε c − γ bn,k+1 = 0
(5.1.7)
for k = n − 1, n − 2, n − 3, . . . , 0, provided that c satisfies ec2 − 2 f c + g = 0. In section 2.6 we found that the monic polynomial solutions {yn }∞ n=0 satisfy the three-term recurrence relation yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
n = 1, 2, 3, . . . ,
(5.1.8)
with initial values y0 (x) = 1 and y1 (x) = x − c0 , where (cf. (2.6.15)) cn =
n (e(n − 1) + 2ε ) (2(e − f ) + ε ) + (e − ε )(γ − 2ε ) , 2 (e(n − 1) + ε ) (en + ε )
n = 0, 1, 2, . . .
and (cf. (2.6.16)) dn = −
n (e(n − 2) + 2ε ) 4 (e(2n − 3) + 2ε ) (e(n − 1) + ε )2 (e(2n − 1) + 2ε ) × e(n − 1)2 (e(n − 1) + 2ε )2 + 2(n − 1) (e(n − 1) + 2ε ) (2eg + 2 f (ε − f ) − eγ ) + 4ε (gε − f γ ) + eγ 2 , n = 1, 2, 3, . . . .
In order to rewrite the difference equation (5.1.1) in self-adjoint form, we need the product rule (3.2.2) with q = 1 and ω = 1
Δ ( f1 (x) f2 (x)) = f1 (x + 1)Δ f2 (x) + f2 (x)Δ f1 (x).
(5.1.9)
By using (5.1.9) the difference equation (5.1.1) multiplied by w(x + 1) can be written in the self-adjoint form (cf. (3.2.5))
5.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions
97
Δ (w(x)ϕ (x − 1)Δ yn (x)) = λn w(x + 1)yn (x + 1), provided that w(x) satisfies the Pearson operator equation (cf. (3.2.6))
Δ (w(x)ϕ (x − 1)) = w(x + 1)ψ (x), where (cf. (2.2.2)) ϕ (x) = ex2 + 2 f x + g and ψ (x) = 2ε x + γ . This equation can be rewritten as w(x + 1) (ϕ (x) − ψ (x)) = w(x)ϕ (x − 1), which leads, by using (5.1.2), to the Pearson difference equation w(x) ϕ (x) − ψ (x) D(x + 1) = = , w(x + 1) ϕ (x − 1) C(x)
(5.1.10)
where C(x) and D(x) are given by (5.1.2). The orthogonality relation (3.2.18) reads N
∑ w(A + ν )ym (A + ν )yn (A + ν ) = 0,
ν =0
m = n,
m, n ∈ {0, 1, 2, . . . , N} (5.1.11)
with boundary conditions w(A − 1)ϕ (A − 2) = 0 and w(A + N)ϕ (A + N − 1) = 0
(5.1.12)
for N ∈ {1, 2, 3, . . .} or N → ∞. For N → ∞ we must have that the moments ∞
∑ w(x)xn ,
n = 0, 1, 2, . . .
x=0
are all finite since we must be able to compute the norms of all polynomials.
5.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions Again we use Favard’s theorem (theorem 3.1) to conclude from (5.1.8) that we have positive-definite orthogonality if cn ∈ R for all n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. Again we consider three different cases depending on the form of ϕ (x) = ex2 + 2 f x + g: Case I. Degree[ϕ ] = 0: e = f = 0 and we may choose g = 1. Then we have cn =
2(n + 1)ε − γ , 2ε
n = 0, 1, 2, . . .
and dn = −
n , 2ε
n = 1, 2, 3, . . . .
98
5 Orthogonal Polynomial Solutions of Real Difference Equations
Hence positive-definite orthogonality occurs for 2ε < 0. Case II. Degree[ϕ ] = 1: e = 0 and we may choose 2 f = 1. Then we have cn =
2n(ε − 1) + 2ε − γ , 2ε
n = 0, 1, 2, . . .
and
n ((n − 1)(2ε − 1) + 2gε − γ ) , 4ε 2 Hence we have positive-definite orthogonality for dn = −
n = 1, 2, 3, . . . .
Case IIa. 2ε ≤ 1 and 2gε − γ < 0. Note that the regularity condition (2.3.3) requires that 2ε = 0. For 2ε > 1 we only have a finite orthogonal polynomial system with N + 1 polynomials. Note that 2gε − γ 4ε 2 2gε − γ 2(2ε − 1) 1+ d2 = − 4ε 2 2ε − 1 .. . 2gε − γ N(2ε − 1) dN = − N −1+ 4ε 2 2ε − 1 2gε − γ (N + 1)(2ε − 1) . N+ dN+1 = − 4ε 2 2ε − 1 d1 = −
Hence for 2ε > 1 and −N ≤ and dN+1 ≤ 0. Case IIb. 2ε > 1 and −N ≤
2gε − γ < −N + 1 we have d1 , d2 , d3 , . . . , dN positive 2ε − 1
2gε − γ < −N + 1. 2ε − 1
Case III. Degree[ϕ ] = 2: we may choose e = 1. Then we have cn = and
n(n − 1 + 2ε ) (2(1 − f ) + ε ) + (1 − ε )(γ − 2ε ) , 2(n − 1 + ε )(n + ε )
n = 0, 1, 2, . . .
5.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions
dn = −
99
n(n − 2 + 2ε ) 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε ) × (n − 1)2 (n − 1 + 2ε )2 + 2(n − 1)(n − 1 + 2ε )(2g + 2 f (ε − f ) − γ ) + 4ε (gε − f γ ) + γ 2 , n = 1, 2, 3, . . . .
The latter formula can also be rewritten in the form dn = −
n(n − 2 + 2ε ) Dn , 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . ,
where 2 Dn = (n − 1 + ε )2 − δ 2 − η 2 − 4δ 2 η 2 , with
δ 2 = ( f − ε )2 − g + γ
n = 1, 2, 3, . . .
(5.2.1)
and η 2 = f 2 − g.
Hence δ and η are either real or pure imaginary. The role of δ and η can be explained as follows: (c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ and
(c + 1 − f + ε )2 = δ 2
⇐⇒
ϕ (x) = x2 + 2 f x + g = (x + f )2 − η 2 .
For ε > 0 we have −
n(n − 2 + 2ε ) < 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . .
From (5.2.1) it follows that Dn = (n − 1 + ε )4 − 2(δ 2 + η 2 )(n − 1 + ε )2 + (δ 2 − η 2 )2 ,
n = 1, 2, 3, . . . . (5.2.2)
Hence for dn > 0 for n = 1, 2, 3, . . . we must have Dn < 0 for n = 1, 2, 3, . . ., which implies, by using (5.2.1) and (5.2.2), that both δ and η must be real. In that case we have for n = 1, 2, 3, . . . (5.2.3) Dn = (n − 1 + ε )2 − (δ + η )2 (n − 1 + ε )2 − (δ − η )2 , with
δ=
( f − ε )2 − g + γ
and η =
Finally, we remark that Dn can also be written in the form
f 2 − g.
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5 Orthogonal Polynomial Solutions of Real Difference Equations
Dn = (n − 1 + ε + δ + η )(n − 1 + ε + δ − η ) × (n − 1 + ε − δ + η )(n − 1 + ε − δ − η )
(5.2.4)
for n = 1, 2, 3, . . .. Now we conclude that we have positive-definite orthogonality for Case IIIa. ε > 0 and (δ − η )2 < (n − 1 + ε )2 < (δ + η )2 . Note that this cannot be true for all n = 1, 2, 3, . . .. However, for |δ − η | < ε and N − 1 < δ + η − ε (≤ N) we have dn > 0 for n = 1, 2, 3, . . . , N. Further we also find finite orthogonal polynomial systems in the following cases. First we define 2ε = −2N − t
with N ∈ {1, 2, 3, . . .}
− 1 < t ≤ 1.
and
Then we have −
n(n − 2 + 2ε ) > 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . , N
and for n = N + 1 this is not longer true. So in that case, for dn > 0 for n = 1, 2, 3, . . . , N we must have Dn > 0 for n = 1, 2, 3, . . . , N. This implies that we have positive-definite orthogonality in the following two finite cases: Case IIIb. δ 2 > 0 and η 2 > 0 (the polynomial ϕ has two different real zeros). In that case we must have (δ + η )2 < (n − 1 + ε )2 or (n − 1 + ε )2 < (δ − η )2 . Hence
δ + η < |n − 1 + ε | or |n − 1 + ε | < |δ − η |. t Since n − 1 + ε = n − 1 − N − , this leads to 1 2 |t| t t ≤ |δ − η | ≤ δ + η < 1 + or N + < |δ − η |. 2 2 2 Case IIIc. δ 2 ≤ 0 and/or η 2 ≤ 0 (the polynomial ϕ has two equal real zeros or two nonreal, id est complex conjugate, zeros). In that case we must have δ pure imaginary (including zero) and/or η pure imaginary (including zero) and2
ε ± δ ± η = 0, −1, −2, . . . , −N + 1.
Now we have proved: 1 2
The extra condition |δ − η | ≥ |t|/2 implies that dn > 0 does no longer hold for n = N + 1. This condition prevents that dn = 0 for n ∈ {1, 2, 3, . . . , N} in view of (5.2.4).
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
101
Theorem 5.1. For the polynomial solutions yn of the difference equation (5.1.1) (ex2 + 2 f x + g) Δ 2 yn (x) + (2ε x + γ ) (Δ yn ) (x) = n(e(n − 1) + 2ε )yn (x + 1) with e, f , g, ε , γ ∈ R and n = 0, 1, 2, . . . we only have positive-definite orthogonality in the following cases: Case I. e = f = 0, g = 1 and 2ε < 0. An infinite system of orthogonal polynomials. Case IIa. e = 0, 2 f = 1, 2ε ≤ 1 and 2gε − γ < 0. An infinite system of orthogonal polynomials. Case IIb. e = 0, 2 f = 1, 2ε > 1 and −N ≤ N + 1 orthogonal polynomials.
2gε − γ < −N + 1. A finite system of 2ε − 1
Case IIIa. e = 1, ε > 0, |δ − η | < ε and N − 1 < δ + η − ε (≤ N). A finite system of N + 1 orthogonal polynomials. Case IIIb. e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1. A finite system of N + 1 orthogonal polynomials if t t |t| ≤ |δ − η | ≤ δ + η < 1 + or N + < |δ − η |. 2 2 2 Case IIIc. e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, δ 2 ≤ 0 and/or η 2 ≤ 0 and ε ± δ ± η = 0, −1, −2, . . . , −N + 1. A finite system of N + 1 orthogonal polynomials.
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions Case I. We have e = f = 0, g = 1 and 2ε < 0. Then the Pearson difference equation (5.1.10) reads w(x) = −2ε x − γ + 1. w(x + 1) In order to get A = 0 (and N → ∞) in the boundary conditions (5.1.12), we set γ = 2ε + 1, which leads to a positive-definite weight function of the form w(x) =
1 , (−2ε )x Γ(x + 1)
x = 0, 1, 2, . . . ,
2ε < 0
for the Charlier polynomials. This weight function for the Charlier polynomials is connected with the Poisson distribution in stochastics. The difference equation (5.1.1) reads
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5 Orthogonal Polynomial Solutions of Real Difference Equations
Δ 2 yn (x) + (2ε (x + 1) + 1) (Δ yn ) (x) = 2nε yn (x + 1),
n = 0, 1, 2, . . .
yn (x + 1) + (2ε x − 1)yn (x) − 2ε xyn (x − 1) = 2nε yn (x),
n = 0, 1, 2, . . .
or
and the three-term recurrence relation (5.1.8) can be written as 1 n yn (x) + yn−1 (x), n = 1, 2, 3, . . . yn+1 (x) = x − n + 2ε 2ε with y0 (x) = 1 and y1 (x) = x + 1/2ε . For the coefficients of the representation (5.1.3) we obtain from (5.1.4) 2ε (n − k)an,k = an,k+1 ,
k = n − 1, n − 2, n − 3, . . . 0
if c satisfies (5.1.5), which means that 1 = −2ε (c + 1) + γ . Since an,n = n!, this leads to n! 1 (−1)k (−n)k (2ε )k , k = 0, 1, 2, . . . , n. an,k = = (2ε )n−k (n − k)! (2ε )n Since γ = 2ε + 1, we have c = 0. Hence by using (5.1.3), we obtain for 2ε < 0 n 1 −n, −x 1 k k x = ; 2 (−1) (−n) (2 ε ) F ε yn (x) = 2 0 k ∑ k − (2ε )n k=0 (2ε )n for n = 0, 1, 2, . . .. The Rodrigues formula (3.4.28) reads (−1)n Γ(x + 1) n 1 , Δ yn (x) = (−2ε )n−x (−2ε )x−n Γ(x − n + 1)
n = 0, 1, 2, . . . .
Now we have ∞ 1 (−2ε )−x (−2ε )−x =∑ = exp − > 0. d0 := ∑ w(x) = ∑ x! 2ε x=0 x=0 Γ(x + 1) x=0 ∞
∞
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as ∞ ym (x)yn (x) exp − 21ε n! = δmn , m, n = 0, 1, 2, . . . . ∑ x (−2ε )n x=0 (−2ε ) x! Case IIa. We have e = 0, 2 f = 1, 2ε ≤ 1, 2ε = 0 and 2gε < γ . Now we have to distinguish between the cases that 2ε < 0, 0 < 2ε < 1 and 2ε = 1. Case IIa1. For 2ε < 0 the two-term recurrence relation (5.1.4) can be used. In that case we have for (5.1.5)
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
−(c + 1) + g = −2ε (c + 1) + γ
⇐⇒
c+1 =
103
g−γ . 1 − 2ε
The Pearson difference equation (5.1.10) reads (1 − 2ε )x + g − γ w(x) = . w(x + 1) x+g−1 In order to get A = 0 (and N → ∞) in the boundary conditions (5.1.12), we set g = γ − 2ε + 1. This implies that c = 0. With γ > 2ε this leads to a positive-definite weight function of the form w(x) =
Γ(x + γ − 2ε ) , (1 − 2ε )x Γ(x + 1)
x = 0, 1, 2, . . .
for the Meixner polynomials. This weight function for the Meixner polynomials is connected with the negative binomial distribution in stochastics. The difference equation (5.1.1) reads (x + γ − 2ε + 1) Δ 2 yn (x) + (2ε x + γ ) (Δ yn ) (x) = 2nε yn (x + 1), n = 0, 1, 2, . . . or (x + γ − 2ε )yn (x + 1) − (2(1 − ε )x + γ − 2ε ) yn (x) + (1 − 2ε )xyn (x − 1) = 2nε yn (x), n = 0, 1, 2, . . . and the three-term recurrence relation (5.1.8) can be written as 2n(ε − 1) + 2ε − γ n(1 − 2ε )(n − 1 + γ − 2ε ) yn (x) − yn+1 (x) = x − yn−1 (x) 2ε 4ε 2 for n = 1, 2, 3, . . . with y0 (x) = 1 and y1 (x) = x − 1 + γ /2ε . For the coefficients of the representation (5.1.3) we obtain from (5.1.4) 2ε (n − k)an,k = (k + γ − 2ε )an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
Since an,n = n!, this leads to an,k =
(k + γ − 2ε )n−k n! (γ − 2ε )n (−n)k (−2ε )k , = (2ε )n (γ − 2ε )k (2ε )n−k (n − k)!
k = 0, 1, 2, . . . , n.
Hence n
yn (x) =
∑ an,k
k=0
(γ − 2ε )n x −n, −x = , F ; 2 ε 2 1 k (2ε )n γ − 2ε
The Rodrigues formula (3.4.28) reads
n = 0, 1, 2, . . . ,
γ > 2ε .
104
5 Orthogonal Polynomial Solutions of Real Difference Equations
(1 − 2ε )x Γ(x + 1) n yn (x) = Δ (2ε )n Γ(x + γ − 2ε )
Γ(x + γ − 2ε ) , (1 − 2ε )x−n Γ(x − n + 1)
n = 0, 1, 2, . . . .
Now we have 1 Γ(x + γ − 2ε ) γ − 2ε = Γ( ; γ − 2 ε ) · F 1 0 ∑ x − 1 − 2ε x=0 (1 − 2ε ) x! γ −2ε 1 − 2ε > 0. = Γ(γ − 2ε ) −2ε ∞
d0 :=
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as ∞
(1 − 2ε )n+γ −2ε Γ(n + γ − 2ε )n! Γ(x + γ − 2ε ) y (x)y (x) = δmn m n x (−2ε )2n+γ −2ε x=0 (1 − 2ε ) x!
∑
for 2ε < 0, γ > 2ε and m, n = 0, 1, 2, . . .. Case IIa2. For 0 < 2ε < 1 the two-term recurrence relation (5.1.4) is still valid with c+1 =
g−γ . 1 − 2ε
The Pearson difference equation (5.1.10) reads (1 − 2ε )x + g − γ w(x) = . w(x + 1) x+g−1 In order to satisfy the boundary conditions (5.1.12), we have to take A → −∞. Then we may choose A + N = 0, which implies that g = 1. With γ > 2ε this leads to a positive-definite weight function of the form w(x) =
Γ(r − x) , (1 − 2ε )x Γ(1 − x)
r=
γ − 2ε , 1 − 2ε
x = 0, −1, −2, . . .
for the Meixner polynomials. Note that this implies that c = −r. The difference equation (5.1.1) reads (x + 1) Δ 2 yn (x) + (2ε x + γ ) (Δ yn ) (x) = 2nε yn (x + 1), n = 0, 1, 2, . . . or xyn (x + 1) − (2(1 − ε )x + 2ε − γ ) yn (x) + ((1 − 2ε )x + 2ε − γ ) yn (x − 1) = 2nε yn (x),
n = 0, 1, 2, . . .
and the three-term recurrence relation (5.1.8) can be written as 2n(ε − 1) + 2ε − γ n ((n − 1)(1 − 2ε ) − 2ε + γ ) yn (x) − yn+1 (x) = x − yn−1 (x) 2ε 4ε 2
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
105
for n = 1, 2, 3, . . . with y0 (x) = 1 and y1 (x) = x − 1 + γ /2ε . For the coefficients of the representation (5.1.3) we obtain from (5.1.4) 2ε (n − k)an,k = (k + r)an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
Since an,n = n!, this leads to an,k =
(k + r)n−k n! (r)n (−n)k (−2ε )k , = (2ε )n−k (n − k)! (2ε )n (r)k
k = 0, 1, 2, . . . , n.
Hence by using c = −r, we obtain for r > 0 n (r)n x−r −n, r − x = ; 2 , F ε yn (x) = ∑ an,k 2 1 k r (2ε )n k=0 The Rodrigues formula (3.4.28) reads Γ(r − x + n) (1 − 2ε )x Γ(1 − x) n , yn (x) = Δ (−2ε )n Γ(r − x) (1 − 2ε )x−n Γ(1 − x)
n = 0, 1, 2, . . . .
n = 0, 1, 2, . . . .
Now we have 0
d0 :=
∑
x=−∞
∞
(1 − 2ε )x Γ(x + r) Γ(x + 1) x=0 r Γ(r) ; 1 − 2ε = = Γ(r) · 1 F0 > 0. − (2ε )r
w(x) =
∑
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as ∞
(1 − 2ε )n Γ(n + r)n! (1 − 2ε )x Γ(x + r) ym (−x)yn (−x) = δmn x! (2ε )2n+r x=0
∑
for 0 < 2ε < 1, γ > 2ε and m, n = 0, 1, 2, . . .. The connection between these Meixner polynomials and those found in the previous case can be described as follows. If we define −n, −x (r)n (1) ; 2 , n = 0, 1, 2, . . . F ε Mn (x) = 2 1 r (2ε )n and (2) Mn (x) =
(r)n 2 F1 (2ε )n
−n, r − x ; 2ε , r (1)
n = 0, 1, 2, . . . , (2)
then the orthogonality relations for Mn (x) and for Mn (−x) are equal. Moreover we have by using (1.7.2)
106
5 Orthogonal Polynomial Solutions of Real Difference Equations n
(−1)
(2) Mn (−x)
−n, r + x (−1)n (r)n ; 2ε = 2 F1 r (2ε )n −n, −x (r)n (1) ; 2ε ∗ = Mn (x), = 2 F1 r (2ε ∗ )n
where 2ε ∗ =
2ε 2ε − 1
(note that this implies 0 < 2ε ∗ < 1
⇐⇒
2ε < 0).
Case IIa3. For 2ε = 1 the two-term recurrence relation (5.1.4) does not hold. In that case we use the representation (5.1.6) with c = g. The Pearson difference equation (5.1.10) reads g−γ w(x) = . w(x + 1) x + g − 1 In order to satisfy the boundary conditions (5.1.12), we have to take A → −∞. Then we may choose A + N = 0, which implies that g = 1. With γ > 1 this leads to a positive-definite weight function of the form w(x) =
1 , (γ − 1)x Γ(1 − x)
x = 0, −1, −2, . . .
for the Charlier polynomials. The difference equation (5.1.1) reads (x + 1) Δ 2 yn (x) + (x + γ ) (Δ yn ) (x) = nyn (x + 1),
n = 0, 1, 2, . . .
or xyn (x + 1) − (x + 1 − γ )yn (x) + (1 − γ )yn (x − 1) = nyn (x),
n = 0, 1, 2, . . .
and the three-term recurrence relation (5.1.8) can be written as yn+1 (x) = (x + n − 1 + γ )yn (x) + n(1 − γ )yn−1 (x),
n = 1, 2, 3, . . .
with y0 (x) = 1 and y1 (x) = x − 1 + γ . For the coefficients of the representation (5.1.6) we obtain from (5.1.7) (n − k)bn,k = (γ − 1)bn,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
Since bn,n = n!, this leads to bn,k =
(γ − 1)n−k n! = (γ − 1)n (−n)k (n − k)!
1 1−γ
k ,
k = 0, 1, 2, . . . , n.
Hence by using (5.1.6) with c = 1, we obtain for n = 0, 1, 2, . . .
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
107
k x+k−1 k k=0 1 −n, x ; = (γ − 1)n 2 F0 . − 1−γ n
yn (x) = (γ − 1)n ∑ (−n)k
1 1−γ
The Rodrigues formula (3.4.28) reads yn (x) = (−1) (γ − 1) Γ(1 − x) Δ n
x
n
(γ − 1)n−x Γ(1 − x)
n = 0, 1, 2, . . . .
,
Now we have ∞
0
d0 :=
∑
w(x) =
x=−∞
(γ − 1)x = eγ −1 > 0. x! x=0
∑
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as ∞
(γ − 1)x ym (−x)yn (−x) = eγ −1 (γ − 1)n n! δmn , x! x=0
∑
m, n = 0, 1, 2, . . . .
The connection between these Charlier polynomials and those obtained in Case I can be described as follows. If we define −n, −x 1 (1) ; 2 , n = 0, 1, 2, . . . F ε Cn (x) = 2 0 − (2ε )n
and (2) Cn (x) = (γ − 1)n 2 F0 (1)
1 −n, x ; − 1−γ
,
n = 0, 1, 2, . . . ,
(2)
then we simply have (−1)nCn (−x) = Cn (x), where 1−γ =
1 2ε
(note that this implies γ > 1
⇐⇒
2ε < 0).
2gε − γ < −N + 1. Again the 2ε − 1 two-term recurrence relation (5.1.4) can be used with (cf. (5.1.5))
Case IIb. We have e = 0, 2 f = 1, 2ε > 1 and −N ≤
c+1 =
g−γ . 1 − 2ε
Again we set g = γ − 2ε + 1 in order to get A = 0 in the boundary conditions (5.1.12). This implies that c = 0 and −N ≤ γ − 2ε < −N + 1. The Pearson difference equation (5.1.10) now reads (1 − 2ε )(x + 1) w(x) = . w(x + 1) x + γ − 2ε
108
5 Orthogonal Polynomial Solutions of Real Difference Equations
Now we have to distinguish between two different cases. Case IIb1. For γ − 2ε = −N this leads to a positive-definite weight function of the form 1 , x = 0, 1, 2, . . . , N w(x) = x (2ε − 1) Γ(x + 1)Γ(N + 1 − x) for the Krawtchouk polynomials. This weight function for the Krawtchouk polynomials is connected with the binomial distribution in stochastics. The difference equation (5.1.1) reads (x − N + 1) Δ 2 yn (x) + (2ε (x + 1) − N) (Δ yn ) (x) = 2nε yn (x + 1) for n = 0, 1, 2, . . . , N, or (x − N)yn (x + 1) − (2(1 − ε )x − N) yn (x) + (1 − 2ε )xyn (x − 1) = 2nε yn (x) for n = 0, 1, 2, . . . , N, and the three-term recurrence relation (5.1.8) can be written as 2n(ε − 1) + N n(1 − 2ε )(n − 1 − N) yn (x) − yn−1 (x) yn+1 (x) = x − 2ε 4ε 2 for n = 1, 2, 3, . . . , N − 1 with y0 (x) = 1 and y1 (x) = x − N/2ε . For the coefficients of the representation (5.1.3) we obtain from (5.1.4) 2ε (n − k)an,k = (k − N)an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
Since an,n = n!, this leads to an,k =
(k − N)n−k n! (−N)n (−n)k (−2ε )k , = (2ε )n (−N)k (2ε )n−k (n − k)!
k = 0, 1, 2, . . . , n.
Hence (−N)n x −n, −x = ; 2ε , yn (x) = ∑ an,k 2 F1 k −N (2ε )n k=0 n
n = 0, 1, 2, . . . , N.
The Rodrigues formula (3.4.28) reads yn (x) =
(2ε − 1)x Γ(x + 1)Γ(N + 1 − x) (−2ε )n 1 n , ×Δ (2ε − 1)x−n Γ(x − n + 1)Γ(N + 1 − x)
n = 0, 1, 2, . . . , N.
Now we have N
d0 :=
1
(2ε )N
∑ (2ε − 1)x x! (N − x)! = (2ε − 1)N N! > 0.
x=0
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
109
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as (2ε )N−2n n! 1 y (x)y (x) = δmn m n ∑ x (2ε − 1)N−n (N − n)! x=0 (2ε − 1) x! (N − x)! N
for 2ε > 1, γ − 2ε = −N and m, n = 0, 1, 2, . . . , N. Case IIb2. In the case that −N < γ −2ε < −N +1 we were not able to find a solution for the Pearson difference equation such that the boundary conditions (5.1.12) can be fulfilled. However, the concept of orthogonality in section 3.2 can be generalized to an integration on the (possibly deformed) imaginary axis in the complex plane. Then we must have (cf. (3.2.14)) ( λn − λm ) where
i∞ −i∞
i∞
w(x)ym (x)yn (x) dx =
−i∞
{sn,m (x) − sn,m (x − 1)} dx
(5.3.1)
sn,m (x) = w(x)ϕ (x) {yn (x + 1)ym (x) − ym (x + 1)yn (x)} .
If the regularity condition (2.3.3) holds, id est en + 2ε = 0 for n = 0, 1, 2, . . . then this leads to an orthogonality relation if the right-hand side of (5.3.1) cancels. Hereby all moments of the form i∞
−i∞
w(x)xn dx,
n = 0, 1, 2, . . .
should exist (this is usually guaranteed by the asymptotic behaviour of the gamma functions involved). By using Cauchy’s integral theorem it can be shown that the right-hand side of (5.3.1) cancels. See [375]. With g = γ − 2ε + 1 we obtain a positive-definite weight function of the form w(x) = (−1)N
Γ(x + γ − 2ε )Γ(−x) (2ε − 1)x
for the Krawtchouk polynomials. Then the Rodrigues formula (3.4.28) reads (2ε − 1)x n Γ(x + γ − 2ε )Γ(n − x) , Δ yn (x) = (2ε )n Γ(x + γ − 2ε )Γ(−x) (2ε − 1)x−n
n = 0, 1, 2, . . . .
Now we may use Barnes’ integral representation (1.6.2) to find that 1 2π i
i∞ Γ(x + γ − 2ε )Γ(−x)
dx (2ε − 1)x 1 2ε − 1 γ −2ε γ − 2ε ; = Γ(γ − 2ε ) . = Γ(γ − 2ε ) 1 F0 − 1 − 2ε 2ε −i∞
110
5 Orthogonal Polynomial Solutions of Real Difference Equations
The latter integrand has poles in {0, 1, 2, . . .} and {2ε − γ , 2ε − γ − 1, 2ε − γ − 2, . . .}. Since N − 1 < 2ε − γ < N, these poles are all separated and the path can be taken as in (1.6.2) and we have d0 :=
(−1)N 2π i
i∞ Γ(x + γ − 2ε )Γ(−x)
dx (2ε − 1)x 2ε − 1 γ −2ε = (−1)N Γ(γ − 2ε ) 2ε N (−1) Γ(γ − 2ε + N) 2ε − 1 γ −2ε = > 0. (γ − 2ε )N 2ε −i∞
If (5.1.11) and (3.1.4) are used, this leads to the orthogonality relation (−1)N 2π i
i∞ Γ(x + γ − 2ε )Γ(−x) −i∞
(2ε − 1)x
ym (x)yn (x) dx
n! (−1)N+n Γ(γ − 2ε + N)(γ − 2ε )n = (γ − 2ε )N
2ε − 1 2ε
n+γ −2ε
δmn
for m, n = 0, 1, 2, . . . , N. Case III. We have e = 1, δ = that
( f − ε )2 − g + γ and η =
f 2 − g. Then we find
ϕ (x) = x2 + 2 f x + g = (x + f )2 − ( f 2 − g) = (x + f )2 − η 2 = (x + f + η )(x + f − η ) and
ψ (x) = 2ε x + γ = 2ε x + ( f − ε )2 − g + γ − ( f 2 − g) + 2ε f − ε 2 = 2ε (x + f ) + δ 2 − η 2 − ε 2 = 2ε (x + f − η ) + δ 2 − (η − ε )2 = 2ε (x + f − η ) + (δ + η − ε )(δ − η + ε ). This implies that the difference equation (5.1.1) can be written as (x + f + η )(x + f − η ) Δ 2 yn (x) + {2ε (x + f − η ) + (δ + η − ε )(δ − η + ε )} Δ yn (x) = n(n − 1 + 2ε )yn (x + 1), n = 0, 1, 2, . . . . or equivalently C(x)yn (x + 1) − {C(x) + D(x)} yn (x) + D(x)yn (x − 1) = n(n − 1 + 2ε )yn (x) for n = 0, 1, 2, . . ., where
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
111
C(x) = ϕ (x − 1) = (x − 1 + f + η )(x − 1 + f − η ) and D(x) = C(x) − ψ (x − 1) = (x − 1 + f + η )(x − 1 + f − η ) − 2ε (x − 1 + f ) − δ 2 + η 2 + ε 2 = (x − 1 + f )2 − 2ε (x − 1 + f ) + ε 2 − δ 2 = (x − 1 + f − ε )2 − δ 2 = (x − 1 + f − ε + δ )(x − 1 + f − ε − δ ). For the hypergeometric representation we use (5.1.3) and (5.1.4) provided that (cf. (5.1.5)) (c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ ⇐⇒
(c + 1)2 − 2( f − ε )(c + 1) + g − γ = 0.
This implies that
c + 1 = f − ε ± ( f − ε )2 − g + γ = f − ε ± δ . In that case we have (k − 1 − c)2 + 2 f (k − 1 − c) + g = (k − 1 − c)2 + 2 f (k − 1 − c) + g + ( f 2 − g) − η 2 = (k − 1 − c + f )2 − η 2 = (k − 1 − c + f + η )(k − 1 − c + f − η ) = (k + ε ∓ δ + η )(k + ε ∓ δ − η ). Therefore, the two-term recurrence relation (5.1.4) reads (n − k)(n + k − 1 + 2ε )an,k = (k + ε ∓ δ + η )(k + ε ∓ δ − η )an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. By using an,n = n!, we obtain (k + ε ∓ δ + η )n−k (k + ε ∓ δ − η )n−k n! (n + k − 1 + 2ε )n−k (n − k)! (ε ∓ δ + η )n (ε ∓ δ − η )n (−n)k (n − 1 + 2ε )k (−1)k = (n − 1 + 2ε )n (ε ∓ δ + η )k (ε ∓ δ − η )k
an,k =
for k = 0, 1, 2, . . . , n, which implies that n n x+c (−x − c)k = ∑ an,k (−1)k yn (x) = ∑ an,k k k! k=0 k=0 −n, n − 1 + 2ε , ε ∓ δ + 1 − f − x (ε ∓ δ + η )n (ε ∓ δ − η )n ;1 = 3 F2 (n − 1 + 2ε )n ε ∓ δ + η,ε ∓ δ − η for n = 0, 1, 2, . . .. These polynomials are called Hahn polynomials.
112
5 Orthogonal Polynomial Solutions of Real Difference Equations
In order to obtain an orthogonality relation, we use the Pearson difference equation (5.1.10), which can be written as D(x + 1) (x + f − ε + δ )(x + f − ε − δ ) w(x) = = w(x + 1) C(x) (x − 1 + f + η )(x − 1 + f − η ) (x + f − ε + δ )(ε + δ − f − x) = (x − 1 + f + η )(1 + η − f − x) (ε − δ − f − x)(ε + δ − f − x) = (x − 1 + f + η )(x − 1 + f − η ) (x + f − ε + δ )(x + f − ε − δ ) = (1 − η − f − x)(1 + η − f − x) (ε − δ − f − x)(ε + δ − f − x) = (1 − η − f − x)(1 + η − f − x)
(5.3.2) (5.3.3) (5.3.4) (5.3.5) (5.3.6)
First we consider positive-definite orthogonality for the Hahn polynomials in the case that ε > 0. Case IIIa. We have ε > 0, |δ − η | < ε and N − 1 < δ + η − ε (≤ N). In this case we use (5.3.3) to obtain a solution of the form w(x) =
Γ(x − 1 + f + η )Γ(1 + ε + δ − f − x) Γ(x + f − ε + δ )Γ(2 + η − f − x)
for the Pearson difference equation. Now we have to distinguish between two different cases. Case IIIa1. We have ε > 0 and |δ − η | < ε . In order to get A = 0 in the boundary conditions (5.1.12), we set f − ε + δ = 1 and 1 − f + η = N. Note that this implies that δ + η − ε = N. If we make the substitutions
ε + δ − η = α + 1 and ε − δ + η = β + 1, then we have 2ε = α + β + 2, 2δ = α + N + 1, 2η = β + N + 1 and we obtain that ϕ (x) = (x + β + 2)(x − N + 1) and ψ (x) = (α + β + 2)(x − N + 1) + (α + 1)N = (α + β + 2)(x + 1) − (β + 1)N. We also have C(x) = (x + β + 1)(x − N) and D(x) = x(x − α − N − 1). This implies that the difference equation can be written as (x + β + 2)(x − N + 1) Δ 2 yn (x) + {(α + β + 2)(x + 1) − (β + 1)N} Δ yn (x) = n(n + α + β + 1)yn (x + 1), n = 0, 1, 2, . . . or equivalently
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
113
(x + β + 1)(x − N)yn (x + 1) − {(x + β + 1)(x − N) + x(x − α − N − 1)} yn (x) + x(x − α − N − 1)yn (x − 1) = n(n + α + β + 1)yn (x), n = 0, 1, 2, . . . . Further we have yn (x) =
(β + 1)n (−N)n 3 F2 (n + α + β + 1)n
−n, n + α + β + 1, −x ;1 , β + 1, −N
n = 0, 1, 2, . . .
and the weight function reads w(x) =
Γ(x + β + 1)Γ(α + N + 1 − x) , Γ(x + 1)Γ(N + 1 − x)
x = 0, 1, 2, . . . , N.
For positivity we must have α + 1 > 0 and β + 1 > 0. This weight function for the Hahn polynomials is connected with the hypergeometric or P´olya distribution in stochastics. The Rodrigues formula (3.4.28) can be written as yn (x) =
(−1)n Γ(x + 1)Γ(N + 1 − x) (n + α + β + 1)n Γ(x + β + 1)Γ(α + N + 1 − x) n Γ(x + β + 1)Γ(α + N + n + 1 − x) ×Δ , Γ(N + 1 − x)Γ(x − n + 1)
n = 0, 1, 2, . . . .
Now we have N
d0 :=
Γ(x + β + 1)Γ(α + N + 1 − x) x! (N − x)! x=0 Γ(β + 1)Γ(α + N + 1) −N, β + 1 = ;1 2 F1 −α − N Γ(N + 1) Γ(β + 1)Γ(α + N + 1) (α + β + 2)N = Γ(N + 1) (α + 1)N Γ(α + 1)Γ(β + 1)Γ(α + β + N + 2) > 0. = Γ(α + β + 2)N! N
∑ w(x) = ∑
x=0
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as Γ(x + β + 1)Γ(α + N + 1 − x) ym (x)yn (x) x! (N − x)! x=0 N
∑ =
Γ(n + α + 1)Γ(n + β + 1)Γ(n + α + β + 1)Γ(n + α + β + N + 2)n! δmn Γ(2n + α + β + 1)Γ(2n + α + β + 2)(N − n)!
for m, n = 0, 1, 2, . . . , N. Case IIIa2. We have ε > 0, |δ − η | < ε and N − 1 < δ + η − ε < N. In this case we were not able to find a solution for the Pearson difference equation such that the boundary conditions (5.1.12) can be fulfilled. However we might proceed as on
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5 Orthogonal Polynomial Solutions of Real Difference Equations
page 109. By using (5.3.4) we obtain w(x) = Γ(x − 1 + f + η )Γ(x − 1 + f − η )Γ(1 + ε − δ − f − x)Γ(1 + ε + δ − f − x) as a possible solution for the Pearson difference equation. Here we also set f − ε + δ = 1 for simplicity. Then the Rodrigues formula (3.4.28) reads yn (x) =
Δ n (Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(n − x)Γ(n + 2δ − x)) (n − 1 + 2ε )n Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(−x)Γ(2δ − x)
for n = 0, 1, 2, . . . , N. Then we may use the Mellin-Barnes integral (1.6.3) again to find that 1 2π i
i∞
−i∞
i∞ 1 Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(−x)Γ(2δ − x) dx 2π i −i∞ Γ(ε + δ + η )Γ(ε + δ − η )Γ(ε − δ + η )Γ(ε − δ − η ) . = Γ(2ε )
w(x)dx =
Note that the increasing poles are {0, 1, 2, . . .} and {2δ , 2δ + 1, 2δ + 2, . . .} and the decreasing poles are {δ ± η − ε , δ ± η − ε − 1, δ ± η − ε − 2, . . .}. Hence for |δ − η | < ε and N − 1 < δ + η − ε < N these increasing and decreasing poles stay separated. Since −N < ε − δ − η < −N + 1, we have (−1)N Γ(ε − δ − η ) =
Γ(ε − δ − η + N) > 0. (−1)N (ε − δ − η )N
This implies that
(−1)N i∞ Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(−x)Γ(2δ − x) dx 2π i −i∞ (−1)N Γ(ε + δ + η )Γ(ε + δ − η )Γ(ε − δ + η )Γ(ε − δ − η ) > 0. = Γ(2ε )
d0 :=
If (5.1.11) and (3.1.4) are used, this leads to the orthogonality relation (cf. page 109)
(−1)N i∞ Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(−x)Γ(2δ − x)ym (x)yn (x) dx 2π i −i∞ n! (−1)N+n Γ(n + 2ε − 1) = Γ(2n + 2ε − 1)Γ(2n + 2ε ) × Γ(n + ε + δ + η )Γ(n + ε + δ − η )Γ(n + ε − δ + η )Γ(n + ε − δ − η ) δmn for m, n = 0, 1, 2, . . . , N. Now we consider positive-definite orthogonality for the Hahn polynomials in the case that ε < 0.
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
115
Case IIIb. We have 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 and |t| t t ≤ |δ − η | ≤ δ + η < 1 + or N + < |δ − η |. 2 2 2 Now we have to distinguish between three different cases. Case IIIb1. First we consider the case that |t| t ≤ |δ − η | ≤ δ + η < 1 + 2 2
and ε ± δ ± η = −N.
In this case we use (5.3.5) to obtain a solution of the form w(x) =
1 Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x)
for the Pearson difference equation. In order to get A = 0 in the boundary conditions (5.1.12), we set ⎧ ⎨ f − ε + δ = 1 and 1 − η − f = N =⇒ δ − η − ε = N ⎩ or
⎧ ⎨ f − ε + δ = 1 and 1 + η − f = N ⎩
or
δ +η −ε = N
and 1 − η − f = N
=⇒
−δ − η − ε = N
and we define ε − δ + η = α + 1 and ε + δ − η = β + 1
⎧ ⎨ f −ε −δ = 1 ⎩
=⇒
and we define ε + δ − η = α + 1 and ε − δ + η = β + 1
⎧ ⎨ f −ε −δ = 1 ⎩
or
and we define ε + δ + η = α + 1 and ε − δ − η = β + 1
and 1 + η − f = N
=⇒
−δ + η − ε = N
and we define ε − δ − η = α + 1 and ε + δ + η = β + 1.
In all cases this implies that 2ε = α + β + 2. Further we have 2δ = α + N + 1 (first and second case) or −2δ = α + N + 1 (third and fourth case) and 2η = β + N + 1 (second and fourth case) or −2η = β + N + 1 (first and third case). In all cases we find the same difference equation and hypergeometric representation as in Case IIIa. The weight function now reads w(x) =
1 , Γ(x + 1)Γ(N + 1 − x)Γ(x − N − α )Γ(−β − x)
x = 0, 1, 2, . . . , N.
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5 Orthogonal Polynomial Solutions of Real Difference Equations
For positivity we must have α < −N and β < −N. The Rodrigues formula (3.4.28) can be written as yn (x) =
Γ(x + 1)Γ(N + 1 − x)Γ(x − N − α )Γ(−β − x) (n + α + β + 1)n 1 n ×Δ Γ(N + 1 − x)Γ(x − n + 1)Γ(x − n − N − α )Γ(−β − x)
for n = 0, 1, 2, . . .. By using Dougall’s bilateral sum (1.5.9), we find that ∞
∑
w(x)
x=−∞ ∞
=
1 x=−∞ Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x)
=
Γ(1 − 2ε ) , Γ(1 − ε + δ − η )Γ(1 − ε − δ − η )Γ(1 − ε + δ + η )Γ(1 − ε − δ + η )
∑
provided that 2 f − 2 + 1 < 2 f − 2ε or equivalently 2ε < 1. For f − ε ± δ = 1 and 1 ± η − f = N this sum reduces to a finite sum over {0, 1, 2, . . . , N}, id est N
d0 :=
1
∑ x! (N − x)! Γ(x − N − α )Γ(−β − x)
x=0
=
Γ(−α − β − 1) > 0. N! Γ(−N − α − β − 1)Γ(−α )Γ(−β )
Hence with (5.1.11) and (3.1.4), the orthogonality relation can be written as N
1
∑ x! (N − x)! Γ(x − N − α )Γ(−β − x) ym (x)yn (x)
x=0
=
Γ(−2n − α − β )Γ(−2n − α − β − 1)n! δmn Γ(−n − α − β )Γ(−n − α )Γ(−n − β )Γ(−n − N − α − β − 1)(N − n)!
for m, n = 0, 1, 2, . . . , N. Case IIIb2. Now we consider the case that t |t| ≤ |δ − η | ≤ δ + η < 1 + 2 2
and ε ± δ ± η = −N.
In this case we use (5.3.5) to obtain a solution of the form w(x) =
1 Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x)
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
117
for the Pearson difference equation. For simplicity we now take f − ε − δ = 1. Then we have w(x) =
1 Γ(x + 1)Γ(x + 2δ + 1)Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x)
and the Rodrigues formula (3.4.28) reads yn (x) =
Γ(x + 1)Γ(x + 2δ + 1)Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x) (n − 1 + 2ε )n 1 ×Δn Γ(x − n + 1)Γ(x − n + 2δ + 1) 1 × Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x)
for n = 0, 1, 2, . . . , N. By using Dougall’s bilateral sum (1.5.8), we obtain ∞
d0 :=
∑
w(x)
x=−∞ ∞
=
1
∑ Γ(x + 1)Γ(x + 2δ + 1)Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x)
x=0
=
Γ(1 − 2ε ) > 0, Γ(1 − ε + δ + η )Γ(1 − ε + δ − η )Γ(1 − ε − δ + η )Γ(1 − ε − δ − η )
and if (5.1.11) and (3.1.4) are used, this leads to the orthogonality relation ∞
1
∑ x! Γ(x + 2δ + 1)Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x) ym (x)yn (x)
x=0
=
Γ(1 − 2ε − 2n)Γ(2 − 2ε − 2n)n! Γ(2 − 2ε − n)Γ(1 − ε + δ + η − n)Γ(1 − ε − δ − η − n) 1 × δmn , m, n = 0, 1, 2, . . . , N. Γ(1 − ε + δ − η − n)Γ(1 − ε − δ + η − n)
Case IIIb3. Further we consider the case that t |δ − η | > N + . 2 For δ > η > 0 this implies that δ − η > N + t/2. In that case we use (5.3.2) to obtain Γ(x − 1 + f + η )Γ(x − 1 + f − η ) w(x) = Γ(x + f − ε + δ )Γ(x + f − ε − δ ) as a possible solution for the Pearson difference equation. Again we take f − ε − δ = 1 for simplicity. Then the Rodrigues formula (3.4.28) reads
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5 Orthogonal Polynomial Solutions of Real Difference Equations
yn (x) =
Γ(x + 1)Γ(x + 2δ + 1) (n − 1 + 2ε )n Γ(x + ε + δ + η )Γ(x + ε + δ − η ) n Γ(x + ε + δ + η )Γ(x + ε + δ − η ) ×Δ , Γ(x − n + 1)Γ(x − n + 2δ + 1)
n = 0, 1, 2, . . . , N.
By using Gauss’s summation formula (1.5.3), we obtain ∞
d0 :=
∑
∞
w(x) =
x=−∞
=
Γ(x + ε + δ + η )Γ(x + ε + δ − η ) Γ(x + 1)Γ(x + 2δ + 1) x=0
∑
Γ(1 − 2ε )Γ(ε + δ + η )Γ(ε + δ − η ) > 0, Γ(1 − ε + δ + η )Γ(1 − ε + δ − η )
which, if (5.1.11) and (3.1.4) are used, leads to the orthogonality relation ∞
Γ(x + ε + δ + η )Γ(x + ε + δ − η ) ym (x)yn (x) x! Γ(x + 2δ + 1) x=0
∑
=
Γ(1 − 2ε − 2n)Γ(2 − 2ε − 2n)Γ(n + ε + δ + η )Γ(n + ε + δ − η )n! δmn Γ(2 − 2ε − n)Γ(1 − ε + δ + η − n)Γ(1 − ε + δ − η − n)
for m, n = 0, 1, 2, . . . , N. For η > δ > 0 this implies that η − δ > N + t/2. In that case we use (5.3.6) to obtain Γ(1 + ε − δ − f − x)Γ(1 + ε + δ − f − x) w(x) = Γ(2 + η − f − x)Γ(2 − η − f − x) as a possible solution for the Pearson difference equation. For simplicity we now take f + η = 1. Then the Rodrigues formula (3.4.28) reads yn (x) =
Γ(1 − x)Γ(2η + 1 − x) (n − 1 + 2ε )n Γ(ε + δ + η − x)Γ(ε − δ + η − x) Γ(n + ε + δ + η − x)Γ(n + ε − δ + η − x) ×Δn Γ(1 − x)Γ(2η + 1 − x)
for n = 0, 1, 2, . . . , N. By using Gauss’s summation formula (1.5.3), we obtain ∞
d0 :=
∑
x=−∞
∞
0
w(x) =
∑
w(x) =
x=−∞
=
Γ(x + ε + δ + η )Γ(x + ε − δ + η ) Γ(x + 1)Γ(x + 2η + 1) x=0
∑
Γ(1 − 2ε )Γ(ε + δ + η )Γ(ε − δ + η ) > 0, Γ(1 − ε + δ + η )Γ(1 − ε − δ + η )
which, if (5.1.11) and (3.1.4) are used, leads to the orthogonality relation
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
119
∞
Γ(x + ε + δ + η )Γ(x + ε − δ + η ) ym (−x)yn (−x) x! Γ(x + 2η + 1) x=0
∑
=
Γ(1 − 2ε − 2n)Γ(2 − 2ε − 2n)Γ(n + ε + δ + η )Γ(n + ε − δ + η )n! δmn Γ(2 − 2ε − n)Γ(1 − ε + δ + η − n)Γ(1 − ε − δ + η − n)
for m, n = 0, 1, 2, . . . , N. Case IIIc. We have 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 and δ pure imaginary (including zero) and/or η pure imaginary (including zero) and
ε ± δ ± η = 0, −1, −2, . . . , −N + 1. In this case we use (5.3.5) to obtain a solution of the form w(x) =
1 Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x)
for the Pearson difference equation. In this case we may write δ = δ1 + iδ2 and η = η1 + iη2 with δ1 δ2 = 0 = η1 η2 and δ1 , δ2 , η1 , η2 ∈ [0, ∞). Then the Rodrigues formula (3.4.28) reads yn (x) =
Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x) (n − 1 + 2ε )n 1 ×Δn Γ(x − n + f − ε + δ )Γ(x − n + f − ε − δ ) 1 × Γ(2 + η − f − x)Γ(2 − η − f − x)
for n = 0, 1, 2, . . . , N. By using Dougall’s bilateral sum (1.5.9), we obtain ∞
d0 :=
∑
w(x)
x=−∞
=
Γ(1 − 2ε ) > 0, Γ(1 − ε + δ + η )Γ(1 − ε + δ − η )Γ(1 − ε − δ + η )Γ(1 − ε − δ − η )
which, if (5.1.11) with A → −∞ and N → ∞ and (3.1.4) are used, leads to the orthogonality relation
120
5 Orthogonal Polynomial Solutions of Real Difference Equations ∞
1 x=−∞ Γ(x + f − ε + δ )Γ(x + f − ε − δ )
∑
1 ym (x)yn (x) Γ(2 − η − f − x)Γ(2 + η − f − x) Γ(1 − 2ε − 2n)Γ(2 − 2ε − 2n)n! = Γ(2 − 2ε − n)Γ(1 − ε + δ + η − n)Γ(1 − ε − δ − η − n) 1 × δmn , m, n = 0, 1, 2, . . . , N. Γ(1 − ε + δ − η − n)Γ(1 − ε − δ + η − n) ×
In this chapter we have proved:
Theorem 5.2. The positive-definite orthogonal polynomial solutions yn (x) of the difference equation (5.1.1) (ex2 + 2 f x + g) Δ 2 yn (x) + (2ε x + γ ) (Δ yn ) (x) = n(e(n − 1) + 2ε )yn (x + 1) for n = 0, 1, 2, . . . with e, f , g, ε , γ ∈ R consist of four infinite systems Case I. Charlier polynomials with orthogonality on {0, 1, 2, . . .} with respect to w(x) = 1/(−2ε )x Γ(x + 1) ; e = f = 0, g = 1, 2ε < 0, γ = 2ε + 1 Case IIa1. Meixner polynomials with orthogonality on {0, 1, 2, . . .} with respect to w(x) = Γ(x + γ − 2ε )/(1 − 2ε )x Γ(x + 1) ; e = 0, 2 f = 1, 2ε < 0, g = γ − 2ε + 1 and γ > 2ε Case IIa2. Meixner polynomials with orthogonality on {. . . , −2, −1, 0} with respect to w(x) = Γ(r − x)/(1 − 2ε )x Γ(1 − x) with r = (γ − 2ε )/(1 − 2ε ) ; e = 0, 2 f = 1, 0 < 2ε < 1, g = 1 and γ > 2ε Case IIa3. Charlier polynomials with orthogonality on {. . . , −2, −1, 0} with respect to w(x) = 1/(γ − 1)x Γ(1 − x) ; e = 0, 2 f = 1, 2ε = 1, g = 1 and γ > 1 and eight finite systems of N + 1 polynomials Case IIb1. Krawtchouk polynomials with orthogonality on {0, 1, 2, . . . , N} with respect to w(x) = 1/(2ε − 1)x Γ(x + 1)Γ(N + 1 − x) ; e = 0, 2 f = 1, 2ε > 1, g = γ − 2ε + 1 and γ − 2ε = −N Case IIb2. Krawtchouk polynomials with orthogonality on (−i∞, i∞) with respect to w(x) = (−1)N Γ(x + γ − 2ε )Γ(−x)/(2ε − 1)x ; e = 0, 2 f = 1, 2ε > 1, g = γ − 2ε + 1 and −N < γ − 2ε < −N + 1
5.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
121
Case IIIa1. Hahn polynomials with orthogonality on {0, 1, 2, . . . , N} with + 1)Γ(N + 1 − x) ; e = respect to w(x) = Γ(x + β + 1)Γ(α + N + 1 − x)/Γ(x ε > 0, α > −1, β > −1, ε − δ − η = −N ( δ = ( f − ε )2 − g + γ , η = 1, 2 f − g and 2ε = α + β + 2) Case IIIa2. Hahn polynomials with orthogonality on (−i∞, i∞) with respect to w(x) = (−1)N Γ(x + ε − δ + η )Γ(x + ε − δ − η )Γ(−x)Γ(−β − x) ; e = 1, ε > 0, |δ − η | < ε and −N < ε − δ − η < −N + 1 Case IIIb1. Hahn polynomials with orthogonality on {0, 1, 2, . . . , N} with respect to w(x) = 1/Γ(x + 1)Γ(N + 1 − x)Γ(x − N − α )Γ(−β − x) ; e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, α < −N, β < −N, δ + η < 1 + t/2 and ε ± δ ± η = −N Case IIIb2. Hahn polynomials with orthogonality on {0, 1, 2, . . .} with respect to w(x) = 1/Γ(x + 1)Γ(x + 2δ + 1)Γ(1 − ε − δ − η − x)Γ(1 − ε − δ + η − x) ; e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, δ + η < 1 + t/2 and ε ± δ ± η = −N Case IIIb3. Hahn polynomials with orthogonality on {. . . , −2, −1, 0} with respect to w(x) = Γ(x + ε + δ + η )Γ(x + ε + |δ − η |)/Γ(x + 1)Γ(x + 2 max(δ , η ) + 1) ; e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 and |δ − η | > N + t/2 Case IIIc. Hahn polynomials with orthogonality on {. . . , −1, 0, 1, . . .} with respect to w(x) = 1/Γ(x + f − ε + δ )Γ(x + f − ε − δ )Γ(2 − η − f − x)Γ(2 + η − f − x) ; e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, ε ± δ ± η = 0, −1, −2, . . . , −N + 1, δ = δ1 + iδ2 , η = η1 + iη2 , δ1 δ2 = 0 = η1 η2 and δ1 , δ2 , η1 , η2 ∈ [0, ∞).
Chapter 6
Orthogonal Polynomial Solutions of Complex Difference Equations Discrete Classical Orthogonal Polynomials II
6.1 Real Polynomial Solutions of Complex Difference Equations We consider the difference equation (5.1.1) with complex coefficients, id est (ez2 + 2 f z + g) Δ 2 yn (z) + (2ε z + γ ) (Δ yn ) (z) = n(e(n − 1) + 2ε )yn (z + 1) (6.1.1) for n = 0, 1, 2, . . .. This difference equation can also be written in the form e(z − 1)2 + 2 f (z − 1) + g (Δ (∇yn )) (z) + (2ε (z − 1) + γ ) (∇yn ) (z) = n(e(n − 1) + 2ε )yn (z) or in the form C(z)yn (z + 1) − {C(z) + D(z)} yn (z) + D(z)yn (z − 1) = n(e(n − 1) + 2ε )yn (z) for n = 0, 1, 2, . . ., where C(z) = e(z − 1)2 + 2 f (z − 1) + g and
D(z) = C(z) − 2ε (z − 1) − γ .
We look for monic polynomial solutions of the form (cf. (2.4.14)) n z+c , c ∈ C, an,n = n!, n = 0, 1, 2, . . . , yn (z) = ∑ an,k k k=0
(6.1.2)
(6.1.3)
where the coefficients satisfy the two-term recurrence relation (cf. (2.4.15)) (n − k) (e(n + k − 1) + 2ε ) an,k − e(k − 1 − c)2 + 2 f (k − 1 − c) + g an,k+1 = 0
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 6, © Springer-Verlag Berlin Heidelberg 2010
(6.1.4)
123
124
6 Orthogonal Polynomial Solutions of Complex Difference Equations
for k = n − 1, n − 2, n − 3, . . . , 0 provided that c ∈ C satisfies (cf. (5.1.5)) e(c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ . In section 2.6 we found that the monic polynomial solutions {yn }∞ n=0 satisfy the three-term recurrence relation yn+1 (z) = (z − cn )yn (z) − dn yn−1 (z),
n = 1, 2, 3, . . . ,
(6.1.5)
with initial values y0 (z) = 1 and y1 (z) = z − c0 , where (cf. (2.6.15)) cn =
n (e(n − 1) + 2ε ) (2(e − f ) + ε ) + (e − ε )(γ − 2ε ) , 2 (e(n − 1) + ε ) (en + ε )
n = 0, 1, 2, . . .
and (cf. (2.6.16)) dn = −
n (e(n − 2) + 2ε ) 4 (e(2n − 3) + 2ε ) (e(n − 1) + ε )2 (e(2n − 1) + 2ε ) × e(n − 1)2 (e(n − 1) + 2ε )2 + 2(n − 1) (e(n − 1) + 2ε ) (2eg + 2 f (ε − f ) − eγ ) + 4ε (gε − f γ ) + eγ 2 , n = 1, 2, 3, . . . .
In order to obtain real polynomial solutions, we set z = a + ix with a ∈ R and x ∈ R. Then we define yn (z) = yn (a + ix) = in yn (x). Substitution into (6.1.5) leads to yn+1 (x) = (x + i(cn − a)) yn (x) + dn yn−1 (x),
n = 1, 2, 3, . . .
(6.1.6)
with y0 (x) = 1 and y1 (x) = x + i(c0 − a). Since the coefficients have to be real, we conclude that cn − a must be pure imaginary (including zero) for all n = 0, 1, 2, . . . and dn must be real for all n = 1, 2, 3, . . .. Again we consider three different cases as in section 5.2. Case I. Degree[ϕ ] = 0: e = f = 0 and we may choose g = 1. Then we have cn =
2(n + 1)ε − γ , 2ε
n = 0, 1, 2, . . .
and dn = −
n , 2ε
n = 1, 2, 3, . . . .
Since dn has to be real, we must have ε ∈ R. However, it is impossible to choose γ = γ1 + iγ2 with γ1 , γ2 ∈ R such that cn − a is pure imaginary for all n = 0, 1, 2, . . .. Case II. Degree[ϕ ] = 1: e = 0 and we may choose 2 f = 1. Then we have
6.1 Real Polynomial Solutions of Complex Difference Equations
cn =
2n(ε − 1) + 2ε − γ , 2ε
125
n = 0, 1, 2, . . .
and
n ((n − 1)(2ε − 1) + 2gε − γ ) , n = 1, 2, 3, . . . . 4ε 2 If we set g = g1 + ig2 , ε = ε1 + iε2 and γ = γ1 + iγ2 with g1 , g2 , ε1 , ε2 , γ1 , γ2 ∈ R, we find that dn = −
cn − a = =
2n(ε1 − 1 + iε2 ) + 2(1 − a)(ε1 + iε2 ) − (γ1 + iγ2 ) 2(ε1 + iε2 ) 2n(ε12 + ε22 ) − 2n(ε1 − iε2 ) + 2(1 − a)(ε12 + ε22 ) − (γ1 + iγ2 )(ε1 − iε2 ) . 2(ε12 + ε22 )
Since cn − a must be pure imaginary for all n = 0, 1, 2, . . ., we conclude that
ε12 + ε22 = ε1
and 2(1 − a)ε1 = γ1 ε1 + γ2 ε2 .
(6.1.7)
Further we have n ((n − 1)(2ε1 − 1 + 2iε2 ) + 2(g1 + ig2 )(ε1 + iε2 ) − (γ1 + iγ2 )) 4(ε1 + iε2 )2 2 2 n ε1 − ε2 − 2iε1 ε2 =− 4(ε12 + ε22 )2 × ((n − 1)(2ε1 − 1 + 2iε2 ) + 2(g1 + ig2 )(ε1 + iε2 ) − (γ1 + iγ2 )) .
dn = −
Since dn has to be real for all n = 1, 2, 3, . . ., we conclude that the imaginary part must be zero. Hence by using (6.1.7), we find that 0 = 2(n − 1)ε2 (ε12 − ε22 ) − 2(n − 1)ε1 ε2 (2ε1 − 1) + 2(ε12 − ε22 )(g1 ε2 + g2 ε1 ) − 4ε1 ε2 (g1 ε1 − g2 ε2 ) − γ2 (ε12 − ε22 ) + 2ε1 ε2 γ1 = 2(n − 1)ε2 ε1 − ε12 − ε22 − 2g1 ε2 (ε12 + ε22 ) + 2g2 ε1 (ε12 + ε22 ) − γ2 (ε12 + ε22 ) + 2γ2 ε22 + 2ε1 ε2 γ1 = (ε12 + ε22 ) (2g2 ε1 − 2g1 ε2 − γ2 ) + 2ε2 (ε1 γ1 + ε2 γ2 ) = ε1 {2(g2 ε1 − g1 ε2 ) − γ2 + 4(1 − a)ε2 } . Since ε = 0 in view of the regularity condition (2.3.3), we have ε1 = ε12 + ε22 > 0 and therefore γ2 = 4(1 − a)ε2 + 2(g2 ε1 − g1 ε2 ). (6.1.8) This implies that i(cn − a) = −
2nε2 + γ1 ε2 − γ2 ε1 , 2 ε1
and by using (6.1.7) and (6.1.8), we obtain
n = 0, 1, 2, . . .
(6.1.9)
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6 Orthogonal Polynomial Solutions of Complex Difference Equations
4ε12 dn = 4(ε12 + ε22 )2 dn = −n (n − 1)(2ε1 − 1)(ε12 − ε22 ) + 4(n − 1)ε1 ε22 + 2(g1 ε1 − g2 ε2 )(ε12 − ε22 )
+4ε1 ε2 (g1 ε2 + g2 ε1 ) − γ1 (ε12 − ε22 ) − 2γ2 ε1 ε2 = −n 2(n − 1)ε1 (ε12 + ε22 ) − (n − 1)(ε12 − ε22 ) + 2g1 ε1 (ε12 + ε22 )
+2g2 ε2 (ε12 + ε22 ) − γ1 (ε12 + ε22 − 2ε22 ) − 2γ2 ε1 ε2 = −n {(n − 1)ε1 + ε1 (2g1 ε1 + 2g2 ε2 − γ1 ) + 2ε2 (γ1 ε2 − γ2 ε1 )} = −n {(n − 1)ε1 + 2g1 ε1 + 2ε2 (g2 ε1 − g1 ε2 ) + γ1 ε1 − 2ε1 (γ1 ε1 + γ2 ε2 )}
= −n (n − 1)ε1 + 2g1 ε1 + (1 − 2ε1 )(γ1 ε1 + γ2 ε2 ) − 4(1 − a)ε22
= −n (n − 1)ε1 + 2g1 ε1 + 2(1 − a)ε1 − 4(1 − a)(ε12 + ε22 ) = −nε1 {n − 1 + 2g1 − 2(1 − a)} . Since ε1 > 0, this implies that dn = −
n(n + 2g1 + 2a − 3) , 4 ε1
n = 1, 2, 3, . . . .
(6.1.10)
Case III. Degree[ϕ ] = 2: we may choose e = 1. Then we have cn =
n(n − 1 + 2ε ) (2(1 − f ) + ε ) + (1 − ε )(γ − 2ε ) , 2(n − 1 + ε )(n + ε )
n = 0, 1, 2, . . .
and dn = −
n(n − 2 + 2ε ) 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε ) × (n − 1)2 (n − 1 + 2ε )2 + 2(n − 1)(n − 1 + 2ε )(2g + 2 f (ε − f ) − γ ) + 4ε (gε − f γ ) + γ 2 , n = 1, 2, 3, . . . .
The latter formula can also be written in the form dn = − where
n(n − 2 + 2ε ) Dn , 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . ,
2 Dn = (n − 1 + ε )2 − δ 2 − η 2 − 4δ 2 η 2 ,
n = 1, 2, 3, . . .
with Note that we have
δ 2 = ( f − ε )2 − g + γ
and η 2 = f 2 − g.
(6.1.11)
(6.1.12)
6.1 Real Polynomial Solutions of Complex Difference Equations
cn =
127
(n − 1 + ε )(n + ε ) (2(1 − f ) + ε ) + (1 − ε )(γ − 2 f ε + ε 2 ) , 2(n − 1 + ε )(n + ε )
n = 0, 1, 2, . . . ,
which implies that cn − a =
(n − 1 + ε )(n + ε ) (2(1 − f − a) + ε ) + (1 − ε )(γ − 2 f ε + ε 2 ) 2(n − 1 + ε )(n + ε )
for n = 0, 1, 2, . . .. Since ε occurs in both cn − a and dn in combination with n, we conclude that cn − a can only be pure imaginary for all n = 0, 1, 2, . . . and dn can only be real for all n = 1, 2, 3, . . . if ε is real. If we now set f = f1 + i f2 , g = g1 + ig2 and γ = γ1 + iγ2 with f1 , f2 , g1 , g2 , ε , γ1 , γ2 ∈ R, we obtain cn − a =
(n − 1 + ε )(n + ε ) (2(1 − f1 − a) + ε ) + (1 − ε )(γ1 − 2 f1 ε + ε 2 ) 2(n − 1 + ε )(n + ε ) 2(n − 1 + ε )(n + ε ) f2 − (1 − ε )(γ2 − 2 f2 ε ) . −i 2(n − 1 + ε )(n + ε )
Since cn − a must be pure imaginary for all n = 0, 1, 2, . . ., we conclude that
ε = 2 f1 + 2a − 2 and (1 − ε )(γ1 − 2 f1 ε + ε 2 ) = 0.
(6.1.13)
Then we have i(cn − a) = f2 −
(1 − ε )(γ2 − 2 f2 ε ) , 2(n − 1 + ε )(n + ε )
n = 0, 1, 2, . . . .
As before, we may write
Dn = (n − 1 + ε )2 − (δ + η )2 (n − 1 + ε )2 − (δ − η )2 = (n − 1 + ε ) − 2(δ + η )(n − 1 + ε ) + (δ − η ) 4
2
2
2
2
2 2
(6.1.14) (6.1.15)
for n = 1, 2, 3, . . .. If we set δ = δ1 + iδ2 and η = η1 + iη2 with δ1 , δ2 , η1 , η2 ∈ R, we find that (δ + η )2 = (δ1 + η1 )2 − (δ2 + η2 )2 + 2i(δ1 + η1 )(δ2 + η2 ) and
(δ − η )2 = (δ1 − η1 )2 − (δ2 − η2 )2 + 2i(δ1 − η1 )(δ2 − η2 ).
Since dn must be real for all n = 1, 2, 3, . . . this implies, by using (6.1.11) and (6.1.14), that (δ1 + η1 )(δ2 + η2 ) = 0 and (δ1 − η1 )(δ2 − η2 ) = 0. Note that both δ and η are not uniquely determined in view of (6.1.12). Without loss of generality we may choose δ1 + iδ2 = η1 − iη2 with δ1 = η1 ≥ 0. Hence δ and η are complex conjugates, id est δ = η . Then we have
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6 Orthogonal Polynomial Solutions of Complex Difference Equations
(δ + η )2 = 4δ12
and
(δ − η )2 = −4δ22 .
Now we use (6.1.12) again to find that
δ 2 + η 2 = ( f − ε )2 − g + γ + f 2 − g = 2 f 2 − 2 f ε + ε 2 − 2g + γ = 2( f1 + i f2 )2 − 2( f1 + i f2 )ε + ε 2 − 2(g1 + ig2 ) + γ1 + iγ2 = 2 f12 − 2 f22 − 2 f1 ε + ε 2 − 2g1 + γ1 + i {4 f1 f2 − 2 f2 ε − 2g2 + γ2 } and
δ 2 − η 2 = ( f − ε )2 − g + γ − f 2 + g = γ − 2 f ε + ε 2 = γ1 + iγ2 − 2( f1 + i f2 )ε + ε 2 = γ1 − 2 f1 ε + ε 2 + i(γ2 − 2 f2 ε ). Hence we have (δ 2 − η 2 )2 = (γ1 − 2 f1 ε + ε 2 )2 − (γ2 − 2 f2 ε )2 + 2i(γ1 − 2 f1 ε + ε 2 )(γ2 − 2 f2 ε ). Since dn must be real for all n = 1, 2, 3, . . ., this implies that, by using (6.1.11) and (6.1.15)
γ2 = 2 f2 ε − 4 f1 f2 + 2g2
and (γ1 − 2 f1 ε + ε 2 )(γ2 − 2 f2 ε ) = 0.
For γ2 = 2 f2 ε we obtain (δ 2 − η 2 )2 = (γ1 − 2 f1 ε + ε 2 )2 ≥ 0. On the other hand we also have
δ 2 − η 2 = (δ1 + iδ2 )2 − (δ1 − iδ2 )2 = 4iδ1 δ2
=⇒
(δ 2 − η 2 )2 = −16δ12 δ22 ≤ 0.
Hence dn can only be real for all n = 1, 2, 3, . . . for
γ1 = 2 f1 ε − ε 2
and γ2 = 2 f2 ε − 4 f1 f2 + 2g2 .
(6.1.16)
Combining (6.1.13) and (6.1.16), we conclude, since cn − a must be pure imaginary for all n = 0, 1, 2, . . . and dn must be real for all n = 1, 2, 3, . . ., that we must have
ε = 2 f1 + 2a − 2, γ1 = 2 f1 ε − ε 2 = 2ε (1 − a) and γ2 = 2 f2 ε − 4 f1 f2 + 2g2 . Note that we have 4(δ12 − δ22 ) = (δ + η )2 + (δ − η )2 = 2(δ 2 + η 2 ) = 4 f12 − 4 f22 − 4g1 and
−16δ12 δ22 = (δ 2 − η 2 )2 = −(γ2 − 2 f2 ε )2 = −(2g2 − 4 f1 f2 )2 ,
which implies that
(6.1.17)
6.1 Real Polynomial Solutions of Complex Difference Equations
δ12 − δ22 = f12 − f22 − g1
129
and 4δ12 δ22 = (g2 − 2 f1 f2 )2 .
Finally, we write the difference equation (6.1.1) in a different form. Note that we have
Δ yn (z) = yn (z + 1) − yn (z) = yn (a + ix + 1) − yn (a + ix) = in ( yn (x − i) − yn (x)) and 2 Δ yn (z) = yn (z + 2) − 2yn (z + 1) + yn (z) = in ( yn (x − 2i) − 2 yn (x − i) + yn (x)) . The difference equation (6.1.1) can be written as ϕ (z) Δ 2 yn (z) + ψ (z)Δ yn (z) = λn yn (z + 1),
n = 0, 1, 2, . . .
which is equivalent to
ϕ (a + ix) ( yn (x − 2i) − 2 yn (x − i) + yn (x)) yn (x − i) − yn (x)) = λn yn (x − i) + ψ (a + ix) ( for n = 0, 1, 2, . . .. By shifting x to x + i the latter formula can also be written in the form y(x − i) − C(x) + D(x) yn (x + i) = λn yn (x) (6.1.18) C(x) yn (x) + D(x) for n = 0, 1, 2, . . ., where − ψ (a + ix − 1). = ϕ (a + ix − 1) and D(x) = C(x) C(x)
(6.1.19)
Note that the difference equation (6.1.1) can be written in the selfadjoint form
Δ (w(z)ϕ (z + 1)Δ yn (z)) = λn w(z + 1)yn (z + 1), where
ϕ (z + 2) = ez2 + 2 f z + g,
ψ (z + 1) = 2ε z + γ
and λn = n (e(n − 1) + 2ε ) ,
provided that w(z) satisfies the Pearson difference equation
Δ (w(z)ϕ (z + 1)) = w(z + 1)ψ (z + 1). This implies that the Pearson difference equation (cf. (5.1.10)) can now be written in the form ϕ (z + 2) − ψ (z + 1) D(z + 1) w(z) = = , (6.1.20) w(z + 1) ϕ (z + 1) C(z) where C(z) and D(z) are given by (6.1.2).
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6 Orthogonal Polynomial Solutions of Complex Difference Equations
6.2 Classification of the Real Positive-Definite Orthogonal Polynomial Solutions Again we use Favard’s theorem (theorem 3.1) to conclude from (6.1.6) that we have positive-definite orthogonality if i(cn − a) ∈ R for all n = 0, 1, 2, . . . and dn < 0 for all n = 1, 2, 3, . . .. Again we consider three different cases depending on the form of ϕ (x) = ex2 + 2 f x + g: Case I. Degree[ϕ ] = 0: e = f = 0 and we may choose g = 1. In this case we do not have real polynomial solutions and therefore we cannot have orthogonal polynomial solutions. Case II. Degree[ϕ ] = 1: e = 0 and we may choose 2 f = 1. Further we have (6.1.7), (6.1.8) and n(n + 2g1 + 2a − 3) , n = 1, 2, 3, . . . . dn = − 4 ε1 Hence positive-definite orthogonality occurs for ε1 = ε12 + ε22 > 0 and g1 > 1 − a. Case III. Degree[ϕ ] = 2: we may choose e = 1. Further we have (6.1.17) and dn = −
n(n − 2 + 2ε ) Dn , 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . .
where
Dn = (n − 1 + ε )2 − 4δ12 (n − 1 + ε )2 + 4δ22 , and
δ12 − δ22 = f12 − f22 − g1
n = 1, 2, 3, . . .
and 4δ12 δ22 = (g2 − 2 f1 f2 )2 .
For ε > 0 we have −
n(n − 2 + 2ε ) < 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . . .
This implies that positive-definite orthogonality occurs for Case IIIa. ε > 0 and 2δ1 < ε . We also have a finite orthogonal polynomial system in this case. First we define 2ε = −2N − t Then we have
with N ∈ {1, 2, 3, . . .}
and
− 1 < t ≤ 1.
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
−
n(n − 2 + 2ε ) > 0, 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
131
n = 1, 2, 3, . . . , N
and for n = N + 1 this is not longer true. So in that case we have dn < 0 for
n = 1, 2, 3, . . . , N
⇐⇒
Dn < 0 for
n = 1, 2, 3, . . . , N.
This implies that we have positive-definite orthogonality for Case IIIb. 2δ1 > −ε .
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions Case I. We have e = f = 0 and g = 1. In this case we do not have real orthogonal polynomial solutions. Case II. We have e = 0, 2 f = 1, ε1 > 0 and g1 > 1 − a. From (6.1.7) and (6.1.8) it follows that
γ1 ε1 = 2(1 − a)ε1 − γ2 ε2 = 2(1 − a)ε1 − 4(1 − a)ε22 − 2ε2 (g2 ε1 − g1 ε2 ) = 2(1 − a)ε1 − 4(1 − a)(ε1 − ε12 ) − 2g2 ε1 ε2 + 2g1 (ε1 − ε12 ) = 2ε1 {g1 (1 − ε1 ) − g2 ε2 − (1 − a)(1 − 2ε1 )} , which implies that
γ1 = 2 {g1 (1 − ε1 ) − g2 ε2 − (1 − a)(1 − 2ε1 )} .
(6.3.1)
The Pearson difference equation (6.1.20) reads (1 − 2ε )z + g − γ w(z) = , w(z + 1) z+g−1 where z = a + ix,
g = g1 + ig2 ,
ε = ε1 + i ε 2
and γ = γ1 + iγ2
with a, x, g1 , g2 , ε1 , ε2 , γ1 , γ2 ∈ R. By using (6.1.8) and (6.3.1), we obtain g − γ = g1 + ig2 − γ1 − iγ2 = −g1 + 2(g1 ε1 + g2 ε2 ) + 2(1 − a)(1 − 2ε1 ) + i {g2 − 4(1 − a)ε2 − 2(g2 ε1 − g1 ε2 )} = (2ε1 − 1 + 2iε2 ) {g1 − 2(1 − a) − ig2 } .
(6.3.2)
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6 Orthogonal Polynomial Solutions of Complex Difference Equations
Hence we have (2ε1 − 1 + 2iε2 ) {g1 + a − 2 − i(x + g2 )} w(a + ix) = . w(a + ix + 1) g1 + a − 1 + i(x + g2 ) Since ε1 = ε12 + ε22 , we have (2ε1 − 1)2 + 4ε22 = 1. Hence we may define ϕ ∈ R such that eiϕ := 2ε1 − 1 + 2iε2
=⇒
cos ϕ = 2ε1 − 1 and
sin ϕ = 2ε2 .
Then we obtain a positive-definite weight function of the form w(a + ix) = |Γ(g1 + a − 1 + i(x + g2 ))|2 eϕ x
(6.3.3)
for the Meixner-Pollaczek polynomials. For the difference equation (6.1.18) we obtain + D(x) yn (x − i) − C(x) yn (x + i) = 2n(ε1 + iε2 ) yn (x) + D(x) yn (x − i), C(x) where, by using (6.1.19), we get = ϕ (a + ix − 1) = a + ix − 1 + g = g1 + a − 1 + i(x + g2 ) C(x) and − ψ (a + ix − 1) = a + ix − 1 + g − 2ε (a + ix − 1) − γ = C(x) D(x) = (1 − 2ε )(a + ix − 1) + g − γ . Hence by using (6.3.2), we obtain = (2ε1 − 1 + 2iε2 )(g1 + a − 1 − i(x + g2 )). D(x) By using (6.1.9) and (6.1.10), the three-term recurrence relation (6.1.6) reads 2nε2 + γ1 ε2 − γ2 ε1 n(n + 2g1 + 2a − 3) yn (x) − yn−1 (x) yn+1 (x) = x − 2 ε1 4 ε1 for n = 1, 2, 3, . . . with y0 (x) = 1 and y1 (x) = x − (γ1 ε2 − γ2 ε1 )/2ε1 . For the coefficients of the representation (6.1.3) we obtain from (6.1.4) 2ε (n − k)an,k = (k − 1 − c + g)an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0,
where (1 − 2ε )(c + 1) = g − γ . Hence we have −(2ε1 − 1 + 2iε2 )(c + 1) = (2ε1 − 1 + 2iε2 ) {g1 − 2(1 − a) − ig2 } , which implies that c = 1 − 2a − g1 + ig2 and c + 1 − g = 2(1 − a − g1 ), provided that 2ε = 1. However, 2ε = 1 (i.e. 2ε1 = 1 and ε2 = 0) is impossible since ε1 = ε12 + ε22 .
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
133
Hence we have 2ε (n − k)an,k = (k + 2g1 + 2a − 2)an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
Since an,n = n!, this leads to an,k =
(k + 2g1 + 2a − 2)n−k n! (2g1 + 2a − 2)n (−n)k (−2ε )k = (2ε )n (2g1 + 2a − 2)k (2ε )n−k (n − k)!
for k = 0, 1, 2, . . . , n. Hence by using (6.1.3), we obtain n a + ix + 1 − 2a − g1 + ig2 yn (a + ix) = ∑ an,k k k=0 (2g1 + 2a − 2)n −n, g1 + a − 1 − i(x + g2 ) = ; 2 F ε 2 1 2g1 + 2a − 2 (2ε )n for n = 0, 1, 2, . . .. The Rodrigues formula (3.4.28) for n = 0, 1, 2, . . . reads yn (a + ix) =
1 (1 + eiϕ )n eϕ x |Γ(g1 + a − 1 + i(x + g2 ))|2
× Δ n Γ(g1 + a − 1 + i(x + g2 ))Γ(g1 + a + n − 1 − i(x + g2 ))eϕ (x+in) . Now we use (1.6.8) to find
1 ∞ |Γ(g1 + a − 1 + i(x + g2 ))|2 eϕ x dx 2π −∞ = Γ(2g1 + 2a − 2)e−g2 ϕ (4ε1 )1−a−g1 > 0.
d0 :=
Then using (5.1.11) and (3.1.4) the orthogonality relation can be written as
1 ∞ |Γ(g1 + a − 1 + i(x + g2 ))|2 eϕ x ym (x) yn (x) dx 2π −∞ = Γ(2g1 + 2a + n − 2)e−g2 ϕ (4ε1 )1−a−g1 −n n! δmn , m, n = 0, 1, 2, . . . .
Case III. We have e = 1 and (6.1.17). Note that
ϕ (x) = x2 + 2 f x + g = (x + f )2 + g − f 2 which, if (6.1.17) is used, leads to
and ψ (x) = 2ε x + γ ,
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6 Orthogonal Polynomial Solutions of Complex Difference Equations
ϕ (a + ix − 1) = {a + f1 − 1 + i(x + f2 )}2 + g1 + ig2 − ( f1 + i f2 )2 = −(x + f2 )2 + (a + f1 − 1)2 + g1 − f12 + f22 + i {2(a + f1 − 1)(x + f2 ) + g2 − 2 f1 f2 } ε2 = −(x + f2 )2 + + g1 − f12 + f22 + i {ε (x + f2 ) + g2 − 2 f1 f2 } 4 and
ψ (a + ix − 1) = 2ε (a + ix − 1) + γ1 + iγ2 = γ1 − 2ε (1 − a) + i(2ε x + γ2 ) = 2i {ε (x + f2 ) + g2 − 2 f1 f2 } . This implies that the difference equation (6.1.18) can be written as y(x − i) − C(x) + D(x) yn (x + i) = λn yn (x), n = 0, 1, 2, . . . , C(x) yn (x) + D(x) where, by using (6.1.17), we get = ϕ (a + ix − 1) C(x) = −(x + f2 )2 +
ε2 + g1 − f12 + f22 + i {ε (x + f2 ) + g2 − 2 f1 f2 } 4
and − ψ (a + ix − 1) D(x) = C(x) = −(x + f2 )2 +
ε2 + g1 − f12 + f22 − i {ε (x + f2 ) + g2 − 2 f1 f2 } . 4
Since i(cn − a) = f2 −
(1 − ε )(γ2 − 2 f2 ε ) (1 − ε )(g2 − 2 f1 f2 ) = f2 − , 2(n − 1 + ε )(n + ε ) (n − 1 + ε )(n + ε )
n = 0, 1, 2, . . .
and dn = −
n(n − 2 + 2ε ) Dn , 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε )
n = 1, 2, 3, . . .
where Dn = (n − 1 + ε )4 + 4(g1 − f12 + f22 )(n − 1 + ε )2 − 4(g2 − 2 f1 f2 )2 , the three-term recurrence relation (6.1.6) can be written as
n = 1, 2, 3, . . .
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
135
(1 − ε )(g2 − 2 f1 f2 ) yn (x) yn+1 (x) = x + f2 − (n − 1 + ε )(n + ε ) n(n − 2 + 2ε ) − 4(2n − 3 + 2ε )(n − 1 + ε )2 (2n − 1 + 2ε ) × (n − 1 + ε )4 + 4(g1 − f12 + f22 )(n − 1 + ε )2
−4(g2 − 2 f1 f2 )2 yn−1 (x) for n = 1, 2, 3, . . . with y0 (x) = 1 and y1 (x) = x + f2 + (g2 − 2 f1 f2 )/ε . For the hypergeometric representation we use (6.1.3) and (6.1.4), provided that (c + 1)2 − 2 f (c + 1) + g = −2ε (c + 1) + γ ⇐⇒ (c + 1)2 − 2( f − ε )(c + 1) + g − γ = 0. Since g − γ = ( f − ε )2 − δ 2 , this implies that (c + 1 − f + ε )2 = δ 2 . In that case we have (k − 1 − c)2 + 2 f (k − 1 − c) + g = (k − 1 − c)2 + 2 f (k − 1 − c) + g + ( f 2 − g) − η 2 = (k − 1 − c + f )2 − η 2 = (k − 1 − c + f + η )(k − 1 − c + f − η ) = (k + ε ∓ δ + η )(k + ε ∓ δ − η ). Therefore, the two-term recurrence relation (6.1.4) reads (n − k)(n + k − 1 + 2ε )an,k = (k + ε ∓ δ + η )(k + ε ∓ δ − η )an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. By using an,n = n!, we obtain (k + ε ∓ δ + η )n−k (k + ε ∓ δ − η )n−k n! (n + k − 1 + 2ε )n−k (n − k)! (ε ∓ δ + η )n (ε ∓ δ − η )n (−n)k (n − 1 + 2ε )k (−1)k = (n − 1 + 2ε )n (ε ∓ δ + η )k (ε ∓ δ − η )k
an,k =
for k = 0, 1, 2, . . . , n, which implies that n n a + ix + c (−a − ix − c)k = ∑ an,k (−1)k yn (a + ix) = ∑ an,k k k! k=0 k=0 =
(ε ∓ δ + η )n (ε ∓ δ − η )n (n − 1 + 2ε )n −n, n − 1 + 2ε , ε ∓ δ + 1 − f − a − ix × 3 F2 ;1 ε ∓ δ + η,ε ∓ δ − η
for n = 0, 1, 2, . . .. Since δ = δ1 + iδ2 = η1 − iη2 = η , we have
136
6 Orthogonal Polynomial Solutions of Complex Difference Equations
±(δ + η ) = ±(δ + δ ) = ±2δ1
and
± (δ − η ) = ±(δ − δ ) = ±2iδ2 .
Hence by using (6.1.17), we now obtain −n, n − 1 + 2ε , ε /2 ∓ δ1 − i(x + f2 ± δ2 ) (ε ∓ 2δ1 )n (ε ∓ 2iδ2 )n F ; 1 yn (x) = 3 2 (n − 1 + 2ε )n in ε ∓ 2δ1 , ε ∓ 2iδ2 for n = 0, 1, 2, . . .. These polynomials are called continuous Hahn polynomials. The Pearson difference equation (6.1.20) reads z2 + 2( f − ε )z + g − γ (z + f − ε + δ )(z + f − ε − δ ) w(z) = = . 2 w(z + 1) (z − 1) + 2 f (z − 1) + g (z − 1 + f + η )(z − 1 + f − η ) Case IIIa. ε > 0 and 2δ1 < ε . In this case we use w(z) = Γ(z − 1 + f + η )Γ(z − 1 + f − η )Γ(1 + ε − δ − f − z)Γ(1 + ε + δ − f − z) as a possible solution for the Pearson difference equation. First we set z = a + ix and use η = δ and (6.1.17) to obtain w(a + ix) = Γ(a + ix − 1 + f + δ )Γ(a + ix − 1 + f − δ ) × Γ(1 + ε − δ − f − a − ix)Γ(1 + ε + δ − f − a − ix) = Γ(ε /2 + δ1 + i(x + f2 − δ2 ))Γ(ε /2 − δ1 + i(x + f2 + δ2 )) × Γ(ε /2 − δ1 − i(x + f2 + δ2 ))Γ(ε /2 + δ1 − i(x + f2 − δ2 )) = |Γ(ε /2 + δ1 + i(x + f2 − δ2 ))|2 |Γ(ε /2 − δ1 + i(x + f2 + δ2 ))|2 . This leads to the Rodrigues formula yn (x) =
1 (n − 1 + 2ε )n ×
1
|Γ(ε /2 + δ1 + i(x + f2 − δ2 ))| |Γ(ε /2 − δ1 + i(x + f2 + δ2 ))|2 × Δ n (Γ(ε /2 + δ1 + i(x + f2 − δ2 ))Γ(ε /2 − δ1 + i(x + f2 + δ2 )) 2
× Γ(n + ε /2 + δ1 − i(x + f2 + δ2 )) × Γ(n + ε /2 − δ1 − i(x + f2 − δ2 ))) for n = 0, 1, 2, . . .. Further we obtain by using the Mellin-Barnes integral (1.6.3) that
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
1 2π
∞
137
−∞
i∞ 1 w(a + s) ds 2π i −i∞ Γ(ε − δ + η )Γ(ε + δ + η )Γ(ε − δ − η )Γ(ε + δ − η ) = Γ(2ε ) Γ(ε + 2δ1 )Γ(ε − 2δ1 )Γ(ε + 2iδ2 )Γ(ε − 2iδ2 ) , = Γ(2ε )
w(a + ix) dx =
which implies that d0 :=
1 2π
∞ −∞
w(a + ix) dx =
Γ(ε + 2δ1 )Γ(ε − 2δ1 ) |Γ(ε + 2iδ2 )|2 > 0, Γ(2ε )
since 0 < 2δ1 < ε . This leads to the orthogonality relation 1 2π
∞ −∞
|Γ(ε /2 + δ1 + i(x + f2 − δ2 ))|2 × |Γ(ε /2 − δ1 + i(x + f2 + δ2 ))|2 ym (x) yn (x) dx
=
Γ(n − 1 + 2ε )Γ(n + ε + 2δ1 )Γ(n + ε − 2δ1 ) |Γ(n + ε + 2iδ2 )|2 n! δmn Γ(2n − 1 + 2ε )Γ(2n + 2ε )
for m, n = 0, 1, 2, . . .. Case IIIb. We have 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 and 2δ1 > −ε . In this case we use w(z) =
Γ(z − 1 + f + η )Γ(1 + ε + δ − f − z) Γ(z + f − ε + δ )Γ(2 + η − f − z)
as a possible solution for the Pearson difference equation. As before, we set z = a + ix and use η = δ and (6.1.17) to obtain w(a + ix) =
Γ(a + ix − 1 + f + δ )Γ(1 + ε + δ − f − a − ix)
Γ(a + ix + f − ε + δ )Γ(2 + δ − f − a − ix) Γ(ε /2 + δ1 + i(x + f2 − δ2 ))Γ(ε /2 + δ1 − i(x + f2 − δ2 )) = Γ(1 − ε /2 + δ1 + i(x + f2 + δ2 ))Γ(1 − ε /2 + δ1 − i(x + f2 + δ2 )) Γ(ε /2 + δ1 + i(x + f2 − δ2 )) 2 . = Γ(1 − ε /2 + δ1 + i(x + f2 + δ2 ))
This leads to the Rodrigues formula
138
6 Orthogonal Polynomial Solutions of Complex Difference Equations
yn (x) =
Γ(1 − ε /2 + δ1 + i(x + f2 + δ2 )) 2 (−1)n (n − 1 + 2ε )n Γ(ε /2 + δ1 + i(x + f2 − δ2 )) Γ(ε /2 + δ1 + i(x + f2 − δ2 )) ×Δn Γ(1 − ε /2 + δ1 − n + i(x + f2 − δ2 )) Γ(n + ε /2 + δ1 − i(x + f2 − δ2 )) × Γ(1 − ε /2 + δ1 − i(x + f2 − δ2 ))
for n = 0, 1, 2, . . . , N. Now, by using (1.6.9), we obtain 1 2π
∞ −∞
w(a + ix) dx =
1 2π i
i∞ −i∞
w(a + s) ds
Γ(ε + δ + η )Γ(1 − 2ε ) Γ(1 − ε + δ + η )Γ(1 − ε + δ − η )Γ(1 − ε − δ + η ) Γ(ε + 2δ1 )Γ(1 − 2ε ) , = Γ(1 − ε + 2δ1 )Γ(1 − ε + 2iδ2 )Γ(1 − ε − 2iδ2 ) =
which implies that d0 :=
1 2π
∞ −∞
w(a + ix) dx =
Γ(ε + 2δ1 )Γ(1 − 2ε ) Γ(1 − ε + 2δ1 ) |Γ(1 − ε + 2iδ2 )|2
> 0,
since 0 < −ε < 2δ1 . This leads to the orthogonality relation 1 2π =
2
∞ Γ(ε /2 + δ1 + i(x + f2 − δ2 )) −∞
yn (x) dx Γ(1 − ε /2 + δ1 + i(x + f2 + δ2 )) ym (x)
Γ(2 − 2ε − 2n)Γ(1 − 2ε − 2n)Γ(n + ε + 2δ1 )n! Γ(2 − 2ε − n)Γ(1 − ε + 2δ1 − n) |Γ(1 − ε − n + 2iδ2 )|2
for m, n = 0, 1, 2, . . . , N.
δmn
6.3 Properties of the Positive-Definite Orthogonal Polynomial Solutions
139
In this chapter we have proved:
Theorem 6.1. The positive-definite orthogonal polynomial solutions yn (x) of the difference equation (5.1.1) (ex2 + 2 f x + g) Δ 2 yn (x) + (2ε x + γ ) (Δ yn ) (x) = n(e(n − 1) + 2ε )yn (x + 1) for n = 0, 1, 2, . . . with e, f , g, ε , γ ∈ C consist of two infinite systems Case II. Meixner-Pollaczek polynomials with orthogonality on (−∞, ∞) with respect to w(x) = |Γ(g1 + a − 1 + i(x + g2 )|2 eϕ (x) ; e = 0, 2 f = 1, ε1 > 0, g1 > 1 − a and 2ε1 − 1 + 2iε2 = eiϕ Case IIIa. Continuous Hahn polynomials with orthogonality on (−∞, ∞) with respect to w(x) = |Γ(ε /2 + δ1 + i(x + f2 − δ2 ))Γ(ε /2 − δ1 + i(x + f2 + δ2 ))|2 ; e = 1, ε > 0 and 2δ1 < ε and one finite system Case IIIb. Continuous Hahn polynomials with orthogonality on (−∞, ∞) with respect to w(x) = |Γ(ε /2 + δ1 + i(x + f2 − δ2 ))/Γ(1 − ε /2 + δ1 + i(x + f2 + δ2 ))|2 ; e = 1, 2ε = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 and 2δ1 > −ε .
Chapter 7
Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations Discrete Classical Orthogonal Polynomials III
7.1 Motivation for Polynomials in x(x + u) Through Duality In the definition 3.1 we introduced the sequences {κn }Nn=0 with κm = κn for m = n and {λn }Nn=0 with λm = λn for m = n of eigenvalues and the sequences {yn }Nn=0 with degree[yn ] = n and {zn }Nn=0 with degree[zn ] = n of polynomials where N ∈ {1, 2, 3, . . .} or N → ∞. The polynomials yn (x) and zn (x) are called dual polynomials with respect to the sequences of eigenvalues {κn }Nn=0 and {λn }Nn=0 when yn (κm ) = zm (λn ) for all m, n = 0, 1, 2, . . . , N. In section 5.3 we obtained the difference equation C(x)yn (x + 1) − {C(x) + D(x)} yn (x) + D(x)yn (x − 1) = λn yn (x)
(7.1.1)
for n = 0, 1, 2, . . . , N − 1 where λn = n(n − 1 + 2ε ), C(x) = (x − 1 + f + η )(x − 1 + f − η ) and D(x) = (x − 1 + f − ε + δ )(x − 1 + f − ε − δ ) with
δ=
( f − ε )2 − g + γ
and η =
f2 −g
for the Hahn polynomials. If the regularity condition (2.3.3) holds, all eigenvalues are different. This implies because of theorem 3.6 that there exists a sequence of dual polynomials. In this case we have λn = n(n − 1 + 2ε ) and κm = m since q = ω = 1 and x0 = 0. If the conditions (D(0) =) 1 − 2 f + g + 2ε − γ = 0 and
(C(n) =) (n − 1)2 + 2 f (n − 1) + g = 0
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 7, © Springer-Verlag Berlin Heidelberg 2010
141
142
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
hold, the dual polynomials {zm }Nm=0 satisfy the three-term recurrence relation C(m)zm+1 (x) − {C(m) + D(m)} zm (x) + D(m)zm−1 (x) = xzm (x)
(7.1.2)
for m = 0, 1, 2, . . . , N − 1 with the convention that z−1 (x) := 0. If we set x = κm = m into (7.1.1), we find that C(m)yn (m + 1) − {C(m) + D(m)} yn (m) + D(m)yn (m − 1) = λn yn (m)
(7.1.3)
for n = 0, 1, 2, . . . , N − 1. Since yn (κm ) = yn (m) = zm (λn ) for all m, n = 0, 1, 2, . . . , N, this implies that for the Hahn polynomials there exist dual polynomials with argument λn = n(n − 1 + 2ε ). This motivates the study of orthogonal polynomials in x(x + u) with x a real variable and u ∈ R a constant.
7.2 Difference Equations Having Real Polynomial Solutions with Argument x(x + u) In order to build a theory, we must first classify all second-order difference equations with real coefficients which have real polynomial solutions {yn (x(x + u))}Nn=0 with degree[yn ] = n and N ∈ {1, 2, 3, . . .} or N → ∞. Therefore we consider the difference equation
ϕ (x + 2)Δ 2 yn (x(x + u)) + ψ (x + 1)Δ yn (x(x + u)) = λn ρ (x + 1)yn ((x + 1)(x + 1 + u))
(7.2.1)
for n = 0, 1, 2, . . ., where Δ yn (x(x + u)) = yn ((x + 1)(x + 1 + u)) − yn (x(x + u)) and Δ 2 yn (x(x + u)) = Δ (Δ yn (x(x + u))). By replacing x by x − 1, we can write this in the symmetric form C(x)yn ((x + 1)(x + 1 + u)) − {C(x) + D(x)} yn (x(x + u)) + D(x)yn ((x − 1)(x − 1 + u)) = λn ρ (x)yn (x(x + u)) with
(7.2.2)
C(x) = ϕ (x + 1) and D(x) = ϕ (x + 1) − ψ (x).
Now we look for eigenvalues λn and coefficients C(x), D(x) and ρ (x) so that for each eigenvalue λn there exists exactly one polynomial solution yn (x(x + u)) with degree[yn ] = n in x(x + u) up to a constant factor. Since (x(x + u))n can be expressed as a linear combination of (x − k + 1)k (x + u)k for k = 0, 1, 2, . . . , n, we set n
yn (x(x + u)) =
∑ an,k
k=0
Then we have
(x − k + 1)k (x + u)k , k!
an,n = 0,
n = 0, 1, 2, . . . .
(7.2.3)
7.2 Difference Equations Having Real Polynomial Solutions with Argument x(x + u) n
Δ yn (x(x + u)) = (2x + 1 + u) ∑ an,k k=1
143
(x − k + 2)k−1 (x + 1 + u)k−1 . (k − 1)!
This leads to the first simplification C(x) = (2x − 1 + u)C∗ (x), D(x) = (2x + 1 + u)D∗ (x) and ρ (x) = (2x − 1 + u)(2x + 1 + u)ρ ∗ (x).
(7.2.4)
For the moment we assume that 2x = −u ± 1. Then we have n
C∗ (x) ∑ an,k k=1
(x − k + 2)k−1 (x + 1 + u)k−1 (k − 1)! n
− D∗ (x) ∑ an,k k=1
(x − k + 1)k−1 (x + u)k−1 = λn ρ ∗ (x)yn (x(x + u)). (k − 1)!
For a second simplification we note that (x − k + 2)k−1 (x + 1 + u)k−1 − (x − k + 1)k−1 (x + u)k−1 = (k − 1)(2x + u)(x − k + 2)k−2 (x + 1 + u)k−2 for k = 2, 3, 4, . . .. Now we define C∗ (x) − D∗ (x) = (2x + u)B(x) and ρ ∗ (x) = (2x + u)r(x)
(7.2.5)
with the assumption that 2x = −u. Without loss of generality we may choose r(x) = 1 so that we have n
B(x) ∑ an,k k=1 n
n (x − k + 1)k−1 (x + u)k−1 (x − k + 2)k−2 (x + 1 + u)k−2 +C∗ (x) ∑ an,k (k − 1)! (k − 2)! k=2
= λn ∑ an,k k=0
(x − k + 1)k (x + u)k k!
(7.2.6)
for n = 2, 3, 4, . . .. For n = 0 we have y0 (x(x + u)) = a0,0 (= 0) which, if (7.2.1) is used, leads to λ0 = 0 (except for the trivial situation that r(x) = 0). For n = 1 we have y1 (x(x + u)) = a1,0 + a1,1 x(x + u) with a1,1 = 0, which leads to B(x)a1,1 = λ1 {a1,0 + a1,1 x(x + u)} . Since all eigenvalues must be different, we conclude that λ1 = 0 (= λ0 ). Hence B(x) = v + wx(x + u) with v, w ∈ R,
w = 0.
(7.2.7)
This form can be used as a third simplification. For the first term on the left-hand side of (7.2.6) we obtain
144
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
n
∑ wkan,k
k=0
n (x − k + 1)k (x + u)k (x − k + 2)k−1 (x + u)k−1 + ∑ v + w(k − 1)2 an,k k! (k − 1)! k=1
n
− ∑ (v + wx2 )(x + u)an,k k=2
(x − k + 2)k−2 (x + 1 + u)k−2 . (k − 2)!
(7.2.8)
This implies that C∗ (x) must be of the form C∗ (x) = (v + wx2 )(x + u) + σ x(x + u) + τ x(x − 1)(x + u)(x + 1 + u) = (x + u) v + σ x + wx2 + τ x(x − 1)(x + 1 + u) , σ , τ ∈ R. (7.2.9) This leads to the following theorem. Theorem 7.1. The difference equation (7.2.1) only has real polynomial solutions yn (x(x + u)) with degree[yn ] = n in x(x + u) for n = 0, 1, 2, . . . if the coefficients C(x), D(x) and ρ (x) have the form C(x) = ϕ (x + 1) = (x + u)(2x − 1 + u) v + σ x + wx2 + τ x(x − 1)(x + 1 + u) , D(x) = ϕ (x + 1) − ψ (x) = x(2x + 1 + u) −v + σ (x + u) − w(x + u)2 + τ (x − 1)(x + u)(x + 1 + u) and
ρ (x) = (2x − 1 + u)(2x + u)(2x + 1 + u)
with u, v, w, σ , τ ∈ R and w = 0. Note that the assumption that 2x ∈ / {−u − 1, −u, −u + 1} can be dropped.
7.3 The Hypergeometric Representation In order to find the hypergeometric representation of the polynomials in the form (7.2.3), we use (7.2.7), (7.2.9) and (7.2.8) to obtain from (7.2.6) n
∑ {w + (k − 1)τ } an,k
k=0
(x − k + 1)k (x + u)k k k!
n
+ ∑ {v + (k − 1) (σ + (k − 1)w + (k − 2)(u + k)τ )} k=1
(x − k + 2)k−1 (x + u)k−1 (k − 1)! n (x − k + 1)k (x + u)k , n = 1, 2, 3, . . . . = λn ∑ an,k k! k=0 × an,k
By comparing the coefficients of (x − k + 1)k (x + u)k on both sides, we find that
7.3 The Hypergeometric Representation
145
k {w + (k − 1)τ } an,k + {v + k (σ + kw + (k − 1)(u + k + 1)τ )} an,k+1 = λn an,k , which holds for k = 0, 1, 2, . . . , n if we define an,n+1 := 0. This leads to the eigenvalues λn = n ((n − 1)τ + w) , n = 0, 1, 2, . . . (7.3.1) and the two-term recurrence relation (n − k) {(n + k − 1)τ + w} an,k = {v + k (σ + kw + (k − 1)(u + k + 1)τ )} an,k+1
(7.3.2)
for k = n − 1, n − 2, n − 3, . . . , 0. Hence the coefficients {an,k }nk=0 in (7.2.3) are uniquely determined in terms of an,n = 0 if (n + k − 1)τ + w = 0
(7.3.3)
for k = n − 1, n − 2, n − 3, . . . , 0 and n ∈ {1, 2, 3, . . .}. This condition holds if the eigenvalues in (7.3.1) are all different. In the sequel we will always assume that this holds. Since w = 0, we only have to consider two different cases. Case I. τ = 0 and w = 1. Then we write C∗ (x) = (x + u)(x2 + σ x + v) = (x + u)(x + x1 )(x + x2 ),
σ = x1 + x2 ,
v = x 1 x2 .
Then we use (7.2.5) and (7.2.7) to show that D∗ (x) = C∗ (x) − (2x + u)B(x) = (x + u)(x + x1 )(x + x2 ) − (2x + u) (x1 x2 + x(x + u)) = −x(x + u − x1 )(x + u − x2 ). Hence, if (7.2.4), (7.2.5) and (7.3.1) are used, the difference equation (7.2.1) reads (x + 1 + u)(2x + 1 + u)(x + 1 + x1 )(x + 1 + x2 )Δ 2 yn (x(x + u)) + {(x + 1 + u)(2x + 1 + u)(x + 1 + x1 )(x + 1 + x2 ) +(x + 1)(2x + 3 + u) ×(x + 1 + u − x1 )(x + 1 + u − x2 )} Δ yn (x(x + u)) (7.3.4) = n(2x + 1 + u)(2x + 2 + u)(2x + 3 + u)yn ((x + 1)(x + 1 + u)) for n = 0, 1, 2, . . ., and the symmetric form (7.2.2) reads
146
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
(x + u)(2x − 1 + u)(x + x1 )(x + x2 )yn ((x + 1)(x + 1 + u)) − {(x + u)(2x − 1 + u)(x + x1 )(x + x2 ) − x(2x + 1 + u)(x + u − x1 )(x + u − x2 )} yn (x(x + u)) − x(2x + 1 + u)(x + u − x1 )(x + u − x2 )yn ((x − 1)(x − 1 + u)) = n(2x − 1 + u)(2x + u)(2x + 1 + u)yn (x(x + u)),
n = 0, 1, 2, . . . . (7.3.5)
For the two-term recurrence relation (7.3.2) we obtain (n − k)an,k = (k + x1 )(k + x2 )an,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
This implies that for the monic polynomials, id est an,n = n!, we have an,k =
(x1 )n (x2 )n (k + x1 )n−k (k + x2 )n−k n! = (−1)k (−n)k , (n − k)! (x1 )k (x2 )k
k = 0, 1, 2, . . . , n.
By using (7.2.3) and (−1)k (x − k + 1)k = (−x)k , we obtain the representation n
(−n)k (−x)k (x + u)k (x1 )k (x2 )k k! k=0 −n, −x, x + u ;1 , = (x1 )n (x2 )n 3 F2 x1 , x2
yn (x(x + u)) = (x1 )n (x2 )n ∑
n = 0, 1, 2, . . . (7.3.6)
for the dual Hahn polynomials. This is explained by the observation that the (not normalized) Hahn polynomials can be written as (cf. section 5.3) −n, n − 1 + 2ε , ε − δ + 1 − f − x ∗ ; 1 , n = 0, 1, 2, . . . . yn (x) = 3 F2 ε − δ + η,ε − δ − η If we now set f − ε + δ = 1, ε − δ + η = x1 , ε − δ − η = x2 and u = 2ε − 1, then we obtain −n, n + u, −x ; 1 , n = 0, 1, 2, . . . . y∗n (x) = 3 F2 x1 , x2 For the (not normalized) polynomials in (7.3.6) we have −n, −x, x + u ; 1 , n = 0, 1, 2, . . . . z∗n (x(x + u)) = 3 F2 x1 , x2 Apparently we have y∗n (x) = z∗x (n(n + u)) for n, x = 0, 1, 2, . . .; {y∗n (x)} and {z∗n (x(x + u))} are dual with respect to the sequences of eigenvalues {κn } with κn = n and {λn } with λn = n(n + u). Note that we also have yn (κ0 ) = z0 (λn ). Case II. τ = 1. Then we write C∗ (x) = (x + u) x3 + (u + w)x2 + (σ − u − 1)x + v = (x + u)(x + x1 )(x + x2 )(x + x3 )
7.3 The Hypergeometric Representation
147
where w = x1 + x2 + x3 − u,
σ = x1 x2 + x1 x3 + x2 x3 + u + 1 and v = x1 x2 x3 .
Then we also obtain by using (7.2.5) and (7.2.7) D∗ (x) = C∗ (x) − (2x + u)B(x) = (x + u)(x + x1 )(x + x2 )(x + x3 ) − (2x + u) (x1 x2 x3 + x(x + u)(x1 + x2 + x3 − u)) = x(x + u − x1 )(x + u − x2 )(x + u − x3 ). Hence, if (7.2.4), (7.2.5) and (7.3.1) are used, the difference equation (7.2.1) reads (x + 1 + u)(2x + 1 + u)(x + 1 + x1 )(x + 1 + x2 )(x + 1 + x3 )Δ 2 yn (x(x + u)) + {(x + 1 + u)(2x + 1 + u)(x + 1 + x1 )(x + 1 + x2 )(x + 1 + x3 ) −(x + 1)(2x + 3 + u)(x + 1 + u − x1 ) ×(x + 1 + u − x2 )(x + 1 + u − x3 )} Δ yn (x(x + u)) = n(n − 1 + x1 + x2 + x3 − u) × (2x + 1 + u)(2x + 2 + u)(2x + 3 + u)yn ((x + 1)(x + 1 + u))
(7.3.7)
and the symmetric form (7.2.2) reads (x + u)(2x − 1 + u)(x + x1 )(x + x2 )(x + x3 )yn ((x + 1)(x + 1 + u)) − {(x + u)(2x − 1 + u)(x + x1 )(x + x2 )(x + x3 ) + x(2x + 1 + u)(x + u − x1 )(x + u − x2 )(x + u − x3 )} yn (x(x + u)) + x(2x + 1 + u)(x + u − x1 )(x + u − x2 )(x + u − x3 )yn ((x − 1)(x − 1 + u)) = n(n − 1 + x1 + x2 + x3 − u)(2x − 1 + u)(2x + u)(2x + 1 + u)yn (x(x + u)) (7.3.8) for n = 0, 1, 2, . . .. For the two-term recurrence relation (7.3.2) we obtain (n − k)(n + k − 1 + x1 + x2 + x3 − u)an,k = (k + x1 )(k + x2 )(k + x3 )an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. This implies that for the monic polynomials, id est an,n = n!, we have (k + x1 )n−k (k + x2 )n−k (k + x3 )n−k n! (n + k − 1 + x1 + x2 + x3 − u)n−k (n − k)! (x1 )n (x2 )n (x3 )n (n − 1 + x1 + x2 + x3 − u)k = (−1)k (−n)k (n − 1 + x1 + x2 + x3 − u)n (x1 )k (x2 )k (x3 )k
an,k =
for k = 0, 1, 2, . . . , n. If (7.2.3) and (−1)k (x − k + 1)k = (−x)k are used, this leads for n = 0, 1, 2, . . . to the representation
148
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
yn (x(x + u)) =
=
(x1 )n (x2 )n (x3 )n (n − 1 + x1 + x2 + x3 − u)n n (−n)k (n − 1 + x1 + x2 + x3 − u)k (−x)k (x + u)k ×∑ (x1 )k (x2 )k (x3 )k k! k=0 (x1 )n (x2 )n (x3 )n (n − 1 + x1 + x2 + x3 − u)n −n, n − 1 + x1 + x2 + x3 − u, −x, x + u × 4 F3 ; 1 (7.3.9) x1 , x2 , x3
for the Racah polynomials. These polynomials have a certain symmetry in n and x which can be seen as follows. The (not normalized) Racah polynomials given by (7.3.9) can be written as −n, n − 1 + x1 + x2 + x3 − u, −x, x + u ∗ yn (x(x + u)) = 4 F3 ;1 (7.3.10) x1 , x2 , x3 for n = 0, 1, 2, . . .. If we replace u by −1 + x1 + x2 + x3 − u, then we have z∗n (x(x − 1 + x1 + x2 + x3 − u)) −n, n + u, −x, x − 1 + x1 + x2 + x3 − u ;1 = 4 F3 x1 , x2 , x3
(7.3.11)
for n = 0, 1, 2, . . .. Now we have y∗n (x(x + u)) = z∗x (n(n − 1 + x1 + x2 + x3 − u)) for n, x = 0, 1, 2, . . . and {y∗n (x(x + u))} and {z∗n (x(x − 1 + x1 + x2 + x3 − u))} are dual polynomial systems with respect to the sequences of eigenvalues {κn } with κn = n(n + u) and {λn } with λn = n(n − 1 + x1 + x2 + x3 − u). Note that we also have y∗n (κ0 ) = z∗0 (λn ). The polynomials z∗n (x(x − 1 + x1 + x2 + x3 − u)) can be considered as dual Racah polynomials. This fact will be used in the next section.
7.4 The Three-Term Recurrence Relation In order to obtain the three-term recurrence relation, we use the concept of duality. Again we consider the two different cases. Case I. For the dual Hahn polynomials we start with the difference equation (7.1.1) for the Hahn polynomials. The coefficients of this difference equation give us the coefficients of the three-term recurrence relation for the dual Hahn polynomials. We have C(x) = (x − 1 + f + η )(x − 1 + f − η ) = (x + x1 )(x + x2 ) and D(x) = (x − 1 + f − ε + δ )(x − 1 + f − ε − δ ) = x(x − 1 + x1 + x2 − u)
7.4 The Three-Term Recurrence Relation
149
with f − ε + δ = 1,
x1 = f − 1 + η ,
x2 = f − 1 − η
and u = 2ε − 1.
This implies that the three-term recurrence relation for the dual Hahn polynomials zn (λx ) can be written as (n + x1 )(n + x2 )zn+1 (λx ) − {(n + x1 )(n + x2 ) + n(n − 1 + x1 + x2 − u)} zn (λx ) + n(n − 1 + x1 + x2 − u)zn−1 (λx ) = λx zn (λx ), n = 1, 2, 3, . . . with z0 (λx ) = 1, z1 (λx ) = 1 + λx /x1 x2 (x1 x2 = 0) and λx = x(x + u). For the monic dual Hahn polynomials yn (x(x + u)) this can be written as yn+1 (x(x + u)) = {x(x + u) + (n + x1 )(n + x2 ) +n(n − 1 + x1 + x2 − u)} yn (x(x + u)) − n(n − 1 + x1 )(n − 1 + x2 ) × (n − 1 + x1 + x2 − u)yn−1 (x(x + u))
(7.4.1)
for n = 1, 2, 3, . . . with y0 (x(x + u)) = 1 and y1 (x(x + u)) = x(x + u) + x1 x2 . Case II. For the Racah polynomials we start with the difference equation (cf. (7.2.2)) for the (not normalized) Racah polynomials y∗n (x(x + u)) given by (7.3.10): C(x)y∗n (κx+1 ) − {C(x) + D(x)} y∗n (κx ) + D(x)y∗n (κx−1 ) = λn ρ (x)y∗n (κx ). Now we have y∗n (κx ) = z∗x (λn ) with κx = x(x+u) and λn = n(n−1+x1 +x2 +x3 −u), which implies that the difference equation for the dual Racah polynomials z∗x (λn ) given by (7.3.11) can be written as C(x)z∗x+1 (λn ) − {C(x) + D(x)} z∗x (λn ) + D(x)z∗x−1 (λn ) = λn ρ (x)z∗x (λn ). For the coefficients we now have C(x) = (x + u)(2x − 1 + u)(x + x1 )(x + x2 )(x + x3 ) and D(x) = x(2x + 1 + u)(x + u − x1 )(x + u − x2 )(x + u − x3 ) which leads to the three-term recurrence relation (n + u)(2n − 1 + u)(n + x1 )(n + x2 )(n + x3 )z∗n+1 (λx ) − {(n + u)(2n − 1 + u)(n + x1 )(n + x2 )(n + x3 ) + n(2n + 1 + u)(n + u − x1 )(n + u − x2 )(n + u − x3 )} z∗n (λx ) + n(2n + 1 + u)(n + u − x1 )(n + u − x2 )(n + u − x3 )z∗n−1 (λx ) = λx (2n − 1 + u)(2n + u)(2n + 1 + u)z∗n (λx ), n = 1, 2, 3, . . .
150
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
with z∗0 (λx ) = 1 and z∗1 (λx ) = 1 + (u + 1)λx /x1 x2 x3 (x1 x2 x3 = 0). When u is replaced by −1 + x1 + x2 + x3 − u (= −1 + w), the polynomials z∗n (λx ) given by (7.3.11) change into the polynomials y∗n (κx ) given by (7.3.10). This implies that the three-term recurrence relation for the polynomials y∗n (κx ) can be written as (n − 1 + w)(2n − 2 + w)(n + x1 )(n + x2 )(n + x3 )y∗n+1 (κx ) − {(n − 1 + w)(2n − 2 + w)(n + x1 )(n + x2 )(n + x3 ) + n(2n + w)(n − 1 + w − x1 ) ×(n − 1 + w − x2 )(n − 1 + w − x3 )} y∗n (κx ) + n(2n + w)(n − 1 + w − x1 )(n − 1 + w − x2 )(n − 1 + w − x3 )y∗n−1 (κx ) = κx (2n − 2 + w)(2n − 1 + w)(2n + w)y∗n (κx ), n = 1, 2, 3, . . . with y∗0 (κx ) = 1, y∗1 (κx ) = 1 + wκx /x1 x2 x3 (x1 x2 x3 = 0) and κx = x(x + u). The connection with the monic Racah polynomials yn (κx ) is given by y∗n (κx ) =
(n − 1 + w)n yn (κx ), (x1 )n (x2 )n (x3 )n
n = 0, 1, 2, . . . .
Hence the three-term recurrence relation for the monic Racah polynomials yn (κx ) can be written in the form (1) (2) (1) (2) yn+1 (κx ) = κx + cn + cn yn (κx ) − cn−1 cn yn−1 (κx ), n = 1, 2, 3, . . . (7.4.2) with y0 (κx ) = 1 and y1 (x) = κx + x1 x2 x3 /w, where κx = x(x + u) and w = x1 + x2 + x3 − u with w = 0, −1, −2, . . ., (1)
cn =
(n − 1 + w)(n + x1 )(n + x2 )(n + x3 ) , (2n − 1 + w)(2n + w)
n = 0, 1, 2, . . .
(7.4.3)
and (2)
cn =
n(n − 1 + w − x1 )(n − 1 + w − x2 )(n − 1 + w − x3 ) , (2n − 2 + w)(2n − 1 + w)
n = 1, 2, 3, . . . . (7.4.4)
7.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions Favard’s theorem (theorem 3.1) can be extended for monic polynomials yn (x(x + u)) of degree n ∈ {0, 1, 2, . . .} with u ∈ R. The polynomials given by yn+1 (x(x + u)) = {x(x + u) − cn } yn (x(x + u)) − dn yn−1 (x(x + u))
(7.5.1)
for n = 1, 2, 3, . . . with cn , dn ∈ R, y0 (x(x + u)) = 1 and y1 (x(x + u)) = x(x + u) − c0 (c0 ∈ R) are orthogonal with respect to a positive-definite linear functional Λ , id est
7.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions
Λ [ym (x(x + u))yn (x(x + u))] =
n
∏ dk
151
δmn ,
m, n = 0, 1, 2, . . . ,
(7.5.2)
k=0
where Λ [y0 (x(x + u))] = d0 ∈ R and Λ [yn (x(x + u))] = 0 for n = 1, 2, 3, . . . iff d0 , d1 , d2 , . . . , dn are positive. The proof is similar to the proof of theorem 3.1. Case I. For the monic dual Hahn polynomials given by (7.3.6) we have the threeterm recurrence relation (7.4.1) with cn = −(n + x1 )(n + x2 ) − n(n − 1 + x1 + x2 − u),
n = 0, 1, 2, . . .
and dn = n(n − 1 + x1 )(n − 1 + x2 )(n − 1 + x1 + x2 − u),
n = 1, 2, 3, . . . ,
u ∈ R.
Since u ∈ R, we have cn ∈ R for all n = 0, 1, 2, . . . if x1 x2 ∈ R and x1 + x2 ∈ R. This implies that x1 , x2 ∈ R or x1 and x2 are complex conjugates. Positive-definite orthogonality occurs in six infinite and seven finite cases given by table 7.1, where x1 and x2 can be interchanged because of the symmetry.
case
conditions
positive
1 x1,2 > 0, x1 + x2 − u > 0 d1 , d2 , d3 , . . . 2 x1,2 = α ± iβ (α , β ∈ R, β = 0), 2α − u > 0 d1 , d2 , d3 , . . . 3 x1 > 0, −N < x2 < −N + 1, −N < x1 + x2 − u < −N + 1 d1 , d2 , d3 , . . . 4 x1 > 0, x2 ≤ −N, −N < x1 + x2 − u < −N + 1 d1 , d2 , d3 , . . . , dN 5 x1 > 0, −N < x2 < −N + 1, x1 + x2 − u ≤ −N d1 , d2 , d3 , . . . , dN 6 x1 > 0, x2 ≤ −N, x1 + x2 − u ≤ −N −N − j < x2 ≤ −N − j + 1, j ∈ {1, 2, 3, . . .} −N − k < x1 + x2 − u ≤ −N − k + 1, k ∈ {1, 2, 3, . . .} a j = k with x2 = −N − j + 1 and x1 + x2 − u = −N − j + 1 d1 , d2 , d3 , . . . b j = k with x2 = −N − j + 1 or x1 + x2 − u = −N − j + 1 d1 , d2 , d3 , . . . , dN+ j−1 c j = k d1 , d2 , d3 , . . . , dN+min( j,k)−1 7 −N < x1,2 < −N + 1, x1 + x2 − u > 0 d1 , d2 , d3 , . . . 8 −N < x1 < −N + 1, x2 ≤ −N, x1 + x2 − u > 0 d1 , d2 , d3 , . . . , dN 9 x1,2 ≤ −N, x1 + x2 − u > 0 −N − j < x1 ≤ −N − j + 1, j ∈ {1, 2, 3, . . .} −N − k < x2 ≤ −N − k + 1, k ∈ {1, 2, 3, . . .} a j = k with x1 = −N − j + 1 and x2 = −N − j + 1 d1 , d2 , d3 , . . . b j = k with x1 = −N − j + 1 or x2 = −N − j + 1 d1 , d2 , d3 , . . . , dN+ j−1 c j = k d1 , d2 , d3 , . . . , dN+min( j,k)−1 Table 7.1 Case I, N ∈ {1, 2, 3, . . .}.
152
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
Case II. For the monic Racah polynomials given by (7.3.9) we have the three-term recurrence relation (7.4.2) with (1)
c0 = −c0 ,
(1)
(2)
cn = −cn − cn ,
n = 1, 2, 3, . . .
and (1)
(2)
dn = cn−1 cn =
n(n − 2 + w) Dn , (2n − 3 + w)(2n − 2 + w)2 (2n − 1 + w)
n = 1, 2, 3, . . .
and Dn = (n − 1 + x1 )(n − 1 + x2 )(n − 1 + x3 ) × (n − 1 + w − x1 )(n − 1 + w − x2 )(n − 1 + w − x3 )
(7.5.3)
for n = 1, 2, 3, . . ., where w ∈ R. Since w ∈ R, we have cn ∈ R for all n = 0, 1, 2, . . . if x1 x2 x3 ∈ R, x1 + x2 + x3 ∈ R and x1 x2 + x1 x3 + x2 x3 ∈ R. This implies that x1 , x2 , x3 ∈ R or one is real and the other two are complex conjugates. Note the similarity with case III in section 4.2 with 2ε replaced by w. As in section 4.2 we conclude that for w > 0 we have n(n − 2 + w) > 0, (2n − 3 + w)(2n − 2 + w)2 (2n − 1 + w)
n = 1, 2, 3, . . .
and for w = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1 we have n(n − 2 + w) < 0, (2n − 3 + w)(2n − 2 + w)2 (2n − 1 + w)
n = 1, 2, 3, . . . , N
and this does no longer hold for n = N + 1. In this case the range −1 ≤ w < 0 must also be considered since this might also lead to positive-definite orthogonality. For w = −1 this is impossible since d1 is not defined in that case. For −1 < w < 0 we have n(n − 2 + w) > 0, n = 1, 3, 4, . . . (2n − 3 + w)(2n − 2 + w)2 (2n − 1 + w) and for n = 2
2w < 0. (w + 1)(w + 2)2 (w + 3)
Case IIa. w > 0. Then we must have Dn > 0 for n = 1, 2, 3, . . .. For w > 0 and xi < 0 for i = 1, 2, 3 we have w − xi > 0 for i = 1, 2, 3. However, w > 0 and w − xi < 0 for i = 1, 2, 3 implies that xi > 0 for i = 1, 2, 3. Therefore, D1 can only be positive if all six factors in (7.5.3) with n = 1 are positive or four are positive and two are negative. This observation leads to positive-definite orthogonality in eight infinite and ten finite cases given by table 7.2.
7.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions
case 1 2 3 4 5
a b c 6 7 8
a b c 9 10 11 12
a b c
conditions
153
positive
0 < x1,2,3 < w d1 , d2 , d3 , . . . 0 < x1 < w, x2,3 = α ± iβ (α , β ∈ R, β = 0) d1 , d2 , d3 , . . . 0 < x1 < w, −N < x2,3 < −N + 1 d1 , d2 , d3 , . . . 0 < x1 < w, x2 ≤ −N, −N < x3 < −N + 1 d1 , d2 , d3 , . . . , dN 0 < x1 < w, x2,3 ≤ −N −N − j < x2 ≤ −N − j + 1, j ∈ {1, 2, 3, . . .} −N − k < x3 ≤ −N − k + 1, k ∈ {1, 2, 3, . . .} j = k with x2 = −N − j + 1 and x3 = −N − j + 1 d1 , d2 , d3 , . . . j = k with x2 = −N − j + 1 or x3 = −N − j + 1 d1 , d2 , d3 , . . . , dN+ j−1 j = k d1 , d2 , d3 , . . . , dN+min( j,k)−1 0 < x1 < w, −N < w − x2,3 < −N + 1 d1 , d2 , d3 , . . . 0 < x1 < w, −N < w − x2 < −N + 1, w − x3 ≤ −N d1 , d2 , d3 , . . . , dN 0 < x1 < w, w − x2,3 ≤ −N −N − j < w − x2 ≤ −N − j + 1, j ∈ {1, 2, 3, . . .} −N − k < w − x3 ≤ −N − k + 1, k ∈ {1, 2, 3, . . .} j = k with w − x2 = −N − j + 1 and w − x3 = −N − j + 1 d1 , d2 , d3 , . . . j = k with w − x2 = −N − j + 1 or w − x3 = −N − j + 1 d1 , d2 , d3 , . . . , dN+ j−1 j = k d1 , d2 , d3 , . . . , dN+min( j,k)−1 0 < x1 < w, −N < x2 < −N + 1, −N < w − x3 < −N + 1 d1 , d2 , d3 , . . . 0 < x1 < w, x2 ≤ −N, −N < w − x3 < −N + 1 d1 , d2 , d3 , . . . , dN 0 < x1 < w, −N < x2 < −N + 1, w − x3 ≤ −N d1 , d2 , d3 , . . . , dN 0 < x1 < w, x2 ≤ −N, w − x3 ≤ −N −N − j < x2 ≤ −N − j + 1, j ∈ {1, 2, 3, . . .} −N − k < w − x3 ≤ −N − k + 1, k ∈ {1, 2, 3, . . .} j = k with x2 = −N − j + 1 and w − x3 = −N − j + 1 d1 , d2 , d3 , . . . j = k with x2 = −N − j + 1 or w − x3 = −N − j + 1 d1 , d2 , d3 , . . . , dN+ j−1 j = k d1 , d2 , d3 , . . . , dN+min( j,k)−1 Table 7.2 Case IIa, N ∈ {1, 2, 3, . . .}.
Case IIb. −1 < w < 0. In this case we must have D2 < 0 and Dn > 0 for n = 1, 3, 4, . . .. For w < 0 and xi > 0 for i = 1, 2, 3 we have w − xi < 0 for i = 1, 2, 3. However, w < 0 and w − xi > 0 for i = 1, 2, 3 implies that xi < 0 for i = 1, 2, 3. Therefore, D1 can
case
conditions
1 w < x1 < 0, −2 < x2 < −1, −1 < x3 < w 2 w < x1 < 0, −2 < x2 < −1, 0 < x3 < w + 1 3 w < x1 < 0, w + 1 < x2 < w + 2, −1 < x3 < w 4 w < x1 < 0, w + 1 < x2 < w + 2, 0 < x3 < w + 1 Table 7.3 Case IIb.
positive d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . d1 , d2 , d3 , . . .
154
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
only be positive if all six factors in (7.5.3) with n = 1 are negative or four factors are negative and two are positive. For w > −1 and xi < 0 for i = 1, 2, 3 we have w − xi > −1 for i = 1, 2, 3. However, w > −1 and w − xi < 0 for i = 1, 2, 3 implies that xi > −1 for i = 1, 2, 3. We conclude that if all six factors in (7.5.3) with n = 1 are negative, then all six factors for n = 2 are positive. Therefore, D2 can only be negative in the case that four factors in (7.5.3) with n = 2 are negative and two are positive. Indeed, hereby only one negative pair (xi , w − xi ) with i ∈ {1, 2, 3} may occur. This observation leads to positive-definite orthogonality in four infinite cases given by table 7.3. Case IIc. w = −2N − t with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1. Then we must have Dn < 0 for n = 1, 2, 3, . . ..
case
conditions
positive
1 x1,2,3 > 0 d1 , d2 , d3 , . . . , dN 2 x1,2 > 0, x3 < −2N − t d1 , d2 , d3 , . . . , dN 3 x1 > 0, x2,3 = α ± iβ (α , β ∈ R, β = 0) d1 , d2 , d3 , . . . , dN 4 x1 > 0, x2,3 < −2N − t d1 , d2 , d3 , . . . , dN 5 x1 > 0, −N − t − 1 < x2,3 < −N + 1 d1 , d2 , d3 , . . . , dN 6 x1 > 0, −2N − t < x2 ≤ −N − t − 1, − N − t − 1 < x3 < −N + 1 −N − t − j − 1 < x2 ≤ −N − t − j, j ∈ {1, 2, 3, . . . , N − 1} d1 , d2 , d3 , . . . , dN− j 7 x1 > 0, −2N − t < x2,3 ≤ −N − t − 1 −N − t − j − 1 < x2 ≤ −N − t − j, j ∈ {1, 2, 3, . . . , N − 1} −N − t − k − 1 < x3 ≤ −N − t − k, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N − t − j and x3 = −N − t − j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N − t − j or x3 = −N − t − j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) 8 x1 > 0, −N + 1 ≤ x2,3 < 0 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} −N + k ≤ x3 < −N + k + 1, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N + j and x3 = −N + j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N + j or x3 = −N + j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) 9 x1 > 0, −N + 1 ≤ x2 < 0, −N − t − 1 < x3 < −N + 1 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} d1 , d2 , d3 , . . . , dN− j 10 x1 > 0, −N + 1 ≤ x2 < 0, −2N − t < x3 ≤ −N − t − 1 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} −N − t − k − 1 < x3 ≤ −N − t − k, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N + j and x3 = −N − t − j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N + j or x3 = −N − t − j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) Table 7.4 Case IIc1, N ∈ {1, 2, 3, . . .}.
7.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions
155
For w < −1 and xi > 0 for i = 1, 2, 3 we have w − xi < −1 for i = 1, 2, 3. However, w < −1 and w − xi > 0 for i = 1, 2, 3 implies that xi < −1 for i = 1, 2, 3. Therefore, D1 can only be negative if three factors in (7.5.3) with n = 1 are negative and three are positive or one factor is positive and five are negative. This observation leads to positive-definite orthogonality in four infinite and forty finite cases given by table 7.4, table 7.5, table 7.6 and table 7.7. In table 7.4 we list all cases with x1 > 0.
case
conditions
positive
11 x1,2,3 < −2N − t d1 , d2 , d3 , . . . , dN 12 x1 < −2N − t, x2,3 = α ± iβ (α , β ∈ R, β = 0) d1 , d2 , d3 , . . . , dN 13 x1 < −2N − t, −N − t − 1 < x2,3 < −N + 1 d1 , d2 , d3 , . . . , dN 14 x1 < −2N − t, −2N − t < x2 ≤ −N − t − 1, − N − t − 1 < x3 < −N + 1 −N − t − j − 1 < x2 ≤ −N − t − j, j ∈ {1, 2, 3, . . . , N − 1} d1 , d2 , d3 , . . . , dN− j 15 x1 < −2N − t, −2N − t < x2,3 ≤ −N − t − 1 −N − t − j − 1 < x2 ≤ −N − t − j, j ∈ {1, 2, 3, . . . , N − 1} −N − t − k − 1 < x3 ≤ −N − t − k, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N − t − j and x3 = −N − t − j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N − t − j or x3 = −N − t − j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) 16 x1 < −2N − t, −N + 1 ≤ x2,3 < 0 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} −N + k ≤ x3 < −N + k + 1, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N + j and x3 = −N + j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N + j or x3 = −N + j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) 17 x1 < −2N − t, −N + 1 ≤ x2 < 0, −N − t − 1 < x3 < −N + 1 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} d1 , d2 , d3 , . . . , dN− j 18 x1 < −2N − t, −N + 1 ≤ x2 < 0, −2N − t < x3 ≤ −N − t − 1 −N + j ≤ x2 < −N + j + 1, j ∈ {1, 2, 3, . . . , N − 1} −N − t − k − 1 < x3 ≤ −N − t − k, k ∈ {1, 2, 3, . . . , N − 1} a j = k with x2 = −N + j and x3 = −N − t − j d1 , d2 , d3 , . . . , dN b j = k with x2 = −N + j or x3 = −N − t − j d1 , d2 , d3 , . . . , dN− j c j = k d1 , d2 , d3 , . . . , dN−max( j,k) Table 7.5 Case IIc2, N ∈ {1, 2, 3, . . .}.
In all cases, the condition x1 > 0 can be replaced by the condition x1 < −2N − t. This leads to the extra cases listed in table 7.5. Even more cases can be obtained by interchanging the role of x1 and w − x1 in table 7.5. If we distinguish between −1 < t < 0 and 0 < t < 1 we have more cases. For −1 < t < 0 we have the cases listed in table 7.6 and for 0 < t < 1 we have the cases listed in table 7.7.
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
case
conditions
positive
19 −t < x1 < −t + 1, −N − t − 1 < x2 < −N, −N < x3 < −N − t 20 −t < x1 < −t + 1, −N − t < x2 < −N + 1, −N < x3 < −N − t 21 0 < x1 ≤ −t, −N − t − 1 < x2 < −N, −N < x3 < −N − t 22 0 < x1 ≤ −t, −N − t < x2 < −N + 1, −N < x3 < −N − t 23 x1 ≥ −t + 1, −N − t − 1 < x2 < −N, −N < x3 < −N − t 24 x1 ≥ −t + 1, −N − t < x2 < −N + 1, −N < x3 < −N − t
d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . , d2N d1 , d2 , d3 , . . . , d2N d1 , d2 , d3 , . . . , d2N+1 d1 , d2 , d3 , . . . , d2N+1
Table 7.6 Case IIc3, N ∈ {1, 2, 3, . . .} and −1 < t < 0.
case
conditions
positive
25 −t + 1 < x1 < −t + 2, −N − 1 < x2 < −N − t, −N − t < x3 < −N 26 −t + 1 < x1 < −t + 2, −N < x2 < −N − t + 1, −N − t < x3 < −N 27 0 < x1 ≤ −t + 1, −N − 1 < x2 < −N − t, −N − t < x3 < −N 28 0 < x1 ≤ −t + 1, −N < x2 < −N − t + 1, −N − t < x3 < −N 29 x1 ≥ −t + 2, −N − 1 < x2 < −N − t, −N − t < x3 < −N 30 x1 ≥ −t + 2, −N < x2 < −N − t + 1, −N − t < x3 < −N 31 x1 > 0, −N − t + 1 ≤ x2 < −N + 1, −N − t < x3 < −N 32 x1 > 0, −N − t − 1 < x2 ≤ −N − 1, −N − t < x3 < −N
d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . d1 , d2 , d3 , . . . , d2N+1 d1 , d2 , d3 , . . . , d2N+1 d1 , d2 , d3 , . . . , d2N+2 d1 , d2 , d3 , . . . , d2N+2 d1 , d2 , d3 , . . . , dN+1 d1 , d2 , d3 , . . . , dN+1
Table 7.7 Case IIc4, N ∈ {1, 2, 3, . . .} and 0 < t < 1.
7.6 The Self-Adjoint Difference Equation The orthogonality relations can be obtained in a similar way as in section 5.3. Therefore we write the difference equation (7.3.4) for the dual Hahn polynomials and the difference equation (7.3.7) for the Racah polynomials in the form (cf. (7.2.1))
ϕ (x + 2)Δ 2 yn (x(x + u)) + ψ (x + 1)Δ yn (x(x + u)) = λn ρ (x + 1)yn ((x + 1)(x + 1 + u)), n = 0, 1, 2, . . . . If we multiply by w(x + 1), this can also be written in the self-adjoint form
Δ {w(x)ϕ (x + 1)Δ yn (x(x + u))} = λn ρ (x + 1)w(x + 1)yn ((x + 1)(x + 1 + u)),
n = 0, 1, 2, . . . ,
(7.6.1)
where w(x) satisfies the Pearson difference equation
Δ {w(x)ϕ (x + 1)} = w(x + 1)ψ (x + 1).
(7.6.2)
As before, we multiply the difference equation (7.6.1) by ym ((x + 1)(x + 1 + u)), subtract the result with m and n interchanged, and finally replace x by x − 1. Then
7.6 The Self-Adjoint Difference Equation
157
we obtain (λn − λm )ρ (x)w(x)ym (x(x + u))yn (x(x + u)) = sn,m (x) − sn,m (x − 1)
(7.6.3)
sn,m (x) = w(x)ϕ (x + 1) {yn ((x + 1)(x + 1 + u))ym (x(x + u)) − ym ((x + 1)(x + 1 + u))yn (x(x + u))}
(7.6.4)
with
for m, n ∈ {0, 1, 2, . . .}. If the eigenvalues given by (7.3.1) are distinct, id est if (7.3.3) holds, then (7.6.3) and (7.6.4) lead to several different kinds of orthogonality relations with respect to the weight function w∗ (x) := ρ (x)w(x) similar to the case of chapter 5. For the dual Hahn polynomials given by (7.3.6) and the Racah polynomials given by (7.3.9) we obtain an orthogonality relation of the form (cf. (5.1.11)) A+N
∑ w∗ (x)ym (x(x + u))yn (x(x + u)) = σn δmn ,
m, n = 0, 1, 2, . . . , N
(7.6.5)
x=A
with boundary conditions (cf. (5.1.12)) w∗ (A − 1)ϕ (A) = 0
and w∗ (A + N)ϕ (A + N + 1) = 0.
(7.6.6)
Compare with theorem 3.3, where we have ϕ (x + 1) in (7.6.1) instead of ϕ (x − 1) in (3.2.5). By using the theorem of Favard (cf. theorem 3.1, (3.1.4), (4.2.5) and (4.2.6)), we obtain A+N
n
σn = ∏ dk ,
n = 0, 1, 2, . . . , N
with
d0 =
∑ w∗ (x)
(7.6.7)
x=A
k=0
and d1 , d2 , d3 , . . . , dN given by the three-term recurrence relation (7.4.1) or (7.4.2). If the second boundary condition in (7.6.6) cannot be satisfied for N ∈ {1, 2, 3, . . .}, then it might be possible to use the condition lim w∗ (x)xk = 0,
k = 0, 1, 2, . . .
x→∞
(7.6.8)
instead. In that case all moments of the form ∞
∑ w∗ (x)xk ,
k = 0, 1, 2, . . .
(7.6.9)
x=A
should exist. As a third possibility the sum in (7.6.5) can be replaced by an improper integral over the (possibly deformed) imaginary axis (cf. page 109):
i∞ 1 w∗ (x)ym (x(x + u))yn (x(x + u)) dx 2π i −i∞ = σn δmn , m, n = 0, 1, 2, . . . , N.
(7.6.10)
158
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
Also in this case all moments of the form 1 2π i
i∞ −i∞
w∗ (x)xk dx,
k = 0, 1, 2, . . .
(7.6.11)
should exist. In the latter two cases we may also find infinite orthogonal polynomial systems. For positive-definite orthogonality we must have d0 > 0.
7.7 Orthogonality Relations for Dual Hahn Polynomials In the case of the dual Hahn polynomials given by (7.3.6) we obtain by using (7.6.2) and (7.3.4) for the Pearson equation w∗ (x) w∗ (x + 1)
ρ (x) {ϕ (x + 2) − ψ (x + 1)} ρ (x + 1)ϕ (x + 1) (2x + u)(x + 1)(x + 1 + u − x1 )(x + 1 + u − x2 ) . =− (2x + 2 + u)(x + u)(x + x1 )(x + x2 )
=
(7.7.1)
It will turn out that the seven finite and six infinite cases of table 7.1 for the dual Hahn polynomials can be treated (with only a few exceptions) by using two weight functions w∗1 (x) and w∗2 (x).
The Finite Cases Note that (7.7.1) can be written in the form (2x + u)(x + 1)(x + 1 + u − x1 )(x + 1 + u − x2 ) w∗ (x) = , w∗ (x + 1) (2x + 2 + u)(x + u)(x + x1 )(−x − x2 ) which leads (up to a factor of period 1 in x) to the solution w∗1 (x) =
(2x + u)Γ(x + u)Γ(x + x1 ) . Γ(x + 1)Γ(x + 1 + u − x1 )Γ(x + 1 + u − x2 )Γ(1 − x2 − x)
By using this weight function for the dual Hahn polynomials given by (7.3.6) the boundary conditions (7.6.6) can be fulfilled if we set A = 0 and x2 = −N. By using (1.5.7), we obtain
7.7 Orthogonality Relations for Dual Hahn Polynomials
159
N
(1)
d0 =
∑ w∗1 (x)
x=0 N
=
2 Γ(x + u/2 + 1)Γ(x + u)Γ(x + x1 )
∑ x! (N − x)! Γ(x + u/2)Γ(x + 1 + u − x1)Γ(x + 1 + u + N)
x=0
N (−1)x (u/2 + 1)x (u)x (x1 )x (−N)x Γ(1 + u)Γ(x1 ) ∑ N! Γ(1 + u − x1 )Γ(1 + u + N) x=0 x! (u/2)x (1 + u − x1 )x (1 + u + N)x u/2 + 1, u, x1 , −N Γ(1 + u)Γ(x1 ) ; −1 = 4 F3 u/2, 1 + u − x1 , 1 + u + N N! Γ(1 + u − x1 )Γ(1 + u + N) Γ(1 + u)Γ(x1 ) Γ(1 + u − x1 )Γ(1 + u + N) = · N! Γ(1 + u − x1 )Γ(1 + u + N) Γ(1 + u)Γ(1 + u − x1 + N) Γ(x1 ) . = N! Γ(1 + u − x1 + N)
=
Further we obtain by using (7.4.1) and (7.6.7) (1)
σn =
Γ(n + x1 )n! , Γ(1 + u − x1 + N − n)(N − n)!
n = 0, 1, 2, . . . , N.
This leads to the orthogonality relation N
∑ w∗1 (x)ym (x(x + u))yn (x(x + u)) = σn
(1)
δmn ,
m, n = 0, 1, 2, . . . , N
(7.7.2)
x=0
for the dual Hahn polynomials yn (x(x + u)) given by (7.3.6). In cases where it is not possible to satisfy the second boundary condition in (7.6.6) because x2 = −N, one can try to use infinite sums instead. In that case we have (1)
σn =
Γ(n + x1 )n! , Γ(1 + u − x1 − x2 − n)Γ(1 − x2 − n)
n = 0, 1, 2, . . . , N
and the orthogonality relation ∞
∑ w∗1 (x)ym (x(x + u))yn (x(x + u)) = σn
(1)
δmn ,
m, n = 0, 1, 2, . . . , N.
(7.7.3)
x=0
For the convergence of this sum we need the existence of the moments ∞
∑ w∗1 (x)xk ,
k = 0, 1, 2, . . . , 4N.
x=0
By using the weight function w∗1 (x) for the dual Hahn polynomials, we can treat the seven finite cases of table 7.1 with only one exception (see the remark below). We only need the positivity of
160
7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations (1)
d0 =
Γ(x1 ) . Γ(1 − x2 )Γ(1 + u − x1 − x2 ) (1)
Case I. We study the sign of d0 for the seven finite cases of table 7.1. In case 4 we have x1 > 0, x2 ≤ −N and −N < x1 + x2 − u < −N + 1. This implies (1) that Γ(x1 ) > 0, Γ(1 − x2 ) > 0 and Γ(1 + u − x1 − x2 ) > 0. Hence we have d0 > 0. In case 5 we have x1 > 0, −N < x2 < −N + 1 and x1 + x2 − u ≤ −N. This implies (1) that Γ(x1 ) > 0, Γ(1 − x2 ) > 0 and Γ(1 + u − x1 − x2 ) > 0. Hence we have d0 > 0. In this case x2 = −N is not possible. In case 6 we have x1 > 0, x2 ≤ −N and x1 + x2 − u ≤ −N. This implies that (1) Γ(x1 ) > 0, Γ(1 − x2 ) > 0 and Γ(1 + u − x1 − x2 ) > 0. Hence we have d0 > 0. In case 8 we have −N < x1 < −N + 1, x2 ≤ −N and x1 + x2 − u > 0. Then the sign of Γ(x1 ) equals (−1)N and Γ(1 − x2 ) > 0. Further we have 1 + u − x1 − x2 < 1, hence if −N − j < 1 + u − x1 − x2 < −N − j + 1 for j ∈ {−N, −N + 1, −N + 2, . . .}, (1) then the sign of Γ(1 + u − x1 − x2 ) equals (−1)N+ j . For the positivity of d0 we need j a factor (−1) . In case 9 we have x1 ≤ −N, x2 ≤ −N and x1 + x2 − u > 0. Then we have Γ(1 − x2 ) > 0. If −N − j < x1 < −N − j + 1 and −N − k < 1 + u − x1 − x2 < −N − k + 1 for j, k ∈ {1, 2, 3, . . .}, then the sign of Γ(x1 ) equals (−1)N+ j and the sign of Γ(1 + (1) u − x1 − x2 ) equals (−1)N+k . For the positivity of d0 we need a factor (−1) j+k . Remark. The cases where x1 ∈ {−N, −N − 1, −N − 2, . . .} cannot be treated by using the weight function w∗1 (x). A separate treatment of these cases is left out here.
The Infinite Cases In order to deal with the six infinite cases of table 7.1 for the dual Hahn polynomials, we write instead of (7.7.1) (x + u/2)(x + 1/2 + u/2)(−x − 1)(−x − 1 − u + x1 )(−x − 1 − u + x2 ) w∗ (x) = w∗ (x + 1) (−x − 1 − u/2)(−x − 1/2 − u/2)(x + u)(x + x1 )(x + x2 ) which leads (up to a factor of period 1 in x) to the solution w∗2 (x) =
Γ(x + u)Γ(x + x1 )Γ(x + x2 )Γ(−x)Γ(x1 − u − x)Γ(x2 − u − x) . Γ(x + u/2)Γ(x + 1/2 + u/2)Γ(−u/2 − x)Γ(1/2 − u/2 − x)
For this weight function it is impossible to satisfy the boundary conditions (7.6.6). However, the concept of orthogonality of section 3.2 can be generalized to integration over the (possibly deformed) imaginary axis (cf. page 109). In that case we need the existence of the required moments, which is in general taken care of by the asymptotic properties of the gamma functions involved.
7.7 Orthogonality Relations for Dual Hahn Polynomials
161
By using (1.6.10), we obtain (2) d0
i∞ 1 = w∗ (x) dx 2π i −i∞ 2
i∞ 1 Γ(x + u)Γ(x + x1 )Γ(x + x2 )Γ(−x)Γ(x1 − u − x)Γ(x2 − u − x) dx = 2π i −i∞ Γ(x + u/2)Γ(x + 1/2 + u/2)Γ(−u/2 − x)Γ(1/2 − u/2 − x) 1 = Γ(x1 )Γ(x2 )Γ(x1 + x2 − u). (7.7.4) 2π
Further we obtain by using (7.4.1) and (7.6.7) (2)
σn =
1 Γ(n + x1 )Γ(n + x2 )Γ(n + x1 + x2 − u)n!, 2π
n = 0, 1, 2, . . . ,
which leads to the orthogonality relation 1 2π i =
i∞ −i∞
w∗2 (x)ym (x(x + u))yn (x(x + u)) dx
1 Γ(n + x1 )Γ(n + x2 )Γ(n + x1 + x2 − u)n! δmn , 2π
m, n = 0, 1, 2, . . . .(7.7.5)
By using the weight function w∗2 (x) for the dual Hahn polynomials, we can treat the six infinite cases of table 7.1 with only one exception (see the remark below). We only need the positivity of (2)
d0 =
1 Γ(x1 )Γ(x2 )Γ(x1 + x2 − u). 2π (2)
Case I. We study the sign of d0 for the six infinite cases of table 7.1. In case 1 we have x1 > 0, x2 > 0 and x1 + x2 − u > 0. This implies that Γ(x1 ) > 0, (2) Γ(x2 ) > 0 and Γ(x1 + x2 − u) > 0. Hence we have d0 > 0. In case 2 we have x1,2 = α ± iβ with α , β ∈ R and β = 0 and 2α − u > 0. This implies that Γ(α + iβ )Γ(α − iβ ) = |Γ(α + iβ )|2 > 0 and Γ(x1 + x2 − u) = Γ(2α − (2) u) > 0. Hence we have d0 > 0. In case 3 we have x1 > 0, −N < x2 < −N + 1 and −N < x1 + x2 − u < −N + 1. This implies that Γ(x1 ) > 0 and the sign of both Γ(x2 ) and Γ(x1 + x2 − u) equals (2) (−1)N . Hence we have d0 > 0. In case 6 we have x1 > 0, x2 ≤ −N and x1 + x2 − u ≤ −N. This implies that Γ(x1 ) > 0 and if −N − j < x2 < −N − j + 1 and −N − j < x1 + x2 − u < −N − j + 1 for j ∈ {−N + 1, −N + 2, −N + 3, . . .} the sign of both Γ(x2 ) and Γ(x1 + x2 − u) (2) equals (−1)N+ j . Hence we have d0 > 0.
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
In case 7 we have −N < x1 < −N + 1, −N < x2 < −N + 1 and x1 + x2 − u > 0. This implies that Γ(x1 + x2 − u) > 0 and the sign of both Γ(x1 ) and Γ(x2 ) equals (2) (−1)N . Hence we have d0 > 0. In case 9 we have x1 ≤ −N, x2 ≤ −N and x1 + x2 − u > 0. This implies that Γ(x1 + x2 − u) > 0 and if −N − j < x1,2 < −N − j + 1 for j ∈ {−N + 1, −N + 2, −N + 3, . . .} (2) then the sign of both Γ(x1 ) and Γ(x2 ) equals (−1)N+ j . Hence we have d0 > 0. Remark. The cases where x1,2 ∈ {−N, −N − 1, −N − 2, . . .} cannot be treated by using the weight function w∗2 (x). A separate treatment of these cases is left out.
7.8 Orthogonality Relations for Racah Polynomials In the case of the Racah polynomials given by (7.3.9), we obtain by using (7.6.2) and (7.3.7) for the Pearson equation w∗ (x)
ρ (x) {ϕ (x + 2) − ψ (x + 1)} ρ (x + 1)ϕ (x + 1) (2x + u)(x + 1) = (2x + 2 + u) (x + 1 + u − x1 )(x + 1 + u − x2 )(x + 1 + u − x3 ) . (7.8.1) × (x + u)(x + x1 )(x + x2 )(x + x3 )
=
w∗ (x + 1)
It will turn out that the finite and infinite cases of table 7.2 up to table 7.7 for the Racah polynomials can be treated (with only a few exceptions) by using two weight functions w∗3 (x) and w∗4 (x).
The Finite Cases Note that (7.8.1) can be written in the form w∗ (x) w∗ (x + 1)
=
(2x + u)(x + 1)(x + 1 + u − x1 )(x + 1 + u − x2 )(x + 1 + u − x3 ) , (2x + 2 + u)(x + u)(x + x1 )(−x − x2 )(−x − x3 )
which leads (up to a factor of period 1 in x) to the solution w∗3 (x) =
2 Γ(x + u/2 + 1)Γ(x + u) Γ(x + u/2)Γ(x + 1 + u − x1 )Γ(x + 1 + u − x2 )Γ(x + 1 + u − x3 ) Γ(x + x1 ) . × Γ(x + 1)Γ(1 − x2 − x)Γ(1 − x3 − x)
7.8 Orthogonality Relations for Racah Polynomials
163
If this weight function for the Racah polynomials given by (7.3.9) is used, the boundary conditions (7.6.6) can be fulfilled if we set A = 0 and x2 = −N (or x3 = −N). By using (1.5.6), we obtain N
(3)
d0 =
∑ w∗3 (x)
x=0 N
=
2 Γ(x + u/2 + 1)Γ(x + u)
∑ Γ(x + u/2)Γ(x + 1 + u − x1)Γ(x + 1 + u + N)Γ(x + 1 + u − x3)
x=0
Γ(x + x1 ) x! (N − x)! Γ(1 − x3 − x) 2 Γ(u/2 + 1)Γ(u)Γ(x1 ) = N! Γ(u/2)Γ(1 + u − x1 )Γ(1 + u + N)Γ(1 + u − x3 )Γ(1 − x3 ) ×
N
(−1)2x (u/2 + 1)x (u)x (x1 )x (−N)x (x3 )x x=0 x! (u/2)x (1 + u − x1 )x (1 + u + N)x (1 + u − x3 )x
×∑
Γ(1 + u)Γ(x1 ) N! Γ(1 + u − x1 )Γ(1 + u + N)Γ(1 + u − x3 )Γ(1 − x3 ) 1 + u/2, u, x1 , −N, x3 × 5 F4 ;1 u/2, 1 + u − x1 , 1 + u + N, 1 + u − x3 Γ(1 + u)Γ(x1 ) = N! Γ(1 + u − x1 )Γ(1 + u + N)Γ(1 + u − x3 )Γ(1 − x3 ) Γ(1 + u − x1 )Γ(1 + u + N)Γ(1 + u − x3 )Γ(1 + u − x1 − x3 + N) × Γ(1 + u)Γ(1 + u − x1 + N)Γ(1 + u − x1 − x3 )Γ(1 + u − x3 + N) Γ(x1 )Γ(1 + u − x1 − x3 + N) . = N! Γ(1 + u − x1 + N)Γ(1 + u − x3 + N)Γ(1 + u − x1 − x3 )Γ(1 − x3 ) =
Further we get by using (7.4.2), (7.6.7) and w = x1 − N + x3 − u (3)
σn =
Γ(w)Γ(1 − w)Γ(n + w − 1) Γ(2n + w)Γ(2n + w − 1)Γ(1 − w + x1 − n)Γ(1 − w − N − n) Γ(n + x1 )(−1)n n! , n = 0, 1, 2, . . . , N. × Γ(1 − w + x3 − n)Γ(1 − x3 − n)(N − n)!
This leads to the orthogonality relation N
∑ w∗3 (x)ym (x(x + u))yn (x(x + u)) = σn
(3)
δmn ,
m, n = 0, 1, 2, . . . , N
(7.8.2)
x=0
for the Racah polynomials yn (x(x + u)) given by (7.3.9). In cases where it is not possible to satisfy the second boundary condition in (7.6.6) because x2 = −N, one can try to use infinite sums instead. In that case we have
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations (3)
σn =
Γ(w)Γ(1 − w)Γ(n + w − 1) Γ(2n + w)Γ(2n + w − 1)Γ(1 − w + x1 − n)Γ(1 − w + x2 − n) Γ(n + x1 )(−1)n n! , n = 0, 1, 2, . . . , N × Γ(1 − w + x3 − n)Γ(1 − x2 − n)Γ(1 − x3 − n)
with w = x1 + x2 + x3 − u and the orthogonality relation ∞
∑ w∗3 (x)ym (x(x + u))yn (x(x + u)) = σn
(3)
δmn ,
m, n = 0, 1, 2, . . . , N.
(7.8.3)
x=0
For the convergence of this sum we need the existence of the moments ∞
∑ w∗3 (x)xk ,
k = 0, 1, 2, . . . , 4N.
x=0
By using this weight function w∗3 (x) for the Racah polynomials, we can treat almost all finite cases of table 7.2 and table 7.4 up to table 7.7. We only need the positivity of Γ(x1 )Γ(1 − w) (3) . d0 = Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x1 )Γ(1 − w + x2 )Γ(1 − w + x3 ) Case IIa. In table 7.2 we have w > 0. In all cases we have 0 < x1 < w which implies that Γ(x1 ) > 0, Γ(w) > 0 and Γ(w − x1 ) > 0. Further we obtain by using (1.2.3) (3)
d0 =
Γ(x1 )Γ(w − x1 )Γ(w − x2 )Γ(w − x3 ) Γ(w)Γ(1 − x2 )Γ(1 − x3 ) sin π (w − x1 ) sin π (w − x2 ) sin π (w − x3 ) . × π 2 sin π w
In case 4 and case 5 we have x2 < 0 and x3 < 0, which implies that we also have Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. In these cases we need a factor with the same sign as the quotient of sines above. Note that we sometimes need a special treatment for the cases where w, w − x1 , w − x2 , w − x3 ∈ {0, ±1, ±2, . . .}. This will be left out here. By using (1.2.3), we may also write (3)
d0 =
sin π (w − x1 ) sin π x2 sin π x3 Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 ) · . Γ(w)Γ(1 − w + x2 )Γ(1 − w + x3 ) π 2 sin π w
In case 7 and case 8 we have w − x2 < 0 and w − x3 < 0, which implies that Γ(x2 ) > 0, Γ(x3 ) > 0, Γ(1−w+x2 ) > 0 and Γ(1−w+x3 ) > 0. In these cases we need a factor with the same sign as the quotient of sines above. Note that we sometimes need a special treatment for the cases where w, w − x1 , x2 , x3 ∈ {0, ±1, ±2, . . .}. This will be left out here.
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165
Finally, we also obtain by using (1.2.3) (3)
d0 =
Γ(x1 )Γ(x3 )Γ(w − x1 )Γ(w − x2 ) sin π (w − x1 ) sin π (w − x2 ) sin π x3 · . Γ(w)Γ(1 − x2 )Γ(1 − w + x3 ) π 2 sin π w
In case 10, case 11 and case 12 we have x2 < 0 and w − x3 < 0, which implies that Γ(1−x2 ) > 0, Γ(x3 ) > 0, Γ(w−x2 ) > 0 and Γ(1−w+x3 ) > 0. In these cases we need a factor with the same sign as the quotient of sines above. Note that we sometimes need a special treatment for the cases where w, x3 , w − x1 , w − x2 ∈ {0, ±1, ±2, . . .}. This will be left out here. Remark. In the cases 7, 8 and 11 it is impossible to take x2 = −N. Case IIb. In table 7.3 we have no finite cases. Case IIc1. In table 7.4 we have w = −2N −t < 0 with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1, which implies that Γ(1 − w) > 0. Further we have x1 > 0 and therefore Γ(x1 ) > 0 and Γ(1 − w + x1 ) > 0. Again we obtain by using (1.2.3) (3)
d0 =
sin π x2 sin π x3 Γ(x1 )Γ(x2 )Γ(x3 )Γ(1 − w) · . Γ(1 − w + x1 )Γ(1 − w + x2 )Γ(1 − w + x3 ) π2
In case 1 we also have x2 > 0 and x3 > 0, which implies that Γ(x2 ) > 0, Γ(x3 ) > 0, Γ(1 − w + x2 ) > 0 and Γ(1 − w + x3 ) > 0. In this case we need a factor with the same sign as the product of sines above. Note that we sometimes need a special treatment for the cases where x2 , x3 ∈ {0, ±1, ±2, . . .}. This will be left out here. By using (1.2.3), we may also write (3)
d0 =
sin π x2 sin π (w − x3 ) Γ(x1 )Γ(x2 )Γ(w − x3 )Γ(1 − w) · . Γ(1 − x3 )Γ(1 − w + x1 )Γ(1 − w + x2 ) π2
In case 2 we have x2 > 0 and x3 < w < 0, which implies that Γ(x2 ) > 0, Γ(1−x3 ) > 0, Γ(1 − w + x2 ) > 0 and Γ(w − x3 ) > 0. In this case we need a factor with the same sign as the product of sines above. Note that we sometimes need a special treatment for the cases where x2 , w − x3 ∈ {0, ±1, ±2, . . .}. This will be left out here. Again we use (1.2.3) to write (3)
d0 =
Γ(x1 )Γ(w − x2 )Γ(w − x3 )Γ(1 − w) sin π (w − x2 ) sin π (w − x3 ) · . Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x1 ) π2
In case 4 we have x2 < w < 0 and x3 < w < 0, which implies that Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. In this case we need a factor with the same sign as the product of sines above. Note that we sometimes need a special treatment for the cases where w − x2 , w − x3 ∈ {0, ±1, ±2, . . .}. This will be left out here.
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
In case 3 we have Γ(1 − x2 )Γ(1 − x3 ) = Γ(1 − α − iβ )Γ(1 − α + iβ ) = |Γ(1 − α + iβ )|2 > 0 and Γ(1 − w + x2 )Γ(1 − w + x3 ) = Γ(1 − w + α + iβ )Γ(1 − w + α − iβ ) = |Γ(1 − w + α + iβ )|2 > 0. Hence in case 3 and also in case 5 through case 10 we simply have (3)
d0 =
Γ(x1 )Γ(1 − w) > 0. Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x1 )Γ(1 − w + x2 )Γ(1 − w + x3 )
Remark. Only in case 5 it is possible to take x2 = −N. Case IIc2. In table 7.5 we have w = −2N − t < 0 with N ∈ {1, 2, 3, . . .} and −1 < t ≤ 1. Further we have x1 < w < 0 and therefore Γ(1 − x1 ) > 0, Γ(1 − w) > 0 and Γ(w − x1 ) > 0. Again we obtain by using (1.2.3) (3)
d0 =
Γ(w − x1 )Γ(w − x2 )Γ(w − x3 )Γ(1 − w) Γ(1 − x1 )Γ(1 − x2 )Γ(1 − x3 ) sin π (w − x1 ) sin π (w − x2 ) sin π (w − x3 ) . × π 2 sin π x1
In case 11 we also have x2 < w < 0 and x3 < w < 0, which implies that Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. In this case we need a factor with the same sign as the quotient of sines above. Note that we sometimes need a special treatment for the cases where x1 , w − x1 , w − x2 , w − x3 ∈ {0, ±1, ±2, . . .}. This will be left out here. For the other cases we use (3)
d0 =
sin π (w − x1 ) Γ(w − x1 )Γ(1 − w) · . Γ(1 − x1 )Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x2 )Γ(1 − w + x3 ) sin π x1
In case 12 we have x2,3 = α ± iβ , which implies, as before, that both Γ(1 − x2 )Γ(1 − x3 ) > 0 and Γ(1 − w + x2 )Γ(1 − w + x3 ) > 0. In case 13 through case 18 we have w < x2 < 0 and w < x3 < 0, which implies that Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(1 − w + x2 ) > 0 and Γ(1 − w + x3 ) > 0. Hence in case 12 through case 18 we need a factor with the same sign as the quotient of sines above. Note that sometimes we need a special treatment for the cases where x1 , w − x1 ∈ {0, ±1, ±2, . . .}. This will be left out here. Remark. Only in case 13 it is possible to take x2 = −N.
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167
Case IIc3. In table 7.6 we have w = −2N −t < 0 with N ∈ {1, 2, 3, . . .} and −1 < t < 0. In all finite cases we have Γ(x1 ) > 0, Γ(1 − w) > 0, Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(1 − w + x1 ) > 0, Γ(1 − w + x2 ) > 0 and Γ(1 − w + x3 ) > 0, which implies that (3)
d0 =
Γ(x1 )Γ(1 − w) > 0. Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x1 )Γ(1 − w + x2 )Γ(1 − w + x3 )
Remark. In these cases it is impossible to take x2 = −N. Case IIc4. In table 7.7 we have w = −2N −t < 0 with N ∈ {1, 2, 3, . . .} and 0 < t < 1. In all finite cases we have Γ(x1 ) > 0, Γ(1 − w) > 0, Γ(1 − x2 ) > 0, Γ(1 − x3 ) > 0, Γ(1 − w + x1 ) > 0, Γ(1 − w + x2 ) > 0 and Γ(1 − w + x3 ) > 0 which implies that (3)
d0 =
Γ(x1 )Γ(1 − w) > 0. Γ(1 − x2 )Γ(1 − x3 )Γ(1 − w + x1 )Γ(1 − w + x2 )Γ(1 − w + x3 )
Remark. In these cases it is impossible to take x2 = −N.
The Infinite Cases In order to deal with the infinite cases of table 7.2 through table 7.7 for the Racah polynomials, we write instead of (7.8.1) w∗ (x) w∗ (x + 1)
(x + u/2)(x + 1/2 + u/2) (x + u)(x + x1 )(x + x2 )(x + x3 ) (−x − 1)(−x − 1 − u + x1 )(−x − 1 − u + x2 )(−x − 1 − u + x3 ) × (−x − 1 − u/2)(−x − 1/2 − u/2)
=
which leads (up to a factor of period 1 in x) to the solution w∗4 (x) =
Γ(x + u)Γ(x + x1 )Γ(x + x2 )Γ(x + x3 ) Γ(x + u/2)Γ(x + 1/2 + u/2) Γ(−x)Γ(x1 − u − x)Γ(x2 − u − x)Γ(x3 − u − x) . × Γ(−u/2 − x)Γ(1/2 − u/2 − x)
For this weight function it is impossible to satisfy the boundary conditions (7.6.6). However, the concept of orthogonality of section 3.2 can be generalized to integration over the (possibly deformed) imaginary axis (cf. page 109). In that case we need the existence of the required moments, which is in general taken care of by the asymptotic properties of the gamma functions involved. By using (1.6.6), we have
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations (4)
i∞ 1 w∗ (x) dx 2π i −i∞ 4
i∞ Γ(x + u)Γ(x1 − u − x)Γ(x2 − u − x)Γ(x3 − u − x) 1 = 2π i −i∞ Γ(x + u/2)Γ(x + 1/2 + u/2)Γ(−u/2 − x)Γ(1/2 − u/2 − x) × Γ(x + x1 )Γ(x + x2 )Γ(x + x3 )Γ(−x) dx 1 Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 )Γ(w − x2 )Γ(w − x3 ) , (7.8.4) = 2π Γ(w)
d0 =
where w = x1 + x2 + x3 − u. Further we obtain by using (7.4.2) and (7.6.7) (4)
σn =
Γ(n + x1 )Γ(n + x2 )Γ(n + x3 )Γ(n + w − 1) 2π Γ(2n + w)Γ(2n − 1 + w) × Γ(n + w − x1 )Γ(n + w − x2 )Γ(n + w − x3 )n!
for n = 0, 1, 2, . . ., which leads to the orthogonality relation 1 2π i
i∞ −i∞
w∗4 (x)ym (x(x + u))yn (x(x + u)) dx = σn δmn , (4)
m, n = 0, 1, 2, . . . . (7.8.5)
By using the weight function w∗4 (x) for the Racah polynomials, we can treat almost all infinite cases of table 7.2 through table 7.7. We only need the positivity of (4)
d0 =
1 Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 )Γ(w − x2 )Γ(w − x3 ) . 2π Γ(w)
Case IIa. In table 7.2 we have w > 0. In all cases we have 0 < x1 < w and therefore Γ(x1 ) > 0, Γ(w) > 0 and Γ(w − x1 ) > 0. In case 1 we also have 0 < x2 < w and 0 < x3 < w, which implies that Γ(x2 ) > 0, (4) Γ(x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. Hence we have d0 > 0. In case 2 we have x2,3 = α ± iβ , which implies that both Γ(x2 )Γ(x3 ) = Γ(α + iβ )Γ(α − iβ ) = |Γ(α + iβ )|2 > 0 and Γ(w − x2 )Γ(w − x3 ) = Γ(w − α − iβ )Γ(w − α + iβ ) = |Γ(w − α + iβ )|2 > 0. (4)
Hence we have d0 > 0. In case 3 and case 5 we have x2 < 0 and x3 < 0, which implies that Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. If −N − j < x2,3 < −N − j + 1 for j ∈ {0, 1, 2, . . .} then the sign (4) of both Γ(x2 ) and Γ(x3 ) equals (−1)N+ j . Hence we have d0 > 0.
7.8 Orthogonality Relations for Racah Polynomials
169
In case 6 and case 8 we have w − x2 < 0 and w − x3 < 0, which implies that Γ(x2 ) > 0 and Γ(x3 ) > 0. If −N − j < w − x2,3 < −N − j + 1 for j ∈ {0, 1, 2, . . .} then (4) the sign of both Γ(w − x2 ) and Γ(w − x3 ) equals (−1)N+ j . Hence we have d0 > 0. In case 9 and case 12 we have x2 < 0 and w − x3 < 0, which implies that Γ(w − x2 ) > 0 and Γ(x3 ) > 0. If −N − j < x2 < −N − j + 1 and −N − j < w − x3 < −N − j + 1 for j ∈ {0, 1, 2, . . .} then the sign of both Γ(x2 ) and Γ(w − x3 ) equals (−1)N+ j . (4) Hence we have d0 > 0. Case IIb. In table 7.3 we have −1 < w < 0. In all cases we have w < x1 < 0 and therefore Γ(x1 ) < 0, Γ(w) < 0 and Γ(w − x1 ) < 0. In case 1 we have −2 < x2 < −1 and −1 < x3 < w, which implies that Γ(x2 ) > 0, (4) Γ(x3 ) < 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. Hence we have d0 > 0. In case 2 we have −2 < x2 < −1 and 0 < x3 < w+1, which implies that Γ(x2 ) > 0, (4) Γ(x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) < 0. Hence we have d0 > 0. In case 3 we have w + 1 < x2 < w + 2 and −1 < x3 < w, which implies that (4) Γ(x2 ) > 0, Γ(x3 ) < 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) > 0. Hence we have d0 > 0. In case 4 we have w + 1 < x2 < w + 2 and 0 < x3 < w + 1, which implies that (4) Γ(x2 ) > 0, Γ(x3 ) > 0, Γ(w − x2 ) > 0 and Γ(w − x3 ) < 0. Hence we have d0 > 0. Case IIc1. In table 7.4 we have no infinite cases. Case IIc2. In table 7.5 we have no infinite cases. Case IIc3. In table 7.6 we have w = −2N −t < 0 with N ∈ {1, 2, 3, . . .} and −1 < t < 0. In both cases we have Γ(w) > 0 and Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 )Γ(w − x2 )Γ(w − (4) x3 ) > 0, which implies that d0 > 0. Case IIc4. In table 7.7 we have w = −2N −t < 0 with N ∈ {1, 2, 3, . . .} and 0 < t < 1. In both cases we have Γ(w) < 0 and Γ(x1 )Γ(x2 )Γ(x3 )Γ(w−x1 )Γ(w−x2 )Γ(w−x3 ) < (4) 0 which implies that d0 > 0.
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7 Orthogonal Polynomial Solutions in x(x + u) of Real Difference Equations
In this chapter we have proved:
Theorem 7.2. The positive-definite orthogonal polynomial solutions y(x(x + u)) with x ∈ R and u ∈ R of the difference equation (7.2.1)
ϕ (x + 2)Δ 2 yn (x(x + u)) + ψ (x + 1)Δ yn (x(x + u)) = λn ρ (x + 1)yn ((x + 1)(x + 1 + u)), n = 0, 1, 2, . . . , consist of the polynomial solutions of the (real) difference equations (7.3.4) and (7.3.7). In fact this leads to both infinite and finite systems of • dual Hahn polynomials and • Racah polynomials.
Chapter 8
Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations Discrete Classical Orthogonal Polynomials IV
8.1 Real Polynomial Solutions of Complex Difference Equations As in chapter 6, the difference equations (7.3.4) for the dual Hahn polynomials and (7.3.7) for the Racah polynomials can also be considered in the case that the coefficients are complex. In that case we look for polynomial solutions yn (z(z + u)) with z ∈ C and u ∈ R. The three-term recurrence relations (7.4.1) and (7.4.2) still hold with x replaced by z. Again we set z = a + ix with a ∈ R and x ∈ R. Then we have z(z + u) = (a + ix)(a + ix + u) = a(a + u) + i(2a + u)x − x2 . For u = −2a the imaginary part cancels and we have z(z + u) = −a2 − x2 . Now we define yn (z(z + u)) = yn (−a2 − x2 ) = (−1)n yn (x2 ),
n = 0, 1, 2, . . . .
(8.1.1)
Case I. For the (monic) dual Hahn polynomials given by (7.3.6) this leads to the (monic) continuous dual Hahn polynomials −n, −a − ix, −a + ix 2 n ;1 (8.1.2) yn (x ) = (−1) (x1 )n (x2 )n 3 F2 x1 , x2 for n = 0, 1, 2, . . .. These continuous dual Hahn polynomials satisfy the three-term recurrence relation yn+1 (x2 ) = x2 + a2 − (n + x1 )(n + x2 ) − n(n − 1 + 2a + x1 + x2 ) yn (x2 ) yn−1 (x2 ) (8.1.3) − n(n − 1 + x1 )(n − 1 + x2 )(n − 1 + 2a + x1 + x2 )
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 8, © Springer-Verlag Berlin Heidelberg 2010
171
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8 Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations
for n = 1, 2, 3, . . . with y0 (x2 ) = 1 and y1 (x2 ) = x2 + a2 − x1 x2 . Since the coefficients in yn (x2 ) can only be real for x1 and x2 both real or complex conjugates, the positive-definite orthogonality can be obtained in a similar way as in the real case. Case II. For the (monic) Racah polynomials given by (7.3.9), this leads to the (monic) Wilson polynomials (x1 )n (x2 )n (x3 )n −n, n − 1 + w, −a − ix, −a + ix ;1 (8.1.4) yn (x2 ) = (−1)n 4 F3 (n − 1 + w)n x1 , x2 , x3 for n = 0, 1, 2, . . ., where w = 2a + x1 + x2 + x3 . These Wilson polynomials satisfy the three-term recurrence relation (1) (2) (1) (2) (8.1.5) yn+1 (x2 ) = x2 + a2 − cn − cn yn (x2 ) − cn−1 cn yn−1 (x2 ) (1)
(2)
for n = 1, 2, 3, . . . with y0 (x2 ) = 1, y1 (x2 ) = x2 +a2 −x1 x2 x3 /w and cn and cn given by (7.4.3) and (7.4.4) respectively. Note that
2(x1 + 1)(x2 + 1)(x3 + 1) − 2a − 1 (x2 + a2 ) y2 (x2 ) = (x2 + a2 )2 − w+2 x1 x2 x3 (x1 + 1)(x2 + 1)(x3 + 1) . + (w + 1)(w + 2) Since a ∈ R, the coefficients of yn (x2 ) can only be real if x1 x2 x3 ∈ R, w
(x1 + 1)(x2 + 1)(x3 + 1) ∈R w+2
and
x1 x2 x3 ∈ R. w+1
For x1 x2 x3 = 0 this implies that w ∈ R and therefore x1 x2 x3 ∈ R and (x1 + 1)(x2 + 1)(x3 +1) ∈ R. Moreover, d1 ∈ R leads to (w−x1 )(w−x2 )(w−x3 ) ∈ R and therefore x1 + x2 + x3 ∈ R and x1 x2 + x1 x3 + x2 x3 ∈ R. Note that xi = 0 for some i ∈ {1, 2, 3} or x j = w for some j ∈ {1, 2, 3} leads to d1 = 0. This implies that there are only two possibilities, id est x1 , x2 , x3 ∈ R or one is real and the other two are complex (1) (2) conjugates. In either case both cn given by (7.4.3) and cn given by (7.4.4) are real for all n = 1, 2, 3, . . .. This shows that in contrast with chapter 6, the coefficients in the difference equation for yn (x2 ) are real except for the powers of a + ix involved. This implies that the positive-definite orthogonality can be obtained in a similar way as in the real case.
8.2 Orthogonality Relations for Continuous Dual Hahn Polynomials
173
8.2 Orthogonality Relations for Continuous Dual Hahn Polynomials For the dual Hahn polynomials given by (7.3.6) and the Racah polynomials given by (7.3.9) we obtained orthogonality relations of the form (7.6.5) with boundary conditions (7.6.6). We use the transformation (8.1.1). If we set a + ix = t, then we obtain x = i(a −t) and therefore x2 = −(a−t)2 . Hence, for the continuous dual Hahn polynomials given by (8.1.2) and the Wilson polynomials given by (8.1.4) we obtain an orthogonality relation of the form A+N
∑ w∗ (t)ym (−(a − t)2 )yn (−(a − t)2 ) = σn δmn ,
m, n = 0, 1, 2, . . . , N
(8.2.1)
t=A
with the boundary conditions w∗ (A − 1)ϕ (A) = 0
and w∗ (A + N)ϕ (A + N + 1) = 0.
(8.2.2)
Then we have A+N
n
σn = ∏ dk ,
n = 0, 1, 2, . . . , N
with d0 =
k=0
∑ w∗ (t)
(8.2.3)
t=A
and d1 , d2 , d3 , . . . , dN are given by the three-term recurrence relation (8.1.3) or (8.1.5). Also in this case it is possible to use an infinite sum or an improper integral over the (possibly deformed) imaginary axis instead of the finite sum in (8.2.1). For positive-definite orthogonality we must have d0 > 0. For the continuous dual Hahn polynomials yn (x2 ) given by (8.1.2), the Pearson equation (7.7.1) still holds.
The Finite Cases In order to deal with the seven finite cases in table 7.1, we write instead of (7.7.1) (−x − u/2)(−x − 1/2 − u/2)(x + 1)(−x − 1 − u + x1 )(x + 1 + u − x2 ) w∗ (x) = w∗ (x + 1) (x + 1 + u/2)(x + 1/2 + u/2)(−x − u)(x + x1 )(−x − x2 ) which leads (up to a factor of period 1 in x) to the solution w∗5 (x) = Γ(x + 1 + u/2)Γ(x + 1/2 + u/2) Γ(1 − u/2 − x)Γ(1/2 − u/2 − x)Γ(x1 − u − x)Γ(x + x1 ) × . Γ(x + 1)Γ(x + 1 + u − x2 )Γ(1 − u − x)Γ(1 − x2 − x)
174
8 Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations
By using Legendre’s duplication formula (1.2.6), we have Γ(x + 1 + u/2)Γ(x + 1/2 + u/2)Γ(1 − u/2 − x)Γ(1/2 − u/2 − x) = π Γ(2x + 1 + u)Γ(1 − u − 2x) for 2x + u = ±1, ±2, ±3, . . ., which implies that w∗5 (x) =
π Γ(2x + 1 + u)Γ(1 − u − 2x)Γ(x1 − u − x)Γ(x + x1 ) . Γ(x + 1)Γ(x + 1 + u − x2 )Γ(1 − u − x)Γ(1 − x2 − x)
If we now replace x by a + ix and u by −2a, we obtain Γ(1 + 2ix)Γ(x1 + a + ix) 2 . w∗5 (a + ix) = π Γ(1 + a + ix)Γ(1 − a − x2 + ix) In this case the boundary conditions (7.6.6) can be satisfied by taking A = 0 and x2 = −N. By using (1.5.7) and (1.2.6) we obtain N
(5)
d0 =
∑ w∗5 (x)
x=0 N
=
∑ Γ(x + 1 + u/2)Γ(x + 1/2 + u/2)
x=0
Γ(1 − u/2 − x)Γ(1/2 − u/2 − x)Γ(x1 − u − x)Γ(x + x1 ) x! Γ(x + 1 + u + N)Γ(1 − u − x)(N − x)! Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) = N! Γ(1 − u)Γ(1 + u + N) ×
N
(−1)x (−N)x (1 + u/2)x (u)x (x1 )x x=0 x! (u/2)x (1 + u − x1 )x (1 + u + N)x
×∑
Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) N! Γ(1 − u)Γ(1 + u + N) −N, 1 + u/2, u, x1 × 4 F3 ; −1 u/2, 1 + u − x1 , 1 + u + N Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) = N! Γ(1 − u)Γ(1 + u + N) Γ(1 + u − x1 )Γ(1 + u + N) × Γ(1 + u)Γ(1 + u − x1 + N) π Γ(x1 )Γ(x1 − u)Γ(1 + u − x1 ) . = N! Γ(1 + u − x1 + N) =
Further we obtain by using (8.1.3) and (7.6.7) (5)
σn =
π Γ(n + x1 )Γ(x1 − u)Γ(1 + u − x1 )n! , Γ(1 + u − x1 + N − n)(N − n)!
n = 0, 1, 2, . . . , N,
8.2 Orthogonality Relations for Continuous Dual Hahn Polynomials
175
which leads to the orthogonality relation (cf. (8.2.1)) N
Γ(1 − 2a + 2t)Γ(1 + 2a − 2t)Γ(x1 + 2a − t)Γ(x1 + t)
∑ Γ(1 + t)Γ(1 − 2a + N + t)Γ(1 + 2a − t)Γ(1 + N − t)
t=0
× ym (−(a − t)2 ) yn (−(a − t)2 ) Γ(n + x1 )Γ(2a + x1 )Γ(1 − 2a − x1 )n! δmn , m, n = 0, 1, 2, . . . , N. (8.2.4) = Γ(1 − 2a − x1 + N − n)(N − n)! If the boundary conditions (8.2.2) cannot be satisfied, one may use infinite sums. In that case we have (5)
σn =
π Γ(n + x1 )Γ(2a + x1 )Γ(1 − 2a − x1 )n! , Γ(1 − 2a − x1 − x2 − n)Γ(1 − x2 − n)
n = 0, 1, 2, . . . , N,
which leads to the orthogonality relation (cf. (8.2.1)) ∞
Γ(1 − 2a + 2t)Γ(1 + 2a − 2t)Γ(x1 + 2a − t)Γ(x1 + t)
∑ Γ(1 + t)Γ(1 − 2a − x2 + t)Γ(1 + 2a − t)Γ(1 − x2 − t)
t=0
× ym (−(a − t)2 ) yn (−(a − t)2 ) Γ(n + x1 )Γ(2a + x1 )Γ(1 − 2a − x1 )n! δmn , m, n = 0, 1, 2, . . . , N. (8.2.5) = Γ(1 − 2a − x1 − x2 − n)Γ(1 − x2 − n) Again we need the existence of the moments involved in order to have convergent sums. Table 7.1 also holds for the continuous dual Hahn polynomials yn (x2 ) given by (8.1.2). The seven finite cases can be treated by using the weight function w∗5 (x) for the continuous dual Hahn polynomials with only one exception, as in the preceding (5) section. The positivity of d0 is dealt with in the same way as before. The factor Γ(2a + x1 )Γ(1 − 2a − x1 ) with 2a + x1 = 0, ±1, ±2, . . . determines the sign which is necessary for positivity.
The Infinite Cases For the six infinite cases of table 7.1 we may use the weight function w∗2 (x) for the dual Hahn polynomials again. By using Legendre’s duplication formula (1.2.6), we have Γ(x + u/2)Γ(x + 1/2 + u/2)Γ(−u/2 − x)Γ(1/2 − u/2 − x) = 4π Γ(2x + u)Γ(−u − 2x) for 2x + u = 0, ±1, ±2, . . ., which implies that
176
8 Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations
w∗2 (x) =
Γ(x + u)Γ(−x)Γ(x + x1 )Γ(x1 − u − x)Γ(x + x2 )Γ(x2 − u − x) . 4π Γ(2x + u)Γ(−u − 2x)
If we now replace x by a + ix and u by −2a, we obtain w∗2 (a + ix) =
1 Γ(−a + ix)Γ(x1 + a + ix)Γ(x2 + a + ix) 2 . 4π Γ(2ix)
In this case the boundary conditions (8.2.2) cannot be satisfied. However, as before, the concept of orthogonality can be extended to integration over the (possibly deformed) imaginary axis (cf. page 109). Then we need again the existence of the required moments which is in general taken care of by the asymptotic properties of the gamma functions involved. By using z = a + ξ with u = −2a, we have Γ(z − 2a)Γ(−z)Γ(z + x1 )Γ(x1 + 2a − z)Γ(z + x2 )Γ(x2 + 2a − z) Γ(z − a)Γ(z − a + 1/2)Γ(a − z)Γ(1/2 + a − z) Γ(ξ − a)Γ(−a − ξ ) = Γ(ξ )Γ(−ξ ) Γ(ξ + a + x1 )Γ(x1 + a − ξ )Γ(ξ + a + x2 )Γ(x2 + a − ξ ) . × Γ(ξ + 1/2)Γ(1/2 − ξ )
w∗2 (z) =
Analogous to (7.7.4) we obtain by using (1.6.10) 1 2π i
i∞ −i∞
w∗2 (a + ξ ) d ξ =
1 Γ(x1 )Γ(x2 )Γ(x1 + x2 + 2a) 2π
which, if ξ = ix is used, leads to
∞ (2) d0 = w∗2 (a + ix) dx = Γ(x1 )Γ(x2 )Γ(x1 + x2 + 2a). −∞
Then by using (8.1.3), we obtain (2)
σn = Γ(n + x1 )Γ(n + x2 )Γ(n + x1 + x2 + 2a)n!,
n = 0, 1, 2, . . .
and therefore the orthogonality relation 1 ∞ Γ(−a + ix)Γ(x1 + a + ix)Γ(x2 + a + ix) 2 2 yn (x2 ) dx ym (x ) 4π −∞ Γ(2ix) = Γ(n + x1 )Γ(n + x2 )Γ(n + x1 + x2 + 2a)n! δmn , m, n = 0, 1, 2, . . . . (8.2.6) The six infinite cases of table 7.1 can be treated by using this weight function w∗2 (x) (2) with only one exception. The positivity of d0 is treated as before.
8.3 Orthogonality Relations for Wilson Polynomials
177
8.3 Orthogonality Relations for Wilson Polynomials For the Wilson polynomials yn (x2 ) given by (8.1.4), the Pearson equation (7.8.1) still holds.
The Finite Cases In order to deal with the finite cases in table 7.2 through table 7.7, we write instead of (7.8.1) w∗ (x) w∗ (x + 1)
=
(−x − u/2)(−x − 1/2 − u/2)(x + 1)(−x − 1 − u + x1 ) (x + 1 + u/2)(x + 1/2 + u/2)(−x − u)(x + x1 ) (x + 1 + u − x2 )(x + 1 + u − x3 ) × (−x − x2 )(−x − x3 )
which leads (up to a factor of period 1 in x) to the solution w∗6 (x) =
Γ(x + 1 + u/2)Γ(x + 1/2 + u/2)Γ(1 − u/2 − x)Γ(1/2 − u/2 − x) Γ(x + 1)Γ(1 − u − x) Γ(x1 − u − x)Γ(x + x1 ) . × Γ(1 − x2 − x)Γ(1 − x3 − x)Γ(x + 1 + u − x2 )Γ(x + 1 + u − x3 )
By using Legendre’s duplication formula (1.2.6) we obtain Γ(x + 1 + u/2)Γ(x + 1/2 + u/2)Γ(1 − u/2 − x)Γ(1/2 − u/2 − x) = π Γ(2x + 1 + u)Γ(1 − u − 2x) for 2x + u = ±1, ±2, ±3, . . ., which implies that w∗6 (x) =
π Γ(2x + 1 + u)Γ(1 − u − 2x) Γ(x + 1)Γ(1 − u − x) Γ(x1 − u − x)Γ(x + x1 ) . × Γ(1 − x2 − x)Γ(1 − x3 − x)Γ(x + 1 + u − x2 )Γ(x + 1 + u − x3 )
If we now replace x by a + ix and u by −2a then we obtain w∗6 (a + ix) = π
2 Γ(1 + 2ix)Γ(x1 + a + ix) Γ(1 + a + ix)Γ(1 − a − x2 + ix)Γ(1 − a − x3 + ix) .
In this case the boundary conditions (7.6.6) can be satisfied by taking A = 0 and x2 = −N. By using (1.5.6) and (1.2.6) we obtain
178
8 Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations N
(6)
d0 =
∑ w∗6 (x)
x=0 N
=
Γ(x + 1 + u/2)Γ(x + 1/2 + u/2)Γ(1 − u/2 − x)Γ(1/2 − u/2 − x) Γ(x + 1)Γ(1 − u − x) x=0
∑
Γ(x1 − u − x)Γ(x + x1 ) Γ(1 + N − x)Γ(1 − x3 − x)Γ(x + 1 + u + N)Γ(x + 1 + u − x3 ) Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) = N! Γ(1 − u)Γ(1 + u + N)Γ(1 − x3 )Γ(1 + u − x3 ) ×
N
(−1)2x (−N)x (1 + u/2)x (u)x (x1 )x (x3 )x x=0 x! (u/2)x (1 + u − x1 )x (1 + u + N)x (1 + u − x3 )x
×∑
Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) N! Γ(1 − u)Γ(1 + u + N)Γ(1 − x3 )Γ(1 + u − x3 ) 1 + u/2, u, x1 , −N, x3 × 5 F4 ;1 u/2, 1 + u − x1 , 1 + u + N, 1 + u − x3 Γ(1 + u/2)Γ(1/2 + u/2)Γ(1 − u/2)Γ(1/2 − u/2)Γ(x1 )Γ(x1 − u) = N! Γ(1 − u)Γ(1 + u + N)Γ(1 − x3 )Γ(1 + u − x3 ) Γ(1 + u − x1 )Γ(1 + u + N)Γ(1 + u − x3 )Γ(1 + u − x1 − x3 + N) × Γ(1 + u)Γ(1 + u − x1 + N)Γ(1 + u − x1 − x3 )Γ(1 + u − x3 + N) π Γ(x1 )Γ(x1 − u)Γ(1 + u − x1 )Γ(1 − w) , = N! Γ(1 − x3 )Γ(1 + x1 − w)Γ(1 + x3 − w)Γ(1 − w − N) =
where w = x1 − N + x3 − u. Further we get by using (8.1.5) and (7.6.7) (6)
σn =
π Γ(w)Γ(1 − w)Γ(w − 1 + n)Γ(x1 − u)Γ(1 + u − x1 ) Γ(2n + w)Γ(2n − 1 + w)Γ(1 − x3 − n) Γ(n + x1 )(−1)n n! × Γ(1 + x1 − w − n)Γ(1 − N − w − n)Γ(1 + x3 − w − n)(N − n)!
for n = 0, 1, 2, . . . , N, which leads to the orthogonality relation (cf. (8.2.1)) Γ(1 − 2a + 2t)Γ(1 + 2a − 2t)Γ(x1 + 2a − t)Γ(x1 + t) Γ(1 + t)Γ(1 + 2a − t)Γ(1 + N − t)Γ(1 − x3 − t) t=0 N
∑
× =
1 yn (−(a − t)2 ) ym (−(a − t)2 ) Γ(1 − 2a + N + t)Γ(1 − 2a − x3 + t)
1 (6) σn δmn , π
m, n = 0, 1, 2, . . . , N.
(8.3.1)
If the boundary conditions (8.2.2) cannot be satisfied, one may use infinite sums. In that case we have w = x1 + x2 + x3 − u and
8.3 Orthogonality Relations for Wilson Polynomials (6)
σn =
179
π Γ(w)Γ(1 − w)Γ(w − 1 + n)Γ(x1 − u)Γ(1 + u − x1 ) Γ(2n + w)Γ(2n − 1 + w)Γ(1 − x2 − n)Γ(1 − x3 − n) Γ(n + x1 )(−1)n n! × Γ(1 + x1 − w − n)Γ(1 + x2 − w − n)Γ(1 + x3 − w − n)
for n = 0, 1, 2, . . . , N, which leads to the orthogonality relation (cf. (8.2.1)) ∞
Γ(1 − 2a + 2t)Γ(1 + 2a − 2t)Γ(x1 + 2a − t)Γ(x1 + t) Γ(1 + t)Γ(1 + 2a − t)Γ(1 − x2 − t)Γ(1 − x3 − t) t=0
∑
× =
1 yn (−(a − t)2 ) ym (−(a − t)2 ) Γ(1 − 2a − x2 + t)Γ(1 − 2a − x3 + t)
1 (6) σn δmn , π
m, n = 0, 1, 2, . . . , N.
(8.3.2)
Again we need the existence of the moments involved in order to have convergent sums. Table 7.2 through table 7.7 also hold for the Wilson polynomials yn (x2 ) given by (8.1.4). Almost all finite cases can be treated by using the weight function w∗6 (x) for (6) the Wilson polynomials. The positivity of d0 is dealt with in the same way as in the preceding section. The factor Γ(2a + x1 )Γ(1 − 2a − x1 ) with 2a + x1 = 0, ±1, ±2, . . . determines the sign which is necessary for positivity.
The Infinite Cases The infinite cases of table 7.2 through table 7.7 can be treated by using the weight function w∗4 (x) for the Racah polynomials again. By using Legendre’s duplication formula (1.2.6), we obtain Γ(x + u/2)Γ(x + 1/2 + u/2)Γ(−u/2 − x)Γ(1/2 − u/2 − x) = 4π Γ(2x + u)Γ(−u − 2x) for 2x + u = 0, ±1, ±2, . . ., which implies that w∗4 (x) =
Γ(x + u)Γ(−x)Γ(x + x1 )Γ(x1 − u − x) 4π Γ(2x + u)Γ(−u − 2x) × Γ(x + x2 )Γ(x2 − u − x)Γ(x + x3 )Γ(x3 − u − x)
If we now replace x by a + ix and u by −2a, then we obtain w∗4 (a + ix) =
1 4π
Γ(−a + ix)Γ(x1 + a + ix)Γ(x2 + a + ix)Γ(x3 + a + ix) 2 . Γ(2ix)
180
8 Orthogonal Polynomial Solutions in z(z + u) of Complex Difference Equations
In this case the boundary conditions (8.2.2) cannot be satisfied. However, as before, the concept of orthogonality can be extended to integration over the (possibly deformed) imaginary axis (cf. page 109). Then we need again the existence of the required moments which is in general taken care of by the asymptotic properties of the gamma functions involved. By using z = a + ξ with u = −2a, we obtain Γ(z − 2a)Γ(−z)Γ(z + x1 )Γ(x1 + 2a − z) Γ(z − a)Γ(z − a + 1/2)Γ(a − z)Γ(1/2 + a − z) × Γ(z + x2 )Γ(x2 + 2a − z)Γ(z + x3 )Γ(x3 + 2a − z) Γ(ξ − a)Γ(−a − ξ )Γ(ξ + a + x1 )Γ(x1 + a − ξ ) = Γ(ξ )Γ(ξ + 1/2)Γ(−ξ )Γ(1/2 − ξ ) × Γ(ξ + a + x2 )Γ(x2 + a − ξ )Γ(ξ + a + x3 )Γ(x3 + a − ξ ).
w∗4 (z) =
Analogous to (7.7.4) we obtain by using (1.6.6) 1 2π i
i∞ −i∞
w∗4 (a + ξ ) d ξ =
Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 )Γ(w − x2 )Γ(w − x3 ) 2π Γ(w)
which, if ξ = ix is used, leads to
(4) d0 =
∞
−∞
w∗4 (a + ix) dx =
Γ(x1 )Γ(x2 )Γ(x3 )Γ(w − x1 )Γ(w − x2 )Γ(w − x3 ) . Γ(w)
Then we obtain by using (8.1.5) (4)
σn =
Γ(n − 1 + w)Γ(n + x1 )Γ(n + x2 )Γ(n + x3 ) Γ(2n + w)Γ(2n − 1 + w) × Γ(n + w − x1 )Γ(n + w − x2 )Γ(n + w − x3 )n!
for n = 0, 1, 2, . . . and therefore the orthogonality relation 1 4π
2 ∞ Γ(−a + ix)Γ(x1 + a + ix)Γ(x2 + a + ix)Γ(x3 + a + ix) −∞
(4)
= σn δmn ,
Γ(2ix)
m, n = 0, 1, 2, . . . .
2 yn (x2 ) dx ym (x )
(8.3.3)
Almost all infinite cases of table 7.2 through table 7.7 can be treated by using this (4) weight function w∗4 (x). The positivity of d0 is treated as before.
8.3 Orthogonality Relations for Wilson Polynomials
181
In this chapter we have proved:
Theorem 8.1. The positive-definite orthogonal polynomial solutions y(z(z + u)) with z ∈ C and u ∈ R of the difference equation (7.2.1)
ϕ (z + 2)Δ 2 yn (z(z + u)) + ψ (z + 1)Δ yn (z(z + u)) = λn ρ (z + 1)yn ((z + 1)(z + 1 + u)), n = 0, 1, 2, . . . , consist of the polynomial solutions of the (real) difference equations (7.3.4) and (7.3.7). In fact this leads to both infinite and finite systems of • continuous dual Hahn polynomials and • Wilson polynomials.
Chapter 9
Hypergeometric Orthogonal Polynomials
In this chapter we deal with all families of hypergeometric orthogonal polynomials appearing in the Askey scheme on page 183. For each family of orthogonal polynomials we state the most important properties such as a representation as a hypergeometric function, orthogonality relation(s), the three-term recurrence relation, the second-order differential or difference equation, the forward shift (or degree lowering) and backward shift (or degree raising) operator, a Rodrigues-type formula and some generating functions. In each case we use the notation which seems to be most common in the literature. Moreover, in each case we mention the connection between various families by stating the appropriate limit relations. See also [500] for an algebraic approach of this Askey scheme and [498] for a view from asymptotic analysis. For notations the reader is referred to chapter 1.
9.1 Wilson Hypergeometric Representation Wn (x2 ; a, b, c, d) (a + b)n (a + c)n (a + d)n −n, n + a + b + c + d − 1, a + ix, a − ix ;1 . = 4 F3 a + b, a + c, a + d
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 9, © Springer-Verlag Berlin Heidelberg 2010
(9.1.1)
185
186
9 Hypergeometric Orthogonal Polynomials
Orthogonality Relation If Re(a, b, c, d) > 0 and non-real parameters occur in conjugate pairs, then 1 2π
∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 0
Γ(2ix)
×Wm (x2 ; a, b, c, d)Wn (x2 ; a, b, c, d) dx Γ(n + a + b) · · · Γ(n + c + d) (n + a + b + c + d − 1)n n! δmn , = Γ(2n + a + b + c + d)
(9.1.2)
where Γ(n + a + b) · · · Γ(n + c + d) = Γ(n + a + b)Γ(n + a + c)Γ(n + a + d)Γ(n + b + c)Γ(n + b + d)Γ(n + c + d). If a < 0 and a + b, a + c, a + d are positive or a pair of complex conjugates occur with positive real parts, then 1 2π
∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 0
Γ(2ix)
×Wm (x2 ; a, b, c, d)Wn (x2 ; a, b, c, d) dx Γ(a + b)Γ(a + c)Γ(a + d)Γ(b − a)Γ(c − a)Γ(d − a) + Γ(−2a) (2a)k (a + 1)k (a + b)k (a + c)k (a + d)k × ∑ (a)k (a − b + 1)k (a − c + 1)k (a − d + 1)k k! k=0,1,2... a+k<0
×Wm (−(a + k)2 ; a, b, c, d)Wn (−(a + k)2 ; a, b, c, d) Γ(n + a + b) · · · Γ(n + c + d) (n + a + b + c + d − 1)n n! δmn . (9.1.3) = Γ(2n + a + b + c + d)
Recurrence Relation − a2 + x2 W˜ n (x2 ) = AnW˜ n+1 (x2 ) − (An +Cn ) W˜ n (x2 ) +CnW˜ n−1 (x2 ), where W˜ n (x2 ) := W˜ n (x2 ; a, b, c, d) = and
Wn (x2 ; a, b, c, d) (a + b)n (a + c)n (a + d)n
(9.1.4)
9.1 Wilson
187
⎧ (n + a + b + c + d − 1)(n + a + b)(n + a + c)(n + a + d) ⎪ ⎪ ⎪ ⎨ An = (2n + a + b + c + d − 1)(2n + a + b + c + d) ⎪ n(n + b + c − 1)(n + b + d − 1)(n + c + d − 1) ⎪ ⎪ ⎩ Cn = . (2n + a + b + c + d − 2)(2n + a + b + c + d − 1)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn − a2 )pn (x) + An−1Cn pn−1 (x),
(9.1.5)
where Wn (x2 ; a, b, c, d) = (−1)n (n + a + b + c + d − 1)n pn (x2 ).
Difference Equation n(n + a + b + c + d − 1)y(x) = B(x)y(x + i) − [B(x) + D(x)] y(x) + D(x)y(x − i),
(9.1.6)
where y(x) = Wn (x2 ; a, b, c, d) ⎧ (a − ix)(b − ix)(c − ix)(d − ix) ⎪ ⎪ ⎪ B(x) = ⎨ 2ix(2ix − 1)
and
⎪ (a + ix)(b + ix)(c + ix)(d + ix) ⎪ ⎪ ⎩ D(x) = . 2ix(2ix + 1)
Forward Shift Operator Wn ((x + 12 i)2 ; a, b, c, d) −Wn ((x − 12 i)2 ; a, b, c, d) = −2inx(n + a + b + c + d − 1)Wn−1 (x2 ; a + 12 , b + 12 , c + 12 , d + 12 ) (9.1.7) or equivalently
δ Wn (x2 ; a, b, c, d) δ x2 = −n(n + a + b + c + d − 1)Wn−1 (x2 ; a + 12 , b + 12 , c + 12 , d + 12 ). (9.1.8)
188
9 Hypergeometric Orthogonal Polynomials
Backward Shift Operator (a − 12 − ix)(b − 12 − ix)(c − 12 − ix)(d − 12 − ix)Wn ((x + 12 i)2 ; a, b, c, d) − (a − 12 + ix)(b − 12 + ix)(c − 12 + ix)(d − 12 + ix)Wn ((x − 12 i)2 ; a, b, c, d) = −2ixWn+1 (x2 ; a − 12 , b − 12 , c − 12 , d − 12 )
(9.1.9)
or equivalently
δ ω (x; a, b, c, d)Wn (x2 ; a, b, c, d) δ x2 1 = ω (x; a − 2 , b − 12 , c − 12 , d − 12 )Wn+1 (x2 ; a − 12 , b − 12 , c − 12 , d − 12 ), (9.1.10) where
ω (x; a, b, c, d) :=
1 Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 . 2ix Γ(2ix)
Rodrigues-Type Formula ω (x; a, b, c, d)Wn (x2 ; a, b, c, d)
δ n ω (x; a + 12 n, b + 12 n, c + 12 n, d + 12 n) . = 2 δx
(9.1.11)
Generating Functions 2 F1
2 F1
2 F1
a + ix, b + ix ;t a+b a + ix, c + ix ;t a+c
2 F1
a + ix, d + ix ;t a+d
2 F1
2 F1
c − ix, d − ix ;t c+d b − ix, d − ix ;t b+d b − ix, c − ix ;t b+c
∞
=
Wn (x2 ; a, b, c, d)t n
∑ (a + b)n (c + d)n n! .
(9.1.12)
n=0
∞
=
Wn (x2 ; a, b, c, d)t n
∑ (a + c)n (b + d)n n! .
(9.1.13)
n=0
∞
=
Wn (x2 ; a, b, c, d)t n
∑ (a + d)n (b + c)n n! .
n=0
(9.1.14)
9.1 Wilson
189
(1 − t)1−a−b−c−d 1 (a + b + c + d − 1), 12 (a + b + c + d), a + ix, a − ix 4t ;− × 4 F3 2 a + b, a + c, a + d (1 − t)2 ∞ (a + b + c + d − 1)n Wn (x2 ; a, b, c, d)t n . (9.1.15) =∑ (a + b) (a + c) (a + d) n! n n n n=0
Limit Relations Wilson → Continuous Dual Hahn The continuous dual Hahn polynomials given by (9.3.1) can be found from the Wilson polynomials by dividing by (a + d)n and letting d → ∞: Wn (x2 ; a, b, c, d) = Sn (x2 ; a, b, c). d→∞ (a + d)n lim
(9.1.16)
Wilson → Continuous Hahn The continuous Hahn polynomials given by (9.4.1) are obtained from the Wilson polynomials by the substitutions a → a − it, b → b − it, c → c + it, d → d + it and x → x + t and the limit t → ∞ in the following way: Wn ((x + t)2 ; a − it, b − it, c + it, d + it) = pn (x; a, b, c, d). t→∞ (−2t)n n! lim
(9.1.17)
Wilson → Jacobi The Jacobi polynomials given by (9.8.1) can be found from the Wilson polynomials by substituting a = b = 12 (α + 1), c = 12 (β + 1) + it, d = 12 (β + 1) − it and x → t 12 (1 − x) in the definition (9.1.1) of the Wilson polynomials and taking the limit t → ∞. In fact we have Wn ( 12 (1 − x)t 2 ; 12 (α + 1), 12 (α + 1), 12 (β + 1) + it, 12 (β + 1) − it) t→∞ t 2n n! lim
(α ,β )
= Pn
(x).
(9.1.18)
190
9 Hypergeometric Orthogonal Polynomials
Remarks Note that for k < n we have (a + b)n (a + c)n (a + d)n = (a + b + k)n−k (a + c + k)n−k (a + d + k)n−k , (a + b)k (a + c)k (a + d)k which implies that the Wilson polynomials defined by (9.1.1) can also be seen as polynomials in the parameters a, b, c and d. If we set a = 12 (γ + δ + 1), b = 12 (2α − γ − δ + 1), c = 12 (2β − γ + δ + 1), and
d = 12 (γ − δ + 1)
ix → x + 12 (γ + δ + 1)
in W˜ n (x2 ; a, b, c, d) =
Wn (x2 ; a, b, c, d) , (a + b)n (a + c)n (a + d)n
given by (9.1.1) and take
α + 1 = −N
or
β + δ + 1 = −N
or
γ + 1 = −N
with N a nonnegative integer, we obtain the Racah polynomials given by (9.2.1).
References [51], [71], [72], [80], [277], [278], [281], [282], [339], [340], [387], [401], [407], [410], [503], [512], [513].
9.2 Racah Hypergeometric Representation Rn (λ (x); α , β , γ , δ ) −n, n + α + β + 1, −x, x + γ + δ + 1 = 4 F3 ;1 , α + 1, β + δ + 1, γ + 1 where
λ (x) = x(x + γ + δ + 1)
n = 0, 1, 2, . . . , N, (9.2.1)
9.2 Racah
191
and
α + 1 = −N
or
β + δ + 1 = −N
γ + 1 = −N
or
with N a nonnegative integer.
Orthogonality Relation (α + 1)x (β + δ + 1)x (γ + 1)x (γ + δ + 1)x ((γ + δ + 3)/2)x
N
∑ (−α + γ + δ + 1)x (−β + γ + 1)x ((γ + δ + 1)/2)x (δ + 1)x x!
x=0
× Rm (λ (x))Rn (λ (x)) (n + α + β + 1)n (α + β − γ + 1)n (α − δ + 1)n (β + 1)n n! δmn , (9.2.2) =M (α + β + 2)2n (α + 1)n (β + δ + 1)n (γ + 1)n where and
Rn (λ (x)) := Rn (λ (x); α , β , γ , δ ) ⎧ (−β )N (γ + δ + 2)N ⎪ ⎪ ⎪ ⎪ (− β + γ + 1)N (δ + 1)N ⎪ ⎪ ⎪ ⎪ ⎨ (−α + δ ) (γ + δ + 2) N N M= ⎪ (− α + γ + δ + 1) ( δ + 1) N N ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (α + β + 2)N (−δ )N ⎪ ⎩ (α − δ + 1)N (β + 1)N
if
α + 1 = −N
if
β + δ + 1 = −N
if
γ + 1 = −N.
Recurrence Relation λ (x)Rn (λ (x)) = An Rn+1 (λ (x)) − (An +Cn ) Rn (λ (x)) +Cn Rn−1 (λ (x)), where and
Rn (λ (x)) := Rn (λ (x); α , β , γ , δ ) ⎧ (n + α + 1)(n + α + β + 1)(n + β + δ + 1)(n + γ + 1) ⎪ ⎪ ⎪ ⎨ An = (2n + α + β + 1)(2n + α + β + 2) ⎪ n(n + α + β − γ )(n + α − δ )(n + β ) ⎪ ⎪ ⎩ Cn = , (2n + α + β )(2n + α + β + 1)
hence
(9.2.3)
192
9 Hypergeometric Orthogonal Polynomials
⎧ (n + β − N)(n + β + δ + 1)(n + γ + 1)(n − N) ⎪ ⎪ ⎪ ⎪ (2n + β − N)(2n + β − N + 1) ⎪ ⎪ ⎪ ⎪ ⎨ (n + α + 1)(n + α + β + 1)(n + γ + 1)(n − N) An = ⎪ (2n + α + β + 1)(2n + α + β + 2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (n + α + 1)(n + α + β + 1)(n + β + δ + 1)(n − N) ⎪ ⎩ (2n + α + β + 1)(2n + α + β + 2)
if α + 1 = −N if β + δ + 1 = −N if γ + 1 = −N
and ⎧ n(n + β )(n + β − γ − N − 1)(n − δ − N − 1) ⎪ ⎪ ⎪ ⎪ (2n + β − N − 1)(2n + β − N) ⎪ ⎪ ⎪ ⎪ ⎨ n(n + α + β + N + 1)(n + α + β − γ )(n + β ) Cn = ⎪ (2n + α + β )(2n + α + β + 1) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n(n + α + β + N + 1)(n + α − δ )(n + β ) ⎪ ⎩ (2n + α + β )(2n + α + β + 1)
if α + 1 = −N if β + δ + 1 = −N if γ + 1 = −N.
Normalized Recurrence Relation xpn (x) = pn+1 (x) − (An +Cn )pn (x) + An−1Cn pn−1 (x),
(9.2.4)
where Rn (λ (x); α , β , γ , δ ) =
(n + α + β + 1)n pn (λ (x)). (α + 1)n (β + δ + 1)n (γ + 1)n
Difference Equation n(n + α + β + 1)y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), (9.2.5) where and
y(x) = Rn (λ (x); α , β , γ , δ ) ⎧ (x + α + 1)(x + β + δ + 1)(x + γ + 1)(x + γ + δ + 1) ⎪ ⎪ ⎪ ⎨ B(x) = (2x + γ + δ + 1)(2x + γ + δ + 2) ⎪ x(x − α + γ + δ )(x − β + γ )(x + δ ) ⎪ ⎪ ⎩ D(x) = . (2x + γ + δ )(2x + γ + δ + 1)
9.2 Racah
193
Forward Shift Operator Rn (λ (x + 1); α , β , γ , δ ) − Rn (λ (x); α , β , γ , δ ) n(n + α + β + 1) = (α + 1)(β + δ + 1)(γ + 1) × (2x + γ + δ + 2)Rn−1 (λ (x); α + 1, β + 1, γ + 1, δ )
(9.2.6)
or equivalently
Δ Rn (λ (x); α , β , γ , δ ) Δ λ (x) n(n + α + β + 1) Rn−1 (λ (x); α + 1, β + 1, γ + 1, δ ). (9.2.7) = (α + 1)(β + δ + 1)(γ + 1)
Backward Shift Operator (x + α )(x + β + δ )(x + γ )(x + γ + δ )Rn (λ (x); α , β , γ , δ ) − x(x − β + γ )(x − α + γ + δ )(x + δ )Rn (λ (x − 1); α , β , γ , δ ) (9.2.8) = αγ (β + δ )(2x + γ + δ )Rn+1 (λ (x); α − 1, β − 1, γ − 1, δ ) or equivalently ∇ [ω (x; α , β , γ , δ )Rn (λ (x); α , β , γ , δ )] ∇λ (x) 1 = ω (x; α − 1, β − 1, γ − 1, δ )Rn+1 (λ (x); α − 1, β − 1, γ − 1, δ ), (9.2.9) γ +δ where
ω (x; α , β , γ , δ ) =
(α + 1)x (β + δ + 1)x (γ + 1)x (γ + δ + 1)x . (−α + γ + δ + 1)x (−β + γ + 1)x (δ + 1)x x!
Rodrigues-Type Formula ω (x; α , β , γ , δ )Rn (λ (x); α , β , γ , δ ) = (γ + δ + 1)n (∇λ )n [ω (x; α + n, β + n, γ + n, δ )] , where
(9.2.10)
194
9 Hypergeometric Orthogonal Polynomials
∇λ :=
∇ . ∇λ (x)
Generating Functions For x = 0, 1, 2, . . . , N we have −x, −x + α − γ − δ x + β + δ + 1, x + γ + 1 ; t 2 F1 ;t 2 F1 α +1 β +1 (β + δ + 1)n (γ + 1)n Rn (λ (x); α , β , γ , δ )t n , ( β + 1) n! n n=0 N
=
∑
if β + δ + 1 = −N 2 F1
−x, −x + β − γ ;t β +δ +1
2 F1
or γ + 1 = −N.
x + α + 1, x + γ + 1 ;t α −δ +1
(9.2.11)
(α + 1)n (γ + 1)n Rn (λ (x); α , β , γ , δ )t n , ( α − δ + 1) n! n n=0 N
=
∑
if α + 1 = −N 2 F1
−x, −x − δ ;t γ +1
2 F1
or
γ + 1 = −N.
x + α + 1, x + β + δ + 1 ;t α +β −γ +1
(9.2.12)
(α + 1)n (β + δ + 1)n Rn (λ (x); α , β , γ , δ )t n , n=0 (α + β − γ + 1)n n! N
=
∑
if
α + 1 = −N
or β + δ + 1 = −N.
(9.2.13)
(1 − t)−α −β −1 × 4 F3
1
1 4t 2 (α + β + 1), 2 (α + β + 2), −x, x + γ + δ + 1 ; − (1 − t)2 α + 1, β + δ + 1, γ + 1
(α + β + 1)n Rn (λ (x); α , β , γ , δ )t n . =∑ n! n=0
N
N
(9.2.14)
9.2 Racah
195
Limit Relations Racah → Hahn The Hahn polynomials given by (9.5.1) can be obtained from the Racah polynomials by taking γ + 1 = −N and letting δ → ∞: lim Rn (λ (x); α , β , −N − 1, δ ) = Qn (x; α , β , N).
δ →∞
(9.2.15)
The Hahn polynomials given by (9.5.1) can also be obtained from the Racah polynomials by taking δ = −β − N − 1 and letting γ → ∞: lim Rn (λ (x); α , β , γ , −β − N − 1) = Qn (x; α , β , N).
γ →∞
(9.2.16)
Another way to do this is to take α + 1 = −N and β → β + γ + N + 1 and then take the limit δ → ∞. In that case we obtain the Hahn polynomials given by (9.5.1) in the following way: lim Rn (λ (x); −N − 1, β + γ + N + 1, γ , δ ) = Qn (x; γ , β , N).
δ →∞
(9.2.17)
Racah → Dual Hahn The dual Hahn polynomials given by (9.6.1) are obtained from the Racah polynomials if we take α + 1 = −N and let β → ∞: lim Rn (λ (x); −N − 1, β , γ , δ ) = Rn (λ (x); γ , δ , N).
β →∞
(9.2.18)
The dual Hahn polynomials given by (9.6.1) are also obtained from the Racah polynomials if we take β = −δ − N − 1 and let α → ∞: lim Rn (λ (x); α , −δ − N − 1, γ , δ ) = Rn (λ (x); γ , δ , N).
α →∞
(9.2.19)
Finally, the dual Hahn polynomials given by (9.6.1) are also obtained from the Racah polynomials if we take γ + 1 = −N and δ → α + δ + N + 1 and take the limit β → ∞: lim Rn (λ (x); α , β , −N − 1, α + δ + N + 1) = Rn (λ (x); α , δ , N).
β →∞
(9.2.20)
196
9 Hypergeometric Orthogonal Polynomials
Remark If we set α = a + b − 1, β = c + d − 1, γ = a + d − 1, δ = a − d and x → −a + ix in the definition (9.2.1) of the Racah polynomials we obtain the Wilson polynomials given by (9.1.1): Rn (λ (−a + ix); a + b − 1, c + d − 1, a + d − 1, a − d) = W˜ n (x2 ; a, b, c, d) =
Wn (x2 ; a, b, c, d) . (a + b)n (a + c)n (a + d)n
References [51], [70], [72], [80], [82], [187], [340], [376], [381], [387], [416], [417], [426], [437], [439], [512].
9.3 Continuous Dual Hahn Hypergeometric Representation Sn (x2 ; a, b, c) = 3 F2 (a + b)n (a + c)n
−n, a + ix, a − ix ;1 . a + b, a + c
(9.3.1)
Orthogonality Relation If a, b and c are positive except possibly for a pair of complex conjugates with positive real parts, then 1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix) 2 2 2 Sm (x ; a, b, c)Sn (x ; a, b, c) dx 2π 0 Γ(2ix) = Γ(n + a + b)Γ(n + a + c)Γ(n + b + c)n! δmn . (9.3.2) If a < 0 and a + b, a + c are positive or a pair of complex conjugates with positive real parts, then
9.3 Continuous Dual Hahn
1 2π
197
2 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)
2 2 Sm (x ; a, b, c)Sn (x ; a, b, c) dx Γ(2ix) Γ(a + b)Γ(a + c)Γ(b − a)Γ(c − a) + Γ(−2a) (2a)k (a + 1)k (a + b)k (a + c)k (−1)k × ∑ (a)k (a − b + 1)k (a − c + 1)k k!
0
k=0,1,2... a+k<0
× Sm (−(a + k)2 ; a, b, c)Sn (−(a + k)2 ; a, b, c) = Γ(n + a + b)Γ(n + a + c)Γ(n + b + c)n! δmn .
(9.3.3)
Recurrence Relation − a2 + x2 S˜n (x2 ) = An S˜n+1 (x2 ) − (An +Cn ) S˜n (x2 ) +Cn S˜n−1 (x2 ), where
(9.3.4)
Sn (x2 ; a, b, c) S˜n (x2 ) := S˜n (x2 ; a, b, c) = (a + b)n (a + c)n ⎧ ⎨ An = (n + a + b)(n + a + c)
and
⎩
Cn = n(n + b + c − 1).
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn − a2 )pn (x) + An−1Cn pn−1 (x),
(9.3.5)
where Sn (x2 ; a, b, c) = (−1)n pn (x2 ).
Difference Equation ny(x) = B(x)y(x + i) − [B(x) + D(x)] y(x) + D(x)y(x − i), where y(x) = Sn (x2 ; a, b, c) and
(9.3.6)
198
9 Hypergeometric Orthogonal Polynomials
⎧ (a − ix)(b − ix)(c − ix) ⎪ ⎪ ⎪ ⎨ B(x) = 2ix(2ix − 1) ⎪ (a + ix)(b + ix)(c + ix) ⎪ ⎪ ⎩ D(x) = . 2ix(2ix + 1)
Forward Shift Operator Sn ((x + 12 i)2 ; a, b, c) − Sn ((x − 12 i)2 ; a, b, c) = −2inxSn−1 (x2 ; a + 12 , b + 12 , c + 12 )
(9.3.7)
or equivalently
δ Sn (x2 ; a, b, c) = −nSn−1 (x2 ; a + 12 , b + 12 , c + 12 ). δ x2
(9.3.8)
Backward Shift Operator (a − 12 − ix)(b − 12 − ix)(c − 12 − ix)Sn ((x + 12 i)2 ; a, b, c) − (a − 12 + ix)(b − 12 + ix)(c − 12 + ix)Sn ((x − 12 i)2 ; a, b, c) = −2ixSn+1 (x2 ; a − 12 , b − 12 , c − 12 )
(9.3.9)
or equivalently
δ ω (x; a, b, c)Sn (x2 ; a, b, c) δ x2 1 = ω (x; a − 2 , b − 12 , c − 12 )Sn+1 (x2 ; a − 12 , b − 12 , c − 12 ), where
(9.3.10)
1 Γ(a + ix)Γ(b + ix)Γ(c + ix) 2 ω (x; a, b, c) = . 2ix Γ(2ix)
Rodrigues-Type Formula
ω (x; a, b, c)Sn (x ; a, b, c) = 2
δ δ x2
n
ω (x; a + 12 n, b + 12 n, c + 12 n) .
(9.3.11)
9.3 Continuous Dual Hahn
199
Generating Functions (1 − t)−c+ix 2 F1 −b+ix
(1 − t)
2 F1
a + ix, b + ix ;t a+b a + ix, c + ix ;t a+c
∞
=
Sn (x2 ; a, b, c) n t . n=0 (a + b)n n!
=
Sn (x2 ; a, b, c) n t . n=0 (a + c)n n!
∑ ∞
∑
∞ b + ix, c + ix Sn (x2 ; a, b, c) n ;t = ∑ t . b+c n=0 (b + c)n n! ∞ a + ix, a − ix Sn (x2 ; a, b, c) n t ; −t = ∑ e 2 F2 t . a + b, a + c n=0 (a + b)n (a + c)n n!
(1 − t)−a+ix 2 F1
−γ
(1 − t)
3 F2
γ , a + ix, a − ix t ; a + b, a + c t −1
∞
=
(9.3.12)
(γ )n Sn (x2 ; a, b, c) n t , ∑ n=0 (a + b)n (a + c)n n!
(9.3.13)
(9.3.14)
(9.3.15)
γ arbitrary.
(9.3.16)
Limit Relations Wilson → Continuous Dual Hahn The continuous dual Hahn polynomials can be found from the Wilson polynomials given by (9.1.1) by dividing by (a + d)n and letting d → ∞: Wn (x2 ; a, b, c, d) = Sn (x2 ; a, b, c). d→∞ (a + d)n lim
Continuous Dual Hahn → Meixner-Pollaczek The Meixner-Pollaczek polynomials given by (9.7.1) can be obtained from the continuous dual Hahn polynomials by the substitutions x → x − t, a = λ + it, b = λ − it and c = t cot φ and the limit t → ∞: (λ )
Sn ((x − t)2 ; λ + it, λ − it,t cot φ ) Pn (x; φ ) = . t→∞ t n n! (sin φ )n lim
(9.3.17)
200
9 Hypergeometric Orthogonal Polynomials
Remark Since we have for k < n (a + b)n (a + c)n = (a + b + k)n−k (a + c + k)n−k , (a + b)k (a + c)k the continuous dual Hahn polynomials defined by (9.3.1) can also be seen as polynomials in the parameters a, b and c.
References [72], [280], [339], [340], [378], [380], [381], [388], [389], [410], [415], [503].
9.4 Continuous Hahn Hypergeometric Representation pn (x; a, b, c, d) −n, n + a + b + c + d − 1, a + ix n (a + c)n (a + d)n ; 1 . (9.4.1) =i 3 F2 a + c, a + d n!
Orthogonality Relation ¯ then If Re(a, b, c, d) > 0, c = a¯ and d = b, 1 2π =
∞ −∞
Γ(a + ix)Γ(b + ix)Γ(c − ix)Γ(d − ix)pm (x; a, b, c, d)pn (x; a, b, c, d) dx
Γ(n + a + c)Γ(n + a + d)Γ(n + b + c)Γ(n + b + d) δmn . (2n + a + b + c + d − 1)Γ(n + a + b + c + d − 1)n!
(9.4.2)
Recurrence Relation (a + ix) p˜n (x) = An p˜n+1 (x) − (An +Cn ) p˜n (x) +Cn p˜n−1 (x), where
(9.4.3)
9.4 Continuous Hahn
201
p˜n (x) := p˜n (x; a, b, c, d) = and
n! pn (x; a, b, c, d) in (a + c)n (a + d)n
⎧ (n + a + b + c + d − 1)(n + a + c)(n + a + d) ⎪ ⎪ ⎪ An = − ⎨ (2n + a + b + c + d − 1)(2n + a + b + c + d) ⎪ ⎪ ⎪ ⎩ Cn =
n(n + b + c − 1)(n + b + d − 1) . (2n + a + b + c + d − 2)(2n + a + b + c + d − 1)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + i(An +Cn + a)pn (x) − An−1Cn pn−1 (x), where pn (x; a, b, c, d) =
(9.4.4)
(n + a + b + c + d − 1)n pn (x). n!
Difference Equation n(n + a + b + c + d − 1)y(x) = B(x)y(x + i) − [B(x) + D(x)] y(x) + D(x)y(x − i),
(9.4.5)
where y(x) = pn (x; a, b, c, d) ⎧ ⎨ B(x) = (c − ix)(d − ix)
and
⎩
D(x) = (a + ix)(b + ix).
Forward Shift Operator pn (x + 12 i; a, b, c, d) − pn (x − 12 i; a, b, c, d) = i(n + a + b + c + d − 1)pn−1 (x; a + 12 , b + 12 , c + 12 , d + 12 )
(9.4.6)
or equivalently
δ pn (x; a, b, c, d) = (n + a + b + c + d − 1)pn−1 (x; a + 12 , b + 12 , c + 12 , d + 12 ). (9.4.7) δx
202
9 Hypergeometric Orthogonal Polynomials
Backward Shift Operator (c − 12 − ix)(d − 12 − ix)pn (x + 12 i; a, b, c, d) − (a − 12 + ix)(b − 12 + ix)pn (x − 12 i; a, b, c, d) n+1 pn+1 (x; a − 12 , b − 12 , c − 12 , d − 12 ) = i
(9.4.8)
or equivalently
δ [ω (x; a, b, c, d)pn (x; a, b, c, d)] δx = −(n + 1)ω (x; a − 12 , b − 12 , c − 12 , d − 12 ) × pn+1 (x; a − 12 , b − 12 , c − 12 , d − 12 ), where
(9.4.9)
ω (x; a, b, c, d) = Γ(a + ix)Γ(b + ix)Γ(c − ix)Γ(d − ix).
Rodrigues-Type Formula ω (x; a, b, c, d)pn (x; a, b, c, d)
(−1)n δ n ω (x; a + 12 n, b + 12 n, c + 12 n, d + 12 n) . = n! δx
(9.4.10)
Generating Functions 1 F1
1 F1
a + ix ; −it a+c a + ix ; −it a+d
(1 − t)1−a−b−c−d 3 F2 ∞
=
1 F1
1 F1
d − ix ; it b+d c − ix ; it b+c
∞
=
(9.4.11)
n=0
∞
=
pn (x; a, b, c, d)
∑ (a + d)n (b + c)n t n .
(9.4.12)
n=0
1
1 2 (a + b + c + d − 1), 2 (a + b + c + d), a + ix
a + c, a + d
(a + b + c + d − 1)n p (x; a, b, c, d)t n . n n (a + c) (a + d) i n n n=0
∑
pn (x; a, b, c, d)
∑ (a + c)n (b + d)n t n .
;−
4t (1 − t)2
(9.4.13)
9.4 Continuous Hahn
203
Limit Relations Wilson → Continuous Hahn The continuous Hahn polynomials are obtained from the Wilson polynomials given by (9.1.1) by the substitution a → a − it, b → b − it, c → c + it, d → d + it and x → x + t and the limit t → ∞ in the following way: Wn ((x + t)2 ; a − it, b − it, c + it, d + it) = pn (x; a, b, c, d). t→∞ (−2t)n n! lim
Continuous Hahn → Meixner-Pollaczek The Meixner-Pollaczek polynomials given by (9.7.1) can be obtained from the continuous Hahn polynomials by setting x → x + t, a = λ − it, c = λ + it and b = d = t tan φ and taking the limit t → ∞: (λ )
lim
t→∞
pn (x + t; λ − it,t tan φ , λ + it,t tan φ ) Pn (x; φ ) = . tn (cos φ )n
(9.4.14)
Continuous Hahn → Jacobi The Jacobi polynomials given by (9.8.1) follow from the continuous Hahn polynomials by the substitution x → 12 xt, a = 12 (α +1−it), b = 12 (β +1+it), c = 12 (α +1+it) and d = 12 (β + 1 − it), division by t n and the limit t → ∞: lim
t→∞
pn ( 12 xt; 12 (α + 1 − it), 12 (β + 1 + it), 12 (α + 1 + it), 12 (β + 1 − it)) tn (α ,β )
= Pn
(x).
(9.4.15)
Continuous Hahn → Pseudo Jacobi The pseudo Jacobi polynomials given by (9.9.1) follow from the continuous Hahn polynomials by the substitution x → xt, a = 12 (−N + iν − 2t), b = 12 (−N − iν + 2t), c = 12 (−N − iν − 2t) and d = 12 (−N + iν + 2t), division by t n and the limit t → ∞: pn (xt; 12 (−N + iν − 2t), 12 (−N − iν + 2t), 12 (−N + iν − 2t), 12 (−N − iν + 2t)) t→∞ tn (n − 2N − 1)n Pn (x; ν , N). = (9.4.16) n! lim
204
9 Hypergeometric Orthogonal Polynomials
Remark Since we have for k < n (a + b)n (a + c)n = (a + b + k)n−k (a + c + k)n−k , (a + b)k (a + c)k the continuous Hahn polynomials defined by (9.4.1) can also be seen as polynomials in the parameters a, b and c.
References [48], [51], [80], [81], [89], [258], [327], [340], [378], [381], [383].
9.5 Hahn Hypergeometric Representation Qn (x; α , β , N) = 3 F2
−n, n + α + β + 1, −x ;1 , α + 1, −N
n = 0, 1, 2, . . . , N.
(9.5.1)
Orthogonality Relation For α > −1 and β > −1, or for α < −N and β < −N, we have N
∑
x=0
=
α +x x
β +N −x Qm (x; α , β , N)Qn (x; α , β , N) N −x
(−1)n (n + α + β + 1)N+1 (β + 1)n n! δmn . (2n + α + β + 1)(α + 1)n (−N)n N!
(9.5.2)
Recurrence Relation − xQn (x) = An Qn+1 (x) − (An +Cn ) Qn (x) +Cn Qn−1 (x), where
Qn (x) := Qn (x; α , β , N)
(9.5.3)
9.5 Hahn
205
⎧ (n + α + β + 1)(n + α + 1)(N − n) ⎪ ⎪ ⎪ ⎨ An = (2n + α + β + 1)(2n + α + β + 2)
and
⎪ n(n + α + β + N + 1)(n + β ) ⎪ ⎪ ⎩ Cn = . (2n + α + β )(2n + α + β + 1)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn ) pn (x) + An−1Cn pn−1 (x), where Qn (x; α , β , N) =
(9.5.4)
(n + α + β + 1)n pn (x). (α + 1)n (−N)n
Difference Equation n(n + α + β + 1)y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), (9.5.5) where
y(x) = Qn (x; α , β , N) ⎧ ⎨ B(x) = (x + α + 1)(x − N)
and
⎩
D(x) = x(x − β − N − 1).
Forward Shift Operator Qn (x + 1; α , β , N) − Qn (x; α , β , N) n(n + α + β + 1) Qn−1 (x; α + 1, β + 1, N − 1) =− (α + 1)N
(9.5.6)
or equivalently
Δ Qn (x; α , β , N) = −
n(n + α + β + 1) Qn−1 (x; α + 1, β + 1, N − 1). (α + 1)N
(9.5.7)
206
9 Hypergeometric Orthogonal Polynomials
Backward Shift Operator (x + α )(N + 1 − x)Qn (x; α , β , N) − x(β + N + 1 − x)Qn (x − 1; α , β , N) (9.5.8) = α (N + 1)Qn+1 (x; α − 1, β − 1, N + 1) or equivalently ∇ [ω (x; α , β , N)Qn (x; α , β , N)] N +1 ω (x; α − 1, β − 1, N + 1)Qn+1 (x; α − 1, β − 1, N + 1), = β where
ω (x; α , β , N) =
(9.5.9)
α +x β +N −x . x N −x
Rodrigues-Type Formula ω (x; α , β , N)Qn (x; α , β , N) (−1)n (β + 1)n n ∇ [ω (x; α + n, β + n, N − n)] . = (−N)n
(9.5.10)
Generating Functions For x = 0, 1, 2, . . . , N we have 1 F1
−x ; −t α +1 2 F0
1 F1
x−N ;t β +1
N
=
(−N)n
∑ (β + 1)n n! Qn (x; α , β , N)t n .
(9.5.11)
n=0
−x, −x + β + N + 1 ; −t −
2 F0
x − N, x + α + 1 ;t −
(−N)n (α + 1)n Qn (x; α , β , N)t n . n! n=0
N
=
∑
(9.5.12)
9.5 Hahn
207
(1 − t)−α −β −1 3 F2
1
2 (α + β
4t + 1), 12 (α + β + 2), −x ;− α + 1, −N (1 − t)2
(α + β + 1)n Qn (x; α , β , N)t n . =∑ n! n=0
N
N
(9.5.13)
Limit Relations Racah → Hahn If we take γ + 1 = −N and let δ → ∞ in the definition (9.2.1) of the Racah polynomials, we obtain the Hahn polynomials. Hence lim Rn (λ (x); α , β , −N − 1, δ ) = Qn (x; α , β , N).
δ →∞
And if we take δ = −β − N − 1 and let γ → ∞ in the definition (9.2.1) of the Racah polynomials, we also obtain the Hahn polynomials: lim Rn (λ (x); α , β , γ , −β − N − 1) = Qn (x; α , β , N).
γ →∞
Another way to do this is to take α + 1 = −N and β → β + γ + N + 1 in the definition (9.2.1) of the Racah polynomials and then take the limit δ → ∞. In that case we obtain the Hahn polynomials in the following way: lim Rn (λ (x); −N − 1, β + γ + N + 1, γ , δ ) = Qn (x; γ , β , N).
δ →∞
Hahn → Jacobi To find the Jacobi polynomials given by (9.8.1) from the Hahn polynomials we take x → Nx and let N → ∞. In fact we have lim Qn (Nx; α , β , N) =
N→∞
(α ,β )
Pn
(1 − 2x)
(α ,β ) Pn (1)
.
(9.5.14)
Hahn → Meixner The Meixner polynomials given by (9.10.1) can be obtained from the Hahn polynomials by taking α = b − 1, β = N(1 − c)c−1 and letting N → ∞: lim Qn (x; b − 1, N(1 − c)c−1 , N) = Mn (x; b, c).
N→∞
(9.5.15)
208
9 Hypergeometric Orthogonal Polynomials
Hahn → Krawtchouk The Krawtchouk polynomials given by (9.11.1) are obtained from the Hahn polynomials if we take α = pt and β = (1 − p)t and let t → ∞: lim Qn (x; pt, (1 − p)t, N) = Kn (x; p, N).
t→∞
(9.5.16)
Remark If we interchange the role of x and n in (9.5.1) we obtain the dual Hahn polynomials given by (9.6.1). Since Qn (x; α , β , N) = Rx (λ (n); α , β , N) we obtain the dual orthogonality relation for the Hahn polynomials from the orthogonality relation (9.6.2) of the dual Hahn polynomials: N
(2n + α + β + 1)(α + 1)n (−N)n N!
∑ (−1)n (n + α + β + 1)N+1 (β + 1)n n! Qn (x; α , β , N)Qn (y; α , β , N)
n=0
δ xy , = α +x β +N −x x N −x
x, y ∈ {0, 1, 2, . . . , N}.
References [16], [34], [39], [46], [51], [54], [59], [72], [80], [82], [146], [152], [155], [164], [184], [185], [226], [228], [265], [268], [277], [308], [337], [340], [363], [364], [367], [372], [374], [375], [377], [381], [391], [415], [416], [417], [432], [434], [435], [440], [465], [488], [489], [512], [515], [521].
9.6 Dual Hahn Hypergeometric Representation Rn (λ (x); γ , δ , N) = 3 F2 where
−n, −x, x + γ + δ + 1 ;1 , γ + 1, −N
λ (x) = x(x + γ + δ + 1).
n = 0, 1, 2, . . . , N, (9.6.1)
9.6 Dual Hahn
209
Orthogonality Relation For γ > −1 and δ > −1, or for γ < −N and δ < −N, we have (2x + γ + δ + 1)(γ + 1)x (−N)x N!
N
∑ (−1)x (x + γ + δ + 1)N+1 (δ + 1)x x! Rm (λ (x); γ , δ , N)Rn (λ (x); γ , δ , N)
x=0
δ mn . = γ +n δ +N −n n N −n
(9.6.2)
Recurrence Relation λ (x)Rn (λ (x)) = An Rn+1 (λ (x)) − (An +Cn ) Rn (λ (x)) +Cn Rn−1 (λ (x)), where
(9.6.3)
Rn (λ (x)) := Rn (λ (x); γ , δ , N) ⎧ ⎨ An = (n + γ + 1)(n − N)
and
⎩
Cn = n(n − δ − N − 1).
Normalized Recurrence Relation xpn (x) = pn+1 (x) − (An +Cn )pn (x) + An−1Cn pn−1 (x), where Rn (λ (x); γ , δ , N) =
(9.6.4)
1 pn (λ (x)). (γ + 1)n (−N)n
Difference Equation − ny(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where and
y(x) = Rn (λ (x); γ , δ , N)
(9.6.5)
210
9 Hypergeometric Orthogonal Polynomials
⎧ (x + γ + 1)(x + γ + δ + 1)(N − x) ⎪ ⎪ ⎪ ⎨ B(x) = (2x + γ + δ + 1)(2x + γ + δ + 2) ⎪ x(x + γ + δ + N + 1)(x + δ ) ⎪ ⎪ ⎩ D(x) = . (2x + γ + δ )(2x + γ + δ + 1)
Forward Shift Operator Rn (λ (x + 1); γ , δ , N) − Rn (λ (x); γ , δ , N) n(2x + γ + δ + 2) Rn−1 (λ (x); γ + 1, δ , N − 1) =− (γ + 1)N
(9.6.6)
or equivalently
Δ Rn (λ (x); γ , δ , N) n =− Rn−1 (λ (x); γ + 1, δ , N − 1). Δ λ (x) (γ + 1)N
(9.6.7)
Backward Shift Operator (x + γ )(x + γ + δ )(N + 1 − x)Rn (λ (x); γ , δ , N) − x(x + γ + δ + N + 1)(x + δ )Rn (λ (x − 1); γ , δ , N) = γ (N + 1)(2x + γ + δ )Rn+1 (λ (x); γ − 1, δ , N + 1)
(9.6.8)
or equivalently ∇ [ω (x; γ , δ , N)Rn (λ (x); γ , δ , N)] ∇λ (x) 1 = ω (x; γ − 1, δ , N + 1)Rn+1 (λ (x); γ − 1, δ , N + 1), γ +δ where
ω (x; γ , δ , N) =
(−1)x (γ + 1)x (γ + δ + 1)x (−N)x . (γ + δ + N + 2)x (δ + 1)x x!
(9.6.9)
9.6 Dual Hahn
211
Rodrigues-Type Formula ω (x; γ , δ , N)Rn (λ (x); γ , δ , N) = (γ + δ + 1)n (∇λ )n [ω (x; γ + n, δ , N − n)] , (9.6.10) where ∇λ :=
∇ . ∇λ (x)
Generating Functions For x = 0, 1, 2, . . . , N we have (1 − t)N−x 2 F1
−x, −x − δ ;t γ +1
(1 − t)x 2 F1
N
=
(−N)n Rn (λ (x); γ , δ , N)t n . n! n=0
∑
x − N, x + γ + 1 ;t −δ − N
(9.6.11)
(γ + 1)n (−N)n Rn (λ (x); γ , δ , N)t n . n=0 (−δ − N)n n! N
=
∑
(9.6.12)
N −x, x + γ + δ + 1 Rn (λ (x); γ , δ , N) n t ; −t t . =∑ e 2 F2 γ + 1, −N n! N n=0
−ε
(1 − t)
3 F2
ε , −x, x + γ + δ + 1 t ; γ + 1, −N t −1
(ε )n Rn (λ (x); γ , δ , N)t n , n! n=0 N
=
∑
(9.6.13)
N
ε arbitrary.
(9.6.14)
Limit Relations Racah → Dual Hahn If we take α + 1 = −N and let β → ∞ in the definition (9.2.1) of the Racah polynomials, then we obtain the dual Hahn polynomials: lim Rn (λ (x); −N − 1, β , γ , δ ) = Rn (λ (x); γ , δ , N).
β →∞
212
9 Hypergeometric Orthogonal Polynomials
And if we take β = −δ − N − 1 and let α → ∞ in the definition (9.2.1) of the Racah polynomials, then we also obtain the dual Hahn polynomials: lim Rn (λ (x); α , −δ − N − 1, γ , δ ) = Rn (λ (x); γ , δ , N).
α →∞
Finally, if we take γ + 1 = −N and δ → α + δ + N + 1 in the definition (9.2.1) of the Racah polynomials and take the limit β → ∞ we find the dual Hahn polynomials in the following way: lim Rn (λ (x); α , β , −N − 1, α + δ + N + 1) = Rn (λ (x); α , δ , N).
β →∞
Dual Hahn → Meixner The Meixner polynomials given by (9.10.1) are obtained from the dual Hahn polynomials if we take γ = β − 1 and δ = N(1 − c)c−1 and let N → ∞: lim Rn (λ (x); β − 1, N(1 − c)c−1 , N = Mn (x; β , c).
N→∞
(9.6.15)
Dual Hahn → Krawtchouk The Krawtchouk polynomials given by (9.11.1) can be obtained from the dual Hahn polynomials by setting γ = pt, δ = (1 − p)t and letting t → ∞: lim Rn (λ (x); pt, (1 − p)t, N) = Kn (x; p, N).
t→∞
(9.6.16)
Remark If we interchange the role of x and n in the definition (9.6.1) of the dual Hahn polynomials we obtain the Hahn polynomials given by (9.5.1). Since Rn (λ (x); γ , δ , N) = Qx (n; γ , δ , N) we obtain the dual orthogonality relation for the dual Hahn polynomials from the orthogonality relation (9.5.2) for the Hahn polynomials: γ +n δ +N −n Rn (λ (x); γ , δ , N)Rn (λ (y); γ , δ , N) ∑ n N −n n=0 N
=
(−1)x (x + γ + δ + 1)N+1 (δ + 1)x x! δxy , (2x + γ + δ + 1)(γ + 1)x (−N)x N!
x, y ∈ {0, 1, 2, . . . , N}.
9.7 Meixner-Pollaczek
213
References [54], [72], [80], [82], [277], [308], [337], [340], [376], [377], [380], [381], [416], [417], [439], [488], [512].
9.7 Meixner-Pollaczek Hypergeometric Representation (λ ) Pn (x; φ ) =
(2λ )n inφ e 2 F1 n!
−n, λ + ix −2iφ . ; 1−e 2λ
(9.7.1)
Orthogonality Relation 1 2π =
∞ −∞
(λ )
(λ )
e(2φ −π )x |Γ(λ + ix)|2 Pm (x; φ )Pn (x; φ ) dx
Γ(n + 2λ ) δmn , (2 sin φ )2λ n!
λ > 0 and 0 < φ < π .
(9.7.2)
Recurrence Relation (λ )
(λ )
(n + 1)Pn+1 (x; φ ) − 2 [x sin φ + (n + λ ) cos φ ] Pn (x; φ ) (λ )
+ (n + 2λ − 1)Pn−1 (x; φ ) = 0.
(9.7.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) −
n+λ tan φ
pn (x) +
where (λ )
Pn (x; φ ) =
n(n + 2λ − 1) pn−1 (x), 4 sin2 φ
(2 sin φ )n pn (x). n!
(9.7.4)
214
9 Hypergeometric Orthogonal Polynomials
Difference Equation eiφ (λ − ix)y(x + i) + 2i [x cos φ − (n + λ ) sin φ ] y(x) − e−iφ (λ + ix)y(x − i) = 0,
(λ )
y(x) = Pn (x; φ ).
(9.7.5)
Forward Shift Operator (λ )
(λ + 1 )
(λ )
Pn (x + 12 i; φ ) − Pn (x − 12 i; φ ) = (eiφ − e−iφ )Pn−1 2 (x; φ ) or equivalently
(9.7.6)
(λ )
δ Pn (x; φ ) (λ + 1 ) = 2 sin φ Pn−1 2 (x; φ ). δx
(9.7.7)
Backward Shift Operator (λ )
(λ )
eiφ (λ − 12 − ix)Pn (x + 12 i; φ ) + e−iφ (λ − 12 + ix)Pn (x − 12 i; φ ) (λ − 1 )
= (n + 1)Pn+1 2 (x; φ ) or equivalently (λ ) δ ω (x; λ , φ )Pn (x; φ )
δx where
(9.7.8)
(λ − 1 )
= −(n + 1)ω (x; λ − 12 , φ )Pn+1 2 (x; φ ),
(9.7.9)
ω (x; λ , φ ) = Γ(λ + ix)Γ(λ − ix)e(2φ −π )x .
Rodrigues-Type Formula (λ ) ω (x; λ , φ )Pn (x; φ ) =
(−1)n n!
δ δx
n
ω (x; λ + 12 n, φ ) .
(9.7.10)
9.7 Meixner-Pollaczek
215
Generating Functions (1 − eiφ t)−λ +ix (1 − e−iφ t)−λ −ix =
∞
(λ )
∑ Pn
(x; φ )t n .
(9.7.11)
Pn (x; φ ) ∑ (2λ )n einφ t n . n=0
(9.7.12)
n=0
et 1 F1
λ + ix ; (e−2iφ − 1)t 2λ
(1 − t)−γ 2 F1 ∞
=
∞
=
(λ )
γ , λ + ix (1 − e−2iφ )t ; 2λ t −1 (λ )
(γ )n Pn (x; φ ) ∑ (2λ )n einφ t n , n=0
γ arbitrary.
(9.7.13)
Limit Relations Continuous Dual Hahn → Meixner-Pollaczek The Meixner-Pollaczek polynomials can be obtained from the continuous dual Hahn polynomials given by (9.3.1) by the substitutions x → x − t, a = λ + it, b = λ − it and c = t cot φ and the limit t → ∞: (λ )
Sn ((x − t)2 ; λ + it, λ − it,t cot φ ) Pn (x; φ ) = . t→∞ t n n! (sin φ )n lim
Continuous Hahn → Meixner-Pollaczek By setting x → x + t, a = λ − it, c = λ + it and b = d = t tan φ in the definition (9.4.1) of the continuous Hahn polynomials and taking the limit t → ∞ we obtain the Meixner-Pollaczek polynomials: (λ )
lim
t→∞
pn (x + t; λ − it,t tan φ , λ + it,t tan φ ) Pn (x; φ ) = . t n n! (cos φ )n
216
9 Hypergeometric Orthogonal Polynomials
Meixner-Pollaczek → Laguerre The Laguerre polynomials given by (9.12.1) can be obtained from the MeixnerPollaczek polynomials by the substitution λ = 12 (α + 1), x → − 12 φ −1 x and the limit φ → 0: ( 1 α + 12 )
lim Pn 2
φ →0
(α )
(− 12 φ −1 x; φ ) = Ln (x).
(9.7.14)
Meixner-Pollaczek → Hermite The Hermite polynomials given by (9.15.1) are √obtained from the Meixner-Pollaczek polynomials if we substitute x → (sin φ )−1 (x λ − λ cos φ ) and then let λ → ∞: √ 1 Hn (x) (λ ) . lim λ − 2 n Pn ((sin φ )−1 (x λ − λ cos φ ); φ ) = n! λ →∞
(9.7.15)
Remark Since we have for k < n
(2λ )n = (2λ + k)n−k , (2λ )k
the Meixner-Pollaczek polynomials defined by (9.7.1) can also be seen as polynomials in the parameter λ .
References [16], [20], [34], [36], [37], [51], [72], [80], [82], [135], [138], [146], [270], [273], [277], [283], [295], [317], [340], [342], [363], [364], [381], [392], [406], [416], [434], [512], [517].
9.8 Jacobi Hypergeometric Representation (α ,β )
Pn
(x) =
(α + 1)n 2 F1 n!
−n, n + α + β + 1 1 − x ; . α +1 2
(9.8.1)
9.8 Jacobi
217
Orthogonality Relation For α > −1 and β > −1 we have 1
(α ,β )
−1
=
(1 − x)α (1 + x)β Pm
(α ,β )
(x)Pn
(x) dx
Γ(n + α + 1)Γ(n + β + 1) 2α +β +1 δmn . 2n + α + β + 1 Γ(n + α + β + 1)n!
(9.8.2)
For α + β < −2N − 1, β > −1 and m, n ∈ {0, 1, 2, . . . , N} we also have ∞ 1
(α ,β )
(x + 1)α (x − 1)β Pm
=−
(α ,β )
(−x)Pn
(−x) dx
2α +β +1 Γ(−n − α − β )Γ(n + α + β + 1) δmn . 2n + α + β + 1 Γ(−n − α )n!
(9.8.3)
Recurrence Relation (α ,β )
xPn
(x) =
2(n + 1)(n + α + β + 1) (α ,β ) P (x) (2n + α + β + 1)(2n + α + β + 2) n+1 +
β 2 − α2 (α ,β ) Pn (x) (2n + α + β )(2n + α + β + 2) 2(n + α )(n + β ) (α ,β ) + P (x). (9.8.4) (2n + α + β )(2n + α + β + 1) n−1
Normalized Recurrence Relation β 2 − α2 pn (x) (2n + α + β )(2n + α + β + 2) 4n(n + α )(n + β )(n + α + β ) pn−1 (x), (9.8.5) + (2n + α + β − 1)(2n + α + β )2 (2n + α + β + 1)
xpn (x) = pn+1 (x) +
where (α ,β )
Pn
(x) =
(n + α + β + 1)n pn (x). 2n n!
218
9 Hypergeometric Orthogonal Polynomials
Differential Equation (1 − x2 )y (x) + [β − α − (α + β + 2)x] y (x) + n(n + α + β + 1)y(x) = 0,
(α ,β )
y(x) = Pn
(x).
(9.8.6)
Forward Shift Operator n + α + β + 1 (α +1,β +1) d (α ,β ) Pn Pn−1 (x) = (x). dx 2
(9.8.7)
Backward Shift Operator d (α ,β ) (α ,β ) Pn (x) + [(β − α ) − (α + β )x] Pn (x) dx (α −1,β −1) (x) = −2(n + 1)Pn+1
(9.8.8)
d (α ,β ) (1 − x)α (1 + x)β Pn (x) dx (α −1,β −1) (x). = −2(n + 1)(1 − x)α −1 (1 + x)β −1 Pn+1
(9.8.9)
(1 − x2 )
or equivalently
Rodrigues-Type Formula α
(1 − x)
(α ,β ) (1 + x)β Pn (x) =
(−1)n 2n n!
d dx
n
(1 − x)n+α (1 + x)n+β .
(9.8.10)
Generating Functions ∞ 2α + β (α ,β ) = Pn (x)t n , ∑ R(1 + R − t)α (1 + R + t)β n=0
R=
1 − 2xt + t 2 .
(9.8.11)
9.8 Jacobi
219
0 F1
− (x − 1)t ; α +1 2 (α ,β )
∞
=
Pn
0 F1
− (x + 1)t ; β +1 2
(x)
∑ (α + 1)n (β + 1)n t n .
(9.8.12)
n=0
−α −β −1
(1 − t)
1 2 F1
2 (α + β
∞
=
+ 1), 12 (α + β + 2) 2(x − 1)t ; α +1 (1 − t)2
(α + β + 1)n (α ,β ) Pn (x)t n . n=0 (α + 1)n
∑
−α −β −1
(1 + t)
1 2 F1
2 (α + β
+ 1), 12 (α + β + 2) 2(x + 1)t ; (1 + t)2 β +1
(9.8.13)
∞
=
(α + β + 1)n (α ,β ) Pn (x)t n . (β + 1)n n=0
∑
(9.8.14)
γ, α + β + 1 − γ 1 − R − t γ, α + β + 1 − γ 1 − R + t ; ; 2 F1 2 F1 α +1 β +1 2 2 ∞ (γ )n (α + β + 1 − γ )n (α ,β ) Pn (x)t n , R = 1 − 2xt + t 2 (9.8.15) =∑ n=0 (α + 1)n (β + 1)n
with γ arbitrary.
Limit Relations Wilson → Jacobi The Jacobi polynomials can be found from the Wilson polynomials given by (9.1.1) by substituting a = b = 12 (α + 1), c = 12 (β + 1) + it, d = 12 (β + 1) − it and x → t 12 (1 − x) in the definition (9.1.1) of the Wilson polynomials and taking the limit t → ∞. In fact we have Wn ( 12 (1 − x)t 2 ; 12 (α + 1), 12 (α + 1), 12 (β + 1) + it, 12 (β + 1) − it) (α ,β ) = Pn (x). t→∞ t 2n n! lim
220
9 Hypergeometric Orthogonal Polynomials
Continuous Hahn → Jacobi The Jacobi polynomials follow from the continuous Hahn polynomials given by (9.4.1) by using the substitution x → 12 xt, a = 12 (α + 1 − it), b = 12 (β + 1 + it), c = 12 (α + 1 + it) and d = 12 (β + 1 − it) in (9.4.1), division by t n and the limit t → ∞: lim
t→∞
pn ( 12 xt; 12 (α + 1 − it), 12 (β + 1 + it), 12 (α + 1 + it), 12 (β + 1 − it)) (α ,β ) = Pn (x). tn
Hahn → Jacobi To find the Jacobi polynomials from the Hahn polynomials given by (9.5.1) we take x → Nx in (9.5.1) and let N → ∞. In fact we have lim Qn (Nx; α , β , N) =
(α ,β )
Pn
(1 − 2x)
(α ,β ) Pn (1)
N→∞
.
Jacobi → Laguerre The Laguerre polynomials given by (9.12.1) can be obtained from the Jacobi polynomials by setting x → 1 − 2β −1 x and then the limit β → ∞: (α ,β )
lim Pn
β →∞
(α )
(1 − 2β −1 x) = Ln (x).
(9.8.16)
Jacobi → Bessel The Bessel polynomials given by (9.13.1) are obtained from the Jacobi polynomials if we take β = a − α and let α → −∞: (α ,a−α )
lim
α →−∞
Pn
(1 + α x)
(α ,a−α ) Pn (1)
= yn (x; a).
(9.8.17)
Jacobi → Hermite The Hermite polynomials given by (9.15.1) follow from the Jacobi polynomials by taking β = α and letting α → ∞ in the following way: 1
(α ,α )
lim α − 2 n Pn
α →∞
1
(α − 2 x) =
Hn (x) . 2n n!
(9.8.18)
9.8 Jacobi
221
Remarks The definition (9.8.1) of the Jacobi polynomials can also be written as: (α ,β )
Pn
(x) =
1 n (−n)k ∑ k! (n + α + β + 1)k (α + k + 1)n−k n! k=0
1−x 2
k .
In this way the Jacobi polynomials can also be seen as polynomials in the parameters α and β . Therefore they can be defined for all α and β . Then we have the following connection with the Meixner polynomials given by (9.10.1): (β )n (β −1,−n−β −x) Mn (x; β , c) = Pn ((2 − c)c−1 ). n! The Jacobi polynomials are related to the pseudo Jacobi polynomials defined by (9.9.1) in the following way: Pn (x; ν , N) =
(−2i)n n! (−N−1+iν ,−N−1−iν ) Pn (ix). (n − 2N − 1)n
The Jacobi polynomials are also related to the Gegenbauer (or ultraspherical) polynomials given by (9.8.19) by the quadratic transformations: (λ )
C2n (x) = and (λ )
C2n+1 (x) =
(λ )n (λ − 12 ,− 21 ) 2 Pn (2x − 1) ( 12 )n (λ )n+1 (λ − 12 , 12 ) 2 xPn (2x − 1). ( 12 )n+1
References [2], [4], [9], [11], [15], [19], [34], [35], [41], [42], [43], [44], [45], [46], [47], [51], [54], [55], [56], [57], [58], [69], [72], [91], [109], [117], [130], [132], [137], [146], [149], [152], [155], [165], [166], [167], [171], [172], [173], [175], [178], [187], [193], [196], [198], [202], [216], [219], [220], [221], [222], [223], [224], [225], [226], [227], [228], [229], [239], [241], [242], [243], [244], [247], [251], [253], [262], [264], [266], [267], [268], [274], [277], [283], [287], [314], [316], [317], [327], [330], [331], [332], [334], [335], [336], [339], [340], [357], [358], [363], [364], [367], [381], [382], [393], [394], [397], [399], [400], [403], [405], [408], [412], [416], [417], [424], [428], [430], [431], [433], [438], [450], [456], [464], [470], [477], [479], [480], [483], [485], [489], [493], [496], [505], [516], [518], [519], [521], [522].
222
9 Hypergeometric Orthogonal Polynomials
Special Cases 9.8.1 Gegenbauer / Ultraspherical Hypergeometric Representation The Gegenbauer (or ultraspherical) polynomials are Jacobi polynomials with α = β = λ − 12 and another normalization: (2λ )n (λ − 12 ,λ − 12 ) Pn (x) (λ + 12 )n −n, n + 2λ 1 − x (2λ )n , F ; = 2 1 λ + 12 n! 2
(λ )
Cn (x) =
λ = 0.
(9.8.19)
and λ = 0.
(9.8.20)
Orthogonality Relation 1
=
(λ )
(λ )
(1 − x2 )λ − 2 Cm (x)Cn (x) dx 1
−1
π Γ(n + 2λ )21−2λ
{Γ(λ )} (n + λ )n! 2
δmn ,
λ >−
1 2
Recurrence Relation (λ )
(λ )
(λ )
2(n + λ )xCn (x) = (n + 1)Cn+1 (x) + (n + 2λ − 1)Cn−1 (x).
(9.8.21)
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
n(n + 2λ − 1) pn−1 (x), 4(n + λ − 1)(n + λ )
where (λ )
Cn (x) =
2n (λ )n pn (x). n!
(9.8.22)
9.8 Jacobi
223
Differential Equation (1 − x2 )y (x) − (2λ + 1)xy (x) + n(n + 2λ )y(x) = 0,
(λ )
y(x) = Cn (x).
(9.8.23)
Forward Shift Operator d (λ ) (λ +1) Cn (x) = 2λ Cn−1 (x). dx
(9.8.24)
Backward Shift Operator (1 − x2 )
d (λ ) (n + 1)(2λ + n − 1) (λ −1) (λ ) Cn (x) + (1 − 2λ )xCn (x) = − Cn+1 (x) (9.8.25) dx 2(λ − 1)
or equivalently 1 (λ ) d (1 − x2 )λ − 2 Cn (x) dx 3 (λ −1) (n + 1)(2λ + n − 1) (1 − x2 )λ − 2 Cn+1 (x). =− 2(λ − 1)
(9.8.26)
Rodrigues-Type Formula 2 λ − 21
(1 − x )
n
1 (1 − x2 )λ +n− 2 .
(9.8.27)
∑ Cn
(x)t n .
(9.8.28)
(λ + 12 )n (λ ) Cn (x)t n , (2 λ ) n n=0
R=
(λ ) Cn (x) =
(2λ )n (−1)n (λ + 12 )n 2n n!
d dx
Generating Functions (1 − 2xt + t 2 )−λ =
∞
(λ )
n=0
−1
R
1 + R − xt 2
1 −λ
∞
2
=
∑
1 − 2xt + t 2 .
(9.8.29)
224
9 Hypergeometric Orthogonal Polynomials
0 F1
− (x − 1)t ; λ + 12 2
0 F1
e
xt
0 F1
− (x + 1)t ; λ + 12 2
− (x2 − 1)t 2 1 ; λ+2 4
γ , 2λ − γ 1 − R − t ; 2 λ + 12
(λ )
∞
=
Cn (x)
∑ (2λ )n (λ + 1 )n t n .
n=0
∞
=
(λ )
Cn (x) n t . n=0 (2λ )n
∑
γ , 2λ − γ 1 − R + t ; 2 F1 2 F1 2 λ + 12 ∞ (γ )n (2λ − γ )n (λ ) Cn (x)t n , R = 1 − 2xt + t 2 , =∑ 1 n=0 (2λ )n (λ + 2 )n −γ
(1 − xt) ∞
=
2 F1
(γ )n
1 1 1 2 γ, 2 γ + 2 1 λ+2
(λ )
∑ (2λ )n Cn
(x2 − 1)t 2 ; (1 − xt)2
γ arbitrary.
(x)t n ,
(9.8.30)
2
(9.8.31)
γ arbitrary. (9.8.32)
(9.8.33)
n=0
Limit Relation Gegenbauer / Ultraspherical → Hermite The Hermite polynomials given by (9.15.1) follow from the Gegenbauer (or ultraspherical) polynomials by taking λ = α + 12 and letting α → ∞ in the following way: 1
(α + 12 )
lim α − 2 nCn
α →∞
1
(α − 2 x) =
Hn (x) . n!
(9.8.34)
Remarks The case λ = 0 needs another normalization. In that case we have the Chebyshev polynomials of the first kind described in the next subsection. The Gegenbauer (or ultraspherical) polynomials are related to the Jacobi polynomials given by (9.8.1) by the quadratic transformations: (λ )
C2n (x) = and
(λ )n (λ − 12 ,− 21 ) 2 Pn (2x − 1) ( 12 )n
9.8 Jacobi
225 (λ )
C2n+1 (x) =
(λ )n+1 (λ − 12 , 12 ) 2 xPn (2x − 1). ( 12 )n+1
References [2], [5], [35], [38], [40], [45], [46], [51], [55], [66], [99], [104], [107], [109], [110], [117], [119], [120], [121], [123], [124], [130], [146], [154], [156], [163], [167], [168], [169], [170], [174], [183], [190], [191], [194], [198], [200], [219], [225], [231], [253], [267], [277], [317], [340], [366], [390], [398], [403], [413], [416], [417], [450], [456], [458], [467], [479], [493], [496], [508], [522].
9.8.2 Chebyshev Hypergeometric Representation The Chebyshev polynomials of the first kind can be obtained from the Jacobi polynomials by taking α = β = − 12 : (− 12 ,− 12 )
Tn (x) =
Pn
(− 21 ,− 21 )
Pn
(x)
= 2 F1
(1)
−n, n 1 − x ; 1 2 2
(9.8.35)
and the Chebyshev polynomials of the second kind can be obtained from the Jacobi polynomials by taking α = β = 12 : ( 1 , 12 )
Un (x) = (n + 1)
Pn 2
( 1 , 12 )
Pn 2
(x)
= (n + 1) 2 F1
(1)
−n, n + 2 1 − x . ; 3 2 2
(9.8.36)
Orthogonality Relation ⎧ π ⎪ ⎨ δmn , n = 0 1 (1 − x2 )− 2 Tm (x)Tn (x) dx = 2 ⎪ −1 ⎩ π δmn , n = 0.
1
1 −1
1
(1 − x2 ) 2 Um (x)Un (x) dx =
π δmn . 2
(9.8.37)
(9.8.38)
226
9 Hypergeometric Orthogonal Polynomials
Recurrence Relations 2xTn (x) = Tn+1 (x) + Tn−1 (x),
2xUn (x) = Un+1 (x) +Un−1 (x),
T0 (x) = 1 and T1 (x) = x.
(9.8.39)
U0 (x) = 1 and U1 (x) = 2x.
(9.8.40)
Normalized Recurrence Relations 1 xpn (x) = pn+1 (x) + pn−1 (x), 4
(9.8.41)
where T1 (x) = p1 (x) = x
and Tn (x) = 2n pn (x),
n = 1.
1 xpn (x) = pn+1 (x) + pn−1 (x), 4
(9.8.42)
where Un (x) = 2n pn (x).
Differential Equations (1 − x2 )y (x) − xy (x) + n2 y(x) = 0,
y(x) = Tn (x).
(9.8.43)
(1 − x2 )y (x) − 3xy (x) + n(n + 2)y(x) = 0,
y(x) = Un (x).
(9.8.44)
Forward Shift Operator d Tn (x) = nUn−1 (x). dx
(9.8.45)
Backward Shift Operator (1 − x2 ) or equivalently
d Un (x) − xUn (x) = −(n + 1)Tn+1 (x) dx
(9.8.46)
9.8 Jacobi
227
1 − 1 d 2 2 1−x Un (x) = −(n + 1) 1 − x2 2 Tn+1 (x). dx
(9.8.47)
Rodrigues-Type Formulas (−1)n d n 2 n− 21 (1 − x . ) 1 ( 2 )n 2n dx (n + 1)(−1)n d n 2 12 2 n+ 21 (1 − x . (1 − x ) Un (x) = ) dx ( 32 )n 2n 1 (1 − x2 )− 2 Tn (x) =
(9.8.48) (9.8.49)
Generating Functions ∞ 1 − xt = ∑ Tn (x)t n . 1 − 2xt + t 2 n=0
−1
R
1 ∞ 1 (1 + R − xt) = ∑ 2 n Tn (x)t n , 2 n=0 n! − (x − 1)t 1 ; 2 2
0 F1
2 F1 ∞
=
1 − 2xt + t 2 .
∞ − (x + 1)t Tn (x) = ∑ 1 t n. 0 F1 1 ; 2 2 n=0 2 n n! ∞ − (x2 − 1)t 2 Tn (x) n =∑ t . ext 0 F1 1 ; 4 n=0 n! 2
R=
(9.8.50)
γ , −γ 1 − R − t ; 1 2 2
(γ )n (−γ )n 1 Tn (x)t n , n! n=0 2 n −γ
(1 − xt) ∞
2 F1
γ , −γ 1 − R + t ; 1 2 2 R = 1 − 2xt + t 2 , γ arbitrary.
1 1 1 2 γ, 2 γ + 2 1 2
(γ )n Tn (x)t n , n=0 n!
∑
(9.8.52)
(9.8.53)
2 F1
∑
=
(9.8.51)
(x2 − 1)t 2 ; (1 − xt)2
γ arbitrary.
∞ 1 = ∑ Un (x)t n . 1 − 2xt + t 2 n=0
(9.8.54)
(9.8.55)
(9.8.56)
228
9 Hypergeometric Orthogonal Polynomials
R
1 2 (1 + R − xt)
0 F1
− (x − 1)t 3 ; 2 2
n Un (x)t n , ∑ (n +2 1)!
=
0 F1
∞
=
− (x2 − 1)t 2 3 ; 4 2
(γ )n (2 − γ )n Un (x)t n , (n + 1)! 2 n (1 − xt) ∞
=
1 − 2xt + t 2 .
(9.8.57)
2 F1
∞
Un (x) t n. (n + 1)! n
(9.8.58)
∑ 3
=
n=0
∞
2
Un (x)
∑ (n + 1)! t n .
=
(9.8.59)
n=0
2 F1
∑ 3
−γ
− (x + 1)t 3 ; 2 2
γ, 2 − γ 1 − R − t ; 3 2 2
n=0
R=
n=0
ext 0 F1
2 F1
3
∞
1
γ, 2 − γ 1 − R + t ; 3 2 2 R = 1 − 2xt + t 2 ,
1 1 1 2 γ, 2 γ + 2 3 2
(γ )n
∑ (n + 1)! Un (x)t n ,
(x2 − 1)t 2 ; (1 − xt)2
γ arbitrary.
γ arbitrary. (9.8.60)
(9.8.61)
n=0
Remarks The Chebyshev polynomials can also be written as: Tn (x) = cos(nθ ), and Un (x) = Further we have
sin(n + 1)θ , sin θ
x = cos θ x = cos θ .
(1)
Un (x) = Cn (x) (λ )
where Cn (x) denotes the Gegenbauer (or ultraspherical) polynomial given by (9.8.19) in the preceding subsection.
9.8 Jacobi
229
References [2], [55], [60], [61], [94], [146], [156], [168], [198], [253], [264], [277], [317], [399], [403], [416], [417], [456], [457], [459], [466], [493], [496], [514], [522], [525], [526].
9.8.3 Legendre / Spherical Hypergeometric Representation The Legendre (or spherical) polynomials are Jacobi polynomials with α = β = 0: −n, n + 1 1 − x (0,0) ; . (9.8.62) Pn (x) = Pn (x) = 2 F1 1 2
Orthogonality Relation 1 −1
Pm (x)Pn (x) dx =
2 δmn . 2n + 1
(9.8.63)
Recurrence Relation (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x).
(9.8.64)
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
n2 pn−1 (x), (2n − 1)(2n + 1)
where Pn (x) =
(9.8.65)
2n 1 pn (x). n 2n
Differential Equation (1 − x2 )y (x) − 2xy (x) + n(n + 1)y(x) = 0,
y(x) = Pn (x).
(9.8.66)
230
9 Hypergeometric Orthogonal Polynomials
Rodrigues-Type Formula Pn (x) =
(−1)n 2n n!
d dx
n
(1 − x2 )n .
(9.8.67)
Generating Functions ∞ 1 √ = ∑ Pn (x)t n . 1 − 2xt + t 2 n=0 ∞ − (x − 1)t − (x + 1)t Pn (x) n ; ; =∑ F F t . 0 1 0 1 2 1 1 2 2 n=0 (n!) ∞ − (x2 − 1)t 2 Pn (x) n xt ; =∑ t . e 0 F1 1 4 n=0 n!
(9.8.68) (9.8.69)
(9.8.70)
γ, 1 − γ 1 − R − t γ, 1 − γ 1 − R + t ; ; 2 F1 2 F1 1 2 1 2 ∞ (γ )n (1 − γ )n =∑ Pn (x)t n , R = 1 − 2xt + t 2 , γ arbitrary. (9.8.71) (n!)2 n=0
−γ
(1 − xt) ∞
=
1 2 F1
1 1 2 γ, 2 γ + 2
(γ )n Pn (x)t n , n=0 n!
∑
1
(x2 − 1)t 2 ; (1 − xt)2
γ arbitrary.
(9.8.72)
References [2], [6], [16], [103], [109], [127], [146], [156], [158], [168], [195], [198], [253], [403], [416], [417], [424], [456], [493], [496], [522].
9.9 Pseudo Jacobi
231
9.9 Pseudo Jacobi Hypergeometric Representation −n, n − 2N − 1 1 − ix (−2i)n (−N + iν )n (9.9.1) ; 2 F1 −N + iν (n − 2N − 1)n 2 −n, N + 1 − n − iν 2 ; , n = 0, 1, 2, . . . , N. = (x + i)n 2 F1 2N + 2 − 2n 1 − ix
Pn (x; ν , N) =
Orthogonality Relation 1 2π =
∞ −∞
(1 + x2 )−N−1 e2ν arctan x Pm (x; ν , N)Pn (x; ν , N) dx
Γ(2N + 1 − 2n)Γ(2N + 2 − 2n)22n−2N−1 n! Γ(2N + 2 − n) |Γ(N + 1 − n + iν )|2
δmn .
(9.9.2)
Recurrence Relation (N + 1)ν Pn (x; ν , N) (n − N − 1)(n − N) n(n − 2N − 2) − (2n − 2N − 3)(n − N − 1)2 (2n − 2N − 1) × (n − N − 1 − iν )(n − N − 1 + iν )Pn−1 (x; ν , N). (9.9.3)
xPn (x; ν , N) = Pn+1 (x; ν , N) +
Normalized Recurrence Relation (N + 1)ν pn (x) (n − N − 1)(n − N) n(n − 2N − 2)(n − N − 1 − iν )(n − N − 1 + iν ) pn−1 (x), (9.9.4) − (2n − 2N − 3)(n − N − 1)2 (2n − 2N − 1)
xpn (x) = pn+1 (x) +
where
Pn (x; ν , N) = pn (x).
232
9 Hypergeometric Orthogonal Polynomials
Differential Equation (1 + x2 )y (x) + 2 (ν − Nx) y (x) − n(n − 2N − 1)y(x) = 0, where
(9.9.5)
y(x) = Pn (x; ν , N).
Forward Shift Operator d Pn (x; ν , N) = nPn−1 (x; ν , N − 1). dx
(9.9.6)
Backward Shift Operator d Pn (x; ν , N) + 2 [ν − (N + 1)x] Pn (x; ν , N) dx = (n − 2N − 2)Pn+1 (x; ν , N + 1)
(9.9.7)
d (1 + x2 )−N−1 e2ν arctan x Pn (x; ν , N) dx = (n − 2N − 2)(1 + x2 )−N−2 e2ν arctan x Pn+1 (x; ν , N + 1).
(9.9.8)
(1 + x2 )
or equivalently
Rodrigues-Type Formula (1 + x2 )N+1 e−2ν arctan x Pn (x; ν , N) = (n − 2N − 1)n
d dx
n
(1 + x2 )n−N−1 e2ν arctan x .
(9.9.9)
Generating Function
0 F1 N
=
− ; (x + i)t −N + iν
(n − 2N − 1)n
0 F1
− ; (x − i)t −N − iν
∑ (−N + iν )n (−N − iν )n n! Pn (x; ν , N)t n .
n=0
N
(9.9.10)
9.9 Pseudo Jacobi
233
Limit Relation Continuous Hahn → Pseudo Jacobi The pseudo Jacobi polynomials follow from the continuous Hahn polynomials given by (9.4.1) by the substitutions x → xt, a = 12 (−N + iν − 2t), b = 12 (−N − iν + 2t), c = 12 (−N − iν − 2t) and d = 12 (−N + iν + 2t), division by t n and the limit t → ∞: pn (xt; 12 (−N + iν − 2t), 12 (−N − iν + 2t), 12 (−N + iν − 2t), 12 (−N − iν + 2t)) t→∞ tn (n − 2N − 1)n Pn (x; ν , N). = n! lim
Remarks Since we have for k < n (−N + iν )n = (−N + iν + k)n−k , (−N + iν )k the pseudo Jacobi polynomials given by (9.9.1) can also be seen as polynomials in the parameter ν . The weight function for the pseudo Jacobi polynomials can be written as (1 + x2 )−N−1 e2ν arctan x = (1 + ix)−N−1−iν (1 − ix)−N−1+iν . The pseudo Jacobi polynomials are related to the Jacobi polynomials defined by (9.8.1) in the following way: Pn (x; ν , N) =
(−2i)n n! (−N−1+iν ,−N−1−iν ) Pn (ix). (n − 2N − 1)n
If we set x → ν x in the definition (9.9.1) of the pseudo Jacobi polynomials and take the limit ν → ∞ we obtain a special case of the Bessel polynomials given by (9.13.1) in the following way: 2n Pn (ν x; ν , N) = yn (x; −2N − 2). ν →∞ νn (n − 2N − 1)n lim
References [50], [108], [382].
234
9 Hypergeometric Orthogonal Polynomials
9.10 Meixner Hypergeometric Representation Mn (x; β , c) = 2 F1
−n, −x 1 . ; 1− β c
(9.10.1)
Orthogonality Relation ∞
(β )x x c Mm (x; β , c)Mn (x; β , c) x=0 x!
∑ =
c−n n! δmn , (β )n (1 − c)β
β > 0 and 0 < c < 1.
(9.10.2)
Recurrence Relation (c − 1)xMn (x; β , c) = c(n + β )Mn+1 (x; β , c) − [n + (n + β )c] Mn (x; β , c) + nMn−1 (x; β , c). (9.10.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
n + (n + β )c n(n + β − 1)c pn (x) + pn−1 (x), 1−c (1 − c)2
where Mn (x; β , c) =
1 (β )n
c−1 c
(9.10.4)
n pn (x).
Difference Equation n(c − 1)y(x) = c(x + β )y(x + 1) − [x + (x + β )c] y(x) + xy(x − 1), where
y(x) = Mn (x; β , c).
(9.10.5)
9.10 Meixner
235
Forward Shift Operator Mn (x + 1; β , c) − Mn (x; β , c) =
n β
c−1 Mn−1 (x; β + 1, c) c
(9.10.6)
or equivalently n Δ Mn (x; β , c) = β
c−1 Mn−1 (x; β + 1, c). c
(9.10.7)
Backward Shift Operator c(β + x − 1)Mn (x; β , c) − xMn (x − 1; β , c) = c(β − 1)Mn+1 (x; β − 1, c) or equivalently (β )x cx (β − 1)x cx Mn (x; β , c) = Mn+1 (x; β − 1, c). ∇ x! x!
(9.10.8)
(9.10.9)
Rodrigues-Type Formula x (β )x cx n (β + n)x c Mn (x; β , c) = ∇ . x! x!
(9.10.10)
Generating Functions ∞ t x (β )n Mn (x; β , c)t n . (1 − t)−x−β = ∑ c n=0 n! ∞ −x 1−c Mn (x; β , c) n t t =∑ t . ; e 1 F1 β c n! n=0
−γ
(1 − t)
2 F1
1−
γ , −x (1 − c)t ; β c(1 − t)
∞
=
(γ )n Mn (x; β , c)t n , n=0 n!
∑
(9.10.11) (9.10.12)
γ arbitrary. (9.10.13)
236
9 Hypergeometric Orthogonal Polynomials
Limit Relations Hahn → Meixner If we take α = b − 1, β = N(1 − c)c−1 in the definition (9.5.1) of the Hahn polynomials and let N → ∞ we find the Meixner polynomials: lim Qn (x; b − 1, N(1 − c)c−1 , N) = Mn (x; b, c).
N→∞
Dual Hahn → Meixner To obtain the Meixner polynomials from the dual Hahn polynomials we have to take γ = β − 1 and δ = N(1 − c)c−1 in the definition (9.6.1) of the dual Hahn polynomials and let N → ∞: lim Rn (λ (x); β − 1, N(1 − c)c−1 , N) = Mn (x; β , c).
N→∞
Meixner → Laguerre The Laguerre polynomials given by (9.12.1) are obtained from the Meixner polynomials if we take β = α + 1 and x → (1 − c)−1 x and let c → 1: lim Mn ((1 − c)−1 x; α + 1, c) =
c→1
(α )
Ln (x) (α )
.
(9.10.14)
Ln (0)
Meixner → Charlier The Charlier polynomials given by (9.14.1) are obtained from the Meixner polynomials if we take c = (a + β )−1 a and let β → ∞: lim Mn (x; β , (a + β )−1 a) = Cn (x; a).
β →∞
(9.10.15)
Remarks The Meixner polynomials are related to the Jacobi polynomials given by (9.8.1) in the following way: (β )n (β −1,−n−β −x) Mn (x; β , c) = Pn ((2 − c)c−1 ). n!
9.11 Krawtchouk
237
The Meixner polynomials are also related to the Krawtchouk polynomials given by (9.11.1) in the following way: Kn (x; p, N) = Mn (x; −N, (p − 1)−1 p).
References [7], [11], [16], [20], [22], [29], [34], [39], [46], [51], [54], [59], [61], [72], [80], [82], [96], [97], [125], [146], [155], [198], [215], [217], [218], [226], [228], [265], [277], [279], [283], [289], [295], [303], [307], [317], [340], [363], [364], [375], [377], [381], [391], [406], [416], [417], [434], [503], [506], [521], [523].
9.11 Krawtchouk Hypergeometric Representation Kn (x; p, N) = 2 F1
−n, −x 1 ; , −N p
n = 0, 1, 2, . . . , N.
(9.11.1)
Orthogonality Relation N ∑ x px (1 − p)N−x Km (x; p, N)Kn (x; p, N) x=0 (−1)n n! 1 − p n = δmn , 0 < p < 1. (−N)n p N
(9.11.2)
Recurrence Relation − xKn (x; p, N) = p(N − n)Kn+1 (x; p, N) − [p(N − n) + n(1 − p)] Kn (x; p, N) + n(1 − p)Kn−1 (x; p, N).
(9.11.3)
238
9 Hypergeometric Orthogonal Polynomials
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [p(N − n) + n(1 − p)] pn (x) + np(1 − p)(N + 1 − n)pn−1 (x), where Kn (x; p, N) =
(9.11.4)
1 pn (x). (−N)n pn
Difference Equation − ny(x) = p(N − x)y(x + 1) − [p(N − x) + x(1 − p)] y(x) + x(1 − p)y(x − 1), (9.11.5) where y(x) = Kn (x; p, N).
Forward Shift Operator Kn (x + 1; p, N) − Kn (x; p, N) = − or equivalently
Δ Kn (x; p, N) = −
n Kn−1 (x; p, N − 1) Np
n Kn−1 (x; p, N − 1). Np
(9.11.6)
(9.11.7)
Backward Shift Operator 1− p Kn (x − 1; p, N) (N + 1 − x)Kn (x; p, N) − x p = (N + 1)Kn+1 (x; p, N + 1)
(9.11.8)
or equivalently x x p p N +1 N Kn (x; p, N) = Kn+1 (x; p, N + 1). (9.11.9) ∇ x x 1− p 1− p
9.11 Krawtchouk
239
Rodrigues-Type Formula x x p p N N −n . Kn (x; p, N) = ∇n x x 1− p 1− p
(9.11.10)
Generating Functions For x = 0, 1, 2, . . . , N we have N N (1 − p) x N−x Kn (x; p, N)t n . t (1 + t) 1− =∑ p n=0 n
et 1 F1
−x t ;− p −N
N
= N
(9.11.11)
Kn (x; p, N) n t . n! n=0
∑
(9.11.12)
γ , −x t ; (1 − t)−γ 2 F1 −N p(t − 1) N (γ )n Kn (x; p, N)t n , n=0 n! N
=
∑
γ arbitrary.
(9.11.13)
Limit Relations Hahn → Krawtchouk If we take α = pt and β = (1 − p)t in the definition (9.5.1) of the Hahn polynomials and let t → ∞ we obtain the Krawtchouk polynomials: lim Qn (x; pt, (1 − p)t, N) = Kn (x; p, N).
t→∞
Dual Hahn → Krawtchouk The Krawtchouk polynomials follow from the dual Hahn polynomials given by (9.6.1) if we set γ = pt, δ = (1 − p)t and let t → ∞: lim Rn (λ (x); pt, (1 − p)t, N) = Kn (x; p, N).
t→∞
240
9 Hypergeometric Orthogonal Polynomials
Krawtchouk → Charlier The Charlier polynomials given by (9.14.1) can be found from the Krawtchouk polynomials by taking p = N −1 a and letting N → ∞: lim Kn (x; N −1 a, N) = Cn (x; a).
N→∞
(9.11.14)
Krawtchouk → Hermite The Hermite polynomials given by (9.15.1) follow from the Krawtchouk polynomials by setting x → pN + x 2p(1 − p)N and then letting N → ∞: N (−1)n Hn (x) Kn (pN + x 2p(1 − p)N; p, N) = (9.11.15) lim n . N→∞ n p 2n n! 1− p
Remarks The Krawtchouk polynomials are self-dual, which means that Kn (x; p, N) = Kx (n; p, N),
n, x ∈ {0, 1, 2, . . . , N}.
By using this relation we easily obtain the so-called dual orthogonality relation from the orthogonality relation (9.11.2): 1− p x N N p ∑ n pn (1 − p)N−n Kn (x; p, N)Kn (y; p, N) = N δxy , n=0 x where 0 < p < 1 and x, y ∈ {0, 1, 2, . . . , N}. The Krawtchouk polynomials are related to the Meixner polynomials given by (9.10.1) in the following way: Kn (x; p, N) = Mn (x; −N, (p − 1)−1 p).
References [16], [34], [39], [46], [51], [59], [72], [80], [125], [141], [146], [164], [182], [184], [187], [188], [198], [203], [226], [228], [265], [277], [307], [338], [340], [363],
9.12 Laguerre
241
[364], [372], [375], [377], [381], [391], [416], [417], [429], [434], [436], [488], [489], [493], [521], [523].
9.12 Laguerre Hypergeometric Representation (α ) Ln (x) =
(α + 1)n 1 F1 n!
−n ;x . α +1
(9.12.1)
Orthogonality Relation ∞ 0
(α )
(α )
e−x xα Lm (x)Ln (x) dx =
Γ(n + α + 1) δmn , n!
α > −1.
(9.12.2)
Recurrence Relation (α )
(α )
(α )
(n + 1)Ln+1 (x) − (2n + α + 1 − x)Ln (x) + (n + α )Ln−1 (x) = 0.
(9.12.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (2n + α + 1)pn (x) + n(n + α )pn−1 (x), where (α )
Ln (x) =
(9.12.4)
(−1)n pn (x). n!
Differential Equation xy (x) + (α + 1 − x)y (x) + ny(x) = 0,
(α )
y(x) = Ln (x).
(9.12.5)
242
9 Hypergeometric Orthogonal Polynomials
Forward Shift Operator d (α ) (α +1) Ln (x) = −Ln−1 (x). dx
(9.12.6)
Backward Shift Operator d (α ) (α ) (α −1) Ln (x) + (α − x)Ln (x) = (n + 1)Ln+1 (x) dx
(9.12.7)
d −x α (α ) (α −1) e x Ln (x) = (n + 1)e−x xα −1 Ln+1 (x). dx
(9.12.8)
x or equivalently
Rodrigues-Type Formula (α ) e−x xα Ln (x) =
1 n!
d dx
n
−x n+α e x .
(9.12.9)
Generating Functions (1 − t)−α −1 exp et 0 F1
(1 − t)−γ 1 F1
xt t −1
− ; −xt α +1
γ xt ; α +1 t −1
∞
=
∞
=
(γ )n
(x)t n .
(9.12.10)
n=0
(α )
∞
=
(α )
∑ Ln
Ln (x) ∑ (α + 1)n t n . n=0 (α )
∑ (α + 1)n Ln
(x)t n ,
γ arbitrary.
(9.12.11)
(9.12.12)
n=0
Limit Relations Meixner-Pollaczek → Laguerre The Laguerre polynomials can be obtained from the Meixner-Pollaczek polynomials given by (9.7.1) by the substitution λ = 12 (α + 1), x → − 12 φ −1 x and the limit φ → 0:
9.12 Laguerre
243 ( 1 α + 12 )
lim Pn 2
φ →0
(α )
(− 12 φ −1 x; φ ) = Ln (x).
Jacobi → Laguerre The Laguerre polynomials are obtained from the Jacobi polynomials given by (9.8.1) if we set x → 1 − 2β −1 x and then take the limit β → ∞: (α ,β )
lim Pn
β →∞
(α )
(1 − 2β −1 x) = Ln (x).
Meixner → Laguerre If we take β = α + 1 and x → (1 − c)−1 x in the definition (9.10.1) of the Meixner polynomials and let c → 1 we obtain the Laguerre polynomials: lim Mn ((1 − c)−1 x; α + 1, c) =
c→1
(α )
Ln (x) (α )
.
Ln (0)
Laguerre → Hermite The Hermite polynomials given by (9.15.1) can be obtained from the Laguerre polynomials by taking the limit α → ∞ in the following way: 1n 1 2 2 (α ) (−1)n Hn (x). Ln ((2α ) 2 x + α ) = lim α →∞ α n!
(9.12.13)
Remarks The definition (9.12.1) of the Laguerre polynomials can also be written as: (α )
Ln (x) =
1 n (−n)k ∑ k! (α + k + 1)n−k xk . n! k=0
In this way the Laguerre polynomials can also be seen as polynomials in the parameter α . Therefore they can be defined for all α . The Laguerre polynomials are related to the Bessel polynomials given by (9.13.1) in the following way: (α )
Ln (x) =
(−x)n yn (2x−1 ; −2n − α − 1). n!
244
9 Hypergeometric Orthogonal Polynomials
The Laguerre polynomials are related to the Charlier polynomials given by (9.14.1) in the following way: (−a)n (x−n) Cn (x; a) = Ln (a). n! The Laguerre polynomials and the Hermite polynomials given by (9.15.1) are also connected by the following quadratic transformations: (− 21 )
H2n (x) = (−1)n n! 22n Ln and
(x2 )
(1)
H2n+1 (x) = (−1)n n! 22n+1 xLn 2 (x2 ). In combinatorics the Laguerre polynomials with α = 0 are often called Rook polynomials.
References [1], [2], [4], [7], [10], [11], [15], [16], [19], [20], [34], [35], [41], [46], [51], [54], [58], [59], [61], [65], [72], [93], [95], [109], [111], [112], [117], [123], [124], [128], [129], [130], [131], [133], [136], [137], [144], [146], [153], [155], [158], [165], [166], [170], [176], [192], [197], [198], [201], [202], [227], [229], [240], [245], [246], [248], [249], [251], [253], [263], [267], [268], [277], [279], [283], [287], [289], [295], [297], [300], [307], [313], [315], [317], [333], [335], [339], [340], [354], [356], [359], [360], [363], [364], [365], [368], [369], [370], [381], [382], [390], [397], [403], [406], [416], [417], [424], [425], [427], [456], [466], [475], [476], [477], [478], [479], [493], [496], [501], [502], [506].
9.13 Bessel Hypergeometric Representation x −n, n + a + 1 ;− − 2 x n 2 −n ; = (n + a + 1)n , F 1 1 −2n − a x 2
yn (x; a) = 2 F0
(9.13.1) n = 0, 1, 2, . . . , N.
9.13 Bessel
245
Orthogonality Relation ∞ 0
2
xa e− x ym (x; a)yn (x; a) dx
=−
2a+1 Γ(−n − a)n! δmn , 2n + a + 1
a < −2N − 1.
(9.13.2)
Recurrence Relation 2(n + a + 1)(2n + a)yn+1 (x; a) = (2n + a + 1) [2a + (2n + a)(2n + a + 2)x] yn (x; a) + 2n(2n + a + 2)yn−1 (x; a).
(9.13.3)
Normalized Recurrence Relation 2a pn (x) (2n + a)(2n + a + 2) 4n(n + a) pn−1 (x), − (2n + a − 1)(2n + a)2 (2n + a + 1)
xpn (x) = pn+1 (x) −
where yn (x; a) =
(9.13.4)
(n + a + 1)n pn (x). 2n
Differential Equation x2 y (x) + [(a + 2)x + 2] y (x) − n(n + a + 1)y(x) = 0,
y(x) = yn (x; a).
(9.13.5)
Forward Shift Operator n(n + a + 1) d yn (x; a) = yn−1 (x; a + 2). dx 2
(9.13.6)
246
9 Hypergeometric Orthogonal Polynomials
Backward Shift Operator x2
d yn (x; a) + (ax + 2)yn (x; a) = 2yn+1 (x; a − 2) dx
(9.13.7)
or equivalently 2 d a −2 x e x yn (x; a) = 2xa−2 e− x yn+1 (x; a − 2). dx
(9.13.8)
Rodrigues-Type Formula 2 2 yn (x; a) = 2−n x−a e x Dn x2n+a e− x .
(9.13.9)
Generating Function − 12
(1 − 2xt)
2 √ 1 + 1 − 2xt
a
2t √ exp 1 + 1 − 2xt
∞
=
tn
∑ yn (x; a) n! .
(9.13.10)
n=0
Limit Relation Jacobi → Bessel If we take β = a − α in the definition (9.8.1) of the Jacobi polynomials and let α → −∞ we find the Bessel polynomials: (α ,a−α )
lim
α →−∞
Pn
(1 + α x)
(α ,a−α ) Pn (1)
= yn (x; a).
Remarks The following notations are also used for the Bessel polynomials: yn (x; a, b) = yn (2b−1 x; a) and θn (x; a, b) = xn yn (x−1 ; a, b). However, the Bessel polynomials essentially depend on only one parameter. The Bessel polynomials are related to the Laguerre polynomials given by (9.12.1) in the following way:
9.14 Charlier
247 (α )
Ln (x) =
(−x)n yn (2x−1 ; −2n − α − 1). n!
The special case a = −2N − 2 of the Bessel polynomials can be obtained from the pseudo Jacobi polynomials by setting x → ν x in the definition (9.9.1) of the pseudo Jacobi polynomials and taking the limit ν → ∞ in the following way: 2n Pn (ν x; ν , N) = yn (x; −2N − 2). ν →∞ νn (n − 2N − 1)n lim
References [32], [102], [126], [159], [179], [181], [255], [277], [352], [384], [417].
9.14 Charlier Hypergeometric Representation Cn (x; a) = 2 F0
1 −n, −x ;− . − a
(9.14.1)
Orthogonality Relation ∞
ax
∑ x! Cm (x; a)Cn (x; a) = a−n ea n! δmn ,
a > 0.
(9.14.2)
x=0
Recurrence Relation − xCn (x; a) = aCn+1 (x; a) − (n + a)Cn (x; a) + nCn−1 (x; a).
(9.14.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (n + a)pn (x) + napn−1 (x), where
(9.14.4)
248
9 Hypergeometric Orthogonal Polynomials
1 n Cn (x; a) = − pn (x). a
Difference Equation − ny(x) = ay(x + 1) − (x + a)y(x) + xy(x − 1),
y(x) = Cn (x; a).
(9.14.5)
Forward Shift Operator n Cn (x + 1; a) −Cn (x; a) = − Cn−1 (x; a) a
(9.14.6)
n Δ Cn (x; a) = − Cn−1 (x; a). a
(9.14.7)
or equivalently
Backward Shift Operator x Cn (x; a) − Cn (x − 1; a) = Cn+1 (x; a) a or equivalently
∇
ax ax Cn (x; a) = Cn+1 (x; a). x! x!
(9.14.8)
(9.14.9)
Rodrigues-Type Formula x ax n a Cn (x; a) = ∇ . x! x!
(9.14.10)
∞ Cn (x; a) n t x t . et 1 − =∑ a n! n=0
(9.14.11)
Generating Function
9.14 Charlier
249
Limit Relations Meixner → Charlier If we take c = (a + β )−1 a in the definition (9.10.1) of the Meixner polynomials and let β → ∞ we find the Charlier polynomials: lim Mn (x; β , (a + β )−1 a) = Cn (x; a).
β →∞
Krawtchouk → Charlier The Charlier polynomials can be found from the Krawtchouk polynomials given by (9.11.1) by taking p = N −1 a and letting N → ∞: lim Kn (x; N −1 a, N) = Cn (x; a).
N→∞
Charlier → Hermite The Hermite polynomials given by (9.15.1) are obtained from the Charlier polynomials if we set x → (2a)1/2 x + a and let a → ∞. In fact we have 1
1
lim (2a) 2 nCn ((2a) 2 x + a; a) = (−1)n Hn (x).
a→∞
(9.14.12)
Remark The Charlier polynomials are related to the Laguerre polynomials given by (9.12.1) in the following way: (−a)n (x−n) Cn (x; a) = Ln (a). n!
References [7], [11], [16], [20], [22], [34], [39], [46], [59], [72], [80], [96], [98], [146], [147], [184], [198], [226], [228], [252], [265], [279], [317], [340], [353], [363], [364], [365], [372], [375], [377], [381], [391], [397], [406], [416], [417], [492], [493], [506], [521], [523].
250
9 Hypergeometric Orthogonal Polynomials
9.15 Hermite Hypergeometric Representation n
Hn (x) = (2x) 2 F0
−n/2, −(n − 1)/2 1 ;− 2 − x
.
(9.15.1)
Orthogonality Relation 1 √ π
∞ −∞
e−x Hm (x)Hn (x) dx = 2n n! δmn . 2
(9.15.2)
Recurrence Relation Hn+1 (x) − 2xHn (x) + 2nHn−1 (x) = 0.
(9.15.3)
Normalized Recurrence Relation n xpn (x) = pn+1 (x) + pn−1 (x), 2
(9.15.4)
where Hn (x) = 2n pn (x).
Differential Equation y (x) − 2xy (x) + 2ny(x) = 0,
y(x) = Hn (x).
(9.15.5)
Forward Shift Operator d Hn (x) = 2nHn−1 (x). dx
(9.15.6)
9.15 Hermite
251
Backward Shift Operator
or equivalently
d Hn (x) − 2xHn (x) = −Hn+1 (x) dx
(9.15.7)
2 d −x2 e Hn (x) = −e−x Hn+1 (x). dx
(9.15.8)
Rodrigues-Type Formula e−x Hn (x) = (−1)n 2
d dx
n
2 e−x .
(9.15.9)
Generating Functions exp 2xt − t 2 = ⎧ √ ⎪ ⎪ et cos(2x t) = ⎪ ⎨
∞
Hn (x) n t . n=0 n!
∑
(9.15.10)
∞
(−1)n H2n (x)t n (2n)! n=0
∑
∞ ⎪ √ et (−1)n ⎪ ⎪ H2n+1 (x)t n . ⎩ √ sin(2x t) = ∑ (2n + 1)! t n=0 ⎧ ∞ H2n (x) 2n −t 2 ⎪ ⎪ t cosh(2xt) = e ∑ ⎪ ⎨ (2n)! n=0
⎪ 2 ⎪ ⎪ ⎩ e−t sinh(2xt) =
∞
H2n+1 (x) ∑ (2n + 1)! t 2n+1 . n=0
⎧ ∞ γ x2 t 2 (γ )n ⎪ 2 −γ ⎪ H2n (x)t 2n =∑ ) F ; (1 + t ⎪ 1 1 1 2 ⎪ 1 + t (2n)! ⎨ n=0 2 1 ∞ (γ + 1 ) 2 2 ⎪ γ+2 x t ⎪ xt n ⎪ ⎪√ = ; 1 F1 ⎩ ∑ (2n +21)! H2n+1 (x)t 2n+1 3 2 1 + t 1 + t2 n=0 2
(9.15.11)
(9.15.12)
(9.15.13)
with γ arbitrary. 1 + 2xt + 4t 2 (1 + 4t 2 )
3 2
4x2t 2 exp 1 + 4t 2
∞
=
Hn (x)
∑ n/2 ! t n ,
n=0
where n/2 denotes the largest integer smaller than or equal to n/2.
(9.15.14)
252
9 Hypergeometric Orthogonal Polynomials
Limit Relations Meixner-Pollaczek → Hermite √ If we take x → (sin φ )−1 (x λ − λ cos φ ) in the definition (9.7.1) of the MeixnerPollaczek polynomials and then let λ → ∞ we obtain the Hermite polynomials: √ 1 Hn (x) (λ ) . lim λ − 2 n Pn ((sin φ )−1 (x λ − λ cos φ ); φ ) = n! λ →∞ Jacobi → Hermite The Hermite polynomials follow from the Jacobi polynomials given by (9.8.1) by taking β = α and letting α → ∞ in the following way: 1
(α ,α )
lim α − 2 n Pn
α →∞
1
(α − 2 x) =
Hn (x) . 2n n!
Gegenbauer / Ultraspherical → Hermite The Hermite polynomials follow from the Gegenbauer (or ultraspherical) polynomials given by (9.8.19) by taking λ = α + 12 and letting α → ∞ in the following way: 1 1 Hn (x) (α + 1 ) lim α − 2 nCn 2 (α − 2 x) = . α →∞ n! Krawtchouk → Hermite The Hermite polynomials follow from the Krawtchouk polynomials given by (9.11.1) by setting x → pN + x 2p(1 − p)N and then letting N → ∞: N (−1)n Hn (x) Kn (pN + x 2p(1 − p)N; p, N) = lim n . N→∞ n p 2n n! 1− p Laguerre → Hermite The Hermite polynomials can be obtained from the Laguerre polynomials given by (9.12.1) by taking the limit α → ∞ in the following way:
9.15 Hermite
253
1n 1 2 2 (α ) (−1)n Hn (x). lim Ln ((2α ) 2 x + α ) = α →∞ α n! Charlier → Hermite If we set x → (2a)1/2 x + a in the definition (9.14.1) of the Charlier polynomials and let a → ∞ we find the Hermite polynomials. In fact we have 1
1
lim (2a) 2 nCn ((2a) 2 x + a; a) = (−1)n Hn (x).
a→∞
Remarks The Hermite polynomials can also be written as: Hn (x) n/2 (−1)k (2x)n−2k = ∑ , n! k! (n − 2k)! k=0 where n/2 denotes the largest integer smaller than or equal to n/2. The Hermite polynomials and the Laguerre polynomials given by (9.12.1) are also connected by the following quadratic transformations: (− 21 )
H2n (x) = (−1)n n! 22n Ln and
(x2 )
(1)
H2n+1 (x) = (−1)n n! 22n+1 xLn 2 (x2 ).
References [2], [11], [16], [19], [20], [34], [35], [38], [41], [46], [51], [58], [72], [88], [99], [105], [109], [111], [112], [134], [136], [146], [153], [156], [165], [166], [177], [198], [202], [245], [246], [250], [251], [253], [267], [277], [295], [317], [340], [355], [361], [365], [381], [382], [390], [396], [403], [406], [416], [417], [424], [427], [456], [466], [479], [484], [493], [496], [497], [501], [506], [510], [520].
Part II
Classical q-Orthogonal Polynomials
Chapter 10
Orthogonal Polynomial Solutions of q-Difference Equations Classical q-Orthogonal Polynomials I
10.1 Polynomial Solutions of q-Difference Equations In the case of the q-derivative operator Dq := Aq,0 , we have to deal with (cf. (2.2.7)): (ex2 + 2 f x + g) Dq2 yn (x) + (2ε x + γ ) (Dq yn ) (x) =
[n] (e[n − 1] + 2ε )yn (qx), qn
(10.1.1)
for n = 0, 1, 2, . . . and q ∈ R \ {−1, 0, 1}. In the symmetric form (cf. (2.2.12)) this reads C(x)yn (qx) − {C(x) + D(x)} yn (x) + D(x)yn (q−1 x) q−n − 1 e(1 − qn−1 ) + 2ε (1 − q) yn (x) = 2 (q − 1) with (cf. (2.2.13)) C(x) =
ex2 + 2 f qx + gq2 q(q − 1)2 x2 and
D(x) =
{e + 2ε (1 − q)} x2 + {2 f + γ (1 − q)} qx + gq2 . (q − 1)2 x2
Hence
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 10, © Springer-Verlag Berlin Heidelberg 2010
257
258
10 Orthogonal Polynomial Solutions of q-Difference Equations
ex2 + 2 f qx + gq2 yn (qx) − {e(1 + q) + 2ε q(1 − q)} x2
+ {2 f (1 + q) + γ q(1 − q)} qx + gq2 (1 + q) yn (x) + {e + 2ε (1 − q)} x2 + {2 f + γ (1 − q)} qx + gq2 qyn (q−1 x) = (q−n − 1) e(1 − qn−1 ) + 2ε (1 − q) qx2 yn (x), n = 0, 1, 2, . . . . (10.1.2) The regularity condition (2.3.3) implies that ε = 0. It will turn out to be convenient to introduce
α := e + 2ε (1 − q) and β := 2 f + γ (1 − q).
(10.1.3)
Then (10.1.2) can be written as 2 ex + 2 f qx + gq2 yn (qx) − (e + α q)x2 + (2 f + β q)qx + gq2 (1 + q) yn (x) + α x2 + β qx + gq2 qyn (q−1 x) (10.1.4) = (q−n − 1) α − eqn−1 qx2 yn (x), n = 0, 1, 2, . . . with e, f , g, α , β ∈ C. Note that, in view of the homogeneity, one of the coefficients can be chosen arbitrarily. Furthermore, without loss of generality we may assume that e ∈ R. In section 10.4 we will see that this implies that all coefficients e, f , g, α and β must be real.
10.2 The Basic Hypergeometric Representation Since ω = 0, we have (cf. (2.4.2)) k x; c x; c x + cqi−1 := ∏ := 1 and , k 0 [i] i=1
k = 1, 2, 3, . . . .
Now we have k (−cx−1 ; q)k x + cqi−1 1 + cx−1 qi−1 = (1 − q)k xk ∏ = (1 − q)k xk . i [i] 1 − q (q; q) k i=1 i=1 k
∏
Therefore we try to find polynomial solutions of the form (cf. (2.4.3)) n
yn (x) =
∑ an,k
k=0
(−cx−1 ; q)k (1 − q)k xk , (q; q)k
an,n = 0,
n = 0, 1, 2, . . . .
(10.2.1)
Substitution of (10.2.1) into (10.1.4) leads to a three-term recurrence relation for the coefficients {an,k }nk=0 (cf. (2.4.4)). If c satisfies the relation (cf. (2.4.5))
10.2 The Basic Hypergeometric Representation
259
α c2 − β qc + gq2 = 0
(10.2.2)
the recurrence relation reduces to the two-term recurrence relation (cf. (2.4.6))
[n − k] α − eqn+k−1 an,k = − c(α − eq2k ) − q(β − 2 f qk ) qn−k−1 an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. For c = 0 (and therefore g = 0) this can be written as [n − k] α − eqn+k−1 an,k = β − 2 f qk qn−k an,k+1 (10.2.3) for k = n − 1, n − 2, n − 3, . . . , 0, and for c = 0 this can be written as [n − k] α − eqn+k−1 can,k = eq2k−2 c2 − 2 f qk−1 c + g qn−k+1 an,k+1
(10.2.4)
for k = n − 1, n − 2, n − 3, . . . , 0. Now we have by using (10.1.3) e[n] + 2ε =
α − eqn , 1−q
n = 0, 1, 2, . . . .
The regularity condition (2.3.3) holds if and only if all eigenvalues in (2.2.6) are different. In that case we have α − eqn+k−1 = 0 which implies that the coefficients {an,k }nk=0 are determined uniquely in terms of an,n = 0. Now we will distinguish between two different cases: g = 0 and g = 0 (cf. [385]). For g = 0 the condition (10.2.2) can be satisfied for c = 0. Then the representation (10.2.1) reads n
yn (x) =
∑ an,k
k=0
n (1 − q)k k xk , x = ∑ an,k (q; q)k [k]! k=0
an,n = 0,
n = 0, 1, 2, . . . .
Note that this is equivalent to the second approach (2.4.12). In that case the coefficients {an,k }nk=0 satisfy the two-term recurrence relation (10.2.3). Hence the coefficients {an,k }nk=0 are determined uniquely in terms of an,n = 0. In fact we have n−k qi β − 2 f qn−i an,k = ∏ an,n , k = 0, 1, 2, . . . , n − 1. 2n−i−1 ) i=1 [i] (α − eq Now we have
n−k
∏ qi = q(
n−k+1 2
) = q(n−k)(n−k+1)/2
i=1
and by using (1.8.16) n−k
1
∏ [i] = i=1
k (1 − q)n−k (q−n ; q)k = (−1)k (1 − q)n−k q−(2)+nk . (q; q)n−k (q; q)n
(10.2.5)
260
10 Orthogonal Polynomial Solutions of q-Difference Equations
In order to find monic polynomials we choose an,n = [n]! = (q; q)n /(1 − q)n . Then we have from (10.2.1)
n (q−n ; q)k n−k β − 2 f qn−i n(n+1)/2 (10.2.6) yn (x) = q ∑ (q; q)k ∏ α − eq2n−i−1 (−x)k i=1 k=0 for n = 0, 1, 2, . . .. Case I. g = 0, β = 0 and α = 0. Since β = 0 we have f = 0 and since α = 0 we have e = 0. Hence by using (10.2.6), this leads to the basic hypergeometric representation k (q−n ; q)k −eqn−1 x = q ∑ 2f k=0 (q; q)k −n n eqn−1 x q 2f −n(n−3)/2 ; q, − , n = 0, 1, 2, . . . . = q 1 φ0 e − 2f
(I) yn (x; q)
2f e
n
−n(n−3)/2
n
The q-polynomials in this class have, besides q, in fact only one free parameter e/ f or f /e. Later we will see that there are no orthogonal polynomial solutions in this case. Case II. g = 0, β = 0 and α = 0. Since β = 0 we have f = 0. Hence by using (10.2.6), this leads to the basic hypergeometric representation (II) yn (x; q)
n 1 (q−n ; q)k (eα −1 qn−1 ; q)k −(k) α x k 2 f qn n 2 = − q ∑ α (eα −1 qn−1 ; q)n k=0 (q; q)k 2f n q−n , eα −1 qn−1 1 αx 2 f qn ; q, − = − 2 φ0 −1 n−1 α (eα q ; q)n − 2f
for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, two free parameters e/α and f /α . A special case of q-polynomials in this class is the family of Stieltjes-Wigert polynomials without free parameters. Case III. g = 0, β = 0 and α = 0. Since α = 0 we have e = 0. Hence by using (10.2.6), this leads to the basic hypergeometric representation (III)
yn
n−1 k n β n −n(n−2) (q−n ; q)k (2k) eq x − q (2 f β −1 ; q)n ∑ q −1 ; q) (q; q) e β k k k=0 (2 f β n −n n−1 q β −eq x = − q−n(n−2) (2 f β −1 ; q)n 1 φ1 ; q, −1 e 2fβ β
(x; q) =
for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, two free parameters e/β and f /β . A special case of q-polynomials in this class is the family of q-Laguerre polynomials with only one free parameter.
10.2 The Basic Hypergeometric Representation
261
Another special case of q-polynomials in this class is again the family of StieltjesWigert polynomials without free parameters. Sometimes families of q-orthogonal polynomials belong to different classes for 0 < q < 1 or q > 1. For instance, in the case of the Stieltjes-Wigert polynomials we have by using (1.8.7): if −n q 1 n+1 ; q, −q x , 0 < q < 1, Sn (x; q) = 1 φ1 (q; q)n 0 then
n
Sn (x; q−1 ) =
(−1)n qn+(2) 2 φ0 (q; q)n
q−n , 0 ; q, −x , −
q > 1.
Case IV. g = 0, β = 0 and α = 0. By using (10.2.6), this leads to the basic hypergeometric representation n β (2 f β −1 ; q)n (IV ) qn(n+1)/2 yn (x; q) = α (eα −1 qn−1 ; q)n n (q−n ; q)k (eα −1 qn−1 ; q)k αx k − ×∑ (2 f β −1 ; q)k (q; q)k β k=0 −n n −1 q , eα −1 qn−1 β ; q)n αx n(n+1)/2 (2 f β = q ; q, − 2 φ1 α (eα −1 qn−1 ; q)n 2 f β −1 β for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, three free parameters e/α , f /β and β /α . Special cases of q-polynomials in this class are the little q-Jacobi polynomials with two free parameters and the little q-Laguerre and the q-Bessel polynomials with only one free parameter. If g = 0 we cannot use (10.2.2) in the case that both α = 0 and β = 0. In that case we use the second approach (2.4.12). Since ω = 0, we have v = c(1 − q), and therefore (cf. (2.4.11)) k x + c + [k − i − 1]vqi+1−k x ; c + [k − 2]vq2−k =∏ k [i] i=1 x + c + c(1 − qk−i−1 )qi+1−k [i] i=1 k
=∏
(−cx−1 q2−k ; q)k x + cqi+1−k = (1 − q)k xk . [i] (q; q)k i=1 k
=∏
Hence, the representation (2.4.12) reads n
yn (x) =
∑ bn,k
k=0
(−cx−1 q2−k ; q)k (1 − q)k xk , (q; q)k
bn,n = 0,
n = 0, 1, 2, . . . .
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10 Orthogonal Polynomial Solutions of q-Difference Equations
Since α = 0 we have e = 0, which implies that ec2 − 2 f c + g = 0
(10.2.7)
can be solved for c = 0. This implies that we have the two-term recurrence relation (cf. (2.4.13))
[n − k] (e[n + k − 1] + 2ε ) bn,k + (e[k] + 2ε )([k + 1]q−k − 1)v + (2[k](ec − f ) + 2ε c − γ )} qn−k bn,k+1 = 0,
k = n − 1, n − 2, n − 3, . . . , 0.
Now we use (10.1.3) and the fact that α = 0, β = 0 and v = c(1 − q) to obtain eqn+k−1 (1 − qn−k )bn,k = (2 f − ec)(1 − q)qn bn,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0
or equivalently, by using (10.2.7), ecqk−1 (1 − qn−k )bn,k = g(1 − q)bn,k+1 ,
k = n − 1, n − 2, n − 3, . . . , 0.
This implies that g n−k bn,k = ec
n−k
1
∏ [i]qn−i−1
bn,n ,
k = 0, 1, 2, . . . , n − 1.
i=1
Now we have as before by using (1.8.16) n−k
1
∏ [i]qn−i−1 i=1
qi (1 − q)n−k q(n−k)(n−k+1)/2 = q−(n−1)(n−k) (q; q)n−k i=1 [i]
n−k
= q−(n−1)(n−k) ∏ =
(q−n ; q)k (−1)k (1 − q)n−k q−n(n−3)/2+(n−1)k . (q; q)n
In order to find monic polynomials we choose bn,n = (q; q)n /(1 − q)n . Hence we have g n−k (q−n ; q) k (−1)k q−n(n−3)/2+(n−1)k , k = 0, 1, 2, . . . , n. bn,k = ec (1 − q)k By using (1.8.14) this leads to the representation g n yn (x) = q−n(n−3)/2 ec 2 n k n (q−n ; q)k (−c−1 xq−1 ; q)k k −(2k ) ec q ×∑ (−1) q (q; q)k g k=0 for n = 0, 1, 2, . . ..
(10.2.8)
10.2 The Basic Hypergeometric Representation
263
Case V. g = 0, α = 0 and β = 0. Since α = 0 we have e = 0. We also have c = 0 and by using (10.2.8) this leads to the basic hypergeometric representation (V )
2 n k g n n k ec q (q−n ; q)k (−c−1 xq−1 ; q)k q−n(n−3)/2 ∑ (−1)k q−(2) ec (q; q) g k k=0 −n g n −1 −1 2 n ec q q , −c xq ; q, , n = 0, 1, 2, . . . . = q−n(n−3)/2 2 φ0 ec − g
yn (x; q) =
The q-polynomials in this class have, besides q, two free parameters g/e (or e/g) and c(= 0), a solution of (10.2.7). A special case of q-polynomials in this class is the family of discrete q-Hermite II polynomials with no free parameters. If g = 0 and (α , β ) = (0, 0) we use the representation (10.2.1). In that case we also have c = 0 in view of (10.2.2). Then we have by using (10.2.4)
n−k eq2n−i−1 c2 − 2 f qn c + gqi+1 k−n an,n , k = 0, 1, 2, . . . , n − 1. an,k = c ∏ [i] (α − eq2n−i−1 ) i=1 Now we have n−k
∏
eq2n−i−1 c2 − 2 f qn c + gqi+1
i=1
=g
n−k (n−k)(n−k+3)/2
q
n−k
∏ i=1
2f e 1 − qn−i−1 c + q2n−2i−2 c2 g g
n−k
= gn−k q(n−k)(n−k+3)/2 ∏
1 + cγ1 qn−i−1
1 + cγ2 qn−i−1
,
i=1
where γ1 γ2 = e/g and γ1 + γ2 = −2 f /g. Hence we have gγi2 + 2 f γi + e = 0,
i = 1, 2.
Later we will see that e, f and g must be real , which implies that γ1 and γ2 are both real in the case that f 2 ≥ eg, id est − f − f 2 − eg − f + f 2 − eg and γ2 = , γ1 = g g and that they are complex conjugates in the case that f 2 < eg. As before we choose an,n = [n]! = (q; q)n /(1 − q)n in order to get monic polynomials. Then we have, by using (10.2.5), from (10.2.1)
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10 Orthogonal Polynomial Solutions of q-Difference Equations
yn (x) =
g n
(q−n ; q)k (−cx−1 ; q)k c (q; q)k k=0 n−k 1 + cγ qn−i−1 1 + cγ qn−i−1 cx k 1 2 − × ∏ α − eq2n−i−1 gq i=1 n
qn(n+3)/2 ∑
(10.2.9)
for n = 0, 1, 2, . . .. Case VI. g = 0, α = 0 and β = 0. Since α = 0 we have e = 0. Note that in this case (10.2.2) implies that c = gq/β . Hence by using (10.2.9), this leads to the basic hypergeometric representation g n (V I) yn (x; q) = − q−n(n−3) −cγ1 q−1 , −cγ2 q−1 ; q n ec k n k ecqn−2 x (q−n ; q)k (−cx−1 ; q)k ( ) 2 ×∑ q −1 −1 g k=0 (−cγ1 q , −cγ2 q ; q)k (q; q)k n β = − q−n(n−3) −gβ −1 γ1 , −gβ −1 γ2 ; q n eq −n q , −gβ −1 qx−1 eqn−1 x × 2 φ2 ; q, − −gβ −1 γ1 , −gβ −1 γ2 β for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, three free parameters e/β , f /β and g/β . By using the fact that γ1 γ2 = e/g we may apply (1.13.5) to find that n −n q , γ1 q−1 x gγ2 qn β (V I) −n(n−2) −1 −gβ γ1 ; q n 2 φ1 q ; q, − yn (x; q) = − −gβ −1 γ1 γ1 γ2 g β for n = 0, 1, 2, . . .. Case VII. g = 0 and α = 0. By using (10.2.9), this leads to the basic hypergeometric representation g n −cγ1 q−1 , −cγ2 q−1 ; q n (V II) qn(n+3)/2 yn (x; q) = cα (eα −1 qn−1 ; q)n n cα x k (q−n ; q)k (−cx−1 ; q)k (eα −1 qn−1 ; q)k − ×∑ (−cγ1 q−1 , −cγ2 q−1 ; q)k (q; q)k gq k=0 −1 −1 g n −cγ1 q , −cγ2 q ; q n = qn(n+3)/2 cα (eα −1 qn−1 ; q)n −n q , −cx−1 , eα −1 qn−1 cα x × 3 φ2 ; q, − −cγ1 q−1 , −cγ2 q−1 gq
10.2 The Basic Hypergeometric Representation
265
for n = 0, 1, 2, . . ., where c satisfies (10.2.2). The q-polynomials in this class have, besides q, four free parameters e/α , f /α , g/α and β /α . Special cases of qpolynomials in this class are the big q-Jacobi polynomials with three free parameters, the big q-Laguerre polynomials with two free parameters, the Al-SalamCarlitz I and II polynomials with one free parameter and the discrete q-Hermite I and II polynomials with no free parameters. We will see that we have:
Theorem 10.1. All orthogonal polynomial solutions yn (x) of the q-difference equation (10.1.4) 2 ex + 2 f qx + gq2 yn (qx) − (e + α q)x2 + (2 f + β q) qx + gq2 (1 + q) yn (x) + α x2 + β qx + gq2 qyn (q−1 x) = (q−n − 1) α − eqn−1 qx2 yn (x), n = 0, 1, 2, . . . , where α = e + 2ε (1 − q) and β = 2 f + γ (1 − q), can be divided into six different cases: Case II. g = 0, β = 0 and α = 0 Case III. g = 0, β = 0 and α = 0 Case IV. g = 0, β = 0 and α = 0 Case V. g = 0, α = 0 and β = 0 Case VI. g = 0, α = 0 and β = 0 Case VII. g = 0 and α = 0.
In chapters 4 and 5 we have indicated that weight functions for classical orthogonal polynomials are connected to certain probability distributions in stochastics. In the case of classical q-orthogonal polynomials weight functions are connected to certain (discrete) q-distributions in probability and statistics as well. Examples of qdistributions of this kind are the Euler, Heine and Kemp distributions, which appear in models of specific processes in physics, biology and mathematical economy. See for instance [309], [310] and [311].
266
10 Orthogonal Polynomial Solutions of q-Difference Equations
10.3 The Three-Term Recurrence Relation In section 2.6 we have seen that the monic polynomial solutions {yn }∞ n=0 of (10.1.1) satisfy the three-term recurrence relation yn+1 (x) = (x − cn )yn (x) − dn yn−1 (x),
n = 1, 2, 3, . . . ,
(10.3.1)
with initial values y0 (x) = 1 and y1 (x) = x − c0 , where (cf. (2.6.11)) c0 = −γ q/2ε , qn (e[n − 1] + 2ε ) (2 f [n](1 + q) + γ q) − eγ [n + 1]qn−1 , n = 1, 2, 3, . . . cn = − (e[2n − 2] + 2ε ) (e[2n] + 2ε ) and (cf. (2.6.12)) dn =
qn+1 [n] (e[n − 2] + 2ε ) (e[2n − 3] + 2ε ) (e[2n − 2] + 2ε )2 (e[2n − 1] + 2ε ) × qn−1 (2 f [n − 1] + γ ) 2 f {e[n − 1] + 2ε } − qn−1 eγ −g (e[2n − 2] + 2ε )2 , n = 1, 2, 3, . . . .
By using (10.1.3) we obtain c0 = (2 f − β )q/(α − e), 2 f qn α − α (1 + q)qn + eq2n−1 cn = − (α − eq2n−2 ) (α − eq2n ) β qn+1 α − e(1 + q)qn−2 + eq2n−1 , − (α − eq2n−2 ) (α − eq2n ) and dn =
n = 1, 2, 3, . . . (10.3.2)
qn+1 (1 − qn ) α − eqn−2 (α − eq2n−3 ) (α − eq2n−2 )2 (α − eq2n−1 ) × qn−1 (β − 2 f qn−1 )(2 f α − eβ qn−1 ) − g(α − eq2n−2 )2 (10.3.3)
for n = 1, 2, 3, . . .. In the case that g = 0 we have c0 = A0 , cn = An + Cn and dn = An−1Cn for n = 1, 2, 3, . . . where qn α − eqn−2 β − 2 f qn−1 , n = 1, 2, 3, . . . An−1 = − (α − eq2n−3 ) (α − eq2n−2 ) and
qn (1 − qn ) 2 f α − eβ qn−1 Cn = − , (α − eq2n−2 ) (α − eq2n−1 )
n = 1, 2, 3, . . . .
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
267
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions Again we will use Favard’s theorem (theorem 3.1) to conclude that there exist positive definite orthogonal polynomial solutions of (10.1.1) if and only if cn ∈ R for all n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. First of all we write instead of (10.3.2) cn = −
qn (α − eq2n−2 )(α − eq2n ) × (α + eq2n−1 )(2 f + β q) − qn−1 (1 + q)(2 f α q + eβ ) ,
which holds for n = 0, 1, 2, . . .. Note that (2 f − β )q c0 = α −e
and
2f −β c1 = α −e
α (1 + q)(1 − q2 ) β q(1 − q2 ) 1− − . α − eq2 α − eq2
Further we write for (10.3.3) (1)
(2)
dn = qn+1 (1 − qn )Dn Dn ,
n = 1, 2, 3, . . .
(10.4.1)
with
α − eqn−2
(1)
Dn =
(α − eq2n−3 ) (α − eq2n−2 )2 (α − eq2n−1 )
,
n = 1, 2, 3, . . .
(10.4.2)
and (2)
Dn = qn−1 (β − 2 f qn−1 )(2 f α − eβ qn−1 ) − g(α − eq2n−2 )2 = −α (β − 2 f qn−1 )2 + β (α − eq2n−2 )(β − 2 f qn−1 ) − g(α − eq2n−2 )2 ,
n = 1, 2, 3, . . . .
(10.4.3)
Note that d1 = −
q2 (1 − q) α (β − 2 f )2 − β (α − e)(β − 2 f ) + g(α − e)2 . 2 (α − e) (α − eq)
For α = 0 we may write (2) Dn = −α β − 2 f qn−1 + δ1 (α − eq2n−2 ) β − 2 f qn−1 + δ2 (α − eq2n−2 ) (10.4.4) = −α 2 f qn−1 + δ1 eq2n−2 + δ2 α 2 f qn−1 + δ1 α + δ2 eq2n−2 for n = 1, 2, 3, . . ., where δ1 δ2 = g/α and δ1 + δ2 = −β /α . Hence we have
268
10 Orthogonal Polynomial Solutions of q-Difference Equations
αδi2 + β δi + g = 0,
i = 1, 2.
Note that δ1 , δ2 ∈ R in the case that β 2 ≥ 4gα and that δ1 , δ2 ∈ C with δ2 = δ1 in the case that β 2 < 4gα . We also distinguish between the cases α = 0 and α = 0 in the q-difference equation (10.1.4). For α = 0 we have 2 ex + 2 f qx + gq2 yn (qx) − ex2 + (2 f + β q)qx + gq2 (1 + q) yn (x) + β qx + gq2 qyn (q−1 x) = eqn (1 − q−n )x2 yn (x),
n = 0, 1, 2, . . . .
(10.4.5)
In view of the homogeneity, one of the coefficients e, f , g and β can be chosen arbitrarily. We remark that we have e = 0 in view of (10.1.3) and the fact that ε = 0. Further we have (1)
Dn = −
q−7n+6 e3
and
(2) Dn = −eq2n−2 β (β − 2 f qn−1 ) + egq2n−2 ,
which implies that (1)
(2)
dn = qn+1 (1 − qn )Dn Dn =
qn+1 (1 − qn ) β (β − 2 f qn−1 ) + egq2n−2 2 5n−4 e q
for n = 1, 2, 3, . . .. Note that
β d1 = q(1 − q) e
β 2f − e e
g + . e
Hence for α = 0 we conclude that c0 ∈ R implies that (β − 2 f )/e ∈ R and c1 ∈ R further implies that also β /e ∈ R. Therefore we also have 2 f /e ∈ R. Finally, we conclude that d1 > 0 further implies that g/e ∈ R. In view of the homogeneity of (10.4.5) we may assume that e ∈ R without loss of generality. This implies that in (10.4.5) all coefficients e, f , g and β are real. For α = 0 the q-difference equation (10.1.4) can be divided by α . Then we obtain 2 e x + 2 f qx + g q2 yn (qx) − (e + q)x2 + (2 f + β q)qx + g q2 (1 + q) yn (x) + x2 + β qx + g q2 qyn (q−1 x) (10.4.6) = (q−n − 1) 1 − e qn−1 qx2 yn (x), n = 0, 1, 2, . . . with e =
e , α
f =
f , α
g =
g α
and β =
β . α
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
269
Note that this q-difference equation is equivalent to (10.1.4) with α = 1 and e, f , g and β replaced by e , f , g and β , respectively. In the sequel we will avoid the notation e , f , g , β by simply setting α = 1 and using e, f , g, β instead of e , f , g , β . We remark that through α = 1 the arbitrary choice of one of the parameters in (10.1.4) has been made. Then we have cn = −
qn (1 − eq2n−2 )(1 − eq2n )
(1 + eq2n−1 )(2 f + β q) − qn−1 (1 + q)(2 f q + eβ )
for n = 0, 1, 2, . . .. Without loss of generality, we may assume that e ∈ R. Note that β q(1 − q2 ) (1 + q)(1 − q2 ) (2 f − β )q 2f −β and c1 = 1− − . c0 = 2 1−e 1−e 1 − eq 1 − eq2 Further we have q2 (1 − q) d1 = − 1 − eq
β −2f 1−e
2
β (β − 2 f ) +g . − 1−e
Hence c0 ∈ R implies that 2 f − β ∈ R and c1 ∈ R further implies that also β ∈ R. Therefore we also have f ∈ R. Finally, we conclude that d1 > 0 further implies that g ∈ R. The sign of qn+1 (1 − qn ) for n = 1, 2, 3, . . . is given in table 10.1 and for α = 1 the (1) sign of Dn for n = 1, 2, 3, . . . is given in table 10.2.
q qn+1 (1 − qn )
q < −1 −1 < q < 0 0 < q < 1 q > 1 +
(−1)n+1
+
−
Table 10.1 sign of qn+1 (1 − qn ), n = 1, 2, 3, . . .
Case I. g = 0, β = 0 and α = 0. In this case we have dn = 0 for all n = 1, 2, 3, . . ., which implies that there exists no positive-definite orthogonal polynomials. Case II. g = 0, β = 0 and α = 0. In this case we set α = 1 and by using (10.4.3) we conclude that (2) Dn = −4 f 2 q2n−2 , n = 1, 2, 3, . . . . This implies, once more, that we must have f = 0. Hence by using table 10.1 and table 10.2, we conclude that we have positive-definite orthogonality for an infinite system of polynomials only in one case: Case IIa1. q > 1, f = 0 and e ≤ 0.
270
10 Orthogonal Polynomial Solutions of q-Difference Equations (1)
q
extra conditions
Dn
q < −1
eq > 1
(−1)n
n = 1, 2, 3, . . .
0 < eq < 1 with q < eq2N ≤ q−1
+
n = 1, 2, 3, . . . , N
e=0
+
n = 1, 2, 3, . . .
0<e
−1 < q < 0
< 1 with eq2N
+
n = 1, 2, 3, . . . , N
0 < e < 1 with 1 < eq2N < q2
+
n = 1, 2, 3, . . . , 2N + 1
e>1
(−1)n+1
n = 1, 2, 3, . . .
(−1)n
n = 1, 2, 3, . . . , N
+
n = 1, 2, 3, . . .
(−1)n+1
n = 1, 2, 3, . . . , N
eq > 1
with q−1
=1
for
≤ eq2N
q < eq < 1 e
> 1 with eq2N
=1
e > 1 with q2 < eq2N < 1 0
e<1 e > 1 with q
< eq2N
≤ q−1
(−1)n+1 n = 1, 2, 3, . . . , 2N + 1 +
n = 1, 2, 3, . . .
−
n = 1, 2, 3, . . . , N
e≤0
+
n = 1, 2, 3, . . .
0 < e < 1 with q−1 ≤ eq2N < q
+
n = 1, 2, 3, . . . , N
e>1
−
n = 1, 2, 3, . . .
q>1
(1)
Table 10.2 sign of Dn , α = 1 and N ∈ {1, 2, 3, . . .}
It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in the following three cases: Case IIb1. −1 < q < 0, f = 0 and eq > 1 with q−1 ≤ eq2N < q. Case IIb2. 0 < q < 1, f = 0 and e > 1 with q < eq2N ≤ q−1 . Case IIb3. q > 1, f = 0 and 0 < e < 1 with q−1 ≤ eq2N < q. Case III. g = 0, β = 0 and α = 0. In this case we have e = 0 and dn =
qn+1 (1 − qn ) q5n−4
2 β 2f 1 − qn−1 , e β
n = 1, 2, 3, . . . .
By using table 10.1 we conclude that for q < −1 we must have 2 f n−1 n > 0. (−1) 1 − q β For n = 1 this reads
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
− 1−
2f β
⇐⇒
>0
271
2f > 1. β
Since q < −1 this implies that 2f 2f q < q < −1 < 1 =⇒ 1 − q > 0. β β 2f Hence, (−1)n 1 − qn−1 > 0 also holds for n = 2. Further we have q2 > 1 which β implies that 2f 2 2f 4 2f < q < q < ... 1< β β β and
2f 5 2f 3 2f q < q < q < q < −1 < 1. β β β 2f Hence, if (−1)n 1 − qn−1 > 0 is true for n = 1, then it holds for all n = β 1, 2, 3, . . .. ... <
For −1 < q < 0 we must have 1 − and n = 2, since
2 f n−1 q < 0. This cannot be true for both n = 1 β
2f 2f > 1 and q > 1 cannot hold simultaneously if −1 < q < 0. β β
For 0 < q < 1 we must have 1 − 1− In the case that 0 ≤
2f >0 β
⇐⇒
2f < 1. β
2f < 1 we then have β ... ≤
and in the case that
2 f n−1 q > 0. For n = 1 this reads β
2f 2f 2 2f q ≤ q≤ <1 β β β
2f ≤ 0 we conclude that β 2f 2f 2f 2 ≤ q≤ q ≤ . . . ≤ 0 < 1. β β β
Hence if 1 −
2 f n−1 q > 0 is true for n = 1, then it holds for all n = 1, 2, 3, . . .. β
272
10 Orthogonal Polynomial Solutions of q-Difference Equations
For q > 1 we must have 1 −
2 f n−1 q < 0. For n = 1 this reads β 2f > 1, β
1−
2f <0 β
1<
2f 2f 2 2f < q< q < .... β β β
⇐⇒
which implies that
Hence if 1 −
2 f n−1 q < 0 is true for n = 1, then it holds for all n = 1, 2, 3, . . .. β
Hence, we conclude that we have positive-definite orthogonality in the following three infinite cases: Case IIIa1. q < −1 and
2f > 1. β
Case IIIa2. 0 < q < 1 and
Case IIIa3. q > 1 and
2f < 1. β
2f > 1. β
We also conclude that it is impossible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .}. Case IV. g = 0, β = 0 and α = 0. In this case we set α = 1 and by using (10.4.3) we have 2f 2 f n−1 (2) 2 n−1 n−1 1− q , n = 1, 2, 3, . . . . (10.4.7) − eq Dn = β q β β For f = 0 we must have e = 0 and (2)
Dn = −eβ 2 q2n−2 ,
n = 1, 2, 3, . . . .
Then we conclude by using (10.4.1), table 10.1 and table 10.2 that dn > 0 for all n = 1, 2, 3, . . . is only possible in the case that 0 < q < 1 and e < 0. Now we assume that f = 0. In order to study the positivity of dn we need the signs 2 f n−1 2f of the factors 1 − q and − eqn−1 . Therefore the sign of A qn−1 for n = β β 1, 2, 3, . . . is given in table 10.3.
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions q q < −1
273
A qn−1
A
A > 0 . . . < A q5 < A q3 < A q < 0 < A < A q2 < A q4 < . . . A < 0 . . . < A q4 < A q2 < A < 0 < A q < A q3 < A q5 < . . .
−1 < q < 0 A > 0 A q < A q3 < A q5 < . . . < 0 < . . . < A q4 < A q2 < A A < 0 A < A q2 < A q4 < . . . < 0 < . . . < A q5 < A q3 < A q 0
0
0 < . . . < A q2 < A q < A
A<0
A < A q < A q2 < . . . < 0
A>0
0 < A < A q < A q2 < . . .
A<0
. . . < A q2 < A q < A < 0
q>1
Table 10.3 sign of A qn−1 , n = 1, 2, 3, . . .
By using (10.4.1), (10.4.2), (10.4.7), table 10.1 and table 10.2 we conclude that in (2) order to have dn > 0 for all n = 1, 2, 3, . . . we must have that the sign of Dn for n = 1, 2, 3, . . . should be as in table 10.4. (2)
q
extra conditions
Dn
q < −1
eq > 1
(−1)n
e=0
+
e>1
(−1)n+1
−1 < q < 0
q < eq < 1
(−1)n+1
0
e<1
+
q>1
e≤0
−
e>1
+
(2)
Table 10.4 sign of Dn needed for dn > 0 for all n = 1, 2, 3, . . .
In the case q < −1 and e = 0 we have 2 f n−1 2 f n−1 n+1 n 2 q , dn = q (1 − q )β 1 − q β β
n = 1, 2, 3, . . . .
274
10 Orthogonal Polynomial Solutions of q-Difference Equations
2f 2f 2f > 0 it follows from d1 > 0 that 1 − > 0 or equivalently < 1. Hence β β β 2f 2f 2f 0< < 1. However, d2 > 0 then requires that 1− q < 0 or equivalently q > 1, β β β which cannot be true. For
2f 2f 2f < 0 it follows from d1 > 0 that 1 − < 0 or equivalently > 1, which β β β cannot be true.
For
By using (10.4.7) and table 10.4 we conclude that for all other cases the product 2f 2f 1 − qn−1 − eqn−1 β β should be either positive or negative for all n = 1, 2, 3, . . .. Now we first consider the sign of the factor 1 −
2 f n−1 q for n = 1, 2, 3, . . . with the β
aid of table 10.3. 2f 2f > 0 we cannot have that 1− qn−1 > 0 for all n = 1, 2, 3, . . . β β since q2n → ∞ for n → ∞.
1. For q < −1 and
2f 2f < 0 we cannot have that 1− qn−1 > 0 for all n = 1, 2, 3, . . . β β since q2n+1 → −∞ for n → ∞.
2. For q < −1 and
2f 2f > 0 we cannot have that 1− qn−1 < 0 for all n = 1, 2, 3, . . . β β since q2n+1 → −∞ for n → ∞.
3. For q < −1 and
2f 2f < 0 we cannot have that 1− qn−1 < 0 for all n = 1, 2, 3, . . . β β since q2n → ∞ for n → ∞.
4. For q < −1 and
2f 2f > 0 it is possible to have 1 − qn−1 > 0 for all n = β β 2f 1, 2, 3, . . . provided that < 1 since qn → 0 for n → ∞. Hence we need that β 2f < 1 in this case. 0< β
5. For −1 < q < 0 and
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
275
2f 2f < 0 it is possible to have 1 − qn−1 > 0 for all n = β β 2f 1, 2, 3, . . . provided that q < 1 since qn → 0 for n → ∞. Hence we need that β 2f < 0 in this case. q−1 < β
6. For −1 < q < 0 and
2f 2f > 0 we cannot have that 1 − qn−1 < 0 for all n = β β 1, 2, 3, . . . since qn → 0 for n → ∞.
7. For −1 < q < 0 and
2f 2f < 0 we cannot have that 1 − qn−1 < 0 for all n = β β 1, 2, 3, . . . since qn → 0 for n → ∞.
8. For −1 < q < 0 and
2f 2f > 0 it is possible to have 1 − qn−1 > 0 for all n = β β 2f 1, 2, 3, . . . provided that < 1 since qn → 0 for n → ∞. Hence we need that β 2f < 1 in this case. 0< β
9. For 0 < q < 1 and
2f 2f < 0 it is possible to have 1 − qn−1 > 0 for all n = β β 2f 1, 2, 3, . . .. In fact, for < 0 this is always true since qn → 0 for n → ∞. β
10. For 0 < q < 1 and
2f 2f > 0 we cannot have that 1 − qn−1 < 0 for all n = β β 1, 2, 3, . . . since qn → 0 for n → ∞.
11. For 0 < q < 1 and
2f 2f < 0 we cannot have that 1 − qn−1 < 0 for all n = β β 1, 2, 3, . . . since qn → 0 for n → ∞.
12. For 0 < q < 1 and
2f 2f > 0 we cannot have that 1 − qn−1 > 0 for all n = 1, 2, 3, . . . β β since qn → ∞ for n → ∞.
13. For q > 1 and
14. For q > 1 and In fact, for
2f 2f < 0 it is possible to have 1 − qn−1 > 0 for all n = 1, 2, 3, . . .. β β
2f < 0 this is always true since qn → ∞ for n → ∞. β
276
10 Orthogonal Polynomial Solutions of q-Difference Equations
2f 2f > 0 it is possible to have 1 − qn−1 < 0 for all n = 1, 2, 3, . . . β β 2f 2f provided that > 1 since qn → ∞ for n → ∞. Hence we need that > 1 in this β β case.
15. For q > 1 and
2f 2f < 0 we cannot have that 1 − qn−1 < 0 for all n = 1, 2, 3, . . . β β since qn → ∞ for n → ∞.
16. For q > 1 and
2f − eqn−1 we only need to consider the six possible cases 5, 6, 9, 10, β 14 and 15 above. Again we use table 10.3. For the factor
2f 2f < 1 and e > 0 we must have − eqn−1 > 0 for all β β 2f > e. Hence we need that 0 < n = 1, 2, 3, . . . which is possible provided that β 2f e< < 1 in this case. β
1. For −1 < q < 0, 0 <
2f 2f < 1 and e ≤ 0 we must have − eqn−1 > 0 for all β β 2f > eq. Hence we need that 0 ≤ n = 1, 2, 3, . . . which is possible provided that β 2f eq < < 1 in this case. β
2. For −1 < q < 0, 0 <
2f 2f < 0 and e > 0 we must have − eqn−1 > 0 for all β β n = 1, 2, 3, . . . which is not possible since qn → 0 for n → ∞.
3. For −1 < q < 0, q−1 <
2f 2f < 0 and e ≤ 0 we must have − eqn−1 > 0 for all β β n = 1, 2, 3, . . . which is not possible since qn → 0 for n → ∞.
4. For −1 < q < 0, q−1 <
2f 2f < 1 and e > 0 we must have − eqn−1 > 0 for all β β 2f n = 1, 2, 3, . . . which is possible provided that > e. Hence we need that 0 < β 2f < 1 in this case. e< β
5. For 0 < q < 1, 0 <
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
277
2f 2f < 1 and e ≤ 0 we must have − eqn−1 > 0 for all β β 2f n = 1, 2, 3, . . . which is possible. In fact, for 0 < < 1 and e ≤ 0 this is always β true.
6. For 0 < q < 1, 0 <
2f 2f < 0 and e > 0 we must have − eqn−1 > 0 for all n = β β 1, 2, 3, . . . which is not possible since qn → 0 for n → ∞.
7. For 0 < q < 1,
2f 2f < 0 and e ≤ 0 we must have − eqn−1 > 0 for all n = β β 1, 2, 3, . . . which is not possible since qn → 0 for n → ∞.
8. For 0 < q < 1,
2f 2f < 0 and e > 0 we must have − eqn−1 > 0 for all n = 1, 2, 3, . . . β β which is not possible since qn → ∞ for n → ∞.
9. For q > 1,
2f 2f < 0 and e ≤ 0 we must have − eqn−1 < 0 for all n = 1, 2, 3, . . . β β which is not possible since qn → ∞ for n → ∞.
10. For q > 1,
2f 2f > 1 and e > 0 we must have − eqn−1 < 0 for all n = 1, 2, 3, . . . β β 2f 2f which is possible provided that < e. Hence we need that 1 < < e in this β β case.
11. For q > 1,
2f 2f > 1 and e ≤ 0 we must have − eqn−1 > 0 for all n = 1, 2, 3, . . . β β 2f which is possible. In fact, for > 1 and e ≤ 0 this is always true. β
12. For q > 1,
Hence, we conclude that we have positive-definite orthogonality in the following seven infinite cases: Case IVa1. −1 < q < 0 and 0 < e <
2f < 1. β
Case IVa2. −1 < q < 0 and 0 ≤ eq <
Case IVa3. 0 < q < 1 and 0 < e <
2f < 1. β
2f < 1. β
278
10 Orthogonal Polynomial Solutions of q-Difference Equations
2f < 1. β
Case IVa4. 0 < q < 1, e ≤ 0 and 0 <
Case IVa5. 0 < q < 1, e < 0 and f = 0. Case IVa6. q > 1 and 1 <
2f < e. β
Case IVa7. q > 1, e ≤ 0 and
2f > 1. β
It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .}. By using (10.4.1), (10.4.2), (10.4.7), table 10.1 and table 10.2 we conclude that in (2) order to have dn > 0 we must have that the sign of Dn should be as in table 10.5.
q q < −1
0 < eq < 1 with q 0<e<1
=1
< eq2N
< q2
for
+
n = 1, 2, 3, . . . , N
+
n = 1, 2, 3, . . . , N
+ n = 1, 2, 3, . . . , 2N + 1 −
n = 1, 2, 3, . . . , N
with eq2N
+
n = 1, 2, 3, . . . , N
e>1
q>1
≤ q−1
Dn
eq > 1 with q−1 ≤ eq2N < q e>1
0
< eq2N
with eq2N
0 < e < 1 with 1 −1 < q < 0
(2)
extra conditions
with q2
<
e > 1 with q < 0<e<1
Table 10.5 sign of
eq2N
with q−1 (2) Dn
=1
eq2N
<1
≤ q−1
≤ eq2N
+ n = 1, 2, 3, . . . , 2N + 1 −
n = 1, 2, 3, . . . , N
−
n = 1, 2, 3, . . . , N
needed for dn > 0 with N ∈ {1, 2, 3, . . .}
By using (10.4.7) and table 10.5 we conclude that the sign of the product 2f 2 f n−1 n−1 1− q − eq β β should alternate in the cases that q < −1 or −1 < q < 0 and should be negative in the cases that 0 < q < 1 or q > 1. First of all we note that for q < −1 and for −1 < q < 0 it is possible that the factor 2f 1 − qn−1 keeps the same sign for n = 1, 2, 3, . . . , N for certain N ∈ {1, 2, 3, . . .} β
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
and that the factor
279
2f − eqn−1 alternates for these values of n (in the case that f = 0 β
for instance). As before, for f = 0 we must have e = 0 and (2)
Dn = −eβ 2 q2n−2 ,
n = 1, 2, 3, . . . .
By using table 10.5 we conclude that we only have dn > 0 for n = 1, 2, 3, . . . , N in the cases that q < −1 and 0 < eq < 1 with q < eq2N ≤ q−1 , or 0 < q < 1 and e > 1 with q < eq2N ≤ q−1 , or q > 1 and 0 < e < 1 with q−1 ≤ eq2N < q. Now we assume that f = 0 and we consider the sign of the factor 1 −
2 f n−1 q for β
n = 1, 2, 3, . . . , N or n = 1, 2, 3, . . . , 2N + 1 with the aid of table 10.3. 2 f n−1 2f > 0 it is possible to have 1 − q > 0 for all n = β β 2 f N−2 1, 2, 3, . . . , N provided that q < 1 and N even. β
1. For q < −1 and
2f 2 f n−1 < 0 it is possible to have 1 − q > 0 for all n = β β 2 f N−2 q < 1 and N odd. 1, 2, 3, . . . , N provided that β
2. For q < −1 and
2f 2 f n−1 > 0 we cannot have that 1 − q < 0 for all n = β β 2f 2f 1, 2, 3, . . . , N with N ≥ 2 since 1 − q < 0 then implies that < q−1 < 0. β β
3. For q < −1 and
2 f n−1 2f < 0 we cannot have that 1 − q < 0 for all n = β β 2f 1, 2, 3, . . . , N since then we have 1 − > 1 > 0. β
4. For q < −1 and
5. For −1 < q < 0 and implies that 0 <
2f 2f > 0 we have: if 1 − qn−1 > 0 is true for n = 1, which β β
2f < 1, then it holds for all n = 1, 2, 3, . . .. β
280
10 Orthogonal Polynomial Solutions of q-Difference Equations
2f 2f < 0 we have: if 1 − qn−1 > 0 is true for n = 1 and β β 2 f n = 2, which implies that q−1 < < 0, then it holds for all n = 1, 2, 3, . . .. β
6. For −1 < q < 0 and
2f 2f > 0 we cannot have that 1 − qn−1 < 0 for all n = β β 2f 2f 1, 2, 3, . . . , N with N ≥ 2 since 1 − q < 0 then implies that < q−1 < 0. β β
7. For −1 < q < 0 and
2f 2f < 0 we cannot have that 1 − qn−1 < 0 for all n = β β 2f > 1 > 0. 1, 2, 3, . . . , N since then we have 1 − β
8. For −1 < q < 0 and
2f 2f > 0 we have: if 1 − qn−1 > 0 is true for n = 1, which β β
9. For 0 < q < 1 and implies that 0 <
2f < 1, then it holds for all n = 1, 2, 3, . . .. β
10. For 0 < q < 1 and
2f 2f < 0 we have: 1 − qn−1 > 0 is true for all n = 1, 2, 3, . . .. β β
2f 2f > 0 it is possible to have 1 − qn−1 < 0 for all n = β β 2 f N−1 1, 2, 3, . . . , N provided that q > 1. β
11. For 0 < q < 1 and
12. For 0 < q < 1 and
2f 2f < 0 we have: 1 − qn−1 < 0 is false for all n = 1, 2, 3, . . .. β β
2f 2f > 0 it is possible to have 1− qn−1 > 0 for all n = 1, 2, 3, . . . , N β β 2 f N−1 q < 1. provided that β
13. For q > 1 and
14. For q > 1 and
2f 2f < 0 we have: 1 − qn−1 > 0 is true for all n = 1, 2, 3, . . .. β β
15. For q > 1 and
2f 2f > 0 we have: if 1− qn−1 < 0 is true for n = 1, which implies β β
that
2f > 1, then it holds for all n = 1, 2, 3, . . .. β
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
16. For q > 1 and
281
2f 2f < 0 we have: 1 − qn−1 < 0 is false for all n = 1, 2, 3, . . .. β β
2f − eqn−1 we only need to consider the ten possible cases 1, 2, 5, 6, β 9, 10, 11, 13, 14 and 15 above. Again we use table 10.3.
For the factor
1. For q < −1, of
2f 2 f N−2 > 0 and q < 1 with N even, we conclude that the sign β β
2f − eqn−1 equals (−1)n for all n = 1, 2, 3, . . . , N provided that e > 0 and β
2f < e and that it equals (−1)n+1 for all n = 1, 2, 3, . . . , N provided that e < 0 β 2f and < eq. β 2 f N−2 2f < 0 and q < 1 with N odd, we conclude that the sign of β β 2f 2f − eqn−1 equals (−1)n for all n = 1, 2, 3, . . . , N provided that e > 0 and > eq β β 2f > e. and that it equals (−1)n+1 for all n = 1, 2, 3, . . . , N provided that e < 0 and β
2. For q < −1,
2f 2f < 1, we conclude that the sign of − eqn−1 equals β β 2f (−1)n for n = 1 provided that e > > 0. Then we use table 10.3 to conclude β 2f − eqn−1 > 0 for all n = 1, 2, 3, . . . , N we must have that for (−1)n β
3. For −1 < q < 0 and 0 <
eq < eq3 < eq5 < . . . < 0 < . . . <
2f < eqN−1 < . . . < eq4 < eq2 < e β
with N odd. 2f 2f < 0, we conclude that the sign of − eqn−1 β β 2f equals (−1)n+1 for n = 1 provided that e < < 0. Then we use table 10.3 to β 2f − eqn−1 > 0 for all n = 1, 2, 3, . . . , N we must conclude that for (−1)n+1 β have
4. For −1 < q < 0 and q−1 <
e < eq2 < eq4 < . . . < eqN−1 <
2f < . . . < 0 < . . . < eq5 < eq3 < eq β
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10 Orthogonal Polynomial Solutions of q-Difference Equations
with N even. 2f 2f < 1, we conclude that − eqn−1 < 0 for all n = β β 2f < eqN−1 . 1, 2, 3, . . . , N provided that β
5. For 0 < q < 1 and 0 <
2f 2f < 0, we conclude that − eqn−1 < 0 for all n = β β 2f 2f 1, 2, 3, . . . , N provided that < e. In fact, then we have that − eqn−1 < 0 β β for all n = 1, 2, 3, . . ..
6. For 0 < q < 1 and
2f 2 f N−1 2f > 0 and q > 1, we conclude that − eqn−1 > 0 for all β β β 2f 2f > e. In fact, then we have that −eqn−1 > 0 n = 1, 2, 3, . . . , N provided that β β for all n = 1, 2, 3, . . ..
7. For 0 < q < 1,
2 f N−1 2f 2f > 0 and q < 1, we conclude that − eqn−1 < 0 for all β β β 2f 2f n = 1, 2, 3, . . . , N provided that < e. In fact, then we have that −eqn−1 < 0 β β for all n = 1, 2, 3, . . ..
8. For q > 1,
2f 2f < 0, we conclude that − eqn−1 < 0 for n = 1 provided that β β 2f 2f 2f < e. If e > 0, which implies that < 0 < e, then we have that −eqn−1 < 0 β β β 2f for all n = 1, 2, 3, . . .. If e < 0, which implies that < e < 0, then we have that β 2f 2f − eqn−1 < 0 for all n = 1, 2, 3, . . . , N provided that < eqN−1 . β β
9. For q > 1 and
2f 2f > 1, we conclude that − eqn−1 > 0 for all n = 1, 2, 3, . . . , N β β 2f provided that > eqN−1 . β
10. For q > 1 and
Hence, we conclude that we have positive-definite orthogonality in the following eight finite cases: 2 f N−2 2f < eq < 1 with q < eq2N ≤ q−1 and 0 < q <1 β β with N even. In that case we have a finite system of N + 1 polynomials.
Case IVb1. q < −1, 0 <
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
283
2f 2 f N−2 < 0 with q < eq2N ≤ q−1 and 0 < q <1 β β with N odd. In that case we have a finite system of N + 1 polynomials.
Case IVb2. q < −1, q−1 < e <
Case IVb3. 0 < q < 1, e > 1 with q < eq2N ≤ q−1 , 0 <
2f 2f < 1 and < eqN−1 . In β β
that case we have a finite system of N + 1 polynomials. Case IVb4. 0 < q < 1, e > 1 with q < eq2N ≤ q−1 and
2f < 0. In that case we have β
a finite system of N + 1 polynomials. Case IVb5. 0 < q < 1, e > 1 with q < eq2N ≤ q−1 ,
2f 2 f N−1 > e and q > 1. In that β β
case we have a finite system of N + 1 polynomials. Case IVb6. q > 1, 0 < e < 1 with q−1 ≤ eq2N < q, 0 <
2f 2 f N−1 < e and q < 1. In β β
that case we have a finite system of N + 1 polynomials. Case IVb7. q > 1, 0 < e < 1 with q−1 ≤ eq2N < q and
2f < 0. In that case we have β
a finite system of N + 1 polynomials. Case IVb8. q > 1, 0 < e < 1 with q−1 ≤ eq2N < q,
2f 2f > 1 and > eqN−1 . In that β β
case we have a finite system of N + 1 polynomials. Case V. g = 0, α = 0 and β = 0. In this case we have e = 0 and g qn+1 (1 − qn ) , e q3n−2
dn =
n = 1, 2, 3, . . . .
By using table 10.1 we conclude that we have positive-definite orthogonality in the following three cases: Case Va1. −1 < q < 0 and Case Va2. 0 < q < 1 and Case Va3. q > 1 and
g < 0. e
g > 0. e
g < 0. e
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10 Orthogonal Polynomial Solutions of q-Difference Equations
We also conclude that it is impossible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .}. Case VI. g = 0, α = 0 and β = 0. In this case we have e = 0 and g 2n−2 qn+1 (1 − qn ) β β 2 f n−1 − q q dn = + q5n−4 e e e e 2 n+1 n q (1 − q ) β 2 f n−1 eg 2n−2 = 1− q + 2q q5n−4 e β β 2 n+1 n γ1 g n−1 γ2 g n−1 q (1 − q ) β = 1 + 1 + , n = 1, 2, 3, . . . , q q q5n−4 e β β with γ1 + γ2 = −2 f /g and γ1 γ2 = e/g. We remark that for f 2 < eg we have 1−
2 2 f n−1 eg 2n−2 f eg − f 2 2n−2 q + 2q = 1 − qn−1 + q >0 β β β β2
for all n = 1, 2, 3, . . .. In that case γ1 and γ2 are complex conjugates. In the case that f 2 ≥ eg they are real: − f − f 2 − eg − f + f 2 − eg and γ2 = . γ1 = g g By using table 10.1 we conclude that for q < −1 we must have 2f eg (−1)n 1 − qn−1 + 2 q2n−2 > 0, β β which cannot be true for all n = 1, 2, 3, . . ., since (−1)n q2n−2 → −∞ for odd n → ∞ and (−1)n q2n−2 → ∞ for even n → ∞. Moreover, the constant eg is either positive or negative. For −1 < q < 0 we must have 1−
2 f n−1 eg 2n−2 q + 2q < 0, β β
which cannot be true for all n = 1, 2, 3, . . ., since qn−1 → 0 for n → ∞. For 0 < q < 1 we must have 1−
2 f n−1 eg 2n−2 q + 2q > 0, β β
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
285
which is true for all n = 1, 2, 3, . . . in the case that f 2 < eg. In the case that f 2 ≥ eg we must have γ1 g n−1 γ2 g n−1 eg 2f 1+ > 0. q q 1 − qn−1 + 2 q2n−2 = 1 + β β β β Since qn → 0 for n → ∞ this can only be true for all n = 1, 2, 3, . . . if both factors are positive, which implies that we must have
γ1 g > −1 and β
γ2 g > −1. β
For q > 1 we must have 1−
2 f n−1 eg 2n−2 q + 2q < 0, β β
which is false in the case f 2 < eg. In the case that f 2 ≥ eg we must have 2 f n−1 eg 2n−2 γ1 g n−1 γ2 g n−1 1+ < 0. + 2q = 1+ q q 1− q β β β β Since qn → ∞ for n → ∞ this can only be true for all n = 1, 2, 3, . . . if γ1 γ2 =
e < 0. g
We conclude that we must have either
γ1 g < −1 and β or
γ1 g > 0 and β
γ2 g > 0, β γ2 g < −1. β
Hence, we conclude that we have positive-definite orthogonality in the following four infinite cases: Case VIa1. 0 < q < 1 and f 2 < eg. Case VIa2. 0 < q < 1, f 2 ≥ eg,
γ1 g γ2 g > −1 and > −1. β β
Case VIa3. q > 1,
γ1 g γ2 g < −1 and > 0. β β
Case VIa4. q > 1,
γ1 g γ2 g > 0 and < −1. β β
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10 Orthogonal Polynomial Solutions of q-Difference Equations
It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .}. For q < −1 we must have that 2f eg (−1)n 1 − qn−1 + 2 q2n−2 > 0 β β is true for all n ≤ N and false for n = N + 1. For −1 < q < 0 we must have that 1−
2 f n−1 eg 2n−2 q + 2q <0 β β
is true for all n ≤ N and false for n = N + 1. We recall that both are impossible in the case that f 2 < eg. In the case that f 2 ≥ eg we have 2 f n−1 eg 2n−2 γ1 g n−1 γ2 g n−1 1− q 1+ + 2q = 1+ q q β β β β with
γ1 =
−f −
f 2 − eg
g
and γ2 =
Without loss of generality we will assume that
γ2 g < 0 we must have β γ1 g γ2 g 1+ >0 − 1+ β β
−f +
f 2 − eg
g
.
γ1 g γ2 g ≤ (since γ1 and γ2 are real). β β
For q < −1 and
=⇒
γ1 g γ2 g < −1 < < 0. β β
Now we use table 10.3 to conclude that ... < −
γ2 g 5 γ2 g 3 γ2 g γ2 g γ2 g 2 γ2 g 4 q <− q <− q<0<− <− q <− q < ..., β β β β β β
which implies that ... < 1+ and 1 < 1+
γ2 g 4 γ2 g 2 γ2 g q < 1+ q < 1+ <1 β β β
γ2 g γ2 g 3 γ2 g 5 q < 1+ q < 1+ q < .... β β β
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
287
γ2 g n−1 n > 0 for all n = 1, 2, 3, . . .. Now we have Hence we have that (−1) 1 + q β γ1 g n−1 γ1 g N γ1 g N q > 0 for n = 1, 2, 3, . . . , N and 1 + q ≤ 0 provided that q ≤ 1+ β β β γ1 g N−2 q and N even. −1 < β
γ2 g γ2 g n−1 > 0 it is possible to have 1 + q > 0 for n = 1, 2, 3, . . . , N β β γ2 g N γ2 g N γ2 g N−2 q ≤ 0 provided that q ≤ −1 < q with N odd. Combined and 1 + β β β γ1 g n−1 > 0 for all n = 1, 2, 3, . . . as above this leads to another q with (−1)n 1 + β γ1 g γ2 g <0< . finite case with −1 < β β
For q < −1 and
γ2 g < 0 we must have β γ1 g γ2 g 1+ < 0 =⇒ 1+ β β
For −1 < q < 0 and
γ1 g γ2 g < −1 < < 0. β β
Now we use table 10.3 to conclude that −
γ2 g γ2 g 3 γ2 g 5 γ2 g 4 γ2 g 2 γ2 g q<− q <− q < ... < 0 < ... < − q <− q <− < 1, β β β β β β
γ2 g n−1 q > 0 for all n = 1, 2, 3, . . .. However, then we cannot β γ1 g n−1 q < 0 for n = 1, 2, 3, . . . , N. have that 1 + β which implies that 1 +
γ2 g > 0 we must have β γ1 g γ2 g 1+ < 0 =⇒ 1+ β β
For −1 < q < 0 and
γ1 g γ2 g < −1 < 0 < . β β
Now we use table 10.3 to conclude that −
γ2 g γ2 g 2 γ2 g 4 γ2 g 5 γ2 g 3 γ2 g <− q <− q < ... < 0 < ... < − q <− q <− q, β β β β β β
which implies that 1+ and
γ2 g γ2 g 3 γ2 g 5 q < 1+ q < 1+ q < ... < 1 β β β
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10 Orthogonal Polynomial Solutions of q-Difference Equations
1 < ... < 1+
γ2 g 4 γ2 g 2 γ2 g q < 1+ q < 1+ . β β β
Furthermore we have −
γ1 g γ1 g 3 γ1 g 5 γ1 g 4 γ1 g 2 γ1 g q<− q <− q < ... < 0 < ... < − q <− q <− , β β β β β β
which implies that
γ1 g γ1 g 2 γ1 g 4 < 1+ q < 1+ q < ... < 1 β β β
1+ and
1 < ... < 1+
γ1 g 5 γ1 g 3 γ1 g q < 1+ q < 1+ q. β β β
Now we want that γ1 g n−1 γ2 g n−1 2f eg 1+ = 1 − qn−1 + 2 q2n−2 < 0, q q 1+ β β β β and 1 −
n = 1, 2, 3, . . . , N
2 f N eg 2N q + 2 q ≥ 0. This implies that we must have β β 1+
γ1 g N−2 γ1 g N q < 0 ≤ 1+ q β β
with N even
or
γ2 g N γ2 g N−2 q ≤ 0 < 1+ q with N odd. β β γ1 g N−2 γ1 g N γ2 g N Hence, we must have either q < −1 ≤ q with N even, or q ≤ −1 < β β β γ2 g N−2 q with N odd. β 1+
For 0 < q < 1 we must have γ1 g n−1 γ2 g n−1 1+ > 0, q q 1+ β β with both
n = 1, 2, 3, . . . , N
γ1 g γ2 g γ1 g γ2 g < −1 and < −1. With the assumption that ≤ we must β β β β
have that 1+
γ2 g N−1 γ2 g N q < 0 ≤ 1+ q β β
For q > 1 we must have
=⇒
γ2 g N−1 γ2 g N q < −1 ≤ q . β β
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
289
γ1 g n−1 γ2 g n−1 1+ 1+ < 0, n = 1, 2, 3, . . . , N q q β β γ1 g N γ2 g N γ1 g γ1 g n−1 1+ > 0. If q q < −1, then we have 1 + q <0 and 1 + β β β β for all n = 1, 2, 3, . . .. Hence, then we must have that −1 <
γ2 g < 0 with β
γ2 g N γ2 g N−1 q ≤ −1 < q . β β
γ2 g γ2 g n−1 > 0, then we have 1 + q > 0 for all n = 1, 2, 3, . . .. Then it is impossible β β γ1 g n−1 γ1 g N q < 0 for n = 1, 2, 3, . . . , N and 1 + q > 0. to have 1 + β β
If
Hence, we conclude that we have positive-definite orthogonality in the following six finite cases:
γ1 g γ2 g γ1 g N γ1 g N−2 < −1 < < 0 with q ≤ −1 < q and N Case VIb1. q < −1 and β β β β even. Case VIb2. q < −1 and −1 <
γ1 g γ2 g γ2 g N γ2 g N−2 <0< with q ≤ −1 < q and N β β β β
odd.
γ1 g γ2 g γ1 g N−2 γ1 g N < −1 < 0 < with q < −1 ≤ q and N Case VIb3. −1 < q < 0, β β β β even. Case VIb4. −1 < q < 0,
γ1 g γ2 g γ2 g N γ2 g N−2 < −1 < 0 < with q ≤ −1 < q and N β β β β
odd. Case VIb5. 0 < q < 1 and
Case VIb6. q > 1 and
γ1 g γ2 g γ2 g N−1 γ2 g N ≤ < −1 with q < −1 ≤ q . β β β β
γ1 g γ2 g γ2 g N γ2 g N−1 < −1 < < 0 with q ≤ −1 < q . β β β β
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10 Orthogonal Polynomial Solutions of q-Difference Equations
Case VII. g = 0 and α = 0. In this case we set α = 1 and by using (10.4.4) we have (2) Dn = − 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 ,
n = 1, 2, 3, . . . .
In the case that β 2 < 4g we have δ2 = δ1 which implies that (2) Dn = − 2 f qn−1 + δ1 eq2n−2 + δ1 2 f qn−1 + δ1 + δ1 eq2n−2 = − 2 f qn−1 + δ1 eq2n−2 + δ1 2 f qn−1 + δ1 + δ1 eq2n−2 2 = − 2 f qn−1 + δ1 eq2n−2 + δ1 < 0, n = 1, 2, 3, . . . . By using γ1 γ2 = e/g, γ1 + γ2 = −2 f /g and δ1 δ2 = g, we obtain that e = γ1 γ2 δ1 δ2
and 2 f = −(γ1 + γ2 )δ1 δ2 ,
which implies that (2) Dn = −δ1 δ2 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 1 − γ2 δ1 qn−1 1 − γ2 δ2 qn−1 for n = 1, 2, 3, . . .. In the case that β 2 ≥ 4g both δ1 and δ2 are real. Moreover, in the case that f 2 < eg we have γ2 = γ1 which implies that (2) Dn = −δ1 δ2 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 = −δ1 δ2 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 (1 − γ1 δ1 qn−1 ) (1 − γ1 δ2 qn−1 ) 2 2 = −δ1 δ2 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 , n = 1, 2, 3, . . . . (2)
This implies that Dn has the same sign as −δ1 δ2 = −g in that case. Finally, we consider the case that γ1 , γ2 , δ1 , δ2 ∈ R. Again we use table 10.1 and (2) table 10.2 to conclude that for q < −1 we must have either eq > 1 and (−1)n Dn > 0, (2) (2) or e = 0 and Dn > 0, or e > 1 and (−1)n Dn < 0, in order to have dn > 0 for all (2) (2) n = 1, 2, 3, . . .. However, note that (−1)n Dn > 0 or (−1)n Dn < 0 cannot be true (2) for all n = 1, 2, 3, . . .. Hence we must have e = 0 and Dn > 0 or 2 f qn−1 + δ2 2 f qn−1 + δ1 < 0, which is also impossible for all n = 1, 2, 3, . . .. (2)
For −1 < q < 0 we must have q < eq < 1 and (−1)n+1 Dn > 0. Also in this case (2) (2) (−1)n+1 Dn > 0 cannot be true for all n = 1, 2, 3, . . ., since Dn → −δ1 δ2 = −g for n → ∞.
10.4 Classification of the Positive-Definite Orthogonal Polynomial Solutions
291
(2)
For 0 < q < 1 we must have e < 1 and Dn > 0, which implies that 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 < 0, n = 1, 2, 3, . . . . For n = 1 this reads (2 f + δ1 e + δ2 ) (2 f + δ1 + δ2 e) < 0, which implies that
δ1 e + δ2 < −2 f < δ1 + δ2 e or δ1 + δ2 e < −2 f < δ1 e + δ2 . The first inequality implies that (1 − e)δ2 < (1 − e)δ1
=⇒
δ2 < δ1 ,
since 1 − e > 0. This contradicts the fact that δ1 < δ2 . So we must have
δ1 + δ2 e < −2 f < δ1 e + δ2 . Further we must have that g = δ1 δ2 < 0, which implies that δ1 < 0 < δ2 . We conclude that it is possible to have orthogonality for an infinite system of polynomials in that case. (2)
(2)
For q > 1 we must have either e ≤ 0 and Dn < 0, or e > 1 and Dn > 0. So, for e ≤ 0 we must have 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 > 0,
n = 1, 2, 3, . . . .
For n = 1 this reads (2 f + δ1 e + δ2 ) (2 f + δ1 + δ2 e) > 0, which implies that both factors are either positive or negative. Further we must have that g = δ1 δ2 > 0, which implies that either δ1 ≤ δ2 < 0 or 0 < δ1 ≤ δ2 . We conclude that it is possible to have orthogonality for an infinite system of polynomials in this case too. For e > 1 we must have 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 < 0,
n = 1, 2, 3, . . . .
In this case we must have
δ1 e + δ2 < −2 f < δ1 + δ2 e and g = δ1 δ2 < 0.
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10 Orthogonal Polynomial Solutions of q-Difference Equations
Hence, we have δ1 < 0 < δ2 , which implies that δ1 e < 0 and δ2 e > 0. Also in this case it is possible to have orthogonality for an infinite system of polynomials. Hence, we conclude that we have positive-definite orthogonality in the following three infinite cases: Case VIIa1. 0 < q < 1, e < 1, δ1 < 0 < δ2 and 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 < 0,
n = 1, 2, 3, . . . .
Case VIIa2. q > 1, e ≤ 0, δ1 δ2 > 0 with δ1 , δ2 ∈ R or δ2 = δ1 , and 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 > 0, n = 1, 2, 3, . . . . Case VIIa3. q > 1, e > 1, δ1 < 0 < δ2 and 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 < 0,
n = 1, 2, 3, . . . .
It is also possible to have positive-definite orthogonality for a finite system of N + 1 or 2N + 2 polynomials with N ∈ {1, 2, 3, . . .}. We only consider the cases where (1) Dn has opposite sign for n = N and n = N + 1 according to table 10.5. Skipping further details, we conclude that we have positive-definite orthogonality, at least in the following eight finite cases: Case VIIb1. q < −1, 0 < eq < 1 with q < eq2N ≤ q−1 and Dn > 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials. (2)
(2)
Case VIIb2. q < −1, 0 < e < 1 with eq2N = 1 and Dn > 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials. (2)
Case VIIb3. q < −1, 0 < e < 1 with 1 < eq2N < q2 and Dn > 0 for n = 1, 2, 3, . . . , 2N + 1. In that case we have a finite system of 2N + 2 polynomials. Case VIIb4. −1 < q < 0, eq > 1 with q−1 ≤ eq2N < q and Dn < 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials. (2)
(2)
Case VIIb5. −1 < q < 0, e > 1 with eq2N = 1 and Dn > 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials.
10.5 Solutions of the q-Pearson Equation
293 (2)
Case VIIb6. −1 < q < 0, e > 1 with q2 < eq2N < 1 and Dn > 0 for n = 1, 2, 3, . . . , 2N + 1. In that case we have a finite system of 2N + 2 polynomials. Case VIIb7. 0 < q < 1, e > 1 with q < eq2N ≤ q−1 and Dn < 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials. (2)
Case VIIb8. q > 1, 0 < e < 1 with q−1 ≤ eq2N < q and Dn < 0 for n = 1, 2, 3, . . . , N. In that case we have a finite system of N + 1 polynomials. (2)
We remark that it is also possible to have positive-definite orthogonality for finite (2) systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in cases where Dn has opposite sign for n = N and n = N + 1.
10.5 Solutions of the q-Pearson Equation We look for solutions of the q-Pearson equation (cf. (3.2.8)) w(x)C(x) = qw(qx)D(qx), with C(x) =
0 < |q| < 1
ex2 + 2 f qx + gq2 q(q − 1)2 x2
and D(x) =
{e + 2ε (1 − q)} x2 + {2 f + γ (1 − q)} qx + gq2 α x2 + β qx + gq2 = . 2 2 (q − 1) x (q − 1)2 x2
Hence we have by using (10.1.3) q2 {e + 2ε (1 − q)} x2 + {2 f + γ (1 − q)} x + g w(x) = w(qx) ex2 + 2 f qx + gq2 =
q2 (α x2 + β x + g) . ex2 + 2 f qx + gq2
(10.5.1)
We consider two types of solutions: A. Continuous solutions for x ∈ R in terms of (convergent) infinite products. For the convergence of these infinite products we refer to the book [471] by L.J. Slater. ν B. Discrete solutions for {xν }Nν =0 with N → ∞ or {xν }∞ ν =−∞ , id est xν = Aq with A ∈ C and ν = 0, ±1, ±2, . . ., in terms of finite products. Note that xν +1 = qxν . Without loss of generality we can choose A = 1 in each case.
294
10 Orthogonal Polynomial Solutions of q-Difference Equations
In order to find solutions of the q-Pearson equation (10.5.1), we distinguish between 0 < |q| < 1 and |q| > 1. For 0 < |q| < 1 we make the following observations. For ∞
w(x) = (−rx; q)∞ = ∏ (1 + rxqk ),
r∈C
(10.5.2)
k=0
we have
∞
w(qx) = (−rqx; q)∞ = ∏ (1 + rxqk+1 ), k=0
which implies that w(x) = 1 + rx. w(qx)
(10.5.3)
For ∞ q qk+1 , w(x) = −rx, − ; q = ∏ 1 + rxqk 1+ rx rx ∞ k=0 we have
∞
w(qx) = ∏ 1 + rxq k=0
k+1
r ∈ C,
r = 0 (10.5.4)
qk , 1+ rx
which implies that 1 + rx w(x) = = rx. 1 w(qx) 1 + rx
(10.5.5)
Finally, for ∞ qk+1 q k , 1+ w(x) = −sx, − ; q = ∏ 1 + sxq tx tx ∞ k=0
we have
s,t ∈ C,
t = 0 (10.5.6)
∞ qk w(qx) = ∏ 1 + sxqk+1 1 + , tx k=0
which implies that w(x) 1 + sx 1 + sx = · tx. = w(qx) 1 + tx1 1 + tx Combining the latter two results, we conclude that −sx, − txq ; q ∞ , r, s,t ∈ C, r = 0, w(x) = −rx, − rxq ; q ∞ implies that
(10.5.7)
t = 0
(10.5.8)
10.5 Solutions of the q-Pearson Equation
295
1 + sx tx 1 + sx t w(x) = · = · . w(qx) 1 + tx rx 1 + tx r
(10.5.9)
For |q| > 1 we rewrite the q-Pearson equation (10.5.1) in the following way. If we set q = p−1 , then we have p−2 e + 2ε (1 − p−1 ) x2 + 2 f + γ (1 − p−1 ) x + g w(x) = w(p−1 x) ex2 + 2 f p−1 x + gp−2
α x2 + β x + g p−2 (α x2 + β x + g) , = ex2 + 2 f p−1 x + gp−2 ep2 x2 + 2 f px + g
(10.5.10)
α := e + 2ε (1 − p−1 ) and β := 2 f + γ (1 − p−1 ).
(10.5.11)
= where
For |q| > 1 this implies that 0 < |p| < 1. Now we replace x by px to obtain ep4 x2 + 2 f p2 x + g w(x) = 2 2 , w(px) α p x + β px + g
0 < |p| < 1.
(10.5.12)
Note that if α = e + 2ε (1 − q) does not depend on q, for instance in the case that α = 0 or α = 1, then α = α . The same holds for β . Case II-A. g = 0, β = 0 = β , α = 0 respectively α = 0 and f = 0. From (10.5.1) we have 1 w(x) α q2 x 2 α qx = 2 = ex · w(qx) ex + 2 f qx 1 + 2 f q 2 f and by using (10.5.2) and (10.5.3) with r = e/2 f q and by using (10.5.4) and (10.5.5) with r = α q/2 f , we obtain the solution 2f , − ; q − α2qx f αx ∞ , 0 < |q| < 1. w(II) (x; q) = ex −2fq;q ∞
From (10.5.12) we have ep4 x2 + 2 f p2 x 2f ep2 x w(x) = · = 1+ w(px) α p2 x 2 2f αx and by using (10.5.2) and (10.5.3) with r = ep2 /2 f and by using (10.5.4) and (10.5.5) with r = α /2 f , we obtain the solution
296
10 Orthogonal Polynomial Solutions of q-Difference Equations
w(II) (x; p) =
2 − ep2 f x ; p
∞
− α2 fx , − 2αf xp ; p
,
0 < |p| < 1.
∞
The special case 0 < p < 1, e = 0 and α = 2 f leads to the weight function w(x; p) =
1 −x, − xp ; p ∞
for the Stieltjes-Wigert polynomials. Case II-B. g = 0, β = 0 = β , α = 0 respectively α = 0 and f = 0. Then we have from (10.5.1) with xν = qν w(xν ) αq ν 1 = ·q · w(xν +1 ) 1 + eqν 2 f 2fq with possible solution ν ν 2f e ;q q− ( 2 ) . w(II) (xν ; q) = − 2fq α q ν From (10.5.10) we have with xν = p−ν w(xν ) α 1 = · p−ν · w(xν +1 ) 1 + ep1−ν 2 f p 2f with possible solution 2 f p ν (ν ) ep2−ν ;p p 2 . w(II) (xν ; p) = − 2f α ν Case III-A. g = 0, β = 0 respectively β = 0, α = 0 = α and e = 0. From (10.5.1) we have for f = 0 w(x) β q2 x βq 1 = 2 = · w(qx) ex + 2 f qx 1 + 2exf q 2 f and by using (10.5.8) and (10.5.9) with r = e/β q2 , s = 0 and t = e/2 f q, we obtain the solution f q2 − 2 ex ;q ∞ , 0 < |q| < 1. w(III) (x; q) = β q3 ex − β q2 , − ex ; q ∞
10.5 Solutions of the q-Pearson Equation
For f = 0 we have
297
w(x) β q2 = w(qx) ex
and by using (10.5.4) and (10.5.5) with r = e/β q2 , we obtain the solution w(III) (x; q)
f =0
1 , = 3 ex − β q2 , − βexq ; q
0 < |q| < 1,
∞
which implies that the solution above is also valid for f = 0. The special case 0 < q < 1, f = 0 and e = β q2 leads to the weight function w(x; q) =
1 −x, − qx ; q ∞
for the Stieltjes-Wigert polynomials. ln 2βf 2f 2f ∈ R. Then we > 0 we may write = qv with q > 0, q = 1 and v = β β ln q have q1−v w(x) = w(qx) 1 + 2exf q
For
with possible solution w(III) (x; q) =
xv−1
,
− 2exf q ; q
0 < |q| < 1.
∞
The special case e = 2 f q leads to the weight function w(x; q) =
xv−1 (−x; q)∞
for the q-Laguerre polynomials. From (10.5.12) we have for f = 0 w(x) ep4 x2 + 2 f p2 x ep2 x 2f p = = 1+ · w(px) β px 2f β and by using (10.5.8) and (10.5.9) with r = ep3 /β , s = 0 and t = ep2 /2 f , we obtain the solution
298
10 Orthogonal Polynomial Solutions of q-Difference Equations
3 − epβ x , − epβ2 x ; p ∞ w(III) (x; p) = , 2f − epx ;p
0 < |p| < 1.
∞
For f = 0 we have w(x) ep4 x2 ep3 x = = . w(px) β px β By using (10.5.4) and (10.5.5) with r = ep3 /β , we obtain the solution β ep3 x (III) w (x; p) = − , − 2 ; p , 0 < |p| < 1, f =0 β ep x ∞ which implies that the solution above is also valid for f = 0. Case III-B. g = 0, β = 0 respectively β = 0, α = 0 = α and e = 0. For f = 0 we have from (10.5.1) with xν = qν 1 w(xν ) βq = · w(xν +1 ) 1 + eqν 2 f 2fq with possible solution ν 2f e ;q . w(III) (xν ; q) = − 2fq β q ν For f = 0 we have
w(xν ) β q2 − ν = ·q w(xν +1 ) e
with possible solution w(III) (xν ; q)
f =0
=
e β q2
ν
ν
q( 2 ) .
For f = 0 we have from (10.5.10) with xν = p−ν w(xν ) β 1 = · 1− ν w(xν +1 ) 1 + ep 2f p 2f with possible solution (III)
w
2f p ν ep2−ν ;p (xν ; p) = − . 2f β ν
10.5 Solutions of the q-Pearson Equation
For f = 0 we have
299
β w(xν ) = 2 · pν w(xν +1 ) ep
with possible solution w(III) (xν ; p)
f =0
=
ep2 β
ν
ν
p− ( 2 ) .
Case IV-A. g = 0, β = 0 respectively β = 0 and α = 0 respectively α = 0. From (10.5.1) we have for f = 0 1 + αβx β q w(x) α q2 x 2 + β q2 x = = · w(qx) ex2 + 2 f qx 1 + 2exf q 2 f and by using (10.5.8) and (10.5.9) with r = e/β q2 , s = α /β and t = e/2 f q, we obtain the solution f q2 − αβx , − 2 ex ;q ∞ , e = 0, 0 < |q| < 1. w(IV ) (x; q) = β q3 ex − β q2 , − ex ; q ∞
For f = 0 we have e = 0 and w(x) α q2 x 2 + β q2 x αx β q2 = . · = 1 + w(qx) ex2 β ex By using (10.5.2) and (10.5.3) with r = α /β and (10.5.4) and (10.5.5) with r = e/β q2 , we obtain the solution − αβx ; q ∞ w(IV ) (x; q) = , 0 < |q| < 1, 3 f =0 − βexq2 , − βexq ; q ∞
which implies that the solution above is also valid for f = 0. For e = 0 we have f = 0 and w(x) αx βq = 1+ · w(qx) β 2f By using (10.5.8) and (10.5.9) with r = α q/2 f , s = 0 and t = α /β , we obtain the solution
300
10 Orthogonal Polynomial Solutions of q-Difference Equations
2f − α2qx , − ; q f αx ∞, = βq e=0 − αx ; q
w(IV ) (x; q)
0 < |q| < 1.
∞
From (10.5.12) we have for f = 0 2
ep x ep4 x2 + 2 f p2 x 1 + 2 f 2 f p w(x) = 2 2 = · w(px) α p x + β px 1 + α px β β
and by using (10.5.8) and (10.5.9) with r = α /2 f , s = ep2 /2 f and t = α p/β , we obtain the solution 2 − ep2 f x , − αβ x ; p ∞ , 0 < |p| < 1. w(IV ) (x; p) = 2f p αx − 2 f , − αx ; p ∞
For f = 0 we have e = 0 and w(x) ep3 x 1 ep3 x = = · . w(px) α px + β β 1 + αβpx By using (10.5.2) and (10.5.3) with r = α p/β and by using (10.5.4) and (10.5.5) with r = ep3 /β , we obtain the solution 3 − epβ x , − epβ2 x ; p ∞ , 0 < |p| < 1. w(IV ) (x; p) = f =0 − αβpx ; p ∞
Case IV-B. g = 0, β = 0 respectively β = 0 and α = 0 respectively α = 0. For f = 0 we have from (10.5.1) with xν = qν ν
1 + αβq β q w(xν ) = ν · w(xν +1 ) 1 + 2eqf q 2 f with possible solution − 2 ef q ; q 2 f ν ν . w(IV ) (xν ; q) = βq − αβ ; q ν
10.5 Solutions of the q-Pearson Equation
301
The special case e = abq2 , 2 f = −a, α = 1 and β = −q−1 leads to the weight function (bq; q)ν ν w(xν ; q) = a (q; q)ν for the little q-Jacobi polynomials. The special case e = 0, 2 f = −a, α = 1 and β = −q−1 leads to the weight function w(xν ; q) =
1 aν (q; q)ν
for the little q-Laguerre polynomials. For f = 0 we have e = 0 and w(xν ) α qν β q2 − ν = 1+ ·q · w(xν +1 ) β e with possible solution (IV )
w
(xν ; q)
1
= f =0 − αβ ; q
ν
e β q2
ν
ν
q( 2 ) .
The special case e = −aq, α = 1 and β = −q−1 leads to the weight function w(xν ; q) =
ν 1 aν q( 2 ) (q; q)ν
for the q-Bessel polynomials. For f = 0 we have from (10.5.10) with xν = p−ν −ν
1 + α βp w(xν ) β = · 1− ν w(xν +1 ) 2f p 1 + ep2 f with possible solution (IV )
w
2−ν
− ep2 f ; p
(xν ; p) = 1−ν − α pβ ; p
ν
For f = 0 we have e = 0 and
ν
2f p β
ν .
302
10 Orthogonal Polynomial Solutions of q-Difference Equations
w(xν ) α p− ν β = 1+ · · pν w(xν +1 ) β ep2 with possible solution (IV )
w
(xν ; p)
f =0
=
1
1−ν − α pβ ; p ν
ep2 β
ν
ν
p− ( 2 ) .
Case V-A. g = 0, α = 0 = α and β = 0 = β . From (10.5.1) we have 1 gq2 w(x) 1 = , = 2 = γ1 x 2fx ex2 w(qx) ex + 2 f qx + gq2 1 + gq + gq2 1− q 1 − γ2qx where γ1 γ2 = e/g and γ1 + γ2 = −2 f /g, which implies that for f 2 ≥ eg − f − f 2 − eg − f + f 2 − eg and γ2 = . γ1 = g g In the case that f 2 < eg we have γ2 = γ1 . By using (10.5.2) and (10.5.3), once with r = −γ1 /q and once with r = −γ2 /q, we obtain the solution w(V ) (x; q) =
1
,
γ1 x γ2 x q , q ;q ∞
0 < |q| < 1.
The special case e = q2 , f = 0 and g = 1, which implies that γ1 = iq and γ2 = −iq, leads to the weight function w(xν ; q) =
1 1 = 2 2 (ix, −ix; q)∞ (x ; q )∞
for the discrete q-Hermite II polynomials. Note that γ1 and γ2 are non-real in this case. From (10.5.12) we have ep4 x2 + 2 f p2 x + g 2 f p2 x ep4 x2 w(x) = = 1+ + = 1 − γ1 p2 x 1 − γ2 p2 x . w(px) g g g By using (10.5.2) and (10.5.3), once with r = −γ1 p2 and once with r = −γ2 p2 , we obtain the solution w(V ) (x; p) = γ1 p2 x, γ2 p2 x; p ∞ , 0 < |p| < 1.
10.5 Solutions of the q-Pearson Equation
303
Case V-B. g = 0, α = 0 = α and β = 0 = β . Then we have from (10.5.1) with xν = qν w(xν ) 1 = ν −1 w(xν +1 ) (1 − γ1 q ) (1 − γ2 qν −1 ) with possible solution
w(V ) (xν ; q) =
γ1 γ2 , ;q q q
. ν
From (10.5.10) we have with xν = p−ν 1 w(xν ) = w(xν +1 ) (1 − γ1 p1−ν )(1 − γ2 p1−ν ) with possible solution w(V ) (xν ; p) = γ1 p2−ν , γ2 p2−ν ; p ν . Case VI-A. g = 0, α = 0 = α and β = 0 respectively β = 0. From (10.5.1) we have 2 1 + βx gq 2 1 + βgx g w(x) q (β x + g) = , = = ex2 w(qx) ex2 + 2 f qx + gq2 1 − γ1qx 1 − γ2qx gq2 1 + 2gqf x + gq 2 where, as before, γ1 γ2 = e/g and γ1 + γ2 = −2 f /g, which implies that for f 2 ≥ eg − f − f 2 − eg − f + f 2 − eg and γ2 = . γ1 = g g In the case that f 2 < eg we have γ2 = γ1 . By using (10.5.2) and (10.5.3), once with r = β /g, once with r = −γ1 /q and once with r = −γ2 /q, we obtain the solution − βgx ; q ∞ , 0 < |q| < 1. w(V I) (x; q) = γ1 x γ2 x q , q ;q ∞
From (10.5.12) we have 2 f p2 x ep4 x2 1 − γ1 p2 x 1 − γ2 p2 x ep4 x2 + 2 f p2 x + g 1 + g + g w(x) = = = . w(px) β px + g 1 + β px 1 + β px g
g
By using (10.5.2) and (10.5.3), once with r = −γ1 p2 , once with r = −γ2 p2 and once with r = β p/g, we obtain the solution
304
10 Orthogonal Polynomial Solutions of q-Difference Equations
(V I)
w
γ1 p2 x, γ2 p2 x; p (x; p) = − β gpx ; p
∞
0 < |p| < 1.
,
∞
Case VI-B. g = 0, α = 0 = α and β = 0 respectively β = 0. Then we have from (10.5.1) with xν = qν ν
1 + β gq w(xν ) = w(xν +1 ) (1 − γ1 qν −1 ) (1 − γ2 qν −1 ) with possible solution
w(V I) (xν ; q) =
γ1 γ2 q , q ;q ν − βg ; q ν
.
From (10.5.10) we have with xν = p−ν −ν
1 + β pg w(xν ) = w(xν +1 ) (1 − γ1 p1−ν )(1 − γ2 p1−ν ) with possible solution (V I)
w
2−ν γ1 p , γ2 p2−ν ; p ν 1−ν (xν ; p) = . − β pg ; p ν
Case VII-A. g = 0 and α = 0 respectively α = 0. From (10.5.1) we have 2 1 − δx 1 − δx 1 + βgx + αgx w(x) q2 (α x2 + β x + g) 1 2 , = 2 = = ex2 γ1 x γ2 x w(qx) ex + 2 f qx + gq2 1 + 2gqf x + gq 1− q 1− q 2 where, as before, in the case that f 2 ≥ eg − f − f 2 − eg − f + f 2 − eg and γ2 = γ1 = g g and δ1 δ2 = g/α and δ1 + δ2 = −β /α , which implies that for β 2 ≥ 4gα −β − β 2 − 4gα −β + β 2 − 4gα δ1 = and δ2 = . 2α 2α
10.5 Solutions of the q-Pearson Equation
305
In the case that f 2 < eg we have γ2 = γ1 and in the case that β 2 < 4gα we have δ2 = δ1 . By using (10.5.2) and (10.5.3), we obtain the solution x x , ; q δ1 δ2 ∞ , 0 < |q| < 1. w(V II) (x; q) = γ1 x γ2 x , ; q q q ∞
The special case e = abq2 , 2 f = −a(b + c)q, g = ac, α = 1 and β = −(a + c) leads to γ1 = q, γ2 = bq/c, δ1 = c and δ2 = a. Then we obtain the weight function x x , ;q w(x; q) = a bxc ∞ x, c ; q ∞ for the big q-Jacobi polynomials. The special case e = 0, 2 f = −abq, g = ab, α = 1 and β = −(a + b) leads to γ1 = q, γ2 = 0, δ1 = b and δ2 = a. Then we obtain the weight function x x , ;q w(x; q) = a b ∞ (x; q)∞ for the big q-Laguerre polynomials. The special case e = f = 0, g = a/q2 , α = 1 and β = −(a+1)/q leads to γ1 = γ2 = 0, δ1 = a/q and δ2 = 1/q. Then we obtain the weight function qx w(x; q) = qx, ; q a ∞ for the Al-Salam-Carlitz I polynomials. The special case e = f = 0, g = −1/q2 , α = 1 and β = 0 leads to γ1 = γ2 = 0, δ1 = −1/q and δ2 = 1/q. Then we obtain by using (1.8.24) the weight function w(x; q) = (qx, −qx; q)∞ = q2 x2 ; q2 ∞ for the discrete q-Hermite I polynomials. From (10.5.12) we have 2
4 2
2f p x ep x ep4 x2 + 2 f p2 x + g 1 + g + g (1 − γ1 p2 x)(1 − γ2 p2 x) w(x) , = 2 2 = = w(px) α p x + β px + g 1 + β px + α p2 x2 1 − δpx 1 − δpx g g 1
2
where δ1 δ2 = g/α and δ1 + δ2 = −β /α , which implies that for (β )2 ≥ 4gα
306
10 Orthogonal Polynomial Solutions of q-Difference Equations
δ1
=
−β −
(β )2 − 4gα 2α
δ2
and
=
−β +
(β )2 − 4gα . 2α
In the case that (β )2 < 4gα we have δ2 = δ1 . By using (10.5.2) and (10.5.3), we obtain the solution 2 γ1 p x, γ2 p2 x; p ∞ (V II) (x; p) = , 0 < |p| < 1. w px px , ;p δ δ 1
∞
2
The special case e = f = 0, g = p2 , α ∗ = 1 and β ∗ = 0 leads to γ1 = γ2 = 0, δ1 = −ip and δ2 = ip. Then we obtain by using (1.8.24) the weight function w(x; p) =
1 1 = 2 2 (ix, −ix; p)∞ (x ; p )∞
for the discrete q-Hermite II polynomials. Note that δ1 and δ2 are non-real in this case. Case VII-B. g = 0 and α = 0. Then we have from (10.5.1) with xν = qν qν qν 1 − 1 − δ1 δ2 w(xν ) = w(xν +1 ) (1 − γ1 qν −1 ) (1 − γ2 qν −1 ) with possible solution
w(V II) (xν ; q) =
γ1 γ2 q , q ;q ν 1 1 δ1 , δ2 ; q ν
.
From (10.5.10) we obtain with xν = p−ν −ν −ν 1 − pδ 1 − pδ w(xν ) 1 2 = w(xν +1 ) (1 − γ1 p1−ν )(1 − γ2 p1−ν ) with possible solution w(V II) (xν ; p) =
(γ1 p2−ν , γ2 p2−ν ; p)ν 1−ν 1−ν . p , pδ ; p δ 1
2
ν
The special case e = f = 0, g = ap2 , α = 1 and β = −(a + 1)p leads to γ1 = γ2 = 0, δ1 = ap and δ2 = p. Then we obtain by using (1.8.14) the weight function
10.6 Orthogonality Relations
w(xν ; p) =
307
1 p− ν ,
p−ν a
= ;p
ν
ν 1 aν pν (ν +1) aν p2ν +2(2 ) = (p, ap; p)ν (p, ap; p)ν
for the Al-Salam-Carlitz II polynomials.
10.6 Orthogonality Relations In the preceding section we have obtained both continuous and discrete solutions of the q-Pearson equation (10.5.1) and the p−1 -Pearson equation (10.5.10) or (10.5.12). In this section we will derive orthogonality relations for several cases obtained in section 10.4. We will not give explicit orthogonality relations for each different case, but we will restrict to the most important cases (either continuous or discrete). We remark that for each different case the appropriate boundary conditions (3.2.12) or (3.2.19) should be satisfied. Therefore we have to consider w(q−1 x; q)ϕ (q−2 x) with ϕ (x) = ex2 + 2 f x + g and w(x; q) the involving weight function which satisfies the q-Pearson equation. This product should vanish at both ends of the interval of orthogonality. Case II. In this case we have g = 0, β = 0 = β and α = 0. In section 10.4 we have seen that it is only possible to have positive-definite orthogonality for an infinite system of polynomials in the case that q > 1, f = 0 and e ≤ 0. Further we have seen that it is possible to have positive-definite orthogonality for a finite system of polynomials in three different cases. Here we will only treat the infinite case. Case IIa1. q > 1, f = 0 and e ≤ 0. If we write q = p−1 , then we have ⇐⇒
q>1
0 < p < 1.
Now we use the weight function w(II) (x; p) =
2 − ep2 f x ; p
∞
− α2 fx , − 2αf xp ; p
Since α = 0 we set α = 1. Then we have
, ∞
0 < |p| < 1.
308
10 Orthogonal Polynomial Solutions of q-Difference Equations
3
− ep2 f x ; p
∞ w(II) (px; p)ϕ (p2 x) = · (ep4 x2 + 2 f p2 x) − 2pxf , − 2xf ; p ∞ ep2 x − 2f ; p ∞ · 2 f p2 x, = − 2pxf , − 2xf ; p ∞
which vanishes for x → 0. It also vanishes for x → ∞ in the case that e = 0. Then we have orthogonality on the interval (0, ∞) and by using (1.12.3) we have for f > 0 ∞
d0 :=
0
dx = 2f − 2xf , − 2 xf p ; p
∞
∞
0
dt = −2 f ln p (p; p)∞ > 0. −t, − pt ; p ∞
Further we find by using (10.4.1), (10.4.2) and (10.4.3) with e = 0 and q = p−1 dn = −4 f 2 q3n−1 (1 − qn ) = 4 f 2 p−4n+1 (1 − pn ),
n = 1, 2, 3, . . .
which implies that n
∏ dk = (4 f 2 )n p−n(2n+1) (p; p)n ,
n = 1, 2, 3, . . . .
k=1
This leads to the orthogonality relation ∞ (II) (II) ym (x; p)yn (x; p) 0
− 2xf , − 2 xf p ; p
dx = −2 f ln p (p; p)∞ (4 f 2 )n p−n(2n+1) (p; p)n δmn
∞
for f > 0 and m, n = 0, 1, 2, . . .. The special case 2 f = 1 leads to the orthogonality relation ∞ ym (x; p)yn (x; p) 0
dx = − ln p (p; p)∞ p−n(2n+1) (p; p)n δmn , −x, − xp ; p ∞
m, n = 0, 1, 2, . . .
for the Stieltjes-Wigert polynomials. Case III. In this case we have g = 0, β = 0 respectively β = 0 and α = 0 = α . In section 10.4 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in three different cases and that it is impossible to have positive-definite orthogonality for a finite system of polynomials. Here we will only treat one case.
10.6 Orthogonality Relations
309
2f < 1. We use the weight function β f q2 ;q − 2 ex ∞ , 0 < |q| < 1. w(III) (x; q) = β q3 ex − β q2 , − ex ; q
Case IIIa2. 0 < q < 1 and
∞
Then we have
w(III) (q−1 x; q)ϕ (q−2 x) =
=
3
fq − 2 ex ;q
β q4
· eq−4 x2 + 2 f q−2 x
∞
− βexq3 , − ex ; q f q2 ;q − 2 ex
∞
∞ β q4 ex − β q3 , − ex ; q ∞
· eq−4 x2 ,
which vanishes for x → 0 if f = 0. Then it also vanishes for x → ∞. Hence we have orthogonality on the interval (0, ∞) if f = 0 and by using (1.12.3) we obtain for β >0 e ∞
d0 :=
0
1 β q2 dx = 3 e − βexq2 , − βexq ; q
∞ 0
∞
1 dt −t, − qt ; q ∞
β q2 ln q (q; q)∞ > 0. = − e In this case we have for f = 0 dn =
qn+1 (1 − qn ) β 2 −4n+5 n−1 β ( β − 2 f q ) = q (1 − qn ), e2 q5n−4 e2
n = 1, 2, 3, . . . ,
which implies that 2n β ∏ dk = e q−n(2n−3) (q; q)n , k=1 n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation ∞ 0
=− for
β > 0. e
1
(III) y(III) m (x; q)yn (x; q) dx
3 − βexq2 , − βexq ; q ∞ 2 βq β
e
ln q (q; q)∞
e
2n
q−n(2n−3) (q; q)n δmn ,
m, n = 0, 1, 2, . . .
310
10 Orthogonal Polynomial Solutions of q-Difference Equations
The special case e = β q2 leads to the orthogonality relation ∞ ym (x; q)yn (x; q) 0
dx = − ln q (q; q)∞ q−n(2n+1) (q; q)n δmn , −x, − qx ; q ∞
m, n = 0, 1, 2, . . .
for the Stieltjes-Wigert polynomials. For 0 <
2f < 1 we may use the weight function β w(III) (x; q) =
xv−1 − 2exf q ; q
,
v=
ln 2βf ln q
> 0.
∞
Then we have q−v+1 xv−1 −4 2 2 f q−v−1 xv · eq x + 2 f q−2 x = , w(III) (q−1 x; q)ϕ (q−2 x) = ex − 2 ex − ; q ; q 2fq f q2 ∞
∞
which vanishes for x = 0, since v > 0, and also for x → ∞. Hence we have orthogof nality on the interval (0, ∞) and by using (1.12.2) we obtain for > 0 e 1−v ∞ q ;q ∞ 2 f q v ∞ t v−1 2fq v π xv−1 dx = d0 := dt = . e e sin π v (q; q)∞ 0 0 (−t; q)∞ − ex ; q 2fq
In this case we have n
∏ dk =
k=1
∞
2f = qv , which implies that β
2f eqv
2n
q−n(2n−3) (q; q)n (qv ; q)n ,
n = 1, 2, 3, . . . .
This leads to the orthogonality relation ∞
xv−1 (III) y(III) m (x; q)yn (x; q) dx 0 − 2exf q ; q ∞ 1−v v q ; q ∞ 2 f 2n −n(2n−3) π 2fq = q (q; q)n (qv ; q)n δmn e sin π v (q; q)∞ eqv
for v > 0,
f > 0 and m, n = 0, 1, 2, . . .. e
The special case e = 2 f q leads to the orthogonality relation (for v > 0)
10.6 Orthogonality Relations
311
∞
xv−1 ym (x; q)yn (x; q) dx 0 (−x; q)∞ 1−v q ; q ∞ −n(2n+2v−1) π = q (q; q)n (qv ; q)n δmn , sin π v (q; q)∞
m, n = 0, 1, 2, . . .
for the q-Laguerre polynomials. Case IV. g = 0, β = 0 respectively β = 0 and α = 0 respectively α = 0. In section 10.4 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in seven different cases and for a finite system of polynomials in eight different cases. Here we will only treat three infinite cases. 2f < 1. We use the weight function β − 2 ef q ; q 2 f ν ν . w(IV ) (xν ; q) = α βq −β ;q
Case IVa3. 0 < q < 1 and 0 < e <
ν
Since α = 0 we set α = 1. Then we have by using xν = qν ν −1 − 2 ef q ; q 2f ν −1 (IV ) −1 −2 w (q xν ; q)ϕ (q xν ) = · eq2ν −4 + 2 f qν −2 1 βq − ;q
=
β
ν −1
− 2 ef q ; q ν − β1 ; q ν −1
2f β
ν
β , q
which should vanish for ν = 0. Since β = 0 this implies by using (1.8.6) that 1 1 1 = − ;q = 1+ = 0 =⇒ β = −q−1 . 1 βq βq 1 − ;q β
−1
Note that it also vanishes for ν → ∞ since 0 <
2f < 1. Then we have by using β
(1.11.1) for xν = qν , 0 < e < 1 and 0 < −2 f q < 1 − 2 ef q ; q ∞ ∞ (e; q)∞ ν (−2 f q)ν = > 0. d0 := ∑ w(IV ) (xν ; q)xν = ∑ (q; q) (−2 f q; q)∞ ν ν =0 ν =0 Further we have by using (10.4.1), (10.4.2) and (10.4.3) with β = −q−1
312
10 Orthogonal Polynomial Solutions of q-Difference Equations
dn = qn+1 (1 − qn )
1 − eqn−2 (1 − eq2n−3 )(1 − eq2n−2 )2 (1 − eq2n−1 )
× qn−1 (β − 2 f qn−1 )(2 f − eβ qn−1 ) = −2 f q2n−1
(1 − qn )(1 − eqn−2 ) e n−2 n (1 + 2 f q q ) 1 + (1 − eq2n−3 )(1 − eq2n−2 )2 (1 − eq2n−1 ) 2f
for n = 1, 2, 3, . . ., which implies that (q, eq−1 ; q)n ∏ dk = (−2 f ) q (eq−1 , e; q)2n k=1 n
n n2
e ;q , −2 f q, − 2fq n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation − 2 ef q ; q ∞ (IV ) (IV ) ∑ (q; q)ν ν (−2 f q)ν ym (qν ; q)yn (qν ; q) ν =0 −1 e (e; q)∞ n n2 (q, eq ; q)n ; q δmn −2 f q, − (−2 f ) q = (−2 f q; q)∞ (eq−1 , e; q)2n 2fq n for 0 < e < 1 and 0 < −2 f q < 1 and m, n = 0, 1, 2, . . .. 2f < 1. In this case we can use the same β weight function as above leading to the same orthogonality relation for e ≤ 0 and 0 < −2 f q < 1. This implies that this orthogonality relation holds for e < 1 and 0 < −2 f q < 1. Case IVa4. 0 < q < 1, e ≤ 0 and 0 <
Then the special case e = abq2 and 2 f = −a leads to the orthogonality relation (for 0 < aq < 1 and bq < 1) ∞
(bq; q)ν (aq)ν ym (qν ; q)yn (qν ; q) (q; q) ν ν =0
∑ =
(abq2 ; q)∞ n n2 (q, abq; q)n a q (aq, bq; q)n δmn , (aq; q)∞ (abq, abq2 ; q)2n
m, n = 0, 1, 2, . . .
for the little q-Jacobi polynomials. The special case e = 0 and 2 f = −a now leads to the orthogonality relation (for 0 < aq < 1) ∞
(aq)ν an q n ∑ (q; q)ν ym (qν ; q)yn (qν ; q) = (aq; q)∞ (q; q)n (aq; q)n δmn , ν =0 for the little q-Laguerre polynomials.
2
m, n = 0, 1, 2, . . .
10.6 Orthogonality Relations
313
Case IVa5. 0 < q < 1, e < 0 and f = 0. Now we use the weight function e ν (ν ) 1 (IV ) w (xν ; q) = q 2 . β q2 f =0 −α ;q β
ν
Since α = 0 we set α = 1. Then we have (IV )
w
(q xν ; q) −1
1
−2
f =0
ϕ (q xν ) =
− β1 ; q ν −1
e β q2
ν −1
which vanishes for ν = 0 if as before by using (1.8.6) 1 1 1 = − ;q = 1+ = 0 =⇒ 1 β q β q 1 −β ;q
ν −1 2
q(
) · eq2ν −4 ,
β = −q−1 .
−1
Then we have by using (1.14.2) since e < 0
∞
d0 :=
∞
(−e)ν
ν
∑ w(IV ) (xν ; q) f =0 xν = ∑ (q; q)ν q(2) = (e; q)∞ > 0.
ν =0
ν =0
Further we have by using (10.4.1), (10.4.2) and (10.4.3) with f = 0 and β = −q−1 dn = qn+1 (1 − qn )
1 − eqn−2 (1 − eq2n−3 )(1 − eq2n−2 )2 (1 − eq2n−1 )
× qn−1 (β − 2 f qn−1 )(2 f − eβ qn−1 ) = −eq3n−3
(1 − qn )(1 − eqn−2 ) (1 − eq2n−3 )(1 − eq2n−2 )2 (1 − eq2n−1 )
,
n = 0, 1, 2, . . . ,
which implies that (q, eq−1 ; q)n
n
∏ dk = (−e)n q3n(n−1)/2 (eq−1 , e; q)2n ,
n = 1, 2, 3, . . . .
k=1
This leads to the orthogonality relation ∞
(−e)ν
ν
∑ (q; q)ν q(2) ym
ν =0
(IV )
(qν ; q)yn
(IV )
(qν ; q) = (e; q)∞ (−e)n q3n(n−1)/2
(q, eq−1 ; q)n δmn (eq−1 , e; q)2n
for e < 0 and m, n = 0, 1, 2, . . .. The special case e = −aq leads to the orthogonality relation (for a > 0)
314
10 Orthogonal Polynomial Solutions of q-Difference Equations ∞
(aq)ν
ν
∑ (q; q)ν q(2) ym (qν ; q)yn (qν ; q)
ν =0
= (−aq; q)∞ an qn(3n−1)/2
(q, −a; q)n δmn , (−a, −aq; q)2n
m, n = 0, 1, 2, . . .
for the q-Bessel polynomials. Case V. g = 0, α = 0 = α and β = 0 = β . In section 10.4 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in three different cases and that it is impossible to have positive-definite orthogonality for a finite system of polynomials. Here we will only treat one case. Case Va2. 0 < q < 1 and
g > 0. We use the weight function e
w(V ) (x; q) =
1
,
0 < |q| < 1.
γ1 x γ2 x q , q ;q ∞
Then we have w(V ) (q−1 x; q)ϕ (q−2 x) =
eq−4 x2 + 2 f q−2 x + g
=
γ1 x γ2 x , ;q q2 q2 ∞
g
g 1 − γq12x 1 − γq22x = γ1 x γ2 x , ; q q2 q2 ∞
,
γ1 x γ2 x q , q ;q ∞
which clearly vanishes for x → ±∞ since γ1 γ2 = 0. Hence we have for q2 < |γ1 | < q, / R) by using q2 < |γ2 | < q and γ1 γ2 > 0 (if γ1 , γ2 ∈ R), or for γ2 = γ1 (if γ1 , γ2 ∈ (1.15.13) γ1 γ2 q3 ∞ , − ; q q, −q, −1, − 2 γ1 γ2 1 q ∞ dq x = (1 − q) > 0. d0 := γ1 γ1 q2 γ2 q2 q2 γ2 q2 −∞ γ1 x , γ2 x ; q , − , , − , , − , , − ; q q q q q γ1 γ1 q q γ2 γ2 ∞
∞
Further we have by using e = γ1 γ2 g dn =
1 −2n+3 q (1 − qn ), γ1 γ2
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
1 γ1 γ2
n
q−n(n−2) (q; q)n ,
This leads to the orthogonality relation
n = 1, 2, 3, . . . .
10.6 Orthogonality Relations
∞ −∞
1
315
(V ) (V ) ym (x; q)yn (x; q) dq x
γ1 x γ2 x q , q ;q ∞
= (1 − q)
q, −q, −1, − γq1 γ22 , − γq1 γ2 ; q 3
∞ γ1 γ1 q2 γ2 q2 q2 γ2 q2 q , − q , γ1 , − γ1 , q , − q , γ2 , − γ2 ; q ∞
1 γ1 γ2
n
q−n(n−2) (q; q)n δmn
for q2 < |γ1 | < q, q2 < |γ2 | < q, γ1 γ2 > 0 (if γ1 , γ2 ∈ R), or for γ2 = γ1 (if γ1 , γ2 ∈ / R) and m, n = 0, 1, 2, . . .. The special case e = q2 , f = 0 and g = 1, which implies that γ1 = iq and γ2 = −iq, leads to the orthogonality relation ∞
1 ym (x; q)yn (x; q) dq x (ix, −ix; q)∞ −∞ 2 (q, −q, −1, −1, −q; q)∞ q−n (q; q)n δmn , = (1 − q) (i, −i, −iq, iq, −i, i, iq, −iq; q)∞
m, n = 0, 1, 2, . . .
for the discrete q-Hermite II polynomials. Case VI. g = 0, α = 0 = α and β = 0 respectively β = 0. In section 10.4 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in four different cases and for a finite system of polynomials in six different cases. Here we will only treat one infinite and one finite case. Case VIa1/2. 0 < q < 1, f 2 < eg or f 2 ≥ eg with
γ1 g γ2 g > −1 and > −1. We use β β
the weight function − βgx ; q ∞ , w(V I) (x; q) = γ1 x γ2 x q , q ;q
0 < |q| < 1.
∞
Then we have
w(V I) (q−1 x; q)ϕ (q−2 x) =
− βgqx ; q
∞ · eq−4 x2 + 2 f q−2 x + g
γ1 x γ2 x , ;q q2 q2 ∞ − βgqx ; q ∞
γ x γ2 x · g 1 − 12 1− 2 γ1 x γ2 x q q , ;q q2 q2 ∞ − βgqx ; q ∞ · g, = γ1 x γ2 x , ; q q q
=
∞
316
10 Orthogonal Polynomial Solutions of q-Difference Equations
which vanishes for x → ∞ since γ1 γ2 = 0. Since β = 0 it also vanishes for x = −gq/β . g Hence we have for > 0, q2 < |γ1 | < q, q2 < |γ2 | < q and γ1 γ2 > 0 (if γ1 γ2 ∈ R), or β g / R) by using (1.15.12) for > 0 and γ2 = γ1 (if γ1 , γ2 ∈ β 2 ∞ − βgx ; q q, − gq , − βg , − γ1βγq2 g , − γβ1 γq2 g ; q β ∞ dq x = (1 − q) ∞ > 0. d0 := gq γ1 x γ2 x γ1 g γ2 g γ1 q2 γ2 q2 −β − β , − β , q , γ1 , q , γ2 ; q q , q ;q ∞
∞
Further we have by using e = γ1 γ2 g dn =
β γ1 γ2 g
2
γ1 gqn−1 γ2 gqn−1 1+ , q−4n+5 (1 − qn ) 1 + β β
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
β γ1 γ2 g
2n q
−n(2n−3)
γ1 g γ2 g q, − ,− ;q β β
,
n = 1, 2, 3, . . . .
n
This leads to the orthogonality relation ∞ − βgx ; q (V I) (V I) ∞ ym (x; q)yn (x; q) dq x gq γ x γ x 1 2 −β q , q ;q ∞ β γ1 γ2 g β q2 q, − gq , − , − , − ; q g γ1 γ2 g β βq ∞ = (1 − q) γ1 g γ2 g γ1 q2 γ2 q2 − β , − β , q , γ1 , q , γ2 ; q ∞ 2n β γ g γ2 g 1 −n(2n−3) q, − × q ,− ; q δmn , γ1 γ2 g β β n
m, n = 0, 1, 2, . . .
g g > 0, q2 < |γ1 | < q, q2 < |γ2 | < q and γ1 γ2 > 0 (if γ1 , γ2 ∈ R), or for > 0 and β β γ2 = γ1 (if γ1 , γ2 ∈ / R).
for
Case VIb5. 0 < q < 1 and the weight function
γ1 g γ2 g γ2 g N−1 γ2 g N ≤ < −1 with q < −1 ≤ q . We use β β β β γ1 γ2 , q ;q q ν. w(V I) (xν ; q) = β − g ;q ν
Then we have
10.6 Orthogonality Relations
317
−1
−2
(q xν ; q)ϕ (q xν ) =
(V I)
w
=
γ1 γ2 q , q ; q ν −1 − βg ; q ν
· eq2ν −4 + 2 f qν −2 + g
γ1 γ2 q , q ; q ν −1 − βg ; q ν −1
=
· g 1 − γ1 qν −2 1 − γ2 qν −2
γ1 γ2 q , q ;q ν − βg ; q ν −1
· g,
which vanishes for ν = 0 if we have by using (1.8.6) 1 β β = − ;q = 1+ =0 β g q g q 1 − g ;q
β = −g q.
=⇒
−1
γ2 = q−N . Hence we take γ2 = q−N+1 and obtain by q using (1.11.5) with xν = qν for γ1 > 0 γ1 −N N ,q ;q N N q γ1 ν ν q = > 0. d0 := ∑ w(V I) (xν ; q)xν = ∑ (q; q)ν q ν =0 ν =0
It also vanishes for ν = N + 1 if
Further we have N 2 q q−4n+5 (1 − qn ) 1 − γ1 qn−2 1 − qn−N−1 , dn = γ1
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
qN γ1
2n q
−n(2n−3)
γ1 −N q, , q ; q , q n
This leads to the orthogonality relation γ1 −N , q ; q N q (V I) (V I) ∑ (q; q)ν ν qν ym (qν ; q)yn (qν ; q) ν =0 N N 2n q γ1 γ1 q−n(2n−3) q, , q−N ; q δmn , = q γ1 q n for γ1 > 0.
n = 1, 2, 3, . . . .
m, n = 0, 1, 2, . . . , N
318
10 Orthogonal Polynomial Solutions of q-Difference Equations
Case VII. g = 0 and α = 0 respectively α = 0. In section 10.4 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in three different cases and for a finite system of polynomials in eight different cases. Here we will only treat two infinite cases. Case VIIa1. 0 < q < 1, e < 1, δ1 , δ2 ∈ R and 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 < 0,
n = 1, 2, 3, . . . .
We use the weight function
w(V II) (x; q) =
Then we have
x x δ1 , δ2 ; q ∞ , γ1 x γ2 x q , q ;q ∞
w(V II) (q−1 x; q)ϕ (q−2 x) =
0 < |q| < 1.
x x δ1 q , δ2 q ; q ∞ γ1 x γ2 x , ; q 2 2 q q ∞
· eq−4 x2 + 2 f q−2 x + g
x , ; q δ1 q δ2 q γ1 x γ2 x ∞ ·g 1− 2 1− 2 = γ1 x γ2 x q q , ; q 2 2 q q ∞ x x δ1 q , δ2 q ; q ∞ · g, = γ1 x γ2 x q , q ;q
x
∞
which clearly vanishes for both x = δ1 q and x = δ2 q. Hence we have for δ1 < δ2 , δ2 q δ1 q < 1, < 1, γ1 δ1 < 1, γ1 δ2 < 1, γ2 δ1 < 1 and γ2 δ2 < 1 by using (1.15.11) δ1 δ2 δ2 q δ1 q x x δ2 q q, , ; q , , γ γ δ δ ; q 1 2 1 2 δ δ δ1 δ2 ∞ 1 2 ∞ dq x = (δ2 − δ1 )q(1 − q) > 0. d0 := γ1 x γ2 x ( γ δ , γ δ , γ δ , γ δ ; q) δ1 q 1 1 1 2 2 1 2 2 ∞ , ; q q q ∞
Further we have for δ1 δ2 = g < 0 by using e = γ1 γ2 δ1 δ2 , (10.4.1), (10.4.2) and (10.4.3) dn = −δ1 δ2 qn+1 (1 − qn ) ×
1 − γ1 γ2 δ1 δ2 qn−2 2n−3 (1 − γ1 γ2 δ1 δ2 q )(1 − γ1 γ2 δ1 δ2 q2n−2 )2 (1 − γ1 γ2 δ1 δ2 q2n−1 ) × 1 − γ1 δ1 qn−1 1 − γ1 δ2 qn−1 1 − γ2 δ1 qn−1 1 − γ2 δ2 qn−1
for n = 1, 2, 3, . . ., which implies that
10.6 Orthogonality Relations n
∏ dk = (−δ1 δ2 )n qn(n+3)/2
k=1
319
(q, γ1 γ2 δ1 δ2 q−1 ; q)n (γ1 δ1 , γ1 δ2 , γ2 δ1 , γ2 δ2 ; q)n (γ1 γ2 δ1 δ2 q−1 , γ1 γ2 δ1 δ2 ; q)2n
for n = 1, 2, 3, . . .. This leads to the orthogonality relation x x δ2 q δ1 , δ2 ; q ∞ (V II) ym (x; q)yn(V II) (x; q) dq x γ1 x γ2 x δ1 q q , q ;q ∞ q, δδ2 q , δδ1 q , γ1 γ2 δ1 δ2 ; q 1 2 ∞ = (δ2 − δ1 )q(1 − q) (−δ1 δ2 )n qn(n+3)/2 (γ1 δ1 , γ1 δ2 , γ2 δ1 , γ2 δ2 ; q)∞ × for δ1 < 0 < δ2 , m, n = 0, 1, 2, . . ..
(q, γ1 γ2 δ1 δ2 q−1 ; q)n (γ1 δ1 , γ1 δ2 , γ2 δ1 , γ2 δ2 ; q)n δmn (γ1 γ2 δ1 δ2 q−1 , γ1 γ2 δ1 δ2 ; q)2n
δ2 q δ1 q < 1, < 1, γ1 δ1 < 1, γ1 δ2 < 1, γ2 δ1 < 1, γ2 δ2 < 1 and δ1 δ2
The special case γ1 = q, γ2 = bq/c, δ1 = c and δ2 = a leads to the orthogonality relation (for 0 < aq < 1, 0 ≤ bq < 1 and c < 0) aq x x , ;q a bxc ∞ ym (x; q)yn (x; q) dq x cq x, c ; q ∞ q, ac−1 q, a−1 cq, abq2 ; q ∞ (−ac)n qn(n+3)/2 = (a − c)q(1 − q) (aq, bq, cq, abc−1 q; q)∞ (q, abq; q)n aq, bq, cq, abc−1 q; q n δmn , m, n = 0, 1, 2, . . . × (abq, abq2 ; q)2n for the big q-Jacobi polynomials. The special case γ1 = q, γ2 = 0, δ1 = b and δ2 = a leads to the orthogonality relation (for 0 < aq < 1 and b < 0) aq x x a, b;q ∞ ym (x; q)yn (x; q) dq x (x; q)∞ bq q, ab−1 q, a−1 bq; q ∞ = (a − b)q(1 − q) (aq, bq; q)∞ × (−ab)n qn(n+3)/2 (q; q)n (aq, bq; q)n δmn ,
m, n = 0, 1, 2, . . .
for the big q-Laguerre polynomials. The special case γ1 = γ2 = 0, δ1 = a/q and δ2 = 1/q leads to the orthogonality relation (for a < 0)
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10 Orthogonal Polynomial Solutions of q-Difference Equations
1
qx, a
qx ; q ym (x; q)yn (x; q) dq x a ∞ n
= (1 − a)(1 − q)(q, aq, a−1 q; q)∞ (−a)n q(2) (q; q)n δmn ,
m, n = 0, 1, 2, . . .
for the Al-Salam-Carlitz I polynomials. The special case γ1 = γ2 = 0, δ1 = −1/q and δ2 = 1/q leads to the orthogonality relation 1 −1
(qx, −qx; q)∞ ym (x; q)yn (x; q) dq x n
= 2(1 − q)(q, −q, −q; q)∞ q(2) (q; q)n δmn ,
m, n = 0, 1, 2, . . .
for the discrete q-Hermite I polynomials. Case VIIa2. q > 1, e ≤ 0, δ1 , δ2 ∈ R or δ2 = δ1 and 2 f qn−1 + δ1 eq2n−2 + δ2 2 f qn−1 + δ1 + δ2 eq2n−2 > 0,
n = 1, 2, 3, . . . .
We write q = p−1 , which implies that 0 < p < 1, and use the weight function 2 γ1 p x, γ2 p2 x; p ∞ (V II) w (x; p) = , 0 < |p| < 1. px px , ; p δ δ 1
2
∞
Then we have (V II)
w
3 γ1 p x, γ2 p3 x; p ∞ (px; p)ϕ (p x) = 2 2 p x p x , ;p δ δ 31 2 3 ∞ γ1 p x, γ2 p x; p = 2 2 ∞ p x p x , ;p δ δ 21 2 2 ∞ γ1 p x, γ2 p x; p = 2 2 ∞ p x p x , ;p δ δ 2
1
2
· ep4 x2 + 2 f p2 x + g · g 1 − γ1 p2 x 1 − γ2 p2 x · g,
∞
which clearly vanishes for x → ±∞ if γ1 = γ2 = 0, which implies that e = f = 0. Then we have for p < |δ1 | < 1, p < |δ2 | < 1 and δ1 δ2 = g > 0 by using (1.15.13) δ1 δ2 p2 ∞ , − ; p p, −p, −1, − p δ1 δ2 1 ∞ d p x = (1 − p) d0 := > 0. px px p p p p , −δ ; p −∞ , ; p , − , δ , − δ , , − , δ 1 δ 2 δ δ δ δ 1 δ 2 1
2
∞
1
Further we have for γ1 = γ2 = 0 and q = p−1
1
2
2
∞
10.6 Orthogonality Relations
321
dn = −δ1 δ2 qn+1 (1 − qn ) = δ1 δ2 p−2n−1 (1 − pn ),
n = 1, 2, 3, . . . ,
which implies that n
∏ dk = (δ1 δ2 )n p−n(n+2) (p; p)n ,
n = 1, 2, 3, . . . .
k=1
This leads to the orthogonality relation ∞ −∞
1
II) (V II) y(V (x; p)yn (x; p) d p x m
px px , ;p δ1 δ2 ∞
δ1 δ2 p ;p ∞ 1 2 (δ1 δ2 )n p−n(n+2) (p; p)n δmn = (1 − p) p p p p , − δ , δ1 , −δ1 , δ , − δ , δ2 , −δ2 ; p δ1 1 2 2 ∞ 2
p, −p, −1, − δp δ , −
for p < |δ1 | < 1, p < |δ2 | < 1, δ1 δ2 = g > 0 and m, n = 0, 1, 2, . . .. The special case δ1 = −ip and δ2 = ip, which implies that δ1 δ2 = p2 > 0, leads to the orthogonality relation ∞
1 ym (x; p)yn (x; p) d p x −∞ (ix, −ix; p)∞ 2 (p, −p, −1, −1, −p; p)∞ p−n (p; p)n δmn , = (1 − p) (i, −i, −ip, ip, −i, i, ip, −ip; p)∞
m, n = 0, 1, 2, . . .
for the discrete q-Hermite II polynomials. Alternatively, we use the weight function w(V II) (xν ; p) =
(γ1 p2−ν , γ2 p2−ν ; p)ν 1−ν 1−ν . p , pδ ; p δ 1
2
ν
For γ1 = γ2 = 0, which implies that e = f = 0, we have by using (1.8.14) w
(V II)
ν (δ1 δ2 )ν p2( 2 ) (xν ; p) = 1−ν 1−ν = . p δ1 , δ2 ; p ν , pδ ; p δ
1
1
2
ν
Then we have w(V II) (pxν ; p)ϕ (p2 xν ) =
ν −1 (δ1 δ2 )ν −1 p2( 2 ) · g, δ1 , δ2 ; p ν −1
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10 Orthogonal Polynomial Solutions of q-Difference Equations
which vanishes for ν = 0 if δ1 = p or δ2 = p. Hence we have by using (1.11.8) with δ2 = p and xν = p−ν for 0 < δ1 < 1 ∞
d0 :=
∑w
(V II)
ν =0
ν (δ1 )ν p2(2 ) 1 (xν ; p)xν = ∑ = > 0. , p; p ( δ ; δ 1 p)∞ ν =0 1 ν
∞
Further we have for γ1 = γ2 = 0, δ2 = p, q = p−1 dn = −δ1 δ2 qn+1 (1 − qn ) = δ1 p−2n (1 − pn ),
n = 1, 2, 3, . . . ,
which implies that n
∏ dk = (δ1 )n p−n(n+1) (p; p)n ,
n = 1, 2, 3, . . . .
k=1
This leads to the orthogonality relation ν (δ1 )ν p2(2 ) (V II) −ν 1 (V II) ∑ δ , p; p ym (p ; p)yn (p−ν ; p) = (δ ; p)∞ (δ1 )n p−n(n+1)(p; p)n δmn 1 ν =0 1 ν
∞
for 0 < δ1 < 1 and m, n = 0, 1, 2, . . .. The special case δ1 = ap leads to the orthogonality relation (for 0 < ap < 1) ∞
ν
2 1 (ap)ν p2(2 ) ∑ (p, ap; p) ym (p−ν ; p)yn (p−ν ; p) = (ap; p)∞ an p−n (p; p)n δmn ν ν =0
with m, n = 0, 1, 2, . . ., for the Al-Salam-Carlitz II polynomials. We remark that we cannot take δ1 = −p in order to obtain an orthogonality relation for the discrete q-Hermite II polynomials, since then we have δ1 δ2 = −p2 < 0.
Chapter 11
Orthogonal Polynomial Solutions in q−x of q-Difference Equations Classical q-Orthogonal Polynomials II
11.1 Polynomial Solutions in q−x of q-Difference Equations In the case that ω = 0, q > 0 and q = 1 we might replace q by q−1 in (3.2.1) and then replace x by q−x . Then we have f (q−x−1 ) − f (q−x ) f (q−x−1 ) − f (q−x ) , = D1/q f (q−x ) = −x−1 −x q −q q−x−1 (1 − q)
q > 0,
q = 1.
In this case the eigenvalue problem reads (cf. (10.1.1)) 2 (eq−2x + 2 f q−x + g) D1/q yn (q−x ) + (2ε q−x + γ ) D1/q yn (q−x ) =
q(qn − 1) eq(1 − q−n+1 ) − 2ε (1 − q) yn (q−x−1 ) 2 (q − 1)
(11.1.1)
for n = 0, 1, 2, . . .. This can also be written in the symmetric form A∗ (x)yn (q−x−1 ) − {A∗ (x) + B∗ (x)} yn (q−x ) + B∗ (x)yn (q−x+1 ) = (qn − 1) eq(1 − q−n+1 ) − 2ε (1 − q) yn (q−x )
(11.1.2)
for n = 0, 1, 2, . . ., with A∗ (x) = eq2 + 2 f qx+1 + gq2x and
B∗ (x) = eq − 2ε (1 − q) + {2 f q − γ (1 − q)} qx−1 + gq2x−1 .
The regularity condition (2.3.3) implies that ε = 0.
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 11, © Springer-Verlag Berlin Heidelberg 2010
323
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
324
It will turn out to be convenient to introduce (cf. (10.1.3))
α ∗ := eq − 2ε (1 − q) and β ∗ := 2 f q − γ (1 − q).
(11.1.3)
Then we have A∗ (x) = eq2 + 2 f qx+1 + gq2x
B∗ (x) = α ∗ + β ∗ qx−1 + gq2x−1 ,
and
(11.1.4)
where e, f , g, α ∗ , β ∗ ∈ C. In view of the homogeneity, one of the coefficients can be chosen arbitrarily. In this chapter, without loss of generality we may assume that α ∗ ∈ R. In section 11.6 we will see that this implies that all coefficients e, f , g, α ∗ and β ∗ must be real.
11.2 The Basic Hypergeometric Representation We try to find solutions of the form yn (q−x ) =
n
∑ an,k
k=0
(−c−1 q−x ; q)k k c (1 − q)k , (q; q)k
an,n = 0,
n = 0, 1, 2, . . . (11.2.1)
for (11.1.2) with c = 0. Substitution of (11.2.1) into (11.1.2) leads to a three-term recurrence relation (cf. (2.4.4)) for the coefficients {an,k }nk=0 . If c satisfies the relation (cf. (2.4.5)) α ∗ qc2 − β ∗ c + g = 0 (11.2.2) the recurrence relation reduces to the two-term recurrence relation (cf. (2.4.7)) [n − k] α ∗ − eq−n−k+2 can,k = q−k ec2 q−k+1 − 2 f c + gqk−1 an,k+1 (11.2.3) for k = n − 1, n − 2, n − 3, . . . , 0 and n = 1, 2, 3, . . . for an,k . In this case the regularity condition (2.3.3) implies that
α ∗ − eq−n+1 = 0,
n = 0, 1, 2, . . . .
(11.2.4)
This implies that all coefficients an,k are uniquely determined in terms of an,n (= 0) provided that c = 0. In fact we have
n−k ec2 q−n+i+1 − 2 f c + gqn−i−1 an,n . an,k = ∏ [i] (α ∗ qn−i − eq−n+2 ) c i=1 In order to have monic polynomials we choose n
an,n = [n]! q−(2) =
(q; q)n −(n) q 2 . (1 − q)n
11.2 The Basic Hypergeometric Representation
325
Hence, by using (10.2.5) we obtain
n−k n k (q−n ; q)k ec2 q−n+i+1 − 2 f c + gqn−i−1 an,k = ∏ (−1)k q−(2)−(2)+nk . ∗ n−i −n+2 k n−k α q − eq (1 − q) c i=1 If e = 0 we have α ∗ = −2ε (1 − q) = 0 and n−k
∏
α ∗ qn−i − eq−n+2 = (α ∗ )n−k q(n−k)(n+k−1)/2 .
i=1
If e = 0 we have n−k
∏
α ∗ qn−2 k q ;q e n−k ∗ α n−2 n−k e q ; q n α∗ . = −eq−n+2 n−2 ; q e q k
n−k α ∗ qn−i − eq−n+2 = −eq−n+2
i=1
Case I. If e = f = 0 and g = 0 we have α ∗ = −2ε (1 − q) = 0 and n−k
∏
ec2 q−n+i+1 − 2 f c + gqn−i−1 = gn−k q(n−k)(n+k−3)/2 .
i=1
Hence for e = f = 0 and g = 0 the monic polynomials can be written as
k (q−n ; q)k (−c−1 q−x ; q)k −(k) α ∗ c2 qn+1 2 q − ∑ (q; q)k g k=0
n
−n −1 −x ∗ 2 n+1 n q , −c q g α c q ; q, = q−(2) 2 φ0 ∗ α cq − g
yn (q−x ; q) = (I)
g α ∗ cq
n
n
q−(2)
n
for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, one free parameter α ∗ /g (g = 0) or g/α ∗ (α ∗ = 0). Special cases of q-polynomials in this class are the Al-Salam-Carlitz I and II polynomials with one free parameter and the discrete q-Hermite I and II polynomials with no free parameters. Case II. If e = 0 and f = 0 we have α ∗ = −2ε (1 − q) = 0 and by using (1.8.12)
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
326 n−k
∏ i=1
n−k g n−i−1 ec2 q−n+i+1 − 2 f c + gqn−i−1 = (−2 f c)n−k ∏ 1 − q 2fc i=1
g k n−k q ;q = (−2 f c) 2 f cq n−k g 2 f cq ; q n . = (−2 f c)n−k g 2 f cq ; q k
Hence for e = 0 and f = 0 the monic polynomials can be written as
n g 2f n (II) −x yn (q ; q) = − ∗ ; q q−2(2) α 2 f cq n
n −n (q ; q)k (−c−1 q−x ; q)k α ∗ cqn k ×∑ g 2f k=0 2 f cq ; q k (q; q)k
n g α ∗ cqn 2f q−n , −c−1 q−x ; q 2 φ1 = − ∗ n−1 ; q, g α q 2 f cq 2f n 2 f cq for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, two free parameters α ∗ / f and g/ f ( f = 0) or f /α ∗ and g/α ∗ (α ∗ = 0). Special cases of q-polynomials in this class are the q-Meixner polynomials and the quantum qKrawtchouk polynomials with two free parameters each and the q-Charlier polynomials with only one free parameter. Case III. If e = 0 we have n−k
∏
ec2 q−n+i+1 − 2 f c + gqn−i−1
i=1
n−k 2f g = ∏ ec2 q−n+i+1 1 − qn−i−1 + 2 q2n−2i−2 ec ec i=1 n−k −(n−k)(n+k−3)/2 n−k = ec2 q ∏ 1 + c−1 ξ1 qn−i−1 1 + c−1 ξ2 qn−i−1 , i=1
where ξ1 ξ2 = g/e and ξ1 + ξ2 = −2 f /e. This implies that for f 2 ≥ eg − f − f 2 − eg − f + f 2 − eg and ξ2 = . ξ1 = e e Note that for g = 0 we have that ξ1 = γ2−1 and ξ2 = γ1−1 with γ1 γ2 = e/g and γ1 + γ2 = −2 f /g and therefore for f 2 ≥ eg
11.2 The Basic Hypergeometric Representation
γ1 =
−f −
f 2 − eg
g
and γ2 =
327
−f +
f 2 − eg
g
as in the previous chapter. In the case that f 2 < eg we have both γ2 = γ1 and ξ2 = ξ1 . Hence for e = 0 the monic polynomials can be written as (III)
yn
n (−c)n (q−x ; q) = α ∗ n−2 ∑ −c−1 ξ1 qk−1 , −c−1 ξ2 qk−1 ; q n−k ; q n k=0 e q ∗
−n −1 −x α n−2 (q ; q)k (−c q ; q)k k q ;q × q e (q; q)k k −1 −1 −c ξ1 q , −c−1 ξ2 q−1 ; q n α∗ = (−c)n qn−2 ; q n
e ∗ q−n , −c−1 q−x , αe qn−2 × 3 φ2 ; q, q −c−1 ξ1 q−1 , −c−1 ξ2 q−1
for n = 0, 1, 2, . . .. The q-polynomials in this class have, besides q, three free parameters ξ1 , ξ2 and α ∗ /e (e = 0). Special cases of q-polynomials in this class are the q-Hahn polynomials with three free parameters and the q-Krawtchouk and the affine q-Krawtchouk polynomials with two free parameters each. Now we have:
Theorem 11.1. All orthogonal polynomial solutions yn (q−x ) of the qdifference equation (11.1.2) 2 eq + 2 f qx+1 + gq2x yn (q−x−1 ) − eq2 + 2 f qx+1 + gq2x + α ∗ + β ∗ qx−1 + gq2x−1 yn (q−x ) + α ∗ + β ∗ qx−1 + gq2x−1 yn (q−x+1 ) = (qn − 1) α ∗ − eq−n+2 yn (q−x ), n = 0, 1, 2, . . . , where α ∗ := eq − 2ε (1 − q) and β ∗ := 2 f q − γ (1 − q), can be divided into three different cases: Case I. e = f = 0 and g = 0 Case II. e = 0 and f = 0 Case III. e = 0.
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
328
11.3 The Three-Term Recurrence Relation Analogous to section 10.3 we obtain for the monic polynomial solutions of (11.1.1) the three-term recurrence relation yn+1 (q−x ) = (q−x − cn )yn (q−x ) − dn yn−1 (q−x ),
n = 1, 2, 3, . . .
(11.3.1)
with initial values y0 (q−x ) = 1 and y1 (q−x ) = q−x − c0 where c0 = −γ /2ε q. If we replace q by q−1 we find by using (10.1.3) and (11.1.3) that
α = e + 2ε (1 − q)
−→
e + 2ε (1 − q−1 ) = (eq − 2ε (1 − q))q−1 = α ∗ q−1
−→
2 f + γ (1 − q−1 ) = (2 f q − γ (1 − q))q−1 = β ∗ q−1 .
and
β = 2 f + γ (1 − q)
Further we obtain from (10.3.2) and (10.3.3) that c0 = −(2 f q − β ∗ )/(eq − α ∗ )q, 2 f α ∗ q−1 − α ∗ q−1 (1 + q−1 )q−n + eq−2n+1 cn = − qn (α ∗ q−1 − eq−2n+2 ) (α ∗ q−1 − eq−2n ) β ∗ q−2 α ∗ q−1 − e(1 + q−1 )q−n+2 + eq−2n+1 − qn (α ∗ q−1 − eq−2n+2 ) (α ∗ q−1 − eq−2n ) ∗ 2 f q α q − α ∗ (1 + q)q−n + eq−2n+3 = − n+1 ∗ q (α − eq−2n+3 ) (α ∗ − eq−2n+1 ) β ∗ α ∗ − e(1 + q)q−n+2 + eq−2n+2 (11.3.2) − n+1 ∗ q (α − eq−2n+3 ) (α ∗ − eq−2n+1 ) for n = 1, 2, 3, . . . and dn =
=
q−n−1 (1 − q−n ) α ∗ q−1 − eq−n+2 (α ∗ q−1 − eq−2n+3 ) (α ∗ q−1 − eq−2n+2 )2 (α ∗ q−1 − eq−2n+1 ) × q−n+1 (β ∗ q−1 − 2 f q−n+1 )(2 f α ∗ q−1 − eβ ∗ q−1 q−n+1 ) −g(α ∗ q−1 − eq−2n+2 )2 q−n (1 − q−n ) α ∗ − eq−n+3 (α ∗ − eq−2n+4 ) (α ∗ − eq−2n+3 )2 (α ∗ − eq−2n+2 ) × q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1 2 −g α ∗ − eq−2n+3
for n = 1, 2, 3, . . ..
(11.3.3)
11.4 Orthogonality and the Self-Adjoint Operator Equation
329
11.4 Orthogonality and the Self-Adjoint Operator Equation In this section we will use the following form of Hahn’s q-operator (cf. (3.2.1)) D1/q f (q−x ) := A1/q,0 f (q−x ) =
f (q−x−1 ) − f (q−x ) , q−x−1 − q−x
q > 0,
q = 1,
(11.4.1)
where f is a complex-valued function in q−x whose domain contains both q−x and q−x−1 for each x ∈ R. For two such functions f1 and f2 , the product rule (3.2.2) reads D1/q ( f1 f2 ) (q−x ) = D1/q f1 (q−x ) f2 (q−x ) + f1 (q−x−1 ) D1/q f2 (q−x ). (11.4.2) Analogous to (3.2.3) and (3.2.4) we now define S1/q f (q−x ) = f (q−x−1 ) −1 and f(q−x ) = S1/q f (q−x ) = f (q−x+1 ).
(11.4.3)
As in section 3.2 2 S1/q w (q−x )ϕ (q−x ) D1/q yn (q−x ) + S1/q w (q−x )ψ (q−x ) D1/q yn (q−x ) = λn S1/q w (q−x ) S1/q yn (q−x ) where λn = q(qn − 1) eq(1 − q−n+1 ) − 2ε (1 − q) /(q − 1)2 ,
ϕ (q−x ) = eq−2x + 2 f q−x + g and ψ (q−x ) = 2ε q−x + γ leads to the self-adjoint form D1/q wϕD1/q yn (q−x ) = λn S1/q w (q−x ) S1/q yn (q−x )
(11.4.4)
if the Pearson operator equation D1/q (wϕ) (q−x ) = S1/q w (q−x )ψ (q−x )
(11.4.5)
holds. By using the product rule (11.4.2), we find that w(q−x−1 ) − w(q−x ) D1/q (wϕ) (q−x ) = ϕ(q−x ) q−x−1 − q−x ϕ(q−x−1 ) − ϕ(q−x ) + w(q−x−1 ) q−x−1 − q−x −x−1 −x w(q )ϕ (q ) − w(q−x )ϕ(q−x ) = . q−x−1 − q−x
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
330
Note that the right-hand side of (11.4.5) can be written in the form w(q−x−1 )ψ (q−x ). Hence (11.4.5) becomes (11.4.6) w(q−x )ϕ(q−x ) = w(q−x−1 ) ϕ (q−x ) − ψ (q−x )(q−x−1 − q−x ) . Further we have from (2.2.13) with ω = 0 q ex2 /q2 + 2 f x/q + g ex2 + 2 f qx + gq2 qϕ (x/q) A(x) C(x) = = = = , q(q − 1)2 x2 (q − 1)2 x2 (q − 1)2 x2 q(q − 1)2 x2 where A(x) = ex2 + 2 f qx + gq2 . If we replace q by q−1 and then x by q−x , we obtain that q2x+1 q q 2 x+1 2x (eq + 2 f q + gq ) = ϕ (q−x+1 ) = A∗ (x). (1 − q)2 (1 − q)2 (1 − q)2 Moreover, we also find from (2.2.13) with ω = 0 that D(qx) = qC(qx) −
1 ψ (x) {ϕ (x) + (1 − q)xψ (x)} . = (q − 1)x (q − 1)2 x2
If we replace q by q−1 and then x by q−x , the latter becomes q2x+2 ϕ (q−x ) + (1 − q−1 )q−x ψ (q−x ) 2 (1 − q) q2x+2 ϕ (q−x ) − (1 − q)q−x−1 ψ (q−x ) . = (1 − q)2 Note that B∗ (x) = eq + 2 f qx + gq2x−1 − (1 − q)(2ε + γ qx−1 ) = q2x−1 ϕ (q−x+1 ) − (1 − q)qx−1 ψ (q−x+1 ). Hence
qB∗ (x + 1) = q2x+2 ϕ (q−x ) − (1 − q)q−x−1 ψ (q−x ) .
This implies that q q2x+2 ϕ (q−x ) − (1 − q)q−x−1 ψ (q−x ) = B∗ (x + 1). (1 − q)2 (1 − q)2 Hence the Pearson equation (11.4.5) is equivalent to (11.4.7) qw(q−x )A∗ (x) = w(q−x−1 )B∗ (x + 1). −x Now we multiply (11.4.4) by S1/q ym (q ) and subtract from the resulting −1 to the equation the same equation with m and n exchanged. Then we apply S1/q
11.4 Orthogonality and the Self-Adjoint Operator Equation
331
result and use the commutation relation (cf. (2.5.3)) −1 −1 D1/q = q−1 D1/q S1/q S1/q
(11.4.8)
to find that (λn − λm )w(q−x )ym (q−x )yn (q−x ) −1 wϕD1/q yn (q−x )ym (q−x ) = q−1 D1/q S1/q −1 wϕD1/q ym (q−x )yn (q−x ). − q−1 D1/q S1/q As before this leads to two kinds of orthogonality. A. Consider the interval (a, b) on the real line and the q-integration by parts formula (cf. (3.2.10)) b
a
D1/q f1 (q−x ) f2 (q−x ) dq x b
= f1 (q−x ) f2 (q−x )
a
−
b a
S1/q f1 (q−x ) D1/q f2 (q−x ) dq x
(11.4.9)
for arbitrary complex-valued functions f1 and f2 which are q-integrable on the interval (a, b). Then we have b
w(q−x )ym (q−x )yn (q−x ) dq x ( λn − λm ) a −1 −1 D1/q yn (q−x )ym (q−x ) = q−1 S1/q (wϕ) (q−x ) S1/q b −1 D1/q ym (q−x )yn (q−x ) . − S1/q a
In view of the regularity condition (11.2.4), we have λm = λn for m = n. So we have Theorem 11.2. Let {yn }∞ n=0 denote the polynomial solutions of the eigenvalue problem (11.1.1), let the regularity condition (11.2.4) hold for n = 0, 1, 2, . . . and let w denote a complex valued function which is q-integrable on the interval (a, b) on the real line and which satisfies the Pearson operator equation (11.4.5). Then we have the orthogonality relation b a
w(q−x )ym (q−x )yn (q−x ) dq x = 0,
m = n,
m, n ∈ {0, 1, 2, . . .}
with weight function w if the boundary conditions −1 −1 S1/q (wϕ) (q−a ) = 0 and S1/q (wϕ) (q−b ) = 0 hold. Here a continuous extension of wϕ might be necessary.
(11.4.10)
(11.4.11)
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
332
If the necessary convergence conditions hold, the integral in (11.4.10) can also be taken over (a, ∞), (−∞, b) or (−∞, ∞) with appropriate boundary conditions. B. For N ∈ {1, 2, 3, . . .} we consider the set of points xν := Aq−ν ,
A ∈ C,
ν = 0, 1, 2, . . . , N + 1.
(11.4.12)
Then we have the summation by parts formula N
∑
ν =0
D1/q f1 (xν ) f2 (xν ) q−1 (1 − q)xν
N+1 = f1 (xν ) f2 (xν ) ν =0
−
N
∑
ν =0
S1/q f1 (xν ) D1/q f2 (xν ) q−1 (1 − q)xν
(11.4.13)
for arbitrary complex-valued functions f1 and f2 in q−x whose domain contains the set of points {xν }Nν =0 . Hence we have N (λn − λm ) ∑ w(xν )ym (xν )yn (xν ) q−1 (1 − q)xν
ν =0
−1 −1 =q D1/q yn (xν )ym (xν ) S1/q (wϕ) (xν ) S1/q N+1 −1 D1/q ym (xν )yn (xν ) − S1/q . −1
ν =0
In view of the regularity condition (11.2.4), we have λm = λn for m = n. So we have Theorem 11.3. Let {yn }∞ n=0 denote the polynomial solutions of the eigenvalue problem (11.1.1), let the regularity condition (11.2.4) hold for n = 0, 1, 2, . . . and let w denote a function whose domain contains the set of points {xν }N+1 ν =0 and that satisfies the Pearson operator equation (11.4.5). Then we have the orthogonality relation N
∑ w(xν )ym (xν )yn (xν )xν = 0,
ν =0
if the boundary conditions −1 S1/q (wϕ) (x0 ) = 0
m = n,
and
m, n ∈ {0, 1, 2, . . . , N}
−1 S1/q (wϕ) (xN+1 ) = 0
hold. Here a continuous extension of wϕ might be necessary.
(11.4.14)
(11.4.15)
11.5 Rodrigues Formulas
333
11.5 Rodrigues Formulas We start with the self-adjoint operator equation (cf. (11.4.4)) −x −1 (q ) D1/q w S1/q ϕ D1/q yn −x −x = λn S1/q w (q ) S1/q yn (q ).
(11.5.1)
−1 We apply the operator S1/q on both sides of (11.5.1) and use (11.4.8) and the fact that λ0 = 0 to obtain −x −1 −2 −1 D1/q yn (q ) D1/q S1/q w S1/q ϕ S1/q
= q(λn − λ0 )w(q−x )yn (q−x ). This formula can be generalized to
k −k −i−1 −k k (q−x ) S1/q w D1/q ∏ S1/q ϕ S1/q D1/q yn i=1
= q(λn − λk−1 )
k−1 −k+1 −i−1 −k+1 k−1 × S1/q S1/q D1/q (q−x ) w ϕ yn ∏ S1/q i=1
for k = 1, 2, 3, . . .. This can be proved by using induction similar to the situation in section 3.4. n −x If we assume that the polynomials {yn }∞ n=0 are monic, id est D1/q yn (q ) = n q−(2) [n]!, and the regularity condition (11.2.4) holds, then we have the Rodrigues formula
n Kn n −n −k−1 D1/q (q−x ) S1/q w ∏ S1/q ϕ yn (q ) = w(q−x ) k=1 −x
(11.5.2)
for n = 1, 2, 3, . . ., where Kn =
1 q(
n
[k]
, ) ∏ λn − λn−k
n+1 2
k=1
n = 1, 2, 3, . . . .
(11.5.3)
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
334
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions Again we will use Favard’s theorem (theorem 3.1) to conclude that there exist positive definite orthogonal polynomial solutions iff cn ∈ R for n = 0, 1, 2, . . . and dn > 0 for all n = 1, 2, 3, . . .. In section 11.3 we have obtained 2 f q α ∗ q − α ∗ (1 + q)q−n + eq−2n+3 cn = − n+1 ∗ q (α − eq−2n+3 ) (α ∗ − eq−2n+1 ) β ∗ α ∗ − e(1 + q)q−n+2 + eq−2n+2 − n+1 ∗ q (α − eq−2n+3 ) (α ∗ − eq−2n+1 ) for n = 0, 1, 2, . . . and dn =
q−n (1 − q−n ) α ∗ − eq−n+3 (α ∗ − eq−2n+4 ) (α ∗ − eq−2n+3 )2 (α ∗ − eq−2n+2 ) 2 × q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1 − g α ∗ − eq−2n+3
for n = 1, 2, 3, . . .. Now we write (1) (2) (1) (2) dn = q−n 1 − q−n Dn Dn = −q−2n (1 − qn )Dn Dn ,
n = 1, 2, 3, . . . (11.6.1)
with
α ∗ − eq−n+3
(1)
Dn =
(α ∗ − eq−2n+4 ) (α ∗ − eq−2n+3 )2 (α ∗ − eq−2n+2 )
,
n = 1, 2, 3, . . . (11.6.2)
and 2 (2) (11.6.3) Dn = q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1 − g α ∗ − eq−2n+3 for n = 1, 2, 3, . . .. Case I. e = f = 0 and g = 0. Then we have α ∗ = −2ε (1 − q) = 0 and cn = − Further we have
β∗ , α ∗ qn+1
n = 0, 1, 2, . . . .
g , n = 1, 2, 3, . . . . (11.6.4) α∗ Note that cn ∈ R for n = 0, 1, 2, . . . implies that we must have β ∗ /α ∗ ∈ R and that positive-definite orthogonality for an infinite system of polynomials occurs for g(1 − qn )/α ∗ > 0. Since α ∗ ∈ R this implies that β ∗ ∈ R and g ∈ R. Now we have: dn = q−2n (1 − qn )
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions
Case Ia1. 0 < q < 1, e = f = 0, g = 0 and Case Ia2. q > 1, e = f = 0, g = 0 and
335
g > 0. α∗
g < 0. α∗
In this case we have no finite systems of positive-definite orthogonal polynomials. Case II. e = 0 and f = 0. Then we have α ∗ = −2ε (1 − q) = 0 and cn =
2 f (1 + q − qn+1 ) − β ∗ qn−1 , α ∗ q2n
n = 0, 1, 2, . . .
and for n = 1, 2, 3, . . . q−2n (1 − qn ) 2 f q−n+1 (β ∗ − 2 f q−n+2 ) − gα ∗ ∗ 2 (α ) β ∗ n−2 gα ∗ 2n−3 4 f 2 −4n+3 n q = q (1 − q ) 1 − + q (α ∗ )2 2f 4f2 2
2f η1 n−1 η2 g n−2 −4n+3 n q = 1− , q (1 − q ) 1 − q α∗ 2f 2f
dn = −
where
η1 η2 = α ∗
Note that we have c0 = and c1 =
(11.6.5)
and η1 q + η2 g = β ∗ .
2 f − β ∗ q−1 2fq−β∗ = α∗ α ∗q
2 f (1 + q − q2 ) − β ∗ 2 f q − β ∗ 2 f (1 − q2 ) = + . α ∗ q2 α ∗ q2 α ∗ q2
It can be shown that cn =
2 f q − β ∗ 2 f (1 + q)(1 − qn ) + , α ∗ qn+1 α ∗ q2n
n = 0, 1, 2, . . . .
Now c0 ∈ R and c1 ∈ R implies that 2 f /α ∗ ∈ R and also β ∗ /α ∗ ∈ R. Since α ∗ ∈ R this implies that we also have f ∈ R and β ∗ ∈ R. Then we must also have g ∈ R since dn ∈ R. In the case that (β ∗ )2 < 4gα ∗ q we have
β ∗ n−2 gα ∗ 2n−3 β ∗ n−2 2 4gα ∗ q − (β ∗ )2 2n−4 1− q + 2q = 1− q + q >0 2f 4f 4f 16 f 2
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
336
for all n = 1, 2, 3, . . .. This implies that dn > 0 for all n = 1, 2, 3, . . . in the case that 0 < q < 1 and (β ∗ )2 < 4gα ∗ q and that dn < 0 for all n = 1, 2, 3, . . . in the case that q > 1 and (β ∗ )2 < 4gα ∗ q. In the case that (β ∗ )2 ≥ 4gα ∗ q we have β ∗ ± (β ∗ )2 − 4gα ∗ q β ∗ ∓ (β ∗ )2 − 4gα ∗ q and η2 = . η1 = 2q 2g Note that for 0 < q < 1 we have q−4n+3 (1 − qn ) > 0 for n = 1, 2, 3, . . . and for q > 1 we have q−4n+3 (1 − qn ) < 0 for n = 1, 2, 3, . . .. For 0 < q < 1 we must have
η1 n−1 η2 g n−2 q 1− > 0, 1− q 2f 2f
n = 1, 2, 3, . . . ,
which implies that both 1−
η1 n−1 η2 g n−2 q q > 0 and 1 − > 0, 2f 2f
n = 1, 2, 3, . . . ,
η1 η2 g < 1 and < 1. And, as before, 2f 2fq
η1 η2 g n−2 q 1 − qn−1 1− >0 2f 2f
since qn → 0 for n → ∞. For n = 1 this reads this implies that
holds for all n = 1, 2, 3, . . .. For q > 1 we must have
η1 η2 g n−2 q 1 − qn−1 1− < 0, 2f 2f
n = 1, 2, 3, . . . ,
which implies that either 1− or
η1 n−1 η2 g n−2 q q < 0 and 1 − > 0, 2f 2f
n = 1, 2, 3, . . . ,
η1 n−1 η2 g n−2 q q > 0 and 1 − < 0, n = 1, 2, 3, . . . . 2f 2f η1 η2 g η1 η2 g > 1 and < 0, or < 0 and > 1. This implies that we must either have 2f 2fq 2f 2fq 1−
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions
337
Hence we have positive-definite orthogonality for an infinite system of polynomials in the following four cases: Case IIa1. 0 < q < 1, e = 0, f = 0, (β ∗ )2 < 4gα ∗ q. Case IIa2. 0 < q < 1, e = 0, f = 0, (β ∗ )2 ≥ 4gα ∗ q,
η1 η2 g < 1 and < 1. 2f 2fq
Case IIa3. q > 1, e = 0, f = 0, (β ∗ )2 ≥ 4gα ∗ q,
η1 η2 g > 1 and < 0. 2f 2f
Case IIa4. q > 1, e = 0, f = 0, (β ∗ )2 ≥ 4gα ∗ q,
η1 η2 g < 0 and > 1. 2f 2fq
In an analogous way we can show that it is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in the following three cases:
η1 η1 N1 η1 N1 −1 η2 g > 1 with q ≤1< q >1 , 2f 2f 2f 2fq η2 g N2 −1 η2 g N2 −2 q q with ≤1< and N = min(N1 , N2 ). 2f 2f
Case IIb1. 0 < q < 1, e = 0, f = 0,
Case IIb2. q > 1, e = 0, f = 0, 0 <
η1 η1 N−1 η1 N η2 g < 1 with q q and > 1. <1≤ 2f 2f 2f 2fq
Case IIb3. q > 1, e = 0, f = 0, 0 <
η1 > 1. 2f
η2 g η2 g N−2 η2 g N−1 < 1 with q q <1≤ and 2fq 2f 2f
Case III. e = 0. Note that we have c0 =
2fq−β∗ , q(α ∗ − eq)
and c2 =
c1 =
2fq−β∗
q2 (α ∗ − eq−1 )
+
(1 − q2 )(2 f α ∗ − eβ ∗ )
q2 (α ∗ − eq) (α ∗ − eq−1 )
(1 + q)(1 − q2 )(2 f qα ∗ − eβ ∗ ) 2fq−β∗ + . q3 (α ∗ − eq−3 ) q5 (α ∗ − eq−1 ) (α ∗ − eq−3 )
In fact, in general we have (1 + q)(1 − qn ) 2 f qn−1 α ∗ − eβ ∗ 2fq−β∗ cn = n+1 ∗ + q (α − eq−2n+1 ) q3n−1 (α ∗ − eq−2n+1 ) (α ∗ − eq−2n+3 ) for n = 0, 1, 2, . . .. Further we have
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
338
dn = −q−2n (1 − qn )Dn Dn , (1)
(2)
n = 1, 2, 3, . . .
(11.6.6)
with
α ∗ − eq−n+3
(1)
Dn =
(α ∗ − eq−2n+4 ) (α ∗ − eq−2n+3 )2 (α ∗ − eq−2n+2 ) ∗
1 − αe qn−3 q7n−9 =− 3 · 2 ∗ ∗ ∗ e 1 − αe q2n−4 1 − αe q2n−3 1 − αe q2n−2
(11.6.7)
for n = 1, 2, 3, . . ., and 2 (2) Dn = q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1 − g α ∗ − eq−2n+3 (11.6.8) for n = 1, 2, 3, . . .. For g = 0 we have for n = 1, 2, 3, . . . (2) Dn = q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1
∗ 2 f α ∗ n−1 −2n+2 −n+2 ∗ q = −eq β −2fq β − e
(11.6.9)
and for g = 0 we have for n = 1, 2, 3, . . . 2 (2) Dn = q−n+1 β ∗ − 2 f q−n+2 2 f α ∗ − eβ ∗ q−n+1 − g α ∗ − eq−2n+3 = −eq−2n+2 β ∗ − 2 f q−n+2 + ξ1 qn−1 α ∗ − eq−2n+3 × β ∗ − 2 f q−n+2 + ξ2 qn−1 α ∗ − eq−2n+3 = −eq−2n+2 β ∗ + ξ2 eq−n+2 + ξ1 α ∗ qn−1 × β ∗ + ξ1 eq−n+2 + ξ2 α ∗ qn−1 = −eq−2n+2 1 + ξ1 η2 qn−2 η1 q + ξ2 eq−n+2 (11.6.10) × 1 + ξ2 η2 qn−2 η1 q + ξ1 eq−n+2 , where as before η1 η2 = α ∗ , η1 q + η2 g = β ∗ , ξ1 ξ2 = g/e and ξ1 + ξ2 = −2 f /e, which implies that for f 2 ≥ eg − f − f 2 − eg − f + f 2 − eg and ξ2 = . ξ1 = e e In the case that f 2 < eg we have ξ1 , ξ2 ∈ C with ξ2 = ξ1 . Finally, we remark that for g = 0 and α ∗ = 0 we have 2 (2) Dn = −eq−4n+6 β ∗ qn−2 − f + eg − f 2 , n = 1, 2, 3, . . . . Hence, for g = 0 and α ∗ = 0 we obtain that
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions
dn = −
2 qn−3 n ∗ n−2 2 , (1 − q ) β q − f + eg − f e2
339
n = 1, 2, 3, . . . .
This implies that in the case that f 2 < eg we have dn < 0 for all n = 1, 2, 3, . . . for 0 < q < 1 and dn > 0 for all n = 1, 2, 3, . . . for q > 1. In the case that f 2 ≥ eg we have ξ1 , ξ2 ∈ R and
β ∗ n−2 β ∗ n−2 n−3 n q q dn = −ξ1 ξ2 q (1 − q ) 1 + 1+ ξ1 e ξ2 e
g ξ1 β ∗ n−2 ξ2 β ∗ n−2 q q 1+ , n = 1, 2, 3, . . . . = − qn−3 (1 − qn ) 1 + e g g This eventually leads to some finite systems of orthogonal polynomials in the case that 0 < q < 1, g = 0 and α ∗ = 0. Note that for 0 < q < 1 we have q−2n (1 − qn ) > 0 for n = 1, 2, 3, . . . and for q > 1 we have q−2n (1 − qn ) < 0 for n = 1, 2, 3, . . .. Analogous to the situation in the previous chapter (cf. table 10.2 on page 270) the (1) sign of Dn is given in table 11.1. (1)
q
extra conditions
Dn
for
0
e < 0 and α ∗ > eq
+
n = 1, 2, 3, . . .
e > 0 and α ∗ < eq
−
n = 1, 2, 3, . . .
α∗ 0 q>1
< eq < 0
< eq < α ∗
with e ≤ α ∗ q2N with eq2
< eq2
< α ∗ q2N
− n = 1, 2, 3, . . . , N
≤ e + n = 1, 2, 3, . . . , N
0 ≤ α∗
+
n = 1, 2, 3, . . .
α∗ ≤ 0 < e
−
n = 1, 2, 3, . . .
e<
eq < α ∗ < 0 with eq2 < α ∗ q2N ≤ e + n = 1, 2, 3, . . . , N 0 < α ∗ < eq with e ≤ α ∗ q2N < eq2 − n = 1, 2, 3, . . . , N
α ∗ < eq < 0
−
n = 1, 2, 3, . . .
0 < eq < α ∗
+
n = 1, 2, 3, . . .
(1)
Table 11.1 sign of Dn , N ∈ {1, 2, 3, . . .}
Hence, for 0 < q < 1, e < 0 and α ∗ > eq we must have Dn < 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 −n+2 ∗ q < 0, β −2fq β − e (2)
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11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
which is true for β ∗ = 0 provided that α ∗ > 0, and for β ∗ = 0 if
2 f −n+2 2 f α ∗ n−1 1− ∗q 1− < 0, q β eβ ∗ which is true for all n = 1, 2, 3, . . . if and only if 2fq > 1 and β∗
2 f α∗ < 1. eβ ∗
And for 0 < q < 1, e > 0 and α ∗ < eq we must have Dn > 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 −n+2 ∗ q < 0, β −2fq β − e (2)
which is true for β ∗ = 0 provided that α ∗ < 0, and for β ∗ = 0 if
2 f −n+2 2 f α ∗ n−1 1− ∗q 1− < 0, q β eβ ∗ which is true for all n = 1, 2, 3, . . . if and only if 2fq > 1 and β∗
2 f α∗ < 1. eβ ∗
Similarly, for q > 1, e < 0 ≤ α ∗ we must have Dn > 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 q > 0, β − 2 f q−n+2 β ∗ − e (2)
which is false for β ∗ = 0. For β ∗ = 0 we must have
2f 2 f α ∗ n−1 1 − ∗ q−n+2 1− > 0, q β eβ ∗ which is true for all n = 1, 2, 3 . . . if and only if 2fq < 1 and β∗
2 f α∗ ≤ 0. eβ ∗
For q > 1, 0 < eq < α ∗ we must have Dn > 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 −n+2 ∗ q < 0, β −2fq β − e (2)
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions
341
which is false for β ∗ = 0. For β ∗ = 0 we must have
2 f −n+2 2 f α ∗ n−1 1− ∗q 1− < 0, q β eβ ∗ which is true for all n = 1, 2, 3 . . . if and only if 2 f α∗ > 1. eβ ∗
2fq < 1 and β∗
And for q > 1, α ∗ < eq < 0 we must have Dn < 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 −n+2 ∗ q < 0, β −2fq β − e (2)
which is false for β ∗ = 0. For β ∗ = 0 we must have
2 f −n+2 2 f α ∗ n−1 1− ∗q 1− < 0, q β eβ ∗ which is true for all n = 1, 2, 3 . . . if and only if 2fq < 1 and β∗
2 f α∗ > 1. eβ ∗
For q > 1, α ∗ ≤ 0 < e we must have Dn < 0. In the case that g = 0 this implies that we must have
∗ 2 f α ∗ n−1 −n+2 ∗ q > 0, β −2fq β − e (2)
which is false for β ∗ = 0. For β ∗ = 0 we must have
2 f −n+2 2 f α ∗ n−1 1− ∗q 1− > 0, q β eβ ∗ which is true for all n = 1, 2, 3 . . . if and only if 2fq < 1 and β∗
2 f α∗ ≤ 0. eβ ∗
Finally, we remark that in the case that 0 < q < 1 and g = 0 we have Dn ∼ −ge2 q−4n+6 (2)
for
n → ∞,
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11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations (2)
which implies that Dn > 0 for all n = 1, 2, 3, . . . can only be true if g < 0 and that (2) Dn < 0 for all n = 1, 2, 3, . . . can only be true if g > 0. Similarly, in the case that q > 1 and g = 0 we have Dn ∼ −g (α ∗ )2 (2)
for n → ∞,
(2)
which implies that Dn > 0 for all n = 1, 2, 3, . . . can only be true if g < 0 and that (2) Dn < 0 for all n = 1, 2, 3, . . . can only be true if g > 0. Hence we conclude that we have positive-definite orthogonality for an infinite system of polynomials, at least in the following nine cases: Case IIIa1. 0 < q < 1, g = 0, β ∗ = 0 and e < 0 < α ∗ . Case IIIa2. 0 < q < 1, g = 0, β ∗ = 0 and α ∗ < 0 < e. Case IIIa3. 0 < q < 1, g = 0, β ∗ = 0, e < 0, α ∗ > eq,
2fq 2 f α∗ > 1 and < 1. ∗ β eβ ∗
Case IIIa4. 0 < q < 1, g = 0, β ∗ = 0, e > 0, α ∗ < eq,
2fq 2 f α∗ > 1 and < 1. β∗ eβ ∗
Case IIIa5. q > 1, g = 0, β ∗ = 0, e < 0 ≤ α ∗ ,
2fq 2 f α∗ < 1 and ≤ 0. ∗ β eβ ∗
Case IIIa6. q > 1, g = 0, β ∗ = 0, 0 < eq < α ∗ ,
2fq 2 f α∗ < 1 and > 1. β∗ eβ ∗
Case IIIa7. q > 1, g = 0, β ∗ = 0, α ∗ < eq < 0,
2fq 2 f α∗ < 1 and > 1. ∗ β eβ ∗
Case IIIa8. q > 1, g = 0, β ∗ = 0, α ∗ ≤ 0 < e, Case IIIa9. q > 1, g = 0, α ∗ = 0 and f 2 < eg.
2fq 2 f α∗ < 1 and ≤ 0. β∗ eβ ∗
11.6 Classification of the Positive-Definite Orthogonal Polynomial Solutions
343
It is also possible to have positive-definite orthogonality for finite systems of polynomials. We only consider the cases where g = 0 and α ∗ = 0, and, for α ∗ = 0, the cases (1) where Dn has opposite sign for n = N and n = N + 1 according to table 11.1. Skipping the details, we conclude that we have positive-definite orthogonality, at least in the following finite cases:
ξ1 β ∗ ξ1 β ∗ N1 −2 g < 0, α ∗ = 0, < −1 with q < −1 ≤ e gq g ξ1 β ∗ N1 −1 ξ2 β ∗ ξ2 β ∗ N2 −2 ξ2 β ∗ N2 −1 q < −1 with q q , < −1 ≤ and N = min(N1 , N2 ), g gq g g
Case IIIb1. 0 < q < 1,
Case IIIb2. 0 < q < 1,
ξ1 β ∗ ξ1 β ∗ N−2 g > 0, α ∗ = 0, < −1 with q < −1 ≤ e gq g
ξ1 β ∗ N−1 ξ2 β ∗ q > −1. and g gq Case IIIb3. 0 < q < 1,
g ξ2 β ∗ ξ2 β ∗ N−2 > 0, α ∗ = 0, < −1 with q < −1 ≤ e gq g
ξ2 β ∗ N−1 ξ1 β ∗ q > −1. and g gq
Case IIIb4. 0 < q < 1, α ∗ < eq < 0 with e ≤ α ∗ q2N < eq2 and Dn > 0. (2)
Case IIIb5. 0 < q < 1, 0 < eq < α ∗ with eq2 < α ∗ q2N ≤ e and Dn < 0. (2)
Case IIIb6. q > 1, eq < α ∗ < 0 with eq2 < α ∗ q2N ≤ e and Dn > 0. (2)
Case IIIb7. q > 1, 0 < α ∗ < eq with e ≤ α ∗ q2N < eq2 and Dn < 0. (2)
We remark that it is also possible to have positive-definite orthogonality for finite (2) systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in cases where Dn has opposite sign for n = N and n = N + 1. As an example we mention Case IIIb8. q > 1, g = 0, β ∗ = 0, α ∗ ≤ 0 < e, 2 f α∗ N 2 f α ∗ N−1 q ≤ 1 < q . eβ ∗ eβ ∗
2fq 2 f α∗ < 1 and 0 < < 1 with β∗ eβ ∗
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344
11.7 Solutions of the q−1 -Pearson Equation We look for solutions of the q−1 -Pearson equation (11.4.7) qw(q−x )A∗ (x) = w(q−x−1 )B∗ (x + 1) with A∗ (x) = eq2 + 2 f qx+1 + gq2x
and B∗ (x + 1) = α ∗ + β ∗ qx + gq2x+1 .
Hence we have B∗ (x + 1) α ∗ + β ∗ qx + gq2x+1 w(q−x ) = = . −x−1 ∗ w(q ) qA (x) q (eq2 + 2 f qx+1 + gq2x ) For simplicity we set y = q−x , which leads to w(y) α ∗ + β ∗ y−1 + gqy−2 α ∗ y2 + β ∗ y + gq = = . −1 2 −1 −2 w(q y) q (eq + 2 f qy + gy ) q (eq2 y2 + 2 f qy + g)
(11.7.1)
In order to find solutions of the q−1 -Pearson equation (11.7.1), we distinguish between 0 < q < 1 and q > 1. For q > 1 we set q = p−1 , which implies that 0 < p < 1. Then we have p α ∗ + β ∗ p−x + gp−2x−1 w(px ) α ∗ p2x+1 + β ∗ px+1 + g = . = w(px+1 ) ep−2 + 2 f p−x−1 + gp−2x ep2x−2 + 2 f px−1 + g Now we set z = px for simplicity, which leads to w(z) α ∗ pz2 + β ∗ pz + g = −2 2 . w(pz) ep z + 2 f p−1 z + g
(11.7.2)
We consider two types of solutions: A. Continuous solutions for y ∈ R in terms of (convergent) infinite products. For the convergence of these infinite products we refer to the book [471] by L.J. Slater. −ν with B. Discrete solutions for {yν }Nν =0 with N → ∞ or {yν }∞ ν =−∞ , id est yν = Aq A ∈ C and ν = 0, ±1, ±2, . . ., in terms of finite products. Note that yν +1 = q−1 yν . Without loss of generality we can choose A = 1 in each case.
The solutions are obtained in a similar way as in the previous chapter. In this chapter some of the details (analogous to the previous chapter) are left for the reader. Case I-A. e = f = 0 and g = 0. From (11.7.1) we have
11.7 Solutions of the q−1 -Pearson Equation
345
w(y) α ∗ y2 + β ∗ y + gq β ∗ y α ∗ y2 η1 y η2 y = = 1 + + = 1 + 1 + , w(q−1 y) gq gq gq g q where η1 η2 = α ∗ and η1 q + η2 g = β ∗ as before, which implies that β ∗ ± (β ∗ )2 − 4gα ∗ q β ∗ ∓ (β ∗ )2 − 4gα ∗ q and η2 = , η1 = 2q 2g provided that (β ∗ )2 ≥ 4gα ∗ q. Then we easily find the solution w(I) (y; q−1 ) =
1 − η1gqy , −η2 y; q
. ∞
The special case e = f = 0, g = q, α ∗ = 1 and β ∗ = 0, which implies that η1 = −i and η2 = i, leads by using (1.8.24) to the weight function w(y; q) =
1 1 = 2 2 (iy, −iy; q)∞ (y ; q )∞
for the discrete q-Hermite II polynomials. Note that η1 and η2 are non-real in this case. From (11.7.2) we have
α ∗ pz2 + β ∗ pz + g β ∗ pz α ∗ pz2 ε1 z w(z) = = 1+ + = 1+ (1 + ε2 pz) , w(pz) g g g g where ε1 ε2 = α ∗ and ε1 p−1 + ε2 g = β ∗ , which implies that β ∗ ± (β ∗ )2 − 4gα ∗ p−1 β ∗ ∓ (β ∗ )2 − 4gα ∗ p−1 , ε1 = and ε2 = 2p−1 2g provided that (β ∗ )2 ≥ 4gα ∗ p−1 . Then we easily find the solution
ε1 z w(I) (z; p) = − , −ε2 pz; p . g ∞ The special case e = f = 0, g = a/p, α ∗ = 1 and β ∗ = −(a + 1)/p, which implies that ε1 = ε2 = −1, leads to the weight function pz , pz; p w(z; p) = a ∞ for the Al-Salam-Carlitz I polynomials.
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The special case e = f = 0, g = −1/p, α ∗ = 1 and β ∗ = 0, which implies that ε1 = ε2 = −1, leads to the weight function w(z; p) = (−pz, pz; p)∞ for the discrete q-Hermite I polynomials. Note that for q → p−1 we have that η1 → ε1 and η2 → ε2 . Case I-B. e = f = 0 and g = 0. Note that e = 0 implies that η1 η2 = α ∗ = −2ε (1 − q) = 0. Then we have from (11.7.1) with yν = q−ν
w(yν ) α ∗ q−2ν + β ∗ q−ν + gq η1 η2 −2ν gqν qν +1 = = 1+ ·q 1+ · w(yν +1 ) gq η1 η2 gq with possible solution −1
1
w (yν ; q ) = − ηg1 , − ηq2 ; q (I)
ν
gq η1 η2
ν
ν
q2( 2 ) .
The special case e = f = 0, g = aq, α ∗ = 1 and β ∗ = −(a + 1)q, which implies that η1 = η2 = −1, leads to the weight function w(yν ; q−1 ) =
1 aν qν (ν +1) (aq, q; q)ν
for the Al-Salam-Carlitz II polynomials. From (11.7.2) we have with zν = pν
w(zν ) α ∗ p2ν +1 + β ∗ pν +1 + g ε1 p ν = = 1+ 1 + ε2 pν +1 w(zν +1 ) g g with possible solution 1 . w(I) (zν ; p) = ε1 − g , −ε2 p; p ν
Case II-A. e = 0 and f = 0. From (11.7.1) we have for g = 0 η1 y η2 y 1 + 1 + ∗ 2 ∗ g q α y + β y + gq w(y) = = 2 f qy w(q−1 y) q(2 f qy + g) 1+ g
11.7 Solutions of the q−1 -Pearson Equation
347
with possible solution 2 − 2 f gq y ; q
∞ w(II) (y; q−1 ) = . − η1gqy , −η2 y; q ∞
For g = 0 we use the fact that α ∗ = −2ε (1 − q) = 0 to obtain
α ∗ y2 + β ∗ y β∗ α ∗y w(y) = = 1 + · −1 2 ∗ w(q y) 2fq y α y 2 f q2 with possible solution w(II) (y; q−1 )
g=0
∗ − αβ∗ y ; q
∞ = ∗ . 2 − α2 f qy , − 2αf∗qy ; q ∞
From (11.7.2) we have for g = 0 w(z) = w(pz)
α ∗ pz2 + β ∗ pz + g 2 f p−1 z + g
=
1 + εg1 z (1 + ε2 pz) 1 + 2gpf z
with possible solution − εg1 z , −ε2 pz; p ∞. w(II) (z; p) = 2fz − gp ; p ∞
For g = 0 we use the fact that α ∗ = −2ε (1 − q) = 0 to obtain
∗ 2 α ∗ pz2 + β ∗ pz β∗ α p z w(z) = = 1 + · −1 ∗ w(pz) 2f p z α z 2f with possible solution ∗ 2 − α 2pf z , − α2∗ fpz ; p ∞ ∗ = . g=0 − βα ∗pz ; p
w(II) (z; p)
∞
Case II-B. e = 0 and f = 0. Note that e = 0 implies that η1 η2 = α ∗ = −2ε (1 − q) = 0. Then we have from (11.7.1) with yν = q−ν
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
348
1 + η1−1 gqν 1 + η2−1 qν +1 η1 η2 −ν w(yν ) α ∗ q−2ν + β ∗ q−ν + gq = = · ·q ν −1 w(yν +1 ) 2 f q−ν +2 + gq 2 f q2 1 + gq 2f
with possible solution − 2gf q ; q
ν w(II) (yν ; q−1 ) = −1 −η1 g, −η2−1 q; q ν
2 f q2 η1 η2
ν
ν
q( 2 ) .
The special case e = 0, 2 f = cq−1 , g = −bcq, α ∗ = 1 and β ∗ = −q(1 − bc), which implies that η1 = η2 = −1, leads to the weight function w(yν ; q−1 ) =
ν +1 (bq; q)ν cν q( 2 ) (−bcq, q; q)ν
for the q-Meixner polynomials. The special case e = 0, 2 f = −b−1 q−1 , g = b−1 q−N , α ∗ = 1 and β ∗ = −q − b−1 q−N , which also implies that η1 = η2 = −1, leads to the weight function −1 ν (ν +1) (q−N ; q)ν −b q 2 . −1 −N (b q , q; q)ν
w(yν ; q−1 ) =
Then we use (1.8.17) to obtain the weight function w(yν ; q−1 ) =
(q; q)N (bq; q)N−ν (−1)ν (ν +1) q 2 , (bq; q)N (q; q)N−ν (q; q)ν
for the quantum q-Krawtchouk polynomials. The special case e = 0, 2 f = aq−1 , g = 0, α ∗ = 1 and β ∗ = −q, which again implies that η1 = η2 = −1, leads to the weight function w(yν ; q−1 ) =
ν +1 aν q( 2 ) (q; q)ν
for the q-Charlier polynomials. From (11.7.2) we have with zν = pν for g = 0 w(zν ) = w(zν +1 ) with possible solution
α ∗ p2ν +1 + β ∗ pν +1 + g 2 f pν −1 + g
=
ν 1 + ε1gp 1 + ε2 pν +1 ν −1
1 + 2 f pg
11.7 Solutions of the q−1 -Pearson Equation
349
w(II) (zν ; p) =
2f − gp ;p
ν
− εg1 , −ε2 p; p
. ν
For g = 0 we use the fact that α ∗ = −2ε (1 − q) = 0 to obtain
∗ 2 α ∗ p2ν +1 + β ∗ pν +1 β ∗ p− ν α p w(zν ) = · pν · = 1 + w(zν +1 ) 2 f pν −1 α∗ 2f with possible solution w(II) (zν ; p)
g=0
=
1 −
β ∗ p−ν +1 α∗
;p
ν
2f α ∗ p2
ν
ν
p− ( 2 ) .
Case III-A. e = 0. From (11.7.1) we have for g = 0 β ∗y α ∗ y2 η1 y η2 y ∗ 2 ∗ 1 + gq 1 + 1 + + gq gq g q w(y) α y + β y + gq = , = = qy qy w(q−1 y) q (eq2 y2 + 2 f qy + g) gq 1 + 2 f qy + eq2 y2 1 − 1 − g g ξ ξ 1
2
where as before η1 η2 = α ∗ , η1 q + η2 g = β ∗ , ξ1 ξ2 = g/e and ξ1 + ξ2 = −2 f /e, which implies that for (β ∗ )2 ≥ 4gα ∗ q β ∗ ± (β ∗ )2 − 4gα ∗ q β ∗ ∓ (β ∗ )2 − 4gα ∗ q and η2 = , η1 = 2q 2g and for f 2 ≥ eg
ξ1 =
−f −
f 2 − eg
e
and
ξ2 =
−f +
f 2 − eg
e
As before we easily find the solution
w(III) (y; q−1 ) =
q2 y q2 y ξ1 , ξ2 ; q ∞ . − η1gqy , −η2 y; q ∞
In the case that g = 0 we distinguish between β ∗ = 0 and β ∗ = 0. For g = 0 and β ∗ = 0 we obtain
.
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
350
∗
1 + αβ ∗y β ∗ w(y) α ∗ y2 + β ∗ y = = · 3 2f w(q−1 y) q (eq2 y2 + 2 f qy) eq y 1 + eqy with possible solution ∗ 4 − eqβ3 y , − eqβ ∗y ; q ∞ . = ∗ 2f g=0,β ∗ =0 − eqy , − αβ ∗qy ; q
w(III) (y; q−1 )
∞
For g = 0 and β ∗ = 0 we must have α ∗ = 0. Assuming that f = 0 we obtain
α ∗ y2 α ∗y 1 w(y) = = eqy · −1 2 2 w(q y) q (eq y + 2 f qy) 1 + 2 f 2 f q2 with possible solution w(III) (y; q−1 )
g=0,β ∗ =0
=
2 − eq2 f y ; q
∞ . 2 f q2 α∗y − 2 f q , − α∗y ; q ∞
For f = g = 0, β ∗ = 0 and α ∗ = 0 we obtain
α∗ w(y) = w(q−1 y) eq3 with possible solution w(III) (y; q−1 )
−eq3 y, − eq12 y ; q ∞. = q f =g=0,β ∗ =0 ∗ −α y, − α ∗ y ; q
∞
From (11.7.2) we have for g = 0 ε1 z 1 + (1 + ε2 pz) g w(z) = −2 2 = −1 w(pz) ep z + 2 f p z + g 1 − ξzp 1 − ξzp
α ∗ pz2 + β ∗ pz + g
1
with possible solution − εg1 z , −ε2 pz; p ∞. w(III) (z; p) = z z , ; p ξ p ξ p 1
2
∞
2
11.7 Solutions of the q−1 -Pearson Equation
351
In the case that g = 0 we distinguish between β ∗ = 0 and β ∗ = 0. For g = 0 and β ∗ = 0 we obtain ∗
1 + αβ ∗z β ∗ p3 α ∗ pz2 + β ∗ pz w(z) = −2 2 = · w(pz) ep z + 2 f p−1 z 1 + 2 f p ez ez with possible solution w(III) (z; p)
g=0,β ∗ =0
∗
2
fp − αβ ∗z , − 2 ez ;p
∞ . = ∗ 4 − β ∗ezp3 , − β ezp ; p ∞
For g = 0 and β ∗ = 0 we must have α ∗ = 0. Assuming that f = 0 we obtain 1 w(z) α ∗ pz2 α ∗ p2 z = −2 2 = ez · −1 w(pz) ep z + 2 f p z 1 + 2 f p 2f with possible solution ∗ 2 − α 2pf z , − α2∗ fpz ; p ∞ = . ez g=0,β ∗ =0 −2f p; p
w(III) (z; p)
∞
For f = g = 0, β ∗ = 0 and α ∗ = 0 we obtain w(z) α ∗ p3 = w(pz) e with possible solution (III)
w
(z; p)
∗ −α z, − αp∗ z ; p ∞ . = 4 f =g=0,β ∗ =0 − pez3 , − pez ; p ∞
Case III-B. e = 0. From (11.7.1) we have with yν = q−ν for α ∗ = 0 and g = 0 gqν qν +1 1 + 1 + ∗ −2 ν ∗ − ν η1 η2 w(yν ) α q + β q + gq η 1 η 2 ξ1 ξ2 = = · w(yν +1 ) q(eq−2ν +2 + 2 f q−ν +1 + g) (1 − ξ1 qν −1 )(1 − ξ2 qν −1 ) gq3 with possible solution
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352
(III)
w
−1
(yν ; q ) =
ξ1 ξ2 q , q ;q ν − ηg1 , − ηq2 ; q ν
gq3 η 1 η 2 ξ1 ξ2
ν .
The special case e = q−2 , 2 f = −α − q−N−1 , g = α q−N+1 , α ∗ = αβ q and β ∗ = −αβ q2 − α q−N+1 , which implies that ξ1 = α q2 , ξ2 = q−N+1 , η1 = −αβ q and η2 = −1, leads to the weight function
1 ν (α q, q−N ; q)ν −1 w(yν ; q ) = −1 −N (β q , q; q)ν αβ for the q-Hahn polynomials. For α ∗ = 0 and g = 0 we obtain ∗ ν
1 + βαq∗ α ∗ q−2ν + β ∗ q−ν α∗ w(yν ) = = · ν −1 w(yν +1 ) q(eq−2ν +2 + 2 f q−ν +1 ) 1 + 2 f q eq3 e with possible solution − 2eqf ; q eq3 ν = ∗ ν . g=0 α∗ − αβ ∗ ; q
w(III) (yν ; q−1 )
ν
For α ∗ = 0 we distinguish between β ∗ = 0 and β ∗ = 0. For α ∗ = 0 and β ∗ = 0 we obtain ν +1
1 + gqβ ∗ w(yν ) β ∗ q−ν + gq β ∗ ξ1 ξ2 ν = = · ·q w(yν +1 ) q (eq−2ν +2 + 2 f q−ν +1 + g) (1 − ξ1 qν −1 )(1 − ξ2 qν −1 ) gq3 with possible solution w(III) (yν ; q−1 )
α ∗ =0,β ∗ =0
=
ξ1 ξ2 q , q ;q ν − βgq∗ ; q ν
gq3 ∗ β ξ1 ξ2
ν
ν
q− ( 2 ) .
The special case e = q−2 , 2 f = −b − q−N−1 , g = bq−N+1 , α ∗ = 0 and β ∗ = −bq−N+1 , which implies that ξ1 = bq2 and ξ2 = q−N+1 , leads by using (1.8.18) to the weight function
11.7 Solutions of the q−1 -Pearson Equation
w(yν ; q−1 ) =
353
ν (bq, q−N ; q)ν (bq; q)ν (q; q)N −ν (−b−1 qN )ν q−(2 ) = b (q; q)ν (q; q)ν (q; q)N−ν
for the affine q-Krawtchouk polynomials. For α ∗ = 0 and β ∗ = 0 we must have g = 0 and we obtain gq 1 w(yν ) = = · ξ1 ξ2 q−2 · q2ν −2 ν +2 − ν +1 ν −1 w(yν +1 ) q (eq +2fq + g) (1 − ξ1 q )(1 − ξ2 qν −1 ) with possible solution w
(III)
(yν ; q ) −1
α ∗ =0,β ∗ =0
=
ξ1 ξ2 , ;q q q
ν
−ν −2(ν ) ξ1 ξ2 q−2 q 2 .
From (11.7.2) we have with zν = pν for g = 0 ν 1 + ε1gp 1 + ε2 pν +1 α ∗ p2ν +1 + β ∗ pν +1 + g w(zν ) = = ν −1 ν −1 w(zν +1 ) ep2ν −2 + 2 f pν −1 + g 1 − pξ 1 − pξ 1
2
with possible solution ξ
1
, 1 ;p p ξ p
1 2 ν . w(III) (zν ; p) = ε1 − g , −ε2 p; p
ν
For g = 0 we distinguish between β ∗ = 0 and β ∗ = 0. For g = 0 and β ∗ = 0 we obtain ∗
1 + αβ ∗ pν α ∗ p2ν +1 + β ∗ pν +1 β ∗ p3 − ν w(zν ) = ·p = · w(zν +1 ) ep2ν −2 + 2 f pν −1 e 1 + 2ef p−ν +1 with possible solution w(III) (zν ; p)
− 2ef p−ν +2 ; p = ∗ g=0,β ∗ =0 − αβ ∗ ; p ν
Note that for f = 0 we have by using (1.8.14)
ν
e ∗ β p3
ν
ν
p( 2 ) .
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354
2 f p ν −(ν ) 2f e ;p − p−ν +2 ; p = − p 2 . e 2f p e ν ν Then the special case 2 f = −epN−1 , g = 0, α ∗ = −p−1 and β ∗ = 1 leads to the weight function (p−N ; p)ν w(zν ; p) = (−b−1 q)ν (p; p)ν for the q-Krawtchouk polynomials. For g = 0 and β ∗ = 0 we must have α ∗ = 0 and we obtain
α ∗ p3 1 w(zν ) = · 2 f w(zν +1 ) 1 + p−ν +1 e e with possible solution (III)
w
ν
e 2 f −ν +2 = − p ;p . ∗ 3 e g=0,β ∗ =0 ν α p
(zν ; p)
11.8 Orthogonality Relations In the preceding section we have obtained both continuous and discrete solutions of the q−1 -Pearson equation (11.7.1) and the p-Pearson equation (11.7.2). In this section we will derive orthogonality relations for several cases obtained in section 11.6. We will not give explicit orthogonality relations for each different case, but we will restrict to the most important cases (either continuous or discrete). In each different case the boundary conditions (11.4.11) or (11.4.15) should be satisfied. Therefore we have to consider w(qy; q−1 )ϕ (q2 y) with ϕ (y) = ey2 + 2 f y + g and w(y; q−1 ) the involving weight function which satisfies the q−1 -Pearson equation (11.7.1) or equivalently w(p−1 z; p)ϕ (p−2 z)
with ϕ (z) = ez2 + 2 f z + g
and w(z; p) the involving weight function which satisfies the p-Pearson equation (11.7.2). These products should vanish at both ends of the interval of orthogonality. Case I. In this case we have e = f = 0 and g = 0. In section 11.6 we have seen that we have positive-definite orthogonality for an infinite system of polynomials in two different cases and that it is impossible to have positive-definite orthogonality for a finite system of polynomials. We will treat both cases here.
11.8 Orthogonality Relations
355
Case Ia1. 0 < q < 1, e = f = 0 and
g > 0. We use the weight function α∗
w(I) (y; q−1 ) =
1
,
− η1gqy , −η2 y; q ∞
where η1 η2 = α ∗ and η1 q + η2 g = β ∗ . Then we have w(I) (qy; q−1 )ϕ (q2 y) =
g
,
2 − η1gq y , −η2 qy; q ∞
g which vanishes for y → ±∞. Now we use (1.15.13) to obtain for q < < 1, η1 g g = >0 q < |η2 | < 1 and η1 η2 α ∗ ∞
1 dq y −∞ − η1 qy , −η y; q 2 g ∞ q, −q, −1, − η1 ηg 2 q , − η1gη2 ; q ∞ > 0. = (1 − q) η1 q η1 q q q g g − g , g , − η1 , η1 , −η2 , η2 , − η2 , η2 ; q
d0 :=
∞
Further we have by using (11.6.4) dn = q−2n (1 − qn )
g , η1 η2
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
g η1 η2
n
q−n(n+1) (q; q)n ,
n = 1, 2, 3, . . . .
This leads to the orthogonality relation ∞
1 (I) y(I) m (y; q)yn (y; q) dq y η1 qy − g , −η2 y; q ∞ q, −q, −1, − η1 ηg 2 q , − η1gη2 ; q ∞ = (1 − q) η1 q η1 q q q g g − g , g , − η1 , η1 , −η2 , η2 , − η2 , η2 ; q ∞
n g × q−n(n+1) (q; q)n δmn , m, n = 0, 1, 2, . . . , η1 η2 −∞
g g > 0. for q < < 1, q < |η2 | < 1 and η1 η1 η2
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356
The special case e = f = 0, g = q, α ∗ = 1 and β ∗ = 0, which implies that η1 = −i and η2 = i, leads to the orthogonality relation ∞
1 ym (y; q)yn (y; q) dq y (iy, −iy; q)∞ 2 (q, −q, −1, −1, −q; q)∞ q−n (q; q)n δmn , = (1 − q) (i, −i, −iq, iq, −i, i, iq, −iq; q)∞ −∞
m, n = 0, 1, 2, . . .
for the discrete q-Hermite II polynomials. Alternatively, we use the weight function
1
−1
w (yν ; q ) = − ηg1 , − ηq2 ; q (I)
ν
gq η1 η2
ν
ν
q2( 2 ) ,
where η1 η2 = α ∗ and η1 q + η2 g = β ∗ . Then we have with yν = q−ν w(I) (qyν ; q−1 )ϕ (q2 yν ) =
g
− ηg1 , − ηq2 ; q
ν −1
g η1 η2
ν −1
q(ν −1) , 2
which vanishes for ν = 0 if we have by using (1.8.6)
q −1 1 1 = − q ;q = 1+ = 0. q η2 η2 1 − η2 ; q −1
This implies that η2 = −1 and therefore η1 = −α ∗ . Then we use (1.11.8) to obtain g with yν = q−ν for 0 < ∗ < 1 α d0 :=
∞
∞
ν =0
ν =0
1 qν (ν −1) g ν = g > 0. ∗ α α α∗ ; q ∞ ν
∑ w(I) (yν ; q−1 )yν = ∑ g∗ , q; q
Further we have by using (11.6.4) dn = q−2n (1 − qn )
g , α∗
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
g n q−n(n+1) (q; q)n , α∗
This leads to the orthogonality relation
n = 1, 2, 3, . . . .
11.8 Orthogonality Relations
357
qν (ν −1) g ν (I) −ν (I) ∑ g∗ , q; q α ∗ ym (q ; q)yn (q−ν ; q) ν =0 α ν g n 1 q−n(n+1) (q; q)n δmn , m, n = 0, 1, 2, . . . =g ∗ α ; q α∗ ∞ ∞
for 0 <
g < 1. α∗
The special case 0 < q < 1, e = f = 0, g = aq, α ∗ = 1 and β ∗ = −(a + 1)q, which implies that η1 = η2 = −1, leads to the orthogonality relation (for 0 < aq < 1) ∞
2 1 aν qν ∑ (aq, q; q)ν ym (q−ν ; q)yn (q−ν ; q) = (aq; q)∞ an q−n (q; q)n δmn ν =0 2
with m, n = 0, 1, 2, . . . for the Al-Salam-Carlitz II polynomials. We remark that the special case 0 < q < 1, e = f = 0, g = −q, α ∗ = 1 and β ∗ = 0, which implies that η1 = η2 = −1, cannot be used here in order to obtain an orthogonality relation for the discrete q-Hermite II polynomials, since then we have g = −q < 0. α∗ g Case Ia2. q > 1, e = f = 0 and ∗ < 0. Since q > 1 we set q = p−1 with 0 < p < 1 α and we use the weight function
ε1 z (I) w (z; p) = − , −ε2 pz; p , g ∞ where ε1 ε2 = α ∗ and ε1 p−1 + ε2 g = β ∗ . Then we have
ε1 z w(I) (p−1 z; p)ϕ (p−2 z) = − , −ε2 z; p · g, gp ∞ which vanishes for z = −gp/ε1 and for z = −1/ε2 . By using (1.15.11) we obtain for gp ε1 gp2 ε2 1 < , < 1 and <1 ε2 ε 1 g ε2 ε1 d0 :=
−1 ε2 ε1 z − gp ε
=
−
1
gp 1 − ε1 ε2
g
, −ε2 pz; p dpz
∞
ε1 gp2 ε2 (1 − p) p, , ;p g ε2 ε 1
Further we have by using (11.6.4) with q = p−1
> 0. ∞
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
358
dn = q−2n (1 − qn )
g g = −pn (1 − pn ) , ε1 ε2 ε1 ε2
n = 1, 2, 3, . . . ,
which implies that
g n (n+1) d = − p 2 (p; p)n , ∏ k ε1 ε2 k=1 n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation −1 ε2 ε1 z
−
− gp ε
1
=
for
g
gp 1 − ε1 ε2
(I) (I) , −ε2 pz; p ym (z; p−1 )yn (z; p−1 ) d p z
∞
(1 − p) p,
g 1 gp ε1 < 0, < , < 1, ε1 ε2 ε2 ε 1 g ε2
ε1 gp2 ε2 , ;p g ε2 ε 1 gp2 ε
2
ε1
∞
−
g ε1 ε2
n
n+1 2
p(
) (p; p) δ n mn
< 1 and m, n = 0, 1, 2, . . ..
The special case 0 < p < 1, e = f = 0, g = a/p, α ∗ = 1 and β ∗ = −(a + 1)/p, which implies that ε1 = ε2 = −1, leads to the orthogonality relation (for a < 0) 1 pz
ym (z; p−1 )yn (z; p−1 ) d p z p n = (1 − a)(1 − p) p, , ap; p (−a)n p(2) (p; p)n δmn , a ∞ a
a
, pz; p
∞
m, n = 0, 1, 2, . . .
for the Al-Salam-Carlitz I polynomials. The special case 0 < q < 1, e = f = 0, g = −1/p, α ∗ = 1 and β ∗ = 0, which implies that ε1 = ε2 = −1, leads to the orthogonality relation 1 −1
(−pz, pz; p)∞ ym (z; p−1 )yn (z; p−1 ) d p z n
= 2(1 − p) (p, −p, −p; p)∞ p(2) (p; p)n δmn ,
m, n = 0, 1, 2, . . .
for the discrete q-Hermite I polynomials. Case II. In this case we have e = 0 and f = 0. In section 11.6 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in four different cases and for a finite system of polynomials in three different cases. We will only treat two infinite cases here and also one finite case. Case IIa2. 0 < q < 1, e = 0, f = 0, (β ∗ )2 ≥ 4gα ∗ q, the weight function
η1 η2 g < 1 and < 1. We use 2f 2fq
11.8 Orthogonality Relations
(II)
w
359
−1
− 2gf q ; q
ν
(yν ; q ) = − ηg1 , − ηq2 ; q
ν
2 f q2 η1 η2
ν
ν
q( 2 ) .
Then we have with yν = q−ν (II)
w
ν (qyν ; q )ϕ (q yν ) = − ηg1 , − ηq2 ; q −1
2
− 2gf q ; q
ν −1
2f η1 η2
ν −1
ν −1 2
· 2 f q ν +(
),
which vanishes for ν = 0 if we have by using (1.8.6)
1 q 1 = − q−1 ; q = 1 + = 0. q η η 2 2 1 − η2 ; q −1
This implies that η2 = −1 and η1 = −α ∗ . Then we have by using (1.11.6) with 2f g yν = q−ν for ∗ > 0 and ∗ < 1 α α − 2gf q ; q 2 f q ν ν − 2αf∗q ; q ∞ ∞ ν q(2 ) = g ∞ > 0. d0 := ∑ w(II) (yν ; q−1 )yν = ∑ g α∗ ν =0 ν =0 α ∗ , q; q ν α∗ ; q ∞ Further we have by using (11.6.5) dn =
2f α∗
2
g α ∗ n−1 q 1 + qn−2 , q−4n+3 (1 − qn ) 1 + 2f 2f
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
2f α∗
2n q
−n(2n−1)
g α∗ ;q , q, − , − 2f 2fq n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation − 2gf q ; q 2 f q ν ν ∞ (II) (II) ∑ g∗ , q; q ν α ∗ q(2) ym (q−ν ; q)yn (q−ν ; q) ν =0 α ν 2fq
− α ∗ ; q 2 f 2n g α∗ −n(2n−1) , − ; q q, − q δmn , = g ∞ α∗ 2f 2fq n α∗ ; q ∞ for −
2f g g < 1, ∗ > 0 and ∗ < 1. 2fq α α
m, n = 0, 1, 2, . . . ,
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
360
The special case 2 f = cq−1 , g = −bcq, α ∗ = 1 and β ∗ = −q + bcq, which implies that η1 = η2 = −1, leads to the orthogonality relation (for 0 ≤ bq < 1 and c > 0) ∞
(bq; q)ν
ν
∑ (−bcq, q; q)ν cν q(2) ym (q−ν ; q)yn (q−ν ; q)
ν =0
=
(−c; q)∞ 2n −n(2n+1) c q (q, −c−1 q, bq; q)n δmn , (−bcq; q)∞
m, n = 0, 1, 2, . . .
for the q-Meixner polynomials. The special case 2 f = aq−1 , g = 0, α ∗ = 1 and β ∗ = −q, which implies that η1 = η2 = −1, leads to the orthogonality relation (for a > 0) ∞
aν
ν
∑ (q; q)ν q(2) ym (q−ν ; q)yn (q−ν ; q)
ν =0
= (−a; q)∞ a2n q−n(2n+1) (q, −a−1 q; q)n δmn ,
m, n = 0, 1, 2, . . .
for the q-Charlier polynomials.
η1 η2 g > 1 and < 0. Since q > 1 2f 2f ε1 ε2 g > 1 and < 0. This implies we set q = p−1 with 0 < p < 1. Then we have 2f 2f that g = 0 and we might use the weight function − εg1 z , −ε2 pz; p ∞, w(II) (z; p) = − 2gpf z ; p Case IIa3. q > 1, e = 0, f = 0, (β ∗ )2 ≥ 4gα ∗ q,
∞
where ε1 ε2 = α ∗ = 0 and ε1 p−1 + ε2 g = β ∗ . Then we have ε1 z 1z − εgp − , −ε2 z; p , − ε z; p 2 gp ∞ · 2 f p−2 z + g = ∞ · g, w(II) (p−1 z; p)ϕ (p−2 z) = 2fz − gp2 ; p − 2gpf z ; p ∞
∞
which vanishes for z = −gp/ε1 and for z = −1/ε2 . Now we use (1.15.11) to obtain gp ε1 gp2 ε2 1 for < , < 1 and <1 ε2 ε 1 g ε2 ε1 ε1 ε2 gp2
− 1 − ε1 z , −ε2 pz; p p, , ; p g ε2 g ε1 ε2 gp 1 ∞ dpz = ∞ > 0. (1 − p) d0 := gp − 2 f z 2 f 2 f ε1 ε2 −ε − ; p , ; p 1 gp ε1 ε2 gp ∞
Further we have by using (11.6.5) with q = p−1
∞
11.8 Orthogonality Relations
361
4 f 2 −4n+3 β ∗ n−2 gα ∗ 2n−3 n q q (1 − q ) 1 − + q (α ∗ )2 2f 4f2
2 f n−2 2 f n−1 g n n p 1− , p (1 − p ) 1 − p =− ε1 ε2 ε1 ε2 g
dn =
which implies that
n 2f 2f g n n(n+1)/2 ;p , p, , ∏ d k = − ε1 ε2 p ε1 ε2 gp n k=1
n = 1, 2, 3, . . . .
This leads to the orthogonality relation − 1 − ε1 z , −ε2 pz; p g ε2 (II) ∞ y(II) m (z; p)yn (z; p) d p z gp 2fz −ε − ; p 1 gp ∞ ε1 ε2 gp2
p, , ; p ε2 g ε1 gp 1 ∞ = (1 − p) − 2f 2f ε1 ε2 , ; p ε1 ε2 gp ∞
n 2 f g 2f n(n+1)/2 ; p δmn , p, , × − p ε1 ε2 ε1 ε2 gp n for
n = 1, 2, 3, . . . ,
m, n = 0, 1, 2, . . .
1 gp ε1 gp2 ε2 ε1 ε2 g > 1, < 0, < , < 1 and < 1. 2f 2f ε2 ε 1 g ε2 ε1
η1 η1 N1 η1 N1 −1 η2 g > 1 with q ≤1< q >1 , 2f 2f 2f 2fq and N = min(N1 , N2 ). Again we use the weight
Case IIb1. 0 < q < 1, e = 0, f = 0,
η2 g N2 −1 η2 g N2 −2 q q ≤1< 2f 2f function
with
− 2gf q ; q
ν w(II) (yν ; q−1 ) = q g − η1 , − η2 ; q
Then we have as before with yν
ν
2 f q2 η1 η2
ν
ν
q( 2 ) .
= q−ν
− 2gf q ; q
ν w(II) (qyν ; q−1 )ϕ (q2 yν ) = q g − η1 , − η2 ; q
ν −1
2f η1 η2
ν −1
ν −1 2
· 2 f q ν +(
),
which vanishes for ν = 0 if η2 = −1, which implies that η1 = −α ∗ . It also vanishes g = q−N . Then we have by using (1.11.6) with yν = q−ν and for ν = N + 1 if − 2fq 2fq g = −2 f q−N+1 for − ∗ < qN α
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
362 N
d0 :=
∑ w(II) (yν ; q−1 )yν =
ν =0
N
∑
ν =0
(q−N ; q)ν −
2 f q−N+1 α∗
, q; q
ν
2fq α∗
ν
ν
q(2 )
1
> 0. = −N+1 − 2 f qα ∗ ; q N
Now we have by using (11.6.5) dn =
2f α∗
2
α ∗ n−1 q 1 − qn−N−1 , q−4n+3 (1 − qn ) 1 + 2f
n = 1, 2, 3, . . . ,
which implies that n
∏ dk =
k=1
2f α∗
2n
α∗ q−n(2n−1) q, − , q−N ; q , 2f n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation
2 f q ν (ν ) (II) −ν (q−N ; q)ν (II) q 2 ym (q ; q)yn (q−ν ; q) ∑ 2 f q−N+1 ∗ α ν =0 − α ∗ , q; q ν 2n
2f α ∗ −N 1 −n(2n−1) q, − , q ; q δmn q = −N+1 α∗ 2f n − 2 f qα ∗ ; q N
N
for −
2fq < qN and m, n = 0, 1, 2, . . . , N. α∗
Since we have w(yν ; q−1 ) =
−1 ν (ν ) (q−N ; q)ν (q; q)N (bq; q)N−ν (−1)ν (ν ) 2 = −b q q 2 , (b−1 q−N , q; q)ν (bq; q)N (q; q)N−ν (q; q)ν
the special case 2 f = −b−1 q−1 , g = b−1 q−N , α ∗ = 1 and β ∗ = −q − b−1 q−N , which implies that η1 = η2 = −1, leads to the orthogonality relation (for b > q−N ) (q; q)N (bq; q)N−ν (−1)ν (ν ) q 2 ym (q−ν ; q)yn (q−ν ; q) (q; q) (q; q) N− ν ν ν =0 1 = −1 −N b−2n q−n(2n+1) q, bq, q−N ; q n δmn , m, n = 0, 1, 2, . . . , N (b q ; q)N N
∑ (bq; q)N
for the quantum q-Krawtchouk polynomials. Case III. In this case we have e = 0. In section 11.6 we have seen that it is possible to have positive-definite orthogonality for an infinite system of polynomials in at
11.8 Orthogonality Relations
363
least nine different cases. It is also possible to have positive-definite orthogonality for finite systems of polynomials. We will only treat three finite cases here. g ξ2 β ∗ ξ2 β ∗ N−2 > 0, α ∗ = 0, < −1 with q Case IIIb3. 0 < q < 1, < −1 ≤ e gq g ξ2 β ∗ N−1 ξ1 β ∗ q > −1. We use the weight function and g gq ξ1 ξ2
ν , ; q ν q q gq3 (III) −1 ν w (yν ; q ) ∗ = q− ( 2 ) . ∗ξ ξ gq α =0,β ∗ =0 β 1 2 − β∗ ;q ν
Then we have by using e = g/ξ1 ξ2 with yν = q−ν w(III) (qyν ; q−1 ) ∗ ϕ (q2 yν ) α =0,β ∗ =0 ξ1 ξ2
ν −1 , ; q q q ν −1 gq3 ν −1 = q−( 2 ) · eq−2ν +4 + 2 f q−ν +2 + g ∗ξ ξ gq β 1 2 − β∗ ;q ν −1 ξ1 ξ2
ν −1 q , q ; q ν −1 ν −1 gq3 q−( 2 ) · eq−2ν +4 1 − ξ1 qν −2 1 − ξ2 qν −2 = ∗ gq β ξ1 ξ2 − ;q
=
β∗
ν −1
ξ1 ξ2 q , q ;q ν − βgq∗ ; q ν −1
g ξ1 ξ2
ν
q β∗
ν −1
ν −1 2
q2−(
),
which vanishes for ν = 0 if we have by using (1.8.6)
1 gq −1 g = − ∗ q ; q = 1 + ∗ = 0. gq β β 1 − β∗ ;q −1
This implies that g = −β ∗ . It also vanishes for ν = N + 1 if ξ2 = q−N+1 . Then we have by using (1.11.7) with yν = q−ν ξ1 −N N+1 ν , q ; q N N q q qN −(ν2 ) (III) −1 ν − d0 := ∑ w (yν ; q ) ∗ y = q . = ν ∑ α =0,β ∗ =0 (q; q)ν ξ1 ξ1N ν =0 ν =0 Note that this is positive for ξ1 > 0. Further we have by using (11.6.6), (11.6.7) and (11.6.8) dn = −ξ1 ξ2 qn−3 (1 − qn ) 1 − ξ1 qn−2 1 − ξ2 qn−2 , n = 1, 2, 3, . . . with ξ2 = q−N+1 , which implies that
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
364 n
∏ dk =
−ξ1 q
−N n n(n−3)/2
k=1
q
ξ1 −N q, , q ; q , q n
n = 1, 2, 3, . . . .
This leads to the orthogonality relation ξ1 −N ; q qN+1 ν N ν q ,q (III) (III) ∑ (q; q)ν ν − ξ1 q−(2) ym (q−ν ; q)yn (q−ν ; q) ν =0
qN ξ1 −N −N n n(n−3)/2 = N −ξ1 q q, , q ; q δmn , m, n = 0, 1, 2, . . . , N q q ξ1 n for ξ1 > 0. The special case e = q−2 , 2 f = −b − q−N−1 , g = bq−N+1 , α ∗ = 0 and β ∗ = −bq−N+1 , which implies that ξ1 = bq2 and ξ2 = q−N+1 , leads by using (1.8.18) and ν (bq; q)ν (q; q)N −ν (bq, q−N ; q)ν (−b−1 qN )ν q−( 2 ) = b (q; q)ν (q; q)ν (q; q)N−ν to the orthogonality relation (for 0 < bq < 1) N
(bq; q)ν (q; q)N
∑ (q; q)ν (q; q)N−ν (bq)−ν ym (q−ν ; q)yn (q−ν ; q)
ν =0
= (−1)n bn−N q−N(n+1) qn(n+1)/2 (q, bq, q−N ; q)n δmn ,
m, n = 0, 1, 2, . . .
for the affine q-Krawtchouk polynomials. Case IIIb8. q > 1, g = 0, β ∗ = 0, α ∗ ≤ 0 < e,
2fq 2 f α∗ < 1 and 0 < < 1 with ∗ β eβ ∗
2 f α ∗ N−1 2 f α∗ N q ≤1< q . We use the weight function ∗ eβ eβ ∗ − 2ef p−ν +2 ; p e ν ν ν w(III) (zν ; p) = p( 2 ) . ∗ β ∗ p3 g=0,β ∗ =0 − αβ ∗ ; p ν
Then we have with zν = pν
11.8 Orthogonality Relations
365
ν −1 − 2ef p−ν +3 ; p ν −1 e ν −1 (III) −1 −2 w (p zν ; p) ϕ (p zν ) = p( 2 ) ∗ 3 ∗ ∗ α β p g=0,β =0 − β∗ ; p ν −1 × ep2ν −4 + 2 f pν −2 − 2ef p−ν +2 ; p e ν −1 ν −1 ν · ep−ν −1+( 2 ) , = ∗ ∗ β − αβ ∗ ; p ν −1
which vanishes for ν = 0 if we have by using (1.8.6)
α ∗ −1 α∗ 1 = − ∗ p ; p = 1 + ∗ = 0. ∗ β β p 1 −α ; p β∗
−1
This implies that β ∗ = −α ∗ p−1 . Hence α ∗ = 0, so that we must have α ∗ < 0 < e. Note that for f = 0 we have by using (1.8.14)
2f 2 f p ν −(ν ) e ;p − p−ν +2 ; p = − p 2 , e 2f p e ν ν e = p−N . Therefore we take 2 f = −epN−1 and 2f p use (1.11.2) with zν = pν to obtain −N
N N p ; p ν epN−1 ν e (III) ;p d0 := ∑ w (zν ; p) zν = ∑ = > 0, α∗ α∗ p g=0,α ∗ =0 N ν =0 ν =0 (p; p)ν
which vanishes for ν = N + 1 if −
since α ∗ < 0 < e. Further we have by using (11.6.6), (11.6.7) and (11.6.9) for 2 f = −epN−1 ∗ ∗ 2 1 − αe p−n+3 1 − αe∗ pn+N−2 1 − p−n+N+1 α dn = − p−3n+5 (1 − p−n ) 2 ∗ ∗ ∗ e 1 − αe p−2n+4 1 − αe p−2n+3 1 − αe p−2n+2 1 − αe∗ pn−3 1 − αe∗ pn+N−2 1 − pn−N−1 e 2n+N−3 n (1 − p ) = ∗p 2 α 1 − e∗ p2n−4 1 − e∗ p2n−3 1 − e p2n−2 α
α
α∗
for n = 1, 2, 3, . . ., which implies that epN−1 e −N ; p N−1 n , , p p, n ∗ n α ep α ∗ p2 n p2(2) , ∏ dk = α ∗ e k=1 , e∗ ; p α ∗ p2 α p
This leads to the orthogonality relation
2n
n = 1, 2, 3, . . . .
11 Orthogonal Polynomial Solutions in q−x of q-Difference Equations
366
p−N ; p ν epN−1 ν (III) ν (III) ym (p ; p)yn (pν ; p) ∑ (p; p) ∗ α ν ν =0 epN−1 e −N ; p
N−1 n p, , , p ∗ α ep e α ∗ p2 n ;p pn(n−1) δmn , = ∗ e e α∗ p α N , ; p α ∗ p2 α ∗ p N
n = 1, 2, 3, . . .
2n
for α ∗ < 0 < e. The special case e > 0, 2 f = −epN−1 , g = 0, α ∗ = −p−1 and β ∗ = 1 leads to the orthogonality relation (for e > 0) ν (p−N ; p)ν −epN ym (pν ; p)yn (pν ; p) ν =0 (p; p)ν p, −epN , p−N ; p n N n n(n−1) = (−e; p)N −ep p δmn , (−ep−1 , −e; p)2n N
∑
m, n = 0, 1, 2, . . . , N
for the q-Krawtchouk polynomials. Case IIIb9. For 0 < q < 1, e = 0, α ∗ = 0 and g = 0 we use the weight function ξ1 ξ2
ν , ; q q q gq3 (III) −1 ν . w (yν ; q ) = η 1 η 2 ξ1 ξ2 − ηg1 , − ηq2 ; q ν
Then we have as before with yν = q−ν (III)
w
−1
(qyν ; q )ϕ (q yν ) = 2
ξ1 ξ2 q , q ; q ν −1 − ηg1 , − ηq2 ; q ν −1
g ξ1 ξ2
ν
qν +1 , (η1 η2 )ν −1
which vanishes for ν = 0 if η2 = −1, which implies that η1 = −α ∗ . It also vanishes ξ2 for ν = N + 1 if = q−N . Then we have by using (1.11.3) with yν = q−ν and q ξ2 = q−N+1 ξ1 −N eξ1 q eq−N+1 2 ν , q ; q , ; q N N ∗ ∗ q α α eq ν ∞ . d0 := ∑ w(III) (yν ; q−1 )yν = ∑ = ∗ eξ1 q−N+1 eξ1 q−N+1 eq2 α ν =0 ν =0 , q; q , α∗ ; q α∗ α∗ ν
The parameters e, ξ1 and have as before
α∗
∞
should be chosen in such a way that d0 > 0. Now we
11.8 Orthogonality Relations
367 ∗
1 − αe qn−3
dn = −ξ1 qn−N−2 (1 − qn )
2 ∗ ∗ ∗ 1 − αe q2n−4 1 − αe q2n−3 1 − αe q2n−2
α ∗ n−1 α ∗ n+N−2 q 1− q × 1 − ξ1 qn−2 1 − qn−N−1 1 − e ξ1 e
for n = 1, 2, 3, . . ., which implies that α ∗ ξ1 −N α ∗ α ∗ qN−1 q, eq , eξ , e ; q n 2 , q ,q n 1 n , ∏ dk = −ξ1 q−N+1 qn(n−5)/2 α∗ α∗ k=1 , ; q 2 eq eq
n = 1, 2, 3, . . . .
2n
This leads to the orthogonality relation ξ1 −N 2 ν , q ; q N q eq (III) (III) ν ym (q−ν ; q)yn (q−ν ; q) ∑ eξ1 q−N+1 ∗ α ν =0 , q; q α∗ ν eξ1 q eq−N+2 n α∗ , α∗ ; q ∞ −ξ1 q−N+1 qn(n−5)/2 = −N+1 2 e ξ1 q , eq α∗ α∗ ; q ∞ α ∗ ξ1 −N α ∗ α ∗ qN−1 q, eq , eξ , e ; q 2 , q ,q 2 n × δmn , m, n = 0, 1, 2, . . . , N α∗ α∗ , ; q 2 eq eq 2n
for suitable conditions on the parameters e, ξ1 and α ∗ . The special case e = q−2 , 2 f = −α − q−N−1 , g = α q−N+1 , α ∗ = αβ q and β ∗ = −αβ q2 − α q−N+1 , which implies that ξ1 = α q2 , ξ2 = q−N+1 , η1 = −αβ q and η2 = −1, leads to the orthogonality relation (for 0 < α q < 1 and 0 < β q < 1)
ν 1 (α q, q−N ; q)ν ∑ (β −1 q−N , q; q)ν αβ q ym (q−ν ; q)yn (q−ν ; q) ν =0 −1 −1 −1 −N−1 β ,α β q ;q = −1 −N −1 −1 −1 ∞ (−α q−N )n qn(n+1)/2 (β q , α β q ; q)∞ q, αβ q, α q, q−N , β q, αβ qN+2 ; q n × δmn , m, n = 0, 1, 2, . . . , N (αβ q, αβ q2 ; q)2n N
for the q-Hahn polynomials.
Chapter 12
Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations Classical q-Orthogonal Polynomials III
12.1 Motivation for Polynomials in q−x + uqx Through Duality As in chapter 7 we have the concept of duality introduced in definition 3.1. In chapter 11 we obtained a q-difference equation of the form (cf. (11.1.2)) A∗ (x)yn (q−x−1 ) − {A∗ (x) + B∗ (x)} yn (q−x ) + B∗ (x)yn (q−x+1 ) = λn∗ yn (q−x ), n = 0, 1, 2, . . . , N − 1 with N ∈ {1, 2, 3, . . .} or N → ∞, where λn∗ = (qn − 1) eq(1 − q1−n ) − 2ε (1 − q) , A∗ (x) = eq2 + 2 f qx+1 + gq2x with
and
(12.1.1)
n = 0, 1, 2, . . . , N − 1,
B∗ (x) = α ∗ + β ∗ qx−1 + gq2x−1
α ∗ := eq − 2ε (1 − q) and β ∗ := 2 f q − γ (1 − q),
where e, f , g, α ∗ , β ∗ ∈ R, q > 0, q = 1 and ε = 0. If the regularity condition (11.2.4) holds all eigenvalues λn∗ are different. This implies by using theorem 3.7 that there exists a sequence of dual polynomials. In this case we have λn∗ = (qn − 1) eq(1 − q1−n ) − 2ε (1 − q) and κn = q−n with ω = 0 and x0 = 1 = q−0 . Furthermore we have by using (11.2.2) B∗ (0) = α ∗ + β ∗ q−1 + gq−1 = 0 if we choose c = −1 in (11.2.1). Hence if
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 12, © Springer-Verlag Berlin Heidelberg 2010
369
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
370
(A∗ (n) =) eq2 + 2 f qn+1 + gq2n = 0,
n = 0, 1, 2, . . . , N − 1
hold, the dual polynomials {zm }Nm=0 satisfy the three-term recurrence relation A∗ (m)zm+1 (q−x ) − {A∗ (m) + B∗ (m)} zm (q−x ) + B∗ (m)zm−1 (q−x ) = q−x zm (q−x ),
m = 0, 1, 2, . . . , N − 1
(12.1.2)
with the convention that z−1 (q−x ) := 0. If we restrict x in (12.1.1) to x = m for m = 0, 1, 2, . . ., then we have A∗ (m)yn (q−m−1 ) − {A∗ (m) + B∗ (m)} yn (q−m ) + B∗ (m)yn (q−m+1 ) (12.1.3) = λn∗ yn (q−m ), n = 0, 1, 2, . . . , N − 1. Since yn (κm ) = yn (q−m ) = zm (λn ) for all m, n = 0, 1, 2, . . ., this implies that there exist dual polynomials with argument λn∗ = (qn − 1) eq(1 − q1−n ) − 2ε (1 − q) = (qn − 1) α ∗ − eq2−n . This motivates the study of orthogonal polynomials in q−x + uqx with q > 0 and q = 1, x a real variable and u ∈ R \ {0} a constant.
12.2 Difference Equations Having Real Polynomial Solutions with Argument q−x + uqx We start with eigenvalue problems of the form (qz) (Dq yn ) (z) = λn ρ(qz) ϕ(qz) Dq2 yn (z) + ψ yn (qz),
n = 0, 1, 2, . . .
(12.2.1)
with Dq := Aq,0 (cf. (3.2.1)), z ∈ C, q > 0 and q = 1. By using (Dq yn ) (z) =
yn (qz) − yn (z) (q − 1)z
and
yn (qz) + q yn (z) yn (q2 z) − (1 + q) Dq2 yn (z) = q(q − 1)2 z2
this can be written in the symmetric form yn (qz) − C(z) + D(z) yn (q−1 z) = λn ρ(z) C(z) yn (z), yn (z) + D(z) where = C(z)
qϕ(z) (q − 1)2 z2
(12.2.2)
2 = q ϕ (z) + q(1 − q)zψ (z) . and D(z) (q − 1)2 z2
For z ∈ R with z > 0 we may write z = qx with x ∈ R, q > 0 and q = 1. By setting
12.2 Difference Equations Having Real Polynomial Solutions with Argument q−x + uqx
371
yn (z) = yn (qx ) = yn (x), yn (qz) = yn (qx+1 ) = yn (x + 1), yn (q−1 z) = yn (qx−1 ) = yn (x − 1) and = C(q x ) = C(x), C(z)
= D(q x ) = D(x) and ρ(z) = ρ(qx ) = ρ (x), D(z)
we conclude that (12.2.1) can be written in the form (cf. (2.2.12)) C(x)yn (x + 1) − {C(x) + D(x)} yn (x) + D(x)yn (x − 1) = λn ρ (x)yn (x)
(12.2.3)
for n = 0, 1, 2, . . .. Now we look for eigenvalues λn and coefficients C(x), D(x) and ρ (x) so that for each eigenvalue λn there exists exactly one real polynomial solution yn with degree[yn ] = n in q−x + uqx up to a constant factor. Since (q−x + uqx )n can be expressed as a linear combination of (q−x ; q)k (uqx ; q)k for k = 0, 1, 2, . . . , n, we set yn (q−x + uqx ) =
n
∑ an,k
k=0
(q−x ; q)k (uqx ; q)k , (q; q)k
an,n = 0,
n = 0, 1, 2, . . . . (12.2.4)
Now we have (q−x−1 ; q)k (uqx+1 ; q)k − (q−x ; q)k (uqx ; q)k = q−x−1 (1 − qk )(uq2x+1 − 1)(q−x ; q)k−1 (uqx+1 ; q)k−1 . This leads to the first simplification C(x) = q(uq2x−1 − 1)C∗ (x),
D(x) = (uq2x+1 − 1)D∗ (x)
and ρ (x) = q−x (uq2x−1 − 1)(uq2x+1 − 1)ρ ∗ (x).
(12.2.5)
For the moment we assume that uq2x±1 = 1. Then we have n
C∗ (x) ∑ an,k k=1
(q−x ; q)k−1 (uqx+1 ; q)k−1 (q; q)k−1 n
− D∗ (x) ∑ an,k k=1
(q−x+1 ; q)k−1 (uqx ; q)k−1 = λn ρ ∗ (x)yn (x). (q; q)k−1
For a second simplification we note that (q−x ; q)k−1 (uqx+1 ; q)k−1 − (q−x+1 ; q)k−1 (uqx ; q)k−1 = q−x (1 − qk−1 )(uq2x − 1)(q−x+1 ; q)k−2 (uqx+1 ; q)k−2 ,
k = 2, 3, 4, . . . .
Now we define C∗ (x) − D∗ (x) = q−x (uq2x − 1)B(x) and ρ ∗ (x) = q−x (uq2x − 1)ρ ∗∗ (x) (12.2.6)
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
372
with the assumption that uq2x = 1. Without loss of generality we may choose ρ ∗∗ (x) = q so that we have n
B(x) ∑ an,k k=1
n (q−x+1 ; q)k−1 (uqx ; q)k−1 (q−x+1 ; q)k−2 (uqx+1 ; q)k−2 +C∗ (x) ∑ an,k (q; q)k−1 (q; q)k−2 k=2
n
= λn q ∑ an,k k=0
(q−x ; q)k (uqx ; q)k , (q; q)k
n = 2, 3, 4, . . . .
(12.2.7)
For n = 0 we have y0 (q−x + uqx ) = a0,0 (= 0) which leads by using (12.2.3) to λ0 = 0 (except for the trivial situation that ρ ∗∗ (x) = 0). For n = 1 we have y1 (q−x + uqx ) = a1,0 + a1,1 (1 − q−x )(1 − uqx )/(1 − q) with a1,1 = 0, which leads to
(1 − q−x )(1 − uqx ) . B(x)a1,1 = λ1 q a1,0 + a1,1 1−q Since all eigenvalues must be different, we conclude that λ1 = 0 (= λ0 ). Hence B(x) = v + w(1 − q−x )(1 − uqx ) with
v, w ∈ R,
w = 0.
(12.2.8)
This form can be used as a third simplification. For the first term on the left-hand side of (12.2.7) we obtain n
∑
vqk−1 + w (1 − u)(1 − qk−1 ) + (1 − q−x )(1 − uqx )
k=1
(q−x ; q)k−1 (uqx ; q)k−1 (q; q)k−1 n + ∑ v − w(1 − q−x )2 (1 − uqx ) × an,k
k=2
× an,k
(q−x+1 ; q)k−2 (uqx+1 ; q)k−2 . (q; q)k−2
(12.2.9)
This implies that C∗ (x) must be of the form C∗ (x) = (1 − uqx ) −v + w(1 − q−x )2 + σ (1 − q−x )
+ τ (1 − q−x )(1 − q−x+1 )(1 − uqx+1 ) ,
σ , τ ∈ R. (12.2.10)
This leads to the following theorem. Theorem 12.1. The q-difference equation (12.2.3) only has real polynomial solutions yn (x∗ ) with degree[yn ] = n in x∗ := q−x +uqx for n = 0, 1, 2, . . . if the coefficients C(x), D(x) and ρ (x) have the form
12.3 The Basic Hypergeometric Representation
373
C(x) = q(1 − uqx )(1 − uq2x−1 ) × v − w(1 − q−x )2 − σ (1 − q−x )
− τ (1 − q−x )(1 − q−x+1 )(1 − uqx+1 ) ,
D(x) = (1 − q−x )(1 − uq2x+1 ) × v − w(1 − uqx )2 − σ (1 − uqx )
− τ (1 − q−x+1 )(1 − uqx )(1 − uqx+1 ) and
ρ (x) = −q−2x+1 (1 − uq2x−1 )(1 − uq2x )(1 − uq2x+1 )
with u, v, w, σ , τ ∈ R, w = 0, q > 0 and q = 1. Note that the assumptions that uq2x±1 = 1 and uq2x = 1 can be dropped.
12.3 The Basic Hypergeometric Representation In order to find the hypergeometric representation of the polynomials in the form (12.2.4), we use (12.2.8), (12.2.10) and (12.2.9) to obtain from (12.2.7) n
∑
vqk−1 + w (1 − u)(1 − qk−1 ) + (1 − q−x )(1 − uqx )
k=1
× an,k
(q−x ; q)k−1 (uqx ; q)k−1 (q; q)k−1
n + ∑ σ + τ (1 − q−x+1 )(1 − uqx+1 ) (1 − q−x )(1 − uqx ) k=2
(q−x+1 ; q)k−2 (uqx+1 ; q)k−2 (q; q)k−2 n (q−x ; q)k (uqx ; q)k = λn q ∑ an,k , n = 1, 2, 3, . . . . (q; q)k k=0 × an,k
Note that this can also be written as
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
374 n
∑
k=0
(q−x ; q)k (uqx ; q)k q−k+1 (1 − qk ) w + τ q(1 − qk−1 ) − λn q an,k (q; q)k n + ∑ vqk − wq−k (1 − qk )2 + σ (1 − qk ) k=0
−τ q−k+1 (1 − qk−1 )(1 − qk )(1 − uqk+1 ) × an,k+1
(q−x ; q)k (uqx ; q)k =0 (q; q)k
for n = 1, 2, 3, . . . with an,n+1 := 0. By comparing the coefficients of (q−x ; q)k (uqx ; q)k on both sides, we find that λn q − q−k+1 (1 − qk ) w + τ q(1 − qk−1 ) an,k = vqk − wq−k (1 − qk )2 + σ (1 − qk ) − τ q−k+1 (1 − qk−1 )(1 − qk )(1 − uqk+1 ) an,k+1 , which holds for k = 0, 1, 2, . . . , n with an,n+1 = 0. This leads to eigenvalues of the form λn = (q−n − 1) w + τ q(1 − qn−1 ) , n = 0, 1, 2, . . . (12.3.1) and the two-term recurrence relation qk+1 w(q−k − q−n ) + τ q q−k (1 + q2k−1 ) − q−n (1 + q2n−1 ) an,k = −vq2k + (1 − qk ) (12.3.2) × w(1 − qk ) − σ qk + τ q(1 − qk−1 )(1 − uqk+1 ) an,k+1 for k = n − 1, n − 2, n − 3, . . . , 0. Hence the coefficients {an,k }nk=0 in (12.2.4) are uniquely determined in terms of an,n = 0 if (q−k − q−n )(w + τ q) + τ (qk − qn ) = 0
(12.3.3)
for k = n − 1, n − 2, n − 3, . . . , 0 and n ∈ {1, 2, 3, . . .}. This condition holds if the eigenvalues in (12.3.1) are all different, since (q−k − q−n )(w + τ q) + τ (qk − qn ) = λk − λn . In the sequel we will always assume that this holds. Note that the coefficients C(x) and D(x) can also be written as
12.3 The Basic Hypergeometric Representation
375
q2xC(x) = q(1 − uqx )(1 − uq2x−1 )
× −w − τ q + 2w + σ + τ (1 + q + uq2 ) qx
+ v − w − σ − τ (1 + uq + uq2 ) q2x + τ uq3x+1
(12.3.4)
q2x D(x) = (1 − qx )(1 − uq2x+1 )
× −τ q − v − w − σ − τ (1 + uq + uq2 ) qx
− 2w + σ + τ (1 + q + uq2 ) uq2x + (w + τ q)u2 q3x
(12.3.5)
and
respectively. Then we have for u = 0 q2xC(x) = −q(1 − uqx )(1 − uq2x−1 )(1 − x1 qx )(1 − x2 qx )(1 − x3 qx ) and q2x D(x) = −u−1 (1 − qx )(1 − uq2x+1 )(x1 − uqx )(x2 − uqx )(x3 − uqx ), where
τ=
x 1 x2 x3 , uq
w = 1−
x1 x2 x3 , u
σ = x1 + x2 + x3 − 2 −
x1 x 2 x 3 1 − q + uq2 uq
and v = −(1 − x1 )(1 − x2 )(1 − x3 ), which leads to the following theorem: Theorem 12.2. The q-difference equation q2xC(x)yn (q−x−1 + uqx+1 ) − q2x {C(x) + D(x)} yn (q−x + uqx ) + q2x D(x)yn (q−x+1 + uqx−1 ) = −q(1 − uq2x−1 )(1 − uq2x )(1 − uq2x+1 )λn yn (q−x + uqx )
(12.3.6)
only has real polynomial solutions yn (x∗ ) with degree[yn ] = n in x∗ = q−x + uqx for n = 0, 1, 2, . . . if the coefficients q2xC(x) and q2x D(x) and the eigenvalues λn have the form
and
q2xC(x) = −q(1 − uqx )(1 − uq2x−1 )(1 − x1 qx )(1 − x2 qx )(1 − x3 qx ),
(12.3.7)
q2x D(x) = −u−1 (1 − qx )(1 − uq2x+1 )(x1 − uqx )(x2 − uqx )(x3 − uqx )
(12.3.8)
x1 x2 x3 n−1 q , λn = (q−n − 1) 1 − u
n = 0, 1, 2, . . . ,
(12.3.9)
376
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
where u ∈ R \ {0}, q > 0, q = 1 and x1 , x2 , x3 ∈ R or one is real and the other two are complex conjugates. For the two-term recurrence relation (12.3.2) we obtain for u = 0 x1 x2 x3 n+k−1 q an,k = (1 − x1 qk )(1 − x2 qk )(1 − x3 qk )an,k+1 q(1 − qk−n ) 1 − u for k = n − 1, n − 2, n − 3, . . . , 0. This implies that we have (x1 qk ; q)n−k (x2 qk ; q)n−k (x3 qk ; q)n−k an,n (qk−n ; q)n−k x1 xu2 x3 qn+k−1 ; q n−k qn−k (q−n ; q)k x1 xu2 x3 qn−1 ; q k k (x1 ; q)n (x2 ; q)n (x3 ; q)n = −n an,n q (x1 ; q)k (x2 ; q)k (x3 ; q)k (q ; q)n x1 xu2 x3 qn−1 ; q n qn
an,k =
for k = 0, 1, 2, . . . , n. By using (12.2.4) this leads to the representation yn (q−x + uqx ) =
(x1 ; q)n (x2 ; q)n (x3 ; q)n an,n −n (q ; q)n x1 xu2 x3 qn−1 ; q n qn n (q−n ; q) x1 x2 x3 qn−1 ; q (q−x ; q) (uqx ; q) k k k k u k ×∑
k=0
=
(x1 ; q)k (x2 ; q)k (x3 ; q)k (q; q)k
(x1 ; q)n (x2 ; q)n (x3 ; q)n an,n (q−n ; q)n x1 xu2 x3 qn−1 ; q n qn −n x1 x2 x3 n−1 −x x q , u q , q , uq × 4 φ3 ; q, q x1 , x2 , x3
q
(12.3.10)
for the q-Racah polynomials with n = 0, 1, 2, . . .. Special cases are the dual q-Hahn polynomials (for x3 = 0) and the dual qKrawtchouk polynomials (for x2 = x3 = 0). Another special case is the family of dual q-Charlier polynomials (for x1 = x2 = x3 = 0) (see [386]). We remark that we have to choose an,n = (q−n ; q)n qn in order to get monic polynomials. Remark. Note that the q-Racah, the dual q-Hahn and the dual q-Krawtchouk polynomials not only appear as finite systems of orthogonal polynomials. Usually one gets finite systems by setting x1 = q−N , for instance. However, this is not necessary. The q-Racah polynomials have a certain symmetry in n and x which leads to duality. The (not normalized) q-Racah polynomials given by (12.3.10) can be written as −n x1 x2 x3 n−1 −x x q , u q , q , uq ∗ −x x (yn (κx ) =) yn (q + uq ) = 4 φ3 ; q, q (12.3.11) x1 , x2 , x3
12.4 The Three-Term Recurrence Relation
377
for n = 0, 1, 2, . . . with κx = (q−x − 1)(1 − uqx ). If we replace u by x1 x2 x3 /uq, then we have −n n −x x1 x2 x3 x−1 q , uq , q , u q x 1 x2 x3 x q ) = 4 φ3 ; q, q (12.3.12) (z∗n (λx ) =) zn (q−x + uq x1 , x2 , x3 for n = 0, 1, 2, 3, . . . with λx = (q−x − 1)(1 − x1 x2 x3 qx−1 /u). Now we have y∗n (κm ) = z∗m (λn ) for m, n = 0, 1, 2, . . .. In view of definition 3.1, {yn (q−x + uqx )} and {zn (q−x + x1 x2 x3 x uq q )} are dual polynomial systems with respect to the sequences of eigenvalues {κn } and {λn }. Note that we also have y∗n (κ0 ) = z∗0 (λn ). The polynomials zn (q−x + x1 x2 x3 x uq q ) can be considered as dual q-Racah polynomials. This fact will be used in the next section.
12.4 The Three-Term Recurrence Relation In order to obtain the three-term recurrence relation, we use the concept of duality. For the q-Racah polynomials we start with the difference equation (cf. (12.2.3)) for the (not normalized) q-Racah polynomials yn (q−x + uqx ) given by (12.3.11): C(x)y∗n (κx+1 ) − {C(x) + D(x)} y∗n (κx ) + D(x)y∗n (κx−1 ) = λn ρ (x)y∗n (κx ). Now we have y∗n (κx ) = z∗x (λn ) with κx = (q−x − 1)(1 − uqx ) and λn = (q−n − 1)(1 − x1 x2 x3 qn−1 /u) which implies that the difference equation for the dual q-Racah polynomials z∗x (λn ) given by (12.3.12) can be written as C(x)z∗x+1 (λn ) − {C(x) + D(x)} z∗x (λn ) + D(x)z∗x−1 (λn ) = λn ρ (x)z∗x (λn ). For the coefficients we have (cf. (12.3.7)) q2xC(x) = −q(1 − uqx )(1 − uq2x−1 )(1 − x1 qx )(1 − x2 qx )(1 − x3 qx ) and (cf. (12.3.8)) q2x D(x) = −u−1 (1 − qx )(1 − uq2x+1 )(x1 − uqx )(x2 − uqx )(x3 − uqx ) which leads to the three-term recurrence relation q(1 − uqn )(1 − uq2n−1 )(1 − x1 qn )(1 − x2 qn )(1 − x3 qn )z∗n+1 (λx ) − q(1 − uqn )(1 − uq2n−1 )(1 − x1 qn )(1 − x2 qn )(1 − x3 qn )
+ u−1 (1 − qn )(1 − uq2n+1 )(x1 − uqn )(x2 − uqn )(x3 − uqn ) z∗n (λx )
+ u−1 (1 − qn )(1 − uq2n+1 )(x1 − uqn )(x2 − uqn )(x3 − uqn )z∗n−1 (λx ) = q(1 − uq2n−1 )(1 − uq2n )(1 − uq2n+1 )λx z∗n (λx ),
n = 1, 2, 3, . . .
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
378
with z∗0 (λx ) = 1 and z∗1 (λx ) = 1 + (1 − uq)λx /(1 − x1 )(1 − x2 )(1 − x3 ). By replacing u by x1 x2 x3 /uq (= τ ) the polynomials z∗n (λn ) given by (12.3.12) change into the polynomials y∗n (κx ) given by (12.3.11). This implies that the threeterm recurrence relation for the polynomials y∗n (κx ) can be written as u(u − x1 x2 x3 qn−1 )(u − x1 x2 x3 q2n−2 )(1 − x1 qn )(1 − x2 qn )(1 − x3 qn )y∗n+1 (κx ) − u(u − x1 x2 x3 qn−1 )(u − x1 x2 x3 q2n−2 )(1 − x1 qn )(1 − x2 qn )(1 − x3 qn ) + (1 − qn )(u − x1 x2 qn−1 )(u − x1 x3 qn−1 )(u − x2 x3 qn−1 ) ×(u − x1 x2 x3 q2n ) y∗n (κx ) + (1 − qn )(u − x1 x2 qn−1 )(u − x1 x3 qn−1 )(u − x2 x3 qn−1 ) × (u − x1 x2 x3 q2n )y∗n−1 (κx ) = (u − x1 x2 x3 q2n−2 )(u − x1 x2 x3 q2n−1 )(u − x1 x2 x3 q2n )κx y∗n (κx ),
n = 1, 2, 3, . . .
with y∗0 (κx ) = 1, y∗1 (κx ) = 1 + (u − x1 x2 x3 )κx /u(1 − x1 )(1 − x2 )(1 − x3 ) and κx = (q−x − 1)(1 − uqx ). The connection with the monic q-Racah polynomials yn (κx ) is given by x1 x2 x3 n−1 ;q n u q ∗ yn (κx ), n = 0, 1, 2, . . . . yn (κx ) = (x1 ; q)n (x2 ; q)n (x3 ; q)n Hence the three-term recurrence relation for the monic q-Racah polynomials yn (κx ) can be written in the form (1) (2) (1) (2) yn+1 (κx ) = κx + cn + cn yn (κx ) − cn−1 cn yn−1 (κx ), n = 1, 2, 3, . . . (12.4.1) with y0 (κx ) = 1 and y1 (x) = κx + u(1 − x1 )(1 − x2 )(1 − x3 )/(u − x1 x2 x3 ), where (1)
cn =
u(u − x1 x2 x3 qn−1 )(1 − x1 qn )(1 − x2 qn )(1 − x3 qn ) , (u − x1 x2 x3 q2n−1 )(u − x1 x2 x3 q2n )
n = 0, 1, 2, . . . (12.4.2)
and (2)
cn =
(1 − qn )(u − x1 x2 qn−1 )(u − x1 x3 qn−1 )(u − x2 x3 qn−1 ) (u − x1 x2 x3 q2n−2 )(u − x1 x2 x3 q2n−1 )
for n = 1, 2, 3, . . ..
(12.4.3)
12.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions
379
12.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions Favard’s theorem (theorem 3.1) can be extended for monic polynomials yn (q−x + uqx ) in q−x + uqx of degree n ∈ {0, 1, 2, . . .} with u ∈ R \ {0}. The polynomials given by yn+1 (q−x + uqx ) = (q−x − 1)(1 − uqx ) − cn yn (q−x + uqx ) (12.5.1) − dn yn−1 (q−x + uqx ) for n = 1, 2, 3, . . . with y0 (q−x +uqx ) = 1 and y1 (q−x +uqx ) = (q−x −1)(1−uqx )−c0 are orthogonal with respect to a positive-definite linear functional Λ , id est n
−x x −x x Λ ym (q + uq )yn (q + uq ) = ∏ dk δmn , m, n = 0, 1, 2, . . . , (12.5.2) k=0
where Λ [y0 (q−x + uqx )] = d0 ∈ R and Λ [yn (q−x + uqx )] = 0 for n = 1, 2, 3, . . . iff cn ∈ R for all n = 0, 1, 2, . . . and d0 , d1 , d2 , . . . , dn are positive. The proof is similar to the proof of theorem 3.1. For the monic q-Racah polynomials given by (12.3.10) we have the three-term recurrence relation (12.4.1) with (1)
c0 = −c0 , (1)
(2)
dn = cn−1 cn =
(1)
(2)
cn = −cn − cn ,
n = 1, 2, 3, . . . ,
u(1 − qn )(u − x1 x2 x3 qn−2 ) Dn 2n−3 (u − x1 x2 x3 q )(u − x1 x2 x3 q2n−2 )2 (u − x1 x2 x3 q2n−1 )
for n = 1, 2, 3, . . ., and Dn = (1 − x1 qn−1 )(1 − x2 qn−1 )(1 − x3 qn−1 ) × (u − x1 x2 qn−1 )(u − x1 x3 qn−1 )(u − x2 x3 qn−1 )
(12.5.3)
for n = 1, 2, 3, . . ., where u ∈ R \ {0}. Hence we have cn ∈ R for all n = 0, 1, 2, . . . if x1 x2 x3 ∈ R, x1 + x2 + x3 ∈ R and x1 x2 + x1 x3 + x2 x3 ∈ R. This implies that x1 , x2 , x3 ∈ R or one is real and the other two are complex conjugates. To study the positivity of dn for n = 1, 2, 3, . . ., we only need to consider the cases 0 < q < 1 and q > 1 in view of the argument q−x + uqx . Further we have u ∈ R \ {0}. In all cases where x1 x2 x3 = 0 we have dn =
u(1 − qn )(u − x1 x2 x3 qn−2 ) 1 − qn D = Dn n (u − x1 x2 x3 q2n−3 )(u − x1 x2 x3 q2n−2 )2 (u − x1 x2 x3 q2n−1 ) u2
for n = 1, 2, 3, . . .. Case I. x1 = x2 = x3 = 0. Then we have
380
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
dn = u(1 − qn ),
n = 1, 2, 3, . . . .
This implies that we have positive-definite orthogonality in the following two infinite cases: Case Ia1. x1 = x2 = x3 = 0, 0 < q < 1 and u > 0. Case Ia2. x1 = x2 = x3 = 0, q > 1 and u < 0. In this case we have no finite systems of positive-definite orthogonal polynomials. Case II. x1 = 0 and x2 = x3 = 0. Then we have dn = u(1 − qn )(1 − x1 qn−1 ),
n = 1, 2, 3, . . . .
This leads to positive-definite orthogonality in the following three infinite cases: Case IIa1. x1 = 0, x2 = x3 = 0, 0 < q < 1, u > 0 and x1 < 1. Case IIa2. x1 = 0, x2 = x3 = 0, q > 1, u > 0 and x1 > 1. Case IIa3. x1 = 0, x2 = x3 = 0, q > 1, u < 0 and x1 < 0. It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in the following two cases: Case IIb1. x1 = 0, x2 = x3 = 0, 0 < q < 1, u < 0 and x1 > 1 with x1 qN ≤ 1 < x1 qN−1 . Case IIb2. x1 = 0, x2 = x3 = 0, q > 1, u < 0 and 0 < x1 < 1 with x1 qN−1 < 1 ≤ x1 qN . Case III. x1 x2 = 0 and x3 = 0. Then we have dn = (1 − qn )(1 − x1 qn−1 )(1 − x2 qn−1 )(u − x1 x2 qn−1 ),
n = 1, 2, 3, . . . .
Without loss of generality we assume that x1 ≤ x2 . Then we have positive-definite orthogonality in the following six infinite cases: Case IIIa1. x1 x2 = 0, x3 = 0, 0 < q < 1, u > 0, x1 < x2 < 1 and u > x1 x2 . Case IIIa2. x1 x2 = 0, x3 = 0, 0 < q < 1, x1 = x2 and u > x1 x2 . Case IIIa3. x1 x2 = 0, x3 = 0, q > 1, x1 < x2 < 0 and u < x1 x2 .
12.5 Classification of the Positive-Definite Orthogonal Polynomial Solutions
381
Case IIIa4. x1 x2 = 0, x3 = 0, q > 1, x1 < 0, x2 > 1 and u > x1 x2 . Case IIIa5. x1 x2 = 0, x3 = 0, q > 1, 1 < x1 < x2 and u < x1 x2 . Case IIIa6. x1 x2 = 0, x3 = 0, q > 1, x1 = x2 and u < x1 x2 . It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in the following seven cases: Case IIIb1. x1 x2 = 0, x3 = 0, 0 < q < 1, x1 < 0 < x2 < 1 and x1 x2 qN−1 < u ≤ x1 x2 qN . Case IIIb2. x1 x2 = 0, x3 = 0, 0 < q < 1, x1 < 0, x2 > 1 with x2 qN ≤ 1 < x2 qN−1 and u < x1 x2 . Case IIIb3. x1 x2 = 0, x3 = 0, 0 < q < 1, 0 < x1 < 1 < x2 with x2 qN1 ≤ 1 < x2 qN1 −1 and x1 x2 qN2 ≤ u < x1 x2 qN2 −1 and N = min(N1 , N2 ). Case IIIb4. x1 x2 = 0, x3 = 0, 0 < q < 1, 1 < x1 ≤ x2 with x1 qN1 ≤ 1 < x1 qN1 −1 , x2 qN2 ≤ 1 < x2 qN2 −1 , N = min(N1 , N2 ) and u > x1 x2 . Since x1 ≤ x2 we have N = N1 in this case. Case IIIb5. x1 x2 = 0, x3 = 0, q > 1, x1 < 0 < x2 < 1 with x2 qN1 −1 < 1 ≤ x2 qN1 , x1 x2 qN2 ≤ u < x1 x2 qN2 and N = min(N1 , N2 ). Case IIIb6. x1 x2 = 0, x3 = 0, q > 1, 0 < x1 ≤ x2 < 1 with x1 qN1 −1 < 1 ≤ x1 qN1 , x2 qN2 −1 < 1 ≤ x2 qN2 , N = min(N1 , N2 ) and u < x1 x2 . Since x1 ≤ x2 we have N = N2 in this case. Case IIIb7. x1 x2 = 0, x3 = 0, q > 1, 0 < x1 < 1 < x2 with x1 qN1 −1 < 1 ≤ x1 qN1 , x1 x2 qN2 −1 < u ≤ x1 x2 qN2 and N = min(N1 , N2 ). Case IV. x1 x2 x3 = 0. Now we write (1)
(2)
dn = u(1 − qn )Dn Dn ,
n = 1, 2, 3, . . .
(12.5.4)
with (1)
Dn = and
1 − τ qn−1
(1 − τ q2n−2 )(1 − τ q2n−1 )2 (1 − τ q2n )
,
n = 1, 2, 3, . . .
(12.5.5)
382
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations (2)
Dn = (1 − x1 qn−1 )(1 − x2 qn−1 )(1 − x3 qn−1 ) τ τ τ × 1 − qn 1 − qn 1 − qn x1 x2 x3 for n = 1, 2, 3, . . ., where
τ=
(12.5.6)
x 1 x2 x3 . uq
Note that dn ∼ u for n → ∞ both for 0 < q < 1 and q > 1. This implies that dn > 0 can only be true for all n = 1, 2, 3, . . . if u > 0. (1)
q
extra conditions
Dn
for
0
τ <0
+
n = 1, 2, 3, . . .
0 < τq < 1
+
n = 1, 2, 3, . . .
τ q > 1 with 1 < τ q2N ≤ q−2
− n = 1, 2, 3, . . . , N
τ <0
q>1 0 < τq < 1
with q−2
n = 1, 2, 3, . . .
+ ≤ τ q2N
< 1 + n = 1, 2, 3, . . . , N
τq > 1
−
n = 1, 2, 3, . . .
(1)
Table 12.1 sign of Dn , q > 0 and N ∈ {1, 2, 3, . . .}
(1)
Without loss of generality we assume that x1 ≤ x2 ≤ x3 . The sign of Dn for n = 1, 2, 3, . . . is given by table 12.1 (cf. table 10.2). By using (12.5.4), (12.5.6) and table 12.1, we conclude that we have positive-definite orthogonality in the following four infinite cases: (2)
Case IVa1. x1 x2 x3 = 0, 0 < q < 1, u > 0, τ < 0 and Dn > 0. (2)
Case IVa2. x1 x2 x3 = 0, 0 < q < 1, u > 0, 0 < τ q < 1 and Dn > 0. (2)
Case IVa3. x1 x2 x3 = 0, q > 1, u > 0, τ < 0 and Dn < 0. (2)
Case IVa4. x1 x2 x3 = 0, q > 1, u > 0, τ q > 1 and Dn > 0. It is also possible to have positive-definite orthogonality for finite systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in the following four cases: Case IVb1. x1 x2 x3 = 0, 0 < q < 1, u > 0, τ q > 1 with 1 < τ q2N ≤ q−2 and Dn < 0. (2)
Case IVb2. x1 x2 x3 = 0, 0 < q < 1, u < 0, τ q > 1 with 1 < τ q2N ≤ q−2 and Dn > 0. (2)
12.6 Solutions of the q-Pearson Equation
383
Case IVb3. x1 x2 x3 = 0, q > 1, u > 0, 0 < τ q < 1 with q−2 ≤ τ q2N < 1 and Dn < 0. (2)
Case IVb4. x1 x2 x3 = 0, q > 1, u < 0, 0 < τ q < 1 with q−2 ≤ τ q2N < 1 and Dn > 0. (2)
We remark that it is also possible to have positive-definite orthogonality for finite (2) systems of N + 1 polynomials with N ∈ {1, 2, 3, . . .} in cases where Dn has opposite sign for n = N and n = N + 1.
12.6 Solutions of the q-Pearson Equation In this section we use the following operator1 (cf. (2.1.1) with ω = 0) (A f ) (q−x + uqx ) =
f (q−x−1 + uqx+1 ) − f (q−x + uqx ) , q−x−1 + uqx+1 − q−x − uqx
q > 0,
q = 1, (12.6.1)
where f is a complex-valued function in q−x + uqx whose domain contains both q−x + uqx and q−x−1 + uqx+1 for each x ∈ R. For two such functions f1 and f2 , a product rule similar to (11.4.2) applies. In fact we have (A ( f1 f2 )) (q−x + uqx ) = (A f1 ) (q−x + uqx ) f2 (q−x + uqx ) + (S f1 ) (q−x + uqx ) (A f2 ) (q−x + uqx ), (12.6.2) where, analogous to (11.4.3), we now define (S f ) (q−x + uqx ) = f (q−x−1 + uqx+1 ) −1 −x and S f (q + uqx ) = f (q−x+1 + uqx−1 ).
(12.6.3)
We start with a general second-order operator equation of the form ϕ (q−x + uqx ) A 2 yn (q−x + uqx ) + ψ (q−x + uqx ) (A yn ) (q−x + uqx ) = λn (S yn ) (q−x + uqx ) in terms of the operator (12.6.1) and later we will compare this to the q-difference equation (12.3.6) in theorem 12.2. If we multiply both sides of this equation by (S w) (q−x + uqx ) we obtain (S w) (q−x + uqx )ϕ (q−x + uqx ) A 2 yn (q−x + uqx ) + (S w) (q−x + uqx )ψ (q−x + uqx ) (A yn ) (q−x + uqx ) (12.6.4) = λn (S w) (q−x + uqx ) (S yn ) (q−x + uqx ), 1 It turns out that this form makes sense. However, this form no longer has the property that its action on a polynomial in q−x + uqx of degree n leads to a polynomial in q−x + uqx of degree n − 1 for n = 1, 2, 3, . . ..
384
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
which leads to the self-adjoint form −1 A w S ϕ (A yn ) (q−x + uqx ) = λn (S w) (q−x + uqx ) (S yn ) (q−x + uqx )
(12.6.5)
if the Pearson operator equation −1 −x A w S ϕ (q + uqx ) = (S w) (q−x + uqx )ψ (q−x + uqx ) holds. Furthermore, the productrule (12.6.2) leads to −1 −x A w S ϕ (q + uqx ) w(q−x−1 + uqx+1 ) − w(q−x + uqx ) = S −1 ϕ (q−x + uqx ) q−x−1 + uqx+1 − q−x − uqx −1 −x −x x ϕ (q + uqx ) −x−1 x+1 ϕ (q + uq ) − S + w(q + uq ) . q−x−1 + uqx+1 − q−x − uqx Combining the latter two equations we obtain (cf. (3.2.7) and (11.4.6)) w(q−x + uqx )ϕ (q−x+1 + uqx−1 ) = w(q−x−1 + uqx+1 ) × ϕ (q−x + uqx ) − q−x−1 + uqx+1 − q−x − uqx ψ (q−x + uqx ) . (12.6.6) Note that the q-difference equation (12.3.6) in theorem 12.2 can be written in the form −x−1 −x+1 + D(x) C(x)y + uqx+1 ) − C(x) yn (q−x + uqx ) + D(x)y + uqx−1 ) n (q n (q = λn yn (q−x + uqx ),
(12.6.7)
where, by using (12.3.7), (12.3.8) and (12.3.9), q2xC(x) q(1 − uq2x−1 )(1 − uq2x )(1 − uq2x+1 ) (1 − uqx )(1 − x1 qx )(1 − x2 qx )(1 − x3 qx ) , = (1 − uq2x )(1 − uq2x+1 )
=− C(x)
q2x D(x) q(1 − uq2x−1 )(1 − uq2x )(1 − uq2x+1 ) (1 − qx )(x1 − uqx )(x2 − uqx )(x3 − uqx ) . = uq(1 − uq2x−1 )(1 − uq2x )
(12.6.8)
D(x) =−
and
x1 x2 x3 n−1 q , λn = (q−n − 1) 1 − u
n = 0, 1, 2, . . . .
(12.6.9)
12.6 Solutions of the q-Pearson Equation
385
Now we have by using (12.6.1) (A yn ) (q−x + uqx ) =
yn (q−x−1 + uqx+1 ) − yn (q−x + uqx ) q−x−1 + uqx+1 − q−x − uqx
(12.6.10)
and 2 −x A yn (q + uqx ) 1 = −x−1 x+1 q + uq − q−x − uqx yn (q−x−2 + uqx+2 ) − yn (q−x−1 + uqx+1 ) × q−x−2 + uqx+2 − q−x−1 − uqx+1 −
yn (q−x−1 + uqx+1 ) − yn (q−x + uqx ) . q−x−1 + uqx+1 − q−x − uqx
(12.6.11)
If we apply the operator S −1 to the left of (12.6.4) we obtain w(q−x + uqx )ϕ (q−x+1 + uqx−1 ) A 2 yn (q−x+1 + uqx−1 ) + w(q−x + uqx )ψ (q−x+1 + uqx−1 ) (A yn ) (q−x+1 + uqx−1 ) = λn w(q−x + uqx )yn (q−x + uqx ). Now we divide by w(q−x + uqx ) and use (12.6.10) and (12.6.11) to find yn (q−x−1 + uqx+1 )ϕ (q−x+1 + uqx−1 ) (q−x + uqx − q−x+1 − uqx−1 )(q−x−1 + uqx+1 − q−x − uqx ) yn (q−x + uqx ) − −x q + uqx − q−x+1 − uqx−1 ϕ (q−x+1 + uqx−1 ) × q−x−1 + uqx+1 − q−x − uqx
ϕ (q−x+1 + uqx−1 ) + −x − ψ (q−x+1 + uqx−1 ) q + uqx − q−x+1 − uqx−1 +
yn (q−x+1 + uqx−1 ) q−x + uqx − q−x+1 − uqx−1
ϕ (q−x+1 + uqx−1 ) −x+1 x−1 × − ψ (q + uq ) q−x + uqx − q−x+1 − uqx−1
= λn yn (q−x + uqx ). If we compare this with the q-difference equation (12.6.7) we conclude that
ϕ (q−x+1 + uqx−1 ) (12.6.12) = q−x + uqx − q−x+1 − uqx−1 q−x−1 + uqx+1 − q−x − uqx C(x)
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
386
and
ψ (q−x+1 + uqx−1 ) ϕ (q−x+1 + uqx−1 ) + = − q−x + uqx − q−x+1 − uqx−1 D(x) q−x + uqx − q−x+1 − uqx−1 −x x −x+1 x−1 D(x) = − q + uq − q − uq −x−1 x+1 −x (12.6.13) + q + uq − q − uqx C(x). This implies that the q-Pearson equation (12.6.6) can be written in the form w(q−x + uqx ) q−x + uqx − q−x+1 − uqx−1 C(x) + 1). = w(q−x−1 + uqx+1 ) q−x−1 + uqx+1 − q−x − uqx D(x Hence we have the q-Pearson equation = w(q + 1), −x−1 + uqx+1 )D(x −x + uqx )C(x) w(q
(12.6.14)
−x + uqx ) := q−x + uqx − q−x+1 − uqx−1 w(q−x + uqx ). w(q
(12.6.15)
where
Finally we use (12.6.8) and (12.6.9) to conclude that we look for solutions of the q-Pearson equation + 1) −x + uqx ) D(x w(q = −x−1 x+1 + uq ) w(q C(x) =
(1 − qx+1 )(1 − uq2x )(x1 − uqx+1 )(x2 − uqx+1 )(x3 − uqx+1 ) . (12.6.16) uq(1 − uqx )(1 − uq2x+2 )(1 − x1 qx )(1 − x2 qx )(1 − x3 qx )
As before, for q > 1 we set q = p−1 . Then we obtain x + up−x ) w(p x+1 + up−x−1 ) w(p =
(1 − p−x−1 )(1 − up−2x )(x1 − up−x−1 )(x2 − up−x−1 )(x3 − up−x−1 ) . (12.6.17) up−1 (1 − up−x )(1 − up−2x−2 )(1 − x1 p−x )(1 − x2 p−x )(1 − x3 p−x )
Case I. x1 = x2 = x3 = 0. From (12.6.16) we have −x + uqx ) (1 − qx+1 )(1 − uq2x ) w(q = −u2 q3x+2 −x−1 x+1 + uq ) (1 − uqx )(1 − uq2x+2 ) w(q with possible solution
12.6 Solutions of the q-Pearson Equation
w(I) (q−x + uqx ) = −
1 u2
x
387
q−x(3x+1)/2 (1 − uq2x )
(u; q)x , (q; q)x
0 < q < 1.
From (12.6.17) we have x + up−x ) u(1 − px+1 )(1 − u−1 p2x ) w(p = − 3x+1 x+1 −x−1 + up ) p (1 − u−1 px )(1 − u−1 p2x+2 ) w(p with possible solution 1 x x(3x−1)/2 (u−1 ; p)x p (1 − u−1 p2x ) , w(I) (px + up−x ) = − u (p; p)x
0 < p < 1.
Case II. x1 = 0 and x2 = x3 = 0. From (12.6.16) we have x+1 )(1 − uq2x )(x − uqx+1 ) −x + uqx ) w(q 1 2x+1 (1 − q = uq −x−1 + uqx+1 ) (1 − uqx )(1 − uq2x+2 )(1 − x1 qx ) w(q
with possible solution
w(II) (q−x + uqx ) =
1 x1 uq
x
x
q−2(2) (1 − uq2x )
(u, x1 ; q)x , (q, x1−1 uq; q)x
0 < q < 1.
From (12.6.17) we have x + up−x ) (1 − px+1 )(1 − u−1 p2x )(1 − x1 u−1 px+1 ) u w(p = x+1 + up−x−1 ) x1 p2x+1 (1 − u−1 px )(1 − u−1 p2x+2 )(1 − x1−1 px ) w(p with possible solution w(II) (px + up−x ) =
x p x x (u−1 , x1−1 ; p)x 1 p2(2) (1 − u−1 p2x ) , u (p, x1 u−1 p; p)x
0 < p < 1.
Case III. x1 x2 = 0 and x3 = 0. From (12.6.16) we have x+1 )(1 − uq2x )(x − uqx+1 )(x − uqx+1 ) −x + uqx ) w(q 1 2 x (1 − q = −q −x−1 + uqx+1 ) (1 − uqx )(1 − uq2x+2 )(1 − x1 qx )(1 − x2 qx ) w(q
with possible solution 1 x −( x ) (u, x1 , x2 ; q)x (III) −x x q 2 (1 − uq2x ) , w (q + uq ) = − x1 x2 (q, x1−1 uq, x2−1 uq; q)x
0 < q < 1.
388
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
From (12.6.17) we have x + up−x ) (1 − px+1 )(1 − u−1 p2x ) u w(p = − x+1 + up−x−1 ) x1 x2 px+1 (1 − u−1 px )(1 − u−1 p2x+2 ) w(p ×
(1 − x1 u−1 px+1 )(1 − x2 u−1 px+1 ) (1 − x1−1 px )(1 − x2−1 px )
with possible solution w(III) (px + up−x ) x x p x x (u−1 , x1−1 , x2−1 ; p)x 1 2 = − p(2) (1 − u−1 p2x ) , u (p, x1 u−1 p, x2 u−1 p; p)x
0 < p < 1.
Case IV. x1 x2 x3 = 0. From (12.6.16) we have −x + uqx ) x1 x2 x3 (1 − qx+1 )(1 − uq2x ) w(q = −x−1 x+1 + uq ) uq (1 − uqx )(1 − uq2x+2 ) w(q ×
(1 − x1−1 uqx+1 )(1 − x2−1 uqx+1 )(1 − x3−1 uqx+1 ) (1 − x1 qx )(1 − x2 qx )(1 − x3 qx )
with possible solution x uq (u, x1 , x2 , x3 ; q)x (IV ) −x x (1 − uq2x ) , w (q + uq ) = −1 x1 x2 x3 (q, x1 uq, x2−1 uq, x3−1 uq; q)x
0 < q < 1.
From (12.6.17) we have x + up−x ) w(p u (1 − px+1 )(1 − u−1 p2x ) = x+1 −x−1 + up ) x1 x2 x3 p (1 − u−1 px )(1 − u−1 p2x+2 ) w(p ×
(1 − x1 u−1 px+1 )(1 − x2 u−1 px+1 )(1 − x3 u−1 px+1 ) (1 − x1−1 px )(1 − x2−1 px )(1 − x3−1 px )
with possible solution w(IV ) (px + up−x ) =
x x x p x 1 2 3 (1 − u−1 p2x ) u (u−1 , x1−1 , x2−1 , x3−1 ; p)x × , (p, x1 u−1 p, x2 u−1 p, x3 u−1 p; p)x
0 < p < 1.
12.7 Orthogonality Relations
389
12.7 Orthogonality Relations In the preceding section we have obtained solutions of the q-Pearson equation (12.6.16) and the p-Pearson equation (12.6.17). In this section we will derive orthogonality relations for several cases obtained in section 12.5. We will not give explicit orthogonality relations for each different case, but we will restrict to the most important cases. As in section 3.2 we now multiply (12.6.5) by (S ym ) (q−x + uqx ) and subtract the same equation with m and n interchanged to obtain (λn − λm ) (S w) (q−x + uqx ) (S yn ) (q−x + uqx ) (S ym ) (q−x + uqx ) = A w S −1 ϕ (A yn ) (q−x + uqx ) (S ym ) (q−x + uqx ) − A w S −1 ϕ (A ym ) (q−x + uqx ) (S yn ) (q−x + uqx ). Now we apply the operator S −1 and use the commutation rule (cf. (2.5.3) and (11.4.8)) −x q + uqx − q−x+1 − uqx−1 S −1 A = q−x−1 + uqx+1 − q−x − uqx A S −1 , which easily follows from (12.6.1) and (12.6.3), to obtain (cf. (3.2.9)) −x + uqx )ym (q−x + uqx )yn (q−x + uqx ) (λn − λm )w(q −x−1 = q + uqx+1 − q−x − uqx × A S −1 w S −1 ϕ (A yn ) (q−x + uqx )ym (q−x + uqx ) − A S −1 w S −1 ϕ (A ym ) (q−x + uqx )yn (q−x + uqx ) , (12.7.1) −x + uqx ) is given by (12.6.15). Now the product rule (12.6.2) leads to the where w(q summations by parts formula (cf. (3.2.17)) N
∑
q−x−1 + uqx+1 − q−x − uqx (A f1 ) (q−x + uqx ) f2 (q−x + uqx )
x=0
N+1 = f1 (q−x + uqx ) f2 (q−x + uqx ) x=0
− ∑ q−x−1 + uqx+1 − q−x − uqx (S f1 ) (q−x + uqx ) (A f2 ) (q−x + uqx ). N
x=0
Applying this to (12.7.1) we obtain
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
390
N
−x + uqx )ym (q−x + uqx )yn (q−x + uqx ) (λn − λm ) ∑ w(q x=0
−x (q + uqx ) = S −1 w S −1 ϕ × S −1 (A yn ) (q−x + uqx )ym (q−x + uqx ) N+1 . − S −1 (A ym ) (q−x + uqx )yn (q−x + uqx ) x=0
As in theorem 3.8 we now obtain orthogonality relations of the form N
−x + uqx )ym (q−x + uqx )yn (q−x + uqx ) = σn δmn , ∑ w(q
m, n = 0, 1, 2, . . . , N
x=0
with
n
σn = ∏ dk ,
n = 1, 2, 3, . . . , N
k=1
and N ∈ {1, 2, 3, . . .} or possibly N → ∞. In each different case the appropriate boundary conditions (cf. (3.2.19)) should be satisfied, id est: −1 −1 0 S w S ϕ (q + uq0 ) = 0 (12.7.2) and
−1 −1 −N−1 w S ϕ (q S + uqN+1 ) = 0,
(12.7.3)
is given by (12.6.15). where the relation between w and w (2)
Case IVa1. x1 x2 x3 = 0, 0 < q < 1, u > 0, τ < 0 and Dn > 0. We use the weight function x uq (u, x1 , x2 , x3 ; q)x w(IV ) (q−x + uqx ) = (1 − uq2x ) , 0 < q < 1. x1 x2 x3 (q, x1−1 uq, x2−1 uq, x3−1 uq; q)x Then we have by using (1.8.25) and Jackson’s summation formula (1.11.10) x ∞ uq (u, x1 , x2 , x3 ; q)x (1 − uq2x ) d0 := ∑ −1 −1 −1 x x x (q, x 1 2 3 x=0 1 uq, x2 uq, x3 uq; q)x √ √ q u, −q u, u, x1 , x2 , x3 uq = (1 − u) 6 φ5 √ √ −1 ; q, x1 x2 x3 u, − u, x1 uq, x2−1 uq, x3−1 uq −1 −1 −1 −1 −1 −1 uq (uq, x1 x2 uq, x x3 uq, x2 x3 uq; q)∞ < 1. , = (1 − u) −1 x1 x2 x3 (x1 uq, x2−1 uq, x3−1 uq, x1−1 x2−1 x3−1 uq; q)∞ Further we have
12.7 Orthogonality Relations
dn =
391
u(1 − qn )(1 − x1 x2 x3 u−1 qn−2 )(1 − x1 qn−1 )(1 − x2 qn−1 )(1 − x3 qn−1 ) (1 − x1 x2 x3 u−1 q2n−3 )(1 − x1 x2 x3 u−1 q2n−2 )2 (1 − x1 x2 x3 u−1 q2n−1 ) × (1 − x1 x2 u−1 qn−1 )(1 − x1 x3 u−1 qn−1 )(1 − x2 x3 u−1 qn−1 )
for n = 1, 2, 3, . . ., which implies that n q, x1 x2 x3 u−1 q−1 , x1 , x2 , x3 , x1 x2 u−1 , x1 x3 u−1 , x2 x3 u−1 ; q n n σn = ∏ dk = u (x1 x2 x3 u−1 q−1 , x1 x2 x3 u−1 ; q)2n k=1 for n = 1, 2, 3, . . .. This leads to the orthogonality relation x ∞ uq (u, x1 , x2 , x3 ; q)x ∑ x1 x2 x3 (1 − uq2x ) (q, x−1 uq, x−1 uq, x−1 uq; q) ym (q−x + uqx )yn (q−x + uqx ) x x=0 1 2 3 = (1 − u)
(uq, x1−1 x2−1 uq, x−1 x3−1 uq, x2−1 x3−1 uq; q)∞
(x−1 uq, x2−1 uq, x3−1 uq, x1−1 x2−1 x3−1 uq; q)∞ 1 q, x1 x2 x3 u−1 q−1 , x1 , x2 , x3 , x1 x2 u−1 , x1 x3 u−1 , x2 x3 u−1 ; q n n ×u δmn (x1 x2 x3 u−1 q−1 , x1 x2 x3 u−1 ; q)2n
for m, n = 0, 1, 2, . . .. Note that this is an orthogonality relation for an infinite system of q-Racah polynomials given by (12.3.10). However, note that we have by using (12.6.12) and (12.6.15) −1 −1 −x −x+1 + uqx−1 )ϕ (q−x+2 + uqx−2 ) w(q S w S ϕ (q + uqx ) = q−x+1 + uqx−1 − q−x+2 − uqx−2 x−1 u (u, x1 , x2 , x3 ; q)x = q−1 (1 − q) . −1 x1 x2 x3 (q, x1 uq, x2−1 uq, x3−1 uq; q)x−1 Now we use (1.8.5) to obtain 1 = (1; q)1 = 0 and (q; q)−1
1 u = (xi−1 u; q)1 = 1 − , xi (xi−1 uq; q)−1
xi = 0.
This implies that the boundary condition (12.7.2) holds. If we choose x1 , x2 or x3 equal to q−N , then the other boundary condition (12.7.3) also holds. So, in order to find an orthogonality relation for a finite system of q-Racah polynomials, one of the parameters x1 , x2 or x3 should be set equal to q−N . For instance, if we set x3 = q−N , then we obtain N+1 x uq (u, x1 , x2 , q−N ; q)x (1 − uq2x ) , 0 < q < 1. w(IV ) (q−x + uqx ) = −1 x1 x2 (q, x1 uq, x2−1 uq, uqN+1 ; q)x Then we have by using (1.8.25) and Jackson’s summation formula (1.11.11)
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
392
x uqN+1 (u, x1 , x2 , q−N ; q)x ∑ x1 x2 (1 − uq2x ) (q, x−1 uq, x−1 uq, uqN+1 ; q) x x=0 1 2 √ √ q u, −q u, u, x1 , x2 , q−N uqN+1 √ = (1 − u) 6 φ5 √ ; q, x1 x2 u, − u, x1−1 uq, x2−1 uq, uqN+1 N
d0 :=
= (1 − u)
(uq, x1−1 x2−1 uq; q)N . (x1−1 uq, x2−1 uq; q)N
Further we have
q, x1 x2 u−1 q−N−1 , q−N , x1 , x2 , x1 x2 u−1 , x1 u−1 q−N , x2 u−1 q−N ; q n σn = ∏ dk = u (x1 x2 u−1 q−N−1 , x1 x2 u−1 q−N ; q)2n k=1 n
n
for n = 1, 2, 3, . . . , N. This leads to the orthogonality relation N
∑
x=0
uqN+1 x1 x2
x (1 − uq2x )
(u, x1 , x2 , q−N ; q)x ym (q−x + uqx )yn (q−x + uqx ) −1 (q, x1 uq, x2−1 uq, uqN+1 ; q)x
(uq, x1−1 x2−1 uq; q)N (x−1 uq, x2−1 uq; q)N 1 q, x1 x2 u−1 q−N−1 , q−N , x1 , x2 , x1 x2 u−1 , x1 u−1 q−N , x2 u−1 q−N ; q n n ×u δmn (x1 x2 u−1 q−N−1 , x1 x2 u−1 q−N ; q)2n
= (1 − u)
for m, n = 0, 1, 2, . . . , N. Note that this is an orthogonality relation for a finite system of q-Racah polynomials. From this general case involving the q-Racah polynomials we obtain orthogonality relations for the dual q-Hahn polynomials by taking the limit x2 → 0. By using k+1 2
x2k (x2−1 uq; q)k = (x2 − uq)(x2 − uq2 ) · · · (x2 − uqk ) → (−1)k uk q(
),
x2 → 0
for k ∈ {0, 1, 2, . . .}, we find that lim w(IV ) (q−x + uqx ) = w(III) (q−x + uqx ).
x2 →0
We remark that this limit can only be used in the finite case. Now we have for instance Case IIIb3. x1 x2 = 0, x3 = 0, 0 < q < 1, 0 < x1 < 1 < x2 with x2 qN1 ≤ 1 < x2 qN1 −1 and x1 x2 qN2 ≤ u < x1 x2 qN2 −1 and N = min(N1 , N2 ). We use the weight function 1 x −( x ) (u, x1 , x2 ; q)x q 2 (1 − uq2x ) , 0 < q < 1. w(III) (q−x + uqx ) = − x1 x2 (q, x1−1 uq, x2−1 uq; q)x Choosing x2 = q−N we have by using (1.8.25) and the summation formula (1.11.12)
12.7 Orthogonality Relations
393
N x x q (u, x1 , q−N ; q)x ∑ − x1 q−(2) (1 − uq2x ) (q, x−1 uq, uqN+1 ; q) x x=0 1 √ √ q u, −q u, u, x1 , 0, q−N (uq; q)N qN √ −1 = (1 − u) N −1 = (1 − u) 6 φ4 √ ; q, . N+1 x1 u, − u, x1 uq, uq x1 (x1 uq; q)N N
d0 :=
Further we have n σn = ∏ dk = un q, q−N , x1 , x1 u−1 q−N ; q n ,
n = 1, 2, 3, . . . , N.
k=1
This leads to the orthogonality relation N
∑
x=0
−
qN x1
= (1 − u)
x
x
(u, x1 , q−N ; q)x ym (q−x + uqx )yn (q−x + uqx ) (q, x1−1 uq, uqN+1 ; q)x un q, q−N , x1 , x1 u−1 q−N ; q n δmn , m, n = 0, 1, 2, . . . , N
q−(2) (1 − uq2x )
(uq; q)N N x1 (x1−1 uq; q)N
for a finite system of dual q-Hahn polynomials. In the same way as above this orthogonality relation for the dual q-Hahn polynomials leads to an orthogonality relation for the dual q-Krawtchouk polynomials by taking the limit x1 → 0. Now we have lim w(III) (q−x + uqx ) = w(II) (q−x + uqx ).
x1 →0
Then we have for instance Case IIb1. x1 = 0, x2 = x3 = 0, 0 < q < 1, u < 0 and x1 > 1 with x1 qN ≤ 1 < x1 qN−1 . We use the weight function x x 1 (u, x1 ; q)x q−2(2) (1 − uq2x ) , 0 < q < 1. w(II) (q−x + uqx ) = x1 uq (q, x1−1 uq; q)x Choosing x1 = q−N we have by using (1.8.25) and the summation formula (1.11.13) N
d0 :=
∑
x=0
qN−1 u
x
x
q−2(2) (1 − uq2x )
(u, q−N ; q)x (q, uqN+1 ; q)x
√ √ q u, −q u, u, 0, 0, q−N qN−1 √ √ = (1 − u) 6 φ3 ; q, u, − u, uqN+1 u N+1 = (1 − u) (−1)N u−N q−( 2 ) (uq; q)N . Further we have
12 Orthogonal Polynomial Solutions in q−x + uqx of Real q-Difference Equations
394
n σn = ∏ dk = un q, q−N ; q n ,
n = 1, 2, 3, . . . , N.
k=1
This leads to the orthogonality relation
x
(u, q−N ; q)x ym (q−x + uqx )yn (q−x + uqx ) (q, uqN+1 ; q)x x=0 N+1 = (1 − u) (−1)N un−N q−( 2 ) (uq; q)N q, q−N ; q n δmn , m, n = 0, 1, 2, . . . , N N
∑
qN−1 u
x
q−2(2) (1 − uq2x )
for a finite system of dual q-Krawtchouk polynomials. This process cannot be continued in order to obtain an orthogonality relation for the dual q-Charlier polynomials. In that case we only have orthogonality in the case of an infinite family of polynomials. However, we don’t know how to derive an orthogonality relation for such an infinite family of dual q-Charlier polynomials.
Chapter 13
Orthogonal Polynomial Solutions in az + uz a of Complex q-Difference Equations Classical q-Orthogonal Polynomials IV
13.1 Real Polynomial Solutions in az + uz a with u ∈ R \ {0} and a, z ∈ C \ {0} It is also possible to obtain real polynomial solutions of the (complex) q-difference equation (12.2.1) (qz∗ ) (Dq yn ) (z∗ ) = λn ρ(qz∗ ) ϕ(qz∗ ) Dq2 yn (z∗ ) + ψ yn (qz∗ ),
n = 0, 1, 2, . . .
with argument z∗ := az + uz a where u ∈ R \ {0} and a, z ∈ C \ {0}. By using z = x + iy, a = α + iβ with x, y, α , β ∈ R, we find that the imaginary part of a uz α + iβ u(x + iy) (α + iβ )(x − iy) u(x + iy)(α − iβ ) + = + = + z a x + iy α + iβ x 2 + y2 α2 + β 2 equals (β x − α y)(α 2 + β 2 ) + u(α y − β x)(x2 + y2 ) (β x − α y) α 2 + β 2 − u(x2 + y2 ) = . (x2 + y2 )(α 2 + β 2 ) (x2 + y2 )(α 2 + β 2 ) This is equal to zero for all x ∈ R and y ∈ R if x 2 + y2 = r 2
and
ur2 = α 2 + β 2 (= aa).
Without loss of generality we set r = 1, which implies that z is on the unit circle. So if we define z = e iθ ,
a = |a|e−iφ ,
θ , φ ∈ R,
|a| > 0
and u = aa = |a|2 ,
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 13, © Springer-Verlag Berlin Heidelberg 2010
395
396
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
then we find that a uz a + = + az = |a|e−i(θ +φ ) + |a|ei(θ +φ ) = 2|a| cos(θ + φ ). z a z a uz a by q−x , then we have + −→ q−x + uqx . So, similar z z a to the situation in the previous chapter we now look for polynomial solutions of the form Note that if we replace
yn (z∗ ) =
n
∑ bn,k
k=0
( az ; q)k ( uz a ; q)k , (q; q)k
in
z∗ =
bn,n = 0,
n = 0, 1, 2, . . .
(13.1.1)
a uz + , z a
of the q-difference equation (12.2.2) ∗ ) ∗ ) + D(z ∗ ) yn (z∗ ) + D(z ∗ ) C(z yn (qz∗ ) − C(z yn (q−1 z∗ ) = λn ρ(z∗ ) yn (z∗ ), with q ∈ R \ {−1, 0, 1}. Note that yn (z∗ ) is a polynomial with degree[yn ] = n in z∗ = az + uz a = 2|a| cos(θ + φ ). In a similar way as in the previous chapter (cf. theorem 12.2) we now have the following theorem:
Theorem 13.1. The q-difference equation
z2 z2 z2 a uqz a uz aq uz + − 2 {C(z) + D(z)} yn + + 2 D(z)yn + C(z)yn a2 qz a a z a a z aq
a uz a 2 a 2 aq 2 + (13.1.2) 1− z 1 − z λn y n = −q 1 − z aq a a z a only has polynomial solutions yn (z∗ ) with degree[yn ] = n in z∗ =
a uz + (= 2|a| cos(θ + φ )) z a 2
2
for n = 0, 1, 2, . . . if the coefficients az 2 C(z) and az 2 D(z) and the eigenvalues λn have the form
z2 z3 z1 a 2 z2 z (13.1.3) 1− z 1− z 1− z , C(z) = −q (1 − az) 1 − 2 a aq a a a
z z2 1 aq 2 1 − 1 − z (z1 − az) (z2 − az) (z3 − az) D(z) = − (13.1.4) a2 aa a a and
z1 z2 z3 n−1 q , λn = (q−n − 1) 1 − aa
n = 0, 1, 2, . . . ,
(13.1.5)
13.1 Real Polynomial Solutions in
a z
+ uz a with u ∈ R \ {0} and a, z ∈ C \ {0}
397
where a, z1 , z2 , z3 ∈ C, a = 0 and q ∈ R \ {−1, 0, 1}. By using az = |a|e−i(θ +φ ) and representation (cf. (12.3.10)) yn (z∗ ) =
uz a
= az = |a|ei(θ +φ ) , we obtain for n = 0, 1, 2, . . . the
(z1 ; q)n (z2 ; q)n (z3 ; q)n z2 z3 bn,n (q−n ; q)n z1aa qn−1 ; q n qn
z2 z3 n−1 q−n , z1aa q , |a|e−i(θ +φ ) , |a|ei(θ +φ ) × 4 φ3 ; q, q (13.1.6) z1 , z2 , z3
with
a uz a + = + az = 2|a| cos(θ + φ ) (13.1.7) z a z for the Askey-Wilson polynomials, the continuous q-Hahn polynomials or the continuous q-Jacobi polynomials. Special cases are the continuous dual q-Hahn polynomials (z3 = 0), the qMeixner-Pollaczek polynomials, the Al-Salam-Chihara polynomials or the continuous q-Laguerre polynomials (z2 = z3 = 0) and the continuous big q-Hermite polynomials (z1 = z2 = z3 = 0). We remark that we have to choose bn,n = (q−n ; q)n qn in order to get monic polynomials. z∗ =
In view of (12.4.1), (12.4.2) and (12.4.3), we conclude that the monic Askey-Wilson, continuous q-Hahn or continuous q-Jacobi polynomials satisfy the three-term recurrence relation (1) (2) yn+1 (z∗ ) = 2|a| cos(θ + φ ) − |a|2 − 1 + cn + cn yn (z∗ ) − cn−1 cn yn−1 (z∗ ), (1)
(2)
n = 1, 2, 3, . . .
(13.1.8)
with y0 (z∗ ) = 1 and y1 (z∗ ) = 2|a| cos(θ + φ ) − |a|2 − 1 + |a|2 (1 − z1 )(1 − z2 )(1 − z3 )/(|a|2 − z1 z2 z3 ), where (1)
cn =
|a|2 (|a|2 − z1 z2 z3 qn−1 )(1 − z1 qn )(1 − z2 qn )(1 − z3 qn ) (|a|2 − z1 z2 z3 q2n−1 )(|a|2 − z1 z2 z3 q2n )
(13.1.9)
for n = 0, 1, 2, . . ., and (2)
cn =
(1 − qn )(|a|2 − z1 z2 qn−1 )(|a|2 − z1 z3 qn−1 )(|a|2 − z2 z3 qn−1 ) (|a|2 − z1 z2 z3 q2n−2 )(|a|2 − z1 z2 z3 q2n−1 )
for n = 1, 2, 3, . . ..
(13.1.10)
398
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
13.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions The classification of positive-definite orthogonal polynomial solutions can be done as in chapter 12. However, in this case we have q ∈ R \ {−1, 0, 1} and u = aa = |a|2 > 0. Now we have (1)
c0 = −c0 , (1)
(2)
dn = cn−1 cn =
(1)
(2)
cn = −cn − cn ,
n = 1, 2, 3, . . . ,
|a|2 (1 − qn )(|a|2 − z1 z2 z3 qn−2 ) Dn (|a|2 − z1 z2 z3 q2n−3 )(|a|2 − z1 z2 z3 q2n−2 )2 (|a|2 − z1 z2 z3 q2n−1 )
for n = 1, 2, 3, . . . and Dn = (1 − z1 qn−1 )(1 − z2 qn−1 )(1 − z3 qn−1 ) × (|a|2 − z1 z2 qn−1 )(|a|2 − z1 z3 qn−1 )(|a|2 − z2 z3 qn−1 ) (13.2.1) for n = 1, 2, 3, . . ., where q ∈ R \ {−1, 0, 1} and a ∈ C \ {0}. Hence we have cn ∈ R for all n = 0, 1, 2, . . . if z1 z2 z3 ∈ R, z1 + z2 + z3 ∈ R and z1 z2 + z1 z3 + z2 z3 ∈ R. This implies that z1 , z2 , z3 ∈ R or one is real and the other two are complex conjugates. As before, we start the study of the positivity of dn for n = 1, 2, 3, . . . with the cases where z1 z2 z3 = 0. Then we have dn =
|a|2 (1 − qn )(|a|2 − z1 z2 z3 qn−2 ) 1 − qn D = Dn . n (|a|2 − z1 z2 z3 q2n−3 )(|a|2 − z1 z2 z3 q2n−2 )2 (|a|2 − z1 z2 z3 q2n−1 ) |a|4
The sign of 1 − qn is given in table 13.1. q 1 − qn
q < −1 −1 < q < 0 0 < q < 1 q > 1 (−1)n+1
+
+
−
Table 13.1 sign of 1 − qn , n = 1, 2, 3, . . .
Case I. z1 = z2 = z3 = 0. Then we have dn = |a|2 (1 − qn ),
n = 1, 2, 3, . . . .
This implies that we have positive-definite orthogonality in the following two infinite cases: Case Ia1. z1 = z2 = z3 = 0 and −1 < q < 0. Case Ia2. z1 = z2 = z3 = 0 and 0 < q < 1.
13.2 Classification of the Positive-Definite Orthogonal Polynomial Solutions
399
In this case we have no finite systems of positive-definite orthogonal polynomials. Case II. z1 = 0 and z2 = z3 = 0. Then we have dn = |a|2 (1 − qn )(1 − z1 qn−1 ),
n = 1, 2, 3, . . . .
In this case we must have z1 ∈ R. This leads to positive-definite orthogonality in the following four infinite cases: Case IIa1. z1 ∈ R \ {0}, z2 = z3 = 0, q < −1 and the sign of 1 − z1 qn−1 equal to (−1)n+1 . This implies that z1 < q−1 . Case IIa2. z1 ∈ R \ {0}, z2 = z3 = 0, −1 < q < 0 and 1 − z1 qn−1 > 0. This implies that q−1 < z1 < 1. Case IIa3. z1 ∈ R \ {0}, z2 = z3 = 0, 0 < q < 1 and 1 − z1 qn−1 > 0. This implies that z1 < 1. Case IIa4. z1 ∈ R \ {0}, z2 = z3 = 0, q > 1 and 1 − z1 qn−1 < 0. This implies that z1 > 1. Also in this case we have no finite systems of positive-definite orthogonal polynomials. Case III. z1 z2 = 0 and z3 = 0. Then we have dn = (1 − qn )(1 − z1 qn−1 )(1 − z2 qn−1 )(|a|2 − z1 z2 qn−1 ),
n = 1, 2, 3, . . . .
This leads to positive-definite orthogonality in the following four infinite cases: Case IIIa1. z1 z2 = 0, z3 = 0, q < −1 and the sign of (1 − z1 qn−1 )(1 − z2 qn−1 )(|a|2 − z1 z2 qn−1 ) equal to (−1)n+1 . Case IIIa2. z1 z2 = 0, z3 = 0, −1 < q < 0 and (1 − z1 qn−1 )(1 − z2 qn−1 )(|a|2 − z1 z2 qn−1 ) > 0. Case IIIa3. z1 z2 = 0, z3 = 0, 0 < q < 1 and (1 − z1 qn−1 )(1 − z2 qn−1 )(|a|2 − z1 z2 qn−1 ) > 0. Case IIIa4. z1 z2 = 0, z3 = 0, q > 1 and (1 − z1 qn−1 )(1 − z2 qn−1 )(|a|2 − z1 z2 qn−1 ) < 0.
400
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
It is also possible to have positive-definite orthogonality for finite systems of polynomials. However, we will not treat these finite cases. Case IV. z1 z2 z3 = 0. Now we write (cf. (12.5.4)) (1)
(2)
dn = |a|2 (1 − qn )Dn Dn ,
n = 1, 2, 3, . . .
(13.2.2)
with (cf. (12.5.5)) (1)
Dn =
1 − τ qn−1 , (1 − τ q2n−2 )(1 − τ q2n−1 )2 (1 − τ q2n )
n = 1, 2, 3, . . .
(13.2.3)
and (cf. (12.5.6)) (2)
Dn = (1 − z1 qn−1 )(1 − z2 qn−1 )(1 − z3 qn−1 )
τ τ τ × 1 − qn 1 − qn 1 − qn z1 z2 z3 for n = 1, 2, 3, . . ., where
τ=
(13.2.4)
z1 z2 z3 . |a|2 q (1)
The sign of 1 − qn for n = 1, 2, 3, . . . is given in table 13.1. The sign of Dn n = 1, 2, 3, . . . is given in table 12.1 (for q > 0) and in table 13.2 (for q < 0).
q
(1)
extra conditions
τ q2 > 1
q < −1 0
< τ q2
< 1 with 1
0 < τq < 1
≤ τ q2N
with τ q2N+1
0 < τ q < 1 with q
< τ q2N
−1 < q < 0
> 1 with 1
< τ q2N
0 < τ q2
for
(−1)n
n = 1, 2, 3, . . .
< q−2
+
n = 1, 2, 3, . . . , N
=1
+
n = 1, 2, 3, . . . , N
< q−1
τq > 1 τ q2
Dn
≤ q−2
<1
+
n = 1, 2, 3, . . . , 2N + 1
(−1)n+1
n = 1, 2, 3, . . .
(−1)n
n = 1, 2, 3, . . . , N
+
n = 1, 2, 3, . . .
0 < τq < 1
+
n = 1, 2, 3, . . .
τ q > 1 with τ q2N+1 = 1
(−1)n+1
n = 1, 2, 3, . . . , N
τ q > 1 with q−1 < τ q2N < q
for
(−1)n+1 n = 1, 2, 3, . . . , 2N + 1
(1)
Table 13.2 sign of Dn , q < 0 and N ∈ {1, 2, 3, . . .}
By using table 12.1, table 13.1 and table 13.2, we conclude that we have positivedefinite orthogonality in the following eight infinite cases:
13.3 Solutions of the q-Pearson Equation
401 (2)
Case IVa1. z1 z2 z3 = 0, q < −1, τ q2 > 1 and Dn < 0. (2)
Case IVa2. z1 z2 z3 = 0, q < −1, τ q > 1 and Dn > 0. (2)
Case IVa3. z1 z2 z3 = 0, −1 < q < 0, 0 < τ q2 < 1 and Dn > 0. (2)
Case IVa4. z1 z2 z3 = 0, −1 < q < 0, 0 < τ q < 1 and Dn > 0. (2)
Case IVa5. z1 z2 z3 = 0, 0 < q < 1, τ < 0 and Dn > 0. (2)
Case IVa6. z1 z2 z3 = 0, 0 < q < 1, 0 < τ q < 1 and Dn > 0. (2)
Case IVa7. z1 z2 z3 = 0, q > 1, τ < 0 and Dn < 0. (2)
Case IVa8. z1 z2 z3 = 0, q > 1, τ q > 1 and Dn > 0. It is also possible to have positive-definite orthogonality for finite systems of polynomials. However, we will not treat these finite cases.
13.3 Solutions of the q-Pearson Equation In this section we use the following operator1 (cf. (12.6.1))
f a + uqz − f a + uz qz a z a a uz + = , q ∈ R \ {−1, 0, 1}, (A f ) uqz a a uz z a qz + a − z − a
(13.3.1)
a uz where f is a complex-valued function in az + uz a whose domain contains both z + a uqz a and qz + a for each z ∈ C. For two such functions f1 and f2 , a product rule similar to (12.6.2) applies. In fact we have
a uz a uz a uz + = (A f1 ) + f2 + (A ( f1 f2 )) z a z a z a
a uz a uz + (A f2 ) + , (13.3.2) + (S f1 ) z a z a
where, analogous to (12.6.3), we now define 1 It turns out that this form makes sense. However, this form no longer has the property that its a uz action on a polynomial in az + uz a of degree n leads to a polynomial in z + a of degree n − 1 for n = 1, 2, 3, . . ..
402
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
a uqz a uz + =f + (S f ) z a qz a
−1 a uz aq uz + =f + . and S f z a z aq
(13.3.3)
Now we start with a general second-order operator equation of the form
a uz 2 a uz + A yn + ϕ z a z a
a uz a uz a uz + (A yn ) + = λn (S yn ) + +ψ z a z a z a in terms of the operator (13.3.1) and later we will compare this to the q-difference equation
in theorem 13.1. If we multiply both sides of this equation by (13.1.2) a uz + we obtain (S w) z a
a uz a uz 2 a uz + + A yn + (S w) ϕ z a z a z a
a uz a uz a uz + + (A yn ) + + (S w) ψ z a z a z a
a uz a uz + (S yn ) + , (13.3.4) = λn (S w) z a z a which leads to the self-adjoint form
−1 a uz + A w S ϕ (A yn ) z a
a uz a uz + (S yn ) + = λn (S w) z a z a if the Pearson operator equation
−1 a uz a uz a uz + = (S w) + + A w S ϕ ψ z a z a z a holds. Furthermore, the productrule (13.3.2) leads to
(13.3.5)
13.3 Solutions of the q-Pearson Equation
403
−1 a uz + A w S ϕ z a
w a + uqz − w a + uz a uz qz a z a = S −1 ϕ + uqz a a uz z a + − − qz a z a a uz a uqz ϕ z + a − S −1 ϕ az + uz a + +w . uqz a a uz qz a qz + a − z − a Combining the latter two equations we obtain (cf. (3.2.7) and (11.4.6))
aq uz a uz + + ϕ w z a z aq
a uqz + =w qz a
a uqz a uz a uz a uz + − + − − + . × ϕ ψ z a qz a z a z a
(13.3.6)
The q-difference equation (13.1.2) in theorem 13.1 can now be written in the form
a uz
a uqz aq uz + − C(z) + D(z) yn + + D(z)yn + C(z)yn qz a z a z aq
a uz + , (13.3.7) = λn y n z a where, by using (13.1.3), (13.1.4) and (13.1.5), z2C(z) a 2 2 1 − aa z2 1 − aq a2 q 1 − aq z a z (1 − az) 1 − za1 z 1 − za2 z 1 − za3 z = , 2 1 − aa z2 1 − aq a z
=− C(z)
z2 D(z) a 2 2 1 − aa z2 1 − aq a2 q 1 − aq z z a z 1 − a (z1 − az) (z2 − az) (z3 − az) = a 2 1 − aa z2 aa q 1 − aq z
=− D(z)
(13.3.8)
z1 z2 z3 n−1 q , λn = (q−n − 1) 1 − aa Note that we have by using (13.3.1) and
n = 0, 1, 2, . . . .
(13.3.9)
404
13 Orthogonal Polynomial Solutions in
(A yn )
a uz + z a
and
a uz A 2 yn + = z a
yn =
a qz
a z
+ uz a of Complex q-Difference Equations
a uz − y + uqz + n a z a
a qz
a uz + uqz a − z − a
⎧ ⎨ yn
a qz
a q2 z
(13.3.10)
2 a + uqa z − yn qz + uqz a
1 a uz ⎩ 2 a a + uqz + uqa z − qz − uqz a − z − a a q2 z ⎫ uqz a a uz ⎬ yn qz + a − yn z + a − . (13.3.11) uqz a a uz ⎭ qz + a − z − a
If we apply the operator S −1 to the left of (13.3.4) we obtain
aq uz 2 aq uz a uz + + A yn + w ϕ z a z aq z aq
aq uz aq uz a uz + + (A yn ) + ψ +w z a z aq z aq
a uz a uz + yn + . = λn w z a z a
a uz + and use (13.3.10) and (13.3.11) to find Now we divide by w z a aq a uz yn qz + uqz ϕ + a z aq aq uqz a uz uz a a uz + − − + − − z a z aq qz a z a ⎧ aq uz ⎨ ϕ z + aq yn az + uz a − a uz aq uz a a uz ⎩ qz + uqz z + a − z − aq a − z − a ⎫ uz
⎬ ϕ aq + z aq aq uz + + a uz aq uz − ψ + − − z aq ⎭ z a z aq ⎧ ⎫ uz uz
⎨ ϕ aq yn aq z + aq z + aq aq uz ⎬ + + a uz aq uz aq uz − ψ ⎩ az + uz z aq ⎭ z + a − z − aq a − z − aq
a uz = λn y n + . z a If we compare this with the q-difference equation (13.3.7) we conclude that
13.3 Solutions of the q-Pearson Equation
405
aq uz + z aq
a uqz a uz a uz aq uz + − − + − − C(z) = z a z aq qz a z a
ϕ
and
(13.3.12)
aq uz + ψ z aq aq uz
ϕ + z aq a uz aq uz + − − =− D(z) + a uz aq uz z a z aq + − z a z − aq
a uqz a uz a uz aq uz + − − D(z) + + − − C(z). (13.3.13) =− z a z aq qz a z a
This implies that the q-Pearson equation (13.3.6) can be written in the form
a uz aq uz a uz + + − − C(z) w z a z a z aq
a uqz a uz a uqz + + − − D(qz). =w qz a qz a z a Hence we have the q-Pearson equation
a uqz a uz + C(z) = w + D(qz), w z a qz a where
a uz + w z a
:=
(13.3.14)
a uz aq uz a uz + − − w + . z a z aq z a
(13.3.15)
Finally we use (13.3.8) and (13.3.9) to conclude that we look for solutions of the q-Pearson equation az + uz w 1 − aa z2 1 − qz a D(qz) a (z1 − aqz)(z2 − aqz)(z3 − aqz) = = 2 C(z) a + uqz aa q 1 − aq z2 (1 − az) 1 − z1 z 1 − z2 z 1 − z3 z w qz
where
a
a
2 1 − aq 1− 1− a z = a (1 − az) 1 − aqa2 z2 1 − aqz 2 z3 z1 z2 1 − aqz 1 − aqz 1 − aqz , × 1 − za1 z 1 − za2 z 1 − za3 z
a qz
a
a
a
a 2 az
(13.3.16)
406
z = eiθ ,
13 Orthogonal Polynomial Solutions in
a = |a|e−iφ ,
θ , φ ∈ R,
a z
+ uz a of Complex q-Difference Equations
a uz + = 2|a| cos(θ + φ ). z a
|a| > 0 and
Note that this is equivalent to (12.6.16) with u = |a|2 = aa, qx replaced by z/a and xi replaced by zi for i = 1, 2, 3. Now we make the following observations: 2
az a w(z) = , 2 ;q a az ∞ 2 aq 2 az a a 2 , ; q 1 − 1 − z z 2 a az a a w(z) ∞ , = = 22 =⇒ aq z a a a w(qz) 1 − 1 − , ; q 2 2 2 2 2 a aq z aq z aqz ∞
1 w(z) = a az, z ; q ∞
=⇒
a a aqz, qz ;q 1 − qz w(z) = a ∞= w(qz) 1 − az az, z ; q ∞
and
w(z) = zi z a
1
zi , az ;q
zi qz a , zi z a ,
w(z) = w(qz)
=⇒ ∞
zi aqz ; q ∞ zi az ; q ∞
=
zi 1 − aqz
1 − zai z
,
i = 1, 2, 3.
Case I. z1 = z2 = z3 = 0. From (13.3.16) we have a 1 − qz 1 − aa z2 1 − aq z2 az + uz ) w( a a = a a qz + uqz w( a ) (1 − az) 1 − aqa2 z2 1 − aqz 2 with possible solution w(I)
a uz + z a
=
az2 a a , az2 ; q a ∞, az, z ; q ∞
z = eiθ ,
a = |a|e−iφ .
Case II. z1 ∈ R \ {0} and z2 = z3 = 0. From (13.3.16) we have aq 2 z1 a a 2 a uz 1 − 1 − 1 − 1 − z z z+ a) w( qz a a aqz = a a a qz + uqz ) w( a 1− 1 − z1 z (1 − az) 1 − aq2 z2
with possible solution
aqz2
a
13.4 Orthogonality Relations
w(II)
a uz + z a
407
=
az2 a a , az2 ; q a z z z ∞ , az, z , a1 , az1 ; q ∞
z = eiθ ,
a = |a|e−iφ .
Case III. z1 z2 = 0 and z3 = 0. From (13.3.16) we have z1 z2 a 2 1 − qz 1 − aa z2 1 − aq 1 − 1 − z az + uz ) w( a aqz aqz a = a z1 z2 a a qz + uqz ) w( a (1 − az) 1 − aq2 z2 1 − aqz2 1 − a z 1 − a z with possible solution w(III)
a uz + z a
= az,
az2 a a , az2 ; q ∞ , a z1 z z1 z2 z z2 z , a , az , a , az ; q ∞
z = eiθ ,
a = |a|e−iφ .
Case IV. z1 z2 z3 = 0. From (13.3.16) we have z3 aq 2 z1 z2 a a 2 a uz 1 − 1 − 1 − 1 − 1 − 1 − z z w( z + a ) qz a a aqz aqz aqz = a z3 z1 z2 a a qz + uqz ) w( a 1− 1− z 1− z 1− z (1 − az) 1 − aq2 z2
aqz2
a
a
a
with possible solution w(IV )
a uz + z a
= az,
az2 a a , az2 ; q ∞ , a z1 z z1 z2 z z2 z3 z z3 z , a , az , a , az , a , az ; q ∞
z = eiθ ,
a = |a|e−iφ .
13.4 Orthogonality Relations In the preceding section we have obtained solutions of the q-Pearson equation (13.3.16). In this section we will derive orthogonality relations for several cases obtained in section 13.2. We will not give explicit orthogonality relations for each different case, but we will restrict to the most important cases. In order to find orthogonality relations we cannot use a similar method as in the previous chapter. However, as in theorem 3.8 we now obtain orthogonality relations of the form
β a uz a uz a uz + ym + yn + dz = σn δmn , m, n = 0, 1, 2, . . . , w z a z a z a α with
408
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
n
σn = ∏ dk ,
n = 1, 2, 3, . . . .
k=1
In each different case the appropriate boundary conditions should be satisfied. This implies (cf. (12.7.2) and (12.7.3)) that 2 aq uz uz aq w z + aq ϕ z + aq2 aq z
2
uz uz + aq − aqz − aq 2
should vanish for both z = α and z = β , the ends of the interval of orthogonality (α , β ) with possibly α → −∞ and/or β → ∞. Case Ia2. z1 = z2 = z3 = 0 and 0 < q < 1. We use the weight function 2 az a
, ; q a az2 a uz + = a ∞ , z = eiθ , a = |a|e−iφ . w(I) z a az, z ; q ∞ For the boundary conditions we should have that 2 aq uz uz aq + ϕ + w z aq z aq2 aq z
2
uz uz + aq − aqz − aq 2 2 2 aq az az
, ; q 1 − 2 2 q a uz aq uz aq az ∞ + − − = az aq az2 az2 z a z aq 1 − 1 − , ; q 2 q z aq aq 2 2 ∞ az aq aq , az2 ; q a = aq ∞ (1 − q) az, z ; q ∞ z
vanishes at both ends of the interval of orthogonality. Note that this is true for z=±
a |a|
=⇒
cos(θ + φ ) = ±1.
This implies that we should have 0 ≤ θ + φ ≤ π . Note that, since θ ∈ R and φ ∈ R, the weight function can be written in the form e2i(θ +φ ) , e−2i(θ +φ ) ; q
e2i(θ +φ ) ; q 2 ∞ = ∞ . w(I) (2|a| cos(θ + φ )) = i(θ +φ ) |a|e , |a|e−i(θ +φ ) ; q ∞ |a|ei(θ +φ ) ; q ∞ Further we have
13.4 Orthogonality Relations
409
2 2 1 π (e2iθ ; q)∞ ∞ d θ = dθ . |a|ei(θ +φ ) ; q 2π 0 (|a|eiθ ; q)∞ ∞
d0 :=
1 −φ +π
2π
−φ
e2i(θ +φ ) ; q
Hence we have by using the Askey-Wilson q-beta integral (1.12.4) that d0 =
1 2π
=
1 2π
2 π 2iθ (e ; q)∞ 0
(|a|eiθ ; q)∞ d θ
0
1 (e2iθ , e−2iθ ; q)∞ dθ = , (q; q)∞ (|a|eiθ , |a|e−iθ ; q)∞
π
|a| < 1.
Further we have dn = |a|2 (1 − qn ) which implies that n
σn = ∏ dk = |a|2n (q; q)n ,
n = 1, 2, 3, . . . .
k=1
This leads, for 0 < |a| < 1, to the orthogonality relation 1 2π
π 2iθ (e ; q)∞ 0
2 2n ym (2|a| cos θ )yn (2|a| cos θ ) d θ = |a| (q; q)n δmn (|a|eiθ ; q)∞ (q; q)∞
with m, n = 0, 1, 2, . . . for the continuous big q-Hermite polynomials. Since |a| > 0 this orthogonality relation can be normalized to read 1 2π
2 π 2iθ (e ; q)∞
(q; q)n yn (cos θ ) d θ = δmn , (|a|eiθ ; q)∞ ym (cos θ ) (q; q)∞
0
m, n = 0, 1, 2, . . . .
Then the limit case a → 0 leads to the orthogonality relation 1 2π
π 2 (q; q)n 2iθ δmn , (e ; q)∞ ym (cos θ )yn (cos θ ) d θ =
(q; q)∞
0
m, n = 0, 1, 2, . . .
for the continuous q-Hermite polynomials. (2)
Case IVa5. z1 z2 z3 = 0, 0 < q < 1, τ < 0 and Dn > 0. We use the weight function 2 az a
, ; q a az2 a uz + = a z1 z z1 z2 z z2 z∞3 z z3 , z = eiθ , a = |a|e−iφ . w(IV ) z a az, z , a , az , a , az , a , az ; q ∞ Also in this case the boundary conditions hold for z=±
a |a|
=⇒
cos(θ + φ ) = ±1,
which implies that we should have 0 ≤ θ + φ ≤ π .
410
13 Orthogonal Polynomial Solutions in
a z
+ uz a of Complex q-Difference Equations
Note that, since θ ∈ R and φ ∈ R, the weight function can be written in the form e2i(θ +φ ) ; q ∞ w(IV ) (2|a| cos(θ + φ )) = z1 i(θ +φ ) z2 i(θ +φ ) z3 i(θ +φ ) |a|ei(θ +φ ) , |a| e , |a| e , |a| e ;q
∞
2 .
Further we have e2i(θ +φ ) ; q
d0 :=
2 ∞ dθ . z3 i(θ +φ ) z z 1 2 i( θ + φ ) i( θ + φ ) i( θ + φ ) |a|e , |a| e , |a| e , |a| e ;q
1 −φ +π
2π
−φ
∞
Hence we have as before by using the Askey-Wilson q-beta integral (1.12.4) that
2 ∞ dθ d0 = 2π 0 |a|eiθ , z1 eiθ , z2 eiθ , z3 eiθ ; q |a| |a| |a| ∞ z1 z2 z3 ; q |a|2 ∞ . = z1 z2 z1 z3 z2 z3 z1 , z2 , z3 , |a|2 , |a|2 , |a|2 , q; q
1 π
e2iθ ; q
∞
Further we have dn =
|a|2 (1 − qn )(1 − τ qn−1 )(1 − z1 qn−1 )(1 − z2 qn−1 )(1 − z3 qn−1 ) (1 − τ q2n−2 )(1 − τ q2n−1 )2 (1 − τ q2n )
z1 z2 z3 τ τ τ × 1 − qn 1 − qn 1 − qn , τ = z1 z2 z3 |a|2 q
for n = 1, 2, 3, . . ., which implies that q, τ , z1 , z2 , z3 , τz1q , τz2q , τz3q ; q n n σn = ∏ dk = |a|2n , ( τ , τ q; q) 2n k=1
τ=
z1 z2 z3 , |a|2 q
n = 1, 2, 3, . . . .
This leads to the orthogonality relation
2 ym (2|a| cos θ )yn (2|a| cos θ ) d θ 2π 0 |a|eiθ , z1 eiθ , z2 eiθ , z3 eiθ ; q |a| |a| |a| ∞ z1 z2 z3 z1 z2 z1 z3 z2 z3 1 z2 z3 q, z|a| ; q 2 q , z1 , z2 , z3 , |a|2 , |a|2 , |a|2 ; q |a|2 n ∞ |a|2n = δmn z1 z2 z3 z1 z2 z3 z1 z2 z1 z3 z2 z3 z1 , z2 , z3 , |a|2 , |a|2 , |a|2 , q; q , ; q 2 2 |a| q |a| 1 π
2iθ e ;q ∞
∞
2n
for m, n = 0, 1, 2, . . .. This is an orthogonality relation for the Askey-Wilson polynomials.
13.4 Orthogonality Relations
411
Similar to the situation in the preceding chapter this orthogonality relation for the Askey-Wilson polynomials leads to orthogonality relations for the special cases of continuous dual q-Hahn polynomials (z3 = 0), the q-Meixner-Pollaczek polynomials, the Al-Salam-Chihara polynomials or the continuous q-Laguerre polynomials (z2 = z3 = 0) and the continuous big q-Hermite polynomials (z1 = z2 = z3 = 0).
Chapter 14
Basic Hypergeometric Orthogonal Polynomials
In this chapter we deal with all families of basic hypergeometric orthogonal polynomials appearing in the q-analogue of the Askey scheme on the page 413. For each family of orthogonal polynomials we state the most important properties such as a representation as a basic hypergeometric function, orthogonality relation(s), the three-term recurrence relation, the second-order q-difference equation, the forward shift (or degree lowering) and backward shift (or degree raising) operator, a Rodrigues-type formula and some generating functions. Throughout this chapter we assume that 0 < q < 1. In each case we use the notation which seems to be most common in the literature. Moreover, in each case we also state the limit relations between various families of q-orthogonal polynomials and the limit relations (q → 1) to the classical hypergeometric orthogonal polynomials belonging to the Askey scheme on page 183. For notations the reader is referred to chapter 1.
14.1 Askey-Wilson Basic Hypergeometric Representation an pn (x; a, b, c, d|q) (ab, ac, ad; q)n −n q , abcdqn−1 , aeiθ , ae−iθ ; q, q , = 4 φ3 ab, ac, ad
x = cos θ .
(14.1.1)
The Askey-Wilson polynomials are q-analogues of the Wilson polynomials given by (9.1.1).
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5 14, © Springer-Verlag Berlin Heidelberg 2010
415
416
14 Basic Hypergeometric Orthogonal Polynomials
Orthogonality Relation If a, b, c, d are real, or occur in complex conjugate pairs if complex, and max(|a|, |b|, |c|, |d|) < 1, then we have the following orthogonality relation 1 2π
1
w(x) √ pm (x; a, b, c, d|q)pn (x; a, b, c, d|q) dx = hn δmn , −1 1 − x2
(14.1.2)
where 2 (e2iθ ; q)∞ w(x) := w(x; a, b, c, d|q) = iθ iθ iθ (ae , be , ce , deiθ ; q)∞ 1
=
1
h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) , h(x, a)h(x, b)h(x, c)h(x, d)
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
and hn =
x = cos θ
(abcdqn−1 ; q)n (abcdq2n ; q)∞ . (qn+1 , abqn , acqn , adqn , bcqn , bdqn , cdqn ; q)∞
If a > 1 and b, c, d are real or one is real and the other two are complex conjugates, max(|b|, |c|, |d|) < 1 and the pairwise products of a, b, c and d have absolute value less than 1, then we have another orthogonality relation given by: 1 2π
1
w(x) √ pm (x; a, b, c, d|q)pn (x; a, b, c, d|q) dx 1 − x2 + ∑ wk pm (xk ; a, b, c, d|q)pn (xk ; a, b, c, d|q) = hn δmn , (14.1.3)
−1
k 1
where w(x) and hn are as before, −1 aqk + aqk xk = 2 and wk =
(a−2 ; q)∞ (q, ab, ac, ad, a−1 b, a−1 c, a−1 d; q)∞ ×
q k (1 − a2 q2k )(a2 , ab, ac, ad; q)k . (1 − a2 )(q, ab−1 q, ac−1 q, ad −1 q; q)k abcd
14.1 Askey-Wilson
417
Recurrence Relation
2x p˜n (x) = An p˜n+1 (x) + a + a−1 − (An +Cn ) p˜n (x) +Cn p˜n−1 (x), where p˜n (x) := p˜n (x; a, b, c, d|q) =
(14.1.4)
an pn (x; a, b, c, d|q) (ab, ac, ad; q)n
⎧ (1 − abqn )(1 − acqn )(1 − adqn )(1 − abcdqn−1 ) ⎪ ⎪ = A ⎪ n ⎪ ⎨ a(1 − abcdq2n−1 )(1 − abcdq2n )
and
⎪ ⎪ a(1 − qn )(1 − bcqn−1 )(1 − bdqn−1 )(1 − cdqn−1 ) ⎪ ⎪ ⎩ Cn = . (1 − abcdq2n−2 )(1 − abcdq2n−1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
1
1 a + a−1 − (An +Cn ) pn (x) + An−1Cn pn−1 (x), 2 4
(14.1.5)
where pn (x; a, b, c, d|q) = 2n (abcdqn−1 ; q)n pn (x).
q-Difference Equation 1 1 1 1 ˜ aq 2 , bq 2 , cq 2 , dq 2 |q)Dq y(x) (1 − q)2 Dq w(x; ˜ a, b, c, d|q)y(x) = 0, + λn w(x; where w(x; ˜ a, b, c, d|q) := and
y(x) = pn (x; a, b, c, d|q), (14.1.6)
w(x; a, b, c, d|q) √ 1 − x2
λn = 4q−n+1 (1 − qn )(1 − abcdqn−1 ).
If we define (ab, ac, ad; q)n Pn (z) := 4 φ3 an
q−n , abcdqn−1 , az, az−1 ; q, q ab, ac, ad
then the q-difference equation can also be written in the form
418
14 Basic Hypergeometric Orthogonal Polynomials
q−n (1 − qn )(1 − abcdqn−1 )Pn (z)
= A(z)Pn (qz) − A(z) + A(z−1 ) Pn (z) + A(z−1 )Pn (q−1 z), where A(z) =
(14.1.7)
(1 − az)(1 − bz)(1 − cz)(1 − dz) . (1 − z2 )(1 − qz2 )
Forward Shift Operator δq pn (x; a, b, c, d|q) = −q− 2 n (1 − qn )(1 − abcdqn−1 )(eiθ − e−iθ ) 1
1
1
1
1
× pn−1 (x; aq 2 , bq 2 , cq 2 , dq 2 |q),
(14.1.8)
where x = cos θ , or equivalently 1
Dq pn (x; a, b, c, d|q) = 2q− 2 (n−1)
(1 − qn )(1 − abcdqn−1 ) 1−q 1
1
1
1
× pn−1 (x; aq 2 , bq 2 , cq 2 , dq 2 |q).
(14.1.9)
Backward Shift Operator δq [w(x; ˜ a, b, c, d|q)pn (x; a, b, c, d|q)] = q− 2 (n+1) (eiθ − e−iθ )w(x; ˜ aq− 2 , bq− 2 , cq− 2 , dq− 2 |q) 1
1
1
1
1
1
1
1
× pn+1 (x; aq− 2 , bq− 2 , cq− 2 , dq− 2 |q),
1
x = cos θ
(14.1.10)
or equivalently ˜ a, b, c, d|q)pn (x; a, b, c, d|q)] Dq [w(x; 1
=−
1 1 1 1 2q− 2 n w(x; ˜ aq− 2 , bq− 2 , cq− 2 , dq− 2 |q) 1−q 1
1
1
1
× pn+1 (x; aq− 2 , bq− 2 , cq− 2 , dq− 2 |q).
(14.1.11)
14.1 Askey-Wilson
419
Rodrigues-Type Formula w(x; ˜ a, b, c, d|q)pn (x; a, b, c, d|q) 1 1 1 1 q − 1 n 1 n(n−1) ˜ aq 2 n , bq 2 n , cq 2 n , dq 2 n |q) . (14.1.12) q4 (Dq )n w(x; = 2
Generating Functions
−iθ aeiθ , beiθ ce , de−iθ −iθ iθ ; q, e t 2 φ1 ; q, e t 2 φ1 ab cd ∞ pn (x; a, b, c, d|q) n t , x = cos θ . =∑ n=0 (ab, cd, q; q)n −iθ aeiθ , ceiθ be , de−iθ −iθ iθ ; q, e t 2 φ1 ; q, e t 2 φ1 ac bd ∞ pn (x; a, b, c, d|q) n t , x = cos θ . =∑ n=0 (ac, bd, q; q)n
(14.1.13)
−iθ −iθ aeiθ , deiθ be , ce −iθ iθ ; q, e t 2 φ1 ; q, e t 2 φ1 ad bc ∞ pn (x; a, b, c, d|q) n t , x = cos θ . =∑ n=0 (ad, bc, q; q)n
(14.1.14)
(14.1.15)
Limit Relations Askey-Wilson → Continuous Dual q-Hahn The continuous dual q-Hahn polynomials given by (14.3.1) simply follow from the Askey-Wilson polynomials given by (14.1.1) by setting d = 0 in (14.1.1): pn (x; a, b, c, 0|q) = pn (x; a, b, c|q).
(14.1.16)
420
14 Basic Hypergeometric Orthogonal Polynomials
Askey-Wilson → Continuous q-Hahn The continuous q-Hahn polynomials given by (14.4.1) can be obtained from the Askey-Wilson polynomials given by (14.1.1) by the substitutions θ → θ + φ , a → aeiφ , b → beiφ , c → ce−iφ and d → de−iφ : pn (cos(θ + φ ); aeiφ , beiφ , ce−iφ , de−iφ |q) = pn (cos(θ + φ ); a, b, c, d; q). (14.1.17) Askey-Wilson → Big q-Jacobi The big q-Jacobi polynomials given by (14.5.1) can be obtained from the AskeyWilson polynomials by setting x → 12 a−1 x, b = a−1 α q, c = a−1 γ q and d = aβ γ −1 in an pn (x; a, b, c, d|q) p˜n (x; a, b, c, d|q) = (ab, ac, ad; q)n given by (14.1.1) and then taking the limit a → 0: lim p˜n ( 12 a−1 x; a, a−1 α q, a−1 γ q, aβ γ −1 |q) = Pn (x; α , β , γ ; q).
a→0
(14.1.18)
Askey-Wilson → Continuous q-Jacobi If we take a = q 2 α + 4 , b = q 2 α + 4 , c = −q 2 β + 4 and d = −q 2 β + 4 in the definition (14.1.1) of the Askey-Wilson polynomials and change the normalization we find the continuous q-Jacobi polynomials given by (14.10.1): 1
1
1
3
1
1
1
q( 2 α + 4 )n pn (x; q 2 α + 4 , q 2 α + 4 , −q 2 β + 4 , −q 2 β + 4 |q) 1
1
1
1
1
3
1
1
1
3
(q, −q 2 (α +β +1) , −q 2 (α +β +2) ; q)n 1
1
(α ,β )
= Pn
3
(x|q).
(14.1.19)
Askey-Wilson → Continuous q-ultraspherical / Rogers 1
1
1
1
1
1
If we set a = β 2 , b = β 2 q 2 , c = −β 2 and d = −β 2 q 2 in the definition (14.1.1) of the Askey-Wilson polynomials and change the normalization we obtain the continuous q-ultraspherical (or Rogers) polynomials given by (14.10.17). In fact we have: 1
1
1
1
1
1
(β 2 ; q)n pn (x; β 2 , β 2 q 2 , −β 2 , −β 2 q 2 |q) 1
1
(β q 2 , −β , −β q 2 , q; q)n
= Cn (x; β |q).
(14.1.20)
14.1 Askey-Wilson
421
Askey-Wilson → Wilson To find the Wilson polynomials given by (9.1.1) from the Askey-Wilson polynomials we set a → qa , b → qb , c → qc , d → qd and eiθ = qix (or θ = ln qx ) in the definition (14.1.1) and take the limit q → 1: pn ( 12 qix + q−ix ; qa , qb , qc , qd |q) = Wn (x2 ; a, b, c, d). (14.1.21) lim q→1 (1 − q)3n
Remarks The q-Racah polynomials given by (14.2.1) and the Askey-Wilson polynomials given by (14.1.1) are related in the following way. If we substitute a2 = γδ q, b2 = α 2 γ −1 δ −1 q, c2 = β 2 γ −1 δ q, d 2 = γδ −1 q and e2iθ = γδ q2x+1 in the definition (14.1.1) of the Askey-Wilson polynomials we find: Rn (μ (x); α , β , γ , δ |q) 1
=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
(γδ q) 2 n pn (ν (x); γ 2 δ 2 q 2 , αγ − 2 δ − 2 q 2 , β γ − 2 δ 2 q 2 , γ 2 δ − 2 q 2 |q) , (α q, β δ q, γ q; q)n
where
1
1
1
1
ν (x) = 12 γ 2 δ 2 qx+ 2 + 12 γ − 2 δ − 2 q−x− 2 .
If we replace q by q−1 we find p˜n (x; a, b, c, d|q−1 ) = p˜n (x; a−1 , b−1 , c−1 , d −1 |q).
References [3], [16], [34], [51], [72], [73], [77], [80], [82], [83], [92], [115], [116], [236], [238], [256], [257], [271], [275], [278], [282], [286], [287], [290], [294], [298], [306], [323], [326], [347], [348], [351], [371], [408], [409], [414], [416], [418], [422], [442], [443], [445], [446], [448], [449], [451], [455], [474], [495], [507], [513].
422
14 Basic Hypergeometric Orthogonal Polynomials
14.2 q-Racah Basic Hypergeometric Representation Rn (μ (x); α , β , γ , δ |q) −n q , αβ qn+1 , q−x , γδ qx+1 ; q, q , = 4 φ3 α q, β δ q, γ q where
n = 0, 1, 2, . . . , N, (14.2.1)
μ (x) := q−x + γδ qx+1
and
α q = q−N
or
β δ q = q−N
or γ q = q−N ,
with N a nonnegative integer. Since k−1 (q−x , γδ qx+1 ; q)k = ∏ 1 − μ (x)q j + γδ q2 j+1 , j=0
it is clear that Rn (μ (x); α , β , γ , δ |q) is a polynomial of degree n in μ (x).
Orthogonality Relation (α q, β δ q, γ q, γδ q; q)x
N
∑ (q, α −1 γδ q, β −1 γ q, δ q; q)x
x=0
× where
(1 − γδ q2x+1 ) Rm (μ (x))Rn (μ (x)) = hn δmn , (αβ q)x (1 − γδ q) Rn (μ (x)) := Rn (μ (x); α , β , γ , δ |q)
and hn =
(α −1 β −1 γ , α −1 δ , β −1 , γδ q2 ; q)∞ (α −1 β −1 q−1 , α −1 γδ q, β −1 γ q, δ q; q)∞ ×
This implies
(1 − αβ q)(γδ q)n (q, αβ γ −1 q, αδ −1 q, β q; q)n . (1 − αβ q2n+1 ) (α q, αβ q, β δ q, γ q; q)n
(14.2.2)
14.2 q-Racah
423
⎧ −1 (β , γδ q2 ; q)N (1 − β q−N )(γδ q)n ⎪ ⎪ ⎪ ⎪ (β −1 γ q, δ q; q)N (1 − β q2n−N ) ⎪ ⎪ ⎪ ⎪ (q, β q, β γ −1 q−N , δ −1 q−N ; q)n ⎪ ⎪ × ⎪ ⎪ (β q−N , β δ q, γ q, q−N ; q)n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (αβ q2 , β γ −1 ; q)N (1 − αβ q)(β −1 γ q−N )n ⎪ ⎪ ⎨ (αβ γ −1 q, β q; q)N (1 − αβ q2n+1 ) hn = N+2 (q, αβ q , αβ γ −1 q, β q; q)n ⎪ ⎪ ⎪ × ⎪ ⎪ (α q, αβ q, γ q, q−N ; q)n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (αβ q2 , δ −1 ; q)N (1 − αβ q)(δ q−N )n ⎪ ⎪ ⎪ ⎪ (αδ −1 q, β q; q)N (1 − αβ q2n+1 ) ⎪ ⎪ ⎪ ⎪ (q, αβ qN+2 , αδ −1 q, β q; q)n ⎪ ⎩ × (α q, αβ q, β δ q, q−N ; q)n
if α q = q−N
if β δ q = q−N
if γ q = q−N .
Recurrence Relation − 1 − q−x 1 − γδ qx+1 Rn (μ (x)) = An Rn+1 (μ (x)) − (An +Cn ) Rn (μ (x)) +Cn Rn−1 (μ (x)), where
(14.2.3)
⎧ (1 − α qn+1 )(1 − αβ qn+1 )(1 − β δ qn+1 )(1 − γ qn+1 ) ⎪ ⎪ ⎪ An = ⎪ ⎨ (1 − αβ q2n+1 )(1 − αβ q2n+2 ) ⎪ ⎪ q(1 − qn )(1 − β qn )(γ − αβ qn )(δ − α qn ) ⎪ ⎪ . ⎩ Cn = (1 − αβ q2n )(1 − αβ q2n+1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 + γδ q − (An +Cn )] pn (x) + An−1Cn pn−1 (x), where Rn (μ (x); α , β , γ , δ |q) =
(αβ qn+1 ; q)n pn (μ (x)). (α q, β δ q, γ q; q)n
(14.2.4)
424
14 Basic Hypergeometric Orthogonal Polynomials
q-Difference Equation Δ [w(x − 1)B(x − 1)Δ y(x − 1)] − q−n (1 − qn )(1 − αβ qn+1 )w(x)y(x) = 0, where
(14.2.5)
y(x) = Rn (μ (x); α , β , γ , δ |q)
and w(x) := w(x; α , β , γ , δ |q) =
(α q, β δ q, γ q, γδ q; q)x (1 − γδ q2x+1 ) −1 −1 (q, α γδ q, β γ q, δ q; q)x (αβ q)x (1 − γδ q)
and B(x) as below. This q-difference equation can also be written in the form q−n (1 − qn )(1 − αβ qn+1 )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where and
(14.2.6)
y(x) = Rn (μ (x); α , β , γ , δ |q) ⎧ (1 − α qx+1 )(1 − β δ qx+1 )(1 − γ qx+1 )(1 − γδ qx+1 ) ⎪ ⎪ ⎪ B(x) = ⎪ ⎨ (1 − γδ q2x+1 )(1 − γδ q2x+2 ) ⎪ ⎪ q(1 − qx )(1 − δ qx )(β − γ qx )(α − γδ qx ) ⎪ ⎪ . ⎩ D(x) = (1 − γδ q2x )(1 − γδ q2x+1 )
Forward Shift Operator Rn (μ (x + 1); α , β , γ , δ |q) − Rn (μ (x); α , β , γ , δ |q) =
q−n−x (1 − qn )(1 − αβ qn+1 )(1 − γδ q2x+2 ) (1 − α q)(1 − β δ q)(1 − γ q) × Rn−1 (μ (x); α q, β q, γ q, δ |q)
(14.2.7)
Δ Rn (μ (x); α , β , γ , δ |q) Δ μ (x) −n+1 (1 − qn )(1 − αβ qn+1 ) q = (1 − q)(1 − α q)(1 − β δ q)(1 − γ q) × Rn−1 (μ (x); α q, β q, γ q, δ |q).
(14.2.8)
or equivalently
14.2 q-Racah
425
Backward Shift Operator (1 − α qx )(1 − β δ qx )(1 − γ qx )(1 − γδ qx )Rn (μ (x); α , β , γ , δ |q) − (1 − qx )(1 − δ qx )(α − γδ qx )(β − γ qx )Rn (μ (x − 1); α , β , γ , δ |q) = qx (1 − α )(1 − β δ )(1 − γ )(1 − γδ q2x ) × Rn+1 (μ (x); α q−1 , β q−1 , γ q−1 , δ |q)
(14.2.9)
or equivalently ∇ [w(x; ˜ α , β , γ , δ |q)Rn (μ (x); α , β , γ , δ |q)] ∇μ (x) 1 w(x; ˜ α q−1 , β q−1 , γ q−1 , δ |q) = (1 − q)(1 − γδ ) × Rn+1 (μ (x); α q−1 , β q−1 , γ q−1 , δ |q), where w(x; ˜ α , β , γ , δ |q) =
(14.2.10)
(α q, β δ q, γ q, γδ q; q)x . (q, α −1 γδ q, β −1 γ q, δ q; q)x (αβ )x
Rodrigues-Type Formula w(x; ˜ α , β , γ , δ |q)Rn (μ (x); α , β , γ , δ |q) n ˜ α qn , β qn , γ qn , δ |q)] , = (1 − q)n (γδ q; q)n ∇μ [w(x; where ∇μ :=
(14.2.11)
∇ . ∇μ (x)
Generating Functions For x = 0, 1, 2, . . . , N we have −x q , αγ −1 δ −1 q−x β δ qx+1 , γ qx+1 x+1 −x ; q, γδ q t 2 φ1 ; q, q t 2 φ1 αq βq (β δ q, γ q; q)n Rn (μ (x); α , β , γ , δ |q)t n , ( β q, q; q) n n=0 N
=
∑
if
β δ q = q−N
or γ q = q−N .
(14.2.12)
426
14 Basic Hypergeometric Orthogonal Polynomials
2 φ1
q−x , β γ −1 q−x ; q, γδ qx+1t βδq
2 φ1
α qx+1 , γ qx+1 ; q, q−xt αδ −1 q
(α q, γ q; q)n
N
∑ (αδ −1 q, q; q)n Rn (μ (x); α , β , γ , δ |q)t n ,
=
n=0
if 2 φ1
q−x , δ −1 q−x ; q, γδ qx+1t γq
N
=
α q = q−N
or
2 φ1
γ q = q−N .
α qx+1 , β δ qx+1 ; q, q−xt αβ γ −1 q
(14.2.13)
(α q, β δ q; q)n
∑ (αβ γ −1 q, q; q)n Rn (μ (x); α , β , γ , δ |q)t n ,
n=0
if
α q = q−N
or
β δ q = q−N .
(14.2.14)
Limit Relations q-Racah → Big q-Jacobi The big q-Jacobi polynomials given by (14.5.1) can be obtained from the q-Racah polynomials by setting δ = 0 in the definition (14.2.1): Rn (μ (x); a, b, c, 0|q) = Pn (q−x ; a, b, c; q).
(14.2.15)
q-Racah → q-Hahn The q-Hahn polynomials follow from the q-Racah polynomials by the substitution δ = 0 and γ q = q−N in the definition (14.2.1) of the q-Racah polynomials: Rn (μ (x); α , β , q−N−1 , 0|q) = Qn (q−x ; α , β , N|q).
(14.2.16)
Another way to obtain the q-Hahn polynomials from the q-Racah polynomials is by setting γ = 0 and δ = β −1 q−N−1 in the definition (14.2.1): Rn (μ (x); α , β , 0, β −1 q−N−1 |q) = Qn (q−x ; α , β , N|q).
(14.2.17)
And if we take α q = q−N , β → β γ qN+1 and δ = 0 in the definition (14.2.1) of the q-Racah polynomials we find the q-Hahn polynomials given by (14.6.1) in the following way: Rn (μ (x); q−N−1 , β γ qN+1 , γ , 0|q) = Qn (q−x ; γ , β , N|q).
(14.2.18)
14.2 q-Racah
427
Note that μ (x) = q−x in each case. q-Racah → Dual q-Hahn To obtain the dual q-Hahn polynomials from the q-Racah polynomials we have to take β = 0 and α q = q−N in (14.2.1): Rn (μ (x); q−N−1 , 0, γ , δ |q) = Rn (μ (x); γ , δ , N|q), with
(14.2.19)
μ (x) = q−x + γδ qx+1 .
We may also take α = 0 and β = δ −1 q−N−1 in (14.2.1) to obtain the dual q-Hahn polynomials from the q-Racah polynomials: Rn (μ (x); 0, δ −1 q−N−1 , γ , δ |q) = Rn (μ (x); γ , δ , N|q), with
(14.2.20)
μ (x) = q−x + γδ qx+1 .
And if we take γ q = q−N , δ → αδ qN+1 and β = 0 in the definition (14.2.1) of the q-Racah polynomials we find the dual q-Hahn polynomials given by (14.7.1) in the following way: Rn (μ (x); α , 0, q−N−1 , αδ qN+1 |q) = Rn (μ (x); α , δ , N|q), with
(14.2.21)
μ (x) = q−x + αδ qx+1 .
q-Racah → q-Krawtchouk The q-Krawtchouk polynomials given by (14.15.1) can be obtained from the q-Racah polynomials by setting α q = q−N , β = −pqN and γ = δ = 0 in the definition (14.2.1) of the q-Racah polynomials: Rn (q−x ; q−N−1 , −pqN , 0, 0|q) = Kn (q−x ; p, N; q).
(14.2.22)
Note that μ (x) = q−x in this case. q-Racah → Dual q-Krawtchouk The dual q-Krawtchouk polynomials given by (14.17.1) easily follow from the qRacah polynomials given by (14.2.1) by using the substitutions α = β = 0, γ q = q−N and δ = c:
428
14 Basic Hypergeometric Orthogonal Polynomials
Rn (μ (x); 0, 0, q−N−1 , c|q) = Kn (λ (x); c, N|q). Note that
(14.2.23)
μ (x) = λ (x) = q−x + cqx−N .
q-Racah → Racah If we set α → qα , β → qβ , γ → qγ , δ → qδ in the definition (14.2.1) of the q-Racah polynomials and let q → 1 we easily obtain the Racah polynomials given by (9.2.1): lim Rn (μ (x); qα , qβ , qγ , qδ |q) = Rn (λ (x); α , β , γ , δ ),
q→1
where
(14.2.24)
⎧ −x x+γ +δ +1 ⎨ μ (x) = q + q ⎩
λ (x) = x(x + γ + δ + 1).
Remarks The Askey-Wilson polynomials given by (14.1.1) and the q-Racah polynomials given by (14.2.1) are related in the following way. If we substitute α = abq−1 , β = cdq−1 , γ = adq−1 , δ = ad −1 and qx = a−1 e−iθ in the definition (14.2.1) of the q-Racah polynomials we find:
μ (x) = 2a cos θ and Rn (2a cos θ ; abq−1 , cdq−1 , adq−1 , ad −1 |q) =
an pn (x; a, b, c, d|q) . (ab, ac, ad; q)n
If we replace q by q−1 we find Rn (μ (x); α , β , γ , δ |q−1 ) = Rn (μ˜ (x); α −1 , β −1 , γ −1 , δ −1 |q), where
μ˜ (x) := q−x + γ −1 δ −1 qx+1 .
References [16], [27], [34], [70], [72], [80], [140], [142], [204], [233], [235], [238], [271], [301], [345], [416], [426], [442].
14.3 Continuous Dual q-Hahn
429
14.3 Continuous Dual q-Hahn Basic Hypergeometric Representation an pn (x; a, b, c|q) = 3 φ2 (ab, ac; q)n
q−n , aeiθ , ae−iθ ; q, q , ab, ac
x = cos θ .
(14.3.1)
Orthogonality Relation If a, b, c are real, or one is real and the other two are complex conjugates, and max(|a|, |b|, |c|) < 1, then we have the following orthogonality relation 1 2π
1
w(x) √ pm (x; a, b, c|q)pn (x; a, b, c|q) dx = hn δmn , −1 1 − x2
(14.3.2)
where 2 1 1 2 2 (e2iθ ; q)∞ = h(x, 1)h(x, −1)h(x, q )h(x, −q ) , w(x) := w(x; a, b, c|q) = iθ iθ iθ h(x, a)h(x, b)h(x, c) (ae , be , ce ; q)∞ with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
and hn =
1 (qn+1 , abqn , acqn , bcqn ; q)∞
x = cos θ
.
If a > 1 and b and c are real or complex conjugates, max(|b|, |c|) < 1 and the pairwise products of a, b and c have absolute value less than 1, then we have another orthogonality relation given by: 1 2π
1
w(x) √ pm (x; a, b, c|q)pn (x; a, b, c|q) dx −1 1 − x2 + ∑ wk pm (xk ; a, b, c|q)pn (xk ; a, b, c|q) = hn δmn , k 1
where w(x) and hn are as before, xk =
−1 aqk + aqk 2
(14.3.3)
430
14 Basic Hypergeometric Orthogonal Polynomials
and k (a−2 ; q)∞ (1 − a2 q2k )(a2 , ab, ac; q)k wk = (−1)k q−(2) (q, ab, ac, a−1 b, a−1 c; q)∞ (1 − a2 )(q, ab−1 q, ac−1 q; q)k
1 a2 bc
k .
Recurrence Relation
2x p˜n (x) = An p˜n+1 (x) + a + a−1 − (An +Cn ) p˜n (x) +Cn p˜n−1 (x), where p˜n (x) := and
(14.3.4)
an pn (x; a, b, c|q) (ab, ac; q)n
⎧ ⎨ An = a−1 (1 − abqn )(1 − acqn ) ⎩
Cn = a(1 − qn )(1 − bcqn−1 ).
Normalized Recurrence Relation
xpn (x) = pn+1 (x) +
1
a + a−1 − (An +Cn ) pn (x) 2
1 + (1 − qn )(1 − abqn−1 ) 4 × (1 − acqn−1 )(1 − bcqn−1 )pn−1 (x),
(14.3.5)
where pn (x; a, b, c|q) = 2n pn (x).
q-Difference Equation 1 1 1 ˜ aq 2 , bq 2 , cq 2 |q)Dq y(x) (1 − q)2 Dq w(x; ˜ a, b, c|q)y(x) = 0, + 4q−n+1 (1 − qn )w(x; where y(x) = pn (x; a, b, c|q) and
(14.3.6)
14.3 Continuous Dual q-Hahn
431
w(x; ˜ a, b, c|q) := If we define (ab, ac; q)n Pn (z) := 3 φ2 an
w(x; a, b, c|q) √ . 1 − x2
q−n , az, az−1 ; q, q ab, ac
then the q-difference equation can also be written in the form
q−n (1−qn )Pn (z) = A(z)Pn (qz)− A(z) + A(z−1 ) Pn (z)+A(z−1 )Pn (q−1 z), (14.3.7) where A(z) =
(1 − az)(1 − bz)(1 − cz) . (1 − z2 )(1 − qz2 )
Forward Shift Operator δq pn (x; a, b, c|q) = −q− 2 n (1 − qn )(eiθ − e−iθ ) 1
1
1
1
x = cos θ (14.3.8)
× pn−1 (x; aq 2 , bq 2 , cq 2 |q), or equivalently 1
Dq pn (x; a, b, c|q) = 2q− 2 (n−1)
1 1 1 1 − qn pn−1 (x; aq 2 , bq 2 , cq 2 |q). 1−q
(14.3.9)
Backward Shift Operator δq [w(x; ˜ a, b, c|q)pn (x; a, b, c|q)] = q− 2 (n+1) (eiθ − e−iθ )w(x; ˜ aq− 2 , bq− 2 , cq− 2 |q) 1
1
1
1
1
× pn+1 (x; aq− 2 , bq− 2 , cq− 2 |q),
1
1
x = cos θ
(14.3.10)
or equivalently ˜ a, b, c|q)pn (x; a, b, c|q)] Dq [w(x; 1
1 1 1 1 1 1 2q− 2 n w(x; ˜ aq− 2 , bq− 2 , cq− 2 |q)pn+1 (x; aq− 2 , bq− 2 , cq− 2 |q). (14.3.11) =− 1−q
432
14 Basic Hypergeometric Orthogonal Polynomials
Rodrigues-Type Formula w(x; ˜ a, b, c|q)pn (x; a, b, c|q) 1 1 1 q − 1 n 1 n(n−1) ˜ aq 2 n , bq 2 n , cq 2 n |q) . q4 (Dq )n w(x; = 2
(14.3.12)
Generating Functions iθ iθ ae , be (ct; q)∞ −iθ ; q, e t 2 φ1 ab (eiθ t; q)∞ ∞ pn (x; a, b, c|q) n t , x = cos θ . =∑ n=0 (ab, q; q)n iθ iθ ae , ce (bt; q)∞ −iθ ; q, e t 2 φ1 ac (eiθ t; q)∞ ∞ pn (x; a, b, c|q) n t , x = cos θ . =∑ n=0 (ac, q; q)n iθ iθ be , ce (at; q)∞ −iθ ; q, e t 2 φ1 bc (eiθ t; q)∞ ∞ pn (x; a, b, c|q) n t , x = cos θ . =∑ n=0 (bc, q; q)n aeiθ , ae−iθ , 0 ; q, t ab, ac n ∞ n n ( (−1) a q 2) =∑ pn (x; a, b, c|q)t n , n=0 (ab, ac, q; q)n
(14.3.13)
(14.3.14)
(14.3.15)
(t; q)∞ · 3 φ2
x = cos θ .
(14.3.16)
Limit Relations Askey-Wilson → Continuous Dual q-Hahn The continuous dual q-Hahn polynomials given by (14.3.1) simply follow from the Askey-Wilson polynomials given by (14.1.1) by setting d = 0 in (14.1.1):
14.4 Continuous q-Hahn
433
pn (x; a, b, c, 0|q) = pn (x; a, b, c|q). Continuous Dual q-Hahn → Al-Salam-Chihara The Al-Salam-Chihara polynomials given by (14.8.1) simply follow from the continuous dual q-Hahn polynomials by taking c = 0 in the definition (14.3.1) of the continuous dual q-Hahn polynomials: pn (x; a, b, 0|q) = Qn (x; a, b|q).
(14.3.17)
Continuous Dual q-Hahn → Continuous Dual Hahn To find the continuous dual Hahn polynomials given by (9.3.1) from the continuous dual q-Hahn polynomials we set a → qa , b → qb , c → qc and eiθ = qix (or θ = ln qx ) in the definition (14.3.1) and take the limit q → 1: pn ( 12 qix + q−ix ; qa , qb , qc |q) = Sn (x2 ; a, b, c). (14.3.18) lim q→1 (1 − q)2n
References [73], [260].
14.4 Continuous q-Hahn Basic Hypergeometric Representation (aeiφ )n pn (x; a, b, c, d; q) (abe2iφ , ac, ad; q)n q−n , abcdqn−1 , aei(θ +2φ ) , ae−iθ ; q, q , = 4 φ3 abe2iφ , ac, ad
x = cos(θ + φ ). (14.4.1)
434
14 Basic Hypergeometric Orthogonal Polynomials
Orthogonality Relation If c = a and d = b then we have, if a and b are real and max(|a|, |b|) < 1, or if b = a and |a| < 1: 1 4π =
π −π
w(cos(θ + φ ))pm (cos(θ + φ ); a, b, c, d; q)pn (cos(θ + φ ); a, b, c, d; q) d θ
(abcdqn−1 ; q)n (abcdq2n ; q)∞ n+1 n 2i (q , abq e φ , acqn , adqn , bcqn , bdqn , cdqn e−2iφ ; q)∞
δmn ,
(14.4.2)
where 2 (e2i(θ +φ ) ; q)∞ w(x) := w(x; a, b, c, d; q) = i(θ +2φ ) i(θ +2φ ) iθ i θ (ae , be , ce , de ; q)∞ 1
=
1
h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) , h(x, aeiφ )h(x, beiφ )h(x, ce−iφ )h(x, de−iφ )
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α ei(θ +φ ) , α e−i(θ +φ ) ; q , ∞
k=0
x = cos(θ + φ ).
Recurrence Relation
2x p˜n (x) = An p˜n+1 (x) + aeiφ + a−1 e−iφ − (An +Cn ) p˜n (x) +Cn p˜n−1 (x), (14.4.3) where p˜n (x) := p˜n (x; a, b, c, d; q) = and
(aeiφ )n pn (x; a, b, c, d; q) (abe2iφ , ac, ad; q)n
⎧ (1 − abe2iφ qn )(1 − acqn )(1 − adqn )(1 − abcdqn−1 ) ⎪ ⎪ ⎪ An = ⎪ ⎨ aeiφ (1 − abcdq2n−1 )(1 − abcdq2n ) ⎪ ⎪ aeiφ (1 − qn )(1 − bcqn−1 )(1 − bdqn−1 )(1 − cde−2iφ qn−1 ) ⎪ ⎪ ⎩ Cn = . (1 − abcdq2n−2 )(1 − abcdq2n−1 )
14.4 Continuous q-Hahn
435
Normalized Recurrence Relation
xpn (x) = pn+1 (x) +
1 iφ ae + a−1 e−iφ − (An +Cn ) pn (x) 2
1 + An−1Cn pn−1 (x), 4
(14.4.4)
where pn (x; a, b, c, d; q) = 2n (abcdqn−1 ; q)n pn (x).
q-Difference Equation 1 1 1 1 ˜ aq 2 , bq 2 , cq 2 , dq 2 ; q)Dq y(x) (1 − q)2 Dq w(x; ˜ a, b, c, d; q)y(x) = 0, + λn w(x; where w(x; ˜ a, b, c, d; q) := and
y(x) = pn (x; a, b, c, d; q), (14.4.5)
w(x; a, b, c, d; q) √ 1 − x2
λn = 4q−n+1 (1 − qn )(1 − abcdqn−1 ).
Forward Shift Operator δq pn (x; a, b, c, d; q) = −q− 2 n (1 − qn )(1 − abcdqn−1 )(ei(θ +φ ) − e−i(θ +φ ) ) 1
1
1
1
1
× pn−1 (x; aq 2 , bq 2 , cq 2 , dq 2 ; q),
x = cos(θ + φ )
(14.4.6)
or equivalently 1
Dq pn (x; a, b, c, d; q) = 2q− 2 (n−1)
(1 − qn )(1 − abcdqn−1 ) 1−q 1
1
1
1
× pn−1 (x; aq 2 , bq 2 , cq 2 , dq 2 ; q).
(14.4.7)
436
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator δq [w(x; ˜ a, b, c, d; q)pn (x; a, b, c, d; q)] = q− 2 (n+1) (ei(θ +φ ) − e−i(θ +φ ) )w(x; ˜ aq− 2 , bq− 2 , cq− 2 , dq− 2 ; q) 1
1
1
1
1
1
1
× pn+1 (x; aq− 2 , bq− 2 , cq− 2 , dq− 2 ; q),
1
1
x = cos(θ + φ )
(14.4.8)
or equivalently ˜ a, b, c, d; q)pn (x; a, b, c, d; q)] Dq [w(x; 1
=−
1 1 1 1 2q− 2 n w(x; ˜ aq− 2 , bq− 2 , cq− 2 , dq− 2 ; q) 1−q 1
1
1
1
× pn+1 (x; aq− 2 , bq− 2 , cq− 2 , dq− 2 ; q).
(14.4.9)
Rodrigues-Type Formula w(x; ˜ a, b, c, d; q)pn (x; a, b, c, d; q) 1 1 1 1 q − 1 n 1 n(n−1) ˜ aq 2 n , bq 2 n , cq 2 n , dq 2 n ; q) . (14.4.10) q4 (Dq )n w(x; = 2
Generating Functions 2 φ1
aei(θ +2φ ) , bei(θ +2φ ) ; q, e−i(θ +φ )t abe2iφ ce−i(θ +2φ ) , de−i(θ +2φ ) i(θ +φ ) × 2 φ1 ; q, e t cde−2iφ
∞
=
pn (x; a, b, c, d; q)t n
∑ (abe2iφ , cde−2iφ , q; q)n ,
x = cos(θ + φ ).
(14.4.11)
n=0
2 φ1
aei(θ +2φ ) , ceiθ ; q, e−i(θ +φ )t ac
∞
=
∑
n=0
pn (x; a, b, c, d; q) n t , (ac, bd, q; q)n
2 φ1
be−iθ , de−i(θ +2φ ) ; q, ei(θ +φ )t bd
x = cos(θ + φ ).
(14.4.12)
14.4 Continuous q-Hahn
2 φ1
aei(θ +2φ ) , deiθ ; q, e−i(θ +φ )t ad
∞
=
437
∑
n=0
pn (x; a, b, c, d; q) n t , (ad, bc, q; q)n
2 φ1
be−iθ , ce−i(θ +2φ ) ; q, ei(θ +φ )t bc
x = cos(θ + φ ).
(14.4.13)
Limit Relations Askey-Wilson → Continuous q-Hahn The continuous q-Hahn polynomials given by (14.4.1) can be obtained from the Askey-Wilson polynomials given by (14.1.1) by the substitutions θ → θ + φ , a → aeiφ , b → beiφ , c → ce−iφ and d → de−iφ : pn (cos(θ + φ ); aeiφ , beiφ , ce−iφ , de−iφ |q) = pn (cos(θ + φ ); a, b, c, d; q). Continuous q-Hahn → q-Meixner-Pollaczek The q-Meixner-Pollaczek polynomials given by (14.9.1) simply follow from the continuous q-Hahn polynomials if we set d = a and b = c = 0 in the definition (14.4.1) of the continuous q-Hahn polynomials: pn (cos(θ + φ ); a, 0, 0, a; q) = Pn (cos(θ + φ ); a|q). (q; q)n
(14.4.14)
Continuous q-Hahn → Continuous Hahn If we set a → qa , b → qb , c → qc , d → qd and e−iθ = qix (or θ = ln q−x ) in the definition (14.4.1) of the continuous q-Hahn polynomials and take the limit q → 1 we find the continuous Hahn polynomials given by (9.4.1) in the following way: lim
q→1
pn (cos(ln q−x + φ ); qa , qb , qc , qd ; q) = (−2 sin φ )n pn (x; a, b, c, d). (1 − q)n (q; q)n
(14.4.15)
438
14 Basic Hypergeometric Orthogonal Polynomials
Remark If we replace q by q−1 we find p˜n (x; a, b, c, d; q−1 ) = p˜n (x; a−1 , b−1 , c−1 , d −1 ; q).
References [34], [72], [238].
14.5 Big q-Jacobi Basic Hypergeometric Representation Pn (x; a, b, c; q) = 3 φ2
q−n , abqn+1 , x ; q, q . aq, cq
(14.5.1)
Orthogonality Relation For 0 < aq < 1, 0 ≤ bq < 1 and c < 0 we have aq −1 −1 (a x, c x; q)∞ cq
(x, bc−1 x; q)∞
= aq(1 − q) ×
Pm (x; a, b, c; q)Pn (x; a, b, c; q) dq x
(q, abq2 , a−1 c, ac−1 q; q)∞ (aq, bq, cq, abc−1 q; q)∞
n (1 − abq) (q, bq, abc−1 q; q)n (−acq2 )n q(2) δmn . (1 − abq2n+1 ) (aq, abq, cq; q)n
(14.5.2)
Recurrence Relation (x − 1)Pn (x; a, b, c; q) = An Pn+1 (x; a, b, c; q) − (An +Cn ) Pn (x; a, b, c; q) (14.5.3) +Cn Pn−1 (x; a, b, c; q), where
14.5 Big q-Jacobi
439
⎧ (1 − aqn+1 )(1 − abqn+1 )(1 − cqn+1 ) ⎪ ⎪ = A ⎪ n ⎪ ⎨ (1 − abq2n+1 )(1 − abq2n+2 ) ⎪ ⎪ (1 − qn )(1 − abc−1 qn )(1 − bqn ) ⎪ ⎪ ⎩ Cn = −acqn+1 . (1 − abq2n )(1 − abq2n+1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 − (An +Cn )] pn (x) + An−1Cn pn−1 (x), where Pn (x; a, b, c; q) =
(14.5.4)
(abqn+1 ; q)n pn (x). (aq, cq; q)n
q-Difference Equation q−n (1 − qn )(1 − abqn+1 )x2 y(x) = B(x)y(qx) − [B(x) + D(x)] y(x) + D(x)y(q−1 x),
(14.5.5)
where y(x) = Pn (x; a, b, c; q) ⎧ ⎨ B(x) = aq(x − 1)(bx − c)
and
⎩
D(x) = (x − aq)(x − cq).
Forward Shift Operator Pn (x; a, b, c; q) − Pn (qx; a, b, c; q) =
q−n+1 (1 − qn )(1 − abqn+1 ) xPn−1 (qx; aq, bq, cq; q) (1 − aq)(1 − cq)
(14.5.6)
or equivalently Dq Pn (x; a, b, c; q) =
q−n+1 (1 − qn )(1 − abqn+1 ) Pn−1 (qx; aq, bq, cq; q). (1 − q)(1 − aq)(1 − cq)
(14.5.7)
440
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator (x − a)(x − c)Pn (x; a, b, c; q) − a(x − 1)(bx − c)Pn (qx; a, b, c; q) (14.5.8) = (1 − a)(1 − c)xPn+1 (x; aq−1 , bq−1 , cq−1 ; q) or equivalently Dq [w(x; a, b, c; q)Pn (x; a, b, c; q)] (1 − a)(1 − c) w(x; aq−1 , bq−1 , cq−1 ; q) = ac(1 − q) × Pn+1 (x; aq−1 , bq−1 , cq−1 ; q), where w(x; a, b, c; q) =
(14.5.9)
(a−1 x, c−1 x; q)∞ . (x, bc−1 x; q)∞
Rodrigues-Type Formula w(x; a, b, c; q)Pn (x; a, b, c; q) =
an cn qn(n+1) (1 − q)n (Dq )n [w(x; aqn , bqn , cqn ; q)] . (aq, cq; q)n
(14.5.10)
Generating Functions −1 aqx−1 , 0 bc x ; q, xt 1 φ1 ; q, cqt 2 φ1 aq bq ∞ (cq; q)n =∑ Pn (x; a, b, c; q)t n . (bq, q; q)n n=0
−1 cqx−1 , 0 bc x ; q, xt 1 φ1 ; q, aqt 2 φ1 cq abc−1 q ∞ (aq; q)n =∑ Pn (x; a, b, c; q)t n . −1 n=0 (abc q, q; q)n
(14.5.11)
(14.5.12)
14.5 Big q-Jacobi
441
Limit Relations Askey-Wilson → Big q-Jacobi The big q-Jacobi polynomials given by (14.5.1) can be obtained from the AskeyWilson polynomials by setting x → 12 a−1 x, b = a−1 α q, c = a−1 γ q and d = aβ γ −1 in an pn (x; a, b, c, d|q) p˜n (x; a, b, c, d|q) = (ab, ac, ad; q)n given by (14.1.1) and then taking the limit a → 0: lim p˜n ( 12 a−1 x; a, a−1 α q, a−1 γ q, aβ γ −1 |q) = Pn (x; α , β , γ ; q).
a→0
q-Racah → Big q-Jacobi The big q-Jacobi polynomials given by (14.5.1) can be obtained from the q-Racah polynomials by setting δ = 0 in the definition (14.2.1): Rn (μ (x); a, b, c, 0|q) = Pn (q−x ; a, b, c; q). Big q-Jacobi → Big q-Laguerre If we set b = 0 in the definition (14.5.1) of the big q-Jacobi polynomials we obtain the big q-Laguerre polynomials given by (14.11.1): Pn (x; a, 0, c; q) = Pn (x; a, c; q).
(14.5.13)
Big q-Jacobi → Little q-Jacobi The little q-Jacobi polynomials given by (14.12.1) can be obtained from the big qJacobi polynomials by the substitution x → cqx in the definition (14.5.1) and then by the limit c → −∞: lim Pn (cqx; a, b, c; q) = pn (x; a, b|q).
c→−∞
(14.5.14)
Big q-Jacobi → q-Meixner If we set b = −a−1 cd −1 (with d > 0) in the definition (14.5.1) of the big q-Jacobi polynomials and take the limit c → −∞ we obtain the q-Meixner polynomials given by (14.13.1):
442
14 Basic Hypergeometric Orthogonal Polynomials
lim Pn (q−x ; a, −a−1 cd −1 , c; q) = Mn (q−x ; a, d; q).
c→−∞
(14.5.15)
Big q-Jacobi → Jacobi If we set c = 0, a = qα and b = qβ in the definition (14.5.1) of the big q-Jacobi polynomials and let q → 1 we find the Jacobi polynomials given by (9.8.1): lim Pn (x; qα , qβ , 0; q) =
q→1
(α ,β )
Pn
(2x − 1)
(α ,β ) Pn (1)
.
(14.5.16)
If we take c = −qγ for arbitrary real γ instead of c = 0 we find lim Pn (x; qα , qβ , −qγ ; q) =
q→1
(α ,β )
Pn
(x)
(α ,β ) Pn (1)
.
(14.5.17)
Remarks The big q-Jacobi polynomials with c = 0 and the little q-Jacobi polynomials given by (14.12.1) are related in the following way: Pn (x; a, b, 0; q) =
n (bq; q)n (−1)n an qn+(2) pn (a−1 q−1 x; b, a|q). (aq; q)n
Sometimes the big q-Jacobi polynomials are defined in terms of four parameters instead of three. In fact the polynomials given by the definition −n q , abqn+1 , ac−1 qx ; q, q Pn (x; a, b, c, d; q) = 3 φ2 aq, −ac−1 dq are orthogonal on the interval [−d, c] with respect to the weight function (c−1 qx, −d −1 qx; q)∞ dq x. (ac−1 qx, −bd −1 qx; q)∞ These polynomials are not really different from those given by (14.5.1) since we have Pn (x; a, b, c, d; q) = Pn (ac−1 qx; a, b, −ac−1 d; q) and Pn (x; a, b, c; q) = Pn (x; a, b, aq, −cq; q).
14.5 Big q-Jacobi
443
References [12], [16], [34], [79], [80], [157], [211], [238], [256], [259], [261], [271], [295], [298], [305], [320], [345], [348], [351], [408], [416], [419], [420], [421], [474], [482].
Special Case 14.5.1 Big q-Legendre Basic Hypergeometric Representation The big q-Legendre polynomials are big q-Jacobi polynomials with a = b = 1: −n n+1 q ,q ,x ; q, q . (14.5.18) Pn (x; c; q) = 3 φ2 q, cq
Orthogonality Relation q cq
Pm (x; c; q)Pn (x; c; q) dq x
= q(1 − c)
n (1 − q) (c−1 q; q)n (−cq2 )n q(2) δmn , (1 − q2n+1 ) (cq; q)n
c < 0. (14.5.19)
Recurrence Relation (x − 1)Pn (x; c; q) = An Pn+1 (x; c; q) − (An +Cn ) Pn (x; c; q) +Cn Pn−1 (x; c; q), where
⎧ (1 − qn+1 )(1 − cqn+1 ) ⎪ ⎪ An = ⎪ ⎪ ⎨ (1 + qn+1 )(1 − q2n+1 ) ⎪ ⎪ (1 − qn )(1 − c−1 qn ) ⎪ ⎪ ⎩ Cn = −cqn+1 . (1 + qn )(1 − q2n+1 )
(14.5.20)
444
14 Basic Hypergeometric Orthogonal Polynomials
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 − (An +Cn )] pn (x) + An−1Cn pn−1 (x), where Pn (x; c; q) =
(14.5.21)
(qn+1 ; q)n pn (x). (q, cq; q)n
q-Difference Equation q−n (1 − qn )(1 − qn+1 )x2 y(x) = B(x)y(qx) − [B(x) + D(x)] y(x) + D(x)y(q−1 x),
(14.5.22)
where y(x) = Pn (x; c; q) ⎧ ⎨ B(x) = q(x − 1)(x − c)
and
⎩
D(x) = (x − q)(x − cq).
Rodrigues-Type Formula
cn qn(n+1) (1 − q)n (Dq )n (q−n x, c−1 q−n x; q)n (q, cq; q)n
(1 − q)n (Dq )n (qx−1 , cqx−1 ; q)n x2n . = (q, cq; q)n
Pn (x; c; q) =
(14.5.23)
Generating Functions 2 φ1
qx−1 , 0 ; q, xt q
2 φ1
1 φ1
cqx−1 , 0 ; q, xt cq
c−1 x ; q, cqt q
1 φ1
∞
=
c−1 x ; q, qt c−1 q
(cq; q)n
∑ (q, q; q)n Pn (x; c; q)t n .
(14.5.24)
n=0
∞
=
Pn (x; c; q)
∑ (c−1 q; q)n t n .
n=0
(14.5.25)
14.6 q-Hahn
445
Limit Relations Big q-Legendre → Legendre / Spherical If we set c = 0 in the definition (14.5.18) of the big q-Legendre polynomials and let q → 1 we simply obtain the Legendre (or spherical) polynomials given by (9.8.62): lim Pn (x; 0; q) = Pn (2x − 1).
(14.5.26)
q→1
If we take c = −qγ for arbitrary real γ instead of c = 0 we find lim Pn (x; −qγ ; q) = Pn (x).
(14.5.27)
q→1
References [324], [345].
14.6 q-Hahn Basic Hypergeometric Representation
−x
Qn (q ; α , β , N|q) = 3 φ2
q−n , αβ qn+1 , q−x ; q, q , α q, q−N
n = 0, 1, 2, . . . , N. (14.6.1)
Orthogonality Relation For 0 < α q < 1 and 0 < β q < 1, or for α > q−N and β > q−N , we have N
(α q, q−N ; q)x
∑ (q, β −1 q−N ; q)x (αβ q)−x Qm (q−x ; α , β , N|q)Qn (q−x ; α , β , N|q)
x=0
=
(αβ q2 ; q)N (q, αβ qN+2 , β q; q)n (1 − αβ q)(−α q)n (n)−Nn q2 δmn . (14.6.2) (β q; q)N (α q)N (α q, αβ q, q−N ; q)n (1 − αβ q2n+1 )
446
14 Basic Hypergeometric Orthogonal Polynomials
Recurrence Relation − 1 − q−x Qn (q−x ) = An Qn+1 (q−x ) − (An +Cn ) Qn (q−x ) +Cn Qn−1 (q−x ), where and
(14.6.3)
Qn (q−x ) := Qn (q−x ; α , β , N|q) ⎧ (1 − qn−N )(1 − α qn+1 )(1 − αβ qn+1 ) ⎪ ⎪ = A ⎪ n ⎪ ⎨ (1 − αβ q2n+1 )(1 − αβ q2n+2 ) ⎪ ⎪ α qn−N (1 − qn )(1 − αβ qn+N+1 )(1 − β qn ) ⎪ ⎪ ⎩ Cn = − . (1 − αβ q2n )(1 − αβ q2n+1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 − (An +Cn )] pn (x) + An−1Cn pn−1 (x), where Qn (q−x ; α , β , N|q) =
(14.6.4)
(αβ qn+1 ; q)n pn (q−x ). (α q, q−N ; q)n
q-Difference Equation q−n (1 − qn )(1 − αβ qn+1 )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where and
y(x) = Qn (q−x ; α , β , N|q) ⎧ ⎨ B(x) = (1 − qx−N )(1 − α qx+1 ) ⎩
D(x) = α q(1 − qx )(β − qx−N−1 ).
(14.6.5)
14.6 q-Hahn
447
Forward Shift Operator Qn (q−x−1 ; α , β , N|q) − Qn (q−x ; α , β , N|q) =
q−n−x (1 − qn )(1 − αβ qn+1 ) Qn−1 (q−x ; α q, β q, N − 1|q) (1 − α q)(1 − q−N )
(14.6.6)
or equivalently
Δ Qn (q−x ; α , β , N|q) q−n+1 (1 − qn )(1 − αβ qn+1 ) = −x Δq (1 − q)(1 − α q)(1 − q−N ) × Qn−1 (q−x ; α q, β q, N − 1|q).
(14.6.7)
Backward Shift Operator (1 − α qx )(1 − qx−N−1 )Qn (q−x ; α , β , N|q) − α (1 − qx )(β − qx−N−1 )Qn (q−x+1 ; α , β , N|q) = qx (1 − α )(1 − q−N−1 )Qn+1 (q−x ; α q−1 , β q−1 , N + 1|q)
(14.6.8)
or equivalently ∇ [w(x; α , β , N|q)Qn (q−x ; α , β , N|q)] ∇q−x 1 w(x; α q−1 , β q−1 , N + 1|q)Qn+1 (q−x ; α q−1 , β q−1 , N + 1|q), (14.6.9) = 1−q where w(x; α , β , N|q) =
(α q, q−N ; q)x (αβ )−x . (q, β −1 q−N ; q)x
Rodrigues-Type Formula w(x; α , β , N|q)Qn (q−x ; α , β , N|q) = (1 − q)n (∇q )n [w(x; α qn , β qn , N − n|q)] , where ∇q :=
∇ . ∇q−x
(14.6.10)
448
14 Basic Hypergeometric Orthogonal Polynomials
Generating Functions For x = 0, 1, 2, . . . , N we have −x x−N q q ,0 −x ; q, α qt 2 φ1 ; q, q t 1 φ1 αq βq N
=
(q−N ; q)n
∑ (β q, q; q)n Qn (q−x ; α , β , N|q)t n .
(14.6.11)
n=0
2 φ1
q−x , β qN+1−x ; q, −α qx−N+1t 0
2 φ0
qx−N , α qx+1 ; q, −q−xt −
(α q, q−N ; q)n −(n) q 2 Qn (q−x ; α , β , N|q)t n . (q; q)n n=0
N
=
∑
(14.6.12)
Limit Relations q-Racah → q-Hahn The q-Hahn polynomials follow from the q-Racah polynomials by the substitution δ = 0 and γ q = q−N in the definition (14.2.1) of the q-Racah polynomials: Rn (μ (x); α , β , q−N−1 , 0|q) = Qn (q−x ; α , β , N|q). Another way to obtain the q-Hahn polynomials from the q-Racah polynomials is by setting γ = 0 and δ = β −1 q−N−1 in the definition (14.2.1): Rn (μ (x); α , β , 0, β −1 q−N−1 |q) = Qn (q−x ; α , β , N|q). And if we take α q = q−N , β → β γ qN+1 and δ = 0 in the definition (14.2.1) of the q-Racah polynomials we find the q-Hahn polynomials given by (14.6.1) in the following way: Rn (μ (x); q−N−1 , β γ qN+1 , γ , 0|q) = Qn (q−x ; γ , β , N|q). Note that μ (x) = q−x in each case. q-Hahn → Little q-Jacobi If we set x → N − x in the definition (14.6.1) of the q-Hahn polynomials and take the limit N → ∞ we find the little q-Jacobi polynomials:
14.6 q-Hahn
449
lim Qn (qx−N ; α , β , N|q) = pn (qx ; α , β |q),
N→∞
(14.6.13)
where pn (qx ; α , β |q) is given by (14.12.1). q-Hahn → q-Meixner The q-Meixner polynomials given by (14.13.1) can be obtained from the q-Hahn polynomials by setting α = b and β = −b−1 c−1 q−N−1 in the definition (14.6.1) of the q-Hahn polynomials and letting N → ∞: lim Qn (q−x ; b, −b−1 c−1 q−N−1 , N|q) = Mn (q−x ; b, c; q).
N→∞
(14.6.14)
q-Hahn → Quantum q-Krawtchouk The quantum q-Krawtchouk polynomials given by (14.14.1) simply follow from the q-Hahn polynomials by setting β = p in the definition (14.6.1) of the q-Hahn polynomials and taking the limit α → ∞: lim Qn (q−x ; α , p, N|q) = Knqtm (q−x ; p, N; q).
α →∞
(14.6.15)
q-Hahn → q-Krawtchouk If we set β = −α −1 q−1 p in the definition (14.6.1) of the q-Hahn polynomials and then let α → 0 we obtain the q-Krawtchouk polynomials given by (14.15.1): lim Qn (q−x ; α , −α −1 q−1 p, N|q) = Kn (q−x ; p, N; q).
α →0
(14.6.16)
q-Hahn → Affine q-Krawtchouk The affine q-Krawtchouk polynomials given by (14.16.1) can be obtained from the q-Hahn polynomials by the substitution α = p and β = 0 in (14.6.1): Qn (q−x ; p, 0, N|q) = KnA f f (q−x ; p, N; q).
(14.6.17)
q-Hahn → Hahn The Hahn polynomials given by (9.5.1) simply follow from the q-Hahn polynomials given by (14.6.1), after setting α → qα and β → qβ , in the following way:
450
14 Basic Hypergeometric Orthogonal Polynomials
lim Qn (q−x ; qα , qβ , N|q) = Qn (x; α , β , N).
q→1
(14.6.18)
Remark The q-Hahn polynomials given by (14.6.1) and the dual q-Hahn polynomials given by (14.7.1) are related in the following way: Qn (q−x ; α , β , N|q) = Rx (μ (n); α , β , N|q), with or
μ (n) = q−n + αβ qn+1 Rn (μ (x); γ , δ , N|q) = Qx (q−n ; γ , δ , N|q),
where
μ (x) = q−x + γδ qx+1 .
References [16], [30], [34], [70], [72], [80], [186], [205], [235], [238], [261], [305], [326], [329], [343], [345], [416], [419], [442], [486], [488], [489].
14.7 Dual q-Hahn Basic Hypergeometric Representation Rn (μ (x); γ , δ , N|q) = 3 φ2 where
q−n , q−x , γδ qx+1 ; q, q , γ q, q−N
μ (x) := q−x + γδ qx+1 .
n = 0, 1, 2, . . . , N, (14.7.1)
14.7 Dual q-Hahn
451
Orthogonality Relation For 0 < γ q < 1 and 0 < δ q < 1, or for γ > q−N and δ > q−N , we have (γ q, γδ q, q−N ; q)x (1 − γδ q2x+1 ) Nx−(x) 2 q ∑ N+2 , δ q; q) (1 − γδ q)(−γ q)x x x=0 (q, γδ q N
× Rm (μ (x); γ , δ , N|q)Rn (μ (x); γ , δ , N|q) =
(q, δ −1 q−N ; q)n (γδ q2 ; q)N (γ q)−N (γδ q)n δmn . (δ q; q)N (γ q, q−N ; q)n
(14.7.2)
Recurrence Relation − 1 − q−x 1 − γδ qx+1 Rn (μ (x)) = An Rn+1 (μ (x)) − (An +Cn ) Rn (μ (x)) +Cn Rn−1 (μ (x)), where
(14.7.3)
Rn (μ (x)) := Rn (μ (x); γ , δ , N|q) ⎧ ⎨ An = 1 − qn−N 1 − γ qn+1
and
⎩
Cn = γ q (1 − qn ) δ − qn−N−1 .
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 + γδ q − (An +Cn )] pn (x) + γ q(1 − qn )(1 − γ qn ) × (1 − qn−N−1 )(δ − qn−N−1 )pn−1 (x), where Rn (μ (x); γ , δ , N|q) =
1 (γ q, q−N ; q)n
(14.7.4)
pn (μ (x)).
q-Difference Equation q−n (1 − qn )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where
(14.7.5)
452
14 Basic Hypergeometric Orthogonal Polynomials
y(x) = Rn (μ (x); γ , δ , N|q) and
⎧ (1 − qx−N )(1 − γ qx+1 )(1 − γδ qx+1 ) ⎪ ⎪ B(x) = ⎪ ⎪ ⎨ (1 − γδ q2x+1 )(1 − γδ q2x+2 ) ⎪ ⎪ γ qx−N (1 − qx )(1 − γδ qx+N+1 )(1 − δ qx ) ⎪ ⎪ ⎩ D(x) = − . (1 − γδ q2x )(1 − γδ q2x+1 )
Forward Shift Operator Rn (μ (x + 1); γ , δ , N|q) − Rn (μ (x); γ , δ , N|q) =
q−n−x (1 − qn )(1 − γδ q2x+2 ) Rn−1 (μ (x); γ q, δ , N − 1|q) (1 − γ q)(1 − q−N )
(14.7.6)
or equivalently
Δ Rn (μ (x); γ , δ , N|q) Δ μ (x) q−n+1 (1 − qn ) Rn−1 (μ (x); γ q, δ , N − 1|q). = (1 − q)(1 − γ q)(1 − q−N )
(14.7.7)
Backward Shift Operator (1 − γ qx )(1 − γδ qx )(1 − qx−N−1 )Rn (μ (x); γ , δ , N|q) + γ qx−N−1 (1 − qx )(1 − γδ qx+N+1 )(1 − δ qx )Rn (μ (x − 1); γ , δ , N|q) x (14.7.8) = q (1 − γ )(1 − q−N−1 )(1 − γδ q2x )Rn+1 (μ (x); γ q−1 , δ , N + 1|q) or equivalently ∇ [w(x; γ , δ , N|q)Rn (μ (x); γ , δ , N|q)] ∇μ (x) 1 = w(x; γ q−1 , δ , N + 1|q) (1 − q)(1 − γδ ) × Rn+1 (μ (x); γ q−1 , δ , N + 1|q), where w(x; γ , δ , N|q) =
x (γ q, γδ q, q−N ; q)x (−γ −1 )x qNx−(2) . (q, γδ qN+2 , δ q; q)x
(14.7.9)
14.7 Dual q-Hahn
453
Rodrigues-Type Formula w(x; γ , δ , N|q)Rn (μ (x); γ , δ , N|q) n = (1 − q)n (γδ q; q)n ∇μ [w(x; γ qn , δ , N − n|q)] , where ∇μ :=
(14.7.10)
∇ . ∇μ (x)
Generating Functions For x = 0, 1, 2, . . . , N we have (q−N t; q)N−x · 2 φ1
q−x , δ −1 q−x ; q, γδ qx+1t γq
(q−N ; q)n Rn (μ (x); γ , δ , N|q)t n . n=0 (q; q)n N
=
∑
(γδ qt; q)x · 2 φ1 N
=
qx−N , γ qx+1 ; q, q−xt δ −1 q−N
(14.7.11)
(q−N , γ q; q)n
∑ (δ −1 q−N , q; q)n Rn (μ (x); γ , δ , N|q)t n .
(14.7.12)
n=0
Limit Relations q-Racah → Dual q-Hahn To obtain the dual q-Hahn polynomials from the q-Racah polynomials we have to take β = 0 and α q = q−N in (14.2.1): Rn (μ (x); q−N−1 , 0, γ , δ |q) = Rn (μ (x); γ , δ , N|q), with
μ (x) = q−x + γδ qx+1 .
We may also take α = 0 and β = δ −1 q−N−1 in (14.2.1) to obtain the dual q-Hahn polynomials from the q-Racah polynomials: Rn (μ (x); 0, δ −1 q−N−1 , γ , δ |q) = Rn (μ (x); γ , δ , N|q),
454
14 Basic Hypergeometric Orthogonal Polynomials
with
μ (x) = q−x + γδ qx+1 .
And if we take γ q = q−N , δ → αδ qN+1 and β = 0 in the definition (14.2.1) of the q-Racah polynomials we find the dual q-Hahn polynomials given by (14.7.1) in the following way: Rn (μ (x); α , 0, q−N−1 , αδ qN+1 |q) = Rn (μ (x); α , δ , N|q), with
μ (x) = q−x + αδ qx+1 .
Dual q-Hahn → Affine q-Krawtchouk The affine q-Krawtchouk polynomials given by (14.16.1) can be obtained from the dual q-Hahn polynomials by the substitution γ = p and δ = 0 in (14.7.1): Rn (μ (x); p, 0, N|q) = KnA f f (q−x ; p, N; q).
(14.7.13)
Note that μ (x) = q−x in this case. Dual q-Hahn → Dual q-Krawtchouk The dual q-Krawtchouk polynomials given by (14.17.1) can be obtained from the dual q-Hahn polynomials by setting δ = cγ −1 q−N−1 in (14.7.1) and letting γ → 0: lim Rn (μ (x); γ , cγ −1 q−N−1 , N|q) = Kn (λ (x); c, N|q).
γ →0
(14.7.14)
Dual q-Hahn → Dual Hahn The dual Hahn polynomials given by (9.6.1) follow from the dual q-Hahn polynomials by simply taking the limit q → 1 in the definition (14.7.1) of the dual q-Hahn polynomials after applying the substitution γ → qγ and δ → qδ : lim Rn (μ (x); qγ , qδ , N|q) = Rn (λ (x); γ , δ , N),
q→1
where
⎧ −x x+γ +δ +1 ⎨ μ (x) = q + q ⎩
λ (x) = x(x + γ + δ + 1).
(14.7.15)
14.8 Al-Salam-Chihara
455
Remark The dual q-Hahn polynomials given by (14.7.1) and the q-Hahn polynomials given by (14.6.1) are related in the following way: Qn (q−x ; α , β , N|q) = Rx (μ (n); α , β , N|q), with
μ (n) = q−n + αβ qn+1
or
Rn (μ (x); γ , δ , N|q) = Qx (q−n ; γ , δ , N|q),
where
μ (x) = q−x + γδ qx+1 .
References [31], [34], [70], [72], [80], [238], [329], [416], [488].
14.8 Al-Salam-Chihara Basic Hypergeometric Representation q−n , aeiθ , ae−iθ ; q, q ab, 0 −n −iθ q , be −1 iθ = (aeiθ ; q)n e−inθ 2 φ1 ; q, a qe a−1 q−n+1 e−iθ −n iθ q , ae −iθ inθ −1 −iθ , ; q, b qe = (be ; q)n e 2 φ1 b−1 q−n+1 eiθ
Qn (x; a, b|q) =
(ab; q)n 3 φ2 an
(14.8.1)
x = cos θ .
Orthogonality Relation If a and b are real or complex conjugates and max(|a|, |b|) < 1, then we have the following orthogonality relation 1 2π
1
w(x) δmn √ Qm (x; a, b|q)Qn (x; a, b|q) dx = n+1 , 2 (q , abqn ; q)∞ −1 1−x
(14.8.2)
456
14 Basic Hypergeometric Orthogonal Polynomials
where (e2iθ ; q)∞ 2 h(x, 1)h(x, −1)h(x, q 21 )h(x, −q 21 ) = , w(x) := w(x; a, b|q) = iθ iθ h(x, a)h(x, b) (ae , be ; q)∞ with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
x = cos θ .
If a > 1 and |ab| < 1, then we have another orthogonality relation given by: 1 2π
1 −1
+
w(x) √ Qm (x; a, b|q)Qn (x; a, b|q) dx 1 − x2
∑
wk Qm (xk ; a, b|q)Qn (xk ; a, b|q) =
k
δmn , (14.8.3) (qn+1 , abqn ; q)∞
1
where w(x) is as before,
and
−1 aqk + aqk xk = 2
(1 − a2 q2k )(a2 , ab; q)k −k2 (a−2 ; q)∞ wk = q (q, ab, a−1 b; q)∞ (1 − a2 )(q, ab−1 q; q)k
1 a3 b
k .
Recurrence Relation 2xQn (x) = Qn+1 (x) + (a + b)qn Qn (x) + (1 − qn )(1 − abqn−1 )Qn−1 (x).
(14.8.4)
Normalized Recurrence Relation 1 1 xpn (x) = pn+1 (x) + (a + b)qn pn (x) + (1 − qn )(1 − abqn−1 )pn−1 (x), (14.8.5) 2 4 where Qn (x; a, b|q) = 2n pn (x).
14.8 Al-Salam-Chihara
457
q-Difference Equation 1 1 ˜ aq 2 , bq 2 |q)Dq y(x) (1 − q)2 Dq w(x; ˜ a, b|q)y(x) = 0, + 4q−n+1 (1 − qn )w(x;
(14.8.6)
where y(x) = Qn (x; a, b|q) and
w(x; a, b|q) . w(x; ˜ a, b|q) := √ 1 − x2
If we define (ab; q)n 3 φ2 an
Pn (z) :=
q−n , az, az−1 ; q, q ab, 0
then the q-difference equation can also be written in the form
q−n (1−qn )Pn (z) = A(z)Pn (qz)− A(z) + A(z−1 ) Pn (z)+A(z−1 )Pn (q−1 z), (14.8.7) where A(z) =
(1 − az)(1 − bz) . (1 − z2 )(1 − qz2 )
Forward Shift Operator δq Qn (x; a, b|q) = −q− 2 n (1 − qn )(eiθ − e−iθ )Qn−1 (x; aq 2 , bq 2 |q), 1
1
1
x = cos θ (14.8.8)
or equivalently 1
Dq Qn (x; a, b|q) = 2q− 2 (n−1)
1 1 1 − qn Qn−1 (x; aq 2 , bq 2 |q). 1−q
(14.8.9)
Backward Shift Operator δq [w(x; ˜ a, b|q)Qn (x; a, b|q)] = q− 2 (n+1) (eiθ − e−iθ )w(x; ˜ aq− 2 , bq− 2 |q) 1
1
1
1
× Qn+1 (x; aq− 2 , bq− 2 |q),
1
x = cos θ
(14.8.10)
458
14 Basic Hypergeometric Orthogonal Polynomials
or equivalently ˜ a, b|q)Qn (x; a, b|q)] Dq [w(x; 1
=−
1 1 1 1 2q− 2 n w(x; ˜ aq− 2 , bq− 2 |q)Qn+1 (x; aq− 2 , bq− 2 |q). 1−q
(14.8.11)
Rodrigues-Type Formula w(x; ˜ a, b|q)Qn (x; a, b|q) 1 1 q − 1 n 1 n(n−1) ˜ aq 2 n , bq 2 n |q) . q4 (Dq )n w(x; = 2
(14.8.12)
Generating Functions ∞ (at, bt; q)∞ Qn (x; a, b|q) n = t , ∑ (eiθ t, e−iθ t; q)∞ n=0 (q; q)n
1 2 φ1 (eiθ t; q)∞
aeiθ , beiθ ; q, e−iθ t ab
x = cos θ .
∞
=
Qn (x; a, b|q) n t , n=0 (ab, q; q)n
∑
aeiθ , ae−iθ ; q, t (t; q)∞ · 2 φ1 ab n ∞ (−1)n an q(2) =∑ Qn (x; a, b|q)t n , n=0 (ab, q; q)n
(14.8.13)
x = cos θ . (14.8.14)
x = cos θ .
γ , aeiθ , beiθ (γ eiθ t; q)∞ −iθ ; q, e t 3 φ2 ab, γ eiθ t (eiθ t; q)∞ ∞ (γ ; q)n Qn (x; a, b|q)t n , x = cos θ , =∑ (ab, q; q)n n=0
γ arbitrary.
(14.8.15)
(14.8.16)
14.8 Al-Salam-Chihara
459
Limit Relations Continuous Dual q-Hahn → Al-Salam-Chihara The Al-Salam-Chihara polynomials given by (14.8.1) simply follow from the continuous dual q-Hahn polynomials by taking c = 0 in the definition (14.3.1) of the continuous dual q-Hahn polynomials: pn (x; a, b, 0|q) = Qn (x; a, b|q). Al-Salam-Chihara → Continuous Big q-Hermite If we take b = 0 in the definition (14.8.1) of the Al-Salam-Chihara polynomials we simply obtain the continuous big q-Hermite polynomials given by (14.18.1): Qn (x; a, 0|q) = Hn (x; a|q).
(14.8.17)
Al-Salam-Chihara → Continuous q-Laguerre The continuous q-Laguerre polynomials given by (14.19.1) can be obtained from 1 1 the Al-Salam-Chihara polynomials given by (14.8.1) by taking a = q 2 α + 4 and b = 1 3 q 2 α+ 4 : 1 1 1 3 (q; q)n (α ) Qn (x; q 2 α + 4 , q 2 α + 4 |q) = 1 1 Pn (x|q). (14.8.18) ( q 2 α + 4 )n Al-Salam-Chihara → Meixner-Pollaczek If we set a = qλ e−iφ , b = qλ eiφ and eiθ = qix eiφ in the definition (14.8.1) of the Al-Salam-Chihara polynomials and take the limit q → 1 we obtain the MeixnerPollaczek polynomials given by (9.7.1) in the following way: Qn (cos(ln qx + φ ); qλ eiφ , qλ e−iφ |q) (λ ) = Pn (x; φ ). q→1 (q; q)n lim
(14.8.19)
References [16], [20], [21], [64], [73], [86], [118], [148], [151], [160], [209], [292], [295], [320], [328].
460
14 Basic Hypergeometric Orthogonal Polynomials
14.9 q-Meixner-Pollaczek Basic Hypergeometric Representation q−n , aei(θ +2φ ) , ae−iθ ; q, q (14.9.1) Pn (x; a|q) = a e a2 , 0 −n iθ q , ae (ae−iθ ; q)n in(θ +φ ) −1 −i(θ +2φ ) = e ; q, qa e 2 φ1 (q; q)n a−1 q−n+1 eiθ −n −inφ
(a2 ; q)n 3 φ2 (q; q)n
with x = cos(θ + φ ).
Orthogonality Relation 1 2π =
π −π
w(cos(θ + φ ); a|q)Pm (cos(θ + φ ); a|q)Pn (cos(θ + φ ); a|q) d θ
δmn , (q; q)n (q, a2 qn ; q)∞
0 < a < 1,
(14.9.2)
where 2 1 1 (e2i(θ +φ ) ; q) h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) ∞ w(x; a|q) = i(θ +2φ ) iθ , = (ae h(x, aeiφ )h(x, ae−iφ ) , ae ; q)∞ with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α ei(θ +φ ) , α e−i(θ +φ ) ; q , k=0
∞
x = cos(θ + φ ).
Recurrence Relation 2xPn (x; a|q) = (1 − qn+1 )Pn+1 (x; a|q) + 2aqn cos φ Pn (x; a|q) + (1 − a2 qn−1 )Pn−1 (x; a|q).
(14.9.3)
14.9 q-Meixner-Pollaczek
461
Normalized Recurrence Relation 1 xpn (x) = pn+1 (x) + aqn cos φ pn (x) + (1 − qn )(1 − a2 qn−1 )pn−1 (x), 4 where 2n Pn (x; a|q) = pn (x). (q; q)n
(14.9.4)
q-Difference Equation 1 ˜ aq 2 |q)Dq y(x) + 4q−n+1 (1 − qn )w(x; (1 − q)2 Dq w(x; ˜ a|q)y(x) = 0,
(14.9.5)
where y(x) = Pn (x; a|q) and
w(x; a|q) . w(x; ˜ a|q) := √ 1 − x2
Forward Shift Operator δq Pn (x; a|q) = −q− 2 n (ei(θ +φ ) − e−i(θ +φ ) )Pn−1 (x; aq 2 |q), 1
1
x = cos θ
(14.9.6)
or equivalently 1
Dq Pn (x; a|q) =
1 2q− 2 (n−1) Pn−1 (x; aq 2 |q). 1−q
(14.9.7)
Backward Shift Operator δq [w(x; ˜ a|q)Pn (x; a|q)] = q− 2 (n+1) (1 − qn+1 )(eiθ − e−iθ ) 1
1
1
× w(x; ˜ aq− 2 |q)Pn+1 (x; aq− 2 |q),
x = cos θ
(14.9.8)
or equivalently 1
˜ a|q)Pn (x; a|q)] = −2q− 2 n Dq [w(x;
1 1 1 − qn+1 w(x; ˜ aq− 2 |q)Pn+1 (x; aq− 2 |q). (14.9.9) 1−q
462
14 Basic Hypergeometric Orthogonal Polynomials
Rodrigues-Type Formula w(x; ˜ a|q)Pn (x; a|q) =
q−1 2
n
1
q 4 n(n−1)
1 1 ˜ aq 2 n |q) . (14.9.10) (Dq )n w(x; (q; q)n
Generating Functions (aeiφ t; q)∞ (ei(θ +φ )t; q) =
2
∞ i φ (ae t, ae−iφ t; q)∞ i( (e θ +φ )t, e−i(θ +φ )t; q)∞
1 (ei(θ +φ )t; q)∞ ∞
=
2 φ1
Pn (x; a|q) n t , 2 n=0 (a ; q)n
∑
∞
=
∑ Pn (x; a|q)t n ,
x = cos(θ + φ ). (14.9.11)
n=0
aei(θ +2φ ) , aeiθ ; q, e−i(θ +φ )t a2
x = cos(θ + φ ).
(14.9.12)
Limit Relations Continuous q-Hahn → q-Meixner-Pollaczek The q-Meixner-Pollaczek polynomials given by (14.9.1) simply follow from the continuous q-Hahn polynomials if we set d = a and b = c = 0 in the definition (14.4.1) of the continuous q-Hahn polynomials: pn (cos(θ + φ ); a, 0, 0, a; q) = Pn (cos(θ + φ ); a|q). (q; q)n q-Meixner-Pollaczek → Continuous q-ultraspherical / Rogers If we take θ = 0 and a = β in the definition (14.9.1) of the q-Meixner-Pollaczek polynomials we obtain the continuous q-ultraspherical (or Rogers) polynomials given by (14.10.17): (14.9.13) Pn (cos φ ; β |q) = Cn (cos φ ; β |q).
14.10 Continuous q-Jacobi
463
q-Meixner-Pollaczek → Continuous q-Laguerre If we take eiφ = q− 4 , a = q 2 α + 2 and eiθ → q 4 eiθ in the definition (14.9.1) of the q-Meixner-Pollaczek polynomials we obtain the continuous q-Laguerre polynomials given by (14.19.1): 1
1
1
1
(α )
Pn (cos(θ + φ ); q 2 α + 2 |q) = q−( 2 α + 4 )n Pn (cos θ |q). 1
1
1
1
(14.9.14)
q-Meixner-Pollaczek → Meixner-Pollaczek To find the Meixner-Pollaczek polynomials given by (9.7.1) from the q-MeixnerPollaczek polynomials we substitute a = qλ and eiθ = q−ix (or θ = ln q−x ) in the definition (14.9.1) of the q-Meixner-Pollaczek polynomials and take the limit q → 1 to find: (λ ) (14.9.15) lim Pn (cos(ln q−x + φ ); qλ |q) = Pn (x; −φ ). q→1
References [16], [21], [64], [72], [135], [270].
14.10 Continuous q-Jacobi Basic Hypergeometric Representation If we take a = q 2 α + 4 , b = q 2 α + 4 , c = −q 2 β + 4 and d = −q 2 β + 4 in the definition (14.1.1) of the Askey-Wilson polynomials we find after renormalizing 1
(α ,β )
Pn
1
(x|q)
(qα +1 ; q)n = 4 φ3 (q; q)n with x = cos θ .
1
3
1
1
1
3
q−n , qn+α +β +1 , q 2 α + 4 eiθ , q 2 α + 4 e−iθ 1
1
1
qα +1 , −q 2 (α +β +1) , −q 2 (α +β +2) 1
1
1
; q, q
(14.10.1)
464
14 Basic Hypergeometric Orthogonal Polynomials
Orthogonality Relation For α ≥ − 12 and β ≥ − 12 we have 1 2π
1
w(x) (α ,β ) (α ,β ) √ Pm (x|q)Pn (x|q) dx −1 1 − x2 (q 2 (α +β +2) , q 2 (α +β +3) ; q)∞ 1
=
1
(q, qα +1 , qβ +1 , −q ×
1 (α +β +1) 2
, −q
1 (α +β +2) 2
1 − qα +β +1 2n+α +β +1 ; q)∞ 1 − q
1 (qα +1 , qβ +1 , −q 2 (α +β +3) ; q)n (α + 1 )n 2 q δmn , 1 (q, qα +β +1 , −q 2 (α +β +1) ; q)n
(14.10.2)
where 2 (e2iθ ; q)∞ w(x) := w(x; q , q |q) = 1 1 (q 2 α + 4 eiθ , q 12 α + 34 eiθ , −q 12 β + 14 eiθ , −q 12 β + 34 eiθ ; q)∞ 2 1 (eiθ , −eiθ ; q 2 )∞ = 1 1 (q 2 α + 4 eiθ , −q 12 β + 14 eiθ ; q 12 )∞ α
β
1
=
1
h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) h(x, q 2 α + 4 )h(x, q 2 α + 4 )h(x, −q 2 β + 4 )h(x, −q 2 β + 4 ) 1
1
1
3
1
1
1
3
,
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
x = cos θ .
Recurrence Relation 1 1 1 1 2xP˜n (x|q) = An P˜n+1 (x|q) + q 2 α + 4 + q− 2 α − 4 − (An +Cn ) P˜n (x|q) +Cn P˜n−1 (x|q), where P˜n (x|q) := and
(q; q)n (α ,β ) Pn (x|q) (qα +1 ; q)n
(14.10.3)
14.10 Continuous q-Jacobi
465
⎧ 1 1 ⎪ (1 − qn+α +1 )(1 − qn+α +β +1 )(1 + qn+ 2 (α +β +1) )(1 + qn+ 2 (α +β +2) ) ⎪ ⎪ An = ⎪ 1 1 ⎪ ⎨ q 2 α + 4 (1 − q2n+α +β +1 )(1 − q2n+α +β +2 ) ⎪ 1 1 1 1 ⎪ ⎪ q 2 α + 4 (1 − qn )(1 − qn+β )(1 + qn+ 2 (α +β ) )(1 + qn+ 2 (α +β +1) ) ⎪ ⎪ ⎩ Cn = . (1 − q2n+α +β )(1 − q2n+α +β +1 )
Normalized Recurrence Relation
xpn (x) = pn+1 (x) +
1 1 1 1 α+ 1 q 2 4 + q− 2 α − 4 − (An +Cn ) pn (x) 2
1 + An−1Cn pn−1 (x), 4 where (α ,β )
Pn
2n q( 2 α + 4 )n (qn+α +β +1 ; q)n 1
(x|q) =
(14.10.4)
1
(q, −q 2 (α +β +1) , −q 2 (α +β +2) ; q)n 1
1
pn (x).
q-Difference Equation ˜ qα +1 , qβ +1 |q)Dq y(x) + λn w(x; (1 − q)2 Dq w(x; ˜ qα , qβ |q)y(x) = 0, where
(α ,β )
y(x) = Pn and
w(x; ˜ qα , qβ |q) :=
(14.10.5)
(x|q)
w(x; qα , qβ |q) √ , 1 − x2
λn = 4q−n+1 (1 − qn )(1 − qn+α +β +1 ).
Forward Shift Operator
(α ,β ) δq Pn (x|q)
q−n+ 2 α + 4 (1 − qn+α +β +1 )(eiθ − e−iθ ) 1
=−
3
(1 + q 2 (α +β +1) )(1 + q 2 (α +β +2) ) 1
(α +1,β +1)
× Pn−1 or equivalently
1
(x|q),
x = cos θ
(14.10.6)
466
14 Basic Hypergeometric Orthogonal Polynomials (α ,β ) Dq Pn (x|q)
2q−n+ 2 α + 4 (1 − qn+α +β +1 ) 1
=
5
(1 − q)(1 + q 2 (α +β +1) )(1 + q 2 (α +β +2) ) 1
1
(α +1,β +1)
× Pn−1
(14.10.7)
(x|q).
Backward Shift Operator (α ,β ) ˜ qα , qβ |q)Pn δq w(x; (x|q) = q− 2 α − 4 (1 − qn+1 )(1 + q 2 (α +β −1) )(1 + q 2 (α +β ) )(eiθ − e−iθ ) 1
1
1
1
(α −1,β −1)
× w(x; ˜ qα −1 , qβ −1 |q)Pn+1
(x|q),
x = cos θ
(14.10.8)
or equivalently (α ,β ) ˜ qα , qβ |q)Pn Dq w(x; (x|q) (1 − qn+1 )(1 + q 2 (α +β −1) )(1 + q 2 (α +β ) ) 1−q 1
= −2q
− 21 α + 14
1
(α −1,β −1)
× w(x; ˜ qα −1 , qβ −1 |q)Pn+1
(14.10.9)
(x|q).
Rodrigues-Type Formula (α ,β )
w(x; ˜ qα , qβ |q)Pn (x|q) n 1 2 1 q 4 n + 2 nα q−1 = 1 1 2 (q, −q 2 (α +β +1) , −q 2 (α +β +2) ; q)n ˜ qα +n , qβ +n |q) . × (Dq )n w(x;
(14.10.10)
Generating Functions 2 φ1 ∞
=
q 2 α + 4 eiθ , q 2 α + 4 eiθ ; q, e−iθ t qα +1 1
1
1
3
2 φ1
−q 2 β + 4 e−iθ , −q 2 β + 4 e−iθ ; q, eiθ t qβ +1
(α ,β )
1
1
(−q 2 (α +β +1) , −q 2 (α +β +2) ; q)n Pn (x|q) n 1 α + 1 )n t , α +1 β +1 ( (q , q ; q)n q 2 4 n=0
∑
1
1
1
x = cos θ .
3
(14.10.11)
14.10 Continuous q-Jacobi
2 φ1
q 2 α + 4 eiθ , −q 2 β + 4 eiθ 1
1
∑
2 φ1
(α ,β )
1 (α +β +3) 2
; q)n q
1
1
1
1
n=0
; q, e
(α ,β )
(−q
1 (α +β +2) 2
; q)n
−iθ
(x|q)
t
1
3
1
1
2 φ1
; q, eiθ t (14.10.12)
q 2 α + 4 e−iθ , −q 2 β + 4 e−iθ 1
3
−q 2 (α +β +3)
3
1
1
−q 2 (α +β +2) 1
x = cos θ .
t n,
1 1 q( 2 α + 4 )n
q 2 α + 4 e−iθ , −q 2 β + 4 e−iθ
x = cos θ .
(−q 2 (α +β +1) ; q)n Pn 1
2 φ1
t n,
3
−q 2 (α +β +2)
∑
(x|q)
( 12 α + 14 )n
q 2 α + 4 eiθ , −q 2 β + 4 eiθ
∞
; q, e−iθ t
(−q 2 (α +β +2) ; q)n Pn 1
1
1
n=0 (−q
=
1
−q 2 (α +β +1) ∞
=
467
iθ
; q, e t (14.10.13)
Limit Relations Askey-Wilson → Continuous q-Jacobi If we take a = q 2 α + 4 , b = q 2 α + 4 , c = −q 2 β + 4 and d = −q 2 β + 4 in the definition (14.1.1) of the Askey-Wilson polynomials and change the normalization we find the continuous q-Jacobi polynomials given by (14.10.1): 1
1
1
3
1
1
1
q( 2 α + 4 )n pn (x; q 2 α + 4 , q 2 α + 4 , −q 2 β + 4 , −q 2 β + 4 |q) 1
1
1
(q, −q
1
1
1 2 (α +β +1)
3
, −q
1
1
1 (α +β +2) 2
1
3
; q)n
3
(α ,β )
= Pn
(x|q).
Continuous q-Jacobi → Continuous q-Laguerre The continuous q-Laguerre polynomials given by (14.19.1) follow simply from the continuous q-Jacobi polynomials given by (14.10.1) by taking the limit β → ∞: (α ,β )
lim Pn
β →∞
(α )
(x|q) = Pn (x|q).
(14.10.14)
Continuous q-Jacobi → Jacobi If we take the limit q → 1 in the definitions (14.10.1) of the continuous q-Jacobi polynomials we simply find the Jacobi polynomials given by (9.8.1): (α ,β )
lim Pn
q→1
(α ,β )
(x|q) = Pn
(x).
(14.10.15)
468
14 Basic Hypergeometric Orthogonal Polynomials
Remarks In [441] M. Rahman takes a = q 2 , b = qα + 2 , c = −qβ + 2 and d = −q 2 in the definition (14.1.1) of the Askey-Wilson polynomials to obtain after renormalizing 1
(α ,β )
Pn
(x; q) =
1
1
1
(qα +1 , −qβ +1 ; q)n (q, −q; q)n 1 1 q−n , qn+α +β +1 , q 2 eiθ , q 2 e−iθ ; q, q (14.10.16) × 4 φ3 qα +1 , −qβ +1 , −q
with x = cos θ . These two q-analogues of the Jacobi polynomials are not really different, since they are connected by the quadratic transformation: (α ,β )
Pn
(x|q2 ) =
(−q; q)n (α ,β ) qnα Pn (x; q). (−qα +β +1 ; q)n
The continuous q-Jacobi polynomials given by (14.10.16) and the continuous qultraspherical (or Rogers) polynomials given by (14.10.17) are connected by the quadratic transformations: C2n (x; qλ |q) = and C2n+1 (x; qλ |q) =
(qλ , −q; q)n 1 2
1 2
(q , −q ; q)n
(qλ , −1; q)n+1 1 2
1 2
(λ − 12 ,− 21 )
1
q− 2 n Pn
(q , −q ; q)n+1
1
(2x2 − 1; q)
(λ − 21 , 12 )
q− 2 n xPn
(2x2 − 1; q).
If we replace q by q−1 we find (α ,β )
Pn
(α ,β )
(x|q−1 ) = q−nα Pn
(α ,β )
(x|q) and Pn
(α ,β )
(x; q−1 ) = q−n(α +β ) Pn
(x; q).
References [72], [208], [236], [238], [288], [290], [293], [351], [414], [416], [441], [443], [444], [446], [474], [495].
14.10 Continuous q-Jacobi
469
Special Cases 14.10.1 Continuous q-Ultraspherical / Rogers Basic Hypergeometric Representation 1
1
1
1
1
1
If we set a = β 2 , b = β 2 q 2 , c = −β 2 and d = −β 2 q 2 in the definition (14.1.1) of the Askey-Wilson polynomials and change the normalization we obtain the continuous q-ultraspherical (or Rogers) polynomials: 1 1 q−n , β 2 qn , β 2 eiθ , β 2 e−iθ (β 2 ; q)n − 1 n β 2 4 φ3 ; q, q (14.10.17) Cn (x; β |q) = 1 1 (q; q)n β q 2 , −β , −β q 2 −n q , β , β e2iθ (β 2 ; q)n −n −inθ ; q, q β e = 3 φ2 (q; q)n β 2, 0 −n q ,β (β ; q)n inθ −1 −2iθ = , x = cos θ . e 2 φ1 ; q, β qe β −1 q−n+1 (q; q)n
Orthogonality Relation 1 2π
1
w(x) √ Cm (x; β |q)Cn (x; β |q) dx −1 1 − x2 (β , β q; q)∞ (β 2 ; q)n (1 − β ) δmn , |β | < 1, = (β 2 , q; q)∞ (q; q)n (1 − β qn )
(14.10.18)
where 2 (e2iθ ; q)∞ w(x) := w(x; β |q) = 1 1 1 1 1 1 (β 2 eiθ , β 2 q 2 eiθ , −β 2 eiθ , −β 2 q 2 eiθ ; q)∞ 2iθ 1 1 (e ; q)∞ 2 h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) = = 1 1 1 1 1 1 , (β e2iθ ; q)∞ h(x, β 2 )h(x, β 2 q 2 )h(x, −β 2 )h(x, −β 2 q 2 ) with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , k=0
∞
x = cos θ .
470
14 Basic Hypergeometric Orthogonal Polynomials
Recurrence Relation 2(1 − β qn )xCn (x; β |q) = (1 − qn+1 )Cn+1 (x; β |q) + (1 − β 2 qn−1 )Cn−1 (x; β |q). (14.10.19)
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
(1 − qn )(1 − β 2 qn−1 ) pn−1 (x), 4(1 − β qn−1 )(1 − β qn )
where Cn (x; β |q) =
(14.10.20)
2n (β ; q)n pn (x). (q; q)n
q-Difference Equation (1 − q)2 Dq [w(x; ˜ β q|q)Dq y(x)] + λn w(x; ˜ β |q)y(x) = 0, where
(14.10.21)
y(x) = Cn (x; β |q), w(x; β |q) w(x; ˜ β |q) := √ 1 − x2
and
λn = 4q−n+1 (1 − qn )(1 − β 2 qn ).
Forward Shift Operator δqCn (x; β |q) = −q− 2 n (1 − β )(eiθ − e−iθ )Cn−1 (x; β q|q), 1
x = cos θ
(14.10.22)
or equivalently 1
DqCn (x; β |q) = 2q− 2 (n−1)
1−β Cn−1 (x; β q|q). 1−q
(14.10.23)
14.10 Continuous q-Jacobi
471
Backward Shift Operator δq [w(x; ˜ β |q)Cn (x; β |q)] 1
= q− 2 (n+1)
(1 − qn+1 )(1 − β 2 qn−1 ) iθ (e − e−iθ ) (1 − β q−1 )
× w(x; ˜ β q−1 |q)Cn+1 (x; β q−1 |q),
x = cos θ
(14.10.24)
or equivalently ˜ β |q)Cn (x; β |q)] Dq [w(x; 1
=−
2q− 2 n (1 − qn+1 )(1 − β 2 qn−1 ) (1 − q)(1 − β q−1 ) × w(x; ˜ β q−1 |q)Cn+1 (x; β q−1 |q).
(14.10.25)
Rodrigues-Type Formula w(x; ˜ β |q)Cn (x; β |q) q − 1 n 1 n(n−1) (β ; q)n q4 (Dq )n [w(x; ˜ β qn |q)] . = 2 (q, β 2 qn ; q)n
(14.10.26)
Generating Functions ∞ (β eiθ t; q)∞ 2 (β eiθ t, β e−iθ t; q)∞ = = ∑ Cn (x; β |q)t n , (eiθ t; q)∞ (eiθ t, e−iθ t; q)∞ n=0 1 2 φ1 i θ (e t; q)∞
β , β e2iθ ; q, e−iθ t β2
∞
=
Cn (x; β |q) n t , 2 n=0 (β ; q)n
∑
x = cos θ . (14.10.27)
x = cos θ .
β , β e2iθ −iθ ; q, e t β2 n ∞ (−1)n β n q(2) Cn (x; β |q)t n , x = cos θ . =∑ 2 ; q) ( β n n=0
(e−iθ t; q)∞ · 2 φ1
(14.10.28)
(14.10.29)
472
14 Basic Hypergeometric Orthogonal Polynomials
2 φ1
β 2 e iθ , β 2 q 2 e iθ 1
1
1
βq2
(−β , −β q 2 ; q)n
n=0
(β 2 , β q 2 ; q)n
∑
2 φ1
−iθ
2 φ1
t
2 φ1
1
1
1
1
Cn (x; β |q)t n ,
β 2 eiθ , −β 2 eiθ ; q, e−iθ t −β 1
1
1
iθ
; q, e t
x = cos θ .
2 φ1
1
(14.10.30)
β 2 q 2 e−iθ , −β 2 q 2 e−iθ ; q, eiθ t −β q 1
1
1
1
1
1
1
; q, e
1
−β q 2 ∞
(−β , β q 2 ; q)n
n=0
(β 2 , −β q 2 ; q)n
x = cos θ .
β 2 eiθ , −β 2 q 2 eiθ
∑
1
βq2
(β q 2 , −β q 2 ; q)n ∑ (β 2 , −β q; q)n Cn (x; β |q)t n , n=0
−β 2 e−iθ , −β 2 q 2 e−iθ
1
∞
=
; q, e
∞
=
=
1
−iθ
t
2 φ1
(14.10.31)
β 2 q 2 e−iθ , −β 2 e−iθ 1
1
1
1
−β q 2
iθ
; q, e t
1
1
Cn (x; β |q)t n ,
x = cos θ .
(14.10.32)
γ , β , β e2iθ (γ eiθ t; q)∞ −iθ ; q, e φ t 3 2 β 2 , γ e iθ t (eiθ t; q)∞ ∞ (γ ; q)n Cn (x; β |q)t n , x = cos θ , =∑ 2 ( β ; q) n n=0
γ arbitrary.
(14.10.33)
Limit Relations Askey-Wilson → Continuous q-ultraspherical / Rogers 1
1
1
1
1
1
If we set a = β 2 , b = β 2 q 2 , c = −β 2 and d = −β 2 q 2 in the definition (14.1.1) of the Askey-Wilson polynomials and change the normalization we obtain the continuous q-ultraspherical (or Rogers) polynomials given by (14.10.17). In fact we have: 1
1
1
1
1
1
(β 2 ; q)n pn (x; β 2 , β 2 q 2 , −β 2 , −β 2 q 2 |q) 1
1
(β q 2 , −β , −β q 2 , q; q)n
= Cn (x; β |q).
q-Meixner-Pollaczek → Continuous q-ultraspherical / Rogers If we take θ = 0 and a = β in the definition (14.9.1) of the q-Meixner-Pollaczek polynomials we obtain the continuous q-ultraspherical (or Rogers) polynomials given by
14.10 Continuous q-Jacobi
(14.10.17):
473
Pn (cos φ ; β |q) = Cn (cos φ ; β |q).
Continuous q-ultraspherical / Rogers → Continuous q-Hermite The continuous q-Hermite polynomials given by (14.26.1) can be obtained from the continuous q-ultraspherical (or Rogers) polynomials given by (14.10.17) by taking the limit β → 0. In fact we have lim Cn (x; β |q) =
β →0
Hn (x|q) . (q; q)n
(14.10.34)
Continuous q-ultraspherical / Rogers → Gegenbauer / Ultraspherical If we set β = qλ in the definition (14.10.17) of the continuous q-ultraspherical (or Rogers) polynomials and let q tend to 1 we obtain the Gegenbauer (or ultraspherical) polynomials given by (9.8.19): (λ )
lim Cn (x; qλ |q) = Cn (x).
(14.10.35)
q→1
Remarks The continuous q-ultraspherical (or Rogers) polynomials can also be written as: Cn (x; β |q) =
(β ; q)k (β ; q)n−k i(n−2k)θ e , k=0 (q; q)k (q; q)n−k n
∑
x = cos θ .
They can be obtained from the continuous q-Jacobi polynomials given by (14.10.1) 1 in the following way. Set β = α in the definition (14.10.1) and replace qα + 2 by β and we find the continuous q-ultraspherical (or Rogers) polynomials with a different normalization. We have q (α ,α ) (x|q) Pn
α + 21
→β
−→
1
(β q 2 ; q)n 1 n β 2 Cn (x; β |q). (β 2 ; q)n
If we set β = qα + 2 in the definition (14.10.17) of the q-ultraspherical (or Rogers) polynomials we find the continuous q-Jacobi polynomials given by (14.10.1) with β = α . In fact we have 1
Cn (x; qα + 2 |q) = 1
(q2α +1 ; q)n 1 1 (qα +1 ; q)n q( 2 α + 4 )n
(α ,α )
Pn
(x|q).
474
14 Basic Hypergeometric Orthogonal Polynomials
If we replace q by q−1 we find Cn (x; β |q−1 ) = (β q)nCn (x; β −1 |q). The special case β = q of the continuous q-ultraspherical (or Rogers) polynomials equals the Chebyshev polynomials of the second kind given by (9.8.36). In fact we have sin(n + 1)θ Cn (x; q|q) = = Un (x), x = cos θ . sin θ The limit case β → 1 leads to the Chebyshev polynomials of the first kind given by (9.8.35) in the following way: 1 − qn Cn (x; β |q) = cos nθ = Tn (x), β →1 2(1 − β )
x = cos θ ,
lim
n = 1, 2, 3, . . . .
The continuous q-Jacobi polynomials given by (14.10.16) and the continuous qultraspherical (or Rogers) polynomials given by (14.10.17) are connected by the quadratic transformations: C2n (x; qλ |q) = and C2n+1 (x; qλ |q) =
(qλ , −q; q)n 1 2
1 2
(q , −q ; q)n
(qλ , −1; q)n+1 1 2
1 2
(λ − 12 ,− 21 )
1
q− 2 n Pn
(q , −q ; q)n+1
1
(2x2 − 1; q)
(λ − 21 , 12 )
q− 2 n xPn
(2x2 − 1; q).
Finally we remark that the continuous q-ultraspherical (or Rogers) polynomials are related to the continuous q-Legendre polynomials given by (14.10.36) in the following way: 1 1 Cn (x; q 2 |q) = q− 4 n Pn (x|q).
References [13], [14], [16], [34], [51], [52], [53], [62], [63], [64], [66], [72], [80], [114], [119], [120], [210], [230], [231], [232], [234], [236], [237], [238], [271], [288], [294], [295], [299], [325], [326], [344], [350], [351], [395], [414], [416], [421], [446], [448], [452], [453], [454], [460], [461], [462], [473].
14.10 Continuous q-Jacobi
475
14.10.2 Continuous q-Legendre Basic Hypergeometric Representation The continuous q-Legendre polynomials are continuous q-Jacobi polynomials with α = β = 0: 1 1 q−n , qn+1 , q 4 eiθ , q 4 e−iθ ; q, q , x = cos θ . Pn (x|q) = 4 φ3 (14.10.36) 1 q, −q 2 , −q
Orthogonality Relation 1 2π
1 w(x; 1|q) −1
√
1 − x2 1
=
(q 2 ; q)∞
Pm (x|q)Pn (x|q) dx 1
q2n
1
1
(q, q, −q 2 , −q; q)∞ 1 − qn+ 2
δmn ,
(14.10.37)
where 2 (e2iθ ; q)∞ w(x; a|q) = 1 3 1 3 (aq 4 eiθ , aq 4 eiθ , −aq 4 eiθ , −aq 4 eiθ ; q)∞ 2 2 1 (e2iθ ; q) (eiθ , −eiθ ; q 2 )∞ ∞ = = 1 1 1 1 (aq 4 eiθ , −aq 4 eiθ ; q 2 )∞ (a2 q 2 e2iθ ; q)∞ 1
=
1
h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) 1
3
1
3
h(x, aq 4 )h(x, aq 4 )h(x, −aq 4 )h(x, −aq 4 )
,
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
x = cos θ .
Recurrence Relation 1
1
1
2(1 − qn+ 2 )xPn (x|q) = q− 4 (1 − qn+1 )Pn+1 (x|q) + q 4 (1 − qn )Pn−1 (x|q). (14.10.38)
476
14 Basic Hypergeometric Orthogonal Polynomials
Normalized Recurrence Relation xpn (x) = pn+1 (x) +
(1 − qn )2 1
pn−1 (x),
1
4(1 − qn− 2 )(1 − qn+ 2 )
where
1
Pn (x|q) =
(14.10.39)
1
2n q 4 n (q 2 ; q)n pn (x). (q; q)n
q-Difference Equation 1 ˜ q 2 |q)Dq y(x) + λn w(x; (1−q)2 Dq w(x; ˜ 1|q)y(x) = 0, where
y(x) = Pn (x|q), (14.10.40)
λn = 4q−n+1 (1 − qn )(1 − qn+1 )
and
w(x; a|q) . w(x; ˜ a|q) := √ 1 − x2
Rodrigues-Type Formula w(x; ˜ 1|q)Pn (x|q) =
q−1 2
n
1 2
q4n 1 2
(q, −q , −q; q)n
1 ˜ q 2 n |q) . (Dq )n w(x;
(14.10.41)
Generating Functions 1 (q 2 eiθ t; q) 2 (q 12 eiθ t, q 12 e−iθ t; q) ∞ Pn (x|q) n ∞ ∞ = iθ = ∑ 1n t , i θ −i θ (e t; q)∞ (e t, e t; q)∞ n=0 q 4 1 2 φ1 i θ (e t; q)∞
q 2 , q 2 e2iθ ; q, e−iθ t q 1
1
∞
=
∑
Pn (x|q) 1
n n=0 (q; q)n q 4
t n,
x = cos θ . (14.10.42)
x = cos θ . (14.10.43)
14.10 Continuous q-Jacobi
477
−iθ
(e
t; q)∞ · 2 φ1
∞
= 2 φ1 ∞
=
1
n
1
3
2 φ1
2 φ1
q 4 eiθ , −q 4 eiθ 1
1
∞
n=0
2 φ1
(14.10.44)
−q 4 e−iθ , −q 4 e−iθ ; q, eiθ t q 1
3
; q, e
(−q; q)n Pn (x|q) 3
(−q 2 ; q)n
1
q4n
−iθ
∞
2 φ1
t
(14.10.45)
q 4 e−iθ , −q 4 e−iθ 3
3
3
−q 2
iθ
; q, e t
x = cos θ .
t n,
q 4 eiθ , −q 4 eiθ ; q, e−iθ t −q 1
x = cos θ .
1
−q 2
∑
=
x = cos θ .
1
(−q 2 , −q; q)n Pn (x|q) n t , ∑ (q, q; q)n 1 q4n n=0
=
1
(−1)n q 4 n+(2) ∑ (q; q)n Pn (x|q)t n , n=0
q 4 e iθ , q 4 e iθ ; q, e−iθ t q 1
q 2 , q 2 e2iθ ; q, e−iθ t q
3
2 φ1
(14.10.46)
q 4 e−iθ , −q 4 e−iθ ; q, eiθ t −q 3
1
1
(−q 2 ; q)n Pn (x|q) n t , 1 q4n n=0 (−q; q)n
∑
(γ eiθ t; q)∞ 3 φ2 (eiθ t; q)∞ ∞
=
γ , q 2 , q 2 e2iθ ; q, e−iθ t q, γ eiθ t 1
(γ ; q)n Pn (x|q)
∑ (q; q)n
n=0
x = cos θ .
q
1n 4
t n,
1
x = cos θ ,
(14.10.47)
γ arbitrary.
(14.10.48)
Limit Relations Continuous q-Legendre → Legendre / Spherical The Legendre (or spherical) polynomials given by (9.8.62) easily follow from the continuous q-Legendre polynomials given by (14.10.36) by taking the limit q → 1: lim Pn (x|q) = Pn (x).
q→1
(14.10.49)
478
14 Basic Hypergeometric Orthogonal Polynomials
Remarks The continuous q-Legendre polynomials can also be written as: 1
n
1
1
(q 2 ; q)k (q 2 ; q)n−k i(n−2k)θ e , k=0 (q; q)k (q; q)n−k
Pn (x|q) = q 4 n ∑
If we set α = β = 0 in (14.10.16) we find 1 1 q−n , qn+1 , q 2 eiθ , q 2 e−iθ ; q, q , Pn (x; q) = 4 φ3 q, −q, −q
x = cos θ .
x = cos θ ,
but these are not really different from those given by (14.10.36) in view of the quadratic transformation Pn (x|q2 ) = Pn (x; q). If we replace q by q−1 we find Pn (x|q−1 ) = Pn (x|q). The continuous q-Legendre polynomials are related to the continuous q-ultraspherical (or Rogers) polynomials given by (14.10.17) in the following way: 1
1
Pn (x|q) = q 4 nCn (x; q 2 |q).
References [320], [323], [341], [345], [348].
14.11 Big q-Laguerre Basic Hypergeometric Representation q−n , 0, x ; q, q Pn (x; a, b; q) = 3 φ2 aq, bq −n 1 x q , aqx−1 = −1 −n ; q, . 2 φ1 (b q ; q)n aq b
(14.11.1)
14.11 Big q-Laguerre
479
Orthogonality Relation For 0 < aq < 1 and b < 0 we have aq −1 (a x, b−1 x; q)∞ bq
(x; q)∞
= aq(1 − q)
Pm (x; a, b; q)Pn (x; a, b; q) dq x
n (q, a−1 b, ab−1 q; q)∞ (q; q)n (−abq2 )n q(2) δmn . (14.11.2) (aq, bq; q)∞ (aq, bq; q)n
Recurrence Relation (x − 1)Pn (x; a, b; q) = An Pn+1 (x; a, b; q) − (An +Cn ) Pn (x; a, b; q) (14.11.3) +Cn Pn−1 (x; a, b; q), where
⎧ ⎨ An = (1 − aqn+1 )(1 − bqn+1 ) ⎩
Cn = −abqn+1 (1 − qn ).
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 − (An +Cn )] pn (x) − abqn+1 (1 − qn )(1 − aqn )(1 − bqn )pn−1 (x), (14.11.4) where Pn (x; a, b; q) =
1 pn (x). (aq, bq; q)n
q-Difference Equation q−n (1 − qn )x2 y(x) = B(x)y(qx) − [B(x) + D(x)] y(x) + D(x)y(q−1 x), where y(x) = Pn (x; a, b; q) and
⎧ ⎨ B(x) = abq(1 − x) ⎩
D(x) = (x − aq)(x − bq).
(14.11.5)
480
14 Basic Hypergeometric Orthogonal Polynomials
Forward Shift Operator Pn (x; a, b; q) − Pn (qx; a, b; q) =
q−n+1 (1 − qn ) xPn−1 (qx; aq, bq; q) (1 − aq)(1 − bq)
(14.11.6)
or equivalently Dq Pn (x; a, b; q) =
q−n+1 (1 − qn ) Pn−1 (qx; aq, bq; q). (1 − q)(1 − aq)(1 − bq)
(14.11.7)
Backward Shift Operator (x − a)(x − b)Pn (x; a, b; q) − ab(1 − x)Pn (qx; a, b; q) = (1 − a)(1 − b)xPn+1 (x; aq−1 , bq−1 ; q)
(14.11.8)
or equivalently Dq [w(x; a, b; q)Pn (x; a, b; q)] (1 − a)(1 − b) w(x; aq−1 , bq−1 ; q)Pn+1 (x; aq−1 , bq−1 ; q), (14.11.9) = ab(1 − q) where w(x; a, b; q) =
(a−1 x, b−1 x; q)∞ . (x; q)∞
Rodrigues-Type Formula w(x; a, b; q)Pn (x; a, b; q) =
an bn qn(n+1) (1 − q)n (Dq )n [w(x; aqn , bqn ; q)] . (aq, bq; q)n
(14.11.10)
Generating Functions (bqt; q)∞ · 2 φ1
aqx−1 , 0 ; q, xt aq
∞
=
(bq; q)n Pn (x; a, b; q)t n . n=0 (q; q)n
∑
(14.11.11)
14.11 Big q-Laguerre
481
(aqt; q)∞ · 2 φ1 (t; q)∞ · 3 φ2
bqx−1 , 0 ; q, xt bq
0, 0, x ; q, t aq, bq
∞
=
(aq; q)n Pn (x; a, b; q)t n . (q; q) n n=0
∑
∞
(14.11.12)
n
(−1)n q(2) =∑ Pn (x; a, b; q)t n . n=0 (q; q)n
(14.11.13)
Limit Relations Big q-Jacobi → Big q-Laguerre If we set b = 0 in the definition (14.5.1) of the big q-Jacobi polynomials we obtain the big q-Laguerre polynomials given by (14.11.1): Pn (x; a, 0, c; q) = Pn (x; a, c; q). Big q-Laguerre → Little q-Laguerre / Wall The little q-Laguerre (or Wall) polynomials given by (14.20.1) can be obtained from the big q-Laguerre polynomials by taking x → bqx in (14.11.1) and letting b → −∞: lim Pn (bqx; a, b; q) = pn (x; a|q).
b→−∞
(14.11.14)
Big q-Laguerre → Al-Salam-Carlitz I If we set x → aqx and b → ab in the definition (14.11.1) of the big q-Laguerre polynomials and take the limit a → 0 we obtain the Al-Salam-Carlitz I polynomials given by (14.24.1): Pn (aqx; a, ab; q) (b) lim = qnUn (x; q). (14.11.15) a→0 an Big q-Laguerre → Laguerre The Laguerre polynomials given by (9.12.1) can be obtained from the big q-Laguerre polynomials by the substitution a = qα and b = (1 − q)−1 qβ in the definition (14.11.1) of the big q-Laguerre polynomials and the limit q → 1: lim Pn (x; qα , (1 − q)−1 qβ ; q) =
q→1
(α )
Ln (x − 1) (α )
Ln (0)
.
(14.11.16)
482
14 Basic Hypergeometric Orthogonal Polynomials
Remark The big q-Laguerre polynomials given by (14.11.1) and the affine q-Krawtchouk polynomials given by (14.16.1) are related in the following way: KnA f f (q−x ; p, N; q) = Pn (q−x ; p, q−N−1 ; q).
References [16], [28], [74], [157], [284].
14.12 Little q-Jacobi Basic Hypergeometric Representation pn (x; a, b|q) = 2 φ1
q−n , abqn+1 ; q, qx . aq
(14.12.1)
Orthogonality Relation For 0 < aq < 1 and bq < 1 we have ∞
(bq; q)k (aq)k pm (qk ; a, b|q)pn (qk ; a, b|q) (q; q) k k=0
∑ =
(abq2 ; q)∞ (1 − abq)(aq)n (q, bq; q)n δmn . (aq; q)∞ (1 − abq2n+1 ) (aq, abq; q)n
(14.12.2)
Recurrence Relation − xpn (x; a, b|q) = An pn+1 (x; a, b|q) − (An +Cn ) pn (x; a, b|q) +Cn pn−1 (x; a, b|q), where
(14.12.3)
14.12 Little q-Jacobi
483
⎧ n+1 )(1 − abqn+1 ) ⎪ n (1 − aq ⎪ ⎪ = q A n ⎪ ⎨ (1 − abq2n+1 )(1 − abq2n+2 ) ⎪ ⎪ ⎪ ⎪ ⎩ Cn = aqn
(1 − qn )(1 − bqn ) . (1 − abq2n )(1 − abq2n+1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn )pn (x) + An−1Cn pn−1 (x), where
(14.12.4)
n
pn (x; a, b|q) =
(−1)n q−(2) (abqn+1 ; q)n pn (x). (aq; q)n
q-Difference Equation q−n (1 − qn )(1 − abqn+1 )xy(x) = B(x)y(qx) − [B(x) + D(x)] y(x) + D(x)y(q−1 x),
(14.12.5)
where y(x) = pn (x; a, b|q) ⎧ ⎨ B(x) = a(bqx − 1)
and
⎩
D(x) = x − 1.
Forward Shift Operator pn (x; a, b|q) − pn (qx; a, b|q) =−
q−n+1 (1 − qn )(1 − abqn+1 ) xpn−1 (x; aq, bq|q) (1 − aq)
(14.12.6)
or equivalently Dq pn (x; a, b|q) = −
q−n+1 (1 − qn )(1 − abqn+1 ) pn−1 (x; aq, bq|q). (1 − q)(1 − aq)
(14.12.7)
484
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator a(bx − 1)pn (x; a, b|q) − (x − 1)pn (q−1 x; a, b|q) = (1 − a)pn+1 (x; aq−1 , bq−1 |q)
(14.12.8)
or equivalently Dq−1 w(x; α , β |q)pn (x; qα , qβ |q) =
1 − qα w(x; α − 1, β − 1|q)pn+1 (x; qα −1 , qβ −1 |q), qα −1 (1 − q)
where w(x; α , β |q) =
(14.12.9)
(qx; q)∞ α x . (qβ +1 x; q)∞
Rodrigues-Type Formula w(x; α , β |q)pn (x; qα , qβ |q) n n qnα +(2) (1 − q)n Dq−1 [w(x; α + n, β + n|q)] . = α +1 (q ; q)n
(14.12.10)
Generating Function 0 φ1
− ; q, aqxt aq
2 φ1
x−1 , 0 ; q, xt bq
n (−1)n q(2) =∑ pn (x; a, b|q)t n . (14.12.11) n=0 (bq, q; q)n
∞
Limit Relations Big q-Jacobi → Little q-Jacobi The little q-Jacobi polynomials given by (14.12.1) can be obtained from the big qJacobi polynomials by the substitution x → cqx in the definition (14.5.1) and then by the limit c → −∞: lim Pn (cqx; a, b, c; q) = pn (x; a, b|q).
c→−∞
14.12 Little q-Jacobi
485
q-Hahn → Little q-Jacobi If we set x → N − x in the definition (14.6.1) of the q-Hahn polynomials and take the limit N → ∞ we find the little q-Jacobi polynomials: lim Qn (qx−N ; α , β , N|q) = pn (qx ; α , β |q),
N→∞
where pn (qx ; α , β |q) is given by (14.12.1). Little q-Jacobi → Little q-Laguerre / Wall The little q-Laguerre (or Wall) polynomials given by (14.20.1) are little q-Jacobi polynomials with b = 0. So if we set b = 0 in the definition (14.12.1) of the little q-Jacobi polynomials we obtain the little q-Laguerre (or Wall) polynomials: pn (x; a, 0|q) = pn (x; a|q).
(14.12.12)
Little q-Jacobi → q-Laguerre If we substitute a = qα and x → −b−1 q−1 x in the definition (14.12.1) of the little q-Jacobi polynomials and then take the limit b → −∞ we find the q-Laguerre polynomials given by (14.21.1): lim pn (−b−1 q−1 x; qα , b|q) =
b→−∞
(q; q)n (α ) Ln (x; q). (qα +1 ; q)n
(14.12.13)
Little q-Jacobi → q-Bessel If we set b → −a−1 q−1 b in the definition (14.12.1) of the little q-Jacobi polynomials and then take the limit a → 0 we obtain the q-Bessel polynomials given by (14.22.1): lim pn (x; a, −a−1 q−1 b|q) = yn (x; b; q).
a→0
(14.12.14)
Little q-Jacobi → Jacobi / Laguerre The Jacobi polynomials given by (9.8.1) simply follow from the little q-Jacobi polynomials given by (14.12.1) in the following way: lim pn (x; qα , qβ |q) =
q→1
(α ,β )
Pn
(1 − 2x)
(α ,β ) Pn (1)
.
(14.12.15)
486
14 Basic Hypergeometric Orthogonal Polynomials
If we take a = qα , b = −qβ for arbitrary real β and x → 12 (1 − q)x in the definition (14.12.1) of the little q-Jacobi polynomials and then take the limit q → 1 we obtain the Laguerre polynomials given by (9.12.1): lim pn ( 12 (1 − q)x; qα , −qβ |q) =
q→1
(α )
Ln (x)
(α ) Ln (0)
.
(14.12.16)
Remarks The little q-Jacobi polynomials given by (14.12.1) and the big q-Jacobi polynomials given by (14.5.1) are related in the following way: pn (x; a, b|q) =
n (bq; q)n (−1)n b−n q−n−(2) Pn (bqx; b, a, 0; q). (aq; q)n
The little q-Jacobi polynomials and the q-Meixner polynomials given by (14.13.1) are related in the following way: Mn (q−x ; b, c; q) = pn (−c−1 qn ; b, b−1 q−n−x−1 |q).
References [12], [16], [23], [24], [33], [34], [51], [79], [80], [157], [212], [213], [214], [235], [238], [256], [261], [271], [287], [298], [304], [326], [329], [343], [345], [346], [348], [402], [408], [416], [442], [480], [482], [485].
Special Case 14.12.1 Little q-Legendre Basic Hypergeometric Representation The little q-Legendre polynomials are little q-Jacobi polynomials with a = b = 1: −n n+1 q ,q ; q, qx . (14.12.17) pn (x|q) = 2 φ1 q
14.12 Little q-Jacobi
487
Orthogonality Relation 1 0
∞
pm (x|q)pn (x|q) dq x = (1 − q) ∑ qk pm (qk |q)pn (qk |q) k=0
(1 − q)qn δmn . = (1 − q2n+1 )
(14.12.18)
Recurrence Relation − xpn (x|q) = An pn+1 (x|q) − (An +Cn ) pn (x|q) +Cn pn−1 (x|q), where
(14.12.19)
⎧ (1 − qn+1 ) ⎪ n ⎪ ⎪ = q A n ⎪ ⎨ (1 + qn+1 )(1 − q2n+1 ) ⎪ ⎪ ⎪ ⎪ ⎩ Cn = qn
(1 − qn ) . (1 + qn )(1 − q2n+1 )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn )pn (x) + An−1Cn pn−1 (x), where
(14.12.20)
n
(−1)n q−(2) (qn+1 ; q)n pn (x|q) = pn (x). (q; q)n
q-Difference Equation q−n (1 − qn )(1 − qn+1 )xy(x) = B(x)y(qx) − [B(x) + D(x)] y(x) + D(x)y(q−1 x), where y(x) = pn (x|q) and
⎧ ⎨ B(x) = qx − 1 ⎩
D(x) = x − 1.
(14.12.21)
488
14 Basic Hypergeometric Orthogonal Polynomials
Rodrigues-Type Formula n n q(2) (1 − q)n Dq−1 [(qx; q)n xn ] . pn (x|q) = (q; q)n
(14.12.22)
Generating Function 0 φ1
− ; q, qxt q
2 φ1
x−1 , 0 ; q, xt q
n (−1)n q(2) ∑ (q, q; q)n pn (x|q)t n . n=0
∞
=
(14.12.23)
Limit Relation Little q-Legendre → Legendre / Spherical If we take the limit q → 1 in the definition (14.12.17) of the little q-Legendre polynomials we simply find the Legendre (or spherical) polynomials given by (9.8.62): lim pn (x|q) = Pn (1 − 2x).
(14.12.24)
q→1
References [345], [346], [447], [504].
14.13 q-Meixner Basic Hypergeometric Representation −x
Mn (q ; b, c; q) = 2 φ1
qn+1 q−n , q−x ; q, − bq c
.
(14.13.1)
14.13 q-Meixner
489
Orthogonality Relation ∞
(bq; q)x
∑ (q, −bcq; q)x cx q(2) Mm (q−x ; b, c; q)Mn (q−x ; b, c; q) x
x=0
=
(−c; q)∞ (q, −c−1 q; q)n −n q δmn , (−bcq; q)∞ (bq; q)n
0 ≤ bq < 1,
c > 0. (14.13.2)
Recurrence Relation q2n+1 (1 − q−x )Mn (q−x ) = c(1 − bqn+1 )Mn+1 (q−x )
− c(1 − bqn+1 ) + q(1 − qn )(c + qn ) Mn (q−x ) + q(1 − qn )(c + qn )Mn−1 (q−x ), where
(14.13.3)
Mn (q−x ) := Mn (q−x ; b, c; q).
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + 1 + q−2n−1 c(1 − bqn+1 ) + q(1 − qn )(c + qn ) pn (x) + cq−4n+1 (1 − qn )(1 − bqn )(c + qn )pn−1 (x), where
(14.13.4)
2
Mn (q−x ; b, c; q) =
(−1)n qn pn (q−x ). (bq; q)n cn
q-Difference Equation − (1 − qn )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where and
y(x) = Mn (q−x ; b, c; q)
(14.13.5)
490
14 Basic Hypergeometric Orthogonal Polynomials
⎧ ⎨ B(x) = cqx (1 − bqx+1 ) ⎩
D(x) = (1 − qx )(1 + bcqx ).
Forward Shift Operator Mn (q−x−1 ; b, c; q) − Mn (q−x ; b, c; q) q−x (1 − qn ) Mn−1 (q−x ; bq, cq−1 ; q) =− c(1 − bq)
(14.13.6)
or equivalently
Δ Mn (q−x ; b, c; q) q(1 − qn ) Mn−1 (q−x ; bq, cq−1 ; q). = − Δ q−x c(1 − q)(1 − bq)
(14.13.7)
Backward Shift Operator cqx (1 − bqx )Mn (q−x ; b, c; q) − (1 − qx )(1 + bcqx )Mn (q−x+1 ; b, c; q) (14.13.8) = cqx (1 − b)Mn+1 (q−x ; bq−1 , cq; q) or equivalently ∇ [w(x; b, c; q)Mn (q−x ; b, c; q)] ∇q−x 1 w(x; bq−1 , cq; q)Mn+1 (q−x ; bq−1 , cq; q), = 1−q where w(x; b, c; q) =
(14.13.9)
x+1 (bq; q)x cx q( 2 ) . (q, −bcq; q)x
Rodrigues-Type Formula
w(x; b, c; q)Mn (q−x ; b, c; q) = (1 − q)n (∇q )n w(x; bqn , cq−n ; q) , where ∇q :=
∇ . ∇q−x
(14.13.10)
14.13 q-Meixner
491
Generating Functions 1 1 φ1 (t; q)∞
q−x ; q, −c−1 qt bq
∞
=
Mn (q−x ; b, c; q) n t . (q; q)n n=0
∑
−1 −1 −x −b c q 1 ; q, bqt φ 1 1 (t; q)∞ −c−1 q ∞ (bq; q)n Mn (q−x ; b, c; q)t n . =∑ −1 q, q; q) (−c n n=0
(14.13.11)
(14.13.12)
Limit Relations Big q-Jacobi → q-Meixner If we set b = −a−1 cd −1 (with d > 0) in the definition (14.5.1) of the big q-Jacobi polynomials and take the limit c → −∞ we obtain the q-Meixner polynomials given by (14.13.1): lim Pn (q−x ; a, −a−1 cd −1 , c; q) = Mn (q−x ; a, d; q).
c→−∞
(14.13.13)
q-Hahn → q-Meixner The q-Meixner polynomials given by (14.13.1) can be obtained from the q-Hahn polynomials by setting α = b and β = −b−1 c−1 q−N−1 in the definition (14.6.1) of the q-Hahn polynomials and letting N → ∞: lim Qn (q−x ; b, −b−1 c−1 q−N−1 , N|q) = Mn (q−x ; b, c; q).
N→∞
q-Meixner → q-Laguerre The q-Laguerre polynomials given by (14.21.1) can be obtained from the q-Meixner polynomials given by (14.13.1) by setting b = qα and q−x → cqα x in the definition (14.13.1) of the q-Meixner polynomials and then taking the limit c → ∞: lim Mn (cqα x; qα , c; q) =
c→∞
(q; q)n (α ) Ln (x; q). (qα +1 ; q)n
(14.13.14)
492
14 Basic Hypergeometric Orthogonal Polynomials
q-Meixner → q-Charlier The q-Charlier polynomials given by (14.23.1) can easily be obtained from the qMeixner given by (14.13.1) by setting b = 0 in the definition (14.13.1) of the qMeixner polynomials: Mn (x; 0, c; q) = Cn (x; c; q). (14.13.15) q-Meixner → Al-Salam-Carlitz II The Al-Salam-Carlitz II polynomials given by (14.25.1) can be obtained from the q-Meixner polynomials given by (14.13.1) by setting b = −ac−1 in the definition (14.13.1) of the q-Meixner polynomials and then taking the limit c → 0: 1 n (n) (a) −1 q 2 Vn (x; q). (14.13.16) lim Mn (x; −ac , c; q) = − c→0 a q-Meixner → Meixner To find the Meixner polynomials given by (9.10.1) from the q-Meixner polynomials given by (14.13.1) we set b = qβ −1 and c → (1 − c)−1 c and let q → 1: lim Mn (q−x ; qβ −1 , (1 − c)−1 c; q) = Mn (x; β , c).
q→1
(14.13.17)
Remarks The q-Meixner polynomials given by (14.13.1) and the little q-Jacobi polynomials given by (14.12.1) are related in the following way: Mn (q−x ; b, c; q) = pn (−c−1 qn ; b, b−1 q−n−x−1 |q). The q-Meixner polynomials and the quantum q-Krawtchouk polynomials given by (14.14.1) are related in the following way: Knqtm (q−x ; p, N; q) = Mn (q−x ; q−N−1 , −p−1 ; q).
References [16], [27], [28], [30], [74], [80], [125], [238], [261], [276], [416], [472].
14.14 Quantum q-Krawtchouk
493
14.14 Quantum q-Krawtchouk Basic Hypergeometric Representation Knqtm (q−x ; p, N; q) = 2 φ1
q−n , q−x n+1 , ; q, pq q−N
n = 0, 1, 2, . . . , N.
(14.14.1)
Orthogonality Relation N
(pq; q)N−x
∑ (q; q)x (q; q)N−x (−1)N−x q(2) Kmqtm (q−x ; p, N; q)Knqtm (q−x ; p, N; q) x
x=0
=
(−1)n pN (q; q)N−n (q, pq; q)n (N+1)−(n+1)+Nn 2 q 2 δmn , (q, q; q)N
p > q−N . (14.14.2)
Recurrence Relation −pq2n+1 (1 − q−x )Knqtm (q−x ) qtm −x (q ) = (1 − qn−N )Kn+1
n−N − (1 − q ) + q(1 − qn )(1 − pqn ) Knqtm (q−x ) qtm −x (q ), + q(1 − qn )(1 − pqn )Kn−1
where
(14.14.3)
Knqtm (q−x ) := Knqtm (q−x ; p, N; q).
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + 1 − p−1 q−2n−1 (1 − qn−N ) + q(1 − qn )(1 − pqn ) pn (x) + p−2 q−4n+1 (1 − qn )(1 − pqn )(1 − qn−N−1 )pn−1 (x), where
2
Knqtm (q−x ; p, N; q) =
pn qn pn (q−x ). (q−N ; q)n
(14.14.4)
494
14 Basic Hypergeometric Orthogonal Polynomials
q-Difference Equation − p(1 − qn )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where
(14.14.5)
y(x) = Knqtm (q−x ; p, N; q) ⎧ ⎨ B(x) = −qx (1 − qx−N )
and
⎩
D(x) = (1 − qx )(p − qx−N−1 ).
Forward Shift Operator Knqtm (q−x−1 ; p, N; q) − Knqtm (q−x ; p, N; q) pq−x (1 − qn ) qtm −x Kn−1 (q ; pq, N − 1; q) = 1 − q−N
(14.14.6)
or equivalently
Δ Knqtm (q−x ; p, N; q) pq(1 − qn ) K qtm (q−x ; pq, N − 1; q). = Δ q−x (1 − q)(1 − q−N ) n−1
(14.14.7)
Backward Shift Operator (1 − qx−N−1 )Knqtm (q−x ; p, N; q) + q−x (1 − qx )(p − qx−N−1 )Knqtm (q−x+1 ; p, N; q) qtm −x (q ; pq−1 , N + 1; q) = (1 − q−N−1 )Kn+1
(14.14.8)
or equivalently
∇ w(x; p, N; q)Knqtm (q−x ; p, N; q) ∇q−x 1 qtm −x w(x; pq−1 , N + 1; q)Kn+1 = (q ; pq−1 , N + 1; q), 1−q where w(x; p, N; q) =
x+1 (q−N ; q)x (−p)−x q( 2 ) . −1 −N (q, p q ; q)x
(14.14.9)
14.14 Quantum q-Krawtchouk
495
Rodrigues-Type Formula w(x; p, N; q)Knqtm (q−x ; p, N; q) = (1 − q)n (∇q )n [w(x; pqn , N − n; q)] , where ∇q :=
(14.14.10)
∇ . ∇q−x
Generating Functions For x = 0, 1, 2, . . . , N we have (q
x−N
t; q)N−x · 2 φ1
q−x , pqN+1−x ; q, qx−N t 0
(q−N ; q)n qtm −x Kn (q ; p, N; q)t n . (q; q) n n=0 N
=
∑
(q−xt; q)x · 2 φ1 N
=
qx−N , 0 ; q, q−xt pq
(14.14.11)
(q−N ; q)n
∑ (pq, q; q)n Knqtm (q−x ; p, N; q)t n .
(14.14.12)
n=0
Limit Relations q-Hahn → Quantum q-Krawtchouk The quantum q-Krawtchouk polynomials given by (14.14.1) simply follow from the q-Hahn polynomials by setting β = p in the definition (14.6.1) of the q-Hahn polynomials and taking the limit α → ∞: lim Qn (q−x ; α , p, N|q) = Knqtm (q−x ; p, N; q).
α →∞
Quantum q-Krawtchouk → Al-Salam-Carlitz II If we set p = a−1 q−N−1 in the definition (14.14.1) of the quantum q-Krawtchouk polynomials and let N → ∞ we obtain the Al-Salam-Carlitz II polynomials given by (14.25.1). In fact we have
496
14 Basic Hypergeometric Orthogonal Polynomials
lim
N→∞
Knqtm (x; a−1 q−N−1 , N; q) =
1 n (n) (a) − q 2 Vn (x; q). a
(14.14.13)
Quantum q-Krawtchouk → Krawtchouk The Krawtchouk polynomials given by (9.11.1) easily follow from the quantum qKrawtchouk polynomials given by (14.14.1) in the following way: lim Knqtm (q−x ; p, N; q) = Kn (x; p−1 , N).
q→1
(14.14.14)
Remarks The quantum q-Krawtchouk polynomials given by (14.14.1) and the q-Meixner polynomials given by (14.13.1) are related in the following way: Knqtm (q−x ; p, N; q) = Mn (q−x ; q−N−1 , −p−1 ; q). The quantum q-Krawtchouk polynomials are related to the affine q-Krawtchouk polynomials given by (14.16.1) by the transformation q ↔ q−1 in the following way: p n −(n) A f f x−N −1 qtm x −1 −1 Kn (q ; p, N; q ) = (p q; q)n − q 2 Kn (q ; p , N; q). q
References [238], [343], [345], [472].
14.15 q-Krawtchouk Basic Hypergeometric Representation For n = 0, 1, 2, . . . , N we have q−n , q−x , −pqn ; q, q (14.15.1) q−N , 0 −n −x (qx−N ; q)n q ,q n+N+1 = −N . φ ; q, −pq 2 1 (q ; q)n qnx qN−x−n+1
Kn (q−x ; p, N; q) = 3 φ2
14.15 q-Krawtchouk
497
Orthogonality Relation (q−N ; q)x (−p)−x Km (q−x ; p, N; q)Kn (q−x ; p, N; q) (q; q) x x=0 N
∑ =
(q, −pqN+1 ; q)n (1 + p) (−p, q−N ; q)n (1 + pq2n ) n 2 N+1 × (−pq; q)N p−N q−( 2 ) −pq−N qn δmn ,
p > 0.
(14.15.2)
Recurrence Relation − 1 − q−x Kn (q−x ) = An Kn+1 (q−x ) − (An +Cn ) Kn (q−x ) +Cn Kn−1 (q−x ), where and
(14.15.3)
Kn (q−x ) := Kn (q−x ; p, N; q) ⎧ (1 − qn−N )(1 + pqn ) ⎪ ⎪ An = ⎪ ⎪ ⎨ (1 + pq2n )(1 + pq2n+1 ) ⎪ ⎪ (1 + pqn+N )(1 − qn ) ⎪ ⎪ ⎩ Cn = −pq2n−N−1 . (1 + pq2n−1 )(1 + pq2n )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + [1 − (An +Cn )] pn (x) + An−1Cn pn−1 (x), where Kn (q−x ; p, N; q) =
(14.15.4)
(−pqn ; q)n pn (q−x ). (q−N ; q)n
q-Difference Equation q−n (1 − qn )(1 + pqn )y(x)
= (1 − qx−N )y(x + 1) − (1 − qx−N ) − p(1 − qx ) y(x) − p(1 − qx )y(x − 1),
(14.15.5)
498
14 Basic Hypergeometric Orthogonal Polynomials
where
y(x) = Kn (q−x ; p, N; q).
Forward Shift Operator Kn (q−x−1 ; p, N; q) − Kn (q−x ; p, N; q) q−n−x (1 − qn )(1 + pqn ) = Kn−1 (q−x ; pq2 , N − 1; q) 1 − q−N
(14.15.6)
or equivalently
Δ Kn (q−x ; p, N; q) q−n+1 (1 − qn )(1 + pqn ) Kn−1 (q−x ; pq2 , N − 1; q). (14.15.7) = Δ q−x (1 − q)(1 − q−N )
Backward Shift Operator (1 − qx−N−1 )Kn (q−x ; p, N; q) + pq−1 (1 − qx )Kn (q−x+1 ; p, N; q) (14.15.8) = qx (1 − q−N−1 )Kn+1 (q−x ; pq−2 , N + 1; q) or equivalently ∇ [w(x; p, N; q)Kn (q−x ; p, N; q)] ∇q−x 1 w(x; pq−2 , N + 1; q)Kn+1 (q−x ; pq−2 , N + 1; q), = 1−q where w(x; p, N; q) =
(q−N ; q)x (q; q)x
(14.15.9)
q x − . p
Rodrigues-Type Formula
w(x; p, N; q)Kn (q−x ; p, N; q) = (1 − q)n (∇q )n w(x; pq2n , N − n; q) , where ∇q :=
∇ . ∇q−x
(14.15.10)
14.15 q-Krawtchouk
499
Generating Function For x = 0, 1, 2, . . . , N we have −x x−N q q ,0 −x ; q, −q t ; q, pqt 2 φ0 1 φ1 − 0 (q−N ; q)n −(n) q 2 Kn (q−x ; p, N; q)t n . (q; q) n n=0 N
=
∑
(14.15.11)
Limit Relations q-Racah → q-Krawtchouk The q-Krawtchouk polynomials given by (14.15.1) can be obtained from the q-Racah polynomials by setting α q = q−N , β = −pqN and γ = δ = 0 in the definition (14.2.1) of the q-Racah polynomials: Rn (q−x ; q−N−1 , −pqN , 0, 0|q) = Kn (q−x ; p, N; q). Note that μ (x) = q−x in this case. q-Hahn → q-Krawtchouk If we set β = −α −1 q−1 p in the definition (14.6.1) of the q-Hahn polynomials and then let α → 0 we obtain the q-Krawtchouk polynomials given by (14.15.1): lim Qn (q−x ; α , −α −1 q−1 p, N|q) = Kn (q−x ; p, N; q).
α →0
q-Krawtchouk → q-Bessel If we set x → N − x in the definition (14.15.1) of the q-Krawtchouk polynomials and then take the limit N → ∞ we obtain the q-Bessel polynomials given by (14.22.1): lim Kn (qx−N ; p, N; q) = yn (qx ; p; q).
N→∞
(14.15.12)
q-Krawtchouk → q-Charlier By setting p = a−1 q−N in the definition (14.15.1) of the q-Krawtchouk polynomials and then taking the limit N → ∞ we obtain the q-Charlier polynomials given by
500
(14.23.1):
14 Basic Hypergeometric Orthogonal Polynomials
lim Kn (q−x ; a−1 q−N , N; q) = Cn (q−x ; a; q).
N→∞
(14.15.13)
q-Krawtchouk → Krawtchouk If we take the limit q → 1 in the definition (14.15.1) of the q-Krawtchouk polynomials we simply find the Krawtchouk polynomials given by (9.11.1) in the following way: (14.15.14) lim Kn (q−x ; p, N; q) = Kn (x; (p + 1)−1 , N). q→1
Remark The q-Krawtchouk polynomials given by (14.15.1) and the dual q-Krawtchouk polynomials given by (14.17.1) are related in the following way: Kn (q−x ; p, N; q) = Kx (λ (n); −pqN , N|q) with or with
λ (n) = q−n − pqn Kn (λ (x); c, N|q) = Kx (q−n ; −cq−N , N; q)
λ (x) = q−x + cqx−N .
References [30], [70], [80], [125], [238], [416], [421], [487], [488].
14.16 Affine q-Krawtchouk
501
14.16 Affine q-Krawtchouk Basic Hypergeometric Representation
q−n , 0, q−x ; q, q pq, q−N n −n x−N q ,q q−x (−pq)n q(2) , ; q, = 2 φ1 (pq; q)n q−N p
KnA f f (q−x ; p, N; q) = 3 φ2
(14.16.1) n = 0, 1, 2, . . . , N.
Orthogonality Relation N
(pq; q)x (q; q)N
∑ (q; q)x (q; q)N−x (pq)−x KmA f f (q−x ; p, N; q)KnA f f (q−x ; p, N; q)
x=0
= (pq)n−N
(q; q)n (q; q)N−n δmn , (pq; q)n (q; q)N
0 < pq < 1.
(14.16.2)
Recurrence Relation −(1 − q−x )KnA f f (q−x ) A f f −x = (1 − qn−N )(1 − pqn+1 )Kn+1 (q )
n−N n+1 − (1 − q )(1 − pq ) − pqn−N (1 − qn ) KnA f f (q−x ) A f f −x − pqn−N (1 − qn )Kn−1 (q ),
where
(14.16.3)
KnA f f (q−x ) := KnA f f (q−x ; p, N; q).
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + 1 − (1 − qn−N )(1 − pqn+1 ) − pqn−N (1 − qn ) pn (x) − pqn−N (1 − qn )(1 − pqn )(1 − qn−N−1 )pn−1 (x), where
(14.16.4)
502
14 Basic Hypergeometric Orthogonal Polynomials
KnA f f (q−x ; p, N; q) =
1 pn (q−x ). (pq, q−N ; q)n
q-Difference Equation q−n (1 − qn )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where
(14.16.5)
y(x) = KnA f f (q−x ; p, N; q) ⎧ ⎨ B(x) = (1 − qx−N )(1 − pqx+1 )
and
⎩
D(x) = −p(1 − qx )qx−N .
Forward Shift Operator KnA f f (q−x−1 ; p, N; q) − KnA f f (q−x ; p, N; q) q−n−x (1 − qn ) K A f f (q−x ; pq, N − 1; q) = (1 − pq)(1 − q−N ) n−1
(14.16.6)
or equivalently
Δ KnA f f (q−x ; p, N; q) Δ q−x q−n+1 (1 − qn ) K A f f (q−x ; pq, N − 1; q). = (1 − q)(1 − pq)(1 − q−N ) n−1
(14.16.7)
Backward Shift Operator (1 − pqx )(1 − q−x+N+1 )KnA f f (q−x ; p, N; q) − p(1 − qx )KnA f f (q−x+1 ; p, N; q) A f f −x = (1 − p)(1 − qN+1 )Kn+1 (q ; pq−1 , N + 1; q)
or equivalently
(14.16.8)
14.16 Affine q-Krawtchouk
503
∇ w(x; p, N; q)KnA f f (q−x ; p, N; q) ∇q−x =
1 − qN+1 1−q
A f f −x w(x; pq−1 , N + 1; q)Kn+1 (q ; pq−1 , N + 1; q),
where
(14.16.9)
(pq; q)x p−x . (q; q)x (q; q)N−x
w(x; p, N; q) =
Rodrigues-Type Formula w(x; p, N; q)KnA f f (q−x ; p, N; q) n (−1)n q−Nn+(2) (1 − q)n = (∇q )n [w(x; pqn , N − n; q)] , (q−N ; q)n where ∇q :=
(14.16.10)
∇ . ∇q−x
Generating Functions For x = 0, 1, 2, . . . , N we have (q
−N
t; q)N−x · 1 φ1
q−x ; q, pqt pq
(−pq
−N+1
(q−N ; q)n A f f −x Kn (q ; p, N; q)t n . (14.16.11) n=0 (q; q)n N
=
∑
t; q)x · 2 φ0
qx−N , pqx+1 ; q, −q−xt −
(pq, q−N ; q)n −(n) A f f −x q 2 Kn (q ; p, N; q)t n . (q; q) n n=0 N
=
∑
(14.16.12)
504
14 Basic Hypergeometric Orthogonal Polynomials
Limit Relations q-Hahn → Affine q-Krawtchouk The affine q-Krawtchouk polynomials given by (14.16.1) can be obtained from the q-Hahn polynomials by the substitution α = p and β = 0 in (14.6.1): Qn (q−x ; p, 0, N|q) = KnA f f (q−x ; p, N; q). Dual q-Hahn → Affine q-Krawtchouk The affine q-Krawtchouk polynomials given by (14.16.1) can be obtained from the dual q-Hahn polynomials by the substitution γ = p and δ = 0 in (14.7.1): Rn (μ (x); p, 0, N|q) = KnA f f (q−x ; p, N; q). Note that μ (x) = q−x in this case. Affine q-Krawtchouk → Little q-Laguerre / Wall If we set x → N − x in the definition (14.16.1) of the affine q-Krawtchouk polynomials and take the limit N → ∞ we simply obtain the little q-Laguerre (or Wall) polynomials given by (14.20.1): lim KnA f f (qx−N ; p, N; q) = pn (qx ; p; q).
N→∞
(14.16.13)
Affine q-Krawtchouk → Krawtchouk If we let q → 1 in the definition (14.16.1) of the affine q-Krawtchouk polynomials we obtain: (14.16.14) lim KnA f f (q−x ; p, N|q) = Kn (x; 1 − p, N), q→1
where Kn (x; 1 − p, N) is the Krawtchouk polynomial given by (9.11.1).
Remarks The affine q-Krawtchouk polynomials given by (14.16.1) and the big q-Laguerre polynomials given by (14.11.1) are related in the following way: KnA f f (q−x ; p, N; q) = Pn (q−x ; p, q−N−1 ; q).
14.17 Dual q-Krawtchouk
505
The affine q-Krawtchouk polynomials are related to the quantum q-Krawtchouk polynomials given by (14.14.1) by the transformation q ↔ q−1 in the following way: KnA f f (qx ; p, N; q−1 ) =
1 K qtm (qx−N ; p−1 , N; q). (p−1 q; q)n n
References [80], [141], [161], [162], [186], [214], [238], [488].
14.17 Dual q-Krawtchouk Basic Hypergeometric Representation q−n , q−x , cqx−N ; q, q Kn (λ (x); c, N|q) = 3 φ2 q−N , 0 −n −x (qx−N ; q)n q ,q x+1 = −N , φ ; q, cq 2 1 (q ; q)n qnx qN−x−n+1
where
(14.17.1) n = 0, 1, 2, . . . , N,
λ (x) := q−x + cqx−N .
Orthogonality Relation (cq−N , q−N ; q)x (1 − cq2x−N ) −x x(2N−x) c q Km (λ (x))Kn (λ (x)) (q, cq; q)x (1 − cq−N ) x=0 N
∑
= (c−1 ; q)N where
(q; q)n (cq−N )n δmn , (q−N ; q)n
c < 0,
Kn (λ (x)) := Kn (λ (x); c, N|q).
(14.17.2)
506
14 Basic Hypergeometric Orthogonal Polynomials
Recurrence Relation −(1 − q−x )(1 − cqx−N )Kn (λ (x)) = (1 − qn−N )Kn+1 (λ (x))
− (1 − qn−N ) + cq−N (1 − qn ) Kn (λ (x)) + cq−N (1 − qn )Kn−1 (λ (x)), where
(14.17.3)
Kn (λ (x)) := Kn (λ (x); c, N|q).
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (1 + c)qn−N pn (x) + cq−N (1 − qn )(1 − qn−N−1 )pn−1 (x), where Kn (λ (x); c, N|q) =
(14.17.4)
1 pn (λ (x)). (q−N ; q)n
q-Difference Equation q−n (1 − qn )y(x) = B(x)y(x + 1) − [B(x) + D(x)] y(x) + D(x)y(x − 1), where and
y(x) = Kn (λ (x); c, N|q) ⎧ (1 − qx−N )(1 − cqx−N ) ⎪ ⎪ ⎪ B(x) = ⎪ ⎨ (1 − cq2x−N )(1 − cq2x−N+1 ) ⎪ ⎪ ⎪ ⎪ ⎩ D(x) = cq2x−2N−1
(1 − qx )(1 − cqx ) . (1 − cq2x−N−1 )(1 − cq2x−N )
(14.17.5)
14.17 Dual q-Krawtchouk
507
Forward Shift Operator Kn (λ (x + 1); c, N|q) − Kn (λ (x); c, N|q) =
q−n−x (1 − qn )(1 − cq2x−N+1 ) Kn−1 (λ (x); c, N − 1|q) 1 − q−N
(14.17.6)
or equivalently
Δ Kn (λ (x); c, N|q) q−n+1 (1 − qn ) = Kn−1 (λ (x); c, N − 1|q). Δ λ (x) (1 − q)(1 − q−N )
(14.17.7)
Backward Shift Operator (1 − qx−N−1 )(1 − cqx−N−1 )Kn (λ (x); c, N|q) − cq2(x−N−1) (1 − qx )(1 − cqx )Kn (λ (x − 1); c, N|q) = qx (1 − q−N−1 )(1 − cq2x−N−1 )Kn+1 (λ (x); c, N + 1|q)
(14.17.8)
or equivalently ∇ [w(x; c, N|q)Kn (λ (x); c, N|q)] ∇λ (x) 1 = w(x; c, N + 1|q)Kn+1 (λ (x); c, N + 1|q), (14.17.9) (1 − q)(1 − cq−N−1 ) where w(x; c, N|q) =
(q−N , cq−N ; q)x −x 2Nx−x(x−1) c q . (q, cq; q)x
Rodrigues-Type Formula w(x; c, N|q)Kn (λ (x); c, N|q) = (1 − q)n (cq−N ; q)n (∇λ )n [w(x; c, N − n|q)] , where ∇λ :=
∇ . ∇λ (x)
(14.17.10)
508
14 Basic Hypergeometric Orthogonal Polynomials
Generating Function For x = 0, 1, 2, . . . , N we have (cq−N t; q)x · (q−N t; q)N−x =
(q−N ; q)n ∑ (q; q)n Kn (λ (x); c, N|q)t n . n=0 N
(14.17.11)
Limit Relations q-Racah → Dual q-Krawtchouk The dual q-Krawtchouk polynomials given by (14.17.1) easily follow from the qRacah polynomials given by (14.2.1) by using the substitutions α = β = 0, γ q = q−N and δ = c: Rn (μ (x); 0, 0, q−N−1 , c|q) = Kn (λ (x); c, N|q). Note that
μ (x) = λ (x) = q−x + cqx−N .
Dual q-Hahn → Dual q-Krawtchouk The dual q-Krawtchouk polynomials given by (14.17.1) can be obtained from the dual q-Hahn polynomials by setting δ = cγ −1 q−N−1 in (14.7.1) and letting γ → 0: lim Rn (μ (x); γ , cγ −1 q−N−1 , N|q) = Kn (λ (x); c, N|q).
γ →0
Dual q-Krawtchouk → Al-Salam-Carlitz I If we set c = a−1 in the definition (14.17.1) of the dual q-Krawtchouk polynomials and take the limit N → ∞ we simply obtain the Al-Salam-Carlitz I polynomials given by (14.24.1): 1 n −(n) (a) x −1 q 2 Un (q ; q). (14.17.12) lim Kn (λ (x); a , N|q) = − N→∞ a Note that λ (x) = q−x + a−1 qx−N .
14.18 Continuous Big q-Hermite
509
Dual q-Krawtchouk → Krawtchouk If we set c = 1 − p−1 in the definition (14.17.1) of the dual q-Krawtchouk polynomials and take the limit q → 1 we simply find the Krawtchouk polynomials given by (9.11.1): (14.17.13) lim Kn (λ (x); 1 − p−1 , N|q) = Kn (x; p, N). q→1
Remark The dual q-Krawtchouk polynomials given by (14.17.1) and the q-Krawtchouk polynomials given by (14.15.1) are related in the following way: Kn (q−x ; p, N; q) = Kx (λ (n); −pqN , N|q) with
λ (n) = q−n − pqn
or
Kn (λ (x); c, N|q) = Kx (q−n ; −cq−N , N; q)
with
λ (x) = q−x + cqx−N .
References [141], [326], [345], [348].
14.18 Continuous Big q-Hermite Basic Hypergeometric Representation q−n , aeiθ , ae−iθ ; q, q 0, 0 −n iθ q , ae ; q, qn e−2iθ , = einθ 2 φ0 −
Hn (x; a|q) = a−n 3 φ2
(14.18.1) x = cos θ .
510
14 Basic Hypergeometric Orthogonal Polynomials
Orthogonality Relation If a is real and |a| < 1, then we have the following orthogonality relation 1 2π
1
w(x) δmn √ Hm (x; a|q)Hn (x; a|q) dx = n+1 , 2 (q ; q)∞ −1 1−x
(14.18.2)
where 2iθ (e ; q)∞ 2 h(x, 1)h(x, −1)h(x, q 21 )h(x, −q 21 ) = , w(x) := w(x; a|q) = iθ h(x, a) (ae ; q)∞ with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
x = cos θ .
If a > 1, then we have another orthogonality relation given by: 1 2π
1
w(x) √ Hm (x; a|q)Hn (x; a|q) dx −1 1 − x2
∑
+
wk Hm (xk ; a|q)Hn (xk ; a|q) =
k
δmn , n+1 (q ; q)∞
(14.18.3)
1
where w(x) is as before, xk = and wk =
−1 aqk + aqk 2
1 k (a−2 ; q)∞ (1 − a2 q2k )(a2 ; q)k − 3 k2 − 1 k 2 2 − q . (q; q)∞ (1 − a2 )(q; q)k a4
Recurrence Relation 2xHn (x; a|q) = Hn+1 (x; a|q) + aqn Hn (x; a|q) + (1 − qn )Hn−1 (x; a|q).
(14.18.4)
Normalized Recurrence Relation 1 1 xpn (x) = pn+1 (x) + aqn pn (x) + (1 − qn )pn−1 (x), 2 4 where
(14.18.5)
14.18 Continuous Big q-Hermite
511
Hn (x; a|q) = 2n pn (x).
q-Difference Equations 1 ˜ aq 2 |q)Dq y(x) + 4q−n+1 (1 − qn )w(x; (1 − q)2 Dq w(x; ˜ a|q)y(x) = 0,
(14.18.6)
where y(x) = Hn (x; a|q) and
w(x; a|q) . w(x; ˜ a|q) := √ 1 − x2
If we define −n
Pn (z) := a
3 φ2
q−n , az, az−1 ; q, q 0, 0
then the q-difference equation can also be written in the form
q−n (1 − qn )Pn (z) = A(z)Pn (qz) − A(z) + A(z−1 ) Pn (z) + A(z−1 )Pn (q−1 z), where A(z) =
(14.18.7)
(1 − az) . (1 − z2 )(1 − qz2 )
Forward Shift Operator δq Hn (x; a|q) = −q− 2 n (1 − qn )(eiθ − e−iθ )Hn−1 (x; aq 2 |q), 1
1
x = cos θ
(14.18.8)
or equivalently 1
Dq Hn (x; a|q) =
1 2q− 2 (n−1) (1 − qn ) Hn−1 (x; aq 2 |q). 1−q
(14.18.9)
Backward Shift Operator δq [w(x; ˜ a|q)Hn (x; a|q)] = q− 2 (n+1) (eiθ − e−iθ ) 1
1
1
× w(x; ˜ aq− 2 |q)Hn+1 (x; aq− 2 |q),
x = cos θ
(14.18.10)
512
14 Basic Hypergeometric Orthogonal Polynomials
or equivalently 1
Dq [w(x; ˜ a|q)Hn (x; a|q)] = −
1 1 2q− 2 n w(x; ˜ aq− 2 |q)Hn+1 (x; aq− 2 |q). 1−q
(14.18.11)
Rodrigues-Type Formula w(x; a|q)Hn (x; a|q) =
q−1 2
n
1 1 q 4 n(n−1) (Dq )n w(x; aq 2 n |q) .
(14.18.12)
Generating Functions (at; q)∞ (eiθ t, e−iθ t; q)∞
∞
=
Hn (x; a|q) n t , n=0 (q; q)n
∑
x = cos θ .
aeiθ ; q, e−iθ t (e t; q)∞ · 1 φ1 e iθ t n ∞ (−1)n q(2) Hn (x; a|q)t n , =∑ (q; q) n n=0 iθ
(14.18.13)
x = cos θ .
γ , aeiθ (γ eiθ t; q)∞ −iθ ; q, e φ t 2 1 γ e iθ t (eiθ t; q)∞ ∞ (γ ; q)n Hn (x; a|q)t n , x = cos θ , =∑ (q; q) n n=0
γ arbitrary.
(14.18.14)
(14.18.15)
Limit Relations Al-Salam-Chihara → Continuous Big q-Hermite If we take b = 0 in the definition (14.8.1) of the Al-Salam-Chihara polynomials we simply obtain the continuous big q-Hermite polynomials given by (14.18.1): Qn (x; a, 0|q) = Hn (x; a|q).
14.18 Continuous Big q-Hermite
513
Continuous Big q-Hermite → Continuous q-Hermite The continuous q-Hermite polynomials given by (14.26.1) can easily be obtained from the continuous big q-Hermite polynomials given by (14.18.1) by taking a = 0: Hn (x; 0|q) = Hn (x|q).
(14.18.16)
Continuous Big q-Hermite → Hermite If we set a = 0 and x → x 12 (1 − q) in the definition (14.18.1) of the continuous big q-Hermite polynomials and let q tend to 1, we obtain the Hermite polynomials given by (9.15.1) in the following way: Hn (x 12 (1 − q); 0|q) = Hn (x). (14.18.17) lim n q→1 1−q 2
2
If we take a → a 2(1 − q) and x → x 12 (1 − q) in the definition (14.18.1) of the continuous big q-Hermite polynomials and take the limit q → 1 we find the Hermite polynomials given by (9.15.1) with shifted argument: Hn (x 12 (1 − q); a 2(1 − q)|q) = Hn (x − a). (14.18.18) lim n q→1 1−q 2
References [73], [85], [86], [207].
2
514
14 Basic Hypergeometric Orthogonal Polynomials
14.19 Continuous q-Laguerre Basic Hypergeometric Representation The continuous q-Laguerre polynomials can be obtained from the continuous qJacobi polynomials given by (14.10.1) by taking the limit β → ∞: 1 1 1 1 (qα +1 ; q)n q−n , q 2 α + 4 eiθ , q 2 α + 4 e−iθ (α ) ; q, q (14.19.1) Pn (x|q) = 3 φ2 (q; q)n qα +1 , 0 (q 2 α + 4 e−iθ ; q)n ( 1 α + 1 )n inθ q 2 4 e (q; q)n 1 1 q−n , q 2 α + 4 eiθ − 12 α + 14 −iθ × 2 φ1 ; q, q e , 1 1 q− 2 α + 4 −n eiθ 1
=
3
x = cos θ .
Orthogonality Relation For α ≥ − 12 we have 1
w(x) (α ) (α ) √ Pm (x|q)Pn (x|q) dx 1 − x2 (qα +1 ; q)n (α + 1 )n 1 2 q δmn , = α +1 (q, q ; q)∞ (q; q)n
1 2π
−1
(14.19.2)
where 2 (eiθ , −eiθ ; q 12 ) 2 2iθ ; q) (e ∞ ∞ w(x) := w(x; qα |q) = 1 1 = 1 1 (q 2 α + 4 eiθ , q 12 α + 34 eiθ ; q)∞ (q 2 α + 4 eiθ ; q 12 )∞ 1
=
1
h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ) h(x, q 2 α + 4 )h(x, q 2 α + 4 ) 1
1
1
3
,
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , k=0
∞
x = cos θ .
14.19 Continuous q-Laguerre
515
Recurrence Relation (α )
(α )
2xPn (x|q) = q− 2 α − 4 (1 − qn+1 )Pn+1 (x|q) 1
1
(α )
+ qn+ 2 α + 4 (1 + q 2 )Pn (x|q) 1
1
+q
1
1 α+ 1 2 4
(α )
(1 − qn+α )Pn−1 (x|q).
(14.19.3)
Normalized Recurrence Relation 1 1 1 1 xpn (x) = pn+1 (x) + qn+ 2 α + 4 (1 + q 2 )pn (x) 2 1 + (1 − qn )(1 − qn+α )pn−1 (x), 4
where
2n q( 2 α + 4 )n pn (x). (q; q)n 1
(α )
Pn (x|q) =
(14.19.4)
1
q-Difference Equations
˜ qα +1 |q)Dq y(x) + 4q−n+1 (1 − qn )w(x; (1 − q)2 Dq w(x; ˜ qα |q)y(x) = 0, (14.19.5) where
(α )
y(x) = Pn (x|q) and
w(x; qα |q) . w(x; ˜ qα |q) := √ 1 − x2
Forward Shift Operator (α )
(α +1)
δq Pn (x|q) = −q−n+ 2 α + 4 (eiθ − e−iθ )Pn−1 (x|q), 1
3
x = cos θ
(14.19.6)
or equivalently (α ) Dq Pn (x|q) =
2q−n+ 2 α + 4 (α +1) Pn−1 (x|q). 1−q 1
5
(14.19.7)
516
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator (α ) ˜ qα |q)Pn (x|q) δq w(x; = q− 2 α − 4 (1 − qn+1 )(eiθ − e−iθ ) 1
1
(α −1)
× w(x; ˜ qα −1 |q)Pn+1 (x|q),
x = cos θ
(14.19.8)
or equivalently (α ) ˜ qα |q)Pn (x|q) Dq w(x; = −2q− 2 α + 4 1
1
1 − qn+1 (α −1) w(x; ˜ qα −1 |q)Pn+1 (x|q). 1−q
(14.19.9)
Rodrigues-Type Formula α
w(x; ˜ q
(α ) |q)Pn (x|q) =
q−1 2
n
q 4 n + 2 nα ˜ qα +n |q) . (Dq )n w(x; (q; q)n 1 2
1
(14.19.10)
Generating Functions ∞
(qα + 2 t, qα +1t; q)∞ 1
(q 2 α + 4 eiθ t, q 2 α + 4 e−iθ t; q)∞ 1
1
1
1
1 2 φ1 i θ (e t; q)∞ =
=
(α )
∑ Pn
q 2 α + 4 eiθ , q 2 α + 4 eiθ ; q, e−iθ t qα +1 1
1
1
(α )
Pn (x|q)t n
n=0
(qα +1 ; q)n q( 2 α + 4 )n 1
(t; q)∞ · 2 φ1
1
3
x = cos θ .
,
q 2 α + 4 eiθ , q 2 α + 4 e−iθ ; q, t qα +1 1
1
1
n (−1)n q(2) (α ) Pn (x|q)t n , = ∑ α +1 ; q)n n=0 (q
∞
x = cos θ .
(14.19.11)
n=0
∞
∑
(x|q)t n ,
1
(14.19.12)
x = cos θ .
(14.19.13)
14.19 Continuous q-Laguerre
517
(γ eiθ t; q)∞ 3 φ2 (eiθ t; q)∞ ∞
=
γ , q 2 α + 4 e iθ , q 2 α + 4 e iθ ; q, e−iθ t qα +1 , γ eiθ t 1
(γ ; q)n
∑ (qα +1 ; q)n
n=0
1
Pn (x|q) 1 1 q( 2 α + 4 )n
1
t n,
3
x = cos θ ,
γ arbitrary. (14.19.14)
Limit Relations Al-Salam-Chihara → Continuous q-Laguerre The continuous q-Laguerre polynomials given by (14.19.1) can be obtained from 1 1 the Al-Salam-Chihara polynomials given by (14.8.1) by taking a = q 2 α + 4 and b = 1 α+ 3 q2 4 : 1 1 1 3 (q; q)n (α ) Qn (x; q 2 α + 4 , q 2 α + 4 |q) = 1 1 Pn (x|q). q( 2 α + 4 )n q-Meixner-Pollaczek → Continuous q-Laguerre If we take eiφ = q− 4 , a = q 2 α + 2 and eiθ → q 4 eiθ in the definition (14.9.1) of the q-Meixner-Pollaczek polynomials we obtain the continuous q-Laguerre polynomials given by (14.19.1): 1
1
1
1
(α )
Pn (cos(θ + φ ); q 2 α + 2 |q) = q−( 2 α + 4 )n Pn (cos θ |q). 1
1
1
1
Continuous q-Jacobi → Continuous q-Laguerre The continuous q-Laguerre polynomials given by (14.19.1) follow simply from the continuous q-Jacobi polynomials given by (14.10.1) by taking the limit β → ∞: (α ,β )
lim Pn
β →∞
(α )
(x|q) = Pn (x|q).
Continuous q-Laguerre → Continuous q-Hermite The continuous q-Hermite polynomials given by (14.26.1) can be obtained from the continuous q-Laguerre polynomials given by (14.19.1) by taking the limit α → ∞ in the following way: (α ) Pn (x|q) Hn (x|q) . (14.19.15) lim 1 1 = α →∞ q( 2 α + 4 )n (q; q)n
518
14 Basic Hypergeometric Orthogonal Polynomials
Continuous q-Laguerre → Laguerre If we set x → qx in the definitions (14.19.1) of the continuous q-Laguerre polynomials and take the limit q → 1 we find the Laguerre polynomials given by (9.12.1). In fact we have: (α ) (α ) (14.19.16) lim Pn (qx |q) = Ln (2x). q→1
Remark If we let β tend to infinity in (14.10.16) and renormalize we obtain 1 1 (qα +1 ; q)n q−n , q 2 eiθ , q 2 e−iθ (α ) ; q, q , x = cos θ . (14.19.17) Pn (x; q) = 3 φ2 (q; q)n qα +1 , −q These two q-analogues of the Laguerre polynomials are connected by the following quadratic transformation: (α )
(α )
Pn (x|q2 ) = qnα Pn (x; q).
Reference [72].
14.20 Little q-Laguerre / Wall Basic Hypergeometric Representation q−n , 0 ; q, qx aq −n −1 1 x q ,x = −1 −n ; q, . φ 2 0 (a q ; q)n − a
pn (x; a|q) = 2 φ1
(14.20.1)
14.20 Little q-Laguerre / Wall
519
Orthogonality Relation ∞
(aq)k
(aq)n
(q; q)n
∑ (q; q)k pm (qk ; a|q)pn (qk ; a|q) = (aq; q)∞ (aq; q)n δmn ,
0 < aq < 1. (14.20.2)
k=0
Recurrence Relation − xpn (x; a|q) = An pn+1 (x; a|q) − (An +Cn ) pn (x; a|q) +Cn pn−1 (x; a|q), (14.20.3) ⎧ ⎨ An = qn (1 − aqn+1 )
where
⎩
Cn = aqn (1 − qn ).
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn )pn (x) + aq2n−1 (1 − qn )(1 − aqn )pn−1 (x), (14.20.4) where
n (−1)n q−(2) pn (x). pn (x; a|q) = (aq; q)n
q-Difference Equation − q−n (1 − qn )xy(x) = ay(qx) + (x − a − 1)y(x) + (1 − x)y(q−1 x),
(14.20.5)
where y(x) = pn (x; a|q).
Forward Shift Operator pn (x; a|q) − pn (qx; a|q) = −
q−n+1 (1 − qn ) xpn−1 (x; aq|q) 1 − aq
(14.20.6)
or equivalently Dq pn (x; a|q) = −
q−n+1 (1 − qn ) pn−1 (x; aq|q). (1 − q)(1 − aq)
(14.20.7)
520
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator apn (x; a|q) − (1 − x)pn (q−1 x; a|q) = (a − 1)pn+1 (x; q−1 a|q)
(14.20.8)
or equivalently Dq−1 [w(x; α |q)pn (x; qα |q)] =
1 − qα
qα −1 (1 − q)
where
w(x; α − 1|q)pn+1 (x; qα −1 |q),
(14.20.9)
w(x; α |q) = (qx; q)∞ xα .
Rodrigues-Type Formula n qnα +(2) (1 − q)n D [w(x; α + n|q)] . −1 q (qα +1 ; q)n n
w(x; α |q)pn (x; qα |q) =
(14.20.10)
Generating Function (t; q)∞ 0 φ1 (xt; q)∞
− ; q, aqxt aq
∞
n
(−1)n q(2) =∑ pn (x; a|q)t n . n=0 (q; q)n
(14.20.11)
Limit Relations Big q-Laguerre → Little q-Laguerre / Wall The little q-Laguerre (or Wall) polynomials given by (14.20.1) can be obtained from the big q-Laguerre polynomials by taking x → bqx in (14.11.1) and letting b → −∞: lim Pn (bqx; a, b; q) = pn (x; a|q).
b→−∞
Little q-Jacobi → Little q-Laguerre / Wall The little q-Laguerre (or Wall) polynomials given by (14.20.1) are little q-Jacobi polynomials with b = 0. So if we set b = 0 in the definition (14.12.1) of the little q-Jacobi polynomials we obtain the little q-Laguerre (or Wall) polynomials:
14.20 Little q-Laguerre / Wall
521
pn (x; a, 0|q) = pn (x; a|q). Affine q-Krawtchouk → Little q-Laguerre / Wall If we set x → N − x in the definition (14.16.1) of the affine q-Krawtchouk polynomials and take the limit N → ∞ we simply obtain the little q-Laguerre (or Wall) polynomials given by (14.20.1): lim KnA f f (qx−N ; p, N; q) = pn (qx ; p; q).
N→∞
Little q-Laguerre / Wall → Laguerre / Charlier If we set a = qα and x → (1 − q)x in the definition (14.20.1) of the little q-Laguerre (or Wall) polynomials and let q tend to 1, we obtain the Laguerre polynomials given by (9.12.1): (α ) Ln (x) (14.20.12) lim pn ((1 − q)x; qα |q) = (α ) . q→1 Ln (0) If we set a → (q − 1)a and x → qx in the definition (14.20.1) of the little qLaguerre (or Wall) polynomials and take the limit q → 1 we obtain the Charlier polynomials given by (9.14.1) in the following way: lim
q→1
pn (qx ; (q − 1)a|q) Cn (x; a) = . (1 − q)n an
(14.20.13)
Remark If we set a = qα and replace q by q−1 we find the q-Laguerre polynomials given by (14.21.1) in the following way: pn (x; q−α |q−1 ) =
(q; q)n (α ) Ln (−x; q). (qα +1 ; q)n
References [16], [26], [84], [144], [146], [147], [157], [214], [238], [345], [346], [416], [504], [509].
522
14 Basic Hypergeometric Orthogonal Polynomials
14.21 q-Laguerre Basic Hypergeometric Representation (α )
−n q (qα +1 ; q)n n+α +1 φ ; q, −q x 1 1 (q; q)n qα +1 −n 1 q , −x n+α +1 ; q, q = . φ 2 1 0 (q; q)n
Ln (x; q) =
(14.21.1)
Orthogonality Relation The q-Laguerre polynomials satisfy two kinds of orthogonality relations, an absolutely continuous one and a discrete one. These orthogonality relations are given by, respectively: ∞ 0
=
xα (α ) (α ) Lm (x; q)Ln (x; q) dx (−x; q)∞
(q−α ; q)∞ (qα +1 ; q)n Γ(−α )Γ(α + 1) δmn , (q; q)∞ (q; q)n qn
α > −1
(14.21.2)
and ∞
qkα +k (α ) (α ) Lm (cqk ; q)Ln (cqk ; q) k ; q) (−cq ∞ k=−∞
∑
=
(q, −cqα +1 , −c−1 q−α ; q)∞ (qα +1 ; q)n δmn , (qα +1 , −c, −c−1 q; q)∞ (q; q)n qn
α > −1,
c > 0. (14.21.3)
For c = 1 the latter orthogonality relation can also be written as ∞ 0
=
xα (α ) (α ) Lm (x; q)Ln (x; q) dq x (−x; q)∞
1 − q (q, −qα +1 , −q−α ; q)∞ (qα +1 ; q)n δmn , 2 (qα +1 , −q, −q; q)∞ (q; q)n qn
α > −1.
(14.21.4)
14.21 q-Laguerre
523
Recurrence Relation (α )
(α )
− q2n+α +1 xLn (x; q) = (1 − qn+1 )Ln+1 (x; q)
(α ) − (1 − qn+1 ) + q(1 − qn+α ) Ln (x; q) (α )
+ q(1 − qn+α )Ln−1 (x; q).
(14.21.5)
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + q−2n−α −1 (1 − qn+1 ) + q(1 − qn+α ) pn (x) + q−4n−2α +1 (1 − qn )(1 − qn+α )pn−1 (x), where (α )
Ln (x; q) =
(14.21.6)
(−1)n qn(n+α ) pn (x). (q; q)n
q-Difference Equation − qα (1 − qn )xy(x) = qα (1 + x)y(qx) − [1 + qα (1 + x)] y(x) + y(q−1 x), (14.21.7) where
(α )
y(x) = Ln (x; q).
Forward Shift Operator (α )
(α )
(α +1)
Ln (x; q) − Ln (qx; q) = −qα +1 xLn−1 (qx; q) or equivalently (α )
Dq Ln (x; q) = −
qα +1 (α +1) L (qx; q). 1 − q n−1
(14.21.8)
(14.21.9)
Backward Shift Operator (α )
(α )
(α −1)
Ln (x; q) − qα (1 + x)Ln (qx; q) = (1 − qn+1 )Ln+1 (x; q) or equivalently
(14.21.10)
524
14 Basic Hypergeometric Orthogonal Polynomials
1 − qn+1 (α ) (α −1) w(x; α − 1; q)Ln+1 (x; q), Dq w(x; α ; q)Ln (x; q) = 1−q where w(x; α ; q) =
(14.21.11)
xα . (−x; q)∞
Rodrigues-Type Formula (α )
w(x; α ; q)Ln (x; q) =
(1 − q)n (Dq )n [w(x; α + n; q)] . (q; q)n
(14.21.12)
Generating Functions 1 1 φ1 (t; q)∞ 1 0 φ1 (t; q)∞ (t; q)∞ · 0 φ2
−x ; q, qα +1t 0
−
; q, −q qα +1
−
qα +1 ,t
α +1
; q, −qα +1 xt
∞
=
(α )
∑ Ln
(x; q)t n .
(14.21.13)
n=0
xt
∞
=
(α )
Ln (x; q)
∑ (qα +1 ; q)n t n .
(14.21.14)
n=0
n (−1)n q(2) (α ) ∑ (qα +1 ; q)n Ln (x; q)t n . n=0
∞
=
(γ t; q)∞ γ α +1 ; q, −q φ xt 1 2 (t; q)∞ qα +1 , γ t ∞ (γ ; q)n (α ) Ln (x; q)t n , γ arbitrary. = ∑ α +1 (q ; q) n n=0
(14.21.15)
(14.21.16)
Limit Relations Little q-Jacobi → q-Laguerre If we substitute a = qα and x → −b−1 q−1 x in the definition (14.12.1) of the little q-Jacobi polynomials and then take the limit b → −∞ we find the q-Laguerre polynomials given by (14.21.1): lim pn (−b−1 q−1 x; qα , b|q) =
b→−∞
(q; q)n (α ) Ln (x; q). (qα +1 ; q)n
14.21 q-Laguerre
525
q-Meixner → q-Laguerre The q-Laguerre polynomials given by (14.21.1) can be obtained from the q-Meixner polynomials given by (14.13.1) by setting b = qα and q−x → cqα x in the definition (14.13.1) of the q-Meixner polynomials and then taking the limit c → ∞: lim Mn (cqα x; qα , c; q) =
c→∞
(q; q)n (α ) Ln (x; q). (qα +1 ; q)n
q-Laguerre → Stieltjes-Wigert If we set x → xq−α in the definition (14.21.1) of the q-Laguerre polynomials and take the limit α → ∞ we simply obtain the Stieltjes-Wigert polynomials given by (14.27.1): (α ) (14.21.17) lim Ln xq−α ; q = Sn (x; q). α →∞
q-Laguerre → Laguerre / Charlier If we set x → (1 − q)x in the definition (14.21.1) of the q-Laguerre polynomials and take the limit q → 1 we obtain the Laguerre polynomials given by (9.12.1): (α )
(α )
lim Ln ((1 − q)x; q) = Ln (x).
q→1
(14.21.18)
If we set x → −q−x and qα = a−1 (q − 1)−1 (or α = −(ln q)−1 ln(q − 1)a) in the definition (14.21.1) of the q-Laguerre polynomials, multiply by (q; q)n , and take the limit q → 1 we obtain the Charlier polynomials given by (9.14.1): (α )
lim (q; q)n Ln (−q−x ; q) = Cn (x; a),
q→1
where qα =
1 a(q − 1)
or
α =−
(14.21.19)
ln(q − 1)a . ln q
Remarks The q-Laguerre polynomials are sometimes called the generalized Stieltjes-Wigert polynomials. If we replace q by q−1 we obtain the little q-Laguerre (or Wall) polynomials given by (14.20.1) in the following way:
526
14 Basic Hypergeometric Orthogonal Polynomials (α )
Ln (x; q−1 ) =
(qα +1 ; q)n pn (−x; qα |q). (q; q)n qnα
The q-Laguerre polynomials given by (14.21.1) and the q-Bessel polynomials given by (14.22.1) are related in the following way: yn (qx ; a; q) (x−n) = Ln (aqn ; q). (q; q)n The q-Laguerre polynomials given by (14.21.1) and the q-Charlier polynomials given by (14.23.1) are related in the following way: Cn (−x; −q−α ; q) (α ) = Ln (x; q). (q; q)n Since the Stieltjes and Hamburger moment problems corresponding to the q-Laguerre polynomials are indeterminate there exist many different weight functions.
References [12], [16], [49], [51], [72], [84], [139], [144], [146], [147], [150], [199], [238], [256], [276], [291], [302], [411].
14.22 q-Bessel Basic Hypergeometric Representation q−n , −aqn ; q, qx 0 −n q n+1 = (q−n+1 x; q)n · 1 φ1 ; q, −aq x q−n+1 x −n −1 q−n+1 q ,x n n ; q, − . = (−aq x) · 2 φ1 0 a
yn (x; a; q) = 2 φ1
(14.22.1)
14.22 q-Bessel
527
Orthogonality Relation ∞
ak
∑ (q; q)k q( 2 ) ym (qk ; a; q)yn (qk ; a; q) k+1
k=0
= (q; q)n (−aqn ; q)∞
n+1 an q( 2 ) δmn , (1 + aq2n )
a > 0.
(14.22.2)
Recurrence Relation − xyn (x; a; q) = An yn+1 (x; a; q) − (An +Cn )yn (x; a; q) +Cn yn−1 (x; a; q), (14.22.3) where
⎧ (1 + aqn ) ⎪ n ⎪ = q A ⎪ n ⎪ ⎨ (1 + aq2n )(1 + aq2n+1 ) ⎪ ⎪ ⎪ ⎪ ⎩ Cn = aq2n−1
(1 − qn ) . (1 + aq2n−1 )(1 + aq2n )
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (An +Cn )pn (x) + An−1Cn pn−1 (x), where
(14.22.4)
n
yn (x; a; q) = (−1)n q−(2) (−aqn ; q)n pn (x).
q-Difference Equation −q−n (1 − qn )(1 + aqn )xy(x) = axy(qx) − (ax + 1 − x)y(x) + (1 − x)y(q−1 x), where y(x) = yn (x; a; q).
(14.22.5)
528
14 Basic Hypergeometric Orthogonal Polynomials
Forward Shift Operator yn (x; a; q) − yn (qx; a; q) = −q−n+1 (1 − qn )(1 + aqn )xyn−1 (x; aq2 ; q)
(14.22.6)
or equivalently Dq yn (x; a; q) = −
q−n+1 (1 − qn )(1 + aqn ) yn−1 (x; aq2 ; q). 1−q
(14.22.7)
Backward Shift Operator aqx−1 yn (qx ; a; q) − (1 − qx )yn (qx−1 x; a; q) = −yn+1 (qx ; aq−2 ; q)
(14.22.8)
or equivalently ∇ [w(x; a; q)yn (qx ; a; q)] q2 w(x; aq−2 ; q)yn+1 (qx ; aq−2 ; q), = ∇qx a(1 − q) where
(14.22.9)
x
w(x; a; q) =
ax q(2) . (q; q)x
Rodrigues-Type Formula
w(x; a; q)yn (qx ; a; q) = an (1 − q)n qn(n−1) (∇q )n w(x; aq2n ; q) , where ∇q :=
(14.22.10)
∇ . ∇qx
Generating Functions
− ; q, −aqx+1t 0 φ1 0 ∞ yn (qx ; a; q) n t , =∑ n=0 (q; q)n
(t; q)∞ 1 φ3 (xt; q)∞
xt ; q, −aqxt 0, 0,t
2 φ0
q−x , 0 ; q, qxt −
x = 0, 1, 2, . . . .
∞
(14.22.11)
n
(−1)n q(2) =∑ yn (x; a; q)t n . n=0 (q; q)n
(14.22.12)
14.22 q-Bessel
529
Limit Relations Little q-Jacobi → q-Bessel If we set b → −a−1 q−1 b in the definition (14.12.1) of the little q-Jacobi polynomials and then take the limit a → 0 we obtain the q-Bessel polynomials given by (14.22.1): lim pn (x; a, −a−1 q−1 b|q) = yn (x; b; q).
a→0
q-Krawtchouk → q-Bessel If we set x → N − x in the definition (14.15.1) of the q-Krawtchouk polynomials and then take the limit N → ∞ we obtain the q-Bessel polynomials given by (14.22.1): lim Kn (qx−N ; p, N; q) = yn (qx ; p; q).
N→∞
q-Bessel → Stieltjes-Wigert The Stieltjes-Wigert polynomials given by (14.27.1) can be obtained from the qBessel polynomials by setting x → a−1 x in the definition (14.22.1) of the q-Bessel polynomials and then taking the limit a → ∞. In fact we have lim yn (a−1 x; a; q) = (q; q)n Sn (x; q).
a→∞
(14.22.13)
q-Bessel → Bessel If we set x → − 12 (1 − q)−1 x and a → −qa+1 in the definition (14.22.1) of the qBessel polynomials and take the limit q → 1 we find the Bessel polynomials given by (9.13.1): lim yn (− 12 (1 − q)−1 x; −qa+1 ; q) = yn (x; a).
q→1
(14.22.14)
q-Bessel → Charlier If we set x → qx and a → a(1 − q) in the definition (14.22.1) of the q-Bessel polynomials and take the limit q → 1 we find the Charlier polynomials given by (9.14.1): yn (qx ; a(1 − q); q) = anCn (x; a). q→1 (q − 1)n lim
(14.22.15)
530
14 Basic Hypergeometric Orthogonal Polynomials
Remark In [318] and [319] these q-Bessel polynomials were called alternative q-Charlier polynomials. The q-Bessel polynomials given by (14.22.1) and the q-Laguerre polynomials given by (14.21.1) are related in the following way: yn (qx ; a; q) (x−n) = Ln (aqn ; q). (q; q)n
Reference [157].
14.23 q-Charlier Basic Hypergeometric Representation qn+1 q−n , q−x ; q, − 0 a −n qn+1−x q = (−a−1 q; q)n · 1 φ1 ; q, − . −a−1 q a
Cn (q−x ; a; q) = 2 φ1
(14.23.1)
Orthogonality Relation ∞
ax
∑ (q; q)x q(2)Cm (q−x ; a; q)Cn (q−x ; a; q) x
x=0
= q−n (−a; q)∞ (−a−1 q, q; q)n δmn ,
a > 0.
(14.23.2)
14.23 q-Charlier
531
Recurrence Relation q2n+1 (1 − q−x )Cn (q−x ) = aCn+1 (q−x ) − [a + q(1 − qn )(a + qn )]Cn (q−x ) + q(1 − qn )(a + qn )Cn−1 (q−x ), where
(14.23.3)
Cn (q−x ) := Cn (q−x ; a; q).
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + 1 + q−2n−1 {a + q(1 − qn )(a + qn )} pn (x) + aq−4n+1 (1 − qn )(a + qn )pn−1 (x), where
(14.23.4)
2
(−1)n qn pn (q−x ). Cn (q ; a; q) = an −x
q-Difference Equation qn y(x) = aqx y(x + 1) − qx (a − 1)y(x) + (1 − qx )y(x − 1), where
(14.23.5)
y(x) = Cn (q−x ; a; q).
Forward Shift Operator Cn (q−x−1 ; a; q) −Cn (q−x ; a; q) = −a−1 q−x (1 − qn )Cn−1 (q−x ; aq−1 ; q)
(14.23.6)
or equivalently
Δ Cn (q−x ; a; q) q(1 − qn ) Cn−1 (q−x ; aq−1 ; q). = − Δ q−x a(1 − q)
(14.23.7)
532
14 Basic Hypergeometric Orthogonal Polynomials
Backward Shift Operator Cn (q−x ; a; q) − a−1 q−x (1 − qx )Cn (q−x+1 ; a; q) = Cn+1 (q−x ; aq; q)
(14.23.8)
or equivalently ∇ [w(x; a; q)Cn (q−x ; a; q)] 1 w(x; aq; q)Cn+1 (q−x ; aq; q), = −x ∇q 1−q where
(14.23.9)
x+1 ax q( 2 ) . w(x; a; q) = (q; q)x
Rodrigues-Type Formula
w(x; a; q)Cn (q−x ; a; q) = (1 − q)n (∇q )n w(x; aq−n ; q) , where
(14.23.10)
∇ . ∇q−x
∇q :=
Generating Functions 1 1 φ1 (t; q)∞ 1 0 φ1 (t; q)∞
q−x ; q, −a−1 qt 0
∞
=
− ; q, −a−1 q−x+1t −a−1 q
Cn (q−x ; a; q) n t . (q; q)n n=0
∑
∞
=
Cn (q−x ; a; q)
∑ (−a−1 q, q; q)n t n .
(14.23.11)
(14.23.12)
n=0
Limit Relations q-Meixner → q-Charlier The q-Charlier polynomials given by (14.23.1) can easily be obtained from the qMeixner given by (14.13.1) by setting b = 0 in the definition (14.13.1) of the qMeixner polynomials: (14.23.13) Mn (x; 0, c; q) = Cn (x; c; q).
14.23 q-Charlier
533
q-Krawtchouk → q-Charlier By setting p = a−1 q−N in the definition (14.15.1) of the q-Krawtchouk polynomials and then taking the limit N → ∞ we obtain the q-Charlier polynomials given by (14.23.1): lim Kn (q−x ; a−1 q−N , N; q) = Cn (q−x ; a; q). N→∞
q-Charlier → Stieltjes-Wigert If we set q−x → ax in the definition (14.23.1) of the q-Charlier polynomials and take the limit a → ∞ we obtain the Stieltjes-Wigert polynomials given by (14.27.1) in the following way: (14.23.14) lim Cn (ax; a; q) = (q; q)n Sn (x; q). a→∞
q-Charlier → Charlier If we set a → a(1 − q) in the definition (14.23.1) of the q-Charlier polynomials and take the limit q → 1 we obtain the Charlier polynomials given by (9.14.1): lim Cn (q−x ; a(1 − q); q) = Cn (x; a).
q→1
(14.23.15)
Remark The q-Charlier polynomials given by (14.23.1) and the q-Laguerre polynomials given by (14.21.1) are related in the following way: Cn (−x; −q−α ; q) (α ) = Ln (x; q). (q; q)n
References [30], [80], [238], [261], [328], [416], [524].
534
14 Basic Hypergeometric Orthogonal Polynomials
14.24 Al-Salam-Carlitz I Basic Hypergeometric Representation n (a) Un (x; q) = (−a)n q(2) 2 φ1
qx q−n , x−1 ; q, . 0 a
(14.24.1)
Orthogonality Relation 1 a
(qx, a−1 qx; q)∞Um (x; q)Un (x; q) dq x (a)
(a)
n
= (−a)n (1 − q)(q; q)n (q, a, a−1 q; q)∞ q(2) δmn ,
a < 0.
(14.24.2)
Recurrence Relation (a)
(a)
(a)
(a)
xUn (x; q) = Un+1 (x; q)+(a+1)qnUn (x; q)−aqn−1 (1−qn )Un−1 (x; q). (14.24.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (a + 1)qn pn (x) − aqn−1 (1 − qn )pn−1 (x), where
(14.24.4)
(a)
Un (x; q) = pn (x).
q-Difference Equation
(1 − qn )x2 y(x) = aqn−1 y(qx) − aqn−1 + qn (1 − x)(a − x) y(x) + qn (1 − x)(a − x)y(q−1 x), where
(a)
y(x) = Un (x; q).
(14.24.5)
14.24 Al-Salam-Carlitz I
535
Forward Shift Operator (a)
(a)
(a)
Un (x; q) −Un (qx; q) = (1 − qn )xUn−1 (x; q) or equivalently (a)
DqUn (x; q) =
1 − qn (a) U (x; q). 1 − q n−1
(14.24.6)
(14.24.7)
Backward Shift Operator aUn (x; q) − (1 − x)(a − x)Un (q−1 x; q) = −q−n xUn+1 (x; q) (a)
(a)
(a)
(14.24.8)
or equivalently q−n+1 (a) (a) w(x; a; q)Un+1 (x; q), Dq−1 w(x; a; q)Un (x; q) = a(1 − q) where
(14.24.9)
w(x; a; q) = (qx, a−1 qx; q)∞ .
Rodrigues-Type Formula n 1 (a) w(x; a; q)Un (x; q) = an q 2 n(n−3) (1 − q)n Dq−1 [w(x; a; q)] .
(14.24.10)
Generating Function ∞ (t, at; q)∞ Un (x; q) n =∑ t . (xt; q)∞ n=0 (q; q)n (a)
(14.24.11)
Limit Relations Big q-Laguerre → Al-Salam-Carlitz I If we set x → aqx and b → ab in the definition (14.11.1) of the big q-Laguerre polynomials and take the limit a → 0 we obtain the Al-Salam-Carlitz I polynomials given by (14.24.1):
536
14 Basic Hypergeometric Orthogonal Polynomials
lim
a→0
Pn (aqx; a, ab; q) (b) = qnUn (x; q). an
Dual q-Krawtchouk → Al-Salam-Carlitz I If we set c = a−1 in the definition (14.17.1) of the dual q-Krawtchouk polynomials and take the limit N → ∞ we simply obtain the Al-Salam-Carlitz I polynomials given by (14.24.1): 1 n −(n) (a) x −1 q 2 Un (q ; q). lim Kn (λ (x); a , N|q) = − N→∞ a Note that λ (x) = q−x + a−1 qx−N . Al-Salam-Carlitz I → Discrete q-Hermite I The discrete q-Hermite I polynomials given by (14.28.1) can easily be obtained from the Al-Salam-Carlitz I polynomials given by (14.24.1) by the substitution a = −1: (−1)
Un
(x; q) = hn (x; q).
(14.24.12)
Al-Salam-Carlitz I → Charlier / Hermite If we set a → a(q−1) and x → qx in the definition (14.24.1) of the Al-Salam-Carlitz I polynomials and take the limit q → 1 after dividing by an (1 − q)n we obtain the Charlier polynomials given by (9.14.1): (a(q−1))
(qx ; q) = anCn (x; a). (14.24.13) q→1 (1 − q)n If we set x → x 1 − q2 and a → a 1 − q2 − 1 in the definition (14.24.1) of the n Al-Salam-Carlitz I polynomials, divide by (1 − q2 ) 2 , and let q tend to 1 we obtain the Hermite polynomials given by (9.15.1) with shifted argument. In fact we have √ (a 1−q2 −1) Un (x 1 − q2 ; q) Hn (x − a) = . (14.24.14) lim n q→1 2n (1 − q2 ) 2 lim
Un
14.25 Al-Salam-Carlitz II
537
Remark The Al-Salam-Carlitz I polynomials are related to the Al-Salam-Carlitz II polynomials given by (14.25.1) in the following way: Un (x; q−1 ) = Vn (x; q). (a)
(a)
References [16], [18], [20], [68], [80], [144], [146], [157], [160], [180], [238], [269], [289], [312], [524].
14.25 Al-Salam-Carlitz II Basic Hypergeometric Representation n
Vn (x; q) = (−a)n q−(2) 2 φ0 (a)
qn q−n , x ; q, − a
.
(14.25.1)
Orthogonality Relation ∞
2
qk ak (a) (a) ∑ (q; q)k (aq; q)k Vm (q−k ; q)Vn (q−k ; q) k=0 =
(q; q)n an 2 δmn , (aq; q)∞ qn
0 < aq < 1.
(14.25.2)
Recurrence Relation xVn (x; q) = Vn+1 (x; q) + (a + 1)q−nVn (x; q) (a)
(a)
(a)
+ aq−2n+1 (1 − qn )Vn−1 (x; q). (a)
(14.25.3)
538
14 Basic Hypergeometric Orthogonal Polynomials
Normalized Recurrence Relation xpn (x) = pn+1 (x) + (a + 1)q−n pn (x) + aq−2n+1 (1 − qn )pn−1 (x), where
(14.25.4)
(a)
Vn (x; q) = pn (x).
q-Difference Equation −(1 − qn )x2 y(x) = (1 − x)(a − x)y(qx) − [(1 − x)(a − x) + aq] y(x) + aqy(q−1 x), (14.25.5) where
(a)
y(x) = Vn (x; q).
Forward Shift Operator Vn (x; q) −Vn (qx; q) = q−n+1 (1 − qn )xVn−1 (qx; q) (a)
(a)
(a)
or equivalently (a)
DqVn (x; q) =
q−n+1 (1 − qn ) (a) Vn−1 (qx; q). 1−q
(14.25.6)
(14.25.7)
Backward Shift Operator (a)
(a)
(a)
aVn (x; q) − (1 − x)(a − x)Vn (qx; q) = −qn xVn+1 (x; q)
(14.25.8)
or equivalently (a) Dq w(x; a; q)Vn (x; q) = −
qn (a) w(x; a; q)Vn+1 (x; q), a(1 − q)
where w(x; a; q) =
1 . (x, a−1 x; q)∞
(14.25.9)
14.25 Al-Salam-Carlitz II
539
Rodrigues-Type Formula n
w(x; a; q)Vn (x; q) = an (q − 1)n q−(2) (Dq )n [w(x; a; q)] . (a)
(14.25.10)
Generating Functions n ∞ (xt; q)∞ (−1)n q(2) (a) =∑ Vn (x; q)t n . (t, at; q)∞ n=0 (q; q)n
(14.25.11)
x ∞ qn(n−1) (a) ; q, t = ∑ Vn (x; q)t n . (q; q) at n n=0
(14.25.12)
(at; q)∞ · 1 φ1
Limit Relations q-Meixner → Al-Salam-Carlitz II The Al-Salam-Carlitz II polynomials given by (14.25.1) can be obtained from the q-Meixner polynomials given by (14.13.1) by setting b = −ac−1 in the definition (14.13.1) of the q-Meixner polynomials and then taking the limit c → 0: 1 n (n) (a) −1 q 2 Vn (x; q). lim Mn (x; −ac , c; q) = − c→0 a Quantum q-Krawtchouk → Al-Salam-Carlitz II If we set p = a−1 q−N−1 in the definition (14.14.1) of the quantum q-Krawtchouk polynomials and let N → ∞ we obtain the Al-Salam-Carlitz II polynomials given by (14.25.1). In fact we have 1 n (n) (a) qtm −1 −N−1 , N; q) = − q 2 Vn (x; q). lim Kn (x; a q N→∞ a Al-Salam-Carlitz II → Discrete q-Hermite II The discrete q-Hermite II polynomials given by (14.29.1) follow from the Al-SalamCarlitz II polynomials given by (14.25.1) by the substitution a = −1 in the following way: (−1) i−nVn (ix; q) = h˜ n (x; q). (14.25.13)
540
14 Basic Hypergeometric Orthogonal Polynomials
Al-Salam-Carlitz II → Charlier / Hermite If we set a → a(1 − q) and x → q−x in the definition (14.25.1) of the Al-SalamCarlitz II polynomials and taking the limit q → 1 we find (q−x ; q) = anCn (x; a). (14.25.14) q→1 (q − 1)n If we set x → ix 1 − q2 and a → ia 1 − q2 − 1 in the definition (14.25.1) of the n Al-Salam-Carlitz II polynomials, divide by in (1 − q2 ) 2 , and let q tend to 1 we obtain the Hermite polynomials given by (9.15.1) with shifted argument. In fact we have √ (ia 1−q2 −1) Vn (ix 1 − q2 ; q) Hn (x − a) = . (14.25.15) lim n q→1 2n in (1 − q2 ) 2 (a(1−q))
lim
Vn
Remark The Al-Salam-Carlitz II polynomials are related to the Al-Salam-Carlitz I polynomials given by (14.24.1) in the following way: Vn (x; q−1 ) = Un (x; q). (a)
(a)
References [16], [18], [20], [67], [101], [143], [144], [146], [160], [214], [269], [276].
14.26 Continuous q-Hermite Basic Hypergeometric Representation Hn (x|q) = einθ 2 φ0
q−n , 0 ; q, qn e−2iθ , −
x = cos θ .
(14.26.1)
14.26 Continuous q-Hermite
541
Orthogonality Relation 1 2π where
1 w(x|q)
δmn √ Hm (x|q)Hn (x|q) dx = n+1 , 2 (q ; q)∞ −1 1−x
(14.26.2)
2 1 1 w(x|q) = e2iθ ; q = h(x, 1)h(x, −1)h(x, q 2 )h(x, −q 2 ), ∞
with ∞ h(x, α ) := ∏ 1 − 2α xqk + α 2 q2k = α eiθ , α e−iθ ; q , ∞
k=0
x = cos θ .
Recurrence Relation 2xHn (x|q) = Hn+1 (x|q) + (1 − qn )Hn−1 (x|q).
(14.26.3)
Normalized Recurrence Relation 1 xpn (x) = pn+1 (x) + (1 − qn )pn−1 (x), 4
(14.26.4)
where Hn (x|q) = 2n pn (x).
q-Difference Equation −n+1 (1 − q)2 Dq [w(x|q)D ˜ (1 − qn )w(x|q)y(x) ˜ = 0, q y(x)] + 4q
(14.26.5)
where y(x) = Hn (x|q) and
w(x|q) . w(x|q) ˜ := √ 1 − x2
Forward Shift Operator δq Hn (x|q) = −q− 2 n (1 − qn )(eiθ − e−iθ )Hn−1 (x|q), 1
x = cos θ
(14.26.6)
542
14 Basic Hypergeometric Orthogonal Polynomials
or equivalently 1
2q− 2 (n−1) (1 − qn ) Hn−1 (x|q). Dq Hn (x|q) = 1−q
(14.26.7)
Backward Shift Operator − (n+1) iθ δq [w(x|q)H ˜ (e − e−iθ )w(x|q)H ˜ n (x|q)] = q 2 n+1 (x|q), 1
x = cos θ (14.26.8)
or equivalently 1
2q− 2 n w(x|q)H ˜ ˜ Dq [w(x|q)H n (x|q)] = − n+1 (x|q). 1−q
(14.26.9)
Rodrigues-Type Formula w(x|q)H ˜ n (x|q) =
q−1 2
n
1
q 4 n(n−1) (Dq )n [w(x|q)] ˜ .
(14.26.10)
Generating Functions 1 2 |(eiθ t; q)∞ |
=
∞
1 (eiθ t, e−iθ t; q)∞ iθ
(e t; q)∞ · 1 φ1
=
Hn (x|q) n t , n=0 (q; q)n
∑
0
−iθ
x = cos θ .
(14.26.11)
; q, e t e iθ t n ∞ (−1)n q(2) Hn (x|q)t n , x = cos θ . =∑ (q; q) n n=0
γ, 0 (γ eiθ t; q)∞ −iθ ; q, e φ t 2 1 γ e iθ t (eiθ t; q)∞ ∞ (γ ; q)n Hn (x|q)t n , x = cos θ , =∑ (q; q) n n=0
γ arbitrary.
(14.26.12)
(14.26.13)
14.26 Continuous q-Hermite
543
Limit Relations Continuous Big q-Hermite → Continuous q-Hermite The continuous q-Hermite polynomials given by (14.26.1) can easily be obtained from the continuous big q-Hermite polynomials given by (14.18.1) by taking a = 0: Hn (x; 0|q) = Hn (x|q). Continuous q-Laguerre → Continuous q-Hermite The continuous q-Hermite polynomials given by (14.26.1) can be obtained from the continuous q-Laguerre polynomials given by (14.19.1) by taking the limit α → ∞ in the following way: (α ) Pn (x|q) Hn (x|q) . lim 1 1 = α →∞ q( 2 α + 4 )n (q; q)n Continuous q-Hermite → Hermite The Hermite polynomials given by (9.15.1) can be obtained from the continuous q-Hermite polynomials given by (14.26.1) by setting x → x 12 (1 − q). In fact we have Hn (x 12 (1 − q)|q) = Hn (x). (14.26.14) lim n q→1 1−q 2
2
Remark The continuous q-Hermite polynomials can also be written as: n
Hn (x|q) =
(q; q)n
∑ (q; q)k (q; q)n−k ei(n−2k)θ ,
x = cos θ .
k=0
References [8], [16], [17], [25], [34], [51], [52], [62], [63], [72], [73], [75], [76], [78], [80], [86], [87], [100], [113], [119], [122], [210], [234], [238], [272], [276], [285], [294], [296], [416], [460], [461], [462], [481].
544
14 Basic Hypergeometric Orthogonal Polynomials
14.27 Stieltjes-Wigert Basic Hypergeometric Representation 1 Sn (x; q) = 1 φ1 (q; q)n
q−n n+1 ; q, −q x . 0
(14.27.1)
Orthogonality Relation ∞ Sm (x; q)Sn (x; q) 0
(−x, −qx−1 ; q)∞
dx = −
ln q (q; q)∞ δmn . qn (q; q)n
(14.27.2)
Recurrence Relation −q2n+1 xSn (x; q) = (1 − qn+1 )Sn+1 (x; q) − [1 + q − qn+1 ]Sn (x; q) + qSn−1 (x; q). (14.27.3)
Normalized Recurrence Relation
xpn (x) = pn+1 (x) + q−2n−1 1 + q − qn+1 pn (x) + q−4n+1 (1 − qn )pn−1 (x), where
(14.27.4)
2
Sn (x; q) =
(−1)n qn pn (x). (q; q)n
q-Difference Equation − x(1 − qn )y(x) = xy(qx) − (x + 1)y(x) + y(q−1 x),
y(x) = Sn (x; q).
(14.27.5)
14.27 Stieltjes-Wigert
545
Forward Shift Operator Sn (x; q) − Sn (qx; q) = −qxSn−1 (q2 x; q) or equivalently Dq Sn (x; q) = −
q Sn−1 (q2 x; q). 1−q
(14.27.6) (14.27.7)
Backward Shift Operator Sn (x; q) − xSn (qx; q) = (1 − qn+1 )Sn+1 (q−1 x; q),
(14.27.8)
or equivalently Dq [w(x; q)Sn (x; q)] =
1 − qn+1 −1 q w(q−1 x; q)Sn+1 (q−1 x; q), 1−q
where w(x; q) =
(14.27.9)
1 . (−x, −qx−1 ; q)∞
Rodrigues-Type Formula w(x; q)Sn (x; q) =
qn (1 − q)n ((Dq )n w) (qn x; q). (q; q)n
(14.27.10)
Generating Functions ∞ − 1 ; q, −qxt = ∑ Sn (x; q)t n . 0 φ1 0 (t; q)∞ n=0 ∞ n − ; q, −qxt = ∑ (−1)n q(2) Sn (x; q)t n . (t; q)∞ · 0 φ2 0,t n=0 (γ t; q)∞ 1 φ2 (t; q)∞
γ ; q, −qxt 0, γ t
∞
=
∑ (γ ; q)n Sn (x; q)t n ,
n=0
γ arbitrary.
(14.27.11) (14.27.12)
(14.27.13)
546
14 Basic Hypergeometric Orthogonal Polynomials
Limit Relations q-Laguerre → Stieltjes-Wigert If we set x → xq−α in the definition (14.21.1) of the q-Laguerre polynomials and take the limit α → ∞ we simply obtain the Stieltjes-Wigert polynomials given by (14.27.1): (α ) lim Ln xq−α ; q = Sn (x; q). α →∞
q-Bessel → Stieltjes-Wigert The Stieltjes-Wigert polynomials given by (14.27.1) can be obtained from the qBessel polynomials by setting x → a−1 x in the definition (14.22.1) of the q-Bessel polynomials and then taking the limit a → ∞. In fact we have lim yn (a−1 x; a; q) = (q; q)n Sn (x; q).
a→∞
q-Charlier → Stieltjes-Wigert If we set q−x → ax in the definition (14.23.1) of the q-Charlier polynomials and take the limit a → ∞ we obtain the Stieltjes-Wigert polynomials given by (14.27.1) in the following way: lim Cn (ax; a; q) = (q; q)n Sn (x; q). a→∞
Stieltjes-Wigert → Hermite The Hermite polynomials given by (9.15.1) can be obtained from the Stieltjes-Wigert polynomials given by (14.27.1) by setting x → q−1 x 2(1 − q) + 1 and taking the limit q → 1 in the following way: (q; q)n Sn (q−1 x 2(1 − q) + 1; q) = (−1)n Hn (x). (14.27.14) lim n q→1
1−q 2
2
Remark Since the Stieltjes and Hamburger moment problems corresponding to the StieltjesWigert polynomials are indeterminate there exist many different weight functions. For instance, they are also orthogonal with respect to the weight function
14.28 Discrete q-Hermite I
547
γ w(x) = √ exp −γ 2 ln2 x , π
x > 0,
with γ 2 = −
1 . 2 ln q
References [49], [51], [86], [145], [146], [160], [276], [416], [490], [493], [503], [511].
14.28 Discrete q-Hermite I Basic Hypergeometric Representation The discrete q-Hermite I polynomials are Al-Salam-Carlitz I polynomials with a = −1: −n −1 n q ,x (−1) ( ) 2 ; q, −qx (14.28.1) hn (x; q) = Un (x; q) = q 2 φ1 0 −n −n+1 2 q2n−1 q ,q q ; . = xn 2 φ0 − x2
Orthogonality Relation 1 −1
(qx, −qx; q)∞ hm (x; q)hn (x; q) dq x n
= (1 − q)(q; q)n (q, −1, −q; q)∞ q(2) δmn .
(14.28.2)
Recurrence Relation xhn (x; q) = hn+1 (x; q) + qn−1 (1 − qn )hn−1 (x; q).
(14.28.3)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + qn−1 (1 − qn )pn−1 (x), where
(14.28.4)
548
14 Basic Hypergeometric Orthogonal Polynomials
hn (x; q) = pn (x).
q-Difference Equation − q−n+1 x2 y(x) = y(qx) − (1 + q)y(x) + q(1 − x2 )y(q−1 x),
(14.28.5)
where y(x) = hn (x; q).
Forward Shift Operator hn (x; q) − hn (qx; q) = (1 − qn )xhn−1 (x; q) or equivalently Dq hn (x; q) =
1 − qn hn−1 (x; q). 1−q
(14.28.6)
(14.28.7)
Backward Shift Operator hn (x; q) − (1 − x2 )hn (q−1 x; q) = q−n xhn+1 (x; q)
(14.28.8)
or equivalently Dq−1 [w(x; q)hn (x; q)] = −
q−n+1 w(x; q)hn+1 (x; q), 1−q
(14.28.9)
where w(x; q) = (qx, −qx; q)∞ .
Rodrigues-Type Formula n 1 w(x; q)hn (x; q) = (q − 1)n q 2 n(n−3) Dq−1 [w(x; q)] .
(14.28.10)
Generating function ∞ (t 2 ; q2 )∞ hn (x; q) n =∑ t . (xt; q)∞ n=0 (q; q)n
(14.28.11)
14.28 Discrete q-Hermite I
549
Limit Relations Al-Salam-Carlitz I → Discrete q-Hermite I The discrete q-Hermite I polynomials given by (14.28.1) can easily be obtained from the Al-Salam-Carlitz I polynomials given by (14.24.1) by the substitution a = −1: (−1)
Un
(x; q) = hn (x; q).
Discrete q-Hermite I → Hermite The Hermite polynomials given by (9.15.1) can be found from the discrete qHermite I polynomials given by (14.28.1) in the following way: hn (x 1 − q2 ; q) Hn (x) lim = . (14.28.12) n q→1 2n (1 − q2 ) 2
Remark The discrete q-Hermite I polynomials are related to the discrete q-Hermite II polynomials given by (14.29.1) in the following way: hn (ix; q−1 ) = in h˜ n (x; q).
References [16], [18], [80], [100], [119], [238], [261], [349].
550
14 Basic Hypergeometric Orthogonal Polynomials
14.29 Discrete q-Hermite II Basic Hypergeometric Representation The discrete q-Hermite II polynomials are Al-Salam-Carlitz II polynomials with a = −1: −n n ˜hn (x; q) = i−nVn(−1) (ix; q) = i−n q−(2) 2 φ0 q , ix ; q, −qn (14.29.1) − −n −n+1 2 q2 q ,q q ;− = xn 2 φ1 . 0 x2
Orthogonality Relation
∞
∑
h˜ m (cqk ; q)h˜ n (cqk ; q) + h˜ m (−cqk ; q)h˜ n (−cqk ; q) w(cqk ; q)qk
k=−∞
=2
(q2 , −c2 q, −c−2 q; q2 )∞ (q; q)n δmn , (q, −c2 , −c−2 q2 ; q2 )∞ qn2
where w(x; q) =
c > 0,
(14.29.2)
1 1 = . (ix, −ix; q)∞ (−x2 ; q2 )∞
For c = 1 this orthogonality relation can also be written as 2 ∞ ˜ q , −q, −q; q2 ∞ (q; q)n hm (x; q)h˜ n (x; q) dq x = 3 δmn . (−x2 ; q2 )∞ (q , −q2 , −q2 ; q2 )∞ qn2 −∞
(14.29.3)
Recurrence Relation xh˜ n (x; q) = h˜ n+1 (x; q) + q−2n+1 (1 − qn )h˜ n−1 (x; q).
(14.29.4)
Normalized Recurrence Relation xpn (x) = pn+1 (x) + q−2n+1 (1 − qn )pn−1 (x), where h˜ n (x; q) = pn (x).
(14.29.5)
14.29 Discrete q-Hermite II
551
q-Difference Equation −(1 − qn )x2 h˜ n (x; q) = (1 + x2 )h˜ n (qx; q) − (1 + x2 + q)h˜ n (x; q) + qh˜ n (q−1 x; q).
(14.29.6)
Forward Shift Operator h˜ n (x; q) − h˜ n (qx; q) = q−n+1 (1 − qn )xh˜ n−1 (qx; q) or equivalently
q−n+1 (1 − qn ) ˜ Dq h˜ n (x; q) = hn−1 (qx; q). 1−q
(14.29.7)
(14.29.8)
Backward Shift Operator h˜ n (x; q) − (1 + x2 )h˜ n (qx; q) = −qn xh˜ n+1 (x; q)
(14.29.9)
or equivalently
qn w(x; q)h˜ n+1 (x; q). Dq w(x; q)h˜ n (x; q) = − 1−q
(14.29.10)
Rodrigues-Type Formula n
w(x; q)h˜ n (x; q) = (q − 1)n q−(2) (Dq )n [w(x; q)] .
(14.29.11)
Generating Functions n ∞ (−xt; q)∞ q(2) ˜ hn (x; q)t n . =∑ (−t 2 ; q2 )∞ n=0 (q; q)n
(−it; q)∞ · 1 φ1
ix ; q, it −it
(14.29.12)
∞
=
(−1)n qn(n−1) ˜ hn (x; q)t n . (q; q) n n=0
∑
(14.29.13)
552
14 Basic Hypergeometric Orthogonal Polynomials
Limit Relations Al-Salam-Carlitz II → Discrete q-Hermite II The discrete q-Hermite II polynomials given by (14.29.1) follow from the Al-SalamCarlitz II polynomials given by (14.25.1) by the substitution a = −1 in the following way: (−1) i−nVn (ix; q) = h˜ n (x; q). Discrete q-Hermite II → Hermite The Hermite polynomials given by (9.15.1) can be found from the discrete qHermite II polynomials given by (14.29.1) in the following way: h˜ n (x 1 − q2 ; q) Hn (x) = . (14.29.14) lim n q→1 2n (1 − q2 ) 2
Remark The discrete q-Hermite II polynomials are related to the discrete q-Hermite I polynomials given by (14.28.1) in the following way: h˜ n (x; q−1 ) = i−n hn (ix; q).
References [100], [349].
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Index
A Affine q-Krawtchouk polynomials, 327, 353, 364, 501 Al-Salam-Carlitz I polynomials, 265, 305, 320, 325, 345, 358, 534 Al-Salam-Carlitz II polynomials, 265, 307, 322, 325, 346, 357, 537 Al-Salam-Chihara polynomials, 397, 455 Alternative q-Charlier polynomials, see q-Bessel polynomials Askey scheme, 183 q-analogue of, 413 Askey-Wilson integral, 18 Askey-Wilson polynomials, 397, 410, 415 Askey-Wilson q-beta integral, 18
B Bailey’s integral, 9 Barnes’ first lemma, 8 Barnes’ integral representation, 8 Barnes’ second lemma, 8 Basic hypergeometric function, 15 Balanced, 15 Saalsch¨utzian, 15 Bessel function, 6 Bessel polynomials, 89, 244 Beta distribution, 86 Beta function, 3 Big q-Hermite polynomials Continuous, 397, 409, 509 Big q-Jacobi polynomials, 265, 305, 319, 438 Big q-Laguerre polynomials, 265, 305, 319, 478 Big q-Legendre polynomials, 443 Binomial coefficient, 4
Binomial distribution, 108 Negative, 103 Binomial theorem, 7 C Charlier polynomials, 101, 106, 247 Chebyshev polynomials, 225 Chu-Vandermonde summation formula, 7 Classical orthogonal polynomials Continuous, 79 Discrete, 95, 123, 141, 171 Classical q-orthogonal polynomials, 257, 323, 369, 395 Continuous big q-Hermite polynomials, 397, 409, 509 Continuous classical orthogonal polynomials, 79 Continuous dual Hahn polynomials, 171, 196 Orthogonality relations for, 173 Continuous dual q-Hahn polynomials, 397, 429 Continuous Hahn polynomials, 136, 200 Continuous q-Hahn polynomials, 397, 433 Continuous q-Hermite polynomials, 409, 540 Continuous q-Jacobi polynomials, 397, 463 Continuous q-Laguerre polynomials, 397, 514 Continuous q-Legendre polynomials, 475 Continuous q-ultraspherical polynomials, 469 D Difference equations Polynomial solutions of, 95, 123, 142, 171 Differential equations Polynomial solutions of, 79 Discrete classical orthogonal polynomials, 95, 123, 141, 171 Discrete q-Hermite I polynomials, 265, 305, 320, 325, 346, 358, 547
R. Koekoek et al., Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-05014-5, © Springer-Verlag Berlin Heidelberg 2010
575
576 Discrete q-Hermite II polynomials, 263, 265, 302, 306, 315, 321, 322, 325, 345, 356, 357, 550 Distribution Beta, 86 Binomial, 108 Euler, 265 Gamma, 85 Heine, 265 Hypergeometric, 113 Kemp, 265 Negative binomial, 103 Normal, 84 Poisson, 101 P´olya, 113 Student’s t-, 90 Dougall’s bilateral sum, 7 Dual Hahn polynomials, 146, 208 Continuous, 171, 196 Orthogonality relations for, 158 Dual q-Charlier polynomials, 376, 394 Dual q-Hahn polynomials, 376, 393, 450 Continuous, 397, 429 Dual q-Krawtchouk polynomials, 376, 394, 505 Dual q-Racah polynomials, 377 Duality, 71, 141, 369 E Eigenvalue problems, 30 Euler distribution, 265 Euler’s integral representation for a 2 F1 , 8 Euler’s transformation formula, 10 F Favard’s theorem, 53 G Gamma distribution, 85 Gamma function, 3 Gauss summation formula, 7 Gegenbauer polynomials, 222 H Hahn polynomials, 111, 204 Continuous, 136, 200 Continuous dual, 171, 196 Dual, 146, 208 Hahn’s q-operator, 29 Hahn-Exton q-Bessel function, 24 Heine distribution, 265 Heine’s series, 15 Heine’s transformation formula, 19 Hermite polynomials, 84, 250
Index Hypergeometric distribution, 113 Hypergeometric function, 5 Balanced, 5 Barnes’ integral representation, 8 Euler’s integral representation, 8 Saalsch¨utzian, 5 J Jackson’s q-Bessel function, 23 Jackson’s summation formula, 17 Jackson’s transformation formula, 21 Jackson-Thomae q-integral, 25, 59 Jacobi polynomials, 86, 88, 216 Pseudo, 90, 231 K Kemp distribution, 265 Kravchuk polynomials, see Krawtchouk polynomials Krawtchouk polynomials, 108, 109, 237 Kummer’s transformation formula, 10 L Laguerre polynomials, 85, 241 Legendre polynomials, 229 Little q-Jacobi polynomials, 261, 301, 312, 482 Little q-Laguerre polynomials, 261, 301, 312, 518 Little q-Legendre polynomials, 486 M Meixner polynomials, 103, 104, 234 Meixner-Pollaczek polynomials, 132, 213 Mellin-Barnes integral, 8 N Negative binomial distribution, 103 Normal distribution, 84 O Orthogonal polynomials, 1 Continuous classical, 79 Discrete classical, 95, 123, 141, 171 Orthogonality, 53 Orthogonality relations for continuous dual Hahn polynomials, 173 for dual Hahn polynomials, 158 for Racah polynomials, 162 for Wilson polynomials, 177 P Pfaff-Kummer transformation formula, 10 Pfaff-Saalsch¨utz summation formula, 7 Pochhammer symbol, see Shifted factorial
Index Poisson distribution, 101 P´olya distribution, 113 Polynomial solutions of complex difference equations, 123, 171 of complex q-difference equations, 395 of differential equations, 79 of q-difference equations, 257, 323 of real difference equations, 95, 142 of real q-difference equations, 370 Pseudo Jacobi polynomials, 90, 231 Q q-Askey scheme, 413 q-Bessel function, 23 q-Bessel polynomials, 261, 301, 314, 526 q-Beta integral, 18 q-Binomial coefficient, 14 q-Binomial theorem, 16 q-Charlier polynomials, 326, 348, 360, 530 Alternative, see q-Bessel polynomials Dual, 376, 394 q-Chu-Vandermonde summation formula, 17 q-Derivative operator, 24 q-Difference equations Polynomial solutions of, 257, 323, 370, 395 q-Distribution, 265 q-Euler transformation formula, 19 q-Exponential function, 22 q-Gamma function, 13 q-Gauss summation formula, 17 q-Hahn polynomials, 327, 352, 367, 445 Continuous, 397, 433 Continuous dual, 397, 429 Dual, 376, 393, 450 q-Hermite polynomials Continuous, 409, 540 Continuous big, 397, 409, 509 q-Hermite I polynomials Discrete, 265, 305, 320, 325, 346, 358, 547 q-Hermite II polynomials Discrete, 263, 265, 302, 306, 315, 321, 322, 325, 345, 356, 357, 550 q-Hypergeometric function, see Basic hypergeometric function q-Integral, 25 q-Jacobi polynomials Big, 265, 305, 319, 438 Continuous, 397, 463 Little, 261, 301, 312, 482 q-Kravchuk polynomials, see q-Krawtchouk polynomials q-Krawtchouk polynomials, 327, 354, 366, 496 Affine, 327, 353, 364, 501 Dual, 376, 394, 505
577 Quantum, 326, 348, 362, 493 q-Laguerre polynomials, 260, 297, 311, 522 Big, 265, 305, 319, 478 Continuous, 397, 514 Little, 261, 301, 312, 518 q-Legendre polynomials Big, 443 Continuous, 475 Little, 486 q-Meixner polynomials, 326, 348, 360, 488 q-Meixner-Pollaczek polynomials, 397, 460 q-Orthogonal polynomials Classical, 257, 323, 369, 395 q-Pfaff-Kummer transformation formula, 19 q-Pfaff-Saalsch¨utz summation formula, 17 q-Pochhammer symbol, see q-Shifted factorial q-Racah polynomials, 376, 391, 392, 422 Dual, 377 q-Saalsch¨utz summation formula, 17 q-Shifted factorial, 11 q-Ultraspherical polynomials Continuous, 469 q-Vandermonde summation formula, 17 Quantum q-Krawtchouk polynomials, 326, 348, 362, 493 R Racah polynomials, 148, 190 Orthogonality relations for, 162 Recurrence relation, 40 Regularity condition, 33 Rodrigues formulas, 62 Rogers polynomials, 469 S Saalsch¨utz summation formula, 7 Sears’ transformation formula, 22 Self-adjoint operator equation, 55 Shifted factorial, 4 Singh’s transformation formula, 22 Spherical polynomials, 229 Stieltjes-Wigert polynomials, 260, 261, 296, 297, 308, 310, 544 Stochastic distribution, see Distribution Student’s t-distribution, 90 Summation formula Binomial theorem, 7 Chu-Vandermonde, 7 Dougall, 7 for a 0 φ1 , 17 for a 1 φ1 , 17 for a terminating 2 φ0 , 17 for a very-well-poised 5 F4 , 7 for a very-well-poised 6 φ5 , 17
578 Gauss, 7 Jackson, 17 Pfaff-Saalsch¨utz, 7 q-Binomial theorem, 16 q-Chu-Vandermonde, 17 q-Gauss, 17 q-Pfaff-Saalsch¨utz, 17 q-Saalsch¨utz, 17 q-Vandermonde, 17 Saalsch¨utz, 7 Vandermonde, 7 T Tchebichef/Tchebycheff polynomials, see Chebyshev polynomials Three-term recurrence relation, 40 Transformation formula Euler, 10 Heine, 19 Jackson, 21
Index Kummer, 10 Pfaff-Kummer, 10 q-Euler, 19 q-Pfaff-Kummer, 19 Sears, 22 Singh, 22 Whipple, 11 U Ultraspherical polynomials, 222 V Vandermonde summation formula, 7 W Wall polynomials, 518 Whipple’s transformation formula, 11 Wilson polynomials, 172, 185 Orthogonality relations for, 177 Wilson’s integral, 10
