# The Rational rank of the support of generalized power series solutions of
differential and $q$-difference equations

J. Cano and P. Fortuny Ayuso  J. Cano: Universidad de Valladolid, Spain
<EMAIL_ADDRESS>P. Fortuny Ayuso: Universidad de Oviedo, Spain
<EMAIL_ADDRESS>
###### Abstract.

Given a differential or $q$-difference equation $P$ of order $n$, we prove
that the set of exponents of a generalized power series solution has its
rational rank bounded by the rational rank of the support of $P$ plus $n$. We
also prove that when the support of the solution has maximum rational rank, it
is convergent. Using the Newton polygon technique, we show also that in the
maximum rational rank case, an initial segment can always be completed to a
true solution. The techniques are the same for the differential and the
$q$-difference case.

This work is part of action “PID2022-139631NB-I00”, supported by the
MICIU/AEI, with ID number 10.13039/501100011033, and by FEDER, UE

##  1\. Introduction

An important class of solutions of differential and $q$-difference equations
are formal power series. Their relevance comes from algebraic to geometric,
analytical, combinatorial and logical point of views. Denef and Lipschitz’s
paper [9] deals with the logical questions; Singer’s and Grigoriev’s works
study the possibility of finding these solutions, and some of their
convergence properties [19, 13]; the excellent book on combinatorics by
Flajolet and Sedgewick [11] contains a complete section devoted to the
relation between solutions of polynomial differential equations and
combinatorics; and one cannot forget the relation between power series,
derivations, Hardy fields, o-minimal structures and related topics [16, 3,
17]. Finally, any survey of the bibliography would be incomplete without
citing van der Hoeven’s works on transseries [10] and the _magnum opus_ [1].

Grigoriev and Singer in [13] introduce the field $\Omega$ of generalized power
series with complex coefficients and real exponents: series
$\sum_{i=0}^{\infty}a_{i}\,x^{\nu_{i}}$, where $a_{i}\in\mathbb{C}$,
$\nu_{i}\in\mathbb{R}$, $v_{0}<\nu_{1}<\cdots$ and $\lim\nu_{i}=\infty$, and
prove that if $y(x)=\sum_{i=0}^{\infty}a_{i}\,x^{\nu_{i}}\in\Omega$ is a
solution of a non trivial polynomial ordinary differential equation
$P(y,y^{\prime},\ldots,y^{(n)})=0$ with coefficients in the field of formal
Laurent series $\mathbb{C}((x))$, then its support
$\operatorname{supp}y(x)=\\{\nu_{i}\mid a_{i}\neq 0\\}$ (the set of exponents
with non-zero coefficient), generates a finite $\mathbb{Z}$-module but they
provide no information on its rational rank. Our first aim (Theorem 1) is to
prove in an elementary way that, modulo the support of the equation, this
rational rank is at most the order of the equation. Our arguments work almost
word by word for $q$-difference equations. Our proof of Theorem 1 allows us to
show that in both cases, when the rational rank of the solution reaches its
maximum possible value, then the solution is necessarily convergent (assuming
$P$ is a polynomial), which is Theorem 2. We make use of the convergence
results of Gontsov, Goryuchkina and Lastra [14, 12]. Theorem 3 addresses the
possibility of completing an initial segment $s_{0}(x)=\sum_{1\leq i\leq
k}c_{i}\,x^{\mu_{i}}$ to an actual power series $s(x)$ solution of the
differential or $q$-difference equation $P=0$. Finally, Theorem 4 is the
consequence of applying the previous results to the case of autonoumous first
order differential equations, providing new proofs about the existence and
convergence of Puiseux solutions of these equations given in [8].

Other authors have proved related finiteness results in other contexts [16, 3,
17, 6, 7], but among these, none gives effective bounds for the rational rank
of the support of the solutions. The unique effective bound we have knowledge
of is by van der Hoeven, in [10] (Corollary 8.38), who shows in the language
of transseries that if $P(y,y^{\prime},\ldots,y^{(n)})=0$ is an algebraic
differential equation and $f$ is a transseries solution, then _essentially_
the rational rank of the semigroup of monomials of $f$ can only increase by
the order $n$ of $P$. This is the same result as our Theorem 1 in the case of
differential equations, and over a field which neither contains nor is
contained in $\Omega$. Our arguments have the advantage of simplicity and
conciseness.

Rosenlicht, in the context of Hardy fields [18] shows that the rank (the
number of comparability classes) of the differential field generated by a
solution of the differential equation is bounded by the order of the equation.
We, however, work over $\Omega$ which has rank $1$, and his results do not
give insight about the rational rank of the support of the solution.

In this paper we work indistinctly with ordinary differential equations and
$q$-difference equations. Given a polynomial
$P\in\Omega[y_{0},y_{1},\ldots,y_{n}]$, the equation $P=0$ means
$P(y(x),y^{\prime}(x),\ldots,y^{(n)}(x))=0$, with the assumption that
$y^{\prime}(x)$ refers to either the usual derivation operator
$y^{\prime}(x)=\frac{d\,y(x)}{d\,x}$, the Euler derivation
$y^{\prime}(x)=x\,\frac{d\,y(x)}{d\,x}$ or $y^{\prime}(x)=y(q\,x)$, for some
$q\in\mathbb{C}$, with $|q|\neq 1$. We do not consider mixed equations.

Given $P\in\Omega[y_{0},y_{1},\ldots,y_{n}]$ we denote by
$\operatorname{supp}P$ the union of the supports of the coefficients of $P$,
and if $E$ is a subset of $\mathbb{R}$ then $\left\langle E\right\rangle$ is
the $\mathbb{Q}$-linear subspace of $\mathbb{R}$ generated by $E$. Our first
result is Theorem 1 which states that if $s(x)\in\Omega$ is a solution of
$P=0$ then

(1)
$\dim_{\mathbb{Q}}\frac{\left\langle\operatorname{supp}s(x)\cup\operatorname{supp}P\right\rangle}{\left\langle\operatorname{supp}P\right\rangle}\leq
n.$

In particular, if the coefficients of $P$ are formal power series with
integral exponents, then the maximal number of rationally independent
irrational exponents that can appear in $\operatorname{supp}s(x)$ is the order
$n$ of the equation $P=0$.

The second result of this paper address the question of the convergence of the
solution $s(x)$ provided $P$ is a polynomial in $x,y_{0},\ldots,y_{n}$. It is
well known that the convergence of a solution is not guaranteed: Euler’s
example $y-x^{2}\,y^{\prime}-x=0$ has as solution the formal power series
$\sum_{n\geq 0}n!\,x^{n+1}$. B. Malgrange in [15] gives a sufficient condition
for a formal power series solution $s(x)$ of a polynomial differential
equations $P=0$ to be convergent in terms of the linearized differential
operator along the solution, $L=\sum_{i=0}^{n}\frac{\partial P}{\partial
y_{i}}(s(x),\ldots,s^{(n)}(x))\,\delta^{i}$: if $L$ has a regular singularity
at the origin then the solution $s(x)$ is convergent. This criterion has been
extended in the differential case by R. Gontsov and I. Goryunchkina in [14]
and in the $q$-difference case by R. Gontsov and I. Goryunchkina and A. Lastra
in [12] for power series solutions with complex exponents. In Theorem 2 of
this paper we provide a new sufficient condition for the convergence of the
solution $s(x)$ which only depends on the order $n$ of the differential or
$q$-difference equation $P=0$, and on the support of the solution.
Specifically, if $\langle\operatorname{supp}s(x)\rangle$ has the maximal
possible rational rank (Equation (1) is an equality) then $s(x)$ is
necessarily convergent. In particular, if $P$ is of order one and with
constant coefficients, any solution of the equation $P=0$ is convergent.

Our third main result addresses the question of the possibility of completing
an initial segment $s_{0}(x)=\sum_{1\leq i\leq k}c_{i}\,x^{\mu_{i}}$ to an
actual power series solution $s(x)$ solution of the equation $P=0$. Given an
equation $P=0$, the Newton polygon method ([13, 6, 7]), provides, for any
positive integer $k$, a finite family of necessary initial conditions
$\operatorname{NIC}_{k}(P)$, such that the first $k$ coefficients and
exponents of any solution $\sum_{i=1}^{\infty}a_{i}\,x^{\nu_{i}}$ of $P=0$
satisfy $\operatorname{NIC}_{k}(P)$. We say that $s_{0}(x)$ is an _admissible
initial segment_ for $P=0$ if its coefficients and exponents satisfy
$\operatorname{NIC}_{k}(P)$.

In the algebraic case (i.e. $n=0$), Puiseux’s Theorem shows that any
admissible initial segment for $P=0$ is the truncation of an actual solution.
This is no longer true for differential and $q$-difference equations: even in
order and degree one, there are examples [4, 2] of admissible initial segments
which cannot be completed to a solution. Section 3 of [13] introduces the
concept of _stabilization_ as a criterion which guarantees the existence (and
uniqueness) of an actual solution with initial segment $s_{0}(x)$. In Theorem
3 we introduce a criterion of a different nature: if the support of an
admissible polynomial $s_{0}(x)$ for $P=0$ has $n$ linearly independent
irrational numbers over $\mathbb{Q}$ (modulo
$\langle\operatorname{supp}P\rangle$) then $s_{0}(x)$ is in fact the
truncation of an actual solution of $P=0$. Indeed, there can be more than one
solution $s(x)$ with the same truncation $s_{0}(x)$, unlike the stabilization
criterion, for which the solution is unique (see example in Section 5). In
particular, if $P=P(x,y)=0$ is a polynomial in $x,y$ (i.e. the equation of an
algebraic curve), then the hyphothesis of Thereom 3 always hold, so that it
may be understood as a kind of generalization of Puiseux’s Theorem to
differential and $q$-difference equations.

Section 2 introduces the context and a technical result from which the two
main theorems follow in Section 3. Then, in Section 4 we recall the essential
notions and results on the Newton polygon process (see [13, 5, 4, 6, 7, 2]),
which we use to prove the technical Proposition 1, from which Theorem 3
follows straightforwardly. In subsection 4.3 we apply the previous results to
the case of autonomous ordinary differential equations. In Section 5 we give a
detailed example to illustrate our main results.

Notice that instead of generalized power series in $x$, we could have worked
with generalized power series in $x^{-1}$. The results and proves are the same
with the obvious modifications.

##  2\. Notation and a Technical Lemma

Following [13], we denote by $\Omega$ the field of formal power series
$s(x)=\sum c_{i}x^{\mu_{i}}$ with complex coefficients and real exponents,
such that $\lim_{i\rightarrow\infty}\mu_{i}=\infty$, where we always assume
that $\mu_{1}<\mu_{2}<\cdots$. The _support_ of a power series $s(x)$ is the
set of exponents with non-zero coefficient:
$\operatorname{supp}s(x)=\\{\mu_{i}\ :\ c_{i}\neq 0\\}$. Given an additive
subgroup $\Gamma$ of $\mathbb{R}$ we denote by
$\mathbb{C}((x^{\Gamma}))\subset\Omega$ the subfield of series $s(x)\in\Omega$
whose support is contained in $\Gamma$. We also denote by
$\hat{K}=\mathbb{C}((x^{\mathbb{Z}}))$ the field of fractions of the ring of
formal power series.

We are going to deal with differential and $q$-difference equations
simultaneously, hence it should be more convenient to use the Euler derivative
in the case of differential equations, but some of our results are specific
for the case of differential equations with the ordinary derivation, in
particular, the case of autonomous equations. Hence we deal simultaneously
with differential equations in terms of the ordinary differential operator, in
terms of the Euler differential operator and with $q$-difference equations,
with $|q|\neq 1$.

Thorought this paper $P$ will denote a polynomial $P(y_{0},\ldots,y_{n})$ in
the indeterminates $y_{0},y_{1},\ldots,y_{n}$ with coefficients in
$\mathbb{C}((x^{\Gamma}))$ for $\Gamma\subset\mathbb{R}$ as above. Given
$s(x)\in\Omega$, we set $P(s(x))=P(s(x),s^{\prime}(x),\ldots,s^{(n)}(x))$,
where $s^{\prime}(x)$ is either the ordinary derivative
$\frac{d\,s(x)}{d\,x}$, the Euler derivative $x\,\frac{d\,s(x)}{d\,x}$ or the
$q$-difference operator $s(q\,x)$, and $s^{(\kappa)}(x)$ is the $\kappa$-th
iteration. We do not consider mixed equations.

If $s(x)=\sum_{i=0}^{\infty}c_{i}\,x^{\mu_{i}}$, and $\kappa=0,\ldots,n$ then
we write the $\kappa$-th iteration of the operator ′ as:

$s^{(\kappa)}(x)=\sum_{i=0}^{\infty}c_{i}\,\delta_{\mu_{i}}^{(\kappa)}\,x^{\mu_{i}-\varepsilon\kappa},$

where $\varepsilon=1$ in the case of differential equations with the ordinary
differential operator $\frac{d\,}{d\,x}$ or $\varepsilon=0$ for differential
equations with the Euler differential operator, and $q$-difference equations;
the symbol $\delta_{\mu}^{(\kappa)}$ is defined as
$\delta_{\mu}^{(\kappa)}=\mu\,(\mu-1)\cdots(\mu-\kappa+1)$ for differential
equations with the operator $\frac{d\,}{d\,x}$,
$\delta_{\mu}^{(\kappa)}=\mu^{\kappa}$ for differential equations with the
Euler differential operator $x\,\frac{d\,}{d\,x}$, and
$\delta_{\mu}^{(\kappa)}=q^{\kappa\,\mu}$ for $q$-difference equations. In
particular $\delta_{\mu}^{(0)}=1$ in all cases, and we denote
$\delta_{\mu}=\delta_{\mu}^{(1)}$.

The expression $P=0$ denotes the corresponding equation of order $n$. A
_solution_ of $P=0$ is a power series $s(x)$ such that
$P(s(x),s^{\prime}(x),\ldots,s^{(n)}(x))=0$. The _support_ of $P$ is the union
of the supports of its coefficients.

Our first two results rely on the following elementary fact about derivations
with respect to _transcendent monomials_. Fix $\Gamma$ an additive subgroup of
$\mathbb{R}$ and consider a family of exponents
$\mu_{1},\ldots,\mu_{m}\in\mathbb{R}$ whose classes modulo
$\left\langle\Gamma\right\rangle$,
$\bar{\mu}_{1},\ldots,\bar{\mu}_{m}\in\mathbb{R}/\langle\Gamma\rangle$ are
$\mathbb{Q}$-linearly independent. Let
$\Gamma^{\prime}=\langle\Gamma\cup\\{\mu_{1},\ldots,\mu_{m}\\}\rangle$. By
definition, any $\alpha\in\Gamma^{\prime}$ can be written uniquely as

$\alpha=[\alpha]_{\Gamma}+\sum_{j=1}^{m}[\alpha]_{\mu_{j}}\,\mu_{j},$

where $[\alpha]_{\Gamma}\in\left\langle\Gamma\right\rangle$ and
$[\alpha]_{\mu_{j}}\in\mathbb{Q}$.

######  Lemma 1.

With the above notation and conditions, for any $j=1,\ldots,m$, there exists
derivation $\mathcal{D}_{j}$ on $\mathbb{C}((x^{\Gamma^{\prime}}))$ such that

$\mathcal{D}_{j}\left(\sum_{i=1}^{\infty}c_{i}\,x^{\alpha_{i}}\right)=\sum_{i=1}^{\infty}c_{i}\,[\alpha_{i}]_{\mu_{j}}\,x^{\alpha_{i}}.$

###### Proof.

The expression above is obviously a well-defined map in
$\mathbb{C}((x^{\Gamma^{\prime}}))$, due to the linear independence of the
classes. A straightforward computation shows that it is linear and satisfies
the Leibniz rule. ∎

Let us remark that the above derivation $\mathcal{D}_{j}$ depends on $\Gamma$
and $\mu_{1},\ldots,\mu_{m}$ although we do not stress this fact in the
notation. Notice also that $\mathcal{D}_{j}$ vanishes on
$\mathbb{C}((x^{\Gamma}))$ and on $\mathbb{C}((x^{\Gamma_{j-1}}))$, for
$\Gamma_{j-1}=\langle\Gamma\cup\\{\mu_{1},\ldots,\mu_{j-1}\\}\rangle$. Let us
give some examples: Assume that $\Gamma=\\{0\\}$, $\mu_{1}=1$ and
$\mu_{2}=\pi$, hence $\mathcal{D}_{1}(x^{2+3\,\pi})=2\,x^{2+3\,\pi}$ and
$\mathcal{D}_{2}(x^{2+3\,\pi})=3\,x^{2+3\,\pi}$; now assume that
$\Gamma=\pi\,\mathbb{Z}$, $\mu_{1}=2$ and $\mu_{2}=e$, hence
$\mathcal{D}_{1}(x^{2+3\pi/2})=x^{2+3\pi/2}$,
$\mathcal{D}_{2}(x^{2+3\pi/2})=0$, and
$\mathcal{D}_{2}(x^{\pi+5\,e})=5\,x^{\pi+5\,e}$.

In what follows, we shall not usually work with the substitution $P(s(x))$ for
$s(x)\in\Omega$, but with the more useful “full substitution”:

$P[s(x)]=P(s(x)+y_{0},s^{\prime}(x)+y_{1},\ldots,s^{(n)}(x)+y_{n})\in\Omega[y_{0},y_{1},\ldots,y_{n}].$

By the chain rule:

$\frac{\partial P}{\partial y_{\kappa}}[s(x)]=\frac{\partial}{\partial
y_{\kappa}}(P[s(x)]),\quad\kappa=0,1,\ldots,n.$

For $P\in\mathbb{C}((x^{\Gamma}))[y_{0},\ldots,y_{n}]$ and
$\mu_{1},\ldots,\mu_{m}$ and $\Gamma^{\prime}$ as in Lemma 1, take
$s(x)\in\mathbb{C}((x^{\Gamma^{\prime}}))$. Extending the derivation
$\mathcal{D}_{j}$ to the ring
$\mathbb{C}((x^{\Gamma^{\prime}}))[y_{0},\ldots,y_{n}]$ by acting trivially on
the indeterminates $y_{0},y_{1},\ldots,y_{n}$ we get, as the $\mathcal{D}_{j}$
are derivations vanishing on $\mathbb{C}((x^{\Gamma}))$, the equality:

(2)
$\mathcal{D}_{j}\left(P[s(x)]\right)=\sum_{\kappa=0}^{n}\frac{\partial\,}{\partial
y_{\kappa}}(P[s(x)])\,\mathcal{D}_{j}(s^{(\kappa)}(x)).$

We introduce some notation useful for studying the coefficients of $P[s(x)]$.
Given $\rho=(\rho_{0},\ldots,\rho_{n})\in\mathbb{Z}_{\geq 0}^{n+1}$, the
expression $P[s(x)]_{\rho}$ will denote the coefficient of
$y_{0}^{\rho_{0}}y_{1}^{\rho_{1}}\cdots y_{n}^{\rho_{n}}$ in $P[s(x)]$. We
shall use the basic vectors
$e_{0}=(1,0,\ldots,0),\ldots,e_{n}=(0,\ldots,0,1)$, each having $n+1$
components (we start at $0$ because $e_{i}$ will be related to $y_{i}$).
Examining each term on both sides of (2), we obtain the key equality:

(3)
$\mathcal{D}_{j}(P[s(x)]_{\rho})=\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)(P[s(x)]_{\rho+e_{\kappa}})\,\mathcal{D}_{j}(s^{(\kappa)}(x)).$

##  3\. Solutions and Rational Rank

Our first theorem follows mainly from Lemma 1 and Equality (3).

######  Theorem 1.

Let $P\in\Omega[y_{0},\ldots,y_{n}]$ be a non-zero polynomial, and
$s(x)\in\Omega$ a solution of the equation $P=0$. Then:

$\dim_{\mathbb{Q}}\frac{\left\langle\operatorname{supp}s(x)\cup\operatorname{supp}P\right\rangle}{\left\langle\operatorname{supp}P\right\rangle}\leq
n.$

Thus we obtain a bound for the rational rank of solutions of equations with
coefficients Puiseux series.

######  Corollary 1.

If $s(x)\in\Omega$ is a generalized power series with

$\dim_{\mathbb{Q}}\frac{\langle\operatorname{supp}s(x)\cup\mathbb{Q}\rangle}{\mathbb{Q}}>n$

then $s(x)$ is not the solution of a nontrivial equation of order $n$ over the
fraction field of the ring of Puiseux series.

######  Corollary 2.

If $s(x)\in\Omega$ is a generalized power series with

$\dim_{\mathbb{Q}}\langle\operatorname{supp}s(x)\rangle>n$

then $s(x)$ is not the solution of a nontrivial equation $P=0$, of order $n$
with constant coefficients.

For instance $x+x^{\pi}$ is not a solution of a non-trivial equation $P=0$ of
order one and constant coefficients, that is, with
$P(y_{0},y_{1})\in\mathbb{C}[y_{0},y_{1}]$. We will improve this result in the
case of autonomous first order differential equations (see Theorem 4).

Our second main result deals with the case when $P$ is a polynomial in $x$
also. Notice that in this case $\langle\operatorname{supp}P\rangle$ is either
$\\{0\\}$ if $P$ has constant coefficients or $\mathbb{Q}$ otherwise.

######  Theorem 2.

Take $P\in\mathbb{C}[x,y_{0},\dots,y_{n}]$ with $P\neq 0$ and assume
$s(x)\in\Omega$ is a solution of $P=0$ whose support has maximal rational
rank, that is

$\dim_{\mathbb{Q}}\frac{\left\langle\operatorname{supp}s(x)\cup\operatorname{supp}P\right\rangle}{\langle\operatorname{supp}P\rangle}=n.$

Then $s(x)$ converges uniformly in any sector $S$ of sufficiently small radius
with vertex at the origin and of the opening less than $2\,\pi$.

The following result covers Theorem 11 in [8] and is a consequence of the
previous Theorem for the case of first order equations with constant
coefficients:

######  Corollary 3.

Let $P(y_{0},y_{1})$ a polynomial with constant coefficients. Let
$s(x)\in\Omega$ be a solution of the equation $P=0$. Then $s(x)$ is convergent
in the sense of Theorem 2. Moreover, if $s(x)$ is a Puiseux series then it is
convergent in a neighbourhood of the origin.

###### Proof.

Either $s(x)$ is a constant or
$\dim_{\mathbb{Q}}\langle\operatorname{supp}s(x)\rangle\geq 1$. Since
$\langle\operatorname{supp}P\rangle=\\{0\\}$, by Theorem 1,
$\dim_{\mathbb{Q}}\langle\operatorname{supp}s(x)\rangle=1$ and the result
follows from Theorem 2. ∎

We divide the proofs of Theorems 1 and 2 into a shared initial part and their
respective ends.

_Common part of the proof of Theorems 1 and 2_. Let
$s(x)=\sum_{i=1}^{\infty}a_{i}\,x^{\alpha_{i}}\in\Omega$ be a solution of the
non-trivial equation $P=0$. We may assume without loss of generality that for
some $0\leq\kappa\leq n$, $\frac{\partial P}{\partial y_{\kappa}}(s(x))\neq
0$: otherwise we proceed by induction on the total degree of $P$, as
$\frac{\partial P}{\partial y_{\kappa}}$ and $s(x)$ would satisfy the
hypotheses of both results. Thus, we can define:

$\lambda=\min\\{\operatorname{ord}_{x}\frac{\partial P}{\partial
y_{\kappa}}(s(x))-\varepsilon\,\kappa\mid\,0\leq\kappa\leq
n\\},\,\,\Lambda=\min\\{\operatorname{ord}_{x}\frac{\partial^{\rho}P}{\partial
y^{\rho}}(s(x))\mid\,\rho\in\mathbb{Z}_{\geq 0}^{n+1}\\},$

This $\lambda$ corresponds to the one defined in [12] and $\Lambda$ is just an
auxiliary constant to be used in the forthcoming proof.

Consider

$m_{0}=\dim_{\mathbb{Q}}\frac{\langle\operatorname{supp}s(x)\cup\operatorname{supp}P\rangle}{\langle\operatorname{supp}P\rangle}\in\mathbb{Z}_{\geq
0}\cup\\{\infty\\},$

which in theory could be infinity. Part of our argument will consist in
proving that it cannot be. In the proof of Theorem 1 we shall assume that
$m_{0}\geq n+1$ and argue by contradiction, whereas in the proof of Theorem 2,
we shall assume that $m_{0}=n$, as this is equivalent to the hypothesis.

Let $\Gamma_{0}=\left\langle{\operatorname{supp}P}\right\rangle$. For $1\leq
j\leq m_{0}$ (if $m_{0}=\infty$ the last inequality is strict), define
$\mu_{j}$ and $\Gamma_{j}$ inductively as follows: Assume that
$\Gamma_{0},\Gamma_{1},\ldots,\Gamma_{j-1}$ and,
$\mu_{1},\mu_{2},\ldots,\mu_{j-1}$ have been defined (so that for $j=1$ only
$\Gamma_{0}$ exists). By the definition of $m_{0}$, there exists the minimum
$k$ of the set of indexes $i$ such that $a_{i}\neq 0$ and
$\alpha_{i}\not\in\Gamma_{j-1}$. We set $\mu_{j}=\alpha_{k}$ and
$\Gamma_{j}=\left\langle{\Gamma_{j-1}\cup\\{\mu_{j}\\}}\right\rangle$.

As $P(s(x))=0$, we can choose $N$ large enough such that the truncation
$\bar{s}(x)=\sum_{i=1}^{N}a_{i}\,x^{\alpha_{i}}$ of $s(x)$ satisfies the
following property:

(4) $\operatorname{ord}(s(x)-\bar{s}(x))>|\lambda|+2\,n+|\Lambda|,$

which is possible because either $s(x)$ has a finite number of terms or its
exponents tend to infinity. Notice that $\bar{s}(x)$ could be equal to $s(x)$
if the latter has a finite number of terms.

Once $N$ is set, there exists a finite $m$ such that we can rewrite
$\bar{s}(x)$ as follows:

(5)
$\bar{s}(x)=\sum_{j=0}^{m}\big{(}c_{j,0}\,x^{\mu_{j,0}}+c_{j,1}\,x^{\mu_{j,1}}+\cdots+c_{j,t_{j}}\,x^{\mu_{j,t_{j}}}\big{)},$

where the following properties hold:

  1. (i)

The sequence of exponents $(\mu_{j,i})$ is increasing with respect the
lexicographical order of their indices $(j,i)$.

  2. (ii)

The exponents $\mu_{j,i}\in\Gamma_{j}$, for $i=0,\ldots,t_{j}$ and
$j=0,\ldots,m$.

  3. (iii)

All the coefficients $c_{j,i}$ are non-zero for $j=1,\ldots,m$ and $0\leq
i\leq t_{j}$.

  4. (iv)

The exponent $\mu_{j,0}=\mu_{j}\not\in\Gamma_{j-1}$ for $j=1,..,m$, that is,
it adds one to the rational rank.

By properties (i)–(iv), the set $\Gamma=\Gamma_{0}$, the exponents
$\mu_{1},\ldots,\mu_{m}$ and $\Gamma_{m}$ satisfy the hypothesis of Lemma 1.
In order to make $\mathbb{C}((x^{\Gamma_{m}}))$ a differential field with the
operator $\frac{d\,}{d\,x}$ we do the following: If $1\not\in\Gamma_{m}$, we
set $\Gamma^{\prime}=\langle\Gamma_{m}\cup\\{1\\}\rangle$ otherwise we set
$\Gamma^{\prime}=\Gamma_{m}$. Hence $\Gamma$ and $\mu_{1},\ldots,\mu_{m}$, and
eventually $1$, satisty hyphotesis of Lemma 1 and
$\mathbb{C}((x^{\Gamma^{\prime}}))$ is closed under the three operators that
we use. Any coefficient of $P[\bar{s}(x)]$ belongs also to
$\mathbb{C}((x^{\Gamma^{\prime}}))$, and we may apply the derivations
$\mathcal{D}_{j}$, $j=1,\ldots,m$, to these coefficients (if
$1\not\in\Gamma_{m}$, we will not use the operator $\mathcal{D}_{m+1}$
corresponding to $1$).

Noticing that $P[\bar{s}(x)]_{e_{k}}=\frac{\partial P}{\partial
y_{\kappa}}(\bar{s}(x))$, Equation (3) applied to $\rho=(0,0,\ldots,0)$ gives:

(6) $\mathcal{D}_{j}(P(\bar{s}(x)))=\sum_{\kappa=0}^{n}\frac{\partial
P}{\partial y_{\kappa}}(\bar{s}(x))\,\mathcal{D}_{j}(\bar{s}^{(\kappa)}(x))$

From now on, the notation $t(x)=ax^{\mu}+\cdots$ will mean that
$a\in\mathbb{C}$ ($a$ may be $0$), and
$\operatorname{ord}_{x}(t(x)-a\,x^{\mu})>\mu$. By definition of $\lambda$, we
can write, for $\kappa\in 0,\ldots,n$:

$\frac{\partial P}{\partial
y_{\kappa}}(s(x))=d_{\kappa}\,x^{\lambda+\varepsilon\kappa}+\cdots$

and we know that there is at least one $\kappa\in\\{0,\ldots,n\\}$ with
$d_{\kappa}\neq 0$. Using the Taylor expansion of the left hand side of this
equality and property (4), we also obtain

$\frac{\partial P}{\partial
y_{\kappa}}(\bar{s}(x))=d_{\kappa}\,x^{\lambda+\varepsilon\,\kappa}+\cdots.$

From the properties of the derivations $\mathcal{D}_{j}$, and because
$[\mu_{j,0}]_{\mu_{j}}=1$ and $[\mu_{j^{\prime},i}]_{\mu_{j}}=0$ for every
$(j^{\prime},i)<(j,0)$, for $j=1,\ldots,m$, we have:

$\mathcal{D}_{j}(\bar{s}^{(\kappa)}(x))=c_{j,0}\,\delta_{\mu_{j}}^{(\kappa)}\,x^{\mu_{j}-\varepsilon\kappa}+\cdots.$

Hence, the right hand side of Equation (6) is

(7)
$\sum_{\kappa=0}^{n}(d_{\kappa}\,x^{\lambda+\varepsilon\,\kappa}+\cdots)(c_{j,0}\,\delta_{\mu_{j}}^{(\kappa)}\,x^{\mu_{j}-\varepsilon\,\kappa}+\cdots)=c_{j,0}\left(\sum_{\kappa=0}^{n}d_{\kappa}\,\delta_{\mu_{j}}^{(\kappa)}\right)x^{\lambda+\mu_{j}}+\cdots.$

This finishes the common arguments.

###### End of proof of Theorem 1.

In order to obtain a contradiction, we assume that $m_{0}>n$ so that
$\mu_{n+1}$ is defined. This allows us to choose $N$ large enough such that
$m\geq n+1$ and $\operatorname{ord}_{x}P(\bar{s}(x))>\lambda+\mu_{n+1}$. Now
$\mu_{n+1,0}=\mu_{n+1}$, and all this properties imply that the left hand side
of equation (6) has order greater than $\lambda+\mu_{n+1,0}$. Since
$c_{j,0}\neq 0$, using (7) we obtain the $n+1$ equalities:

(8) $\sum_{\kappa=0}^{n}d_{\kappa}\,\delta_{\mu_{j}}^{(\kappa)}=0,\quad
j=1,2,\ldots,n+1,$

which is a square linear system with coefficient matrix
$(\delta_{\mu_{j}}^{(\kappa)})$. Because $|q|\neq 1$ and $\mu_{j}$ is an
increasing sequence of real numbers, then $q^{\mu_{j}}\neq q^{\mu_{i}}$ for
$i\neq j$. Thus the coefficient matrix is a Vandermonde matrix in the case of
$q$-difference equations and in the case of differential equations with the
Euler operator. In the case of differential equations with the ordinary
differential operator, the matrix $(\delta_{\mu_{j}}^{(\kappa)})$ is reduced
by elementary column operations to the Vandermonde matrix
$(\mu_{j}^{\kappa})$. Hence, system (8) has the unique solution
$d_{0}=d_{1}=\cdots=d_{n}=0$, contradicting the fact that at least one
$d_{\kappa}$ is non-zero. Thus, $m_{0}\leq n$ which gives Theorem 1. ∎

###### End of proof of Theorem 2.

As $m_{0}=n$ in this case, $\mu_{n}$ is defined and we can choose $N$ such
that $\operatorname{ord}_{x}P(\bar{s}(x))>\lambda+\mu_{n}$, and $m=n$. Thus
$\mu_{n,0}=\mu_{n}$ is also defined. The left hand side of equation (6) has
order greater than $\lambda+\mu_{n,0}$ and we obtain the system of equations
(8) for $j=1,2,\ldots,n$. If $d_{n}=0$, this system of equations becomes
another Vandermonde system for $d_{0},d_{1},\ldots,d_{n-1}$, so that
$d_{0}=d_{1}=\cdots=d_{n-1}=d_{n}=0$, again contradicting the existence of one
non-zero $d_{k}$. As a consequence, $d_{n}\neq 0$, so that
$\operatorname{ord}\frac{\partial P}{\partial
y_{n}}(s(x))=\lambda+\varepsilon\,n$, and we get:

(9) $\operatorname{ord}\frac{\partial P}{\partial
y_{n}}(s(x))-\varepsilon\,n=\lambda\leq\operatorname{ord}\frac{\partial
P}{\partial y_{\kappa}}(s(x))-\varepsilon\,\kappa,\quad\text{for
}\kappa=0,1,\ldots,n.$

This means that the linearized differential operator
$\sum_{\kappa=0}^{n}\frac{\partial P}{\partial y_{\kappa}}(s(x))\,y_{\kappa}$
along $s(x)$ has a regular singularity at $x=0$. If $s(x)$ is a Puisuex formal
power series, we apply the main result in [15] to guarantee that $s(x)$ is
convergent in both differential cases. If $s(x)\in\Omega$, Condition (9) is
the hypothesis of Theorem 1 of [14], which proves the Euler differential case.
To prove the ordinary differential case we only need to rewrite $P$ in terms
of the Euler operator. Condition (9) becomes then the hypothesis of Theorem 1
of [14] again.

For the case of $q$-difference equations we need to distinguish the cases
$|q|>1$ and $|q|<1$, as in [12]. If $|q|>1$, the fact that $d_{n}\neq 0$ and
that $\mu_{j}$ is an increasing sequence of positive real numbers let us apply
Theorem 1 of [12] straightforwardly. If $|q|<1$, in order to apply the same
Theorem we need to show that $d_{0}\neq 0$. As in the previous argument, if
$d_{0}=0$, then (8), for $j=1,\ldots,n$ becomes a Vandermonde system for
$d_{1},\ldots,d_{n}$, whose only solution is $d_{1}=\cdots=d_{n}=0$, getting
the same contradiction. Thus $d_{0}\neq 0$ and we can apply Theorem 1 of [12]
again to obtain the convergence. ∎

##  4\. The Newton-Puiseux Polygon

In this section we give a short description of the well known method of the
Newton polygon (or Newton-Puiseux) applied to polynomial differential and
$q$-difference equations, which we shall use to prove Theorem 3.

Let us introduce some notation: For
$\rho=(\rho_{0},\ldots,\rho_{n})\in\mathbb{Z}_{\geq 0}^{n+1}$ we denote
$|\rho|=\rho_{0}+\rho_{1}+\cdots+\rho_{n}$ and
$\omega(\rho)=\rho_{1}+2\,\rho_{2}+\cdots+n\,\rho_{n}$. As before, we set
$\varepsilon=0$ in the cases of differential equations with respect the Euler
derivative or $q$-difference equations, and $\varepsilon=1$ in the case of
differential equations with respect the derivative $\frac{d\,}{d\,x}$.

Fix a non-zero polynomial $P\in\Omega[y_{0},\ldots,y_{n}]$, and write it
uniquely as

$P=\sum_{\rho\in\mathbb{Z}_{\geq
0}^{n+1}}\sum_{\alpha\in\operatorname{supp}(P)-\varepsilon\,\omega(\rho)}P_{(\alpha,\rho)}\,x^{\alpha+\varepsilon\,\omega(\rho)}y_{0}^{\rho_{0}}\cdots
y_{n}^{\rho_{n}}$

with $P_{(\alpha,\rho)}\in\mathbb{C}$. Given
$V=(\alpha,r)\in\mathbb{R}\times\mathbb{Z}_{\geq 0}$, $P_{V}$ will denote the
sum of all the terms of $P$ corresponding to the point $V$, that is

$P_{V}=P_{(\alpha,r)}=\sum_{(\alpha,|\rho|)=V}P_{(\alpha,\rho)}x^{\alpha+\varepsilon\,\omega(\rho)}y_{0}^{\rho_{0}}\cdots
y_{n}^{\rho_{n}}.$

There is no confusion possible between $P_{(\alpha,r)}$ and
$P_{(\alpha,\rho)}$ except when $n=0$, in which case we shall abuse the
notation as the context will clarify what value we are using. The _cloud of
points_ $\mathcal{C}(P)$ of $P$ is the set of points in the plane:

$\mathcal{C}(P)=\left\\{(\alpha,r)\in\mathbb{R}\times\mathbb{Z}_{\geq
0}:P_{(\alpha,r)}\neq 0\right\\}.$

We are interested in generalized power series with exponents in increasing
order. In this setting, the _Newton-Puiseux polygon_ (Newton polygon for
short) of $P$, denoted $\mathcal{N}(P)$, is the convex envelope in
$\mathbb{R}^{2}$ of the set obtained by adjoining the half-line
$\mathbb{R}_{\geq 0}\times\\{0\\}$ to each point in $\mathcal{C}(P)$:

$\mathcal{N}(P)=\mathrm{conv.env.}\left\\{(\alpha,r)+(\mathbb{R}_{\geq
0}\times\\{0\\}):(\alpha,r)\in\mathcal{C}(P)\right\\}.$

Its border is composed of a sequence of points and segments. Given a positive
number $\mu\in\mathbb{R}$, the _supporting line_ of co-slope $\mu$, is the
unique line with equation $L_{\mu}(P)\equiv\mu\,r+\alpha=\alpha_{0}$ with
$\alpha_{0}$ minimum and $L_{\mu}(P)\cap\mathcal{N}(P)\neq\emptyset$. The
_element of co-slope $\mu$ of $\mathcal{N}(P)$_, $E_{\mu}(P)$ is that
intersection $L_{\mu}(P)\cap\mathcal{N}(P)$, which can be either a segment
(called a _side_ of $\mathcal{N}(P)$) or a point (a _vertex_). In both cases
we denote by $\operatorname{Top}(E_{\mu}(P))$ and
$\operatorname{Bot}(E_{\mu}(P))$ the highest and lowest points of
$E_{\mu}(P)$, respectively (if $E$ is a vertex, they coincide).

Given $V=(\alpha,r)\in\mathbb{R}\times\mathbb{Z}_{\geq 0}$, the _indicial
polynomial_ of $P$ at $V$ is

$\Psi_{(P;V)}(T)=\sum_{(\alpha,|\rho|)=V}P_{(\alpha,\rho)}\,T^{\langle\rho\rangle}\in\mathbb{C}[T],$

where $T^{\langle\rho\rangle}$ is equal to $T^{\omega(\rho)}$ for the Euler
differential operator or the $q$-difference operator, and
$T^{\langle\rho\rangle}=\prod_{\kappa=1}^{n}\big{(}T\,(T-1)\cdots(T-\kappa+1)\big{)}^{\rho_{\kappa}}$
for the differential operator $\frac{d\,}{d\,x}$.

The _characteristic polynomial_ of $P$ with respect to a co-slope
$\mu\in\mathbb{R}_{\geq 0}$ is

$\Phi_{(P;\mu)}(C)=\sum_{(\alpha,|\rho|)\in
E_{\mu}(P)}\delta_{\mu}^{\langle\rho\rangle}\,P_{(\alpha,\rho)}\,C^{|\rho|}=\sum_{V\in
E_{\mu}(P)}\Psi_{(P;V)}(\delta_{\mu})\,C^{\operatorname{ht}(V)}\in\mathbb{C}[C],$

where $\operatorname{ht}(V)$ is the ordinate of $V$. The key Lemma of the
Newton polygon process gives the following necessary condition, which shows
the importance of the characteristic polynomial (see [6, 7] for a short
proof):

######  Lemma 2.

Let $s(x)=c\,x^{\mu}+\sum_{\alpha>\mu}c_{\alpha}\,x^{\alpha}\in\Omega$ be a
solution of $P=0$. Then

$\Phi_{(P;\mu)}(c)=0.$

In fact, this lemma translates into a sequence of necessary conditions for a
power series to be a solution of $P=0$. Given
$s(x)=\sum_{i=0}^{\infty}c_{i}\,x^{\nu_{i}}\in\Omega$, with
$\nu_{i}<\nu_{i+1}$ for all $i\geq 0$, denote, for the sake of simplicity:

$P_{0}=P,\quad\text{and}\quad
P_{i+1}=P_{i}[c_{i}\,x^{\nu_{i}}]=P[c_{0}x^{\nu_{0}}+\cdots+c_{i}\,x^{\nu_{i}}],\quad
i=1,2,\ldots.$

If $s(x)$ is a solution of $P=0$, then for each $i\in\mathbb{Z}_{\geq 0}$, the
series $\sum_{l=i}^{\infty}c_{l}\,x^{\nu_{l}}$ must be a solution of
$P_{i}=0$. Thus, we obtain the sequence of necessary initial conditions on the
coefficients of $s(x)$ when it is a solution of $P=0$:

(10) $\Phi_{(P_{i};\nu_{i})}(c_{i})=0,\quad i=0,1,\ldots.$

We need, however, to check whether a finite power series is effectively the
truncation of a solution. To this end, we introduce the following concepts: a
finite sum $r(x)=c_{0}x^{\nu_{0}}+\cdots+c_{k}x^{\nu_{k}}$ with
$c_{i}\in\mathbb{C}^{\ast}$ (i.e. non-zero) and $\nu_{i}<\nu_{i+1}$, with
$\nu_{i}\in\mathbb{R}$ will be called a _generalized polynomial_. Such a
generalized polynomial $r(x)$ is _admissible for $P$, or for $P=0$_ (or simply
_admissible_) if the necessary initial conditions (10) are fulfilled for
$i=0,1,\ldots,k$. In particular, any truncation $r(x)$ of a solution
$s(x)\in\Omega$ is an admissible generalized polynomial, but the converse is
not true, even for linear equations: the generalized polynomial $r(x)=x$ is
admissible for the differential equation $P=2\,y_{0}-y_{1}-x+x^{2}$ (in terms
of the Euler differential operator) but there is no solution $s(x)\in\Omega$
of $P=0$ having $r(x)=x$ as a truncation. Notice that the same holds
considering $P=0$ as a $q$-difference equation with $q=\sqrt{2}$ and
$r(x)=(1+\sqrt{2}/2)\,x$.

However, the converse statement holds when $\operatorname{supp}r(x)$ has
maximum rational rank. This is our third main result:

######  Theorem 3.

Assume that the equation $P=0$ of order $n$ admits a generalized admissible
polynomial $r(x)$ with

(11)
$\dim_{\mathbb{Q}}\frac{\left\langle\operatorname{supp}r(x)\cup\operatorname{supp}P\right\rangle}{\left\langle\operatorname{supp}P\right\rangle}\geq
n.$

Then $r(x)$ is the truncation of a solution $s(x)$ of $P=0$, and the above
inequality is actually an equality.

Before proving this result, we need to study the behaviour of the Newton
polygon under changes of variables of the form $y=c\,x^{\nu}+y$. (See [6, 7]).

######  Lemma 3.

Let $Q=P[c\,x^{\nu}]$. Then:

  1. (1)

$\mathcal{N}(Q)$ is contained in the closed right half-plane defined by the
supporting line $L_{\nu}(P)$.

  2. (2)

Let $h$ the height of the point $\operatorname{Top}(E_{\nu}(P))$. Then
$\mathcal{N}(P)$ and $\mathcal{N}(Q)$ are equal above $h$. In particular
$\operatorname{Top}(E_{\nu}(P))=\operatorname{Top}(E_{\nu}(Q))$ and
$L_{\nu}(P)=L_{\nu}(Q)$. Moreover, $P_{(\alpha,\rho)}=Q_{(\alpha,\rho)}$ for
all $(\alpha,\rho)$ with $(\alpha,|\rho|)$ in the border of $\mathcal{N}(P)$
and $|\rho|\geq h$.

  3. (3)

The height of the point $\operatorname{Bot}(E_{\nu}(Q))$ is zero if and only
if $\Phi_{(P;\nu)}(c)\neq 0$.

  4. (4)

The following sequence of inequalities holds:

$\operatorname{ht}(\operatorname{Top}(E_{\nu}(P)))\geq\operatorname{ht}(\operatorname{Bot}(E_{\nu}(Q)))\geq\operatorname{ht}(\operatorname{Top}(E_{\mu}(Q))),$

where $\mu>\nu$ in the last expression.

We have the following characterization of generalized admissible polynomials
in terms of the Newton polygon:

######  Lemma 4.

Let $r(x)=c_{0}x^{\nu_{0}}+\cdots+c_{k}\,x^{\nu_{k}}$ with
$\nu_{0}<\cdots<\nu_{k}$, and, as above, $P_{0}=P$ and
$P_{i+1}=P_{i}[c_{i}x^{\nu_{i}}]$ for $i=0,\ldots,k$. Let us denote
$Q=P[r(x)]=P_{k+1}$. Then $r(x)$ is an admissible generalized polynomial for
$P$ if and only the bottom vertex of $E_{\nu_{k}}(Q)$ has height greater than
o equal to one.

###### Proof.

Assume that $r(x)$ is admissible. Applying Lemma 3 iteratively, we obtain the
inequality $\operatorname{ht}(\operatorname{Bot}(E_{\nu_{k}}(P_{k+1})))\geq
1$.

Conversely, assume that $r(x)$ is not admissible for $P$. Let $i$ be the
minimum index for which equation (10) does not hold. By part (3) of Lemma 3,
$\operatorname{ht}(\operatorname{Bot}(E_{\nu_{i}}(P_{i+1}))=0$ and by part (4)
of the same Lemma,

$0=\operatorname{ht}(\operatorname{Bot}(E_{\nu_{i}}(P_{i+1}))\geq\operatorname{ht}(\operatorname{Top}(E_{\nu_{i+1}}(P_{i+1}))\geq\operatorname{ht}(\operatorname{Bot}(E_{\nu_{i+1}}(P_{i+2}))\geq
0.$

Applying iteratively this argument we get
$\operatorname{ht}(\operatorname{Bot}(E_{\nu_{k}}(P_{k+1}))=0$. ∎

The following result is also well-known but we include its proof for the
convenience of the reader.

######  Corollary 4.

Let $r(x)$ be an admissible generalized polynomial for $P=0$, and let
$Q=P[r(x)]$. Then either $r(x)$ is already a solution of the equation $P=0$,
or $\mathcal{N}(Q)$ has a side $E_{\nu}(Q)$ with
$\nu>\max(\operatorname{supp}(r(x)))$.

###### Proof.

By Lemma 4, we know that
$\operatorname{ht}(\operatorname{Bot}(E_{\nu_{k}}(Q)))\geq 1$. If
$\mathcal{N}(Q)$ does not intersect the horizontal axis of coordinates then
the null power series is a solution of $Q=0$ and therefore $r(x)$ is already a
solution of the equation $P=0$. Otherwise, there is a vertex of
$\mathcal{N}(Q)$ of the form $(\beta,0)$. Let $(\beta^{\prime},0)$ be the
intersection of $L_{\nu_{k}}(Q)$ with the horizontal axis. Since
$\operatorname{ht}(\operatorname{Bot}(E_{\nu_{k}}(Q)))\geq 1$, the point
$(0,\beta^{\prime})\not\in\mathcal{C}(Q)$, and by (1) in Lemma 3 we get
$\beta^{\prime}<\beta$, which implies the existence of a side $E_{\nu}(Q)$ of
$\mathcal{N}(Q)$ with co-slope $\nu>\nu_{k}$. ∎

The next Lemma provides several relations between $P=0$ and some of the
equations $\frac{\partial P}{\partial y_{i}}=0$, which will be useful in the
proof of the main Proposition of this section.

######  Lemma 5.

Let $r(x)$ be an admissible generalized polynomial for $P$, and set
$Q=P[r(x)]$. Let $\nu>\max(\operatorname{supp}r(x))$ and assume there exists a
point $V=(\alpha,r)\in E_{\nu}(Q)\cap\mathcal{C}(Q)$, with $r\geq 2$. Take
$\rho$ with $|\rho|=r$ such that $Q_{(\alpha,\rho)}\neq 0$, which exists
because $V\in\mathcal{C}(Q)$. Let $i$ be an index $0\leq i\leq n$ such that
$\rho_{i}\geq 1$, and set $\overline{P}=\frac{\partial P}{\partial y_{i}}$ and
$\overline{Q}=\overline{P}[r(x)]$.

Then $r(x)$ is an admissible generalized polynomial for $\overline{P}$, and
the point $\overline{V}=(\alpha,r-1)$ belongs to
$E_{\nu}(\overline{Q})\cap\mathcal{C}(\overline{Q})$. Moreover, if $V$ is a
vertex of $\mathcal{N}(Q)$ then $\overline{V}$ is a vertex of
$\mathcal{N}(\overline{Q})$.

###### Proof.

The cloud of points of $\overline{P}$ is a subset of the image of the points
of $\mathcal{C}(P)$ with height at least $1$ under the map
$\Delta(\alpha,r)=(\alpha,r-1)$. By the chain rule
$\overline{Q}=\frac{\partial Q}{\partial y_{i}}$, hence
$\mathcal{C}(\overline{Q})$ is a subset of the image of the points of
$\mathcal{C}(Q)$ with height at least one under $\Delta$. Moreover
$\overline{Q}_{(\alpha,\rho-e_{i})}=\rho_{i}\,Q_{(\alpha,\rho)}$, so that the
point $\overline{V}=(\alpha,r-1)$ belongs to
$\mathcal{C}(\overline{Q})\cap\Delta(L_{\nu}(Q))$. As $\Delta$ is an affine
translation, the existence of $\overline{V}$ implies that $\Delta(L_{\nu}(Q))$
is the supporting line of co-slope $\nu$ of $\mathcal{C}(\overline{Q})$.
Therefore $\overline{V}\in E_{\nu}(\overline{Q})$.

As $\operatorname{ht}\overline{V}=r-1\geq 1$, and $\overline{V}\in
E_{\nu}(\overline{Q})$, then
$\operatorname{ht}(\operatorname{Top}(E_{\nu}(\overline{Q})))\geq 1$. Let
$\nu_{k}=\max\operatorname{supp}r(x)$. Since $\nu>\nu_{k}$, by (4) of Lemma 3
we get
$\operatorname{ht}\operatorname{Bot}(E_{\nu_{k}}(\overline{Q}))\geq\operatorname{Top}(E_{\nu}(\overline{Q}))\geq
1$. Hence by Lemma 4, the generalized polynomial $r(x)$ is admissible for
$\overline{P}$.

Assume now that $V$ is a vertex of $\mathcal{N}(Q)$, so that there exists
$\nu^{\prime}>\max\operatorname{supp}r(x)$ with $\nu\neq\nu^{\prime}$ and
$V\in E_{\nu}(Q)\cap E_{\nu^{\prime}}(Q)$. We conclude that $\overline{V}\in
E_{\nu}(\overline{Q})\cap E_{\nu^{\prime}}(\overline{Q})$ which proves that
$\overline{V}$ is a vertex of $\mathcal{N}(\overline{Q})$. ∎

###  4.1. Admissible polynomials with maximal rational rank and characteristic
polynomials

The goal of this subsection is to prove that if $r(x)$ is an admissible
generalized polynomial for the equation $P=0$ which has maximum rational rank
then the characteristic polynomials of the relevant sides of the Newton
polygon of $P[r(x)]$ have always non-zero roots. That is, after the last
exponent of $r(x)$, the equation $P[r(x)]=0$ behaves as an algebraic curve and
$r(x)$ can always be completed to a solution of $P=0$.

Throughout this subsection, $r(x)$ will be denote an admissible generalized
polynomial for $P=0$ such that

(12)
$\dim_{\mathbb{Q}}\frac{\langle\operatorname{supp}P\cup\operatorname{supp}r(x)\rangle}{\langle\operatorname{supp}P\rangle}=m\geq
n,$

where $n$ is the order of $P=0$. Let
$\Gamma=\langle\operatorname{supp}P\rangle$ and write $r(x)$ as in Equation
(5), satisfying properties (i)–(iv). Thus, we have:

(13)
$r(x)=\sum_{j=0}^{m}\big{(}c_{j,0}\,x^{\mu_{j,0}}+c_{j,1}\,x^{\mu_{j,1}}+\cdots+c_{j,t_{j}}\,x^{\mu_{j,t_{j}}}\big{)},$

and we set, from now on, $Q=P[r(x)]$.

As in the previous section, we denote $\mu_{j}=\mu_{j,0}$ and either
$\Gamma^{\prime}=\langle\Gamma\cup\\{\mu_{1},\ldots,\mu_{m}\\}\rangle$ or
$\Gamma^{\prime}=\langle\Gamma\cup\\{\mu_{1},\ldots,\mu_{m},1\\}\rangle$
according as $1\in\langle\Gamma\cup\\{\mu_{1},\ldots,\mu_{m}\\}$ or not. The
derivations $\mathcal{D}_{j}$, $j=1,\ldots,m$, on
$\mathbb{C}((x^{\Gamma^{\prime}}))$ are defined as in Lemma. Notice that
$Q\in\mathbb{C}((x^{\Gamma^{\prime}}))[y_{0},\ldots,y_{n}]$. Given
$\rho=(\rho_{0},\rho_{1},\ldots,\rho_{n})\in\mathbb{Z}_{\geq 0}^{n+1}$ and
$\alpha\in\Gamma^{\prime}$, we write
$Q_{\rho}\in\mathbb{C}((x^{\Gamma^{\prime}}))$ for the coefficient of
$y_{0}^{\rho_{0}}y_{1}^{\rho_{1}}\cdots y_{n}^{\rho_{n}}$ in $Q$, and
$Q_{(\alpha,\rho)}\in\mathbb{C}$ for the coefficient of
$x^{\alpha+\varepsilon\,\omega(\rho)}$ in $Q_{\rho}$. Recall that $e_{i}$ is a
vector with $n+1$ components, all zero except the $i+1-th$, which is $1$.

The existence of the admissible generalized polynomial $r(x)$ generates
conditions on the coefficients of $Q$ at elements of co-slope greater than
$\max\operatorname{supp}r(x)=\mu_{m,t_{m}}$.

######  Lemma 6.

Let $\nu>\max\operatorname{supp}r(x)=\mu_{m,t_{m}}$ and consider the element
$E_{\nu}(Q)$ of co-slope $\nu$ of $\mathcal{N}(Q)$. Assume there is a point
$V=(\alpha,r)\in E_{\nu}(Q)$, with $r=\operatorname{ht}(V)\geq 1$. Then for
all $\rho=(\rho_{0},\rho_{1},\ldots,\rho_{n})\in\mathbb{Z}_{\geq 0}^{n+1}$
with $|\rho|=r-1$, and for all $j=1,2,\ldots,m$, the following equality holds:

(14)
$\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)\,\delta_{\mu_{j}}^{(\kappa)}\,Q_{(\alpha,\rho+e_{\kappa})}=0.$

######  Remark 1.

Notice that each $\rho=(\rho_{0},\ldots,\rho_{n})$ with $|\rho|=r-1$ gives
rise to $m$ conditions (14) on the coefficients of $Q_{V}$.

$\mathcal{N}(Q)$$\alpha$$|\rho|$$L_{\nu}(Q)$$E_{\nu}(Q)$$V=(\alpha,r)$$V_{{}_{j}}$$r-1$$\alpha+\mu_{j}$$\alpha+\nu$$\eta$
Figure 1. Newton Polygon of $Q=P[r(x)]$ and some elements that appear in the
proof of Lemma 6.

###### Proof.

Fix for the all this proof a $\rho\in\mathbb{Z}_{\geq 0}^{n+1}$ with
$|\rho|=r-1$. Let $V_{j}=(\alpha+\mu_{j},r-1)$ for $j=1\,\ldots,m$. Notice
that $V_{j}\not\in\mathcal{N}(Q)$ because $V\in E_{\nu}(Q)$ and
$\nu>\mu_{m,t_{m}}\geq\mu_{j}$. Thus, $Q_{(\alpha+\mu_{j},\rho)}=0$.

Since $\mu_{1},\ldots,\mu_{m}$ and $\Gamma$ fulfill the hypothesis of Lemma 1,
we may apply the derivations $\mathcal{D}_{j}$ to $Q$, and for each
$j=1,\ldots,m$, Equation (3) becomes

(15)
$\mathcal{D}_{j}(Q_{\rho})=\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)(Q_{\rho+e_{\kappa}})\,\mathcal{D}_{j}(r^{(\kappa)}(x)).$

Since $V=(\alpha,r)$ is on the border of $\mathcal{N}(Q)$ and
$|\rho+e_{\kappa}|=r$, we have:

$Q_{\rho+e_{\kappa}}=Q_{(\alpha,\rho+e_{\kappa})}x^{\alpha+\varepsilon\,\omega(\rho+e_{\kappa})}+\cdots=Q_{(\alpha,\rho+e_{\kappa})}x^{\alpha+\varepsilon\,\omega(\rho)+\varepsilon\,\kappa}+\cdots$

where the last equality holds because
$\omega(\rho+e_{\kappa})=\omega(\rho)+\kappa$. Because of Properties (i)–(iv)
in the expression of $r(x)$ the following equality holds:

$\mathcal{D}_{j}(r^{(\kappa)}(x))=c_{j,0}\,\delta_{\mu_{j}}^{(\kappa)}\,x^{\mu_{j}-\varepsilon\,\kappa}+\cdots,$

where recall that $c_{j,0}\neq 0$ for each $j=1,\ldots,m$. Hence, the right
hand side of Equation (15) is equal to

(16)
$c_{j,0}\left(\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)\,\delta_{\mu_{j}}^{(\kappa)}\,Q_{(\alpha,\rho+e_{\kappa})}\right)x^{\alpha+\mu_{j}+\varepsilon\,\omega(\rho)}+\cdots,$

and the left hand side of Equation (15) has, by definition of
$\mathcal{D}_{j}$, order at least $\operatorname{ord}_{x}Q_{\rho}$. Thus, if
$Q_{\rho}=0$ we are done. So we assume $Q_{\rho}\neq 0$ from now on. Set then
$\operatorname{ord}_{x}Q_{\rho}=\eta+\varepsilon\,\omega(\rho)$.

It is enough to prove that $\eta>\alpha+\mu_{j}$ because this implies that the
first term of (16) is zero which combined with (15) gives Equation (14). By
definition of $L_{\nu}(Q)$, any point in $\mathcal{C}(Q)$ belongs to the
closed right halfplane defined by $L_{\nu}(Q)$. Since $V=(\alpha,r)\in
E_{\nu}(Q)\subset L_{\nu}(Q)$, the point $(\alpha+\nu,r-1)\in L_{\nu}(Q)$
because $L_{\nu}(Q)$ is a line with co-slope $\nu$. Therefore, any point with
ordinate $r-1$ in $\mathcal{N}(Q)$ has abscissa greater than or equal to
$\alpha+\nu$. As $Q_{\rho}\neq 0$ then
$Q_{(\eta+\varepsilon\,\omega(\rho),\rho)}\neq 0$ so that
$(\eta,r-1)\in\mathcal{C}(Q)$ and hence $\eta\geq\alpha+\nu>\alpha+\mu_{j}$. ∎

The key consequence of Lemma 6 is the following result.

######  Proposition 1.

Let $\nu>\max\operatorname{supp}r(x)$ and $V=(\alpha,r)\in E_{\nu}(Q)$ with
$Q_{V}\neq 0$. The indicial polynomial $\Psi_{(Q;V)}(T)$ is a power of the
polynomial $(T-\delta_{\mu_{1}})\cdots(T-\delta_{\mu_{n}})$ up to a non-zero
constant $c$:

$\Psi_{(Q;V)}(T)=c\,\bigg{(}\prod_{j=1}^{n}(T-\delta_{\mu_{j}})\bigg{)}^{r}.$

As a consequence, $\Psi_{(Q;V)}(\delta_{\nu})\neq 0$ because
$\nu>\max\operatorname{supp}r(x)>\mu_{j}$ for all $j=1,\ldots,n$, so that if
$E_{\nu}(Q)$ is a side of $\mathcal{N}(Q)$, then the characteristic polynomial
$\Phi_{(Q,\nu)}(C)$ has at least one non-zero root.

###### Proof.

We proceed by induction on $r$. The case $r=0$ is trivial because
$\Psi_{(Q;V)}(T)=Q_{(\alpha,(0,\ldots,0))}$ is a non-zero constant under our
hypothesis $Q_{V}\neq 0$.

We need to prove the case $r=1$ because our induction argument requires $r\geq
2$.

Assume $r=1$. The indicial polynomial of $Q$ at $V=(\alpha,1)$ is, by
definition,

$\Psi_{(Q;V)}(T)=\sum_{k=0}^{n}Q_{(\alpha,e_{\kappa})}\,T^{\langle
e_{\kappa}\rangle}.$

Applying Lemma 6 to $\rho=(0,\ldots,0)$ we get
$\Psi_{(Q;V)}(\delta_{\mu_{j}})=0$ for $j=1,\ldots,n$ (i.e. each
$\delta_{\mu_{j}}$ is a root of $\Psi_{(Q;V)}(T)$). Since the polynomial
$\Psi_{(Q;V)}(T)$ has degree at most $n$, we must have:

$\Psi_{(Q;V)}(T)=\sum_{k=0}^{n}Q_{(\alpha,e_{\kappa})}\,T^{\langle
e_{\kappa}\rangle}=Q_{(\alpha,e_{n})}\,\prod_{j=1}^{n}(T-\delta_{\mu_{j}}),$

and we need to show that $Q_{(\alpha,e_{n})}\neq 0$. If it were not so, then
$\Psi_{(Q;V)}(T)=0$ and all the coefficients $Q_{(\alpha,e_{\kappa})}$ should
be zero for $\kappa=0,\ldots,n$, which contradicts the fact that $Q_{V}\neq
0$. This finishes case $r=1$.

Assume the result holds for $r-1$ and consider the general case. Let
$V=(\alpha,r)$, with $r\geq 2$ and all other notations as before. For any
$\rho\in\mathbb{Z}_{\geq 0}^{n+1}$ with $|\rho|=r-1$, Equation (14) for
$j=1,\ldots,n$ implies the following equality of polynomials in the variable
$\mu$ of degree at most $n$:

(17) $\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)\,T^{\langle
e_{\kappa}\rangle}\,Q_{(\alpha,\rho+e_{\kappa})}=(\rho_{n}+1)\,Q_{(\alpha,\rho+e_{n})}\,\prod_{j=1}^{n}(T-\delta_{\mu_{j}}),$

because $\delta_{\mu_{j}}$, for $j=1,\ldots,n$, are different roots of the
left hand side.

We are going to prove that $Q_{(\alpha,(0,\ldots,0,r))}\neq 0$ and apply Lemma
5 to use the induction hypothesis. Arguing by contradiction assume
$Q_{(\alpha,(0,\ldots,0,r))}=0$. Take $\gamma\in\mathbb{Z}_{\geq 0}^{n+1}$
with $|\gamma|=r$ maximal with respect the reverse lexicographical order such
that $Q_{(\alpha,\gamma)}\neq 0$. This $\gamma$ exists because $Q_{V}\neq 0$,
and $\gamma\neq(0,0,\ldots,r)$ by the assumption. Let $\rho=\gamma-e_{\kappa}$
for some $\kappa<n$ with $\gamma_{\kappa}>0$, which exists too because
$\gamma_{n}<r$. Equation (17) applied to this $\rho$ implies that if
$Q_{(\alpha,\rho+e_{n})}=0$, then the left hand side of (17) is the null
polynomial so that $Q_{(\alpha,\gamma)}=0$ because $\gamma=\rho+e_{\kappa}$.
Hence $Q_{(\alpha,\rho+e_{n})}\neq 0$. This contradicts the definition of
$\gamma$. Thus our assumption is false and $Q_{(\alpha,(0,\ldots,0,r))}\neq
0$.

Equation (17) is important because it will allow us to show that
$\Psi_{(Q;V)}(T)$ is a polynomial divisible by
$\prod_{j=1}^{n}(T-\delta_{\mu_{j}})$ from which we shall be able to apply the
induction hypothesis.

Multiplying both sides of Equation (17) by $T^{\langle\rho\rangle}$, as
$T^{\langle\rho\rangle}T^{\langle
e_{\kappa}\rangle}=T^{\langle\rho+e_{\kappa}\rangle}$, we get

(18)
$\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)\,T^{\langle\rho+e_{\kappa}\rangle}\,Q_{(\alpha,\rho+e_{\kappa})}=(\rho_{n}+1)\,Q_{(\alpha,\rho+e_{n})}\,T^{\langle\rho\rangle}\,\prod_{j=1}^{n}(T-\delta_{\mu_{j}}).$

The sum of all the left hand sides of (18) for $\rho$ with $|\rho|=r-1$ gives:

$\sum_{|\rho|=r-1}\sum_{\kappa=0}^{n}(\rho_{\kappa}+1)\,T^{\langle\rho+e_{\kappa}\rangle}\,Q_{(\alpha,\rho+e_{\kappa})}=r\,\sum_{|\gamma|=r}T^{\langle\gamma\rangle}\,Q_{(\alpha,\gamma)}=r\,\Psi_{(Q;V)}(T),$

where the second equality is the definition of the indicial polynomial, and
the first one follows from collapsing the preimages of the map
$(\rho,\kappa)\mapsto\gamma=\rho+e_{\kappa}$. As a consequence, summing all
the corresponding right hand sides of (18) we get

(19)
$r\,\Psi_{(Q;V)}(T)=\prod_{j=1}^{n}(T-\delta_{\mu_{j}})\sum_{|\rho|=r-1}(\rho_{n}+1)\,Q_{(\alpha,\rho+e_{n})}\,T^{\langle\rho\rangle}.$

We are now going to apply the induction hypothesis in order to prove that the
summation on the right hand side is, up to a non-constant factor, an $r-1$
power of $\prod_{j=1}^{n}(T-\delta_{\mu_{j}})$. To this end, recall that
$Q_{(\alpha,(0,0,\ldots,r))}\neq 0$ and consider $\overline{P}=\frac{\partial
P}{\partial y_{n}}$. By the chain rule,
$\overline{Q}=\overline{P}[r(x)]=\frac{\partial Q}{\partial y_{n}}$, and in
particular,
$\overline{Q}_{(\alpha,(0,0,\ldots,r-1)}=r\,Q_{(\alpha,(0,0,\ldots,r))}\neq
0$. Lemma 5 guarantees that $r(x)$ is an admissible generalized polynomial for
$\overline{P}$. Let $\overline{V}=(\alpha,r-1)$. We know that then
$\overline{Q}_{\overline{V}}\neq 0$, and by Lemma 5 again, $\overline{V}$
belongs to an element $E_{\nu}(\overline{Q})$ of co-slope $\nu$. These
properties allow us to apply the induction hypothesis to $\overline{P}$,
$r(x)$ and $\overline{V}$ which means that the indicial polynomial
$\Psi_{(\overline{Q},\overline{V})}(T)$ is, up to a non-zero constant, an
$r-1$-th power of $\prod_{j=1}^{n}(T-\delta_{\mu_{j}})$. Thus, on one hand, by
definition and because $\overline{Q}=\frac{\partial Q}{\partial y_{n}}$ we
have

(20)
$\Psi_{(\overline{Q},\overline{V})}(T)=\sum_{|\rho|=r-1}\overline{Q}_{(\alpha,\rho)}\,T^{\langle\rho\rangle}=\sum_{|\rho|=r-1}(\rho_{n}+1)\,Q_{(\alpha,\rho+e_{n})}\,T^{\langle\rho\rangle},$

and, on the other, by the induction hypothesis, there is a non-zero constant
$c$ with

(21)
$\Psi_{(\overline{Q},\overline{V})}(T)=c\,\bigg{(}\prod_{j=1}^{n}(T-\delta_{\mu_{j}})\bigg{)}^{r-1}.$

The proof of the first statement in the Proposition finishes by connecting
(20) and (21), and inserting the result into (19).

It only remains to prove that the characteristic polynomial
$\Phi_{(Q,\nu)}(C)$ has at least one non-zero root. By definition:

$\Phi_{(Q,\nu)}(C)=\sum_{W\in
E_{\nu}(Q)}\Psi_{(Q;W)}(\delta_{\nu})\,C^{\operatorname{ht}(W)}.$

We have already proved that for all $W\in E_{\nu}(Q)\cap\mathcal{C}(Q)$,
$\Psi_{(Q;W)}(\delta_{\nu})\neq 0$, because $\nu>\mu_{j}$, for all
$j=1,\ldots,n$. Since $E_{\nu}(Q)$ is a side, it has at least two vertices,
whence $\Psi_{(Q,\nu)}(C)$ has at least two non-zero monomials, and thus it
has at least one non-zero root. ∎

###### Proof of Theorem 3.

Let $Q=P[r(x)]$. Consider the following procedure, analogue to the Newton-
Puiseux construction for algebraic curves:

######  Procedure 1.

Completion of $r(x)$ to a solution of $P=0$.

  1. Setup.

Set $r_{0}(x)=r(x)$, $Q_{1}=Q$, $\eta_{0}=\max(\operatorname{supp}(r(x))$, and
$i=1$.

  2. 1.

If $y=0$ is a solution of $Q_{i}=0$ Then Return $s(x)=r_{i-1}(x)$ Else

  3. 2.

Choose a side of $\mathcal{N}(Q_{i})$ with co-slope $\eta_{i}>\eta_{i-1}$ and

  4. 3.

a non-zero root $d_{i}$ of the characteristic polynomial
$\Phi_{(Q_{i},\eta_{i})}(C)$.

  5. 4.

Set $r_{i}(x)=r_{i-1}(x)+d_{i}\,x^{\eta_{i}}$ and
$Q_{i+1}=Q_{i}[d_{i}\,x^{\eta_{i}}]$.

  6. 5.

Set $i=i+1$ and Goto 1.

  7. 6.

Return $s(x)=r(x)+\sum_{i=1}^{\infty}d_{i}\,x^{\eta_{i}}$.

We shall show that in each iteration, the main loop of the above procedure can
be performed, that the output $s(x)$ belongs to $\Omega$, and it is a solution
of the equation $P=0$. From this follows, by Theorem 1, that $m\leq n$, hence
$m=n$.

Let us prove that each step can be performed. Assume $r_{i-1}(x)$, $Q_{i}$,
$\eta_{i-1}$ are defined for some $i\geq 1$ and $r_{i-1}(x)$ is admissible for
$P=0$. If the condition of line 1 holds the procedure finished. Otherwise, as
$r_{i-1}(x)$ is admissible for $P=0$ there exists $\eta_{i}>\eta_{i-1}$ by
Corollary 4 such that $\mathcal{N}(Q_{i})$ has a side of co-slope $\eta_{i}$.
This allows line 2 of Procedure 1 to be performed. Proposition 1 guarantees
the existence of a non-zero root $d_{i}$ of the characteristic polynomial
$\Phi_{(Q_{i},\eta_{i})}(C)$ which gives line 3.

We have proved that the above method either returns a generalized polynomial
$r_{i-1}(x)$ which is a solution of $P=0$ or, after infinitely many steps, it
returns a formal power series
$s(x)=r(x)+\sum_{i=1}^{\infty}d_{i}\,x^{\eta_{i}}$. From the definition of
$\eta_{0}$ and the fact that $\eta_{0}<\eta_{j}$, it is obvious that $r(x)$ is
a truncation of $s(x)$ but it remains to prove that $s(x)\in\Omega$ and that
$s(x)$ is a solution of $P=0$. These facts follow from the following
Proposition, which can be found in the literature under slightly different
hypotheses: In [13] the equation $P=0$ is differential and has coefficients in
the field $\mathbb{C}((x^{\mathbb{Z}}))$, in [5] the series $s(x)$ has
rational exponents, and in [7] $P=0$ is a $q$-difference equation whose
coefficients are grid-based series. The arguments presented in those proofs
are equally valid in our setting with minimal modification. For the
convenience of the reader we sketch the proof of the specific statement we
need, thus ending the proof of Theorem 3.

∎

######  Proposition 2.

Let $P\in\Omega[y_{0},y_{1},\ldots,y_{n}]$ and
$s(x)=\sum_{i=0}^{\infty}c_{i}\,x^{\nu_{i}}$ be a series with
$\nu_{0}<\nu_{1}<\dots$ such that $s(x)$ satisfies the necessary initial
conditions (10) for any $i\geq 0$. Then $s(x)\in\Omega$ and it is solution of
the equation $P=0$.

###### Proof.

First, we are going to define a special point which we call the pivot point of
$P$ along the series $\sum_{i=0}^{\infty}c_{i}\,x^{\nu_{i}}$ and will be
central to our arguments.

As previously, we denote $P_{0}=P$ and $P_{i+1}=P_{i}[c_{i}\,x^{\nu_{i}}]$. By
Lemma 3 we know that
$\operatorname{ht}\operatorname{Top}(E_{\nu_{i}}(P_{i}))\geq\operatorname{ht}\operatorname{Top}(E_{\nu_{i+1}}(P_{i+1}))\geq
1$, for any $i\geq 0$. Hence there exists an index $i_{0}$ such that this
sequence of heights stabilizes, and therefore the points
$\operatorname{Top}(E_{\nu_{i}}(P_{i}))$ stabilize too: there is a point
$V=(\alpha,r)$ such that
$\operatorname{Top}(E_{\nu_{i_{0}}}(P_{i_{0}})=\operatorname{Top}(E_{\nu_{i}}(P_{i}))=V$
for all $i\geq i_{0}$. This $V$ is the _pivot point_ of $P$ along $s(x)$.

We know that $r\geq 1$ by Lemma 3. Notice that for any $k\geq i_{0}$, setting
$\alpha_{k}=\operatorname{ord}_{x}P(\sum_{i=0}^{k-1}c_{i}\,x^{\nu_{i}})$ one
has $\alpha_{k}\geq\alpha+r\,\nu_{k}$ because the point
$(\alpha_{k},0)\in\mathcal{C}(P_{k})$ and the supporting line
$L_{\nu_{k}}(P_{k})$ passes through $V$. Hence, in order to prove the
Proposition it is enough to show that $s(x)\in\Omega$, that is
$\lim\nu_{i}=\infty$, because then automatically $\lim\alpha_{k}=\infty$ which
makes $s(x)$ a solution of $P=0$.

We may assume that $r=1$, because if $r\geq 2$ we can apply $r-1$ times Lemma
5 to find some derivative $P^{\prime}=\frac{\partial^{r-1}P}{\partial
y_{0}^{j_{0}}\dots\partial y_{n}^{j_{n}}}$ of $P$, such that, the height of
the pivot point of $P^{\prime}$ along $s(x)$ is one. Thus from now on we
assume that $r=1$.

We may also assume that $\alpha=0$ so that $V=(0,1)$. Let us sketch why (a
detailed argument can be found in [7]). Performing the change of variable
$\bar{y}=x^{\nu_{i_{0}}}\,y$ on $P=0$, and multiplying the resulting equation
by $x^{-\beta}$, with $\beta=\alpha+\nu_{i_{0}}$, we obtain a new equation
$\overline{P}=0$ whose pivot point along $\bar{s}(x)=x^{-\nu_{i_{0}}}\,s(x)$
is $(0,1)$ and such that the coefficients of $\overline{P}_{i_{0}}$ are
elements of $\Omega$ with non-negative order in $x$. Using
$\overline{P}_{i_{0}}$ instead of $P$, we may assume that $P=0$ is in _quasi-
solved form_ : $\mathcal{N}(P)$ is contained in the first quadrant, and in
particular the coefficients of $P$ have non-negative order in $x$, and
$V=(0,1)$ is a vertex of $\mathcal{N}(P)$, and $\nu_{0}\geq 0$.

We first prove the Proposition assuming that $P$ is in quasi-solved form and
with the additional hypothesis that its coefficients are grid-based, that is:
there exists a finitely generated semigroup $G$ of $\mathbb{R}_{\geq 0}$ such
that $\operatorname{supp}P\subset G$. In this case we can prove that $s(x)$ is
also grid-based. The general case will follow from this one.

By Lemma 3, the point $V=(0,1)$ is a vertex of $\mathcal{N}(P_{i})$ for all
$i\geq 0$. Also by Lemma 3, the equality
$P_{(0,e_{\kappa})}=({P_{i}})_{(0,e_{\kappa})}$ holds for all $i\geq 0$ and
$\kappa=0,1,\ldots,n$. Hence the indicial polynomials $\Psi_{(P;V)}(T)$ and
$\Psi_{(P_{i};V)}(T)$ are equal for all $i\geq 0$. Let
$\Sigma=\\{\mu\in\mathbb{R}_{\geq 0}\mid\Psi_{(P;V)}(\delta_{\mu})=0\\}$. This
set is finite because the set of roots of the polynomial
$\Psi_{(P;V)}(T)=\sum_{\kappa=0}^{n}P_{(0,e_{\kappa})}\,T^{\kappa}$ is finite
(it is not identically null since $(0,1)$ is a vertex), and $|q|\neq 1$ in the
case of $q$-difference equations. Let $G^{\prime}$ be the semigroup generated
by $G$ and $\Sigma$, which is finitely generated.

Let us prove inductively that $\operatorname{supp}P_{i}\subset G^{\prime}$ and
that $\nu_{i}\in G^{\prime}$. The case $i=0$ holds by hypothesis. Let $i\geq
0$ and assume that $\operatorname{supp}P_{i}\subset G^{\prime}$; let us first
prove that $\nu_{i}\in G^{\prime}$. By hypothesis,
$\Phi_{(P_{i};\nu_{i})}(c_{i})=0$. If the element $E_{\nu_{i}}(P_{i})$ is a
side then its vertices are $(0,1)$ and $(\alpha_{i},0)$, so that
$\nu_{i}=\alpha_{i}\in\operatorname{supp}P_{i}\subset G^{\prime}$. Otherwise
$E_{\nu_{i}}(P_{i})$ is the vertex $\\{V\\}$, which implies that
$\Phi_{(P_{i};\nu_{i})}(c_{i})=\Psi_{(P_{i},V)}(\delta_{\nu_{i}})c_{i}$, so
that $\nu_{i}\in\Sigma$. In both cases, $\nu_{i}\in G^{\prime}$. Since
$P_{i+1}=P_{i}[c_{i}\,x^{\nu_{i}}]$ then $\operatorname{supp}P_{i+1}\subset
G^{\prime}$. This shows that $\operatorname{supp}s(x)\subset G^{\prime}$ which
implies that $\lim\nu_{i}=\infty$ as desired.

It only remains to prove the Proposition when $P$ has coefficients in
$\Omega$, not necessarily grid-based. Take $N>0$ and denote by $R$ the
polynomial obtained by truncating in the variable $x$ the coefficients of $P$
up to orden $N$. Define as usually $R_{0}=R$ and
$R_{i+1}=R_{i}[c_{i}\,x^{\nu_{i}}]$ for $i\geq 0$. Since the coefficients of
$P$ have non-negative order in $x$ and $\nu_{i}\geq 0$, then the truncation up
to order $N$ of $R_{i}$ and that of $P_{i}$ coincide for all $i\geq 0$. In
particular, if $\nu_{k}\leq N$ then, $\Phi_{(R_{i},\nu_{i})}(c_{i})=0$ for
$0\leq i\leq k$. Since $\operatorname{supp}(R)$ is finite, it generates a
finitely generated semigroup $G$ of $\mathbb{R}_{\geq 0}$. As before, we may
prove by induction that $\nu_{i}\in G^{\prime}$ for $0\leq i\leq k$. Since the
set $G^{\prime}\cap\mathbb{R}_{\leq N}$ finite, then the set
$\operatorname{supp}s(x)\cap\mathbb{R}_{\leq N}$ is finite for any $N$ and
therefore either $\lim\nu_{i}=\infty$ or $\operatorname{supp}s(x)$ is finite;
in both cases $s(x)\in\Omega$ and we are done. ∎

###  4.2. On the completion of admissible polynomials to solutions

Grigoriev and Singer in [13] provide, in the case of differential equations, a
criterion for an admissible polynomial to be the truncation of an actual
solution of $P=0$, based on the following definition. As before,
$r(x)=\sum_{i=0}^{k-1}c_{i}\,x^{\nu_{i}}$ is an admissible generalized
polynomial for the equation $P=0$, and we denote $P_{0}=P$ and
$P_{i+1}=P_{i}[c_{i}\,x^{\nu_{i}}]$, for $0\leq i\leq k-1$. Let
$V=\operatorname{Bot}(E_{\nu_{k-1}}(P_{k}))$ be the bottom vertex of
$E_{\nu_{k-1}}(P_{k})$.

######  Definition 1.

We say that _$r(x)$ stabilizes $P$_ if the height $\operatorname{ht}(V)$ is
$1$, and $\nu_{k-1}$ is greater than the maximum of the finite set
$\Sigma=\\{\mu\in\mathbb{R}_{\geq 0}\mid\Psi_{(P;V)}(\delta_{\mu})=0\\}$.

In Section 3 of [13] the authors prove the following _stabilization criterion_
(for differential equations): if $r(x)$ stabilizes $P$ then it is the
truncation of a unique solution $s(x)$ of $P=0$. For the benefit of the reader
we include a proof which covers also the $q$-difference case.

###### Proof of the stabilization criterion of [13].

We need to show the existence and uniqueness of a solution
$s(x)=r(x)+\sum_{i=k}^{\infty}c_{i}\,x^{\nu_{i}}\in\Omega$, with
$\nu_{i}<\nu_{i+1}$ for all $i$. By Proposition 2, it is enough to prove that
$r(x)$ can be completed uniquely to an $s(x)$ such that $s(x)$ satisfies the
necessary initial conditions (10).

Notice that whenever $\nu_{k}>\nu_{k-1}$ and $c_{k}$ satisfy the necessary
initial condition $\Phi_{(P_{k};\nu_{k})}(c_{k})=0$ then the generalized
polynomial $r(x)+c_{k}\,x^{\nu_{k}}$ stabilizes $P$. Thus, by recurrence, we
only need to prove that: either (I) $P(r(x))=0$ and the condition
$\Phi_{(P_{k};\nu_{k})}(c_{k})=0$ does not hold for any $\nu_{k}>\nu_{k-1}$
and $c_{k}\neq 0$; or (II) $P(r(x))\neq 0$ and there only exist one
$\nu_{k}>\nu_{k-1}$ and one $c_{k}\neq 0$ with
$\Phi_{(P_{k};\nu_{k})}(c_{k})=0$.

Let $\nu_{k}>\nu_{k-1}$. Because $\operatorname{ht}V=1$, there are only three
cases for the element $E_{\nu_{k}}(P_{k})$: (a) $E_{\nu_{k}}(P_{k})$ is a
vertex of $\mathcal{N}(P_{k})$ with ordinate zero; (b) $E_{\nu_{k}}(P_{k})$ is
the vertex $V$; (c) $E_{\nu_{k}}(P_{k})$ is the side joining $V$ and the
vertex of $\mathcal{N}(P_{k})$ on the horizontal axis, say $(\beta,0)$.

In case (a), $\Phi_{(P_{k};\nu_{k})}(C)$ is a non-zero constant polynomial, so
that it has no roots. In case (b),
$\Phi_{(P_{k};\nu_{k})}(C)=\Psi_{(P_{k},V)}(\delta_{\nu_{k}})\,C$ whose only
root is $C=0$ as
$\Psi_{(P_{k},V)}(\delta_{\nu_{k}})=\Psi_{(P_{k-1},V)}(\delta_{\nu_{k}})\neq
0$ because $r(x)$ stabilizes $P$. In case (c) we have

$\Phi_{(P_{k};\nu_{k})}(C)=\Psi_{(P_{k},V)}(\delta_{\nu_{k}})\,C+{P_{k}}_{(\beta,(0,\ldots,0))},$

both of whose coefficients are non-zero by assumption so that it has a single
root.

If $P(r(x))=0$ then we are necessarily in case (b) because
$\mathcal{C}(P_{k})$ has no points with ordinate zero, whence (I) holds.

Assume that $P(r(x))=d\,x^{\gamma}+\cdots$ with $d\neq 0$, hence $(\gamma,0)$
is a vertex of $\mathcal{N}(P_{k})$. As $V=(\alpha,1)$ for some $\alpha$ and
$V\in L_{\nu_{k-1}}(P_{k})$, then $\gamma>\alpha+\nu_{k-1}$. If
$\nu_{k}\neq\gamma-\alpha$ then we are either in case (a) or case (b), so that
$\Phi_{(P_{k};\nu_{k})}(C)$ has no non-zero roots. As
$\gamma-\alpha>\nu_{k-1}$, there only remains the possibility
$\nu_{k}=\gamma-\alpha$, giving case (c) with $\beta=\gamma$, and
$\Phi_{(P_{k};\nu_{k})}(C)$ has a single non-zero root, which is (II). ∎

The criterion presented in Theorem 3 is of different nature that this one: it
only depends on the dimension of the $\mathbb{Q}$-vector space spanned by the
support of $r(x)$ and the order of the equation, while the above criterion
required knowing the whole equation $P$, $r(x)$, and the substitution
$P_{k}=P[r(x)]$.

In Section 5 we present an example of an admissible generalized polynomial
that satisfies the criterion of Theorem 3 and that can be completed in two
different ways to a solution of the differential equation so that the
stabilization criterion of Grigoriev and Singer does not apply.

###  4.3. Autonomous ordinary differential equations of first order

In this subsection we will apply our previous results to the case of first
order autonomous ordinary differential equations. In the next Theorem we prove
that any solution $s(x)\in\Omega$ of that kind of equation is a formal Puiseux
series (statement (1)). We also provide alternative proofs of two known
results: Theorems 11 and 12 in [8] (statements (2) and (3)).

Let $P=0$ be a non-trivial an autonomous ordinary differential equation of
first order, that is, a first order differential equation invariant by the
translations $x\mapsto x+c$, $c\in\mathbb{C}$. Hence $P$ can be written as a
non-zero polynomial with constant coefficients in $y_{0}$ and $y_{1}$, where
$y_{1}$ refers to to the derivative $\frac{d\,}{d\,x}$. Let us denote by
$\hat{K}^{*}=\cup_{d\geq 1}\mathbb{C}((x^{\frac{1}{d}\mathbb{Z}}))$ the field
of formal Puiseux power series.

######  Theorem 4.

Let $P(y_{0},y_{1})=0$ be non-trivial an autonomous first order ordinary
differential equation. Then

  1. (1)

If $s(x)\in\Omega$ be a solution of the equation $P=0$, then
$s(x)\in\hat{K}^{*}$.

  2. (2)

If $s(x)\in\hat{K}^{*}$ is a solution of $P=0$, then $s(x)$ is a convergent
Puiseux power series.

  3. (3)

For any point $(x_{0},c_{0})\in\mathbb{C}^{2}$, there exists an analytic
solution $s(x)$ of $P=0$ passing through the point $(x_{0},c_{0})$.

###### Proof.

Part (2) is consequence of our Theorem 2, see Corollary 3. In order to prove
parts (1) and (3), let us note some specific properties of the Newton polygon
of $P$ in this case.

Since $P$ is a polynomial with constant coefficients, we can write

(22)
$P=\sum_{i=0}^{d_{0}}\sum_{j=0}^{d_{1}}P_{i,j}\,y_{0}^{i}\,y_{1}^{j},\quad
P_{i,j}\in\mathbb{C}.$

Then any term $P_{i,j}\,y_{0}^{i}\,y_{1}^{j}$ corresponds to the point
$(-j,i+j)$ in $\mathcal{C}(P)$, which provides a bijection between the non-
zero terms of $P$ and $\mathcal{C}(P)$. Let $V\in\mathcal{C}(P)$ be a point
lying in some side or vertex of $\mathcal{N}(P)$. There exist unique integers
$0\leq i\leq d_{1}$ and $0\leq j\leq d_{2}$ such that $V=(-j,i+j)$. Then, the
indicial polynomial $\Psi_{(P,V)}(T)=P_{i,j}\,T^{j}$, with $P_{i,j}\neq 0$
because $V\in\mathcal{C}(P)$. We are assuming that $P$ is effectively a
differential equation, that is, $y_{1}$ appears in $P$, so that there is a
point with strictly negative abscissa in $\mathcal{C}(P)$. This implies that
for any $\mu\in\mathbb{R}$, the characteristic polynomial $\Phi_{(P;\mu)}(C)$
has a non-zero root if and only if $\mu=0$ or $\mu$ is the co-slope of a side
of $\mathcal{N}(P)$.

Let us prove part (1). Let $s(x)=c_{0}\,x^{\mu_{0}}+\cdots\in\Omega$, with
$c_{0}\neq 0$, be a non-constant solution of $P=0$. We may assume that
$\mu_{0}\neq 0$, otherwise, we perform the change of variable
$P[c_{0}]=P(y_{0}+c_{0},y_{1})$ to obtain a polynomial with constant
coefficients which has $s(x)-c_{0}=c_{1}\,x^{\mu_{1}}+\cdots$, with
$\mu_{1}>0$, as solution. By Lemma 2, $c_{0}$ is a non-zero root of the
characteristic polynomial $\Phi_{(P;\mu_{0})}(C)$. By the above properties of
$\mathcal{N}(P)$, either $\mu_{0}=0$ or $\mu_{0}$ is the co-slope of a side of
$\mathcal{N}(P)$. Since the sides of $\mathcal{N}(P)$ are rational,
necessarily $\mu_{0}$ is a non-zero rational number. Hence
$r(x)=c_{0}\,x^{\mu_{0}}$ is an admissible polynomial for $P=0$. Since
$\operatorname{supp}P=\\{0\\}$ and
$\langle\operatorname{supp}r(x)\rangle=\mathbb{Q}$, then $r(x)$ satisfies the
hypothesis of Theorem 3. Hence, $r(x)$ is the truncation of a solution $s(x)$
of $P=0$. By Theorem 1, $\langle\operatorname{supp}s(x)\rangle=\mathbb{Q}$,
hence $s(x)\in\mathbb{C}((x^{\mathbb{Q}}))$. Now, the arguments given in
Proposition 2 for the grid-based case, shows that $s(x)$ is grid-based, hence
$s(x)\in\hat{K}^{*}$.

Let us prove part (3). Since $P$ is autonomous, we may assume that $x_{0}=0$.
Performing the transformation $P(y_{0}+c_{0},y_{1})$, we may assume that
$c_{0}=0$. Write $P$ as in (22). If $P_{0,0}=0$, then $s(x)=0$ is a solution
and we are done. Assume that $P_{0,0}\neq 0$. Since $P$ is of order one, the
Newton polygon $\mathcal{N}(P)$ has a vertex of the form $(-j,i+j)$, with
$j\geq 1$. Hence $\mathcal{N}(P)$ has at least one side with co-slope
$0<\mu_{0}\in\mathbb{Q}$. By the properties of $\mathcal{N}(P)$, the
characteristic polynomial $\Phi_{(P;\mu_{0})}(C)$ has a non-zero root $c_{0}$,
and as in part (2), there exists a power series
$s(x)=c_{0}\,x^{\mu_{0}}+\cdots\in\hat{K}^{*}$ solution of $P=0$. By Corollary
3, $s(x)$ is convergent. ∎

##  5\. Example

Consider the irrational number $\tau=\pi/2$ and the linear differential
equations

$\displaystyle L_{1}$
$\displaystyle=\tau\,y_{0}-y_{1}-\big{(}(\tau-1)\,x+(\tau-2)\,x^{2}+(\tau-3)\,x^{3}\big{)},$
$\displaystyle L_{2}$
$\displaystyle=\tau\,y_{0}-y_{1}-\big{(}(\tau-1)\,x+(\tau-2)\,x^{2}+(\tau-5)\,x^{5}\big{)},$

where $y_{1}$ refers to the Euler derivative $x\frac{d\,y(x)}{d\,x}$ and
$y_{0}=y(x)$, for $y(x)\in\Omega$. Set $P=L_{1}\,L_{2}+x^{6}\,y_{0}\,y_{1}$
and $r(x)=x+x^{\tau}$. Here $L_{1}\,L_{2}$ is considered as a product of
polynomials, not as a composition of differential operators. More precisely

$P=y_{1}^{2}+\tau^{2}\,y_{0}^{2}+\left(-2\,\tau+x^{6}\right)\,y_{0}\,y_{1}\\\
+\left((2\,\tau-2)\,x+(2\,\tau-4)\,x^{2}+(\tau-3)\,x^{3}+(\tau-5)\,x^{5}\right)\,y_{1}\\\
+\left(2\,\tau\,(1-\tau)x+2\,\tau(2-\tau)\,x^{2}+\tau\,(3-\tau)\,x^{3}+\tau\,(5-\tau)\,x^{5}\right)\,y_{0}\\\
+(1-2\,\tau+\tau^{2})\,x^{2}+(4-6\tau+2\,\tau^{2})\,x^{3}+\cdots+(15-8\,\tau+\tau^{2})\,x^{8}.$

Let us verify that $r(x)$ satisfies the criterion of Theorem 3 for $P=0$ but
it can be continued to two different solutions, so that $r(x)$ does not
satisfy the stabilization criterion. Recall that Theorem 3 guarantees also the
convergence of those continuations.

As the order of the equation $P=0$ is one and $\tau$ is irrational we only
need to show that $r(x)$ is admissible. To this end, define $P_{0}=P$,
$P_{1}=P_{0}[x]$, $P_{2}=P_{1}[x^{\tau}]$. Writing only the terms
corresponding to the border of each Newton polygon, we have

$\displaystyle P_{1}$
$\displaystyle=(\tau\,y_{0}-y_{1})^{2}+2\,\left(\tau-2\right)\,x^{2}\,\left(y_{1}-\tau\,y_{0}\right)+\left(\tau-2\right)^{2}\,x^{4}+\cdots,$
$\displaystyle P_{2}$
$\displaystyle=(\tau\,y_{0}-y_{1})^{2}+2\,\left(\tau-2\right)\,x^{2}\,\left(y_{1}-\tau\,y_{0}\right)+\left(\tau-2\right)^{2}\,x^{4}+\cdots.$

From now on we refer the reader to Figure LABEL:fig:example-A1 for the
structure and elements of the Newton polygon of each the equations we shall
compute.

The element $E_{1}(P_{0})$ of co-slope $\nu_{1}=1$ is the side of
$\mathcal{N}(P_{0})$ with vertices $(0,2)$ and $(2,2)$. Its characteristic
polynomial is $\Phi_{(P;1)}(C)=\tau^{2}(\tau^{-1}-1)(C-1)^{2}$, whose only
root is $c_{1}=1$. Hence the polynomial $r_{1}(x)=x$ is admissible for $P$ but
does not satisfy either the criterion of Theorem 3 (it has no irrational
exponents), or the stabilization criterion (because
$\operatorname{ht}\operatorname{Bot}(E_{1}(P_{1}))=2$).

Now for $P_{1}$ the indicial polynomial for the vertex $V=(0,2)$ of
$\mathcal{N}(P_{1})$ is $\Psi_{(P_{1},V)}(\mu)=\tau-\mu$, whose root is
$\mu=\tau$. Set $\nu_{2}=\tau$. Since $1<\tau<2$, the element
$E_{\tau}(P_{1})$ is $V$, and its characteristic polynomial is
$\Phi_{(P_{1};\tau)}(C)=0$ so that any $c_{2}\in\mathbb{C}$ is a root. We
choose $c_{2}=1$. Thus $r_{2}(x)=x+x^{\tau}$ is an admissible polynomial for
$P=0$. We have not reached the stabilization step because
$E_{\tau}(P_{2})=\\{V\\}$ and the bottom vertex is $V$ which has height $2$
(see $\mathcal{N}(P_{2})$ in Figure LABEL:fig:example-A1). However, we can
already apply Theorem 3 so that $x+x^{\tau}$ can be continued to at least one
solution of $P=0$. Moreover, by Theorem 2, any solution which is a
continuation of $x+x^{\tau}$ is convergent.

Let us try and continue the process to see when we reach the stabilization
step.

As $\mathcal{N}(P_{2})$ has a unique side, of co-slope $2$, $E_{2}(P_{2})$, we
need to compute the corresponding characteristic polynomial

$\Phi_{(P_{2};2)}(C)=(\tau-2)^{2}+2\,(\tau-2)(2-\tau)\,C+(\tau-2)^{2}=(\tau-2)^{2}(C-1)^{2},$

whose only root is $c_{3}=1$ which makes $r_{3}(x)=x+x^{\tau}+x^{2}$ an
admissible polynomial. Let $P_{3}=P_{2}[x^{2}]$:

$P_{3}=(\tau\,y_{0}-y_{1})^{2}+\left(\tau-3\right)\,x^{3}\,\left(y_{1}-\tau\,y_{0}\right)+\left(\tau-4\right)^{2}\,x^{8}+\cdots,$

whose Newton polygon $\mathcal{N}(P_{3})$ has $E_{2}(P_{3})=\\{V\\}$. Again,
we still have not reached the stabilization step.

The Newton polygon $\mathcal{N}(P_{3})$ has two sides: setting $W=(3,1)$, they
are $E_{3}(P_{3})=[V,W]$ and $E_{5}(P_{3})=[W,(8,0)]$ of co-slopes $3$ and $5$
respectively. Notice that Proposition 1 applies to the points $V$, $W$ and
$(8,0)$ and as a consequence their corresponding indicial polynomials are
powers of $(\mu-\tau)$ up to a constant factor:

$\Psi_{(P_{3};V)}(\mu)=(\tau-\mu)^{2},\,\,\,\Psi_{(P_{3};W)}(\mu)=(\tau-3)\,(\mu-\tau),\,\,\,\Psi_{(P_{3};(8,0)}(\mu)=(\tau-4)^{2}\,(\mu-\tau)^{0}.$

This guarantees that the characteristic polynomials $\Phi_{(P_{3};3)}(C)$ and
$\Phi_{(P_{3};5)}(C)$ have non-zero roots. For co-slope $3$ we get

$\Phi_{(P_{3};3)}(C)=(\tau-3)^{2}\,C^{2}+(\tau-3)\,(3-\tau)\,C=(\tau-3)^{2}\,C\,(C-1),$

and for co-slope $5$,

$\Phi_{(P_{3};5)}(C)=(\tau-3)\,(5-\tau)\,C+(\tau-4)^{2},$

whose single root is $d_{5}=\frac{(\tau-4)^{2}}{(\tau-3)(\tau-5)}$. Since
$\Phi_{(P_{3};3)}(1)=0$ and $\Phi_{(P_{3};5)}(d_{5})=0$, we can continue
$r_{3}(x)$ with either $x^{3}$ or $d_{5}\,x^{5}$ to get admissible
polynomials. Let $P_{4}=P_{3}[x^{3}]$ and $Q_{4}=P_{3}[d_{5}\,x^{5}]$:

$\displaystyle P_{4}$
$\displaystyle=(\tau\,y_{0}-y_{1})^{2}-\left(\tau-3\right)\,x^{3}\,\left(y_{1}-\tau\,y_{0}\right)+x^{8}+\cdots,$
$\displaystyle Q_{4}$
$\displaystyle=(\tau\,y_{0}-y_{1})^{2}+\left(\tau-3\right)\,x^{3}\,\left(y_{1}-\tau\,y_{0}\right)+(1+\tau)\,x^{7+\tau}+\cdots.$

The structure of $\mathcal{N}(P_{4})$ guarantees that
$z(x)=x+x^{\tau}+x^{2}+x^{3}$ stabilize $P$ because
$\operatorname{Bot}(E_{3}(P_{4}))=W$ has height $1$ and its indicial
polynomial $\Psi_{(P_{4},W)}(\mu)=(\tau-3)(\mu-\tau)$ has no roots greater
than $3$. The stabilization criterion now implies that $z(x)$ can be completed
to an actual solution $\bar{z}(x)$, necessarily convergent. Similar arguments
apply to the other choice $Q_{4}=P_{3}[d_{5}\,x^{5}]$ and
$w(x)=x+x^{\tau}+x^{2}+d_{5}\,x^{5}$, which can be continued to a convergent
solution $\bar{w}(x)$. For the sake of the reader we include some of the first
terms of $\bar{z}(x)$ and $\bar{w}(x)$:

$\displaystyle\bar{z}(x)$
$\displaystyle=x+x^{\tau}+x^{2}+x^{3}+\frac{1}{(\tau-5)(3-\tau)}\,x^{5}+\frac{\tau+1}{4(\tau-3)}\,x^{4+\tau}+\cdots,$
$\displaystyle\bar{w}(x)$
$\displaystyle=x+x^{\tau}+x^{2}+d_{5}\,x^{5}-\frac{\tau+1}{4(\tau-3)}\,x^{4+\tau}+\frac{3}{(\tau-6)(\tau-3)}x^{6}+\cdots$

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