appell polynomials and their zero attractorsrboyer/papers/appell_poly.pdf · appell polynomials and...

Appell Polynomials and Their Zero Attractors

Robert P. Boyer and William M. Y. Goh

Abstract. A polynomial family {pn(x)} is Appell if it is given by ext

g(t)=

P

∞

n=0 pn(x)tn or, equivalently, p′n(x) = pn−1(x). If g(t) is an entire function,g(0) 6= 0, with at least one zero, the asymptotics of linearly scaled polynomials{pn(nx)} are described by means of finitely zeros of g, including those ofminimal modulus. As a consequence, we determine the limiting behavior oftheir zeros as well as their density. The techniques and results extend ourearlier work on Euler polynomials.

1. Introduction

Let g(t) be an entire function such that g(0) 6= 0.

Definition 1.1. The Appell polynomials {pn(x)} associated with generatingfunction g(t) are given by

(1.1)ext

g(t)=

∞∑

n=0

pn(x)tn.

Some important examples are: the Taylor polynomials of ex, with g(t) = 1− t;the Euler polynomials, with g(t) = (et + 1)/2; and the Bernoulli polynomials, withg(t) = (et − 1)/t; and their higher order analogues.

The asymptotics and limiting behavior of the zeros of these families have beeninvestigated by many people; for example, [2], [6], and so on.

In this paper, we obtained the asymptotics and the limiting behavior of the ze-ros for all Appell families provided the generating function g(t) satisfies one furthercondition: that g must have at least one zero. We use the ideas in our earlier paper[2]; furthermore, we simultaneously simplify and generalize some of the techniquesthere.

We found that the asymptotics in the general case are built from the basicexample g(t) = 1 − t which coincides with the classical work of Szego on theTaylor polynomials of the exponential function. In our paper [2], we found that theasymptotics for the Euler and the Bernoulli polynomials are controlled by certainroots of g(t), the ones of minimal modulus. In the general situation, as expected,the minimal modulus roots of g(t) are needed to describe the asymptotics but theremay be finitely many other roots needed to determine the asymptotics. These

1991 Mathematics Subject Classification. Primary 05C38, 15A15; Secondary 05A15, 15A180.Key words and phrases. Appell polynomials, zeros of polynomials, asymptotics.

1

2 ROBERT P. BOYER AND WILLIAM M. Y. GOH

additional roots are determined through a geometric condition described in termsof rotated and scaled versions of the Szego curve: |xe1−x| = 1, |x| ≤ 1, x ∈ C (seeFigure 3).

We frequently use the following notations. Let Z(g) denote the set of all zeros ofg and let r0 < r1 < r2 < . . . denote the distinct moduli of these zeros in increasingorder.

Recall that if K1 and K2 are two non-empty compact subsets of C, then theirHausdorff distance is the larger of sup{d(x,K1) : x ∈ K2} and sup{d(x,K2) : x ∈K1}.

Definition 1.2. For a family {qn(x)} of polynomials whose degrees are in-creasing to infinity, their zero attractor is the limit of their set of zeros Z(qn) inthe Hausdorff metric on the space of all non-empty compact subsets of the complexplane C.

In the appendix, we discuss how the zero attractor is found in terms of thelimsup and liminf of the zero sets.

’0.2-0.2

-0.1

0.15

0.1

-0.15

0.40

0.05

-0.4

-0.05

0

Figure 1. Zeros for degree 1000 polynomial, with generating func-tion g(t) = J0(t)

There is a related work on the asymptotics and zeros of the Taylor polynomialsfor linear combinations of exponentials

∑

cjeλjx where the parameters λj satisfy

a geometric constraint [1]. The techniques of proof are very different from ourapproach.

2. The Generalized Szego Approximations

It is convenient to collect together several results from [2] and some extensionsof them concerning the asymptotics of Sn(x) =

∑nk=0 x

k/k!. The domains of wheretheir asymptotics hold are critical in understanding the behavior for the Appellpolynomials.

Proposition 2.1. (Left-Half Plane) Let 1/3 < α < 1/2 and 1 ≤ j. Onany compact subset K of {w : ℜw < 1}, we have

APPELL POLYNOMIALS AND THEIR ZERO ATTRACTORS 3

(1)Sn−1(nw)

enw= 1 − (we1−w)n

√2πn(1 − w)

(

1 +O(n1−3α))

,

(2) Dj−1w (w−nSn−1(nw)) = Dj−1

w (w−nenw)− (j − 1)!√2πn

en

(1 − w)j

(

1 +O(n1−3α))

,

where the big O constant holds uniformly for x ∈ K and Dw is the usualdifferential operator.

The proof of part (1) is in [2]. Part (2) follows from an application of the saddlepoint method.

The following Proposition is also from [2]:

Proposition 2.2. (Outside Disk) Let S be a compact subset contained in|w| > 1 with distance δ > 0 from the unit circle, and let α be chosen so 1/3 < α <1/2. Then

Sn−1(nw)

enw=

(we1−w)n

√2πn(w − 1)

(

1 +O(n1−3α))

,

where the big O term holds uniformly for w ∈ S.

Proposition 2.3. (Evaluations of Integrals) If ǫ < |w| and j ≥ 1, thenwe have

(1)1

2πi

∮

|t|=ǫ

(

ext

t

)n1

t− wdt = −w−nSn−1(wxn).

(2)1

2πi

∮

|t|=ǫ

(

ext

t

)n1

(t− w)jdt =

−1

(j − 1)!Dj−1

w (w−nSn−1(wxn)),

where Dw is the differentiation operator ddw .

Proof. (1) By expanding 1/(t− z) into an infinite geometric series and per-forming a term-by-term integration, we obtain

1

2πi

∮

|t|=ε

(

ext

t

)n1

t− zdt =

−1

z2πi

∮

|t|=ε

(

ext

t

)n1

1 − tz

dt

=−1

z2πi

∮

|t|=ε

(

ext

t

)n

∑

m≥0

(

t

z

)m

dt.

By the Cauchy integral theorem the terms correspond to m ≥ n vanish. Hence

1

2πi

∮

|t|=ε

(

ext

t

)n1

t− zdt =

−1

z

∑

n−1≥m≥0

1

zm

(

1

2πi

∮

|t|=ε

extnt−n+m dt

)

=−1

z

∑

n−1≥m≥0

1

zm

(xn)n−m−1

(n−m− 1)!

=−1

zz−n+1

∑

n−1≥m≥0

(xnz)n−m−1

(n−m− 1)!

= −z−n∑

n−1≥j≥0

(xnz)j

j!= (−1)z−nSn−1(zxn).

Part (2) follows from differentiating (1) j − 1 times with respect to z. �


3. Asymptotics Outside the Disk D(0; 1/r0)

Theorem 3.1. Given the Appell family {pn(x)} with generating function g(t)we have

pn(nx)

(xe)n/√

2πn=

1

g(1/x)(1 +O(1/n))

uniformly for x ∈ K where K is a compact subset of the annulus A(1/r0;∞).

Proof. We shall find an asymptotic approximation to pn(nx) in the region

A(1/r0;∞) ={

x : |x| > 1r0

}

. Use the generating relation equation (1.1) to get

pn(x) =1

2πi

∮

|t|=ǫ

ext

g(t)tn+1dt,

where ǫ < r0. Since both sides of the above equation are entire functions of x, byanalytic continuation this representation for pn(x) is valid for all x ∈ C. Hence wecan replace x by nx to get

(3.1) pn(nx) =1

2πi

∮

|t|=ǫ

(

ext

t

)ndt

tg(t).

The above expression is valid for 0 < ǫ < r0 and is the starting point of the analysisin the sequel.

Let K be an arbitrary compact subset ⊆ {x : |x| > 1r0} and let x ∈ K. We

can certainly choose ǫ small enough so that for all x ∈ K, |ǫx| < 1. By a change ofvariables, we get

pn(nx) =xn

2πi

∮

|t|=ǫ|x|

(

et

t

)ndt

tg(t/x).

Observe that the zeros of g(t/x) have the form ax where a ∈ Z(g). Moreover,they must lie outside the closed unit disk since |x| > 1/r0, so we can deform theintegration path from the circle with radius ǫ |x| to the unit circumference. Thus

pn(nx) =xn

2πi

∮

|t|=1

(

et

t

)ndt

tg(t/x)

=xn

2πi

∮

|t|=1

en(t−ln t) dt

tg(t/x).

It can be easily seen that t = 1 is the saddle point of the integral and the classicalsaddle point method is applicable here [3]. Hence

pn(nx) =(ex)n

√2πng( 1

x)

(

1 +O(1

n)

)

,

where the implied O constant holds uniformly for x ∈ K. �

The last equation can be written as

pn(nx)

(ex)n/√

2πn=

1

g( 1x )

(

1 +O(1

n)

)

, |x| > 1/r0.

We have the:


Corollary 3.2. (1) On the complement of the disk D(0; 1/r0),

limn→∞

1

nln

∣

∣

∣

∣

pn(nx)

(ex)n/√

2πn

∣

∣

∣

∣

= 0

where the limit holds uniformly on compact subsets.(2) The zero attractor must be contained in the closed disk D(0; 1/r0).

Note that part (2) follows easily from (1) since g(t) never vanishes outside thedisk D(0; 1/r0).

4. Asymptotics on the Basic Regions Rℓ

Let r0, r1, . . . denote the distinct moduli of the zeros of the generating functiong for the Appell family {pn(x)}. Fix a positive integer ℓ. We fix a large ρ > 0 so itis not equal to any zero modulus {r0, r1, . . . }. For each zero a ∈ Z(g) with |a| = rℓ,we consider the circle |x| = 1/|a| and the disk D(1/a; δa).

Now the tangent line Ta to the circle |x| = 1/rℓ at the point 1/a determines thehalf-plane Ha, which contains 0; that is, ℜ(ax) < 1. We choose ǫℓ > 0 to be lessthan the distance from the portion of the tangent line Ta that lies outside the diskD(1/a; δa) to the circle |x| = 1/|a| for any |a| = rℓ; that is, ǫℓ <

√

1/r2ℓ + δ2a −1/rℓ.This has the effect that the circle |x| = 1/|a|+ ǫℓ never intersects the portion of thetangent line Ta outside the disk D(1/a; δa). Finally, we make the requirement thedisks D(1/a; δa) be mutually disjoint for all a ∈ Z(g) with |a| < ρ.

Definition 4.1. With these conventions, the region Rℓ is described in termsof the half-planes Ha and disks as

(4.1) Rℓ =⋂

{

Ha \D( 1a ; δa) : |a| = rℓ

}

∩D(0; 1/rℓ + ǫℓ} \D(0; 1rℓ+1

+ ǫℓ+1)

We note that the regions Rℓ are not disjoint; in fact, by construction, its innerboundary which consists of the portion of the circle |x| = 1

rℓ+1+ǫℓ+1 that lie outside

the disks D(1/a; δa), |a| = rℓ+1, actually lies inside the region Rℓ+1.Note the order of dependence: first we have the cut-off modulus ρ > 0 for the

moduli of the zeros; next, δa > 0 for each a ∈ Z(g) is given and is a function of ρ(described later in this section), then finally, ǫℓ is determined relative to each zeromoduli rℓ which is a function of δa.

It is convenient to introduce a region that contains all of the Rℓ’s:

Definition 4.2. Let Rρ be the domain given by(4.2)

Rρ =⋂

{Ha : a ∈ Z(g), |a| = r0} \[

⋃

{D(1/a; δa : a ∈ Z(g), |a| < ρ} ∪D(0; 1ρ)]

For any a ∈ Z(g) with r0 ≤ |a| < ρ, let sa(t) be the singular part of

1

tg(t)

at its pole a. Next let g1(t) be a normalized version of the generating function g(t)given as

(4.3) g1(t) =1

tg(t)−∑

{sa(t) : a ∈ Z(g), r0 ≤ |a| < ρ}

so g1(t) is analytic in the disk: |t| < ρ.


We develop the asymptotics for {pn(nx)} on the regions Rℓ where r0 ≤ rℓ < ρ.Now we saw already that we can write pn(nx) as

pn(nx) =1

2πi

∫

|t|=ǫ

(

ext

t

)n

g1(t) dt+1

2πi

∫

|t|=ǫ

(

ext

t

)n

s(t) dt,

where s(t) =∑{sa(t) : a ∈ Z(g), r0 ≤ |a| ≤ ρ}.

Lemma 4.3. With g1(t) given above in equation (4.3), we have

1

2πi

∫

|t|=ǫ

(

ext

t

)n

g1(t) dt =xn−1en

√2πn

g1(1/x) (1 +O(1/n))

uniformly on compact subsets of the annulus A(1/ρ;∞).

Proof. Let x ∈ K ⊂ A(1/ρ,∞). By a change of variables, we write

1

2πi

∫

|t|=ǫ

(

ext

t

)n

g1(t) dt =xn−1

2πi

∫

|t|=ǫ|x|

(

et

t

)n

g1(t/x) dt.

By construction, g1(t/x) is analytic on a disk of radius greater than 1. So thecontour in the last integral can be deformed to the unit circle |t| = 1 withoutchanging its value. Finally, by an application of the saddle point method we findthat

xn−1

2πi

∫

|t|=1

(

et

t

)n

g1(t/x) dt =xn−1en

√2πn

g1(1/x)(

1 +O( 1n ))

.

�

’

K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0

Figure 2. Generic Plot of Polynomial Zeros and Zero AttractorWhen g Has Two Roots; Tangent Lines and Circles Displayed

To state the next two lemmas, we need to introduce special polynomials In(z)in z−1 and J(a; z) in z.


The polynomial In(z) comes from expanding the derivative of Dm−1z (z−nenz).

Consider

Dm−1z (z−nenz) =

m−1∑

p=0

(

m− 1

p

)

(Dpzz

−n)(Dm−1−pz enz)

=

m−1∑

p=0

(

m− 1

p

)

(−n)(−n− 1) · · · (−n− p+ 1)z−n−p(nm−1−penz)

= z−nenznm−1m−1∑

p=0

(

m− 1

p

)

(−n)(−n− 1) · · · (−n− p+ 1)(nz)−p

= z−nenznm−1m−1∑

p=0

(−1)pp!

(

m− 1

p

)(

n+ p− 1

p

)

(nz)−p

= z−nenznm−1Im−1(nz),

where Im−1(z) is given in

Definition 4.4.

(4.4) Im−1(z) =

m−1∑

p=0

(−1)pp!

(

m− 1

p

)(

n+ p− 1

p

)

z−p.

For a ∈ Z(g), we define J(a; z) which are also polynomials in z. We write outthe singular part sa(t) of the function 1

tg(t) at its nonzero pole a by

(4.5) sa(t) :=

βa∑

m=1

ba,m

(t− a)m,

where βa is the order of a as a zero of g(t) so ba,βa6= 0.

Definition 4.5. For a ∈ Z(g), let J(a; z) be the polynomial in z given as

(4.6) J(a; z) =

βa∑

m=1

ba,m

(m− 1)!zm−1Im−1(az).

Lemma 4.6. Let a ∈ Z(g) and let x ∈ K, a compact subset of the half-planeHa, ℜ(ax) < 1. Then

1

2πi

∫

|t|=ǫ

(

ext

t

)n

sa(t) dt = −a−nenaxJ(a;nx) +enxn−1

√2πn

sa(1/x)(

1 +O(n1−3α))

where sa(t) is the singular part of 1/(tg(t)) at the zero a of g(t).

Proof. We first write out the integral in terms of the singular part sa(t)

1

2πi

∫

|t|=ǫ

(

ext

t

)n

sa(t) dt = −βa∑

m=1

ba,m

(m− 1)!Dm−1

a

(

a−nSn−1(nax))

where the coefficients ba,m are given in equation (4.5). We now study the asymp-totics of the typical term Dm−1

a (a−nSn−1(nax)).


We may use the generalized half-plane Szego asymptotics with 13 < α < 1

2because of the restriction that a ∈ Z(g) with ℜ(ax) < 1 to obtain

Dm−1a (a−nSn−1(nax)) = xn+m−1Dm−1

ax ((ax)−nSn−1(nax))

= xn+m−1

{

Dm−1z (z−nenz)

∣

∣

z=ax− (m− 1)!√

2πn

en

(1 − ax)m

(

1 +O(n1−3α))

}

.

Combining these estimates we obtain

Dm−1a (a−nSn−1(nax)) = xn+m−1{(ax)−nenaxnm−1Im−1(nax)

− (m− 1)!√2πn

en

(1 − ax)m

(

1 +O(n1−3α))

}

= a−nenax(nx)m−1Im−1(nax) −(m− 1)!√

2πn

enxn+m−1

(1 − ax)m

(

1 + O(n1−3α))

.(4.7)

Hence after summation we obtain

1

2πi

∮

|t|=ǫ

(

ext

t

)n


m=1

ba,m

(m− 1)!Dm−1

a (a−nSn−1(nax))

= −a−nenaxJ(a;nx) +enxn−1

√2πn

sa(1

x)(

1 +O(n1−3α))

.(4.8)

�

Corollary 4.7. For a ∈ Z(g), |a| ≤ rℓ, we have

1

2πi

∫

|t|=ǫ

(

ext

t

)n

sa(t) dt = −a−nenaxJ(a;nx) +enxn−1

√2πn

sa(1/x)(

1 +O(n1−3α))

uniformly on the compact subsets of Rℓ, where sa(t) is the singular part of 1/(tg(t))at the zero a of g(t).

Proof. By the definition of Rℓ, when x ∈ Rℓ and |a| = rℓ, we have ℜ(ax) <1− c(δ). When |a| < rℓ, we have |xa| < 1− c(δ). So in both cases, the asymptoticsstated in Proposition 2.1 applies. �

Lemma 4.8. Let a ∈ Z(g) and let x ∈ K, where K compact subset of thedisk-complement A(1/|a|;∞). Then

1

2πi

∫

|t|=ǫ

(

ext

t

)n

sa(t) dt =enxn−1

√2πn

sa(1/x)(

1 +O(n1−3α))

.

Proof. We will use the disk-complement generalized Szego asymptotics. Forz in the annulus A(1 + c,∞), for any c > 0, we have

Sn−1(nz) = − zn

2πi

∮

|ζ|=1

en(ζ−ln ζ)

ζ − zdζ

By Dividing zn and taking derivatives up to order m− 1, we get

Dm−1z (z−nSn−1(nz)) = − (m− 1)!

2πi

∮

|ζ|=1

en(ζ−ln ζ)

(ζ − z)mdζ

= − (m− 1)!√2πn

en

(1 − z)m

(

1 +O(n1−3α))

.


In the above, replace z by ax to obtain

Dm−1a (a−nSn−1(nax)) = xn+m−1Dm−1

ax ((ax)−nSn−1(nax))

= − (m− 1)!√2πn

enxn+m−1

(1 − ax)m

(

1 +O(n1−3α))

.(4.9)

By summation, we obtain the asymptotics for the original integral:

1

2πi

∮

|t|=ǫ

(

ext

t

)n


m=1

ba,m

(m− 1)!Dm−1

a (a−nSn−1(nx))

=enxn−1

√2πn

sa(1

x)(

1 +O(n1−3α))

.(4.10)

�

Corollary 4.9. For a ∈ Z(g) with rℓ+1 ≤ |a| < ρ, we have

1

2πi

∫

|t|=ǫ

(

ext

t

)n

sa(t) dt =enxn−1

√2πn

sa(1/x)(

1 +O(n1−3α))

,

uniformly on the compact subsets of Rℓ.

Proof. When x ∈ Rℓ, we have rℓ+1 ≤ |a| < ρ. By definition of Rℓ, we have|xa| ≥ 1 + c(δ). Hence the asymptotics in Proposition 2.2 applies. �

The remaining case for the above integration involving sa(t) on the diskD(1/a; δ)will be handled in a later section.

Proposition 4.10. For x ∈ Rℓ, we have

pn(nx)

(xe)n/√

2πn=

1

x

1

g1(1/x)(1 +O(1/n))

−√

2πn∑

{

J(a;nx)1

φ(ax)n: a ∈ Z(g), |a| ≤ rℓ

}

+∑

{

1

xsa

(

1

x

)

: a ∈ Z(g), |a| < ρ

}

(1 +O(n1−3α))

uniformly on the compact subsets of Rℓ, where φ(x) = xe1−x and 1/3 < α < 1/2.

Proof. Putting the last two corollaries into equation (3.1) and using Lemma4.3 to simplify, we have the result. �

Proposition 4.11. For x ∈ Rℓ, we have

pn(nx)

(ex)n/√

2πn=

1

g(1/x)−√

2πn∑

{J(a;nx)

φn(ax): a ∈ Z(g), |a| ≤ rℓ} +O(n1−3α)

uniformly on the compact subsets of Rℓ, where 1/3 < α < 1/2.

Proof. By the definition of normalized version of the generating function g1(t)(see equation (4.3)), we see that

(4.11)1

xg1

(

1

x

)

=1

g(1/x)−∑

{

1

xsa(

1

x) : a ∈ Z(g), |a| < ρ

}

.


We insert this into the expression for pn(nx)

(ex)n/√

2πnin Proposition 4.11. Since the

sa(1/x) term cancels, we have uniformly for x ∈ Rℓ:

pn(nx)

(ex)n/√

2πn=

1

g(1/x)(4.12)

−√

2πn∑

{

(axe1−ax)−nJ(a;nx) : a ∈ Z(g), |a| < rℓ}

+O(

n1−3α)

.

�

Lemma 4.12. If a ∈ Z(g) with |a| < ρ and x ∈ Rℓ, then

J(a;nx) =ba,βa

(βa − 1)!(nx)βa−1

(

ax− 1

ax

)βa−1

(1 + o(1)).

Proof. Recall that

J(a;nx) =

βa∑

m=1

ba,mIm−1(nax)

(m− 1)!, Im−1(nax) =

m−1∑

p=0

(−1)p

(

m− 1

p

)(

n+ p− 1

p

)

p! (nax)−p.

It is easy to see that(

n+ p− 1

p

)

(nax)−p =(ax)−p

p!(1 + o(1)) ,

that is, as n→ ∞

Im−1(nax) →m−1∑

p=0

(−1)p

(

m− 1

p

)

(ax)−p =

(

ax− 1

ax

)m−1

.

Hence

(4.13) J(a;nx) =ba,βa

(βa − 1)!(nx)βa−1

(

ax− 1

ax

)βa−1

(1 + o(1)).

Since the coefficient ba,βain the definition of the singular part sa(t) is nonzero, we

find for fixed x that the precise order of J(a;nx) as a polynomial in n is nβa−1. �

We note the following

Corollary 4.13. limn→∞

pn(nx)

(xe)n/√

2πn=

1

g(1/x), x ∈ Rℓ provided |φ(ax)| > 1

for all a ∈ Z(g) with |a| ≤ rℓ.

We can summarize this section in the following

Theorem 4.14. On Rρ, we have the following uniform asymptotics

pn(nx)

(ex)n/√

2πn=

1

g(1/x)(1 +O(1/n))

−√

2πn∑

{φ(ax)−nJ(a;nx) : a ∈ Z(g), |a| < ρ} +O(n1−3α)

where 1/3 < α < 1/2.

It remains to develop the asymptotics in the disks D(1/a; δa) and well as de-termining domination among a ∈ Z(g) of |φ(ax)|.


5. Geometry of Szego curves

Recall that φ(x) = xe1−x is an entire function conformal on the open unit disk.The standard Szego curve S is the portion of the level curve |φ(x)| = 1 that liesinside the closed unit disk or equivalently, inside the closed left-hand plane ℜ(x) ≤ 1.

S is a simple closed convex curve; in fact, it has the form t = ±√e2(s−1) − s2 where

x = s + it and s ∈ [−W (e−1), 1] and W is the principal branch of the LambertW−function.

Definition 5.1. Let a be a nonzero complex number. We call any curve ofthe form 1

aS a Szego curve.

xK0.5 0 0.5 1.0

K0.4

K0.3

K0.2

K0.1

0.1

0.2

0.3

0.4

Figure 3. Szego Curve: |ze1−z| = 1 and |z| ≤ 1

Remark 5.2. Note that the full curve |φ(x)| = 1 divides the complex x-planeinto three domains, one bounded (the interior of the standard Szego curve S) andtwo unbounded. The inequality

∣

∣φ−1(x)∣

∣ > 1consists of two domains: the interior ofS and the unbounded domain that contains the real axis where x > 1. Furthermore,the deleted circumference {x : |x| = 1, x 6= 1} lies in the domain where

∣

∣φ−1(x)∣

∣ < 1.

For brevity denote the interior of 1aS by Ga so the interior of S is denoted by

G1. Of course, if x ∈ G1, then∣

∣φ−1(x)∣

∣ > 1. Let

G+1 :=

{

x : x /∈ G1,∣

∣φ−1(x)∣

∣ > 1}

, G−1 :=

{

x : x /∈ G1,∣

∣φ−1(x)∣

∣ < 1}

G+1 is the unbounded domain that contains the real axis where x > 1 while G−

1 isthe remaining domain where

∣

∣φ−1(x)∣

∣ < 1. Since the difference between G1 and

a typical Ga is a matter of rotation and stretching, the domains G+a and G−

a aresimilarly defined. In terms of these notations, the above remarks can be equivalentlyphrased as

(5.1) {x : |x| ≤ 1} \G1 ⊂ G−1 .

In general, if |a| > 0, then

(5.2)

{

x : |x| ≤ 1

|a|

}

\Ga ⊂ G−a .

Lemma 5.3. The image of S \ {x = 1} under the inversion map x → 1x lies

G−1 .


Proof. We saw that the level curve |φ(x)| = 1 divides the complex plane intothree connected components whose boundaries are described in terms of

f(t) =√

e2(t−1) − t2, t ≥ −W (e−1) ≃ −0.2784645428.

For example, S is given by the two graphs of ±f(t), with t ∈ [−W (e−1), 1]. Wewant to show that the inverted Szego curve lies outside the standard Szego curve Sin the half plane ℜ(x) < 1 and either above or below the the graph of ±f(t) whent > 1.

For convenience, let G denote the two domains Ga and G+a where |φ(x)| < 1.

Since S is symmetric about the real axis, it is enough to show that the portion ofS with positive real part lies inside G under the map T : w 7→ 1/w.

Now S \ {1} lies inside the open unit disk. So the portion of the image of Sthat lies outside the unit disk with real part < 1 will lie inside the desired set G.

Given the point p(t) = (t, f(t)) on the upper portion of the Szego curve, itsimage under T is given as

(

t

t2 + f(t)2,

f(t)

t2 + f(t)2

)

=(

te−2(t−1), e−2(t−1)f(t))

, −W (e−1) ≤ t ≤ 1

since t2 + f(t)2 = e2(t−1). Now ℜ(T (p(t))) = te−2(t−1) < 1 if t < −W (−2e−2)/2 ≃0.2031878700 and the modulus of T (p(t)) is

√

t2 + f(t)2/(t2+f(t)2) = 1/√

t2 + f(t)2

which reduces to 1/√e2(t−1) = e−(t−1) > 1 for t < 1. This shows that T (p(t)) lies

inside the region G provided t < −W (−2e−2)/2.It remains to examine the location of T (p(t)) for −W (−2e−2)/2 ≤ t < 1. Of

course, for such points, we know that their real part is greater than 1. Now thefunction t/(t2 + f(t)2) = te−2(t−1) is increasing on the interval [−W (−2e−2)/2, 1/2]and decreasing on [1/2, 1] and is ≥ 1 on both intervals. It will be enough to showthe following inequality:

f(t)

t2 + f(t)2= e−2(t−1)f(t) > f

(

t

t2 + f(t)2

)

= f(te−2(t−1))

which is straightforward to verify. �

Lemma 5.4. Let a, b be two distinct non-zero complex numbers. Then the in-tersection 1

aS ∩ 1bS has at most two points.

Proof. The intersection of the two curves 1aS ∩ 1

bS must satisfy |φ(ax)| =|φ(bx)|. This modulus condition determines a line; in fact, It is easy to give anexplicit form for this line. Write x = s+it and b−a = α+iβ. Then |φ(ax)| = |φ(bx)|reduces to the line: |ae−ax| =

∣

∣be−bx∣

∣ ·∣

∣e(b−a)x∣

∣ = |b/a|; that is, ℜ((b − a)x) =ln |b/a| · αs− βt = ln |b/a|. Since the Szego curves are both convex, the number ofintersection points is bounded above by 2. �

We need to determine exactly the size of this intersection.

Lemma 5.5. Choose |a| > 1 so that 1/a lies on the Szego curve; that is,|φ(1/a)| = 1. Then the equation |φ(ax)| = |φ(x)| has a unique solution: 1/a.In this case, 1

aS is properly contained inside S except at the point 1a . Conversely,

if |a| > 1 and 1aS ∩S consists of just one point, then this common point must be 1

a .


Proof. We use the form of the equation for |φ(ax)| = |φ(x)| from the aboveproof for Lemma 5.4 where we set b = 1. The slope of this line is α/β. Recall that

the upper portion of S is the graph of y =√e2(x−1) − x2 with derivative

y′ =e2(x−1) − x

y=x2 + y2 − x

y

We set 1/a = x0 + iy0 where y0 =√

e2(x0−1) − x20 so 1/a lies on S. Write a as

a =x0

x20 + y2

0

− y0x2

0 + y20

i.

The slope of the line |φ(ax)| = |φ(x)| is

α

β=

1 − x0/(x20 + y2

0)

y0/(x20 + y2

0)=x2

0 + y20 − x0

y0

Hence the slope of the tangent line at 1/a agrees with the slope of the line |φ(ax)| =|φ(x)|. Since S is convex, there is just one intersection point with the tangentline. �

The following two corollaries are immediate consequences of this lemma:

Corollary 5.6. Let |a| > |b| > 0. Assume 1a /∈ Gb, the closure of Gb, then

∣

∣

∣

∣

1

aS ∩ 1

bS∣

∣

∣

∣

= 2.

Corollary 5.7. Let |a| > |b| > 0. If 1a ∈ Gb, then 1

aS is properly containedin Gb.

We now introduce the definition of dominant zero of the function g. Since g isan entire function, the zeros of g can be quite general. In fact, any discrete pointset with a possible limit point at infinity is qualified as the zero set of g.

Definition 5.8. Let a ∈ Z(g). The definition of dominant zero is inductiveon the magnitude of a. First every zero α with |α| = r0 is dominant. Secondly, azero α with |α| = r1 is dominant if

1

α/∈⋃

{

Ga : |a| = r0}

.

A zero α, with |α| = r2, is dominant if

1

α/∈⋃

{

Ga : a dominant , |a| ≤ r1}

This procedure is carried out inductively.

Let W denote the principal value of the Lambert W -function.

Lemma 5.9. If a′ ∈ Z(g) such that |a′| > r0/W (e−1), then a′ must be a non-dominant zero. Hence there are at most finitely many dominant zeros.

Proof. The proof follows from the fact that the radius of the largest opencircular disk centered 0 that lies in the interior of the standard Szego curve S isW (e−1). �

Lemma 5.10. Let a and b be two dominant zeros of g. Then 1aS ∩ 1

bS consistsof two points.


Proof. If |a| = |b|, then equation of the line of intersection is reduced to the

line argx = − arg a+arg b2 . It is easy to verify that, indeed, we have exactly two

points of intersection. When |a| > |b|, by definition

1

a/∈⋃

{

Gβ : β, β dominant, |β| < |a|}

Since Gb ⊂ ⋃ {Gβ : β, β dominant, |β| < |a|}, we have 1a /∈ Gb. Again by

Lemma 5.6 we get the result. The case where |a| < |b| is proved similarly. �

Definition 5.11. If a and b are two dominant zeros of g, then by Lemma 5.10the intersection line La,b |φ(ax)| = |φ(bx)| always exists. Of the two half planesthis line determines, let Ea+,b denote the one that contains 1

a .

Lemma 5.12. For two dominant zeros a 6= b of the generating function g(t),we have

Ea+,b ={

x :∣

∣φ−1(ax)∣

∣ >∣

∣φ−1(bx)∣

∣

}

Proof. Let x = 1a in the inequality to get 1 >

∣

∣φ−1(b/a)∣

∣. It is equivalent to

showing that 1 >∣

∣φ−1(b/a)∣

∣ is true for all distinct dominant zeros a and b. Wedivide the situation into three cases:

Case 1: |a| = |b|. In this case we have |b/a| = 1 and b/a 6= 1, by Remark 5.2the number b/a lies in G−

1 . Hence we have∣

∣φ−1(b/a)∣

∣ < 1.

Case 2: |a| > |b|. By definition of dominant zero we have 1a /∈ Gb. Since

∣

∣

1a

∣

∣ <∣

∣

1b

∣

∣, by equation (5.1) we see that 1a lies G−

b . Hence∣

∣φ−1(b/a)∣

∣ < 1.

Case 3: |a| < |b|. To see∣

∣φ−1(b/a)∣

∣ < 1, we invoke Lemma 5.3 to get 1a ∈ G−

b .

Hence∣

∣φ−1(b/a)∣

∣ < 1. �

We introduce two key domains needed to describe the Appell polynomial asymp-totics.

Definition 5.13. Let D0 be the domain given as

D0 :=⋃

{Ga : a dominant zero of g}so D0 is a domain that contains 0.

Definition 5.14. For a dominant zero a, let

Da := Ga ∩⋂

{

Ea+,b : b dominant , b 6= a}

Note that by Lemma 5.10 for all dominant b 6= a, Ea+,b is a non-empty domain

that contains 1a . Hence

⋂{

Ea+,b : b, b dominant , b 6= a}

is a domain containing 1a .

Therefore, Da is a non-empty domain.

Lemma 5.15. (1) Let a′ be a non-dominant zero of the generating function g.We have

1

a′∈⋃

{

Ga : a, |a| < |a′| , a dominant}

(2) For all zeros a of g, 1a ∈ D0.

(3) For all dominant zeros a of g, we have Da ⊂ Ga ⊂ D0.

(4)⋃

{Da : a dominant} ⊂ D0

(5) For all dominant zero a, we have

Da ={

x : x ∈ Ga,∣

∣φ−1(ax)∣

∣ >∣

∣φ−1(bx)∣

∣ for all dominant b 6= a}


(6) Let a and b are two distinct dominant zeros of g. We have

Da ∩Db = ∅.Proof. The proof of these statements follows mostly from definitions. We do

not prove all of them.Part (1) follows directly from definition. For (2), note that if a is dominant,

then, of course, we have 1a ∈ Ga and for b dominant, 1

a ∈ Ea+,b. Hence by Definition

5.14, 1a ∈ Da. If a is non-dominant, by (1) and Definition 5.13 we still have 1

a ∈ Da.Hence (2) follows.

Part (5) follows from Definition 5.14 and Lemma 5.12. For (6), assume x0 ∈Da ∩ Db. Since x0 ∈ Da, by (5) we have

∣

∣φ−1(ax0)∣

∣ >∣

∣φ−1(bx0)∣

∣. Similarly, we

have∣

∣φ−1(ax0)∣

∣ <∣

∣φ−1(bx0)∣

∣. A contradiction thus arises. Hence (6) follows. �

Lemma 5.16.⋃

{Da : a dominant zero of g(t)} ⊂ D0.

Proof. We prove a claim first.Claim: If x0 ∈ D0 and x0 /∈

⋃ {La,b : a, b dominant zeros, a 6= b}, then x0 ∈ Dα

for some dominant α. For the notation for the line segment La,b, see Definition 5.11.

Proof. Since x0 /∈ ⋃ {La,b : a, b dominant zero}, the set {∣

∣φ−1(ax0)∣

∣ : a is

dominant} consists of distinct numbers. Let∣

∣φ−1(αx0)∣

∣ be the unique maximum

of the set. Hence for all dominant b, b 6= α, we have∣

∣φ−1(αx0)∣

∣ >∣

∣φ−1(bx0)∣

∣.Next, since x0 ∈ D0, by Definition 5.13 there exists a dominant zero β such that

x0 ∈ Gβ . So∣

∣φ−1(αx0)∣

∣ >∣

∣φ−1(βx0)∣

∣ > 1 and |x0| <∣

∣

∣

1β

∣

∣

∣. First, it is easy to see

that x0 /∈ 1αS, the boundary of Gα (otherwise it would contradict to

∣

∣φ−1(αx0)∣

∣ > 1).

Assume that x0 /∈ Gα.

Case 1:∣

∣

1α

∣

∣ ≥∣

∣

∣

1β

∣

∣

∣. Since |x0| <∣

∣

∣

1β

∣

∣

∣, then∣

∣

1α

∣

∣ > |x0|. By equation (5.2)

x0 ∈ G−α . Therefore,

∣

∣φ−1(αx0)∣

∣ < 1. This contradicts to∣

∣φ−1(αx0)∣

∣ > 1.

Case 2:∣

∣

1α

∣

∣ <∣

∣

∣

1β

∣

∣

∣. Now both α and β are dominant. We have 1α /∈ Gβ . By

Lemma 5.3, Gβ \Gα ⊂ G−α . Because x0 ∈ Gβ \Gα, we have x0 ∈ G−

α , which implies∣

∣φ−1(αx0)∣

∣ < 1. This is still a contradiction. Thus x0 ∈ Gα. By (5) x0 ∈ Dα. �

For the proof of the lemma, we note that the set D0 \ ⋃{

Da : a dominant}

is an open set which we will assume is non-empty. Then there exists a disk ∆ ⊂D0 \ ⋃

{

Da : a dominant}

. Observe that ∆ \ ⋃ {La,b : a, b dominant } is neverempty. Thus exists x0 ∈ ∆\⋃ {La,b : a, b dominant }. By the above claim x0 ∈ Dβ

for some dominant β. This is a contradiction since x0 /∈{

Da : a dominant}

. �

According to Lemma 5.16, the general picture for Da is now clear. Roughly,the set {Dα : α is a dominant zero of g} partitions D0 so that the borders betweentwo adjacent D′

as are segments of the lines La,b.

Lemma 5.17. Uniformly on the compact subsets of D(0; 1r0

) \D0, we have

(5.3) limn→∞

pn(nx)

(ex)n/√

2πn=

1

g( 1x ).


Proof. Let K be a compact subset of D(0; 1r0

) \ D0. By part 2 of Lemma5.15, K contains no zeros of g. Therefore, we can choose δ small enough so thatK does not intersect any disk D( 1

a ; δ), where a ∈ Z(g). Recall the definition of theset Rℓ, ℓ ≥ 0. Let

Kℓ := K ∩Rℓ.

Note that by definition of Rℓ we know, for all large ℓ, Kℓ = ∅. Since⋃

l≥0Rℓ ⊃ K,we have

⋃

ℓ≥0

Kℓ = K.

There are at most finitely many Kℓ in the above union. Consider a typical Kℓ. Letx ∈ Kℓ, so x also lies in Rℓ. By the way Rℓ is defined and a variant of equation(5.1) x lies in G−

a for all a ∈ Z(g), with |a| ≤ Rℓ, we get∣

∣φ−1(ax)∣

∣ < 1. Now weinvoke Proposition 4.11 to obtain

pn(nx)

(ex)n/√

2πn=

1

g( 1x )

+O(n1−3α).

Note in the above equation, the exponentially small terms corresponding to J(a;nx)φn(ax)

are absorbed in O(n1−3α). Hence limn→∞pn(nx)

(ex)n/√

2πn= 1

g( 1x)

for x ∈ Kℓ. Since the

number of Kℓ is finite, proof of the lemma follows. �

We close this section with a strengthening of Theorem :

Theorem 5.18. Let ρ be chosen greater than 1/|a| where a is any dominantzero of the generating function g(t). Then on Rρ, we have the following uniformasymptotics

pn(nx)

(ex)n/√

2πn=

1

g(1/x)(1 +O(1/n))

−√

2πn∑

{

φ(ax)−nJ(a;nx) : a ∈ Z(g) and dominant}

+O(

n1−3α)

+ o(Φ(x)),

where 1/3 < α < 1/2 and Φ(x) = max{|φ(ax)|−1 : a ∈ Z(g) and dominant }.

6. Asymptotics for Other Domains

6.1. Asymptotics Inside the Disk D(1/a′; δ) Where a′ Is a Non-DominantZero.

Proposition 6.1. Let a′ ∈ Z(g). Then on the disk D(1/a′; δa′), the normalizedAppell polynomials have the asymptotics

pn(nx)

(ex)n/√

2πn=

(

1

g(1/x)− 1

xsa′(

1

x)

) (

1 +O(1

n)

)

−√

2πn∑

{

J(a;nx)

φn(ax): a ∈ Z(g), |a| ≤ |a′| , a 6= a′

}

−√

2πn

(ex)nσa′(x) +O(n1−3α),

where

σa′(x) =

βa′∑

m=1

ba′,m

(m− 1)!Dm−1

a′ ((a′)−nSn−1(na′x)).

Proof. The proof is very similar to that of Proposition 4.11. We shall notrepeat it here. �


This proposition shows that we still need to estimate√

2πn(ex)n σa′(x). Since x ∈

D(1/a′; δ), the approximations in Propositions 1 and 2 do not work. We handlethis in the following proposition.

Proposition 6.2. Let a′ be a non-dominant zero of g with |a′| < ρ. Thenthere exists a choice of δ such that

√2πn

(ex)nσa′(x) = O(e6nδρ)

Proof. To estimate σa, we make use of the elementary estimate: If f(z) isanalytic function of z, then for any ǫ > 0, we have

∣

∣Dj−1z f(z)

∣

∣ ≤ (j − 1)!

ǫj−1max

|ζ−z|=ǫ|f(ζ)| .

By the definition of σa′(x), we find

|σa′ | ≤

∣

∣

∣

∣

∣

∣

βa′∑

m=1

ba′,m

(m− 1)!Dm−1

a′ (|a′|−nSn−1(na′x))

∣

∣

∣

∣

∣

∣

≤βa′∑

m=1

|ba′,m|δm−1a′

max|ζ−a′|=δa′

∣

∣ζ−nSn−1(nζx)∣

∣

≤ Kδa′max

|ζ−a′|=δa′

(|ζ|−n Sn−1(|ζx|n))

where Kδa′> 0 is a constant that depends on the zero a′ and the radius δa′ .

To go further we observe for x ∈ D( 1a , δa′) and |ζ − a′| = δa′ :

|ζx| ≤ (|a′| + δa′) |x| ≤ |a′| |x| + |x| δa′

≤ 1 + |a′| δa′ + |x| δa′ = 1 + δa′(|a′| + |x|).Since |a′| < ρ by assumption, |ζx| ≤ 1+2ρδa′. But |ζ| ≥ |a′|−δa′ and |x| ≥ 1

|a′|−δa′ ,so we get

|ζx| ≥ (|a′| − δa′)

(

1

|a′| − δa′

)

≥ 1 − δa′

(

1

|a′| + |a′|)

≥ 1 − 2δa′ρ.

Collecting these two inequalities, we get

1 − 2δa′ρ ≤ |ζx| ≤ 1 + 2δa′ρ.

Now use that |Sn−1(nt)| ≤ ent:


|eζx|−nSn−1(|ζx|n) ≤ e−n |1 − 2δa′ρ|−n

en(1+2δa′ρ)

= |1 − 2δa′ρ|−ne2nδa′ρ

For 0 ≤ x ≤ 1/2, 1/(1 − x) ≤ e2x; if we choose δa′ such that 2δa′ρ ≤ 1/2, then we

have |1 − 2δa′ρ|−n ≤ e4δa′ρ. With this choice of δ, we obtain the desired bound


(|eζx|−n Sn−1(|ζx|n)) ≤ e4δa′ρe2nδa′ρ = e6nδa′ρ

�


6.2. Asymptotics Inside the Domain Dβ Where β Is a Dominant Zero.Let β1, β2, · · · , βk be the dominant zeros of g. For each βi, 1 ≤ i ≤ k define set Ai

as

Ai :=

{

1

α: α ∈ Z(g),

1

α∈ Dβi

, |a| < ρ

}

So Ai consists of reciprocals of zeros that fall intoDβi. Finally let the remaining

part of reciprocals be denoted by B, namely,

B :=

1

α: α ∈ Z(g), α non-dominant,

1

α/∈

k⋃

j=1

Aj , |a| < ρ

The set B consists of the reciprocals of those zeros lying on the border linesamong {Dβj

}. Note that each Ai ∪B is a finite set. If 1α ∈ Ai ∪B, then α is non-

dominant. We now investigate the asymptotics of pn(nx)

(ex)n/√

2πnfor x ∈ Dβi

, 1 ≤ i ≤ k.

We remind the readers that there could be many zeros a of g such that 1a ∈ Dβi

.This fact prevents the situation given in equation (5.3) from occuring. We need alemma for estimation.

Lemma 6.3. If 0 < |a| < |b| and 1b ∈ Ga, then for all x ∈ Gb, we have

∣

∣φ−1(bx)∣

∣ ≤∣

∣φ−1(ax)∣

∣ .

For x ∈ Gb, we have∣

∣φ−1(bx)∣

∣ <∣

∣φ−1(ax)∣

∣ .

Proof. Apply Lemma 5.7 and maximum modulus principle to the harmonicfunction ln

∣

∣φ−1(ax)∣

∣− ln∣

∣φ−1(bx)∣

∣ for x ∈ 1bS. �

We note that this result can be sharpen as: 1b ∈ Ga, then there exists δ > 0

such that for all x ∈ Gb, we have

eδ∣

∣φ−1(bx)∣

∣ ≤∣

∣φ−1(ax)∣

∣

Proposition 6.4. For x ∈ Dβi\D(0; 1

ρ), there exists δ = δ(ρ) > 0 such that

pn(nx)

(ex)n/√

2πn= −

√2πn

J(βi, nx)

φn(βix)+ o(φ−n(βix)) +O(e6nρδ).

Note that when x lies in a compact subset of Dβi\D(0; 1

ρ), the term O(e6nρδ)

can be absorbed in o(φ−n(βix)).

Proof. Let K be a compact subset of Dβi\D(0, 1

ρ). We choose δ small enough

so that for all 1a ∈ B \D(0; 1

ρ ) we have D( 1a ; δ) ∩K = φ. Let

rℓ = |βi| .Note that Rℓ−1∩Dβi

= φ. The first Rj which possibly has a non-empty overlapwith K is Rℓ. Hence we define Kj := Rl+j ∩K, j ≥ 0. Note that

⋃

j≥0

Kj ⊂ K.

The left-hand side of the above is a finite union and equality of sets does nothold in general. What is missing in

⋃

j≥0Kj is that many small disks centered


at some 1a where a ∈ Z(g) are not included in

⋃

j≥0Kj. To see the pattern ofestimation, we apply Proposition 4.11 to K0 which is Rℓ. Thus for x ∈ K0 we have

pn(nx)

(ex)n/√

2πn=

1

g(1/x)−√

2πn∑

{

J(a;nx)

φn(ax): a ∈ Z(g), |a| ≤ rℓ

}

+O(n1−3α)

Next, the summation is broken into three parts σ1 + σ2 −√

2πnJ(βi,nx)φn(βix) , where

σ1 := −√

2πn∑

{

J(α;nx)

φn(αx): α dominant, α 6= βi, |α| ≤ rℓ

}

and σ2 is the summation over the remaining part of it. Thus for x ∈ K0,

(6.1)pn(nx)

(ex)n/√

2πn= σ1 + σ2 −

√2πn

J(βi, nx)

φn(βix)+O(n1−3α)

By part 5 of Lemma 5.15 each term in σ1 is of o(φ−n(βix)). Hence

(6.2) σ1 = o(φ−n(βix)).

Let α be a zero that corresponds to a summand in σ2. So α is non-dominant.

By definition of σ2, we get |α| ≤ rℓ. Since x ∈ Dβi, we have |x| <

∣

∣

∣

1βi

∣

∣

∣ which equals1rℓ

. Hence |x| <∣

∣

1α

∣

∣.

Case 1: x /∈ Gα. By equation (5.2) x ∈ G−α . Hence

∣

∣φ−1(αx)∣

∣ ≤ 1.

Case 2: x ∈ Gα. Since α is non-dominant, α lies in Gβjfor some dominant βj .

By Lemma 6.3 we get∣

∣φ−1(αx)∣

∣ <∣

∣φ−1(βjx)∣

∣

We know that when x ∈ Dβi,

∣

∣φ−1(βix)∣

∣ = max{∣

∣φ−1(βmx)∣

∣ : 1 ≤ m ≤ k}So we still get

∣

∣φ−1(αx)∣

∣ ≤∣

∣φ−1(βix)∣

∣. Combining these two cases, we always

have∣

∣φ−1(αx)∣

∣ <∣

∣φ−1(βix)∣

∣. Therefore,

(6.3) σ2 = o(φ−n(βix)).

Putting the results from equations (6.2) and (6.3) into equation (6.1) we get,for x ∈ K0,

pn(nx)

(ex)n/√

2πn= −

√2πn

J(βi, nx)

φn(βix)+ o(φ−n(βix))

The argument works similarly for x ∈ Kj , j ≥ 1. Hence for x ∈ ⋃j≥0Kj, we have

pn(nx)

(ex)n/√

2πn= −

√2πn

J(βi, nx)

φn(βix)+ o(φ−n(βix)).

Since K \ ⋃j≥0Kj may possibly consists of small disks. It remains to study the

behavior of pn(nx)

(ex)n/√

2πnon any such disk. To this end we note that the number of

zeros of g contained in Dβi\D(0; 1

ρ) is obviously finite. Let D(1/a′; δ) be any such

disk contained in K \ ⋃j≥0Kj . To apply Proposition 6.1 for x ∈ D(1/a′; δ), we

write pn(nx)

(ex)n/√

2πnas

pn(nx)

(ex)n/√

2πn= τ1(x) + τ2(x) −

√2πn

(ex)nσa′(x) +O(n1−3α),


where

τ1(x) =1

x(

1

(1/x)g(1/x)− sa′(

1

x))(1 +O(

1

n))

τ2(x) = −√

2πn∑

{

J(a;nx)

φn(ax): a ∈ Z(g), |a| ≤ |a′| , a 6= a′

}

.

Now τ1(x) is obviously bounded in D(1/a′; δ) since sa′( 1x) is the singular part

of 1(1/x)g(1/x) . Let α correspond to a summand in τ2(x).

Case 1: a′ /∈ Gα. Since |α| ≤ |a′| and α 6= a′,∣

∣

1α

∣

∣ ≥∣

∣

1a′

∣

∣. Using equation(5.2) we can obviously choose δ small enough so that for all x ∈ D(1/a′; δ) we havex ∈ G−

α . So∣

∣φ−1(αx)∣

∣ < 1.

Case 2: a′ ∈ Gα. Now α in non-dominant, there exists a dominant βj such

that 1α ∈ Gβj

. Choosing δ small enough and carrying out a careful reasoning using

Lemma 6.3 and the maximality of∣

∣φ−1(xβi)∣

∣ we can show that for all x ∈ D(1/a′, δ), we have

∣

∣φ−1(αx)∣

∣ <∣

∣φ−1(xβi)∣

∣.

Combining these two cases we get∣

∣φ−1(αx)∣

∣ <∣

∣φ−1(xβi)∣

∣. As a result, weobtain for x ∈ D(1/a′; δ),

τ2(x) = o(∣

∣φ−n(xβi)∣

∣).

Finally the term√

2πn(ex)n σa′(x) is O(e6nρδ) by Proposition 6.2. �

7. Zero Attractor and the Density of the Zeros

In our paper [2], we determined the limit points of the zeros of the Eulerpolynomials by means of the asymptotics and the zero density. Here, we separateout first the question of find the support of the zero density measure, which is, ofcourse, the zero attractor. Then we determine the zero density by applying ourgeneral result in the appendix.

Proposition 7.1. Let fn(x) =√

2πnpn(nx)/(xe)n. Then the following limitshold uniformly on compact subsets of the indicated domains:

(1) On the domain A(1/r0;∞), limn→∞

1

nln[fn(x)] = 0.

(2) On the domain Da ∩A(1/ρ;∞) where a is any dominant zero of g,

limn→∞

1

nln[fn(x)] = − lnφ(ax).

Proof. We use the asymptotic expansions for pn(nx) developed in the previ-ous sections. �

To describe the zero attractor requires a closer examination of the boundary ofeach domain Da where a is a dominant zero.

The boundary ∂Da where a is a dominant zero of g has several natural families:∂Da ∩ ∂D0 which is an “outer boundary” and a polygonal curve consisting of theline segments contained in La,b where b is another dominant zero of g. Note that

∂Da ∩ ∂Db is a subset of D0. It will be useful to subdivide ∂Da ∩ ∂D0 intotwo connected components denoted by ∂D±

a that come from deleting {1/a} from[∂Da ∩ ∂D0].


Lemma 7.2. The zero attractor of the Appell polynomials {pn(nx)} must lieinside the compact set

⋃

{∂Da : a is a dominant zero of g} .

Proof. First, we let x∗ let in the infinite exterior of D0. Recall that

limn→∞

√2πnpn(nx)/(xe)n = 1/g(1/x)

uniformly on compact subsets. If xnkis a zero of pnk

(nkx) and xnk→ x∗, then

appealing to this limit we find that the limit must be 0 while the right-handside is 1/g(1/x∗) 6= 0. Secondly, suppose x∗ lies in the interior of D0 but noton any boundary set ∂Da, where a is a dominant zero. By construction, x willlie in the interior of one of the domains Db, where b is a dominant zero. Thenlimn→∞ |

√2πnpn(nx)/(xe)n|1/n = |φ(bx)| uniformly on compacta in the interior

of Db. By the same reasoning as before, x∗ cannot be a limit of zeros. �

The following Theorem is an immediate consequence of the above lemma to-gether with the result of Sokal in section A.1 of the Appendix.

Theorem 7.3. Let {pn(x)} be an Appell family with generating function g(t).Then the zero attractor of the normalized family {pn(nx)} is given by

⋃

{∂Da : a is a dominant zero of g} .where Da is the domain given in Definition 5.8.

Proof. Let a be any dominant zero of g and let x∗ ∈ ∂D±a . Let ǫ > 0 be

given. Then we find that

limn→∞

1

nln

∣

∣

∣

∣

pn(nx)

(xe)n/√

2πn

∣

∣

∣

∣

=

{

0, x ∈ D(x∗; ǫ) \D0,− ln |φ(ax)|, x ∈ D(x∗; ǫ) ∩D0

holds uniformly on compact subsets. Next suppose that x∗ lies on the line segmentof the form ∂Da ∩ ∂Db where Db is a bordering domain of Da. Again, we find that

limn→∞

1

nln

∣

∣

∣

∣

pn(nx)

(xe)n/√

2πn

∣

∣

∣

∣

=

{

− ln |φ(ax)|, x ∈ D(x∗; ǫ) ∩Da,− ln |φ(bx)|, x ∈ D(x∗; ǫ) ∩Db

which also holds uniformly on compact subsets. By Sokal’s result [5] describedin the appendix, we conclude that x∗ is in lim supZ(pn) since there can be noharmonic function v(x) on the disk D(x∗; ǫ) that satisfies the inequalities

lim infn→∞

ln

∣

∣

∣

∣

pn(nx)

(xe)n/√

2πn

∣

∣

∣

∣

≤ v(x) ≤ lim supn→∞

ln

∣

∣

∣

∣

pn(nx)

(xe)n/√

2πn

∣

∣

∣

∣

.

This reasoning handles all 1/a where a is a dominant zero of g. However, sincethe zero attractor must be a compact set and points in D(1/a; ǫ)∩ [∂Da ∩ ∂D0] liein the zero attractor, we conclude that 1/a also lie in the attractor. �

Theorem 7.4. Let g(t) be the generating function of the Appell family {pn(x)}.Suppose a and b are distinct dominant zeros of g.(1) The zero density measure on any proper subcurve of ∂Da ∩∂D0 is the pull-backof the normalized Lebesgue measure on the unit circle under the conformal mapφ(ax) where D0 is the domain given in Definition 5.13.(2) The zero density measure on any proper line segment of ∂Da∩∂Db is a multipleof Lebesgue measure.


Proof. For both parts, we can use the asymptotics given in Theorem 5.18.For part (1), let fn(x) =

√2πng(1/x)pn(nx)/(xe)n. Let a be a dominant zero of

g, and let C be a proper subcurve of ∂D0∩∂D±a . Then there exists a neighborhood

U of C such that U ⊂ Rρ ∩ [(C \ D0) ∪ Da] so that the asymptotics in Theorem5.18 can be written as

pn(nx)

(xe)n/√

2πn=

1

g(1/x)(1 +O(1/n)) −

√2πn

J(a;nx)

φ(ax)n+O(n1−3α) + o(Φn

1,a(x)),

where Φ1,a(x) = max{1, |φ−1(ax)|}. Hence, by multiplying by g(1/x)), we find thatfn(x) has the form:

fn(x) = 1 + an(x)φ(ax)−n + en(x), an(x) = −√

2πng(1/x)J(a;nx),

where

en(x) =

{

o(1), x ∈ U ∩ (C \D0),o(φ(ax)−n), x ∈ U ∩ (Da ∩Rρ).

Since φ(ax) is conformal in the disk D(0; 1/|a|), we may apply Theorem A.1 fromthe Appendix Section A.1 on the density of zeros.

Let a and b be two distinct dominant zeros of g such that ∂Da∩∂Db is nonempty.On Da ∩Db ∩Rρ, the asymptotics in Theorem 5.18 can be written as

pn(nx)

(xe)n/√

2πn=

1

g(1/x)(1 +O(1/n)) −

√2πn

(

J(a;nx)1

φ(ax)n+ J(b;nx)

1

φ(bx)n

+∑

{J(a′;nx)1

φ(a′x)n: a′ dominant zero, a′ 6= a, b }

)

+ O(n1−3α) + o(Φ(x)n)

1

g(1/x)(1 +O(1/n)) −

√2πn

(

J(a;nx)1

φ(ax)n+ J(b;nx)

1

φ(bx)n

)

+O(n1−3α) + o(Φna,b(x)),

where Ψa,b(x) = max{1/|φ(ax)|, 1/|φ(bx)|}.Let L be a proper line segment of the intersection ∂Da ∩ ∂Db. Let U be

a neighborhood of L so both |φ(ax)| < 1 and |φ(bx)| < 1 for x ∈ U . On theintersection U ∩Rρ, we work with a different normalization than before:

Tn(x) = − φ(ax)n

√2πn(xe)nJ(a;nx)

pn(nx).

Note that in this normalization the term that contains φ(ax)−n becomes the con-stant 1 for Tn(x). Of course, this new normalization has exactly the same zeros aspn(nx) in U so the zero density is unchanged. Then we find that

Tn(x) = 1 + an(x)ψ(x)n + en(x),

where

ψ(x) =φ(ax)

φ(bx)=a

be(b−a)x, an(x) =

J(b;nx)

J(a;nx),

and

en(x) = − φ(ax)n

√2πnJ(a;nx)

(

O(n1−3α) + o(Φna,b(x))

)

.


On U , we have that |φ(ax)|n ≤ Φna,b(x) = max{1, |ψ(x)|n}; while on Da ∩ U ,

|ψ(x)| < 1 and on Db ∩ U , |ψ(x)| > 1. This allows us to write en(x) as

en(x) =

o(ψ(x)n), x ∈ Da ∩ U,

o(1), x ∈ Db ∩ U..

By construction, φ(ax)/φ(bx) = ab e

(b−a)x is a conformal map on U ∩Rρ that mapsL onto an arc of the unit circle. By Corollary A.3 in the Appendix section A.1, theresult follows. �

We close with several examples that illustrate the main constructions in thepaper.

Example 7.5. Let g(t) be an entire function whose minimal modulus zeroa1 = 1 such that all its other zeros a satisfy 1/|a| < W (e−1) ≃ 0.27846. Thenthe zero attractor for the associated Appell polynomials coincide with the classicalSzego curve in Figure 3.

’

K0.4 K0.3 K0.2 K0.1 0 0.1 0.2 0.3 0.4

K1.0

K0.5

0.5

1.0

Figure 4. Zero Attractor for Taylor polynomials of cos(x)

Example 7.6. The higher order Euler polynomials E(m)n (x), where m ∈ Z+,

have generating function g(t) = (et + 1)m/2m; while the higher order Bernoulli

polynomials B(m)n (t) have generating function g(t) = (et − 1)m/tm. Then their

zero attractors are independent of m and coincide with a scaled version of the zeroattractor for the Taylor polynomials for cos(x), see Figure 4.

Example 7.7. The zero attractor for the Appell polynomials associated withgenerating function g(t) = J0(t), where J0(t) is the zero-th order Bessel function,is a scaled version of the zero attractor for the Taylor polynomials for cosh(x), seeFigure 1. Here the minimal modulus zeros of J0(t), a = ±2.404825558, are the onlydominant zeros and all the zeros of J0(t) lie on the real axis.

Example 7.8. Let g(t) = (t − 1) (t2 + 2). See Figure 5 for its zero attractorand zeros for degree 400.


’

K0.2 0 0.2 0.4 0.6 0.8

K0.6

K0.4

K0.2

0.2

0.4

0.6

’

K0.4 K0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

K0.6

K0.4

K0.2

0.2

0.4

0.6

0.8

1.0

1.2

Figure 5. (a) Zeros for degree 400 polynomial with generatingfunction g(t) = (t−1) (t2 +2); (b) Zero Attractor with polynomialzeros

Example 7.9. Consider the Appell polynomials with generating function

g(t) = (t− 1/a)(t− 1/b)(t− 1/c), with a = 1.2ei3π/16, b = 1.3ei7π/16, c = 1.5.

In this case, all three roots of g(t) are dominant. See Figures 6 and 7.

These last two examples both illustrate the following general fact. We assumethat the generating function g(t) has exactly three dominant zeros a, b, and c. Thenthe three lines determined by |φ(ax)| = |φ(bx)|, |φ(ax)| = |φ(cx)|, and |φ(bx)| =|φ(cx)| have a common intersection point, a so-called “triple point.” This followsby interpreting the lines as the boundary between the change of asymptotics of theAppell polynomial family; that is, the boundaries of the domains Da, Db, and Dc.


K0.4 K0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

K0.6

K0.4

K0.2

0.2

0.4

0.6

0.8

1.0

1.2

’

K0.4 K0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

K0.6

K0.4

K0.2

0.2

0.4

0.6

0.8

1.0

1.2

Figure 6. (a) Zero Attractor only, for generating function g(t) =(t − 1/a)(t− 1/b)(t− 1/c), a = 1.2ei3π/16, b = 1.3ei7π/16, c = 1.5;(b) Boundary of the Domain D0.

Appendix A. Density of Zeros

A.1. Introduction. We generalize the density result for the zeros of the Eulerpolynomials in [2] to highlight how the asymptotic structure of the polynomialfamily may determine the density of its zeros.

Let ψ(x) be an analytic function on a domain D ⊂ C that is conformal on D.We write ζ = ψ(x). We sometimes write x(ζ) for x = ψ−1(ζ).

We assume that there exists ǫ0 > 0 and 0 ≤ α < β ≤ 2π so that the annularsector

(A.1) S = {ρeiθ : ρ ∈ [1 − ǫ0, 1 + ǫ0], θ ∈ [α, β]}


’

K0.4 K0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

K0.6

K0.4

K0.2

0.2

0.4

0.6

0.8

1.0

1.2

Figure 7. Zeros for degree 400 polynomial together with the ZeroAttractor, for generating function g(t) = (t−1/a)(t−1/b)(t−1/c),a = 1.2ei3π/16, b = 1.3ei7π/16, c = 1.5

lies in the image ψ(D). Next we define two subsectors of S as

S+ = {ρeiθ : ρ ∈ [1 − ǫ0, 1), θ ∈ [α, β]}S− = {ρeiθ : ρ ∈ (1, 1 + ǫ0], θ ∈ [α, β]}.

Let C be the unimodular curve ψ−1({eiθ : θ ∈ [α, β]}), so |φ(x)| = 1 for x ∈ C.By construction, C is smoothly parametrized as x(eiθ) for θ ∈ [α, β]. Of course, wehave ψ−1(S) = ψ−1(S−) ∪ C ∪ ψ−1(S+) as a disjoint union.

Let {Tn(x)} be a sequence of analytic functions on ψ−1(S) where we assumethat the analytic functions satisfy the basic asymptotic relation:

(A.2) Tn(x) = 1 + an(x)ψ(x)cn + en(x),

where {cn} is a increasing unbounded sequence of positive numbers, δ > 0 is aconstant so that |an(x)| ≥ δ, and |an(x)| = exp[o(cn)], uniformly on ψ−1(S). Theterm en(x) satisfy the following estimates uniformly:

en(x) =

{

o(ψ(x)cn), x ∈ S+,o(1), x ∈ S−.

In the sequel, we may assume either form for en(x) if x lies on the common boundaryC of the two regions S± that is, |ψ(x)| = 1.

Let Zn be the set of all zeros of Tn that lie in ψ−1(S), which we assume is finitefor all n. For [γ1, γ2] ⊂ (α, β), let

(A.3) Nn(γ1, γ2) = #{x ∈ Zn : argx ∈ [γ1, γ2]}.

Choose ǫ > 0 so 3ǫ < ǫ0. By the Argument Principle, we find that

Nn(γ1, γ2) =1

2πi

∫

Γ

ddζTn(x(ζ))

Tn(x(ζ))dζ


where Γ is the boundary of the sector {ρeiθ : ρ ∈ [1 − ǫ, 1 + ǫ], θ ∈ [γ1, γ2]}. Theclosed contour Γ naturally has four parts of the form Γ1±ǫ and Γγj

, j = 1, 2 where

Γ1±ǫ = {(1 ± ǫ)eiθ : θ ∈ [γ1, γ2]},Γγ = {ρeiγ : ρ ∈ [1 − ǫ, 1 + ǫ]}.

Theorem A.1. Let α < γ1 < γ2 < β, and let Nn(γ1, γ2) denote the number ofzeros of Tn(x) whose arguments lie in [γ1, γ2], given in equation (A.3). Then

limn→∞

Nn(γ1, γ2)

cn=γ2 − γ1

2π;

that is, the image of the zero density under ψ is Lebesgue measure on an arc of theunit circle.

We need to recall the notions of lim sup and lim inf of a sequence {Xn} ofcompact sets in the complex plane. Now x∗ ∈ lim supXn if for every neighborhoodU of x∗, there exists a sequence xnk

∈ Xnk∩ U that converges to x∗ while x∗ ∈

lim inf Xn if for every neighborhood U of x∗, there exists an index n∗ and a sequencexn ∈ Xn∩U , for n ≥ n∗ that converges to x∗. It is known that if the lim infXn andlim supXn agree and are uniformly bounded, then the sequence {Xn} convergesin the Hausdorff metric. Consequently, when the density result holds, then thelim inf Z(Tn) must agree with lim supZ(Tn). Hence, we have the following:

Corollary A.2. As compact subsets of ψ−1(S), Z(Tn) converges to the uni-modular curve C in the Hausdorff metric.

Although we can determine the zero attractor and the zero density completelyin the above framework, it is conceptually useful to have the result of Sokal thatgives a description of the support of the zero density measure.

[Sokal] [5]: Let D be a domain in C, and let z0 ∈ D. Let {gn} be analytic

functions on D, and let {an} be positive real constants such that {|gn|an} are

uniformly bounded on the compact subsets of D. Suppose that there does notexist a neighborhood V of z0 and a function v on V that is either harmonic or else

identically −∞ such that lim infn→∞

an ln |gn(z)| ≤ v(z) ≤ lim supn→∞

an ln |gn(z)| for all

z ∈ V . Then z0 ∈ lim inf Z(gn).We can state the asymptotic form for Tn(x) in a more symmetric form:

Tn(x) = ψ0(x) +

N∑

k=1

an,k(x)ψk(x)cn + en(x)

where N is fixed and the error term has the form

en(x) = o(max{ψk(x)cn , 0 ≤ k ≤ n})This version explains the asymmetry in the above setup where we have ψ0(x) = 1and the zeros accumulate along the curve |ψ0(x)| = |ψ(x)|.

When the analytic arc C is a straight line segment and ψ(x) has the form eax+b,where a and b are constants, the Density Theorem yields:

Corollary A.3. If the analytic arc C is a straight line segment and ψ(x) isof the form eax+b, where a and b are constants, then the zero density along the linesegment C is a multiple of Lebesgue measure.


References

[1] Pavel Bleher and Robert Mallison, Jr., Zeros of Sections of Exponential Sums, Int. Math.Res. Not. 2006, Art. ID 38937, 49 pp.

[2] Robert P. Boyer and William M.Y. Goh, On the Zero Attractor of the Euler Polynomials,Advances in Applied Math. 38 (2007), no. 1, 97–132.

[3] E. T. Copson, Asymptotic Expansions, Cambridge University Press 1965.[4] William M. Y. Goh, Matthew X. He, Paolo E. Ricci, On the universal zero attractor of the

Tribonacci-related polynomials, Calcolo 46, no. 2 (2009), 95–129.[5] Alan Sokal, Chromatic roots are dense in the whole complex plane, Combin. Probab. Com-

puting. 13 (2004), 221–261.

[6] G. Szego, Uber eine Eigenschaft der Exponentialreihe, Sitzungsber. Berl. Math. Ges., 23

(1924), 50–64.

Department of Mathematics, Drexel University, Philadelphia, PA 19104E-mail address: [email protected]

Department of Statistics and Finance, University of Science and Technology ofChina (USTC), Hefei 230026,China

E-mail address: [email protected]

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