A Generalised Sextic Freud Weight
AA Generalised Sextic Freud Weight
Peter A. Clarkson and Kerstin Jordaan School of Mathematics, Statistics and Actuarial Science,University of Kent, Canterbury, CT2 7FS, UK
[email protected] Department of Decision Sciences,University of South Africa, Pretoria, 0003, South Africa [email protected]
15 April 2020
Abstract
We discuss the recurrence coefficients of orthogonal polynomials with respect to a generalised sextic Freud weight ω ( x ; t, λ ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , x ∈ R , with parameters λ > − and t ∈ R . We show that the coefficients in these recurrence relations can be expressedin terms of Wronskians of generalised hypergeometric functions F ( a ; b , b ; z ) . We derive a nonlinear discreteas well as a system of differential equations satisfied by the recurrence coefficients and use these to investigate theirasymptotic behaviour. We also study properties of the generalised sextic Freud polynomials and their zeros. Weconclude by highlighting a fascinating connection between generalised quartic, sextic, octic and decic Freud weightswhen expressing their first moments in terms of generalised hypergeometric functions. In this paper we are concerned with semi-classical polynomials which are orthogonal with respect to a symmetricweight on an unbounded interval. Orthogonal polynomials find application in various branches of mathematics such asapproximation theory, special functions, continued fractions and integral equations. Examples of classical orthogonalpolynomals on infinite intervals include Hermite and Laguerre polynomials.Although Szeg¨o pioneered much of what is known on the theory of orthogonal polynomials on finite intervals, hedid not carry his ideas over to infinite intervals, despite there being significant differences. It was only in the secondhalf of the 20th century, starting with the work of G´eza Freud on orthogonal polynomials on R , that the study ofFreud-type polynomials, and their generalisations, flourished. One of Freud’s original aims was to extend the theoryof best approximations using Jackson-Bernstein type estimates to the real line and, since a realistic expectation wasthat orthogonal expansions could serve as near-best approximations, a natural approach was to explore properties oforthogonal polynomials [51, 46].Freud [23] studied polynomials { P n ( x ) } ∞ n =0 orthogonal on the real line with respect to a class of exponential-typeweights, known as Freud weights, given by w ( x ) = | x | ρ exp( −| x | m ) , m ∈ N , with ρ > − , for more details see [22, 50, 51, 35, 46]. Freud conjectured that the asymptotic behaviour of recurrencecoefficients β n in the recurrence relation P n +1 ( x ) = xP n ( x ) − β n P n − ( x ) , (1.1)with P − ( x ) = 0 and P ( x ) = 1 , are given by lim n →∞ β n n /m = (cid:20) Γ( m )Γ(1 + m )Γ( m + 1) (cid:21) /m . (1.2)1 a r X i v : . [ n li n . S I] A p r reud [23] showed that if the limit exists for m ∈ Z , then it is equal to the expression in (1.2) and proved existenceof the limit (1.2) for m = 2 , , using a technique that gives rise to an infinite system of nonlinear equations, calledFreud equations. A general proof of Freud’s conjecture was given by Lubinsky, Mhaskar and Saff [37]; see also[18, 22, 23, 38, 51]. Freud explored other properties, such as the asymptotic behavior of the polynomials using therecurrence coefficients and the asymptotic behavior of the greatest zero [24]. Recent contributions on the asymptoticbehavior of the recurrence coefficients associated with Freud-type exponential weights and zeros of the associatedpolynomials can be found in [2, 35, 37, 41, 43, 44, 45, 50, 54]Magnus [40] showed that the coefficients in the three-term recurrence relation for the Freud weight ω ( x ; t ) = exp (cid:0) − x + tx (cid:1) , x, t ∈ R , with t ∈ R a parameter, can be expressed in terms of simultaneous solutions, q n , of the discrete equation q n ( q n − + q n + q n +1 ) + 2 tq n = n, (1.3)which is discrete P I (dP I ) – see equation (1.7) below for a more general version – as earlier shown by Bonan and Nevai[5, p. 135], and the fourth Painlev´e equation (P IV ) d q n d z = 12 q n (cid:18) d q n d z (cid:19) + 32 q n + 4 zq n + 2( z − A n ) q n + B n q n , (1.4)where A n = − n and B n = − n , with n ∈ Z + . The relationship between solutions of P IV (1.4) and dP I (1.3) isreflected in the striking similarity of the results for P IV (1.4) in [4, 48, 53] and those for dP I (1.3) in [25]. Bonan andNevai [5] proved that there is a unique positive solution of the discrete equation (1.3) with initial conditions β = 0 , β = (cid:82) ∞−∞ x exp( − x + tx ) d x (cid:82) ∞−∞ exp( − x + tx ) d x . In [13, 14], we considered the generalised quartic Freud weight ω ( x ; t ) = | x | λ +1 exp( − x + tx ) , x, t ∈ R , (1.5)with λ > − and t ∈ R parameters and gave explicit expressions for the moments of this weight (1.5). The firstmoment is µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp (cid:0) − x + tx (cid:1) d x = Γ( λ + 1)2 ( λ +1) / exp (cid:0) t (cid:1) D − λ − (cid:0) − √ t (cid:1) , (1.6)where D v ( ξ ) is the parabolic cylinder function with integral representation, cf. [52, § D ν ( ξ ) = exp( − ξ )Γ( − ν ) (cid:90) ∞ s − ν − exp (cid:0) − s − ξs (cid:1) d s, Re( ν ) < . Several sequences of monic orthogonal polynomials related to the weight (1.5) have been studied in the literature. Forinstance, for t = 0 , λ = − , the asymptotic and analytic properties of the corresponding orthogonal polynomialswere studied in [49], while the case when t > and λ = − is discussed in [3]. The recurrence coefficients in thethree-term recurrence relations associated with semiclassical orthogonal polynomials can often be expressed in termsof solutions of the Painlev´e equations and associated discrete Painlev´e equations. As shown in [14], the recurrencecoefficients β n in the three-term recurrence relation (1.1) are related to solutions of P IV (1.4) and satisfy the equation d β n d t = 12 β n (cid:18) d β n d t (cid:19) + β n − tβ n + ( t − A n ) β n + B n β n , where the parameters A n and B n are given by (cid:18) A n B n (cid:19) = (cid:18) − λ − n − − n (cid:19) , (cid:18) A n +1 B n +1 (cid:19) = (cid:18) λ − n − λ + n + 1) (cid:19) as well as the nonlinear discrete equation β n (cid:0) β n − + β n + β n +1 − t (cid:1) = n + ( λ + )[1 − ( − n ] , (1.7)2hich is the general discrete P I (dP I ). We remark that the nonlinear discrete equation (1.7) appears in the paper byFreud [23, equation (23), p. 5]; see also [2, §
2] for a historical review of the origin and study of equation (1.7).Iserles and Webb [27] discuss orthogonal systems in L ( R ) which give rise to a real skew-symmetric, tridiagonal,irreducible differentiation matrix. Such systems are important since they are stable by design and, if necessary, pre-serve Euclidean energy for a variety of time-dependent partial differential equations. Iserles and Webb [27] prove thatthere is a one-to-one correspondence between such an orthogonal system { φ n ( x ) } ∞ n =0 and a sequence of polynomials { P n ( x ) } ∞ n =0 which are orthogonal with respect to a symmetric weight.In this paper we consider polynomials orthogonal with respect to the generalised sextic Freud weight ω ( x ; t, λ ) = | x | λ +1 exp( − x + tx ) , x, t ∈ R , with λ > − and t ∈ R . The paper is organised as follows: in §
2, we review some properties of orthogonal polyno-mials with symmetric weight while we prove that the first moment of the generalised sextic Freud weight is a linearcombination of generalised hypergeometric functions F ( a ; b , b ; z ) and use this to derive an expression for therecurrence coefficents in terms of Wronskians of such generalised hypergeometric functions in §
3. In § n and parameter t tend to infinity. In § §
6. Finally,in § Let P n ( x ) , n ∈ N , be the monic orthogonal polynomial of degree n in x with respect to a positive weight ω ( x ) on thereal line R , such that (cid:90) ∞−∞ P m ( x ) P n ( x ) ω ( x ) d x = h n δ m,n , h n > , where δ m,n denotes the Kronekar delta. One of the most important properties of orthogonal polynomials is that theysatisfy a three-term recurrence relationship of the form P n +1 ( x ) = xP n ( x ) − α n P n ( x ) − β n P n − ( x ) , where the coefficients α n and β n are given by the integrals α n = 1 h n (cid:90) ∞−∞ xP n ( x ) ω ( x ) d x, β n = 1 h n − (cid:90) ∞−∞ xP n − ( x ) P n ( x ) ω ( x ) d x, with P − ( x ) = 0 and P ( x ) = 1 . For symmetric weights, i.e. ω ( x ) = ω ( − x ) , then clearly α n ≡ . Hence forsymmetric weights, the monic orthogonal polynomials P n ( x ) , n ∈ N , satisfy the three-term recurrence relation P n +1 ( x ) = xP n ( x ) − β n P n − ( x ) . (2.1)The relationship between the recurrence coefficient β n and the normalisation constants h n is given by h n = β n h n − . (2.2)The coefficient β n in the recurrence relation (2.1) can be expressed in terms of a determinant whose entries are givenin terms of the moments associated with the weight ω ( x ) . Specifically β n = ∆ n +1 ∆ n − ∆ n , (2.3)where ∆ n is the Hankel determinant ∆ n = det (cid:2) µ j + k (cid:3) n − j,k =0 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n − µ µ . . . µ n ... ... . . . ... µ n − µ n . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , n ≥ , (2.4)3ith ∆ = 1 , ∆ − = 0 , and µ k , the k th moment, is given by the integral µ k = (cid:90) ∞−∞ x k ω ( x ) d x. For symmetric weights then clearly µ k − ≡ , for k = 1 , , . . . .For symmetric weights it is possible to write the Hankel determinant ∆ n in terms of the product of two Hankeldeterminants, as given in the following lemma. The decomposition depends on whether n is even or odd. Lemma 2.1.
Suppose that A n and B n are the Hankel determinants given by A n = det (cid:2) µ j +2 k (cid:3) n − j,k =0 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n − µ µ . . . µ n ... ... . . . ... µ n − µ n . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , (2.5a) B n = det (cid:2) µ j +2 k +2 (cid:3) n − j,k =0 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n µ µ . . . µ n +2 ... ... . . . ... µ n µ n +2 . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.5b) Then the determinant ∆ n (2.4) is given by ∆ n = A n B n , ∆ n +1 = A n +1 B n . (2.6) Proof.
The result is obtained by matrix manipulation interchanging rows and columns ∆ n = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n − µ . . . µ n µ µ . . . µ n ... ... ... . . . ... ... µ n − µ n . . . µ n − µ n . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n − µ µ . . . µ n ... ... . . . ... µ n − µ n . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) × (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n µ µ . . . µ n +2 ... ... . . . ... µ n µ n +2 . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = A n B n , and ∆ n +1 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n µ . . . µ n µ µ . . . µ n +2 ... ... ... . . . ... ... µ n . . . µ n − µ n µ n +2 . . . µ n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n µ µ . . . µ n +2 ... ... . . . ... µ n µ n +2 . . . µ n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) × (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ µ . . . µ n µ µ . . . µ n +2 ... ... . . . ... µ n µ n +2 . . . µ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = A n +1 B n , as required. Corollary 2.2.
For a symmetric weight, the recurrence coefficient β n is given by β n = A n +1 B n − A n B n , β n +1 = A n B n +1 A n +1 B n , (2.7) where A n and B n are the Hankel determinants given by (2.5) , with A = B = 1 . roof. From (2.3), the recurrence coefficient β n is given by β n = ∆ n +1 ∆ n − ∆ n = A n +1 B n A n B n − A n B n = A n +1 B n − A n B n ,β n +1 = ∆ n +2 ∆ n ∆ n +1 = A n +1 B n +1 A n B n A n +1 B n = A n B n +1 A n +1 B n , as required. Lemma 2.3.
Suppose that ω ( x ) is a symmetric positive weight on the real line for which all the moments exist and ω ( x ; t ) = exp( tx ) ω ( x ) , with t ∈ R , is a weight such that all the moments also exist. Then the Hankel determinants A n and B n given by (2.5) can be written in terms of Wronskians, as follows A n = W (cid:18) µ , d µ d t , . . . , d n − µ d t n − (cid:19) , B n = W (cid:18) d µ d t , d µ d t , . . . , d n µ d t n (cid:19) , (2.8) where µ ( t ; λ ) = (cid:90) ∞−∞ exp( tx ) ω ( x ) d x. Proof. If ω ( x ; t ) = exp( tx ) ω ( x ) , with t ∈ R then µ n = (cid:90) ∞−∞ x n exp( tx ) ω ( x ) d x = d n d t n (cid:90) ∞−∞ exp( tx ) ω ( x ) d x = d n µ d t n , n = 1 , , . . . , and so it follows from (2.5) that A n and B n are given by (2.8). Lemma 2.4. If A n and B n are Wronskians given by (2.8) , with A = B = 1 , then A n d B n d t − B n d A n d t = A n +1 B n − , B n d A n +1 d t − A n +1 d B n d t = A n +1 B n . (2.9) Proof.
See, for example, Vein and Dale [61, § Corollary 2.5.
For a symmetric weight, if A n and B n are Wronskians given by (2.8) then the recurrence coefficient β n is given by β n = dd t ln B n A n , β n +1 = dd t ln A n +1 B n . Proof.
From (2.7), using (2.9), we have β n = A n +1 B n − A n B n = 1 B n d B n d t − A n d A n d t = dd t ln B n A n ,β n +1 = A n B n +1 A n +1 B n = 1 A n +1 d A n +1 d t − B n d B n d t = dd t ln A n +1 B n , as required. Lemma 2.6.
For a symmetric weight ω ( x ) , the normalisation constants h n are given by h n = A n +1 / A n , h n +1 = B n +1 / B n . Proof.
In terms of the determinants ∆ n , it is well-known that the normalisation constants are given by h n = ∆ n +1 / ∆ n , h = ∆ = µ , and so the result follows immediately from Lemma 2.1. Lemma 2.7.
Let ω ( x ) be a symmetric positive weight on the real line for which all the moments exist and let ω ( x ; t ) = exp( tx ) ω ( x ) , with t ∈ R , is a weight such that all the moments of exist. Then the recurrence coefficient β n ( t ) satisfies the Volterra, or the Langmuir lattice, equation d β n d t = β n ( β n +1 − β n − ) . (2.10)5e remark that the differential-difference equation (2.10) is also known as the discrete KdV equation, or the Kac-vanMoerbeke lattice [33]. Proof.
From (2.1) we have h n ( t ) = (cid:90) ∞−∞ P n ( x ; t ) ω ( x ; t ) d x, which on differentiation gives d h n d t = 2 (cid:90) ∞−∞ P n ( x ; t ) ∂ P n ∂t ( x ; t ) ω ( x ; t ) d x + (cid:90) ∞−∞ P n ( x ; t ) x ω ( x ; t ) d x. (2.11)Since P n ( x ; t ) is a monic polynomial of degree n in x , then ∂ P n ∂t ( x ; t ) is a polynomial of degree less than n andtherefore (cid:90) ∞−∞ P n ( x ; t ) ∂ P n ∂t ( x ; t ) ω ( x ; t ) d x = 0 . Using this and the recurrence relation (2.1) gives d h n d t = (cid:90) ∞−∞ [ P n +1 ( x ; t ) + β n ( t ) P n − ( x ; t )] ω ( x ; t ) d x = (cid:90) ∞−∞ (cid:2) P n +1 ( x ; t ) + β n ( t ) P n − ( x ; t ) (cid:3) ω ( x ; t ) d x = h n +1 + β n h n − and so it follows from (2.2) that d h n d t = h n +1 + β n h n . (2.12)Differentiating (2.2) with respect to t gives d h n d t = d β n d t h n − + β n d h n − d t , and using (2.12) gives h n +1 + β n h n = d β n d t h n − + β n ( h n + β n − h n − ) and so, since h n +1 = β n +1 h n = β n +1 β n h n − , we obtain d β n d t = β n ( β n +1 − β n − ) , as required.The weights of classical orthogonal polynomials satisfy a first-order ordinary differential equation, the Pearson equa-tion dd x [ σ ( x ) ω ( x )] = τ ( x ) ω ( x ) , (2.13)where σ ( x ) is a monic polynomials of degree at most and τ ( x ) is a polynomial with degree . However for semi-classical orthogonal polynomials, the weight function ω ( x ) satisfies the Pearson equation (2.13) with either deg ( σ ) > or deg ( τ ) (cid:54) = 1 (cf. [26, 42]).For example, the weight | x | λ +1 exp (cid:0) − x + tx (cid:1) , λ > − , x, t ∈ R of generalised sextic Freud polynomials, satisfies the Pearson equation (2.13) with σ ( x ) = x, τ ( x ) = 2 λ + 2 + 2 tx − x . For further information about orthogonal polynomials see, for example [10, 29, 58].6
Generalised sextic Freud weight
In this section we are concerned with the generalised sextic Freud weight ω ( x ; t, λ ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , λ > − , x, t ∈ R . (3.1) Lemma 3.1.
For the generalised sextic Freud weight (3.1) , the first moment is given by µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp( − x + tx ) d x = (cid:90) ∞ s λ exp( ts − s ) d s = Γ( λ + ) F ( λ + ; , ; ( t ) ) + t Γ( λ + ) F ( λ + ; , ; ( t ) )+ t Γ( λ + 1) F ( λ + 1; , ; ( t ) ) , (3.2) where F ( a ; b , b ; z ) is the generalised hypergeometric function.Proof. First we shall show that µ ( t ; λ ) = (cid:90) ∞ s λ exp( ts − s ) d s, is a solution of the third order equation d ϕ d t − t d ϕ d t − ( λ + 1) ϕ = 0 . (3.3)Following Muldoon [47], if we seek a solution of (3.3) in the form ϕ ( t ) = (cid:90) ∞ e st v ( s ) d s, then d ϕ d t − t d ϕ d t − ( λ + 1) ϕ = (cid:90) ∞ e st (cid:8) s v ( s ) − tsv ( s ) − ( λ + 1) v ( s ) (cid:9) d s = (cid:90) ∞ e st (cid:26) s v ( s ) + v ( s ) + s d v d s − ( λ + 1) v ( s ) (cid:27) d s = 0 , using integration by parts and assuming that lim s →∞ sv ( s )e st = 0 . Therefore for ϕ ( t ) to be a solution of (3.3) then v ( s ) necessarily satisfies s d v d s + (3 s − λ ) v = 0 ⇒ v ( s ) = s λ exp( − s ) . The general solution of equation (3.3) is given by ϕ ( t ) = c F ( λ + ; , ; ( t ) ) + c t F ( λ + ; , ; ( t ) ) + c t F ( λ + 1; , ; ( t ) ) , with c , c and c arbitrary constants. This can be derived from the third order equation satisfied by F ( a ; b , b ; z ) given in § z d w d z + z ( b + b + 1) d w d z + ( b b − z ) d w d z − a w = 0 , (3.4)which has general solution w ( z ) = c F ( a ; b , b ; z ) + c z − b F (1 + a − b ; 2 − b , − b + b ; z )+ c z − b F (1 + a − b ; 1 + b − b , − b ; z ) , with c , c and c constants. Note that making the transformation w ( z ) = ϕ ( t ) , with z = ( t ) , in (3.4) gives t d ϕ d t + 3 t ( b + b −
1) d ϕ d t + (cid:2) (3 b − b − − t (cid:3) d ϕ d t − a t ϕ = 0 . a = ( λ + 1) , b = and b = we have µ ( t ; λ ) = (cid:90) ∞ s λ exp( ts − s ) d s = c F ( λ + ; , ; ( t ) ) + c t F ( λ + ; , ; ( t ) ) + c t F ( λ + 1; , ; ( t ) ) , where c , c and c are constants to be determined. Since F ( a ; b , b ; 0) = 1 and µ (0; λ ) = (cid:90) ∞ s λ exp( − s ) d s = Γ( λ + ) , d µ d t (0; λ ) = Γ( λ + ) , d µ d t (0; λ ) = Γ( λ + 1) then it follows that c = Γ( λ + ) , c = Γ( λ + ) , c = Γ( λ + 1) , which gives (3.2), as required. Remarks 3.2.
1. If λ = − then µ ( t ; − ) = (cid:90) ∞−∞ exp( − x + tx ) d x = π / − / [Ai ( τ ) + Bi ( τ )] , τ = 12 − / t, where Ai( τ ) and Bi( τ ) are the Airy functions. This result is equation 9.11.4 in the DLMF [52], which is due toMuldoon [47, p32], see also [55].2. The generalised sextic Freud weight (3.1) is an example of a semi-classical weight for which the first moment µ ( t ; λ ) satisfies a third order equation . In our earlier studies of semi-classical weights [12, 13, 14], the first mo-ment has satisfied a second order equation. For example, for the quartic Freud weight (1.5), the first moment isexpressed in terms of parabolic cylinder functions (1.6), or equivalently in terms of the confluent hypergeometricfunction F ( a ; b ; z ) , see (7.2). These are classical special functions that satisfy second order equations.3. Equation (3.3) arises in association with threefold symmetric Hahn-classical multiple orthogonal polynomials[36] and in connection with Yablonskii–Vorob’ev polynomials associated with rational solutions of the secondPainlev´e equation [15].The higher moment µ k ( t ; λ ) is given by µ k ( t ; λ ) = (cid:90) ∞−∞ x k | x | λ +1 exp( − x + tx ) d x, k = 0 , , , . . . , and so µ k ( t ; λ ) = d k d t k µ ( t ; λ ) , µ k +1 ( t ; λ ) = 0 , (3.5)with µ ( t ; λ ) given by (3.2). Lemma 3.3.
Suppose that ∆ n ( t ; λ ) is the Hankel determinant given by ∆ n ( t ; λ ) = det (cid:2) µ j + k ( t ; λ ) (cid:3) n − j,k =0 , and A n ( t ; λ ) and B n ( t ; λ ) are the Hankel determinants given by A n ( t ; λ ) = det (cid:2) µ j +2 k ( t ; λ ) (cid:3) n − j,k =0 , B n ( t ; λ ) = det (cid:2) µ j +2 k +2 ( t ; λ ) (cid:3) n − j,k =0 , (3.6) then ∆ n ( t ; λ ) = A n ( t ; λ ) B n ( t ; λ ) , ∆ n +1 ( t ; λ ) = A n +1 ( t ; λ ) B n ( t ; λ ) . Lemma 3.4. If A n ( t ; λ ) and B n ( t ; λ ) are given by (3.6) then B n ( t ; λ ) = A n ( t ; λ + 1) .Proof. Since µ k +2 ( t ; λ ) = (cid:90) ∞ s λ + k +1 exp( ts − s ) d s = µ k ( t ; λ + 1) , then the result immediately follows. 8 emma 3.5. For the generalised sextic Freud weight (3.1) , the associated monic polynomials P n ( x ) satisfy the recur-rence relation P n +1 ( x ) = xP n ( x ) − β n ( t ; λ ) P n − ( x ) , n = 0 , , , . . . , (3.7) with P − ( x ) = 0 and P ( x ) = 1 , where β n ( t ; λ ) = A n +1 ( t ; λ ) A n − ( t ; λ + 1) A n ( t ; λ ) A n ( t ; λ + 1) = dd t ln A n ( t ; λ + 1) A n ( t ; λ ) ,β n +1 ( t ; λ ) = A n ( t ; λ ) A n +1 ( t ; λ + 1) A n +1 ( t ; λ ) A n ( t ; λ + 1) = dd t ln A n +1 ( t ; λ ) A n ( t ; λ + 1) . where A n ( t ; λ ) is the Wronskian given by A n ( t ; λ ) = W (cid:18) µ , d µ d t , . . . , d n − µ d t n − (cid:19) , with µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp (cid:0) − x + tx (cid:1) d x = Γ( λ + ) F (cid:0) λ + ; , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , ; ( t ) (cid:1) + t Γ( λ + 1) F (cid:0) λ + 1; , ; ( t ) (cid:1) . Plots of the polynomials P n ( x ; t ) , n = 3 , , . . . , , with λ = , for t = 0 , , . . . , , are given in Figure 3.1. Plotsof the polynomials P n ( x ; t ) , n = 3 , , . . . , , with λ = , at times t = 0 , , . . . , are given in Figure 3.2, whichillustrate the interlacing of the roots of successive polynomials. In this section we derive a discrete equation and a differential equation satisfied by recurrence coefficient β n ( t ; λ ) anddiscuss this asymptotics of β n ( t ; λ ) as n → ∞ and t → ±∞ . Lemma 4.1.
The recurrence coefficient β n ( t ; λ ) satisfies the nonlinear discrete equation β n (cid:0) β n +2 β n +1 + β n +1 + 2 β n +1 β n + β n +1 β n − + β n + 2 β n β n − + β n − + β n − β n − (cid:1) − tβ n = n + γ n , (4.1) with γ n = ( λ + )[1 − ( − n ] .Proof. Following Freud [23] we compute the integral (cid:90) ∞−∞ dd x (cid:8) P n ( x ) P n − ( x ) | x | λ +1 (cid:9) ω ( x ) d x (4.2)in two different ways. The first way is simply working out the derivative in the integrand and to use the orthogonalityto evaluate the resulting terms. This gives (cid:90) ∞−∞ dd x (cid:8) P n ( x ) P n − ( x ) | x | λ +1 (cid:9) ω ( x ) d x = (cid:90) ∞−∞ d P n d x ( x ) P n − ( x ) ω ( x ) d x + (cid:90) ∞−∞ P n ( x ) d P n − d x ( x ) ω ( x ) d x + (2 λ + 1) (cid:90) ∞−∞ P n ( x ) P n − ( x ) x ω ( x ) d x, with ω ( x ; t ) = exp( − x + tx ) and ω ( x ; t ) = | x | λ +1 exp( − x + tx ) . Since d P n d x ( x ) = nP n − ( x ) + lower order terms , then (cid:90) ∞−∞ d P n d x ( x ) P n − ( x ) ω ( x ) d x = n (cid:90) ∞−∞ P n − ( x ) ω ( x ) d x = nh n − . (4.3)9 ( x ; t ; − ) P ( x ; t ; − ) P ( x ; t ; − ) P ( x ; t ; − ) P ( x ; t ; − ) P ( x ; t ; − ) Figure 3.1: Plots of the polynomials P n ( x ; t ; ) , n = 3 , , . . . , for t = 0 (black), t = 1 (red), t = 2 (blue), t = 3 (green) and t = 4 (purple).As d P n − d x is a polynomial of degree n − then (cid:90) ∞−∞ P n ( x ) d P n − d x ( x ) ω ( x ) d x = 0 . (4.4)If n is even then P n − ( x ) /x is a polynomial of degree n − so that (cid:90) ∞−∞ P n ( x ) P n − ( x ) x ω ( x ) d x = 0 , whilst if n is odd then P n ( x ) /x = P n − ( x ) + lower order terms , so that (cid:90) ∞−∞ P n ( x ) P n − ( x ) x ω ( x ) d x = h n − . Consequently (cid:90) ∞−∞ P n ( x ) P n − ( x ) x ω ( x ) d x = [1 − ( − n ] h n − , (4.5)and therefore combining (4.3)–(4.5) we obtain (cid:90) ∞−∞ dd x (cid:8) P n ( x ) P n − ( x ) | x | λ +1 (cid:9) ω ( x ) d x = ( n + γ n ) h n − , (4.6)10igure 3.2: Plots of the polynomials P ( x ; t ; ) (black), P ( x ; t ; ) (red), P ( x ; t ; ) (blue), P ( x ; t ; ) (green) for t = 0 , , . . . , .with γ n = ( λ + )[1 − ( − n . We can also evaluate the integral (4.2) using integration by parts (cid:90) ∞−∞ dd x (cid:8) P n ( x ) P n − ( x ) | x | λ +1 (cid:9) ω ( x ) d x = − (cid:90) ∞−∞ (cid:8) P n ( x ) P n − ( x ) | x | λ +1 (cid:9) d ω d x d x = (cid:90) ∞−∞ (6 x − tx ) P n ( x ) P n − ( x ) ω ( x ) d x. From the three-term recurrence relation (3.7) we have xP n ( x ) = P n +1 ( x ) + β n P n − ( x ) , which on iteration gives x P n ( x ) = xP n +1 ( x ) + β n xP n − ( x )= P n +2 ( x ) + ( β n +1 + β n ) P n ( x ) + β n β n − P n − ( x ) , and x P n ( x ) = xP n +2 ( x ) + ( β n +1 + β n ) xP n ( x ) + β n β n − xP n − ( x )= P n +3 ( x ) + ( β n +2 + β n +1 + β n ) P n +1 ( x ) + β n ( β n +1 + β n + β n − ) P n − ( x ) + β n β n − β n − P n − ( x ) . (cid:90) ∞−∞ x P n ( x ) P n − ( x ) ω ( x ) d x = (cid:90) ∞−∞ x P n ( x ) x P n − ( x ) ω ( x ) d x = (cid:90) ∞−∞ (cid:2) P n +3 ( x ) + ( β n +2 + β n +1 + β n ) P n +1 ( x ) + β n ( β n +1 + β n + β n − ) P n − ( x ) + β n β n − β n − P n − ( x ) (cid:3) × (cid:2) P n +1 ( x ) + ( β n + β n − ) P n − ( x ) + β n − β n − P n − ( x ) (cid:3) ω ( x ) d x = ( β n +2 + β n +1 + β n ) h n +1 + β n ( β n + β n − )( β n +1 + β n + β n − ) h n − + β n β n − β n − h n − = ( β n +2 + β n +1 + β n ) β n +1 β n h n − + β n ( β n + β n − )( β n +1 + β n + β n − ) h n − + β n β n − β n − h n − = β n (cid:0) β n +2 β n +1 + β n +1 + 2 β n +1 β n + β n +1 β n − + β n + 2 β n β n − + β n − + β n − β n − (cid:1) h n − , and (cid:90) ∞−∞ xP n ( x ) P n − ( x ) ω ( x ) d x = (cid:90) ∞−∞ [ P n +1 ( x ) + β n P n − ( x )] P n − ( x ) ω ( x ) d x = β n h n − , so that (cid:90) ∞−∞ (6 x − tx ) P n ( x ) P n − ( x ) ω ( x ) d x = 6 β n (cid:0) β n +2 β n +1 + β n +1 + 2 β n +1 β n + β n +1 β n − + β n + 2 β n β n − + β n − + β n − β n − (cid:1) h n − − tβ n h n − . (4.7)By equating (4.6) and (4.7) we obtain the discrete equation (4.1), as required. Remark 4.2.
1. The fourth order nonlinear discrete equation (4.1) when t = 0 was derived by Freud [23]; see also Van Assche[59, § (2)I , a special case of the second member of the discrete Painlev´e I hierarchywhich is given by c β n (cid:0) β n +2 β n +1 + β n +1 + 2 β n +1 β n + β n +1 β n − + β n + 2 β n β n − + β n − + β n − β n − (cid:1) + c β n (cid:0) β n +1 + β n + β n − ) + c β n = c + c ( − n + n, (4.8)with c j , j = 0 , , . . . , constants. Cresswell and Joshi [16, 17] show that if c = 0 then the continuum limit of(4.8) is equivalent to d w d z = 10 w d w d z + 5 (cid:18) d w d z (cid:19) − w + z, which is P (2)I , the second member of the first Painlev´e hierarchy [34], see also [6, 20], and that if c (cid:54) = 0 thenthe continuum limit of (4.8) is equivalent to d w d z = 10 w d w d z + 10 w (cid:18) d w d z (cid:19) − w + zw + α, where α is a constant, which is P (2)II , the second member of the second Painlev´e hierarchy [1, 19]. This isanalogous to the situation for the general discrete Painlev´e I equation c β n (cid:0) β n +1 + β n + β n − ) + c β n = c + c ( − n + n, (4.9)with c j , j = 0 , , , constants. If c = 0 then the continuum limit of (4.9) is equivalent to the first Painlev´eequation d w d z = 6 w + z, whilst if c (cid:54) = 0 then the continuum limit of (4.9) is equivalent to the second Painlev´e equation d w d z = 2 w + zw + α, where α is a constant, see [16, 17] for details. 12 emma 4.3. The recurrence coefficient β n ( t ; λ ) satisfies the system d β n d t − β n + β n +1 ) d β n d t + β n + 6 β n β n +1 + 3 β n β n +1 − tβ n = ( n + γ n ) , (4.10a) d β n +1 d t + 3( β n + β n +1 ) d β n +1 d t + β n +1 + 6 β n +1 β n + 3 β n +1 β n − tβ n +1 = ( n + 1 + γ n +1 ) , (4.10b) with γ n = ( λ + )[1 − ( − n ] .Proof. Following Magnus [40, Example 5], from the Langmuir lattice (2.10) we have d β n − d t = β n − ( β n − β n − )= β n − + 3 β n − β n + β n − β n +1 + β n + 2 β n β n +1 + β n +1 + β n +1 β n +2 − n + γ n β n − t, (4.11a) d β n d t = β n ( β n +1 − β n − ) , (4.11b) d β n +1 d t = β n +1 ( β n +2 − β n ) , (4.11c) d β n +2 d t = β n +2 ( β n +3 − β n +1 )= − β n − β n − β n − β n β n +1 − β n β n +2 − β n +1 − β n +1 β n +2 − β n +2 + n + 1 + γ n +1 β n +1 + t. (4.11d)where we have used the discrete equation (4.1) to eliminate β n +3 and β n − . Solving (4.11b) and (4.11c) for β n +2 and β n − gives β n +2 = β n + 1 β n +1 d β n +1 d t , β n − = β n +1 − β n d β n d t , and substitution into (4.11a) and (4.11d) yields the system (4.10) as required. Lemma 4.4.
The recurrence coefficient β n ( t ; λ ) has the asymptotics, as t → ∞ β n ( t ; λ ) = n t + 3 √ n (2 n − λ − t / + O ( t − ) , (4.12a) β n +1 ( t ; λ ) = √ t − n − λ + 14 t − √ n − λn + 12 λ − λ + 5)32 t / + O ( t − ) . (4.12b) and as t → −∞ β n ( t ; λ ) = − nt − n [10 n + 6(2 λ + 1) n + 3 λ + 3 λ + 2] t + O ( t − ) , (4.13a) β n +1 ( t ; λ ) = − n + λ + 1 t − n + λ + 1)[10 n + (8 λ + 14) n + ( λ + 3)( λ + 2)] t + O ( t − ) . (4.13b) Proof.
First we consider β ( t ; λ ) which is given by β ( t ; λ ) = µ ( t ; λ ) µ ( t ; λ ) = (cid:82) ∞ s λ +1 exp( ts − s ) d s (cid:82) ∞ s λ exp( ts − s ) d s , and satisfies the equation d β d t + 3 β d β d t + β − tβ = ( λ + 1) . (4.14)Since µ ( t ; λ ) given by (3.2) involves the sum of three generalised hypergeometric functions then its asymptotics arenot as straightforward as for a classical special function, as was the case for the generalised quartic Freud weight wediscussed in [13, 14] which involved parabolic cylinder functions, recall (1.6). Using Laplace’s method it follows thatas t → ∞ µ ( t ; λ ) = (cid:90) ∞ s λ exp( ts − s ) d s = t ( λ +1) / (cid:90) ∞ ξ λ exp { t / ξ (1 − ξ ) } d ξ ∼ − / − λ/ t λ/ − / √ π exp (cid:16) √ t / (cid:17) µ ( t ; λ ) = µ ( t ; λ + 1) ∼ − / − λ/ t λ/ / √ π exp (cid:16) √ t / (cid:17) β ( t ; λ ) = µ ( t λ ) µ ( t ; λ ) ∼ √ t, as t → ∞ . Hence we suppose that as t → ∞ β ( t ; λ ) = √ t + a t + a t / + O ( t − ) . Substituting this into (4.14) and equating coefficients of powers of t gives a = (2 λ − , a = −√ λ − λ + 5) / , and so β ( t ; λ ) = √ t + 2 λ − t − √ λ − λ + 5)32 t / + O ( t − ) . Also using Watson’s Lemma it follows that as t → −∞ µ ( t ; λ ) ∼ Γ( λ + 1)( − t ) − λ − , µ ( t ; λ ) ∼ Γ( λ + 2)( − t ) − λ − , and so β ( t ; λ ) ∼ − λ + 1 t , as t → −∞ . Hence we suppose that as t → −∞ β ( t ; λ ) = − λ + 1 t + b t + b t + O ( t − ) . Substituting this into (4.14) and equating coefficients of powers of t gives b = − λ + 1)( λ + 2)( λ + 3) , b = − λ + 1)( λ + 2)( λ + 3)(3 λ + 21 λ + 38) . Then using the Langmuir lattice (2.10) it can be shown that as t → ∞ β ( t ; λ ) = 12 t − √ λ − t / + O ( t − ) , β ( t ; λ ) = √ t + 2 λ − t − √ λ − λ + 41)32 t / + O ( t − ) ,β ( t ; λ ) = 1 t − √ λ − t / + O ( t − ) , β ( t ; λ ) = √ t + 2 λ − t − √ λ − λ + 149)32 t / + O ( t − ) , and as t → −∞ β ( t ; λ ) = − t − λ + 2)( λ + 3) t + O ( t − ) , β ( t ; λ ) = − λ + 2 t − λ + 2)( λ + 3)( λ + 10) t + O ( t − ) ,β ( t ; λ ) = − t − λ + 3)( λ + 6) t + O ( t − ) , β ( t ; λ ) = − λ + 3 t − λ + 3)( λ + 21 λ + 74) t + O ( t − ) . From these we can see a pattern emerging for the asymptotics of β n ( t ; λ ) as t → ±∞ , which are different dependingon whether n is even or odd.Now suppose that u n ( t ; λ ) = β n ( t ; λ ) and v n ( t ; λ ) = β n +1 ( t ; λ ) , which from (4.10) satisfy d u n d t − u n + v n ) d u n d t + u n + 6 u n v n + 3 u n v n − tu n = n, (4.15a) d v n d t + 3( u n + v n ) d v n d t + v n + 6 v n u n + 3 v n u n − tv n = ( n + 1 + λ ) . (4.15b)If we suppose that as t → ∞ u n = a t + a t / + O ( t − ) , v n = √ t + b t + b t / + O ( t − ) , with a , a , b and b constants, then substituting into (4.15) and equating coefficients of powers of t gives (4.12).Also if we suppose that as t → −∞ u n = c t + c t + O ( t − ) , v n = d t + d t + O ( t − ) , with c , c , d and d constants, then substituting into (4.15) and equating coefficients of powers of t gives (4.13).14 n − ( t ; − ) β n − ( t ; ) β n − ( t ; ) β n ( t ; − ) β n ( t ; ) β n ( t ; ) Figure 4.1: Plots of the recurrence coefficients β n − ( t ; λ ) and β n ( t ; λ ) , n = 1 , , . . . , , with λ = − , , , for n = 1 (black), n = 2 (red), n = 3 (blue), n = 4 (green) and n = 5 (purple).Plots of β n ( t ; λ ) , for n = 1 , , . . . , , with λ = − , , are given in Figure 4.1. We see that there is completelydifferent behaviour for β n ( t ; λ ) as t → ∞ , depending on whether n is even or odd, which is reflected in Lemma 4.4.From these plots we make the following conjecture. Conjecture 4.5.
1. The recurrence coefficient β n +1 ( t ; λ ) is a monotonically increasing function of t .2. β n +2 ( t ; λ ) > β n ( t ; λ ) , for all t .3. The recurrence coefficient β n ( t ; λ ) has one maximum at t = t ∗ n , with t ∗ n +2 > t ∗ n . Remarks 4.6.
1. From the Langmuir lattice (2.10) we have β n +1 d β n +1 d t = β n +2 − β n , and so β n +2 ( t ; λ ) > β n ( t ; λ ) if and only if β n +1 ( t ; λ ) is a monotonically increasing function of t since β n +1 ( t ; λ ) > .2. Also from the Langmuir lattice we have β n d β n d t = β n +1 − β n − . and so β n ( t ; λ ) has a maximum when β n +1 ( t ; λ ) = β n − ( t ; λ ) . Since β n ( t ; λ ) → as t → ±∞ and β n ( t ; λ ) > then it is a maximum rather than a minimum.15reud [23] proved the following result, see also [59, § Lemma 4.7.
For the weight w ( x ) = | x | ρ exp( −| x | ) , m ∈ N , the recurrence coefficient β n ( ρ ) has the following asymptotic behaviour as n → ∞ lim n →∞ β n ( ρ ) n / = 1 √ . Theorem 4.8.
The recurrence coefficient β n ( t ; λ ) in the three-term recurrence relation for the sextic Freud weight ω ( x ; t ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , has the asymptotic expansions, if n is even β n ( t ; λ ) = n / κ + tκ n / − (2 λ + 1) κ n / − t (2 λ + 1) κ n / − [ t − λ + 7 λ + 2)] κ n / + O (cid:0) n − (cid:1) (4.16a) and if n is odd β n ( t ; λ ) = n / κ + tκ n / + (2 λ + 1) κ n / + 7 t (2 λ + 1) κ n / − [2 t + 135(6 λ + 6 λ + 1)] κ n / + O (cid:0) n − (cid:1) , (4.16b) with κ = √ , as n → ∞ .Proof. The recurrence coefficient β n satisfies the nonlinear discrete equation (4.1), which for λ (cid:54) = − has a ( − n term which suggests an even-odd dependence in β n ( t ; λ ) . This dependence needs to be taken into account to obtainan asymptotic approximation. Therefore we suppose that β n = (cid:40) u n , if n even ,v n , if n odd , (4.17)where from Freud’s Lemma 4.7 lim n →∞ u n √ n = 1 √ , lim n →∞ v n √ n = 1 √ , then ( u n , v n ) satisfy u n (cid:0) u n +2 v n +1 + v n +1 + 2 v n +1 u n + v n +1 v n − + u n + 2 u n v n − + v n − + v n − u n − (cid:1) − tu n = n, (4.18a) v n (cid:0) v n +2 u n +1 + u n +1 + 2 u n +1 v n + u n +1 u n − + v n + 2 v n u n − + u n − + u n − v n − (cid:1) − tv n = n + 2 λ + 1 . (4.18b)We remark that the transformation (4.17) was used by Cresswell and Joshi [16, 17] when they derived the continuumlimit of (4.8). Now suppose that u n = n / / + (cid:88) j =0 a j n j/ + O (cid:0) n − (cid:1) , v n = n / / + (cid:88) j =0 b j n j/ + O (cid:0) n − (cid:1) , (4.19a)where a j , b j , j = 0 , , . . . , , are constants to be determined. Then u n ± = n / κ + a + a n / + a κ ± κn / + a n + a ∓ a n / + ( a ∓ a ) κ − κn / + O (cid:0) n − (cid:1) , (4.19b) v n ± = n / κ + b + b n / + b κ ± κn / + b n + b ∓ b n / + ( b ∓ b ) κ − κn / + O (cid:0) n − (cid:1) , (4.19c) u n ± = n / κ + a + a n / + a κ ± κn / + a n + a ∓ a n / + ( a ∓ a ) κ − κn / + O (cid:0) n − (cid:1) , (4.19d) v n ± = n / κ + b + b n / + b κ ± κn / + b n + b ∓ b n / + ( b ∓ b ) κ − κn / + O (cid:0) n − (cid:1) , (4.19e)16ith κ = √ . Substituting (4.19) into (4.18) and equating powers of n gives a = b = 0 , a = b = tκ , a = − (2 λ + 1) κ , b = (2 λ + 1) κ ,a = b = 0 , a = − t (2 λ + 1) κ , b = 7 t (2 λ + 1) κ ,a = − [ t − λ + 7 λ + 2)] κ , b = − [2 t + 135(6 λ + 6 λ + 1)] κ , with κ = √ , and so if n is even then β n = n / κ + tκ n / − (2 λ + 1) κ n / − t (2 λ + 1) κ n / − [ t − λ + 7 λ + 2)] κ n / + O (cid:0) n − (cid:1) , (4.20a)whilst if n is odd then β n = n / κ + tκ n / + (2 λ + 1) κ n / + 7 t (2 λ + 1) κ n / − [2 t + 135(6 λ + 6 λ + 1)] κ n / + O (cid:0) n − (cid:1) , (4.20b)as required.Plots of β n ( t ; ) , for n = 1 , , . . . , , with t = 0 , , , , , are given in Figure 4.2 and plots of β n (2; λ ) , for n = 1 , , . . . , , with λ = 0 , , , , , are given in Figure 4.3. In these plots, the blue dots • are β n ( t ; λ ) for n even and the red dots • are β n ( t ; λ ) for n odd. The solid lines are the asymptotics (4.20) and the dashed line is n / κ + tκ n / , (4.21)with κ = √ . Remark 4.9.
In [62], Wang, Zhu and Chen state that β n ( t ; λ ) ∼ / t Θ n ( t ; λ ) + Θ n ( t ; λ )45 × / , as n → ∞ , where Θ n ( t ; λ ) = 48600 (cid:34) n + 2 λ + 1 + (cid:114) (2 n + 2 λ + 1) − t (cid:35) . From this it can be shown that as n → ∞ β n ( t ; λ ) = n / κ + tκ n / + (2 λ + 1) κ n / − t (2 λ + 1) κ n / + O (cid:0) n − / (cid:1) , with κ = √ , though this is not given in [62], which is the average of the asymptotic expressions for β n ( t ; λ ) for n even and odd given by (4.16). In this section we derive a mixed recurrence relation, a differential-difference equation and a differential equationsatisfied by generalised sextic Freud polynomials.The coefficients A n ( x ) and B n ( x ) in the relation d P n d x ( x ) = A n ( x ) P n − ( x ) − B n ( x ) P n ( x ) , (5.1)satisfied by semi-classical orthogonal polynomials can be derived using a technique introduced by Shohat [56] forweights ω ( x ) such that w (cid:48) ( x ) /ω ( x ) is a rational function. The method of ladder operators was introduced by Chenand Ismail in [8], see also [29, Theorem 3.2.1] and adapted in [7] for the situation where the weight function vanishesat one point. Explicit expressions for the coefficients in the differential-difference equation (5.1) when the weightfunction is positive on the real line except for one point are provided in [14].17 n (0; ) β n (1; ) β n (2; ) β n (3; ) β n (4; ) β n (10; ) Figure 4.2: Plots of β n ( t ; ) , for n = 1 , , . . . , , with t = 0 , , , , , . The blue dots • are β n ( t ; ) for n evenand the red dots • are β n ( t ; ) for n odd. The solid lines are the asymptotics (4.20) and the dashed line is (4.21). Theorem 5.1.
Let ω ( x ) = | x − k | ρ exp {− v ( x ) } , x, t, k ∈ R , (5.2) where v ( x ) is a continuously differentiable function on R . Assume that the polynomials { P n ( x ) } ∞ n =0 satisfy theorthogonality relation (cid:90) ∞−∞ P n ( x ) P m ( x ) ω ( x ) d x = h n δ mn . Then, for ρ ≥ , P n ( x ) satisfy the differential-difference equation ( x − k ) d P n d x ( x ) = A n ( x ) P n − ( x ) − B n ( x ) P n ( x ) , where A n ( x ) = x − kh n − (cid:90) ∞−∞ P n ( y ) K ( x, y ) ω ( y ) d y + a n ( x ) ,B n ( x ) = x − kh n − (cid:90) ∞−∞ P n ( y ) P n − ( y ) K ( x, y ) ω ( y ) d y + b n ( x ) , with K ( x, y ) = v (cid:48) ( x ) − v (cid:48) ( y ) x − y , and a n ( x ) = ρh n − (cid:90) ∞−∞ P n ( y ) y − k ω ( y ) d y, b n ( x ) = ρh n − (cid:90) ∞−∞ P n ( y ) P n − ( y ) y − k ω ( y ) d y. n (2; 0) β n (2; ) β n (2; 1) β n (2; 2) β n (2; 3) β n (2; 5) Figure 4.3: Plots of β n (2; λ ) , for n = 1 , , . . . , , with λ = 0 , , , , , . The blue dots • are β n (2; λ ) for n evenand the red dots • are β n (2; λ ) for n odd. The solid lines are the asymptotics (4.20) and the dashed line is (4.21). Proof.
See [14, Theorem 2].
Lemma 5.2.
Consider the weight defined by (5.2) and assume that v ( x ; t ) is an even, continuously differentiablefunction on R . Assume that the polynomials { P n ( x ) } ∞ n =0 satisfy the orthogonality relation (cid:90) ∞−∞ P n ( x ) P m ( x ) ω ( x ) d x = h n δ mn , and the three-term recurrence relation P n +1 ( x ) = xP n ( x ) − β n ( t ) P n − ( x ) , with P = 1 and P = x . Then the polynomials P n ( x ) satisfy (cid:90) ∞−∞ P n ( y ) y − k ω ( y ) d y = 0 , (cid:90) ∞−∞ P n ( y ) P n − ( y ) y − k ω ( y ) d y = [1 − ( − n ] h n − , where n ∈ N and h n = (cid:90) ∞−∞ P n ( y ) ω ( y ) d y. Proof.
See [14, Lemma 1].
Corollary 5.3.
Let ω ( x ) = | x | ρ exp {− v ( x ) } , x, t, k ∈ R , here v ( x ) is an even, continuously differentiable function on R . Assume that the polynomials { P n ( x ) } ∞ n =0 satisfy theorthogonality relation (cid:90) ∞−∞ P n ( x ) P m ( x ) ω ( x ) d x = h n δ mn . Then, for ρ ≥ , P n ( x ) satisfy the differential-difference equation x d P n d x ( x ) = A n ( x ) P n − ( x ) − B n ( x ) P n ( x ) , where A n ( x ) = xh n − (cid:90) ∞−∞ P n ( y ) K ( x, y ) ω ( y ) d y,B n ( x ) = xh n − (cid:90) ∞−∞ P n ( y ) P n − ( y ) K ( x, y ) ω ( y ) d y + ρ [1 − ( − n ] . Proof.
The result is an immediate consequence of Theorem 5.1 and Lemma 5.2.
Lemma 5.4.
For the generalised sextic Freud weight (3.1) ω ( x ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , the monic orthogonal polynomials P n ( x ) with respect to ω ( x ) satisfy (cid:90) ∞−∞ K ( x, y ) P n ( y ) ω ( y ) d y = 6 (cid:2) x − t + x (cid:0) β n + β n +1 (cid:1) + β n +2 β n +1 + (cid:0) β n +1 + β n (cid:1) + β n − β n (cid:3) h n , (5.3a) (cid:90) ∞−∞ K ( x, y ) P n ( y ) P n − ( y ) ω ( y ) d y = 6 x (cid:0) x + β n +1 + β n + β n − (cid:1) h n , (5.3b) where K ( x, y ) = v (cid:48) ( x ) − v (cid:48) ( y ) x − y , with v ( x ) = x − tx and h n = (cid:90) ∞−∞ P n ( y ) ω ( y ) d y. Proof.
For the generalised sextic Freud weight (3.1) we have v ( x ) = x − tx , and so K ( x, y ) = 6( x + x y + x y + xy + y ) − t. Hence for (5.3a) (cid:90) ∞−∞ K ( x, y ) P n ( y ) ω ( y ) d y = (6 x − t ) (cid:90) ∞−∞ P n ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ yP n ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ y P n ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ y P n ( y ) ω ( y ) d y + 6 (cid:90) ∞−∞ y P n ( y ) ω ( y ) d y = (6 x − t ) h n + 6 x (cid:90) ∞−∞ (cid:2) P n +1 ( y ) + β n P n − ( y ) (cid:3) ω ( y ) d y + 6 (cid:90) ∞−∞ (cid:2) P n +2 ( y ) + ( β n +1 + β n ) P n ( y ) + β n β n − P n − ( y ) (cid:3) ω ( y ) d y = (6 x − t ) h n + 6 x ( h n +1 + β n h n − ) + 6( h n +2 + ( β n +1 + β n ) h n + β n β n − h n − )= 6 (cid:2) x − t + x (cid:0) β n + β n +1 (cid:1) + β n +2 β n +1 + (cid:0) β n +1 + β n (cid:1) + β n − β n (cid:3) h n ,
20s required, since (cid:90) ∞−∞ yP n ( y ) ω ( y ) d y = (cid:90) ∞−∞ y P n ( y ) ω ( y ) d y = 0 , as these have odd integrands, β n = h n /h n − , the monic orthogonal polynomials P n ( x ) satisfy the three-term recur-rence relation (3.7), and are orthogonal, i.e. (cid:90) ∞−∞ P m ( y ) P n ( y ) ω ( y ) d y = 0 , if m (cid:54) = n. (5.4)Also for (5.3b) (cid:90) ∞−∞ K ( x, y ) P n ( y ) P n − ( y ) ω ( y ) d y = (6 x − t ) (cid:90) ∞−∞ P n ( y ) P n − ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ yP n ( y ) P n − ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ y P n ( y ) P n − ( y ) ω ( y ) d y + 6 x (cid:90) ∞−∞ y P n ( y ) P n − ( y ) ω ( y ) d y + 6 (cid:90) ∞−∞ y P n ( y ) P n − ( y ) ω ( y ) d y = 6 x (cid:90) ∞−∞ P n ( y ) (cid:2) P n ( y ) + β n − P n − ( y ) (cid:3) ω ( y ) d y + 6 x (cid:90) ∞−∞ (cid:2) P n +2 ( y ) + (cid:0) β n +1 + β n (cid:1) P n ( y ) + β n β n − P n − ( y ) (cid:3)(cid:2) P n ( y ) + β n − P n − ( y ) (cid:3) ω ( y ) d y = 6 x h n + 6 x (cid:2)(cid:0) β n +1 + β n (cid:1) h n + β n β n − h n − (cid:3) = 6 x (cid:0) x + β n +1 + β n + β n − (cid:1) h n , as required, since (cid:90) ∞−∞ P n ( y ) P n − ( y ) ω ( y ) d y = (cid:90) ∞−∞ y P n ( y ) P n − ( y ) ω ( y ) d y = (cid:90) ∞−∞ y P n ( y ) P n − ( y ) ω ( y ) d y = 0 , as these have odd integrands, using the recurrence relation (3.7) and orthogonality (5.4). Theorem 5.5.
For the generalised sextic Freud weight (3.1) the monic orthogonal polynomials P n ( x ) with repsect tothis weight satisfy the differential-difference equation x d P n d x ( x ) = A n ( x ) P n − ( x ) − B n ( x ) P n ( x ) , (5.5) where A n ( x ) = 6 xβ n (cid:2) x − t + x (cid:0) β n + β n +1 (cid:1) + β n +2 β n +1 + (cid:0) β n +1 + β n (cid:1) + β n − β n (cid:3) , (5.6a) B n ( x ) = 6 x β n (cid:0) x + β n +1 + β n + β n − (cid:1) + ( λ + )[1 − ( − n ] , (5.6b) with β n the recurrence coefficient in the three-term recurrence relation (3.7) .Proof. Corollary 5.3 shows that monic orthogonal polynomials P n ( x ) with respect to the weight ω ( x ) = | x | λ +1 exp {− v ( x ) } , satisfy the differential-difference equation (5.5), where A n ( x ) = xh n − (cid:90) ∞−∞ K ( x, y ) P n ( y ) ω ( y ) d y,B n ( x ) = xh n − (cid:90) ∞−∞ K ( x, y ) P n ( y ) P n − ( y ) ω ( y ) d y + ( λ + )[1 + ( − n ] . For the generalised sextic Freud weight (3.1), using Lemma 5.4 yields the result.21 emark 5.6.
In [62], the technique due to Shohat [56] using quasi-orthogonality, was applied to obtain the coefficients A n and B n in (5.1) for the weight (3.1). Note that, in their notation, the expression for B n (cf. [62, eqn. (39)]) shouldbe corrected to read B n ( z ) = 6 z β n + 6 zβ n (cid:0) β n +1 + β n + β n − (cid:1) + α [1 − ( − n ]2 z Now we derive a differential equation satisfied by generalised sextic Freud polynomials.
Theorem 5.7.
Let ω ( x ) = | x | λ +1 exp( − x + tx ) , for x, t ∈ R , then the monic orthogonal polynomials P n ( x ) withrespect to ω ( x ) satisfy x d P n d x + R n ( x ) d P n d x ( x ) + T n ( x ) P n ( x ) = 0 , where R n ( x ) = 2 tx − x + 2 λ + 1 − x (cid:0) x + β n + β n +1 (cid:1) C n ( x ) ,T n ( x ) = 36 xβ n C n − ( x ) C n ( x ) + 12 x β n + (2 λ +1) x (cid:0) x β n D n ( x ) − ( λ + ) (( − n − (cid:1) + 12 xβ n D n ( x ) − (cid:0) x β n D n ( x ) − ( λ + ) (( − n − (cid:1) (cid:16) xβ n D n ( x ) − (2 λ +1)2 x (( − n − − tx + 6 x (cid:17) − (cid:0) C n ( x ) + 4 x + 2 x ( β n + β n +1 ) (cid:1) (cid:0) x β n D n ( x ) − ( λ + ) (( − n − (cid:1) xC n ( x ) , with C n ( x ) = x − t + x (cid:0) β n + β n +1 (cid:1) + β n +2 β n +1 + (cid:0) β n +1 + β n (cid:1) + β n − β n D n ( x ) = x + β n − + β n + β n +1 . Proof.
In [14, Theorem 3] we proved that the coefficients in the differential equation x d P n d x ( x ) + R n ( x ) d P n d x ( x ) + T n ( x ) P n ( x ) = 0 , satisfied by polynomials orthogonal with respect to the weight ω ( x ) = | x | ρ exp {− v ( x ; t ) } , are given by R n ( x ) = ρ − x d v d x − xA n ( x ) d A n d x , (5.7a) T n ( x ) = A n ( x ) A n − ( x ) xβ n − + d B n d x − B n ( x ) (cid:20) d v d x + B n ( x ) − ρx (cid:21) − B n ( x ) A n ( x ) d A n d x , (5.7b)with A n ( x ) = xh n − (cid:90) ∞−∞ P n ( y ) K ( x, y ) ω ( y ) d y,B n ( x ) = xh n − (cid:90) ∞−∞ P n ( y ) P n − ( y ) K ( x, y ) ω ( y ) d y + ρ [1 − ( − n ] . For the generalised sextic Freud weight (3.1) we use (5.7) with k = 0 , ρ = 2 λ + 1 and v ( x ) = x − tx to obtain R n ( x ) = 2 λ + 2 − x + 2 tx − xA n ( x ) d A n d x , (5.8a) T n ( x ) = A n ( x ) A n − ( x ) xβ n − + d B n d x − B n ( x ) (cid:20) x − tx + B n ( x ) − (2 λ + 1) x (cid:21) − B n ( x ) A n ( x ) d A n d x . (5.8b)Substituting the expressions for A n ( x ) and B n ( x ) given by (5.6) and their derivatives into (5.8a) and (5.8b), we obtainthe stated result on simplification. 22ext, we consider a mixed recurrence relation connecting generalised sextic Freud polynomial associated with dif-ferent weight functions. Mixed recurrence relations such as these are typically used to prove interlacing and Stieltjesinterlacing of the zeros of two polynomials from different sequences and also provide a set of points that can be appliedas inner bounds for the extreme zeros of polynomials. The relation derived here will be used in § Lemma 5.8.
Let { P n ( x ; t, λ ) } ∞ n =0 be the sequence of monic generalised sextic Freud polynomials orthogonal withrespect to the weight (3.1) , then, for n fixed, x P n ( x ; t, λ + 1) = xP n +1 ( x ; t, λ ) − ( β n +1 + a n ) P n ( x ; t, λ ) (5.9) where a n = P n +2 (0; t, λ ) P n (0; t, λ ) , if n even ,P (cid:48) n +2 (0; t, λ ) P (cid:48) n (0; t, λ ) , if n odd , Proof.
The weight function associated with the polynomials P n ( x ; t, λ + 1) is ω ( x ; t, λ + 1) = | x | λ +3 exp( − x + tx ) = x ω ( x ; t, λ ) , and therefore Christoffel’s formula (cf. [58, Theorem 2.5]), applied to the monic polynomials P n ( x ; t, λ + 1) , is x P n ( x ; t, λ + 1) = 1 P n (0; t, λ ) P (cid:48) n +1 (0; t, λ ) − P (cid:48) n (0; t, λ ) P n +1 (0; t, λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P n ( x ; t, λ ) P n +1 ( x ; t, λ ) P n +2 ( x ; t, λ ) P n (0; t, λ ) P n +1 (0; t, λ ) P n +2 (0; t, λ ) P (cid:48) n (0; t, λ ) P (cid:48) n +1 (0; t, λ ) P (cid:48) n +2 (0; t, λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Since the weight ω ( x ; t, λ ) is even, we have that P n +1 (0; t, λ ) = P (cid:48) n (0; t, λ ) while P n (0; t, λ ) (cid:54) = 0 and P (cid:48) n +1 (0; t, λ ) (cid:54) =0 , hence x P n ( x ; t, λ + 1) = − P (cid:48) n (0; t, λ ) P n +1 (0; t, λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P n ( x ; t, λ ) P n +1 ( x ; t, λ ) P n +2 ( x ; t, λ )0 P n +1 (0; t, λ ) 0 P (cid:48) n (0; t, λ ) 0 P (cid:48) n +2 (0; t, λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) for n odd, while, for n even, x P n ( x ; t, λ + 1) = − P n (0; t, λ ) P (cid:48) n +1 (0; t, λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P n ( x ; t, λ ) P n +1 ( x ; t, λ ) P n +2 ( x ; t, λ ) P n (0; t, λ ) 0 P n +2 (0; t, λ )0 P (cid:48) n +1 (0; t, λ ) 0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) This yields x P n ( x ; t, λ + 1) = P n +2 ( x ; t, λ ) − a n P n ( x ; t, λ ) (5.10)and the result follows by using the three-term recurrence relation (3.7) to eliminate P n +2 ( x ; t, λ ) in (5.10). In what follows we investigate properties of the zeros of semiclassical orthogonal polynomials with respect to the evenweight (3.1).
Theorem 6.1.
Let { P n ( x ) } ∞ n =0 be the sequence of monic generalised sextic Freud polynomials orthogonal with re-spect to the weight (3.1) and let x k,n , k ∈ { , . . . , n } , denote the n zeros of P n in ascending order. Then, for λ > − and t ∈ R , the zeros are real, distinct and interlace as follows: x ,n < x ,n − < x ,n < · · · < x n − ,n < x n − ,n − < x n,n with x n +12 ,n = 0 when n is odd.Proof. The proofs for classical weights (see, for example, [58, Thm 3.3.1 and 3.3.2]) use the three-term recurrencerelation and definition of orthogonality and therefore the results also hold for semiclassical weights.23hen the weight is even, the zeros of the corresponding orthogonal polynomials are symmetric about the origin andtherefore we only need to consider the monotonicity of the positive zeros since the positive and the negative zeros haveopposing monotonicity.
Theorem 6.2.
Let { P n ( x ) } ∞ n =0 be the sequence of monic generalised sextic Freud polynomials orthogonal with re-spect to the weight (3.1) and let < x (cid:98) n/ (cid:99) ,n < · · · < x ,n < x ,n denote the positive zeros of P n ( x ) where (cid:98) m (cid:99) isthe largest integer smaller than m . Then, for λ > − and t ∈ R and for a fixed value of ν , ν ∈ { , , . . . , (cid:98) n/ (cid:99)} , the ν -th zero x n,ν increases when(i) λ increases;(ii) t increases.Proof. (i) Since for the generalised sextic Freud weight (3.1) ∂∂λ ln w ( x ; t, λ ) = ln | x | is an increasing function of x , it follows from a generalised version of Markov’s monotonicity theorem (cf. [31,Theorem 2.1] that the positive zeros of P n ( x ) increase as λ increases.(ii) Similarly, since ∂∂t ln w ( x ; t, λ ) = x , increases when x > increases, it follows that the zeros of P n ( x ) increase as t increases.Mixed recurrence relations involving polynomials from different orthogonal sequences, such as the relation derivedin Lemma 5.8, provide information on the relative positioning of zeros of the polynomials in the relation. In thenext theorem we prove that the zeros of P n ( x ; t, λ ) , the monic generalised sextic Freud polynomials orthogonal withrespect to the weight (3.1), and the zeros of P n − ( x ; t, λ + k ) interlace for λ > − , t ∈ R and k ∈ (0 , fixed. Theorem 6.3.
Let λ > − , t ∈ R and k ∈ (0 , . Let { P n ( x ; t, λ ) } be the monic generalised sextic Freud polynomialsorthogonal with respect to the weight (3.1) . Denote the positive zeros of P n ( x ; t, λ + k ) by < x ( t,λ + k ) (cid:98) n (cid:99) ,n < x ( t,λ + k ) (cid:98) n (cid:99)− ,n < · · · < x ( t,λ + k )2 ,n < x ( t,λ + k )1 ,n . If n is even, then In Theorem 6.2 we proved that the positive zeros of P n − ( x ; t, λ ) monotonically increase as λ increases. Thisimplies that, for each fixed (cid:96) ∈ { , , . . . , (cid:98) n − (cid:99)} , x ( t,λ ) (cid:96),n − < x ( t,λ + k ) (cid:96),n − < x ( t,λ +1) (cid:96),n − . (6.3)On the other hand, it was shown in Theorem 6.1 that the zeros of P n ( x ; t, λ ) and P n − ( x ; t, λ ) are interlacing, that is,when n is even, < x ( t,λ ) (cid:98) n (cid:99) ,n < x ( t,λ ) (cid:98) n − (cid:99) ,n − < x ( t,λ ) (cid:98) n (cid:99)− ,n < · · · < x ( t,λ )2 ,n < x ( t,λ )1 ,n − < x ( t,λ )1 ,n . (6.4)Next, we prove that the zeros of P n ( x ; t, λ )( x ) interlace with those of P n − ( x ; t, λ + 1) : Replacing n by n − in (5.9)yields P n − ( x ; t, λ + 1) = 1 x ( xP n ( x ; t, λ ) − ( β n + a n − ) P n − ( x ; t, λ )) . (6.5)24valuating (6.5) at consecutive zeros x (cid:96) = x ( t,λ ) (cid:96),n and x (cid:96) +1 = x ( t,λ ) (cid:96) +1 ,n , (cid:96) = 1 , , · · · , (cid:98) n (cid:99) − , of P n ( x ; t, λ )( x ) , weobtain P n − ( x (cid:96) ; t, λ + 1) P n − ( x (cid:96) +1 ; t, λ + 1) = 1 x (cid:96) x (cid:96) +1 ( β n + a n − ) P n − ( x (cid:96) ; t, λ ) P n − ( x (cid:96) +1 ; t, λ ) < since the zeros of P n ( x ; t, λ ) and P n − ( x ; t, λ ) seperate each other. So there is at least one positive zero of P n ( x ; t, λ +1) in the interval ( x (cid:96) , x (cid:96) +1 ) for each (cid:96) = 1 , , · · · , (cid:98) n (cid:99) − and this implies that < x ( t,λ ) (cid:98) n (cid:99) ,n < x ( t,λ +1) (cid:98) n − (cid:99) ,n − < x ( t,λ ) (cid:98) n (cid:99)− ,n < x ( t,λ +1) (cid:98) n − (cid:99)− ,n − < · · · < x ( t,λ +1)2 ,n − < x ( t,λ )2 ,n < x ( t,λ +1)1 ,n − < x ( t,λ )1 ,n (6.6)(6.6), (6.3) and (6.4) yield (6.1). The proof of (6.2) follows along the same lines.Considering that when the weight function is even, the zeros of P n ( x ; t, λ ) are symmetric about the origin with a zeroat the origin when n is odd, we have the following corollary. Corollary 6.4. With the same symbols as Theorem 6.3, we have for n odd that x ( t,λ ) n,n < x ( t,λ ) n − ,n − < x ( t,λ + k ) n − ,n − < x ( t,λ +1) n − ,n − < x ( t,λ ) n − ,n < · · · < x ( t,λ )2 ,n < x ( t,λ )1 ,n − < x ( t,λ + k )1 ,n − < x ( t,λ +1)1 ,n − < x ( t,λ )1 ,n while for n even x ( t,λ ) n,n < x ( t,λ ) n − ,n − < x ( t,λ + k ) n − ,n − < x ( t,λ +1) n − ,n − < x ( t,λ ) n − ,n < · · · < x ( t,λ +1) (cid:98) n − (cid:99) +2 ,n − < x ( t,λ ) (cid:98) n (cid:99) +1 ,n < and < x ( t,λ ) (cid:98) n (cid:99) ,n < x ( t,λ ) (cid:98) n − (cid:99) ,n − < x ( t,λ + k ) (cid:98) n − (cid:99) ,n − < x ( t,λ +1) (cid:98) n − (cid:99) ,n − < x ( t,λ ) (cid:98) n (cid:99)− ,n < · · · < x ( t,λ +1)1 ,n − < x ( t,λ )1 ,n with x ( t,λ ) (cid:98) n − (cid:99) +1 ,n − = x ( t,λ + k ) (cid:98) n − (cid:99) +1 ,n − = x ( t,λ +1) (cid:98) n − (cid:99) +1 ,n − = 0 . The three-term recurrence relation yields information on bounds of the extreme zeros of polynomials. Theorem 6.5. Let { P n ( x ; t, λ ) } ∞ n =0 be the sequence of monic generalised sextic Freud polynomials orthogonal withrespect to the weight (3.1) . For each n = 2 , , . . . , the largest zero, x ,n , of P n ( x ; t, λ ) , satisfies < x ,n < max ≤ k ≤ n − (cid:112) c n β k ( t ; λ ) , where c n = 4 cos (cid:16) πn +1 (cid:17) + ε , ε > .Proof. The upper bound for the largest zero x ,n follows by applying [30, Theorem 2 and 3] follows from the approachbased on the Wall-Wetzel Theorem, introduced by Ismail and Li [30] (see also [29]) to the three-term recurrencerelation (3.7).The Sturm Convexity Theorem (cf. [57]) on the monotonicity of the distances between consecutive zeros, applies tothe zeros of solutions of second-order differential equations in the normal form y (cid:48)(cid:48) ( x ) + F ( x ) y ( x ) = 0 . Next weconsider the implications of the convexity theorem of Sturm for the zeros of generalised sextic Freud polynomialswhen λ = − . We begin by considering the differential equation in normal form satisfied by generalised sextic Freudpolynomials for λ = − proved by Wang, Zhu and Chen in [62]. Theorem 6.6. Let w ( x ) = exp( − x + tx ) , x, t ∈ R , (6.7) and denote the monic orthogonal polynomials with respect to w ( x ) by P n ( x ) . Then, for t < , the polynomials S n ( x ) = P n ( x ) (cid:115) w ( x ) A n ( x ) (6.8)25 atisfy d S n d x + F ( x ) S n ( x ) = 0 , (6.9) where F ( x ) = β n A n − ( x ) A n ( x ) − w (cid:48)(cid:48) ( x )2 w ( x ) − B n ( x )( B n − x + 2 tx ) + 6 x − tx − (cid:18) A (cid:48) n ( x )2 A n ( x ) (cid:19) (6.10) + B (cid:48) n ( x ) − (2 B n ( x ) + 6 x − tx ) A (cid:48) n ( x ) − A (cid:48)(cid:48) n ( x )2 A n ( x ) , A n ( x ) = A n ( x ) xβ n = 6 (cid:2) x − t + x (cid:0) β n + β n +1 (cid:1) + β n +2 β n +1 + (cid:0) β n +1 + β n (cid:1) + β n − β n (cid:3) , B n ( x ) = B n ( x ) x = 6 xβ n (cid:0) x + β n +1 + β n + β n − (cid:1) + ( λ + )[1 − ( − n ] . Proof. See [62, Theorem 4] and note that A n > when t < . Theorem 6.7. Let {P n ( x ) } ∞ n =0 be the monic generalised sextic Freud polynomials orthogonal with respect to theweight (6.7) and let x k , k ∈ { , . . . , n } , denote the n zeros of P n in ascending order. Then, for t > ,(i) if F ( x ) given in (6.10) is strictly increasing on ( a, b ) , then, for the zeros x k ∈ ( a, b ) , we have x k +2 − x k +1 26s given by (cf. [14]) µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp( − x + tx ) d x = Γ( λ )2 ( λ +1) / exp( t ) D − λ − ( − √ t )= Γ( λ + ) F ( λ + ; ; t ) + t Γ( λ + 1) F ( λ + 1; ; t ) , (7.2)where F ( a ; b ; z ) is the confluent hypergeometric function, which is equivalent to the Kummer function M ( a, b, z ) .The relationship between the parabolic cylinder function D ν ( ζ ) and the Kummer function M ( a, b, z ) is given in [52, § µ ( t ; λ ) given by (7.2) satisfies the second-order equation d ϕ d t − t d ϕ d t − ( λ + 1) ϕ = 0 , and β ( t ; λ ) = dd t µ ( t ; λ ) satisfies the Riccati equation d β d t + β − tβ = ( λ + 1) . Lemma 7.1. For the generalised octic Freud weight ω ( x ; t ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , λ > − , x, t ∈ R , (7.3) then the first moment is given by µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp( − x + tx ) d x = (cid:90) ∞ s λ exp( − s + ts ) d s = Γ( λ + ) F (cid:0) λ + ; , , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , , ; ( t ) (cid:1) + t Γ( λ + 1) F (cid:0) λ + 1; , , ; ( t ) (cid:1) , where F ( a ; b , b , b ; z ) is the generalised hypergeometric function. Further µ ( t ; λ ) satisfies the fourth-orderequation d ϕ d t − t d ϕ d t − ( λ + 1) ϕ = 0 , and the first recurrence coefficient β ( t ; λ ) = dd t ln µ ( t ; λ ) satisfies the third-order equation d β d t + 4 β d β d t + 3 (cid:18) d β d t (cid:19) + 6 β d β d t + β − tβ = ( λ + 1) . (7.4) Proof. The proof is analogous that for the generalised sextic Freud weight (3.1) in Lemma 3.1. Lemma 7.2. The recurrence coefficient β n ( t ; λ ) for the generalised octic Freud weight (7.3) has the asymptotics as t → ∞ β n ( t ; λ ) = n t + 2 / n (3 n − − λ )9 t / + O ( t − / ) ,β n +1 ( t ; λ ) = ( t ) / − n − λ + 13 t − / [36 n − n (4 λ − 1) + 6 λ − λ + 7]27 t / + O ( t − / ) , and as t → −∞ β n ( t ; λ ) = − nt + O ( t − ) , β n +1 ( t ; λ ) = − n + λ + 1 t + O ( t − ) . Proof. Using Laplace’s method it follows that as t → ∞ µ ( t ; λ ) = (cid:90) ∞ s λ exp( ts − s ) d s = t ( λ +1) / (cid:90) ∞ ξ λ exp { t / ξ (1 − ξ ) } d ξ ∼ ( π ) / ( t ) ( λ − / exp (cid:110) t ) / (cid:111) µ ( t ; λ ) = µ ( t ; λ + 1) ∼ ( π ) / ( t ) λ/ exp (cid:110) t ) / (cid:111) β ( t ; λ ) = µ ( t ; λ ) µ ( t ; λ ) ∼ ( t ) / , as t → ∞ . Hence we suppose that as t → ∞ β ( t ; λ ) = ( t ) / + a t + a t / + O ( t − / ) . Substituting this into (7.4) and equating coefficients of powers of t gives a = ( λ − , a = − / (6 λ − λ + 7) / , and so β ( t ; λ ) = √ t + λ − t − / (6 λ − λ + 7)27 t / + O ( t − / ) . For the analogous result for the generalised sextic weight (3.1), in Lemma 4.4 we were able to use the system (4.15)to derive the leading asymptotics for β n ( t ; λ ) as t → ±∞ . However for the generalised octic Freud weight (7.3) wedon’t have the analog of (4.15). Instead we can use the Langmuir lattice (2.10). Now suppose that u n ( t ; λ ) = β n ( t ; λ ) and v n ( t ; λ ) = β n +1 ( t ; λ ) , then from the Langmuir lattice (2.10) we obtain u n +1 = u n + dd t ln v n , v n +1 = v n + dd t ln u n +1 , (7.5)with u = 0 , v ( t ; λ ) = β ( t ; λ ) = √ t + λ − t − / (6 λ − λ + 7)27 t / + O ( t − / ) . Next we assume that as t → ∞ u n = a n, t + a n, t / + O ( t / ) , v n = √ t + b n, t + b n, t / + O ( t / ) . (7.6)Substituting this into (7.5) and equating powers of t gives the recurrence relations a n +1 , = a n, + , b n +1 , = b n, − , a n +1 , = a n, − / b n, , b n +1 , = b n, − a n +1 , a n +1 , . Solving these with initial conditions a , = a , = 0 , b , = ( λ − , b , = − / (6 λ − λ + 7)27 , gives a n, = n, a n, = 2 / n (3 n − − λ )9 ,b n, = ( λ − − n ) , b n, = − / [36 n − n (4 λ − 1) + 6 λ − λ + 7]27 , as required.Using Watson’s Lemma it follows that as t → −∞ µ ( t ; λ ) = Γ( λ + 1)( − t ) − λ − (cid:2) O ( t − ) (cid:3) , µ ( t ; λ ) = Γ( λ + 2)( − t ) − λ − (cid:2) O ( t − ) (cid:3) , and so β ( t ; λ ) = − λ + 1 t + O ( t − ) , as t → −∞ . Consequently, we now assume that u n = c n t + O ( t − ) , v n = d n t + O ( t − ) , as t → −∞ . Substituting these into (7.5) and equating powers of t gives the recurrence relations c n +1 = c n − , d n +1 = d n − . Solving these with initial conditions c = 0 , d = − ( λ + 1) , gives c n = − n, d n = − ( λ + n + 1) , as required. 28 .2 The generalised decic Freud weight Lemma 7.3. For the generalised decic Freud weight ω ( x ; t ) = | x | λ +1 exp (cid:0) − x + tx (cid:1) , λ > − , x, t ∈ R , then the first moment is given by µ ( t ; λ ) = (cid:90) ∞−∞ | x | λ +1 exp( − x + tx ) d x = (cid:90) ∞ s λ exp( − s + ts ) d s = Γ( λ + ) F (cid:0) λ + ; , , , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , , , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , , , ; ( t ) (cid:1) + t Γ( λ + ) F (cid:0) λ + ; , , , ; ( t ) (cid:1) + t Γ( λ + 1) F (cid:0) λ + 1; , , , ; ( t ) (cid:1) where F ( a ; b , b , b , b ; z ) is the generalised hypergeometric function. Further µ ( t ; λ ) satisfies the fifth-orderequation d ϕ d t − t d ϕ d t − ( λ + 1) ϕ = 0 , and the first recurrence coefficient β ( t ; λ ) = dd t ln µ ( t ; λ ) satisfies the fourth-order equation d β d t + 5 β d β d t + 10 (cid:18) d β d t + β (cid:19) d β d t + 15 β (cid:18) d β d t (cid:19) + 10 β d β d t + β − tβ = ( λ + 1) . Proof. As for Lemma 7.1 above, the proof is analogous to that for the generalised sextic Freud weight (3.1) in Lemma3.1. Lemma 7.4. The recurrence coefficient β n ( t ; λ ) for the generalised decic Freud weight (7.3) has the asymptotics as t → ∞ β n ( t ; λ ) = n t + 5 / n (4 n − − λ )32 t / + O ( t − / ) ,β n +1 ( t ; λ ) = ( t ) / − n − λ + 38 t − / [40 n − λ − 1) + 4 λ − λ + 9]128 t / + O ( t − / ) , and as t → −∞ β n ( t ; λ ) = − nt + O ( t − ) , β n +1 ( t ; λ ) = − n + λ + 1 t + O ( t − ) . Proof. The proof is very similar to the proof of Lemma 7.2 above. In this case using Laplace’s method µ ( t ; λ ) ∼ ( π ) / ( t ) (2 λ − / exp (cid:110) t ) / (cid:111) , as t → ∞ and from Watson’s lemma µ ( t ; λ ) = Γ( λ + 1)( − t ) − λ − (cid:2) O ( t − ) (cid:3) , as t → −∞ . In this paper we studied generalised sextic Freud weights, the associated orthogonal polynomials and the recurrencecoefficients. We also investigated the interesting structural connections between the moments of the weight when theorder of the polynomial in the exponential factor of the weight is increased. Further analysis of this interesting class ofgeneralised higher order Freud polynomials and their properties, such as asymptotic expressions for the polynomialsand their greatest zeros, is currently in progress. It is important to note that our technique of expressing the Hankeldeterminants ∆ n and ∆ n +1 , for symmetric weights such as the generalised Freud weights, in terms of smallerHankel determinants A n and B n , as was done in § 2, had several benefits. The method resulted in expressions for β n and β n +1 in terms of A n that allowed the derivation of nonlinear discrete and nonlinear differential equationsfor β n which do not appear to exist when using ∆ n . 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