NUMERICALLY SATISFACTORY SOLUTIONS OF HYPERGEOMETRIC RECURSIONS

MATHEMATICS OF COMPUTATION Volume 76, Number 259, July 2007, Pages 1449 1468 S 0025-5718(07)01918-7 Artile eletronially published on January 31, 2007 NUMERICALLY SATISFACTORY SOLUTIONS OF HYPERGEOMETRIC RECURSIONS AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME Abstrat. Eah family of Gauss hypergeometri funtions f n = 2 F 1 (a + ε 1 n, b + ε 2 n; + ε 3 n; ), n Z, for fixed ε j =0, ±1 (notallε j equal to ero) satisfies a seond order linear differene equation of the form A n f n 1 + B n f n + C n f n+1 =0. Beause of symmetry relations and funtional relations for the Gauss funtions, many of the 26 ases (for different ε j values) an be transformed into eah other. In this way, only with four basi differene equations an all other ases be obtained. For eah of these reurrenes, we give pairs of numerially satisfatory solutions in the regions in the omplex plane where t 1 t 2, t 1 and t 2 being the roots of the harateristi equation. 1. Introdution The families of Gauss hypergeometri funtions a + ε1 n, b + ε (1.1) y n = 2 F 2 n 1 ;, n Z, + ε 3 n where a, b (1.2) 2F 1 ; (a) n (b) n = () n=0 n n! n, < 1, satisfy seond order linear differene equations (three-term reurrene relations) of the form (1.3) A n y n 1 + B n y n + C n y n+1 =0. Here, ε j Z are fixed and not all ε j are equal to ero. In this paper, we onsider the ases ɛ j =0, ±1. That these reurrenes exist and an be obtained from relations between ontiguous funtions (see [1, pp. 557-558]) are well-known fats. However, the ondition of these reurrenes is essentially an unexplored issue. For a numerial use of a reurrene relation, it is of ruial importane to know whether a reurrene admits a minimal solution and to identify suh a solution Reeived by the editor Otober 18, 2005 and, in revised form, February 2, 2006. 2000 Mathematis Subjet Classifiation. Primary 33C05, 39A11, 41A60, 65D20. Key words and phrases. Gauss hypergeometri funtions, reursion relations, differene equations, stability of reursion relations, numerial evaluation of speial funtions, asymptoti analysis. 1449 2007 Amerian Mathematial Soiety Reverts to publi domain 28 years from publiation

1450 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME when it exists. A solution f n of a three-term reurrene relation (TTRR) is said to be minimal as n + when, for any other solution independent of f n, g n, lim f n /g n =0;g n is said to be a dominant solution. The minimal solution is unique up to a onstant fator (not depending on n). The omputation of values of f n for large n by a forward (inreasing n) appliation of the TTRR is a bad onditioned proess, while the bakward appliation of the reurrene is generally well onditioned. The opposite happens for the dominant solutions. When a TTRR admits a minimal solution, an independent pair of solutions of this TTRR {f n,g n } is said to be numerially satisfatory for large n when it inludes the minimal solution. The goal of this artile is to obtain numerially satisfatory solutions for the hypergeometri reursions in the omplex plane. As shown in [3], of the 26 possible reursions (when ɛ j 1) in priniple only five have to be studied, beause of symmetry relations and onnetion formulas. That is, beause of the relations 2F 1 a, b ; = 2 F 1 b, a ;, 2F 1 a, b ; =(1 ) a 2F 1 a, b ;, 1 (1.4) 2F 1 a, b ; =(1 ) b 2F 1 a, b ;, 1 2F 1 a, b ; =(1 ) a b 2F 1 a, b ;. Using these relations, it follows that we need to study the reursions for the following five basi forms [3]: (1.5) a + n, b + n 2F 1 ; a + n, b + n, 2F 1 ; n 2F 1 ( a + n, b n ; ), 2F 1 ( a, b + n ; ). a + n, b, 2F 1 ;, However, as we will see when expliitly building solutions for these reurrenes, the third and the last ases in Eq. (1.5) are also related, and only four reurrenes need to be studied. In [3] we have desribed the domains in the omplex -plane where minimal and dominant solutions of the differene equations have to be determined. In this paper, we determine the minimal solutions in eah of these domains whih, together with any dominant solution, forms a numerially satisfatory pair of solutions of the orresponding three term reurrene relation. In order to find the minimal solutions in eah domain, it will be important to build solutions of the reurrene relations based on expansions around the three singular regular points of the differential equation: = 0, 1,. The following set of funtions, solutions of the Gauss hypergeometri equation, provides the starting

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1451 point for finding suh solutions of the TTRR: w 1 = 2 F 1 a, b ;, 1+a, 1+b w 2 = 1 2F 1 ;, 2 a, b w 3 = 2 F 1 ;1, a + b +1 (1.6) w 4 =(1 ) a b 2F 1 a, b ;1, +1 a b a, a +1 w 5 =( 1 e iπ ) a 2F 1 ; 1, a +1 b b, b +1 w 6 =( 1 e iπ ) b 2F 1 ; 1. b +1 a This set of solutions plays an important role in defining pairs of linearly independent solutions of the hypergeometri differential equation. With the relations in (1.4) Kummer s 24 solutions of the differential equation an be given; see [4, p. 67]. Eah element of the 24 solutions an be written as a linear ombination of two other elements. The six funtions in (1.6) are also important in the theory of the differene equations for the Gauss hypergeometri funtion. When any of these funtions satisfies a seond order reursion relation, the remaining five funtions w j satisfy the same relation provided we multiply eah of these solutions by an appropriate fator. These fators follow from the linear relations between the funtions w j. For our purposes we mention four relations (see [4, pp. 70 71]) in whih w 1 is written in terms of the other w j, j =2, 3, 4, 5, 6. There are six more relations, but we need only 4 relations. These are [ ] Γ(a +1 )Γ(b +1 ) w 3 Γ( 1) w 1 = Γ(1 ) Γ(a + b +1 ) Γ(a)Γ(b) w 2 [ ] Γ(1 a)γ(1 b) w 4 = Γ(1 ) Γ( +1 a b) Γ( 1) Γ( a)γ( b) w 2 (1.7) [ ] Γ(1 b)γ(1 + a ) w 5 Γ( 1)eiπ( 1) = w 2 Γ(1 ) Γ(1 + a b) Γ(a)Γ( b) [ ] Γ(1 a)γ(1 + b ) w 6 Γ( 1)eiπ( 1) = w 2. Γ(1 ) Γ(1 + b a) Γ(b)Γ( a) When, for example the funtion w 1 of (1.6) is in the form 2 F 1 (a + n, b + n; ; ) and satisfies a ertain three-term reursion relation, any of the w j at the right-hand sides of (1.7) (with a and b replaed by a + n and b + n, and with the given ratios of gamma funtions) are seen to satisfy the same reursion relation. This an also be heked by diret substitution.

1452 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME We will use these relations to build 6 different solutions for eah reurrene relation. Of ourse, only two solutions will be independent but, as we will see, one of them will be minimal when the roots of the harateristi equation have different moduli. When building the six solutions, when neessary we will also use the relation ( 1) n π (1.8) Γ(p n) = sin(πp)γ(n +1 p) in order to avoid front fators with gamma funtions of the form Γ(p n), for p any ombination of the a, b, parameters. Also, the relations (1.4) will be used to express some of the solutions in a more symmetri form. Finally, we will neglet all front fators not depending on n. In the following setions we give the unique minimal solution (up to a multipliative onstant) and five dominant solutions for eah reursion relation orresponding with the 5 basi forms in (1.5). Eah reursion relation has one or more different domains in the omplex plane for whih the minimal and dominant solutions have to be given. 2. Minimal and dominant solutions Consider the differene equation (1.3) and define (2.1) α n := A n C n, β n := B n C n. Then for all equations onsidered in this paper α n and β n tend to well-defined limits (exept, perhaps, at the singular points of the hypergeometri differential equation). We write (2.2) α := lim n α n, β := lim n β n. Perron s theorem (see [6, Appendix B]) gives in the ase of finite limits the following result, whih is also known as Poinaré s theorem. Theorem 1 (Poinaré). Let t 1 and t 2 denote the eros of the harateristi equation t 2 + βt + α =0. Then, if t 1 t 2 the differene equation ( 1.3) has two linear independent solutions f n and g n with the properties f n (2.3) lim = t 1, f n 1 If t 1 = t 2,then lim g n g n 1 = t 2. (2.4) lim sup y n 1 n = t1 n for any non-trivial solution y n of ( 1.3). When t 1 t 2 the minimal solution is the one whose ratio (f n /f n 1 or g n /g n 1 ) onverges to the root with the smallest modulus. When t 1 = t 2 the theorem is inonlusive with respet to the existene of a minimal solution. The same type of analysis an be onsidered as n. By defining ŷ n = y n, we an use Theorem 1 for studying the behaviour of ŷ n and n +. Beause for the hypergeometri ase lim n ± α n = α and lim n ± β n = β the harateristi equation for ŷ n turns out to be reiproal to the previous ase, with roots 1/t 1 and 1/t 2. A simple way to analye the behaviour as n is the following.

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1453 Theorem 2. Suppose that lim n ± α n = α and lim n ± β n = β, andt 1 and t 2 are the roots of t 2 + βt + α =0. Let us assume that, for instane, t 1 < t 2. Then the differene equation has a minimal solution both as n + and as n. The minimal solution when n + satifies lim = t 1, y n 1 while the minimal solution as n is suh that lim n y n y n = t 2. y n 1 We will use this theorem for identifying the minimal solutions in eah region in the omplex plane for whih t 1 t 2. For this, we first need to determine solutions of the reurrene and then to obtain the asymptoti behaviour of ratios of these solutions as n ±. In [3] we announed that all information on minimal and dominant solutions, for all reursions and for all domains, ould be obtained by using asymptoti estimates of the Gauss hypergeometri funtions for large parameters, as given in [5]; some of these estimates are given in [4, 235 242]. Fortunately, the study beomes more straightforward by onsidering the asymptotis of the ratios y n /y n 1,whihisthe only information needed in Theorems 1 and 2. In addition, the study beomes muh more simple by taking into aount the following result: Theorem 3. The harater of the solutions of the hypergeometri TTRR annot hange in a onneted domain where the harateristi roots t 1, t 2 satisfy t 1 t 2 for all in the domain. This, together with the different behaviour of the roots t 1 and t 2 near the singular points of the hypergeometri ODE (0, 1, ) allows for a simple identifiation by studying the behaviour of the solutions around these points. For proving Theorem 3 we only need to prove that the solutions of the reurrene relation, y n, are suh that the ratio (2.5) H n = y n /y n 1 has, for fixed, a bounded derivative as n. Indeed, beause for eah, these ratios onverge to one of the roots of the harateristi equation, when there exists C suh that H n() <Cfor large n (and this is true for eah of the six solutions we provide for eah reurrene), we an guarantee that the limit may only hange (from t 1 to t 2 ) when rossing the urves t 1 = t 2 for any of the solutions. A first step for proving the boundedness of the derivative is the following result: Theorem 4. Under the onditions of Theorem 1 with α 0,let{y k }, k N, be a solution of the hypergeometri reurrene y n+1 + β n y n + α n y n 1 =0and let t 1 t 2 (and therefore β 0and β 2 4α 0). Then y n H n = t(1 + h n ) y n 1 where t is a root of the harateristi equation and h n = O(1/n) as n.

1454 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME Proof. From Poinaré s theorem, we know that H n = t(1 + h n )whereh n 0. In addition, beause β n and α n are rational funtions of n with finite limits as n, then β n = β(1 + γ n )andα n = α(1 + δ n )withδ n,γ n = O(1/n). Substituting these estimates into the differene equation satisfied by H n : α n + β n H n + H n H n+1 =0, and using the harateristi equation, we have: βt(γ n h n+1 )(1 + h n )+α(δ n h n h n+1 (1 + h n )) = 0. Now, supposing that h n goes to ero slower that 1/n, weanwrite: βth n+1 + α(h n+1 + h n )+o(h n )=0; therefore βt + α 0 and we an write: h n+1 lim = α n h n α + βt K. But this is in ontradition with the hypothesis, for all values of K. Indeed, if K < 1, then h n goes to ero as K n (faster that 1/n) while, if K > 1, h n annot go to ero; finally, if K =1,then α = α + βt, but beause α + βt + t 2 =0then t 2 = α ; this is in ontradition with the fat that t 1 t 2, beause t is either t 1 or t 2 and α = t 1 t 2. Now we an prove the boundedness of the derivative of H n (Theorem 5), using Theorem 4 together with some properties of the funtions y n = 2 F 1 (a + ε 1 n, b + ε 2 n; + ε 3 n; ), namely: 1. The three-term reurrene relations f n+1 + β n f n + α n f n 1 = 0 have oeffiients α n and β n with finite limits as n (see Eq. (2.1) and (2.2)). 2. The oeffiients a n,b n,d n,e n of the differene-differential (DDE) equations f n (2.6) = a nf n + d n f n 1, f n 1 = b n f n 1 + e n f n, are of order n as n. 3. The oeffiients a n, b n, d n, e n, α n and β n are rational funtions of n. These properties have been verified for all the reurrenes appearing in the present paper, that is for ɛ j 1, j = 1, 2, 3, and for the ases (ɛ 1,ɛ 2,ɛ 3 ) = (1, 0, 2), (1, 1, 2), (1, 1, 3) (1, 2, 0), whih, as we will see, appear when building the solutions for the basi reurrenes (ɛ 1,ɛ 2,ɛ 3 )=(1, 1, 0), (1, 0, 1), (1, 1, 1). Then, in a strit sense, Theorem 5 is only proved for these ases, although we onjeture that it is true in general for ɛ j Z. a + ε1 n, b + ε Theorem 5. Let y n = 2 F 2 n 1 ;, ɛ + ε 3 n 1,ɛ 2,ɛ 3 Z. For any 0, 1, there exist N>0and C>0 (independent of n), suh that H n() <C for all n>n. a + Thesameistruefory n = g(n)f() n ε1 n, b + ε 2 n 2F 1 ; ζ(), f() and + ε 3 n ζ() being differentiable in C \{0, 1, } and g(n +1)/g(n) bounded for large n.

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1455 Proof. The seond part of the ( theorem follows immediately ) from the first part. a + ε1 n, b + ε The funtions y n = 2 F 2 n 1 ; satisfy DDEs (Eq. (2.6)) and a + ε 3 n TTRR. The oeffiients of the TTRR and the DDE are related. Indeed, replaing n by n + 1 in the seond equation in (2.6) and equating to the first equation we get the TTRR: f n+1 + b n+1 a n f n d n f n 1 =0. e n+1 e n+1 Therefore, given one of the solutions (for instane any of the 6 solutions we provide) of the TTRR y n+1 + β n y n + α n y n 1 = 0, the oeffiients of the system of DDEs satisfied by set of funtions {y k } is related to the oeffiients of the TTRR through: (2.7) β n = b n+1 a n e n+1 = b n a n (1 + O(1/n)), e n α n = d n = d n (1 + O(1/n)). e n+1 e n H n, being a ratio of differentiable funtions (exept at the singular points), is differentiable. Using the DDEs we get: ( (2.8) H n = e n Hn 2 + b n a n H n d ) n = e n (Hn 2 + β n H n + α n + O(1/n)). e n e n Hene, beause α n = α(1 + O(1/n)) and β n = β(1 + O(1/n)), (2.9) H n = e n [(H n t 1 )(H n t 2 )+O(1/n)]. Now, if is suh that t 1 t 2,wehavethateither H n = y n /y n 1 t 1 (1 + O(1/n)) or H n t 2 (1 + O(1/n)) and then, beause e n = O(n) thenh n = O(1), whih means that there exists C>0suh that H n() <C for large n. Therefore, for large n and when t 1 t 2, H n an only be singular on the singular points of the differential equation. Then, if for instane we have that, at = 0 (not a singular point), lim n H n ( 0 )=t 1 ( 0 )and t 1 ( 0 ) t 2 ( 0 ), this limit holds around 0 and, in fat, it will be satisfied in any onneted set not ontaining the singular points of the ODE or the urves t 1 = t 2. This proves Theorem 3. Notie that, if is suh that t 1 = t 2, the argument is not true and the derivative may beome unbounded as n. This is onsistent with the fat that the ratios H n may onverge to t 1 at one side of a urve t 1 = t 2 and to t 2 at the other side. This theorem is ruial in simplifying our study beause is states that a loal analysis in the neighborhood of ertain points (the singular points of the ODE) will suffie to obtain global information. The seond fundamental simplifiation omes from the fat that, as we will see, the urves t 1 = t 2 divide the omplex plane into disjoint regions suh that in the interior of eah of these disjoint regions there is only one singular point of the ODE. Besides, at eah of these interior singular points, the harateristi roots behave differently, allowing for a simple identifiation of the minimal solution. Furthermore, we will see that for any of these interior singular points, the minimal

1456 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME solution around this point will be expressed in terms of an expansion around this same point. Let us give an example for larifiation. The reurrene relation for the ase (+ + +) (not taken as a basi ase but related to (0 0 +)) has as roots t 1 =1/(1 ) and t 2 = 1/. Theurve t 1 = t 2 is the vertial line R =1/2, and there are two singular points ( =0, 1) that are away from this line, whih we will all interior singular points. The minimal solution in the region R < 1/2 is an expansion around the interior singular point inside this region, namely: a + n, b + n y n = 2 F 1 ;. + n For proving this fat, by virtue of Theorem 3, we only need to know the behavior lose to =0,wheret 1 and t 2 behave very differently. Similarly, we an prove that the minimal solution for R >1/2 an be expressed in terms of an expansion around the interior singular point in this region ( =1), namely: Γ( + n) a + n, b + n y n =( 1) n Γ(1 + a + b + n) 2 F 1 1+a + b + n ;1. For proving this we only need to study the behaviour of the solution around 1 =0. For studying the behaviour of the hypergeometri funtions around = 0,the following result is enough: a + ɛ1 n, b + ɛ Theorem 6. Let y n = 2 F 2 n 1 ;,withɛ + ɛ 3 n j =0, ±1, ɛ 3 0, n = y 0, 1, 2..., and/ Z if ɛ 3 < 0, then in a dis around the origin lim n n =t(), y n 1 where t() is the root of the harateristi equation suh that lim 0 t() =1. This result is an immediate onsequene of the following fats: 1. H n (0) = 1. 2. There exists C>0suh that H n(0) <Cfor all n. Indeed, it is easy to hek that (2.10) H n(0) ɛ 1 ɛ 2 +4 abɛ 2 3 + 2 ɛ 1 ɛ 2 aɛ 2 ɛ 3 bɛ 1 ɛ 3. ɛ 3 3. One of the roots of the harateristi equation (t()) tends to 1 as 0 while the other one goes to 0 or. Both are ontinuous funtions of in a dis around = 0 (exluding = 0) where the roots have different modulus. The third ondition is in fat a onsequene of the fat that the harateristi roots are algebrai funtions of, that neessarily one of the roots tends to 1 as 1 (beause of the first ondition) and that α() = lim n α n either has a pole at = 0 or vanishes at = 0 (let us notie that the produts of the harateristi roots is α()). This ondition has been tested expliitly for all the ases onsidered in this artile. ɛ 3 3

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1457 3. Analysis of the four basi ases 3.1. The (+ + 0) reursion. The reursion relation reads (3.1) A(a + n, b + n)y n 1 + B(a + n, b + n)y n + C(a + n, b + n)y n+1 =0, where (3.2) A(a, b) =( a)( b)( a b 1), B(a, b) =( a b){(a + b )+ 2ab +[(a + b)( a b)+2ab +1 ]}, C(a, b) =ab( a b + 1)(1 ) 2. The oeffiients of harateristi equation λ 2 + βλ + α =0are with roots (3.3) t 1 = α =1/(1 ) 2, β = 2(1 + )/(1 ) 2 1 (1 ) 2, t 2 = 1 (1 + ) 2. The equation t 1 = t 2 holds when 0, otherwise t 1 > t 2. In this ase, the region t 1 t 2 is one onneted region. The only singular point away from t 1 = t 2 is = 1, and, as we will see, the minimal solution an be written in terms of an expansion around = 1 (both in the diretion n + and n ). We provide six solutions of the reurrene relation using the method desribed in Setion 1: (3.4) y 1,n = 2 F 1 a + n, b + n ;, Γ(1 + a + n)γ(1 + b + n) 1+a + n, 1+b + n y 2,n = 2F 1 ;, Γ(a + n)γ(b + n) 2 Γ(1 + a + n)γ(1 + b + n) y 3,n = 2F 1 a + n, b + n ;1, Γ(1 + a + b +2n) 1+a + b +2n 2n Γ(a + b +2n) a + n, b + n y 4,n =(1 ) Γ(a + n)γ(b + n) 2 F 1 ;1, 1 a b + 2n n Γ(1 + a + n) a + n, 1+a + n y 5,n =( ) 2F 1 ; 1, Γ(b + n) 1+a b n Γ(1 + b + n) b + n, 1+b + n y 6,n =( ) 2F 1 ; 1. Γ(a + n) 1 a + b

1458 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME The minimal solution as n is y 3,n. Indeed, by Theorem 6 we have that, in a dis around =1, (3.5) lim y 3,n = 1 t(ζ), ζ =1. y 3,n 1 4 The fator 1/4 omes from the limit of the ratio of gamma funtions, while t(ζ) = lim y n /y n 1, y n 2 F 1 (a + n, b + n;1+a + b +2n; ζ), ζ =1. Aording to Theorem 6, lim ζ 0 t(ζ) =1. The limit (3.5) orresponds to the root t 2 () of the (+ + 0) reurrene, beause t 2 () 1/4 as 1. Also, beause t 2 () is the smallest root outside the negative real axis, y 3,n is minimal in a dis around = 1 and therefore, by Theorem 3, is minimal in the omplex plane exept possibly on the negative real axis. An expliit verifiation of the exat limit omes from the omputation of t(ζ) for the (+ + 2+) reurrene relation. We indeed verify that (3.6) lim y 3,n y 3,n 1 = t 2 (), C \{ 0}. Notie that the definition of y 3,n indiates that the reurrene (+ + 0) and (+ + 2+) are in fat related. Although it is not neessary to study the ( 0), beause it an be related to the (+ + 0) by using the last relation in Eq. (1.4), it is also easy to obtain the minimal solution for the negative n diretion. The minimal solution in this ase is y 4,n ; indeed, from Theorems 3 and 6 we have that lim n y 4,n y 4,n 1 = 4 (1 ) 2 1 t(ζ) = t 1(), C \{ 0}. All the six solutions solve the same reurrene relation, eah of them an be expressed as a linear ombination of two others with oeffiients not depending on n. Any other solutions different from the minimal solution found annot be minimal, beause the minimal solution is unique (up to multipliative fators not depending on n). In summary, in C \{ 0}: (3.7) y 1,n y 2,n y 3,n y 4,n y 5,n y 6,n dominant dominant minimal/dominant dominant/minimal dominant dominant where, when two possibilities appear, the first one orresponds to n + and the seond one to n. 3.2. The (00+) reursion. The (0 0 +) reursion relation reads (3.8) A( + n)y n 1 + B( + n)y n + C( + n)y n+1 =0,

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1459 where (3.9) A() =( 1)( 1), B() =[ 1 (2 a b 1)], C() =( a)( b). The harateristi equation has oeffiients α =1 1/ and β = 2+1/. The harateristi roots are (3.10) t 1 =1, t 2 = 1. The urve t 1 = t 2 is the straight line R =1/2 and t 1 < t 2 when R <1/2. There are two interior singular points, that is, singular points whih lie away from the urve t 1 = t 2 : =0, 1. Studying the solutions around these singular points will suffie for obtaining the minimal solutions at both sides of the line R =1/2. Solutions of the reurrene are: y 1,n = 2 F 1 a, b ;, + n ( 1 )n Γ( 1+ + n)γ( + n) y 2,n = Γ( b + + n)γ( a + + n) 2 F 1 1 a, 1 b ;, 2 n Γ( + n)γ( a b + + n) a, b y 3,n = Γ( b + + n)γ( a + + n) 2 F 1 ;1, 1+a + b n (3.11) ( 1 )n Γ( + n) y 4,n = Γ(1 a b + + n) 2 F 1 1 a, 1 b ;1, 1 a b + + n Γ( + n) a, 1+a n y 5,n = Γ( a + + n) 2 F 1 ; 1, 1+a b Γ( + n) b, 1+b n y 6,n = Γ( b + + n) 2 F 1 ; 1. 1 a + b We have followed the method desribed in Setion 1 and applied the last relation in Eq. (1.4) to the seond and fourth solutions (in order to express the solutions in a more symmetrial form). Proeeding as in the previous ase (using Theorems 2, 3 and 6), it is lear that, when R <1/2 y 1,n (3.12) lim = t 1, y 1,n 1 lim n y 2,n y 2,n 1 = t 2. Then, when R <1/2,y 1,n is minimal as n + and y 2,n is minimal as n.

1460 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME Similarly, for R >1/2 we may onsider the expansions around the orresponding interior singular point. It is immediate to see that y 3,n (3.13) lim = t 1, n y 3,n 1 lim y 4,n y 4,n 1 = t 2, and beause t 1 > t 2 in this region, then y 4,n is minimal as n + and y 3,n is minimal as n. Summariing, we have: (3.14) R < 1 2 R > 1 2 y 1,n minimal/dominant dominant y 2,n dominant/minimal dominant y 3,n dominant dominant/minimal y 4,n dominant minimal/dominant y 5,n dominant dominant y 6,n dominant dominant where, when two possibilities appear, the first one orresponds to n + and the seond one to n. It is important to realie that the last two solutions y 5,n and y 6,n are revealing that the reurrenes (0 0 +) (+ 0 0) are related. There are several possible ways to express suh a relation; for instane, by using the last relation in (1.4) and negleting fators not depending on n, we an write the sixth solution: ( (3.15) y 6,n () = 1 1 )n ( Γ( + n) 1 a, a + + n Γ( b + + n) 2 F 1 ; 1 ). 1 a + b This means that, with some parameter substitutions and multiplying by some fators, we an transform a solution of the (0 0 +) reurrene (right side of the equation) into a solution of the (+ + 0) reurrene. The relation an also be inverted in order to build the solutions of the (+ 0 0) relation from the solutions of the (0 0 +) relation. We postpone the analysis to Setion 4. 3.3. The (+ + ) reursion.the reursion relation reads (3.16) where (3.17) A(a + n, b + n, n)y n 1 + B(a + n, b + n, n)y n +C(a + n, b + n, n)y n+1 =0, A(a, b, ) = (a )(a 1)(b 1 )(b )U, B(a, b, ) =[ 1 U + 2 V + 3 UV], 1 =(1 )(b )(b 1)[a 1+(b 1)], 2 = b(b +1 )(1 )(a + (b +2)), 3 = 2b (a b), C(a, b, ) =ab( 1)(1 ) 3 V, U = (a + b +1)(a + b +2)+ab(1 ), V =(1 )(1 a b + ab)+(a + b 1)(a + b 2).

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1461 The oeffiients of the harateristi equation are α = 16/(1 ) 3 and β = (8 2 +20 1)/(1 ) 3. The eros of the harateristi polynomial of the reursion relation (3.16) are 32(1 + w) (3.18) t 1 = (3 + w) 3, t 32(1 w) 2 = (3 w) 3, w = 8 +1, where w = 8 + 1. Observe that lim 0 t 1 () =1. The urve t 1 = t 2 in the omplex w-plane is desribed by (we write w = re iθ ) (3.19) r = 9+6 3osθ, 1 6 π θ 1 6π and Rw =0, together with the half-line 1/8. This is shown in Figure 1. In the domain interior to the urve we have t 1 > t 2. There are two singular points for whih t 1 t 2,whihare =0, 1; on the other hand, lim t 1 /t 2 = 1. As before, one an expet that the minimal solutions an be built by onsidering series around =0, 1. We write the six solutions in the following way: (3.20) y 1,n = 2 F 1 a + n, b + n ;, n n Γ(b +1+2n)Γ(a +1+2n) y 2,n = ( 1) 3 Γ(a + n)γ(b + n)γ(1 + n)γ(2 + n) 2 F 1 1 a n, 1 b n ;, 2 + n Γ(1 + b +2n)Γ(1 + a +2n) y 3,n = Γ(1 + n)γ(1 + a + b +3n) 2 F 1 a + n, b + n ;1, 1+a + b +3n n Γ(a + b +3n) y 4,n = ( 1) 3 Γ(a + n)γ(b + n)γ(1 + n) 2 F 1 1 a n, 1 b n ;1, 1 a b + 3n y 5,n = n Γ(1 + a +2n) a + n, 1+a +2n Γ(b + n)γ(1 + n) 2 F 1 ; 1, 1+a b y 6,n = n Γ(1 + b +2n) b + n, 1+b +2n Γ(n + a)γ(1 + n) 2 F 1 ; 1. 1+b a Proeeding like in the previous ases, it is evident that inside the urve (3.21) lim n y 1,n y 1,n 1 = t 1 and therefore y 1,n is minimal as n beause t 1 > t 2,whereas (3.22) lim y 2,n y 2,n 1 = and hene y 2,n is minimal as n +. 16 ( 1) 3 1 t 1 = α t 1 = t 2,

1462 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME 0.1 1 8 0.1 0.1 Figure 1. The urve t 1 = t 2 for the ase (+ + ). Outside the urve, we have to onsider the series around =1,thatis,the solutions y 3,n and y 4,n.Wehavethat (3.23) lim y 3,n = 16 y 3,n 1 27 t(ζ), ζ =1, where t(ζ) is the harateristi root for the TTRR satisfied by a + n, b + n y n = 2 F 1 ; ζ, +3n suh that lim ζ 0 t(ζ) = 1. Therefore, beause lim 1 t 1 () 16 27 (whereas t 2 is singular at =1)itislearthat y 3,n (3.24) lim = t 1, y 3,n 1 and therefore y 3,n is minimal as n +. The limit an be expliitly heked by omputing t(); for brevity, we don t provide the details of suh a alulation. On the other hand (3.25) lim n y 4,n y 4,n 1 = and hene y 4,n is minimal as n. Summariing: 27 1 ( 1) 3 t(ζ) = 27α 16t(ζ) = t 2, inside the urve outside the urve y 1,n dominant/minimal dominant y 2,n minimal/dominant dominant (3.26) y 3,n dominant minimal/dominant y 4,n dominant dominant/minimal y 5,n dominant dominant y 6,n dominant dominant where, when two possibilities appear, the first one orresponds to n + and the seond one to n.

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1463 Observe that the above results only hold inside the prinipal setor where π < phase(8 +1)<π, that is, in the ut plane, with branh ut from 1 8 to. 3.4. The (+ 0 ) reursion.the (+ 0 ) reursion relation reads (3.27) A(a + n, n)y n 1 + B(a + n, n)y n + C(a + n, n)y n+1 =0, where (3.28) A(a, ) =(a )(a 1)(b )[a + (b +1 )], B(a, ) =[a(a 1)( 1) + a(a 1)(a +3b 4 +2) +(b )(b +1 )(4a 1) 2 (a b)(b )(b +1 ) 3 ], C(a, ) = a( 1)[a 1+(b )](1 ) 2. The oeffiients of the harateristi equation are α = 4/(1 ) 2 and β = ( 2 6 +1)/(1 ) 2. The eros of the harateristi polynomial of the reursion relation (3.27) are (3.29) t 1 =1, t 2 = 4 (1 ) 2. The urve defined by t 1 = t 2 is desribed by (3.30) r =2+osθ ± os 2 θ +4osθ +3, π θ π, = re iθ. Both signs give a losed loop with ommon point 1. In Figure 2 we show this urve in the plane. In the domain interior to the inner urve we have t 1 > t 2 ; between the inner urve and the outer urve we have t 1 < t 2, and outside the outer urve t 1 > t 2. It is important to notie that the three singular points =0, 1, are interior in this ase, eah one inside a different region of the three regions into whih the plane is divided. Then, expansions around eah of the singular points will be needed in order to identify the minimal solution in eah region. 5.0 2.5 2.5 1.0 2.5 5.0 2.5 5.0 Figure 2. The urve t 1 = t 2 for the (+ 0 ) reursion.

1464 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME The six solutions for this ase an be written: (3.31) y 1,n = 2 F 1 a + n, b ;, n n Γ(1 + b + n)γ(1 + a +2n) y 2,n = (1 ) 2 Γ(a + n)γ(1 + n)γ(2 + n) 2 F 1 1 a n, 1 b ;, 2 + n Γ(1 + b + n)γ(1 + a +2n) a + n, b y 3,n = Γ(1 + n)γ(1 + a + b +2n) 2 F 1 ;1, 1+a + b +2n n Γ(a + b +2n) y 4,n = (1 ) 2 Γ(a + n)γ(1 + n) 2 F 1 1 a n, 1 b ;1, 1 a b + 2n n Γ(1 + a +2n) y 5,n = (1 ) 2 Γ(1 + n)γ(1 + a b + n) 2 F 1 1 b, b + n ; 1, 1+a b + n Γ(a b + n)γ(1 + b + n) b, 1+b + n y 6,n = 2F 1 ; 1. Γ(a + n)γ(1 + n) 1 a + b n The situation, as we next prove, an be summaried in the following way: inside inner urve between urves outside outer urve y 1 dominant/minimal dominant dominant y 2 minimal/dominant dominant dominant (3.32) y 3 dominant minimal/dominant dominant y 4 dominant dominant/minimal dominant y 5 dominant dominant minimal/dominant y 6 dominant dominant dominant/minimal where, when two possibilities appear, the first one orresponds to n + and the seond one to n. Let us now searh for the minimal solutions. 3.4.1. Minimal solutions inside the inner urve. This region ontains the point = 0, and we should onsider expansions around this point. In this region t 2 < t 1. Proeeding as before, it is obvious that (3.33) lim n y 1,n y 1,n 1 = t 1. Therefore, y 1,n is minimal as n. On the other hand y 2,n (3.34) lim = 4 1 = α = t 2, y 2,n 1 (1 ) 2 t 1 t 1 and hene y 2,n is minimal as n +.

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1465 3.4.2. Minimal solutions between urves. For the y 3,n solution, we have: y 3,n (3.35) lim = t(ζ), y 3,n 1 where ζ =1 and t(ζ) is the harateristi root for the TTRR satisfied by a + n, b y n = 2 F 1 +2n ; ζ, suh that lim ζ 0 t(ζ) = 1. Therefore, beause t 1 () = 1 (while t 2 is singular at =1)itislearthat y 3,n (3.36) lim = t 1, y 3,n 1 and therefore y 3,n is minimal as n +. The limit an be expliitly heked by omputing t(ζ), whih is equal to t 1 =1. On the other hand y 4,n (3.37) lim = α() n y 4,n 1 t(ζ) = t 2, and therefore y 4,n is minimal as n. 3.4.3. Minimal solutions outside the outer urve. In this region we have the singular point =, and we onsider a series around this point. We have (3.38) lim y 5,n y 5,n 1 = α t 1 = t 2 and y 6,n (3.39) lim = t 1. n y 6,n 1 Therefore y 5,n is minimal as n + while y 6,n is minimal as n. 4. Obtaining information for the remaining 22 + 4 reurrenes As disussed in [3], the 26 reurrene relations with ɛ j 1 an be redued to 5 ases by means of Eq. (1.4). In addition, as we disussed before, when building the six solutions by onsidering linear relations between solutions, new relations appear, as is the ase of the relation between the (0 0 +) reurrene and the (+ 0 0); this relation redues the number of basi reurrene relations to 4. We also saw that 4 other reurrenes are related with the 4 remaining basi reurrenes: (+ + 2+) with (+ + 0), (+ + 3+) and (+ 2+ 0) with (+ + ), and (+ 0 2+) with (+ 0 ). All these ases an be related to the four basi ases by using simple transformations. Let us for instane onsider obtaining all the information for the (+ 0 0) from the (0 0 +) reurrene. From Eq. (3.15) we see that the solutions y (00+) α,β;γ+n () of the reurrene satisfied by the hypergeometri funtions 2 F 1 (α, β; γ + n; ) anbe built from the solutions y (+00) α+n,β;γ () of the reurrene satisfied by the funtions 2F 1 (α + n, β; γ; ) by means of the following transformation: ( (4.1) y (00+) α,β;γ+n () = 1 1 )n Γ(γ + n) Γ( β + γ + n) y(+00) α+γ+n,1 α;1 α+β (1/).

1466 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME y (00+) α,β;γ+n () denotes a generi solution of the (0 0 +) reurrene, and not only the 2 F 1 (α, β; γ + n; ) funtion; also y (+00) α+n,β;γ () is a generi solution of the (+ 0 0) reurrene. Notie that the relation (3.15) provides not only a onnetion between two Gauss hypergeometri funtions, but also a transformation of a reurrene into another one ((+ 0 0) into (0 0 +)). This transformation an be onsidered for any solution (in partiular for the six solutions given for eah reurrene). Denoting a = α+γ, b =1 α, =1 α+β, we an invert this general relation to get: Γ(1 + a + n) (4.2) y (+00) a+n,b; () =(1 ) n Γ(1 + a b + n) y(00+) 1 b, b+;1+a b+n (1/). Therefore, we an build the six solutions for the (+ 0 0) reurrene by performing the following transformations over the solutions of the (0 0 +) reurrene: 1. Perform the substitutions: a 1 b, b b +, 1+a b, 1/. n Γ(1 + a + n) 2. Multiply the resulting funtions by (1 ) Γ(1 + a b + n). With these transformations, we get the following set of solutions for the (+ 0 0) ase: n Γ(1 + a + n) y 1,n =(1 ) Γ(1 + a b + n) 2 F 1 1 b, b + ; 1, 1+a b + n Γ(a b + n) b, 1+b y 2,n = 2F 1 ; 1, Γ(a + n) 1 a + b n n Γ(a + b + n) a + n, b + y 3,n =(1 ) 2F 1 ;1, Γ(a + n) 1 a b + n (4.3) Γ(1 + a + n) a + n, b y 4,n = Γ(1 + a + b + n) 2 F 1 ;1, 1+a + b + n Γ(1 + a + n) 1+a + n, 1+b y 5,n = 2F 1 ;, Γ(a + n) 2 y 6,n = 2 F 1 a + n, b ;. Here, apart from the presribed substitutions, the last transformation of Eq. (1.4) has also been applied in the last 4 ases (and the fators not depending on n removed). ( Given ) that we make the substitution 1/ and we multiply the solutions by n 1, the harateristi roots are the roots for (0 0 +) with replaed by 1/ 1 and then multiplied by 1/(1 ), that is (4.4) t 1 = 1 1,t 2 =1.

SOLUTIONS OF HYPERGEOMETRIC RECURSIONS 1467 Of ourse, the harater of the solutions is maintained in the transformed domains given by the hange 1/. We therefore have: (4.5) R( 1 ) < 1 2 R( 1 ) > 1 2 y 1,n minimal/dominant dominant y 2,n dominant/minimal dominant y 3,n dominant dominant/minimal y 4,n dominant minimal/dominant y 5,n dominant dominant y 6,n dominant dominant Observe that the equation R( 1 )=1/2 is the equation of the irle 1 =1, with R( 1 ) > 1 2 defining the interior of the disk and R( 1 ) < 1 2 the exterior. For this transformation hosen we have the most general situation possible: the parameters and the variable are replaed, and the funtions are multiplied by gamma funtions and by a -dependent fator. A more simple example is provided by the onnetion of (0 0 +) with the (+ + +) ase. Using the last relation in Eq. (1.4), we have: a + n, b + n a, b 2F 1 ; =(1 ) + n a b n 2F 1 ;. + n Therefore, the solutions of the (+ + +) reurrene an be obtained from the solutions of the (0 0 +) by multiplying by (1 ) n and by substituting a by a and b by b. 5. Conlusions The problem of obtaining the pairs of satisfatory numerial ( solutions of the hypergeometri reursions satisfied by the funtions y n = 2 F 2 n ) a + ɛ1 n, b + ɛ 1 ;, + ɛ 3 ɛ j 1has been solved in the omplex plane, exept on the ritial urves t 1 = t 2 (t 1 and t 2 being the roots of the harateristi polynomial) where the Poinaré theorem does not provide information regarding the existene of minimal solutions. The study of the behaviour on the ritial urves needs a separate analysis and is beyond the sope of the present paper. In [2], the real ase for the reurrene (+ + +) is disussed in detail. In this ase it is easy to hek that for =1/2 there is a minimal solution when 1 + a + b 2 0; this fat has important onsequenes on the stability of the reurrenes lose to =1/2. A omprehensive desription of the ondition of the hypergeometri reursions in the omplex plane is now available, exept on the ritial urves. This is an essential piee of information for the omputation of hypergeometri funtions by means of reurrene relations. Aknowledgments The authors wish to thank the referee for omments and suggestions whih have resulted in an improved version of the paper. A. Gil aknowledges finanial support from Ministerio de Eduaión y Cienia (programa Ramón y Cajal). J. Segura aknowledges finanial support from projet BFM2003-06335-C03-02 and from the

1468 AMPARO GIL, JAVIER SEGURA, AND NICO M. TEMME Seretaría de Estado de Universidades e Investigaión (Programa de Movilidad, Ref. PR2005-0333). The authors aknowledge finanial support from Ministerio de Eduaión y Cienia (Projet MTM2004 01367). Referenes [1] M. Abramowit, I. Stegun (Eds). Handbook of Mathematial Funtions. National Bureau of Standards. Applied Mathematis Series, no. 55. U.S. Government Printing Offie, Washington DC (1964). MR0167642 (29:4914) [2] A. Deaño, J. Segura. Transitory minimal solutions of hypergeometri reursions and pseudoonvergene of assoiated ontinued frations. Aepted for publiation in Mathematis of Computation. [3] A. Gil, J. Segura, N. M. Temme. The ABC of hyper reursions. J. Comput. Appl. Math. [4] Y. L. Luke. The speial funtions and their approximations, Vol. I. Mathematis in Siene and Engineering, Vol. 53., Aademi Press, New York, 1969. MR0241700 (39:3039) [5] G.N. Watson. Asymptoti expansions of hypergeometri funtions. Trans. Cambridge Philos. So., 22:277 308, 1918. [6] J. Wimp. Computation with reurrene relations. Appliable Mathematis Series. Pitman (Advaned Publishing Program). Boston, MA. 1984. MR0727118 (85f:65001) Departamento de Matemátias, Estadístia y Computaión, Univ. Cantabria, 39005- Santander, Spain E-mail address: amparo.gil@unian.es Departamento de Matemátias, Estadístia y Computaión, Univ. Cantabria, 39005- Santander, Spain E-mail address: javier.segura@unian.es CWI, P.O. Box 94079, 1090 GB Amsterdam, The Netherlands E-mail address: niot@wi.nl