Asymptotic analysis of the Askey-scheme I: fromKrawtchouk to Charlier
Diego Dominici ∗
Department of Mathematics
State University of New York at New Paltz75 S. Manheim Blvd. Suite 9
New Paltz, NY 12561-2443USA
Phone: (845) 257-2607
Fax: (845) 257-3571
January 5, 2005
∗e-mail: [email protected]
1
Abstract
We analyze the Charlier polynomials Cn(x) and their zeros asymptotically as n →∞. We obtain asymptotic approximations, using the limit relation between the Krawtchoukand Charlier polynomials, involving some special functions. We give numerical exam-ples showing the accuracy of our formulas.
Keywords: Charlier polynomials, Askey-scheme, asymptotic analysis, orthogonal poly-nomials, hypergeometric polynomials, special functions.
MSC-class: 33C45 (Primary) 34E05, 33C10 (Secondary)
2
1 Introduction
The Charlier polynomials Cn(x) [7] are defined by
Cn(x) = 2F0
(−n,−x
−
∣∣∣∣ −1
a
)(1)
where x ≥ 0, n = 0, 1, . . . and a > 0. They satisfy the discrete orthogonality condition [37]
∞∑
j=0
aj
j!Cn(j)Cm(j) = a−nean!δnm.
They are part of the Askey-scheme [18] of hypergeometric orthogonal polynomials:
4F3 Wilson Racah↓ ↘ ↓ ↘
3F2Continuousdual Hahn
ContinuousHahn
Hahn Dual Hahn
↓ ↙ ↓ ↙ ↓ ↘ ↙ ↓
2F1MeixnerPollaczek
Jacobi Meixner Krawtchouk
↘ ↓ ↙ ↘ ↓1F1 Laguerre Charlier 2F0
↘ ↙2F0 Hermite
where the arrows indicate limit relations between the polynomials.The Charlier polynomials have applications in quantum mechanics [28], [31], [36], [39],
difference equations [5], [23], teletraffic theory [16], [27], generating functions [4], [22], [26],and probability theory [3], [29], [30], [32]. The q-analogue of the Charlier polynomials werestudied in [2], [8], [19] and [41]. The generalized Charlier polynomials were analyzed in [14],[17], [33], [34] and [40].
Asymptotics for the Lp-norms and information entropies of Charlier polynomials werederived in [21]. Bounds for their zeros were obtained in [20]. Asymptotic representationswere established in [11] in terms of Hermite polynomials and in [24] in terms of Gammafunctions. Some asymptotic estimates were computed in [15] from a representation of Cn(x)in terms of Bell polynomials. An asymptotic formula when x < 0 was derived in [25] usingprobabilistic methods.
In [13], Goh studied the asymptotic behavior of Cn(x) for large n using an approximationof the Plancharel-Rotach type. A uniform asymptotic expansion was derived in [6] using thesaddle-point method. Asymptotic expansions were obtained in [10] from a second order lineardifferential satisfied by Cn(x) in which a is the independent variable and x is a parameter.
3
In this paper we shall take a different approach and investigate the asymptotic behaviorof Cn(x) as n→ ∞, by using the limit relation between the Krawtchouk polynomials Kn(x)defined by
Kn(x) = Kn(x, p,N) = 2F1
(−n,−x−N
∣∣∣∣1
p
), n = 0, 1, . . . , N, 0 ≤ x ≤ N, 0 ≤ p ≤ 1 (2)
and the Charlier polynomials, namely
limN→∞
Kn
(x,
a
N,N
)= Cn(x). (3)
We shall use the asymptotic expansions derived in [9] for the scaled Krawtchouk polynomialskn(x), with
kn(x) = kn(x, p,N) = (−p)n
(N
n
)Kn(x, p,N). (4)
A similar idea has been used in [12] and [38] to obtain asymptotic approximations ofseveral orthogonal polynomials of the Askey-scheme in terms of Hermite and Laguerre poly-nomials.
2 Preliminaries
The following is the main result derived in [9].
Theorem 1 As N → ∞, kn(x, p,N) admits the following asymptotic approximations (seeFigure1).
1. n = O(1), 0 ≤ y ≤ 1, y 6≈ p.
kn(x) ∼ k(1)n (y) =
ε−n
n!(y − p)n (5)
whereε = N−1, x =
y
ε, n =
z
ε0 ≤ y, z ≤ 1.
2. n = O(1), y ≈ p, y = p + η√
2pqε, η = O(1).
kn(x) ∼ k(2)n (η) =
ε−n2
n!
(pq2
)n2
Hn (η) , (6)
where q = 1 − p and Hn (η) is the Hermite polynomial.
4
X
VII
V
VI
III
III IVVIII
IX
XI
Y-
Y+
p
p
q
q
y
z
Figure 1: A sketch of the different asymptotic regions for kn(x).
5
3. 0 ≤ y < Y −(z), 0 < z < p, where
Y ±(z) = p+ (q − p) z ± 2zU0, U0(z) =
√pq(1 − z)
z. (7)
kn(x) ∼ k(3)(y, z) =
√ε√2π
exp[ψ(y, z, U−)ε−1
]L(z, U−), (8)
withψ(y, z) = (z − 1) ln(U) + (1 − y) ln(U − p) + y ln(U + q), (9)
L(y, z) =
√(U − p)(U + q)
z[U2 − (U0)
2] (10)
and
U±(y, z) = −1
2
(p− y
z+ q − p
)± 1
2
√(p− y
z+ q − p
)2
− 4 (U0)2. (11)
4. Y +(z) < y ≤ 1, 0 < z < q.
kn(x) ∼ k(4)(y, z) =
√ε√2π
exp[ψ(y, z, U+)ε−1
]L(z, U+). (12)
5. x = O(1), p < z < 1.
kn(x) ∼ k(5) (x, z) =
√ε√
2π√z (1 − z)
cos(πx)
(z − p
p
)x
exp[φ0(z)ε
−1]
(13)
− ε
π
x
z − pΓ(x) sin(πx)
(qε
z − p
)x
exp
[(z − 1) ln(q) + πiz
ε
]
whereφ0(z) = (z − 1) ln(1 − z) − z ln(z) + z ln(−p)
and Γ(x) is the Gamma function.
6. x = O(1), z ≈ p, z = p− u√pqε, u = O(1).
kn(x) ∼ k(6)(x, u) =
√ε√
2πpq
[√qε
p
]x
Dx(u) (14)
× exp
[πip− q ln (q)
ε+u√pqπi−u√pq ln (q)
√ε
− u2
4
],
where Dx(u) is the parabolic cylinder function.
6
7. 0 � y < Y −(z), p < z < 1.
kn(x) ∼ k(7)(y, z) = exp
(πiy
ε
) [cos
(πyε
)k(4)(y, z) + 2i sin
(πyε
)k(3)(y, z)
](15)
8. y ≈ Y −(z), 0 < z < p, y = Y −(z) − βε2/3, β = O(1).
kn(x) ∼ k(8)(β, z) = ε13 exp
[ψ0(z)ε
−1 + ln
(U0 + p
U0 − q
)βε−
13
]Ai
(Θ
23 β
) Θ− 13
√zU0
, (16)
where
ψ0(z) = zπi + (z − 1) ln (U0) + Y −(z) ln (U0 − q) +[1 − Y −(z)
]ln (U0 + p) , (17)
Θ(z) =
√U0
z
1
(U0 + p) (U0 − q)(18)
and Ai (·) is the Airy function.
9. y ≈ Y −(z), p < z < 1, y = Y −(z) − βε23 , β = O(1).
kn(x) ∼ k(9)(β, z) = ε13 exp
[ψ0(z)ε
−1 + ln
(U0 + p
U0 − q
)βε−
13
](19)
× 1
2
ϑ−13
√zU0
[λ+(β, z)Ai
(ϑ
23 β
)+ iλ−(β, z)Bi
(ϑ
23 β
)],
where ϑ(z) = −Θ(z),
λ±(β, z) = exp{
2πi[Y − (z) − βε
23
]ε−1
}± 1. (20)
and Ai (·) ,Bi (·) are the Airy functions.
10. Y −(z) < y < Y +(z), 0 < z < 1.
kn(x) ∼ k(10)(y, z) = k(3)(y, z) + k(4)(y, z). (21)
11. y ≈ Y +(z), 0 < z < q, y = Y +(z) + αε23 , α = O(1).
kn(x) ∼ k(11)(α, z) = ε13 exp
[ψ1(z)ε
−1 + ln
(U0 + q
U0 − p
)αε−
13
]Ai
[(Θ1)
23 α
] (Θ1)− 1
3
√zU0
,
(22)where
ψ1(z) = (z − 1) ln (U0) + Y +(z) ln (U0 + q) +[1 − Y +(z)
]ln (U0 − p) , (23)
Θ1(z) =
√U0
z
1
(U0 − p) (U0 + q)(24)
7
In order to obtain the corresponding asymptotic expansions for Kn(x), we need to deriveasymptotic formulas for (−p)n (
Nn
)in the different regions of Theorem 1.
The following lemma follows immediately from Stirling’s formula [1]
Γ(x) ∼√
2π
xxxe−x, x→ ∞. (25)
Lemma 2 As N → ∞, we have the following asymptotic approximations:
1.
(−p)n
(N
n
)∼ 1
n!(−p)n ε−n, n = O(1). (26)
2.
(−p)n
(N
n
)∼
√ε√
2π√z (1 − z)
exp[φ(z)ε−1
], n = zε−1 (27)
whereφ(z) = z ln(p) + zπi + (z − 1) ln(1 − z) − z ln(z). (28)
3.
(−p)n
(N
n
)∼
√ε√
2πpqexp [φ1(u)] , n = pε−1 − u
√pq
ε, u = O(1) (29)
where
φ1(u) = [πip− q ln(q)] ε−1 − u√pq [ln(q) + πi] ε−
12 − 1
2u2. (30)
3 Limit analysis
Settingy = xε, z = nε, q = 1 − p, p = aε (31)
in (7) and letting ε→ 0, we obtain
Y ±(z) → X±(a, n) =(√
n±√a)2, U0(z) →
√a
n. (32)
Hence, the eleven regions of Theorem 1 transform into the following regions (see Figure 2).
1. Region I
From (5) and (26) we have for n = O(1)
Kn(x) ∼(
1 − y
p
)n
. (33)
8
III
III IV
V
VI
VII
VIII
IX
X
XI
X-
X+
a
a
x
n
Figure 2: A sketch of the different asymptotic regions for Cn(x).
9
Thus,
Cn(x) '(1 − x
a
)n
. (34)
The formula above is exact for n = 0, 1 and in the limit as a→ ∞ we have
lima→∞
Cn(av) = (1 − v)n = 1F0
(−n−
∣∣∣∣ v).
2. Region II
From (6) and (26) we have for x = pε−1 + η√
2pqε, η = O(1)
Kn(x) ∼ (−1)n
(εq
2p
)n2
Hn (η) . (35)
Hence,Cn(x) ' (−1)n (2a)−
n2 Hn (η) , x = a+ η
√2a. (36)
Equation (36) is exact for n = 0, 1 and in the limit as a→ ∞ we have
lima→∞
(−1)n (2a)n2 Cn(a+ η
√2a) = Hn (η)
which is equation 2.12.1 in.
3. Region III
From (8), (27) and (28) we have for 0 ≤ y < Y −(z), 0 < z < p
Kn(x) ∼ K(3)(y, z) = exp
[ψ (y, z, U−) − φ(z)
ε
]G
(z, U−)
, (37)
where
G(z, U) =
√(1 − z)(U − p)(U + q)
U2 − Uo2. (38)
From (9), (11) and (38) we obtain
ψ (y, z, U−) − φ(z)
ε→ Ψ3(x),
where
Ψ3(x) = x ln
(a+ x− n+ ∆
2a
)+ n ln
(a− x + n+ ∆
2a
)+
1
2(a− x− n + ∆) (39)
and
G(z, U−)
→ L3(x) ≡√a− x− n+ ∆
2∆, (40)
10
2
4
6
8
10
–3 –2 –1 1 2
x
Figure 3: A comparison of Cn(x) (solid curve) and F3(x) (ooo) for n = 30 with a = 50.165184.
for 0 ≤ x < X−, 0 < n < a, with
∆(a, n, x) =√a2 − 2a(x + n) + (x− n)2. (41)
Thus,Cn(x) ∼ F3(x) = exp [Ψ3(x)]L3(x), 0 ≤ x < X−, 0 < n < a. (42)
Figure 3 shows the accuracy of the approximation (42) with n = 30 and a = 50.165184in the range −3 < x < X−.
Remark 3 Although we have not shown proof of it, the approximation (42) is in factalso valid for x < 0 and n ≥ 0 (see Figure 4).
4. Region IV
11
0
2e+25
4e+25
6e+25
8e+25
–3 –2.5 –2 –1.5 –1 –0.5
x
Figure 4: A comparison of Cn(x) (solid curve) and F3(x) (ooo) for n = 30 with a = 2.165184.
12
From (12), (27) and (28) we have for Y +(z) < y ≤ 1, 0 < z < q
Kn(x) ∼ K(4)(y, z) = exp
[ψ (y, z, U+) − φ(z)
ε
]G
(z, U+
). (43)
From (9), (11) and (38) we obtain
ψ (y, z, U+) − φ(z)
ε→ Ψ4(x) − nπi,
where
Ψ4(x) = x ln
(a+ x− n− ∆
2a
)+ n ln
(x− a− n + ∆
2a
)+
1
2(a− x− n+ ∆) (44)
and
G(z, U−)
→ L4(x) ≡√x− a+ n+ ∆
2∆, (45)
for X+ < x. Therefore,
Cn(x) ∼ F4(x) = (−1)n exp [Ψ4(x)]L4(x), X+ < x. (46)
Figure 5 shows the accuracy of the approximation (46) with n = 30 and a = 2.165184in the range X+ < x <∞.
5. Region V
From (13), (27) and (28) we have for x = O(1), p < z < 1
Kn(x) ∼ exp
[x ln
(z
p− 1
)]cos (πx) −
√ε
z − p
√2
πz (1 − z)Γ (x+ 1) sin (πx) (47)
× exp
(1 − z) ln(
1−zq
)+ z ln
(zp
)
ε+ x ln
(εq
z − p
) .
Hence,
Cn(x) ∼ F5(x) = exp[x ln
(na− 1
)]cos (πx) −
√n
√2
πΓ (x+ 1) sin (πx) (48)
× exp[n ln
(na
)− (x + 1) ln(n− a) + a− n
], x ≈ 0, n > a.
Figure 6 shows the accuracy of the approximation (48) with n = 30, a = 2.165184 andx ≈ 0.
13
0
1e+36
2e+36
3e+36
4e+36
48 49 50 51 52 53 54 55
x
Figure 5: A comparison of Cn(x) (solid curve) and F4(x) (ooo) for n = 30 with a = 2.165184.
14
–6e+20
–5e+20
–4e+20
–3e+20
–2e+20
–1e+20
00.5 1 1.5 2 2.5 3
x
Figure 6: A comparison of Cn(x) (solid curve) and F5(x) (ooo) for n = 30 with a = 2.165184.
15
0
0.2
0.4
0.6
0.8
1
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
x
Figure 7: A comparison of Cn(x) (solid curve) and F6(x) (ooo) for n = 30 with a = 30.165184.
6. Region VI
From (14) and (29)-(30) we have for x = O(1), z = p− u√pqε, u = O(1)
Kn(x) ∼ exp
[x
2ln
(qε
p
)+u2
4
]Dx (u) . (49)
Thus,
Cn(x) ∼ F6(x) = exp
[−x
2ln (a) +
u2
4
]Dx (u) , x ≈ 0, n = a− u
√a. (50)
Figure 7 shows the accuracy of the approximation (50) with n = 30, a = 30.165184(n ≈ a) and x ≈ 0.
7. Region VII
16
From (15), (37) and (43) we have for 0 � y < Y −(z), p < z < 1
Kn(x) ∼ exp
(πiy
ε
)[cos
(πyε
)K(4)(y, z) + 2i sin
(πyε
)K(3)(y, z)
]. (51)
ThereforeCn(x) ∼ exp (πix)
[cos (πx)C(4)
n (x) + 2i sin (πx)C(3)n (x)
]
for 0 � x < X−, n > a, which we can rewrite as
Cn(x) ∼ F7(x) = exp
[x ln
(n− a− x + ∆
2a
)+ n ln
(a+ n− x− ∆
2a
)+
1
2(a− x− n + ∆)
]
× cos (πx)
√x + n− a+ ∆
2∆− 2 sin (πx)
√x + n− a− ∆
2∆(52)
× exp
[x ln
(n− a− x− ∆
2a
)+ n ln
(a+ n− x + ∆
2a
)+
1
2(a− x− n + ∆)
].
Figure 8 shows the accuracy of the approximation (52) with n = 30 and a = 2.165184in the range 0 � x < X−.
8. Region VIII
From (16), (27) and (28) we have for y ≈ Y −(z), 0 < z < p, y = Y −(z) − βε2/3,β = O(1)
Kn(x) ∼ ε−16 exp
[ψ0(z) − φ(z)
ε+ ln
(U0 + p
U0 − q
)βε−
13
](53)
×√
2π
[z (1 − z)
pq
] 14
Ai(Θ
23 β
)Θ− 1
3 .
Sinceβ =
[Y −(z) − y
]ε−
23 (54)
we have from (32)
βε−13 → X− − x. (55)
From (17), (28) and (55) we get
ψ0(z) − φ(z)
ε+ ln
(U0 + p
U0 − q
)βε−
13 → 1
2n ln
(na
)+ x ln
(1 −
√n
a
)+√an−
√n.
From (18) and (55) we obtain
ε−16
√2π
[z (1 − z)
pq
] 14
Θ− 13 →
√2π
(na
) 16 (√
a−√n)1
3 (56)
17
–3
–2
–1
0
1
2
3
2 4 6 8 10 12 14 16
x
Figure 8: A sketch of Cn(x)/F7(x) for n = 30 with a = 2.165184. The vertical lines are dueto the discontinuities at x ' 13, 14 and 15.
18
and
Θ23 β →
(na
) 16 (X− − x)
(√a−
√n)
23
. (57)
Therefore,
Cn(x) ∼ F8(x) =√
2π(na
) 16 (√
a−√n) 1
3 Ai
[(na
) 16 (X− − x)
(√a−
√n)
23
](58)
× exp
[1
2n ln
(na
)+ x ln
(1 −
√n
a
)+√an−
√n
]
for x ≈ X−, 0 < n < a.
9. Region IX
From (19), (20), (27) and (28) we have for y ≈ Y −(z), p < z < 1, y = Y −(z) −βε
23 , β = O(1)
Kn(x) ∼ ε−16 exp
[ψ0(z) − φ(z)
ε+ ln
(U0 + p
U0 − q
)βε−
13
]
×√
2π
[z (1 − z)
pq
] 14
ϑ−13
[λ+(β, z)Ai
(ϑ
23 β
)+ iλ−(β, z)Bi
(ϑ
23 β
)],
which can be written as
Kn(x) ∼ ε−16
√2π exp
(z − 1) ln(
U0
1−z
)+ Y − ln (q − U0) + (1 − Y −) ln (p+ U0) + z ln
(zp
)
ε
(59)
× exp
[ln
(U0 + p
q − U0
)βε−
13
] [z (1 − z)
pq
] 14
ϑ−13
[cos (πx) Ai
(ϑ
23 β
)− sin(πx)Bi
(ϑ
23 β
)].
Using (56) and (57) in (59) with ϑ = −Θ, we have
Cn(x) ∼ F9(x) =√
2π(na
) 16 (√
n−√a) 1
3 exp
[1
2n ln
(na
)+ x ln
(√n
a− 1
)+√an− n
]
(60)
×
{cos (πx) Ai
[(na
) 16 (X− − x)
(√n−
√a)
23
]− sin(πx)Bi
[(na
) 16 (X− − x)
(√n−
√a)
23
]}
for x ≈ X−, n > a.
19
–1.2e+25
–1e+25
–8e+24
–6e+24
–4e+24
–2e+24
031 32 33 34 35 36
x
Figure 9: A comparison of Cn(x) (solid curve) and F10(x) (ooo) for n = 30 with a = 2.165184.
10. Region X
From (21), (37) and (43) we have for Y −(z) < y < Y +(z), 0 < z < 1
Kn(x) ∼ K(3)(y, z) +K(4)(y, z). (61)
Thus,Cn(x) ∼ F10(x) = F3(x) + F4(x), X− < x < X+. (62)
Figure 9 shows the accuracy of the approximation (62) with n = 30, a = 2.165184 inthe range X− < x < X+.
11. Region XI
From (22), (27) and (28) we have for y ≈ Y +(z), 0 < z < q, y = Y +(z)+αε23 , α =
20
O(1),
Kn(x) ∼ ε−16 exp
[ψ1(z) − φ(z)
ε+ ln
(U0 + q
U0 − p
)αε−
13
](63)
×√
2π
[z (1 − z)
pq
] 14
Ai[(Θ1)
23 β
](Θ1)
− 13 .
Sinceα =
[y − Y +(z)
]ε−
23 (64)
we have from (32)
αε−13 → x−X+. (65)
From (23), (28) and (65) we get
ψ1(z) − φ(z)
ε+ ln
(U0 + q
U0 − p
)αε−
13 → 1
2n ln
(na
)+ x ln
(1 +
√n
a
)−√an−
√n− nπi.
From (24) and (65) we obtain
ε−16
√2π
[z (1 − z)
pq
] 14
(Θ1)− 1
3 →√
2π(na
) 16 (√
a+√n)1
3 (66)
and
(Θ1)23 α→
(na
) 16 (x−X+)
(√a +
√n)
23
. (67)
Therefore,
Cn(x) ∼ F11(x) =√
2π(na
) 16 (√
a +√n) 1
3 Ai
[(na
) 16 (x−X+)
(√a+
√n)
23
](68)
× (−1)n exp
[1
2n ln
(na
)+ x ln
(1 +
√n
a
)−√an−
√n
]
for x ≈ X+.
4 Comparison with previous results
We shall now compare our results with those obtained previously in [6] and [13].
1. Region VII: 0 ≤ x < X−, n > a.
Setting x = un, with
u = O(1), 0 ≤ u < 1 − 2
√a
n+a
n< 1
21
in (52) we have, as n→ ∞
F7(x) ∼ g7(u) =cos(unπ)√
1 − uexp
{[u ln
(na
)+ (u− 1) ln (1 − u) − u
]n +
au
u− 1
}(69)
−2 sin (unπ)
√u
1 − uexp
{[ln
(na
)+ (1 − u) ln (1 − u) + u ln(u) − 1
]n+
a
1 − u
}.
The second term of equation (69) is the same as the equation before (5.3) in [6] andequation (84) in [13]. However, the first term is absent in previous works, although itis necessary in the asymptotic approximation, especially when u ' 0, 1, 2, . . . .
2. Region IX: x ≈ X−, n > a.
We now set x = X− + tn16 , t = O(1) in (60) and obtain, as n→ ∞
F9(x) ∼ g9(t) =√
2πa−16n
13 exp
[1
2
(X− + tn
16 + n
)ln
(na
)− n+
3
2a
](70)
×{
cos[(X− + tn
16
)π]Ai
(−ta−
16
)− sin
[(X− + tn
16
)π]Bi
(−ta−
16
)}.
Equation (70) agrees with equation (5.13) in [6] and equation (51) in [13].
3. Region X: X− < x < X+.
Setting x = n+ a+ 2 sin(θ)√an, with −π
2< θ < π
2in (42) we have, as n→ ∞
F3(x) ∼ g3(θ) = (−1)n a−14n
14√
2 cos (θ)exp
{[ln
(na
)− 1
]n+
π
4i}
× exp{√
an[sin (θ) ln
(na
)− sin (θ) (2θ − π) i − 2 cos (θ) i
]}(71)
× exp
{a
[1 − 1
2cos (2θ) +
1
2ln
(na
)− 1
2sin (2θ) i − θi +
π
2i
]}.
Similarly from (46) we get
F4(x) ∼ g4(θ) = (−1)n a−14n
14√
2 cos (θ)exp
{[ln
(na
)− 1
]n− π
4i}
× exp{√
an[sin (θ) ln
(na
)+ sin (θ) (2θ − π) i + 2 cos (θ) i
]}(72)
× exp
{a
[1 − 1
2cos (2θ) +
1
2ln
(na
)+
1
2sin (2θ) i + θi − π
2i
]}.
22
Using (71) and (72) in (62) we have
F10(x) ∼ g10(θ) = (−1)n
√2a−
14n
14√
cos (θ)exp
{[ln
(na
)− 1
]n}
× exp
{√an
[sin (θ) ln
(na
)]+ a
[1 − 1
2cos (2θ) +
1
2ln
(na
)]}(73)
× cos
{√an [sin (θ) (2θ − π) + 2 cos (θ)] + a
[1
2sin (2θ) + θ − π
2
]− π
4
}.
Equation (73) is equivalent to equation (44) in [13].
4. Region XI: x ≈ X+.
We now set x = X+ + sn16 , s = O(1) in (68) and obtain, as n→ ∞
F11(x) ∼ g11(s) =√
2πa−16n
13 exp
[1
2
(X+ + sn
16 + n
)ln
(na
)− n +
3
2a
]Ai
(sa−
16
).
(74)Equation (74) is equation (5.12) in [6] and equation (30) in [13].
5 Zeros
Using the formulas from the previous sections we can obtain approximations to the zeros ofthe Charlier polynomials.
1. x ' 0, n > a.
The first zero is exponentially small. From (48) we have, as x→ 0
C5(x) ∼ 1 +
[ln
(na− 1
)−
√2πn
n− aa−nnnea−n
]x.
Solving for x we obtain
x0 '[ln
(na− 1
)−
√2πn
n− aa−nnnea−n
]−1
∼ en−aann−n
√2πn
, n→ ∞ (75)
where x0 denotes the smallest zero.
2. 0 < x < X−, n > a.
In this range of x, the zeros are exponentially close to 1, 2, . . . , bX−c . Using
t =x− n− a + 2
√an
n16
23
and the asymptotic formulas [35]
Ai(x) ∼exp
[−2
3x
32
]
2√πx
14
, x→ ∞
Bi(x) ∼exp
[23x
32
]
√πx
14
, x → ∞
we have, as n→ ∞
Ai(−ta 1
6
)
Bi(−ta 1
6
) ∼ 1
2exp
[−4
3a−
14n− 1
4
(X− − x
) 32
]. (76)
Using (76) in (70) we have
g9(t) ' 0 ⇔ 1
2exp
[−4
3a−
14n− 1
4
(X− − x
) 32
]' tan (πx) .
Since xj ' j, j = 1, 2, . . . , bX−c , we get
1
2exp
[−4
3a−
14n− 1
4
(X− − j
)32
]' π (xj − j)
which we can solve to obtain
xj ' j +π
2exp
[−4
3a−
14n− 1
4
(X− − j
)32
], j = 1, 2, . . . ,
⌊X−⌋
. (77)
3. X− < x < X+.
Finally, the non-trivial zeros of the Charlier polynomials can be approximated using(73). We have g10(θ) = 0 if and only if
cos
{√an [sin (θ) (2θ − π) + 2 cos (θ)] + a
[1
2sin (2θ) + θ − π
2
]− π
4
}= 0
or equivalently if
√an [sin (θ) (2θ − π) + 2 cos (θ)] + a
[1
2sin (2θ) + θ − π
2
]− π
4=π
2+ πl, l ∈ Z
or
√an [sin (θ) (2θ − π) + 2 cos (θ)] + a
[1
2sin (2θ) + θ − π
2
]− 3π
4− πl = 0, (78)
24
with −π2< θ < π
2. Recalling that
x = n+ a + 2 sin(θ)√an, (79)
we see that the condition X− < x < X+ implies
0 ≤ l ≤ 2√an− a− 3
4. (80)
Equation (78) cannot be solved exactly. However, it can be easily solved numerically toany desired accuracy and using (79) gives very good approximations for the nontrivialzeros.
In Table 1 we computed the exact and approximate zeros of C25(x) with a = 2.16564899using (75), (77) and (78)-(80).
Conclusion 4 We analyzed the asymptotic behavior of the Charlier polynomials in the range0 ≤ x as n → ∞. We also obtained approximations for their zeros. We intend to extendour method to the other polynomials of the Askey-scheme to obtain asymptotic expansions ofthem.
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Table 1: Comparison of the exact and approximate zeros of C25(x) with a = 2.16564899.
l x (exact) x (approximate)− 0.41229323× 10−16 0.41549221 × 10−16
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26
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29