arXiv:1507.05261v1 [math.CA] 19 Jul 2015
Perturbed Hankel determinant, correlation functions and Painlev´ e equations Min Chen1∗, Yang Chen1†and Engui Fan2‡ 1
2
Department of Mathematics, University of Macau, Avenida da Universidade, Taipa, Macau, P.R. China
School of Mathematical Science, Fudan University, Shanghai 200433, P.R. China
Abstract We continue with the study of the Hankel determinant, Dn (t, α, β) := det
Z
1
x
j+k
w(x; t, α, β)dx
0
n−1
,
j,k=0
generated by a Pollaczek-Jacobi type weight, w(x; t, α, β) := xα (1 − x)β e−t/x ,
x ∈ [0, 1],
α > 0,
β > 0,
t ≥ 0.
This reduces to the “pure” Jacobi weight at t = 0. We may take α ∈ R , in the situation while t is strictly greater than 0. It was shown in Chen and Dai (2010), that the logarithmic derivative of this Hankel determinant satisfies a Jimbo-Miwa-Okamoto σ -form of Painlev´e V ( PV ). In fact the logarithmic of the Hankel determinant has an integral representation in terms of a particular PV . In this paper, we show that, under a double scaling, where n the dimension of the Hankel matrix tends to ∞ , and t tends to 0+ , such that s := 2n2 t is finite, the double scaled Hankel determinant (effectively an operator determinant) has an integral representation in terms of a particular PIII′ . Expansions of the scaled Hankel determinant for small and large s are found. A further double scaling with α = −2n + λ, where n → ∞ and t, tends to 0+ , such that s := nt is finite. In this situation the scaled Hankel determinant has an integral representation in terms of a particular PV , and its small and large s asymptotic expansions are also found. The reproducing kernel in terms of monic polynomials orthogonal with respect to the Pollaczek-Jacobi type weight, under the origin (or hard edge) scaling may be expressed in terms of the solutions of a second order linear ordinary differential equation (ODE). ∗
[email protected] Corresponding author(Yang Chen),
[email protected] and
[email protected] ‡
[email protected]
†
1
With special choices of the parameters, the limiting (double scaled) kernel and the second order ODE degenerate to Bessel kernel and the Bessel differential equation, respectively. We also applied this method to polynomials orthogonal with respect to the perturbed Laguerre weight; w(x; t, α) := xα e−x e−t/x , 0 ≤ x < ∞, α > 0, t > 0. The scaled kernel at origin of this perturbed Laguerre ensemble has the same behavior with the above limiting kernel, although difference scaled schemes are adopted on these two kernels.
2
1
Introduction
The determinant of the n × n Hankel matrix, Z j+k w(x)x dx L
,
0≤j,k≤n−1
has an equivalent representation as the multiple integral [39], 1 Dn [w] = n!
Z
Ln
Y
1≤j 0,
(1.9)
is the Pollaczek-Jocobi type weight. For t > 0, e−t/x → 0, as x → 0, with far greater speed than xα tends to 0, if α > 0. The same can be said, for α < 0, as long as t > 0. The Pollaczek-Jacobi type weight violates the Szeg¨o condition (see [43] and [10]), which reads, Z 1 | ln w(x)| p dx < ∞. x(1 − x) 0 5
Any monic polynomial orthogonal with respect to some weight can be represented by Heine’s multiple-integral, and in our case, Z n n Y Y Y 1 2 w(xℓ ; t, α, β)dxℓ. Pn (z; t, α, β) = (z − xm ) (xj − xk ) n!Dn (t, α, β) (0,1)n m=1 ℓ=1 1≤j 0 is finite. We note that our Hankel determinant under this double scaling scheme can be reproduced by another perturbed Laguerre weight, namely, w(x; t) = (x + t)−(λ+β) xβ e−x−t , λ < 0, β > 0, t ≥ 0, x ∈ (0, ∞), studied in [20]. In this case, the Hankel determinant has an integral representation in terms of a particular PV , (equivalent to a PIII ). From which we determine its small and large s expansion. Moreover, the constant which appear in the asymptotic expansion of the doublescaled Hankel determinant, for large s, is found. In Section 4, combining the ladder operator relations in x obtained in [10] and further ladder operator relations in t obtained here, satisfied by the monic orthogonal polynomials 6
Pn (x; t, α, β) (see theorem 12), we construct a Lax pair, involving derivative in x and derivative in t . The natural compatibility condition, reproduces certain results of [10]. In order to analyse the limiting behavior of the kernel arising form the Pollaczek-Jacobi type weight, a scaling scheme is introduced. Here, t → 0+ , n → ∞, such that s = 2n2 t is ζ∗ finite. The “coordinates”, x and y have been re-scaled to x = 4nζ 2 , y = 4n 2 . Ultimately, this shows that the limiting kernel may be characterized by solutions a second order ODE. If s = 0 , this essentially reduces to the Jacobi weight, and the limit kernel and the second order ODE reduce to the Bessel kernel and Bessel differential equation, respectively. In Section 5, we adopt the method in Section 4 to study the kernel arising form the singularly perturbed Laguerre weight. We adopt another scaling scheme, where t → 0+ , n → ∞, such ζ that s = (2n + 1 + α)t is finite. The “coordinates”, x and y have been re-scaled to x = 4n ∗ ζ . and y = 4n
2
Double scaling analysis I.
Chen and Dai [10] applied the ladder operator method to investigate the Hankel determinant obtained from the Pollacaek-Jacobi type weight. It was found that the logarithmic derivative of the Hankel determinant satisfies a particular Jimbo-Miwa-Okamoto σ -form of Painlev´e equation. An immediate consequence of the relationships obtained in [10], is that the Hankel determinant has an integral representation in terms of a PV transcendent in the variable y(t, α, β) . See Lemma 1. We shall be concerned with the behavior of the Hankel determinant, as n , the dimension of the Hankel matrix tends to infinity. For this purpose, a double scaling scheme is introduced, namely, sending n → ∞, t → 0+ , and such that s := 2n2 t remain fixed. We recall theorem 5.4 in [10]. Theorem 1. The logarithmic derivative of the Hankel determinant with respect to t , Hn (t, α, β) := t
d Dn (t, α, β) ln = (2n + α + β)(rn∗ (t) − rn (t)), dt Dn (0, α, β)
(2.10)
satisfies the following ordinary differential equation: ′′
′
′
′
′
(tHn )2 = [n(n + α + β) − Hn + (α + t)Hn ]2 + 4Hn (tHn − Hn )(β − Hn ),
(2.11)
with the initial data Hn (0, α, β) = 0. Here rn∗ (t) and rn (t) variables defined in [10]. By a change of variable, the above ODE turns out to be a particular Jimbo-Miwa-Okamoto σ -from of PV . See [10] for more details. We now recall the Theorem 7.2 in [10] and replace Sn (t) in that paper by y(t, α, β). Theorem 2. Let y(t, α, β) :=
Rn (t) . 2n + 1 + α + β 7
(2.12)
Then y(t, α, β) satisfies the following differential equation: y ′′ =
3y − 1 y ′ (2n + 1 + α + β)2 (y − 1)2 y (y − 1)2 β 2 αy y(y + 1) (y ′)2 − + − + − , 2y(y − 1) t 2t2 2t2 y t 2(y − 1) (2.13)
which is a PV ((2n + 1 + α + β)2 /2, −β 2 /2, α, −1/2). The boundary condition is y(0, α, β) = 1. From results obtained in [10], we show that the Hankel determinant has an integral representation in terms of PV . See the Lemma below. Lemma 1. The logarithmic derivative of the Hankel determinant Hn (t, α, β), and rn∗ , can be expressed in terms of y(t, α, β) and y ′ (t, α, β) as follow, Dn (t, α, β) d ln dt Dn (0, α, β) h 1 2 =− β 2 + 2y 3 − 2(2n + α + β)2 y 3 − 2(t + β)(2n + α + β)y 3 − y 4 − t2 y ′ 2 4y(y − 1) +(2n + α + β)2 y 4 − 4n(t + β)y − 2β(t + α + 2β)y − 2ty ′ y + 2ty ′y 2 + 4n2 y 2 +y 2 (t + α)2 + 4tβ + 6αβ + 6β 2 − 1 + 4n(2t + α + 3β)y 2 + n(n + α − t), (2.14)
Hn (t, α, β) = t
with the initial data Hn (0, α, β) = 0. Proof. Recalling identities (5.8) , (5.9) and (7.2) in [10] as, (2n + 1 + α + β)[2rn2 + (t + 2β − 2rn∗ )rn + (2n + α)rn∗ − nt − trn′ ] Rn (t) = , 2[(rn∗ − rn )2 + (2n + α − t)rn∗ + (β + t)rn − nt] 2r 2 + (t + 2β − 2rn∗ )rn + (2n + α)rn∗ − nt + trn′ 1 , = n Rn (t) 2(2n + 1 + α + β)(β + rn )rn and rn∗ =
Rn − β − t 1 [tRn′ − (2n + 1 + α + β)(2rn − Rn + β)] + rn − . 2Rn 2
With the aid of above equations and the definition of y(t, α, β) in (2.12), the variables rn (t) and rn∗ (t) may be expressed in terms of y and y ′. Straightforward computation produces (2.14). Hence the Hankel determinant has an integral representation in terms of y and y ′ , and that y(t, α, β) satisfies the PV given by (2.13). 8
2.1
Scaling limit of the Hankel determinant in terms of Painlev´ e equations.
Carrying out the double scaling and combining with Theorem 1, Theorem 2 and Lemma 1, we find that the (effectively) infinite dimensional Hankel determinant has an integral representation in terms of ( 2.16 ). Moreover, the logarithmic derivatives of such a Hankel determinant satisfies another σ -form of the corresponding Painlev´e equation. Theorem 3. Let y(t, α, β) := 1 +
f (t, α, β) , n2
and
s := 2n2 t.
(2.15)
t → 0+ and n → ∞ , such that s ∈ (0, ∞). If s , α, β , g(s, α, β) := lim f n→∞ 2n2 then g(s, α, β) satisfies g ′′ =
g ′2 g ′ 2g 2 1 α − + 2 + − , g s s 2s 4g
(2.16)
1 , The equation (2.16) is PIII′ (8, 2α, 0, −1) . with the initial data g(0, α, β) = 0, g ′(0, α, β) = 2α If s H(s, α, β) := lim Hn , α, β , n→∞ 2n2 then H(s, α, β) satisfies,
2 1 ′ (sH ) + 4 (H ) (sH − H) − αH + = 0, 2 ′′ 2
′ 2
′
(2.17)
1 . with the initial conditions H(0, α, β) = 0 , H′ (0, α, β) = − 2α Furthermore, if Dn (s/2n2 , α, β) , ∆(s, α, β) := lim n→∞ Dn (0, α, β)
then H=s
d (s g ′ − g)2 4αsg − s2 α2 ln ∆(s, α, β) = + − g − . ds 4g 2 16g 2 4
(2.18)
Proof. Substituting the definition (2.15) into ( 2.13 ) , then we see that g(s, α, β) satisfied PIII′ (8, 2α, 0, −1) . See ( 2.16 ). Plugging s = 2n2 t into ( 2.11 ), we see that the limit of Hn (t, α, β) satisfies a σ -from Painlev´e equation ( 2.17 ). Moreover, substituting ( 2.15 ) into ( 2.14 ), then the equation ( 2.18 ) is found.
9
Note that the differential equation is satisfied by g(s, α, β) is a PIII′ (8, 2α, 0, −1) , see [30, 42]. From PIII′ (8, 2α, 0, −1) and the σ -form of the Painlev´e equation, together with the boundary conditions, g(s, α, β) and H(s, α, β) are independent of β . We use these notations to distinguish this result from those previously obtained [9]. The σ -from Painlev´e equation of ( 2.17 ) has the same double scaling limit of the logarithmic derivative of the Hankel determinant generated by the singularly perturbed Laguerre weight, xα e−x−t/x , where x ≥ 0, t > 0, and real α, however, the double scaling scheme is different from that in [9]. Remark 1: By the change variables 2 8 x F (x, α, β) = g , α, β , x 8 it is seen that the PIII′ (8, 2α, 0, −1) satsified by g becomes, F ′′ =
1 F ′2 F ′ F 2 2α − + + − , F x x x F
a PIII (1, 2α, 0, −1) . See [30]. The C potential (2.22) introduced in [9], after a change of variable, satisfies the above equation. There are three algebraic solutions of PIII′ (8, 2α, 0, −1), 1 2 g(s, α, β) = s 3 , 2 1 2 1 1 g(s, α, β) = s 3 ∓ s 3 , 2 6
2.2
for α = 0.
for α = ±1.
Asymptotic expansions of the scaled Hankel determinant.
We assume solution of PIII′ (8, 2α, 0, −1) for s → 0+ has the power series exP∞that the 1 j ′ pansion j=0 aj s , with g(0) = 0, g (0) = 2α , and substitute this into (2.16), by some straightforward computations, one finds, g(s, α, β) =
1 1 3 9 − 6α2 3 s− 2 2 s2 + 3 2 s + s4 2α 2α (α − 1) 2α (α − 4)(α2 − 1) α4 (α2 − 1)2 (α2 − 4)(α2 − 9) 5(−36 + 11α2) + 5 2 s5 + O(s6 ), where α 6= Z. (2.19) 2α (α − 1)2 (α2 − 4)(α2 − 9)(α2 − 16)
For large and positive s , we assume that the solution of (2.16) has the following expansion P∞ − 3j . The first term of the expansion is s2/3 /2 . A straight forward computation j=−2 bj s gives, 1 2 α 1 α(α2 − 1) − 1 α2 (α2 − 1) − 2 α(α2 − 1) −1 g(s, α, β) = s 3 − s 3 + s 3+ s 3+ s 2 6 162 486 486 5 α2 (α2 − 1)(2α2 − 11) − 4 s 3 + O(s− 3 ). − 6561 10
(2.20)
Note that this solution becomes the algebraic solutions mentioned in the Remark 1, for α = 0 , and α = ±1 In the next Theorem we obtain asymptotic expressions of the scaled Hankel determinant for small s and large s . Theorem 4. Under the double scaling scheme, the asymptotic expansions of the scaled Hankel determinant generated by the Pollaczek-Jacobi type weight, has the following small and large s expansions: For small s , s s2 s3 3(2α2 − 3)s4 ∆(s, α, β) = exp − + 2 2 − 3 2 + 2α 8α (α − 1) 6α (α − 1)(α2 − 4) 16α4 (α2 − 1)2 (α2 − 4)(α2 − 9) (36 − 11α2 )s5 6 + O(s ) , (2.21) + 10α5 (α2 − 1)2 (α2 − 4)(α2 − 9)(α2 − 16) where α ∈ / Z. For large s ,
9 2 3α 1 1 − 6α2 α(1 − α2 ) − 1 α2 (1 − α2 ) − 2 3 3 ∆(s, α, β) = exp c − s + s + ln s + s 3+ s 3 8 2 36 18 216 α(1 − α2 ) −1 − 34 s + O(s ) , + (2.22) 486 where c = c(α) is an integration constant, independent of s . Proof. By ( 2.18 ) , we see that ln ∆(s, α, β) =
Z
0
s
(ξg ′ − g)2 4αξg − ξ 2 4g + α2 + − 4ξg 2 16ξg 2 4ξ
!
dξ.
For small s , with g(s, α, β) given by ( 2.19 ) the asymptotic expansion ∆(s, α, β) , ( 2.21 ) follows immediately. Similarly, for large s , with g given by ( 2.20 ) , the equation ( 2.22 ) is obtained, following straightforward computations. Note that the scaling limit of Hankel determinant via from the Pollaczek-Jacobi type weight is independent of β . In what follows, we give an account which will ultimately determine c(α). We start from, (−1)n Pn (0; t, α, β) =
Dn (t, α + 1, β) . Dn (t, α, β)
Corollary 1. Sending n → ∞, t → 0+ and such that s = 2n2 t, is finite, then (−1)n Pn (0; 2ns 2 , α, β) ∆(s, α + 1, β) 3 1 1 + 2α α(α + 1) − 1 3 − lim = = exp c + s ln s − s 3 1 n→∞ (−1)n Pn (0; 0, α, β) ∆(s, α, β) 2 6 6 4 α(α + 1)(2α + 1) − 2 α(α + 1) −1 − (2.23) s 3− s + O(s 3 ) , − 108 162 11
where c1 = c1 (α) is a constant, independent of s, and c1 (α) = c(α + 1) − c(α)
(2.24)
and c(α) is the constant in (2.22). Proof. From the fact (−1)n Pn (0; 2ns 2 , α, β) ∆(s, α + 1, β) Dn (s/(2n2 ), α + 1, β) Dn (0, α, β) = lim = , n→∞ n→∞ (−1)n Pn (0; 0, α, β) Dn (s/(2n2 ), α, β) Dn (0, α + 1, β) ∆(s, α, β) lim
the equation (2.23) and (2.24) follow from (2.22). A computation that produces the constant c1 (α) can be found in Section 2.4.
2.3
The Hankel determinant for large β .
In this subsection, we are interested in the behavior of the Hankel determinant for large β e s := 2(1 + β)n e 2 t, and introduce a different scaling process. Let n → ∞, t → 0+ , β := nβ, such that s and βe are fixed, we then obtain the same Painlev´e equations in the theorem 3; e e . We state these just replace f (t, α, β) of ( 2.15 ) in the theorem 3 with f (t, α, β)/(1 + β) results in the theorem below. Theorem 5. Let
e β := nβ,
e 2 t, s := 2(1 + β)n
and
y(t) := 1 +
f (t, α, β) , e 2 (1 + β)n
(2.25)
t → 0+ and n → ∞ such that s and βe are finite, s ∈ (0, ∞) and βe ∈ (−1, ∞). If ! s e := lim f g(s, α, β) , α, nβe , n→∞ e 2 2(1 + β)n
e satisfies the following PIII′ (8, 2α, 0, −1), then g(s, α, β) g ′′ =
α 1 g ′2 g ′ 2g 2 − + 2 + − , g s s 2s 4g
e = 0 , g ′(0, α, β) e = with initial conditions g(0, α, β) If e := lim Hn H(s, α, β) n→∞
e satisfies, then H(s, α, β)
1 2α
, s
e 2 2(1 + β)n
! , α, nβe ,
2 1 ′ = 0, (sH ) + 4 (H ) (sH − H) − αH + 2 ′′ 2
′ 2
′
12
(2.26)
(2.27)
e = 0 , H′ (0, α, β) e = − 1 . Moreover, if with initial conditions H(0, α, β) 2α e Dn 2(1+sβ)n , α, n β e 2 e := lim , ∆(s, α, β) n→∞ Dn 0, α, nβe then
e =s H(s, α, β)
2 ′ 2 2 d e = (sg − g) + 4αsg − s − g − α . ln ∆(s, α, β) ds 4g 2 16g 2 4
(2.28)
e is found to satisfy ( 2.26 ). Plugging Proof. Substituting ( 2.25 ) into the ( 2.13 ), g(s, α, β) 2 e e β = nβ and s = 2(1 + β)n t into the σ -from Painlev´e equation is satisfied by Hn (t, α, β) e satisfies ( 2.27 ) . Moreover, substituting ( 2.25 ) in the theorem 1, we see that H(s, α, β) into ( 2.14 ) , the equation ( 2.28 ) follows. Remark 2: Comparing theorem 3 with theorem 5, one finds that the Painlev´e equations are the same, although we emphasize that their scaling scheme are different from each other.
2.4
Large n behavior.
In order to find the constant c1 (α) in (2.23), we need to determine the large n behavior of the constant terms of the monic orthogonal polynomial, namely, Pn (0; t, α, β). The Szeg¨o limit theorem which computes the determinants of the finite section of Toeplitz matrix with nice symbols, can be adapted to orthogonal polynomials on the line. The large n computation for the orthogonal polynomials where the potential v satisfies the convexity condition [15], can be found in [15, 19] and also in [34]. In fact, as n → ∞ , Pn (z) is approximated by Pn (z) ∼ exp [−S1 (z) − S2 (z)] ,
(2.29)
valid for z ∈ / [a, b]. Here S1 (z) and S2 (z) are given by ( (4.6) and (4.7) in [19]). These formulas are " 1 1 # 1 z−b 4 z−a 4 exp(−S1 (z)) = + , z∈ / [a, b]. (2.30) 2 z−a z−b and √ 2 z−a+ z−b S2 (z) = −n ln 2 "p # Z b (z − a)(z − b) 1 v(x) p + + 1 dx, 2π a x−z (b − x)(x − a) √
13
z∈ / [a, b].
(2.31)
For the problem at hand, the Pollaczek-Jacobi type weight of (1.9), then v(x) and v′ (x) are given by, v(x) = − ln w(x) =
t − αln x − βln(1 − x), x
and v′ (x) = −
Substituting v(x) and v′ (x) into the following identities, Z b v′ (x) p dx = 0, (b − x)(x − a) a
t α β − − . x2 x x−1
(2.32)
and
Z
a
b
xv′ (x) p dx = 2πn. (b − x)(x − a)
(2.33)
These can be found, for instance, in [15, 19]. From (2.32), (2.33) and integral formulas in the Appendix A, it is found that a and b satisfy the following algebraic equations, α β t+ √ − p = 0, ab (1 − a)(1 − b) 2(ab) a+b
(2.34)
3 2
and
t β √ −p + 2n + α + β = 0. ab (1 − a)(1 − b)
(2.35)
Here the parameters a and b determines the end points of the support of the equilibrium density. √ Let X := 1/ ab, and eliminating a + b from (2.34) and (2.35), then X satisfies the quintic, X 3t β 2X 3t Xt − + αX − − (2n + α + β) = 0. 2 2 2(tX + 2n + α + β) 2 We now state a theorem which describes the large n asymptotic of Pn (0; t, α, β) , without displaying the detail steps involved, since these are quite straightforward. Theorem 6. If v(x) = − ln w(x) = xt − αln x − βln(1 − x) , x ∈ [0, 1] , t ≥ 0, α > 0, β > 0, the evaluation at z = 0 of S1 (z; t, α, β) , S2 (z; t, α, β) , and Pn (z; t, α, β) given by 1
1
exp [−S1 (0; t, α, β)] ∼ 2−1 (2n + α + β) 2 {2−1 t(2n + α + β)2 }− 6 ,
(2.36)
h 1 exp [−S2 (0; t, α, β)] ∼(−1)n 4−n (2n + α + β)α exp {2−1 t(2n + α + β)2 } 3 − (2α + β)ln 2 α 1 −1 2 13 −1 2 , (2.37) + {2 t(2n + α + β) } − ln 2 t(2n + α + β) 2 3 14
and Pn (0; t, α, β) ∼ exp [−S1 (0; t, α, β) − S2 (0; t, α, β)] 1 3 α n −n α+ 12 −(β+α+ 21 ) ∼ (−1) 4 n 2 exp {2−1 t(2n + α + β)2 } 3 − ln {2−1 t(2n + α + β)2 } 2 3 1 − ln {2−1 t(2n + α + β)2 } . (2.38) 6 The above asymptotic estimations are uniform with respect to t ∈ (0, t0 ], 0 < t0 < ∞, α > 0, β > 0, n → ∞ such that n2 t is fixed.
In order to derive the constant c1 (α) in ( 2.23 ) , we still need to obtain Pn (0; 0, α, β) , which is the constant terms of monic polynomials orthogonal with respect to the ”shifted” Jacobi weight w(x; 0, α, β) = xα (1 − x)β , x ∈ [0, 1]. This can be found from the monic polynomial orthogonal with respect to Jacobi weight w(x) = (1 − x)α (1 + x)β , x ∈ [−1, 1]. Taking a result from [17], we find that, 1 √ 1 (−1)n 4−n nα+ 2 2−(α+β+ 2 ) 2π n Γ(n + 1 + α)Γ(n + 1 + α + β) ∼ , Pn (0; 0, α, β) = (−1) Γ(α + 1)Γ(2n + 1 + α + β) Γ(α + 1) 1 √ 1 Γ(n + 1 + β)Γ(n + 1 + α + β) 4−n nβ+ 2 2−(α+β+ 2 ) 2π Pn (1; 0, α, β) = ∼ , Γ(β + 1)Γ(2n + 1 + α + β) Γ(β + 1) where use has been made of the asymptotic formula √ 1 Γ(n + 1 + α) ∼ 2πnn+α+ 2 e−n as, n → ∞. Remark 3: To derive the constant c1 (α) in (2.23), we recall the asymptotic estimation of Pn (0; t, α, β) , (2.38), as
Pn (0; t, α, β) ∼ exp [−S1 (0; t, α, β) − S2 (0; t, α, β)] 1 √ 1 1 (−1)n 4−n nα+ 2 2−(α+β+ 2 ) 2π Γ(α + 1) 3 ∼ · √ exp {2−1t(2n + α + β)2 } 3 Γ(α + 1) 2 2π α 1 −1 2 −1 2 − ln {2 t(2n + α + β) } − ln {2 t(2n + α + β) } 3 6 √ n −n α+ 21 −(α+β+ 12 ) 1 2π Γ(α + 1) 3 −1 (−1) 4 n 2 exp ln √ + {2 t(2n + α + β)2 } 3 ∼ Γ(α + 1) 2 2π α 1 −1 2 −1 2 − ln {2 t(2n + α + β) } − ln {2 t(2n + α + β) } , (2.39) 3 6
Hence,
1 3 −1 α Pn (0; t, α, β) ∼exp {2 t(2n + α + β)2 } 3 − ln {2−1 t(2n + α + β)2 } Pn (0; 0, α, β) 2 3 1 −1 2 − ln {2 t(2n + α + β) } + c1 (α) , 6 15
from which c1 (α) is found to be ln
Γ(α + 1) √ . 2π
Replacing 2−1 t(2n + α + β)2 by s , then (2.23) follows. By (2.24) one finds Γ(α + 1) , c(α + 1) − c(α) = c1 (α) = ln √ 2π giving G(α + 1) c(α) = ln , α (2π) 2 which we recognize to be the Tracy-Widom constant that appeared in the Bessel kernel problem. Here G(z) is the Barnes-G function.
2.5
An algorithm of the Hankel determinant for small s .
In this subsection, we use the formulas derived by Normand in [40] and similar process in [27] to compute the Hankel determinant. We look for expansion of the Hankel determinant around t = 0 . We scale the variable t as s = 2n2 t in the expansion and let n → ∞ , in this way we verify the asymptotic expansion for small s derived in section 2. Recall that n−1 Dn (t, α, β) = det [µi+j (t)]i,j=0 ,
where the moments µi+j (t) are given by Z 1 t µi+j (t) = xi+j xα (1 − x)β e− x dx,
i, j = 0, 1, . . .
0
Although the moments µi+j (t) depending on α and β , we do not display this to lighten notations. Lemma 2. The moments µm (t) are, Z 1 t µm (t) = xm xα (1 − x)β e− x dx = ϕm (t) + tm+α+1 ψm (t), 0
m ∈ N,
where ϕm (t) and ψm (t) are analytic at t = 0 , ϕm (t) =
e−t Γ(β + 1)Γ(m + α + 1) 1 F1 (1 + β, −m − α, t), Γ(m + 2 + α + β)
and, ψm (t) = e−t Γ(−m − α − 1) 1 F1 (m + 2 + α + β, m + 2 + α, t), 16
(2.40)
where 1 F1 (a, b; z) is the confluent hypergeometric function of the first kind. Futrthermore, µm (t) has series expansion around t = 0 ,
where
t2 µm (t) = ϕm (0) + tϕ′m (0) + ϕ′′m (0) + O(t3 ) + tm+α+λ+1 ψm (0)(1 + O(t2 )) 2! 2(m+α+λ+1) +O t , ϕm (0) =
Γ(β + 1)Γ(m + α + 1) , Γ(m + 2 + α + β)
ϕ′′m (0) =
ϕ′m (0) = −
Γ(β + 1)Γ(m + α) , Γ(m + 1 + α + β)
Γ(β + 1)Γ(m + α − 1) , Γ(m + α + β)
(2.41)
(2.42)
(2.43)
and ψm (0) = Γ(−m − α − 1). The expansions are valid for α > 0, β > 0, m ∈ N and | arg t| < π. Proof. From ([29], P367), we find Z 1 m+α t m + 2 + α + 2β m + 1 + α − m α β − xt , ,t , µm (t) = x x (1 − x) e dx = t 2 e 2 Γ(β + 1)W − 2 2 0 where W denotes the Whittaker function W (λ, µ, z) =
1 z 1 Γ(−2µ) z µ+ 2 e− 2 1 F1 (µ − λ + , 2µ + 1, z) 2 − µ − λ) 1 z 1 Γ(2µ) z −µ+ 2 e− 2 1 F1 (−µ − λ + , −2µ + 1, z), + 1 2 Γ( 2 + µ − λ)
Γ( 21
subject to | arg z| < π ([29], P1023). Hence ( 2.40 ) and ( 2.41 ) follow. Theorem 7. The series expansion of the Hankel determinant is given by, n(n + α + β)t n−1 Dn (t, α, β) = det [µk+j (t)]k,j=0 = Dn (0, α, β) 1 − α 2 n(n + α + β)(n(n + α + β)α + β)t + + O t3 2 2α(α − 1)
# Γ(n + α + 1)Γ(n + 1 + α + β)πtα+1 (1 + O(t)) + O t2(α+1) , + Qn−1 (β + j) Γ(β + 1)Γ(α + 1)Γ2 (α + 2)sin(απ)(n − 1)! j=1 (2.44) 17
subject to α ∈ / Z, α > 0 and | arg t| < π. Dn (0, α, β) has a closed form expression [4], Γ( α+β+1 )G2 ( α+β+1 )G2 ( α+β + 1) 2 2 2 G(α + β + 1)G(α + 1)G(β + 1) G(n + 1)G(n + α + 1)G(n + β + 1)G(n + α + β + 1) , (2.45) × 2 (n + 1 + α+β )Γ(n + α+β+1 ) G2 (n + α+β+1 )G 2 2 2
Dn (0, α, β) = 4−n(n+α+β) (2π)n
where G(z) is the Barnes G -function. Proof. By ( 2.41 ) , Dn (t, α, β)
n−1 ϕ′′k+j (0) k+j+α+1 ∼ det ϕk+j (0) + +t +t ψk+j (0) 2! k,j=0 n−1 n−1 ′ ∼ det [ϕk+j (0)]k,j=0 + t < t > det ϕk+j (0) + tϕk+j (0) k,j=0 ϕ′′ (0) n−1 ′ 2 k+j 2 2 + t < t > det ϕk+j (0) + tϕk+j (0) + t 2! k,j=0 n−1 α+1 α+1 k+j+α+1 +t det ϕk+j (0) + t ψk+j (0) k,j=0 . (2.46)
n−1 =det [µk+j (t)]k,j=0
tϕ′k+j (0)
2
Here < tm > f (t) denotes the coefficient of tm in the series expansion of f (t) in t. We note here Normand’s formula n−1 n−1 Y Y (b + (n − 1 − j)(1 − a))j Γ(zj ) Γ(zj + i) det (zj − zi ), = Γ(azj + b + i) i,j=0 j=0 Γ(azj + b + n − 1) 0≤i det ϕk+j (0) + tϕk+j (0) k,j=0 = < t > det Γ(2 + α + β + j + k) n−1 Γ(β + 1)Γ(α + j + k) . −t Γ(1 + α + β + j + k) k,j=0 With ( 2.47 ) , we find, n−1 Γ(β + 1)Γ(α + 1 + j + k) n(n + α + β)t Γ(β + 1)Γ(α + j + k) < t > det =− −t Dn (0, α, β), Γ(2 + α + β + j + k) Γ(1 + α + β + j + k) k,j=0 α The terms < t2 > f (t) and < tα+1 can be similarly derived, although with greater effort. 18
Corollary 2. From the series expansion of Dn (t, α, β) around t = 0 , see ( 2.44 ) , the series expansion of Hn (t, α, β) around t = 0 follows, Dn (t, α, β) n(n + α + β) n(n + α + β)(n(n + α + β) + αβ) 2 d ln =− t+ t dt Dn (0, α, β) α α2 (α2 − 1) Γ(n + α + 1)Γ(n + 1 + α + β)π 2(α+1) + O(t3 ) + tα+1 (1 + O(t)) + O t , (α + 1)2 Γ3 (α + 1)sin(απ)Γ(n)Γ(n + β) (2.48)
Hn (t, α, β) = t
subject to α ∈ / Z, α > 0, t > 0 and | arg t| < π.
Proof. With Hn (t, α, β) given by ( 2.10 ) , the series expansion of Dn (t, α, β) , equation ( 2.44 ) , and the identity n−1 Γ(n + β) Y = (β + j), Γ(β + 1) j=1 the equation ( 2.48 ) is obtained.
Corollary 3. Sending t → 0+ , n → ∞, and s = 2n2 t such that s ∈ (0, ∞) is finite, we have, Dn ( 2ns 2 , α, β) d s s2 d = s ln ∆(s, α, β) = − + 2 2 + O(s3 ) H(s, α, β) = lim s ln n→∞ ds Dn (0, α, β) ds 2α 4α (α − 1) πsα+1 + α+1 (1 + O(s)) + O(s2(α+1) ), (2.49) 2 (α + 1)2 Γ3 (α + 1)sin(απ) subject to α ∈ / Z, α > 0 and | arg s| < π.
Proof. Substituting s = 2n2 t into ( 2.48 ), and taking limit n → ∞ , then ( 2.49 ) is obtained with πsα+1 Γ(n + α + 1)Γ(n + 1 + α + β)π sα+1 = , lim α+1 2(α+1) n→∞ 2 n (α + 1)2 Γ3 (α + 1)sin(απ)Γ(n)Γ(n + β) 2α+1 (α + 1)2 Γ3 (α + 1)sin(απ) where α ∈ / Z, α > 0 and Γ(n + α + 1) ∼ Γ(n + 1)nα , as n → ∞, is used in the proof. Remark. Note that the small s expansion of H(s, α, β) is independent of β and coincides with the characterization of Painlev´e equation in the Theorem 3. In particular, the “first portion” of ( 2.49 ) is the same as the series expansion, d (sg ′ − g)2 4αsg − s2 α2 H(s, α, β) = s ln ∆(s, α, β) = + − g − ds 4g 2 16g 2 4 1 1 1 s2 − 3 2 s3 + O(s4 ), =− s+ 2 2 2α 4α (α − 1) 2α (α − 4)(α2 − 1)
where the small expansion s of g(s, α, β) is given by ( 2.19 ) . We note that the “special function portion” of ( 2.49 ) , cannot be obtained through power series. This method, proposed by Normand, capture both “special functions” and “power series”. T complexity in the computations increases very rapidly, as one attempt to include term with higher powers of s and should be a worthy future project to explore. 19
3
Double scaling analysis II.
In this section, we propose another double scaling scheme by setting α = −2n + λ, s = nt, as n → ∞, t → 0 such that λ and s are finite, the scaled determinant has an integral representation in terms of a particular PV , which can be reduced to a PIII , from which its expansions for small and large s can be found.
3.1
The Painlev´ e equations.
For convenience, we introduce a number of items; d ∆n (t, α, β) d Dn (t, α, β) b n (t, α, β) := Hn (t, α, β) + n2 − nλ, ln := t ln + n2 − nλ, and H dt ∆n (0, α, β) dt Dn (0, α, β) (3.50) where d Dn (t, α, β) Hn (t, α, β) := t ln . dt Dn (0, α, β)
t
We proceed as follows. Substituting α = −2n + λ into (3.50), we find, ∆n (t, λ, β) =
tn
2 −nλ
n!
Dn (t, λ, β) =
With the change of variable z =
and
n b n (t, λ, β) = (−1) D n!
Z
tn
2 −nλ
n! (1−x)t , x
Z
Y
2
(0,1)n 1≤j 0, γ > 0, γ − λ − β > −1, x ∈ (0, ∞), (3.52)
which the reduces to the one above (3.52), if γ = β and degenerates to the Laguerre weight at t = 0. 20
We introduce two quantities in term of the monic polynomials Pn (x) orthogonal with respect to (3.52), over [0, ∞). These are Z ∞ 2 γ Pn (y) b Rn (t) := w(y; t)dy, (3.53) hn 0 y Z ∞ Pn−1 (y)Pn (y) γ w(y; t)dy. (3.54) rbn (t) := hn−1 0 y
bn satisfy the following Riccati equations, Moreover, we find rbn and R
t
d rn + 2n − λ − β) + n(n − λ − β) rbn (b rn − γ) rbn (b − , rbn = (2n − λ − β + γ) rbn +n(n−λ−β)+ bn bn dt R 1−R (3.55) d b bn − 1) R bn . t R rn − γ + 2n + γ − β − λ + t(R (3.56) n = 2b dt Let n−1 X b n (t, γ, λ, β) d D b −t Rj , (3.57) Hn (t, γ, λ, β) := t ln b n (0, γ, λ, β) dt D j=0
where the derivation of −t here. Note that,
Pn−1 j=0
Rj (t), obtained through ladder operators, is not re-produce
(−1)n ∆n (0, γ, λ, β) = and a simply identity
b ′ (t). rbn (t) = H n
Theorem 8. For finite n , let
then yb(t, γ, λ, β) satisfies
bn (t) =: 1 + R
(3b y − 1) (b y ′ )2 yb′ (b y − 1)2 yb = − + 2b y (b y − 1) t 2t2 ′′
with the initial conditions
yb(0, γ, λ, β) =
G(n + 1)G(n + 1 + γ − β − λ) . G(γ − β − λ)
γ , β+λ
(3.58)
1 , yb(t, γ, λ, β) − 1
y + 1) γ2 (2n + 1 + γ − β − λ) yb yb (b − , (3.59) (β + λ) yb − − yb t 2(b y − 1) 2
yb′ (0, γ, λ, β) =
γ(2n + 1 + γ − β − λ) . (β + λ)((β + λ − γ)2 − 1)
b n (t, γ, λ, β) defined in ( 3.57 ), and satisfies the following second order ODE, The quantity H 2 2 b ′′ = 4 n + H b′ b n − tH b′ b ′ − γ + (t − γ + β + λ)H b′ − H b n − nγ , (3.60) tH H H n n n n n 21
with the initial conditions
Moreover, b n (t, γ, λ, β) := H
b n (0, γ, λ, β) = 0, H
b ′ (0, γ, λ, β) = H n
nγ . β+λ−γ
((β + λ)b y − γ)2 t(β + λ − 2n)b y γt t2 yb (tb y ′ )2 − + − − . 4b y(b y − 1)2 4b y 2(b y − 1) 2(b y − 1) 4(b y − 1)2 (3.61)
bn (t) = 1 + 1/(b Proof. Eliminating rbn from ( 3.55 ) and ( 3.56 ) , substituting R y(t) − 1) into the resulting equation, we see that yb(t) satisfies ( 3.59 ). Eliminating rbn from ( 3.58 ) and ( 3.55 ), we obtain the following b n = F (H bn, H b′ , H b ′′), R n n
where F (·, ·, ·) is a function of three variables, which we do not display. Combining the expression for Rn , ( 3.57 ), ( 3.58 ) , and n−1 X
b rn ) − rbn (t + λ + β − γ − 2n) − nγ rbn (b rn − γ) bj = Rn (n(n − λ) + (γ − β)n + tb R + , b b bn ) t(1 − Rn ) tRn (1 − R j=0 (3.62) b we arrive at the non-linear second order ODE which is satisfied by Hn (t, γ, λ, β). Finally, combining ( 3.62 ) and ( 3.57 ) , we see that,
b rn ) − rbn (t + λ + β − γ − 2n) − nγ rbn (b rn − γ) b n (t, γ, λ, β) = − Rn (n(n − λ) + (γ − β)n + tb H − . bn ) bn (1 − R bn ) (1 − R R
bn (t) = 1 + 1/(b The equation ( 3.61 ) is found, with ( 3.56 ) and R y(t) − 1) .
b n (t, γ, λ, β) → H b n (t, γ, λ, β) − nt − n(β + λ) , then the non-linear second Remark 4: If H ODE (3.60) becomes, 2 2 b ′′ = H b n − tH b′ + 2 H b ′ 2 − δn H b ′ − 4H b′ H b′ − n H b′ − β − λ H b′ − n − γ , tH n n n n n n n n
where δn = 2n + γ + β + λ. The equation above is the Jimbo-Miwa-Okamoto [36, 41] σ -form of PV , with parameters v0 = 0,
v1 = −n,
v2 = −β − λ,
v3 = −n − γ.
Similar to the investigation in previous sections, let, s ∆n ns , γ, λ, β b b H (s, γ) := lim Hn , γ, λ, β , ∆(s, γ) = lim , n→∞ n→∞ ∆n (0, γ, λ, β) n we have, 22
g (s, γ) := lim yb b n→∞
s
, γ, λ, β ,
n (3.63)
Theorem 9. Let s := nt, t → 0+ , n → ∞, such that s is finite. Then b g (s, γ) satisfies gb′′ =
3b g−1 gb′ (β + λ)2 b g (b g − 1)2 γ 2 (b g − 1)2 2b g 2 (b g ′) − + − − , 2b g (b g − 1) s 2s2 2s2 b g s
(3.64)
with the initial data
g (0, γ) = b
γ , β+λ
gb′ (0, γ) =
2γ . (β + λ)((β + λ − γ)2 − 1)
b γ) satisfies a non-linear second order ODE, The quantity H(s, 2 2 b′ − γ , b ′′ = 4H b − sH b ′ + (−γ + β + λ) H b′ H b′ 1 + H sH
(3.65)
with the initial conditions
Moreover,
b γ) = 0, H(0, b= H
b ′ (0, γ) = H
γ . β+λ−γ
sb y ((β + λ)b y − γ)2 (sb y ′ )2 − . − 2 4b y yb − 1 4b y (b y − 1)
(3.66)
The equation (3.64) is satisfied by PV ((β + λ)2 /2, −γ 2 /2, −2, 0)).
Proof. Inserting s = nt into ( 3.59 ), ( 3.60 ), ( 3.61 ) and followed by the double scaling limit, the equations ( 3.64 ), ( 3.65 ) and ( 3.66 ) are found. b γ) → H(s, b γ) − s, then H b ′ (s, γ) → H b ′ (s, γ) − 1 and the non-linear Remark 5: If H(s, second order ODE of (3.65) becomes 2 2 b ′′ = 4H b′ H b − sH b′ H b ′ − 1 + (−γ + β + λ) H b′ − β − λ , sH
This turns out to be a σ -form of PIII . b n (t, λ, β) = Hn (t, λ, β)+n2 −nλ and s = nt into the σ -from Remark 6: Substituting H b (s, λ, β) satisfies another σ -form of PIII , Painlev´e equation ( 2.11 ), we see that H
3.2
b ′′ )2 = 4H b ′ (1 + H b ′ )(H b − sH b ′ ) + (λH b ′ − β)2 . (sH
(3.67)
The asymptotic Expansions of PV ((β + λ)2/2, −γ 2/2, −2, 0)
For small and positive s , we find, gb(s) =
2γ γ + s β + λ (β + λ)((β + λ − γ)2 − 1) ((β + λ)2 − 1)(3(β + λ) − 5γ) + γ 2 (β + λ + γ) s2 + O(s3 ), + 2 2 (β + λ)(β + λ − γ)((β + λ − γ) − 1)((β + λ − γ) − 4) 23
(3.68)
where β + λ − γ ∈ / Z. The solution of PV ((β + λ)2 /2, −γ 2 /2, −2, 0) for large s . For large and positive s, we find, 1 2 4γ 2 − 1 − 1 1 − 4γ 2 −1 (4γ 2 − 1)(48(β + λ)2 − 4γ 2 + 25) − 3 s2 + 1 + s 2+ s + s 2 β+λ 16(β + λ) 16 1024(β + λ) 5 (4γ 2 − 1)(4γ 2 − 8(β + λ)2 − 17) −2 (3.69) + s + O(s− 2 ). 256 Theorem 10. The scaled Hankel determinant generated by the Pollaczek-Jacobi type weigh has the following small and large s expansions: For small s , γ γ(β + λ) b γ) = exp ∆(s, s+ s2 β+λ−γ 2(β + λ − γ)2 ((β + λ − γ)2 − 1) 2γ(β + λ)(β + λ + γ) 3 4 s + O(s ) , (3.70) + 3(β + λ − γ)3 ((β + λ − γ)2 − 4)((β + λ − γ)2 − 1)
g (s) = b
where β + λ − γ ∈ / Z. For large s , 2 b γ) = exp c2 − s − 2(β + λ)s 21 + (β + λ)(β + λ − 2γ) ln s + (4γ − 1)(β + λ) s− 12 ∆(s, 4 16 2 2 3 (4γ − 1)(β + λ) −1 − s + O(s− 2 ) , (3.71) 64 where c2 = c2 (γ) is an integration constant independent of s . Proof. From ( 3.57 ), ( 3.63 ) and ( 3.66 ) , we obtain, ! Z s 2 ′ 2 ((β + λ)b g (ξ) − γ) (ξb g (ξ)) ξb g (ξ) dξ b γ) = ln ∆(s, − . 2 − 4b g (ξ) gb(ξ) − 1 ξ 4b g (ξ) (b g (ξ) − 1) 0
Substituting (3.68) and (3.69) into the above formula, the equations (3.70) and (3.71) are found. At the end of this section, we evaluate the large n behavior of Pn (0; t, γ). In fact, (−1)n Pn (0; t, γ) =
Dn (t, γ + 1) . Dn (t, γ)
For the weight, w(x; 0) = xγ−(λ+β) e−x ,
γ − λ − β > −1,
x ∈ (0, ∞),
the constant term has a closed form expression in terms of Gamma functions, Γ(n + 1 + γ − λ − β) √ ∼ 2π nn+γ−α−β+1/2 e−n , n → ∞. (−1)n Pn (0; 0, γ) = Γ(1 + γ − λ − β) 24
(3.72)
Corollary 4. With the double scaling, one finds, b γ + 1) (−1)n Pn (0; t, γ) β+λ ∆(s, (1 + 2γ)(β + λ) − 1 lim = exp c3 (γ) + = ln s + s 2 n n→∞ (−1) Pn (0; 0, γ) b γ) 2 4 ∆(s, (1 + 2γ)(β + λ)2 −1 − 32 (3.73) − s + O(s ) 16 where c3 (γ) is a constant independent of s. Moreover, c3 (γ) = c2 (γ + 1) − c2 (γ),
(3.74)
where c2 (γ) is the constant in ( 3.71 ). Proof. From the fact that b n (0, γ) b γ + 1) b n (s/n, γ + 1) D (−1)n Pn (0; ns , γ) ∆(s, D = , = lim n n→∞ (−1) Pn (0; 0, γ) n→∞ b n (s/n, γ) D b n (0, γ + 1) b γ) D ∆(s, lim
(3.73) and (3.74) follow.
3.3
Large n behavior of Pn (0; t, γ) .
In this subsection, we obtain the constant, c3 (γ), following similar steps in section 2.4. We substitute v(x, t) and v′ (x, t) into ( 2.32 ) and ( 2.33 ). With the aid of the integral formulas in the Appendix A, we find a and b satisfy two algebraic equations, p
λ+β
γ − √ + 1 = 0, ab (t + a)(t + b)
λ+β−γ+
a+b t(λ + β) = 2n. −p 2 (t + a)(t + b)
(3.75) (3.76)
The parameters a and b , are, as before, the end points of the support of the equilibrium density. √ b satisfies a b := ab , eliminating a + b from the above equations, we see that X Let X quintic equation, 2 tγ 2 2 b b −X b 2 (λ + β)2 = 0. X + 2t(2n + γ − λ − β + ) − t γ−X (3.77) b X b asymptotic of X, b given by, Note the large n of X
2 2 2 b ∼ γ − γ(β + λ) (nt)− 12 + γ(β + λ) (nt)−1 + γ(β + λ)(γ − 2(β + λ) ) (nt)− 23 . X 2 4 16
(3.78)
with t ∈ (0, t1 ], 0 < t1 < ∞, n → ∞ such that nt is fixed. We now give the large n behavior of Pn (0; t, γ), and place this in the Theorem below, without displaying the detail steps. 25
Theorem 11. If v(x) = − ln w(x; t) = (λ + β) ln(x + t) − γ ln x + x + t, and λ < 0, β > 0, γ > 0, t > 0, γ − λ − β > −1, x ∈ (0, ∞), the evaluation of the orthogonal polynomials at the x = 0, reads, λ+β 1 n n+γ−α−β+ 21 −n ln γ + γ + ln(nt) (−1) Pn (0; t, γ) ∼ n e exp − γ + 2 2 (1 + 2γ)(β + λ) (1 + 2γ)(β + λ)2 − 21 −1 , (nt) − (nt) + 4 16 (3.79) the asymptotic estimation is uniform with respect to t ∈ (0, t1 ], 0 < t1 < ∞, λ < 0, β > 0, γ > 0, γ − λ − β > −1, n → ∞ such that nt is fixed. Remark 7: To obtain the constant c3 (γ) in (3.73), we rewrite the asymptotic estimation of (−1)n Pn (0; t, γ), the equation (3.79), as λ+β 1 n n+γ−α−β+ 12 −n ln γ + γ + (−1) Pn (0; t, γ) ∼n ln(nt) e exp − γ + 2 2 (1 + 2γ)(β + λ) (1 + 2γ)(β + λ)2 − 21 −1 + (nt) − (nt) 4 16 Taking into account the extreme right of (3.72), we see that, (1 + 2γ)(β + λ)2 λ+β (1 + 2γ)(β + λ) Pn (0; t, γ) − 21 −1 , ∼ exp c3 + ln(nt) + (nt) − (nt) Pn (0; 0, γ) 2 4 16 in accordance to (3.73). From these facts c3 is identify to be Γ(1 + γ − λ − β) 1 1 ln − γ− ln γ + γ − ln (2π). γ 2 2 Consequently, the equation (3.74) becomes, 1 1 Γ(1 + γ − λ − β) − γ− ln γ − γ + ln (2π) . c2 (γ + 1) − c2 (γ) = c3 (γ) = ln γ 2 2 We may identify c2 (γ) = ln
G (1 + γ − λ − β) Γ(γ)G(γ)
+ b(γ),
γ>0
up to another constant b(γ), which satisfies, −γt Z ∞ e 1 1 1 − + t dt. b(γ + 1) − b(γ) = 2 t e −1 t 0 26
Binet’s formula, see ([49], P249), says that, −zt Z ∞ 1 1 e 1 1 1 ln Γ(z) = z − ln(z) − z + ln (2π) + − + t dt, 2 2 2 t e −1 t 0
Re z > 0.
Remark 8: As γ → ∞, b(γ + 1) − b(γ) =
1 1 − + O(1/γ 5 ). 12 γ 360γ 3
We also note that, b(γ) = κ −
1 1 ln γ + + +O(1/γ 4 ). 2 720γ 2
(3.80)
It may be of interest to determine the constant κ .
4
Limiting behavior of the Kernel with the PollaczekJacobi type weight.
The ladder operator formalism for orthogonal polynomials and the associated compatibility condition can be found, for example, in [14]. This is a handy tool regarding orthogonal polynomials on the unit circle [3], construction of Jacobi polynomials [17] and has been adapted to obtain the Painlev´e equations, arising the deformation of classical weight. We refer the Reader to [3, 10, 11, 21, 22, 28]. In particular, this r formalism has been applied to specific orthogonal polynomial ensemble in [18, 28], and its comparison with the isomonodromy approach[37]. For the extension of ladder operators to discrete orthogonal polynomials, and allied discrete Painlev´e equations; to q -orthogonal polynomials and the allied q -Painlev´e equations, we refer the Reader to [6, 8, 7, 16, 31, 32, 12]. In this section, we construct the Lax pair for our problem. First, we recall the ladder operator relations of (2.7) and (2.8) in [10], satisfied by the polynomials Pn (x; t, α, β) orthogonal with respect to the Plooaczek-Jacobi type weight, d + Bn (x; t, α, β) Pn (x; t, α, β) = βn An (x; t, α, β)Pn−1(x; t, α, β), (4.81) dx d ′ − Bn (x; t, α, β) − v (x) Pn−1 (x; t, α, β) = −An−1 (x; t, α, β)Pn (x; t, α, β), (4.82) dx The “coefficients”, An (x; t, α, β) and Bn (x; t, α, β) are given by (2.14), (2.15) in [10], respectively; An (x; t, α, β) =
Rn Rn Rn∗ + − , 2 x x x−1
Bn (x; t, α, β) =
rn∗ n − rn rn − − , 2 x x x−1
and Rn∗ , Rn , rn∗ , rn are defined by (2.16) - (2.19) in [10], all depending on t, α, β. 27
Theorem 12. The monic polynomials Pn (x; t, α, β) orthogonal with respect to the PollaczekJacobi type weight, satisfy the ladder operator relations in t , d ∗ xt − rn Pn (x; t, α, β) = βn Rn∗ Pn−1 (x; t, α, β), (4.83) dt d ∗ ∗ ∗ Pn (x; t, α, β). (4.84) xt + rn − t + xRn−1 Pn−1 (x; t, α, β) = Rn−1 dt By the Christoffel–Darboux formula [43], the reproducing kernel reads, s hn φn (x)φn−1 (y) − φn (y)φn−1(x) Kn (x, y) := , hn−1 x−y
(4.85)
where hn is the square of the L2 norm is given by (1.3) and βn = hn /hn−1 . Here, φn (x) and φn−1 (x) are defined in terms of Pn (x; t, α, β) by φn (x) := φn−1 (x) := Let Ψ ,
β t Pn (x; t, α, β) α Pn (x; t, α, β) 1 √ √ w 2 (x; t, α, β) = x 2 (1 − x) 2 e− 2x , hn hn
β t Pn−1 (x; t, α, β) 1 Pn−1 (x; t, α, β) α p p w 2 (x; t, α, β) = x 2 (1 − x) 2 e− 2x . hn−1 hn−1
Ψ=
φn (x) = φn−1 (x)
We find, Ψx = and
Pn (x;t,α,β) α √ x 2 (1 hn Pn−1 (x;t,α,β) α
√
hn−1
β
t
− x) 2 e− 2x
β t x 2 (1 − x) 2 e− 2x .
C1 C2 C3 + + 2 x−1 x x
Ψ,
(4.86)
√ √ −Rn βn Rn βn rn + β2 n − rn √ + α2 √ , C1 = , C2 = −Rn−1 βn −n + rn − α2 Rn−1 βn −rn − β2 t √ ∗ ∗ − r R βn n n 2 √ . C3 = ∗ −Rn−1 βn − 2t + rn∗
(4.87) (4.88)
Similarly, the ladder operator relationships in t, see (4.83) and (4.84) can be re-written as D1 Ψ t = D0 + Ψ, (4.89) x and D0 =
"
R∗n 2t
0
0
−
R∗n−1 2t
#
,
D1 = 28
"
∗ rn −1 t √2 ∗ Rn−1 βn t
∗
√
− Rn t βn 1 2
−
∗ rn t
#
.
(4.90)
The equations (4.86) and (4.89) give the Lax pair mentioned earlier. The compatibility condition of the Lax pair reads Ψxt = Ψtx , which, when un-pact, re-produces the differential differences obtained in [10]. Writing out (4.86) in component form, followed by eliminating φn−1 (x) in favor of φn (x), we find, 2 x3 − x2 φ′′n (x) + 2 3x2 − 2x φ′n (x) − (2(2n + α + β)x + 2(rn − rn∗ ) + t − 2n − α) φn 2(Rn∗ − Rn )x2 (x − 1) ′ Rn∗ − Rn − φ (x) + (2n + α + β)x2 + (2(rn − rn∗ ) n x(Rn∗ − Rn ) − Rn∗ x(Rn∗ − Rn ) − Rn∗ 1 +t − 2n − α) x + 2rn∗ − t) φn − 2 [(x(2(rn − rn∗ ) + t − 2n − α) + 2rn∗ − t 2x (x − 1) i 2 2 ∗ ∗ ∗ ∗ +x (2n + α + β) − 4 (x(Rn − Rn ) − Rn ) x(Rn−1 − Rn−1 ) − Rn−1 βn φn = 0, (4.91)
∗ where rn , rn∗ , Rn∗ , Rn , Rn−1 , Rn−1 may be expressed by Hn , Hn′ (t) and Hn ”(t), these are (2.56), (2.32), (2.33), (3.7), (5.1), (5.2), (5.8) in [10]. For convenience, we list these relations in Appendix C. We give in the next theorem, the scaled kernel.
Theorem 13. Let 2
ζ := 4n x,
∗
2
ζ := 4n y,
2
s := 2n t,
φ(ζ) := lim φn n→∞
ζ 4n2
,
(4.92)
where n → ∞, t → 0+ , ζ and ζ ∗ in compact subsets of (0, ∞), and s finite, then, 1 ζ ζ∗ lim Kn , n→∞ 2n 4n2 4n2 (ζ − 4sH′ )φ(ζ)ζ ∗2 φ′ (ζ ∗) − (ζ ∗ − 4sH′ )φ(ζ ∗)ζ 2 φ′ (ζ) − 2s2 H′′ (ζ − ζ ∗ )φ(ζ)φ(ζ ∗) = , (4.93) (ζ − ζ ∗ )(ζ − 4sH′ )(ζ ∗ − 4sH′ ) Furthermore, φ(ζ) satisfies the ODE, ζ 4sH′ (s)ζ 2s2 H′′ (s) α2 s2 αs ′ 2 ′′ φ (ζ) + ζ φ (ζ) + ζ − + − − 2− − H φ(ζ) = 0, ζ − 4sH′ (s) 4 ζ − 4sH′ (s) 4 ζ ζ (4.94) with the boundary conditions of φ(0, α, β) = 0,
α > 0.
Proof. φn−1 (x) may be expressed in terms of φ′n (x) and φn (x) via one of the ladder operator relations, ((2n + α + β)x2 + (2(n + rn − rn∗ ) + t − α)x + 2rn∗ − t) φn − 2(x − 1)x2 φ′n (x) √ φn−1 (x) = . 2 βn (xRn + (1 − x)Rn∗ ) 29
Substituting φn−1 , given above, ( 4.92 ) and the large n asymptotic of rn∗ , rn , Rn and Rn∗ presented in the Appendix C, into the kernel given by the equation (4.85), and let n → ∞, then the limiting kernel (4.93) is obtained. We first substitute ζ = 4n2 x and s = 2n2 t into (4.91), satisfied by φn (x). Recall rn , ∗ rn∗ , Rn∗ , Rn , Rn−1 , Rn−1 in [10], listed in the Appendix C. Now let n → ∞ , then φ(ζ) satisfies the second order ODE, namely, (4.94). We list here some properties of the limiting kernel. Let s = 0. The boundary condition, H(0, α, β) = 0 , implies the kernel reduces to 1 φ(ζ)ζ ∗φ′ (ζ ∗) − φ(ζ ∗)ζφ′(ζ) ζ ζ∗ lim = Kn , . n→∞ 2n 4n2 4n2 ζ − ζ∗ In which case, the equation (4.94) becomes the Bessel differential equation, with the boundary condition of φ(0) = 0 . Then the limiting kernel becomes the celebrated Bessel Kernel of Tracy and Widom [47]. For s > 0, we make the transformation, ζ → 4sH′ (s)ζ,
ζ ∗ → 4sH′ (s)ζ ∗,
φ(ζ) → ζ ρ(s) φ(ζ),
then the kernel (4.93) becomes as, 1 A(ζ)B(ζ ∗) − A(ζ ∗)B(ζ) ζ ζ∗ lim = Kn , , n→∞ 2n 4n2 4n2 ζ − ζ∗
(4.95)
where the functions A(ζ) and B(ζ) are given by
and
ζ ρ(s) φ(ζ), A(ζ) := p 2 sH′ (s)
ζ2 ζ ρ(s) B(ζ) := p φ′ (ζ), ′ 2 sH (s) ζ − 1
ρ(s) :=
sH”(s) . 2H′ (s)
The ODE (4.94) transforms to, b4 1 b3 b2 b1 2(1 + b0 ) ′ φ (ζ) + − 4 − 3 + 2 + φ(ζ) = 0, − φ”(ζ) + ζ ζ −1 ζ ζ ζ ζ where b4 =
1 (H′ )2 , 16
b3 =
α , 4H′
b2 =
(sH′′ )2 − 4HH′ 2 − α2 H′ 2 , 4H′ 2
The coefficients b1 , b2 , b3 , b4 satisfy the relation, b1 + b2 − b3 − b4 = 0, 30
b1 = sH′ ,
b0 =
sH′′ . 2H′
which is the σ -form of the Painlev´e equation found in the Theorem 3. 2 1 ′′ 2 ′ 2 ′ ′ (sH ) + 4 (H ) (sH − H) − αH + = 0. 2 The kernel (4.95) has continuity property, a feature of the integrable kernels, since it can be written as, P1 ∗ ζ ζ∗ 1 j=0 fj (ζ)gj (ζ ) Kn , = , lim n→∞ 2n 4n2 4n2 ζ − ζ∗
where
sH′′
fj (ζ) = (−1)
2−j
ζ 2H′ √ 2 sH′
ζ2 d ζ − 1 dζ
j
sH′′
φ(ζ),
ζ 2H′ gj (ζ ) = √ 2 sH′ ∗
ζ2 d ζ − 1 dζ
1−j
φ(ζ),
P1 and j ∈ {0, 1}, with the property j=0 fj (ζ)gj (ζ) = 0 . Using the continuity property, the limiting kernel (4.93) becomes, 1 ζ ζ lim Kn , n→∞ 2n 4n2 4n2 ζ 2 φ′ (ζ) − ζ 2φ(ζ)φ′′(ζ) − 2ζφ(ζ)φ′(ζ) ζ 2φ(ζ)φ′(ζ) − 2s2 H′′ (s)φ2 (ζ) + , = ζ − 4sH′ (s) (ζ − 4sH′ (s))2 2 ζ α4 s2 αs 2 ′ ζ (φ (ζ)) + 4 − 4 − ζ 2 − ζ − H φ2 (ζ) , s > 0, = ζ − 4sH′ 2 ζ 2 (φ′ (ζ))2 + 4ζ − α4 φ2 (ζ) , s = 0, = ζ √ √ √ Jα2 ( ζ) − Jα−1 ζ Jα+1 ζ , =λ 4 where Jα (z) is the Bessel function of order α , λ is a parameter. The first equality above, is found by applying L’Hospital rule to ( 4.93 ) , the second equality is achieved by eliminating φ′′ (ζ) with the aid of ( 4.94 ) and H(0, α, β) = 0 . The Bessel equation√satisfied by φ(ζ) with the boundary condition φ(0) = 0 has regular solution φ(ζ) = λ Jα ( ζ) with a parameter of λ.
5
Limiting Kernel with the perturbed Laguerre weight.
In this section, we make use of the method in the previous section, and apply to the kernel generated by the singularly perturbed Laguerre weight. It is interesting that the limiting kernel via from the perturbed Laguerre weight is the same with the limiting kernel that arises from the Pollaczek Jacobi type weight, although their scaling schemes are quite different. 31
First, the monic orthogonal polynomials Pn (x; t, α) with respect to the singular perturbed Laguerre weight, t
w(x; t, α) = xα e−x e− x ,
0 ≤ x < ∞,
α > 0,
t > 0,
satisfy the ladder operator relations have been derived in [18] , which we restate here, d + Bn (x) Pn (x; t, α) = βn An (x)Pn−1 (x; t, α), dx d ′ − Bn (x) − v (x) Pn−1 (x; t, α) = −An−1 (x)Pn (x; t, α), dx where v(x) = − ln w(x; t, α) . The coefficients An (x), Bn (x) are given by (2.7) and (2.8) in [18] , n rn 1 Rn An (x) = + 2 , Bn (x) = − + 2 . x x x x We use Rn (t), rn (t) and t instead of an (s) , bn (s) and s respectively in [18] ; these are defined by, Z ∞ Z ∞ t 1 2 1 t Rn (t) := Pn (y; t, α)w(y; t, α)dy, rn (t) := Pn (y; t, α)Pn−1(y; t, α)w(y)dy, hn 0 y hn−1 0 y depend on t and α , with the initial conditions Rn (0) = 0, rn (0) = 0 . The monic orthogonal polynomials Pn (x; t, α) satisfy ladder operator relations in t , see (5.56) and (5.57) in [18], βn Rn rn d Pn (x; t, α) = − − Pn−1 (x; t, α), dt xt xt d Rn−1 1 rn Rn−1 Pn−1 (x; t, α) = − + + Pn (x; t, α). dt x xt t xt The reproducing Kernel with respect to w(x; t, α) is given by, Kn (x, y) :=
hn hn−1
21
ϕn (x)ϕn−1 (y) − ϕn (y)ϕn−1(x) , x−y
where, Pn (x; t, α) α − x − t Pn (x; t, α) 1 √ √ w 2 (x) = x 2 e 2 2x , hn hn Pn−1 (x; t, α) α − x − t Pn−1 (x; t, α) 1 p w 2 (x) = x 2 e 2 2x , ϕn−1 (x) := p hn−1 hn−1 ϕn (x) :=
where hn is the square of the L2 norm and βn = hn /hn−1 . 32
(5.96)
Let Φ(x)
" Pn√(x;t,α) x α2 e− x2 − 2xt # ϕn (x) hn Φ := = Pn−1 . (x;t,α) α − x − t √ x 2 e 2 2x ϕn−1 (x)
hn−1
We find, Φx =
E1 E2 + 2 E0 + x x
∂ Φ, where Φx denotes ∂x √ 1 −1 0 n√ + α2 βn E0 = , E1 = , − βn −n − α2 2 0 1
The quantity Φ also satisfies,
where F0 =
Rn 2t
0
Φ,
E2 =
(5.97)
t 2
− r√n
−Rn−1 βn
F1 Φ, Φt = F0 + x 0 − Rn−1 2t
,
F1 =
"
rn −1 t √2 Rn−1 βn t
√ Rn βn . − 2t + rn (5.98)
√
− Rn t βn 1 − rtn 2
#
The equations (5.97) and (5.98) is the Lax pair of this problem. Un-packing the compatibility condition is satisfied by the Lax pair, we find the following set of scalar equations, t
d 2(rn2 − trn ) rn = + (2n + 1 + α)rn − nt, dt Rn t
d Rn = Rn2 + (2n + 1 + α)Rn − t + 2rn , dt
(5.99)
d Rn−1 = t − 2rn − (2n − 1 + α + Rn−1 )Rn−1 . (5.100) dt Note that the above two Riccati equations are the same with (3.10) and (3.11) in [18], the third equation in the above also can be derived by (2.16) and (3.10) in [18]. We write ( 5.97 ) as a set of scalar equations, 1 1 2n + α t − 2rn 1 Rn ′ 2 ϕn (x) = − + ϕn−1 (x), (5.101) + ϕ (x) + + β n n 2 2x 2x2 x x2 1 1 Rn−1 1 2n + α t − 2rn ′ 2 + 2 βn ϕn (x) − − + + ϕn−1 (x). (5.102) ϕn−1 (x) = − x x 2 2x 2x2 t
If we eliminate ϕn−1 (x) The resulting ODE reads, ! d t R − R xR x − R n n n n ϕ′n (x) − ϕn (x) + dt x2 ϕ′′n (x) + x + x + Rn 2 2(x + Rn ) 2 x α2 t2 α t αt − − + + + 2 −x n+1+ + Hn ϕn (x) = 0, 4 4 4x 2 2 2x 33
(5.103)
with the aid of ( 5.100 ), and (2.12), (2.13), (3.21), (3.29) in [18]. In order to investigate the limiting behavior of the kernel, we recall the equations of (2.12), (2.13), (3.21), (3.29) in [18] βn (Rn + Rn−1 ) = −(2n + α)rn + nt, rn = tHn′ ,
βn Rn Rn−1 = rn2 − trn .
βn = n2 + nα + tHn′ − Hn .
The equations, (3.28) in [18] becomes, 2 (t(Hn′ )2 − tHn′ ) αn = 2n + 1 + α + , tHn′′ + n − (2n + α)Hn′ and with (2.9) in [18], we find Rn =
2 (t(Hn′ )2 − tHn′ ) , tHn′′ + n − (2n + α)Hn′
Let H(s, α) := lim Hn n→∞
(5.104)
s ,α . 2n + 1 + α
In the next theorem we describe the scaled kernel. Theorem 14. Let ξ := 4nx,
∗
ξ := 4ny,
s := (2n + 1 + α)t,
and
ϕ(ξ) := lim ϕn n→∞
ξ 4n
,
(5.105)
where t → 0+ , n → ∞ and ξ, ξ ∗ are in compact subsets of (0, ∞), s finite, α > 0, then 1 ξ ξ∗ lim Kn , n→∞ 4n 4n 4n (ξ − 4sH′ (s))ϕ(ξ)ξ ∗2 ϕ′ (ξ ∗ ) − (ξ ∗ − 4sH′ (s))ϕ(ξ ∗)ξ 2 ϕ′ (ξ) − 2s2 H′′ (s)(ξ − ξ ∗ )ϕ(ξ)ϕ(ξ ∗) = , (ξ − ξ ∗ )(ξ − 4sH′ (s))(ξ ∗ − 4sH′ (s)) (5.106) and ϕ(ξ) satisfies the second order ODE, ξ 2s2 H′′ (s) α2 s2 αs 4sH′ (s)ξ ′ 2 ′′ ϕ (ξ) + + − − 2− − H(s) ϕ(ξ) = 0, ξ ϕ (ξ) + ξ − ξ − 4sH′ (s) 4 ξ − 4sH′ (s) 4 ξ ξ (5.107) with the boundary conditions ϕ(0) = 0, 34
Proof. By (5.101), ϕn−1 (x) can be represented by ϕn (x) and its derivative as, 1 1 2 t α 2 ′ √ ϕn−1 (x) = x ϕn x) − − x + n + x + − rn ϕn (x) . 2 2 2 (x + Rn ) βn
Inserting the above equation and (5.105) into the kernel given by (5.96) , let n → ∞ , then we arrive at (5.106) and (5.107). In deriving these results, a number asymptotic relations, are listed here. rn = tHn′ (t) ∼ sH′ (s),
Rn ∼ −
sH′ (s) , n
t
d sH′ (s) s2 H′′ (s) Rn ∼ − − , dt n n
where t → 0, n → ∞ and s = (2n + 1 + α)t is fixed. Hence, we see that the scaling limit of the logarithmic derivative of the Hankel determinant Hn (t, α) generated by the perturbed Laguerre weight is the same with the scaling limit of logarithmic derivative of the Hankel determinant generated by the Pollaczek-Jacobi type kernel. Moreover, the limiting kernel ( 5.106 ) is completely characterized by the second order linear ODE of ( 5.107 ), which is the same with the limiting Pollaczek-Jacobi type kernel in the Theorem 13, both of which are “scaled at the origin” but their scaling scheme are different from each other. We present here another version of the ODE ( 5.107 ). For s > 0 , we make the transformation, ξ → 4sH′ ξ, where H′ denotes dH(s)/ds , into ( 5.107 ); we see that ϕ(ξ) satisfies, a4 a3 a2 a1 1 a0 2 ′ ′′ ϕ (ξ) + − 4 − 3 + 2 + ϕ(ξ) = 0, − + ϕ (ξ) + ξ ξ−1 ξ ξ ξ ξ ξ−1 where, 1 α α2 sH′′ sH′′ ′ , a = , a = −H − − , a = sH − , 3 2 1 16H′2 4H′ 4 2H′ 2H′ The coefficients of a4 , a3 , a2 , a1 and a0 satisfies the relation, a4 =
a0 =
sH′′ , 2H′
a20 + 2a0 + a1 + a2 − a3 − a4 = 0, which is the σ -form of the Painlev´e equation, 2 1 ′ (sH ) + 4 (H ) (sH − H) − αH + = 0. 2 ′′ 2
′ 2
′
We consider an interesting special case, where α = 0 The ODE becomes ! 2 2 2 3 + 5 3 − 1 s 1 1 27s 3s 2 ϕ′ (ξ) + − 4 + + ϕ(ξ) = 0, − + ϕ′′ (ξ) + 2 2 4 ξ ξ + 2s 32 ξ 36ξ 2 12s 3 ξ 12(ξs 3 + 2s 3 ) 35
with the boundary condition of ϕ(0) = 0. The solution of the σ -from Painlev´e equation, in this situation, reads, 1 3 2 H(s) = − s 3 + . 4 36 The region ξ > 1. The approximating equation is ψ”(ξ) +
s2 2 ′ ψ (ξ) − 4 ψ(ξ) = 0, ξ ξ
with boundary condition ψ(0) = 0 . The solution, up to a constant multiplier, is given by s
e− ξ . 2
The region ξ >> 1, ξs− 3 >> 1. The approximating equation is ! 2 3 1 1 3s − 1 + ψ(ξ) = 0, ψ ′′ (ξ) + ψ ′ (ξ) + 2 2 2 ξ 12s 3 ξ 12s 3 (ξ + 2s 3 ) we find that, ϕ(ξ) is asymptotic to 1 2 1 ξ −2 A (ξ + 2s )HeunC 0, 1, 0, s 3 , , s 3 + 1 , 2 3 2 2 3
where A is an arbitrary constant. Acknowledgement. We would like to thank the Macau Science and Technology Development Fund for generous support: FDCT 077/2012/A3, and the National Science Foundation of China (ProjectNo.11271079), Doctoral Programs Foundation of the Ministry of Education of China. Appendix A We list here a selection of integral formulas that are relevant for the computations in the main text. These can also be found in [20] and [13]. For 0 < a < b, we have, Z b 1 p dx = π. (A1) (b − x)(x − a) a Z
b
a
Z
a
Z
(a + b)π x p dx = . 2 (b − x)(x − a)
b a
b
x2
π 1 p dx = √ . ab x (b − x)(x − a)
p
1 (a + b)π dx = 3 . (b − x)(x − a) 2(ab) 2 36
(A2) (A3) (A4)
Z
b
a
Z
b
a
√ ! a+ b . 2 √ 2π 2 ab ln x p √ . dx = √ ln √ x (b − x)(x − a) ab a+ b
ln x p dx = 2π ln (b − x)(x − a)
√
(A5)
(A6)
Appendix B We list here a number of identities involving Rn , Rn ∗, rn , rn ∗, βn , Hn which can be found in [10], Rn∗ = Rn − (2n + 1 + α + β), ∗ (rn∗ )2 − trn∗ = βn Rn∗ Rn−1 ,
rn2 + βrn = βn Rn Rn−1 , βn =
−(rn∗ − rn )2 − (β + t)rn + (t − α − 2n)rn∗ + nt , 1 − (2n + α + β)2 rn∗ = rn =
Rn =
nt + tHn′ , 2n + α + β
n(n + α) + tHn′ − Hn , 2n + α + β
(2n + 1 + α + β)[2rn2 + (t + 2β − 2rn∗ )rn + (2n + α)rn∗ − nt − trn′ ] . 2[(rn∗ − rn )2 + (2n + α − t)rn∗ + (β + t)rn − nt]
The expressions of these for large n , are given below, rn∗ (t) =
rn =
sH′ s − (α + β)sH′ 1 + + O( 3 ), 2 2n n n
1 n α − β β 2 − α2 − 4(H − sH) (α + β)((α2 − β 2 ) + 4(H − sH)) + + + + O( 3 ), 2 2 4 8n 16n n Rn = 2n + 1 + α + β − Rn∗ = −
2sH′ (α + β − 1)sH′ − 2s2 H′′ 1 + + O( 3 ), 2 n n n
1 2sH′ (α + β − 1)sH′ − 2s2 H′′ + + O( 3 ), 2 n n n 37
where H is given by
s H(s, α, β) := lim Hn , α, β , n→∞ 2n2 H′ denotes dH(s)/ds and s = 2n2 t .
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