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Asymptotic analysis of the Krawtchouk polynomials by 5 the WKB method 0 0 2 n Diego Dominici ∗ a J Department of Mathematics 4 State University of New York at New Paltz ] 75 S. Manheim Blvd. Suite 9 A C New Paltz, NY 12561-2443 h. USA t a m February 1, 2008 [ 2 v 2 Abstract 4 0 1 We analyze the Krawtchouk polynomials Kn(x,N,p,q) asymptotically. We use 0 singular perturbation methods to analyze them for N , with appropriate scalings 5 → ∞ ofthetwovariablesxandn. Inparticular,theWKBmethodandasymptoticmatching 0 / are used. We obtain asymptotic approximations valid in the whole domain [0,N] h × t [0,N], involving some special functions. We give numerical examples showing the a accuracy of our formulas. m : v i 1 Introduction X r a Definition 1 Let n,N 0 be integers. The Krawtchouk polynomials K (x) are defined by n ≥ [36], n x N x K (x) = − qk( p)n−k (1) n k n k − k=0(cid:18) (cid:19)(cid:18) − (cid:19) X where 0 < p,q < 1 p+q = 1. (2) The binary Krawtchouk polynomials are the special case with p = 1/2 = q [15], [18]. ∗ e-mail: [email protected] 1 The Krawtchouk polynomials [27] are one of the families of classical orthogonal polynomials of a discrete variable [14]. They satisfy the orthogonality relation N N K (k)K (k)̺(k)= (pq)jδ , i,j = 0,...,N i j ij j k=0 (cid:18) (cid:19) X with weight function N ̺(x) = pxqN−x. x (cid:18) (cid:19) Writing the Krawtchouk polynomials in the extended form K (x) = K (x,N,p,q), we n n have the the symmetry formula K (x,N,p,q) = ( 1)nK (N x,N,q,p). (3) n n − − They also satisfy the three-term recurrence (n+1)Kn+1(x)+pq(N n+1)Kn−1(x)+[p(N n)+nq x]Kn(x) = 0 − − − which is the main object of our analysis. The Krawtchouk polynomials are important in the study of the Hamming scheme of classical coding theory [17], [21], [25], [29], [33], [34]. Lloyd’s theorem [23] states that if a perfect code exists in the Hamming metric, then the Krawtchouk polynomial must have integral zeros [3], [5], [20]. Not surprisingly, these zeros have been the subject of extensive research [4], [7] [9], [10], [11], [16], [35]. The Krawtchouk polynomials also have applications in probability theory [8], queueing models [6], stochastic processes [31], quantum mechanics [2], [24] [39] and biology [13]. The asymptotic behavior of the Krawtchouk polynomials as N was studied by → ∞ Sharapudinov for x Np and n = O N1/3 in [32]. He derived an approximation in ≈ terms of the Hermite polynomials (see also [36] for a similar formula). The general case (cid:0) (cid:1) with x,n = O(N) was investigated by Ismail and Simeonov [12] and a uniform asymptotic expansion was derived by Li and Wong [22], both using the saddle point method. The purpose of this paper is to take a different approach, based on the recurrence formula that the Krawtchouk polynomials satisfy and using singular perturbation techniques [26], [37] to analyze it. We scale x = yN, n = zN and obtain asymptotic approximations to K (x) for (y,z) [0,1] [0,1]. Our results agree and extend those obtained in [12], [22] and n ∈ × [30]. InSection 2we review thebasicproperties ofthe Krawtchouk polynomials. InSections 3- 10 we obtain asymptotic expansions from the recurrence formula by using the WKB method. Wemustconsidertwelve relevantregionsofthetwo-dimensionalstatespace. InSection11we summarize our results and numerically compare our approximations with the exact formula. 2 2 The WKB approximation To analyze the recurrence (n+1)Kn+1(x)+pq(N n+1)Kn−1(x)+[p(N n)+nq x]Kn(x) = 0 (4) − − − subject to the boundary conditions K (x) = 1 (5) 0 x K (x) = (6) N+1 N +1 (cid:18) (cid:19) N K (0) = ( p)n (7) n n − (cid:18) (cid:19) N K (N) = qn (8) n n (cid:18) (cid:19) for large N, we introduce the scaled variables y,z defined by x = yN, n = zN, 0 < y,z < 1. (9) We define the function F(y,z) and the small parameter ε by 1 ε = , K (x) = F(εx,εn) = F(y,z) (10) n N and observe that Kn±1(x) = F(y,z ε). ± Substituting (9)-(10) in (4) we get (z +ε)F(y,z+ε)+pq(1 z +ε)F(y,z ε)+[p(1 z)+zq y]F(y,z) = 0. (11) − − − − To find F(y,z) for ε small, we shall use the WKB method [28]. Thus, we consider solutions which have the asymptotic form F(y,z) ενexp ε−1ψ(y,z) L(y,z). (12) ∼ Using (12) in (11), with (cid:2) (cid:3) 1 ε−1ψ(y,z ε) = ε−1ψ(y,z) ψ (y,z)+ ψ (y,z)ε+O ε2 , z zz ± ± 2 dividing by exp[ε−1ψ(y,z)] and expanding in powers of ε we obtain the(cid:0)eik(cid:1)onal equation zU2 +[p y +z(q p)]U +pq(1 z) = 0 (13) − − − and the transport equation 1 zU2 pq(1 z) L + zU2 +pq(1 z) ψ +U2 +pq L = 0 (14) z zz − − 2 − (cid:26) (cid:27) (cid:2) (cid:3) (cid:2) (cid:3) with U(y,z) = exp[ψ (y,z)]. (15) z 3 2.1 The functions ψ and L From (15) we have U U dz z z ψ(y,z) = ln[U(y,z)]dz = zln(U) z dz = zln(U) z dU − U − U dU Z Z Z and using (13) we get U dz (y p)U pq 1 y 1 y z z dU = − − dU = + − dU. U dU (U +q)(U p) U U − U +q Z Z − Z (cid:20) (cid:21) Hence, ψ(y,z) = ln Uz−1(U p)1−y(U +q)y +A(y) (16) − where the function A(y) is still unk(cid:2)nown. (cid:3) From (14) we have 1 [zU2 +pq(1 z)]ψ +U2 +pq L(y,z) = B(y)exp 2 − zz dz − zU2 pq(1 z) (cid:20) Z − − (cid:21) and from (15) ψ = U /U. After changing variables from z to U, we obtain zz z 1 1 (p y)U +pq L(y,z) = B(y)exp + − dU . 2U U +q − (p y)U2 +2pqU +pq(q y) (cid:26)Z (cid:20) − − (cid:21) (cid:27) Thus, U L(y,z) = B(y)(U +q) (17) (p y)U2 +2pqU +pq(q y) s − − where B(y) is to be determined. 2.2 The function U Rewriting (13) as p y U2 + − +q p U +(U )2 = 0 0 z − (cid:18) (cid:19) and solving for U we get 2 1 p y 1 p y U±(y,z) = − +q p − +q p 4(U )2 (18) 0 −2 z − ± 2s z − − (cid:18) (cid:19) (cid:18) (cid:19) where pq(1 z) U (z) = − . (19) 0 z r 4 The discriminant in (18) vanishes if p−y +q p = 2U , which is equivalent to y = Y±, z − ± 0 with ± Y (z) = p+(q p)z 2zU . (20) 0 − ± Rewriting the equation p−y +q p 2 4(U )2 = 0 as z − − 0 2 2 1 (cid:0) 1 (cid:1) 1 1 y + z +2(p q) y z pq = 0 (21) − 2 − 2 − − 2 − 2 − (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19) we can see that (y,z) [0,1] [0,1] : y = Y±(z) = E (22) ∈ × where E is an ellipse centered at 1, 1 (see Figure 1). After rotation by π and translation E (cid:8) 2 2 (cid:9) ±4 to the origin, reduces to one of the canonical forms (cid:0) (cid:1) y2 z2 + = 1, p < q q/2 p/2 1 1 y2 +z2 = , p = = q 4 2 y2 z2 + = 1, p > q. p/2 q/2 The ellipse E is contained in the square [0,1] [0,1] and intersects the y and z axis at the × points (0,p),(1,q),(p,0) and (q,1). Its left side between the points (q,1) and (p,0) coincides with the curve Y−(z) and its right side with the curve Y+(z). Forpoints(y,z)locatedoutsideE,U±(y,z)arerealandforpoints(y,z)insideEU±(y,z) ± ± are complex conjugates. When y = Y the two values U coalesce and we have U+(Y+,z) = U = U−(Y+,z), U+(Y−,z) = U = U−(Y−,z). (23) 0 0 − Writing the function L(y,z) in terms of U we have from (17) 0 (U p)(U +q) L(y,z) = B(y) − . (24) s z(U2 U2) − 0 In the rest of this paper we shall use the following notation ± ± z−1 ± 1−y ± y ψ (y,z) = ln U U p U +q (25) − h i and (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (U± p)(U± +q) ± L (y,z) = − . (26) s z (U±)2 U2 − 0 Hence, we write (cid:2) (cid:3) K (x) ενB−(y)exp ε−1ψ−(y,z)+ε−1A−(y) L−(y,z) (27) n ∼ +ενB+(y)exp ε−1ψ+(y,z)+ε−1A+(y) L+(y,z) (cid:2) (cid:3) ± ± ± where B (y) and A (y) are functions to be determined. From (26) we see that L (y,z) are (cid:2) (cid:3) ± ± singular when U = U , i.e., for y = Y , and also for z = 0. Therefore, we need to find 0 asymptotic solutions valid in those regions. 5 (q,1) 1 0.8 + Y (z) (0,p) 0.6 0.4 z - (1,q) Y (z) 0.2 0 0.2 0.4 0.6 (p,0) 0.8 1 y Figure 1: A sketch of the ellipse E and the curves Y±(z). 6 3 The boundary n = 0 (Region I) For n = O(1) we have from (1) that K (x) = O(Nn) as N . Thus, we introduce the n → ∞ (1) function R (y) and consider solutions of (4) which have the asymptotic form n x K (x) = NnR(1) . (28) n n N (cid:16) (cid:17) Using (28) in (4) and expanding in powers of N gives, to leading order (n+1)R(1) +(p y)R(1) = 0. n+1 − n Solving the recursion above with the initial condition (5) we get (y p)n R(1)(y) = − n n! and hence (y p)n K (x) K(1)(y) = Nn − , n = O(1). (29) n ∼ n n! 3.1 The corner layer at (p,0) (Region II) We shall now find an asymptotic solution in the neighborhood of the point (p,0). We intro- (2) duce the stretched variable η and the function R (η) defined by n y = p+η 2pqε η = O(1), (30) K (x) = ε−n/2ppq n/2R(2) εx−p . n n! 2 n √2pqε (cid:18) (cid:19) (cid:16) (cid:17) Using (30) in (4) yields, to leading order, the equation R(2) 2ηR(2) +2nR(2) = 0 n+1 − n n−1 which we recognize as the recurrence relation for the Hermite polynomials. Thus, ε−n/2 pq n/2 K (x) K(2)(η) = H (η) (31) n ∼ n n! 2 n (cid:16) (cid:17) for n = O(1) and y p = O ε1/2 , where H (η) is the Hermite polynomial of degree n. n − (cid:0) (cid:1) 4 The lower corners (Regions III and IV) Setting n = z/ε in (29) and letting ε 0 we obtain → 1 K(1)(y) ε1/2 exp ε−1[1 ln(z)+ln(y p)]z (32) n ∼ √2πz − − (cid:8) (cid:9) 7 where we have used Stirling’s formula [1] 2π Γ(x) xxe−x, x . (33) ∼ x → ∞ r From (25)-(26) we have as z 0 → [1 ln(z)+ln(y p)]z, y < p ψ−(y,z) − − (34) ∼ (1 y)ln 1−y +yln y +ln pq z, y > p ( − q p y−p (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) z−1/2 +O(z), y < p − L (y,z) = √(y−1)y (35) ( y−p +O(z), y > p (1 y)ln 1−y +yln y +ln pq z, y < p ψ+(y,z) − q p y−p (36) ∼ ( [1(cid:16) ln(cid:17)(z)+ln((cid:16)y (cid:17)p)]z,(cid:16) y >(cid:17) p − − and √(y−1)y +O(z), y < p L+(y,z) = p−y . (37) ( z−1/2 +O(z), y > p Matching (34)-(37) and (32) we conclude that − − K (y,z), 0 < y < Y (z) K (x) (38) n ∼ K+(y,z), Y+(z) < y < 1 (cid:26) with 1 K−(y,z) = ε1/2 exp ε−1ψ−(y,z) L−(y,z) (39) √2π (cid:2) (cid:3) 1 K+(y,z) = ε1/2 exp ε−1ψ+(y,z) L+(y,z). (40) √2π (cid:2) (cid:3) Remark 2 In the remainderof the paper, we will find asymptotic formulas only in the region 0 y Y+(z), 0 z 1. The corresponding results for Y+(z) y 1 can be obtained ≤ ≤ ≤ ≤ ≤ ≤ by using the symmetry formula (3) and noting that under the transformations y 1 y, → − p q, K ( 1)nK , we obtain n n ↔ → − U− U+ → ψ− ψ+ +zπi, L− L+ → → K− K+. → 8 5 The boundary x = 0 We shall now consider the case x = O(1). Lemma 3 Let x = m, with m an integer, m N. ≪ 1. m N n K (m) = ( p)n 1 N−1 , m = 0,1. (41) n − n − p (cid:18) (cid:19)(cid:18) (cid:19) 2. If n = O(N), then m N n K (m) ( p)n 1 N−1 , N , m 2. (42) n ∼ − n − p → ∞ ≥ (cid:18) (cid:19)(cid:18) (cid:19) Proof. From (1) we have for x = m integer m m N m K (m) = − qk( p)n−k (43) n k n k − k=0(cid:18) (cid:19)(cid:18) − (cid:19) X m N−m k N m q = ( p)n n−k − n k N −p (cid:18) (cid:19)k=0(cid:18) (cid:19)(cid:0) n (cid:1) (cid:18) (cid:19) X (cid:0) (cid:1) and (41) follows for m = 0,1. Setting n = zN and using (33) we get N−m k z Nz−k (1 z)m , N . N ∼ − 1 z → ∞ (cid:0) Nz (cid:1) (cid:18) − (cid:19) Using the above in (43) we(cid:0)ha(cid:1)ve m k N m z q K (m) ( p)n (1 z)m n ∼ − n − k −1 zp (cid:18) (cid:19) k=0(cid:18) (cid:19)(cid:18) − (cid:19) X m N p z = ( p)n (1 z)m − − n − p(1 z) (cid:18) (cid:19) (cid:20) − (cid:21) and (42) follows. Using (33) we have, as N → ∞ N ε1/2 ( p)n exp ε−1φ (z) (44) 0 − n ∼ √2π z(1 z) (cid:18) (cid:19) − (cid:2) (cid:3) where p φ (z) = (z 1)ln(1 z) zln(z)+zln( p). (45) 0 − − − − 9 From (39) we get, as y 0 → − ε1/2 φ0(z)+ln p−pz y K (y,z) exp , 0 < z < p. (46) ∼ √2π z(1 z)  ε (cid:16) (cid:17)  − p   − Hence, as y 0, K (y,z) satisfies the boundary condition (7) for 0 < z < p. → 5.1 The boundary layer at x = 0, p < z < 1 (Region V) (5) Taking (42) into account, we define the function R (x) by n N K (x) = ( p)nR(5)(x). (47) n n − n (cid:18) (cid:19) Using (47) in (4) yields p(n N)R(5) +[pN +(q p)n x]R(5) nqR(5) = 0. (48) − n+1 − − n − n−1 Writing (48) in terms of z,ε and the function G(5)(x,z) defined by R(5)(x) = G(5)(x,εn) (49) n we have p(1 z)G(5)(x,z +ε)+[z(p q)+xε p]G(5)(x,z)+zqG(5)(x,z ε) = 0. (50) − − − − Using the WKB anszat 1 G(5)(x,z) ετ exp φ(x,z) W(x,z) (51) ∼ ε (cid:20) (cid:21) in (50), we obtain the equations [exp(φ ) 1][p(z 1)exp(φ )+zq]W = 0 (52) z z − − and [zq +p(z 1)exp(2φ )]W (53) z z − 1 + [p(z 1)exp(2φ ) zq]φ xexp(φ ) W = 0. z zz z 2 − − − (cid:26) (cid:27) Solving (52)-(53) we get G(5)(x,z) ετ1A(5)(x)exp ε−1ϕ (x) (z p)x (54) 1 ∼ − +ετ2B(5)(x)exp ε−1φ2(x,z) ((cid:2)z p)−x−(cid:3)1 z(1 z) − − (cid:2) (cid:3) p 10