asymptotic analysis of the askey-scheme i: from krawtchouk ...dominicd/charl15.pdf · the...

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


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


14

–6e+20

–5e+20

–4e+20

–3e+20

–2e+20

–1e+20

00.5 1 1.5 2 2.5 3

x


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


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)


[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


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),


[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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25

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

− 1.0000000 1.0000000− 2.0000000 2.0000000− 3.0000000 3.0000001− 4.0000000 4.0000009− 5.0000001 5.0000073− 6.0000015 6.0000507− 7.0000227 7.0003063− 8.0002574 8.0015785− 9.0021153 9.0068260− 10.012329 10.024179− 11.050278 11.067497− 12.147166 12.13724211 13.330606 13.33429510 14.615276 14.5608679 16.007976 15.8997278 17.514470 17.3507927 19.142918 18.9217146 20.905595 20.6261105 22.820702 22.4846004 24.915443 24.5279113 27.232157 26.8035912 29.842164 29.3913941 32.883964 32.4462400 36.717784 36.379078

26

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27

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