{'fraction_non_alphanumeric': 0.09736178218668491, 'fraction_numerical': 0.03296703296703297, 'mean_word_length': 3.475315729047072, 'pattern_counts': {'":': 0, '<': 24, '<?xml version=': 0, '>': 5, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 3, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Global Asymptotics of the Hahn Polynomials

In this paper, we study the asymptotics of the Hahn polynomials Q_n(x; {\alpha}, {\beta}, N) as the degree n grows to infinity, when the parameters {\alpha} and {\beta} are fixed and the ratio of n/N = c is a constant in the interval (0, 1). Uniform asymptotic formulas in terms of Airy functions and elementary functions are obtained for z in three overlapping regions, which together cover the whole complex plane. Our method is based on a modified version of the Riemann-Hilbert approach introduced by Deift and Zhou.

Formula (1.3) tells us that these polynomials are orthogonal on an unbounded interval as n → ∞. We now introduce a rescaling so that these polynomials become orthogonal on a bounded interval. Let X N denote the set defined by , where x N,k := k + 1/2 N .
(1. 5) The x N,k 's are called nodes and they all lie in the interval (0, 1). Also, let It is readily seen that the polynomials P N,n (z) are orthogonal on the nodes x N,k with respect to the weight w N,k := w(x N,k ); that is, where h * N,n = N −α−β α!β!h N −1,n . For further properties and applications of the Hahn polynomials, we refer to Beals and Wong [2] and Karlin and Mcgregor [6]. In 1989, Sharapudinov [13] studied the asymptotic behavior of Q n (x; α, β, N ), when the degree n is substantially smaller than N . His main result is an asymptotic formula for Q n (x; α, β, N ) with n = O(N 1/2 ) as n → ∞; which involves the Jacobi polynomial P (β,α) n (t), where t is related to x via the formula x = (N − 1)(1 + t)/2. Recently, Baik et al. [1] used the Riemann-Hilbert approach to investigate the asymptotics of discrete orthogonal polynomials with respect to a general weight function. Their results are very general, but it is difficult to use them to write out explicit formulas for specific polynomials. With regard to the Hahn polynomials, they only considered the case of varying parameters; that is, α = N A and β = N B, where A and B are fixed positive numbers. Moreover, their results are more local in nature; that is, one needs more asymptotic formulas to describe the behavior of these polynomials in the complex plane.
The purpose of this paper is to study uniform asymptotic behavior of the Hahn polynomials as n → ∞, when the parameters α and β are fixed and the ratio n/N is a constant c ∈ (0, 1). Our approach is based on a modified version of the Riemann-Hilbert method introduced by Deift and Zhou [4]. This version has been used successfully to obtain globally uniform asymptotic expansions of several discrete orthogonal polynomials; see [3], [10] and [15]. The presentation of this paper is arranged as follows: In Section 2, we review some preliminaries, including weak asymptotics of the zero distribution and the formulation of the basic interpolation problem Y (z). In Section 3, we transform this interpolation problem into an equivalent Riemann-Hilbert problem (RHP). In Section 4, we give the matrix transformation T (z) that normalizes the RHP for R(z) presented in Section 3 by using a function g(z), which is related to the logarithmic potential (g-function) of the equilibrium measure. The auxiliary functions φ(z) and φ * (z) are also studied in Section 4. In Section 5, we factorize the jump matrix in the RHP for T (z). The final results are presented in Sections 6 and 7, where we divide the whole complex plane into three regions and in each region we derive a uniformly asymptotic formula. As a special case of our result, we give in Section 8 the asymptotic formula of Q n (x; α, β, N ) for fixed values of x.

The equilibrium measure
It is known that the zero distribution of discrete orthogonal polynomials is related to a constrained equilibrium problem for logarithmic potential with an external field ϕ(x), which is defined by the formula ϕ(x) := V (x) + a 2 a 1 ln |x − y|ρ 0 (y)dy, x ∈ (a 1 , a 2 ), (2.1) where V (x) and ρ 0 (x) are real-analytic functions in a neighborhood of the closed interval [a 1 , a 2 ], and ρ 0 (x) is also a density function of the nodes; see [3] and [1, p.26]. In our case, V ( From (2.2), an explicit formula can be given for V N (x N,n ). Indeed, since x N,n = (n + 1 2 )/N and Since n = N x N,n − 1 2 , by replacing x N,n and n by x and N x − 1 2 , respectively, in the above equation, we obtain for x ∈ (0, 1). Using Stirling's formula, it can be shown that Thus, it follows from (2.3) that as N → ∞. This approximation formula suggests the definition of the function and we have Observe that in our case, we also have ϕ(x) = 0. The measure µ(x)dx that minimizes the following energy functional where µ(x) satisfies the upper and lower constraints and the normalization condition is called the equilibrium measure. It follows from (2.8) that there is an upper bound for µ(x), which does not appear in the continuous case. This fact is a crucial difference between discrete orthogonal polynomials and continuous orthogonal polynomials. The equilibrium measure µ(x)dx divides the interval of orthogonality into subintervals of three possible types: (1) achieving the lower constraint; (2) attaining the upper constraint; (3) not reaching the upper or the lower constraints. We call the open intervals of type (1)-(3), Voids, Saturated Regions and Bands, respectively. For reference, we refer to [1, p.19].
In the existing literature, the equilibrium measure is usually obtained by solving a minimization problem of a certain quadratic functional (cf. [1], [3]). Here, we prefer to use the method introduced by Kuijlaars and Van Assche [7]. Following [7], one can construct the equilibrium measure corresponding to the Hahn polynomials. Recall c := n/N , and let a : The density function is given by (2.11) Observe that the equilibrium measure of the Hahn polynomials corresponds to the saturated-bandsaturated configuration defined in [1], and we also note that

The basic interpolation problem
We begin with the basic interpolation problem for the Hahn polynomials. Note that the leading coefficient of P N,n (z) is Hence, the monic Hahn polynomials π N,n (z) are given by π N,n (z) = 1 k N,n P N,n (z). (2.14) Clearly, they satisfy the new orthogonality relation where γ 2 N,n = k 2 N,n /h * N,n . Following [1], we construct the interpolation problem for a 2 × 2 matrix-value function Y (z) with the properties: By the well-known theorem of Fokas, Its and Kitaev [5] (see also Baik et al. [1]), we have Theorem 2.1. The unique solution to the above interpolation problem is given by (2.17)

Riemann-Hilbert Problem
In this section, we will make a sequence of transformations: Y → H → R, and transform the basic interpolation problem into an equivalent Riemann-Hilbert problem (RHP).
First, we want to remove the saturated regions. Following Baik et al. [1], we introduce the first transformation It is readily verified that H(z) is a solution of the interpolation problem: ( H b ) at each x N,k ∈ X N , the second column of H(z) is analytic and the first column of H(z) has a simple pole where the residue is Next, we shall employ an idea of Ou and Wong [10] to remove the poles in H(z), and transform this interpolation problem into an equivalent RHP; see also [15]. Let δ > 0 be a sufficiently small number, and we define for z ∈ Ω ± , and R(z) := H(z) (3.4) for z / ∈ Ω ± , where the domains Ω ± are shown in Figure 1. Let Σ + be the boundary of Ω + in the upper half-plane, and Σ − be the mirror image of Σ + in the lower half-plane. For the contour Σ = (0, 1) ∪ Σ ± , see also Figure 1. For each x N,k ∈ X N , the singularity of R(z) at x N,k is removable; that is, Res Proof. For z ∈ Ω ± , it follows from (3.3) that Figure 1: The domains Ω ± and the contour Σ.
Since H 12 (z) is analytic by (H b ), the residue of R 12 (z) at x N,k is zero. From (3.2), we also have On the other hand, it is readily seen that (3.7) Thus, applying (3.6) and (3.7) to (3.5) shows that the residue of R 11 (z) at z = x N,k is zero. The entries in the second row of the matrix R(z) can be studied similarly. This completes the proof of the lemma.
Note that this transformation makes R + (x) and R − (x) continuous on the interval (0,1). As a consequence, we have created several jump discontinuities on the contour Σ in the complex plane.
It is easily verified that R(z) is a solution of the following RHP: (R a ) R(z) is analytic for z ∈ C \ Σ; (R b ) the jump conditions on the contour Σ: for x ∈ (0, 1), for z ∈ Σ ± , where r ± (z) = ∓2e ±N πiz cos(N πz)

The Auxiliary Functions
To normalize the condition (R c ) in the RHP for R(z), we need some auxiliary functions.
Definition 4.1. The g-function is the logarithmic potential defined by where µ(x) is the density function given in (2.11), and the so-called φ-function is defined by where l := 2 In (4.1) and (4.2), we view log(x − s) as an analytic function of the variable s with a branch cut along (x, +∞). Since log ± (x − s) = log |x − s| ± πi, s ∈ (x, +∞), we obtain Similarly, we have

From (4.3), it follows that
Moreover, it can be easily verified that the g-function satisfies the jump conditions and On account of (4.1), (4.6) and (4.7), one readily sees that e n g(z) can be analytically extended to C \ [a, b] and Hence, by adopting the convention that σ 3 denotes the Pauli matrix we introduce the transformation T (z) := (c 1 e nl ) −σ 3 /2 R(z)e −n g(z)σ 3 (c 1 e nl ) σ 3 /2 (4.10) to normalize the behavior of R(z) at z = ∞, where c 1 is given in (3.9). As we have mentioned before, the density function µ(x) reaches its upper constraint near the endpoints of the interval of the orthogonality; see (2.11). Furthermore, since µ(x) is not differentiable at the point x = a, the function φ(x) is not analytic in the neighborhood of x = a, and we can not construct our global parametrix (such as Airy parametrix) by using φ(z) in the region which includes the critical point a. Hence, for our future analysis, a few more auxiliary functions are needed.
Note by (4.7) that g(z) is analytic for z ∈ (b, 1). This together with (4.5) evokes us to consider a modified measure µ * (x) = µ(x) − 1 c . Furthermore, we want to find a complex function ν(z) which satisfies the requirement and behaves like z as z → +∞, and where log(z − z 2 ) is an analytic function with branch cuts along (−∞, 0] ∪ [1, +∞). In view of the identities On the other hand, the measure µ(x) − 1/c can be rewritten as From (4.14) and (4.15), it clearly follows that Now, we are ready to introduce the auxiliary function where the path of integration from a to z lies entirely in the region z ∈ C \ (−∞, 0] ∪ [a, +∞), except for the initial point a.
Similarly, we set where the path of integration from b to z lies entirely in the region z ∈ C \ (−∞, b] ∪ [1, +∞), except for the initial point b.
The functions φ(z) and φ * (z) defined above play an important role in our argument, and the following are some of their properties.
A geometric interpretation of the results established above is given in Figure 2. Furthermore, if we let C + = {z : Im z > 0} and C − = {z : Im z < 0}, then for z ∈ C ± we have from (4.16) and (4.17) The results in the following proposition can now be proved either directly from definition (4.18), or by using (4.36) and Proposition 4.2.
Proposition 4.3. The mapping properties of φ * (z) are given by: The images of the upper (lower) half of z-plane under the mapping φ * (z) is depicted in Figure 3. The next proposition shows the relationship between the φ-function and the φ-function (φ * -function). (4.2), the following connection formulas between the φ-function and the φ-function (φ * -function) hold: To establish (4.43) and (4.44), we need some properties of g ′ (z). In fact, we shall derive explicit formulas for g(z) and g ′ (z). From (2.11), we have (4.46) The last integral in (4.46) is given by On account of (2.11) and (2.12), a direct calculation gives A combination of (4.46), (4.48), (4.50) and (4.
Let the integrals on the right-hand side of the equality be denoted by I 1 , I 2 and I 3 , respectively. To evaluate I 1 , we note that from [16] Making the change of variable s = z − x, we obtain from (4.53) and see [16,Lemma 1]. Hence, . (4.57) A straightforward calculation shows that Moreover, by using (4.50), (4.54) and (4.55), we have Proof of Proposition 4.4 . Note that when x ∈ (a, b), ν ± (x) = ±πi(µ(x) − 1 c ); see (4.16). Hence, it follows that On the other hand, if x ∈ (−∞, 0) ∪ (1, +∞), we have with the aid of (4.25) and (4.33) To establish (4.43), we need one more functionφ(z) defined bŷ From (4.1), one can show that for x ∈ (0, 1) and satisfies the jump conditionsφ Comparing (4.64) and (4.65) with (4.60) and (4.61), we find or, equivalently, . On account of this, ν(z) −φ(z) can be analytically extended to the whole complex plane. Note that ν + (x) = −ν − (x), and that from (4.59) it can be shown and soφ Hence, from (4.67), (4.16) and (4.69) we have Using analytic continuation, it follows from (4.62) Hence, it follows from (4.17) that Now, let z ∈ C + ; on account of (4.1) and the definition of l, we have In a similar manner, one can show that for We conclude this section with the mapping properties of φ(z) in (4.2). By using Proposition 4.2 and (4.43), one can prove the following proposition; see also Figure 4. (To establish the second equality in (4.78), we make use of (4.44) instead of (4.43).) Proposition 4.5. The mapping properties of φ(z) are given by: The lower half z-plane under φ(z)

Normalization and Contour Deformation
Recall from (4.10) that (T a ) T (z) is analytic for z ∈ C \ Σ; For the sake of simplicity, we introduce some auxiliary functions. Let for z ∈ C ± , where ϕ(z) is defined in (4.4). In view of the equation preceding (4.5), it is easily verified that On account of this, it can be shown that E + (x) E − (x) −1 = 1 for x ∈ (0, 1). Hence, E(z) can be analytically continued to the interval (0, 1). Moreover, As N → ∞, by applying Stirling's formula to (5.4) and evaluating the integral in (4.4) explicitly, one can show that E(z) ∼ 1 uniformly for z bounded away from the interval [0, 1]. Moreover, from (5.5), we have for z ∈ C ± E(z)/ E(z) = 1 + e ±2N πiz ∼ 1 (5.8) uniformly for z bounded away from the real line.
With the properties of φ(z) and φ * (z) established in Section 4 and Lemma 5.1, the jump matrices for T in condition (T b ) can be written as follows: for 0 < x < a, for b < x < 1, and Note that for the jump matrix on (a, b), we have the following factorization: The first and third matrices on the right-hand side of (5.18) have, respectively, the analytic continuations E w E/e 2nφ * 0 ± denote the contour shown in Figure 5. Define where regions Ω ± = 3 i=1 Ω i ± and Ω ∞ are also depicted in Figure 5. It can be verified that the jump matrices of V (z) are given by Figure 5: Contour Σ V and regions Ω 1 ± , Ω 2 ± , Ω 3 ± , Ω ∞ . and In proving (5.23), (5.24) and (5.25), we note that E is analytic in 0 < x < 1. For (5.23), we also make use of (5.18). In establishing (5.27), we have used (4.44) and (5.8). For (5.28) and (5.29), we note that T − (z) −1 T + (z) = I. Thus, V (z) is a solution of the following RHP: For condition (V c ), we note from (5.4) that E(z) ∼ 1 as z → ∞. From the properties of φ * (z) and φ(z) (see also Figure 2 and Figure 4), the regions Ω 2 ± and Ω ± can be chosen sufficiently small so that A direct calculation shows that Therefore, it is natural to suggest that for large n, the solution of the RHP for V (z) may behave asymptotically like the solution of the following RHP for N (z): To solve the above RHP, we first consider the function Its limit at infinity is given by This problem can be solved explicitly, and its solution is given by with a branch cut along [a, b] and a(z) → 1 as z → ∞. Hence, That is, V (z) and N (z) are not uniformly close to each other near these points, and special attention must be paid to the neighborhoods of these points. Let U (0, ε) := {z ∈ C : |z| < ε} be a neighborhood of the origin, where ε is a small positive number. In view of (5.5), (5.26), (5.30) and (5.43), we first wish to find a matrix-valued function V 0 (z) that satisfies the following RHP: where the functions (V 0 ) + (z) and (V 0 ) − (z) denote the boundary values of V 0 (z) taken from the left and right of the imaginary axis, respectively; (V 0,c ) for z ∈ ∂U (0, ε), as n → ∞.
Next, we consider as in [15] the following D-function By applying Stirling's formula to (5.46), we have for | arg z| < π. We also introduce the function which is actually analytic in (−∞, 0 ). From Euler's reflection formula, it follows that as n → ∞. Set the following matrix-valued function It follows from (5.47), (5.48) and (5.50) that Q 0 is a solution of the RHP: where the functions (Q 0 ) + (z) and (Q 0 ) − (z) denote the boundary values of Q 0 (z) taken from the left and right of the imaginary axis, respectively; (Q 0,c ) for z bounded away from the origin, as n → ∞.
A comparison of conditions (Q 0,b ) and (Q 0,c ) with (5.44) and (5.45) shows that a solution of the RHP for V 0 is given by In a similar manner, for z ∈ U (1, ε) = {z ∈ C : |z − 1| < ε}, we want to find a matrix-valued function V 1 (z) that satisfies the RHP: where the functions (V 1 ) + (z) and (V 1 ) − (z) denote the boundary values of V 1 (z) taken from the right and left of the line Re z = 1, respectively; also see Figure 5; (V 1,c ) for z ∈ ∂U (1, ε), as n → ∞.
As in the case of V 0 , it can be verified that a solution of the RHP for V 1 is given by Since V 0 (z) satisfies the same jump conditions as V (z) on the contour (1, ε). Hence, as n → ∞, we have from (5.43) (5.59) Let x 0 be an arbitrary fixed point in the interval (a, b). From (5.54), (5.58) and (5.59), it follows that for Re z < x 0 ,  Hence, it follows that for Re z < x 0 , Similarly, for Re z > x 0 , we have by using (5.57), (5.58) and (5.59) where we have also made use of the identity D * (1 − z) E(z) = D(1 − z)E(z) obtained from (5.5) and (5.48). Define On account of (4.3), (4.10) and (5.4), one can work backwards to get where l is given in Definition 4.1. Figure 6: The regions I, II, III.

Construction of the Parametrices
In this section, we will construct an approximation R(z) to the solution of the RHP for R(z) stated in Section 3 for large n. Given x 1 ∈ (0, a), define the rectangle K = {z ∈ C : x 1 < Re z < 1−x 1 , |Im z| < δ}, and the line Γ 0 = {z ∈ C : Re z = x 0 , |Im z| < δ}, where x 0 is given in Section 5. We divide the complex plane into three regions I, II, III by using Γ 0 and K; see Figure 6. From (5.66), it is natural to set R(z) = (c 1 e nl ) σ 3 /2 N (z)e −nφ(z)σ 3 D(z) for z ∈ I. For z inside the rectangle K, we will construct the parametrices by using Airy functions as in [3,15]. For z ∈ II, it follows from (5.48), (5.65) and (5.66) R(z) ∼ (c 1 e nl ) σ 3 /2 N (z)e −nφ(z)σ 3 e ±N πizσ 3 2 cos(N πz)D(z) and from (4.43) and (5.47) we obtain Using (4.12) and (4.17), one can derive an expansion for φ(z) for z ∈ U (a, ε) := {z ∈ C : |z − a| < ε}. Indeed, we have for z ∈ U (a, ε). By using (4.29) and the proposition of φ (see Figure 2), it can be verified that the function defined by On account of this and (6.4), we conclude that this function is analytic in C \ (−∞, 0] ∪ [b, +∞). In particular, one can see that for z ∈ II ∩ C + −π < arg f (z) < 0, (6.6) and we obtain from (4.22) Note that N (z) = M −σ 3 ∞ N (z)M (z) σ 3 by (5.40) and (5.42), and N (z) has the factorization Since a(z) = (z − b) 1/4 /(z − a) 1/4 , N (z) has fourth-root singularities at z = a and z = b. In order to remove the singularity at z = a, we introduce the function Thus, Coupling (6.9) and (5.38), we obtain Since f (z) has a simple zero at z = a and a(z) has a fourth-root singularity at a, the last formula shows that E n (z) has a removable singularity at z = a. Furthermore, one can show that E n (z) has no jump across the interval (a, b). Indeed, by (5.33) ∈ (a, b). (6.11) From (6.7), (6.9) and (6.11), it follows that for x ∈ (a, b) Therefore, E n (z) is analytic across (a, b). Also, note that from (5.42), (5.34) and (6.5), it is evident that N (z), h(z) and f (z) are analytic for z ∈ (0, a). Thus, the matrix-valued function E n (z) is analytic in By virtue of (6.9), we can rewrite (6.3) as To find an approximation to R(z), we first look for a matrix which is asymptotic to e n φ(z) . (6.14) Set From the asymptotic expansions of the Airy function [9, p.198], we have where ω = e 2πi/3 . It is immediate that for z ∈ II ∩ C + , Similarly, for z ∈ II ∩ C − , we also have 0 < arg f (z) < π and For z ∈ III, there is a result corresponding to (6.3) with φ(z) replaced by φ * (z). Indeed, it follows from (5.65), (5.66) and (4.44) that Since N (z) has a fourth-root singularity at z = b, we define As in the case of E n (z), it is readily seen that b is a removable singularity of E * n (z). By (6.11) and (6.21), one also has for x ∈ (a, b). Therefore, E * n (z) is analytic in C \ (−∞, a] ∪ [1, +∞). On account of this and (6.19), we have for z ∈ III Hence, as before, it can be shown that the matrix [n 2/3 f * (z)] σ 3 /4 e −nφ * (z) ie nφ * (z) ie −nφ * (z) e nφ * (z) (6.25) is the leading term of the asymptotic expansion of the matrices for z ∈ III ∩ C + , and An appeal to the formulas Ai(z) + ωAi(ωz) + ω 2 Ai(ω 2 z) = 0, Ai ′ (z) + ω 2 Ai ′ (ωz) + ωAi ′ (ω 2 z) = 0 (6.29) shows P (z) satisfies Let E n (z) and E * n (z) be defined as in (6.9) and (6.22), respectively. In view of the heuristic argument leading to (6.1), (6.13) and (6.24), it is reasonable to suggest that R(z) is asymptotically approximated by for z ∈ II and III. From (6.31), it can be shown that R(z) has the same large z behavior as R(z) given in (R c ). To see this, we only consider the case when z ∈ I and Re z > x 0 . The discussion for the case z ∈ I and Re z < x 0 is very similar. Note that when z ∈ I and Re z > x 0 , This together with (4.1) and the asymptotic behavior of N (z) in (N c ) gives (S a ) S(z) is analytic for z ∈ C \ Γ; where the contour associated with the matrix J S (z) is Γ = Σ ∪ Γ 0 ∪ ∂K; see Figure 1 and Figure 6.
To solve this problem, we first derive the asymptotic behavior of the jump matrix J S (z) as n → ∞. For convenience, we consider only the cases when z lies on the contour Γ 0 ∩ C + and Γ 3 ∩ C + , and z in the interval (0, 1), since the other cases can be discussed in a similar manner.

Main Results
Let X ij denote the (i, j) element in the matrix X. for z ∈ I, where δ(n) = 0 for z ∈ I \ Ω ± and δ(n) is exponentially small in comparison with its leading term for z ∈ I ∩ Ω ± ; see Figure 1. More precisely, δ(n) = O(N max(α,β) e nφ(z) ), where Re φ(z) is negative when Re z is bounded away from the interval (a, b); see Figure 4. Let f (z) be defined as in (6.5). We have Proof. Since R(z) = (c 1 e nl ) σ 3 /2 S(z)(c 1 e nl ) −σ 3 /2 R(z) by (6.37), we have R 11 (z) = S 11 (z) R 11 (z) + S 12 (z) R 21 (z)c 1 e nl (7.6) and R 12 (z) = S 11 (z) R 12 (z) + S 12 (z) R 22 (z)c 1 e nl . for z ∈ II ∩ C + . In a similar manner, one can see that for 0 < z < x 1 . A comparison of (8.19) and (8.20) with (8.11) and (8.3) implies that our results agree with those in [8]. In fact, one can easily show that when α = β = 0, we do not need to construct the parametrix in region I. However, R(z) defined in (6.31) for z ∈ II can not be extended to the region I when α, β = 0. This is because the function h(z) involved in R(z) is not analytic in the neighborhoods of 0 and 1.