Painlevé V and time dependent Jacobi polynomials
aa r X i v : . [ m a t h - ph ] M a y Painlev´e V and time dependent Jacobipolynomials
Estelle Basor ∗ American Institute of MathematicsPalo Alto, California 94306,[email protected] Yang ChenDepartment of MathematicsImperial College London180 Queen’s GatesLondon SW7 2BZ [email protected] EhrhardtDepartment of MathematicsPOSTECHPohang 790-784, South [email protected]
Abstract
In this paper we study the simplest deformation on a sequence of orthogonal poly-nomials, namely, replacing the original (or reference) weight w ( x ) (supported on R or subsets of R ) by w ( x )e − tx . It is a well-known fact that under such a deforma-tion the recurrence coefficients denoted as α n and β n evolve in t according to theToda equations, giving rise to the time dependent orthogonal polynomials, using Sogo’sterminology.If w is the Gaussian density e − x , x ∈ R , or the Gamma density x α e − x , x ∈ R + , α > − w is the Beta density (1 − x ) α (1 + x ) β , x ∈ [ − ,
1] , α, β > − , the resulting ”time-dependent” Jacobi polynomials will againsatisfy a linear second order ode, but no longer in the Sturm-Louville form. Thisdeformation induces an irregular point at infinity in addition to three regular singularpoints of the hypergeometric equation satisfied by the Jacobi polynomials.We will show that the coefficients of this ode are intimately related to a particularPainlev´e V. In particular we show that p ( n, t ) , where p ( n, t ) is the coefficient of ∗ Supported in part by NSF Grant DMS-0500892 n − of the monic orthogonal polynomials associated with the ”time-dependent” Jacobiweight, satisfies, up to a translation in t, the Jimbo-Miwa σ -form of the same P V ;while a recurrence coefficient α n ( t ) , is up to a translation in t and a linear fractionaltransformation P V ( α / , − β / , n + 1 + α + β, − / . These results are found fromcombining a pair of non-linear difference equations and a pair of Toda equations.This will in turn allow us to show that a certain Fredholm determinant related toa class of Toeplitz plus Hankel operators has a connection to a Painlev´e equation.The case with α = β = − / The study of Hankel determinants has seen a flurry of activity in recent years in part due toconnections with random matrix theory. In this paper we will investigate Hankel determi-nants for a weight of the form (1 − x ) α (1 + x ) β e − tx on the interval [ − , . Here we take t ∈ R . We will call this a time dependent Jacobiweight. Our ultimate goal is to produce a non-linear second order differential equation thatis satisfied by the logarithmic derivative of D n ( t ) , where D n ( t ) is the determinant of theHankel matrix generated from the moments of the weight: D n ( t ) := det ( µ j + k ( t )) n − j,k =0 := det (cid:18)Z − x j + k (1 − x ) α (1 + x ) β e − tx dx (cid:19) n − j,k =0 , and we shall initially assume α, β > . The moments µ k ( t ) can be evaluated as follows: µ k ( t ) = ( − k d k dt k µ ( t ) , k = 0 , , , ... where µ ( t ) = 2 α + β +1 Γ( α + 1)Γ( β + 1) e t M ( β + 1; α + β + 2; − t ) , and M ( a ; b ; z ) is the Kummer function with parameters a and b. Because d k dz k M ( a ; b ; z ) = ( a ) k ( b ) k M ( a + k ; b + k ; z ) , we find, µ k ( t ) = 2 α + β +1 Γ( α + 1)Γ( β + 1) e t k X r =0 (cid:18) kr (cid:19) ( − r ( β + 1) r ( α + β + 2) r M ( β + r + 1; α + β + r + 2; 2 t ) . D n ( t ) comes from the fact D n ( t ) /D n (0) is the generating functionof the linear statistics n X j =1 x j , and can also be thought of as the partition function for the random matrix ensemble witheigenvalue distribution Y ≤ j The purpose of this section is to derive two coupled Toda equations that involve the recur-sion coefficients of the time-dependent Jacobi polynomials. This is not a new result. The3ather more general Toda-hierarchy, can be found for example, in [30], [23] and [37]. Ourscorresponds to the first of the hierarchy. See [24] for a discussion of this in relation to Sato’stheory. See also [1] for the “multi-time” approach to matrix models.We include the necessary computations here for completeness sake and to set the nota-tions to be used throughout this paper.To begin we consider general orthogonal polynomials P i ( x ) with respect to the weight w ( x )e − tx on [ − , . The weight w will be known as the “reference” weight. The orthog-onality condition is Z − P i ( x ) P j ( x ) w ( x )e − tx dx = h i ( t ) δ i,j , (2.1)and the t dependence through e − tx , induces t dependence on the coefficients. We normalizeour monic polynomials as P n ( z ) = z n + p ( n, t ) z n − + ... + P n (0) , (2.2)although sometime we do not display the t dependence of coefficients of z n − .An immediate consequence of the orthogonality condition is the three terms recurrencerelation zP n ( z ) = P n +1 ( z ) + α n P n ( z ) + β n P n − ( z ) (2.3)with the initial conditions P ( z ) = 1 , β P − ( z ) = 0 . (2.4)An easy consequence of the recurrence relation is α n ( t ) = p ( n, t ) − p ( n + 1 , t ) , (2.5)and a telescopic sum of the above equation (bearing in mind that p (0 , t ) = 0) leaves − n − X j =0 α j ( t ) = p ( n, t ) . (2.6)First let us discuss the derivatives of α n and β n with respect to t, as this yields the simplestequations, where we keep w quite general, as long as the moments µ i ( t ) := Z − x i w ( x )e − tx dx, i = 0 , , ... (2.7)of all orders exist. Taking a derivative of h n with respect to th ′ n ( t ) = − Z − w ( x )e − tx xP n ( x ) dx = − α n h n , (2.8)4.e., (log h n ) ′ = − α n , (2.9)and since β n = h n /h n − , we have the first Toda equation, β ′ n = ( α n − − α n ) β n . (2.10)We define D n ( t ) to be the Hankel determinant D n ( t ) = det( µ i + j ( t )) n − i,j =0 . (2.11)It is well-known that D n ( t ) = Q n − i =0 h i ( t ) . This yields in view of (2.9) that ddt log D n ( t ) = − n − X j =0 α j ( t ) = p ( n, t ) . Also, 0 = ddt Z − P n P n − w e − tx dx = − Z − xP n P n − w e − tx dx + h n − ddt p ( n, t )= − h n + h n − ddt p ( n, t ) , and therefore ddt p ( n, t ) = β n ( t ) . (2.12)But since α n = p ( n ) − p ( n + 1) , we have the second Toda equation, α ′ n = β n − β n +1 . (2.13)To summarize we have the following theorem. Theorem 1 The recursion coefficients α n ( t ) and β n ( t ) satisfy the coupled Toda equations, β ′ n = ( α n − − α n ) β n , (2.14) α ′ n = β n − β n +1 . (2.15)It is also worth pointing out that in view of (2.11) we have the obvious Toda moleculeequation [34], d dt log D n ( t ) = ddt p ( n, t ) = β n ( t ) = D n +1 ( t ) D n − ( t ) D n ( t ) . Ladder operators, compatibility conditions, and dif-ference equations. In this section we give an account for a recursive algorithm for the determination of therecurrence coefficients α n , β n based a pair of ladder operators and the associated supple-mentary conditions. Such operators have been derived by various authors over many years.Here we provide a brief guide to the relevant literature, [7], [8], [9], [10], [11], [13], [15], [20],[21] and [26]. In fact Magnus in [26] traced this back to Laguerre. We find the form of theladder operators set out below convenient to use.For a sufficiently well-behaved weight (see [15] for a precise statement) of the form w ( x ) =e − v ( x ) the lowering and raising operators are P ′ n ( z ) = − B n ( z ) P n ( z ) + β n A n ( z ) P n − ( z ) , (3.1) P ′ n − ( z ) = [ B n ( z ) + v ′ ( z )] P n − ( z ) − A n − ( z ) P n ( z ) , (3.2)where A n ( z ) := 1 h n Z − v ′ ( z ) − v ′ ( y ) z − y P n ( y ) w ( y ) dy, (3.3) B n ( z ) := 1 h n − Z − v ′ ( z ) − v ′ ( y ) z − y P n ( y ) P n − ( y ) w ( y ) dy. (3.4)(3.5)Here we have assumed that w ( ± 1) = 0 . Additional terms would have to be included in thedefinitions of A n ( z ) and B n ( z ) if w ( ± = 0 . See [13] and [15].A direct calculation produces two fundamental supplementary (compatibility) conditionsvalid for all z ; B n +1 ( z ) + B n ( z ) = ( z − α n ) A n ( z ) − v ′ ( z ) ( S )1 + ( z − α n )( B n +1 ( z ) − B n ( z )) = β n +1 A n +1 ( z ) − β n A n − ( z ) . ( S )We note here that ( S ) and ( S ) have been applied to random matrix theory in [36]. Itturns out that there is an equation which gives better insight into the α n and β n if ( S )and ( S ) are suitably combined. See [17].Multiplying ( S ) by A n ( z ) we see that the r.h.s. of the resulting equation is a first orderdifference, while the l.h.s., with ( z − α n ) replaced by B n +1 ( z ) + B n ( z ) + v ′ ( z ) , is a firstorder difference plus A n ( z ) . Taking a telescope sum together with the appropriate “initialcondition”, B ( z ) = A − ( z ) = 0 , produces, B n ( z ) + v ′ ( z ) B n ( z ) + n − X j =0 A j ( z ) = β n A n ( z ) A n − ( z ) . ( S ′ ) . S ) , ( S ) and ( S ′ )were also stated to [26]. See also [22].If w is modified by the multiplication of “singular” factors such as | x − t | a or a + bH ( x − t ) , where H is the unit step function, then the ladder operator relations, ( S ) , ( S )and ( S ′ ) remain valid with appropriate adjustments. See [19], [18], and [4].Let Ψ( z ) = P n ( z ) . Eliminating P n − ( z ) from the raising and lowering operators givesΨ ′′ ( z ) − (cid:18) v ′ ( z ) + A ′ n ( z ) A n ( z ) (cid:19) Ψ ′ ( z ) + B ′ n ( z ) − B n ( z ) A ′ n ( z ) A n ( z ) + n − X j =0 A j ( z ) ! Ψ( z ) = 0 , (3.6)where we have used ( S ′ ) to simplify the coefficient of Ψ in (3.6).For the problem at hand, w ( x ) is the ”time-dependent” Jacobi weight, and now we mustsuppose that α > β > w ( x ) := (1 − x ) α (1 + x ) β e − tx , x ∈ [ − , , (3.7) v ( z ) := − α log(1 − z ) − β log(1 + z ) + tz, v ′ ( z ) = − αz − − βz + 1 + t, v ′ ( z ) − v ′ ( y ) z − y = α ( y − z − 1) + β ( y + 1)( z + 1) . (3.8)Substituting these into the definitions of A n ( z ) and B n ( z ) and integrating by parts, pro-duces, A n ( z ) = − R n ( t ) z − t + R n ( t ) z + 1 ,B n ( z ) = − r n ( t ) z − r n ( t ) − nz + 1 , where R n ( t ) := αh n Z − P n ( y )1 − y (1 − y ) α (1 + y ) β e − ty dy,r n ( t ) := αh n − Z − P n ( y ) P n − ( y )1 − y (1 − y ) α (1 + y ) β e − ty dy. Substituting the expressions for A n ( z ) and B n ( z ) into ( S ) and ( S ′ ) , which are identitiesin z , and equating the residues of the poles at z = ± , we find four distinct differenceequations and one which importantly performs the summation P j R j : − ( r n +1 + r n ) = α − R n (1 − α n ) (3.9) r n +1 + r n = 2 n + 1 + β − ( R n + t )(1 + α n ) (3.10)7 n + αr n = β n R n R n − (3.11)( r n − n ) − β ( r n − n ) = β n ( R n + t )( R n − + t ) (3.12) (cid:18) β − α (cid:19) r n + αn − t r n − r n ( r n − n ) − n − X j =0 R j (3.13)= − β n R n ( R n − + t ) + ( R n + t ) R n − ] . We now manipulate the equations (3.9)–(3.13) with the aim of expressing the recurrencecoefficients α n , β n in terms of r n , R n , and of course n, t. Adding (3.9) and (3.10) yields,2 R n = 2 n + α + β + 1 − t − tα n , (3.14)thus, α n is “easily” expressed in terms of R n . Subtracting (3.9) from (3.10) gives, r n +1 + r n = n + β − α + 1 − t − (cid:18) t R n (cid:19) α n . (3.15)Eliminating β n R n R n − from (3.11) and (3.12) we find, n ( n + β ) − (2 n + α + β ) r n = β n [ t + t ( R n − + R n )] . (3.16)Now with the aid of (3.11), replacing β n R n − by ( r n + αr n ) /R n , in (3.16) we find, t ( t + R n ) β n = n ( n + β ) − (2 n + α + β ) r n − tR n ( r n + αr n ) , (3.17)and this expresses β n in terms of R n , r n , n, t . It is important to note the absence of R n ± , and r n ± . The reader will notice that the above manipulations prove that we have expressed α n and β n the recurrence coefficients in terms of auxiliary quantities r n and R n .This is summarized in: Theorem 2 With r n , R n , α n and β n as defined above and with α, β > tα n = 2 n + 1 + α + β − t − R n , (3.18) t ( t + R n ) β n = n ( n + β ) − (2 n + α + β ) r n − tR n ( r n + αr n ) , (3.19)8n what follows we will find two Riccati equations, one in r n with coefficients involving R n and another with the roles of r n and R n reversed. We will first show that ddt log D n ( t ) = n − r n − β n t = − n − X j =0 α j ( t ) = p ( n, t ) . (3.20)The equation (3.20) can be derived as follows: Replace n by j in (3.18) and sum over j from 0 to n − n − X j =0 R j = n ( n − n ( α + β + 1 − t )2 − t n − X j =0 α j . Now we can obtain from (3.13) another expression for P j R j , n − X j =0 R j = (cid:18) β − α (cid:19) r n + αn − tr n − r n + nr n + tβ n R n + R n − ) + r n + αr n = (cid:18) α + β n − t (cid:19) r n + αn β n t R n + R n − )= n ( n + α + β )2 − t r n − β n t , (3.21)where we have eliminated tβ n ( R n + R n − ) using (3.16). These last two equations yield p ( n, t ) = − n − X j =0 α j = n − r n − tβ n . ¿From this we also deduce (see (2.5)), α n = 2( r n +1 − r n ) + t ( β n +1 − β n ) − . (3.22)Hence, in view of (2.12), α n = 2( r n +1 − r n ) − t dα n dt − , (3.23)i.e., t dα n dt + α n + 1 = 2 (cid:18) n + β − α + 1 − t − (cid:18) t R n (cid:19) α n − r n (cid:19) , where we have replaced r n +1 by r n plus an additional term with (3.15). Finally, t dα n dt + α n = 2 n + β − α − t − ( t + 2 R n ) α n − r n , 9r replacing α n in favor of R n from (3.18), t dR n dt = αt + (2 n + 1 + α + β − t ) R n − R n + 2 tr n . (3.24)This is a Riccati equation in R n , but with r n appearing linearly.Now since − n − X j =0 α j ( t ) = p ( n, t ) and p ′ ( n, t ) = β n ( t ) , see (2.6) and (2.11), we find, upon taking a derivative of (3.20) with respect to t, − dr n dt − β n − t dβ n dt = β n , or − dr n dt = β n + t α n − − α n ) β n , (3.25)where we have replaced β ′ n ( t ) by ( α n − − α n ) β n with the aid of the first Toda equation,(2.13). A simple computation with (3.14) gives, t α n − − α n ) = − R n − R n − , hence − dr n dt = β n ( R n − R n − )= β n R n − r n + αr n R n (3.26)= R n t ( t + R n ) (cid:20) n ( n + β ) − (2 n + α + β ) r n − tR n (cid:0) r n + αr n (cid:1)(cid:21) − r n + αr n R n , which is a Riccati equation in r n . We summarize in the following theorem: Theorem 3 The quantities r n and R n satisfy the coupled Riccati equations: t dR n dt = αt + (2 n + 1 + α + β − t ) R n − R n + 2 r n t, (3.27) − dr n dt = R n t ( t + R n ) (cid:20) n ( n + β ) − (2 n + α + β ) r n − tR n (cid:0) r n + αr n (cid:1)(cid:21) − r n + αr n R n (3.28)We end this section by pointing out that the above equations not only produce differentialequations in our various unknown quantities, but also a pair of coupled non-linear first orderdifference equations in R n and r n . If we substitute β n into (3.11) we obtain the followingresult. 10 heorem 4 The quantities r n and R n satisfy the coupled difference equations t ( r n +1 + r n ) = 4 R n + 2 R n (2 t − n − − α − β ) − αt (3.29) n ( n + β ) − (2 n + α + β ) r n = ( r n + αr n ) (cid:18) t R n R n − + tR n + tR n − (cid:19) (3.30) together with the “initial” conditions r = 0 , (3.31) R = α + β + 12 M (1 + β ; α + β + 1; − t ) M (1 + β ; α + β + 2; − t ) . (3.32)The initial condition for R can be found by direct integration. Observe that the represen-tation of R in terms of a ratio of the Kummer functions allows for the analytic continuationof α, β down to α = β = − / , due to the relation,lim b → b M ( a ; b ; z ) = a zM ( a + 1 , , z ) . Hence lim α →− / ,β →− / R ( t ) = t (cid:18) I (2 t ) I (2 t ) − (cid:19) . Indeed, by formally continuing β so that β + 1 = − k, k = 0 , , ..., we find R ( t ) = α L ( α − − k ) k ( − t ) L ( α − k ) k ( − t ) , expressed as ratio of Laguerre polynomials of degree k. It is clear that iterating (3.30) and(3.31) with the above R and r = 0 will generate rational solutions (in the variable t )of our P V derived in section 4. It is interesting to note that R is also a rational functionof α and t, therefore for the values of the parameters stated above our P V is a rationalfunction in α and t. Also note that the above equations define the quantities r n and R n for all α > − β > − t = 0 , then (3.14)gives, R n = n + α + β + 12 , and is consistent with (3.21) at t = 0 . Now equating (3.11) and (3.12) at t = 0 , gives, r n = n ( n + β ) α + β + 2 n , and β n = r n + αr n R n R n − = 4 n ( n + α )( n + β )( n + α + β )(2 n + α + β + 1)(2 n + α + β − n + α + β ) . With the R n given above, we find α n from (3.14) at t = 0 ,α n = β − α (2 n + α + β )(2 n + α + β + 2) , which are in agreement with those of [15]. 11 Identification of P V The idea is to eliminate r n ( t ) from our coupled Ricatti equations to produce a second orderode in R n ( t ) . This is straightforward and messy and we omit the details. A further changeof variable, R n ( t ) = − t − y ( t ) , leaves, after some simplification, y ′′ = 3 y − y ( y − 1) ( y ′ ) − y ′ t + Rat( y, t ) , where the last term is a particular rational function of two variables defined as follows:Rat( y, t ) = − y ( y + 1) y − n + 1 + α + β ) yt + ( y − t (cid:20) α y − β y (cid:21) . Therefore, y ( t ) := 1 + tR n ( t ) , satisfies, y ′′ = 3 y − y ( y − 1) ( y ′ ) − y ′ t + 2(2 n + 1 + α + β ) yt − y ( y + 1) y − y − t (cid:20) α y − β / y (cid:21) , which is almost a P V . To fit the above to into a P V , we make the replacement t → t/ y ( t/ 2) = Y ( t ) , and find Y ′′ = 3 Y − Y ( Y − 1) ( Y ′ ) − Y ′ t + ( Y − t (cid:20) α Y − β / Y (cid:21) + (2 n + 1 + α + β ) Yt − Y ( Y + 1) Y − , (4.33)which is P V ( α / , − β / , n + 1 + α + β, d = − / . The initial conditions are Y (0) = 1 , Y ′ (0) = 12 n + α + β + 1 . It is well known that there is a Hamiltonian associated with P V . To identify it, we substitute R n ( t ) := − tq (4.34) r n ( t ) := − pq ( q − 1) + ρ q, (4.35)12here p = p ( t ) , q = q ( t ) into (3.20), and choose ρ so that the resulting expression is apolynomial in p and q. There are two possible ρ : ρ = n and ρ = n + β. Case I. ρ = nt p ( n, t ) + n ( n + α + β ) − nt = p ( p + 2 t ) q ( q − − ntq + β pq + α p ( q − 1) (4.36) Case II. ρ = n + βt p ( n, t ) + n ( n + α + β ) − αβ − nt = p ( p + 2 t ) q ( q − 1) + 2( β − n ) qt + βpq + αp ( q − . (4.37)Replacing t by t/ t H and t H for our P V . The Hamiltonian as presented in Okamoto [31] (see also [32]) is t H = p ( p + t ) q ( q − 1) + α qt − α pq − α p ( q − , where a = α , b = − α , c = α − α , d = − , α = 1 − α − α − α . Comparing with our t H , while keeping in mind that the t is in fact t/ , we find α = − n, α = − β, α = − α, α = n + 1 + α + β,a = α , b = − β , c = 2 n + 1 + α + β. (4.38)Comparing with our t H , we find α = β − n, α = − β, α = − α, α = n + 1 + αa = α , b = − β , c = 2 n + 1 + α + β. (4.39)Hence both H and H generates our P V , where Y ( t ) = 1 − q ( t/ . σ -form of P V . Recall from the section 2, that ddt log D n ( t ) = p ( n, t ) , p ′ ( n, t ) = β n ( t ) . (5.1)Now we come the continuous σ -form of P V satisfied by p ( n, t ) , with n fixed and t beingthe variable. The idea is to express β n , r n , r ′ n in terms of p ( n, t ) and its derivative withrespect to t. Let us begin with (3.20), p ( n, t ) = n − r n − tβ n = n − r n − t p ′ ( n, t ) . (5.2)¿From the last equality of (5.2) we have r n ( t ) = 12 (cid:20) n − ddt ( t p ( n, t )) (cid:21) . (5.3)Under some minor rearrangements, equations (3.19) and (3.26) become tβ n R n + tR n ( r n + αr n ) = n ( n + β ) − (2 n + α + β ) r n − t β n (5.4) − tβ n R n + tR n ( r n + αr n ) = tr ′ n , (5.5)respectively, which is a system of linear equations in 1 /R n and R n . Solving for 1 /R n and R n we find, 2 tR n ( r n + αr n ) = tr ′ n + n ( n + β ) − (2 n + α + β ) r n − t β n tβ n R n = − tr ′ n + n ( n + β ) − (2 n + α + β ) r n − t β n . (5.6)Taking the product of the above we arrive at an identity free of R n :4 t β n ( r n + αr n ) = [ n ( n + β ) − (2 n + α + β ) r n − t β n ] − ( tr ′ n ) . (5.7)To identify the σ -function of Jimbo and Miwa [25], we replace t by t/ R n ( t/ 2) = − t − Y ( t )) , and substitute the above in (3.27) in the variable t/ . After a little simplification we find t dYdt = tY − r n ( t/ − Y ) − ( Y − αY + 2 n + β ) . Comparing this with the first equation of (C.40) of [25], we have z ( t ) = − r n ( t/ ,θ − θ + θ ∞ α, θ + θ + θ ∞ − n − β, − θ − θ = 2 n + 1 + α + β = c, consistent with the parameter c of our P V . Furthermore comparing (4.33) with (C.41) of[25], we find a possible identification; α = θ − θ + θ ∞ β = θ − θ − θ ∞ , and consequently, θ = − n, θ = − n − α − β, θ ∞ = α − β. But since ddt σ ( t ) = z ( t ) = − r n ( t/ , and bearing in mind (5.3) we have, upon integration and fixing a constant, σ ( t ) = 12 t p ( n, t/ − nt n ( n + β ) . (5.8)The σ -form of our P V is essentially a second order non-linear ode satisfied by p ( n, t ) , andreads ( tσ ′′ ) = [ σ − tσ ′ + (2 n + α + β ) σ ′ ] + 4[ σ − n ( n + β ) − tσ ′ ][( σ ′ ) − ασ ′ ] , (5.9)with the initial conditions σ (0) = n ( n + β ) , σ ′ (0) = − r n (0) = − n ( n + β ) α + β + 2 n . After some calculations we find that (5.9) in fact the Jimbo-Miwa σ -form (C.45) with ν = 0 , ν = − α, ν = n, ν = n + β. (5.10)To obtain (5.9) we first replace t by t/ r n ( t/ 2) = − σ ′ ( t ) ddt r n ( t/ 2) = − σ ′′ ( t ) β n ( t/ 2) = 2 ddt p ( n, t/ 2) = 4 ddt [ σ ( t ) + nt/ − n ( n + β )] t into (5.7) at t/ . Furthermore, since p ( n, t ) = (log D n ( t )) ′ , we have D n ( t ) = D n (0) exp (cid:20)Z t σ (2 s ) − n ( n + β ) + nss ds (cid:21) , D n (0) given by (1.6) of [3].We expect there exists a discrete analog of the continuous σ -form, namely, a differenceequation in the variable n, satisfied by p ( n, t ) with t fixed. To simplify notations, we donot display the t dependence. The idea is similar to the continuous case namely, we express β n , r n and R n in terms of p ( n ) and p ( n ± , and substitute these into (3.11) that is, r n + α r n = β n R n R n − . To begin with, we note that (3.20) is linear in β n and r n , which we re-write as; tβ n + 2 r n = n − p ( n ) . (5.11)We now find a another linear equation in β n and r n . First note that α n = p ( n ) − p ( n + 1) , and from (3.14) we have,2 R n + t = 2 n + 1 + α + β + t [ p ( n + 1) − p ( n )] . (5.12)The sum (5.12) at “ n ” and the same but at “ n − R n + R n − + t = (2 n + α + β ) + ( t/ p ( n + 1) − p ( n − . Substituting the above into (3.16) results the other linear equation mentioned above; tβ n { n + α + β + ( t/ p ( n + 1) − p ( n − } + (2 n + α + β ) r n = n ( n + β ) . (5.13)Solving for tβ n and 2 r n from the linear system (5.12) and (5.13), we find,2 r n = ( t/ n − p ( n )][ p ( n + 1) − p ( n − − n ( n + β ) n + ( α + β ) / t/ p ( n + 1) − p ( n − tβ n = [ n − p ( n )][ n + ( α + β ) / 2] + n ( n + α + β ) n + ( α + β ) / t/ p ( n + 1) − p ( n − , (5.15)and the discrete σ -form results from substituting (5.12), (5.14) and (5.15) into (3.11).Imagine for a moment that we leave our original problem behind and only considerthe functions Y and σ that satisfy the two Painlev´e equations (4.33) and (5.9) with theappropriate initial conditions. Then our orthogonal polynomials P n ( z, t ) satisfies the linearsecond order ode Ψ ′′ ( z ) + P ( z )Ψ ′ ( z ) + Q ( z )Ψ( z ) = 0 , (5.16)where P ( z ) := 1 + αz − βz − − t − z − [1 + Y (2 t )] / [1 − Y (2 t )] (5.17) Q ( z ) := − z − (cid:20) σ (2 t ) + n ( α + 1) − − Y (2 t )2 ddt σ (2 t ) (cid:21) + 12( z + 1) (cid:20) σ (2 t ) + n ( α + 1 + 2 t ) − (cid:20) ddt σ (2 t ) + n (cid:21) (1 − Y (2 t )) (cid:21) − n + ddt σ (2 t )2 [1 − Y (2 t )] (cid:20) z − (1 + Y (2 t )) / (1 − Y (2 t )) (cid:21) . (5.18)16his is a deformation of the classical ode satisfied by the Jacobi polynomials. When t = 0this ode reduces to a hypergeometric equation. In this section we introduce certain matrices that are combinations of finite Toeplitz andHankel matrices. There are identities that link these matrices directly to the Hankel momentmatrices that appear in the first section of this paper and define our quantity D n ( t ) . Wewill use these identities in some special cases to get exact formulas for D n ( t ) and, as aby-product, find Painlev´e type results for some other interesting determinants. We includethe Toeplitz/Hankel computations because as far as we know they are not written downexplicitly in this form in any other place. However, the Case 2 was established already bytwo of the authors in [5, Sec. 2]. The current derivation follows that in [5].Given a sequence { a k } ∞ k = −∞ of complex numbers, we associate the formal Fourier series a ( e iθ ) = ∞ X k = −∞ a k e ikθ , e iθ ∈ T . (6.1)The n × n Toeplitz and Hankel matrices with the (Fourier) symbol a are defined by T n ( a ) = ( a j − k ) n − j,k =0 , H n ( a ) = ( a j + k +1 ) n − j,k =0 . (6.2)Usually a represents an L -function defined on the unit circle T = { z ∈ C : | z | = 1 } , inwhich case the numbers a k are the Fourier coefficients, a k = 12 π Z π − π a ( e iθ ) e − ikθ dθ, k ∈ Z . (6.3)Notice that while the matrices H n ( a ) are classically referred to as Hankel matrices theyare not the same as the Hankel moment matrices considered in the previous sections of thispaper. To make the connection to Hankel matrices defined by moments we write H n [ b ] = ( b j + k ) n − j,k =0 , b k = 1 π Z − b ( x )(2 x ) j + k dx, (6.4)where b ( x ) be an L -function defined on [ − , 1] . Notice the difference in notation incomparison to (2.7) and (2.11). Our goal in this section is to prove four identities. Let z = e iθ . Then for each n > a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) − / (2 − θ ) − / , thendet (cid:16) T n ( a ) − H n ( z − a ) (cid:17) = det H n [ b ] . 17. If a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) − / (2 − θ ) / , thendet (cid:16) T n ( a ) + H n ( a ) (cid:17) = det H n [ b ] . 3. If a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) / (2 − θ ) − / thendet (cid:16) T n ( a ) − H n ( a ) (cid:17) = det H n [ b ] . 4. If a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) / (2 − θ ) / then14 det (cid:16) T n ( a ) + H n ( za ) (cid:17) = det H n [ b ] . In these identities the function a is always even, which means in terms of its Fourier co-effcients that a k = a − k . Moreover, in these formulas we assume that a ∈ L ( T ) , whichimplies that (and in Case 4 is equivalent to) b ∈ L [ − , 1] . We also remark that Cases 2and 3 can be derived from each other by making the substitutions a ( e iθ ) a ( e i ( θ + π ) ) and b ( x ) b ( − x ) .Our interest in the above formulas stems from the circumstance that they allow us touse existing results [6] on the asymptotics of the Toeplitz+Hankel determinants with well-behaved symbols a in order to derive the asymptotics of the Hankel moment determinants.In the above identities four types of finite symmetric Toeplitz+Hankel matrices as well asa finite Hankel moment matrix occur. These finite matrices can be obtained from their (on-sided) infinite matrix versions by taking the finite sections. It turns out that these infinitematrices are related to each other in a very simply way, namely they can be transformed intoone another by multiplying with appropriate upper and lower triangular (infinite) matricesform the left and right. These identities for the infinite matrices will be established in thenext theorem (and the remarks afterwards) in most general setting, where we do not assumethat the symbols are L -functions.Let us introduce the infinite matrices D + = − − , D − = 11 11 1. . . . . . . These are just the well-known Toeplitz operators D ± = T (1 ∓ z ) and their transposes aredenoted by D T ± . We also need the infinite diagonal matrix R = diag( , , , . . . ) . Theorem 5 For sequences of numbers { a n } ∞ n = −∞ , { a + n } ∞ n = −∞ , { a − n } ∞ n = −∞ , and { a n } ∞ n = −∞ satisfying a n = a − n , a + n = a + − n , a − n = a −− n , a n = a − n , efine A = ( a j − k − a j + k +2 ) ∞ j,k =0 A + = ( a + j − k + a + j + k +1 ) ∞ j,k =0 A − = ( a − j − k − a − j + k +1 ) ∞ j,k =0 A = ( a j − k + a j + k ) ∞ j,k =0 . (6.5) Then the following holds true:(1) If a + k = 2 a k − a k − − a k +1 , then D + AD T + = A + .(2) If a − k = 2 a k + a k − + a k +1 , then D − AD T − = A − .(3) If a k = 2 a + k + a + k − + a + k +1 , then D − A + D T − = RA R. (4) If a k = 2 a − k − a − k − − a − k +1 , then D + A − D T + = RA R. Moreover, if we define a sequence { b n } ∞ n =0 and an infinite Hankel matrix by b n = 12 n X k =0 a n − k (cid:18) nk (cid:19) , B = ( b j + k ) ∞ j,k =0 , (6.6) then B = S RA RS T with S = (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) ... . . . . (6.7) Proof: Before we start with the actual proof, we remark that the various products of theinfinite matrices make sense in terms of the usual matrix multiplication because the left andright factors are always (infinite) band matrices.In order to prove the first statement (1) we consider the ( j, k ) -entries of the following(products of) infinite matrices and compute as follows: (cid:2) D + AD T + (cid:3) j,k = (2 a j − k − a j − k − − a j − k +1 ) − ( a j + k +2 − a j + k +1 + a j + k ) if j, k > a − k − a − k +1 ) − ( a k +2 − a k +1 ) if j = 0 , k > a j − a j − ) − ( a j +2 − a j +1 ) if j > , k = 0 a − a if j = k = 0= a + j − k + a + j + k +1 if j, k > a + − k + a + k +1 if j = 0 , k > a + j + a + j +1 if j > , k = 0 a +0 + a +1 if j = k = 0 = (cid:2) A + (cid:3) j,k a k − a k − − a k +2 + a k +1 = (2 a k − a k − − a k +1 )+(2 a k +1 − a k − a k +2 ) ,a similar identity statement for j , and a − a = (2 a − a ) + (2 a − a − a ) . Moreover,we use the assumption that a n = a − n .Similarly, we compute the ( j, k ) -entry for the product appearing in (2): (cid:2) D − AD T − (cid:3) j,k = (2 a j − k + a j − k − + a j − k +1 ) − ( a j + k +2 + 2 a j + k +1 + a j + k ) if j, k > a − k + a − k +1 ) − ( a k +2 + a k +1 ) if j = 0 , k > a j + a j − ) − ( a j +2 + a j +1 ) if j > , k = 0 a − a if j = k = 0= a − j − k − a − j + k +1 if j, k > a −− k − a − k +1 if j = 0 , k > a − j − a − j +1 if j > , k = 0 a − − a − if j = k = 0 = (cid:2) A − (cid:3) j,k . Here we used a k + a k − − a k +2 − a k +1 = (2 a k + a k − + a k +1 ) − (2 a k +1 + a k + a k +2 ) , a similaridentity for j , and a − a = (2 a + 2 a ) − (2 a + a + a ) , and again a n = a − n .As for statement (3) we consider (cid:2) D − A + D T − (cid:3) j,k = (2 a + j − k + a + j − k − + a + j − k +1 ) + ( a + j + k +1 + 2 a + j + k + a + j + k − ) if j, k > a + − k + a + − k +1 ) + ( a + k +1 + a + k ) if j = 0 , k > a + j + a + j − ) + ( a + j +1 + a + j ) if j > , k = 0 a +0 + a +1 if j = k = 0= a j − k + a j + k if j, k > ( a − k + a k ) if j = 0 , k > ( a j + a j ) if j > , k = 0 ( a + a ) if j = k = 0 = (cid:2) RA R (cid:3) j,k . Again, we used that a + n = a + − n and a n = a − n .Statement (4) can be proven in the same way as statement (3). In fact, if we assumeall the hypotheses in (1)–(4), then (4) follows with a little algebra from the previous threestatements (and the from fact that D + and D − commute).In order to prove formula (6.7) first observe that S = ( ξ ( i, j )) ∞ i,j =0 , ξ ( i, j ) = ( (cid:0) i i − j (cid:1) if i > j and i − j even0 otherwise.Put r = 1 / r n = 1 for n > B = S RA RS T can be rephrased20s 12 i + l X m =0 a i + l − m (cid:18) i + lm (cid:19) = i X j =0 l X k =0 ξ ( i, j ) ξ ( l, k ) r j r k ( a j − k + a j + k ) (6.8)to hold true for each i, l > s > a s = a − s are the same on the left and right hand side.First assume s > a s = a − s is equal to thesum N + N + N , where N = X j i k ls = j − k ξ ( i, j ) ξ ( l, k ) r j r k = X u i/ v l/ s = i − u − l + 2 v (cid:18) iu (cid:19)(cid:18) lv (cid:19) r i − u r l − v ,N = X j i k ls = k − j ξ ( i, j ) ξ ( l, k ) r j r k = X u i/ v l/ s = − i + 2 u + l − v (cid:18) iu (cid:19)(cid:18) lv (cid:19) r i − u r l − v ,N = X j i k ls = j + k ξ ( i, j ) ξ ( l, k ) r j r k = X u i/ v l/ s = i − u + l − v (cid:18) iu (cid:19)(cid:18) lv (cid:19) r i − u r l − v . Therein, we made a change of variables j u = ( i − j ) / k v = ( l − k ) / u, v ) . In the above expressions for N and N we makeanother change of variables v l − v and u i − u to get the expressions X u i/ l/ v ls = i − u + l − v (cid:18) iu (cid:19)(cid:18) lv (cid:19) r i − u r v − l and X i/ u i v l/ s = i − u + l − v (cid:18) iu (cid:19)(cid:18) lv (cid:19) r u − i r l − v . Since there are no indices ( u, v ) satisfying i/ u i , l/ v l , and s = i − u + l − v ,we obtain that N + N + N equals X u i v ls = i − u + l − v (cid:18) iu (cid:19)(cid:18) lv (cid:19) = ξ ( i + l, s ) = ( (cid:0) i + l i + l − s (cid:1) if s i + l and i + l − s even0 otherwise. (6.9)This is the desired result since the coefficient for a s = a − s in the left hand side of (6.8) iszero if i + l − s is odd and 12 (cid:18)(cid:18) i + l i + l − s (cid:19) + (cid:18) i + l i + l + s (cid:19)(cid:19) = (cid:18) i + l i + l − s (cid:19) s = 0 the coefficient for the term a on the right hand side of (6.8) equals N := N + N = N + N , while it equals ξ ( i + l, 0) on the right hand side. The manipulationof the expressions N k can be done in the same way, with the only difference that in the endthere are indices which ( u, v ) satisfying i/ u i , l/ v l , i + l = 2( u + v ) . Thiscorresponds to a term N , which happen to be equal to N . Thus N = ( N + · · · + N )with N + · · · + N equaling (6.9). This settles the case s = 0 .Hence we have shown that the identity (6.8) holds, and this implies formula (6.7). ✷ In regard to the first part of the theorem we remark that the hypotheses in (1)–(4) arecompatible to each other in the sense that the hypotheses in (1) and (3), as well as those in(2) and (4) imply that a k = 2 a k − a k − − a k +2 . Correspondingly, we have DAD T = RA R with D = D + D − = D − D + . Elaborating on formula (6.7) we remark that assuming the hypotheses in (1)–(4), onecan express the coefficients b n in terms of a ± k and a k as well. We record the correspondingresults form completeness sake: b n = n X k =0 (cid:18) nk (cid:19) ( a + n − k + a +2 n +1 − k ) , B = S + A + S T + , S + = S D − (6.10) b n = n X k =0 (cid:18) nk (cid:19) ( a − n − k − a − n +1 − k ) , B = S − A − S T − , S − = S D + (6.11) b n = n X k =0 (cid:18) nk (cid:19) ( a n − k − a n +2 − k ) , B = SAS T , S = S D (6.12)The matrices S ± and S evaluate as follows: S + = (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) ... . . . , S − = (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) ... . . . , = (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) − (cid:0) (cid:1) (cid:0) (cid:1) ... . . . . Finally we remark that the recurrence relation (6.6) allows to express the coefficients a n interms of b n , a = 2 b , a n = a − n = [ n ] X k =0 ( − k b n − k (cid:18)(cid:18) n − kk (cid:19) + (cid:18) n − k − k − (cid:19)(cid:19) , n > . Let us now proceed with establishing the identities for the determinants of the finitematrices. We restrict to the cases where the symbols are L -functions because this is whatis of interest to us. Theorem 6 Let a, a + , a − , a ∈ L ( T ) be even, and let b ∈ L [ − , . Assume that1. a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) − / (2 − θ ) − / ,2. a + ( e iθ ) = b (cos θ )(2 + 2 cos θ ) − / (2 − θ ) / ,3. a − ( e iθ ) = b (cos θ )(2 + 2 cos θ ) / (2 − θ ) − / ,4. a ( e iθ ) = b (cos θ )(2 + 2 cos θ ) / (2 − θ ) / .Then, for each n > , det H n [ b ] = det (cid:16) T n ( a ) − H n ( z − a ) (cid:17) = det (cid:16) T n ( a + ) + H n ( a + ) (cid:17) = det (cid:16) T n ( a − ) − H n ( a − ) (cid:17) = 14 det (cid:16) T n ( a ) + H ( za ) (cid:17) . (6.13) Proof. We first notice that the hypotheses on the coefficients stated in (1)–(4) of Theorem5 can be rephrased in terms of the corresponding generating functions (see (6.1) and (6.3))as follows: a + ( z ) = a ( z )(1 − z )(1 − z − ) , a − ( z ) = a ( z )(1 + z )(1 + z − ) ,a ( z ) = a + ( z )(1 + z )(1 + z − ) , a ( z ) = a − ( z )(1 − z )(1 − z − ) . (6.14)Here z = e iθ ∈ T . Incidentally, the relations between a , a + , a − , and a implied by theassumption 1.–4. above are precisely those in (6.14).23ow assume 4. and compute b n = 1 π Z − b ( x )(2 x ) n dx = 1 π Z π b (cos θ )(2 cos θ ) n sin( θ ) dθ = 14 π Z π a ( e iθ )( e iθ + e − iθ ) n dθ = 12 n X k =0 a n − k (cid:18) nk (cid:19) , which is precisely the condition (6.6).In order to use the results of Theorem 5 we take the finite sections of the various identities(i.e., we consider the n × n upper-left corners of the infinite matrices), D + AD T + = A + , D − AD T − = A − ,D − A + D T − = D + A − D T + = RA R, B = S RA RS T , and then take the determinants. The crucial point is that D + , D − , and S are lower-trinagular and have ones on their diagonals. The diagonal matrices R give a factor in the determinants. Now it just remains to check that the finite sections of the infinitematrices (6.5) and (6.6) are indeed the matrices occuring in (6.13). But this follows fromthe definitions (6.2) and (6.4). ✷ It is apparent from the proof that if we are only interested in an identity between two typesof determinants featuring (6.13), then it is enough to assume that only the correspondingsymbols are L -functions and that the appropriate relationships between these symbols hold(see also (6.14)). For instance, if we assume a + , a − ∈ L ( T ) and a + ( z )(1 + z )(1 + z − ) = a − ( z )(1 − z )(1 − z − )then we can conclude thatdet (cid:16) T n ( a + ) + H n ( a + ) (cid:17) = det (cid:16) T n ( a − ) − H n ( a − ) (cid:17) . (6.15)By the way, this relationship between these two type of determinants is not the trivial onefeaturing the “equivalence” between the Cases 2 and 3, which has been pointed out earlier. The idea for this section is that if the α and β are any combination of ± , then we maychoose an operator of the form T n ( a ) + H n ( b ) from our list of identities 1.–4. that has,in a certain sense, a nice symbol and find an explicit formula for the determinants of theassociated Hankel matrices, H n [ b ] . This is because for these values of the parameters andthe right choice of operator, we lose the square root singularities. Fortunately in [6] exactformulas for the types of Toeplitz plus Hankel determinants that appear in the previous24heorem were found. If we specialize the results to the cases at hand we can state the exactformula of the determinants of the matrices H n [ b ] . The four different determinants all havethe form: G [ a ] n F [ a ] det( I + Q n KQ n ) , (7.1)where F [ a ] and G [ a ] are certain constants that depend on our choice of parameters for α and β. The last operator determinant involves orthogonal projections Q n = I − P n , wherethe projections P n acting on ℓ ( Z + ) , Z + = { , , . . . , } , are defined by P n ( a , a , . . . ) = ( a , a , . . . , a n − , , , . . . ) . The operator K , acting on ℓ ( Z ) , is a certain (trace class) semi-infinite Hankel operator.The precise reference for the result (7.1) is Proposition 4.1 and the remark afterwards in[6]. Propositions 3.1 and 3.3 in [6] also have to be consulted. For sake of clarification weremark that our cases 1.-4. correspond to the cases I-IV in [6] as follows: 1.=III, 2.=I, 3.=II,4.=IV, where in Case 4, the operators differ by a constant.In our case the symbol is (up to a constant) a ( e iθ ) = e − t cos θ whence ψ = a − ( e iθ ) e a + ( e iθ ) = e it sin θ , which occurs in the definition of the operator K . The Fourier coefficients ψ k ( k > J k ( t ) of order j with the argument t . The precise description of K is as follows:1. Let α = , β = , then K = K where K has ( j, k ) -entry − J j + k +2 ( t ) .2. Let α = − , β = , then K = K where K has ( j, k ) -entry J j + k +1 ( t ) .3. Let α = , β = − , then K = K where K has ( j, k ) -entry − J j + k +1 ( t ) .4. Let α = − , β = − , then K = K where K has ( j, k ) -entry J j + k ( t ) .Here j, k ≥ . It is known that the operator K is trace class. This is not hard to see sincefor fixed t the entries in the Hankel matrix tend to zero very rapidly. We state the fourcases below. In all cases our function a in the previous identities is e − t cos θ times a factorof a power of 2 . Theorem 7 Let b ( x ) = (1 − x ) α (1 + x ) β e − tx .1. Let α = , β = , then det H n [ b ] = 2 − n e t / det ( I + Q n K Q n ) . 2. Let α = − , β = , then det H n [ b ] = e t / − t/ det ( I + Q n K Q n ) . . Let α = , β = − , then det H n [ b ] = e t / t/ det ( I + Q n K Q n ) . 4. Let α = − , β = − , then det H n [ b ] = 2 n − e t / det ( I + Q n K Q n ) . This does not quite give us the identity of the original D n ( t ) since the above Hankel wasdefined with some extra constants of π and 2 . So first we adjust for these to yield thefollowing. Theorem 8 Let b ( x ) = (1 − x ) α (1 + x ) β e − tx .1. Let α = , β = , then D n ( t ) = 2 − n ( n +1) (2 π ) n e t / det ( I + Q n K Q n ) . 2. Let α = − , β = , then D n ( t ) = 2 − n (2 π ) n e t / − t/ det ( I + Q n K Q n ) . 3. Let α = , β = − , then D n ( t ) = 2 − n (2 π ) n e t / t/ det ( I + Q n K Q n ) . 4. Let α = − , β = − , then D n ( t ) = 2 − n ( n − − (2 π ) n e t / det ( I + Q n K Q n ) . Since Q n tends to zero strongly and the operator K is trace class, the term det ( I + Q n KQ n )tends to one and the asymptotics are given by the previous factors in each case of the aboveresult.More precisely, we obtain the following result. Theorem 9 Let b ( x ) = (1 − x ) α (1 + x ) β e − tx .1. Let α = , β = , then D n ( t ) ∼ − n ( n +1) (2 π ) n e t / . 2. Let α = − , β = , then D n ( t ) ∼ − n (2 π ) n e t / − t/ . . Let α = , β = − , then D n ( t ) ∼ − n (2 π ) n e t / t/ . 4. Let α = − , β = − , then D n ( t ) ∼ − n ( n − − (2 π ) n e t / . If we expand det( I + Q n K Q n ) using the fact that log det( I + A ) = tr log( I + A ) using justthe first couple of terms it seems reasonable to conjecture that, for example, D n ( t ) ∼ − n ( n +1) (2 π ) n e t / e ( t/ n +2Γ(2 n +3) +O ( n +4) ) . Similar conjectures can be made in the other cases.Before ending this section, we conjecture, with the aid the linear statistics formula in[16] and [3] obtained through the heuristic Coulomb Fluid approach [14], that for ”general”values α and β and for large n log (cid:18) D n ( t ) D n (0) (cid:19) ∼ t α − β ) t, where D n (0) ∼ − n ( n + α + β ) n ( α + β ) / − / (2 π ) n G ( α + β ) G ( α + β ) G ( α + β ) G (1 + α + β ) G (1 + α ) G (1 + β ) . 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