Uniqueness of polynomial canonical representations

# Uniqueness of polynomial canonical representations

## Abstract.

Let and be polynomials of the same degree in the complex variables and , respectively. In this extended abstract we study the non-linear functional equation , where is restricted to be analytic in a neighborhood of . We provide sufficient conditions to ensure that all the roots of are contained within the range of as well as to have as the unique analytic solution of the non-linear equation. Our results are motivated from uniqueness considerations of polynomial canonical representations of the phase or amplitude terms of oscillatory integrals encountered in the asymptotic analysis of the coefficients of mixed powers and multivariable generating functions via saddle-point methods. Uniqueness shall prove important for developing algorithms to determine the Taylor coefficients of the terms appearing in these representations. The uniqueness of Levinson’s polynomial canonical representations of analytic functions in several variables follows as a corollary of our one-complex variables results.

###### Key words and phrases:
Airy phenomena, asymptotic enumeration, analytic combinatorics, large powers of generating functions, discrete random structures, saddle point method, uniform asymptotic expansions.

## 1. Introduction

Unless otherwise stated is a fixed integer and . We use boldface notation to denote vectors in . We reserve the script to refer to the zero vector. The script is reserved for a vector with strictly positive real coordinates. We refer to as a polyradius. The coordinates of a vector are denoted . We define , in particular, . The notation means that for all . Similarly, means that for all .

Problem description. A classical example of a polynomial canonical representation is the Weierstrass preparation theorem [Tay02] which asserts the following. If is a complex-valued analytic function in a neighborhood of and

 k:=min{n≥0:∂nU∂tnd(0)≠0}<∞

then there exists a polyradius and analytic functions such that

 (1.1) U(t)=V(t)⋅{tkd+k−1∑j=0uj(t′)⋅tjd},

for . We refer to as the order of vanishing of about the origin with respect to the variable . The factor within the parenthesis above is called the Weierstrass polynomial of about the origin and it will be denoted as . It satisfies the following important property. For all the polynomial equation in the variable : , with , has exactly solutions repeated according to their multiplicity. Since , and since the roots of a monic polynomial identify it uniquely, the factorization in (1.1) must be unique.

The problem of whether itself can be represented as a polynomial with respect to a possibly auxiliary variable dates back to the investigations of Chester, Friedman and Ursell [CheFriUrs57] who studied this problem for the special case of . Later work by Levinson [Lev60a] provided a way to represent certain analytic functions of complex variables in a canonical way as a polynomial in two auxiliary variables. More generally, for , Levinson proved the following [Lev60b]. If is like before then there exists a polyradius and analytic functions such that

 (1.2) U(t)=k∑j=0vj(t′)⋅{x(t)}j,

for , with for , , and and for .

Unlike the representation in (1.1), it is unclear that the representation in (1.2) is unique. Indeed, the issue of uniqueness was omitted in [Lev60b] and to the best of our knowledge it has not been addressed further. The main issue surrounding the uniqueness of this representation as well as other canonical representations is the introduction of auxiliary variables. Loosely speaking, the problem is how to certify in general the validity of the following implication

 [k∑j=0vj(t′)⋅xj=k∑j=0wj(t′)⋅yj]⟹[vj=wj, for all j, and x=y],

under the assumption that and for all sufficiently close to the origin. Clearly, to assert the uniqueness of the above factorization, it suffices to have in some open neighborhood of the origin in . In fact, since and are as functions of locally invertible about the origin, we shall see at the end of Section 2 that the validity of the above implication is closely related to the uniqueness of as an analytic solution of the non-linear functional equation

 (1.3) ⎧⎪⎨⎪⎩k∏i=1(z−zi)=k∏i=1(y(z)−y(zi)),|z|

where is a given radius and are fixed complex numbers such that .

There is one case where the uniqueness issue of the above equation can be addressed directly. If is an entire function i.e. then, according to (1.3), as . Since , is a bounded entire function. Hence, due to Liouville’s theorem [Rud87], must be constant and therefore because . Unfortunately, the case with is not of much use to address uniqueness issues of polynomial canonical representations because — almost always — they only apply locally.

Connections with mixed powers generating functions. Polynomial canonical representations are pivotal for analyzing the asymptotic behavior of oscillatory integrals [BleHan86]. Integrals of this type arise frequently in the context of asymptotic enumeration or the analysis of discrete random structures [PemWil05].

A mixed power generating function is a generating function of the form , where the factors are complex-valued analytic functions near and are nonnegative integers. The term of mixed power was introduced in [Lla06b] to emphasize the fact that one is usually interested in the coefficient of of as . If one defines , this is equivalent to request that where is any norm in .

Generating functions of the above form occur naturally in the context of the Lagrange inversion formula [GouJac04, Wil90] with . More recent applications include the case of to analyze the core size of various types of random planar maps [BanFlaSchSor01].

Coefficients of mixed powers generating functions have been considered in the literature for factors with nonnegative coefficients by Drmota [Drm94a], for and at a comparable rate. Gardy [Gar95] considered the case of nonnegative coefficients for with or and for . A geometrically based approach, in the lines used by Pemantle and Wilson [PemWil02, PemWil04], was proposed in [Lla06b] to handle factors with possibly negative Taylor coefficients. Given with nonnegative coordinates and such that , say that is a strictly minimal critical point associated with provided that

 t0 = d∑i=1ti⋅xf′i(x)fi(x); d∏i=1|fi(z)|ti < d∏i=1|fi(x)|ti,

for all such that and . If the above conditions hold and some pathological behavior is ruled out, it follows from [Lla06b] that

 (1.4) [zn0]d∏i=1{fi(z)}ni∼x−n02πd∏i=1{fi(x)}ni⋅∫π−πexp{−∥(n0,n)∥⋅F(θ;(t0,t))}dθ,

for such that , as . The function is a computable function that is continuous in its arguments however it is also analytic in the variable . For a fixed , it satisfies that at , and the real-part of is minimized at . Furthermore, the above expansion applies uniformly for all , provided that is a compact set such that for all , is a strictly critical point associated with that depends continuously on . In particular, the asymptotic analysis of the above integral is amenable for the saddle-point method to obtain uniform asymptotic expansions for the coefficients in (1.4) for , as .

It is precisely for the asymptotic analysis of integrals such as the one occurring in (1.4) that polynomial canonical representations of the type in (1.2) play a crucial role. In particular, uniqueness of these representations shall prove important to determine the Taylor coefficients of the various terms and auxiliary variables occurring in these representations. This should aid in automatizing the extraction of asymptotic formulae for coefficients of mixed powers generating functions as well as multivariable generating functions.

The lack of analyticity of in (1.4) with respect to the variable can be over passed by thinking of as a function of . The original function can then be recovered by evaluating this new function at . In order to apply the saddle-point method let be the order of vanishing of about with respect to the variable . Since at points of this type, it follows from [Lla06a] that Levinson’s polynomial cannonical representation takes the form

 F(θ;(s0,s);x)=k∑j=2vj((s0,s);x)⋅{y(θ;(s0,s);x)}j,

with and at points of the form that are near . If the above translates into having the integral appearing in (1.4) to be described asymptotically by the Gamma function. In particular, the integral is of order as . On the other hand, if the integral is described asymptotically by the Airy function. In this case the integral in (1.4) has typically an asymptotic series expansion which is a linear combination of terms of order and also of order . See [BleHan86, Lla03] to follow up on uniform asymptotics for integrals that involve the Gamma and Airy function.

The interested reader is referred to [Lla06b] for concrete applications of the above methodology with . The reader is also referred to [BanFlaSchSor01] for a related yet more specialized discussion with . Although our motivation to study the uniqueness of polynomial canonical representations has been argued in the context of mixed powers generating functions, they also play a fundamental role in the extraction of asymptotics of multivariable generating functions. The reader is referred to [PemWil02, PemWil04, Lla06a] to follow up on this last remark.

## 2. Main results

We first introduce two one-complex variable results. Theorem 1 provides sufficient conditions to ensure that all the roots of a polynomial are contained in the range of an analytic function when there exists another polynomial , of the same degree as , such that in a neighborhood of . Under an appropriate rescaling, this translates into having , where is the degree of and are the roots of repeated according to their multiplicity. Theorem 2 provides sufficient conditions in order to conclude from this that . Both theorems are then used to show the uniqueness of Levinson’s representation in (1.2). The proofs of our main two theorems are presented in Section 3. Our main results and accompanying proofs are refined versions of some of the results obtained by the author in his doctoral dissertation [Lla03].

Auxiliary results. In what follows, is a given radius and we use the notation

 D := {z∈C:|z|

For , we define

 ∥f∥r:=sup|z|≤r|f(z)|=sup|z|=r|f(z)|,

where the last identity is justified by the maximum modulus principle [Rud87].

###### Theorem 1.

Let and be polynomials of the same degree and assume that contains all the roots of repeated according to their multiplicity. If is such that , for all , then

 (2.1) [Q−1{0}⊂y(D)]⟺[∀i:y′(zi)≠0, and ∀i,j:y(zi)=y(zj)⇔zi=zj].

Furthermore, if either of the conditions in (2.1) apply then there exists a constant such that

 (2.2) Q(y)=q⋅k∏i=1(y−y(zi)).
###### Theorem 2.

For all and such that there exists a such that if then is the only solution of the non-linear functional equation

 (2.3) ⎧⎪⎨⎪⎩k∏i=1(z−zi)=k∏i=1(y(z)−y(zi)),y∈H;y(0)=0,

that satisfies the condition .

Proof of uniqueness of Levinson’s representation. We use the stated theorems to show the uniqueness of Levinson’s polynomial canonical representation in (1.2). Thus consider analytic in a neighborhood and assume that

 (2.4) U(t)=k∑j=0vj(t′)⋅sj=k∑j=0wj(t′)⋅tj

where for , , , and and at all points in the domain of and of the form . We show that , for all , and that . For this consider the transformation . Since and the Jacobian matrix of at is lower-triangular with non-zero entries along the diagonal, the inverse mapping theorem [Tay02] implies that is a well-defined analytic function in some open neighborhood of the origin in . Define and . It follows from (2.4) that

 (2.5) V(t′,z)=k∑j=0vj(t′)⋅zj=k∑j=0wj(t′)⋅xj.

Observe that and at all points in the domain of of the form . Furthermore, according to the first identity above, vanishes to degree about the origin in the variable . In particular, the Weierstrass preparation theorem [Tay02] implies that, for all sufficiently close to , the roots of can be listed as , repeated according to their multiplicity. Since for sufficiently close to the origin the transformation is a one-to-one transformation, we may use Theorem 1 in (2.5) to conclude that

 vk(t′)⋅k∏j=1{z−zj(t′)}=wk(t′)⋅k∏j=1{x(t′,z)−x(t′,zj(t′))}.

But observe that, according to (2.5), provided that the appropriate branch for the -th root is selected. With this choice of branch, introduce the auxiliary variable

 y=y(t′,z):=x(t′,z)⋅{vk(t′)wk(t′)}−1/k.

Notice that

 k∏j=1{z−zj(t′)}=k∏j=1{y(t′,z)−y(t′,zj(t′))},

for all sufficiently close to the origin in and such that , where is certain real parameter independent of . But observe that, according to the Weierstrass preparation theorem, if is sufficiently close to the origin then , for all . On the other hand, since and is uniformly continuos over compact subsets of its domain, it follows for that

 limt′→0′∥∥y(t′,z)−z∥∥r=0.

Theorem 2 implies that , for all sufficiently close to the origin and all such that . In particular, . Since at all points in the domain of of the form , we conclude that . This in (2.5) implies that , for all . Furthermore, since , with , we find . This shows that Levinson’s polynomial canonical representations are unique.∎

## 3. Proofs of main results

Proof of Theorem 1. Assume that i.e. that all roots of lie in the range of the function . Then the roots of may be listed as  — repeated according to their multiplicity — in such a way that if and only if . Define and let be the multiplicity of as a root of . Observe that

 (3.1) limz→ξiP(z)(z−ξi)ni=limz→ξi{y(z)−y(ξi)z−ξi}ni⋅Q(y(z))(y(z)−yi)ni={y′(ξi)}ni⋅limy→yiQ(y)(y−yi)ni,

where for the last identity we have used that is in the interior of as asserted by the open mapping theorem [Rud87]. Since the limit on the right-hand side above exists, has to be a root of of multiplicity at least . In particular, using that the multiplicity of in the list is precisely , must divide . Since is of degree , and as a result the sequence is just a reordering of . From this it is immediate that if and only if , and that factorizes as described in (2.2). Furthermore, since must be a root of multiplicity of , it follows from (3.1) that . In particular, for all .

To complete the proof of the theorem it suffices to show that if for all and , , and if and only if , then all roots of lie in the range of . For each let be the multiplicity of as a root of and define . Since, like in (3.1), we have

 limy→yiQ(y)(y−yi)mi={1y′(zi)}mi⋅limz→ziP(z)(z−zi)mi

and the limit on the right-hand side above is nonzero and finite, it follows that is a zero of multiplicity of . Since the multiplicity of in the list is , divides . In particular, all roots of lie in the range of because is of degree . This completes the proof of the theorem.∎

Proof of Theorem 2. To prove this result it would suffice to show that the differential of the operator associated with (2.3) is one-to-one and has a continuous inverse. However, a technical issue with this approach is that it is unclear that the inverse of the differential is continuous at all when considering the natural Banach space with the infinite-norm . Technical difficulties arise even showing that the pre-images of functions in under the differential stay bounded near the boundary of .

Due to the above considerations we consider a weaker topology. We embed with the topology of uniform convergence over compact subsets of . In particular, a sequence converges to provided that , for all . This topology is induced by a metric in under which this space is complete i.e. Cauchy sequences are convergent [Tay02]. We define . Clearly is a closed linear subspace of . In particular, is also complete when endowed with the topology of uniform convergence over subsets of .

###### Lemma 1.

If then the operator defined as

 (Lf)(z):=1kk∑j=1f(z)−f(zj)z−zj

is a linear isomorphism.

Proof. Define . According to the removable singularity theorem [Tay02], is a well-defined linear transformation. Since on the other hand, for and it applies that

 ∥Lf∥r≤2∥f∥r⋅sup|z|=r1kk∑j=1∣∣∣1z−zj∣∣∣≤2∥f∥rr−ρ,

it is immediate that is a continuous linear operator.

To show that is one-to-one consider the polynomial and observe that

 (3.2) p′(z)=p(z)⋅k∑j=11z−zj.

Suppose that is such that . Using (3.2), a simple calculation reveals that

 f(z)⋅p′(z)=k∑j=1f(zj)⋅p(z)z−zj.

Since is a polynomial of degree in the variable and the right-hand side above is a polynomial of degree at most , the division algorithm implies that there is a constant and a polynomial of degree at most such that , for such that . But it is well-known that the roots of are convex linear combinations of . In particular, the roots of lie in the disk of radius centered at the origin. Since is bounded in this disk however is of degree at most , it follows from the above identity that . Hence is constant and therefore because . This shows that is one-to-one.

To show that is an isomorphism we define an operator such that , for all . With the understanding that define . Observe that . In particular, the series is analytic for . Since , is analytic in a neighborhood of the origin. Hence if is defined as the coefficient of of the power series expansion of about the origin, it applies that

 (3.3) n∑j=iβn−j⋅αj−i={0,n>i,1,n=i.

Furthermore, an inductive argument shows that

 (3.4) |βn|≤(2ρ)n.

For with define

 (3.5) (Tf)(z):=z⋅∞∑j=0⎧⎨⎩∞∑l=jfl⋅βl−j⎫⎬⎭zj.

To see that the above transformation is well-defined consider . According to the Cauchy estimates [Rud87], . As a result, using (3.4) we obtain that

 (3.6) ∣∣ ∣∣∞∑l=jfl⋅βl−j∣∣ ∣∣≤∥f∥r11−2ρ/r1⋅r−j1.

Since the above holds for any , is analytic for and . Furthermore, is a continuous linear operator because the above inequality implies that

 ∥Tf∥r0≤r0(1−2ρ/r1)(1−r0/r1)⋅∥f∥r1.

Finally we show that , for all . For this let be the linear subspace of polynomials in the complex variable . Since is continuous and is a dense linear subspace, it suffices to show that , for all . However, due to linearity, this is equivalent to show that , for all . For this observe that

 L(Tzn)=L(n∑j=0βn−jzj+1)=n∑j=0βn−jL(zj+1)=n∑j=0βn−jj∑i=0αj−izi=n∑i=0{n∑j=iβn−jαj−i}zi=zn,

where for the last equality we have used (3.3). This completes the proof of the lemma. ∎

###### Lemma 2.

Let . If then , for all , with

 (3.7) c:=min(z1,…,zk)inf|z|=r∣∣ ∣∣1kk∑j=11z−zj∣∣ ∣∣⋅{1+ρ/r(1−2ρ/r)3}−1>0,

where the minimum is taken over all such that .

Proof. Let be the inverse operator of . According to (3.2), since for all , it applies for that

 (Tf)(z)⋅p′(z)=p(z)⋅{k⋅f(z)+k∑j=1(Tf)(zj)z−zj}.

Define and observe that because . It follows from the above identity that

 Extra open brace or missing close brace

where for the second inequality we have used (3.5) and (3.6) with . The above implies that for all , , with

 0

where for the second identity we have used (3.2). Since for such that and it applies that

 ∣∣ ∣∣inf|z|=r∣∣ ∣∣1kk∑j=11z−zj∣∣ ∣∣−inf|z|=r∣∣ ∣∣1kk∑j=11z−yj∣∣ ∣∣∣∣ ∣∣≤maxj|zj−yj|(r−ρ)2,

depends continuously on . This shows (3.7) and completes the proof of the lemma.∎

We finally prove Theorem 2. Let be such that . To study the uniqueness of the functional equation in (2.3) consider the operator defined as . Given define . Observe that

 F(y)−F(\it Id)=F(f+\it Id)−F(\it Id)=p(z)⋅∑J∏j∈Jf(z)−f(zj)z−zj,

where the index in the summation varies over all possible non-empty subsets of the set . In particular,

 F(y)−F(\it Id)=p(z)⋅{k∑j=1f(z)−f(zj)z−zj+∑I∏i∈If(z)−f(zi)z−zi},

where the index varies over all possible subsets of the set of cardinality two or greater. As a result, we find that

 (3.8) ∥F(y)−F(\it Id)∥r≥inf|z|=r|p(z)|⋅⎧⎨⎩∥∥ ∥∥k∑j=1f(z)−f(zj)z−zj∥∥ ∥∥r−∥∥ ∥∥∑I∏i∈If(z)−f(zi)z−zi∥∥ ∥∥r⎫⎬⎭.

According to Lemma 2, we have that

 ∥∥ ∥∥k∑j=1f(z)−f(zj)z−zj∥∥ ∥∥r≥k⋅c⋅