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Solutions of the equation an+(an1+(a2+(a1+xr1)r2)rn=bxa_{n}+(a_{n-1}+\cdots(a_{2}+(a_{1}+x^{r_{1}})^{r_{2}}\cdots)^{r_{n}}=b\,x

Daniel Panazzolo111This material is based upon work supported by the Agence Nationale de la Recherche under Grant ANR-23-CE40-0028.
Université de Haute-Alsace
Abstract

We establish a novel upper bound for the real solutions of the equation specified in the title, employing a generalized derivation-division algorithm. As a consequence, we also derive a new set of Chebyshev functions adapted specifically for this problem.

Dedicated to Jorge Sotomayor

Introduction

The equation

an+(an1+(a2+(a1+xr1)r2)rn=bxa_{n}+(a_{n-1}+\cdots(a_{2}+(a_{1}+x^{r_{1}})^{r_{2}}\cdots)^{r_{n}}=b\,x (1)

frequently emerges in problems concerning the qualitative theory of vector fields. In these problems, the variables (b,a,r)(b,a,r), belonging to >0×2n\mathbb{R}_{>0}\times\mathbb{R}^{2n}, are treated as free parameters. The goal is to estimate the number of real solutions xx to this equation, uniformly with respect to the parameters.

For instance, consider a smooth planar vector field XX whose phase portrait has a hyperbolic polycycle Γ\Gamma with nn saddle points s1,,sns_{1},\ldots,s_{n} and let XλX_{\lambda} be a smooth family of vector fields which unfolds XX. We fix a local transverse section Σ\Sigma (see picture below) such that ΣΓ={p}\Sigma\cap\Gamma=\{p\} and consider the first return map π:(Σ,p)(Σ,p)\pi:(\Sigma,p)\rightarrow(\Sigma,p), which depends on the parameter λ\lambda. The limit cycles bifurcating from Γ\Gamma are related to the isolated fixed points of π\pi.

[Uncaptioned image]

According a result of Mourtada (see [9], Theorem 1), in the vicinity of the parameter value λ=0\lambda=0, there exists an appropriate domain on the transversal where the fixed-point equation π(x)=x\pi(x)=x has an expansion of the form (1), up to some error term which is innocuous under generic conditions. Here the parameters (a,r)(a,r) depend smoothly on λ\lambda and measure respectively the distance between the saddle separatrices and the ratio of eigenvalues at the saddle points. The additional parameter b>0b>0 is related to the first multiplier of the Poincaré map along the unperturbed polycycle and is only relevant if the product r1rnr_{1}\cdots r_{n} equals one.

In such particular setting, one is interested in finding an upper bound for the cyclicity of the origin. Namely, the number of solutions (1) bifurcating from x=0x=0 under small perturbation of the parameters.

Another situation where (1) appears is related to slow-fast systems. More specifically, it is shown in [4] that such equation appears in the study of limit cycles under bifurcation from multi-layer canard cycles. In contrast to the previous situation, here one is interested in finding a global uniform bound for the number of isolated solutions.

Independently, the structure of the group GG of real functions generated by translations xx+ax\mapsto x+a and power maps xxrx\mapsto x^{r} is a subject of interest in group theory (see e.g. [2]). Here, one is interested in the abstract structure of such group and one natural question is whether GG is isomorphic to a free product. From a dynamical systems point of view, this problem is related to the following question, usually called center problem: Characterise those hyperbolic polycycles which lie on the boundary of a continuum of periodic orbits, i.e. those polycycles for which the Poincaré first return map π\pi is the identity. We refer the reader to [11] for some results on this direction.

In this paper, we will be interested in estimating the maximum number of real solutions of (1). To state the problem precisely, we need some definitions: Given (a,r)2n(a,r)\in\mathbb{R}^{2n}, let In(a,r)I_{n}(a,r)\subset\mathbb{R} denote the subset of all real numbers xx such that the functions g0,,gng_{0},\ldots,g_{n} defined by

g0(x)=x and gi+1(x)=ai+1+gi(x)ri+1, for i=1,,n1.g_{0}(x)=x\quad\mbox{ and }\quad g_{i+1}(x)=a_{i+1}+g_{i}(x)^{r_{i+1}},\mbox{ for }i=1,\ldots,n-1. (2)

are all strictly positive. Since all these functions are strictly monotone (increasing or decreasing), it is easy to see that In(a,r)I_{n}(a,r) is an open (possibly empty) interval of \mathbb{R}. If the rir_{i} are all positive, In(a,r)I_{n}(a,r) is a nonempty neighborhood of infinity.

Remark 0.1.

The set {(a,r,In(a,r)):(a,r)n×n}\{(a,r,I_{n}(a,r)):(a,r)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\} is an open subset of 2n+1\mathbb{R}^{2n+1} which belongs to the o-minimal structure of sets definable in the expansion of the real field by the Pfaffian functions (in the sense of Khovanskii).

So, our goal is to estimate the number of connected components of the solution set

Sn(b,a,r)={xIn(a,r)an+(an1+(a1+xr1)r2)rnbx=0},S_{n}(b,a,r)=\{x\in I_{n}(a,r)\mid a_{n}+(a_{n-1}+\cdots(a_{1}+x^{r_{1}})^{r_{2}}\cdots)^{r_{n}}-bx=0\}, (3)

Let FPn(b,a,r)\mathrm{FP}_{n}(b,a,r) denote the number of such connected components. It follows immediately from the analyticity of the functions in (2) that for each fixed parameter (b,a,r)(b,a,r), either Sn(b,a,r)In(a,r)S_{n}(b,a,r)\equiv I_{n}(a,r) or Sn(b,a,r)S_{n}(b,a,r) is a finite set of points. As a consequence, FPn(b,a,r)<\mathrm{FP}_{n}(b,a,r)<\infty, and we define

fp(n)=sup(b,a,r)FPn(b,a,r)\mathrm{fp}(n)=\sup_{(b,a,r)}\mathrm{FP}_{n}(b,a,r) (4)

The first obvious question is whether fp(n)<\mathrm{fp}(n)<\infty. This is a direct Corollary of Khovanskii’s Fewnomial theory [7]. Indeed, by introducing auxiliary variables (x1,,xn)+n(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}_{+}, equation (1) can be equivalently written as a system,

a1+x1r1=x2\displaystyle a_{1}+x_{1}^{r_{1}}=x_{2}\hfill
a2+x2r2=x3\displaystyle a_{2}+x_{2}^{r_{2}}=x_{3}\hfill
\displaystyle\vdots
an+xnrn=bx1\displaystyle a_{n}+x_{n}^{r_{n}}=b\,x_{1}\hfill

In such form, we have a system of nn equations in nn-variables with 2n+12n+1 distinct monomials, and the theory of Khovanskii provides the bound

fp(n)K(n):=2n(2n1)(n+1)2n\mathrm{fp}(n)\leq\mathrm{K}(n):=2^{n(2n-1)}(n+1)^{2n}

(see [7]). In [1], Sotille and Bihan slightly improved the fewnomial’s bound. Their result implies that fp(n)SB(n):=e2+34 2(n2)nn\mathrm{fp}(n)\leq\mathrm{SB}(n):=\frac{e^{2}+3}{4}\,2^{\binom{n}{2}}n^{n}.

One of the difficulties of the problem is that, in general, the sequence (2) is not a Chebyshev system (see e.g. [3], Section 3.3, for the definition). To the author’s knowledge, this was first observed in Mourtada’s thesis [6]. In that work, Mourtada constructed a specific example of a generic hyperbolic polycycle with four singularities which bifurcates into five limit cycles. In our notation, this result translates to fp(4)5\mathrm{fp}(4)\geq 5. We shall see in Section 2 that fp(4)13\mathrm{fp}(4)\leq 13 and, in Section 4, that in fact fp(3)=5\mathrm{fp}(3)=5. The exact value of fp(n)\mathrm{fp}(n) is unknown for n4n\geq 4.

In the next Sections, we shall describe a simple algorithm which gives a new upper bound for fp(n)\mathrm{fp}(n) by a derivation-division algorithm. The algorithm is somewhat inspired in [8].

Such new upper bound coincides with the exact value of fp(n)\mathrm{fp}(n) for n{1,2,3}n\in\{1,2,3\} and is significantly smaller the Khovanskii’s and Bihan-Sottile’s bound for n5n\leq 5. However, it is largely outperformed by such bounds for n6n\geq 6. We refer to Section 6 for a more detailed comparison.

One of the advantages of our method is that it provides as a by-product a system of linearly independent monomials which allows to expand the left-hand side of (1) uniformly with respect to the parameters (a,r)(a,r). This expansion can be interpreted as a generalization of the compensator-type expansion introduced independently by Roussarie [12] and Ecalle [5]. We refer to Section 5 for the details.

1 Derivation/division for Laurent polynomials

We initially consider a more abstract setting of a certain ring equipped with a derivation.

Given nn\in\mathbb{N}, let R=[R1,..,Rn]R=\mathbb{Z}[R_{1},..,R_{n}] denote the commutative ring of polynomials nn variables R1,,RnR_{1},\ldots,R_{n} with integer coefficients. Our basic object will be the ring

𝕃n=R[x1±,y1±,,xn±,yn±],\mathbb{L}_{n}=R[x_{1}^{\pm},y_{1}^{\pm},\ldots,x_{n}^{\pm},y_{n}^{\pm}],

i.e. the ring of Laurent polynomials in xi,yix_{i},y_{i} with coefficients in RR. We will frequently use the fraction notation and write, for instance, x1y11x_{1}y_{1}^{-1} simply as x1/y1x_{1}/y_{1}. We convention that 𝕃0=R\mathbb{L}_{0}=R.

A monomial in 𝕃n\mathbb{L}_{n} is polynomial of the form 𝔪=cxkyl\mathfrak{m}=c\,x^{k}y^{l}, with a nonzero coefficient cR{0}c\in R\setminus\{0\} and where we note

xkyl:=x1k1xnkny1l1ynlnx^{k}y^{l}:=x_{1}^{k_{1}}\cdots x_{n}^{k_{n}}y_{1}^{l_{1}}\cdots y_{n}^{l_{n}}

The vector (k,l)n×n(k,l)\in\mathbb{Z}^{n}\times\mathbb{Z}^{n} will be called the exponent 𝔪\mathfrak{m} and we say 𝔪\mathfrak{m} unitary if c=1c=1. The (poly)-degree of a such monomial is the integer vector pdeg(𝔪)=k+ln\mathrm{pdeg}(\mathfrak{m})=k+l\in\mathbb{Z}^{n}. We note that all unitary monomials are invertible element of 𝕃n\mathbb{L}_{n}.

Each non-zero polynomial p𝕃np\in\mathbb{L}_{n} can be written as a finite sum of monomials

p=iI𝔪ip=\sum_{i\in I}\mathfrak{m}_{i} (5)

where the set supp(p)={𝔪i:iI}\mathrm{supp}(p)=\{\mathfrak{m}_{i}:i\in I\} will be called the support of pp. We now define successively three subrings of 𝕃n\mathbb{L}_{n}.

𝕃nreg𝕃n1hom[xn,yn]𝕃nhom𝕃n\mathbb{L}_{n}^{\mathrm{reg}}\subset\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}]\subset\mathbb{L}_{n}^{\mathrm{hom}}\subset\mathbb{L}_{n} (6)

Firstly, we denote by 𝕃nhom𝕃n\mathbb{L}_{n}^{\mathrm{hom}}\subset\mathbb{L}_{n} the subring of Laurent polynomials which are homogeneous, i.e. such that all monomials in their support have a same degree. We will note by pdeg(p)n\mathrm{pdeg}(p)\in\mathbb{Z}^{n} the degree of an element pp of 𝕃nhom\mathbb{L}_{n}^{\mathrm{hom}}. For each 1jn1\leq j\leq n, we will denote by pdegj(p)\mathrm{pdeg}_{j}(p)\in\mathbb{Z} the degree of pp with respect to the variables (xj,yj)(x_{j},y_{j}) (i.e. the jthj^{th}-component of pdeg(p)\mathrm{pdeg}(p)).

We now define 𝕃n1hom[xn,yn]𝕃nhom\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}]\subset\mathbb{L}_{n}^{\mathrm{hom}} as the subring of homogeneous Laurent polynomials which are polynomials in the (xn,yn)(x_{n},y_{n})-variables. In other words, we consider the Laurent polynomials pp such that each monomial 𝔪=cxkylsupp(p)\mathfrak{m}=cx^{k}y^{l}\in\mathrm{supp}(p) has an exponent (k,l)(k,l) satisfying kn,ln0k_{n},l_{n}\geq 0. Thus, if we let m=pdegn(p)m=\mathrm{pdeg}_{n}(p) and denote the variables (xn,yn)(x_{n},y_{n}) simply as (x,y)(x,y), we can expand such element pp in the form

p=j=0mqjxjymjp=\sum_{j=0}^{m}q_{j}x^{j}y^{m-j} (7)

with coefficients qjq_{j} in 𝕃n1hom\mathbb{L}_{n-1}^{\mathrm{hom}}.

Finally, we define the smallest subring 𝕃nreg𝕃n1hom[xn,yn]\mathbb{L}_{n}^{\mathrm{reg}}\subset\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}] appearing in (6), whose elements will be called nn-regular polynomials. The definition is by induction on nn:

  • (i)

    For n=0n=0, the 0-regular polynomials are simply L0reg=RL_{0}^{\mathrm{reg}}=R.

  • (ii)

    For n1n\geq 1, we say that a Laurent polynomial p𝕃n1hom[xn,yn]p\in\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}] is nn-regular if the coefficient q0q_{0} in the expansion (7) is (n1)(n-1)-regular.

Note that, in particular, if p𝕃nregp\in\mathbb{L}_{n}^{\mathrm{reg}} then pdeg(p)n\mathrm{pdeg}(p)\in\mathbb{N}^{n} is a positive vector and if pdegn(p)=0\mathrm{pdeg}_{n}(p)=0 then pp is (n1)(n-1)-regular. Inductively, we show that pdeg(p)=0\mathrm{pdeg}(p)=0 if and only if pp belongs to the base ring RR.

Let us give a different characterisation of 𝕃nreg\mathbb{L}_{n}^{\mathrm{reg}}. Consider a homogeneous Laurent polynomial p𝕃nhomp\in\mathbb{L}_{n}^{\mathrm{hom}} of degree pdeg(p)=(m1,..,mn)n\mathrm{pdeg}(p)=(m_{1},..,m_{n})\in\mathbb{N}^{n}. Then pp is nn-regular if and only if there exists a sequence of nonzero Laurent polynomials pj𝕃j1hom[xj,yj]p_{j}\in\mathbb{L}_{j-1}^{\mathrm{hom}}[x_{j},y_{j}] for 0jn0\leq j\leq n such that:

  • (1)

    pn=pp_{n}=p and

  • (2)

    for each 1jn1\leq j\leq n, pdegj(pj)=mj\mathrm{pdeg}_{j}(p_{j})=m_{j} and we can inductively write

    pj=Qj+pj1yjmjp_{j}=Q_{j}+p_{j-1}y_{j}^{m_{j}}

    where Qj𝕃j1hom[xj,yj]Q_{j}\in\mathbb{L}_{j-1}^{\mathrm{hom}}[x_{j},y_{j}] is divisible by xjx_{j} (as a polynomial in (xj,yj)(x_{j},y_{j})). In other words, QjQ_{j} contains only monomials of the form xjαyjβ\star x_{j}^{\alpha}y_{j}^{\beta} with positive exponents α,β\alpha,\beta such that α+β=mj\alpha+\beta=m_{j} and β<mj\beta<m_{j}.

This result leads to the following nested form for pp,

p=Qn+((Q2+(Q1+αy1m1)y2m2))ynmnp=Q_{n}+(\cdots(Q_{2}+(Q_{1}+\alpha y_{1}^{m_{1}})y_{2}^{m_{2}})\cdots)y_{n}^{m_{n}} (8)

for some αR\alpha\in R.

Example 1.1.

For n=2n=2, the Laurent polynomials

p=x1y1x2+y2q=(x12y1+x13y12+x1+y12x1)x2y2+(x1+y1)y22p=\frac{x_{1}}{y_{1}}x_{2}+y_{2}\qquad q=\left(\frac{x_{1}^{2}}{y_{1}}+\frac{x_{1}^{3}}{y_{1}^{2}}+x_{1}+\frac{y_{1}^{2}}{x_{1}}\right)x_{2}y_{2}+(x_{1}+y_{1})y_{2}^{2}

are regular, of respective degrees pdeg(p)=(0,1)\mathrm{pdeg}(p)=(0,1) and pdeg(q)=(1,2)\mathrm{pdeg}(q)=(1,2)

We now turn 𝕃n\mathbb{L}_{n} into a differential ring by introducing a derivation

:𝕃n𝕃n\partial:\mathbb{L}_{n}\rightarrow\mathbb{L}_{n}

which satisfies (R)=0\partial(R)=0 and acts on the variables xj,yjx_{j},y_{j} according to the following formulas

(y1)=(x1)=Δ1x1,(yj)=(xj)=Δjxjx1xj1y1yj1,j2\partial(y_{1})=\partial(x_{1})=\Delta_{1}x_{1},\qquad\partial(y_{j})=\partial(x_{j})=\Delta_{j}x_{j}\,\frac{x_{1}\cdots x_{j-1}}{y_{1}\cdots y_{j-1}},\qquad j\geq 2 (9)

where we define Δj=R1RjR\Delta_{j}=R_{1}\cdots R_{j}\in R. The following result is an obvious consequence of the above expressions for \partial and the Leibniz rule.

Proposition 1.2.

(1) The derivation \partial maps the ring 𝕃j1hom[xj,yj]\mathbb{L}_{j-1}^{\mathrm{hom}}[x_{j},y_{j}] into itself, for each index 1jn1\leq j\leq n.
(2) Given p𝕃j1hom[xj,yj]p\in\mathbb{L}_{j-1}^{\mathrm{hom}}[x_{j},y_{j}], we either have p=0\partial p=0 or else

pdeg(p)=pdeg(p)\mathrm{pdeg}(\partial p)=\mathrm{pdeg}(p)

in other words, \partial preserves the degree.

Proof.

By linearity, it suffices to consider the case of a monomial 𝔪=xkyl\mathfrak{m}=x^{k}y^{l}. Then, from the Leibniz rule, have

𝔪=xkylj=1nkj(xj)xj+li(yj)yj=xkylj=1nΔjPj1(kj+ljxjyj)\partial\,\mathfrak{m}=x^{k}y^{l}\sum_{j=1}^{n}k_{j}\frac{\partial(x_{j})}{x_{j}}+l_{i}\frac{\partial(y_{j})}{y_{j}}=x^{k}y^{l}\sum_{j=1}^{n}\Delta_{j}P_{j-1}(k_{j}+l_{j}\frac{x_{j}}{y_{j}})

where Pj1=x1xj1y1yj1P_{j-1}=\frac{x_{1}\cdots x_{j-1}}{y_{1}\cdots y_{j-1}} and P0=1P_{0}=1.

We now remark that each term Pj1P_{j-1} is a Laurent homogeneous polynomial of degree 0. Therefore, by the additivity of the degree under multiplication, we easily obtain

pdeg(𝔪)=pdeg(𝔪).\mathrm{pdeg}(\partial\,\mathfrak{m})=\mathrm{pdeg}(\mathfrak{m}).

Finally, if we further suppose that mm belongs to 𝕃n1hom[xn,yn]\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}] (i.e.  kn,ln0k_{n},l_{n}\geq 0) then it follows from the above formula that 𝔪\partial\mathfrak{m} is a sum of monomials lying in 𝕃n1hom[xn,yn]\mathbb{L}_{n-1}^{\mathrm{hom}}[x_{n},y_{n}]. The same argument works if 𝔪\mathfrak{m} lies in 𝕃j1hom[xj,yj]\mathbb{L}_{j-1}^{\mathrm{hom}}[x_{j},y_{j}]. ∎

We observe however that the ring 𝕃nreg\mathbb{L}_{n}^{\mathrm{reg}} is not preserved by \partial, as the following simple example shows:

Example 1.3.

Consider the polynomial pp given in Example 1.1. Then,

p=(Δ1x1y1+(2Δ2Δ1)x12y12)x2\partial p=\left(\Delta_{1}\frac{x_{1}}{y_{1}}+(2\Delta_{2}-\Delta_{1})\frac{x_{1}^{2}}{y_{1}^{2}}\right)x_{2}

which is not a Laurent regular polynomial according to our definition.

In order to deal with such phenomena, we observe that any element of 𝕃nhom\mathbb{L}^{\mathrm{hom}}_{n} can be transformed into a regular Laurent polynomial upon division by an appropriate monomial.

More precisely, the regularization monomial associated to a non-zero Laurent polynomial p𝕃nhomp\in\mathbb{L}^{\mathrm{hom}}_{n} is a unitary monomial 𝔪=xkyl\mathfrak{m}=x^{k}y^{l} inductively defined as follows:

  • (i)

    Case n=0n=0: We define 𝔪=1\mathfrak{m}=1.

  • (ii)

    Case n1n\geq 1: If we note m=pdegn(p)m=\mathrm{pdeg}_{n}(p) then we observe that we can write an expansion of pp in the variables (x,y)=(xn,yn)(x,y)=(x_{n},y_{n}) as

    p=j=n0n1qjxjymjp=\sum_{j=n_{0}}^{n_{1}}q_{j}x^{j}y^{m-j}

    for some (possibly negative) integers n0n1n_{0}\leq n_{1} and coefficients qj𝕃n1homq_{j}\in\mathbb{L}_{n-1}^{\mathrm{hom}} such that qn0,qn1q_{n_{0}},q_{n_{1}} are non-zero. We then define 𝔪=xn0ydn1𝔪¯\mathfrak{m}=x^{n_{0}}y^{d-n_{1}}\bar{\mathfrak{m}}, where 𝔪¯\bar{\mathfrak{m}} is the regularization monomial of qn0q_{n_{0}}.

We will denote such monomial reg(p)=𝔪\mathrm{reg}(p)=\mathfrak{m}. The following result is obvious:

Proposition 1.4.

The Laurent polynomial p/reg(p)p/\mathrm{reg}(p) is regular.

Example 1.5.

Consider the polynomial q=pq=\partial p given in Example 1.3. Then, the above definitions gives reg(q)=x2x1y12\mathrm{reg}(q)=\frac{x_{2}}{x_{1}}{y_{1}^{2}}. The Laurent polynomial

r:-qreg(q)=Δ1y1+(2Δ2Δ1)x1+Δ1y1r\coloneq\frac{q}{\mathrm{reg}(q)}=\Delta_{1}y_{1}+(2\Delta_{2}-\Delta_{1})x_{1}+\Delta_{1}y_{1}

is regular, and has degree pdeg(r)=(1,0)\mathrm{pdeg}(r)=(1,0).

Theorem 1.6.

Suppose that p𝕃nregp\in\mathbb{L}_{n}^{\mathrm{reg}}. Then either p=0\partial p=0 or the regular polynomial

q=preg(p)q=\frac{\partial p}{\mathrm{reg}(\partial p)}

is such that pdeg(q)<rlexpdeg(p)\mathrm{pdeg}(q)<_{\mathrm{rlex}}\mathrm{pdeg}(p).

Remark 1.7.

Here we denote by <rlex<_{\mathrm{rlex}} the usual reverse lexicographical ordering in n\mathbb{N}^{n}, namely

(m1,..mn)<rlex(m1,..,mn)i:(mj<mj,i<jn)mi<mi(m_{1},..m_{n})<_{\mathrm{rlex}}(m_{1}^{\prime},..,m_{n}^{\prime})\Longleftrightarrow\exists i:(m_{j}<m_{j}^{\prime},i<j\leq n)\land m_{i}<m_{i}^{\prime}
Proof.

If pdeg(p)=0\mathrm{pdeg}(p)=0 then necessarily pRp\in R and hence p=0\partial p=0. So, let us assume that pdeg(p)=(m1,,mn)\mathrm{pdeg}(p)=(m_{1},\ldots,m_{n}) is such that m1==mj1=0m_{1}=\cdots=m_{j-1}=0 and mj1m_{j}\geqslant 1, for some 1jn1\leq j\leq n. We claim that the polynomial qq defined in the enunciate satisfies

pdeg(q)rlex(0,,0,mj1,mj+1,,mn)\mathrm{pdeg}(q)\leq_{\mathrm{rlex}}(0,\ldots,0,m_{j}-1,m_{j+1},\ldots,m_{n}) (10)

Indeed, it follows from the definition of regularity and the assumption on pdeg(p)\mathrm{pdeg}(p) that the expansion (8) can be rewritten as

p=Qn+((Qj+1+(Qj+αyjmj)yj+1mj+1))ynmnp=Q_{n}+(\cdots(Q_{j+1}+(Q_{j}+\alpha y_{j}^{m_{j}})y^{m_{j+1}}_{j+1})\cdots)y^{m_{n}}_{n} (11)

where αR{0}\alpha\in R\setminus\{0\} and, for each index jinj\leq i\leq n, the coefficient QiQ_{i} is an element 𝕃i1[xi,yi]hom\mathbb{L}_{i-1}[x_{i},y_{i}]^{\mathrm{hom}} containing only exponents in ximiyi0,,xiyimi1x_{i}^{m_{i}}y_{i}^{0},\ldots,x_{i}y_{i}^{m_{i}-1}. We recall that the derivation \partial applied to each monomial yimiy_{i}^{m_{i}} gives

yimi=miΔixiyimi1Pi1\partial y_{i}^{m_{i}}=m_{i}\Delta_{i}x_{i}y_{i}^{m_{i}-1}P_{i-1}

where Pi1=x1xi1y1yi1𝕃i1homP_{i-1}=\frac{x_{1}\cdots x_{i-1}}{y_{1}\ldots y_{i-1}}\in\mathbb{L}_{i-1}^{\mathrm{hom}}. In particular, for the innermost term in the expansion (11), we can write

(Qj+αyjmj)=Qj+αyjmj=Pj~\partial(Q_{j}+\alpha y_{j}^{m_{j}})=\partial Q_{j}+\alpha\partial y_{j}^{m_{j}}=\widetilde{P_{j}}

where P~j\tilde{P}_{j} is a Laurent polynomial in 𝕃j1[xj,yj]hom\mathbb{L}_{j-1}[x_{j},y_{j}]^{\mathrm{hom}} containing only exponents in xjmjyj0,,xjyjmj1x_{j}^{m_{j}}y_{j}^{0},\ldots,x_{j}y_{j}^{m_{j}-1} (and hence is divisible by xjx_{j}).

More generally, applying the derivation \partial to the nested expression of pp given above, we obtain

p=Q~n+((Q~j+1+P~jyj+1mj+1))ynmn\partial p=\tilde{Q}_{n}+(\cdots(\tilde{Q}_{j+1}+\tilde{P}_{j}y^{m_{j+1}}_{j+1})\cdots)y^{m_{n}}_{n}

where, for each index j+1inj+1\leqslant i\leqslant n, the coefficient Q~i\tilde{Q}_{i} is a Laurent polynomial in 𝕃i1[xi,yi]hom\mathbb{L}_{i-1}[x_{i},y_{i}]^{\mathrm{hom}} which contains only exponents in ximiyi0,,xiyimi1x_{i}^{m_{i}}y_{i}^{0},\ldots,x_{i}y_{i}^{m_{i}-1}.

Considering separately the cases Pj=0P_{j}=0 and P~j0\tilde{P}_{j}\neq 0, it now suffices to observe that the division of p\partial p by its regularization monomial reg(p)\mathrm{reg}(\partial p) results into a regular polynomial qq such that (10) holds. ∎

Applying successively the above Theorem, we obtain the following:

Corollary 1.8 (Derivation-division Algorithm).

Let p𝕃nregp\in\mathbb{L}_{n}^{\mathrm{reg}}. Then the sequence (p(n))n0(p^{(n)})_{n\geq 0} of Laurent polynomials in 𝕃nreg\mathbb{L}_{n}^{\mathrm{reg}}, defined inductively as

p(0)=p,p(n+1)=p(n)reg(p(n)),n0p^{(0)}=p,\qquad p^{(n+1)}=\frac{\partial\,p^{(n)}}{\mathrm{reg}(\partial\,p^{(n)})},\quad n\geq 0

eventually gives an element p(n)Rp^{(n)}\in R.

Proof.

It suffices to observe that if p𝕃nregp\in\mathbb{L}_{n}^{\mathrm{reg}} then pdeg(p)n\mathrm{pdeg}(p)\in\mathbb{N}^{n} and moreover pRp\in R if and only if pdeg(p)=0\mathrm{pdeg}(p)=0. ∎

In the above setting, we define the derivation-division complexity of pp as the smallest integer m=DD(p)0m=\mathrm{DD}(p)\geq 0 such that p(m)Rp^{(m)}\in R.

Remark 1.9.

The basic idea behind the derivation-division algorithm is quite simple: The expressions given in (9) show that (yj)\partial(y_{j}) only depends on the variables x1,,xjx_{1},\ldots,x_{j} and y1,..,yj1y_{1},..,y_{j-1}. Hence, we inductively eliminate all monomials in y1,..,yny_{1},..,y_{n} and then divide out by the maximal possible factors in x1,..,xnx_{1},..,x_{n}.

Example 1.10.

Let n=1n=1. Writing the variable (x1,y1)(x_{1},y_{1}) simply as (x,y)(x,y), a polynomial p𝕃regp\in\mathbb{L}^{\mathrm{reg}} of degree mm has the form

p=j=0mαjxjymjp=\sum_{j=0}^{m}\alpha_{j}x^{j}y^{m-j}

where the coefficients αj\alpha_{j} lie on the ring RR. Applying the derivation \partial to each monomial xjymjx^{j}y^{m-j} gives

xjymj=Δ1(jxjymj+(mj)xj+1ymj1)\partial x^{j}y^{m-j}=\Delta_{1}\left(jx^{j}y^{m-j}+(m-j)x^{j+1}y^{m-j-1}\right)

Therefore, we can write p=j=0mβjxjymj\partial p=\sum_{j=0}^{m}\beta_{j}x^{j}y^{m-j} where

βj=Δ1(jαj+(mj1)αj1),j1\beta_{j}=\Delta_{1}\left(j\alpha_{j}+(m-j-1)\alpha_{j-1}\right),\quad j\geq 1

and β0=0\beta_{0}=0. As a consequence, p\partial p is divisible by xx and the polynomial p(1)=p/xp^{(1)}=\partial p/x has degree at most m1m-1. We conclude by induction that DD(p)m\mathrm{DD}(p)\leq m.

Example 1.11.

Let n=2n=2. Denoting the variables (x1,y1)(x_{1},y_{1}) and (x2,y2)(x_{2},y_{2}) simply by (x,y)(x,y) and (x,y)(x^{\prime},y^{\prime}), we consider the special case where p𝕃regp\in\mathbb{L}^{\mathrm{reg}} is such that

pR[x,x,y,y]p\in R[x,x^{\prime},y,y^{\prime}]

In other words, we assume that our initial Laurent polynomial is indeed an homogeneous polynomial. Writing pdeg(p)=(m,m)\mathrm{pdeg}(p)=(m^{\prime},m), we claim that in this case DD(p)m(m+1)+m\mathrm{DD}(p)\leq m(m^{\prime}+1)+m^{\prime}.

If m=0m=0, we fall in the situation of the Example 1.10 and the estimate holds. So, let us suppose that m1m\geq 1 and write the expansion

p=Q1+p0ymp=Q_{1}+p_{0}y^{m}

where Q1=j=1mqjxjymjQ_{1}=\sum_{j=1}^{m}{q_{j}x^{j}y^{m-j}} and p0p_{0} is a an element of 𝕃1reg\mathbb{L}^{\mathrm{reg}}_{1}. In fact, due to our hypothesis, p0,q1,..,qmp_{0},q_{1},..,q_{m} are all homogeneous polynomials of degree mm^{\prime} in variables (x,y)(x^{\prime},y^{\prime}).

After at most mm^{\prime} steps of the derivation-division algorithm, we obtain a new Laurent polynomial of the form

q=R1+αym,q=R_{1}+\alpha y^{m},

of degree pdeg(q)=(0,m)\mathrm{pdeg}(q)=(0,m). Here αR\alpha\in R and we have R1=j=1mrjxjymjR_{1}=\sum_{j=1}^{m}{r_{j}x^{j}y^{m-j}}, where each coefficient qjq_{j} is a Laurent polynomial with support contained in the set of monomials xiyix^{\prime i}y^{\prime-i}, with i{0,..,m}i\in\{0,..,m\}. In the subsequent derivation-division step, we will get r=q/mr=\partial q/m, where 𝔪\mathfrak{m} is the monomial

m=xym+1m=\frac{x^{\prime}}{y^{\prime m^{\prime}+1}}

and such that, rr has degree (m,m1)(m^{\prime},m-1). One can verify that such new polynomial rr satisfies the same hypothesis of pp (i.e. lies in R[x,x,y,y]R[x,x^{\prime},y,y^{\prime}]). Therefore, we conclude by induction hypothesis that DD(r)(m1)(m+1)+m\mathrm{DD}(r)\leq(m-1)(m^{\prime}+1)+m^{\prime}. Hence, since DD(p)m+1+DD(r)=m(m+1)+m\mathrm{DD}(p)\leq m^{\prime}+1+\mathrm{DD}(r)=m(m^{\prime}+1)+m^{\prime}.

2 Studying the number of isolated roots

We are now ready to estimate the the real solutions of equation (1).

For each fixed parameter value (a,r)2n(a,r)\in\mathbb{R}^{2n}, consider the sequence of functions g0,f1,g1,..,fn,gng_{0},f_{1},g_{1},..,f_{n},g_{n} defined inductively as g0(x)=xg_{0}(x)=x,

f1(x)=xr1,g1(x)=a1+f1(x)f_{1}(x)=x^{r_{1}},\qquad g_{1}(x)=a_{1}+f_{1}(x)

and, for each 1in11\leq i\leq n-1,

fi+1(x)=gi(x)ri+1,gi+1(x)=fi+1(x)+ai+1f_{i+1}(x)=g_{i}(x)^{r_{i+1}},\qquad g_{i+1}(x)=f_{i+1}(x)+a_{i+1}

As we have seen in the introduction, the functions fi,gif_{i},g_{i} are strictly positive and analytic on some open (possibly unbounded) interval I(a,r)I(a,r)\subset\mathbb{R}. We therefore can consider the subring GnG_{n} of Cω(I(a,r))C^{\omega}(I(a,r)) given by

Gn=r[f1±,g1±,,fn±,gn±]G_{n}=r[f_{1}^{\pm},g_{1}^{\pm},\ldots,f_{n}^{\pm},g_{n}^{\pm}] (12)

i.e. the ring formed by finite linear combinations of monomials fkglf^{k}g^{l} (with possible negative exponents k,lk,l) and coefficients in the ring r=[r1,..,rn]r=\mathbb{Z}[r_{1},..,r_{n}].

The following result relates this latter ring with the ring of Laurent polynomials defined in the previous Section.

Proposition 2.1.

The derivation xddxx\frac{d}{dx} maps the ring GnG_{n} into itself. Moreover, if we consider the morphism of rings Φ:Gn𝕃n\Phi:G_{n}\rightarrow\mathbb{L}_{n} defined by Φ(ri)=Ri\Phi(r_{i})=R_{i} and

Φ(fi)=xi,Φ(gi)=yi,1in,\Phi(f_{i})=x_{i},\qquad\Phi(g_{i})=y_{i},\qquad 1\leq i\leq n,

Then, we have the relation Φ=Φxddx\partial\circ\Phi=\Phi\circ x\frac{d}{dx}.

Proof.

The proof is immediate. Indeed, we have

xddxf1=xddxxr1=r1xr1=r1f1x\frac{d}{dx}f_{1}=x\frac{d}{dx}x^{r_{1}}=r_{1}x^{r_{1}}=r_{1}f_{1}

Similarly,

xddxf2=xddx(a2+(a1+xr1)r2=r1r2(a2+(a1+xr1)r21xr1=(r1r2)f2(f1g1)x\frac{d}{dx}f_{2}=x\frac{d}{dx}(a_{2}+(a_{1}+x^{r_{1}})^{r_{2}}=r_{1}r_{2}(a_{2}+(a_{1}+x^{r_{1}})^{r_{2}-1}x^{r_{1}}=(r_{1}r_{2})f_{2}\left(\frac{f_{1}}{g_{1}}\right)

And, in general,

xddxgj=xddxfj=(r1rj)fjf1fj1f1fj1x\frac{d}{dx}g_{j}=x\frac{d}{dx}f_{j}=(r_{1}\cdots r_{j})\,f_{j}\,\frac{f_{1}\cdots f_{j-1}}{f_{1}\cdots f_{j-1}}

Therefore, comparing with the expressions given in (9), we immediately conclude that Φ=Φxddx\partial\circ\Phi=\Phi\circ x\frac{d}{dx}. ∎

Consider now the analytic function φ:I(a,r)\varphi:I(a,r)\rightarrow\mathbb{R} given by

φ(x)=an+(an1+(a1+xr1)r2)rnbx\varphi(x)=a_{n}+(a_{n-1}+\cdots(a_{1}+x^{r_{1}})^{r_{2}}\cdots)^{r_{n}}-bx (13)

which we can also write as φ=gnbg0\varphi=g_{n}-b\,g_{0}.

An easy computation shows that ψ=x2(ddx)2φ\psi=x^{2}\left(\frac{d}{dx}\right)^{2}\varphi belongs to GnG_{n}. More specifically, if we consider its image q=Φ(ψ)q=\Phi(\psi) under the morphism Φ\Phi given above and the monomial

𝔪=x1xn(y1yn1)2\mathfrak{m}=\frac{x_{1}\cdots x_{n}}{(y_{1}\cdots y_{n-1})^{2}}

Then, p=q/mp=q/m is assumes the very simple form

p=Δnj=1n(ΔjΔj1)(i1=1j1xi1)(i2=jn1yi2)p=\Delta_{n}\sum_{j=1}^{n}(\Delta_{j}-\Delta_{j-1})\;\left(\prod_{i_{1}=1}^{j-1}x_{i_{1}}\right)\left(\prod_{i_{2}=j}^{n-1}y_{i_{2}}\right) (14)

where we define Δ0=1\Delta_{0}=1 and recall that Δi=R1Ri\Delta_{i}=R_{1}\cdots R_{i} for 1in1\leq i\leq n.

Notation: We denote by DD(n):-DD(p)\mathrm{DD}(n)\coloneq\mathrm{DD}(p) the number of steps in the derivation-division algorithm for the above polynomial pp.

Using the fact that p𝕃n1regp\in\mathbb{L}_{n-1}^{\mathrm{reg}}, we can easily obtain the following result:

Theorem 2.2.

The number of isolated roots of φ\varphi on I(a,r)I(a,r) (counted with multiplicities) is bounded by DD(n)+2\mathrm{DD}(n)+2.

Proof.

By Rolle’s Theorem, the number of roots of φ\varphi on I(a,r)I(a,r) is bounded by the number of roots of ψ\psi plus two. Let us assume that ψ\psi is not identically zero. Then, the derivation-division algorithm described in Corollary 1.8, when to the polynomial applied p=Φ(ψ)mp=\frac{\Phi(\psi)}{m} given by (14), defines a sequence of analytic functions on I(a,r)I(a,r) which eventually (after at most DD(n)\mathrm{DD}(n) steps) ends up into a constant function. Therefore, applying again Rolle’s theorem, we conclude that the number of roots of φ\varphi on I(a,r)I(a,r) is bounded by DD(n)+2\mathrm{DD}(n)+2. ∎

Since the number DD(n)\mathrm{DD}(n) does not depend on any specific choice of parameters (a,r)(a,r), we conclude that:

Corollary 2.3.

The number fp(n)\mathrm{fp}(n) (see (4)) is bounded by DD(n)+2\mathrm{DD}(n)+2.

3 At most three solutions for n=2n=2

As a simple illustration of the algorithm, let us prove that the equation

a2+(a1+xr1)r2bx=0a_{2}+(a_{1}+x^{r_{1}})^{r_{2}}-b\,x=0 (15)

has a maximum of three isolated solutions. If we denote by φ\varphi the right-hand side of the above equation, the function ψ=x2(ddx)2φ\psi=x^{2}\left(\frac{d}{dx}\right)^{2}\varphi has the form

ψ=r1r2(r1(r21)(a1+xr1)r22x2r12+(r11)(a1+xr1)r21xr12)\psi=r_{1}r_{2}\left(r_{1}\left({r_{2}}-1\right)(a_{1}+x^{r_{1}})^{r_{2}-2}x^{2r_{1}-2}+(r_{1}-1)(a_{1}+x^{r_{1}})^{r_{2}-1}x^{r_{1}-2}\right)

Therefore, upon division by xr1(a+xr1)r22x^{r_{1}}(a+x^{r_{1}})^{r_{2}-2} we obtain

r1r2(r1(r21)xr1+(r11)(a1+xr1))r_{1}r_{2}\left(r_{1}(r_{2}-1)x^{r_{1}}+(r_{1}-1)(a_{1}+x^{r_{1}})\right)

which, under the morphism Φ\Phi, gives the regular Laurent polynomial

p=A1x1+B1y1p=A_{1}x_{1}+B_{1}y_{1}

where we set A1=Δ2(Δ2Δ1)A_{1}=\Delta_{2}(\Delta_{2}-\Delta_{1}) and B1=Δ2(Δ1Δ0)B_{1}=\Delta_{2}(\Delta_{1}-\Delta_{0}). Now, applying one step of the derivation-division, we obtain

A1x1+B1y1Δ1(A1+B1)x1/x1Δ1(A1+B1)RA_{1}x_{1}+B_{1}y_{1}\overset{\partial}{\Longrightarrow}\Delta_{1}(A_{1}+B_{1})x_{1}\overset{\cdot/x_{1}}{\Longrightarrow}\Delta_{1}(A_{1}+B_{1})\in R

Therefore, DD(n)=1\mathrm{DD}(n)=1 and we conclude that fp(2)3\mathrm{fp}(2)\leq 3.

The following explicit example shows that indeed fp(2)=3\mathrm{fp}(2)=3. Take a1=0.004259259259a_{1}=0.004259259259, a2=0.1516666667a_{2}=-0.1516666667, r1=2r_{1}=2 and r2=1/3r_{2}=1/3. Numerical computations with SageMath shows that the equation

0.1516666667+(0.004259259259+x2)1/3x=0-0.1516666667+(0.004259259259+x^{2})^{1/3}-x=0

has the three isolated solutions x1=0.0123409x_{1}=0.0123409..., x2=0.1741525x_{2}=0.1741525... and x3=0.3585065x_{3}=0.3585065....

4 At most five solutions for n=3n=3

Let us prove that the equation

a3+(a2+(a1+xr1)r2)r3bx=0a_{3}+(a_{2}+(a_{1}+x^{r_{1}})^{r_{2}})^{r_{3}}-b\,x=0 (16)

has a maximum of five solutions, for all parameter values (b,(a,r))>0×6(b,(a,r))\in\mathbb{R}_{>0}\times\mathbb{R}^{6}.

A similar computation to the previous example leads to the polynomial

p=A2x1x2+B2x1y2+C2y1y2p=A_{2}x_{1}x_{2}+B_{2}x_{1}y_{2}+C_{2}y_{1}y_{2}

where A2=Δ3(Δ3Δ2)A_{2}=\Delta_{3}(\Delta_{3}-\Delta_{2}), B2=Δ3(Δ2Δ1)B_{2}=\Delta_{3}(\Delta_{2}-\Delta_{1}) and C2=Δ3(Δ1Δ0)C_{2}=\Delta_{3}(\Delta_{1}-\Delta_{0}).

The following lines show the derivation-division algorithm ends in at most three steps. We omit the dependence on the coefficients Δi\Delta_{i} to simplify the notation:

p(0)=x1x2+x1y2+y1y2p^{(0)}=x_{1}x_{2}+x_{1}y_{2}+y_{1}y_{2}
p(0)=(x1+x12y1)x2+x1y2p(1)=p(0)x1=(1+x1y1)x2+y2\partial p^{(0)}=\left(x_{1}+\frac{x_{1}^{2}}{y_{1}}\right)x_{2}+x_{1}y_{2}\Longrightarrow p^{(1)}=\frac{\partial p^{(0)}}{x_{1}}=\left(1+\frac{x_{1}}{y_{1}}\right)x_{2}+y_{2}
p(1)=((x1y1)2+x1y1)x2p(2)=p(1)(x1/y12)x2=x1+y1\partial p^{(1)}=\left(\left(\frac{x_{1}}{y_{1}}\right)^{2}+\frac{x_{1}}{y_{1}}\right)x_{2}\Longrightarrow p^{(2)}=\frac{\partial p^{(1)}}{(x_{1}/y_{1}^{2})x_{2}}=x_{1}+y_{1}
p(2)=x1p(3)=p(2)x1=1\partial p^{(2)}=x_{1}\Longrightarrow p^{(3)}=\frac{\partial p^{(2)}}{x_{1}}=1

Therefore, we conclude that fp(3)5\mathrm{fp}(3)\leq 5.

The following explicit example shows that indeed fp(3)=5\mathrm{fp}(3)=5. We take

a3=8.39×106,a2=0.0035836,a1=0.012r1=53150,r2=118,r3=2a_{3}=-8.39\times 10^{-6},\;\;a_{2}=0.0035836,\;\;a_{1}=-0.012\;\;r_{1}=\frac{53}{150},\;\;r_{2}=\frac{11}{8},\;\;r_{3}=2

Numerical computations with 20 digits of precision in SageMath give the following five solutions

x1=0.1270599×104x2=0.1921586×104x3=0.4764392×104,x4=0.7949546×104x5=0.2384109\begin{array}[]{l}x_{1}=0.1270599...\times 10^{-4}\\ x_{2}=0.1921586...\times 10^{-4}\\ x_{3}=0.4764392...\times 10^{-4},\\ x_{4}=0.7949546...\times 10^{-4}\\ x_{5}=0.2384109...\end{array}

5 Disconjugacy and compensator-type expansions

We now show that, as a byproduct of the derivation-division algorithm, the function appearing in (1) can be expanded in an uniform basis of functions satisfying the Chebyshev property.

Firstly, we recall some basic definitions about disconjugate differential operators. For a thoughtful treatment, we refer the reader to the excellent Coppel’s book [3]. A mthm^{th}-order linear differential operator LL defined on an interval II is called disconjugate if it can be written in the form

Lu=1ρmddx1ρm1ddxddx1ρ0uLu=\frac{1}{\rho_{m}}\frac{d}{dx}\frac{1}{\rho_{m-1}}\frac{d}{dx}\cdots\frac{d}{dx}\frac{1}{\rho_{0}}u (17)

where, for each 0km0\leq k\leq m, ρk\rho_{k} is a strictly positive analytic function on II. A well-known result of Polya (see e.g. [3], Chapter 3) states that LL is disconjugate if and only if no solution uu of Lu=0Lu=0 has more than m1m-1 zeros in II, counted with multiplicities.

Assuming that LL is written as in (17), we can easily construct a basis of solutions {ω0,,ωm1}\{\omega_{0},\ldots,\omega_{m-1}\} for LL as follows: fix an arbitrary base point x0Ix_{0}\in I and set

ω0(x)=ρ0(x),ω1(x)=ρ0(x)x0xρ1(x1)𝑑x1\omega_{0}(x)=\rho_{0}(x),\qquad\omega_{1}(x)=\rho_{0}(x)\int_{x_{0}}^{x}\rho_{1}(x_{1})dx_{1}

and, in general, define ωk\omega_{k} by the kthk^{th}-iterated integral

ωk(x)=ρ0(x)x0xρ1(x1)x0x1ρ2(x2)x0xk1ρk(xk)𝑑xk𝑑x2𝑑x1\omega_{k}(x)=\rho_{0}(x)\int_{x_{0}}^{x}\rho_{1}(x_{1})\int_{x_{0}}^{x_{1}}\rho_{2}(x_{2})\cdots\int_{x_{0}}^{x_{k-1}}\rho_{k}(x_{k})\,dx_{k}\,\cdots\,dx_{2}\,dx_{1} (18)

Notice that this collection of functions has the following properties:

  • (i)

    They form a Chebyshev system, i.e. any non-trivial \mathbb{R}-linear combination λkωk(x)\sum\lambda_{k}\omega_{k}(x) has at most m1m-1 zeros on II.

  • (ii)

    For each 0km10\leq k\leq m-1, the subset {ω0,,ωk1}\{\omega_{0},\ldots,\omega_{k-1}\} is a basis of solutions of the truncated operator

    L[k]u=1ρkddx1ρk1ddxddx1ρ0uL^{[k]}u=\frac{1}{\rho_{k}}\frac{d}{dx}\frac{1}{\rho_{k-1}}\frac{d}{dx}\cdots\frac{d}{dx}\frac{1}{\rho_{0}}u

    (with the convention that L[0]=1/ρ0L^{[0]}={1}/{\rho_{0}})

  • (iii)

    ωk(x)\omega_{k}(x) has a zero of multiplicity precisely kk at x0x_{0} and L[k]ωk=1L^{[k]}\omega_{k}=1.

The last two properties imply that any solution uu of Lu=0Lu=0 can be expanded as

u(x)=k=0m1λkωk(x)u(x)=\sum_{k=0}^{m-1}\lambda_{k}\,\omega_{k}(x) (19)

where the coefficients λk\lambda_{k}\in\mathbb{R} are determined inductively as follows

λm1=L[m1]u,λk1=L[k1](uj=km1λkuk),1km1\lambda_{m-1}=L^{[m-1]}u,\qquad\lambda_{k-1}=L^{[k-1]}\,\big{(}u-\sum_{j=k}^{m-1}\lambda_{k}u_{k}\big{)},\quad 1\leq k\leq m-1

Let us now apply these results to our original problem.

For each nn\in\mathbb{N} and each parameter value (b,a,r)>0×2n(b,a,r)\in\mathbb{R}_{>0}\times\mathbb{R}^{2n}, we consider the function φa,r:I(a,r)\varphi_{a,r}:I(a,r)\rightarrow\mathbb{R} given by (13). Since the parameter bb will be innocuous in the following discussion, we suppose from now on that b=1b=1 and will omit it to simplify the notation.

Proposition 5.1.

The function φa,r\varphi_{a,r} is a solution of a disconjugate linear differential operator La,rL_{a,r} of order DD(n)+3\mathrm{DD}(n)+3.

Proof.

We recall from the proof of Theorem 2.2 and Corollary 1.8 that if we set ψ=x2(ddx)2\psi=x^{2}\left(\frac{d}{dx}\right)^{2} and consider the Laurent polynomial q=Φ(ψ)q=\Phi(\psi) then there exists a sequence of monomials (𝔪k)(\mathfrak{m}_{k}) for 0kr10\leq k\leq r-1, such that

1𝔪s11𝔪0q=0\partial\frac{1}{\mathfrak{m}_{s-1}}\partial\cdots\partial\frac{1}{\mathfrak{m}_{0}}q=0 (20)

where s=DD(n)s=\mathrm{DD}(n) and \partial is the derivation on 𝕃n\mathbb{L}_{n} defined in (9). Using the morphism Φ\Phi defined in Proposition 2.1, this relation gives La,rφa,r=0L_{a,r}\varphi_{a,r}=0, where La,rL_{a,r} is the disconjugate operator of order m=s+3m=s+3 given by

La,r=ddx(xhr1)ddx(xh0)ddxx2(ddx)L_{a,r}=\frac{d}{dx}\left(\frac{x}{h_{r-1}}\right)\frac{d}{dx}\cdots\left(\frac{x}{h_{0}}\right)\frac{d}{dx}x^{2}\left(\frac{d}{dx}\right)

In this expression, for each 0jr10\leq j\leq r-1, we choose hj=Φ1(𝔪j)h_{j}=\Phi^{-1}(\mathfrak{m}_{j}) to be a function in ring GnG_{n} (given in (12)) which is mapped to 𝔪j\mathfrak{m}_{j} under Φ\Phi. ∎

In particular, we observe that this result allows to expand φa,r\varphi_{a,r} in terms of a globally defined collection of analytic Chebyshev functions. To enunciate this result precisely, we consider the open subset in the parameter space given by

U={(a,r)2nI(a,r) is nonempty }U=\{(a,r)\in\mathbb{R}^{2n}\mid I(a,r)\text{ is nonempty }\}

Note that, as we have observed in the introduction, UU contains in particular the set of all parameters (a,r)(a,r) such that r>0nr\in\mathbb{R}^{n}_{>0}. Now, we define an open set VV in the total space {(a,r,x)2n×}\{(a,r,x)\in\mathbb{R}^{2n}\times\mathbb{R}\} by

V={(a,r,I(a,r))(a,r)U}V=\{\big{(}a,r,I(a,r)\big{)}\mid(a,r)\in U\}
Proposition 5.2 (Compensator type expansion for φa,r\varphi_{a,r}).

There exists a collection of m=DD(n)+3m=\mathrm{DD}(n)+3 analytic functions {ωk}0km\{\omega_{k}\}_{0\leq k\leq m} defined on VV such that

  • (i)

    For each fixed parameter value (a,r)(a,r) in UU, the functions {xωk(a,r,x)}\{x\mapsto\omega_{k}(a,r,x)\} forms a Chebyshev system on I(a,r)I(a,r).

  • (ii)

    We can write the expansion

    φa,r(x)=k=0m1λk(a,r)ωk(a,r,x)\varphi_{a,r}(x)=\sum_{k=0}^{m-1}\lambda_{k}(a,r)\,\omega_{k}(a,r,x)

    where the coefficients λk(a,r)\lambda_{k}(a,r) are analytic functions on UU.

Proof.

The mthm^{th}-order differential operator La,rL_{a,r} defined in Proposition 17 obviously depends analytically on the parameters (a,r)U(a,r)\in U. In other words, the functions ρ0,..,ρm\rho_{0},..,\rho_{m} which appear when we decompose La,rL_{a,r} as (17), are analytic on VV.

Let {ωk}0km\{\omega_{k}\}_{0\leq k\leq m} be the basis of solutions of La,rL_{a,r} defined according to (18). Note that the base point x0x_{0} used in computation of the iterated integrals must chosen as a point on the interval I(a,r)I(a,r). Such base point, seen as a function of the parameters (a,r)(a,r), can also be chosen to be analytic.

Finally, when we write φa,r\varphi_{a,r} in terms of such basis of solutions {ωk}\{\omega_{k}\}, the coefficients λk(a,r)\lambda_{k}(a,r) can be computed according to the recursive procedure described in (19). This shows that they are analytic functions of (a,r)(a,r). ∎

We will say that ideal a,rCω(U){\cal I}_{a,r}\subset C^{\omega}(U) generated by the functions {λk}0km\{\lambda_{k}\}_{0\leq k\leq m} is the ideal of coefficients of φa,r\varphi_{a,r}.

Remark 5.3.

Although the functions λk(a,r)\lambda_{k}(a,r) obviously depend on the choice of the basis {ωk}\{\omega_{k}\}, the ideal a,r{\cal{I}}_{a,r} itself is independent of this choice. In fact one can prove that a,r{\cal{I}}_{a,r} is indeed a polynomial ideal, i.e. generated by polynomials in [a,r]\mathbb{Z}[a,r]. Let us sketch the proof assuming that r>0r\in\mathbb{R}_{>0}: We consider the Laurent polynomial q=Φ(ψ)𝕃nq=\Phi(\psi)\in\mathbb{L}_{n} defined in the proof of Proposition 5.1. By definition, we can write a finite expansion

q=jcj𝔪jq=\sum_{j}{c_{j}\,\mathfrak{m}_{j}}

where 𝔪j\mathfrak{m}_{j} are Laurent monomials and cjc_{j} are coefficients in R=[R1,..,Rn]R=\mathbb{Z}[R_{1},..,R_{n}]. Under the morphism Φ\Phi, each monomial 𝔪j\mathfrak{m}_{j} corresponds to an analytic function hj=Φ1(mj)h_{j}=\Phi^{-1}(m_{j}) defined on some open interval containing ++\infty on the boundary. Therefore, we can consider the asymptotic expansion of hjh_{j} at x=+x=+\infty. In fact, a simple application of the binomial expansion to each monomial shows that the asymptotic expansion of qq has the form

αSPα(r,a)xα\sum_{\alpha\in S}P_{\alpha}(r,a)x^{\alpha}

where SS is the semigroup rn+rn1rn++r1rn\mathbb{N}r_{n}+\mathbb{N}r_{n-1}r_{n}+\cdots+\mathbb{N}r_{1}\cdots r_{n} and each Pj(r,a)P_{j}(r,a) is a polynomial in [a,r]\mathbb{Z}[a,r]. Now, it is not very hard to see that a,r{\cal{I}}_{a,r} is generated by Pα(r,a)αS\langle P_{\alpha}(r,a)\rangle_{\alpha\in S}.

Example 5.4.

[Classical compensators] Let n=1n=1 and consider the function

φ=a+xrx\varphi=a+x^{r}-x

with r>0r>0 and aa\in\mathbb{R}. This function lies in the kernel of the third order disconjugate linear differential operator

L=ddx1xr(ddx)2L=\frac{d}{dx}\frac{1}{x^{r}}\left(\frac{d}{dx}\right)^{2}

Hence, ρ0=ρ1=1\rho_{0}=\rho_{1}=1 and ρ2=xr2\rho_{2}=x^{r-2}. Let us compute the basis of solutions {ω0,ω1,ω2}\{\omega_{0},\omega_{1},\omega_{2}\} by integrating from the base point x0=1x_{0}=1. We have ω0(x)=1\omega_{0}(x)=1, ω1(x)=x1\omega_{1}(x)=x-1 and

ω2(x)=1xΩ(r,y)𝑑y, where Ω(r,y)={logy, if r=1,yr11r1, otherwise\omega_{2}(x)=\int_{1}^{x}\Omega(r,y)dy,\quad\text{ where }\quad\Omega(r,y)=\begin{cases}\log y,&\mbox{ if }r=1,\vspace{0.2cm}\\ \displaystyle{\frac{y^{r-1}-1}{r-1}},&\mbox{ otherwise}\end{cases}

is the well-known Ecalle-Roussarie compensator. Explicitly, we obtain

ω2(x)={xlog(x)x+1, if r=1,rxrxr+1(r1)r, otherwise\omega_{2}(x)=\begin{cases}x\log\left(x\right)-x+1,&\mbox{ if }r=1,\vspace{0.2cm}\\ \displaystyle{-\frac{rx-r-x^{r}+1}{{\left(r-1\right)}r}},&\mbox{ otherwise}\end{cases}

The ideal of coefficients of a,r{\cal{I}}_{a,r} is generated by the polynomials λ0=a\lambda_{0}=a, λ1=r1\lambda_{1}=r-1 and λ2=r(r1)\lambda_{2}=r(r-1).

Motivated by Roussarie’s result for the saddle loop case [12], we expect that the generalized compensators introduced in the last Proposition will ultimately prove useful in analyzing the first return map of non-generic hyperbolic polycycles.

6 The number of steps in the D-D algorithm

We now give an upper estimate on the number DD(p)\mathrm{DD}(p) of steps of the derivation-division algorithm, when applied on a given Laurent polynomial p𝕃nregp\in\mathbb{L}^{\mathrm{reg}}_{n}. Such bound depends solely on the support supp(p)\mathrm{supp}(p), seen as a subset of 2n\mathbb{Z}^{2n}.

More precisely, denoting by pdeg(p)=(m1,,mn)n\mathrm{pdeg}(p)=(m_{1},\ldots,m_{n})\in\mathbb{N}^{n} the degree of pp, let us suppose that

supp(p)[0,m1]××[0,mn]\mathrm{supp}(p)\subset[0,m_{1}]\times\cdots\times[0,m_{n}] (21)

i.e. that pp is a homogeneous polynomial in the variables xi,yix_{i},y_{i} (with no negative exponents). For each 1kn1\leq k\leq n, we denote by

Hk:k\mathrm{H}_{k}:\mathbb{N}^{k}\rightarrow\mathbb{N}

the collection of functions defined recursively as follows:

  • (1)

    If k=1k=1 then H1(m)=m\mathrm{H}_{1}(m)=m

  • (2)

    If k2k\geq 2 then

    • (2.a)

      If mk=0m_{k}=0 then Hk(m1,,mk1,0)=Hk1(m1,,mk1)\mathrm{H}_{k}(m_{1},\ldots,m_{k-1},0)=\mathrm{H}_{k-1}(m_{1},\ldots,m_{k-1})

    • (2.b)

      If mk1m_{k}\geq 1 then Hk(m1,,mk1,mk)=Hk(m1,,Mk1,mk1)\mathrm{H}_{k}(m_{1},\ldots,m_{k-1},m_{k})=\mathrm{H}_{k}(m_{1},\ldots,M_{k-1},m_{k}-1), where Mk1=mk1+H(m1,,mk1)M_{k-1}=m_{k-1}+\mathrm{H}(m_{1},\ldots,m_{k-1}).

We can now state the following result, whose proof we will omit:

Proposition 6.1.

If supp(p)\mathrm{supp}(p) satisfies the condition (21) then

DD(p)Hn(m1,,mn).\mathrm{DD}(p)\leq\mathrm{H}_{n}(m_{1},\cdots,m_{n}).

Suppose now that pp has the special form given by (14). In this situation, the support of pp is quite special and we conjecture, based on an algorithmic implementation, that the number DD(n)\mathrm{DD}(n) is given by

Conjecture.

DD(n)=A(n1,1)\mathrm{DD}(n)=A(n-1,1).

In the enunciate, A(n,k)A(n,k) denotes the Ackerman function, defined inductively as follows:

A(n,k)={k+1if n=0A(n1,1)if n>0 and k=0A(n1,A(n,k1))if n>0 and k>0.A(n,k)=\begin{cases}k+1&\mbox{if }n=0\\ A(n-1,1)&\mbox{if }n>0\mbox{ and }k=0\\ A(n-1,A(n,k-1))&\mbox{if }n>0\mbox{ and }k>0.\end{cases}

Notice that such function grows extremely fast with respect to nn. Here are the first 5 values:

A(0,1)=2,A(1,1)=3,A(2,1)=5,A(3,1)=13,A(4,1)=65533,A(5,1)=2223A(0,1)=2,A(1,1)=3,A(2,1)=5,A(3,1)=13,A(4,1)=65533,A(5,1)={2^{2}}^{{\cdot}^{{\cdot}^{{\cdot}^{2}}}}-3

Here, the rightmost expression is a tower of powers of height A(4,1) + 3A(4,1)\mbox{ + 3}.

We conclude by providing a summary of the various upper bounds and the exact values known for fp(n)\mathrm{fp}(n):

nn DD(n)\mathrm{DD}(n) K(n)\mathrm{K}(n) SB(n)\mathrm{SB}(n) Exact Value of fp(n)\mathrm{fp}(n)
1 2 8 3 2
2 3 5184 21 3
3 5 1.3×108\sim 1.3\times 10^{8} 562 5
4 13 1.0×1014\sim 1.0\times 10^{14} 42554 ?
5 65533 2.0×1021\sim 2.0\times 10^{21} 8.3×106\sim 8.3\times 10^{6} ?

For n6n\geq 6, the DD(n)\mathrm{DD}(n) bound becomes much larger than both Khovanskii’s bound K(n)\mathrm{K}(n) and Bihan Sotile’s bound SB(n)\mathrm{SB}(n).

Up to the author’s knowledge, the exact value of fp(n)\mathrm{fp}(n) is unknown for n4n\geq 4.

Declaration

The author states that there is no conflict of interest.

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D. Panazzolo, IRIMAS, Université de Haute-Alsace, 68093 Mulhouse, France

E-mail address: daniel.panazzolo@uha.fr