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CITY UNIVERSITY OF HONG KONG

香港城市大學

Uniform Asymptotic Expansions of the

Tricomi-Carlitz Polynomials and the Modified

Lommel Polynomials

Tricomi-Carlitz多項式與 Modified Lommel

多項式的一致漸進展開

Submitted to

Department of Mathematics

數學系 in Partial Fulfillment of the Requirements

for the Degree of Doctor of Philosophy

哲學博士學位

by

Lee Kei Fung

李奇峰

June 2012

二零一二年六月

Abstract

In this thesis, we derive uniform asymptotic expansions of the Tricomi-Carlitz poly-

nomials f (α)n (x) and the modified Lommel polynomials hn,ν(x), as n → ∞, valid for

x in (0,∞). Since these two polynomials do not satisfy a second-order differential

equation, the powerful tools developed for differential equations are not applicable.

Our discussion is divided into three parts.

In the first part, we derive directly from the three-term recurrence relation (n +

1)f(α)n+1(x) − (n + α) x f

(α)n (x) + f

(α)n−1(x) = 0, an asymptotic expansion for f (α)

n (x)

which holds uniformly in regions containing the critical values x = ±2/√ν, where

ν = n + 2α − 1/2. This method is based on the turning-point theory for three-term

recurrence introduced by Wang and Wong [Numer. Math. 91 (2002) and 94 (2003)].

In the second part, the expansion is derived by using the cubic transformation for

the integral∫CJ(s; t) exp[νϕ(s; t)] ds, where J(s; t) and ϕ(s; t) are analytic functions

of s, t is a bounded real parameter and ϕ(s; t) have two saddle points s±(t) which

coalesce as t tends to some real number t0. Then we apply the integration-by-part

technique suggested by Bleistein. As an application, an asymptotic expansion for the

zeros of the Tricomi-Carlitz polynomials is derived. The validity for bounded t can be

extended to unbounded t by using a sequence of rational functions introduced by Olde

Daalhuis and Temme. The expansion involves the Airy functions and their derivatives.

Error bounds are also given for one-term and two-term approximations.

We finally study a asymptotic expansion for the modified Lommel polynomials

hn,ν(t/N) which holds uniformly in regions containing the critical values x = ±1/N ,

where N = n + ν. This method is again based on the turning-point theory for three-

term recurrence; their zeros are also derived.

Acknowledgement

I would like to express my gratitude to those who provided endless support and

encouragement throughout the completion of this thesis. First of all, I am grateful

to my supervisor, Prof. Roderick S. C. Wong, for his valuable time, patience and

guidance during the course of my study. Prof. Wong’s keen mathematical insight has

been enlightening, inspirational and key in helping me cultivate a rigorous approach to

research.

I would like to thank Prof. Zhen Wang and Prof. Yu-qiu Zhao for their useful ad-

vice, criticism and input which contributed significantly to the outcome of this thesis.

I would like to express my thanks to Dr. Dan Dai, Dr. Yutian Li, Dr. Yu Lin and Mr.

Jianhui Pan for their generous help during my study, in particular their discussion and

encouragement, which, in no small measure, also contributed to the outcome of this

thesis. The financial support of City University of Hong Kong is gratefully acknowl-

edged.

Last but not least, I would like to express my deep gratitude to God, my parents

and Miss Nga-chi Wong for their unwavering support and unconditional love.

Contents

Abstract i

Acknowledgement ii

1 Introduction 1

1.1 Tricomi-Carlitz Polynomials . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Second-Order Linear Difference Equation . . . . . . . . . . . . . . . 5

1.3 Method of Steepest Descents . . . . . . . . . . . . . . . . . . . . . . 8

1.4 Trigonometric Solution of a Cubic Equation . . . . . . . . . . . . . . 11

2 Uniform Asymptotic Expansion of f (α)n (x) via Difference Equation 13

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Difference Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 Determination of C1(x) and C2(x) . . . . . . . . . . . . . . . . . . . 17

2.4 Numerical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Uniform Asymptotic Expansion of f (α)n (x) via Method of Steepest Descent 20

3.1 Reduction to a Canonical Integral . . . . . . . . . . . . . . . . . . . 20

3.2 Derivation of Expansion . . . . . . . . . . . . . . . . . . . . . . . . 28

3.3 For Unbounded Values of ζ . . . . . . . . . . . . . . . . . . . . . . . 34

3.4 Error Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.5 Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4 Uniform Asymptotic Expansion of Modified Lommel polynomials 56

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Contents iv

4.2 Difference Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.3 Determination of C1(x) and C2(x) . . . . . . . . . . . . . . . . . . . 60

4.4 Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

A Verification of Some Results in §1 72

A.1 Another Generalized Hypergeometric Representations of t(α)n (x) . . . 76

A.2 Connection Formula of rn(x, a) and f (α)n (x) . . . . . . . . . . . . . . 77

A.3 The Fixed xAsymptotic Behavior of Tricomi-Carlitz Polynomial f (α)n (x) 78

B Verification of Some Results in §2 82

C Verification of Some Results in §3 85

C.1 The Transformation s 7→ u in the Case t > 2 . . . . . . . . . . . . . . 102

C.2 The Transformation s 7→ u in the Case 0 < t < 2 . . . . . . . . . . . 115

C.3 A Comparison of the Results (2.3.9) and (3.2.19) . . . . . . . . . . . 132

C.4 Power series solutions of (3.5.13) . . . . . . . . . . . . . . . . . . . . 135

D Verification of Some Results in §4 136

D.1 Power series solutions of (4.4.13) . . . . . . . . . . . . . . . . . . . . 143

Bibliography 144

Chapter 1 Introduction 2

A tremendous amount of research has been carried out on the asymptotics of vari-

ous discrete orthogonal polynomials. Most of the results of their asymptotic behavior

are obtained by using the steepest descent or the saddle point method for integrals.

For instance, we have [28], [12] and [25] for Charlier polynomials, [16] and [32] for

Meixner polynomials, [20], [27] and [10] for Krawtchouk polynomials, and [26] for

Chebyshev polynomials.

In this thesis, we study two non-classical discrete orthogonal polynomials; Tricomi-

Carlitz polynomials and modified Lommel polynomials. We first review the history

and some common properties of the discrete orthogonal polynomials, discuss the meth-

ods of obtaining their asymptotic behavior, and present some new results on their

asymptotic behavior.

The arrangement of the present paper is as follows. In Chapter 1, we first review

the history and some properties of the Tricomi-Carlitz polynomials. Then we recall a

turning-point theory introduced by Wang and Wong [33, 34] for three-term recurrence

relations, and summarize the key steps in the Chester-Friedman-Ursell method.

In Chapter 2, we introduce the sequence Kn the three-term recurrence relation

of f (α)n (x) can be transformed into the canonical form with the coefficients An and Bn.

The coefficients An and Bn can of course be recast when introducing ν = n+ τ0. We

then find the turning points from the characteristic equation. A pair of solutions Pn(x)

and Qn(x) can be obtained. The crucial step is to determine the functions C1(x) and

C2(x). It can be done when we examine the behavior of f (α)n (x), as n → ∞ by using

the classical steepest decent method. As an illustration, we give a numerical result.

In Chapter 3, we present integral representation for Tricomi-Carlitz polynomials

f(α)n (x), and deduce a contour integral into a canonical form by using a cubic trans-

formation. We prove the cubic transformation is one-to-one and analytic in a domain

containing the path of integration. The Bleistain technique of repeated integration by

parts is used, which lead to Airy-type expansions with explicit remainders. We then

present the uniform asymptotic expansions and compute the leading term coefficients.

We also show how to extend the validity of the uniform asymptotic expansion as the

auxiliary parameter t tends to infinity. Error bounds are also given for one-term and

1.1 Tricomi-Carlitz Polynomials 3

two-term approximations. Finally, we present asymptotic formulas for the zeros of

f(α)n (t/

√ν) under consideration.

In Chapter 4, we review the history and some properties of the modified Lommel

Polynomials polynomials hn,ν(x). We then present the uniform asymptotic expansions

by using a turning-point theory for three-term recurrence relations. The functions

C1(x) and C2(x) can be determined by the definition of modified Lommel polyno-

mials. We present several asymptotic formulas for the least zeros, the largest zeros

near the mass points and the zeros located on both sides of the turning point. Fi-

nally, we make a comparison of the approximations with the true values of the zeros

of hn,ν(t/N).

1.1 Tricomi-Carlitz Polynomials

The Tricomi polynomials are defined by

t(α)n (x) =n∑

k=0

(−1)k(x− αk

)xn−k

(n− k)!, n = 0, 1, 2, · · · , (1.1.1)

which satisfy the recurrence relation

(n+ 1)t(α)n+1(x)− (n+ α)t(α)n (x) + x t

(α)n−1(x) = 0, n ≥ 1, (1.1.2)

with initial values t(α)0 (x) = 1, t(α)1 (x) = α. Note that deg t(α)n (x) = [n/2] and Tricomi

polynomials are connected with the Laguerre polynomials L(a)n (x),

t(α)n (x) = (−1)nL(x−α−n)n (x). (1.1.3)

These polynomials can also be expressed in terms of the generalized hypergeometric

series1

t(α)n (x) =xn

n!2F0(−n,−x+ α;

−1x

) = (−1)n(x− αn

)1F1(−n;x− α− n+ 1;x).

(1.1.4)1Detailed analysis can be found in Appendix A.1.

1.1 Tricomi-Carlitz Polynomials 4

The generating function

exp xw − (α− x) log(1− w) =∞∑n=0

t(α)n (x)wn, |w| < 1. (1.1.5)

can be derived from the recurrence relation (1.1.2), when x = 0. The series in (1.1.5)

is convergent for |w| < 1. Tricomi [31] observed that t(α)n (x) is not a system of

orthogonal polynomials since the recurrence relation (1.1.2) fails to have the required

form [30, p. 135]. However, Carlitz [6] discovered that if one sets

f (α)n (x) = xnt(α)n (x−2), (1.1.6)

thenf(α)n (x)

satisfies

(n+ 1)f(α)n+1(x)− (n+ α)x f (α)

n (x) + f(α)n−1(x) = 0, n ≥ 1, (1.1.7)

with initial values f (α)0 (x) = 1, f (α)

1 (x) = αx. Furthermore, he gave the orthogonal

relation ∫ ∞

−∞f (α)m (x) f (α)

n (x) dψ(α)(x) =2 eα

(n+ α)n!δmn, (1.1.8)

where α > 0 and ψ(α)(x) is the step function whose jumps are

dψ(α)(x) =(k + α)k−1e−k

k!at x = xk = ±(k + α)−1/2, k = 0, 1, 2, · · · .

(1.1.9)

The generating function

exp

w

x+

1− αx2

x2log(1− wx)

=

∞∑n=0

f (α)n (x)wn (1.1.10)

can be derived from the recurrence relation (1.1.7) when x = 0. The series in (1.1.10)

is convergent for |wx| < 1. The values xk play a special role in the generating function

(1.1.10) since for these x values we have

ew/xk(1− xkw)k =∞∑n=0

f (α)n (xk)w

n (1.1.11)

and now the series converges for all values of w. If x = 0, then (1.1.10) reduces to

e−12w2

=∞∑n=0

f (α)n (0)wn. (1.1.12)

1.2 Second-Order Linear Difference Equation 5

By expanding e−12w2

at w = 0, one obtains

f(α)2n (0) =

(−1)n

2nn!, f

(α)2n+1(0) = 0, n = 0, 1, 2, · · · . (1.1.13)

These polynomials can also be expressed in terms of the generalized hypergeometric

series

f (α)n (x) =

1

xnn!2F0(−n, α− x−2;−x2). (1.1.14)

Independently and simultaneously, Karlin and McGregor [17] studied a stochastic

(Markovian) birth-and-death process and discovered the random walk polynomials2

rn(x) derived from an infinitely many server process, which are very useful in various

problems. Therefore, the Tricomi-Carlitz polynomials are also known as the Carlitz-

Karlin-McGregor polynomials [3]. In 1984, Askey and Ismail [3] gave the general-

ization of the Tricomi-Carlitz polynomials, namely, Gn(x; a, b) = rn(x)an(b/a)n/n!

with f (α)n (x) = α−n/2Gn(x

√α; 0, α). Other properties of Gn(x; a, b) can be found in

[3] and [15].

1.2 Second-Order Linear Difference Equation

For more general three-term recurrence relation

pn+1(x) = (anx+ bn)pn(x)− cnpn−1(x), n ≥ 1, (1.2.1)

where an, bn and cn are constants. Refer to Wang and Wong [34], we summarize the

main results here.

Define the sequence Kn recursively by

Kn+1

Kn−1

= cn, (1.2.2)

where K0 and K1 depends on the particular sequence of function or polynomials

pn(x). Furthermore, we set

An = anKn

Kn+1

, Bn = bnKn

Kn+1

, Pn =pnKn

. (1.2.3)

2Detailed analysis can be found in Appendix A.2.


(1.2.1) can be transformed into the canonical form

Pn+1(x)− (Anx+Bn)Pn(x) + Pn−1(x) = 0. (1.2.4)

We assume the coefficients An and Bn have asymptotic expansions of the form

An ∼ n−θ

∞∑s=0

αs

nsand Bn ∼

∞∑s=0

βsns, (1.2.5)

where θ is a real number and α0 = 0.

Let τ0 be a constant and put ν = n + τ0. Clearly, the expansions in (1.2.5) can be

recast in the form

An ∼ ν−θ

∞∑s=0

α′s

νsand Bn ∼

∞∑s=0

β′s

νs. (1.2.6)

In (1.2.4), we now let x = νθt and Pn = λn. Substituting (1.2.6) into (1.2.4) and

letting n→∞ (and hence ν →∞), we obtain the characteristic equation

λ2 − (α′0t+ β′

0)λ+ 1 = 0. (1.2.7)

The roots of this equation are given by

λ =1

2

[(α′

0t+ β′0)±

√(α′

0t+ β′0)

2 − 4]

(1.2.8)

and they coincide when t = t±, where

α′0t± + β′

0 = ±2. (1.2.9)

The values t± play an important role in the asymptotic theory of the three-term

recurrence relation (1.2.4) and they correspond to the transition points (i.e., turning

point and poles) occurring in differential equations; cf. Olver [24, p. 362]. For this

reason, we shall also call them transition points. Since t+ and t− have different values,

we may restrict ourselves to just the case t = t+. In terms of the exponent θ in (1.2.6)

and the transition point t+, we have three cases to consider; namely, (i) θ = 0 and

t+ = 0; (ii) θ = 0 and t+ = 0; and (iii) θ = 0. In this thesis, we will investigate the

case (i) only.


We begin with case (i), and assume for simplicity that θ > 0 and |β′0| < 2 (i.e.,

t+ and t− are of opposite signs). The analysis for the case θ < 0 is very similar.

Furthermore, we choose

τ0 = −(α1t+ + β1)

(2− β0)θ(1.2.10)

so that

α′1t+ + β′

1 = 0, (1.2.11)

and define the function ζ(t) by

2

3[ζ(t)]3/2 := α′

0t1/θ

∫ t

t+

s−1/θ√(α′

0s+ β′0)

2 − 4ds

− logα′0t+ β′

0 +√(α′

0t+ β′0)

2 − 4

2, t ≥ t+,

(1.2.12)

and

2

3[−ζ(t)]3/2 := cos−1 α

′0t+ β′

0

2

− α′0t

1/θ

∫ t+

t

s−1/θ√4− (α′

0s+ β′0)

2ds, t < t+.

(1.2.13)

It should be noted that for θ < 0, the minus sign ’−’ should be added into the left-hand

side of (1.2.12) and (1.2.13). Moreover, we put

H0(ζ) := −

√(α′

0t+ β′0)

2 − 4

4ζ(1.2.14)

and

Φ(t) := − 1

ζ1/2

∫ t

t+

α′1τ + β′

1

2θτζ12H0

dτ. (1.2.15)

Now we are ready to state our first result on uniform asymptotic expansions for

difference equations.

Theorem 1.1. Assume that the coefficientsAn andBn in the recurrence relation (1.2.4)

have asymptotic expansions of the form given in (1.2.6) with θ = 0 and |β0| < 2. Let ζ

and Φ be given as in (1.2.12) and (1.2.15), respectively. Then equation (1.2.4) has, for

each value of ν and each nonnegative integer p, a pair of solutions Pn(x) and Qn(x),

1.3 Method of Steepest Descents 8

given by

Pn(νθt) =

[4ζ

(α′0t+ β′

0)2 − 4

] 14[

Ai(ν

23 ζ +

Φ

ν13

) p∑s=0

As(ζ)

νs−16

+ Ai′(ν

23 ζ +

Φ

ν13

) p∑s=0

Bs(ζ)

νs+16

+ εp(ν, t)

] (1.2.16)

and

Qn(νθt) =

(4ζ

(α′0t+ β′

0)2 − 4

)14[

Bi(ν

23 ζ +

Φ

ν13

) p∑s=0

As(ζ)

νs−16

+ Bi′(ν

23 ζ +

Φ

ν13

) p∑s=0

Bs(ζ)

νs+16

+ δp(ν, t)

],

(1.2.17)

where

|εp(ν, t)| ≤Mp

νp+56

Ai(ν

23 ζ +

Φ

ν13

)(1.2.18)

and

|δp(ν, t)| ≤Mp

νp+56

Bi(ν

23 ζ +

Φ

ν13

)(1.2.19)

for δ ≤ t <∞, 0 < δ < t+ and Mp being a constant. In (1.2.16) and (1.2.17), Ai and

Bi are the Airy functions; in (1.2.18) and (1.2.19), Ai and Bi are the modulus functions

given in [34, (7.10) and (7.11)].

1.3 Method of Steepest Descents

In many practical applications, most of the functions can be expressed as an integral

form I(λ, t). For instance, the solutions of many ODEs or PDEs have this integral

representation. Let us first consider the more standard integral

I(λ, t) =

∫CJ(s; t)eλϕ(s;t)ds, (1.3.1)

where ϕ(s; t) and J(s; t) are analytic functions of the complex variable s and the pa-

rameter t. We are interested in the behavior of I(λ, t) when λ is large and positive.

In this case, asymptotic expansions can usually be found by the method of steepest

descents.


The method of steepest descents is invented by Riemann. He used it to obtain an

asymptotic approximation of I(λ, t). We will now give a brief description of it since it

is of considerable relevance to my thesis. For details, we refer the readers to books by

Copson [9, Chapter 7] and Wong [37, Chapter 7] and [38, Chapter 4].

The essence of this method is to deform the path of integration C into C ′ such that

the following conditions hold:

(i) the path C ′ passes through one or more zeros of ϕs(s; t) and

(ii) the imaginary part of ϕ(s; t) is constant on the path.

This method shows that the principle contributions to the integral (1.3.1) arise from the

saddle points.

The position of the saddle points varies with t and we assume that there exists a

critical value of t, say t = t0, such that for t = t0 there are two distinct saddle points

s+ and s− of multiplicity 1, but when t = t0 these two points coincide and give a single

saddle point s0 of multiplicity 2. Thus

ϕs(s+; t) = ϕs(s−; t) = 0, ϕss(s±; t) = 0 (1.3.2)

and

ϕs(s0; t0) = ϕss(s0; t0) = 0, ϕsss(s0; t0) = 0 (1.3.3)

for t = t0.

In view of the fact that the simplest phase function which exhibits two coalescing

saddle points is a cubic polynomial, Chester, Friedman and Ursell (1957) introduced

what is now regarded as a classic method, a change of variable of the form

ϕ(s; t) =1

3u3 − ζ(t)u+ η(t), (1.3.4)

where the coefficients ζ and η are to be determined. To ensure the mapping s 7→ u

defined by (1.3.4) is one-to-one and analytic, ∂s/∂u can neither vanish nor become

infinite. Since∂s

∂u=

u2 − ζϕs(s; t)

(1.3.5)


and ϕs(s; t) = 0 when s = s±, we must assign the saddle points s± of ϕ(s; t) to

correspond to the saddle points ±√ζ(t), respectively. Accordingly, we have

ϕ(s+; t) =1

3ζ

32 − ζ

32 + η = −2

3ζ

32 + η (1.3.6)

and

ϕ(s−; t) = −1

3ζ

32 + ζ

32 + η = +

2

3ζ

32 + η. (1.3.7)

From (1.3.6) and (1.3.7), we find that the coefficients ζ and η are given by

ζ3/2 =3

4[ϕ(s−; t)− ϕ(s+; t)] (1.3.8)

and

η =1

2[ϕ(s−; t) + ϕ(s+; t)]. (1.3.9)

By rewriting (1.3.5) as

ϕs(s; t)∂s

∂u= u2 − ζ (1.3.10)

and differentiating which respect to s, we have(∂s

∂u

∣∣∣∣u=±√

ζ(t)

)2

= ± 2√ζ

ϕss(s±; t), (1.3.11)

if t = t0. If t = t0, then a further differentiation gives(∂s

∂u

∣∣∣∣u=0

)3

=2

ϕsss(s0; t0). (1.3.12)

The transformation (1.3.4) was shown by Chester, Friedman and Ursell to be one-

to-one and analytic for all s, t in a neighborhood of s0, t0. More precisely, there

is a theorem in Wong [37], which says that with the above values of ζ and η, the

transformation (1.3.4) has exactly one branch u = u(s, t) which can be expanded into

a power series in (s− s0), with coefficients which are continuous in t for t near t0. On

this branch the points s = s± correspond to u = ±√ζ(t), respectively. Furthermore,

for t near t0, the correspondence s ↔ u is one-to-one. The above theorem is only a

local result, but not the global case. For our case, we can prove that the transformation

(1.3.4) is in fact one-to-one and analytic in a domain containing the entire path of

integration.

1.4 Trigonometric Solution of a Cubic Equation 11

1.4 Trigonometric Solution of a Cubic Equation

Due to the importance of the solution of cubic equation, we will reproduce the result

below by considering a cubic equation of the form

w3 +Hw +K = 0, (1.4.1)

whereH and K are complex numbers.

For real variable θ, we have

cos 3θ + i sin 3θ

= e3iθ = (cos θ + i sin θ)3

= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ

= cos3 θ − 3 cos θ sin2 θ + i(3 cos2 θ sin θ − sin3 θ

)= cos3 θ − 3 cos θ (1− cos2 θ) + i

(3(1− sin2 θ) sin θ − sin3 θ

)= 4 cos3 θ − 3 cos θ + i

(3 sin θ − 4 sin3 θ

).

Hence, cos 3θ = 4 cos3 θ− 3 cos θ and sin 3θ = 3 sin θ− 4 sin3 θ. By analytic continu-

ation and the fact that the sine and cosine functions are entire, the above formulae also

hold when θ is a complex variable.

Suppose w = A sin θ (where A, θ are complex numbers) is a solution of the above

cubic equation. Here, A is a fixed number to be chosen later and θ is a variable. Then

0 = A3 sin3 θ +HA sin θ +K

= A3 sin3 θ +HA1

3(sin 3θ + 4 sin3 θ) +K

=HA3

sin 3θ + A(4

3H + A2) sin3 θ +K.

If we choose A =√−4H

3, we have

HA3

sin 3θ +K = 0.

Hence, w = A sin θ is a solution if and only if

sin 3θ = − 3KHA

.

1.4 Trigonometric Solution of a Cubic Equation 12

Thus, the three solutions of the cubic equation are

w1 = A sinφ

3, w2 = A sin

φ+ 2π

3, w3 = A sin

φ+ 4π

3, (1.4.2)

where

A =

√−4H

3(1.4.3)

and φ is the solution of

sinφ = − 3KHA

. (1.4.4)

We remark that any possible values of A and φ satisfying (1.4.3) and (1.4.4) respec-

tively will work and give identical results. This is consistent with the fact that a cubic

equation has exactly three complex roots counting multiplicities.

Chapter 2

Uniform Asymptotic Expansion of

f(α)n (x) via Difference Equation

2.1 Introduction

Asymptotic behavior of f (α)n (x) has been investigated by Goh and Wimp [13, 14]. In

their first paper, they used an elementary approach to show that

f(α)n (y/

√α)

α−n/2ynn−α(1−y2)/y2−1→ eα/y

2

Γ(−α(1− y2)/y2), as n→∞, (2.1.1)

uniformly for all y in a compact set K ⊆ C\[−1, 1]. (There is a minor error in the

statement of their result; the factor α−n/2 in the denominator on the left-hand side of

(2.1.1) is missing in their equation (23) in [13].) The validity of (2.1.1) can be veri-

fied by a direct application of Darboux’s method1 [37, p. 116]. Goh and Wimp also

observed that all zeros of f (α)n (y/

√α) lie in the interval [−1, 1]. In their second pa-

per, they used saddle-point methods to study the asymptotics of f (α)n (z/

√n) for z in

neighborhoods of z = ±i/2. Note that the scales in their two papers are different; in

[13] the scale is y/√α, whereas in [14] the scale is z/

√n. The behavior of f (α)

n (x) as

α → ∞ has been studied by Temme [21]. His result is expressed in terms of Hermite

polynomials.

1Detailed analysis can be found in Appendix B.

2.2 Difference Equation 14

The purpose of this section is to present an asymptotic expansion for large n. The

parameter ν is large, with fixed α. In view of the reflection formula f (α)n (−x) =

(−1)nf (α)n (x), a corresponding result can be given for t in (−∞, 0]. Note that our

result is not covered in the two papers of Goh and Wimp; in fact, it complements

those in [13] and [14]. We also point out that the Tricomi-Carlitz polynomials do not

satisfy a second-order differential equation; hence, the powerful tools developed for

differential equations (see, e.g., [24]) are not applicable. Our approach is to use a

turning-point theory recently introduced by Wang and Wong [33, 34] for three-term

recurrence relations.

2.2 Difference Equation

Returning to (1.1.7), we write

f(α)n+1(x)−

n+ α

n+ 1x f (α)

n (x) +1

n+ 1f(α)n−1(x) = 0 (2.2.1)

and introduce the sequence Kn defined by

(n+ 1)Kn+1 = Kn−1 (2.2.2)

with K0 = 1 and K1 =√

2/π. Induction shows that

Kn =1

2n/2Γ(12n+ 1

) . (2.2.3)

With the notation

F (α)n (x) :=

f(α)n (x)

Kn

, (2.2.4)

equation (2.2.1) can be put in the canonical form considered in [34]:

F(α)n+1(x)− (Anx+Bn)F

(α)n (x) + F

(α)n−1(x) = 0 (2.2.5)

with

An =n+ α

n+ 1

Kn

Kn+1

=n+ α√

2

Γ(12(n+ 1)

)Γ(12n+ 1

) (2.2.6)

and Bn = 0. To find an asymptotic expansion for An, we recall the well-known result

[37, p. 47]Γ(12n+ 1

2

)Γ(12n+ 1

) ∼√ 2

n

[1− 1

4n+

1

32n2+ · · ·

].


Thus,

An ∼ n1/2

[1 + (α− 1

4)1

n+ (

1

32− α

4)1

n2+ · · ·

]. (2.2.7)

In terms of notations

An ∼ n−θ

∞∑s=0

αs

nsand Bn ∼

∞∑s=0

βsns, (2.2.8)

used in [34], we have

θ = −1

2, α0 = 1, α1 = α− 1

4, α2 = (

1

32− α

4), · · · (2.2.9)

and β0 = β1 = β2 = · · · = 0. If these expansions are recast in the form

An ∼ ν−θ

∞∑s=0

α′s

νsand Bn ∼

∞∑s=0

β′s

νs, (2.2.10)

where ν = n+ τ0 and τ0 is some fixed real number to be determined, it is easily found

that

α′0 = 1, α′

1 = (α− 1

4)− τ0

2, · · · (2.2.11)

and β′0 = β′

1 = β′2 = · · · = 0. To apply the result in [34], we first choose τ0 so that

α′1 = 0. From (2.2.11), it is obvious that the choice is

τ0 = 2α− 1

2. (2.2.12)

According to equation (2.4) in [34], the characteristic equation is

λ2 − tλ+ 1 = 0, (2.2.13)

where t is the rescaled variable x = ν−12 t. The two roots of this equation are

λ± =1

2(t±√t2 − 4). (2.2.14)

The points t± = ±2 where the two roots conincide are called the turning points of

equation (2.2.5). In view of the symmetry relation F (α)n (−x) = (−1)nF (α)

n (x), we

may restrict ourselves just to the case 0 < t <∞.


We now define the function ζ(t) introduced in [34, (4.10)]. With t+ = 2, θ = −12,

α′0 = 1 and β′

0 = 0, this function is given by

2

3[ζ(t)]3/2 := log

t+√t2 − 4

2− 1

t2

∫ t

2

s2√s2 − 4

ds, t ≥ 2

and2

3[−ζ(t)]3/2 := t−2

∫ 2

t

s2√4− s2

ds− cos−1 t

2, 0 < t < 2.

By direct calculation, one obtains

2

3[ζ(t)]3/2 = log

t+√t2 − 4

2− 1

t2

[1

2t√t2 − 4 + 2 log |t+

√t2 − 4| − 2 ln 2

],

(2.2.15)

for t ≥ 2 and

2

3[−ζ(t)]3/2 = 1

t2

[−2 sin−1 t

2+ π +

t

2

√4− t2

]− cos−1 t

2, (2.2.16)

for 0 < t < 2. We also define the function H0(ζ) and Φ(ζ) introduced in [34, (4.19)

and (4.28)]. In the present situation, these functions are given by

H0(ζ) = −

√t2 − 4

4ζand Φ(ζ) = 0, (2.2.17)

where ζ is the function defined in (2.2.15) and (2.2.16). Note that in our special case,

α′1 = β′

1 = 0; hence, according to the definition of Φ(ζ) given in (4.28) of [34], the

second equation in (2.2.17) holds for 0 < t <∞, instead of t ≥ δ, 0 < δ < 2.

With this preliminary work done, we can now apply the main result in [34] to

conclude that there are constants C1(x) and C2(x), such that the polynomials F (α)n (x)

in (2.2.4) can be expressed as

F (α)n (x) = C1(x)Pn(x) + C2(x)Qn(x), (2.2.18)

where, with x = ν−12 t, Pn(x) and Qn(x) have asymptotic expansions

Pn(ν− 1

2 t) =

(4ζ

t2 − 4

) 14

[Ai(ν

23 ζ) ∞∑

s=0

As(ζ)

νs−16

+ Ai′(ν

23 ζ) ∞∑

s=0

Bs(ζ)

νs+16

](2.2.19)

2.3 Determination of C1(x) and C2(x) 17

and

Qn(ν− 1

2 t) =

(4ζ

t2 − 4

) 14

[Bi(ν

23 ζ) ∞∑

s=0

As(ζ)

νs−16

+ Bi′(ν

23 ζ) ∞∑

s=0

Bs(ζ)

νs+16

]. (2.2.20)

In (2.2.19) and (2.2.20), Ai(·) and Bi(·) are the Airy functions, the leading coefficients

are given by

A0(ζ) = 1 and B0(ζ) = 0, (2.2.21)

and the expansions hold uniformly for 0 ≤ t <∞.

2.3 Determination of C1(x) and C2(x)

First we examine the behavior of f (α)n (x), as n → ∞. By using the classical steepest

decent method, one can show that to leading order, we have

f (α)n (

t√ν) ∼ ν−

n2 e

ν(

λ+t+( 2

t2−1) log λ+

)√2πν(t2 − 4)

14

cos(πα− νπ

t2)

+ν−

n2 e

ν(

λ−t+( 2

t2−1) log λ−

)√2πν(t2 − 4)

14

2 sin(πα− νπ

t2),

(2.3.1)

when t > 2;

f (α)n (

t√ν) ∼

√2ν−

n2 e

ν2

√πν(4− t2) 1

4

sin

[ν

(1

2t

√4− t2 + π

t2− 2

t2sin−1 t

2− cos−1 t

2

)+ πα− νπ

t2+π

4

],

(2.3.2)

when 0 < t < 2 and

f (α)n (

2√ν) ∼ ν−

n2 e

ν2 3

13ν−

13

(1

3Γ(23)cos(πα− νπ

4) +

√3

3Γ(23)sin(πα− νπ

4)

),

(2.3.3)

when t = 2. Next we recall the well-known asymptotic formulae

Ai(η) ∼ η−14

2√πexp(−2

3η

32 ) (2.3.4)

and

Bi(η) ∼ η−14

√πexp(

2

3η

32 ) (2.3.5)

2.4 Numerical Results 18

as η → +∞. See [24, p. 392]. Furthermore, since ν = n+2α− 1/2, it is readily seen

from (2.3.4) and (2.3.5) that

Pn(ν− 1

2 t)

2n2Γ(n

2+ 1)

∼ ν−n2 ν−

12

√2π(t2 − 4)

14

eν(

λ+t+( 2

t2−1) log λ+

), (2.3.6)

Qn(ν− 1

2 t)

2n2Γ(n

2+ 1)

∼√2ν−

n2 ν−

12

π(t2 − 4)14

eν(

λ−t+( 2

t2−1) log λ−

). (2.3.7)

Comparing the left-hand side with the right-hand side in (2.2.18), one concludes that

C1(x) =√π cos(πα− π

x2), C2(x) =

√π sin(πα− π

x2). (2.3.8)

Note that in obtaining (2.3.8), any one of the three formulae (2.3.1), (2.3.2) or (2.3.3)

could have been used. In summary, we have from (2.2.4), (2.2.18), (2.2.19) and

(2.2.20)

f (α)n (

t√ν) =

√π cos(πα− νπ

t2)

212nΓ(1

2n+ 1)

(4ζ

t2 − 4

) 14

ν16

[Ai(ν

23 ζ)+O(

1

ν)

]

+

√π sin(πα− νπ

t2)

212nΓ(1

2n+ 1)

(4ζ

t2 − 4

) 14

ν16

[Bi(ν

23 ζ)+O(

1

ν)

],

(2.3.9)

where x = ν−12 t and ν = n + 2α − 1/2. This result holds uniformly for 0 ≤ t < ∞.

As a check on the validity of (2.3.9), we note that

f(α)2n (0) =

(−1)n

2nn!, f

(α)2n+1(0) = 0, n = 0, 1, 2, . . . . (2.3.10)

From (2.3.9), we obtain

f (α)n (0) ∼ en/2√

πn(n+1)/2sin(−nπ

2+π

2). (2.3.11)

In view of Stirling’s formula, (2.3.11) agrees with (2.3.10).

2.4 Numerical Results

The expansion (2.3.9) is particularly useful near t = 2, where the two characteristic

roots in (2.2.14) coincide. As an illustration, we take α = 1.9 and n = 100. Table 2.1

provides exact and approximate values of 2n/2Γ(n/2+1)f(α)n (t/

√ν). The last column

of the table shows the percentage error of the approximations.

2.4 Numerical Results 19

Table 2.1: Numerical Results

t Exact values Approximate values Error

1.0 -7.46841313×10−1 -7.37169075×10−1 0.0130

1.2 6.30560302×10−1 6.34644034×10−1 0.0065

1.4 8.46491439×10−1 8.31094179×10−1 0.0182

1.6 -1.30620957 -1.29091632 0.0117

1.8 1.51877275 1.49426911 0.0161

2.0 1.90931070 1.89864389 0.0056

2.2 3.26791053×102 3.16621347×102 0.0311

2.4 -9.21372780×104 -8.83780892×104 0.0408

2.6 3.94171949×109 3.74229636×109 0.0506

2.8 1.54972056×1013 1.45539824×1013 0.0609

3.0 7.21884326×1016 6.70106781×1016 0.0717

Chapter 3


f(α)n (x) via Method of Steepest Descent

The purpose of this section is to confirm that the expansion is valid for unbounded

values of t, and to show that numerically computable bounds can be constructed for

the error terms associated with the expansion. The approach taken in this chapter is

based on the modified steepest-decent method of Chester, Friedman and Ursell [7], and

the use of a sequence of rational functions introduced by Olde Daalhuis and Temme

[23].

3.1 Reduction to a Canonical Integral

From (1.1.10), we have by the Cauchy integral formula

f (α)n (x) =

1

2πi

∫Cexp

w

x+

1− αx2

x2log(1− wx)

dw

wn+1, (3.1.1)

where C is a circle centered at the origin and oriented in the counter-clockwise direction

with a radius less than 1/|x|, x = 0. Rescaling the variables

x = t/√ν, w = s

√ν and ν = n+ 2α− 1/2, (3.1.2)

equation (3.1.1) becomes

f (α)n (t/

√ν) =

ν−n/2

2πi

∫Ceν ϕ(s;t) 1

s−2α+1/2s(1− ts)αds, (3.1.3)

3.1 Reduction to a Canonical Integral 21

where

ϕ(s; t) =s

t+

1

t2log(1− ts)− log s. (3.1.4)

For definiteness, we choose for all logarithmic functions in (3.1.4) the principal branch

log ζ = log |ζ|+ i arg ζ, −π < arg ζ ≤ π. (3.1.5)

The phase function ϕ(s; t) has branch points at s = 0 and s = 1/t, although w = 0

is not a branch point of the integrand in (3.1.1). This function is analytic and single-

valued in the complex s-plane with cuts along the intervals (−∞, 0] and [1/t,∞). For

later discussion, we also need to specify the values of the argument along the upper

and lower edges of the cuts. To this end, we let s± = u+ i 0±, u > 1/t, denote points

on the upper and lower edges of the cut along [1/t,∞). For instance, we have

arg(1− ts+) = −π and arg(1− ts−) = π. (3.1.6)

From (1.1.10), it is easily seen that

∞∑n=0

f (α)n (−x) (−w)n =

∞∑n=0

f (α)n (x)wn,

from which it follows that

f (α)n (−x) = (−1)nf (α)

n (x). (3.1.7)

For this reason, we may restrict x to the interval [0,∞).

The saddle points of ϕ(s; t), i.e., zeros of ∂ϕ/∂s, are at

s = s± =1

2(t±√t2 − 4).

It is convenient to consider two separate cases: (i) 0 < t < 2 and (ii) t ≥ 2. The last

equation is equivalent to

s± =

1

2(t± i

√4− t2) if 0 < t < 2,

1

2(t±√t2 − 4) if t ≥ 2.

(3.1.8)


Clearly, the saddle points coalesce at s = 1 when t = 2. Note that for t ≥ 2, we have

1/t < s− < s+ <∞. The steepest descent (and ascent) paths are given by

ℑϕ(s; t) = ℑϕ(s±; t). (3.1.9)

Case 1: t > 2. Since s+± are on the upper edge of the cut, we have from (3.1.6)

log(1− s+±t) = −iπ + log(s+±t− 1) = −iπ + 2 log s+±, (3.1.10)

where use has been made of the fact that s± are zeros of ∂ϕ/∂s, i.e., s2±− t s±+1 = 0.

Similarly, since s−± are on the lower edge of the cut, we have

log(1− s−±t) = +iπ + log(s−±t− 1) = +iπ + 2 log s−±. (3.1.11)

If we write s = u+ iv, then equation (3.1.9) can be written as

v

t+

1

t2arg(1− tu− itv)− arg(u+ iv) = ℑϕ(s+±; t) = −

π

t2(3.1.12)

and

v

t+

1

t2arg(1− tu− itv)− arg(u+ iv) = ℑϕ(s−±; t) = +

π

t2. (3.1.13)

Using Maple, the steepest paths given by (3.1.12) and (3.1.13) are shown in Figure 3.1,

where arrows are used to indicate the direction of descent ( For a detailed analysis of

directions of descent and ascent, see Appendix C ).

Case 2: 0 < t < 2. For u ∈ (−∞,∞), we take the branch of tan−1 u with values

in (−π2, π2). If z = x+ iy, it is easily seen that

arg(x+ iy) =

tan−1(y/x) if x > 0 and y > 0 or < 0,

π + tan−1(y/x) if x < 0 and y > 0,

−π + tan−1(y/x) if x < 0 and y < 0.

(3.1.14)

By again using the fact that s± are zeros of ∂ϕ/∂s, i.e., s2± − t s± + 1 = 0, equation

(3.1.9) can be written as

v

t+

1

t2arg(1− tu− itv)− arg(u+ iv) = ℑϕ(s±; t) = ±θt, (3.1.15)


Figure 3.1: Steepest paths when t ≥ 2.

where θt = 12t

√4− t2 − π

t2+ ( 2

t2− 1) tan−1

√4−t2

t. The graph of the curves given by

(3.1.15) are shown in Figure 2, with arrows again indicating directions of ascent and

descent. ( Detailed analysis can be found in Appendix C. )

0

Figure 3.2: Steepest path when 0 < t < 2.

Since the two saddle points s± coalesce at s = 1, let us expand ϕ(s; t) into a Taylor


series around this point. The result is

ϕ(s; t) =1

t+

1

t2log(1− t) +

[1

t+

1

t(t− 1)− 1

](s− 1)

+1

2!

(1− 1

(t− 1)2

)(s− 1)2 +

1

3!

[2t

(t− 1)3− 2

](s− 1)3 + · · · .

When t → 2, the linear and quadratic terms in the above expansion both vanish, and

the first nonvanishing term is the cubic. This suggests that we make the cubic transfor-

mation

ϕ(s; t) =u3

3− ζ(t)u+ η(t) (3.1.16)

introduced by Chester, Friedman and Ursell [7]. See also [24, p. 351]. Differentiating

(3.1.16) with respect to s gives

∂s

∂u=u2 − ζ(t)∂ϕ/∂s

= (u2 − ζ(t)) s(ts− 1)

(s− s−)(s− s+).

To ensure that the mapping s 7→ u is one-to-one and analytic, ∂s/∂u can neither

vanish nor become infinite. Hence, we must assign the saddle points s± of ϕ(s; t) to

correspond to the saddle points±√ζ(t) of the cubic polynomial on the right-hand side

of (3.1.16). This gives

−2

3ζ3/2(t) + η(t) =

s+t

+1

t2log(1− ts+)− log s+,

2

3ζ3/2(t) + η(t) =

s−t

+1

t2log(1− ts−)− log s−.

(3.1.17)

Note that, in addition to (3.1.10) and (3.1.11), we also have

s+ + s− = t, s+s− = 1,s+s−

=s2+s−s+

= ts+ − 1,

s+ − s− =

√t2 − 4 if t ≥ 2,

i√4− t2 if 0 < t < 2,

s+t− 1

s−t− 1= (s+t− 1)2 = s4+.


With the aid of these identities, we obtain upon solving the two equations in (3.1.17)

ζ3/2(t) =

3

4

[(2− 4

t2) log

(1

2(t+√t2 − 4)

)− 1

t

√t2 − 4

]if t > 2,

−i34

[1

t

√4− t2 + 2π

t2− 4

t2sin−1 t

2− 2 cos−1 t

2

]if 0 < t < 2

(3.1.18)

and

η(t) =1

2∓ iπ

t2, (3.1.19)

where the minus and plus signs in η(t) correspond to the paths in the upper and lower

half of the plane, respectively ( For detailed calculations, we again refer to Appendix

C ). Note that as t → 2, we have ζ(t) → 0. In fact, near t = 2, we have the Taylor

expansion

ζ(t) = (t− 2)

[1− 29

60(t− 2) +

799

3150(t− 2)2 + · · ·

]. (3.1.20)

The cubic equation (3.1.16) can be solved explicitly by using trigonometric func-

tions; the solution is

u(s, t) = 2ζ1/2(t) sin1

3γ, (3.1.21)

where γ satisfies

sin γ =3

2ζ3/2(t)[η(t)− ϕ(s; t)]. (3.1.22)

See [37, p. 374]. The properties of the transformation s 7→ u are best seen by intro-

ducing an intermediate variable Z defined by

u = 2√ζ(t) sin

γ

3,

2

3ζ3/2(t) sin γ = Z, Z = η(t)− ϕ(s; t). (3.1.23)

Note that here, for convenience, we simply write u = u(s; t), γ = γ(s; t) and Z =

Z(s; t). Since we require s+ ↔ u+ = ζ1/2(t) and s− ↔ u− = −ζ1/2(t), the transfor-

mation s 7→ Z take s+ to 23ζ3/2(t) and s− to −2

3ζ3/2(t).

The mapping s 7→ u defined in (3.1.16) is well studied in the literature. See, e.g.

[7] and [37]. Here, we give only a brief sketch of the argument ( Details are given in

Appendix C ). First, we consider


Case 1: t ≥ 2. To make the function Z = Z(s, t) single-valued, we divide the up-

per s-plane into two regions by using part of the steepest-descent path passing through

s+. See Figure 3.3.

BA C’ ECB’

F

E’ JD D’

G

H

I

Figure 3.3: s-plane (t > 2).

Through the sequence of mappings s 7→ Z 7→ γ 7→ u in (3.1.23), the two regions

bounded by ABB’CC’DD’EFGA and E’FGHIJE’ are mapped into two corresponding

regions in the u-plane. See Figure 3.4.

Note that the function ϕ(s; t) has singularities at s = 0, s = 1/t and saddle points

s = s±. One can show that the mapping s 7→ u is one-to-one and analytic on the

boundaries of these two regions. Hence, it also has these properties in the interior of

the regions. See [38, p. 12]. This establishes the one-to-one and analytic nature of the

mapping in the upper half-plane. The mapping properties of the lower half-plane can

be obtained by using reflection with respect to the real-axis.

Case 2: 0 < t < 2. This case is treated in a similar manner. We again divide the

upper half of the s-plane into two pieces by the steepest-descent path passing through

s+. See Figure 3.5.

The two regions with boundaries ABB’CDEA and DC’HGFED are mapped into

two corresponding regions in the u-plane in a one-to-one analytic manner. See Figure

3.6.


B

A

C’ E

C

B’

F

E’D D’

G

H

I

J

Figure 3.4: u-plane (t > 2).

This property of the mapping s 7→ u in the lower half of the s-plane is again

established by using reflection with respect to the real-axis.

We now return to the integral in (3.1.3)

f (α)n (t/

√ν) =

ν−n/2

2πi

∫Ceν ϕ(s;t) 1

s−2α+1/2s(1− ts)αds,

where C is a circle centered at the origin and with a radius less than 1/t. For t ≥ 2, we

see from Figure 3.1 that C can be deformed into the steepest-descent paths shown in

Figure 3.7.

In the case when 0 < t < 2, we deform C into the steepest-descent paths depicted

in Figure 3.2. In either case, the new contour will consist of two parts C+ and C−; C+is the part in the upper half-plane and C− is the part in the lower half-plane. Making

the change of variable s 7→ u given in (3.1.16), we obtain

f (α)n (t/

√ν) =

ν−n/2

2πieν(

12−iπ/t2)

∫L+

eν ( 13u3−ζu)h(u) du

− ν−n/2

2πieν(

12+iπ/t2)

∫L−

eν ( 13u3−ζu)h(u) du,

(3.1.24)

where L± are the images of C± under the transformation and

h(u) =1

s−2α+1/2s(1− ts)αds

du. (3.1.25)

3.2 Derivation of Expansion 28

BA C’

E

CB’

F

G

H

Figure 3.5: s-plane (0 < t < 2).

In (3.1.24), we have also made use of (3.1.19). Recall that a one-to-one analytic map

takes steepest-descent paths to steepest-descent paths. Thus, the curves L+ and L− are

the steepest descent paths of the phase function 13u3− ζu. Sketches of these curves are

shown in Figure 3.8.

Since (1− ts)α = e−iαπ(ts− 1)α for s ∈ C+, (3.1.24) can be written as

f (α)n (t/

√ν) =

ν−n/2

2πieν/2ei(απ−πν/t2)

∫L+

eν ( 13u3−ζu)h(u) du

− ν−n/2

2πieν/2e−i(απ−πν/t2)

∫L−

eν ( 13u3−ζu)h(u) du,

(3.1.26)

where

h(u) =1

s−2α+1/2s(ts− 1)αds

du. (3.1.27)

3.2 Derivation of Expansion

To derive the asymptotic expansion of f (α)n (t/

√ν) from the integral representation

(3.1.26), we apply the integration-by-parts technique developed by Bleistein [4]. First,

we write

h0(u) = h(u) = α0 + β0u+ (u2 − ζ(t))g0(u). (3.2.1)


B

A

C

B’

F

E

D

G

HC’

Figure 3.6: u-plane (0 < t < 2).

Setting u = ±√ζ(t) on both sides of the equation, we have

h0(√ζ(t) ) = α0 + β0

√ζ(t), h0(−

√ζ(t) ) = α0 − β0

√ζ(t).

Hence,

α0 =1

2[h0(

√ζ(t) ) + h0(−

√ζ(t) )], β0 =

1

2√ζ(t)

[h0(√ζ(t) )− h0(−

√ζ(t) )].

(3.2.2)

To obtain the values of h0(±√ζ(t)), we let u → −

√ζ(t) in the equation following

(3.1.16). This gives

ds

du

∣∣∣∣u=−√

ζ(t)

=

[2√ζ(t) s−(1− ts−)s− − s+

]1/2=

[2√ζ(t) s−(ts− − 1)

s+ − s−

]1/2. (3.2.3)

Similarly, we have

ds

du

∣∣∣∣u=√

ζ(t)

=

[2√ζ(t) s+(1− ts+)s− − s+

]1/2=

[2√ζ(t) s+(ts+ − 1)

s+ − s−

]1/2. (3.2.4)

Note that the above two quantities become indeterminate (i.e. 0/0), when t→ 2. Thus,

we must treat the case t = 2 with care. For t > 2, we have from (3.1.8) and (3.1.18)

ds

du

∣∣∣∣u=±√

ζ(t)

=2

12

(t2 − 4)14

[1

2(t±√t2 − 4)

] 32

×(3

4

) 16[(2− 4

t2) log

(1

2(t+√t2 − 4)

)− 1

t

√t2 − 4

] 16

.

(3.2.5)


Figure 3.7: Steepest-descent (t ≥ 2).

Similarly, for 0 < t < 2, we have

ds

du

∣∣∣∣u=±√

ζ(t)

=2

12

(4− t2) 14

[1

2(t± i

√4− t2)

] 32

×(3

4

) 16[1

t

√4− t2 + 2π

t2− 4

t2sin−1 t

2− 2 cos−1 t

2

] 16

.

(3.2.6)

In the case when t = 2, we apply the l’Hopital’s rule and obtain

ds

du

∣∣∣∣u=±√

ζ(2)

= limt→2

ds

du

∣∣∣∣u=±√

ζ(t)

= 1. (3.2.7)

From (3.1.27), it follows that1

h0(±√ζ(t) ) =

√2

(t2 − 4)1/4

(3

4

) 16[(2− 4

t2) log

(1

2(t+√t2 − 4)

)− 1

t

√t2 − 4

] 16

for t > 2,

h0(±√ζ(t) ) =

√2

(4− t2)1/4

(3

4

) 16[1

t

√4− t2 + 2π

t2− 4

t2sin−1 t

2− 2 cos−1 t

2

] 16

for 0 < t < 2 and

h(±√ζ(t) ) = 1 for t = 2.

1h0(±√

ζ(t) ) can also be written as√2(

ζt2−4

) 14

for t > 2 or 0 < t < 2.


(a) u-plane (t ≥ 2) (b) u-plane (0 < t < 2)

Figure 3.8: Contours L+ and L−.

Note that u = ±√ζ(t) correspond to s = s±. A combination of (3.2.2) and the last

three equations gives2

α0 =

√2

(t2 − 4)1/4

(3

4

) 16[(2− 4

t2) log

(1

2(t+√t2 − 4)

)− 1

t

√t2 − 4

] 16

(3.2.8)

for t > 2,

α0 =

√2

(4− t2)1/4

(3

4

) 16[1

t

√4− t2 + 2π

t2− 4

t2sin−1 t

2− 2 cos−1 t

2

] 16

(3.2.9)

for 0 < t < 2 and

α0 = 1 for t = 2. (3.2.10)

Since h(+√ζ(t) ) = h(−

√ζ(t) ) for all t ∈ (0,∞), we have

β0 = 0 for t ∈ (0,∞). (3.2.11)

Substituting (3.2.1) in (3.1.27), and integrating term-by-term, we encounter inte-

grals which can be expressed in terms of the function

V (λ) =

∫γ

ev3/3−λvdv,

where γ is one of the contours γ0, γ1 and γ2 shown in Figure 3.9.

2From (3.1.18), for both cases in (3.2.8) and (3.2.9), α0 can also be written as√2(

ζt2−4

) 14

.


Figure 3.9: Contours γj, j = 0, 1, 2.

In terms of the Airy function, we have

V (λ) =

∫γj

ev3/3−λvdv = −2πiωjAi(λωj),

where ω is the cube root of unity (i.e. ω = e2πi/3). In view of the connection formulas

Ai(z) + ωAi(ωz) + ω2Ai(ω2z) = 0, (3.2.12)

Bi(z) + iωAi(ωz)− iω2Ai(ω2z) = 0, (3.2.13)

after rearranging the terms, equation (3.1.26) gives

e−ν/2νn/2f (α)n (

t√ν) = cos(πα− πν

t2)

[Ai(ν2/3ζ)

α0

ν1/3− Ai′(ν2/3ζ)

β0ν2/3

]+ sin(πα− πν

t2)

[Bi(ν2/3ζ)

α0

ν1/3− Bi′(ν2/3ζ)

β0ν2/3

]+ ϵ1 + δ1,

(3.2.14)

where

ϵ1 = ei(πα−πν/t2) 1

2πi

∫γ1

eν(u3/3−ζu)(u2 − ζ)g0(u) du,

δ1 = e−i(πα−πν/t2) 1

2πi

∫γ2

eν(u3/3−ζu)(u2 − ζ)g0(u) du.


An integration by parts shows that

ϵ1 = −ei(πα−πν/t2) 1

2πiν

∫γ1

eν(u3/3−ζu)g′0(u) du,

δ1 = −e−i(πα−πν/t2) 1

2πiν

∫γ2

eν(u3/3−ζu)g′0(u) du.

(3.2.15)

The error terms ϵ1 and δ1 now have an extra decaying factor 1/ν.

The above procedure can be repeated, and we define inductively

hm(u) = αm + βmu+ (u2 − ζ)gm(u), (3.2.16)

hm+1(u) = g′m(u) (3.2.17)

for m = 0, 1, 2, · · · . As in (3.2.2), we have

αm =1

2[hm(

√ζ(t) )+hm(−

√ζ(t) )], βm =

1

2√ζ(t)

[hm(√ζ(t) )−hm(−

√ζ(t) )].

(3.2.18)

The final result is

e−ν/2νn/2f (α)n (

t√ν)

= cos(πα− πν

t2

)[Ai(ν2/3ζ)

p−1∑k=0

(−1)kαk

νk+1/3− Ai′(ν2/3ζ)

p−1∑k=0

(−1)kβkνk+2/3

]+ ϵp

+ sin(πα− πν

t2

)[Bi(ν2/3ζ)

p−1∑k=0

(−1)kαk

νk+1/3− Bi′(ν2/3ζ)

p−1∑k=0

(−1)kβkνk+2/3

]+ δp,

(3.2.19)

where

ϵp = ei(πα−πν/t2) (−1)p

2πiνp

∫γ1

hp(u)eν(u3/3−ζu) du (3.2.20)

and

δp = e−i(πα−πν/t2) (−1)p

2πiνp

∫γ2

hp(u)eν(u3/3−ζu) du (3.2.21)

for p = 1, 2, 3, · · · . To establish that (3.2.19) is indeed a uniform asymptotic expansion

in t (or ζ), it still remains to show that there are constants Mp, Np, Pp and Qp such that

|ϵp| ≤Mp

νp+1/3|Ai(ν2/3ζ)|+ Np

νp+2/3|Ai′(ν2/3ζ)| (3.2.22)

3.3 For Unbounded Values of ζ 34

and

|δp| ≤Pp

νp+1/3|Bi(ν2/3ζ)|+ Qp

νp+2/3|Bi′(ν2/3ζ)|. (3.2.23)

For bounded values of ζ, the proofs of (3.2.22) and (3.2.23) are now routine., and we

refer to [5, pp. 374–376] and [37, pp. 371–372] for details. The essential arguments

here are: (i) to find the behavior of the two integrals in (3.2.20) and (3.2.21), (ii) to

compare them with the behavior of the Airy functions Ai, Ai′, Bi and Bi′, and (iii) to

observe that Ai and Ai′ do not have a common zero; neither do Bi and Bi′.

3.3 For Unbounded Values of ζ

The advantage of the above derivation, based on Bleistein’s method of integration by

parts, is that it provides exact expressions for the error terms ϵp and δp, and in addition

that these expressions exhibit the decaying factor 1/νp explicitly. Since the integrand

hp(u) = hp(u; t) in (3.2.20) and (3.2.21) is analytic in t, it is evident from these

expressions that the expansion in (3.2.19) is asymptotic for bounded values of t. To

show that it is an asymptotic expansion even for unbounded values of t, we use a more

recent method of Olde Daalhuis and Temme [23] which was specifically designed for

this kind of purpose. Thus, following [23], we introduce the sequence of rational

functions

R0(w, u,√ζ) =

1

w − u,

Rn+1(w, u,√ζ) =

−1w2 − ζ

d

dwRn(w, u,

√ζ), n = 0, 1, · · · ,

(3.3.1)

where w, u,√ζ ∈ C, w = u and w2 = ζ . It can be shown that the function hn(u)

defined by the recursive formula (3.3.1) has the representation

hn(u) =1

2πi

∫Γ

Rn(w, u,√ζ)h0(w)dw, (3.3.2)

where Γ is any simple closed contour in the domain of analyticity of h0(u), enclosing

the points u and ±√ζ . Furthermore, it can be verified that there are constants Cij ,

independent of w, u and√ζ such that

Rn(w, u,√ζ) =

n−1∑i=0

kn,i∑j=0

Cijwi−j

(w − u)n+1−i−j(w2 − ζ)n+i, n = 1, 2, · · · , (3.3.3)


where kn,i = min(i, n− 1− i). Note that the rational functions Rn(w, u,√ζ) defined

in (3.3.1) are independent of the function h0(u). Hence, (3.3.2) can be considered as

an analogue of the Cauchy integral representation for the remainder in a Taylor expan-

sion. To extend the validity of the asymptotic expansion in (3.2.19) from bounded t to

unbounded t, we first define

ρ0(√ζ) = min|u±

√ζ| : u is a singularity of h0(u). (3.3.4)

For t ≥ 2, the points s± = 12(t ±√t2 − 4) in the s-plane are mapped into u = ±

√ζ,

respectively, under the mapping s↔ u given by

s

t+

1

t2log(1− ts)− log s =

1

3u3 − ζ2(t)u+ η(t), (3.3.5)

where the logarithmic function is restricted to its principal value. See (3.1.4) and

(3.1.16). However, the points s± on other sheets of the Riemann surface of logarithmic

function are singular points of the mapping (3.3.5).

Lemma 3.1. As t→∞ (and hence√ζ →∞), we have

ρ0(√ζ) ∼

√2π

ζ1/4

(1− 2

t2

)1/2

. (3.3.6)

Proof. The cubic transformation (3.3.5) can be solved explicitly in terms of trigono-

metric functions. The solution that takes s± to ±√ζ , respectively, is given by (3.1.21)

and (3.1.22); that is,

u(s, t) = 2ζ1/2(t) sin1

3γ, sin γ =

3

2ζ3/2(t)[η(t)− ϕ(s; t)].

Since s = s+ is mapped into u =√ζ , we obtain

√ζ = 2

√ζ sin 1

3γ and

sin γ =3

2ζ3/2(t)

[1

2− s+

t− (

2

t2− 1) log s+

], (3.3.7)

where we have made use of (3.1.10) and (3.1.19). Let S±k denote the singular points

in the u-plane. When u = S±k, we have S±k = 2√ζ sin 1

3γ±k and

sin γ±k =3

2ζ3/2(t)

[1

2− s+

t− (

2

t2− 1)(log s+ ± 2kπi)

]. (3.3.8)


Note that the left-hand side of (3.3.7) is actually equal to 1. Hence, subtracting (3.3.7)

from (3.3.8) gives

sin γ±k = 1± 3kπi

ζ3/2(t)(1− 2

t2) (3.3.9)

and

γ±k = arcsin

[1± 3kπi

ζ3/2(t)(1− 2

t2)

]∼ π

2−

√∓ 6kπi

ζ3/2(t)(1− 2

t2) (3.3.10)

as t → ∞ (or equivalently ζ(t) → ∞). The last asymptotic equality follows from an

expansion for the arcsine function. See[1, p. 81, (4.4.41)]. Using the addition formula

of the sine function and the Maclaurin expansions of sine and cosine, we have from

the definition of S±k and asymptotic formula (3.3.10)

S±k = 2√ζ(t) sin

1

3γ±k

∼ 2√ζ(t)

[1

2cos

1

3

√∓ 6kπi

ζ3/2(t)(1− 2

t2)−√3

2sin

1

3

√∓ 6kπi

ζ3/2(t)(1− 2

t2)

]

∼√ζ(t)−

√∓ 2kπi

ζ1/2(t)(1− 2

t2).

(3.3.11)

Clearly, the two closest singularities to√ζ(t) are

S1 ∼√ζ(t)−

√− 2πi

ζ1/2(t)(1− 2

t2) and S−1 ∼

√ζ(t)−

√2πi

ζ1/2(t)(1− 2

t2).

(3.3.12)

Similarly, s = s− is mapped into u = −√ζ(t), and the two closest singularities to

−√ζ(t) are

S ′1 ∼ −

√ζ(t) +

√2πi

ζ1/2(t)(1− 2

t2) and S ′

−1 ∼ −√ζ(t) +

√− 2πi

ζ1/2(t)(1− 2

t2);

(3.3.13)

thus, proving the lemma.

To estimate the remainder ϵp in (3.2.20), we first deform the path of integration γ1

into the contour γ1a + γ1b, where γ1b runs from −∞ to√ζ along the real axis and γ1a

continues on from√ζ to∞eiπ/3 along the steepest descent path of the phase function

13u3 − ζu in the first quadrant of the u-plane. See Figure 3.10.


Figure 3.10: Coutour γ1a and γ1b.

Next, we split γ1a into γ′1a and γ′′1a, where γ′1a = u ∈ γ1a : |u− ζ12 | ≤ c ζ

12θ and

γ′′1a is the complement of γ′1a on γ1a (i.e., γ′′1a = γ1a \ γ′1a), and c and θ are constants

satisfying c > 0 and −1/2 < θ ≤ 1. Let L1a be a closed curve embracing the contour

γ′1a such that as√ζ →∞,

length of L1a = O(ζ12θ) and distance (L1a, γ

′1a) ∼ c ζ−

14 . (3.3.14)

Let Ω1a denote the domain bounded by L1a and its closure by Ω1a. See Figure 3.11(a).

Similarly, let γ′1b = u ∈ γ1b : |u + ζ12 | ≤ c ζ

12θ and γ′′1b = γ1b \ γ′1b. Also, let L1b

denote a closed curve embracing the contour γ′1b such that as√ζ →∞,

length of L1b = O(ζ12θ) and distance (L1b, γ

′1b) ∼ c ζ−

14 . (3.3.15)

Let Ω1b denote the domain bounded by L1b and its closure by Ω1b. See Figure 3.11(b).

Put Ω1,θ = Ω1a ∪ Ω1b. On account of Lemma 3.1, h0(u) is analytic in Ω1,θ and hence

bounded on Ω1,θ.

Similar to (3.3.14), we let L2a be a closed curve embracing the contour γ′1a such

that as√ζ →∞,

length of L2a = O(ζ12θ) and distance (L2a, γ

′1a) ∼

1

2c ζ−

14 . (3.3.16)

Let Ω2a denote the domain bounded by L2a and its closure by Ω2a. Note that L2a lies

inside L1a. Furthermore, let L2b be a closed curve embracing the contour γ′1b such that

as√ζ →∞,

length of L2b = O(ζ12θ) and distance (L2b, γ

′1a) ∼

1

2c ζ−

14 . (3.3.17)


(a) Curves L1a and L2a (b) Curves L1b and L2b

Figure 3.11: Curve L.

Let Ω2b denote the domain bounded by L2b and its closure by Ω2b. Note that L2b lies

inside L1b. Put Ω2,θ = Ω2a ∪ Ω2b. and L1,θ = L1a ∪ L1b.

Lemma 3.2. There is a constant c0 independent of√ζ such that as

√ζ →∞,

supu∈Ω1,θ

|h0(u)| ≤ c0|h0(√ζ)|(ζ

12θ+ 1

4 )2α−2. (3.3.18)

Proof. Note that from (3.1.18) and the equations following (3.2.7), we have

h0(±√ζ) =

√2

(ζ

t2 − 4

) 14

. (3.3.19)

To bound h0(u), first we consider u ∈ Ω1a and estimate s− s+. Replacing s by s+ in

(3.3.5) and subtracting the resulting equation from (3.3.5), we obtain

s− s+t

+1

t2log(1+

t(s− s+)s+s+

)−log(s− s+t

+1) =1

3(u+2

√ζ)(u−

√ζ)2, (3.3.20)

where use has been made of two identities following (3.1.17). Since there could be

points u ∈ Ω1a such that |u −√ζ| ∼ c ζ

12θ with −1

2< θ ≤ 1, for these u the right-

hand side of (3.3.20) behaves like c2ζ2θ+12 . If 1

t(s − s+) = O(1), then the left-hand

side of (3.3.20) is bounded, which is a contradiction. Thus, (3.3.20) gives

1

t|s− s+| ∼ c2ζ2θ+

12 as t→∞. (3.3.21)

In view of the first equation following (3.1.17), it is easy to show that

1

t|s− s−| ∼ c2ζ2θ+

12 as t→∞. (3.3.22)


A combination of (3.1.27), (3.3.21), (3.3.22) and the equation following (3.1.16) gives∣∣∣∣ h0(u)h0(√ζ)

∣∣∣∣ ∼ 1

t12 c−2α+1(ζ

12θ+ 1

4 )−2α+1

√2(t2 − 4)

14

c ζ12θ+ 1

4

∼√2c2α−2(ζθ+

12 )α−1, (3.3.23)

which certainly implies (3.3.18).

Returning to (3.2.20), we write

ϵp = ϵ′p(1a) + ϵ′′p(1a) + ϵ′p(1b) + ϵ′′p(1b), (3.3.24)

where

ϵ′p(1a) = ei(πα−πν/t2) (−1)p

2πiνp

∫γ′1a

hp(u)eν(u3/3−ζu) du, (3.3.25)

ϵ′′p(1a) = ei(πα−πν/t2) (−1)p

2πiνp

∫γ′′1a

hp(u)eν(u3/3−ζu) du, (3.3.26)

and ϵ′p(1b), ϵ′′p(1b) are defined in a similar manner. By simple calculation, it can easily

be shown that ϵ′′p(1a) and ϵ′′p(1b) are exponentially small in comparison with ϵ′p(1a) and

ϵ′p(1b). See [23]. Furthermore, for w ∈ L1a and u ∈ Ω2a, we have

|w−u| ≥ 1

2c ζ−

14 , |w−

√ζ| ≥ 1

2c ζ−

14 and |w+

√ζ| ≥

√ζ. (3.3.27)

Thus, from (3.3.3) with n replaced by p, we obtain

|Rp(w, u,√ζ)| ≤

p−1∑i=0

kp,i∑j=0

|Cij|(2√ζ)i−j

|w − u|p+1−i−j|w −√ζ|p+i(

√ζ)p+i

, (3.3.28)

p = 1, 2, · · · . Do the same for w ∈ L1b and u ∈ Ω2b, and obtain an estimate corre-

sponding to (3.3.28). Then, we have

supw∈L1,θ , u∈Ω2,θ

|Rp(w, u,B)| ≤p−1∑i=0

kp,i∑j=0

Cij ζ− 3j

4+ 1

4 ≤ Ap ζ14 (3.3.29)

as√ζ → ∞, where the constants Cij and Ap are constants independent of

√ζ. A

combination of (3.3.18), (3.3.29) and (3.3.14) gives∣∣∣∣∣ 1

2πi

∫L1,θ

Rp(w, u,√ζ)h0(w) dw

∣∣∣∣∣ ≤ A′p(ζ

θ+ 12 )α−

12 |h0(

√ζ)|, (3.3.30)


where A′p is again a constant. By the Cauchy integral formula, and (3.2.17), we have

upon integration by parts

hp(u) = −1

2πi

∫L1,θ

d

dwR0(w, u,

√ζ) gp−1(w) dw. (3.3.31)

Furthermore, from (3.3.1) and (3.2.16) it follows that

hp(u) =1

2πi

∫L1,θ

(w2 − ζ)R1(w, u,√ζ) gp−1(w) dw

=1

2πi

∫L1,θ

R1(w, u,√ζ)hp−1(w) dw

− 1

2πi

∫L1,θ

R1(w, u,√ζ) (αp−1 + βp−1w) dw.

(3.3.32)

The coefficients αp−1 and βp−1 are given in (3.2.18). In [23, (4.6)], it has been shown

that for p = 1, 2, · · · ,1

2πi

∫L1,θ

Rp(w, u,√ζ)hp−1(w) dw = O(ζ−3p/2). (3.3.33)

Since the expression for βp−1 has a factor of ζ−1/2, in view of (3.2.18) and (3.3.33),

the last integral in (3.3.32) is dominated by hpO(ζ−3/2), where we define

hp := supu∈Ω1,θ

|hp(u)|. (3.3.34)

The above process can be repeated, and we obtain

hp(u) =1

2πi

∫L1,θ

Rp(w, u,√ζ)h0 dw + hp−1O(ζ

−3/2) + · · ·+ h0O(ζ−3p/2).

(3.3.35)

From (3.3.30) and (3.3.35), we have by induction

hp ≤ cp(ζθ+ 1

2 )α−12 |h0(

√ζ)|, (3.3.36)

where cp is a constant independent of√ζ .

Returning to (3.3.24) – (3.3.26), we have

|ϵp| ≤ ν−p− 13 (cp + 1)(ζθ+

12 )α−

12 |h0(

√ζ)|(Ai(ν2/3ζ), (3.3.37)

since ϵ′′p(1a) and ϵ′′p(1b) are exponentially small in comparison with ϵ′p(1a) and ϵ′p(1b).

See the statement following (3.3.25) – (3.3.26). A similar estimate, with Ai(ν2/3ζ) in

(3.3.37) replaced by Bi(ν2/3ζ), holds for the remainder δp in (3.2.21). We have thus

established the following result.

3.4 Error Bounds 41

Theorem 3.3. The expansion in (3.2.19) holds with the reminders ϵp and δp satisfying

|ϵp|+ |δp| ≤ ν−p− 13 (cp + 1)(ζθ+

12 )α−

12 |h0(

√ζ)|[Ai(ν2/3ζ) + Bi(ν2/3ζ)] (3.3.38)

where |h0(√ζ)| is given by (3.3.19) and cp is a constant.

We have now extended the validity of the expansion (3.2.19) to all positive values

of t; that is, (3.2.19) is indeed an asymptotic expansion which holds uniformly with

respect to all t ≥ 0.

3.4 Error Bounds

To construct an error bound for the asymptotic expansion in (3.2.19) means to find a

value for the constant cp in (3.3.38), or to find values for the constants Mp, Np, Pp and

Qp in (3.2.22) and (3.2.23). In general, this problem is very difficult if not impossible,

and the desire to construct such bounds has been expressed in a survey article of Wong

[36]. With the advent of powerful computers, the problem has now become more

feasible. In this section, we will provide some basic steps in getting computable values

for the constants mentioned above.

To simplify the matter, and to make the problem more tangible, we take p = 1 so

that we need only deal with the two integrals

ϵ1 = −1

νei(πα−πν/t2) 1

2πi

∫γ1

h1(u)eν(u3/3−ζu) du (3.4.1)

and

δ1 = −1

νe−i(πα−πν/t2) 1

2πi

∫γ2

h1(u)eν(u3/3−ζu) du. (3.4.2)

As in Section 3.3, we deform the path of integration γ1 in (3.4.1) into the contour

γ1a + γ1b. See Figure 3.10. If we write u = σ + iτ , then the parametric equation for

the steepest descent path γ1a passing through√ζ is given by τ 2 = 3σ2 − 3

√ζ. From

(3.2.16) and (3.2.17), we have

h1(u) = g′0(u) =1

(u2 − ζ)2[h′0(u)(u

2 − ζ)− 2u (h0(u)− α0)]

(3.4.3)

3.4 Error Bounds 42

(Note that in our case β0 = 0). Since expansion (3.2.19) was purposely designed to

be uniformly valid in a neighborhood of the critical value t = 2, let us restrict it to the

interval 0 < t < 4. Since α is a fixed number, it is relatively small in comparison to

the large asymptotic variable√ν in the function f (α)

n (t/√ν) under consideration. For

simplicity, we also restrict it to 0 < α < 4. What we shall show in the following is that

numerical evidence indicates that with these restrictions, there is a positive constant

c1 < 2 such that

supu∈γ1a

|h1(u)| ≤ c1|h1(√ζ)| and sup

u∈γ1b|h1(u)| ≤ c1|h1(−

√ζ)|. (3.4.4)

Applying (3.4.4) to (3.4.1) and (3.4.2) will yield the error bound

|ϵ1 + δ1| ≤ |ϵ1|+ |δ1| ≤c1ν4/3

[|h1(

√ζ)||Ai(ν2/3ζ)|+ |h1(−

√ζ)||Bi(ν2/3ζ)|

];

(3.4.5)

cf. (3.2.22) and (3.2.23). Here,

h1(±√ζ) = ±h0(

√ζ)

4ζ

2ζ

s±√t2 − 4

[(2α− 3

2)(2α− 5

2)s2± − 2(2α− 3

2)αts±

+ α(α+ 1)t2]∓ ζ

t2 − 4

1

s2±

[4(t− s3±)

((2α− 3

2)s± − αt

)+ 3(s4± − t2)

]+

10ζ(t− s3±)2

3(t2 − 4)32 s3±− 5

12√ζ

.

(3.4.6)

Now, we proceed to describe how to obtain a value for the constant c1. We only

consider the case 2 ≤ t < 4, since the other case 0 < t ≤ 2 can be handled in a similar

manner. Using a computer, one can calculate the values of h1(u) along the contours

γ1a and γ1b, except possibly at the saddle points u = ±√ζ . Note that the denominator

of h1(u) vanishes at u = ±√ζ . See (3.4.3). Numerical computation indicates that the

supremum of h1(u) on γ1a occurs near u =√ζ , and that the supremum of h1(u) on

γ1b occurs near u = −√ζ . To avoid calculation of indeterminate forms near ±

√ζ, we

draw a small circle with the center at each of these two points. Since h1(u) is analytic

along the contour γ1a and γ1b, by the maximum modulus principle the maximum of

h1(u) inside the circles occurs on the boundaries of these circles. Comparing the max-

imal values of h1(u) on the circles and those on the steepest descent paths γ1a and γ1b

3.4 Error Bounds 43

outside the circles, we find that the former are bigger than the latter. This led us to the

claim stated in (3.4.4). To get an estimate on the constant c1, we examine the ratio∣∣∣∣exact value− leading termerror term in (3.4.5)

∣∣∣∣ , (3.4.7)

where the exact value is f (α)n (t/

√ν), the leading term is

eν/2ν−n/2−1/3α0

[cos(πα− πν

t2)Ai(ν2/3ζ) + sin(πα− πν

t2)Bi(ν2/3ζ)

](3.4.8)

obtained from (3.2.14) with β0 = 0, and the error estimate in (3.4.5) is

eν/2ν−n/2−4/3(|h1(

√ζ)||Ai(ν2/3ζ)|+ |h1(−

√ζ)||Bi(ν2/3ζ)|

). (3.4.9)

Numerical computation shows that for 0 < α < 4 and 0 < t < ∞, the values of the

ratio in (3.4.7) lie between 0 and 2. See Tables 3.1 and 3.2. This confirms the estimate

0 < c1 < 2 mentioned earlier.

Table 3.1: Numerical values of c1 with different values of n and α = 1.

t n = 100 n = 500 n = 1000

0.5 0.3165 N.A. N.A.

1.5 0.3705 0.2487 0.0937

2.0 0.000001543 0.000007818 0.00001915

2.5 0.6845 0.6846 0.6843

3.0 0.7679 0.7666 0.7663

3.5 0.7849 0.1913 0.6955

4.0 0.8866 0.8832 0.9580

In principle, one can construct error bounds for higher order asymptotic approxi-

mations, but the amount of work would be tremendous. With the aid of a computer,

we also find the estimate

|ϵ2 + δ2| ≤c2ν7/3

[|h2(

√ζ)||Ai(ν2/3ζ)|+ |h2(−

√ζ)||Bi(ν2/3ζ)|

]. (3.4.10)

3.4 Error Bounds 44

Table 3.2: Numerical values of c1 with different values of n and α = 3.

t n = 100 n = 500 n = 1000

0.5 0.4685 N.A. N.A.

1.5 0.7641 0.1298 0.6106

2.0 0.00004671 0.00002387 0.000009098

2.5 0.4176 0.3775 0.3727

3.0 0.9348 0.5197 0.7809

3.5 1.2532 0.7817 0.2606

4.0 1.4302 1.0293 0.4887

Table 3.3 provides the numerical evidence that the constant c2 satisfies the estimate

0 < c2 < 2.

To conclude this section, we wish to point out that it is difficult to compute the

exact values of f (α)n (t/

√ν) for large values of n (say, e.g., n = 500, 1000) and small

positive values of t. This is mainly because of the factor x−n in the exact expression

of the Tricomi-Carlitz polynomial

f (α)n (x) =

1

xn

n∑k=0

(−1)k(x−2 − α

k

)x2k

(n− k)!

=1

n!xn

n∑k=0

(−1)k(n

k

)(1− αx2

)· · ·(1− (α+ k − 1)x2

),

(3.4.11)

which can be easily derived from the generalized hypergeometric function in (1.1.14)

or its relation with the Tricomi polynomials (given in [19, (1.1)]). Note that the sec-

ond sum in (3.4.11) is a polynomial of degree 2n, and that its coefficients of xk,

k = 0, 1, · · · , n − 1, are all zero, that is, the right-hand side of (3.4.11) is indeed a

polynomial of degree n. In view of this difficulty, in the construction of Tables 3.1 and

3.2 we have used the three-term recurrence relation (1.1.7) to compute the exact value

of f (α)n (t/

√ν) for t = 0.5 and n = 100. (When n = 500 or n = 1000 and t = 0.5,

3.5 Zeros 45

Table 3.3: Numerical values of c2 when n = 100.

t α = 1 α = 3

0.5 0.2218 0.7571

1.5 0.5866 0.4855

2.0 0.0319 0.9556

2.5 0.6565 0.4162

3.0 0.7193 0.9272

3.5 0.7126 1.2348

4.0 0.7781 1.3975

the computation is just too big for our computer to produce a reasonable figure even

by using (1.1.7)). The uniform asymptotic approximation (3.2.14) now becomes very

useful, since its validity includes neighborhoods of t = 0 and t = 2, and there is no

problem with calculating sin(πα− πν/t2) and cos(πα− πν/t2) even for small values

of t.

3.5 Zeros

It is well known that the zeros of the Tricomi-Carlitz Polynomials f (α)n (x) all lie in the

interval −1/√α ≤ x ≤ 1/

√α. See [14]. After the rescaling x = t/

√ν in (3.1.2), the

zeros of f (α)n (t/

√ν) all lie in the interval −

√ν/α ≤ t ≤

√ν/α. In this section, we

shall use the asymptotic approximation given in (3.2.14) to derive asymptotic formulas

for the zeros of f (α)n (t/

√ν) as n → ∞. We divide our discussion into three cases: (i)

large zeros in 2 < t <√ν/α; (ii) small zeros in 0 < t < 2; and (iii) zeros on both

sides of t = 2. Several tables are provided to give a comparison of the asymptotic

zeros and the true zeros.

3.5 Zeros 46

Case (i). Let tn,m denote them-th zero of f (α)n (t/

√ν), arranged in ascending order

−√ν

α≤ tn,1 < tn,2 < · · · < tn,n−1 < tn,n ≤

√ν

α. (3.5.1)

For large zeros, we have the following result.

Theorem 3.4. The asymptotic behavior of the large zeros of f (α)n (t/

√ν) can be ap-

proximated by

tn,n−k =

√ν

α+ k

[1 +O(

1

ν)

](3.5.2)

as n→∞, for fixed k = 0, 1, 2, · · · , where ν = n+ 2α− 12.

Proof. Returning to (3.2.14), we note that the coefficient β0 is zero. See (3.2.11).

Hence, it follows from (3.2.22) and (3.2.23) that

e−ν/2νn/2f (α)n (t/

√ν) =

[cos(πα− πν

t2)Ai(ν2/3ζ)

+ sin(πα− πν

t2)Bi(ν2/3ζ)

]α0

ν1/3+ ϵ1 + δ1,

(3.5.3)

where

|ϵ1| ≤M1

ν4/3|Ai(ν2/3ζ)|+ N1

ν5/3|Ai′(ν2/3ζ)| (3.5.4)

and

|δ1| ≤P1

ν4/3|Bi(ν2/3ζ)|+ Q1

ν5/3|Bi′(ν2/3ζ)|. (3.5.5)

Next, we recall the well-known asymptotic results

Ai(η) ∼ η−14

2√πexp(−2

3η

32 ), Ai′(η) ∼ − η

14

2√πexp(−2

3η

32 ), (3.5.6)

and

Bi(η) ∼ η−14

√πexp(

2

3η

32 ), Bi′(η) ∼ η

14

√πexp(

2

3η

32 ), (3.5.7)

as η → +∞. See [24, pp. 392–393]. Furthermore, since ν = n + 2α − 12

and ζ > 0

for t > 2, the terms involving Ai(ν2/3ζ) and Ai′(ν2/3ζ) are exponentially small, while

the terms involving Bi(ν2/3ζ) and Bi′(ν2/3ζ) are exponentially large. Thus, (3.5.3) can

be reduced to

e−ν/2νn/2f (α)n (t/

√ν) = sin(πα− πν

t2)Bi(ν2/3ζ)

α0

ν1/3+Oν−

32 exp(

2

3νζ3/2).

3.5 Zeros 47

If t is a zero of f (α)n (t/

√ν), the left-hand side of the equation is equal to zero, and we

have

sin(πα− πν

t2) = O(

1

ν) (3.5.8)

on account of (3.5.7). In view of (3.5.1), the solutions of (3.5.8) are given in (3.5.2).

For n = 100 and α = 1, the numerical values of the large zeros are given in Table

3.4.

Table 3.4: True values and approximate values of the large zeros evaluated with Maple,

when n = 100 and α = 1.

m-th true approx. error %

t100,91 3.1859 3.1859 5.3246×10−9

t100,92 3.3582 3.3582 1.0000×10−8

t100,93 3.5620 3.5620 -1.9025×10−9

t100,94 3.8079 3.8079 2.4413×10−9

t100,95 4.1130 4.1130 -8.8517×10−9

t100,96 4.5056 4.5056 -9.4228×10−12

t100,97 5.0374 5.0374 0.

t100,98 5.8166 5.8167 -1.7967×10−9

t100,99 7.1239 7.1239 -1.9025×10−9

t100,100 10.0747 10.0747 0.

Case (ii). To obtain the small zeros when 0 < t < 2, we recall the asymptotic

formulas

Ai(−η) = η−14

√π

[cos(

2

3η

32 − 1

4π) +O(η−

32 )

], (3.5.9)

Bi(−η) = η−14

√π

[− sin(

2

3η

32 − 1

4π) +O(η−

32 )

], (3.5.10)

3.5 Zeros 48

and the corresponding formulas of Ai′(−η) and Bi′(−η); see [24, pp. 392–393]. From

these formulas, it is evident that if t is a zero of the Tricomi-Carlitz polynomials

f(α)n (t/

√ν), then equation (3.5.3) can be reduced to

cos(πα− πν


t2)Bi(ν2/3ζ) = O(

1

ν). (3.5.11)

Let tn,m denote the m-th zero of f (α)n (t/

√ν), arranged in ascending order

0 ≤ tn,1 < tn,2 < · · · < 2. (3.5.12)

Since ζ(t) < 0 when 0 < t < 2, it is readily seen from (3.5.9) and (3.5.10) that the

left-hand side of (3.5.11) can be written as

cos(πα− πν


t2)Bi(ν2/3ζ)

=1

π1/2ν1/6|ζ(t)|1/4

[cos(πα− πν

t2) cos(

2

3ν|ζ(t)|

32 − 1

4π)

− sin(πα− πν

t2) sin(

2

3ν|ζ(t)|

32 − 1

4π) +O(

1

ν)

]=

1

π1/2ν1/6|ζ(t)|1/4

[cos(πα− πν

t2+

2

3ν|ζ(t)|

32 − 1

4π) +O(

1

ν)

].

Therefore, from (3.5.11) we have

cos(πα− πν

t2+

2

3ν|ζ(t)|

32 − 1

4π) = O(

1

ν). (3.5.13)

Note that near t = 0, |ζ(t)| 32 has the expansion

|ζ(t)|32 =

3π

2t2− 3π

4+

1

2t+

1

80t3 +

9

8960t5 +O

(t6). (3.5.14)

Upon solving (3.5.13) and (3.5.14), we have

− n

2+ (

ν

3πt+

ν

120πt3 +

3ν

4480πt5 + · · · ) = 2j ± 1

2+O(

1

ν). (3.5.15)

As a first approximation, we take just the first two terms on the left-hand side of the

equation and obtain

− n

2+νt

3π= 2j ± 1

2+O(

1

ν). (3.5.16)

The solutions of equation (3.5.16) are presented in two separate cases. In the order

arranged in (3.5.12), when n is even, the small zeros are given by

tn,2k+1 =3π

2ν

[4k + 1 +O(

1

ν)

](3.5.17)

3.5 Zeros 49

and

tn,2k+2 =3π

2ν

[4k + 3 +O(

1

ν)

](3.5.18)

for fixed k = 0, 1, 2 · · · . Note that tn,1 = 0. When n is odd, the small zeros are given

by tn,1 = 0,

tn,2k+2 =3π

2ν

[4k + 2 +O(

1

ν)

](3.5.19)

and

tn,2k+3 =3π

2ν

[4k + 4 +O(

1

ν)

](3.5.20)

for fixed k = 0, 1, 2 · · · .

Theorem 3.5. The asymptotic behavior of the small zeros of f (α)n (t/

√ν), arranged in

ascending order in (3.5.12), are given by the formulas in (3.5.17) – (3.5.20).

The above result can be improved3 if we pick three terms in (3.5.15), in which case

we have

− n

2+

ν

3πt+

ν

120πt3 = 2j ± 1

2+O(

1

ν). (3.5.21)

This is a cubic equation, and we are looking for only the real root. Recall that the real

root of a cubic equation of the form

t3 + p t+ q = 0 (3.5.22)

is given explicitly by

t =

(−q2+

√q2

4+p3

27

) 13

+

(−q2−√q2

4+p3

27

) 13

, (3.5.23)

provided thatq2

4+p3

27> 0. (3.5.24)

The solutions of (3.5.21) are again presented in two separate cases. Let Km = 4k +

m+O(1/ν), m = 1, 2, 3 and 4. When n is even, the small zeros are given by

tn,2k+1 =

[30K1

π

ν+ 30

√(K1π

ν)2 +

640

243

] 13

+

[30K1

π

ν− 30

√(K1π

ν)2 +

640

243

] 13

(3.5.25)3Power series solutions can be obtained; detailed analysis can be found in Appendix C.4.

3.5 Zeros 50

and

tn,2k+2 =

[30K3

π

ν+ 30

√(K3π

ν)2 +

640

243

] 13

+

[30K3

π

ν− 30

√(K3π

ν)2 +

640

243

] 13

(3.5.26)

for fixed k = 0, 1, 2, · · · . Note that, tn,1 = 0. When n is odd, the small zeros are given

by tn,1 = 0,

tn,2k+2 =

[30K2

π

ν+ 30

√(K2π

ν)2 +

640

243

] 13

+

[30K2

π

ν− 30

√(K2π

ν)2 +

640

243

] 13

(3.5.27)

and

tn,2k+3 =

[30K4

π

ν+ 30

√(K4π

ν)2 +

640

243

] 13

+

[30K4

π

ν− 30

√(K4π

ν)2 +

640

243

] 13

(3.5.28)

for fixed k = 0, 1, 2, · · · .

It can be shown that as ν →∞, the quantities on the right-hand sides of equations

(3.5.25) – (3.5.28) are asymptotically equal to the quantities on the right-hand sides of

(3.5.17) – (3.5.20), respectively. More precisely, we have[30Km

π

ν+ 30

√(Kmπ

ν)2 +

640

243

] 13

+

[30Km

π

ν− 30

√(Kmπ

ν)2 +

640

243

] 13

∼ 3π

2ν(4k +m+O(

1

ν)), m = 1, 2, 3 and 4.

(3.5.29)

Note that the O – term in (3.5.29) is still O(1/ν). To improve the order estimate from

O(1/ν) to O(1/ν2), one should use (3.2.19) with p = 2, instead of (3.2.14).

Numerical computation shows that the solutions obtained in (3.5.25) – (3.5.28) are

more accurate than those given in (3.5.17) – (3.5.20). Use of the solutions of equation

(3.5.13) produces even better results. See the numerics in Table 3.5.

Case (iii). Now, we consider the zeros near the turning point t = 2; that is, on both

sides of σ = 0, where σ = ν2/3ζ(t) is the variable inside the Airy functions in (3.5.3).

If t is a zero of f (α)n (t/

√ν) and σ is bounded, then equation (3.5.3) gives

cos(πα− πν

t2)Ai(σ) + sin(πα− πν

t2)Bi(σ) = O(

1

ν) (3.5.30)

3.5 Zeros 51

Table 3.5: True values and approximate values of small zeros evaluated with Maple,

when n = 100 and α = 1.

m-th true (3.5.17) – (3.5.20) (3.5.25) – (3.5.28) (3.5.13)

t100,1 0.0464 0.0464 0.0464 0.0464

t100,2 0.1392 0.1393 0.1393 0.1392

t100,3 0.2318 0.2321 0.2318 0.2318

t100,4 0.3241 0.3250 0.3241 0.3241

t100,5 0.4160 0.4178 0.4160 0.4160

t100,6 0.5074 0.5107 0.5074 0.5074

t100,7 0.5981 0.6036 0.5982 0.5980

t100,8 0.6880 0.6964 0.6883 0.6879

t100,9 0.7770 0.7893 0.7775 0.7769

t100,10 0.8649 0.8821 0.8659 0.8649

t100,11 0.9517 0.9750 0.9533 0.9517

t100,12 1.0372 1.0678 1.0397 1.0372

which can be written as

sin

[πα− πν

t2+ arctan

Ai(σ)Bi(σ)

]= O(

1

ν). (3.5.31)

Since

arctanAi(σ)Bi(σ)

=1

6π − 3

4

313 (Γ (2/3))2 σ

π+

9

16

316 (Γ (2/3))4 σ2

π2+O

(σ4)

(3.5.32)

for small σ, replacing arctan in (3.5.31) by its leading term given in (3.5.32) yields

sin(πα− πν

t2+π

6

)= O(

1

ν). (3.5.33)

3.5 Zeros 52

Recall that here we are only concerned with the zeros of the polynomials near t = 2

(i.e. σ close to 0). From (3.5.33), we obtain

α− ν

t2+

1

6= m+O(

1

ν), (3.5.34)

where m is an integer. Recall that σ = ν2/3ζ(t). Hence, by Lagrange’s inversion

formula, we have from (3.1.20)

t = 2 +∞∑j=1

Aj(σν− 2

3 )j, (3.5.35)

where A1 = 1, A2 = −2960

, A3 =7993150

, · · · . Substituting (3.5.35) in (3.5.34) gives

α− ν

4(1− σν−

23 + · · · ) + 1

6= m+O(

1

ν), (3.5.36)

which in turn gives

1

4σν

13 = m− (α+

1

6− ν

4) +O(

1

ν). (3.5.37)

Now, we choose m to be a large negative integer so that m− (α+ 16− ν

4) is a bounded

quantity. Let us write (3.5.37) in the form

σ =t1ν1/3

+ · · · (3.5.38)

with1

4t1 = m− (α+

1

6− ν

4). (3.5.39)

If we denote by να the fractional part of 14ν− 1

6−α (i.e., να = 1

4ν− 1

6−α−[1

4ν− 1

6−α]),

then (3.5.39) can be expressed as

t1 = 4(να + k), (3.5.40)

where k is an integer. Let t(l)n,k+1 and t(r)n,k+1, k = 0, 1, 2, · · · , denote, respectively, the

(k + 1)th zero on the left-hand and on the right-hand side of the turning point t = 2,

arranged in the order

· · · < t(l)n,2 < t

(l)n,1 < 2 < t

(r)n,1 < t

(r)n,2 < · · · . (3.5.41)

3.5 Zeros 53

We can now write down the asymptotic formulas as ν →∞

t(r)n,k+1 = 2 +

4

ν(k + να) (3.5.42)

and

t(l)n,k+1 = 2 +

4

ν(−k − 1 + να) (3.5.43)

for fixed k = 0, 1, 2, · · · , where να denotes the fractional part of 14ν − 1

6− α (i.e.,

να = 14ν − 1

6− α− [1

4ν − 1

6− α]).

Theorem 3.6. The asymptotic behavior of the zeros of f (α)n (t/

√ν) on two sides of the

turning point t = 2, as arranged in (3.5.41), are given in (3.5.42) and (3.5.43).

For n = 100 and α = 1, the numerical values of the first three zeros on the two

sides of t = 2 are given in Table 3.6. Approximation of the zeros of f (α)n (x) as α→∞

has been studied by Lopez and Temme [21].

Asymptotic zero distribution of the Tricomi-Carlitz polynomials has been investi-

gated by Goh and Wimp [13, 14] and Kuijlaars and Van Assche [18]. Goh and Wimp

investigated the zero distribution of f (α)n (z/

√n) as n→∞, by using a probability ap-

proach; while Kuijlaars and Van Assche used the three-term recurrence relation (1.1.7)

to establish the zero distribution of f (α)n (t/

√N) with n/N → c as n → ∞. The zero

distribution obtained by Kuijlaars and Van Assche [18] is

µc(t) =

4

πc|t|3

[arcsin(

|t|√c

2)− |t|

4

√c(4− t2c)

], |t| < 2√

c,

2

c|t|3, |t| ≥ 2√

c,

(3.5.44)

which agrees with the result

µ(t) =

4

π|t|3

[arcsin(

|t|2)− |t|

4

√4− t2

], |t| < 2,

2

|t|3, |t| ≥ 2,

(3.5.45)

3.5 Zeros 54

of Goh and Wimp [13, 14] if we take c = 1 (i.e., n/N → 1 as n→∞). The graph of

the distribution function (3.5.45) is shown in Figure 3.12. Note that in our caseN = ν.

The area under the curve of µ(t) in (3.5.45) in the interval [a, b] represents the ratio

the number of zeros in [a, b]

total number of zeros. (3.5.46)

As an illustration, we take n = 100 and α = 1. We find that the polynomial f (α)n (t/

√ν)

has 25 zeros in the interval 0 < t < 2, and that the value of the integral of µ(t) on the

interval [0, 2] is 0.25.

Table 3.6: True values and approximate values of zeros near the turning point t = 2

evaluated with Maple, when n = 100 and α = 1.

true approx. error %

t(l)100,3 1.8506 1.8900 -2.1289

t(l)100,2 1.9079 1.9294 -1.1283

t(l)100,1 1.9610 1.9688 -0.3970

2.0 2.0 0.0

t(r)100,1 2.0098 2.0082 0.0812

t(r)100,2 2.0554 2.0476 0.3797

t(r)100,3 2.1006 2.0870 0.6463

3.5 Zeros 55

Figure 3.12: The zero distribution of fαn (t/√ν).

Chapter 4


Modified Lommel polynomials

4.1 Introduction

The Lommel polynomialsRn,ν(x) arise in the theory of Bessel functions. The modified

Lommel polynomials hn,ν(x) are defined by

hn,ν(x) = Rn,ν(1/x). (4.1.1)

The modified Lommel polynomials satisfy the three-term recurrence relation

hn+1,ν(x)− 2(n+ ν)xhn,ν(x) + hn−1,ν(x) = 0, n ≥ 0, (4.1.2)

with initial values h−1,ν(x) = 0, h0,ν(x) = 1. The polynomials have the hypergeomet-

ric representation (for n > 1)

hn,ν(x) = (ν)n(2x)n2F3(−n/2, (−n+ 1)/2; ν,−n, 1− ν − n;−1/x2), (4.1.3)

where

(ν)n =Γ(ν + n)

Γ(ν). (4.1.4)

(4.1.3) implied the reflection formula of Modified Lommel Polynomials; that is

hn,ν(−x) = (−1)n hn,ν(x). (4.1.5)

4.1 Introduction 57

Their generating function

π

2x(1− 2xw)−

ν2

[J1−ν

(1

x

)Y−ν

(√1− 2xw

x

)− J−ν

(√1− 2xw

x

)Y1−ν

(1

x

)]=

∞∑n=0

hn,ν(x)wn

n!

(4.1.6)

can be derived from the recurrence relation (4.1.2). Furthermore, the correct orthogo-

nality relation is

∞∑k=1

hm,ν(±xk)hn,ν(±xk)x2k =1

2(n+ ν)δmn, (4.1.7)

where xk = j−1ν−1,k and jν−1,k are the zeros of the Bessel function Jν−1(x), for k =

1, 2, · · · ; see [15, p. 197, (6.5.17)]. (Here, we wish to point out that the orthogonality

relations given in Dickinson [11] and Chihara [8] are incorrect). From some formulas

involving Lommel Polynomials (cf., Watson [35, §9.6]), we may define the modified

Lommel polynomials hn,ν(x) in terms of Bessel functions by

Jν+n(1/x) = Jν(1/x)hn,ν(x)− Jν−1(1/x)hn−1,ν+1(x), (4.1.8)

where n is an integer and ν is not a nonpositive integer, or by

π−1(2x sinπν)hn,ν(x) = Jν+n(1/x)J−ν+1(1/x) + (−1)nJ−ν−n(1/x)Jν−1(1/x),

(4.1.9)

for n an integers and ν not an even integer. The behavior of hn,ν(x) has been inves-

tigated by Hurwitz (see Watson [35, §9.65]) , and he used an elementary approach to

show that

limn→∞

(2x)1−ν−nhn,ν(x)

Γ(n+ ν)= Jν−1(1/x) (4.1.10)

uniformly for all x in any closed and bounded annular region centered at the origin.

The purpose of this chapter is to present an asymptotic expansion for hn,ν(t/N),

which holds uniformly for t in [0,∞), where N = n + ν. Our approach is to use a

turning-point theory recently introduced by Wang and Wong [33, 34] for three-term re-

currence relations. Asymptotic formulas are also obtained for the zeros of the modified

Lommel polynomials.


4.2 Difference Equation

Equation (4.1.2) is already in the canonical form considered in [34]

hn+1,ν(x)− (Anx+Bn)hn,ν(x) + hn−1,ν(x) = 0 (4.2.1)

with

An = 2(n+ ν) and Bn = 0. (4.2.2)

In terms of the notations

An ∼ n−θ

∞∑s=0

αs

nsand Bn ∼

∞∑s=0

βsns

(4.2.3)

used in [34], we have

θ = −1; α0 = 2, α1 = 2ν, αn = 0 for n = 2, 3, · · · , (4.2.4)

and β0 = β1 = β2 = · · · = 0. If these expansions are recast in the form

An ∼ N−θ

∞∑s=0

α′s

N sand Bn ∼

∞∑s=0

β′s

N s, (4.2.5)

where N = n+τ0 and τ0 is some fixed real number to be determined, it is easily found

that

α′0 = 2, α′

1 = 2ν − 2τ0, · · · , (4.2.6)

and β′0 = β′

1 = β′2 = · · · = 0. To apply the result in [34], we first choose τ0 so that

α′1 = 0. From (4.2.6), it is obvious that the choice is

τ0 = ν. (4.2.7)

According to the equation in [34, (2.4)], the characteristic equation is

λ2 − 2tλ+ 1 = 0, (4.2.8)

where t is the rescaled variable t = Nx. The two roots of this equation are

λ± = t±√t2 − 1. (4.2.9)


The points t± = ±1, where the two roots coincide are called the turning points of

equation (4.2.1).

We now define the function ζ(t) introduced in [34, (4.10)]. With t+ = 1, θ = −1,

α′0 = 2 and β′

0 = 0, this function is given by

2

3[ζ(t)]3/2 := log (t+

√t2 − 1)− 1

t

∫ t

1

s√s2 − 1

ds, t ≥ 1

and2

3[−ζ(t)]3/2 := t−1

∫ 1

t

s√1− s2

ds− cos−1 t, 0 < t < 1.

By direct calculation, one obtains

2

3[ζ(t)]3/2 = log (t+

√t2 − 1)− 1

t(t2 − 1)

12 (4.2.10)

for t ≥ 1 and2

3[−ζ(t)]3/2 := 1

t(1− t2)

12 − cos−1 t (4.2.11)

for 0 < t < 1. Note that as t→ 1, we have ζ(t)→ 0. In fact, near t = 1, we have the

Taylor expansion

ζ(t) = 213 (t− 1)

[1− 7

10(t− 1) +

102

175(t− 1)2 + · · ·

]. (4.2.12)

We also define the functions H0(ζ) and Φ(ζ) introduced in [34, (4.19) and (4.28)].

In the present situation, these functions are given by

H0(ζ) := −

√t2 − 1

ζand Φ(ζ) = 0, (4.2.13)

where ζ is the function defined in (4.2.10) and (4.2.11). Note that in our special case,

α′1 = β′

1 = 0; hence, according to the definition of Φ(ζ) given in [34, (4.28)], the

second equation in (4.2.13) holds for 0 < t <∞, instead of t ≥ δ, 0 < δ < 1.

With this preliminary work done, we can now apply the main result in [34] to

conclude that there are constants C1(x) and C2(x) such that the polynomials hn,ν(x)

in (4.2.1) can be expressed as

hn,ν(x) = C1(x)Pn(x) + C2(x)Qn(x), (4.2.14)


where, with x = N− 12 t, Pn(x) and Qn(x) have the asymptotic expansions

Pn(x) =

(ζ

t2 − 1

) 14

[Ai(N

23 ζ) p−1∑

s=0

As(ζ)

N s− 16

+ Ai′(N

23 ζ) p−1∑

s=0

Bs(ζ)

N s+ 16

+ ϵp(N, t)

](4.2.15)

and

Qn(x) =

(ζ

t2 − 1

) 14

[Bi(N

23 ζ) p−1∑

s=0

As(ζ)

N s− 16

+ Bi′(N

23 ζ) p−1∑

s=0

Bs(ζ)

N s+ 16

+ δp(N, t)

].

(4.2.16)

In (4.2.15) and (4.2.16), Ai(·) and Bi(·) are the Airy functions, the leading coefficients

are given by

A0(ζ) = 1 and B0(ζ) = 0, (4.2.17)

and there exists a positive constant Mp such that

|ϵp(N, t)| ≤Mp

Np− 16

|Ai(N23 ζ)| and |δp(N, t)| ≤

Mp

Np− 16

|Bi(N23 ζ)|, (4.2.18)

where Ai and Bi are the modulus functions given in [34, (7.10) and (7.11)]. The

expansions hold uniformly for 0 ≤ t <∞.

4.3 Determination of C1(x) and C2(x)

First, we examine the behavior of hn,ν(x) as n → ∞. Here, we do not use the result

of (4.1.10) since it will provide only the value of C2(x) and not that of C1(x). We

first note that when N is large, we have from the well-known asymptotic formulas of

Jν(1/x) and Yν(1/x)

Jν+n(1

x) = JN(

1

x) ∼ 1√

2 π N(

e

2N x)N , (4.3.1)

Yν+n(1

x) = YN(

1

x) ∼ −

√2

π N(

e

2N x)−N ; (4.3.2)

see [24, p. 436]. For the negative order of the Bessel function, we know that from [24,

p. 243]

J−ν−n(1

x) = J−N(

1

x) = cos(Nπ)JN(1/x)− sin(Nπ)YN(1/x) (4.3.3)


and

Yν−1(z) =1

sin(νπ)(Jν−1(z) cos(νπ) + J−ν+1(z)) . (4.3.4)

Coupling (4.1.9), (4.3.1), (4.3.2), (4.3.3) and (4.3.4), we have

hn,ν(x) ∼eN(2xN)−N

√2 π N

π

2xYν−1(1/x) + (2x)N−1

√2πNN− 1

2 e−NJν−1(1/x).

(4.3.5)

Next, we recall the well-known asymptotic formulas

Ai(η) ∼ η−14

2√πexp(−2

3η

32 ), Ai′(η) ∼ − η

14

2√πexp(−2

3η

32 ), (4.3.6)

and

Bi(η) ∼ η−14

√πexp(

2

3η

32 ), Bi′(η) ∼ η

14

√πexp(

2

3η

32 ) (4.3.7)

as η → +∞. See [24, pp. 392-393]. Furthermore, from (4.2.10) we have

2

3[ζ(t)]3/2 ∼ log(2t)− 1 = log(2xN)− 1. (4.3.8)

Since N = n+ ν, it is readily seen from (4.3.6), (4.3.7) and (4.3.8) that

Pn(x) ∼1

2√π xN

1

(2xN)NeN ,

Qn(x) ∼2N√π(xN)N− 1

2 e−N .

Comparing the two sides of (4.2.14) gives

C1(x) =π√2x

Yν−1(1/x), C2(x) =π√2x

Jν−1(1/x). (4.3.9)

In summary, we have from (4.2.14), (4.2.15) and (4.2.16)

hn,ν(t

N) =

π√2x

Yν−1(1/x)

(ζ

t2 − 1

) 14

N16

[Ai(N

23 ζ)+O(

1

N)

]+

π√2x

Jν−1(1/x)

(ζ

t2 − 1

) 14

N16

[Bi(N

23 ζ)+O(

1

N)

],

(4.3.10)

where x = N−1t and N = n+ ν. This result holds uniformly for 0 ≤ t <∞.

4.4 Zeros 62

4.4 Zeros

It is well known that the zeros of the modified Lommel polynomials hn,ν(x) all lie in

the interval −j−1ν−1,1 ≤ x ≤ j−1

ν−1,1. This can be seen from (4.1.7). After the rescaling

x = N−1t, the zeros of hn,ν(t/N) all lie in the interval −N/j−1ν−1,1 ≤ t ≤ N/j−1

ν−1,1.

In this section, we shall use the asymptotic approximation given in (4.3.10) to derive

asymptotic formulas for the zeros of hn,ν(t/N) as n → ∞. We divide our discussion

into three cases: (i) large zeros in 1 < t < N/j−1ν−1,1; (ii) small zeros in 0 < t < 1; and

(iii) zeros near the turning point t = 1 and on two sides of the point. Several tables are

provided to give a comparison of the numerical values of the asymptotic zeros and the

true zeros.

Case (i). Let tn,m denote the m-th zero of hn,ν(t/N), arranged in ascending order

− N

jν−1,1

≤ tn,1 < tn,2 < · · · < tn,n−1 < tn,n ≤N

jν−1,1

. (4.4.1)

For large zeros, we have the following result.

Theorem 4.1. The asymptotic behavior of the large zeros tn,m is given by

tn,n−k =N

jν−1,k+1

[1 +O(

1

N)

], n→∞, (4.4.2)

for fixed k = 0, 1, 2, · · · , where N = n+ ν.

Proof. Returning to (4.3.10), we have

hn,ν(t

N) =

π√2x

(ζ

t2 − 1

) 14[N

16

(Yν−1(

1

x)Ai(N

23 ζ) + Jν−1(

1

x)Bi(N

23 ζ)

)+O(

1

N)

].

(4.4.3)

Since N = n+ ν and ζ > 0 for t > 1, the terms involving Ai(N2/3ζ) and Ai′(N2/3ζ)

are exponentially small, while the terms involving Bi(N2/3ζ) and Bi′(N2/3ζ) are ex-

ponentially large; see (4.3.6) and (4.3.7). Thus, (4.4.3) reduces to

hn,ν(t

N) =

π√2x

(ζ

t2 − 1

) 14[N

16Jν−1(

1

x)Bi(N

23 ζ) +ON−1 exp(

2

3Nζ

32 )].

If t is a zero of hn,ν(t/N), the left-hand side of the equation is equal to zero, and we

have

Jν−1(1

x) = O(

1

N) (4.4.4)

4.4 Zeros 63

since Bi(N2/3ζ) has no zero for ζ > 0. In view of (4.4.1), the solutions of (4.4.4) are

given in (4.4.2).

For n = 100 and ν = 1, the numerical values of the large zeros are given in Table

4.1. Note that in this case, the polynomial hn,ν(t/N) has 100 zeros; half of them lie on

the positive real axis, and the other half lie on the negative real axis.

Case (ii). To obtain the small zeros when 0 < t < 1, we recall the asymptotic

formulas

Ai(−η) = η−14

√π

[cos(

2

3η

32 − 1

4π) +O(η−

32 )

], (4.4.5)

Bi(−η) = η−14

√π

[− sin(

2

3η

32 − 1

4π) +O(η−

32 )

], (4.4.6)

Jν(η) =

√2

πη

[cos(η − νπ

2− π

4) +O(η−1)

], (4.4.7)

Yν(η) =

√2

πη

[sin(η − νπ

2− π

4) +O(η−1)

], (4.4.8)

and the corresponding formulas of Ai′(−η) and Bi′(−η). See [24, pp. 392-393]. From

these formulas, it is evident that if t is a zero of the Modified Lommel polynomials

polynomial hn,ν(t/N), then equation (4.4.3) can be reduced to

π√2x

N16

[Yν−1(

1

x)Ai(N

23 ζ) + Jν−1(

1

x)Bi(N

23 ζ)

]= O(

1

N). (4.4.9)

Let tn,m denote the m-th zero of hn,ν(t/N), arranged in ascending order

0 ≤ tn,1 < tn,2 < · · · < 1. (4.4.10)

Since ζ(t) < 0 for 0 < t < 1, it is readily seen from (4.4.5) and (4.4.6) that the

left-hand side of (4.4.9) can be written as

π√2x

N16Yν−1(

1

x)Ai(N

23 ζ) +

π√2x

N16Jν−1(

1

x)Bi(N

23 ζ)

=1

|ζ(t)|1/4

[sin(

N

t− νπ

2+π

4) cos(

2

3N |ζ(t)|

32 − π

4)

− cos(N

t− νπ

2+π

4) sin(

2

3N |ζ(t)|

32 − π

4) +O(

1

N)

]=

1

|ζ(t)|1/4

[sin(

N

t− νπ

2+π

4− 2

3N |ζ(t)|

32 +

π

4) +O(

1

N)

].

4.4 Zeros 64

Table 4.1: True values and approximate values of large zeros when n = 100 and ν = 1

evaluated with Maple.

m-th true (4.4.2) error %

t100,1,80 1.5493 1.5493 0.0

t100,1,81 1.6278 1.6278 0.0

t100,1,82 1.7146 1.7146 0.0

t100,1,83 1.8112 1.8112 0.0

t100,1,84 1.9193 1.9193 0.0

t100,1,85 2.0411 2.0411 0.0

t100,1,86 2.1795 2.1795 0.0

t100,1,87 2.3380 2.3380 0.0

t100,1,88 2.5213 2.5213 0.0

t100,1,89 2.7359 2.7359 0.0

t100,1,90 2.9903 2.9903 0.0

t100,1,91 3.2969 3.2969 0.0

t100,1,92 3.6736 3.6736 0.0

t100,1,93 4.1474 4.1474 0.0

t100,1,94 4.7615 4.7615 0.0

t100,1,95 5.5890 5.5890 0.0

t100,1,96 6.7645 6.7645 0.0

t100,1,97 8.5655 8.5655 0.0

t100,1,98 11.6713 11.6713 0.0

t100,1,99 18.2968 18.2968 0.0

t100,1,100 41.9989 41.9989 0.0

4.4 Zeros 65

Therefore, from (4.4.9) we have

sin(N

t− νπ

2+π

4− 2

3N |ζ(t)|

32 +

π

4) = O(

1

N). (4.4.11)


2

3|ζ(t)|

32 =

1

t− π

2+

1

2t+

1

24t3 +

1

80t5 +O

(t6). (4.4.12)

Upon solving (4.4.11) and (4.4.12), we have

n+ 1

2π − 1

2Nt− 1

24Nt3 − 1

80Nt5 + · · · = kπ +O(

1

N). (4.4.13)

As a first approximation, we take just the first two terms on the left-hand side of the

equation, and obtainn+ 1

2π − 1

2Nt = kπ +O(

1

N). (4.4.14)

The solutions of equation (4.4.14) are presented in two separate cases. In the order

arranged in (4.4.10), when n is even, the small zeros are given by

tn,m+1 =2

N

[π

2+mπ +O(

1

N)

](4.4.15)

for fixed m = 0, 1, 2, · · · . Note that tn,1 = 0. When n is odd, the small zeros are given

by tn,1 = 0 and

tn,m+1 =2

N

[mπ +O(

1

N)

](4.4.16)

for fixed m = 0, 1, 2, · · · .

Theorem 4.2. The asymptotic behavior of the small zeros of hn,ν(t/N), arranged in

ascending order in (4.4.10), are given by the formulas in (4.4.15) − (4.4.16).

The above result can be improved1 if we pick three terms in (4.4.13), in which case

we haven+ 1

2π − 1

2Nt− 1

24Nt3 = kπ +O(

1

N). (4.4.17)

This is a cubic equation, and we are looking for only the real root. Recall that the real

root of the cubic equation

t3 + p t+ q = 0 (4.4.18)1Power series solutions can be obtained; detailed analysis can be found in Appendix D.1.

4.4 Zeros 66

is given explicitly by

t =

(−q2+

√q2

4+p3

27

) 13

+

(−q2−√q2

4+p3

27

) 13

, (4.4.19)

provided thatq2

4+p3

27> 0. (4.4.20)

The solutions of (4.4.17) are again presented in two separate cases. Let K1 = π2+

mπ +O( 1N) and K2 = mπ +O( 1

N). When n is even, the small zeros are given by

tn,m+1 =

[12

NK1 + 12

√(K1

N)2 +

4

9

] 13

+

[12

NK1 − 12

√(K1

N)2 +

4

9

] 13

(4.4.21)

for fixed m = 0, 1, 2, · · · . Note that tn,1 = 0. When n is odd, the small zeros are given

by tn,1 = 0 and

tn,m+1 =

[12

NK2 + 12

√(K2

N)2 +

4

9

] 13

+

[12

NK2 − 12

√(K2

N)2 +

4

9

] 13

(4.4.22)

for fixed m = 0, 1, 2 · · · .

It can be shown that as n→∞, the quantities on the right-hand sides of equations

(4.4.21) − (4.4.22) are asymptotically equal to the quantities on the right-hand sides

of (4.4.15) − (4.4.16), respectively. More precisely, we have[12

NK1 + 12

√(K1

N)2 +

4

9

] 13

+

[12

NK1 − 12

√(K1

N)2 +

4

9

] 13

∼ 2

N

[π

2+mπ +O(

1

N)

] (4.4.23)

and [12

NK2 + 12

√(K2

N)2 +

4

9

] 13

+

[12

NK2 − 12

√(K2

N)2 +

4

9

] 13

∼ 2

N

[mπ +O(

1

N)

],

(4.4.24)

respectively. Note that the O – terms in (4.4.23) and (4.4.24) are still O(1/N). To

improve the order estimate from O(1/N) to O(1/N2), one should use (4.2.14) and

(4.3.9) with p = 2 in (4.2.15) and (4.2.16), instead of (4.3.10).

4.4 Zeros 67

Numerical computation shows that the solutions given in (4.4.21) and (4.4.22) are

more accurate than those presented in (4.4.15) and (4.4.16); see the numerics in Table

4.2. Use of numerical solutions of equation (4.4.11) would produce even better results.

Case (iii). Now, we consider the zeros near the turning point t = 1; that is, on both

sides of σ = 0, where σ = N2/3ζ(t) is the variable inside the Airy functions in (4.4.3).

If t is a zero of hn,ν(t/N) and σ is bounded, then equation (4.4.3) gives

π√2x

Yν−1(1

x)Ai(σ) +

π√2x

Jν−1(1

x)Bi(σ) = O(

1

N), (4.4.25)

which can be written as

sin

[N

t− νπ

2+π

4+ arctan

Bi(σ)Ai(σ)

]= O(

1

N). (4.4.26)

Since

arctanBi(σ)Ai(σ)

=1

3π +

3

4

313

π

(Γ(

2

3)

)3

σ − 9

16

316

π2

(Γ(

2

3)

)4

σ2 +O(σ4)

(4.4.27)

for small σ, replacing arctan in (4.4.26) by its leading term given in (4.4.27) yields

sin

(N

t− νπ

2+π

4+π

3

)= O(

1

N). (4.4.28)

Recall that here we are only concerned with the zeros of the polynomials near t = 1

(i.e., σ close to 0). From (4.4.28), we obtain

N

tπ− ν

2+

1

4+

1

3= k +O(

1

N), (4.4.29)

where k is an integer. Recall that σ = N2/3ζ(t). Hence, by Lagrange’s inversion

formula, we have from (4.2.12)

t = 1 +∞∑j=1

Aj(2− 1

3σN− 23 )j, (4.4.30)

where A1 = 1, A2 =710

, A3 =139350

, · · · . Substituting (4.4.30) in (4.4.29) gives

N

π(1− 2−

13σN− 2

3 + · · · ) = k − 7

12+ν

2+O(

1

N), (4.4.31)

4.4 Zeros 68

Table 4.2: True values and approximate values of zeros evaluated with Maple, when

n = 100 and ν = 1.

m-th true (4.4.15) error % (4.4.21) error %

t100,1,1 0.0311 0.0311 0.0081 0.0311 0.0

t100,1,2 0.0932 0.0933 0.0726 0.0932 0.0001

t100,1,3 0.1552 0.1555 0.2021 0.1552 0.0013

t100,1,4 0.2169 0.2177 0.3973 0.2169 0.0053

t100,1,5 0.2781 0.2799 0.6596 0.2782 0.0148

t100,1,6 0.3388 0.3422 0.9906 0.3389 0.0331

t100,1,7 0.3988 0.4044 1.3925 0.3991 0.0645

t100,1,8 0.458 0.4666 1.8683 0.4585 0.1142

t100,1,9 0.5163 0.5288 2.4215 0.5173 0.1878

t100,1,10 0.5735 0.591 3.0565 0.5751 0.292

t100,1,11 0.6294 0.6532 3.7789 0.6322 0.4344

t100,1,12 0.684 0.7154 4.5956 0.6882 0.6236

t100,1,13 0.737 0.7776 5.5151 0.7434 0.8698

t100,1,14 0.7882 0.8398 6.5487 0.7976 1.1851

t100,1,15 0.8375 0.902 7.7109 0.8507 1.5842

t100,1,16 0.8845 0.9643 9.0216 0.9029 2.0861

t100,1,17 0.9289 1.0265 10.5069 0.9541 2.7152

t100,1,18 0.9703 1.0887 12.1976 1.0043 3.4989

t100,1,19 1.0086 1.1509 14.1023 1.0535 4.4433

4.4 Zeros 69

which in turn gives

2−13σ

πN

13 =

N

π− k + 7

12− ν

2+O(

1

N). (4.4.32)

Now, we choose k to be a large integer so that Nπ− k + 7

12− ν

2is a bounded quantity.

Let us write (4.4.32) in the form

σ =t1N1/3

+ · · · (4.4.33)

with1

213πt1 =

N

π− k + 7

12− ν

2. (4.4.34)

If we denote byNν the fractional part of Nπ+ 7

12− ν

2(i.e.,Nν = N

π+ 7

12− ν

2−[N

π+ 7

12− ν

2]),

then (4.4.34) can be expressed as

t1 = 213π(m+Nν), (4.4.35)

where m is an integer. Let t(l)n,m+1 and t(r)n,m+1, m = 0, 1, 2, · · · , denote respectively the

(m+ 1)th zero on the left-hand and on the right-hand sides of the turning point t = 1,

arranged in the order

· · · < t(l)n,2 < t

(l)n,1 < 1 < t

(r)n,1 < t

(r)n,2 < · · · . (4.4.36)

We can now write down the asymptotic formulas as n→∞

t(r)n,m+1 = 1 +

213π

N(m+Nν) (4.4.37)

and

t(l)n,m+1 = 1 +

213π

N(−m− 1 +Nν) (4.4.38)

for fixed m = 0, 1, 2 · · · , where Nν denotes the fractional part of Nπ+ 7

12− ν

2.

Theorem 4.3. The asymptotic behavior of the zeros of hn,ν(t/N) on two sides of the

turning point t = 1, as arranged in (4.4.36), are given in (4.4.37) and (4.4.38).

For n = 100 and ν = 1, the numerical values of the first two zeros on the two

sides of t = 1 are given in Table 4.3.

4.4 Zeros 70

Table 4.3: True values and approximate values of zeros near the turning point t = 1

evaluated with Maple, when n = 100 and ν = 1.

m-th true approx. error %

t(l)100,1,1 0.9703 0.9761 0.5993

1.0 1.0

t(r)100,1,1 1.0086 1.0072 -0.1391

Asymptotic zero distribution of the modified Lommel polynomials has been inves-

tigated by Kuijlaars and Van Assche [18]. They used the three-term recurrence relation

(4.1.2) to establish the zero distribution of hn,ν(t/N) with n/N → c as n → ∞. The

zero distribution obtained by Kuijlaars and Van Assche [18] is

µc(t) =

1

πct2

(1−√1− t2c2

), |t| < 1

c,

1

πct2, |t| ≥ 1

c.

(4.4.39)

(There is a minor error in the statement of their result; the functions α(t) and β(t)

should be α(t) = −1/t and β(t) = 1/t in [18, p. 195]; therefore the zero distribu-

tion of the modified Lommel polynomials hn,ν(t/N) should be the function given in

(4.4.39), instead of the one in [18, (4.18)]). Note that in our case N = n+ ν, the ratio

n/N → 1 when n→∞, and the corresponding asymptotic zero distribution is

µ(t) =

1

πt2

(1−√1− t2

), |t| < 1,

1

πt2, |t| ≥ 1.

(4.4.40)

The graph of the distribution function (4.4.39) is shown in Figure 4.1. The area under

the curve of µ(t) in (4.4.40) in an interval [a, b] represents the ratio

the number of zeros in [a, b]

total number of zeros. (4.4.41)

4.4 Zeros 71

Figure 4.1: The zero distribution of hn,ν(t/N).

As an illustration, we take n = 100 and ν = 1. We find that the polynomial hn,ν(t/N)

has 18 zeros in the interval 0 < t < 1 and that the value of the integral of µ(t) on the

interval [0, 1] is 0.1817.

Appendix A

Verification of Some Results in §1

Verification of (1.1.2).

(n+ 1)t(α)n+1(x) = (n+ 1)

n+1∑k=0

(−1)k(x− αk

)xn+1−k

(n+ 1− k)!

=n+1∑k=0

(−1)k(x− αk

)xn+1−k

(n− k)!

(n+ 1− k + k

n+ 1− k

)

=n+1∑k=0

(−1)k(x− αk

)xn+1−k

(n− k)!

(1 +

k

n+ 1− k

)

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!+

n+1∑k=1

(−1)k(x− αk

)k xn+1−k

(n+ 1− k)!

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!

+n+1∑k=1

(−1)k (x− α)(x− α− 1)...(x− α− k + 1)

k!

k xn+1−k

(n+ 1− k)!

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!

+n+1∑k=1

(−1)k (x− α)(x− α− 1)...(x− α− k + 1)

(k − 1)!

xn+1−k

(n+ 1− k)!

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!+

n+1∑k=1

(−1)k(x− αk − 1

)(x− α− k + 1)

xn+1−k

(n+ 1− k)!

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!+

n∑k=0

(−1)k+1

(x− αk

)(x− α− k) xn−k

(n− k)!

Appendix A Verification of Some Results in §1 73

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!+

n∑k=0

(−1)k+1

(x− αk

)(x− α− n+ n− k) xn−k

(n− k)!

=n∑

k=0

(−1)k(x− αk

)xn+1−k

(n− k)!+

n∑k=0

(−1)k(x− αk

)(n+ α)

xn−k

(n− k)!

−n∑

k=0

(−1)k(x− αk

)(x+ n− k) xn−k

(n− k)!

=n∑

k=0

(−1)k(x− αk

)(x− x− n+ k)

xn−k

(n− k)!+

n∑k=0

(−1)k(x− αk

)(n+ α)

xn−k

(n− k)!

=−n∑

k=1

(−1)k(x− αk

)xn−k

(n− k − 1)!+

n∑k=0

(−1)k(x− αk

)(n+ α)

xn−k

(n− k)!

=− x t(α)n−1(x) + (n+ α)t(α)n (x), n ≥ 1.


Recall that

(−x)k = (−1)k(x

k

)k! and (−x)k = (−1)k(x− k + 1)k.

Then, by the definition of the Tricomi polynomial, we have

t(α)n (x) =n∑

k=0

(−1)k(x− αk

)xn−k

(n− k)!

=n∑

k=0

(−1)k (−1)k

k!(−x+ α)k

xn−k

(n− k)!

=xn

n!

n∑k=0

n!

(n− k)!(−1)k(−1)k(−x+ α)k

x−k

k!

=xn

n!

n∑k=0

(−n)k(−x+ α)k(−1)kx−k

k!

=xn

n!2F0(−n,−x+ α;

−1x

).



Let T (x,w) =∑∞

n=0 t(α)n (x)wn ⇒ T (x, 0) = t

(α)0 (x) = 1. Using (1.1.2) and differen-

tiating both side of T with respect to w, we have

∂T

∂w=

∞∑n=1

nt(α)n (x)wn−1 = t(α)1 (x) +

∞∑n=2

nt(α)n (x)wn−1

=∞∑n=1

(n+ 1)t(α)n+1(x)w

n + t(α)1 (x)

=∞∑n=1

[(n+ α) t(α)n (x)− x t

(α)n−1(x)(x)

]wn + t

(α)1 (x)

= t(α)1 (x) +

∞∑n=1

nt(α)n (x)wn−1w + α

∞∑n=1

t(α)n (x)wn − x∞∑n=1

t(α)n−1(x)w

n

= t(α)1 (x) + w

∂T

∂w+ αT − αt(α)0 (x)− xwT.

=⇒ ∂T

∂w=

(α− xw)T1− w

= Tx− xw − x+ α

1− w

= T

(x+−x+ α

1− w

)

=⇒ ln |T | = xw − (−x+ α) ln(1− w) + C, |w| < 1.

Putting w = 0, we have C = 0. Hence

exw−(α−x) ln(1−w) =∞∑n=0

t(α)n (x)wn, |w| < 1.



Let F (x,w) =∑∞

n=0 f(α)n (x)wn ⇒ F (x, 0) = f

(α)0 (x) = 1. Using (1.1.7) and differ-

entiating both side of F with respect to w, we have

∂F

∂w=

∞∑n=1

nf (α)n (x)wn−1 = f

(α)1 (x) +

∞∑n=2

nf (α)n (x)wn−1

=∞∑n=1

(n+ 1)f(α)n+1(x)w

n + f(α)1 (x)

=∞∑n=1

[(n+ α) x f (α)

n (x)− f (α)n−1(x)

]wn + f

(α)1 (x)

= f(α)1 (x) + x

∞∑n=1

nf (α)n (x)wn + αx

∞∑n=1

f (α)n (x)wn −

∞∑n=1

f(α)n−1(x)w

n

= f(α)1 (x) + x

∞∑n=1

nf (α)n (x)wn + αx

∞∑n=0

f (α)n (x)wn − αxf (α)

0 (x)

−∞∑n=0

f (α)n (x)wn+1

= αxF + xw∂F

∂w− wF.

=⇒ ∂F

∂w=

(αx− w)F1− xw

=

(1

x

)(αx2 − xw1− xw

)F

= F1

x

1− xw − 1 + αx2

1− xw

= F1

x

(1− 1− αx2

1− xw

)

=⇒ ln |F | = w

x−∫

1

x

1− αx2

1− xwdw

=w

x+

1− αx2

x2ln(1− xw) + C, |wx| < 1.

Putting w = 0, we have C = 0. Hence

ew/x+(1−αx2)/x2 ln(1−xw) =∞∑n=0

f (α)n (x)wn |wx| < 1.

A.1 Another Generalized Hypergeometric Representations of t(α)n (x) 76


By using (1.1.11), L’Hopital rule and considering the following limit,

limx→0

[w/x+ (1− αx2) ln(1− xw)/x2

]= lim

x→0

wx+ (1− αx2) ln(1− xw)x2

= limx→0

w − 2αx ln(1− xw) + (1− αx2) −w1−xw

2x

= limx→0

1

2

[−2α ln(1− xw) + 2αxw

1− xw+ 2αx

w

1− xw+ (1− αx2) −w2

(1− xw)2)

]=

1

2

[0 + 0 + 0 + (1− 0)

−w2

1

]=−w2

2.

Hence, we have

limx→0

∞∑n=0

f (α)n (x)wn = lim

x→0ew/x+(1−αx2)/x2 ln(1−xw) = e

−w2

2 .

=⇒ e−w2

2 =∞∑n=0

f (α)n (0)wn =

∞∑n=0

f(α)2n (0)w2n +

∞∑n=0

f(α)2n+1(0)w

2n+1.

But, e−w2

2 =∞∑n=0

(−w2

2

)n1

n!=

∞∑n=0

(−1)n

2nn!w2n.

Therefore, we obtain

f(α)2n (0) = (−1)n2−n/n!, f

(α)2n+1(0) = 0, n = 0, 1, 2, · · · .

A.1 Another Generalized Hypergeometric Representa-

tions of t(α)n (x)

Tricomi polynomials also be expressed in terms of the Kummer confluent hypergeo-

metric series

t(α)n (x) = (−1)n(x− αn

)ex 1F1(x− α+ 1; x− α− n+ 1;−x). (A.1.1)

Then, the recurrence relation (1.1.2) follows from the recurrence relation of the 1F1(a; c−

n;x). The generating function in (1.1.5) follows from the known relation∞∑n=0

L(α−n)n (x)zn = e−zx(1 + z)α. (A.1.2)

(It is suggested by Prof. N. M. Temme.)

A.2 Connection Formula of rn(x, a) and f (α)n (x) 77

A.2 Connection Formula of rn(x, a) and f (α)n (x)

The random walk polynomials rn(x, a) introduced by Karlin and McGregor [17, p.

117] satisfying the three-term recurrence relation

arn+1(x, a)− (n+ a)xrn(x, a) + nrn−1(x, a) = 0

with r0(x, a) = 1 and r1(x, a) = xr0(x, a) = x. The connection formula of Random

Walk Polynomials rn(x, a) and Tricomi-Carlitz polynomials f (α)n (x) is

f (α)n (x) =

αn2

n!rn(√αx, α).

Proof. To find this connection formula, we can start from the generating function of

rn(x, a) and f (α)n (x), i.e.,

ezx

(1− xz

a

)a 1−x2

x2

=∞∑n=0

rn(x, a)zn

n!

and

exp

w

x+

1− αx2

x2log(1− wx)

=

∞∑n=0

f (α)n (x)wn.

Setting x =√αX , z =

√αW and a = α in the generating function of rn(x, a), then

we have

ezx

(1− xz

a

)a 1−x2

x2

=∞∑n=0

rn(x, a)zn

n!

⇒ e√

αW√αX

(1−√αX√αW

α

)α 1−αX2

αX2

=∞∑n=0

rn(√αX,α)

αn2W n

n!

⇒ eWX (1−WX)

1−αX2

X2 =∞∑n=0

rn(√αX,α)

αn2W n

n!.

Thus, we obtain the connection formula

f (α)n (x) =

αn2

n!rn(√αx, α).

A.3 The Fixed x Asymptotic Behavior of Tricomi-Carlitz Polynomial f (α)n (x) 78

A.3 The Fixed xAsymptotic Behavior of Tricomi-Carlitz

Polynomial f (α)n (x)

From the result of uniform asymptotic expansions of the Tricomi-Carlitz polynomials

(2.3.9), we know that

f (α)n (

t√ν) =


t2)

212nΓ(1

2n+ 1)

(4ζ

t2 − 4

) 14

ν16

[Ai(ν

23 ζ)+O(

1

ν)

]

+


t2)

212nΓ(1

2n+ 1)

(4ζ

t2 − 4

) 14

ν16

[Bi(ν

23 ζ)+O(

1

ν)

] (A.3.1)

where x = ν−12 t and ν = n+ 2α− 1/2. This result holds uniformly for 0 ≤ t <∞.

From the above result, we recall the well-known asymptotic results, the Stirling

formula and Airy functions,

n! = Γ(n+ 1) =√2πnn+ 1

2 e−n(1 +O(1

n)),

Ai(η) ∼ η−14

2√πexp(−2

3η

32 ), Ai′(η) ∼ − η

14

2√πexp(−2

3η

32 ), (A.3.2)

and

Bi(η) ∼ η−14

√πexp(

2

3η

32 ), Bi′(η) ∼ η

14

√πexp(

2

3η

32 ). (A.3.3)

Since x = ν−12 t, then

πα− πν

t2= πα− π

x2.

Thus,

2n2Γ(

n

2+ 1) = 2

n2

√2π(

n

2)n2+ 1

2 e−n2 (1 +O(

1

n))

=√πn

n2+ 1

2 e−n2 (1 +O(

1

n)).

Note that,

t > 2, t = x√ν > 2⇔ x >

2√ν,

and

0 < t < 2, t = x√ν < 2⇔ 0 < x <

2√ν.


First, we consider

logt+√t2 − 4

2

= log1

2(t+ t(1− 4

t2)12 )

= log1

2(t+ t− 2

t− 2

t3− 4

t5)

= log(t− 1

t− 1

t3− 2

t5+ · · · )

= log t(1− 1

t2− 1

t4− 2

t6+ · · · )

= log t+ log(1− 1

t2− 1

t4− 2

t6+ · · · )

= log t− 1

t2− 1

t4− 2

t6+ · · · ,

− 1

t21

2t√t2 − 4

= − 1

t21

2t2(1− 4

t2)12

= −1

2(1− 2

t2− 2

t4− 4

t6+ · · · .)

= −1

2+

1

t2+

1

t4+

2

t6+ · · · ,

and

− 1

t2

(2 log(t+

√t2 − 4)− 2 log(2)

)= − 2

t2log

t+√t2 − 4

2

= − 2

t2

[log t− 1

t2− 1

t4− 2

t6+ · · ·

]= − 2

t2log t+

2

t4+ · · · .

Combining the above results, we have

2

3[ζ(t)]3/2 = ln

t+√t2 − 4

2− 1

t2

[1

2t√t2 − 4 + 2 ln |t+

√t2 − 4| − 2 ln 2

]= log t− 1

t2− 1

t4− 2

t6+ · · ·+−1

2+

1

t2+

1

t4+

2

t6+ · · ·

− 2

t2log t+

2

t4+ · · ·

= log t− 1

2− 2

t2log t+O(

1

t4).

Since x = ν−12 t, we have

2

3[ζ(xν

12 )]3/2 = log(xν

12 )− 1

2− 2

x2νlog(xν

12 ) +O(

1

ν2),


Pn(ν− 1

2 t) ∼(

4ζ

t2 − 4

) 14

Ai(ν23 ζ)ν

16

∼

(4ζ

(xν12 )2 − 4

) 14

ν16 e−

23(ν

23 ζ)

32 1

2√π(ν

23 ζ)−

14

=1√2π

(xν12 )−

12 e−

23νζ

32

∼ 1√2π

(xν12 )−

12 e

−ν[log(xν

12 )− 1

2− 2

x2νlog(xν

12 )+O( 1

ν2)]

=1√2π

(xν12 )−

12 (xν

12 )−νe

ν2 (xν

12 )

2x2

=1√2π

(xν12 )

2x2

− 12−νe

ν2

=1√2πx

2x2

− 12−n−2α+ 1

2

[n(1 +

2α− 12

n)

]( 2x2

− 12−n−2α+ 1

2)/2

en2+α− 1

4

∼ 1√2πx

2x2

−2αx−nn1x2

−n2−αe−α+ 1

4 en2+α− 1

4

=1√2πx

2x2

−2αx−nn1x2

−n2−αe

n2

and

Qn(ν− 1

2 t) ∼(

4ζ

t2 − 4

) 14

Bi(ν23 ζ)ν

16

∼

(4ζ

(xν12 )2 − 4

) 14

ν16 e

23(ν

23 ζ)

32 1√

π(ν

23 ζ)−

14

=

√2

π(xν

12 )−

12 e

23νζ

32

∼√

2

π(xν

12 )−

12 e

ν[log(xν

12 )− 1

2− 2

x2νlog(xν

12 )+O( 1

ν2)]

=

√2

π(xν

12 )−

12 (xν

12 )νe−

ν2 (xν

12 )−

2x2

=

√2

π(xν

12 )ν−

12− 2

x2 e−ν2

=

√2

πx−

2x2

−1+n+2α

[n(1 +

2α− 12

n)

](− 2x2

−1+n+2α)/2

e−n2−α+ 1

4

∼√

2

πx−

2x2

+2α−1xnn− 1x2

+n2+α− 1

2 eα−14 e−

n2−α+ 1

4

=

√2

πx2α−1− 2

x2 xnnn2+α− 1

2− 1

x2 e−n2 .


From the above result, when n→∞, fixed x, we have

f (α)n (x) =

1√2π

cos(πα− π

x2)x−n−2α+ 2

x2 n−n− 12−α+ 1

x2 en[1 +O(n−1)

]+

√2

πsin(πα− π

x2)xn+2α−1− 2

x2 nα−1− 1x2[1 +O(n−1)

].

Using another method introduced by R. Wong and H. Li [39, 40], one can find the fixed

x asymptotic behavior for Tricomi-Carlitz polynomial f (α)n (x). Two linear independent

solutions y1(n) and y2(n) are

y1(n) = xnnα−1− 1x2

∞∑s=0

csns

and

y2(n) =1

(n− 2)!x−nn

1x2

−α−2∞∑s=0

csns

such that

f (α)n (x) = C1(x)y1(n) + C2(x)y2(n),

where C1(x) and C2(x) are functions independent of n, and c1 = 1. Note that, by the

Stirling formula,1

(n− 2)!=n(n− 1)

n!∼ n2

e−nnn+ 12

√2π.

we also have,

y2(n) =1√2πx−nn−n− 1

2−α− 1

x2 en∞∑s=0

csns.

However, in general, it is very difficult to find two functions C1(x) and C2(x).

Appendix B



We recall the Darboux’s Method [37, p. 116]. To obtain the asymptotic behavior for

the coefficients an of the Maclaurin expansion,

f(z) =∞∑n=0

anzn.

We assume that

f(z) = (1− z)αg(z)

and

g(z) =∞∑r=0

cr(1− z)r.

Then, the result is

(−1)nan∞∑r=0

cr

(α+ r

n

), n−α−r−1, as n→∞.

Now, we use the generating function for Tricomi-Carlitz polynomials

exp

w

x+

1− αx2

x2log(1− wx)

=

∞∑n=0

f (α)n (x)wn, |wx| < 1.

Set x→ y√α

, z → yw√α⇐⇒ w = z

√α

y, then the above equation becomes

ezαy2 (1− z)α(

1−y2

y2)=

∞∑n=0

f (α)n (

y√α)znα

n2

yn, |z| < 1.

Appendix B Verification of Some Results in §2 83

Set g(z) = ezαy2 =

∑∞r=0 cr(1 − z)r. Then the first two coefficients are c0 = e

αy2 and

c1 = − αy2e

αy2 . Applying the Darboux’s method, for fixed y, we have

(−1)n f (α)n ( y√

α)α

n2

yn∼

∞∑r=0

cr

(α(1−y2

y2) + r

n

), n−α( 1−y2

y2)−r−1, as n→∞.

=⇒ f (α)n (

y√α)α

n2 y−n ∼

∞∑r=0

crΓ(n− r − α(1−y2

y2))

n!Γ(α(1−y2

y2)− r)

=⇒ f (α)n (

y√α)α

n2 y−n ∼ e

αy2

n−α( 1−y2

y2)−1

Γ(−α(1−y2

y2))

[1 +O(

1

n)

].


Kn+1

Kn−1

=2

n−12 Γ

(n−12

+ 1)

2n+12 Γ

(n+12

+ 1) =

1

2

Γ(n+12)

n+12Γ(n+1

2)=

1

n+ 1.

Verification of (2.2.15).2

3[ζ(t)]3/2

= logα′0t+ β′

0 +√

(α′0t+ β′

0)2 − 4

2− α′

0t1/θ

∫ t

t+

s−1/θ√(α′

0s+ β′0)

2 − 4ds

= logt+√t2 − 4

2− 1

t2

∫ t

2

s2√s2 − 4

ds

= logt+√t2 − 4

2− 1

t2

∫ t

2

√s2 − 4 +

4√s2 − 4

ds

= logt+√t2 − 4

2

− 1

t2

[1

2

(s√s2 − 4− 4 ln |s+

√s2 − 4|

)∣∣∣t2+ 4 ln |s+

√s2 − 4|

∣∣∣t2

]= log

t+√t2 − 4

2

− 1

t2

[1

2t√t2 − 4− 2 ln |t+

√t2 − 4|+ 2 ln 2 + 4 ln |t+

√t2 − 4| − 4 ln 2

]= log

t+√t2 − 4

2− 1

t2

[1

2t√t2 − 4 + 2 ln |t+

√t2 − 4| − 2 ln 2

].

Appendix B Verification of Some Results in §2 84


2

3[−ζ(t)]3/2

= − cos−1 α′0t+ β′

0

2+ α′

0t1/θ

∫ t+

t

s−1/θ√4− (α′

0s+ β′0)

2ds

= − cos−1 t

2+ t−2

∫ 2

t

s2√4− s2

ds

= − cos−1 t

2− 1

t2

∫ 2

t

√4− s2 − 4√

4− s2ds

= − cos−1 t

2− 1

t2

[1

2

(s√4− s2 + 4 sin−1(

s

2))∣∣∣2

t− 4 sin−1(

s

2)∣∣∣2t

]= − cos−1 t

2− 1

t2

[1

2

(0 + 4× π

2− t√4− t2 − 4 sin−1 t

2

)− 4

(π

2− sin−1 t

2

)]= − cos−1 t

2− 1

t2

[π − t

2

√4− t2 − 2 sin−1 t

2− 2π + 4 sin−1 t

2

]= − cos−1 t

2− 1

t2

[2 sin−1 t

2− π − t

2

√4− t2

]= − cos−1 t

2+

1

t2

[−2 sin−1 t

2+ π +

t

2

√4− t2

].

Appendix C



From (1.1.10), it is easily seen that∞∑n=0

f (α)n (−x) (−w)n = e(−w)/(−x)+[1−α(−x)2] log[1−(−x)(−w)]/(−x)2

= ewx+ 1−αx2

x2log(1−wx)

=∞∑n=0

f (α)n (x)wn,

from which it follows that

f (α)n (−x) = (−1)nf (α)

n (x).


To find the saddle points of ϕ(s; t), i.e., zeros of ∂ϕ/∂s, we know that

∂ϕ

∂s=

1

t+

1

t

1

ts− 1− 1

s=

(s2 − ts+ 1)

s(ts− 1)

Then the zeros of ∂ϕ/∂s are at

s = s± =1

2(t±√t2 − 4).

Appendix C Verification of Some Results in §3 86


Case 1: t > 2.

Using Maple, the steepest paths given by (3.1.12) and (3.1.13) are shown in Figure

C.1, where arrows are used to indicate the direction of descent. Near s = s+, letting

s− s+ = reiθ,

ϕ(s; t) = ϕ(s+; t) +1

2ϕ′′(s+; t)r

2ei2θ +O(r3).

For the steepest paths,

sin 2θ +O(r) = 0 (C.0.1)

and

ℜϕ(s; t) = ℜϕ(s+; t) +1

2ϕ′′(s+; t)r

2 cos 2θ +O(r3). (C.0.2)

(C.0.2) gives the directions of the steepest paths at s+ while (C.0.2) gives the values of

ϕ(s; t) along the steepest paths. Now,

ϕ′′(s±; t) = ±1

2

√t2 − 4[t(t2 − 3)∓ (t2 − 1)

√t2 − 4]. (C.0.3)

Therefore,

ϕ′′(s+; t) > 0 and ϕ′′(s−; t) < 0.

As r → 0, (C.0.2) implies that sin 2θ = 0. Therefore, θ = 0,±π/2, π. From (C.0.2),

we see that near s = s+, the value of ϕ(s; t) decreases in the direction θ = ±π/2 and

increases in the direction θ = 0, π. Hence, the branches D5 and D6 of (3.1.12) are the

steepest descent path while D7 and D8 are the steepest ascent path.

Similarly, near s = s−, letting s− s− = reiθ,

ϕ(s; t) = ϕ(s−; t) +1

2ϕ′′(s−; t)r

2ei2θ +O(r3)

= ϕ(s−; t) +1

2|ϕ′′(s−; t)|r2ei(π+2θ) +O(r3)

since ϕ′′(s−) < 0. For the steepest paths,

sin(π + 2θ) +O(r) = 0 (C.0.4)

and

ℜϕ(s; t) = ℜϕ(s−; t) +1

2ϕ′′(s−; t)r

2 cos(π + 2θ) +O(r3). (C.0.5)


(C.0.4) gives the directions of the steepest paths at s− while (C.0.5) gives the values

of ϕ(s; t) along the steepest paths. As r → 0, (C.0.4) implies that sin(π + 2θ) = 0.

Therefore, θ = 0,±π/2, π. From (C.0.5), we see that near s = s−, the value of ϕ(s; t)

decreases in the direction θ = 0, π and increases in the direction θ = ±π/2. Hence,

the branches D3 and D4 of (3.1.12) are the steepest descent path while D1 and D2 are

the steepest ascent path.

D1

D2

D3

D4

D5

D7

D8

D6

Figure C.1: Steepest path when t > 2.

Case 2: 0 < t < 2. For u ∈ (−∞,∞), we take the branch of tan−1 u with values in

(−π2, π2). If z = x+ iy, it is easily seen that

arg(x+ iy) =

tan−1(y/x) if x > 0 and y > 0 or < 0,

π + tan−1(y/x) if x < 0 and y > 0,

−π + tan−1(y/x) if x < 0 and y < 0.

(C.0.6)

By again using the fact that s± are zeros of ∂ϕ/∂s, i.e., s2± − t s± + 1 = 0, equation

(3.1.9) can be written as

v

t+

1

t2arg(1− tu− itv)− arg(u+ iv) = ℑϕ(s±; t) = ±θt, (C.0.7)


where θt = 12t

√4− t2 − π

t2+ ( 2

t2− 1) tan−1

√4−t2

t. The graph of the curves given by

(C.0.7) are shown in Figure C.2 – Figure C.4, with arrows again indicating directions

of ascent and descent. Near s = s±, letting s− s± = reiθ,

ϕ(s; t) = ϕ(s±; t) +1

2ϕ′′(s±; t)r

2ei2θ +O(r3).

Now,

ϕ′′(s±; t) =1

2

√4− t2[(t2 − 1)

√4− t2 ± it(t2 − 3)] =

√4− t2e±i δ(t);

therefore,

ϕ(s; t) = ϕ(s±; t) +1

2

√4− t2r2ei[±δ(t)+2θ] +O(r3).

For the steepest paths,

sin[2θ + δ(t)] +O(r) = 0 (C.0.8)

and

ℜϕ(s±; t) +r2

2

√4− t2 cos[2θ + δ(t)] = ℜϕ(s; t). (C.0.9)

Now, we have to consider three cases and define

δ(t) = tan−1(t(t2 − 3)

(t2 − 1)√4− t2

) (C.0.10)

for simplicity.

Subcase 1, 0 < t < 1.

As r → 0, (C.0.8) implies that

sin[2θ ± (δ(t)− π)] = 0.

The solutions in the interval (−π, π] are given by

θ = ∓δ(t)2,π

2∓ δ(t)

2, π ∓ δ(t)

2,3π

2∓ δ(t)

2.

From (C.0.9), we see that near s = s+, the value of ϕ(s; t) decreases in the direction

θ = − δ(t)2

and π − δ(t)2

and increases in the direction π2− δ(t)

2, 3π

2− δ(t)

2. Hence, the

branches D1 and D2 of (3.1.15) are the steepest descent paths while D3 and D4 are the


steepest ascent paths. The case is similar when near s = s− and we conclude that the

branches D7 and D8 of (3.1.15) are the steepest ascent paths while D5 and D6 are the

steepest descent paths. Note that δ(t) is positive.

D1

D2

D3

D4

D5

D7

D8

D6

0

Figure C.2: Steepest path when 0 < t < 1.

Subcase 2, 1 < t <√3.


sin[2θ ± δ(t)] = 0.


θ = −π2∓ δ(t)

2,∓δ(t)

2,π

2∓ δ(t)

2, π ∓ δ(t)

2.


−π2− δ(t)

2and π

2− δ(t)

2and increases in the direction− δ(t)

2, π− δ(t)

2. Hence, the branches

D1 and D2 of (3.1.15) are the steepest descent paths while D3 and D4 are the steepest

ascent paths. The case is similar when near s = s− and we conclude that the branches

D7 and D8 of (3.1.15) are the steepest ascent paths while D5 and D6 are the steepest

descent paths. Note that δ(t) is negative.


D1

D2

D3

D4

D5

D7

D8

D6

0

Figure C.3: Steepest path when 1 < t <√3.

Subcase 3,√3 < t < 2.


sin[2θ ± δ(t)] = 0.


θ = −π2∓ δ(t)

2,∓δ(t)

2,π

2∓ δ(t)

2, π ∓ δ(t)

2.


−π2− δ(t)

2and π

2− δ(t)

2and increases in the direction− δ(t)

2, π− δ(t)

2. Hence, the branches

D1 and D2 of (3.1.15) are the steepest descent paths while D3 and D4 are the steepest

ascent paths. The case is similar when near s = s− and we conclude that the branches

D7 and D8 of (3.1.15) are the steepest ascent paths while D5 and D6 are the steepest

descent paths. Note that δ(t) is positive.


D1

D5

D2

D3

D4

D7

D8

D6

0

Figure C.4: Steepest path when√3 < t < 2.

Verification of (C.0.3).

First, we calculate ϕ′′(s+; t) when t > 2. Since

ϕ′′(s+) = −1

(1− ts+)2+

1

s2+=

1

s2+− 1

s4+

=1

s4+(s2+ − 1)

=1

s4+(s+ − 1)(s+ + 1)

=1

s4+

[1

2(t+√t2 − 4)− 1

] [1

2(t+√t2 − 4) + 1

]=

1

s4+

1

4

[t− 2 +

√t2 − 4

] [t+ 2 +

√t2 − 4

]=

1

4s4+

√t− 2

√t+ 2

[√t− 2 +

√t+ 2

] [√t+ 2 +

√t− 2

]=

1

4s4+

√t2 − 4

(t− 2 + 2

√t2 − 4 + t+ 2

)=

2

4s4+

√t2 − 4

(t+√t2 − 4

)=s+√t2 − 4

s4+=

√t2 − 4

s3+;


however,

s3+ =

(1

2(t+√t2 − 4)

)3

=1

8

(t3 + 3t2

√t2 − 4 + 3t(t2 − 4) +

√t2 − 4(t2 − 4)

)=

1

8

(4t3 − 12t+ 4(t2 − 1)

√t2 − 4

)=

1

2

((t3 − 3t) + (t2 − 1)

√t2 − 4

).

Thus, we have

ϕ′′(s+) =

√t2 − 4

s3+=

2√t2 − 4

(t3 − 3t) + (t2 − 1)√t2 − 4

=2√t2 − 4

((t3 − 3t)− (t2 − 1)

√t2 − 4

)(t3 − 3t)2 − (t2 − 1)2(t2 − 4)

=2√t2 − 4

((t3 − 3t)− (t2 − 1)

√t2 − 4

)t6 − 6t4 + 9t2 − t6 + 6t4 − 9t2 + 4

=1

2

√t2 − 4

(t(t2 − 3)− (t2 − 1)

√t2 − 4

).

Then, we calculate ϕ′′(s−; t) when t > 2. Since

ϕ′′(s−) = −1

(1− ts−)2+

1

s2−=

1

s2−− 1

s4−

=1

s4−(s2− − 1)

=1

s4−(s− − 1)(s− + 1)

=1

s4−

[1

2(t−√t2 − 4)− 1

] [1

2(t−√t2 − 4) + 1

]=

1

s4+

1

4

[t− 2−

√t2 − 4

] [t+ 2−

√t2 − 4

]=

1

4s4+

√t− 2

√t+ 2

[√t− 2−

√t+ 2

] [√t+ 2−

√t− 2

]=−14s4−

√t2 − 4

(t− 2− 2

√t2 − 4 + t+ 2

)=−24s4−

√t2 − 4

(t−√t2 − 4

)= −s−

√t2 − 4

s4−= −√t2 − 4

s3−;


however,

s3− =

(1

2(t−√t2 − 4)

)3

=1

8

(t3 − 3t2

√t2 − 4 + 3t(t2 − 4)−

√t2 − 4(t2 − 4)

)=

1

8

(4t3 − 12t− 4(t2 − 1)

√t2 − 4

)=

1

2

((t3 − 3t)− (t2 − 1)

√t2 − 4

).

Thus, we have

ϕ′′(s−) = −√t2 − 4

s3−=

−2√t2 − 4

(t3 − 3t)− (t2 − 1)√t2 − 4

=−2√t2 − 4

((t3 − 3t) + (t2 − 1)

√t2 − 4

)(t3 − 3t)2 − (t2 − 1)2(t2 − 4)

=−2√t2 − 4

((t3 − 3t) + (t2 − 1)

√t2 − 4

)t6 − 6t4 + 9t2 − t6 + 6t4 − 9t2 + 4

= −1

2

√t2 − 4

(t(t2 − 3) + (t2 − 1)

√t2 − 4

).

To conclude, we have

ϕ′′(s+; t) =

√t2 − 4

s3+> 0

and

ϕ′′(s−; t) = −√t2 − 4

s3−< 0.

To summarize, for t > 2, we have

ϕ′′(s±; t) = ±1

2

√t2 − 4[t(t2 − 3)∓ (t2 − 1)

√t2 − 4].


When 0 < t < 2, since

ϕ′′(s+) = −1

(1− ts+)2+

1

s2+=

1

s2+− 1

s4+

=1

s4+(s2+ − 1)

=1

s4+(s+ − 1)(s+ + 1)

=1

s4+

1

2(t− 2 + i

√4− t2)1

2(t+ 2 + i

√4− t2)

=1

s4+

−14

√2− t(

√2− t− i

√2 + t)

√2 + t(

√2 + t+ i

√2− t)

=1

s4+

−14

√4− t2(

√4− t2 − i (2 + t) + i (2− t) +

√4− t2)

=1

s4+

−14

√4− t2(−2i)(t+ i

√4− t2)

=i s+√4− t2s4+

=i√4− t2s3+

;

however,

s3+ =1

8(t+ i

√4− t2)2(t+ i

√4− t2)

=1

8(t2 − (4− t2) + 2it

√4− t2)(t+ i

√4− t2)

=1

82(t2 − 2 + it

√4− t2)(t+ i

√4− t2)

=1

4

(t(t2 − 2)− t(4− t2) + it2

√4− t2 + i(t2 − 2)

√4− t2

)=

1

4

(t3 − 2t− 4t+ t3 + 2i

√4− t2(t2 − 1)

)=

1

2

((t3 − 3t) + i(t2 − 1)

√4− t2

).

Thus, we have

ϕ′′(s+) =i√4− t2s3+

=i2√4− t2

(t3 − 3t) + i(t2 − 1)√4− t2

=i2√4− t2

((t3 − 3t)− i(t2 − 1)

√4− t2

)(t3 − 3t)2 + (t2 − 1)2(4− t2)

=i2√4− t2

((t3 − 3t)− i(t2 − 1)

√4− t2

)t6 − 6t4 + 9t2 − t6 + 6t4 − 9t2 + 4

=1

2

√4− t2

((t2 − 1)

√4− t2 + it(t2 − 3)

).


The case for ϕ′′(s−; t) when 0 < t < 2 is similar, the result is

ϕ′′(s−) = −1

2

√4− t2

(−(t2 − 1)

√4− t2 + it(t2 − 3)

).

Verification of (3.1.18) and (3.1.19).

From (3.1.17), we immediately obtain2η(t) =

1

t(s++ + s+−) +

1

t2[log(1− s++t) + log(1− s+−t)]− log(s++s

+−)

−4

3ζ3/2(t) =

1

t(s++ − s+−) +

1

t2[log(1− s++t)− log(1− s+−t)]− log(

s++s+−

)

.

Since s+± is on the upper edge of the cut, we have

log(1− s+±t) = −iπ + log(s+±t− 1) = −iπ + 2 log(s+±).

Combining the above results, we have

η(t) =1

2t(s++ + s+−) +

1

2t2[log(1− s++t) + log(1− s+−t)]−

1

2log(s++s

+−)

=1

2t(t) +

1

2t2[−iπ + log(s++t− 1)− iπ + log(s+−t− 1)]− 1

2log(s++s

+−)

=1

2− iπ

t2,

and

4ζ3/2(t)

3= −1

t(s++ − s+−)−

1

t2(log(1− s++t)− log(1− s+−t)

)+ log(

s++s+−

)

= −1

t(s++ − s+−)−

1

t2(−iπ + log(s++t− 1) + iπ − log(s+−t− 1)

)+ log(

s++s+−

)

= −1

t(s++ − s+−)−

1

t2log s4+ + log s2+

= −1

t

√t2 − 4 + (2− 4

t2) log(

1

2(t+√t2 − 4)).

Similarly, when t ≥ 2, s = s−± are mapped to u = ±√ζ. Note that, since s−± is on the

lower edge of the cut, we have

log(1− s−±t) = +iπ + log(s−±t− 1) = +iπ + 2 log(s−±).


Thus, we have

η(t) =1

2∓ iπ

t2,

and4ζ3/2(t)

3= −1

t

√t2 − 4 + (2− 4

t2) log(

1

2(t+√t2 − 4)).

The case for 0 < t < 2 is very similar. Note that,

4ζ3/2(t)

3= −1

t(s+ − s−)−

1

t2log

s+t− 1

s−t− 1+ log(

s+s−

)

= −2i[1

2t

√4− t2 + 2

t2tan−1

(√4− t2t

)− tan−1

(√4− t2t

)]= −2i

[1

2t

√4− t2 + 2

t2cos−1 t

2− cos−1 t

2

]= −2i

[1

2t

√4− t2 + π

t2− 2

t2sin−1 t

2− cos−1 t

2

].


u3 − 3ζ(t)u+ 3[η(t)− ϕ(s; t)] = 0,

then we obtain u1(s, t) = 2

√ζ(t) sin 1

3φ,

u2(s, t) = 2√ζ(t) sin 1

3(φ+ 2π),

u3(s, t) = 2√ζ(t) sin 1

3(φ+ 4π),

(C.0.11)

where φ is a solution of

sinφ = −3[η(t)− ϕ(s; t)]−ζ(t)(2

√ζ(t))

=3

2ζ32 (t)

[η(t)− ϕ(s; t)]. (C.0.12)

Since s+ ←→ u+ and s− ←→ u−, from

u1(s+; t) =√ζ(t), u1(s−; t) = −

√ζ(t),

u2(s+; t) =√ζ(t), u2(s−; t) = 2

√ζ(t),

u3(s+; t) = −2√ζ(t), u3(s−; t) = −

√ζ(t),

we see that u1(s; t) is the desired solution. We call it u(s; t) from now on.


Verification of (3.2.14). For the upper half plane, we see that the curve in s-plane

is transformed to the curve γ1 in u-plane.

ei(πα−πν/t2)

2πiνn/2

∫upper

eν ϕ(s;t)


= ei(πα−πν/t2)ν−n/2eν/2

2πi

∫γ1

eν(u3/3−ζu)[α0 + β0u+ (u2 − ζ)g0(u)] du

= ei(πα−πν/t2)ν−n/2eν/2

2πi

[α0

∫γ1

eν(u3/3−ζu) du+ β0

∫γ1

eν(u3/3−ζu)u du

+

∫γ1

eν(u3/3−ζu)(u2 − ζ)g0(u) du

].

The last integral can be evaluated as follows:∫γ1

eν(u3/3−ζu)(u2 − ζ)g0(u) du =

1

ν

∫γ1

g0(u) deν(u3/3−ζu)

=1

ν

[g0(u) e

ν(u3/3−ζu)∣∣∣γ1−∫γ1

eν(u3/3−ζu)g′0(u) du

]= −1

ν

∫γ1


If we put z = ν1/3u, then∫γ1

eν(u3/3−ζu) du =

∫γ1

ez3/3−ν2/3ζzν−1/3 dz =

2πi

ν1/3(−ωAi(ων2/3ζ)

)and ∫

γ1

eν(u3/3−ζu)u du =

∫γ1

ez3/3−ν2/3ζzν−2/3z dz =

2πi

ν2/3(ω2Ai′(ων2/3ζ)

).

Similarly, for the lower half-plane, we see that the curve in s-plane is transformed to

the curve γ2 in u-plane.

e−i(πα−πν/t2)

2πiνn/2

∫lower

eν ϕ(s;t)


= e−i(πα−πν/t2)ν−n/2eν/2

2πi

∫γ2

eν(u3/3−ζu)[α0 + β0u+ (u2 − ζ)g0(u)] du

= e−i(πα−πν/t2)ν−n/2eν/2

2πi

[α0

∫γ2

eν(u3/3−ζu) du+ β0

∫γ2

eν(u3/3−ζu)u du

+

∫γ2

eν(u3/3−ζu)(u2 − ζ)g0(u) du

].


The last integral can be evaluated as follows:∫γ2

eν(u3/3−ζu)(u2 − ζ)g0(u) du =

1

ν

∫γ2

g0(u) deν(u3/3−ζu)

=1

ν

[g0(u) e

ν(u3/3−ζu)∣∣∣γ2−∫γ2

eν(u3/3−ζu)g′0(u) du

]= −1

ν

∫γ2


If we put z = ν13u, then∫

γ2

eν(u3/3−ζu) du =

∫γ2

ez3/3−ν2/3ζzν−1/3 dz =

2πi

ν1/3(−ω2Ai(ω2ν2/3ζ)

)and ∫

γ2

eν(u3/3−ζu)u du =

∫γ2

ez3/3−ν2/3ζzν−2/3z dz =

2πi

ν2/3(ωAi′(ω2ν2/3ζ)

).


To conclude,

f (α)n (

t√ν)

=ei(πα−

πνt2

)

2πiνn2

∫upper

eν ϕ(s;t)

s−2α+1/2s(ts− 1)αds− e−i(πα−πν

t2)

2πiνn2

∫lower

eν ϕ(s;t)


=ei(πα−

πνt2

)

2πiνn2

2πi

ν13

α0

(−ωAi(ων

23 ζ))+

2πi

ν23

β0

(ω2Ai′(ων

23 ζ))

−1

ν

∫γ1

eν(u3

3−ζu)g′0(u) du

+e−i(πα−πν

t2)

2πiνn2

2πi

ν13

α0

(−ω2Ai(ω2ν

23 ζ))+

2πi

ν23

β0

(ωAi′(ω2ν

23 ζ))

−1

ν

∫γ2

eν(u3

3−ζu)g′0(u) du

= e

ν2 ν−

n2

ν−13 cos(πα− πν

t2)α0

(−ωAi(ων

23 ζ))+ iν−

13 sin(πα− πν

t2)α0

(−ωAi(ων

23 ζ))

+ν−23 cos(πα− πν

t2)β0

(ω2Ai′(ων

23 ζ))+ iν−

23 sin(πα− πν

t2)β0

(ω2Ai′(ων

23 ζ))

+ eν2 ν−

n2

ν−13 cos(πα− πν

t2)α0

(−ω2Ai(ω2ν

23 ζ))− iν−

13 sin(πα− πν

t2)α0

(−ω2Ai(ω2ν

23 ζ))

+ν−23 cos(πα− πν

t2)β0

(ωAi′(ω2ν

23 ζ))− iν−

23 sin(πα− πν

t2)β0

(ωAi′(ω2ν

23 ζ))

− ei(πα−πνt2

)

2πiνn2

1

ν

∫γ1

eν(u3

3−ζu)g′0(u) du−

e−i(πα−πνt2

)

2πiνn2

1

ν

∫γ2

eν(u3

3−ζu)g′0(u) du

= eν2 ν−

n2

ν−

13α0 cos(πα−

πν

t2)[−ωAi(ων

23 ζ)− ω2Ai(ω2ν

23 ζ)]

+ ν−13α0 sin(πα−

πν

t2)[iω2Ai(ω2ν

23 ζ)− iωAi(ων

23 ζ)]

+ ν−23β0 cos(πα−

πν

t2)[ω2Ai′(ων

23 ζ) + ωAi′(ω2ν

23 ζ)]

+ ν−23β0 sin(πα−

πν

t2)[iω2Ai′(ων

23 ζ)− iωAi′(ω2ν

23 ζ)]


)

2πiνn2

1

ν

∫γ1

eν(u3


e−i(πα−πνt2

)

2πiνn2

1

ν

∫γ2

eν(u3

3−ζu)g′0(u) du

= eν2 ν−

n2

ν−

13α0 cos(πα−

πν

t2)Ai(ν

23 ζ) + ν−

13α0 sin(πα−

πν

t2)Bi(ν

23 ζ)

−ν−23β0 cos(πα−

πν

t2)Ai′(ν

23 ζ)− ν−

23β0 sin(πα−

πν

t2)Bi′(ν

23 ζ)


)

2πiνn2

1

ν

∫γ1

eν(u3


e−i(πα−πνt2

)

2πiνn2

1

ν

∫γ2

eν(u3

3−ζu)g′0(u) du.



To prove (3.2.19), we first introduce the following lemma.

Lemma C.1. For m = 0, 1, 2, · · · , hm(u) is analytic along the contour of integration

L.

Proof. We prove this by induction. We have already shown that h0(u) is analytic.

Assume hm(u) is analytic. Then by noting that

gm(u) =hm(u)− αm − βmu

u2 − ζ(t)

is analytic except possibly at u = ±√ζ(t) and is of the indeterminate form [0/0] ( by

the definition of αm and βm ) as u→√ζ(t). By L’Hopital rule,

limu→√

ζ(t)

gm(u) = limu→√

ζ(t)

h′m(u)− βm2u

=h′m(

√ζ(t))− βm

2√ζ(t)

and

limu→−√

ζ(t)

gm(u) = limu→−√

ζ(t)

h′m(u)− βm2u

=h′m(−

√ζ(t))− βm

−2√ζ(t)

.

If we define gm(±√ζ(t)) as the limit of gm(u) as u → ±

√ζ(t), then gm(u) is con-

tinuous at u = ±√ζ(t). From this, we conclude that gm(u) is analytic and so is

hm+1(u) = g′m(u).

Again we prove this by induction. The case p = 1 has been done. Assume the

result is true for p = q. Observe that


ϵq + δq

= ei(πα−πν/t2) (−1)q

2πiνq

∫γ1

eν(u3/3−ζu)[αq + βqu+ (u2 − ζ)gq(u)] du

+ e−i(πα−πν/t2) (−1)q

2πiνq

∫γ2

eν(u3/3−ζu)[αq + βqu+ (u2 − ζ)gq(u)] du

= ei(πα−πν/t2) (−1)q

2πiνq

[αq

∫γ1

eν(u3/3−ζu) du+ βq

∫γ1

eν(u3/3−ζu)u du

+

∫γ1

eν(u3/3−ζu)(u2 − ζ)gq(u) du

]+ e−i(πα−πν/t2) (−1)q

2πiνq

[αq

∫γ2

eν(u3/3−ζu) du+ βq

∫γ2

eν(u3/3−ζu)u du

+

∫γ2

eν(u3/3−ζu)(u2 − ζ)gq(u) du

]= ei(πα−πν/t2) (−1)q

2πiνq

2πi

ν1/3αq

(−ωAi(ων2/3ζ)

)+

2πi

ν2/3βq(ω2Ai′(ων2/3ζ)

)−1

ν

∫γ1

eν(u3/3−ζu)hq+1(u) du

+ e−i(πα−πν/t2) (−1)q

2πiνq

2πi

ν1/3αq

(−ω2Ai(ω2ν2/3ζ)

)+

2πi

ν2/3βq(ωAi′(ω2ν2/3ζ)

)−1

ν

∫γ2

eν(u3/3−ζu)hq+1(u) du

= (−1)qν−q−1/3αq cos(πα− πν/t2)

[−ωAi(ων2/3ζ)− ω2Ai(ω2ν2/3ζ)

]+ (−1)qν−q−1/3αq sin(πα− πν/t2)

[iω2Ai(ω2ν2/3ζ)− iωAi(ων2/3ζ)

]+ (−1)qν−q−2/3βq cos(πα− πν/t2)

[ω2Ai′(ων2/3ζ) + ωAi′(ω2ν2/3ζ)

]+ (−1)qν−q−2/3βq sin(πα− πν/t2)

[iω2Ai′(ων2/3ζ)− iωAi′(ω2ν2/3ζ)

]+ ϵq+1 + δq+1

= (−1)q cos(πα− πν/t2)[

Ai(ν2/3ζ)αq

νq+1/3− Ai′(ν2/3ζ)

βqνq+2/3

]+ ϵq+1

+ (−1)q sin(πα− πν/t2)[

Bi(ν2/3ζ)αq

νq+1/3− Bi′(ν2/3ζ)

βqνq+2/3

]+ δq+1.

The result for p = q + 1 follows from this immediately.

C.1 The Transformation s 7→ u in the Case t > 2 102

Verification of the transformation s 7→ u in the case t > 2.

C.1 The Transformation s 7→ u in the Case t > 2

Note that the function ϕ(s; t) has singularities at s = 0, s = 1/t and saddle points

s = s±. One can show that the mapping s 7→ u is one-to-one and analytic on the

boundaries of these two regions. Hence, it also has these properties in the interior of

the regions. See [38, p. 12]. This establishes the one-to-one and analytic nature of the

mapping in the upper half-plane. The mapping properties of the lower half-plane can

be obtained by using reflection with respect to the real-axis.

Lemma C.2. There exist real numbers s1, s2, s3, s4 with

0 < s1 < s2 <1

t< s3 < s− < s+ < s4

such that s→ Z maps s = s1 and s = s4 to Z = −23ζ3/2(t) and s = s2 and s = s3 to

Z = 23ζ3/2(t).

Proof. Let us consider the mapping of the real-axis in s-plane under the transformation

s 7→ Z and let

ψ(s) := η(t)− ϕ(s; t) =

ψ1(s)−

iπ

t2,

ψ2(s),

where

ψ1(s) =1

2−[s

t+

1

t2log(1− ts)− log(s)

],

ψ2(s) =1

2−[s

t+

1

t2log(ts− 1)− log(s)

].

It is easily shown that

ψ′1(s) = ψ′

2(s) = −[1

t+

1

t

1

ts− 1− 1

s] = −(s− s−)(s− s+)

s(ts− 1)

and

ψ′′(s) = −[−1ts− 1

+1

s2

].


0

S4

S1

S3

S2

Figure C.5: y = ψ(s).

Therefore, ψ′(s) = 0⇔ s = s±, i.e., ψ′(s±) = 0 and

ψ′′(s±) = −[±1

2

√t2 − 4[t(t2 − 3)∓ (t2 − 1)

√t2 − 4]

which gives

ψ′′(s+) < 0 and ψ′′(s−) > 0.

From the sketch of the graph y = ψ(s) in Figure C.5, the results follow.

In order to make the function s 7→ Z single-valued, we consider the two regions

H’I’J’KLMN’NH’ and ABB’CC’DD’EE’FF’GG’HIJA as shown in Figure C.6. Here,

BB’, CC’, DD’, EE’, FF’, GG’ and NN’ are parts of the circle of radius r centered at

origin, s = s1, s = s2, s = 1t, s = s3, s = s− and s = s4, respectively, where s = s−

are the saddle point of ϕ(s; t). HI and I’H’ are parts of the circle of radius r centered

at s = s+. IJ’ and J’I are parts of the steepest descent path through s+. JA and MJ’ are

parts of the circle of radius R centered at the origin. We will let R→∞ and r → 0 in

our analysis. For the region ABB’CC’DD’EE’FF’GG’HIJA, we divide the boundary

into several portions and discuss the mapping of each of them as follows:


(i) AB: s = x, x ∈ [−R,−r],

Z = η(t)− ϕ(s; t)

= η(t)− [x

t+

1

t2log(1− tx)− log(−x)− iπ]

=: g(x) + iπ − iπ

t2,

where g(x) = 12− [x

t+ 1

t2log(1− tx)− log(−x)]. Thus,

g′(x) = −[−(x2 − tx+ 1)

(1− tx)x

]=

(x2 − tx+ 1)

(1− tx)x,

where 0 < s− < s+. Hence, g′(x) < 0 and g(x) is strictly decreasing in x ∈ [−R,−r].

Also, g(x) → −∞ as x → 0− and g(x) → ∞ as x → −∞. We see that as R → ∞

and r → 0, AB is (asymptotically) mapped injectively to the the line segment AB

(with ℑ(Z) = π − π/t2) in the Z-plane.

(ii) BB’: s = reiθ, θ ∈ [0, π].

Z = η(t)− ϕ(s; t)

= η(t)− [reiθ

t+

1

t2log(1− treiθ)− log(reiθ)]

=1

2− r

tcos θ − 1

2t2log((1− rt cos θ)2 + r2t2 sin2 θ) + log r

+i

[−rtsin θ − 1

t2tan−1

(−rt sin θ

1− rt cos θ

)+ θ − π

t2

].

When r → 0+, ℜ(z) → −∞ and ℑ(Z) → θ − π/t2. We see that BB’ is (asymptoti-

cally) mapped injectively to the line segment BB’ in the Z-plane.

(iii) B’C: s = x, x ∈ [r, s1 − r].

Z = η(t)− ϕ(s; t)

= η(t)−[x

t+

1

t2log(1− tx)− log(x)

]=: g(x)− iπ

t2,


t+ 1

t2log(1− tx)− log(x)]. Note that, log(1− tx) is well-defined

as 1− tx > 0. Since

g′(x) =(x2 − tx+ 1)

(1− tx)x=

(x− s−)(x− s+)(1− tx)x

,


then g(x) is strictly increasing in x ∈ [r, s1 − r]. g(x) → −∞ as x → 0 and

g(x) → η(t) − ϕ(s1; t) = −2ζ3/2(t)/3 as x → s1. We see that as r → 0, B’C is

(asymptotically) mapped injectively to the the line segment B’C (with ℑ(Z) = −π/t2)

in the Z-plane.

(iv) CC’: s = s1 + reiθ, θ ∈ [0, π].

Expanding ϕ(s; t) at s = s1 and making use of lemma C.2, we have

Z = η(t)− [ϕ(s1; t) + ϕ′(s1; t)reiθ +O(r2)]

= −2ζ32

3− iπ

t2− ϕ′(s1; t)re

iθ +O(r2).

If we write Z := X + iY = −2ζ3/2/3− iπ/t2 + ρeiΘ, thenX = −2ζ32

3− ϕ′(s1; t)r cos θ +O(r2)

Y = − iπt2− ϕ′(s1; t)r sin θ +O(r2)

.

So, we have

ρ = −ϕ′(s1; t)r +O(r2)

and

tanΘ =Y − (−iπ/t2)

X − (−2ζ3/2(t)/3)=

sin θ +O(r)

cos θ +O(r).

For fixed t and r,

d

dθtanΘ(θ) =

(cos θ +O(r))(cos θ +O(r))− (sin θ +O(r))(− sin θ +O(r))

(cos θ +O(r))2

=1

(cos θ +O(r))2(1 +O(r)).

Hence, if r is sufficiently small,

d

dθtanΘ(θ) > 0.

This implies that Θ(θ) is strictly increasing from 0 to π (as r → 0) when θ ∈ [0, π].

We see that CC’ is asymptotically mapped injectively to the semi-circle CC’ centered

at Z = −2ζ3/2(t)/3− iπ/t2 in the Z-plane.


(v) C’D: similar to (iii).

(vi) DD’: similar to (iv).

(vii) D’E: similar to (iii).

(viii) EE’: s = 1/t+ reiθ, θ ∈ [0, π].

Z = η(t)− ϕ(s; t)

=1

2− [

s

t+

1

t2log(ts− 1)− log(s)]

=1

2−[1

t(1

t+ r cos θ + ir sin θ) +

1

t2log(rteiθ)− log(

1

t+ r cos θ + ir sin θ)

]=

[1

2− 1

t2− r

tcos θ − 1

t2log rt+ log((

1

t+ r cos θ)2 + r2 sin2 θ)

]+i

[−rtsin θ − 1

t2θ + tan−1

(r sin θ

1t+ r cos θ

)].

As r → 0, ℜ(Z)→∞ and ℑ(Z)→ −θ/t2. Furthermore, we let

ℑ(Z) := ϕ1(θ) = −r

tsin θ − 1

t2θ + tan−1

(r sin θ

1t+ r cos θ

).

Then

ϕ′1(θ) = −

r

tcos θ − 1

t2θ +

r2 + rt(cos θ − sin θ)

((1t+ r cos θ)2 + r2 sin2 θ)

.

So, for fixed t > 2 and sufficiently small r,

ϕ′1(θ) < 0 for θ ∈ [0, π).

From this, we see that EE’ is asymptotically mapped injectively to the line segment

EE’ in Z-plane.

(ix) E’F: similar to (iii).

(x) FF’: similar to (iv).

(xi) F’G: similar to (iii).


(xii) GG’ s = s− + reiθ, θ ∈ [0, π].

Z = η(t)− ϕ(s; t)

=1

2−[ϕ(s−; t) + ϕ′(s−; t)(s− s−) +

1

2!ϕ′′(s−; t)(s− s−)2 +O(|s− s−|3)

]= −2ζ

32

3− 1

2!ϕ′′(s−; t)r

2ei2θ +O(r3),

where ϕ′(s−; t) = 0 and ϕ′′(s−; t) < 0 for t > 2. If we write Z := X + iY =

−2ζ3/2(t)/3 + ρeiΘ, thenX = −2ζ3/2

3− 1

2ϕ′′(s−; t)r

2 cos 2θ +O(r3)

Y = −1

2ϕ′′(s−; t)r

2 sin 2θ +O(r3)

.

So,

ρ = −1

2ϕ′′(s−; t)(r

2 +O(r3))

and

tanΘ =Y

X + 2ζ3/2(t)/3=

sin 2θ +O(r)

cos 2θ +O(r).

So, for fixed t and r, we have

d

dθtanΘ(θ) =

(cos 2θ +O(r))(2 cos 2θ +O(r))− (sin 2θ +O(r))(−2 sin 2θ +O(r))

(cos 2θ +O(r))2

=2

(cos 2θ +O(r))2(1 +O(r)).


d

dθtanΘ(θ) > 0.

This implies that Θ(θ) is strictly increasing from 0 to 2π (as r → 0) when θ ∈ [0, π].

We see that GG’ is asymptotically mapped injectively to a small circle GG’ centered

at Z = −2ζ3/2(t)/3 in the Z-plane.

(xiii) G’H: similar to (iii).

(xiv) HI: similar to (xii).


(xv) IJ: since IJ is a part of the steepest descent path through s = s+, it is asymptoti-

cally mapped injectively to a part of the line ℑ(Z) = 0, i.e., the real-axis in Z-plane.

However, it is not easy to prove that the mapping is one-to-one. Write s = x+ iy and

Z = X + iY . From

Z = η(t)− ϕ(s; t) = 1

2− s

t− 1

t2log(ts− 1) + log s,

we haveX = X(x, y, t) = η(t)− x

t− 1

2t2log((tx− 1)2 + t2y2) +

1

2log(x2 + y2)

Y = Y (x, y, t) = −yt− 1

t2arg(tx− 1 + ity) + arg(x+ iy)

.

Since IJ is a part of the steepest descent path through s = s+, it is defined implicitly

by the equation,

ℑ(ϕ(s; t)) = ℑ(ϕ(s+; t)),

or, equivalently,

y

t+

1

t2arg(tx− 1 + ity)− arg(x+ iy) = 0.

We need to prove that the mapping of IJ in s-plane to IJ in Z-plane is one-to-one. We

consider again the system of equation defining the steepest descent path through s+,X =

1

2− x

t− 1

2t2log((tx− 1)2 + t2y2) +

1

2log(x2 + y2)

0 = −yt− 1

t2arg(tx− 1 + ity) + arg(x+ iy)

,

where

arg(x+ iy) =

tan−1(y/x) if x > 0 and y > 0 or < 0,

π + tan−1(y/x) if x < 0 and y > 0,

−π + tan−1(y/x) if x < 0 and y < 0.


and −π/2 < tan−1 u < π/2 for u ∈ (−∞,∞).

We regard this system of equation as a system defining x and y as functions of X . This

is possible by an application of the implicit function theorem. We defineH(x, y, t) = η(t)− x

t− 1

2t2log((tx− 1)2 + t2y2) +

1

2log(x2 + y2)−X,

I(x, y, t) =y

t+

1

t2arg(tx− 1 + ity)− arg(x+ iy).

By simple computations, we obtain their first partial derivatives as follows,∂H∂x

= −1t+ 1−tx

t[(tx−1)2+t2y2]+ x

x2+y2, ∂H

∂y= −y

(tx−1)2+t2y2+ y

x2+y2, ∂H

∂X= −1,

∂I∂x

= −y(tx−1)2+t2y2

+ yx2+y2

, ∂I∂y

= 1t− 1−tx

t[(tx−1)2+t2y2]− x

x2+y2, ∂I

∂Y= 0.

The Jacobian J of the system is

J =

∣∣∣∣∣∣∣∂H∂x

∂H∂y

∂I∂x

∂I∂y

∣∣∣∣∣∣∣= −

[(−1

t+

1− txt[(tx− 1)2 + t2y2]

+x

x2 + y2)2 + (

−y(tx− 1)2 + t2y2

+y

x2 + y2)2].

Lemma C.3. J < 0.

Proof. Suppose J = 0. Then

−y(tx− 1)2 + t2y2

+y

x2 + y2= 0⇒ (tx− 1)2 + t2y2 = x2 + y2

−1

t+

1− txt[(tx− 1)2 + t2y2]

+x

x2 + y2= 0⇒ (tx−1)2+ t2y2 = 1 = x2+y2 ⇒ x =

t

2.

Hence, we have

y = ±i√t2 − 4

2

which is impossible since y must be real. This contradiction gives J = 0. Therefore,

J < 0.


By the implicit function theorem, for every X0 ∈ (2ζ3/2(t)/3,∞), x and y can be

expressed as C1 functions of X in some small neighborhood of X0. Also, we have

∂y

∂X=

1

J

∣∣∣∣∣∣∣∂H∂x−∂H

∂X

∂I∂x− ∂I

∂X

∣∣∣∣∣∣∣=

1

J

∣∣∣∣∣∣∣−1

t+ 1−tx

t[(tx−1)2+t2y2]+ x

x2+y21

−y(tx−1)2+t2y2

+ yx2+y2

0

∣∣∣∣∣∣∣=

1

J(

y

(tx− 1)2 + t2y2− y

x2 + y2).

∂x

∂X=

1

J

∣∣∣∣∣∣∣−∂H

∂X∂H∂y

− ∂I∂X

∂I∂y

∣∣∣∣∣∣∣=

1

J

∣∣∣∣∣∣∣1 −y

(tx−1)2+t2y2+ y

x2+y2

0 1t− 1−tx

t[(tx−1)2+t2y2]− x

x2+y2

∣∣∣∣∣∣∣=

1

J(1

t− 1− txt[(tx− 1)2 + t2y2]

− x

x2 + y2).

Theorem C.4.∂x

∂X< 0 for x ∈ (−∞, s+).

Proof. If x < 0, then obviously, ∂x∂X

< 0. If x > 0, then we first note that

u

1 + u2< tan−1 u if u > 0 (C.1.1)

andu

1 + u2> tan−1 u if u < 0. (C.1.2)

Putting u = y/x and u = ty/(tx− 1) in (C.1.1) and (C.1.2) respectively, and we have

yx

1 + ( yx)2< tan−1

(yx

)⇔ xy

x2 + y2< tan−1

(yx

)and

tytx−1

1 + ( tytx−1

)2> tan−1

(ty

tx− 1

)⇔ (tx− 1)ty

(tx− 1)2 + t2y2> tan−1

(ty

tx− 1

).


Hence,

1

t− 1− txt[(tx− 1)2 + t2y2]

− x

x2 + y2

>1

t+

1

t2ytan−1

(ty

tx− 1

)− 1

ytan−1

(yx

)= 0

Hence, ∂x/∂X < 0 for x ∈ (−∞, s+), or equivalently, X ∈ (−2ζ3/2(t)/3,∞).

This implies that x is in fact an decreasing function of X in the relevant interval.

We conclude that IJ is asymptotically mapped injectively to the line segment IJ (with

ℑ(Z) = 0) in the Z-plane.

(xvi) JA: s = Reiθ, θ ∈ [θJ , π], where θJ → π as R→∞.

Z =1

2− 1

tReiθ − 1

t2log(tReiθ − 1) + logReiθ

ℜ(Z) = 1

2− R

tcos θ − 1

2t2log((tR cos θ − 1)2 + t2R2 sin2 θ) + logR

ℑ(Z) = −Rtsin θ − 1

t2tan−1

(Rt sin θ

Rt cos θ − 1

)+ θ

.

If we write Z = X + iY = ρeiΘ, then

ρ = R +O(R12 ).

Since

tanΘ =Y

X=

−Rtsin θ − 1

t2tan−1

(Rt sin θ

Rt cos θ−1

)+ θ

C(t)− Rtcos θ − 1

2t2log((tR cos θ − 1)2 + t2R2 sin2 θ) + logR

,

for fixed t and R,d

dθtanΘ =

1

X2(O(R)).

Hence, if R is sufficiently large,

d

dθtanΘ > 0.

Θ(θ) is strictly increasing from Θ1(→ 0) to Θ2(asR → ∞) when θ ∈ [θJ , π]. We

conclude that the mapping of JA in s-plane to JA in Z-plane is injective.


Next, we consider the mapping of the boundary of the region H’I’J’KLMN’NH’.

We again divide it into several portions. Since the discussions are similar to the region

ABB’CC’DD’EE’FF’GG’HIJA, we will omit some of the details.

(xvii) J’I’:similar to (xv).

(xviii) I’H’:similar to (xii).

(xix) H’N:similar to (iii).

(xx) NN’:similar to (iv).

(xxi) N’M:similar to (iii).

(xxii) M’J’:similar to (xvi).

From the above analysis, we see that the transformation s 7→ Z maps the two sim-

ple closed curves ABB’CC’DD’EE’FF’GG’HIJA and H’I’J’KLMN’NH’ in s-plane

into two simple closed curves in Z-plane. Therefore, it maps the two regions injec-

tively to their corresponding images in Z-plane; the mappings of the elementary func-

tions sine and its inverse arcsine are well-known (See, for instance, [2, p. 98] or [22,

p. 273–278]).

Therefore, the transformations Z 7→ γ and γ 7→ u map the indicated regions con-

formally to their corresponding images. See Figures C.6, C.7, C.8, C.9, C.10, C.11,

C.12. The transformation s 7→ u being the composition of conformal mappings is

also one-to-one (The region to be transformed is the upper half-plane (including the

real-axis) with s = 0, s1, s2,1t, s3, s± and s4 excluded). The mapping properties of the

lower half-plane can be obtained by using reflection with respect to the real-axis.


BA C’ ECB’ FF’E’

J

G’DD’ G H H’

I’I

M

L

K

J’

N N’

Figure C.6: s-plane (t > 2).

BA

C’ ECB’

FF’E’

G’

D D’

G

H I J

Figure C.7: Z-plane (t > 2).

NM

L

K

J’H’ I’N’

Figure C.8: Z-plane (t > 2).


-3/2 -/2 /2 3/2-

B A

C’

E

C

B’

F F’

E’

G’

D

D’

G H H’I’I

M

LK

J’J

N

N’

Figure C.9: γ-plane (t > 2).

-/2 -/6 /6 /3 /2-/3

B A

C’

E

C

B’

F F’

E’

G’

D

D’

G H H’I’I

M

LK

J’J

N

N’

Figure C.10: γ/3-plane (t > 2).

-½ ½ 1-1

B

A

C’

E

C

B’

F F’E’ G’

D

D’

GH H’

I’I

M

L

K

J’J

N N’

Figure C.11: sin γ/3-plane (t > 2).

C.2 The Transformation s 7→ u in the Case 0 < t < 2 115

B

A

C’

E

C

B’

F F’E’ G’

D

D’

GH H’

I’I

M

L

K

J’J

N N’

Figure C.12: u-plane (t > 2).

Verification of the transformation s 7→ u in the case 0 < t < 2.

C.2 The Transformation s 7→ u in the Case 0 < t < 2

In order to make the function s 7→ Z single-valued, we consider the two regions

ABB’CDEFGA and C’IHGF’E’D’C’ as shown in Figure C.13. Here, BB’ is a part of

the circle of radius r centered at the origin. CD and D’C’ are parts of the circle of

radius r centered at s = 1/t. JG’ and GA are parts of the circle of radius R centered

at the origin. EF and E’F’ are parts of the circle of radius r centered at the saddle

point s = s+ of ϕ(s; t). DE, FG, D’E’ and F’G’ are parts of the steepest descent path

through s+. We will let R→∞ and r → 0 in our analysis.

(i) AB: s = x ∈ [−R,−r],

Z = η(t)− ϕ(s; t)

=1

2− [

x

t+

1

t2log(1− tx) + iπ

t2− log(−x)− iπ]

=: g(x) + iπ − iπ/t2,


where g(x) = 12− [x/t+ log(1− tx)/t2 − log(−x)].

g′(x) = −[−(x2 − tx+ 1)

(1− tx)x

]=

(x2 − tx+ 1)

(1− tx)x.

As the discriminant ∆ of the numerator is t2 − 4 < 0 and x is negative, g(x) is strictly

decreasing in x ∈ [−R,−r]. Also, g(x) → −∞ as x → 0− and g(x) → ∞ as

x → −∞. We see that as R → ∞ and r → 0, AB is (asymptotically) mapped injec-

tively to the line segment AB (with ℑ(Z) = π − π/t2) in the Z-plane.

(ii) BB’: s = reiθ, θ ∈ [0, π].

Z = η(t)− ϕ(s; t)

=1

2− [

reiθ

t+

1

t2log(1− treiθ) + iπ

t2− log(reiθ)]

=1

2− r

tcos θ − 1

2t2log((1− rt cos θ)2 + r2t2 sin2 θ) + log r

+i

[−rtsin θ − 1

t2tan−1

(−rt sin θ

1− rt cos θ

)+ θ − π

t2

].

When r → 0, ℜ(Z) → −∞ and ℑ(Z) → θ − π/t2. We see that BB’ is (asymptoti-

cally) mapped injectively to the line segment BB’ in the Z-plane.

(iii) B’C: s = x, x ∈ [r, 1t− r].

Z = η(t)− ϕ(s; t)

=1

2− [

x

t+

1

t2log(1− tx) + iπ

t2− log(x)] =: g(x)− iπ

t2


t+ 1

t2log(1− tx)− log(x)]. Note that, log(1− tx) is well-defined

as 1− tx > 0. Since

g′(x) =(x2 − tx+ 1)

(1− tx)x> 0,

then g(x) is strictly increasing in x ∈ [r, 1t− r]. g(x)→ −∞ as x→ 0 and g(x)→∞

as x → 1t− r. We see that r → 0, B’C is (asymptotically) mapped injectively to the

line segment B’C (with ℑ(Z) = −π/t2) in the Z-plane.


(iv) CD: s = 1t+ reiθ, θ ∈ [θD, π], 0 < θD < π,

Z = η(t)− ϕ(s; t)

=1

2− [

1

t(1

t+ reiθ) +

1

t2log(t(

1

t+ reiθ)− 1)− log(

1

t+ reiθ)]

=1

2− 1

t2− r

tcos θ − 1

t2log rt+

1

2log((

1

t+ r cos θ)2 + r2 sin2 θ)

+i

[−rtsin θ − θ

t2+ tan−1

(r sin θ

1t+ r cos θ

)].

When r → 0+, ℜ(Z) → ∞ and ℑ(Z) → −π/t2. We see that CD is (asymptotically)

mapped injectively to the line segment CD in the Z-plane.

(v) DE: similar to (xv) of the case t > 2. To conclude, DE is (asymptotically) mapped

injectively to the line segment DE in the Z-plane.

(vi) EF: s = s+ + reiθ, θ ∈ [θE, θF].

We consider the Taylor series expansion of Z = Z(s, t) about s = s+,

Z = η(t)−[ϕ(s+; t) + ϕ′(s+; t)(s− s+) +

1

2ϕ′′(s+; t)(s− s+)2 +O(|s− s+|3)

].

Now, ϕ′(s+; t) = 0 and

ϕ′′(s+; t) =1

2

√4− t2[

√4− t2(t2 − 1) + it(t2 − 3)].

Therefore,

Z =2

3B3 − 1

4

√4− t2[

√4− t2(t2 − 1) + it(t2 − 3)].

If we write Z := X + iY = 2ζ3/2/3 + ρeiΘ, we haveX = −1

4

√4− t2

[√4− t2(t2 − 1) cos 2θ − t(t2 − 3) sin 2θ

]r2 +O(r3)

Y =2

3iζ

32 − 1

4

√4− t2

[√4− t2(t2 − 1) sin 2θ + t(t2 − 3) cos 2θ

]r2 +O(r3)

.

So,

ρ =

[X2 + (Y − 2

3iζ

32 )2] 1

2

=1

2

√4− t2r2 +O(r3)


and

tanΘ =Y − 2

3iζ

32

X=

√4− t2(t2 − 1) sin 2θ + t(t2 − 3) cos 2θ +O(r)√4− t2(t2 − 1) cos 2θ − t(t2 − 3) sin 2θ +O(r)

.

Therefore, for fixed t and r, we have

d

dθtanΘ(θ)

=1

[√4− t2(t2 − 1) cos 2θ − t(t2 − 3) sin 2θ +O(r)]2

×([√4− t2(t2 − 1) cos 2θ − t(t2 − 3) sin 2θ +O(r)]×

[√4− t2(t2 − 1)2 cos 2θ − 2t(t2 − 3) sin 2θ +O(r)]

−[√4− t2(t2 − 1) sin 2θ + t(t2 − 3) cos 2θ +O(r)]×

[√4− t2(t2 − 1)(−2) sin 2θ − t(t2 − 3)2 cos 2θ +O(r)]

)=

2

[√4− t2(t2 − 1) cos 2θ − t(t2 − 3) sin 2θ +O(r)]2

[4 +O(r)].


d

dθtanΘ(θ) > 0.

This implies that Θ(θ) is strictly increasing from 0 to 2π (as r →∞) when θ ∈ [θE, θF].

From this, we see that EF is asymptotically mapped injectively to a small circle EF cen-

tered at Z = 2ζ3/2(t)/3 in the Z-plane.

(vii) FG: It is similar to (v).

(viii) GA: similar to (xvi) of the case t > 2.

From (i) to (viii), we conclude that the boundary of the region ABB’CDEFGA is

mapped injectively to a simple closed contour in Z-plane. The mapping of the bound-

ary of the region C’JIHGF’E’D’C’ from s-plane to Z-plane is considered in exactly

the same way and we omit the details. It should be noted that we deliberately con-

sider a smaller region in order to exclude the point Z = 2ζ3/2(t)/3 since this is one

of the branch points of the transformation Z → γ. By the same argument as in the


BA C’

E

CB’

F

F’

E’

J

G’

D D’

G

H

I

Figure C.13: s-plane.

BA

CB’

F

E D

G

Figure C.14: Z-plane.

case t > 2, we can prove that both regions are mapped injectively to their images in

Z-plane under the transformation s → Z. To prove that both regions are mapped to

their images in the u-plane, we use the same line of argument as in the case t > 2. The

details of the mapping of the two regions under the transformation s→ 3Z/(2ζ3/2(t)),

3Z/(2ζ3/2(t)) → γ and γ → u are shown in Figure C.14, C.15, C.16, C.17, C.18 and

C.19.

Theorem C.5. Let Ω be a region in the complex plane, f be an analytic function in Ω


J C’

I

F’

H

E’

G’

D’

Figure C.15: Z-plane.

B

AC

B’

F E

DG

1

Figure C.16: 3Z/(2ζ3/2)-plane.


J

C’

I

F’

H

E’

G’

D’

1

Figure C.17: 3Z/(2ζ3/2)-plane.

and C a contour in Ω. If the transformation z 7→ u is analytic and one-to-one, then∫C

f(z)dz =

∫C′f(z(u))

dz

dudu, (C.2.1)

where C ′ is the image of C in the u-plane.

Proof. Let us write z = x + iy, u = X + iY and f(z) = p(x, y) + iq(x, y) and let C

be parameterized by

z(t) = α(t) + iβ(t), t ∈ [0, 1],

u(t) = γ(t) + iγ(t), t ∈ [0, 1].

Then, by the definition of contour integrals,∫C

f(z)dz =

∫ 1

0

[p(α(t), β(t)) + iq(α(t), β(t))][α′(t) + iβ′(t)]dt.

Now if we write

u = u(z) = f(x, y) + g(x, y),


B

A

C B’

F

F’

E

E’

DD’

GH G’

C’

/!

IJ

Figure C.18: γ-plane.

B

A

C

B’

F

F’

E

E’

DD’

G

H

G’

C’

I

J

Figure C.19: u-plane.


then

X = f(x, y) and Y = g(x, y).

Since the inverse transform u→ z is also analytic and one-to-one, we may write

z = z(u) = F (X,Y ) + iG(X, Y ),

and so

x = F (X,Y ) and y = G(X,Y ).

Clearly, we have the relations γ(t) = f(α(t), β(t))

δ(t) = g(α(t), β(t))

,

α(t) = F (γ(t), δ(t))

β(t) = G(γ(t), δ(t))

.

By differentiating with respect to t, we haveα′(t) = ∂F∂X

(γ(t), δ(t))γ′(t) + ∂F∂Y

(γ(t), δ(t))δ′(t)

β′(t) = ∂G∂X

(γ(t), δ(t))γ′(t) + ∂G∂Y

(γ(t), δ(t))δ′(t)

,

Hence,

α′(t) + iβ′(t)

=

[∂F

∂X(γ(t), δ(t))γ′(t) +

∂F

∂Y(γ(t), δ(t))δ′(t)

]+i

[∂G

∂X(γ(t), δ(t))γ′(t) +

∂G

∂Y(γ(t), δ(t))δ′(t)

]=

[∂F

∂X(γ(t), δ(t)) + i

∂G

∂X(γ(t), δ(t))

]γ′(t) +[

∂F

∂Y(γ(t), δ(t))δ′(t) + i

∂G

∂Y(γ(t), δ(t))

]δ′(t)

=

[∂F

∂X(γ(t), δ(t))− i∂F

∂Y(γ(t), δ(t))

]γ′(t)

+

[− ∂G∂X

(γ(t), δ(t))δ′(t) + i∂G

∂Y(γ(t), δ(t))

]δ′(t)

=dz

du(γ(t), δ(t))[γ′(t) + iδ′(t)].


In the above derivation, we have made use of the Cauchy-Riemann equations

FX = GY and FY = −GX .

Also,

p(α(t), β(t)) + iq(α(t), β(t))

= p(F (γ(t), δ(t)), G(γ(t), δ(t))) + iq(F (γ(t), δ(t)), G(γ(t), δ(t))),

∫C

f(z)dz

=

∫ 1

0

[p(F (γ(t), δ(t)), G(γ(t), δ(t))) + iq(F (γ(t), δ(t)), G(γ(t), δ(t)))]

dz

du(γ(t), δ(t))[γ′(t) + iδ′(t)]dt

=

∫C′f(z(u))

dz

dudu.

Hence, the results follow.

Lemma C.6. h0(u) and g0(u) are analytic functions.

Proof.

h0(u) = h(u) =1


du

=1

s−2α+1/2s(ts− 1)αs(ts− 1)(u2 − ζ)(s− s−)(s− s+)

=1

s−2α+1/2s(ts− 1)αs(ts− 1)

1

(u−√ζ)

(s− s+)(u+

√ζ)

(s− s−).

(C.2.2)

From this, we see that h0(u) is analytic except possibly when u = ±√ζ(t), which

correspond to s± respectively.


Recall that

h0(u) = α0 + β0u+ (u2 − ζ)g0(u),

h1(u) = g′0(u),(C.2.3)

from (C.2.3), we have

h1(u) =d

du

h0(u)− α0

u2 − ζ. (C.2.4)


Thus,

h1(u) =1

(u2 − ζ)2[h′0(u)(u

2 − ζ)− 2u (h0(u)− α0)]. (C.2.5)

When u→ ±√ζ, we have

h1(√ζ) =

1

4ζ

[√ζ h′′0(

√ζ)− h′0(

√ζ)]

(C.2.6)

and

h1(−√ζ) =

−14ζ

[√ζ h′′0(−

√ζ) + h′0(−

√ζ)]. (C.2.7)

Proof. Since

h1(u) =d

du

h0(u)− α0

u2 − ζ, (C.2.8)

then we have

limu→

√ζh1(u)

= limu→

√ζ

d

du

h0(u)− h0(√ζ)

u2 − ζ

= limu→

√ζ

1

(u2 − ζ)2[(u2 − ζ)h′0(u)− 2u(h0(u)− h0(

√ζ))]

= limu→

√ζ

1

4u(u2 − ζ)

[2uh′0(u) + (u2 − ζ)h′′0(u)− 2(h0(u)− h0(

√ζ))− 2uh′0(u)

]= lim

u→√ζ

1

4(3u2 − ζ)[2uh′′0(u) + (u2 − ζ)h′′′0 (u)− 2h′0(u)

]=

1

8ζ

[2√ζ h′′0(

√ζ)− 2h′0(

√ζ)]

=1

4ζ

[√ζ h′′0(

√ζ)− h′0(

√ζ)].

(C.2.9)

Similarly, we have

limu→−

√ζh1(u)

=1

8ζ

[−2√ζ h′′0(

√ζ)− 2h′0(

√ζ)]

=−14ζ

[√ζ h′′0(

√ζ) + h′0(

√ζ)].

(C.2.10)


From now on, we consider the below functions, let

J(s) :=1

s−2α+1/2s(ts− 1)α= s2α−

32 (st− 1)−α, (C.2.11)

ϕ(s) :=s

t+

1

t2log(ts− 1)− log s. (C.2.12)

Then, h0(u) can be rewritten as

h0(u) := J(s)ds

du. (C.2.13)

First of all, we consider the higher order derivative (w. r. t. u) of h0(u).

h0(u) = J(s)ds

du(C.2.14)

h′0(u) = J ′(s)

(ds

du

)2

+ J(s)d2s

du2(C.2.15)

and

h′′0(u) = J ′′(s)

(ds

du

)3

+ 3J ′(s)ds

du

(ds

du

)2

+ J(s)d3s

du3. (C.2.16)

The higher order derivatives of J(s) are

J ′(s) = (2α− 3

2)s2α−

52 (st− 1)−α − αts2α−

32 (st− 1)−α−1 (C.2.17)

and

J ′′(s) = (2α− 3

2)

[(2α− 5

2)s2α−

72 (st− 1)−α + s2α−

52 (−α)t(st− 1)−α−1

]− αt

[(2α− 3

2)s2α−

52 (st− 1)−α−1 + s2α−

32 (−α− 1)t(st− 1)−α−2

].

(C.2.18)

The higher order derivatives of J(s) at s = s+ are

J(s+) = s2α− 3

2+ (s+t− 1)−α = s

2α− 32

+ s−2α+ = s

− 32

+ , (C.2.19)

J ′(s+) = (2α− 3

2)s

2α− 52

+ (s+t− 1)−α − αts2α−32

+ (s+t− 1)−α−1

= (2α− 3

2)s

2α− 52

+ s−2α+ − αts2α−

32

+ s−2α−2+

= (2α− 3

2)s

− 52

+ − αts− 7

2+

(C.2.20)


and

J ′′(s+) = (2α− 3

2)

[(2α− 5

2)s

2α− 72

+ (s+t− 1)−α − s2α−52

+ αt(s+t− 1)−α−1

]− αt

[(2α− 3

2)s

2α− 52

+ (s+t− 1)−α−1 − s2α−32

+ (−α− 1)t(s+t− 1)−α−2

]= (2α− 3

2)

[(2α− 5

2)s

− 72

+ − s− 5

2+ αts−2

+

]− αt

[(2α− 3

2)s

− 52

+ s−2+ − s

− 32

+ (α+ 1)ts−4+

]= (2α− 3

2)(2α− 5

2)s

− 72

+ − 2(2α− 3

2)αts

− 92

+ + α(α+ 1)t2s− 11

2+ .

(C.2.21)

The higher order derivatives of ϕ(s) are

ϕ(s) =s

t+

1

t2log(ts− 1)− log s,

ϕ′(s) =1

t+

1

t

1

ts− 1− 1

s,

ϕ′′(s) = − 1

(ts− 1)2+

1

s2,

ϕ′′′(s) =2t

(ts− 1)3− 2

s3

and

ϕ(4)(s) = − 6t2

(ts− 1)4+

6

s4.

Thus, the higher order derivatives of ϕ(s) at s = s+ are

ϕ′(s+) = 0, (C.2.22)

ϕ′′(s+) =s+ − s−s3+

, (C.2.23)

ϕ′′′(s+) =2

s6+(t− s3+) (C.2.24)

and

ϕ(4)(s+) =6

s8+(s4+ − t2). (C.2.25)

To find the quantities ofds

du,d2s

du2, and

d3s

du3at u = ±

√ζ , we consider the cubic trans-

formation


s

t+

1

t2log(ts− 1)− log s =: ϕ(s) = Q(u) :=

u3

3− ζ(t)u+ η(t). (C.2.26)

For simplicity, we depict the variable t in ϕ. To findds

duat u = ±

√ζ, we recall that

ϕ′(s)ds

du= Q′(u), (C.2.27)

i.e.,ds

du

∣∣∣∣s=s+,u=+

√ζ

= lims→s+,u→+

√ζ

Q′(u)

ϕ′(s)=

[2√ζ s3+

s+ − s−

]1/2(C.2.28)

ds

du

∣∣∣∣s=s−,u=−

√ζ

= lims→s−,u→−

√ζ

Q′(u)

ϕ′(s)=

[2√ζ s3−

s+ − s−

]1/2. (C.2.29)

It also confirms the equations (3.2.4) and (3.2.3), i.e.,

ds

du

∣∣∣∣u=

√ζ

=

[2√ζ(t) s3+

s+ − s−

]1/2(C.2.30)

andds

du

∣∣∣∣u=−

√ζ

=

[2√ζ(t) s3−

s+ − s−

]1/2. (C.2.31)

To find ϕ′′(s) at s = s±, we consider

ϕ′′(s)

(ds

du

)2

+ ϕ′(s)d2s

du2= Q′′(u) = 2u, (C.2.32)

i.e.,

ϕ′′(s±) =±2√ζ ds

du

∣∣∣∣ s=s±u=±

√ζ

2 = ±s+ − s−s3±

. (C.2.33)

To findd2s

du2at u =

√ζ , we consider

ϕ′′′(s)

(ds

du

)3

+ 3ϕ′′(s)

(ds

du

)d2s

du2+ ϕ′(s)

d3s

du3= Q′′′(u) = 2, (C.2.34)


i.e.,

d2s

du2

∣∣∣∣u=

√ζ

=

2− ϕ′′′(s)

(ds

du

)3

3ϕ′′(s)

(ds

du

)∣∣∣∣∣∣∣∣∣u=

√ζ

=

2− 2s6+(t− s3+)

[2√

ζ(t) s3+s+−s−

]3/23 s+−s−

s3+

[2√

ζ(t) s3+s+−s−

]1/2

=2

3

1− (t− s3+)(

2√ζ

(s+−s−)s+

) 32√

2√ζ (s+−s−)

s3+

.

(C.2.35)

To findd3s

du3at u =

√ζ , we consider

ϕ(4)(s)

(ds

du

)4

+ 6ϕ′′′(s)

(ds

du

)2d2s

du2+ 3ϕ′′(s)

(d2s

du2

)2

+ 4ϕ′′(s)

(ds

du

)d3s

du3+ ϕ′(s)

d4s

du4= 0,

(C.2.36)

i.e.,

d3s

du3

∣∣∣∣u=

√ζ

= −ϕ(4)(s)

(dsdu

)4+ 6ϕ′′′(s)

(dsdu

)2 d2sdu2 + 3ϕ′′(s)

(d2sdu2

)24ϕ′′(s)

(dsdu

)∣∣∣∣∣∣∣u=

√ζ

=−s

32+

2√2√ζ(s+ − s−)

[12

s2+

(s4+ − t2)ζ(s+ − s−)2

+4(t− s3+)

√2√ζ

(s+)32 (s+ − s−)

32

(1− (t− s3+)

(2√ζ

s+(s+ − s−)

) 32

)

+1

3√ζ

(1− (t− s3+)

(2√ζ

s+(s+ − s−)

) 32

)2],

(C.2.37)


where

ϕ(4)(s)

(ds

du

)4

+ 6ϕ′′′(s)

(ds

du

)2d2s

du2+ 3ϕ′′(s)

(d2s

du2

)2∣∣∣∣∣u=

√ζ

=6

s8+(s4+ − t2)

[2√ζ s3+

s+ − s−

]4/2

+ 6

(2

s6+(t− s3+)

)[2√ζ s3+

s+ − s−

]2/22

3

1− (t− s3+)(

2√ζ

(s+−s−)s+

) 32√

2√ζ (s+−s−)

s3+

+ 3

(s+ − s−s3+

)2

3

1− (t− s3+)(

2√ζ

(s+−s−)s+

) 32√

2√ζ (s+−s−)

s3+

2

.

(C.2.38)

To obtain the value

h1(√ζ) =

1

4ζ

[√ζ h′′0(

√ζ)− h′0(

√ζ)], (C.2.39)

we first consider the two quantities. Since

h′0(√ζ) = J ′(s+)

(ds

du

∣∣∣∣u=

√ζ

)2

+ J(s+)d2s

du2

∣∣∣∣u=

√ζ

=

[(2α− 3

2)s+ − αt

]2√ζ

(s+ − s−)s12+

+2

3

1− (t− s3+)(

2√ζ

(s+−s−)s+

) 32√

2√ζ(s+ − s−)

(C.2.40)


and

h′′0(√ζ)

=J ′′(s+)

(ds

du

∣∣∣∣u=

√ζ

)3

+ 3J ′(s+)ds

du

∣∣∣∣u=

√ζ

(ds

du

∣∣∣∣u=

√ζ

)2

+ J(s+)d3s

du3

∣∣∣∣u=

√ζ

=1

s+

(2√ζ

s+ − s−

) 32[(2α− 3

2)(2α− 5

2)(s+)

2 − 2(2α− 3

2)αts+ + α(α+ 1)t2

]+

2

(s+)12 (s+ − s−)

[(2α− 3

2)s+ − αt

](1− (t− (s+)

3)

(2√ζ

s+(s+ − s−)

) 32

− −s32+

2√

2√ζ(s+ − s−)

[12

s2+

(s4+ − t2)ζ(s+ − s−)2

+4(t− s3+)

√2√ζ

(s+)32 (s+ − s−)

32

(1− (t− s3+)

(2√ζ

s+(s+ − s−)

) 32

)

+1

3√ζ

(1− (t− s3+)

(2√ζ

s+(s+ − s−)

) 32

)2

,

(C.2.41)

then, from the above result, we have

h1(√ζ)

=1

4ζ

(2√ζ

s+ − s−

) 12

√ζ

s+

(2√ζ

s+ − s−

)[(2α− 3

2)(2α− 5

2)s2+ − 2(2α− 3

2)αts+ + α(α+ 1)t2

]− 1

s2+

(2√ζ

s+ − s−

)2 [(2α− 3

2)s+ − αt

](t− s3+)−

3ζ

(s+ − s−)2

(s4+ − t2

s2+

)+

5

6

(2√ζ

s+ − s−

)2(t− s3+)2

s3+(s+ − s−)− 5

12√ζ

.

(C.2.42)

Simplifying the above result, we have

h1(√ζ) =

h0(√ζ)

4ζ

2ζ

s+√t2 − 4

[(2α− 3

2)(2α− 5

2)s2+ − 2(2α− 3

2)αts+

+ α(α+ 1)t2]− ζ

t2 − 4

1

s2+

[4(t− s3+)

((2α− 3

2)s+ − αt

)+ 3(s4+ − t2)

]+

10ζ(t− s3+)2

3(t2 − 4)32 s3+− 5

12√ζ

.

(C.2.43)

C.3 A Comparison of the Results (2.3.9) and (3.2.19) 132

Similarly, from the above procedure, we conclude that

h1(±√ζ) = ±h0(

√ζ)

4ζ

2ζ

s±√t2 − 4

[(2α− 3

2)(2α− 5

2)s2± − 2(2α− 3

2)αts±

+ α(α+ 1)t2]∓ ζ

t2 − 4

1

s2±

[4(t− s3±)

((2α− 3

2)s± − αt

)+ 3(s4± − t2)

]+

10ζ(t− s3±)2

3(t2 − 4)32 s3±− 5

12√ζ

.

(C.2.44)

Remark:

limt→2

h1(√ζ) = lim

t→2h1(−

√ζ) =

5

16− 1

2α, (C.2.45)

where α > 0 and

h0(±√ζ) =

√2ζ

14

(t2 − 4)14

=

(4ζ

t2 − 4

) 14

= α0. (C.2.46)

C.3 A Comparison of the Results (2.3.9) and (3.2.19)

From Stirling formula, we have

Γ(n

2+ 1) ∼ 2−

n2√πν

n2+ 1

2 e−ν2 (C.3.1)

where ν = n+ 2α− 1/2.

Proof.

Γ(n

2+ 1) = Γ(

ν + 2α− n2

2+ 1)

∼√2π

(ν + 2α− n

2

2

) ν+2α−n2

2+ 1

2

e−ν+2α−n

22

=√2πν

n2+ 1

2

2n2+ 1

2

(1 +

12− 2α

ν

) ν+2α−n2

2+ 1

2

e−ν+2α−n

22

∼√πν

n2+ 1

2

2n2

e12( 12−2α)e−

12(ν−2α+ 1

2)

= 2−n2√πν

n2+ 1

2 e−ν2


Applying the Stirling formula to (2.3.9), we have

f (α)n (

t√ν)

=


t2)

212nΓ(1

2n+ 1)

α0

[Ai(ν

23 ζ) p−1∑

s=0

As(ζ)

νs−16

+ Ai′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+16

+ ϵp(N, t)

]

+


t2)

212nΓ(1

2n+ 1)

α0

[Bi(ν

23 ζ) p−1∑

s=0

As(ζ)

νs−16

+ Bi′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+16

+ δp(N, t)

]

∼ cos(πα− νπ

t2)ν−

n2− 1

2 eν2α0

[Ai(ν

23 ζ) p−1∑

s=0

As(ζ)

νs−16

+ Ai′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+16

+ ϵp(ν, t)

]

+ sin(πα− νπ

t2)ν−

n2− 1

2 eν2α0

[Bi(ν

23 ζ) p−1∑

s=0

As(ζ)

νs−16

+ Bi′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+16

+ δp(ν, t)

]=cos(πα− νπ

t2)ν−

n2 e

ν2

α0

[Ai(ν

23 ζ) p−1∑

s=0

As(ζ)

νs+13

+ Ai′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+23

+ ν−12 ϵp(ν, t)

]+ sin(πα− νπ

t2)ν−

n2 e

ν2

α0

[Bi(ν

23 ζ) p−1∑

s=0

As(ζ)

νs+13

+ Bi′(ν

23 ζ) p−1∑

s=0

Bs(ζ)

νs+23

+ ν−12 δp(ν, t)

]=cos(πα− νπ

t2)ν−

n2 e

ν2[

Ai(ν

23 ζ) p−1∑

s=0

α0As(ζ)

νs+13

+ Ai′(ν

23 ζ) p−1∑

s=0

α0Bs(ζ)

νs+23

+ α0ν− 1

2 ϵp(ν, t)

]+ sin(πα− νπ

t2)ν−

n2 e

ν2[

Bi(ν

23 ζ) p−1∑

s=0

α0As(ζ)

νs+13

+ Bi′(ν

23 ζ) p−1∑

s=0

α0Bs(ζ)

νs+23

+ α0ν− 1

2 δp(ν, t)

]

=cos(πα− νπ

t2)ν−

n2 e

ν2

[Ai(ν

23 ζ) p−1∑

s=0

α0As(ζ)

νs+13

+ Ai′(ν

23 ζ) p−1∑

s=0

α0Bs(ζ)

νs+23

]

+ sin(πα− νπ

t2)ν−

n2 e

ν2

[Bi(ν

23 ζ) p−1∑

s=0

α0As(ζ)

νs+13

+ Bi′(ν

23 ζ) p−1∑

s=0

α0Bs(ζ)

νs+23

]+ cos(πα− νπ

t2)ν−

n2 e

ν2α0ν

− 12 ϵp(ν, t) + sin(πα− νπ

t2)ν−

n2 e

ν2α0ν

− 12 δp(ν, t).


From the result of the uniform asymptotic expansion of the Tricomi-Carlitz poly-

nomials via integral approach (3.2.19), we have

f (α)n (

t√ν)

=ν−n2 e

ν2 cos

(πα− πν

t2

)[Ai(ν2/3ζ)

p−1∑k=0

(−1)kαk

νk+1/3− Ai′(ν2/3ζ)

p−1∑k=0

(−1)kβkνk+2/3

]

+ ν−n2 e

ν2 sin

(πα− πν

t2

)[Bi(ν2/3ζ)

p−1∑k=0

(−1)kαk

νk+1/3− Bi′(ν2/3ζ)

p−1∑k=0

(−1)kβkνk+2/3

]+ ν−

n2 e

ν2 ϵp + ν−

n2 e

ν2 δp.

We can conclude that two expansions (2.3.9) and (3.2.19) are the same.

C.4 Power series solutions of (3.5.13) 135

C.4 Power series solutions of (3.5.13)

We can use more efficient inversion methods to obtain higher approximations of the

zeros. Since (3.5.13) can be written as

sin(πα− πν

t2+

2

3ν|ζ(t)|

32 +

1

4π) = O(

1

ν). (C.4.1)


|ζ(t)|32 =

3π

2t2− 3π

4+

1

2t+

1

80t3 +

9

8960t5 +O

(t6). (C.4.2)

Upon solving (C.4.1) and (C.4.2), we have

− n− 1

2+ (

ν

3πt+

ν

120πt3 +

3ν

4480πt5 + · · · ) = m+O(

1

ν). (C.4.3)

When n is even, put (C.4.3) in the form,

t+ c3t3 + c5t

5 + . . . = s, s =3π

ν(k +

1

2+O(

1

ν)),

where c3 = 1/40, c5 = 9/4480, . . .. By Lagrange’s inversion formula, for fixed k =

0, 1, 2, · · · , the small zeros are given by

tn,k+1 = s+ d3s3 + d5s

5 + · · ·

where

d3 = −c3 = −1

40, d5 = 3c23 − c5 = −

3

22400, · · · .

Note that, tn,1 = 0. Similarly, when n is odd, put (C.4.3) in the form

t+ c3t3 + c5t

5 + . . . = s, s =3π

ν(k +O(

1

ν)),

where c3 = 1/40, c5 = 9/4480, . . .. By Lagrange’s inversion formula, for fixed k =

0, 1, 2, · · · , the small zeros are given by tn,1 = 0 and

tn,k+1 = s+ d3s3 + d5s

5 + · · ·

where

d3 = −c3 = −1

40, d5 = 3c23 − c5 = −

3

22400, · · · .

(It is suggested by Prof. N. M. Temme and modified by me.)

Appendix D



The reflection formula of the modified Lommel polynomial hn,ν(x) is

hn,ν(−x) = (−1)nhn,ν(x).

Note that when n = 0, 1, 2, · · · ,

h2n,ν(0) = (−1)n and h2n+1,ν(0) = 0.

Proof. Recall the hypergeometric representation of hn,ν(x) is

hn,ν(x) = (ν)n(2x)n2F3(−n/2, (−n+ 1)/2; ν,−n, 1− ν − n;−1/x2).

hn,ν(−x) = (ν)n(−2x)n 2F3(−n/2, (−n+ 1)/2; ν,−n, 1− ν − n;−1/(−x)2)

= (−1)n(ν)n(2x)n 2F3(−n/2, (−n+ 1)/2; ν,−n, 1− ν − n;−1/x2)

= (−1)nhn,ν(x).

To evaluate hn,ν(x) at x = 0, we consider that h0,ν(x) = 1, h1,ν(x) = 2νx and

h2,ν(x) = 4(ν + 1)νx2 − 1.

Then h0,ν(x) = 1, h1,ν(0) = 0 and h2,ν(0) = −1. It is true when n = 0. Assume

h2k,ν(0) = (−1)k and h2k+1,ν(0) = 0 for some integers k. When n = k + 1, from the

three-term recurrence relation,

h2k+2,ν(x) = 2(2k + 1 + ν)xh2k+1,ν(x)− h2k,ν(x)

Appendix D Verification of Some Results in §4 137

and

h2k+3,ν(x) = 2(2k + 2 + ν) xh2k+2,ν(x)− h2k+1,ν(x),

then h2k+2,ν(0) = (−1)(−1)k = (−1)k+1 and h2k+3,ν(0) = 0. By induction, h2n,ν(0) =

(−1)n and h2n+1,ν(0) = 0 when n = 0, 1, 2, · · · .


Let the generating function of modified Lommel polynomials be F (x,w), and

F (x,w) :=∞∑n=0

hn,ν(x)wn

n!. (D.0.1)

Differentiating both sides (D.0.1) with respect to w, then we have

∂F (x,w)

∂w=

∂

∂w

∞∑n=0

hn,ν(x)wn

n!

=∞∑n=1

hn,ν(x)wn−1

(n− 1)!

=∞∑n=0

hn+1,ν(x)wn

n!

=h1,ν(x) +∞∑n=1

hn+1,ν(x)wn

n!

=h1,ν(x) +∞∑n=1

[2(n+ ν) xhn,ν(x)− hn−1,ν(x)]wn

n!

=h1,ν(x) + 2x∞∑n=1

(n+ ν)hn,ν(x)wn

n!−

∞∑n=0

hn,ν(x)wn+1

(n+ 1)!.

Differentiating both sides again with respect to w, then we have

∂2F (x,w)

∂w2= 2x

∞∑n=1

(n+ ν)hn,ν(x)wn−1

(n− 1)!−

∞∑n=0

hn,ν(x)wn

n!.

Since

∂F (x,w)

∂w=

∞∑n=1

hn,ν(x)wn−1

(n− 1)!,

then, multiplying both sides by w, we have

w∂F (x,w)

∂w=

∞∑n=1

hn,ν(x)wn

(n− 1)!.


Differentiating both sides with respect to w, then we have

∂F (x,w)

∂w+ w

∂2F (x,w)

∂w2=

∞∑n=1

hn,ν(x)nwn−1

(n− 1)!.

Hence,

∂2F (x,w)

∂w2= 2x

∞∑n=1

(n+ ν)hn,ν(x)wn−1

n− 1!−

∞∑n=0

hn,ν(x)wn

n!

= 2x∞∑n=1

nhn,ν(x)wn−1

(n− 1)!+ 2xν

∞∑n=1

hn,ν(x)wn−1

(n− 1)!−

∞∑n=0

hn,ν(x)wn

n!

= 2x

(∂F (x,w)

∂w+ w

∂2F (x,w)

∂w2

)+ 2xν

∂F (x,w)

∂w− F,

i.e.,

(1− 2xw)∂2F (x,w)

∂w2− 2x(1 + ν)

∂F (x,w)

∂w+ F (x,w) = 0.

Putting w = 0 in F (x,w) and Fw(x,w), respectively, we have

F (x, 0) = h0,ν(x) = 1

and

∂F (x, 0)

∂w= h1,ν(x) = 2νx.

On solving the above differential equation with initial conditions, we have the general

solution

F (x,w) = C1(1− 2xw)−ν2J−ν

(√1− 2xw

x

)+ C2(1− 2xw)−

ν2Y−ν

(√1− 2xw

x

).

From the initial conditions, we have

F (x,w)

=(1− 2xw)−ν2

J1−ν

(1x

)Y−ν

(√1−2xwx

)− J−ν

(√1−2xwx

)Y1−ν

(1x

)J1−ν

(1x

)Y−ν

(1x

)− J−ν

(1x

)Y1−ν

(1x

) .

Since the Wronskians of Jν(z) and Yν(z), denoted by W (Jν(z), Yν(z)), is given by

W (Jν(z), Yν(z)) = Jν+1(z)Yν(z)− Jν(z)Yν+1(z) =2

zπ.


Setting ν 7→ −ν, z 7→ 1x

in W (Jν(z), Yν(z)), we obtain

π

2x(1− 2xw)−

ν2

[J1−ν

(1

x

)Y−ν

(√1− 2xw

x

)− J−ν

(√1− 2xw

x

)Y1−ν

(1

x

)]=

∞∑n=0

hn,ν(x)wn

n!.


Since α1 = 2ν, t+ = 1, β0 = 0, β1 = 0, and

τ0 = −α1t+ + β1(2− β0)θ

,

then it implies that τ0 = ν.


Since α′0 = 2, β′

0 = 0 and recall from (1.2.7) and (1.2.8), then we have the character-

istic equation for Modified Lommel polynomials, i.e.,

λ2 − (α′0t+ β′

0)λ+ 1 = 0

⇒ λ2 − 2tλ+ 1 = 0

The two roots of this equation are

λ± = t±√t2 − 1.

Verification of (4.2.10).2

3[ζ(t)]3/2

= logα′0t+ β′

0 +√(α′

0t+ β′0)

2 − 4

2− α′

0t1/θ

∫ t

t+

s−1/θ√(α′

0s+ β′0)

2 − 4ds

= log (t+√t2 − 1)− 1

t

∫ t

1

s√s2 − 1

ds

= log (t+√t2 − 1)− 1

t

1

22(s2 − 1)

12 |t1

= log (t+√t2 − 1)− 1

t(t2 − 1)

12 .



2

3[−ζ(t)]3/2

= − cos−1 α′0t+ β′

0

2+ α′

0t1/θ

∫ t+

t

s−1/θ√4− (α′

0s+ β′0)

2ds

= t−1

∫ 1

t

s√1− s2

ds− cos−1 t

= t−1−122(1− s2)

12 |1t − cos−1 t

=1

t(1− t2)

12 − cos−1 t.


From equations (4.1.9) and (4.3.3), i.e.,

π−1(2x sinπν)hn,ν(x) = Jν+n(1/x)J−ν+1(1/x) + (−1)nJ−ν−n(1/x)Jν−1(1/x),

and

J−ν−n(1

x) = J−N(

1

x) = cos(Nπ)JN(1/x)− sin(Nπ)YN(1/x),

where n is an integer and ν not an even integer. Then we have

π−1(2x sinπν)hn,ν(x) = Jν+n(1/x)J−ν+1(1/x) + (−1)nJ−ν−n(1/x)Jν−1(1/x)

= Jν+n(1/x)J−ν+1(1/x)

+ (−1)n [cos(Nπ)JN(1/x)− sin(Nπ)YN(1/x)] Jν−1(1/x)

∼ 1√2πN

(e

2Nx)NJ−ν+1(1/x)

+ (−1)n cos((n+ ν)π)1√2πN

(e

2Nx)NJν−1(1/x)

− (−1)n sin((n+ ν)π)(−1)√

2

πN(e

2Nx)−NJν−1(1/x).


Hence, we have

hn,ν(x) ∼π

2x

1

sin πν

1√2πN

(e

2Nx)N [J−ν+1(1/x) + cos(νπ)Jν−1(1/x)]

+π

2x

1

sin πν

√2

πN(e

2Nx)−N sin(νπ)Jν−1(1/x)

∼ π

2x

1√2πN

(e

2Nx)NYν−1(1/x)

+π

2x

√2

πN(e

2Nx)−NJν−1(1/x).

Verification of the asymptotic of Pn(x) and Qn(x).

First, we consider the asymptotic behavior of the leading term Pn(N− 1

2 t) andQn(N− 1

2 t),

we have

Pn(N− 1

2 t) ∼(

ζ

t2 − 4

) 14

Ai(N23 ζ)N

16

=

(ζ

t2 − 4

) 14 1

2√π

1

(N23 ζ)

14

e−23(N

23 ζ)

32N

16

=

(ζ

t2 − 4

) 14 1

2√π

1

N16 ζ

14

e−23Nζ

32N

16

=1

(t2 − 4)14

1

2√πe−

23Nζ

32

=1

(t2 − 4)14

1

2√πe−N [log λ+− 1

t(t2−1)

12 ],

and

Qn(N− 1

2 t) ∼(

ζ

t2 − 4

) 14

Bi(N23 ζ)N

16

=

(ζ

t2 − 4

) 14 1√

π

1

(N23 ζ)

14

e23(N

23 ζ)

32N

16

=

(ζ

t2 − 4

) 14 1√

π

1

N16 ζ

14

e23Nζ

32N

16

=1

(t2 − 4)14

1√πe

23Nζ

32

=1

(t2 − 4)14

1√πeN [log λ+− 1

t(t2−1)

12 ].


Since x = N−1t and for t ≥ 1,

2

3[ζ(t)]3/2 = log (t+

√t2 − 1)− 1

t(t2 − 1)

12 ,

then2

3[ζ(xN)]3/2 = log (t+

√t2 − 1)− 1

t(t2 − 1)

12

= log (xN +√xN2 − 1)− 1

xN((xN)2 − 1)

12

∼ log (2xN)− 1.

Hence,

Pn(x) ∼1

(t2 − 4)14

1

2√πe−

23Nζ

32

=1

((xN)2 − 4)14

1

2√πe−

23Nζ

32

∼ 1

2√πxN

e−N(log (2xN)−1)

=1

2√πxN

1

(2xN)NeN

and

Qn(x) ∼1

(t2 − 4)14

1√πe

23Nζ

32

=1

((xN)2 − 4)14

1√πe

23Nζ

32

∼ 1√πxN

eN(log (2xN)−1)

=2N√π(xN)N− 1

2 e−N .


From (4.4.3), we have

hn,ν(t

N) =

π√2x

(ζ

t2 − 1

) 14[N

16

(Yν−1(

1

x)Ai(N

23 ζ) + Jν−1(

1

x)Bi(N

23 ζ)

)+ ϵ1 + δ1

].

Since t > 1, then N23 ζ > 0 and

δ1(N, t) ∼M1

N56

|Bi(N23 ζ)| = M1

N56

Bi(N23 ζ) ∼ M1

N56

1√π

1

N16

ζ−14 e

23Nζ

32

= ON−1 exp(2

3Nζ

32 ).

D.1 Power series solutions of (4.4.13) 143

If t is a zero of hn,ν(t/N), the left-hand side of the first equation is equal to zero, and

we have

N16Jν−1(

1

x)Bi(N

23 ζ) = ON−1 exp(

2

3Nζ

32 )

⇒ Jν−1(1

x) = O(

1

N).

D.1 Power series solutions of (4.4.13)

We can use more efficient inversion methods to obtain higher approximations of the

zeros. When n is even, put (4.4.13) in the form,

t+ c3t3 + c5t

5 + . . . = s, s =2

N(π

2+mπ +O(

1

N)),

where c3 = 1/12, c5 = 1/40, . . .. By Lagrange’s inversion formula, for fixed m =

0, 1, 2, · · · , the small zeros are given by

tn,m+1 = s+ d3s3 + d5s

5 + · · ·

where

d3 = −c3 = −1

12, d5 = 3c23 − c5 = −

1

240, · · · .

Note that, tn,1 = 0. Similarly, when n is odd, put (4.4.13) in the form

t+ c3t3 + c5t

5 + . . . = s, s =2

N(mπ +O(

1

N)),

where c3 = 1/12, c5 = 1/40, . . .. By Lagrange’s inversion formula, for fixed m =

0, 1, 2, · · · , the small zeros are given by tn,1 = 0 and

tn,m+1 = s+ d3s3 + d5s

5 + · · ·

where

d3 = −c3 = −1

12, d5 = 3c23 − c5 = −

1

240, · · · .

(It is suggested by Prof. N. M. Temme and modified by me.)

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