Mehler-Heine type formulas for the Krawtchouk polynomials

Journal Pre-proof Mehler-Heine type formulas for the Krawtchouk polynomials

Diego Dominici

PII:

S0022-247X(20)30039-1

DOI:

https://doi.org/10.1016/j.jmaa.2020.123877

Reference:

YJMAA 123877

To appear in:

Journal of Mathematical Analysis and Applications

Received date:

12 June 2019

Please cite this article as: D. Dominici, Mehler-Heine type formulas for the Krawtchouk polynomials, J. Math. Anal. Appl. (2020), 123877, doi: https://doi.org/10.1016/j.jmaa.2020.123877.

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Mehler-Heine type formulas for the Krawtchouk polynomials. Diego Dominici ∗ Johannes Kepler University Linz Doktoratskolleg “Computational Mathematics” Altenberger Straße 69 4040 Linz Austria Permanent address: Department of Mathematics State University of New York at New Paltz 1 Hawk Dr. New Paltz, NY 12561-2443 USA January 14, 2020

Abstract We derive Mehler–Heine type asymptotic expansions for the Krawtchouk polynomials Kn (x; p, N ) with n = O(N ) and x = o(N ). These formulas provide good approximations for Kn (x; p, N ) and determine the asymptotic limit of their zeros as n → ∞.

Keywords: Mehler-Heine formulas, discrete orthogonal polynomials. MSC-class: 41A30 (Primary), 33A65, 33A15, 44A15 (Secondary) ∗

e-mail: [email protected]

1

1

Introduction

Let N0 denote the set N0 = N ∪ {0} = 0, 1, 2, . . . . The monic Krawtchouk polynomials are defined by [54, 18.20.6]   −n, −x 1 n ; , Kn (x; p, N ) = p (−N )n 2 F1 −N p

(1)

where p ∈ [0, 1] , n, N ∈ N0 , n ≤ N, r Fs denotes the generalized hypergeometric function defined by [54, Chapter 16]  r Fs

a1 , . . . , ar ;z b1 , . . . , b s



∞  (a1 )k · · · (ar )k z k , = (b1 )k · · · (bs )k k! k=0

and (x)k the Pochhammer symbol (or rising factorial) [54, 5.2.4], (x)z =

Γ (x + z) , Γ (x)

x, z ∈ C,

(2)

where in general we need − (x + z) ∈ / N0 . Note that when x = 0, we have Kn (0; p, N ) = pn (−N )n .

(3)

The Krawtchouk polynomials are one of the families of discrete classical orthogonal polynomials [52]. They satisfy the orthogonality relation     N x N −x 2 N [p (1 − p)]n δn,m , p (1 − p) Kn (x; p, N ) Km (x; p, N ) = (n!) n x x=0

N 

the three-term recurrence relation xKn = Kn+1 + (N p + n − 2np) Kn + np (1 − p) (N − n + 1) Kn−1 ,

(4)

and have the generating function ∞  n=0

Kn (x; p, N )

tn = [1 + (1 − p) t]x (1 − pt)N −x , n! 2

(5)

from which we obtain the symmetry relation Kn (x; p, N ) = (−1)n Kn (N − x; 1 − p, N ) . Setting p = 0 in (5) gives ∞    x n tn x t , = (1 + t) = Kn (x; 0, N ) n n! n=0 n=0

∞ 

and we conclude that

  x = (x − n + 1)n = ϕn (x) , Kn (x; 0, N ) = n! n

(6)

where the falling factorial polynomials ϕn (x) are defined by ϕ0 (x) = 1 and ϕn (x) =

n−1 

(x − k) = (−1)n (−x)n ,

n ∈ N.

(7)

k=0

Similarly, when p = 1, we see from (5) that  ∞   N −x tn N −x (−t)n , = (1 − t) Kn (x; 1, N ) = n n! n=0 n=0

∞ 

and obtain



N −x Kn (x; 1, N ) = (−1) n! n n

 = (x − N )n .

(8)

The Krawtchouk polynomials are important in the study of the Hamming scheme of classical coding theory [35], [39], [46], [55], [60], [62]. Lloyd’s theorem [42] states that if a perfect code exists in the Hamming metric, then the Krawtchouk polynomial must have integral zeros [5], [10], [38]. Not surprisingly, the zeros of Kn (x; p, N ) have been the subject of extensive research [3], [8], [15], [26], [27], [29], [32], [36], [66], [72]. The Krawtchouk polynomials also have applications in probability theory [19], queueing models [14], stochastic processes [56], quantum mechanics [4], [43], [71], face recognition systems [1], combinatorics [18], and biology [31]. Multivariate [17], [20], [22], [23], [50], [58], [59], [68], and q extensions [6], [7], [21], [24], [25], [34], [37], [61], [65], [63], [64], have also been considered. 3

Suppose that Pn (x) is a sequence of orthogonal polynomials and let xk,n denote the zeros of Pn (x) Pn (xk,n ) = 0,

x1,n < x2,n < · · · < xn,n .

Two standard approximations describing the asymptotic behavior of the polynomials Pn (x) as the degree n tends to infinity are Mehler–Heine type formulas (in a region around the smallest zero) and Plancherel-Rotach type formulas (in a region around the largest zero) x1,n < x2,n < · · · < xn−1,n < xn,n .     Mehler-Heine

Plancherel-Rotach

Mehler–Heine type formulas were introduced by Heinrich Heine in 1861 [28] and Gustav Mehler [48] in 1868 to analyze the asymptotic behavior of Legendre polynomials. See Watson’s book [69, 5.71] for some historical remarks. They have the general form lim fn (z) Pn (A + un z) = G(z),

n→∞

z ∈ S,

where S is a domain in the complex plane, A is a constant, un is a given sequence, G(z) is analytic in S and the functions fn (z) are analytic and don’t have any zeros in S, for sufficiently large n. The convergence is uniform on compact subsets of S. From Hurwitz’s theorem, we conclude that for a fixed k ζk − A xn,k ∼ , n→∞ (9) un where ζ1 < ζ2 < · · · are the zeros of G(z). Mehler–Heine type formulas are very important in the field of Sobolev orthogonal polynomials, see [2], [41], [44], [45], [51], [47]. In [12], we studied Mehler–Heine type formulas for the Charlier and Meixner polynomials and obtained: (1) If Cn (x; a) denotes the monic Charlier polynomials defined by [33, 9.14.1]   1 −n, −x n , a > 0, (10) ;− Cn (x, a) = (−a) 2 F0 − a 4

then, we have (−1)n ea Cn (z, a) = , n→∞ Γ (n − z) Γ (−z) lim

z ∈ C.

(11)

(2) If Mn (x; β, c) denotes the monic Meixner polynomials defined by [33, 9.10.1] n    c 1 −n, −x ;1 − Mn (x; β, c) = (β)n , β > 0, (12) 2 F1 β c−1 c then, for c ∈ C \ [1, ∞) we have (−1)n (1 − c)n+z (1 − c)−β lim Mn (z; β, c) = , n→∞ Γ (n − z) Γ (−z)

z ∈ C,

(13)

where all functions assume their principal values. In [11], we extended our results to full asymptotic expansions. Although it seems that one could apply these results to the Krawtchouk polynomials using the relation   p Kn (x; p, N ) = Mn x; −N, , p−1 we see that this presents many problems because for the Meixner polynomials (i) c ∈ (0, 1) , (ii) β > 0, and (iii) β = O(1), while for the Krawtchouk polynomials we need (i’) c ∈ (−∞, 0) , (ii’) β < 0, and (iii’) |β| > n. In this paper, we should use a different approach based on the asymptotic analysis of the differential equation satisfied by Kn (x; p, N ) as a function of p. A similar idea was followed by Dunster in [16] to study the asymptotic behavior of the Charlier polynomials. Our objective is to prove the following Mehler–Heine type formulas for the Krawtchouk polynomials with n = rN, for some fixed r ∈ (0, 1) : (i) For 0 < r < p ≤ 1, z

p−r Kn (z; p, N ) lim = , z ∈ C. n→∞ (z − N ) pn (1 − r) p n (ii) For 0 ≤ p < r < 1, (−1)n Kn (z; p, N )

p −z−1 (1 − p)z 1− = , Γ (−z) r

z ∈ C. − z) (1 − p)n−N We shall not attempt to obtain full asymptotic approximations for the Krawtchouk polynomials, since this problem has been studied by many authors, including [9], [13], [30], [40], [49], [57], [67]. lim

n→∞ Γ (n

5

2

Asymptotic approximation

From (1), we see that the Krawtchouk polynomials Kn (x; p, N ) satisfy the hypergeometric differential equation p (1 − p)

d2 f df + n (N − n + 1) f = 0, (14) + [x − n + 1 − (N − 2n + 2)p] 2 dp dp

while (8) and (6) give the boundary conditions f (0) = ϕn (x) ,

f (1) = (x − N )n ,

(15)

where the polynomials ϕn (x) where defined in (7). Setting n = rN, in (14), we have d2 f df + rN (N − N r + 1)f = 0. + [(2pr − r − p) N + x − 2p + 1] 2 dp dp (16) Replacing f (p) = [ψ (p)]N L (p) p (1 − p)

in (16) we obtain, to leading order, as N → ∞ with x = o(N ) the nonlinear ODE [(p − 1) ψ  + (1 − r) ψ] (pψ  − rψ) = 0, with solutions and Writing

ψ0 (p) = C0 (1 − p)r−1 , ψ1 (p) = C1 pr ,

p < 1.

p > 0.

(17) (18)

f0 (p) = [ψ0 (p)]N L0 (p) ,

and using (15) and (17), we obtain f0 (p) = ϕn (x) (1 − p)n−N ξ0 (p; x) ,

(19)

ξ0 (0; x) = 1.

(20)

with Note that since n < N, f0 (p) is not defined at p = 1. Similarly, writing f1 (p) = [ψ1 (p)]N L1 (p) , 6

and using (15) and (18), we get f1 (p) = (x − N )n pn ξ1 (p; x) ,

(21)

ξ1 (1; x) = 1.

(22)

with Note that f1 (p) defined at both p = 0 and p = 1. It is clear from the analysis so far that we shall have two main asymptotic expansions as N → ∞, one valid in a neighborhood of p = 1 Kn (x; p, N ) ∼ (x − N )n pn ξ1 (p; x) ,

0 < r < p ≤ 1,

(23)

and the other valid in a neighborhood of p = 0 Kn (x; p, N ) ∼ ϕn (x) (1 − p)n−N ξ0 (p; x) + k (x) (x − N )n pn ξ1 (p; x) , (24) where 0 ≤ p < r < 1, and k (x) is an unknown function to be determined by asymptotic matching. Using the condition (3) in (23) gives ξ1 (p; 0) = 1.

(25)

Using (3) and (25) in (24), we conclude that k (0) = 1,

(26)

since ϕn (x) = 0,

2.1

x = 0, 1, . . . , n − 1,

n ∈ N.

The function ξ1 (p; x)

In this section, we assume that 0 < r < p ≤ 1. Setting ε = N −1 in (16), we have ε2 p (1 − p) Replacing

d2 f df +ε [2pr − r − p + (x − 2p + 1) ε] +r(1−r +ε)f = 0. (27) 2 dp dp  r ln (p) ξ1 (p; x) f (p) = exp ε

7

in (27), we obtain εp2 (1 − p) ξ1 + [ε (x + 1 − 2p) + r − p] pξ1 + rxξ1 = 0.

(28)

If we let ε → 0 in (28), we get (r − p) pξ1 + rxξ1 = 0. Solving (29) with initial condition (22), we see that x  r −x 1− , ε → 0. ξ1 (p; x) ∼ (1 − r) p To find higher order terms, we set x  r −x ξ1 (p; x) = (1 − r) 1− g (p; x, r, ε) , p

(29)

(30)

(31)

in (28), and obtain the ODE   εp (1 − p) (p − r)2 g  + ε p − r + 2pr + px + rx − 2p2 − 2prx (p − r) g  (32) − (p − r)3 g  + εr (1 − r) x (x − 1) g = 0, while (22) gives the initial condition g (1; x, r, ε) = 1.

(33)

If we define the formal power series g (p; x, r, ε) =

∞ 

gk (p; x, r) εk ,

g0 (p; x, r) = 1,

(34)

k=0

we see from (32) that   gk = A2 gk−1 + A1 gk−1 + A0 gk−1 ,

with p (1 − p) p − r + 2pr + px + rx − 2p2 − 2prx , A1 = , p−r (p − r)2 r (1 − r) x (x − 1) , A0 = (p − r)3 A2 =

8

(35)

and using the initial condition (33) in (34) gives gk (1; x, r) = 0,

k ∈ N.

(36)

In Appendix A, we solve the recurrence (35) with initial conditions g0 (p; x, r) = 1 and (36), and obtain gk (p; x, r) =

2k  ak,j (x, r) j=0

(p − r)j

,

(37)

where the coefficients ak,j (x) satisfy the recurrence (x − j + 1) (x − j + 2) ak−1,j−2 j (j − 1) (x − j + 1) ak−1,j−1 + (j − 1) ak−1,j , j, k ∈ N, − (2r − 1) j ak,j = r (r − 1)

with initial conditions a0,0 = 1,

j∈ / {0, 1, . . . , 2k} ,

ak,j = 0,

and ak,0 (x, r) = −

2k  ak,j (x, r) j=1

(1 − r)j

,

k ∈ N.

For example,

 1 1−r r − x (x − 1) , g1 (p; x, r) = 2 1 − r (p − r)2

(38)

where we see that the condition g1 (p; 0, r) = 0 is satisfied, as required by (25). All the coefficients ak,j (x, r) can be easily computed, and in Appendix A we list the first few values for 0 ≤ k ≤ 3.

9

2.2

The function ξ0 (p; x)

In this section, we assume that 0 ≤ p < r. Replacing 

(r − 1) ln (1 − p) ξ0 (p; x) f (p) = exp ε in (27), we obtain εp (1 − p)2 ξ0 + [ε (x + 1 − 2p) + p − r] (1 − p) ξ0 + (x + 1 − rx − p) ξ0 = 0. (39) If we let ε → 0 in (39), we get (p − r) (1 − p) ξ0 + (x + 1 − rx − p) ξ0 = 0.

(40)

Solving (40) with initial condition (20), we see that p −x−1 , ξ0 (p; x) ∼ (1 − p)x 1 − r

ε → 0.

To find higher order terms, we set p −x−1 h (p; x, r, ε) , ξ0 (p; x) = (1 − p)x 1 − r

(41)

(42)

in (39), and obtain the ODE εp (1 − p) (p − r)2 h + ε (2pr − p − r) (p − r) (x + 1) h   + (p − r)3 h + ε rx + p + r − r2 x − 2pr (x + 1) h = 0,

(43)

while (20) gives the initial condition h (0; x, r, ε) = 1.

(44)

If we define the formal power series h (p; x, r, ε) =

m 

hk (p; x, r) εk ,

h0 (p; x, r) = 1,

(45)

k=0

then we see from (43) that hk = B2 hk−1 + B1 hk−1 + B0 hk−1 , 10

(46)

with p (1 − p) (2pr − p − r) (x + 1) , B1 = − , p−r (p − r)2 (rx + p + r − r2 x − 2pr) (x + 1) B0 = − , (p − r)3

B2 = −

and using the initial condition (44) in (45) gives hk (0; x, r) = 0,

k ∈ N.

(47)

In Appendix B we solve the ODE (46) with initial condition (47), and obtain 2k  bk,j (x, r) hk (p; x, r) = (48) j , (p − r) j=0 where the coefficients bk,j (x, r) satisfy the recurrence r (1 − r) (x + j − 1) (x + j) bk−1,j−2 j + (1 − 2r) (x + j) bk−1,j−1 − (j + 1) bk−1,j , j, k ∈ N, bk,j =

with initial conditions b0,0 = 1,

j∈ / {0, 1, . . . , 2k} ,

bk,j = 0,

and bk,0 (x, r) = −

2k  bk,j (x, r) j=1

(−r)j

,

k ∈ N.

For example,

 r (r − 1) 1 (r − 1) x − 2r 2 (2r − 1) − − (x + 2) (x + 1) . h1 (p; x, r) = 2 r p−r (p − r)2 All the coefficients bk,j (x, r) can be easily computed, and in Appendix B we list the first few values for 0 ≤ k ≤ 3. Remark 1 It follows from (31) and (42) that ξ1 (p; x) and ξ0 (p; x) have branch points at p = r. Therefore, in the next section we shall find a new asymptotic approximation in the neighborhood of p = r. 11

3

The turning point at p = r

If we set

√ p = r + u aε,

a > 0,

u = O (1) ,

(49)

in (27), we obtain   d2 f √ ε εau2 + aε (2r − 1) u + r (r − 1) du2   df √ √ √ √ 3 (50) + ε 2ε 2 au + ε a (2r − x − 1) + εau (1 − 2r) + 2 ar (1 − r) du − (ε + 1 − r)2 arf = 0. Replacing

 φ (u) η (u) f (u) = exp √ ε

(51)

in (50) and letting ε → 0, we get   √ 2 φ − a = 0, with solution

√ φ (u) = u a + C.

Using (51) and (52) in (50) and letting ε → 0, we obtain     r (1 − r) η  − a u (2r − 1) η  + au2 + r − x − 1 η = 0. Setting a = r (1 − r) and η (u) = exp



(52)

(53)

 2r − 1 2 u λ (u) , 4

in (53), we see that λ (u) satisfies   2 u 1  −x− λ, λ = 4 2 whose linearly independent solutions are the parabolic cylinder functions Dx (u) , Dx (−u) [70, 16.5]. Hence, we get   2r − 1 2 u [Cp (x) Dx (u) + Cm (x) Dx (−u)] , (54) η (u) = exp 4 12

where Cp (x) and Cm (x) are functions to be determined. Combining (51), (52), and (54), we conclude that 

u2 Kn (x; p, N ) ∼ [Cp (x) Dx (u) + Cm (x) Dx (−u)] exp τ2 (u) + , 4

(55)



where

r (1 − r) 1 − (1 − r) u2 . (56) ε 2 In the next section, we shall determine Cp (x) and Cm (x) by matching (55) asymptotically with (23) and (24). τ2 (u) = u

3.1

Matching the asymptotic expansions

Since from (2) we have (x − N )n = (−1)n

Γ (N − x + 1) , Γ (N − x − n + 1)

(57)

we can use Stirling’s formula [54, 5.11.1] ln Γ (z) ∼ z [ln (z) − 1] −

1 1 1 ln (z) + ln (2π) + , 2 2 12z

and obtain x− 12

n

(−1) (x − N )n ∼ (1 − r)

z → ∞,

 (r − 1) ln (1 − r) − (1 + ln ε) r . exp ε

(58)



(59)

Using (30) and (59) in (23) gives

 x τ1 (p) r 1 exp 1− , (−1) Kn (x; p, N ) ∼ √ ε p 1−r

(60)

τ1 (p) = (r − 1) ln (1 − r) + [ln (p) − ln (ε) − 1] r.

(61)

n

where Similarly, we see from (2) and (6) that ϕn (x) = (−1)n 13

Γ (n − x) , Γ (−x)

(62)

and using Stirling’s formula (58), we obtain √

   1 (ln r − ln ε − 1) r 2πr−x− 2 1 n exp + x+ ln (ε) . (63) (−1) ϕn (x) ∼ Γ (−x) ε 2 Using (30), (41), (59), and (63) in (24) gives

 √ τ0 (p) [ε (1 − p)]x exp (r − p)−x−1 (−1) Kn (x; p, N ) ∼ 2πεr Γ (−x) ε  x

τ1 (p) r k (x) 1− , +√ exp ε p 1−r n

(64)

where τ0 (p) = (r − 1) ln (1 − p) + [ln (r) − ln (ε) − 1] r.

(65)

Using (49) in (61), we obtain τ1 (p) τ1 (r) ∼ + τ2 (u) , ε ε

p → r+ ,

(66)

where τ2 (u) was defined in (56), and using (49) we also have r p−r u 1− = ∼ r (1 − r) ε, p → r+ . p p r Using (66) and (67) in (60), we conclude that as p → r+   x

 τ1 (r) 1 − r 1 n u ε exp + τ2 (u) . (−1) Kn (x; p, N ) ∼ √ r ε 1−r

(67)

(68)

The parabolic cylinder function Dx (u) has the asymptotic approximations [70, 16.5]  2 u Dx (u) ∼ exp − ux , u → ∞, (69) 4 and [70, 16.52] √ u−x−1 exp Dx (−u) ∼ 2π Γ (−x)



u2 4



 2 u + exp − ux , 4

u → ∞.

Using (69) and (70) in (55), we obtain as u → ∞ 

√ u−x−1 u2 x Kn (x; p, N ) ∼ Cp (x) u + 2πCm (x) e 2 exp [τ2 (u)] , Γ (−x) 14

(70)

and comparing with (68), we conclude that Cm (x) = 0 and 1 Cp (x) = √ 1−r √ Therefore, for p − r = O ( ε) 



x 

1−r 2 τ1 (r) ε exp . r ε

 τ1 (r) u2 + τ2 (u) + Dx (u) . exp ε 4 (71) − On the other hand, using (49) in (65), we obtain as p → r

1 (−1) Kn (x; p, N ) ∼ √ 1−r n

 x2

1−r ε r



τ0 (p) τ0 (r) τ1 (r) u2 u2 ∼ + τ2 (u) + = + τ2 (u) + , ε ε 2 ε 2 since τ1 (r) = τ0 (r). Using (49) we also have

−x−1  , (1 − p)x (r − p)−x−1 ∼ (1 − r)x −u r (1 − r) ε

p → r− .

(72)

(73)

Using (72) and (73) in (64), we see that as p → r− 

x

 1−r 2 τ1 (r) ε exp + τ2 (u) r ε  −x−1 √ u2 (−u) × k (x) ux + 2π e2 . Γ (−x)

1 (−1) Kn (x; p, N ) ∼ √ 1−r  n

(74)

Using the asymptotic approximation (70) in (71) and comparing with (74), we conclude that k (x) = (−1)x . (75) Note that k (x) satisfies the condition (26).

4

Main results

Replacing (75) in (24), we conclude that for 0 ≤ p < r < 1 Kn (x; p, N ) ∼ ϕn (x) (1 − p)n−N ξ0 (p; x) + (−1)x (x − N )n pn ξ1 (p; x) . (76) Using (31) and (42) in (23) and (76), we obtain the following result. 15

Proposition 2 Let Kn (x; p, N ) be defined by (1), with x = o(N ), n = rN, and ε = N −1 . Then, as ε → 0 we have the following asymptotic approximations: (i) For 0 < r < p ≤ 1,



Kn (x; p, N ) ∼ (x − N )n p

n−x

p−r 1−r

x g (p; x, r, ε) ,

(77)

where g (p; x, r, ε) was defined in (34). (ii) For 0 ≤ p < r < 1, x+1 r Kn (x; p, N ) ∼ ϕn (x) (1 − p) h (p; x, r, ε) r−p x  r−p n−x g (p; x, r, ε) , + (x − N )n p 1−r n+x−N



(78)

where ϕn (x) was defined in (7), and h (p; x, r, ε) was defined in (45). We now have all the elements needed to state the Mehler–Heine type formulas for the Krawtchouk polynomials. Corollary 3 Let Kn (x; p, N ) be defined by (1), with n = rN. Then, we have: (i) For 0 < r < p ≤ 1,

z

p−r Kn (z; p, N ) = , lim n→∞ (z − N ) pn (1 − r) p n

z ∈ C.

(79)

(ii) For 0 ≤ p < r < 1, lim

(−1)n Kn (z; p, N )

n→∞ Γ (n

− z) (1 − p)n−N

=

p −z−1 (1 − p)z 1− , Γ (−z) r

z ∈ C.

(80)

Proof. Part (i) is just a rewriting of (77), after using (2) and g (p; x, r, 0) = 1. To see that Part (ii) is true, note that from (59) and (63) we have   1−p   p x− 12 + (1 − r) ln r ln 1 (1 − r) (x − N )n pn r 1−r √ , = Γ (−x) (εr)x+ 2 exp n−N ε 2π ϕn (x) (1 − p) 16

and r ln

p r

 + (1 − r) ln

1−p 1−r

 ≤ 0,

p, r ∈ [0, 1] .

Thus, as long as x ∈ / N, the second term in (78) is exponentially small. To remove the conditions on x, we observe that the rhs of (79) and (80) are entire functions of z, and therefore by analytic continuation the limits are valid on the whole complex plane.

4.1

Zeros

If we set x = 1 in (1), we have  n

Kn (1; p, N ) = p (−N )n

n 1− Np

 ,

and hence Kn (1; p, N ) = 0,

n = N p.

(81)

Note that this agrees with (71), since [54, 12.7.2]  2 u exp D1 (u) = u, 4 and therefore if u = 0 (i.e., n = N p), we get (81). For the special case p = 12 (the so called binary Krawtchouk polynomials), we can use the identity [54, 15.4.30]   √ 21−b πΓ (b) a, 1 − a 1 ; =  a+b   b−a+1  2 F1 b 2 Γ 2 Γ 2 in (1), and obtain     √ 2x−n πΓ (n + 1 − x) 1 1−x n  = (−2) Kn x; , 2n = (x − 2n)n  1−x   . 2 2 Γ 2 Γ n + 1 − 12 x n   Hence, Kn x; 12 , 2n has zeros at the odd integers x = 2k − 1, k = 1, . . . , n. If we use the expansion (77) with g (p; x, r, ε) ∼ 1 + g1 (p; x, r) , we have x   p−r1  Kn (x; p, N ) 1 + g1 (p; x, r) N −1 , N → ∞, ∼ n (x − N )n p 1−rp 17

where g1 (p; x, r) was defined in (38). Solving for x in 2 (1 − r) (p − r)2 N = 0, x (x − 1) − r (p − 2r + 1) (1 − p) we obtain an approximation for the smallest zero of Kn (x; p, N ) in the region 0 < r < p ≤ 1,  1 1 8 (1 − r) (p − r)2 x1 ∼ + N. (82) 1+ 2 2 r (p − 2r + 1) (1 − p) For example, when n = 10, N = 50, and p = 0.347, the approximation √ (82)

gives x1 4.27, while the exact value is x1 = 4.11. Note that x1 = O N , and this is in the region x = o(N ), where our approximations are valid. If we expand (78) as x → 0, with g (p; x, r, ε) ∼ 1 and h (p; x, r, ε) ∼ 1, we have   (−1)n N r x n−N pn , Kn (x; p, N ) ∼ − (1 − p) + n n! n r−p where we have used (57) and the expansion ϕn (x) ∼ (−1)n−1 (n − 1)!x,

x → 0.

Therefore, the smallest zero of Kn (x; p, N ) in the region 0 ≤ p < r < 1 is asymptotically given by n    p N p N (1 − p) x1 ∼ n , N → ∞. (83) 1− n 1−p r For example, when n = 40, N = 50, and p = 0.347, the approximation (83) gives x1 1.346 × 10−9 , while the exact solution is x1 = 1.31 × 10−9 .

4.2

Plots

In order to plot the approximations (77)-(78), we need to rescale the terms. We define −x  p−r Kn (x; p, N ) F1 (x, n, N, p) = , (x − N )n pn−x 1 − r 18

Figure 1: A plot of the scaled polynomial K20 (x; 0.747, 50) and its approximations. and Gm (x, n, N, p) =

m 

gk (p; x, r) N −k .

k=0

It follows from (77) that for 0 < r < p ≤ 1 F1 (x, n, N, p) ∼ Gm (x, n, N, p) ,

N → ∞.

In Figure 1, we plot F1 (x, 20, 50, 0.747) and Gm (x, 20, 50, 0.747) for 1 ≤ m ≤ 3. Note that in this case r = 0.4 < p and ε = 0.02. From (62) and the reflection formula 5.5.3 1 sin (πz) =− Γ (z + 1) , Γ (−z) π we have ϕn (x) = (−1)n

Γ (n − x) sin (πx) = − (−1)n Γ (n − x) Γ (x + 1) , Γ (−x) π 19

and we can write (78) as (−1)n πKn (x; p, N ) x+1 ∼ − sin (πx) Γ (x + 1) h (p; x, r, ε) n+x−N r Γ (n − x) (1 − p) r−p x  r−p pn−x p x+1 n (x − N )n + (−1) π g (p; x, r, ε) . 1 − Γ (n − x) (1 − p)n+x−N 1 − r r We define F0 (x, n, N, p) =

(−1)n πKn (x; p, N ) x+1 , n+x−N r Γ (n − x) (1 − p) r−p

and sin (πx) Γ (x + 1) Hm (x, n, N, p) π x r−p pn−x p x+1 n (x − N )n +π (−1) Gm (x, n, N, p) , 1 − Γ (n − x) (1 − p)n+x−N 1 − r r Um (x, n, N, p) = −

where Hm (x, n, N, p) =

m 

hk (p; x, r) N −k .

k=0

It follows from (78) that for 0 ≤ p < r < 1, F0 (x, n, N, p) ∼ Um (x, n, N, p) ,

N → ∞.

In Figure 2, we plot F0 (x, 20, 50, 0.147) and Um (x, 20, 50, 0.147) for 0 ≤ m ≤ 3. Note that in this case r = 0.4 > p and ε = 0.02.

5

Comparison with previous results

In previous works ([13], [30], [40]), several authors have studied multiples of the monic Krawtchouk polynomials with leading coefficient n!1 . From Stirling’s formula (58) with n = rε−1 , we have 

 1 − ln (r) + ln (ε) 1 ε ∼ exp r , ε → 0. (84) n! 2πr ε 20

Figure 2: A plot of the scaled polynomial K20 (x; 0.147, 50) and its approximations.

21

Using (84) in (60), we get for 0 < r < p ≤ 1  x 

 ω (p) ε r n Kn (x; p, N ) ∼ 1− , exp (−1) n! 2π (1 − r) r p ε

(85)

where ω (p) = (r − 1) ln (1 − r) + r [ln (p) − ln (r)] .

(86)

Using (75) and (84) in (64), we get for 0 ≤ p < r < 1 x+1 

1−p (r − 1) ln (1 − p) ε exp r−p ε (87)  x 

 r ω (p) ε + − 1 exp . 2π (1 − r) r p ε  Finally, using (84) in (71) we get for p − r = u r (1 − r) ε Kn (x; p, N ) 1 1 (−1) ∼ n! 1 − p Γ (−x) n



 x  Kn (x; p, N ) ε 1−r 2 (−1) ∼ ε Dx (u) n! 2π (1 − r) r r    (r − 1) ln (1 − r) r (1 − r) 1 2 +u − u (1 − 2r) . × exp ε ε 4 n

(88)

 Since r (1 − r) ≤ 12 for 0 ≤ r ≤ 1, we can combine (85), (87), and (88) in the following Theorem. Theorem 4 Let Kn (x; p, N ) be defined by (1), with x = o(N ), n = rN, u = O(1), and ε = N −1 . Then, as ε → 0 the asymptotic approximations for (−1)n

Kn (x; p, N ) n!

are given by: (i) (85) for r +

u√ ε 2

< p < 1. √ (ii) (87) for 0 < p < r − u2 ε. √ (iii) (88) for |p − r| < u2 ε. 22

(89)

Let q = 1 − p. In [13], we analyzed the recurrence relation (n + 1)Kn+1 + pq(N − n + 1)Kn−1 + [p(N − n) + nq − x] Kn = 0, ) satisfied by Kn (x;p,N , and obtained 12 relevant asymptotic approximations. n! We summarized our results in Section 10, and comparing with the work presented in this paper, we see that: 1) The approximation (85) is identical to the one we called K − (Region III). The restriction for x was

0≤

 x < p + (q − p) r − 2 pq(1 − r)r, N

(90)

which is sharper than our requirement that x = o (N ) . For example, when p = 0.747, r = 0.4, and N = 50 (as in Figure 1), the upper bound (90) gives x < 6. 172 5. 2) The approximation (87) is identical to the one we called K (5) (Region V). 3) The approximation (88) is equivalent to the one we called K (6) (Region VI). We used a different scaling, namely √ r = p − u pqε, instead of (49). In [30], Ismail and Simeonov studied the integral representation Kn (x; p, N ) 1 = n! 2πi



(1 + qt)x (1 − pt)N −x dt, tn+1

(91)

|t|=R

which follows from the generating function (5). They summarized their results in Theorem 4.1, and assuming that 0 ≤ p < 12 and x 0<


p1−r −1 q r

we see that: 23

2 ,

1) For 1 − p < r < 1 they obtain an approximation (equation 4.32) that can be reduced to (87). 2) For p < r ≤ 1 − p they obtain an approximation (equation 4.32) that can be reduced to (88). 3) For 0 < r ≤ p they obtain an approximation (equation 4.37) that can be reduced to (85). Since they used the saddle point method, their formulas are not explicit. In [40], Li and Wong again used the integral representation (91) and the saddle point method, and derived a uniform expansion (equation 5.14). They showed that their results agree what those of Ismail and Simeonov. In particular, their equation 6.20 can be reduced to (87).

6

Conclusions

We have found asymptotic expansions for the monic Krawtchouk polynomials Kn (x; p, N ) , valid for n = O (N ) , x = o(N ) and all values of p ∈ [0, 1] . We have also obtained asymptotic approximations for the smallest zero. In a sequel, we plan to use similar methods to study the Hahn polynomials.

7

Acknowledgments

This work was done while visiting the Johannes Kepler Universit¨at Linz and ¨ 2010 plus” from the supported by the strategic program ”Innovatives OO– Upper Austrian Government. Finally, we also wish to express our gratitude to the anonymous referees, who provided us with invaluable suggestions and comments that greatly improved previous drafts of the paper.

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8

Appendix A

In this section, we study the sequence of functions gk (p; x, r) defined by the recurrence dgk d2 gk−1 dgk−1 = A2 + A0 gk−1 , + A1 2 dp dp dp

k∈N

(92)

with initial conditions g0 (p; x, r) = 1 and gk (1; x, r) = 0,

k ∈ N,

where p (1 − p) p − r + 2pr + px + rx − 2p2 − 2prx , A1 = , p−r (p − r)2 r (1 − r) x (x − 1) . A0 = (p − r)3

A2 =

If we let

gk (p; x, r) ∼ Ck (p − r)sk , 31

p → r,

(93)

then we see from (92) that as p → r sk Ck (p − r)sk −1 ∼ r (1 − r) Ck−1 sk−1 (sk−1 − 1) (p − r)sk−1 −3 +2r (1 − r) xCk−1 sk−1 (p − r)sk−1 −3 + r (1 − r) x (x − 1) Ck−1 (p − r)sk−1 −3 . Hence, sk − 1 = sk−1 − 3, sk Ck = Ck−1 r (1 − r) (x + sk−1 ) (x + sk−1 − 1) , and since C0 = 1, s0 = 0, we conclude that

k 1 r (r − 1) sk = −2k, Ck = (−x)2k . k! 2

(94)

Thus, we consider a solution of (92)-(93) of the form gk (p; x, r) =

2k  ak,j (x, r) j=0

(p − r)j

.

We shall use Einstein summation convention, and write (95) as ak,j , ak,j = 0, j ∈ / {0, 1, . . . , 2k} . gk = (p − r)j

(95)

(96)

Replacing (96) in (92), we obtain

 r (1 − r) 1 − 2r 1 ak,j = j (j + 1) ak−1,j + − −j (p − r)j+1 (p − r)j+3 (p − r)j+2 (p − r)j+1 (97)

 r (1 − r) x (x − 1) 2r (1 − r) x (1 − 2r) (x + 1) 2 − . −jak−1,j j+3 + j+2 j+1 + ak−1,j (p − r) (p − r) (p − r) (p − r)j+3 Shifting indices in (97), we get r (1 − r) 1 − 2r ak,j j+1 = (j − 2) (j − 1) ak−1,j−2 j+1 + (j − 1) jak−1,j−1 (p − r) (p − r) (p − r)j+1 1 2r (1 − r) x (1 − 2r) (x + 1) −j (j + 1) ak−1,j j+1 − (j − 2) ak−1,j−2 j+1 − (j − 1) ak−1,j−1 (p − r) (p − r) (p − r)j+1 (98) 2 r (1 − r) x (x − 1) +jak−1,j . j+1 + ak−1,j−2 (p − r) (p − r)j+1

−j

32

Simplifying (98), we conclude that (x − j + 1) (x − j + 2) (j − 1) (x − j + 1) ak−1,j−2 −(2r − 1) ak−1,j−1 +(j − 1) ak−1,j . j j (99) To find ak,0 , we use the initial condition (93), and obtain ak,j = r (r − 1)

ak,0 (x, r) = −

2k  ak,j (x, r) j=1

(1 − r)j

.

(100)

Using (99)-(100) with initial conditions a0,j = δj,0 , we get a1,0 = a2,0 a2,2 a2,3

1 r ϕ2 (x) , 21−r

1 a1,2 = r (r − 1) ϕ2 (x) , 2

1 r (2r + 8x + 3rx2 − 7rx − 4) = ϕ2 (x) , 24 (r − 1)2  1  = − r −2r + rx2 − rx + 2 ϕ2 (x) , 4 1 1 = − r (2r − 1) (r − 1) ϕ3 (x) , a2,4 = r2 (r − 1)2 ϕ4 (x) , 3 8

and 1 r (−16r − 5r2 x + r2 x2 + 8rx + 6r2 + 12) [ϕ2 (x)]2 , a3,0 = − 3 48 (r − 1) 2 1 r (−48r + 10r x + 8rx3 − 3r2 x2 − 10r2 x3 + 3r2 x4 − 8rx + 24r2 + 24) a3,2 = ϕ2 (x) , 48 r−1   1 a3,3 = r (2r − 1) −6r + rx2 − rx + 6 ϕ3 (x) , 6   1 a3,4 = − r (r − 1) 24r − r2 x + r2 x2 − 24r2 − 4 ϕ4 (x) , 16 1 2 1 a3,5 = − r (2r − 1) (r − 1)2 ϕ5 (x) , a3,6 = r3 (r − 1)3 ϕ6 (x) , 6 48 where ϕn (x) was defined in (7). Note that in each case we get ak,1 = 0 and

k

k 1 r (r − 1) 1 r (r − 1) ak,2k = ϕ2k (x) = (−x)2k , k! 2 k! 2 in agreement with (94). Also, ak,j (0, r) = 0 for k ≥ 1, in agreement with (25). 33

9

Appendix B

In this section, we study the sequence of functions hk (p; x, r) defined by the recurrence dhk d2 hk−1 dhk−1 = B2 + B0 hk−1 , + B 1 dp dp2 dp

k∈N

(101)

with initial conditions h0 (p; x, r) = 1 and hk (0; x, r) = 0,

k ∈ N,

(102)

where p (p − 1) (p + r − 2pr) (x + 1) , B1 = , p−r (p − r)2 (r2 x + 2pr − rx − p − r) (x + 1) B0 = . (p − r)3

B2 =

A similar argument to the one used in the previous section shows that we shall consider a solution of (101)-(102) of the form hk (p; x, r) =

2k  bk,j (x, r) j=0

(p − r)j

,

which we can write as hk =

bk,j , (p − r)j

bk,j = 0,

j∈ / {0, 1, . . . , 2k} .

(103)

Replacing (103) in (101), we obtain

 r (r − 1) bk,j 2r − 1 1 −j = j (j + 1) bk−1,j + + (p − r)j+1 (p − r)j+3 (p − r)j+2 (p − r)j+1

 2r (1 − r) (x + 1) (1 − 2r) (x + 1) −jbk−1,j (104) + (p − r)j+3 (p − r)j+2

 r (r − 1) (x + 1) (x + 2) (2r − 1) (x + 1) +bk−1,j . + (p − r)j+3 (p − r)j+2

34

Shifting indices in (104), we get r (r − 1) 2r − 1 bk,j j+1 = (j − 2) (j − 1) bk−1,j−2 j+1 + (j − 1) jbk−1,j−1 (p − r) (p − r) (p − r)j+1 2r (1 − r) (x + 1) (1 − 2r) (x + 1) bk−1,j +j (j + 1) − (j − 1) bk−1,j−1 j+1 − (j − 2) bk−1,j−2 j+1 (p − r) (p − r) (p − r)j+1 (105) r (r − 1) (x + 1) (x + 2) (2r − 1) (x + 1) +bk−1,j−2 + bk−1,j−1 . j+1 (p − r) (p − r)j+1 −j

Simplifying (105), we conclude that r (1 − r) (x + j − 1) (x + j) bk−1,j−2 +(1 − 2r) (x + j) bk−1,j−1 −(j + 1) bk−1,j . j (106) To find bk,0 , we use the initial condition (102), and obtain bk,j =

bk,0 (x, r) = −

2k  bk,j (x, r) j=1

(−r)j

.

(107)

Using (106)-(107) with initial conditions b0,j = δj,0 , we get 1 (rx − 2r − x) (x + 1)1 , 2 r 1 = r (1 − r) (x + 1)2 , 2

b1,0 = b1,2

24r2 − 2x + 12rx − 10r2 x − x2 + 6rx2 − 5r2 x2 + 3x3 − 6rx3 + 3r2 x3 (x + 1)1 , 24r2 1 (1 − 2r) (−6r − x + rx2 − rx − x2 ) (x + 1)1 , = 2 r  1 = − 24r + x − 2rx2 − r2 x + r2 x2 − 24r2 + x2 − 4 (x + 1)2 , 4 5 1 = (2r − 1) r (r − 1) (x + 1)3 , b2,4 = r2 (1 − r)2 (x + 1)4 , 6 8

b2,0 = b2,1 b2,2 b2,3

b1,1 = (1 − 2r) (x + 1)1 ,

and 35

1  3 2 12r x + 20r3 x − 48r3 − 14r2 x2 − 24r2 x + 4rx2 + 4rx − 2x2 48r3  +x3 − 7rx3 + 15r2 x3 − 9r3 x3 + 2x4 − 4rx4 + 2r2 x4 − x5 + 3rx5 − 3r2 x5 + r3 x5 (x + 1)1 , 1 2r − 1  168r2 − 2x + 36rx + 38r2 x − 3x2 =− 24 r2  +42rx2 − 39r2 x2 + 2x3 − 2r2 x3 + 3x4 − 6rx4 + 3r2 x4 (x + 1)1 , 1  146r3 x + 1464r3 − 2r2 x − 1464r2 − 177rx2 − 122rx + 288r + 27x2 + 26x =− 48r  +3r3 x4 − 2r3 x3 − 147r3 x2 − 9r2 x4 + 2r2 x3 + 297r2 x2 + 9rx4 + 2rx3 − 3x4 − 2x3 (x + 1)2 ,   1 = (2r − 1) 5r2 x2 − 5r2 x − 124r2 − 10rx2 + 124r + 5x2 + 5x − 12 (x + 1)3 , 12   1 = (r − 1)r 3r2 x2 − 3r2 x − 262r2 − 6rx2 + 262r + 3x2 + 3x − 52 (x + 1)4 , 48 7 1 = − (2r − 1) r2 (r − 1)2 (x + 1)5 , b3,6 = r3 (1 − r)3 (x + 1)6 . 24 48

b3,0 =

b3,1

b3,2

b3,3 b3,4 b3,5

Note that in each case we get bk,2k

k 1 r (1 − r) = (x + 1)2k , k! 2

in agreement with (106), since it is the solution of the leading term recurrence bk,2k =

r (1 − r) (x + 2k − 1) (x + 2k) bk−1,2(k−1) . 2k

36