Asymptotic analysis of the Krawtchouk polynomials by the WKB method

0.37 MB·English

by  Diego Dominici

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 polyno-
mials 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