# Autonomous Operator and Differential Systems of order 1 over Integral
Domains

Ronald Orozco López

###### Abstract

In this paper we introduce the autonomous operator which is a nonlinear map
that sends sequences to sequences in an integral domain $R$, in particular
sequences over the ring of exponential generating functions. Properties of
this operator are studied: its relationship with the derivative of the ring of
sequences, its characterization by groups of $k$-homogeneity, condictions of
injectivity and surjectivity and its restriction to a linear operator.
Autonomous differential dynamical systems over integral domain are defined. It
is also shown that the flow of such differential equations is the exponential
generating function of the sequence obtained via the action of autonomous
operator acting over the Hurwitz expansion of the vectorial field of such
equation. It is shown that the flow is a torsion-free cyclic $R$-module. The
$R^{\times}$-modules associates to the groups of $k$-homogeneity are
constructed. We finish showing some examples of autonomous differential
equations and we find its $R^{\times}$-modules when $R$ is one of the rings:
$\mathbb{Z}$, $\mathbb{C}$, $\mathbb{Z}[i]$, $\mathbb{Z}[\omega]$ and
$\mathbb{Z}[\sqrt{2}]$.

Keywords: autonomous operator, Hurwitz expansion, homogeneity groups,
dynamical systems
Mathematics Subject Classification: 47H99, 16W99, 34A34, 11B83

##  1 Introduction

An important issue in the theory of autonomous differential systems is to find
a general analytic solution for the equation $x^{\prime}=f(x)$ where $f$ is an
arbitrary vector field $f:E\subset\mathbb{R}^{k}\rightarrow\mathbb{R}^{k}$.
One of the main approaches to the solution of this equation is the use of
formal series of rooted trees, introduced by Cayley [3] and developed by
Merson and Butcher [5]. Butcher is best known for showing the existence of a
group structure in the Runge-Kutta methods for ordinary differential equations
when those methods are expanded in power series. All this methods implicitly
use the Faá di Bruno formula for the composition of functions. Finally,
Foissy[4] studied the combinatorial form of the Dyson-Scwinger equation using
Hopf-Faá di Bruno algebras in which the coefficients of the solution use the
Bell polynomials. In [6] an analytical solution to the autonomous equation of
dimension $k$ is found by mean using Bell polynomials. In this paper a
nonlinear operator defined by means of the Bell polynomials is studied and
applied to the study of one-dimensional dynamical systems over integral
domain.

Bell polynomials are a very useful tool in mathematics to represent the $nth$
derivative of the composite of two functions [2]. In effect, let $f$ and $g$
be two analytical functions with power series representations $\sum
a_{n}\frac{x^{n}}{n!}$ and $\sum b_{n}\frac{x^{n}}{n!}$ respectively, with
$a_{n},b_{n}\in\mathbb{C}$. Then

$f(g(x))=f(b_{0})+\sum_{n=1}^{\infty}Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})\frac{x^{n}}{n!},$
(1)

where $Y_{n}$ is the $nth$ Bell polynomial. For instance it is well known that

$\displaystyle Y_{1}(b_{1};a_{1})$ $\displaystyle=$ $\displaystyle
a_{1}b_{1},$ $\displaystyle Y_{2}(b_{1},b_{2};a_{1},a_{2})$ $\displaystyle=$
$\displaystyle a_{1}b_{2}+a_{2}b_{1}^{2},$ $\displaystyle
Y_{3}(b_{1},b_{2},b_{3};a_{1},a_{2},a_{3})$ $\displaystyle=$ $\displaystyle
a_{1}b_{3}+a_{2}(3b_{1}b_{2})+a_{3}b_{1}^{3}.$

Bell polynomials can be expressed implicitly using the Faá di Bruno formula
[2],

$Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})=\sum_{k=1}^{n}B_{n,k}a_{k},$ (2)

where

$B_{n,k}=\sum_{|p(n)|=k}\frac{n!}{j_{1}!j_{2}!\cdots
j_{n}!}\left[\frac{b_{1}}{1!}\right]^{j_{1}}\left[\frac{b_{2}}{2!}\right]^{j_{2}}\cdots\left[\frac{b_{n}}{n!}\right]^{j_{n}},$
(3)

where the sum runs over all the partitions $p(n)$ of $n$, i.e.,
$n=j_{1}+2j_{2}+\cdots+nj_{n}$, $j_{h}$ denotes the number of parts of size
$h$ and $|p(n)|=j_{1}+j_{2}+\cdots+j_{n}$ is the length of partition $p(n)$.
Let $f\in C^{\infty}(S)$, where $S\subset\mathbb{R}$ and let

$\begin{cases}u^{\prime}=f(u)\\\ u(0)=x\end{cases}$ (4)

the autonomous ordinary differential equation with initial value problem.

In [6] the author introduced the autonomous polynomials, which are the
coefficients of the analytical solution of the equation (4) expressed as a
function of the initial value in the next way:

$\begin{cases}A_{1}(f(x))=f(x),\\\
A_{n+1}(f(x),f^{\prime}(x),...,f^{(n)}(x))=Y_{n}(A_{1}(f(x)),...,A_{n}(f(x));f^{\prime}(x),...,f^{(n)}(x))\end{cases}$
(5)

From (5) we can see that every $(n+1)$-th autonomous polynomial is defined
over $f(x)$, $f^{\prime}(x)$,…,$f^{(n)}(x)$ and in general every autonomous
polynomial $A_{n}$ is defined on the sequence $(f^{(n)}(x))_{n\geq 0}$ of
derivatives of $f(x)$. We define an operator by means of autonomous
polynomials defined over the sequences of derivatives of exponential
generating functions with coefficients in an integral domain. In this paper
$R$ or $\mathop{\mbox{{Frac}}}\nolimits(R)$, the fractions field of $R$,
contains to $\mathbb{Q}$. Also, $\mathbb{N}$ will be the monoid of nonnegative
integer and $\mathbb{N}_{1}$ will be the semigroup of positive integers.

This paper is organized as follows. In Section 2, the ring of exponential
generating functions and the ring of sequences over this are introduced.
Section 3 introduces the autonomous operator $\mathfrak{A}$. In Section 4, the
properties of $\mathfrak{A}$ are studied. In Section 5, the results of the
previous section are applied to dynamical systems over integral domain. The
$G$-modules of flows of dynamical systems are constructed. Finally, in Section
6 examples of dynamical systems are showed.

##  2 The $R$-algebra $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$

Let $R$ be an integral domain with identity and consider the power series of
the form $\sum_{n=0}^{\infty}a_{n}\frac{X^{n}}{n!}$ with $a_{n}\in R$, for all
$n\geq 0$ and let $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ denote the set of
all this series. Is is clear that
$(\mathop{\mbox{{Exp}}}\nolimits_{R}[X],+,\cdot)$ is a ring with componentwise
addiction and ordinary product of series, i.e.

$f(X)+g(X)=\sum_{n=0}^{\infty}(a_{n}+b_{n})\frac{X^{n}}{n!},$
$f(X)g(X)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}a_{k}b_{n-k}\frac{X^{n}}{n!},$

with
$f(X)=\sum_{n=0}^{\infty}a_{n}\frac{X^{n}}{n!},g(X)=\sum_{n=0}^{\infty}b_{n}\frac{X^{n}}{n!}\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$.

Let $\delta$ denote the formal derivative over the ring
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then the ring of constant of
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$, i.e. the kernel
$\mathop{\mbox{{Ker}}}\nolimits\delta$, is all the ring $R$.

######  Definition 1.

Let $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. The sequence of
derivatives of $f(X)$, denoted $\Delta f(X)$, is the sequence
$(\delta^{n}f(X))_{n\in\mathbb{N}}$.

######  Proposition 1.

Take $f$ and $g$ of $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then

  1. 1.

$\Delta(f+g)=\Delta f+\Delta g$.

  2. 2.

$\Delta(cf)=c\Delta f$, with $c\in R$.

  3. 3.

$\Delta(fg)=\left(\sum_{k=0}^{n}\binom{n}{k}\delta^{k}f\delta^{n-k}g\right)_{n\in\mathbb{N}}$.

  4. 4.

$\Delta(f\circ g)=(Y_{n}(\delta g,\delta^{2}g,...,\delta^{n}g;(\delta f)\circ
g,(\delta^{2}f)\circ g,...,(\delta^{n}f)\circ g)_{n\in\mathbb{N}}$, where
$Y_{n}$ is the $nth$ Bell polynomial and $Y_{0}=f\circ g$.

###### Proof.

$1$ and $2$ are obtained directly of properties of lineality of $\delta$. $3$
is obtained from the Leibniz formula. And $4$ is obtained from of definition
of Bell polynomials. ∎

Let $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ denote the set of all
sequences $\Delta f$ for all $f\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. From
the previous proposition it is followed that
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ is an abelian group because
$\Delta f+\Delta g=\Delta(f+g)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$.
Let $\ast$ denote the convolution product in
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ given by

$\Delta f(X)\ast\Delta
g(X)=\left(\sum_{k=0}^{n}\binom{n}{k}\delta^{k}f(X)\delta^{n-k}g(X)\right).$

Then from $3$ in previous proposition it is followed that

$\Delta f(X)\ast\Delta
g(X)=\Delta(f(X)g(X))\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X].$

As $c\Delta f=\Delta(cf)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$, then
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ is an $R$-algebra with identity
$\textbf{e}_{\Delta}=\Delta(1)=(1,0,...)$ and zero element
$\textbf{0}=(0,0,...)$. Then
$\Delta:\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$
is an $R$-isomorphism of algebras with
$\mathop{\mbox{{Ker}}}\nolimits\Delta=\\{0\\}$. The ring
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ is a subring of the ring
$H\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ of series of Hurwitz with
coefficients in $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and $\Delta f(X)$ is
known as Hurwitz expansion of $f(X)$. (See [1] for more details).

##  3 Autonomous Operator $\mathfrak{A}$

######  Definition 2.

Let $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$ be the multiplicative
algebra of sequences over $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ with
multiplication defined term to term. We define autonomous operator
$\mathfrak{A}$ as the nonlinear map
$\mathfrak{A}:\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$,
defined by $\mathfrak{A}(\Delta f)=(A_{n}([\delta^{n-1}f]))_{n\geq 1}$, with
$[\delta^{n}f]=(f,\delta f,...,\delta^{n}f)$, in where the $A_{n}$ are
recursively defined in the next way:

$\displaystyle A_{1}([\delta^{0}f])$ $\displaystyle=$ $\displaystyle
A_{1}([f])=f,$ (6) $\displaystyle A_{n+1}([\delta^{n}f])$ $\displaystyle=$
$\displaystyle
Y_{n}(A_{1}([\delta^{0}f]),A_{2}([\delta^{1}f])...,A_{n}([\delta^{n-1}f]);\delta^{1}f,...,\delta^{n}f),$

$n\geq 1$.
The polynomials $A_{n}$ in the indeterminates
$\delta^{0}f(X),\delta^{1}f(X),...,\delta^{n-1}f(X)$ will be called autonomous
polynomial.

Some polynomials $A_{n}$ are

$\displaystyle A_{1}([\delta^{0}f])$ $\displaystyle=$ $\displaystyle f,$
$\displaystyle A_{2}([\delta^{1}f])$ $\displaystyle=$ $\displaystyle
A_{1}([\delta^{0}f])\delta^{1}f=f\delta f,$ $\displaystyle
A_{3}([\delta^{2}f])$ $\displaystyle=$ $\displaystyle
A_{2}([\delta^{1}f])\delta f+A_{1}^{2}([\delta^{0}f])\delta^{2}f,$
$\displaystyle=$ $\displaystyle f(\delta f)^{2}+f^{2}\delta^{2}f,$
$\displaystyle A_{4}([\delta^{3}f])$ $\displaystyle=$ $\displaystyle
A_{3}([\delta^{2}f])\delta^{1}f+3A_{1}([\delta^{0}f])A_{2}([\delta^{1}f])\delta^{2}f+A_{1}^{3}([\delta^{0}f])\delta^{3}f,$
$\displaystyle=$ $\displaystyle f(\delta
f)^{3}+4f^{2}\delta^{1}f\delta^{2}f+f^{3}\delta^{3}f.$

Let $R^{\mathbb{N}}$ denote the ring of sequences over the ring $R$ with
componentwise addiction and product and indentity $\textbf{e}=(1,1,...)$. The
map $\gamma:R^{\mathbb{N}}\rightarrow\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$,
with
$\gamma(\textbf{a})=f_{\textbf{a}}(X)=\sum_{n=0}^{\infty}a_{n}\frac{X^{n}}{n!}$
is obviously an isomorphism of ring. Then the ring
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ is the ring of exponential generating
functions of sequences in $R^{\mathbb{N}}$. Let
$\mathcal{E}_{a}:\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow R$ denote
the evaluation map in the element $a$ of $R$ and extend this map to
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$ in this sense

$\mathcal{E}_{a}\Delta
f(X)=(\mathcal{E}_{a}\delta^{n}f(X))_{n\in\mathbb{N}_{0}}=(\delta^{n}f(a))_{n\in\mathbb{N}}.$

Then

$\mathcal{E}_{0}\Delta f_{\textbf{a}}(X)=(f_{\textbf{a}}(0),\delta
f_{\textbf{a}}(0),\delta^{2}f_{\textbf{a}}(0),...)=(a_{0},a_{1},a_{2},...)=\textbf{a}$
(7)

and $R^{\mathbb{N}}\subset\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Also

$\displaystyle\mathcal{E}_{0}\mathfrak{A}(\Delta f_{\textbf{a}}(X))$
$\displaystyle=$ $\displaystyle(f_{\textbf{a}}(0),f_{\textbf{a}}(0)\delta
f_{\textbf{a}}(0),f_{\textbf{a}}(0)(\delta
f_{\textbf{a}}(0))^{2}+f_{\textbf{a}}^{2}(0)\delta^{2}f_{\textbf{a}}(0),...)$
$\displaystyle=$
$\displaystyle(a_{0},a_{0}a_{1},a_{0}a_{1}^{2}+a_{0}^{2}a_{2},...)$

Thus $\mathfrak{A}$ act over any sequence in $R^{\mathbb{N}}$ although $R$ is
not necessarily a differential ring. On the other hand, is not easy, at least
to the autor, to find closed formulas for the images of $\mathfrak{A}$,
especially with of the sum, product or composition of functions in
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. For this reason, we calculate the
action $\mathfrak{A}$ over $\alpha\Delta f(X)$, $\exp(X)f(X)$ and
$\exp(f(X))$.

######  Proposition 2.

Let $f\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and $\alpha\in R$. Then

$\mathfrak{A}(\alpha\Delta
f(X))=(\alpha^{n})_{n\in\mathbb{N}}\mathfrak{A}(\Delta f(X)),$ (8)

###### Proof.

We shall prove that
$A_{n}([\alpha\delta^{n-1}f])=\alpha^{n}A_{n}([\delta^{n-1}f])$. The proof is
by induction. This is true for $n=1$, since $A_{1}([\alpha\delta^{0}f])=\alpha
f=\alpha A_{1}([f])$. Suppose that the statement it holds for all natural
number smaller than $n$ and let’s consider the case $n+1$. Then

$\displaystyle A_{n+1}([\alpha\delta^{n}f])$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\dfrac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{\alpha
A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{\alpha^{n}A_{n}}{n!}\right]^{j_{n}}\alpha\delta^{k}f$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!\alpha^{j_{1}+2j_{2}+\cdots+nj_{n}+1}}{j_{1}!\cdots
j_{n}!}\left[\frac{A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{A_{n}}{n!}\right]^{j_{n}}\delta^{k}f$
$\displaystyle=$
$\displaystyle\alpha^{n+1}\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{A_{n}}{n!}\right]^{j_{n}}\delta^{k}f$
$\displaystyle=$ $\displaystyle\alpha^{n+1}A_{n+1}([\delta^{n}f])$

and the statement follows. ∎

######  Proposition 3.

Let $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and let $\alpha\in R$, then

$\mathfrak{A}(\Delta(\exp(\alpha X)f(X)))=(\exp(\alpha
nX))_{n\in\mathbb{N}}\mathfrak{A}(F(X))$ (9)

where

$F(X)=(F_{n,\alpha}(f(X)))_{n\geq 0}$

with

$F_{n,\alpha}(f(X))=\sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\delta^{k}f(X).$
(10)

###### Proof.

In the previous proposition was shown that $\mathfrak{A}(\alpha
f(X))=(\alpha^{n})_{n\geq 1}\mathfrak{A}(f(X))$ for all $f(X)$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. We use this fact together with

$\delta^{n}(\exp(\alpha X)f(X))=\exp(\alpha X)F_{n,\alpha}(f(X))$

to obtain

$\displaystyle\mathfrak{A}\left(\Delta(\exp(\alpha X)f(X))\right)$
$\displaystyle=$ $\displaystyle\mathfrak{A}\left(\exp(\alpha
X)(F_{n,\alpha}(f(X)))_{n\in\mathbb{N}}\right)$ $\displaystyle=$
$\displaystyle(\exp(\alpha
nX))_{n\in\mathbb{N}}\mathfrak{A}(A)\left((F_{n,\alpha}(f(X)))_{n\in\mathbb{N}}\right)$

Making $f(X)=1$, we have $\mathfrak{A}(\Delta(\exp(\alpha
X)))=\left(\alpha^{n-1}(n-1)!\exp(\alpha nX)\right)_{n\in\mathbb{N}}$. ∎

######  Proposition 4.

Let $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then

$\mathfrak{A}(\Delta(\exp(f(X))))=\left(\exp(nf(X)))\right)_{n\in\mathbb{N}}\mathfrak{A}(Y(X)),$
(11)

where

$Y(X)=\left(Y_{n}(f^{(1)}(X),f^{(2)}(X),...,f^{(n)}(X))\right)_{n\in\mathbb{N}}$
(12)

with $Y_{n}$ the $n$-th Bell’s polynomial .

###### Proof.

From (2),

$Y_{n}=\sum_{k=1}^{n}B_{n,k}\delta^{k}\exp(f(X))=\exp(f(X))\sum_{k=1}^{n}B_{n,k}.$

where

$B_{n,k}=\sum_{|p(n)|=k}\frac{n!}{j_{1}!j_{2}!\cdots
j_{n}!}\left[\frac{f^{(1)}(0)}{1!}\right]^{j_{1}}\left[\frac{f^{(2)}(0)}{2!}\right]^{j_{2}}\cdots\left[\frac{f^{(n)}(0)}{n!}\right]^{j_{n}}.$

∎

##  4 The structure of
$\mathfrak{A}(\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X])$

In this section properties of $\mathfrak{A}$ are studied. In particular the
relationship between $\mathfrak{A}$ and the derivative $\delta$ of ring
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Also we characterize $\mathfrak{A}$
by groups of $k$-homogeneity in $R^{\mathbb{N}}$. We study condiction for
their injectivity and surjectivity. Their action over $R[X]$ and its
restriction to a linear operator.

###  4.1 Relationship between $\delta$ and $A_{n}$

In the next theorem we will see the relationship between the operators
$\mathfrak{A}$ and $\delta$

######  Theorem 1.

Take $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then

$A_{n+1}([\delta^{n}f(X)])=f(X)\delta(A_{n}([\delta^{n-1}f(X)])),$ (13)

with $A_{0}\equiv 1$.

###### Proof.

By definition $A_{2}([\delta f])=f\delta(f)$. Suppose true
$A_{n}([\delta^{n-1}f])=f\delta(A_{n-1}([\delta^{n-2}f]))$. Set $A_{n}\equiv
A_{n}([\delta^{n-1}f])$. On the one hand we have

$\displaystyle A_{n+1}$ $\displaystyle=$ $\displaystyle\delta
fA_{n}+\delta^{2}f\\{A_{1}A_{n-1}+A_{2}A_{n-2}+\cdots\\}+\cdots$
$\displaystyle=$
$\displaystyle\delta(f)f\delta(A_{n-1})+\delta^{2}(f)\\{ff\delta(A_{n-2})+f\delta(A_{1})A_{n-2}+\cdots\\}+\cdots$
$\displaystyle=$ $\displaystyle
f[\delta(f)\delta(A_{n-1})+\delta^{2}(f)\\{f\delta(A_{n-2})+\delta(A_{1})A_{n-2}+\cdots\\}+\cdots]$

And on the other hand,

$\displaystyle f\delta(A_{n})$ $\displaystyle=$ $\displaystyle
f\delta(\delta(f)A_{n-1}+\delta(f)\\{A_{1}A_{n-2}+A_{2}A_{n-3}+\cdots\\}+\cdots)$
$\displaystyle=$ $\displaystyle
f[\delta(f)\delta(A_{n-1})+\delta^{2}(f)A_{n-1}+\delta^{2}(f)\delta(A_{1})A_{n-2}+\delta^{2}(f)A_{1}\delta(A_{n-2})+\cdots]$
$\displaystyle=$ $\displaystyle
f[\delta(f)\delta(A_{n-1})+\delta^{2}(f)\\{f\delta(A_{n-2})+\delta(A_{1})A_{n-2}+\cdots\\}+\cdots]$

We can see that $A_{n+1}=f\delta(A_{n})$ for comparing both results. ∎

######  Corollary 1.

Take $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then

$A_{n}([\delta^{n-1}f(X)])=\overbrace{f(X)\delta(\cdots
f(X)\delta(f(X)))}^{n}.$ (14)

###### Proof.

From above theorem,
$A_{2}([\delta^{1}f])=f\delta(A_{1}([\delta^{0}f]))=f\delta(f)$. Inductively
the claim is followed. ∎

###  4.2 Characterization by groups of $k$-homogeneity

Let $R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ denote the set

$\\{\textbf{a}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]:\textbf{a}\in
R^{\mathbb{N}}\\}$ (15)

where the product $\textbf{a}\Delta f(X)$ is componentwise.
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ is an
$R^{\mathbb{N}}$-module since

  1. 1.

$\textbf{a}(\Delta f(X)+\Delta g(X))=\textbf{a}\Delta f(X)+\textbf{a}\Delta
g(X)$,

  2. 2.

$(\textbf{a}+\textbf{b})\Delta f(X)=\textbf{a}\Delta f(X)+\textbf{b}\Delta
f(X)$,

  3. 3.

$(\textbf{a}\textbf{b})\Delta f(X)=\textbf{a}(\textbf{b}\Delta f(X))$,

  4. 4.

$\textbf{e}\Delta f(X)=\Delta f(X)$.

If we define product in
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ as

$\textbf{a}\Delta f(X)\ast\textbf{b}\Delta g(X)=(\textbf{a}\textbf{b})(\Delta
f(X)\ast\Delta g(X))$ (16)

then $R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ becomes an
$R^{\mathbb{N}}$-algebra.

We wish to study the action of $\mathfrak{A}$ over
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Let
$\textbf{h}_{\alpha}=(\alpha^{n})_{n\geq 0}\in R^{\mathbb{N}}$, $\alpha\neq
0$. We extend $\textbf{h}_{\alpha}$ for all $k\in\mathbb{N}$ by means of to
define

$\textbf{h}^{k}_{\alpha}=\begin{cases}((\alpha^{n})^{k})_{n\geq 0},&if\ k\neq
0,\\\ e=(1)_{n\geq 0},&if\ k=0.\end{cases}$ (17)

Then $\textbf{h}_{\alpha}^{k}=\textbf{h}_{\alpha^{k}}$ and
$\textbf{h}_{\alpha^{k}}\textbf{h}_{\beta^{l}}=\textbf{h}_{\alpha^{k}\beta^{l}}$.
Let $\mathcal{M}_{1}$ denote the commutative multiplicative monoid

$\mathcal{M}_{1}=\\{\textbf{h}_{\alpha}\in R^{\mathbb{N}}:\alpha\neq 0\\}.$
(18)

If all $\alpha$ are taken to be units in $R$, then $\mathcal{M}_{1}$ is a
group and denote this by $\mathcal{H}_{1}$.

In the following theorem we characterize $\mathfrak{A}$ by the monoid
$\mathcal{M}_{1}$

######  Theorem 2.

$\mathfrak{A}$ is homogeneous under the action of $\mathcal{M}_{1}$, that is,

$\mathfrak{A}(\mathcal{M}_{1}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X])=\mathcal{M}_{1}\mathfrak{A}(\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]).$
(19)

###### Proof.

Take $\textbf{h}_{\alpha}\in\mathcal{M}_{1}$. We wish to show that

$A_{n}([\alpha^{(n-1)}\delta^{n-1}f(X)])=\alpha^{(n-1)}A_{n}([\delta^{n-1}f(X)])$

The proof is by induction on $n$. The case $n=1$ is trivial, since
$A_{1}([\alpha^{0}f(X)])=A_{1}([f(X)])=f(X)=\alpha^{0}f(X)=\alpha^{0}A_{1}([f(X)])$.
Suppose true for all natural number smaller than $n$ and we prove for $n+1$

$\displaystyle A_{n+1}([\alpha^{n}\delta^{n}f(X)])$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\dfrac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{\alpha^{0}A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{\alpha^{(n-1)}A_{n}}{n!}\right]^{j_{n}}\alpha^{k}\delta^{k}f(X)$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!\alpha^{(0j_{1}+j_{2}+\cdots+(n-1)j_{n}+1)}}{j_{1}!\cdots
j_{n}!}\left[\frac{A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{A_{n}}{n!}\right]^{j_{n}}\delta^{k}f(X)$
$\displaystyle=$
$\displaystyle\alpha^{n}\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{A_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{A_{n}}{n!}\right]^{j_{n}}\delta^{k}f(X)$
$\displaystyle=$ $\displaystyle\alpha^{n}A_{n+1}([\delta^{n}f(X)]).$

From which it follows that $\mathfrak{A}(\textbf{h}_{\alpha}\Delta
f(X))=\textbf{h}_{\alpha}\mathfrak{A}(\Delta f(X))$ for all nonzero $\Delta
f(X)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and all
$\textbf{h}_{\alpha}\in\mathcal{M}_{1}$. ∎

If $\alpha$ is a unity, then $\mathfrak{A}$ is homogeneous under the action
the group $\mathcal{H}_{1}$. The subscripts in $\mathcal{M}_{1}$ and
$\mathcal{H}_{1}$ are justified by the order of homogeneity of
$\textbf{h}_{\alpha}$. We call $\mathcal{M}_{1}$ and $\mathcal{H}_{1}$ monoid
and group of 1-homogeneity, respectively. The previous theorem tells us that
the monoid $\mathcal{M}_{1}$ defines a partition in
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ in homogeneity
class $\mathcal{M}_{1}\Delta f(X)$, for all
$f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Then the action of
$\mathfrak{A}$ over $R^{\mathbb{N}}$ as an action of class of 1-homogeneity.
Take also $g(X)$ in $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. If $f(X)=g(\alpha
X)$, then $\Delta f(X)=\textbf{h}_{\alpha}\Delta g(X)\circ(\alpha
X)_{n\in\mathbb{N}}$ implies that $\mathfrak{A}(\Delta
f(X))=\textbf{h}_{\alpha}\mathfrak{A}(\Delta g(X))\circ(\alpha
X)_{n\in\mathbb{N}}$, where the composition is componentwise.

Now we study the groups of $k$-homogeneity in
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ for $k>1$.

######  Definition 3.

For all $k\geq 2$ define

$\mathcal{H}_{k}=\\{\textbf{h}\in R^{\mathbb{N}}:\mathfrak{A}(\textbf{h}\Delta
f(X))=\textbf{h}^{k}\mathfrak{A}(\Delta f(X)),\ 0\neq\Delta
f(X)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X],\ h\neq 0\\}$ (20)

as the $k-$homogeneity set of $\mathfrak{A}$ and we will say that
$\mathfrak{A}$ is $\mathcal{H}_{k}-$homogeneous of degree $k$.

######  Theorem 3.

Let $\mathcal{H}_{k}$ be the $k$-homogeneity set of $\mathfrak{A}$. All
element h of $\mathcal{H}_{k}$ has the next form

$\textbf{h}=\\{a^{1-n}b^{n}\\}_{n\geq 0}$ (21)

where $a^{k-1}=1$ y $b^{k-1}=a$, $k>1$.

###### Proof.

Take $\textbf{h}=\\{h_{n}\\}_{n\geq 0}$ in $\mathcal{H}_{k}$. For $n=1$, we
have that $A_{1}([h_{0}x_{0}])=h_{0}^{k}A_{1}([x_{0}])$ implies that
$h_{0}=h_{0}^{k}$ and therefore $h_{0}^{k-1}=1$. Now for $n=2$, we have that
$A_{2}([h_{1}x_{1}])=h_{1}^{k}A_{2}([x_{1}])$ implies $h_{0}h_{1}=h_{1}^{k}$
and therefore $h_{1}^{k-1}=h_{0}$ and thus we have the desired relations. On
the other hand, it is easy to see that $h_{0}=h_{0}^{-(0-1)}h_{1}^{0}$ and
$h_{1}=h_{0}^{-(1-1)}h_{1}^{1}$. Since $\textbf{h}\in H_{k}$, then
$A_{n+1}([h_{n}x_{n}])=h_{n}^{k}A_{n+1}([x_{n}])$ implies that

$h_{0}^{r}h_{l_{1}}h_{l_{2}}\cdots
h_{l_{n+1-r}}x_{0}^{r}x_{l_{1}}x_{l_{2}}\cdots
x_{l_{n+1-r}}=h_{n}^{k}x_{0}^{r}x_{l_{1}}x_{l_{2}}\cdots x_{l_{n+1-r}}$

with $l_{1}+l_{2}+\cdots+l_{n-r}=n$. Then $h_{0}h_{1}^{n}=h_{0}^{n}h_{n}$ and
$h_{n}=h_{0}^{-(n-1)}h_{1}^{n}$. Hence it is followed that

$\displaystyle h_{n}^{k}$ $\displaystyle=$
$\displaystyle(h_{0}^{-(n-1)})^{k}(h_{1}^{n})^{k}$ $\displaystyle=$
$\displaystyle\left(h_{0}^{k}\right)^{-(n-1)}\left(h_{1}^{k}\right)^{n}$
$\displaystyle=$ $\displaystyle h_{0}^{-(n-1)}(h_{0}h_{1})^{n}$
$\displaystyle=$ $\displaystyle h_{0}^{-(n-1)}h_{0}^{n}h_{1}^{n}$
$\displaystyle=$ $\displaystyle h_{0}h_{1}^{n}.$

We want to show that $h_{0}^{r}h_{l_{1}}h_{l_{2}}\cdots
h_{l_{n+1-r}}=h_{0}h_{1}^{n}$ for all $r$. We have

$\displaystyle h_{0}^{r}h_{l_{1}}h_{l_{2}}\cdots h_{l_{n+1-r}}$
$\displaystyle=$ $\displaystyle
h_{0}^{r}h_{0}^{1-l_{1}}h_{1}^{l_{1}}h_{0}^{1-l_{2}}h_{1}^{l_{2}}\cdots
h_{0}^{1-l_{n+1-r}}h_{1}^{l_{n+1-r}}$ $\displaystyle=$ $\displaystyle
h_{0}^{r}h_{0}^{1-l_{1}}h_{0}^{1-l_{2}}\cdots
h_{0}^{1-l_{n+1-r}}h_{1}^{l_{1}}h_{1}^{l_{2}}\cdots h_{1}^{l_{n+1-r}}$
$\displaystyle=$ $\displaystyle
h_{0}^{n+1-(l_{1}+l_{2}+\cdots+l_{n+1-r})}h_{1}^{l_{1}+l_{2}+\cdots+l_{n+1-r}}$
$\displaystyle=$ $\displaystyle h_{0}h_{1}^{n}.$

The preceding result does not depend on the length $n-r$, so we can deduce
that $h_{n}=h_{0}^{-(n-1)}h_{1}^{n}$ for all $n$. On the other hand, suponga
que $h_{n}=h_{0}^{-(n-1)}h_{1}^{n}$. Then

$\displaystyle A_{n+1}([h_{n}x_{n}])$ $\displaystyle=$ $\displaystyle
A_{n+1}([h_{0}^{-(n-1)}h_{1}^{n}x_{n}])$ $\displaystyle=$ $\displaystyle
h_{0}h_{1}^{n}A_{n+1}([x_{n}])$ $\displaystyle=$ $\displaystyle
h_{n}^{k}A_{n+1}([x_{n}])$

and $h=\\{h_{0}^{-(n-1)}h_{1}^{n}\\}_{n\geq 0}\in\mathcal{H}_{k}$.

∎

######  Theorem 4.

$\mathcal{H}_{k}\subset R^{\mathbb{N}}$ is a torsion group of exponent
$(k-1)^{2}$.

###### Proof.

Take $\textbf{f},\textbf{g}\in\mathcal{H}_{k}$. Then

$\mathfrak{A}((\textbf{f}\textbf{g})\Delta
f(X))=\mathfrak{A}(\textbf{f}(\textbf{g}\Delta
f(X)))=\textbf{f}^{k}\mathfrak{A}(\textbf{g}\Delta
f(X))=\textbf{f}^{k}\textbf{g}^{k}\mathfrak{A}(\Delta
f(X))=(\textbf{f}\textbf{g})^{k}\mathfrak{A}(\Delta f(X)).$

Hence $\textbf{f}\textbf{g}\in\mathcal{H}_{k}$. Obviously
$\textbf{e}\in\mathcal{H}_{k}$. From the previous theorem we know that all
element in $\mathcal{H}_{k}$ is invertible because if
$(a^{1-n}b^{n})_{n\in\mathbb{N}}$ exist in $\mathcal{H}_{k}$, then
$(a^{n-1}b^{-n})_{n\in\mathbb{N}}$ is its inverse and thus $\mathcal{H}_{k}$
is a group. From $(a^{1-n}b^{n})^{(k-1)^{2}}=(a^{k-1})^{n}=1$, is followed
that $\textbf{f}^{(k-1)^{2}}=\textbf{e}$ for all
$\textbf{f}\in\mathcal{H}_{k}$ and it follow that $\mathcal{H}_{k}$ is a
torsion group of exponent $(k-1)^{2}$. ∎

As
$\mathfrak{A}(\mathcal{H}_{k}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X])=\mathcal{H}_{k}\mathfrak{A}(\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X])$,
then the operator $\mathfrak{A}$ is an action over class of $k$-homogeneity in
$R^{\mathbb{N}}\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Also let
$f(X)=ag(a^{-1}bX)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ with $a^{k-1}=1$ y
$b^{k-1}=a$. Then $\Delta f(X)=\textbf{h}\Delta
g(X)\circ(a^{-1}bX)_{n\in\mathbb{N}}$ and $\mathfrak{A}(\Delta
f(X))=\mathfrak{A}(\textbf{h}\Delta
g(X)\circ(a^{-1}bX)_{n\in\mathbb{N}})=\textbf{h}^{k}\mathfrak{A}(\Delta
g(X))\circ(a^{-1}bX)_{n\in\mathbb{N}}$.

As the exponential generating functions of elements in $\mathcal{H}_{k}$ are
exponential functions we call to the groups $\mathcal{H}_{k}$ exponential
groups. We finish this section showing the exponential groups
$\mathcal{H}_{k}$ for some rings.

####  4.2.1 Homogeneity group in $\mathbb{Z}^{\mathbb{N}}$

The units of $\mathbb{Z}$ are $\\{-1,1\\}$. Then
$\mathcal{H}_{1}=\\{\textbf{e},(1,-1,1,-1,...)\\}$ is a binary group and
$\mathcal{H}_{k}=\\{\textbf{e}\\}$ for $k>1$.

####  4.2.2 Homogeneity group in $\mathbb{R}^{\mathbb{N}}$

As $\mathbb{R}$ is a field then $\mathcal{H}_{1}\cong\mathbb{R}^{\times}$.
Also $\mathcal{H}_{k}=\\{\textbf{e}\\}$ for all $k>1$.

####  4.2.3 Homogeneity group in $\mathbb{C}^{\mathbb{N}}$

Since $a^{k-1}=1$, $b^{k-1}=a$, then $a$ is a $(k-1)th$ root of the unity,
i.e., $a\in\mathbb{Z}_{k-1}$ and $b\in\sqrt[k-1]{a}\cdot\mathbb{Z}_{k-1}$.
Then

  1. 1.

$\mathcal{H}_{1}\cong\mathbb{C}^{\times}$

  2. 2.

$\mathcal{H}_{2}=\\{\textbf{e}\\}$.

  3. 3.

$\mathcal{H}_{3}=\left\langle((-1)^{n})_{n\in\mathbb{N}},(-i^{n})_{n\in\mathbb{N}},(-(-i)^{n})_{n\in\mathbb{N}}\right\rangle\cong\mathbb{Z}_{4}$

  4. 4.

$\begin{split}\mathcal{H}_{4}=\left\langle(\omega^{n})_{n\in\mathbb{N}},(\omega^{2n})_{n\in\mathbb{N}},\left(\omega^{\frac{3-2n}{3}}\right)_{n\in\mathbb{N}},\left(\omega^{\frac{3+n}{3}}\right)_{n\in\mathbb{N}},\left(\omega^{\frac{3+4n}{3}}\right)_{n\in\mathbb{N}},\right.\\\
\left(\omega^{\frac{6-4n}{3}}\right)_{n\in\mathbb{N}},\left.\left(\omega^{\frac{6-n}{3}}\right)_{n\in\mathbb{N}},\left(\omega^{\frac{6-2n}{3}}\right)_{n\in\mathbb{N}}\right\rangle,\end{split}$

where $\omega=e^{2\pi i/3}$.

In general

$\mathcal{H}_{k}=\left\langle\left(e^{2\pi
i((j(1-n)+ln)(k-1)+jn)/(k-1)^{2}}\right)_{n\in\mathbb{N}}:j=0,1,...,k-2,l=0,1...,k-2\right\rangle$
(22)

####  4.2.4 Homogeneity group in $\mathbb{Z}[w_{d}]^{\mathbb{N}}$

Let $d$ be a square-free integer. Then
$\mathbb{Z}[w_{d}]=\\{a+bw_{d}:a,b\in\mathbb{Z}\\}$ is the ring known as the
quadratic ring, where

$w_{d}=\begin{cases}(1+\sqrt{d})/2&\ si\ d\equiv 1(\mod 4)\\\ \sqrt{d}&\ si\
d\not\equiv 1(\mod 4)\end{cases}$

We find the groups of homogeneity in $\mathbb{Z}[w_{d}]^{\mathbb{N}}$. We
start with the cases $d=-1,-3$, which correspond to the gaussian and
eisenstein rings, respectively.

  1. 1.

$\mathbb{Z}[i]$.

$\displaystyle\mathcal{H}_{1}$ $\displaystyle=$
$\displaystyle\\{\textbf{e},((-1)^{n})_{n\in\mathbb{N}},(i^{n})_{n\in\mathbb{N}},((-i)^{n})_{n\in\mathbb{N}}\\}\cong\mathbb{Z}_{4},$
$\displaystyle\mathcal{H}_{3}$ $\displaystyle=$
$\displaystyle\\{(a^{1-n}b^{n})_{n\in\mathbb{N}}:(a,b)\in\\{(1,1),(1,-1),(-1,i),(-1,-i)\\}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$
$\displaystyle\mathcal{H}_{5}$ $\displaystyle=$
$\displaystyle\\{(a^{1-n}b^{n})_{n\in\mathbb{N}}:(a,b)\in\\{(1,1),(1,-1),(1,i),(1,-i)\\}=\mathcal{H}_{1}\cong\mathbb{Z}_{4}$
$\displaystyle\mathcal{H}_{2^{l}+1}$ $\displaystyle=$
$\displaystyle\mathcal{H}_{1},\ l\geq 3,$ $\displaystyle\mathcal{H}_{q}$
$\displaystyle=$ $\displaystyle\\{\textbf{e}\\},\ q\neq 2^{l}+1.$

Then only there are two non-isomorphs nontrivial exponential groups in
$\mathbb{Z}[i]^{\mathbb{N}}$: $\mathcal{H}_{1}$ and $\mathcal{H}_{3}$.

  2. 2.

$\mathbb{Z}[\omega]$, $\omega=\frac{-1+\sqrt{3}i}{2}$.

$\displaystyle\mathcal{H}_{1}$ $\displaystyle=$
$\displaystyle\\{\textbf{e},((-1)^{n})_{n\in\mathbb{N}},(\omega^{n})_{n\in\mathbb{N}},((-1)^{n}\omega^{n})_{n\in\mathbb{N}},(\omega^{2n})_{n\in\mathbb{N}},((-1)^{n}\omega^{2n})_{n\in\mathbb{N}}\\}$
$\displaystyle\cong\mathbb{Z}_{6}$ $\displaystyle\mathcal{H}_{3}$
$\displaystyle=$
$\displaystyle\\{(a^{1-n}b^{n})_{n\in\mathbb{N}}:(a,b)\in\\{(1,1),(1,-1)\\}\\}\cong\mathbb{Z}_{2}$
$\displaystyle\mathcal{H}_{4}$ $\displaystyle=$
$\displaystyle\\{(a^{1-n}b^{n})_{n\in\mathbb{N}}:(a,b)\in\\{(1,1),(1,\omega),(1,\omega^{2})\\}\\}\cong\mathbb{Z}_{3}$
$\displaystyle\mathcal{H}_{7}$ $\displaystyle=$
$\displaystyle\\{(a^{1-n}b^{n})_{n\in\mathbb{N}}:(a,b)\in\\{(1,\pm
1),(1,\pm\omega),(1,\pm\omega^{2})\\}\\}=\mathcal{H}_{1}\cong\mathbb{Z}_{6}$
$\displaystyle\mathcal{H}_{3\cdot 2^{l}+1}$ $\displaystyle=$
$\displaystyle\mathcal{H}_{1},\ l\geq 2$ $\displaystyle\mathcal{H}_{2^{l}+1}$
$\displaystyle=$ $\displaystyle\mathcal{H}_{3},\ l\geq 2,$
$\displaystyle\mathcal{H}_{3^{m}\cdot 2^{l}+1}$ $\displaystyle=$
$\displaystyle\mathcal{H}_{4},\ m\geq 2,l\geq 0,$
$\displaystyle\mathcal{H}_{q}$ $\displaystyle=$
$\displaystyle\\{\textbf{e}\\},\ q\neq 2^{l}+1,\ q\neq 3^{m}\cdot 2^{l}+1,\
q\neq 3\cdot 2^{l}+1.$

Then only there are two non-isomorphs nontrivial exponential groups in
$\mathbb{Z}[\omega]^{\mathbb{N}}$: $\mathcal{H}_{1}$, $\mathcal{H}_{3}$ and
$\mathcal{H}_{4}$.

The case $d>1$, correspond to the real quadratic rings. For this ring there
are not nontrivial homogeneity groups $\mathcal{H}_{k}$, $k>1$.

  1. 1.

$\mathbb{Z}[\sqrt{2}]$.
$\mathcal{H}_{1}=\\{((\pm(1+\sqrt{2})^{m})^{n})_{n\in\mathbb{N}}:m\in\mathbb{Z}\\}$.

  2. 2.

$\mathbb{Z}[\sqrt{3}]$.
$\mathcal{H}_{1}=\\{((\pm(2+\sqrt{3})^{m})^{n})_{n\in\mathbb{N}}:m\in\mathbb{Z}\\}$.

  3. 3.

$\mathbb{Z}[\varphi]$, $\varphi$ the golden radio.
$\mathcal{H}_{1}=\\{((\pm\varphi^{m})^{n})_{n\in\mathbb{N}}:m\in\mathbb{Z}\\}$.

  4. 4.

In general for all $d>1$,
$\mathcal{H}_{1}=\\{((\pm\epsilon^{m})^{n})_{n\in\mathbb{N}}:m\in\mathbb{Z}\\}$,
where $\epsilon$ is a fundamental unity of $\mathbb{Z}[\sqrt{d}]$.

###  4.3 Injectivity and Surjectivity of $\mathfrak{A}$

Now we study the injectivity and the surjectivity of $\mathfrak{A}$ when this
is restricted to subset of $R^{\mathbb{N}}$.

######  Theorem 5.

The set of zeros of $\mathfrak{A}$ is the ideal

$R^{\mathbb{N}}_{0}=\\{(x_{n})_{n\geq 0}\in R^{\mathbb{N}}:x_{0}=0\\}\subset
R^{\mathbb{N}},$ (23)

i.e., $\mathop{\mbox{{Null}}}\nolimits(\mathfrak{A})=R^{\mathbb{N}}_{0}$.

###### Proof.

Take $f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and let $x^{*}$ be a zero
of $f(X)$. From (13) if $A_{1}([f(x^{*})])=f(x^{*})=0$, then
$A_{n}([\delta^{n-1}f(x^{*})])=f(x^{*})\delta(A_{n}([\delta^{(n-2)}f(x^{*})]))=0$
for all $n>1$. Hence, all the $\Delta
f(x^{*})=(\delta^{n}f(x^{*}))_{n\in\mathbb{N}}$ with $f(x^{*})=0$ are in
$\mathop{\mbox{{Null}}}\nolimits(\mathfrak{A})$, for all
$f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. On the other hand, let
$y\in\mathop{\mbox{{Null}}}\nolimits(\mathfrak{A})$. Then

$\mathfrak{A}(y)=\left(A_{1}([y_{0}]),A_{2}([y_{1}]),A_{3}([y_{2}])...\right)=(0,0,0,...).$

Since $A_{1}([y_{0}])=0$ implies $y_{0}=0$, then $y\in R^{\mathbb{N}}_{0}$ and
$\mathop{\mbox{{Null}}}\nolimits(\mathfrak{A})=R^{\mathbb{N}}_{0}$. ∎

Consequently $\mathfrak{A}$ is a map from quotient ring
$R^{\mathbb{N}}/R^{\mathbb{N}}_{0}$ to $R^{\mathbb{N}}$.

######  Theorem 6.

$\mathfrak{A}:R^{\mathbb{N}}/R^{\mathbb{N}}_{0}\rightarrow R^{\mathbb{N}}$ is
injective.

###### Proof.

Take $\textbf{x}=(x_{n})_{n\geq 0}$ and $\textbf{y}=(y_{n})_{n\geq 0}$ in
$R^{\mathbb{N}}$ with exponential generating functions $f_{\textbf{x}}(X)$ and
$f_{\textbf{y}}(Y)$, respectively, such that $x_{0}\neq 0$ and $y_{0}\neq 0$
and suppose that $A_{n}([x_{n-1}])=A_{n}([y_{n-1}])$ for all $n\geq 1$. Then
$x_{0}=y_{0}$. From (13), we have that
$f_{\textbf{x}}(0)\delta(A_{n-1}([\delta^{(n-2)}f_{\textbf{x}}(0)]))=f_{\textbf{y}}(0)\delta(A_{n-1}([\delta^{(n-2)}f_{\textbf{y}}(0)]))$.
By induction $x_{n}=y_{n}$ for all $n\geq 0$ and therefore
$\textbf{x}=\textbf{y}$. ∎

Then, we show that the operator $\mathfrak{A}$ is bijective if it is
restricted to those sequences $R^{\mathbb{N}}$ having first element
invertible. Thus

######  Theorem 7.

Let $R^{\mathbb{N}}_{U}$ denote the set $R^{\times}\times R^{\mathbb{N}_{1}}$.
Then $\mathfrak{A}:R^{\mathbb{N}}_{U}\rightarrow R^{\mathbb{N}}_{U}$ is
biyective.

###### Proof.

Indeed we note that $\mathfrak{A}(\textbf{x})$ falls into $R^{\mathbb{N}}_{U}$
when $\textbf{x}=(x_{n})\in R^{\mathbb{N}}_{U}$. Take $\textbf{y}\in
R^{\mathbb{N}}$ and let $\textbf{x}\in R^{\mathbb{N}}$ with
$\mathfrak{A}(\textbf{x})=\textbf{y}$. Then from the above theorem
$\mathfrak{A}$ restricted to $R^{\mathbb{N}}_{U}$ is injective. Let
$\textbf{x}\in R^{\mathbb{N}}$ such that
$\mathfrak{A}(\textbf{x})=\textbf{y}$. Then $x_{0}=y_{0}$,
$x_{1}=y_{0}^{-1}y_{1}$, $x_{2}=y_{0}^{-2}y_{2}-y_{0}^{-3}y_{1}^{2}$ and as
$y_{0}$ is an unity in $R$, therefore all element $x_{n}$ is expressed in
terms of $y_{0}^{-1},y_{1},y_{2},...$. Thus there is an only $\textbf{x}\in
R^{\mathbb{N}}_{U}$ such that $\mathfrak{A}(\textbf{x})=\textbf{y}$. Hence
$\mathfrak{A}$ is surjective when it is restricted to $R^{\mathbb{N}}_{U}$. ∎

In general $\mathfrak{A}:R^{\mathbb{N}}\rightarrow R^{\mathbb{N}}$ is not
surjective. For instance, the set $R^{\mathbb{N}}_{0}\setminus\\{0\\}$ is
never the image the some subset of $R^{\mathbb{N}}$ under the action of
$\mathfrak{A}$. This is noticed because if we take $x\in R^{\mathbb{N}}$ with
$A_{1}([x_{0}])=x_{0}=0$, then $A_{n}([x_{n-1}])=0$ for all $n>1$. Then
$\mathfrak{A}(x)=0$ and $\mathfrak{A}(x)\notin
R^{\mathbb{N}}_{0}\setminus\\{0\\}$.

Take a in $R^{\mathbb{N}}$ and let $f_{\textbf{a}}(X)$ its exponential
generating function in $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. Let
$f_{\textbf{a}}(X)\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$ denote
the subring of $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$ of
sequences in $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ multiplied by a factor
$f_{\textbf{a}}(X)$. Then

######  Proposition 5.

$\mathfrak{A}\left(\Delta(\mathop{\mbox{{Exp}}}\nolimits_{R}[X])\right)\subset\bigcup_{f_{\textbf{a}}(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]}f_{\textbf{a}}(X)\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$.

###### Proof.

From (14) we have

$\mathfrak{A}(\Delta(f_{\textbf{a}}(X)))=f_{\textbf{a}}(X)(1,\delta(A_{n-1}([\delta^{n-2}f_{\textbf{a}}(X)])))_{n\geq
2}\in f_{\textbf{a}}(X)\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}.$

∎

######  Corollary 2.

$\mathfrak{A}(R^{\mathbb{N}})\subset\bigcup_{x\in R}\left\langle
x\right\rangle^{\mathbb{N}}$.

###### Proof.

Follows from above proposition and from to evaluate $f_{\textbf{a}}(X)$ in
$X=0$. ∎

######  Proposition 6.

Let $\left\langle a\right\rangle$ be a principal ideal of $R$. Then
$\mathfrak{A}(\left\langle
a\right\rangle^{\mathbb{N}})\subset\mathcal{M}_{1}\left\langle
a\right\rangle^{\mathbb{N}}$.

###### Proof.

Take $a\textbf{x}\in\left\langle a\right\rangle^{\mathbb{N}}$. From
proposition 2,
$\mathfrak{A}(a\textbf{x})=a\textbf{h}_{a}\mathfrak{A}(\textbf{x})$. As
$\left\langle a\right\rangle^{\mathbb{N}}$ is an ideal in $R^{\mathbb{N}}$,
then $\textbf{h}_{a}(a\mathfrak{A}(\textbf{x}))\in\mathcal{M}_{1}\left\langle
a\right\rangle^{\mathbb{N}}$. ∎

Let $J=(I_{i})_{i\in\mathbb{N}}$ be a sequences of ideals in $R$ with the
$I_{i}$ not necessarily different. It is clear that $J$ is an ideal of
$R^{\mathbb{N}}$. Define

$\mathfrak{F}_{J}(X):=\\{f_{\textbf{a}}(X):\textbf{a}\in J\\}$ (24)

as the class of exponential generating functions in $J$ and

$\Delta(\mathfrak{F}_{J}(X)):=\\{\Delta(f_{\textbf{a}}(X)):f_{\textbf{a}}(X)\in\mathfrak{F}_{J}(X)\\}$
(25)

as the class of Hurwitz expansion of $\mathfrak{F}_{J}(X)$. Then

$\mathfrak{A}(\Delta(\mathfrak{F}_{J}(X)))=(A_{n}([\delta^{n-1}\mathfrak{F}_{J}(X)])_{n\in\mathbb{N}}.$
(26)

As we have seen, not every element in $R^{\mathbb{N}}$ is image of
$\mathfrak{A}$. However, if $F=\mathop{\mbox{{Frac}}}\nolimits(R)$, then from
the theorem 7, $\mathfrak{A}:F^{\mathbb{N}}\rightarrow F^{\mathbb{N}}$ is
bijective.

###  4.4 $\mathfrak{A}$ acting over $R[x]$

######  Definition 4.

For each $k\geq 1$ we define the ring

$R^{\mathbb{N}}_{k}=\\{\textbf{x}\in R^{\mathbb{N}}:\
\textbf{x}=(x_{0},x_{1},...,x_{k-1},0,0,...)\\}$ (27)

such that $R^{\mathbb{N}}_{k}\cap R^{\mathbb{N}}_{0}=\\{\textbf{0}\\}$.

The condiction $R^{\mathbb{N}}_{k}\cap R^{\mathbb{N}}_{0}=\\{\textbf{0}\\}$
implies that $x_{0}\neq 0$ for all
$\textbf{x}=(x_{0},x_{1},...,x_{k-1},0,...)$ in $R^{\mathbb{N}}_{k}$ and
$k\geq 2$. then we have the series of inclusions $\\{\textbf{0}\\}\subset
R^{\mathbb{N}}_{1}\subset R^{\mathbb{N}}_{2}\subset\cdots\subset
R^{\mathbb{N}}_{k}$.

The elements in the ring $R^{\mathbb{N}}_{k}$ correspond to the polynomials
$\sum_{n=0}^{k-1}a_{n}\frac{X^{n}}{n!}$ of grade $k-1$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$.

####  4.4.1 Structure of $\mathfrak{A}(R^{\mathbb{N}}_{1})$

The ring $R^{\mathbb{N}}_{1}$ is isomorph to $R$ and is the simpler ring from
which act $\mathfrak{A}$ since
$\mathfrak{A}(R^{\mathbb{N}}_{1})=R^{\mathbb{N}}_{1}$. Over
$R^{\mathbb{N}}_{1}$ the operator $\mathfrak{A}$ fulfil
$\mathfrak{A}(\textbf{x}+\textbf{y})=\mathfrak{A}(\textbf{x})+\mathfrak{A}(\textbf{y})$
and
$\mathfrak{A}(\textbf{x}\textbf{y})=\mathfrak{A}(\textbf{x})\mathfrak{A}(\textbf{y})$
for all $\textbf{x},\textbf{y}$ in $R^{\mathbb{N}}_{1}$.

####  4.4.2 Estructura de $\mathfrak{A}(R^{\mathbb{N}}_{2})$

We have that $R^{\mathbb{N}}_{2}=(R^{\mathbb{N}}_{2}\setminus
R^{\mathbb{N}}_{1})\cup R^{\mathbb{N}}_{1}$. If $\textbf{x}=(a,b,0,0...)\in
R^{\mathbb{N}}_{2}$, then

$\mathfrak{A}(\textbf{x})=\begin{cases}\textbf{x},&si\ \textbf{x}\in
R^{\mathbb{N}}_{1}\\\ a\textbf{h}_{b},&si\ \textbf{x}\in
R^{\mathbb{N}}_{2}\setminus R^{\mathbb{N}}_{1}.\end{cases}$ (28)

Now take $\textbf{x},\textbf{y}$ en $R^{\mathbb{N}}_{2}\setminus
R^{\mathbb{N}}_{1}$. Then we can say that
$\mathfrak{A}(\textbf{x}\textbf{y})=\mathfrak{A}(\textbf{x})\mathfrak{A}(\textbf{y})$
and that $\mathfrak{A}(R^{\mathbb{N}}_{2}\setminus R^{\mathbb{N}}_{1})$ is a
multiplicative group. Also if $\textbf{x}=(a,b,0,0,...)$,
$\textbf{y}=(c,d,0,0,...)$, then
$\mathfrak{A}(\textbf{x}+\textbf{y})=(a+c)\textbf{h}_{b+d}$ and $\mathfrak{A}$
is no linear in $R^{\mathbb{N}}_{2}\setminus R^{\mathbb{N}}_{1}$.

###  4.5 Linear part of $\mathfrak{A}$

######  Definition 5.

Let $\alpha\in K$. The linear part of the operator $\mathfrak{A}$ is the set

$\mathop{\mbox{{Lin}}}\nolimits\mathfrak{A}=\\{\textbf{x}\in
R^{\mathbb{N}}:\mathfrak{A}(\textbf{x})\in R^{\mathbb{N}}_{1}\\}$ (29)

The set $\mathop{\mbox{{Lin}}}\nolimits\mathfrak{A}$ is a sub-algebra of
$R^{\mathbb{N}}$ and the restriction of $\mathfrak{A}$ to
$\mathop{\mbox{{Lin}}}\nolimits\mathfrak{A}$ is a linear operator, i.e., if
$\textbf{x},\textbf{y}\in\mathop{\mbox{{Lin}}}\nolimits\mathfrak{A}$ and if
$a,b\in R$, then
$\mathfrak{A}(a\textbf{x}+b\textbf{y})=a\mathfrak{A}(\textbf{x})+b\mathfrak{A}(\textbf{y})$.
It is easy to note that
$\mathop{\mbox{{Lin}}}\nolimits\mathfrak{A}=R^{\mathbb{N}}_{1}$.

##  5 Dynamical system associated with $\mathfrak{A}(\Delta f(x))$

We know from [6] that the solution $\phi$ of (4) is given by

$\phi(t,x)=x+\sum_{n=1}^{\infty}A_{n}([f^{(n-1)}(x)])\dfrac{t^{n}}{n!}.$ (30)

This solution is a dynamical system defined on the sequence
$f(x),f^{(1)}(x),f^{(2)}(x),...$. Also, it is well known that the soluction
$\phi(t,x)$ is a flow, i.e, satisfy

  1. 1.

$\phi(0,x)=x$

  2. 2.

$\phi(t,\phi(s,x))=\phi(t+s,x)$

The property 2) tell us that $\phi(t,x)$ for all $t\in\mathbb{R}$ is an
oneparametric group, with identity $\phi(0,x)$ and composition product
$\circ$. From this we can note that the operator $\mathfrak{A}$ lead the
sequence of derivative of $f(x)$ the group $\\{\phi(t,x)\\}_{t\in\mathbb{R}}$.
Then, we will define a dynamical system over the integral domain
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$.

######  Definition 6.

Take $\Delta f(x)$ in $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. We define
the dynamical system $\Phi:R\times
R\times\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow\mathop{\mbox{{Exp}}}\nolimits_{S}[T]$,
$S=\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$, related to $\Delta
f(x)$ as

$\Phi(t,x,\Delta
f(x))=x+\sum_{n=1}^{\infty}A_{n}([\delta^{(n-1)}f(x)])\dfrac{t^{n}}{n!}.$ (31)

From previous definition is concluded that the flow associate to Hurwitz
expansion $\Delta f(x)$ is the exponential generating function of
$\mathfrak{A}(\Delta f(X))$. From the theorem 7 is deduced that if
$\textbf{a}\in R^{\mathbb{N}}_{U}$, then there is a function $f(x)$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ such that
$\epsilon_{0}\mathfrak{A}(\Delta f(x))=\textbf{a}$, i.e, all sequence in
$R^{\mathbb{N}}_{U}$ is mapped to a funcion generadora exponencial
$\epsilon_{0}\Phi(t,x,\Delta f(X))$ in $\mathop{\mbox{{Exp}}}\nolimits_{S}[T]$
for some function $f(x)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. In
particular, the previous it holds for all sequence in
$\mathop{\mbox{{Frac}}}\nolimits(R)^{\mathbb{N}}$. For instance, the sequences
$(1,1,2,5,14,42,...)$ and $(1,1,2,3,5,8,13,...)$ have associates the
exponential generating functions $\epsilon_{0}\Phi(t,x,\Delta f(X))$ and
$\epsilon_{0}\Phi(t,x,\Delta g(X))$ for some functions $f$ and $g$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. An important issue is how to obtain
such functions, if possible, but this will not be studied in this paper. We
will deal with from now on with basic results from the theory of dynamical
systems and we will apply the previous results in this paper.

######  Theorem 8.

$\Phi(t,x,\Delta f(x))$ is a flow.

###### Proof.

When $t=0$ we have from (31) that $\Phi(0,x,\Delta f(x))=x$. Now the property
2) will be proved. On one hand, the Taylor expansion of $\Phi(s+t,x,\Delta
f(x))$ is $\sum_{n=0}^{\infty}\delta_{s}^{n}\Phi(s,x,\Delta
f(x))\frac{t^{n}}{n!}$. On the other hand, making $\Phi_{s}=\Phi(s,x,\Delta
f(x))$ we have

$\displaystyle\Phi(t,\Phi(s,x,\Delta f(x)),\Delta f(x))$ $\displaystyle=$
$\displaystyle\Phi_{s}+\sum_{n=1}^{\infty}A_{n}([\delta^{n-1}f(\Phi_{s})])\frac{t^{n}}{n!}$
$\displaystyle=$
$\displaystyle\Phi_{s}+\sum_{n=1}^{\infty}\delta_{s}^{n}\Phi(s,x,\Delta
f(x))\frac{t^{n}}{n!}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\delta_{s}^{n}\Phi(s,x,\Delta
f(x))\frac{t^{n}}{n!}$ $\displaystyle=$ $\displaystyle\Phi(s+t,x,\Delta
f(x)).$

∎

On the other hand, from (13) we know that the operators $\mathfrak{A}$ and
$\delta$ are related by equation
$A_{n+1}([\delta^{n}f(x)])=f(x)\delta(A_{n}([\delta^{n-1}f(x)]))$. Write
$\delta_{x}=\delta$ and suppose that $\delta_{x}$ extend as derivative to
$\Phi$, i.e., $\delta_{x}$ derives formally to the dynamical system
$\Phi(t,x,\Delta f(x))$ with respect to the variable $x$. Also define
$\delta_{t}$ the formal derivative of $\Phi$ with respect to variable $t$. In
the following theorem we will show how are relationed the partial derivatives
$\delta_{x}$ and $\delta_{t}$ of $\Phi$.

######  Theorem 9.

Take $\Delta f(x)\in R^{\mathbb{N}}$. Then

$f(x)\delta_{x}\Phi(t,x,\Delta f(x))=\delta_{t}\Phi(t,x,\Delta f(x))=f(\Phi)$
(32)

###### Proof.

From (31), we have

$\displaystyle f(x)\delta_{x}\Phi(t,x,\Delta f(x))$ $\displaystyle=$
$\displaystyle
f(x)\delta_{x}\left(x+\sum_{n=1}^{\infty}A_{n}([\delta_{x}^{n-1}f(x)])\dfrac{t^{n}}{n!}\right)$
$\displaystyle=$ $\displaystyle
f(x)+\sum_{n=1}^{\infty}f(x)\delta_{x}\left(A_{n}([\delta_{x}^{n-1}f(x)])\right)\dfrac{t^{n}}{n!}$
$\displaystyle=$ $\displaystyle
A_{1}([f(x)])+\sum_{n=1}^{\infty}A_{n+1}([\delta_{x}^{n}f(x)])\dfrac{t^{n}}{n!}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}A_{n+1}([\delta_{x}^{n}f(x)])\dfrac{t^{n}}{n!}$
$\displaystyle=$ $\displaystyle\delta_{t}\Phi(t,x,\Delta f(x)).$

∎

A result of great importance is that we can construct an $R$-module from the
flow $\Phi(t,x,r\Delta f(x))$ for a fixed $r$ in $R$. First we prove the
following theorem who says that the flows of $\delta_{s}\Phi=f(\Phi)$ and
$\delta_{t}\Phi=rf(\Phi)$ have matching trajectories.

######  Theorem 10.

Fix an $r$ in $R$. Then $\Phi(t,x,r\Delta f(x))=\Phi(rt,x,\Delta f(x))$ for
all $t\in R$.

###### Proof.

From proposition 2 we know that $\mathfrak{A}(r\Delta f(x))=(r^{n})_{n\geq
1}\mathfrak{A}(\Delta f(x))$. Then

$\displaystyle\Phi(t,x,r\Delta f(x))$ $\displaystyle=$ $\displaystyle
x+\sum_{n=0}^{\infty}r^{n}A_{n}([\delta^{(n-1)}f(x)])\frac{t^{n}}{n!}$
$\displaystyle=$ $\displaystyle
x+\sum_{n=0}^{\infty}A_{n}([\delta^{(n-1)}f(x)])\frac{(rt)^{n}}{n!}$
$\displaystyle=$ $\displaystyle\Phi(rt,x,\Delta f(x)),$

as we wanted to show. ∎

Now we use the previous results for to construct an $R$-module over the flow
$\Phi$

######  Theorem 11.

Fix an $r$ and an $x$ in $R$ and let $M_{\Phi}(x)=(\\{\Phi_{t,r}(x)\\}_{t\in
R})$ be an abelian group, where $\Phi_{t,r}(x)\equiv\Phi(t,x,r\Delta f(x))$
and the action of ring $R$ over $M_{\Phi}(x)$ is given by the map
$\star:\\{\Phi_{t,r}(x)\\}_{t\in R}\times
R\rightarrow\\{\Phi_{t,r}(x)\\}_{t\in R}$ defined by $\Phi_{t,r}(x)\star
s=\Phi_{ts,r}(x)$. Then $M_{\Phi}(x)$ is a right $R$-module.

###### Proof.

From above theorem
$\Phi_{t,r}(x)=\Phi_{rt,1}(x)=\Phi_{r,t}(x)=\Phi_{1,rt}(x)$. Then we have

  1. 1.

$(\Phi_{t,r}(x)\circ\Phi_{s,r}(x))\star v=\Phi_{t+s,r}(x)\star
v=\Phi_{tv+sv,r}(x)=\Phi_{tv,r}(x)\circ\Phi_{sv,r}(x)=(\Phi_{t,r}(x)\star
v)\circ(\Phi_{s,r}(x)\star v)$,

  2. 2.

$\Phi_{t,r}(x)\star(v+w)=\Phi_{t(v+w),r}(x)=\Phi_{tv+tw,r}(x)=\Phi_{tv,r}(x)\circ\Phi_{tw,r}(x)=(\Phi_{t,r}(x)\star
v)\circ(\Phi_{t,r}(x)\star w)$,

  3. 3.

$\Phi_{t,r}(x)\star(vw)=\Phi_{tvw,r}(x)=\Phi_{tvr}(x)\star
w=(\Phi_{t,r}(x)\star v)\star w$,

  4. 4.

$\Phi_{t,r}(x)\star 1_{R}=\Phi_{t,r}(x)$,

where $v,w\in R$. ∎

For any $x$ in $R$ defines the map $\sigma:R\rightarrow M_{\Phi}(x)$ by
$\sigma(t)=\Phi_{t,r}(x)$. Then
$\sigma(t+s)=\Phi_{t+s,r}(x)=\sigma(t)\circ\sigma(s)$,
$\sigma(st)=\Phi_{st,r}(x)=\sigma(s)\star t=\sigma(t)\star s$. As the ring $R$
is an $R$-module, then the map $\sigma$ becomes an homomorphism of
$R$-modules.

Let

$\Gamma_{x}:=\\{\epsilon_{x}\Phi(t,y,\Delta f(y)):t\in R\\}$ (33)

denote the orbit or trayectory of $x$. Then, all sequence in
$R^{\mathbb{N}}_{U}$ has to the trajectory $\Gamma_{0}$ of the flow
$\Phi(t,x,\Delta f(x))$ as exponential generating function la trayectoria. If
$\Gamma_{x}=\\{x\\}$, then $x$ is an equilibrium point for $\Phi$. The
equilibrium points are obtained when $A_{n}([\delta^{n-1}f(x)])=0$ for all
$n$, i.e., when $f(x)=0$ for some $x\in R$. Thus a solution of equilibrium of
$\delta_{t}\Phi=f(\Phi)$ correspond with a sequence in the ideal
$R^{\mathbb{N}}_{0}$. If $x$ is not an equilibrium point, then this will be
called a regular point of $\Phi$.

Let

$\displaystyle\mathop{\mbox{{Ker}}}\nolimits\sigma$ $\displaystyle=$
$\displaystyle\\{t\in R:\sigma(t)=1_{M_{\Phi}}\\}$ $\displaystyle=$
$\displaystyle\\{t\in R:\Phi(t,x,\Delta f(x))=x\\}$

be the kernel $\sigma$. If $\overline{x}$ is an equilibrium point,
$\sigma:R\rightarrow\\{\overline{x}\\}$ and
$\mathop{\mbox{{Ker}}}\nolimits\sigma=R$. If $x$ is regular point,
$\mathop{\mbox{{Ker}}}\nolimits\sigma=\\{0\\}$ and $\sigma$ is injective. As
$\sigma$ is surjective, then $\sigma$ is an isomorphisms of $R$-modules. We
shall say that the module $M_{\Phi}(x)$ is a trivial module if $x$ is an
equilibrium point. Otherwise, will be called nontrivial module.

######  Theorem 12.

A non-trivial $R$-module $M_{\Phi}(x)$ is a torsion-free cyclic module.

###### Proof.

Suppose $x$ is a regular point. To show that $M_{\Phi}(x)$ is a cyclic module,
is enough to take an unit $u$ in $R$. Then $\Phi_{u,r}(x)\star
R=\Phi_{uR,r}(x)=M_{\Phi}(x)$ and thus $M_{\Phi}(x)$ is cyclic. On the one
hand, there is an ideal $I\subset R$ such that $M_{\Phi}(x)$ is isomorph to
$R/I$. Yet we are already that $R$ and $M_{\Phi}(x)$ are isomorph. Thus $I$ is
the zero ideal. On the other hand, $M_{\Phi}(x)$ is an cyclic $R$-module is
equivalent to saying that the multiplication homomorphism
$\tau_{s}:R\rightarrow M_{\Phi}(x)$, $\tau_{s}(a)=\Phi_{s,r}(x)\star a$, is a
surjective $R$-module homomorphism. Write
$I=\mathop{\mbox{{Ker}}}\nolimits(\tau_{s})$. By the first isomorphism
theorem, $\overline{\tau_{s}}$ is an isomorphism from $R/I$ to $M_{\Phi}(x)$.
Since $\mathop{\mbox{{Ker}}}\nolimits(\tau_{s})$ is the annihilator
$\mathop{\mbox{{Ann}}}\nolimits(\Phi_{s,r}(x))$ of $\Phi_{s,r}(x)$, then
$\mathop{\mbox{{Ann}}}\nolimits(\Phi_{s,r}(x))=I=0$ for any $\Phi_{s,r}(x)\in
M_{\Phi}(x)$ and $0$ is only torsion element in $M_{\Phi}(x)$. ∎

###  5.1 The flow $\Phi$ as an $R^{\times}$-module

In this section we study the existing relationship between the exponential
groups $\mathcal{H}_{k}$ and the flow $\Phi$ of a differential equation
$\delta_{t}\Phi=f(\Phi)$. Actually, we note that each group $\mathcal{H}_{k}$
defines an action over $\Phi$ as $G$-module. First, we study the $G$-module
induced by $\mathcal{H}_{1}$ and then the $G$-modules induced by
$\mathcal{H}_{k}$ for $k\geq 2$. We define a right $R$-algebra via composition
in this way

  1. 1.

$(\Delta f(x)+\Delta g(x))\circ(ax)_{n\in\mathbb{N}}=\Delta
f(x)\circ(ax)_{n\in\mathbb{N}}+\Delta g(x)\circ(ax)_{n\in\mathbb{N}}$

  2. 2.

$\Delta f(x)\circ(x)_{n\in\mathbb{N}}=\Delta f(x)$,

  3. 3.

$\Delta f(x)\circ(abx)_{n\in\mathbb{N}}=(\Delta
f(x)\circ(ax)_{n\in\mathbb{N}})\circ(bx)_{n\in\mathbb{N}}$,

  4. 4.

$(\Delta f(x)\ast\Delta g(x))\circ(ax)_{n\in\mathbb{N}}=(\Delta
f(x)\circ(ax)_{n\in\mathbb{N}})\ast(\Delta g(x)\circ(ax)_{n\in\mathbb{N}})$.

The composition $\circ$ extend to $\mathop{\mbox{{Exp}}}\nolimits_{S}[T]$ by
to do

$\sum_{n=0}^{\infty}\delta^{n}f(x)\circ(ax)\frac{t^{n}}{n!}=\left(\sum_{n=0}^{\infty}\delta^{n}f(x)\frac{t^{n}}{n!}\right)\circ(ax).$

This fact is used to show the existing relationship between the trajectory
$\Gamma_{ax}$ of the flow associated to $\Delta f(x)$ and the flow associated
to $\Delta f(ax)$.

######  Theorem 13.

Let $a$ be non-zero in $R$ and let $\mathcal{H}_{1}$ be the group of
$1$-homogeneity of $\mathfrak{A}$ and
$\textbf{g}_{a}=(a^{n})_{n\in\mathbb{N}_{0}}\in\mathcal{H}_{1}$. Then for all
sequence $\Delta f(x)$ in $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ we
have

$a\Phi(t,x,\Delta f(ax))=a\Phi(t,x,\textbf{g}_{a}\Delta
f(x)\circ(ax)_{n\in\mathbb{N}_{0}})=\Phi(at,x,\Delta f(x))\circ(ax)$ (34)

where the composition is in the $x$ variable.

###### Proof.

If $\textbf{g}_{a}\in\mathcal{H}_{1}$, we have that

$\displaystyle a\Phi(t,x,\Delta f(ax))$ $\displaystyle=$ $\displaystyle
a\Phi(t,x,\textbf{g}_{a}\Delta f(x)\circ(ax)_{n\in\mathbb{N}_{0}})$
$\displaystyle=$ $\displaystyle
ax+a\sum_{n=1}^{\infty}A_{n}([a^{n-1}\delta^{n-1}f(x)\circ(ax)])\dfrac{t^{n}}{n!}$
$\displaystyle=$ $\displaystyle
ax+a\sum_{n=1}^{\infty}a^{n-1}A_{n}([\delta^{n-1}f(x)])\circ(ax)\dfrac{t^{n}}{n!}$
$\displaystyle=$ $\displaystyle
ax+\sum_{n=1}^{\infty}A_{n}([\delta^{n-1}f(x)])\circ(ax)\dfrac{(at)^{n}}{n!}$
$\displaystyle=$ $\displaystyle\Phi(at,x,\Delta f(x))\circ(ax).$

∎

With the previous result we obtain the first $G$-module induced by the
homogeneity group $\mathcal{H}_{1}$.

######  Theorem 14.

Denote $\mathfrak{M}_{1}(\Phi)$ be the set

$\\{u\Phi(t,x,\Delta f(ux)):u\in R^{\times}\\}$ (35)

and let $\diamond:\mathfrak{M}_{1}(\Phi)\times
R^{\times}\rightarrow\mathfrak{M}_{1}(\Phi)$ be a right group action given by

$\Phi(t,x,\Delta f(x))\diamond u=\Phi(t,x,u\Delta f(x))\circ(ux)$ (36)

Then $\mathfrak{M}_{1}(\Phi)$ is a $R^{\times}$-module.

###### Proof.

From previous theorem, $\Phi(t,x,u\Delta f(x))\circ(ux)=u\Phi(t,x,\Delta
f(ux))$. Then

  1. 1.

$\Phi(t,x,\Delta f(x))\diamond u\in\mathfrak{M}_{1}(\Phi)$,

  2. 2.

$\Phi(t,x,\Delta f(x))\diamond 1_{R^{\times}}=\Phi(t,x,\Delta f(x))$,

  3. 3.

$\Phi(t,x,\Delta f(x))\diamond uv=(\Phi(t,x,\Delta f(x))\diamond u)\diamond
v$,

  4. 4.

$\Phi(t+s,x,\Delta f(x))\diamond u=(\Phi(t,x,\Delta f(x))\diamond
u)\circ(\Phi(s,x,\Delta f(x))\diamond u)$,

∎

######  Definition 7.

The $R^{\times}$-module $\mathfrak{M}_{1}(\Phi)$ will be called
$R^{\times}$-module of homogeneity of order 1.

Equally, we will show the existing relationship between the trajectory
$\Gamma_{a^{-1}bx}$ of the flow associated to $\Delta f(x)$ and the flow
associated to $a\Delta f(a^{-1}bx)$, where $a$ and $b$ hold $a^{k-1}=1$ and
$b^{k-1}=a$, $k\geq 2$.

######  Theorem 15.

Let $\mathcal{H}_{k}$ be $k-$homogeneity group of $\mathfrak{A}$ with $k>1$ in
$R^{\mathbb{N}}$ and let
$\textbf{h}=(a^{1-n}b^{n})_{n\in\mathbb{N}_{0}}\in\mathcal{H}_{k}$ Then for
all sequence $\Delta f(x)=(\delta^{n}f(x))_{n\in\mathbb{N}_{0}}$ we have

$\displaystyle a^{-1}b\Phi(t,x,a\Delta f(a^{-1}bx))$ $\displaystyle=$
$\displaystyle a^{-1}b\Phi(t,x,\textbf{h}\Delta
f(x)\circ(a^{-1}bx))_{n\in\mathbb{N}_{0}}$ (37) $\displaystyle=$
$\displaystyle\Phi(a^{-1}bt,x,\Delta f(x))\circ(a^{-1}bx).$

###### Proof.

Igual que el caso anterior. ∎

Ahora obtendremos los $G$-module asociados a los grupos $\mathcal{H}_{k}$,
$k\geq 2$.

######  Theorem 16.

Denote $\mathfrak{M}_{k}(\Phi)$, $k\geq 2$, be the set

$\\{a^{-1}b\Phi(t,x,a\Delta f(a^{-1}bx)):a^{k-1}=1,\ b^{k-1}=a,\ a,b\in
R^{\times}\\}$ (38)

and let $\bullet:\mathfrak{M}_{k}(\Phi)\times
R^{\times}\rightarrow\mathfrak{M}_{k}(\Phi)$ be a right group action given by

$\Phi(t,x,\Delta f(x))\bullet(a^{-1}b)=\Phi(t,x,a^{-1}b\Delta
f(x))\circ(a^{-1}bx).$ (39)

Then $\mathfrak{M}_{k}(\Phi)$ is a $R^{\times}$-module.

###### Proof.

From previous theorem, $\Phi(t,x,a^{-1}b\Delta
f(x))\circ(a^{-1}bx)=a^{-1}b\Phi(t,x,a\Delta f(a^{-1}bx))$. Then

  1. 1.

$\Phi(t,x,\Delta f(x))\bullet(a^{-1}b)\in\mathfrak{M}_{k}(\Phi)$,

  2. 2.

$\Phi(t,x,\Delta f(x))\bullet 1_{R^{\times}}=\Phi(t,x,\Delta f(x))$,

  3. 3.

$\Phi(t,x,\Delta f(x))\bullet(a^{-1}bc^{-1}d)=(\Phi(t,x,\Delta
f(x))\bullet(a^{-1}b))\bullet(c^{-1}d)$,

  4. 4.

$\Phi(t+s,x,\Delta f(x))\bullet(a^{-1}b)=(\Phi(t,x,\Delta
f(x))\bullet(a^{-1}b))\circ(\Phi(s,x,\Delta f(x))\bullet(a^{-1}b))$.

∎

######  Definition 8.

The $R^{\times}$-module $\mathfrak{M}_{k}(\Phi)$ will be called
$R^{\times}$-module of homogeneity of order $k$, $k\geq 2$.

Using the exponential groups $\mathcal{H}_{k}$ found in the section 4.2 we
have the following theorem.

######  Theorem 17.

From the rings $\mathbb{Z}$, $\mathbb{C}$, $\mathbb{Z}[\omega]$ and
$\mathbb{Z}[\epsilon]$ we have the following $G$-modules

  1. 1.

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm\Phi(t,x,\Delta f(\pm x))\\}.$

  2. 2.

    1. (a)

$\mathbb{C}^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\\{c\Phi(t,x,\Delta
f(cx)):c\in\mathbb{C}^{\times}\\}.$

    2. (b)

$\mathbb{Z}_{(k-1)^{2}}$-module

$\mathfrak{M}_{k}(\Phi)=\\{\zeta\Phi(t,x,\Delta f(\zeta
x)):\zeta^{(k-1)^{2}}=1\\}.$

  3. 3.

    1. (a)

$\mathbb{Z}_{6}$-module

$\begin{split}\mathfrak{M}_{1}(\Phi)=\left\\{\pm\Phi(t,x,\Delta f(\pm
x)),\pm\omega\Phi(t,x,\Delta f(\pm\omega x)),\pm\omega^{2}\Phi(t,x,\Delta
f(\pm\omega^{2}x))\right\\}.\end{split}$

    2. (b)

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm\Phi(t,x,\Delta f(\pm x))\\}.$

    3. (c)

$\mathbb{Z}_{3}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\Phi(t,x,\Delta f(x)),\omega\Phi(t,x,\Delta
f(\omega x)),\omega^{2}\Phi(t,x,\Delta f(\omega^{2}x))\right\\}.$

  4. 4.

$\mathbb{Z}[\epsilon]^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm\epsilon^{m}\Phi(t,x,\Delta
f(\pm\epsilon^{m}x)):m\in\mathbb{Z}\\}.$

######  Theorem 18.

From the gaussian ring $\mathbb{Z}[i]$ all the $G$-modules
$\mathfrak{M}_{k}(\Phi)$ are $\mathbb{Z}_{4}$-modules.

###### Proof.

From section 4.2.4, the $G$-modules $\mathfrak{M}_{1}(\Phi)$ and
$\mathfrak{M}_{3}(\Phi)$ are $\mathbb{Z}_{4}$-modules and
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-module, respectively. From exponential
group $\mathcal{H}_{3}$ we have that $a^{-1}b\in\mathbb{Z}_{4}$. Then
$\mathfrak{M}_{1}(\Phi)=\mathfrak{M}_{3}(\Phi)$. ∎

Let $x^{*}$ be an equilibrium point of $\Phi$. If $G$ is some of the above
groups, then each $g^{-1}x^{*}$ is an equilibrium point of $\Phi(t,x,\Delta
f(gx))$ for all $g\in G$. Then, all equilibrium point $x^{*}$ of $\Phi$ is
invariant under the action of the $G$-module.

##  6 Examples of one-dimensional dynamical systems over integral domain

We finish this paper showing the flos and the $R^{\times}$-modules for the
equations $\delta_{t}\Phi=a$, $\delta_{t}\Phi=a+b\Phi$ and
$\delta_{t}\Phi=a\exp(\Phi)$ defined over the rings $\mathbb{Z}$,
$\mathbb{C}$, $\mathbb{Z}[i]$, $\mathbb{Z}[\omega]$ and
$\mathbb{Z}[\sqrt{2}]$. Also some trajectories for the $G$-modules are showed.

###  6.1 Equation $\delta_{t}\Phi=a$

We have $f(x)=a$. Then

$\Delta f(x)=(a,0,0,0,...)$

and

$\mathfrak{A}(\Delta f(x))=\Delta f(x)=(a,0,0,...).$

Thus the solution is

$\Phi(t,x,\Delta f(x))=x+at.$

####  6.1.1 $R=\mathbb{Z}$

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm(x+at)\\}.$

Figure 1: $\mathbb{Z}_{2}$-module para $\delta_{t}\Phi=2$

####  6.1.2 $R=\mathbb{C}$

  1. 1.

$\mathbb{C}^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\\{c(x+at):c\in\mathbb{C}^{\times}\\},$

  2. 2.

$\mathbb{Z}_{(k-1)^{2}}$-module

$\mathfrak{M}_{k}(\Phi)=\\{\zeta(x+at):\zeta^{(k-1)^{2}}=1\\}.$

####  6.1.3 $R=\mathbb{Z}[i]$

$\mathbb{Z}_{4}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm(x+at),\pm i(x-at)\\},$

Figure 2: $\mathbb{Z}_{4}$-module for $\delta_{t}\Phi=2$ and orbit
$\Gamma_{2}$

####  6.1.4 $R=\mathbb{Z}[\omega]$

  1. 1.

$\mathbb{Z}_{6}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm(x+at),\pm\omega(x+at),\pm\omega^{2}(x+at)\\}.$

  2. 2.

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{3}(\Phi)=\\{\pm(x+at)\\}.$

  3. 3.

$\mathbb{Z}_{3}$-module

$\mathfrak{M}_{4}(\Phi)=\\{x+at,\omega(x+at),\omega^{2}(x+at)\\}.$

Figure 3: $\mathbb{Z}_{3}$-module for $\delta_{t}\Phi=2$ and orbit
$\Gamma_{1}$

####  6.1.5 $R=\mathbb{Z}[\sqrt{2}]$

$\mathbb{Z}[\sqrt{2}]^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm(1+\sqrt{2})^{m}(x+at):m\in\mathbb{Z}\\}.$

Figure 4: $\mathbb{Z}[\sqrt{2}]^{\times}$-module for $\delta_{t}\Phi=2$ and
orbit $\Gamma_{2}$

###  6.2 Equation $\delta_{t}\Phi=a+b\Phi$

We have $f(x)=a+bx$. Then

$\Delta f(x)=(a+bx,b,0,0,...)$ (40)

and

$\mathfrak{A}(\Delta f(x))=([a+bx](b)^{n})_{n\in\mathbb{N}_{0}}.$ (41)

Thus the solution is

$\Phi(t,x,\Delta f(x))=x+\left(\frac{a}{b}+x\right)(\exp(bt)-1).$ (42)

When $a\equiv 0$, then (42) is reduced to

$\Phi(t,x,\Delta f(x))=x\exp(bt)$ (43)

which is a solution to the knowed linear equation $\delta_{t}\Phi(t,x)=b\Phi$.

####  6.2.1 $R=\mathbb{Z}$

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\pm x+\left(\frac{a}{b}\pm
x\right)\left(\exp(\pm bt)-1\right)\right\\}.$

Figure 5: $\mathbb{Z}_{2}$-module para $\delta_{t}\Phi=1+\Phi$

####  6.2.2 $R=\mathbb{C}$

  1. 1.

$\mathbb{C}^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{cx+\left(\frac{a}{b}+cx\right)(\exp(bct)-1):c\in\mathbb{C}^{\times}\right\\}.$

  2. 2.

$\mathbb{Z}_{(k-1)^{2}}$-module

$\mathfrak{M}_{k}(\Phi)=\left\\{\zeta x+\left(\frac{a}{b}+\zeta
x\right)(\exp(b\zeta t)-1):\zeta^{(k-1)^{2}}=1\right\\}.$

####  6.2.3 $R=\mathbb{Z}[i]$

$\mathbb{Z}_{4}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\pm x+\left(\frac{a}{b}\pm x\right)(\exp(\pm
bt)-1),\pm ix+\left(\frac{a}{b}\pm ix\right)(\exp(\pm ibt)-1)\right\\}.$

Figure 6: $\mathbb{Z}_{4}$-module for $\delta_{t}\Phi=1+\Phi$ and orbit
$\Gamma_{2}$

####  6.2.4 $R=\mathbb{Z}[\omega]$

  1. 1.

$\mathbb{Z}_{6}$-module

$\begin{split}\mathfrak{M}_{1}(\Phi)=\left\\{\pm x+\left(\frac{a}{b}\pm
x\right)(\exp(\pm bt)-1),\pm\omega x+\left(\frac{a}{b}\pm\omega
x\right)(\exp(\pm\omega bt)-1),\right.\\\
\left.\pm\omega^{2}x+\left(\frac{a}{b}\pm\omega^{2}x\right)(\exp(\pm\omega^{2}bt)-1)\right\\}.\end{split}$

  2. 2.

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{3}(\Phi)=\left\\{\pm x+\left(\frac{a}{b}\pm x\right)(\exp(\pm
bt)-1)\right\\}.$

  3. 3.

$\mathbb{Z}_{3}$-module

$\begin{split}\mathfrak{M}_{4}(\Phi)=\left\\{x+\left(\frac{a}{b}x\right)(\exp(bt)-1),\omega
x+\left(\frac{a}{b}\omega x\right)(\exp(\omega bt)-1),\right.\\\
\left.\omega^{2}x+\left(\frac{a}{b}\omega^{2}x\right)(\exp(\omega^{2}bt)-1)\right\\}.\end{split}$

Figure 7: $\mathbb{Z}_{3}$-module for $\delta_{t}\Phi=1+\Phi$ and orbit
$\Gamma_{2}$

####  6.2.5 $R=\mathbb{Z}[\sqrt{2}]$

$\mathbb{Z}[\sqrt{2}]$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\pm(1+\sqrt{2})^{m}x+\left(\frac{a}{b}\pm(1+\sqrt{2})^{m}x\right)\left(\exp(\pm(1+\sqrt{2})^{m}bt)-1\right):m\in\mathbb{Z}\right\\}.$

Figure 8: $\mathbb{Z}[\sqrt{2}]^{\times}$-module for $\delta_{t}\Phi=1+\Phi$
and orbit $\Gamma_{2}$

###  6.3 Equation $\delta_{t}\Phi=a\exp(\Phi)$

In this case $f(x)=a\exp(x)$ and $\Delta f=(a\exp(x))_{n\in\mathbb{N}_{0}}$.
Then

$\mathfrak{A}(\Delta
f)=(a^{n}\exp(nx))_{n\in\mathbb{N}}((n-1)!)_{n\in\mathbb{N}}.$ (44)

Thus the solution is

$\Phi(t,x,\Delta F)=x-\ln\left(1-a\exp(x)t\right)$ (45)

####  6.3.1 $R=\mathbb{Z}$

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\pm x-\ln\left(1\mp a\exp(\pm
x)t\right)\right\\}.$

Figure 9: $\mathbb{Z}_{2}$-module for $\delta_{t}\Phi=exp(\Phi)$ and orbit
$\Gamma_{2}$

####  6.3.2 $R=\mathbb{C}$

  1. 1.

$\mathbb{C}^{\times}$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{cx-\ln\left(1-ca\exp(cx)t\right):c\in\mathbb{C}^{\times}\right\\}.$

  2. 2.

$\mathbb{Z}_{(k-1)^{2}}$-module

$\mathfrak{M}_{k}(\Phi)=\left\\{\zeta x-\ln\left(1-a\zeta\exp(\zeta
x)t\right):\zeta^{(k-1)^{2}}=1\right\\}.$

####  6.3.3 $R=\mathbb{Z}[i]$

$\mathbb{Z}_{4}$-module

$\mathfrak{M}_{1}(\Phi)=\\{\pm x-\ln\left(1\mp a\exp(\pm x)t\right),\pm
ix-\ln\left(1\mp ia\exp(\pm ix)t\right)\\}.$

Figure 10: $\mathbb{Z}_{4}$-module for $\delta_{t}\Phi=exp(\Phi)$ and orbit
$\Gamma_{2}$

####  6.3.4 $R=\mathbb{Z}[\omega]$

  1. 1.

$\mathbb{Z}_{6}$-module

$\begin{split}\mathfrak{M}_{1}(\Phi)=\left\\{\pm x-\ln\left(1\mp a\exp(\pm
x)t\right),\pm\omega x-\ln\left(1\mp\omega a\exp(\pm\omega
x)t\right),\right.\\\
\left.\pm\omega^{2}x-\ln\left(1\mp\omega^{2}a\exp(\pm\omega^{2}x)t\right)\right\\}.\end{split}$

  2. 2.

$\mathbb{Z}_{2}$-module

$\mathfrak{M}_{3}(\Phi)=\left\\{\pm x-\ln\left(1\mp a\exp(\pm
x)t\right)\right\\}.$

  3. 3.

$\mathbb{Z}_{3}$-module

$\begin{split}\mathfrak{M}_{4}(\Phi)=\left\\{x-\ln\left(1-a\exp(x)t\right),\omega
x-\ln\left(1-\omega a\exp(\omega x)t\right),\right.\\\
\left.\omega^{2}x-\ln\left(1-\omega^{2}a\exp(\omega^{2}x)t\right)\right\\}.\end{split}$

Figure 11: $\mathbb{Z}_{3}$-module for $\delta_{t}\Phi=exp(\Phi)$ and orbit
$\Gamma_{1}$

####  6.3.5 $R=\mathbb{Z}[\sqrt{2}]$

$\mathbb{Z}[\sqrt{2}]$-module

$\mathfrak{M}_{1}(\Phi)=\left\\{\pm(1+\sqrt{2})^{m}x-\ln\left(1\mp(1+\sqrt{2})^{m}a\exp(\pm(1+\sqrt{2})^{m}x)t\right):m\in\mathbb{Z}\right\\}.$

Figure 12: $\mathbb{Z}[\sqrt{2}]^{\times}$-module for
$\delta_{t}\Phi=exp(\Phi)$ and orbit $\Gamma_{2}$

##  7 Conclusions and Further research

In this paper the operator $\mathfrak{A}$ and its properties were studied. A
gol was can to link sequences in an integral domain with the flow of a
dynamical system. Another result of great importance was to show that
$\mathfrak{A}$ has an inverse. Thus to all sequence with first element in the
group of units of the domain is obtained via the action of $\mathfrak{A}$.
Other very important result was to show that the flow of an onedimensional
dynamical system over an integral domain is a torsion-free cyclic module. This
will be very useful as a part of the construction of a cualitative theory for
this system. Also we construct an structure of $G$-module for the flows and
were classified for some rings.

Some potencial issues for further research are:

  1. 1.

To find a closed formula for $\mathfrak{A}(\Delta f(X)\ast\Delta g(X))$.

  2. 2.

Is $\mathfrak{A}(\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X])$ a ring? or a
module or an algebra?

  3. 3.

To find a closed formula for the inverse of the operator $\mathfrak{A}$ and to
study its properties.

  4. 4.

To classify equilibrium point for dynamical system over integral domains. Also
to study other concepts of the cualitative theory of dynamical systems.

## References

  * [1] Keigher W.F. On the ring of hurwitz series, Communications in Algebra, 25:6, 1845-1859 (1997).
  * [2] Bernardini, A., Ricci P.E. Bell Polinomials and Differential Equations of Freud-Type Polynomials, Mathematical and Computer Modelling 36(2002) 1115–1119.
  * [3] Caley, A.On the theory of analytical forms called trees, Philos. Mag, 13(19)(1857):4–9.
  * [4] Foissy, L. Faá di Bruno subalgebras of the Hopf of planar trees from combinatorial Dyson-Schwinger equations, Advances in Mathematics 218(1)(2008):136–162.
  * [5] McLachlan, R.I., Modin, K., Munthe-Kaas, H., Verdier, O. What are Butcher series, really?,arXiv:1512.00906v2. (2016)
  * [6] Orozco, R. Solución Analítica de la Ecuación Diferencial Autónoma de orden k, Revista De Matemática Teoría y Aplicaciones vol(23) fasc.1: 63–84.

E-mail address, R. Orozco<EMAIL_ADDRESS>


\P\left(  |H_{n}(\param)| \ge \eta \right) &\le \sum_{j=1}^{N(\eta/2L)} \P\left(  \left| H_{n}(\param_{j}) \right| \ge \eta/2 \right) 	\le K\left( \frac{2L \, \textup{diam}(\Param)}{\eta} \right)^{p}  \exp\left( - \frac{n \eta^{2}}{2M^{2}} \right).
\end{align}
The above holds for all $n\ge 1$ and $\eta>0$. This shows the assertion.
\end{proof}

Now we prove Theorem \ref{thm:MLE_local_consistency}.

\begin{proof}[\textbf{Proof of Theorem \ref{thm:MLE_local_consistency}}]
Let $X_{1},\dots,X_{n}$ denote i.i.d. samples from $\pi_{\param^{\star}}$ and write $\X=[X_{1},\dots,X_{n}]\in \R^{d\times n}$. Recall that
\begin{align}
\mathcal{L}(\X;\param) := \mathcal{L}_{0}(\X;\param) + n R_{n}(\param), \qquad \mathcal{L}_{0}(\X;\param) := -\sum_{i=1}^{n} \log \pi_{\param}(X_{i}),
\end{align}
where $R(\cdot)$ is the regularizer used in \eqref{eq:def_MLE_problem}. By the hypothesis, $\mathcal{L}_{0}$ is twice continuously differentiable, so $\E[\nabla \mathcal{L}_{0}]=\nabla \E[\mathcal{L}_{0}]$ and $\E[\nabla^{2} \mathcal{L}_{0}]=\nabla^{2} \E[\mathcal{L}_{0}]$ by the dominated convergence theorem.

Fix a constant $C>0$ and let $\alpha_{n}:=n^{-1/2}+\lVert \nabla R_{n}(\param^{\star}) \rVert$. We wish to show the probability bound in \eqref{eq:local_consistency_thm}. We first introduce two random variables that we will bound to be small by using some concentration inequalities:
\begin{align}\label{eq:def_T_S_pf_consistency}
T_{n}(\param) &:=\E\left[ \left\langle \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}),\, \frac{\param-\param^{\star}}{\lVert\param - \param^{\star} \rVert} \right\rangle \right]- \frac{1}{\sqrt{n}}\sum_{i=1}^{n}  \left\langle \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}),\, \frac{\param-\param^{\star}}{\lVert\param - \param^{\star} \rVert} \right\rangle \\
&=  \left\langle  \frac{1}{\sqrt{n}}\sum_{i=1}^{n} \left(  \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}) - \E\left[  \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}) \right] \right) ,\, \frac{\param-\param^{\star}}{\lVert\param - \param^{\star} \rVert} \right\rangle,  \\
S_{n}(\param) &:=
\frac{(\param-\param_{0})^{T}}{\lVert \param-\param^{\star} \rVert} \left( \frac{1}{n}\nabla_{\param}\nabla_{\param^{T}}
\mathcal{L}(\X \,;\,  \param^{\star})  - \nabla_{\param}\nabla_{\param^{T}}
\left( \E\left[ \mathcal{L}(X_{1} \,;\,  \param^{\star})\right]  \right) \right) \frac{\param-\param_{0}}{\lVert \param-\param_{0}\rVert}.
\end{align}
Fix $\param\in \Param$ such that $\lVert \param-\param^{\star} \rVert=C\alpha_{n}$. Since $\param\mapsto \mathcal{L}(\X\,;\, \param)$  is assumed to be three-times continuously differentiable, the quantity $M\ge 0$ in the assertion is well-defined and is finite. Then using a Taylor expansion, we  may write
\begin{align}\label{eq:thm_local_consistency_taylor}
\mathcal{L}(\X \, ;\,  \param)  -  \mathcal{L}(\X \, ;\,  \param^{\star})   \ge \langle \nabla_{\param} \mathcal{L}(\X;\param^{\star}),\, \param-\param^{\star} \rangle   + \frac{1}{2} (\param-\param_{0})^{T} \nabla_{\param}\nabla_{\param^{T}} \mathcal{L}(\X \,;\,  \param^{\star})   (\param-\param_{0}) -  \frac{M}{6} (C\alpha_{n})^{3}.
\end{align}
We will lower bound the first two terms in the right hand side above. Note that
\begin{align}
\langle \nabla_{\param} \mathcal{L}(\X;\param^{\star}),\, \param-\param^{\star} \rangle &= \left[  \langle \nabla_{\param} \mathcal{L}(\X;\param^{\star}),\, \param-\param^{\star} \rangle- \E\left[ \langle \nabla_{\param} \mathcal{L}(\X;\param^{\star}),\, \param-\param^{\star} \rangle \right] \right] \\
&\hspace{3cm} +n \langle \nabla_{\param} \E[\mathcal{L}_{0}(X_{1};\param^{\star})] ,\, \param-\param^{\star} \rangle  + n\langle \nabla R_{n}(\param^{\star}),\, \param-\param^{\star} \rangle \\
&\overset{(a)}{\ge}  \langle \nabla_{\param} \mathcal{L}_{0}(\X;\param^{\star}),\, \param-\param^{\star} \rangle- \E\left[ \langle \nabla_{\param} \mathcal{L}_{0}(\X;\param^{\star}),\, \param-\param^{\star} \rangle \right] - n \lVert \nabla R_{n}(\param^{\star}) \rVert \, \lVert \param-\param^{\star} \rVert\\
&= -\sqrt{n} \lVert \param-\param^{\star} \rVert T_{n}(\param) - n \lVert \nabla R_{n}(\param^{\star}) \rVert \, \lVert \param-\param^{\star} \rVert \\
&\overset{(b)}{\ge} -C \sqrt{n} \alpha_{n} T_{n}(\param) - Cn\alpha_{n}^{2},
\end{align}
where for (a) we use the fact that $\param^{\star}$ is a stationary point of $\E[\mathcal{L}_{0}(X_{1};\param)]$ over $\Param$ and Cauchy-Schwarz inequality;
for (b) we used that $\max(n^{-1/2},\lVert \nabla R_{n}(\param^{\star})  \rVert) \le \alpha_{n}$.

Next, we turn our attention to the second order term in the Taylor expansion \eqref{eq:thm_local_consistency_taylor}.
Under assuming $\lVert \param-\param^{\star} \rVert=C\alpha_{n}$, note that
\begin{align}
(\param-\param^{\star})^{T} \nabla_{\param}\nabla_{\param^{T}} \mathcal{L}(\X \,;\,  \param^{\star})   (\param-\param^{\star}) &= C^{2}n \alpha^{2} \frac{(\param-\param^{\star})^{T}}{\lVert \param-\param^{\star} \rVert} \left( \frac{1}{n}\nabla_{\param}\nabla_{\param^{T}}
\mathcal{L}(\X \,;\,  \param^{\star})   \right) \frac{\param-\param^{\star}}{\lVert \param-\param^{\star}\rVert} \\
&\ge C^{2}n\alpha_{n}^{2}\left( S_{n}(\param) + \lambda_{*}\right),
\end{align}
where the inequality follows from the hypothesis, which implies
\begin{align}
\E\left[ \frac{1}{n}\nabla_{\param}\nabla_{\param^{T}}
\mathcal{L}(\X \,;\,  \param^{\star}) \right] = \nabla_{\param}\nabla_{\param^{T}}
\left( \E\left[ \mathcal{L}(X_{1} \,;\,  \param^{\star})\right]  \right) \succeq \lambda_{*} \I_{p},
\end{align}
where $\lambda_{*}>0$ is a constant. Combining the above inequalities, we obtain
\begin{align}\label{eq:local_consistency_pf_1}
&\hspace{-0.5cm} \inf_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  	\frac{1}{C^{2}n\alpha_{n}^{2}}\left( \mathcal{L}(\X \, ;\,  \param)  -  \mathcal{L}(\X \, ;\,  \param^{\star})  \right) \\
&\qquad \ge \underbrace{\frac{1}{C}\left(  \inf_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }   \frac{-T_{n}(\param)}{\sqrt{n} \alpha_{n}}  - 1 \right)}_{=:A} + \underbrace{\left(  \inf_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  S_{n}(\param) + \lambda_{*} \right)}_{=:B} - \frac{M C\alpha_{n}}{6n}.
\end{align}

According to the hypothesis, we have
\begin{align}
(MC/6)(n^{-3/2} + n^{-1} \rVert \nabla R_{n}(\param^{\star}) \rVert)  =\frac{M C\alpha_{n}}{6n} \le \lambda_{*}/4.
\end{align}
Then the last expression in \eqref{eq:local_consistency_pf_1} is at least $\lambda_{*}/4$ if $A\ge -\lambda_{*}/4$ and $B\ge 3\lambda_{*}/4$. Moreover, Note that for any two events $E_{1},E_{2}$ defined on the same probability space, $\P(E_{1}\cap E_{2})\ge \P(E_{1})+\P(E_{2})-1$. Notice that $A$ and $B$ above are random variables defined on the same probability space, which are deterministic functions of the common random vectors $X_{1},\dots,X_{n}$. Thus
\begin{align}\label{eq:local_consistency_pf_11}
\P\left( \inf_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  	\frac{1}{C^{2}n\alpha_{n}^{2}}\left( \mathcal{L}(\X \, ;\,  \param)  -  \mathcal{L}(\X \, ;\,  \param^{\star}) \right)  \ge \lambda_{*}/4  \right)  \ge \P(A\ge -\lambda_{*}/4) + \P( B\ge 3\lambda_{*}/4) -1.
\end{align}
Note that by the uniform McDirmid's inequality in Lemma \ref{lem:uniform_McDirmid}, there exists constants $C',C''>0$ such that
\begin{align}\label{eq:local_consistency_pf_111}
\P( B <  3\lambda_{*}/4) &\le \P\left(   \inf_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  S_{n}(\param) < -\lambda_{*}/4  \right) \le  C'\exp(-C'' n).
\end{align}
On the other hand, we will show the following inequalities:
\begin{align}
\P(A< -\lambda_{*}/4) \overset{(c)}{\le}  \P\left(  \sup_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  T_{n}(\param)    > \frac{C\lambda_{*}}{4}-1\right) &\overset{(d)}{\le} M \left( \P\left(Z \ge  p^{-1/2}\left(\frac{C\lambda_{*}}{4}-1\right) \right) +   \frac{K}{\sqrt{n}}\right) \\
&\overset{(e)}{\le} M \left( \exp\left(-\left( \frac{C\lambda_{*}}{4}-1 \right)^{2}\bigg/2\sqrt{p} \right)  +   \frac{K}{\sqrt{n}}\right), \label{eq:local_consistency_pf_2}
\end{align}
where $Z\sim N(0,1)$ is an independent standard normal random variable and  $M, K$ are constants that does not depend on $n$ and $\eta$. Then the assertion will follow by combining  \eqref{eq:local_consistency_pf_1},  \eqref{eq:local_consistency_pf_11},  \eqref{eq:local_consistency_pf_111}, and  \eqref{eq:local_consistency_pf_2}. Note that (c) in \eqref{eq:local_consistency_pf_2} follows from the definition of $A$ in \eqref{eq:local_consistency_pf_1} and the fact that $\sqrt{n}\alpha_{n} = 1+n^{-1/2} \lVert \nabla R_{n}(\param^{\star}) \rVert \ge 1$. Also note that (e) above is a simple consequence of Hoeffding's bound for the standard normal tail.

It remains to verify (d) in \eqref{eq:local_consistency_pf_2}. To this end, we first wright $R_{n}:= \frac{1}{\sqrt{n}}\sum_{i=1}^{n} \left(  \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}) - \E\left[  \nabla_{\param} \mathcal{L}_{0}(X_{i};\param^{\star}) \right] \right)  = [R_{n}^{(1)},\dots, R_{n}^{(p)}]\in \R^{p}$. By Cauchy-Schwarz inequality,
\begin{align}
| T_{n}(\param)  |^{2} \le \lVert R_{n} \rVert^{2}=  \sum_{i=1}^{p}  \left( R_{n}^{(i)} \right)^{2},
\end{align}
where the upper bound  $R_{n}$ does not depend on $\param$. Hence by union bound,
\begin{align}\label{eq:pf_sup_T_union_bd}
\P\left(   \sup_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  T_{n}(\param)   \ge t  \right) & \le \P\left(  \lVert R_{n} \rVert \ge t  \right) \le \sum_{i=1}^{p} \P\left( |R_{n}^{(i)}|  \ge  \frac{t}{\sqrt{p}} \right).
\end{align}
Note that if $R^{(i)}_{n}=0$ a.s., then the coordsponding tail probability in the last expression above is zero whenever $t\ne 0$. Since $\E[ R^{(i)}_{n} ]=0$, this is the case if $\var( R^{(i)}_{n}  )=0$. So we may assume without loss of generality that $\var( R^{(i)}_{n}  )>0$ for $1\le i \le p$. Then, by the definition of $R_{n}^{(i)}$ and the classical Berry-Esseen theorem (see, e.g., \cite[Thm. 3.4.17]{Durrett}), for all $z\in \R$,
\begin{align}
\sup_{z\in \R} \left| \P\left( R_{n}^{(i)} \le  z\right)  - \P\left( Z\le z  \right) \right| \le \frac{3 \, \E[ | R_{n}^{(i)} |^{3} ]  }{ \var (R_{n}^{(i)} )^{3/2} \sqrt{ n}}.
\end{align}
Note that $\E[ | R_{n}^{(i)} |^{3} ]  <\infty$ by the hypothesis and $\var (R_{n}^{(i)} ) >0$  by the assumption we just made above. Combining with \eqref{eq:pf_sup_T_union_bd}, we obtain
\begin{align}\label{eq:pf_sup_T_union_bd}
\P\left(   \sup_{\substack{\param\in \Param \\ \lVert \param-\param^{\star} \rVert =C\alpha_{n}} }  T_{n}(\param)   \ge t  \right) & \le  2p\left(   \P( Z \ge t/\sqrt{p} ) +  \frac{K}{\sqrt{n}} \right),
\end{align}
where $K\in (0,\infty)$ is the maximum of $3 \, \E[ | R_{n}^{(i)} |^{3} ]  \bigg/  \E[ |R_{n}^{(i)}|^{2} ]^{3/2} $ for $i=1,\dots,p$. Thus (d) in  \eqref{eq:local_consistency_pf_2} follows.
\end{proof}

\section{Generative model for strongly constrained feature-based SDL}
\label{section:statistical_estimation_BCD_feat}

In Section \ref{subsection:statistical_estimation_BCD}, we have considered a generative model for the filter-based SDL with strong constraints and derived a local consistency result. In this section, we discuss a parallel model for a generative feature-based SDL with strong constraints. Unlike for the filter-based case, we would need to formulate a latent-variable model where the `code matrix' $\H=[h_{1},\dots,\h_{n}]$ is the latent variable, and our general theory of non-asymptotic local consistency of constrained and regularized MLE applies only approximately.

Suppose that the data, auxiliariy covariate, and label triples $(\x_{i}, \x_{i}', y_{i})$ are drawn i.i.d. according to the following generative model:
\begin{align}\label{eq:SDL_prob_BCD_feat}
&\h_{i}\sim N(\h^{\star}, \sigma_{h}^{2}\I_{r}),\quad \x_{i} \sim  N\left( \W^{\star}\h_{i}, \sigma^{2}\I_{p} \right), \quad \x_{i}'\sim N(\blambda^{\star}, (\sigma')^{2}\I_{q}),\\
& y_{i}\,|\, \x_{i}, \x_{i}' \sim \text{Multinomial}\big(1, \g\left( (\Beta^{\star})^{T} \h_{i}+ (\bGamma^{\star})^{T}\x_{i}'\right) \big), \\
&\textup{where}\quad  \W^{\star}\in \R^{p\times r},\,  \h^{\star}\in \R^{r\times 1},\,  \Beta^{\star}\in  \R^{r\times \kappa},\,  \bGamma^{\star}\in \R^{q\times \kappa},\,\blambda^{\star}\in \R^{q\times 1} \,\, \text{s.t. $[\W^{\star},\h^{\star},\Beta^{\star},\bGamma^{\star}]\in \Param$}.
\end{align}
As before, $\Param= \mathcal{C}^{\textup{dict}}\times \mathcal{C}^{\textup{code}}\times  \mathcal{C}^{\textup{beta}} \times \mathcal{C}^{\textup{aux}}$ is the product of convex constraint sets on individual factors. We assume $(\x_{i},\x_{i}',y_{i})$ for $i=1,\dots,n$ are i.i.d. observed data and $\h_{i}$ for $i=1,\dots,n$ are i.i.d. latent variables that we do not observe. However, we assume that the mean latent variable $\h^{\star}$ is known\footnote{In [52], a similar model was considered with $\h^{\star}=\mathbf{0}$.}. We also assume $\x_{i}$ and $\x_{i}'$ are independent for each $1\le i \le n$. We call the above the \textit{generative feature-based SDL model}. Assuming that $\sigma_{n},\sigma,$ and $\sigma'$ are known, our goal is to estimate the true factors $\W^{\star}$, $\Beta^{\star}, \bGamma^{\star}$, and $\blambda^{\star}$ from an observed sample $(\x_{i},\x'_{i},y_{i})$, $i=1,\dots,n$ of size $n$.

We consider the maximum likelihood estimation framework with $L_{2}$-regularization of the parameters. Namely, define
\begin{align}\label{eq:SDL_likelihood_BCD_feature_cond}
\mathcal{\ell}_{n}(\W,\H,\Beta, \bGamma)&:=-\sum_{j=0}^{\kappa}\mathbf{1}(y_{i}=j)g_{j}(\Beta^{T}\h_{i} +  \bGamma^{T} \x_{i}' ) )   +  \frac{1}{2\sigma^{2}} \lVert \X_{\textup{data}} - \W \H \rVert_{F}^{2},
\end{align}
which is the negative log likelihood of observing the samples conditional on the hidden variable $\H=[h_{1},\dots,\h_{n}]$ and also on the auxiliary covarite $\x_{1}',\dots,\x_{n}'$. Integrating out the hidden variable $\H$, we obtain the negative log likelihood of observing the data conditional on the auxiliary covariate:
\begin{align}\label{eq:feat_SDL_approx_MAP0}
\mathcal{L}_{n}(\W,\Beta, \bGamma) = -\log \left(  \int_{\R^{r\times n}} \exp\left( -\mathcal{\ell}_{n}(\W,\H,\Beta, \bGamma)\right) \exp\left( -\frac{1}{(2\sigma_{h})^{r}} \left( \sum_{i=1}^{n}\lVert \h_{i}-\h^{\star} \rVert^{2}\right)   \right) \,d \H \right).
\end{align}
Note that when the variance of the hidden variable $\sigma_{h}^{2}$ is assumed to be small, the leading contribution to the integral over $\H$ above comes when $\H$ is set to the maximizer of the integrand. Hence in this case, the above can be approximate as
\begin{align}\label{eq:feat_SDL_approx_MAP1}
\mathcal{L}_{n}(\W,\Beta, \bGamma) \propto \min_{\H} \,\,  \ell_{n}(\W,\H,\Beta,\bGamma) +\frac{1}{(2\sigma_{h})^{r}} \left( \sum_{i=1}^{n}\lVert \h_{i}-\h^{\star} \rVert^{2}\right) .
\end{align}
Consequently, we may estimate the unknown true parameters $\W^{\star}, \Beta^{\star},$ and $\bGamma^{\star}$ as the minimizer of the following loss function
\begin{align}\label{eq:SDL_likelihood_BCD_filter_full}
L(\W, \H, \Beta, \bGamma,\blambda) &  :=  \ell_{n}(\W, \H,\Beta, \bGamma) +\frac{1}{(2\sigma_{h})^{r}} \left( \sum_{i=1}^{n}\lVert \h_{i}-\h^{\star} \rVert^{2}\right) + \frac{pn\log \sigma}{2} +   \frac{qn\log \sigma'}{2} +  \frac{1}{(2(\sigma')^{2})^{q}} \sum_{i=1}^{n} \lVert \x_{i}'-\blambda \rVert^{2},
\end{align}
where we also simultaneously estimate the hidden variable $\H$. This is analogous to the `MAP approximation' approach and `generative training' employed in [52].

Note that the first two terms on the right-hand side above are equivalent to the feature-based SDL loss in \eqref{eq:ASDL_1} with an additional quadratic regularizer on $\H$. Hence we can compute its derivate and Hessian using a similar computation as for the filter-based model in Subsection \ref{subsection:statistical_estimation_BCD}. By adding a suitable $L_{2}$-regularizer for $\W,\Beta$, and $\bGamma$ (also possibly for $\H$ with different regularization coefficient), we can obtain a similar non-asymptotic local consistency result as we established in Theorem \ref{thm:SDL_BCD_STAT_filt} for the generative filter-based SDL. However, in order to obtain a precise error bound as in Theorem \ref{thm:SDL_BCD_STAT_filt} for the generative feature-based SDL we consider here, we need to account for the `MAP approximation' of the likelihood function in \eqref{eq:feat_SDL_approx_MAP0}, where we replaced integrating over $\H$ by maximizing over $\H$ assuming the variance of $\h_{i}$'s are small. It should be straightforward to estimate this approximation error and relate the Hessian of $\mathcal{L}_{n}(\W,\Beta,\bGamma)$ in \eqref{eq:feat_SDL_approx_MAP1} with the Hessian of $\ell_{n}(\W,\H,\Beta,\bGamma)$. Then we can apply Theorem \ref{thm:MLE_local_consistency} with an explicit lower bound on the added $L_{2}$-regularization coefficients to make the Fisher information positive definite. We omit the details of this sketch in this work.

\section{Auxiliary lemmas from optimization}

\begin{lemma}\label{lem:gradient_mapping}
Fix a differentiable function $f:\R^{p}\times \R$ and a convex set $\Param\subseteq \R^{p}$. Fix $\tau>0$ and \begin{align}
G(\bZ, \tau):= \frac{1}{\tau}(\bZ - \Pi_{\Param}(\param - \tau\nabla f(\param))).
\end{align}
Then for each $\param\in \Param$, $\lVert G(\param,\tau) \rVert\le \lVert \nabla f(\param) \rVert$. %Furthermore, if there is another differentiable function $\bar{f}:\R^{p}\rightarrow \R$ and if $\param^{\star}$ is a stationary point of $\bar{f}$ over $\Param$, then $\lVert G(\param,\tau) \rVert \le \lVert  \rVert$
\end{lemma}

\begin{proof}
The assertion is clear if $\lVert G(\param,\tau) \rVert=0$, so we may assume $\lVert G(\param,\tau) \rVert>0$.
Denote $\hat{\param}:=\Pi_{\Param}(\param- \tau\nabla f(\param)))$. Note that
\begin{align}
\hat{\param}= \argmin_{\param'} \, \lVert \param - \tau\nabla f(\param) - \param' \rVert^{2},
\end{align}
so by the first-order optimality condition,
\begin{align}
\langle \hat{\param} - \param + \tau\nabla f(\param) ,\, \param' -\hat{\param}  \rangle \ge 0 \quad \forall \param'\in \Param.
\end{align}
Plugging in $\param'=\param$ and using Cauchy-Schwarz inequality,
\begin{align}
\tau^{2} \lVert G(\param,\tau) \rVert^{2} =  \lVert \param - \hat{\param} \rVert^{2} \le \tau \langle \nabla f(\param),\, \param-\hat{\param} \rangle \le \tau \lVert \nabla f(\param) \rVert \, \tau \lVert G(\param, \tau)  \rVert.
\end{align}
Hence the assertion follows by dividing both sides by $\tau^{2}\lVert G(\param,\tau) \rVert>0$.
\end{proof}

\section{Generalized multinomial logistic Regression}\label{sec:MLR}
In this section, we provide some background on a generalized multinomial logistic regression and record some useful computations. (See [16] for backgrounds on multinomial logistic regression.)
Without loss of generality, we can assume that the $\kappa$ classes are the integers in $\{1,2,\dots, \kappa \}$. Say we have training examples $(\bphi(\x_{1}),y_{1}),\dots,(\bphi(\x_{N}),y_{N})$, where
\begin{enumerate}[label={\textbf{$\bullet$}}]
\item $\x_{1},\dots,\x_{N}$: Input data (e.g., collection of all medical records of each patient)
\item $\bphi_{i}:=\bphi(\x_{1}),\dots,\bphi_{N}:=\bphi(\x_{N})\in \R^{p}:$ Features (e.g., some useful (derived) information for each patient)
\item $y_{1},\dots,y_{n}\in \{0, 1,\dots, \kappa \}$: $\kappa$ class labels (e.g., digits from 0 to 9).
\end{enumerate}
The basic idea of multinomial logistic regression is to model the output $y$ as a discrete random variable  $Y$ with probability mass function  $\p=[p_{0}, p_{1},\dots,p_{\kappa}]$ that depends on the observed feature $\bphi(\x)$, link function $h:\R \rightarrow \R$, and a parameter $\W=[\w_{1},\dots,\w_{\kappa}]\in \R^{p\times \kappa}$ through the following relation:
\begin{align}\label{eq:MLR}
p_{0}= \frac{1}{1+\sum_{c=1}^{\kappa} h(\langle \bphi(\x), \w_{c} \rangle)},\qquad 	p_{j}=\frac{h(\langle \bphi(\x), \w_{i} \rangle)}{1+\sum_{c=1}^{\kappa} h(\langle \bphi(\x), \w_{c} \rangle)},\qquad \text{for $j=1,\dots, \kappa$}.
\end{align}
That is, given the feature vector $\bphi(\x)$, the probability $p_{i}$ of $\x$ having label $i$ is proportional to $h$ evaluated at the `linear activation' $\langle \bphi(\x), \w_{i} \rangle$. Note that using $h(x)=\exp(x)$, the above multiclass classification model reduces to the classical multinomial logistic regression. In this case, the corresponding predictive probability distribution $\p$ is called  the \textit{softmax distribution} with activation $\mathbf{a}=[a_{1},\dots,a_{\kappa}]$ with $a_{i}=\langle \bphi(\x), \w_{i} \rangle$ for $i=1,\dots,\kappa$. Notice that this model has parameter vectors $\w_{1},\dots,\w_{\kappa}\in \R^{p}$, one for each of the $\kappa$ nonzero class labels.

Next, we derive the maximum log likelihood formulation for finding optimal parameter $\W$ for the given training set $(\bphi_{i},y_{i})_{i=1,\dots,N}$. For each $1\le i \le N$ and $1\le j \le \kappa$, denote $p_{ij} := h(\langle \bphi_{i}, \w_{j} \rangle) \big/ \sum_{c=1}^{\kappa} h(\langle \bphi_{i}, \w_{c} \rangle)$, the predictive probability of the $y_{i}$ given $\bphi_{i}$ being $j$. We introduce the following matrix notations
\begin{align}
\begin{matrix}
\Y:=
\begin{bmatrix}
\mathbf{1}(y_{1}=1) & \cdots & \mathbf{1}(y_{1}=\kappa) \\ \vdots & & \vdots  \\ \mathbf{1}(y_{N}=1)  & \cdots & \mathbf{1}(y_{N}=\kappa)
\end{bmatrix}
\\ \\
\quad \in \{0,1\}^{N\times \kappa}
\end{matrix}
\begin{matrix}
\bP:=
\begin{bmatrix}
p_{11} & \cdots & p_{1\kappa} \\ \vdots & & \vdots  \\ p_{N 1}  & \cdots & p_{N \kappa}
\end{bmatrix}
\\ \\
\quad \in \{0,1\}^{N\times \kappa}
\end{matrix}
\begin{matrix}
\bPhi:=
\begin{bmatrix}
\uparrow & & \uparrow \\
\bphi(\x_{1})  & \cdots &  \bphi(\x_{N}) \\
\downarrow &  &\downarrow
\end{bmatrix}
\\ \\
\quad \in \R^{p\times N}
\end{matrix}
\begin{matrix}
\W :=
\begin{bmatrix}
\uparrow & & \uparrow \\
\w_{1}  & \cdots &  \w_{\kappa}\\
\downarrow &  &\downarrow
\end{bmatrix}
\\ \\
\quad \in \R^{p\times \kappa}
\end{matrix}
\end{align}
Note that the $s$th row of $\Y$ is a one-hot encoding of the lable $y_{s}$ and the corresponding row of $\Q$ is its predictive probability distribution. Then the joint likelihood function of observing labels $(y_{1},\dots,y_{N})$ given input data $(\x_{1},\dots,\x_{N})$ under the above probabilistic model is
\begin{align}
L(y_{1},\dots,y_{N}\,;\, \W) = \P(Y_{1}=y_{1},\dots,Y_{N}=y_{N}\, ;\, \W)=  \prod_{s=1}^{N} \prod_{j=1}^{\kappa}  (p_{sj})^{\mathbf{1}(y_{s}=j)}.
\end{align}
We can derive the negative log likelihood function $\ell(\bPhi, \W)
:= - \sum_{s=1}^{N} \sum_{j=1}^{\kappa} \mathbf{1}(y_{s}=j) \log p_{sj}$ in a matrix form as follows:
\begin{align}
\ell(\bPhi, \W)
& = \sum_{s=1}^{N}  \log \left( \sum_{c=1}^{\kappa} h(\langle \bphi(\x_{s}), \w_{c} \rangle ) \right) - \sum_{s=1}^{N}  \sum_{j=1}^{\kappa} \mathbf{1}(y_{s}=j) \log h\left(  \langle \bphi(\x_{s}), \w_{j}\rangle \right)  \\
&= \left( \sum_{s=1}^{N} \log \left( \sum_{q=1}^{\kappa} h(\langle \bphi(\x_{s}), \w_{q} \rangle ) \right) \right)- \tr\left( \Y^{T} h(\bPhi^{T} \W) \right).
\end{align}
Then the maximum likelihood estimate $\hat{\W}$ is defined as the minimizer of the above loss function in $\W$ while fixing the feature matrix $\bPhi$.

Both the maps $\W\mapsto \ell(\bPhi, \W)$ and $\bPhi\mapsto \ell(\bPhi, \W)$ are convex and we can compute their gradients as well as the Hessian explicitly as follows. For each $y\in \{0,1,\dots\kappa\}$, $\bphi\in \R^{p}$, and $\W\in \R^{p\times \kappa}$, define vector and matrix functions
\begin{align}\label{eq:MNL_h_def_appendix}
\dot{\h}(y,\bphi, \W)& :=(\dot{h}_{1},\dots, \dot{h}_{\kappa})^{T}\in \R^{\kappa\times 1}, \qquad \dot{h}_{j} :=\left( \frac{h'(\langle \bphi,\w_{j} \rangle)}{1+\sum_{c=1}^{\kappa} h(\langle \bphi,\w_{c} \rangle)} - \mathbf{1}(y=j)\frac{h'(\langle \bphi,\w_{j} \rangle)}{h(\langle \bphi,\w_{j} \rangle)} \right)\\
\ddot{\H}(y,\bphi,\W) & := \left( \ddot{\H}_{ij}  \right)_{i,j} \in \R^{\kappa\times \kappa}, \qquad \ddot{\H}_{ij} = \left( \begin{matrix} \hspace{-2cm} \frac{ h''(\langle \bphi,\w_{j} \rangle) \mathbf{1}(i=j)  }{  1+\sum_{c=1}^{\kappa} h(\langle \bphi,\w_{c} \rangle)  } - \frac{ h'(\langle \bphi,\w_{i} \rangle) h'(\langle \bphi,\w_{j} \rangle)   }{  \left( 1+\sum_{c=1}^{\kappa} h(\langle \bphi,\w_{c} \rangle) \right)^{2} }  \\
\hspace{2cm} - \mathbf{1}(y=i=j) \left( \frac{h''(\langle \bphi,\w_{j} \rangle)}{h(\langle \bphi,\w_{j} \rangle)} - \frac{\left( h'(\langle \bphi,\w_{j} \rangle) \right)^{2} }{\left( h(\langle \bphi,\w_{j} \rangle) \right)^{2}}   \right) \end{matrix}  \right).
\end{align}
For each $\W=[\w_{1},\dots,\w_{\kappa}]\in \R^{p\times \kappa}$, let $\W^{\textup{vec}}:=[\w_{1}^{T},\dots, \w_{\kappa}^{T}]^{T}\in \R^{p\kappa}$ denote its vectorization.
Then a straightforward computation shows
\begin{align}
\nabla_{\vect(\W)}   \ell(\bPhi, \W) = \sum_{s=1}^{N} \dot{\h}(y_{s},\bphi_{i}, \W) \otimes \bphi_{s},\qquad 	\H:= \nabla_{\vect(\W)}  \nabla_{\vect(\W)^{T}}   \ell(\bPhi, \W) = \sum_{s=1}^{N} \ddot{\H}(y_{s},\bphi_{s}, \W) \otimes \bphi_{s}\bphi_{s}^{T},
\end{align}
where $\otimes$ above denotes the Kronecker product. 	Recall that the eigenvalues of $\A\times \B$, where $\A$ and $\B$ are two square matrices, are given by $\lambda_{i}\mu_{j}$, where $\lambda_{i}$ and $\mu_{j}$ run over all eigenvalues of $\A$ and $\B$, respectively. Hence we can deduce
\begin{align}\label{eq:MNL_evals_bounds}
\lambda_{\min}\left( \bPhi \bPhi^{T} \right) 	\min_{1\le s \le N,\, \W} \lambda_{\min}\left( \ddot{\H}(y_{s},\bphi_{s}, \W) \right) 	&\le \lambda_{\min}(\H) \\
&\le  \lambda_{\max}(\H)\le \lambda_{\max}\left( \bPhi \bPhi^{T} \right) 	\max_{1\le s \le N,\, \W} \lambda_{\min}\left( \ddot{\H}(y_{s},\bphi_{s}, \W) \right).
\end{align}

There are some particular cases worth noting. First, suppose binary classification case, $\kappa=1$. Then the Hessian $\H$ above reduces to
\begin{align}
\H = \sum_{s=1}^{N} \ddot{\H}_{11}(y_{s}, \bphi_{s},\W)  \bphi_{s}\bphi_{s}^{T}.
\end{align}
%Hence the eigenvalues of $\H$ are between the smallest and the largest possible values of the coefficients $\ddot{\H}_{11}$ depend on possible values of activation $\langle \bphi_{i}, \w_{1} \rangle$. If we restricted the norms of feature vector $\bphi_{i}$ and the regression coefficients $\w_{1}$, then we can get an explicit lower and upper bounds.
Second, let $h(x)=\exp(x)$ and consider the multinomial logistic regression case. Then $h=h'=h''$ so the above yields the following concise matrix expression
\begin{align}
&\nabla_{\W} \, \ell(\bPhi, \W) = \bPhi (\bP - \Y) \in \R^{p\times \kappa},\qquad \nabla_{\bPhi} \, \ell(\bPhi, \W) = \W (\bP - \Y)^{T} \in \R^{p\times N}, \\
& \H= \sum_{s=1}^{N} \begin{bmatrix}
p_{s1}(1-p_{s1}) & - p_{s1}p_{s2} & \dots & -p_{s1}p_{s\kappa} \\
-p_{s2}p_{s1} & p_{s2}(1-p_{s2}) & \dots & -p_{s2}p_{s\kappa} \\
\vdots & \vdots & \ddots & \vdots \\
-p_{s\kappa}p_{s1} & -p_{s\kappa}p_{s2} & \dots & p_{s\kappa}(1-p_{s\kappa})
\end{bmatrix}
\otimes \bphi_{s}\bphi_{s}^{T}.
\end{align}
It follows that  eigenvalues of $\H$ are bounded above by $1/4$. The lower bound on the eigenvalues depend on the range of linear activation $\langle \bphi_{i}, \w_{j} \rangle$ may take. For instance, if we restrict the norms of the input feature vector $\phi_{i}$ and parameter $\w_{j}$, then we can find a suitable positive uniform lower bound on the eigenvalues of $\H$.

\begin{lemma}\label{lem:MNL}
Supose $h(\cdot)=\exp(\cdot)$. Then
\begin{align}
&\lambda_{\min}\left( \ddot{\H}(\bphi_{s},\W) \right)\ge \min_{1\le i \le \kappa} \frac{ \exp(\langle \bphi_{s}, \w_{i} \rangle) }{1+\sum_{c=1}^{\kappa} \exp(\langle \bphi_{s}, \w_{c} \rangle) }, \\
&\lambda_{\max}\left( \ddot{\H}(\bphi_{s},\W) \right)\le \max_{1\le i \le \kappa} \frac{ \exp(\langle \bphi_{s}, \w_{i} \rangle)  }{\left( 1+\sum_{c=1}^{\kappa} \exp(\langle \bphi_{s}, \w_{c} \rangle) \right)^{2} }\left(1+2\sum_{c=2}^{\kappa}  \exp(\langle \bphi_{s}, \w_{c} \rangle)  \right).
\end{align}
\end{lemma}

\begin{proof}
For the lower bound on the minimum eigenvalue, we note that
\begin{align}
\lambda_{\min}\left( \ddot{\H}(\bphi_{s},\W) \right) \ge \min_{1\le i\le \kappa}  \sum_{j=1}^{\kappa}  \ddot{H}_{ij} = \min_{1\le i \le \kappa} p_{si}p_{s0} =\min_{1\le i \le \kappa}  \frac{ \exp(\langle \bphi_{s}, \w_{i} \rangle) }{1+\sum_{c=1}^{\kappa} \exp(\langle \bphi_{s}, \w_{c} \rangle) }
\end{align}
where the first inequality was shown in  \cite[Sec. 3]{amani2021ucb} using the fact that $\ddot{\H}(\bphi_{s},\W)$ is a diagonally dominant $M$-matrix (see [69]). The following equalities can be verified easily.

For the upper bound on the maximum eigenvalue, we use the Gershgorin circle theorem (see, e.g., [28]) to bound
\begin{align}
\lambda_{\max}\left( \ddot{\H}(\bphi_{s},\W) \right) &\le \max_{1\le i \le \kappa} \left( p_{si}(1-p_{si}) + \sum_{c=2}^{\kappa} p_{si}p_{sc}\right) \le \max_{1\le i \le \kappa}  p_{si} \left(2-p_{s0}-2p_{si}\right).
\end{align}
Then simplifying the last expression gives the assertion.
\end{proof}

\newpage

\section{Additioanl figures and tables}

\begin{table}[h!]
\centering
\includegraphics[width=1\linewidth]{Figures/MNIST_table_full.png}
\caption{ Tables of all results from each of the six methods for the semi-sythetic MNIST data in Section \ref{sec:MNIST_setup}. Mean and standard deviations of five runs are reported. We report the reconstruction error of each method after normalizing it by the reconstruction error from NMF. }
\label{table:MNIST_table}
\end{table}

\begin{table}[h!]
\centering
\includegraphics[width=1\linewidth]{Figures/fakejob_table1.png}
\caption{ Tables of all results from each of the six methods for the fake job postings data in Section \ref{sec:fakejob_setup}. without the auxiliary covariates. Mean and standard deviations of five runs are reported. We report the reconstruction error of each method after normalizing it by the reconstruction error from NMF. }
\label{table:fakejob_table1}
\end{table}

\begin{table}[h!]
\centering
\includegraphics[width=1\linewidth]{Figures/fakejob_table2.png}
\caption{ Tables of all results from each of the six methods for the fake job postings data in Section \ref{sec:fakejob_setup} with the 72 auxiliary covariates. Mean and standard deviations of five runs are reported. We report the reconstruction error of each method after normalizing it by the reconstruction error from NMF. }
\label{table:fakejob_table2}
\end{table}

\end{document}