Positivity of rational functions and their diagonals
aa r X i v : . [ m a t h . N T ] M a y Positivity of rational functions and their diagonals
Armin Straub Wadim ZudilinNovember 9, 2018
To Dick Askey on the occassion of his birthday,with many positive wishes
Abstract
The problem to decide whether a given rational function in severalvariables is positive, in the sense that all its Taylor coefficients are positive,goes back to Szeg˝o as well as Askey and Gasper, who inspired more recentwork. It is well known that the diagonal coefficients of rational functionsare D -finite. This note is motivated by the observation that, for severalof the rational functions whose positivity has received special attention,the diagonal terms in fact have arithmetic significance and arise fromdifferential equations that have modular parametrization. In each of thesecases, this allows us to conclude that the diagonal is positive.Further inspired by a result of Gillis, Reznick and Zeilberger, we in-vestigate the relation between positivity of a rational function and thepositivity of its diagonal. Keywords : positivity; rational function; hypergeometric function; modular function;Ap´ery-like sequence; multivariate asymptotics.
The question to decide whether a given rational function is positive , that is,whether its Taylor coefficients are all positive, goes back to Szeg˝o [25] and hassince been investigated by many authors including Askey and Gasper [3, 4, 5],Koornwinder [17], Ismail and Tamhankar [14], Gillis, Reznick and Zeilberger[11], Kauers [15], Straub [24], Kauers and Zeilberger [16], Scott and Sokal [23].The interested reader will find a nice historical account in [23]. A particularlyinteresting instance is the Askey–Gasper rational function A ( x, y, z ) := 11 − x − y − z + 4 xyz , (1)whose positivity is proved in [5] and [11]. Generalizations to more than threevariables are rarely tractable, with the longstanding conjecture of the positivity1f 11 − x − y − z − w + ( xy + xz + xw + yz + yw + zw ) , (2)also referred to as the Lewy–Askey problem. Very recently, Scott and Sokal[23] succeeded in proving the non-negativity of (2), both in an elementary wayby an explicit Laplace-transform formula and based on more general results onthe basis generating polynomials of certain classes of matroids. Note that by aresult from [16] the positivity of (2) would follow from the positivity of D ( x, y, z, w ) := 11 − x − y − z − w + 2( yzw + xzw + xyw + xyz ) + 4 xyzw , (3)which is still an open problem. In another direction, Gillis, Reznick and Zeil-berger conjecture in [11] that 11 − ( x + x + . . . + x d ) + d ! x x · · · x d (4)has non-negative coefficients for any d > d = 2 , diagonal Taylor coefficients are non-negative. Modulo this claim,the cases d = 4 , , e k ( x , . . . , x d ) the elementary symmetric polynomials defined by d Y j =1 ( x + x j ) = d X k =0 e k ( x , . . . , x d ) x d − k . (5) Question 1.1.
Under what (natural) condition(s) is the positivity of a rationalfunction h ( x , . . . , x d ) of the form h ( x , . . . , x d ) = 1 P dk =0 c k e k ( x , . . . , x d ) (6)implied by the positivity of its diagonal? For example, would the positivity of h ( x , . . . , x d − ,
0) be a sufficient condition?Another motivation for this question is the fact that for several importantrational functions, like the ones reproduced above, the diagonal coefficients arearithmetically interesting sequences. In particular, expressing them in terms ofknown hypergeometric summations sometimes makes their positivity apparent.For instance, the diagonal sequence for A ( x, y, z ) is a n,n,n = n X k =0 (cid:18) nk (cid:19) , (7)2ee Example 3.4, while the diagonal of D ( x, y, z, w ) is given by d n,n,n,n = n X k =0 (cid:18) nk (cid:19) (cid:18) kn (cid:19) , (8)as is shown in Example 4.2. Since these diagonal sequences are manifestlypositive, sufficient progress on Question 1.1 might provide a proof of the conjec-tured positivity of D ( x, y, z, w ). More generally, proving positivity of a singlesequence is much simpler from a practical point of view than proving positivityof all Taylor coefficients of a rational function, and tools such as cylindricalalgebraic decomposition can be used for this task rather successfully in specificexamples, as illustrated by [15] and observed in some of the examples herein.We note that, with no loss of generality, we may assume in Question 1.1 that c = 1.In Section 2, we answer Question 1.1 in the affirmative when d = 2. The threeand four-dimensional cases are discussed in Sections 3 and 4, while Section 5covers an approach to positivity via asymptotics. In particular, we prove thatthe conjectural conditions, given in [24], for positivity of rational functions inthree variables are indeed necessary. In the case of only two variables, we are thus interested in the rational function h ( x, y ) = 11 + c ( x + y ) + c xy . Note that the condition c < h ( x, y ),as − c is a Taylor coefficient of its series expansion. The next examples demon-strate that the positivity of the diagonal coefficients of h ( x, y ) and the positivityof h ( x,
0) are not implied by each other.
Example 2.1.
The rational function h ( x, y ) = 1 / (1+ x + y ) has positive diagonalcoefficients but is not positive; indeed, h ( x,
0) = 1 − x + O ( x ). This illustratesthat some condition is indeed needed in Question 1.1. Example 2.2.
Let h ( x, y ) = 1 / (1 − x − y + 2 xy ). Then h ( x,
0) = 1 / (1 − x ) ispositive, but the diagonal coefficients of h ( x, y ) are not positive. Theorem 2.3.
A two-variable rational function h ( x, y ) = 11 + c ( x + y ) + c xy is positive, if both h ( x, and the diagonal of h ( x, y ) are positive. roof. The positivity of h ( x,
0) implies that c <
0. Upon rescaling the variablesby a positive factor, we may assume that c = − − ( x + y ) + axy = ∞ X n,m =0 u n,m x n y m . (9)As demonstrated in the course of [24, Proposition 4], this rational function ispositive if and only if a
1. Here, we only need the easy observation that itis positive if a
1, which follows directly from the geometric series and thefactorization 1 − ( x + y ) + xy = (1 − x )(1 − y ).On the other hand, the diagonal terms u n := u n,n of the Taylor expansion(9) are given by u n = n X k =0 (2 n − k )! k !( n − k )! ( − a ) k . We observe that the sequence u n is characterized by the generating series ∞ X n =0 u n z n = 1 p − − a ) z + a z . For a >
1, the quadratic polynomial 1 − − a ) z + a z has non-real roots,from which we conclude that u n is (eventually) sign-indefinite. Therefore, theseries (9) is positive if and only if its diagonal terms are positive.Theorem 2.3 answers Question 1.1 in the affirmative when d = 2. Remark 2.4.
The sequence u n satisfies the three-term recurrence( n + 1) u n +1 = (2 − a )(2 n + 1) u n − a nu n − , which has characteristic polynomial x − − a ) x + a = ( x + a ) − x . Notethat, for a >
1, this polynomial has complex roots.The ultimate reduction to d = 1 and d = 2 performed in this section showsthat for d ≥ d -variablerational function (6) to satisfy c = 1, c = − c = a ≤
1. This willbe the canonical form of a rational function in Question 1.1.
A partially conjectural classification of positive rational functions of the form h a,b ( x, y, z ) = 11 − ( x + y + z ) + a ( xy + yz + zx ) + bxyz (10)has been given in [24]. It is conjectured there [24, Conjecture 1] that h a,b ispositive if and only if the three inequalities a b < − a ), b − a +2(1 − a ) / hold. In Theorem 5.7 below we show that all three conditions areindeed necessary for positivity. 4 xample 3.1. It is proven in [24] that the rational function h a,b with a = λ ( λ + 2)( λ + 1) , b = − ( λ − λ + 2) ( λ + 1) is positive for all λ >
0. The conjecture mentioned above predicts that it is, infact, positive as long as λ > (1 + √ / − (1 + √ − / − ≈ − . − ( λ + 1)( x + y + z ) + λ ( λ + 2)( xy + yz + zx ) − ( λ − λ + 2) xyz . (11)Empirically, it appears that the Taylor coefficients of this rational function arepolynomials in λ with positive coefficients. We have verified this for the coeffi-cients of x k y m z n with k, m, n h a,b ( x, y,
0) is positive if and only if a
1. The next conjecture is therefore equivalent to an affirmative answer toQuestion 1.1 for d = 3. Conjecture 3.2.
Suppose that a . Then the rational function (10) is posi-tive if and only if its diagonal is positive. In the case a >
1, which is not covered by Conjecture 3.2, the followingis a conjectural characterization of the rational functions which have positivediagonal coefficients.
Conjecture 3.3.
Let a > . The diagonal of (10) is positive if and only if b − a . That the case b = − a plays a special role can be seen from the characteris-tic polynomial of the recurrence of minimal order for the diagonal coefficients.Namely, the diagonal coefficients u n of h a,b , as defined in (10), satisfy a fourth-order recurrence (the coefficients have degree 4 in n , degree 9 in a and degree 5in b ), whose characteristic polynomial is( a + b )( a + ab − (1 − a ) x )(( x + b ) + 27 x ( a + ab − (1 − a ) x )) . (12)Obviously, the first factor vanishes when b = − a . We further observe that thecubic factor in (12) has discriminant − ( a − a − b ) (4 a − a + 6 ab + b − b ) . The second factor vanishes if and only if b = 2 − a ± − a ) / , which includesthe principal condition of [24, Conjecture 1] as stated at the beginning of thissection; see also Example 5.2.In the remainder of this section, we consider two cases of particular interestin the three-variable case, namely the Askey–Gasper rational function A ( x, y, z )from (1) as well as Szeg˝o’s rational function S ( x, y, z ) = 11 − ( x + y + z ) + ( xy + yz + zx ) . (13)5n both cases, we exhibit the arithmetic nature of the diagonals and demonstratethat positivity follows as a consequence. Example 3.4.
According to [9], the diagonal sequence a n := a n,n,n for A ( x, y, z )as in (1) satisfies the three-term recurrence equation( n + 1) a n +1 = (7 n + 7 n + 2) a n + 8 n a n − , the latter already implying the positivity of the diagonal. The same recursionholds for the sequence a n = P nk =0 (cid:0) nk (cid:1) , also known as Franel numbers, which is aclassical result of Franel [10]. Furthermore, both sequences have the same initialconditions a = 1, a = 2. As indicated in (7), the diagonal of A ( x, y, z ) is thusgiven by the Franel numbers. The sequence a n is an Apery-like sequence, namely[1, sequence (4.8) (a)], and the (Calabi–Yau) differential equation satisfied bythe generating function has modular parametrization. As a consequence, thegenerating function has a hypergeometric form, namely ∞ X n =0 a n z n = 11 − z F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) z (1 − z ) (cid:19) , (14)which we record for comparison with the next example. Here and in what follows F (cid:18) a, bc (cid:12)(cid:12)(cid:12)(cid:12) z (cid:19) = ∞ X n =0 z n n − Y j =0 ( a + j )( b + j )(1 + j )( c + j )is the hypergeometric function. Note that positivity of a n is still apparentfrom (14). Example 3.5.
As shown in [24], the positivity of S ( x, y, z ) can be deducedfrom the positivity of A ( x, y, z ). On the other hand, the diagonals of thetwo are not related to each other in an easy way; the diagonal terms s n =[( xyz ) n ] S (2 x, y, z ) are given by1 , , , , , , . . . and satisfy the recurrence2( n + 1) s n +1 = 3 (cid:0) n + 27 n + 8 (cid:1) s n − n − n + 1) s n − . (15)Denoting by y ( z ) = P n > s n z n the generating function of this sequence, it isroutine to verify that y ( z ) = F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) z (2 − z ) (cid:19) ; (16)indeed, both sides in (16) satisfy the same differential equation and initial values.See Remark 3.7 below on how one can find this expression. We note that (16)implies the binomial formula s n = n X k =0 ( − n − k k − n (3 k )! k ! (cid:18) kn − k (cid:19) , (17)though positivity is not apparent here.6 emma 3.6. The sequence s n in Example is positive.Proof. To deduce positivity of y ( z ) in (16), start with Ramanujan’s cubic trans-formation [7, p. 97] F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) − (cid:18) − x x (cid:19) (cid:19) = (1 + 2 x ) F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) x (cid:19) , which is proven in [8]; see also [18, Corollary 6.2]. With x and z related by27 z (2 − z ) = 1 − (cid:18) − x x (cid:19) , we find that 2 x ( z ) = 31 + 2(1 − z ) / − . The binomial theorem shows that (1 − z ) / = 1 − zg ( z ) for some g ( z ) withpositive Taylor coefficients. It follows that x ( z ) = c z + c z + . . . for positive c j , so that y ( z ) = (1 + 2 x ( z )) F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) x ( z ) (cid:19) is seen to have positive coefficients. Remark 3.7.
Let us briefly indicate how we found (16). First, note that y ( z ) is the analytical solution of the differential equation corresponding to(15) characterized by y (0) = 1. Let y ( z ) be the solution such that y ( z ) − y ( z ) log( z ) ∈ z Q [[ z ]]. Then q ( z ) := exp (cid:18) y ( z ) y ( z ) (cid:19) = z + 33 z z + 12203 z z + O (cid:0) z (cid:1) . Denoting by z ( q ) the inverse function, we observed, by computing the first fewterms of the q -expansion, that y ( z ( q/ X n,m ∈ Z q n + nm + m . The right-hand side is the theta series of the planar hexagonal lattice, also knownas the first cubic theta function a ( q ), and its relation to the hypergeometricfunction in (16) is well known; see, for instance, [8]. For further background onthis approach we refer the interested reader to [1]. We now study positivity of the rational functions in four variables, which are oftype (6). That is, we consider the rational functions h a,b,c ( x ) = 11 − e ( x ) + ae ( x ) + be ( x ) + ce ( x ) , (18)7here x = ( x , x , x , x ) and e k ( x ) are the elementary symmetric functionsdefined in (5). Table 1 summarizes the examples we discuss in this section. a b c h , , ;non-negativity proven in [23]0 −
16 positivity proven in [17]; see also [11] − § h , , − Table 1: Interesting instances of h a,b,c as in (18) Example 4.1 (Lewy–Askey rational function) . In [4], Askey and Gasper men-tion the following four-dimensional generalization of Szeg˝o’s function h / , , ( x, y, z, w ) = 11 − ( x + y + z + w ) + ( xy + xz + xw + yz + yw + zw ) , which is just a rescaled version of 1 /e (1 − x, − y, − z, − w ). As al-ready mentioned in the introduction, the non-negativity of this rational func-tion was recently established in [23]. The scaled initial diagonal terms s n :=9 n [( xyzw ) n ] h / , , ( x, y, z, w ) are1 , , , , , , . . ., and one checks that s n = (cid:0) nn (cid:1) u n , where the sequence u n satisfies the recurrenceequation3( n + 1) u n +1 = 4 (cid:0) n + 28 n + 9 (cid:1) u n − n − n + 1) u n − . As in Example 3.5, the differential equation, of which the generating function y ( z ) = P n > u n z n is the unique analytical solution with value 1 at z = 0,admits modular parametrization. This fact was found and communicated to usby van Straten. As a consequence, we have the hypergeometric representation y ( z ) = 1(1 − z + 12288 z ) / F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) − z (3 − z )(1 − z ) (1 − z + 12288 z ) (cid:19) , (19)which, once found, can be verified by comparing the differential equations satis-fied by both sides. In fact, using hypergeometric transformations, we find that(19) simplifies to y ( z ) = 1 √ − z F (cid:18) , (cid:12)(cid:12)(cid:12)(cid:12) − z (3 − z )(1 − z ) (cid:19) . (20)8s in Example 3.5, we can now use the arithmetic properties of this functionto show that the sequence u n , hence the diagonal terms s n , are indeed positive.To do so, we may proceed as in Lemma 3.6, except now using Ramanujan’squadratic transformation [7, p. 146] (also proven in [18, Corollary 6.2]). Alter-natively, we can prove positivity of the diagonal terms from the above three-termrecurrence using cylindrical algebraic decomposition in the style of [15]. Example 4.2 (Kauers–Zeilberger rational function) . On the other hand, thepositivity of the rational function in the previous example is implied, as shownin [16] using positivity preserving operators, by the (conjectured) positivity ofthe rational function h , , ( x, y, z, w ) = 11 − ( x + y + z + w ) + 2 e ( x, y, z, w ) + 4 xyzw , which we also refer to as D ( x, y, z, w ). This rational function, as mentionedin the introduction, has particularly appealing diagonal coefficients. Namely,expanding D ( x, y, z, w ) = ∞ X n =0 [( x + y + z + w ) − yzw + xzw + xyw + xyz ) − xyzw ] n , and applying the binomial theorem, one obtains a five-fold sum for the diagonalcoefficients d n = d n,n,n,n . With the help of the multivariate Zeilberger algorithm[2] we verify that the sequence d n satisfies, for n = 1 , , . . . ,( n + 1) d n +1 − n + 1)(3 n + 3 n + 1) d n + 16 n d n − = 0 . The same recurrence is satisfied by [1, sequence (4.12) ( ǫ )], so that comparinginitial values proves that d n = n X k =0 (cid:18) nk (cid:19) (cid:18) kn (cid:19) . Since this makes positivity of the diagonal terms obvious, an affirmative an-swer to Question 1.1 when d = 4 would prove the conjectured positivity of D ( x, y, z, w ). Example 4.3 (Examples of Szeg˝o and Koornwinder) . Szeg˝o proved, as a higher-order generalization of the function (13), the positivity of 1 /e (1 − x, − y, − z, − w ); see [25, § h / , − / , . The positivityof the rational function h / , − / , can also be obtained, again via positivitypreserving operators, see [24], from the positivity of Koornwinder’s rationalfunction [17] h , , − ( x, y, z, w ) = 11 − ( x + y + z + w ) + 4 e ( x, y, z, w ) − xyzw , h , , − is given by the,obviously positive, sequence n X k =0 (cid:18) kk (cid:19) (cid:18) n − k ) n − k (cid:19) . This, again, is an Ap´ery-like sequence, namely [1, sequence (4.10) ( β )]. Example 4.4.
As shown in [15], modulo the assertion from [11] with an un-published proof, the rational function h , ,c ( x, y, z, w ) = 11 − ( x + y + z + w ) + cxyzw has non-negative coefficients if and only if c
24. The condition c
24 isnecessary because the (diagonal) coefficient of xyzw in h , ,c is 24 − c . On theother hand, asymptotic considerations, such as in Example 5.3 below, suggestthat all but finitely many diagonal coefficients of h , ,c are positive if c < d = 4 of a general conjecture from [11] mentioned in theintroduction. In the general case, the rational function11 − ( x + x + . . . + x d ) + cx x · · · x d can only be non-negative if c d ! since the coefficient of x x · · · x d is seento be d ! − c . It is conjectured in [11] that, for d >
4, the condition c d ! isindeed sufficient, and it is claimed that the non-negativity follows from the non-negativity of the diagonal. On the other hand, some hypergeometric intuitionsuggests that the diagonal coefficients are eventually positive if c < ( d − d − . Example 4.5.
In [15], the rational function h , / , ( x, y, z, w ) = X k,l,m,n ≥ u k,l,m,n x k y l z m w n is conjectured to be positive. As evidence, it is shown in [15], using cylindricalalgebraic decomposition (CAD), that u k,l,m,n > d = 4 would therefore imply the conjectured positivity of h , / , . Multivariate asymptotics, as developed in [6, 19, 20, 22] and further illustratedin [21], is an approach to determine the asymptotics of the coefficients u n ,...,n d
10f a multivariate generating function h ( x , . . . , x d ) = X n ,...,n d > u n ,...,n d x n · · · x n d d directly from h and its singular points.In the sequel, we write x = ( x , . . . , x d ). In the cases we are presentlyinterested in, h = 1 /p is the reciprocal of a polynomial p ( x ). Denote with V ⊆ C d the singular variety defined by p = 0. A point x ∈ V is smoothif ∇ p ( x ) = ( ∂ p ( x ) , . . . , ∂ d p ( x )) = , where ∂ j := ∂/∂x j for j = 1 , . . . , d .The nonsmooth points can be comfortably computed using Gr¨obner bases, asdetailed in [21, Section 4].The next three examples suggest that rational functions which are on theboundary of positivity (that is, slightly perturbing its coefficients can changewhether the function is positive) are intimately linked with rational functionsthat have nonsmooth points on their singular variety. This echoes the remarkin [21] that, while for generic functions all points of the singular variety aresmooth, “interesting applications tend not to be generic.” Example 5.1.
With d = 2, consider the case h ( x , x ) = 11 − ( x + x ) + ax x , which by Theorem 2.3 is positive if and only if a
1. Then the singular varietyhas nonsmooth points if and only if a = 1. The nonsmooth point in the case a = 1 is x = (1 , a < u n,n ∼ (1 + √ − a ) n +1 p πn √ − a . Example 5.2.
With d = 3, consider the case h ( x , x , x ) = 11 − ( x + x + x ) + a ( x x + x x + x x ) + bx x x . Then the singular variety has nonsmooth points if and only if4 a − a + 6 ab + b − b = 0 . Solving this condition for b , gives b = 2 − a ± − a ) / , which includesprecisely the boundary in [24, Conjecture 1] explicitly describing the transitionbetween positive rational functions and those with negative coefficients. Example 5.3.
With d = 4, consider the case h ( x , x , x , x ) = 11 − e ( x ) + ae ( x ) + be ( x ) + ce ( x ) . a + 2 ab − ac + b + c )(64 b − b + c ) + 6 bc (2 c − b ) + c − a (2 b − c )( b + c ) + 18 a (2 b + 10 bc − c ) − a ( b + c ) + 81 a c ) . (21)We note that all the examples in Table 1, with the exception of (0 , ,
24) whichis not ‘natural’ as pointed out in Example 4.4, have nonsmooth points on thesingular variety. The above factorization of the right-hand side of (21) impliesthat, when c = a + 2 ab + b a − , (22)the rational function h has nonsmooth points on its singular variety. The ex-amples ( a, b, c ) = (0 , , −
16) and (8 / , − / ,
0) from Table 1 are of this form.
Example 5.4.
In the case a = 0 of (22), the rational function is h ,b, − b = 11 − ( x + y + z + w ) + b ( yzw + xzw + xyw + xyz ) − b xyzw . By directly computing the coefficients from their recurrence, we observe that itsTaylor coefficients u k,l,m,n are positive for all 0 k, l, m, n
20 if and only if b < . . . . , which suggests that h ,b, − b is positive if and only if b
4. Notethat positivity was proven in [17] for the case b = 4. On the other hand, uponsetting, say, w = 0, it follows from [24, Proposition 5] that positivity of h ,b, − b requires b
4. However, it appears empirically that the diagonal coefficientsare positive for any b ∈ R . We have verified this for the first 50 coefficients.Note that this fits nicely and further illustrates Question 1.1. We note, however,that, for b <
4, the function h ,b, − b does not appear to be on the boundary ofpositivity.Let a direction n = ( n , . . . , n d ) ∈ Z d> be given. Among points on thesingular variety V , a special role is played by the critical points x for n , whichare characterized [21, Proposition 3.11] by the d equations p ( x ) = 0 and, for all j = 1 , . . . , d − n d x j ∂ j p ( x ) = n j x d ∂ d p ( x ) . (23)We note the following consequence of [21, Theorem 3.16], which is a simplereformulation of [22, Proposition 5.1]; see the remark after [21, Theorem 3.16]for the uniqueness. Proposition 5.5.
Let n ∈ Z d> be such that there is a smooth critical pointfor n . If the rational function h is non-negative, then there is a unique criticalpoint for n in R d> . Example 5.6.
For illustration, we use Proposition 5.5 to give an alternativeproof of [24, Proposition 5], which states that the rational function h ( x, y, z ) = 11 − ( x + y + z ) + bxyz
12s not non-negative if b >
4. Suppose that b >
4, in which case V is smooth.A simple computation shows that h has three critical points for n = (1 , , c, c, c ), where c is a solution of1 − c + bc = 0 . (24)This cubic equation has discriminant ∆ = 27 b (4 − b ). The assumption b > <
0, which in turn implies that the equation (24) has only onereal root. Since this real root is necessarily negative by the intermediate valuetheorem, we conclude that none of the three critical points for (1 , ,
1) lies in R > . By Proposition 5.5 it follows that h is not non-negative if b > Theorem 5.7.
For the rational function h ( x, y, z ) = 11 − ( x + y + z ) + a ( xy + yz + zx ) + bxyz to be non-negative it is necessary that a and b − a + 2(1 − a ) / .Proof. Upon setting z = 0, it follows from Section 2 that a h to be non-negative.The defining equations (23) for critical points for n = (1 , ,
1) are equivalentto ( x − z )( ay −
1) = 0 , ( y − z )( ax −
1) = 0 . There are therefore two kinds of critical points. Firstly, the points ( c, c, c ), where c is a solution to 1 − c + 3 ac + bc = 0; (25)secondly, the points ( x, y, z ) where two coordinates are equal to 1 /a and thethird coordinate is a (1 − a ) / ( a + b ).We observe that the discriminant of the cubic equation (25) is negative if a b > − a + 2(1 − a ) / . In that case, by the same argument as inthe previous example, there are no critical points of the first kind in R > . Onthe other hand, unless b = − a , there are three distinct critical points of thesecond kind, which either all lie in R > or all lie outside R > .Suppose that a b > − a + 2(1 − a ) / . Then the case b = − a occurs only if a < −
3, in which case the critical points of the second kind lieoutside R > . We conclude that h cannot have a unique critical point in R > .Since, by Example 5.2, all points on the singular variety of h are smooth, theclaim therefore follows from Proposition 5.5.Note the special role played by the diagonal direction n = (1 , , D ⊂ C , if either D is convex or the degree of thepolynomial p ( x ) = p ( x , . . . , x d ) = d X k =0 c k e k ( x )is d , then, for any ξ , . . . , ξ n ∈ D , there exists ξ ∈ D such that p ( ξ , . . . , ξ n ) = p ( ξ, . . . , ξ ) . Acknowledgements.
We are indebted to Duco van Straten for providing uswith the explicit formula (19) and also with geometric insights behind his deriva-tion. We thank the referees for their healthy criticism on an earlier version ofthis work.
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Math. Z. (1933), 674–688.Armin Straub: Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7,Bonn D-53111, Germany
URL : http://arminstraub.com/ Current address : Department of Mathematics, University of Illinois,1409 West Green St, Urbana, IL 61801, USA
Wadim Zudilin:
School of Mathematical and Physical Sciences, TheUniversity of Newcastle, Callaghan NSW 2308, Australia
URL : http://wain.mi.ras.ru/http://wain.mi.ras.ru/