text:
In~\cite{2016arXiv160704546B}, the authors laid out the preceding construction technique, and then went on to exploit the existence of maximal sets of MUBs in even prime power dimensions and a special class of affine block designs  to construct an infinite family of maximal OGFFs, emphasizing that their construction is highly rigid due to maximality properties of the MUBs and block designs they used, and highlighted the examples'  {cubature properties}.

In this section, we extend their work, noting that numerous other families of tight OGFFs arise from {\it symmetric block designs} via the final theorem of the preceding section; similarly, we observe examples arising from  other types of affine block designs -- all unmentioned in~\cite{2016arXiv160704546B}.

We begin with symmetric designs, which are  surprising well-suited to this construction principle in a certain sense.
\subsection{Symmetric block designs}
\begin{defn}
	A $t$-$(m,l,\lambda)$-block design is {\bf symmetric} if $m=b$ or, equivalently, if $l=r$.
\end{defn}

As verified in~\cite{cameron_lint_1991}, symmetric block designs have the useful property that the pairwise block intersections is constant.
\begin{thm}\label{th:sym.eq}[\cite{cameron_lint_1991}]
	For a symmetric $t$-$(m,l,\lambda)$ block design, $\left(\INDEX{m},  \BLOCKSET\right)$,  every $\mathcal J_j, \mathcal J_{j'} \in \BLOCKSET$ with $j \neq j'$ satisfies $$\left| \mathcal J_j \cap \mathcal J_{j'} \right| = \lambda.$$
\end{thm}

\subsubsection{Simple ETFFs from all t-block designs}
An immediate -- albeit trivial -- corollary to this observation is that every symmetric block designs yields an ETFF.

\begin{cor}\label{prop:symmequi}
	Given any symmetric $t$-$(m,l,\lambda)$ block design and a tight  $(m, 1)$-fusion frame, $\Pi = \CBRACK{\pi_i}_{i=1}^m$, for $\mathbb F^m$,  then the family of $n=b$ block projections for $\Pi$ forms an ETFF  for $\mathbb F^m$.
\end{cor}
\begin{proof}
	By design, Proposition~\ref{prop_1designs} ensures that the sequence of $n=b$ rank $l$ block projections, $\mathcal P = \CBRACK{P_i := \sum_{j \in \mathcal J_i}  \pi_j}_{i=1}^m$ forms a tight fusion frame for $\mathbb F^m$.  By Theorem~\ref{th:cohbds}, these are chordally equiangular, since, for $i \neq i'$,  we have
	$$
	\TR\PARENTH{P_i P_{i'} } = \sum_{j \in \mathcal J_i}\sum_{j' \in \mathcal J_{i'}} \TR\PARENTH{  \pi_j   \pi_{j'} } = \lambda,
	$$
	where one can confirm that $\lambda$ is the packing constant for this special case.
\end{proof}

As implicated by the numerous examples of symmetric block designs in ~\cite{MR2246267} and the references therein,  these simple examples of ETFFs exist in abundance.

\subsubsection{Tight OGFFs from certain block dseigns}

Of more significance is the multitude of examples of tight OGFFs that arise from the construction outlined in theorem~\ref{th_costruct_OGFF}, depending heavily on the existence of maximal MUBs. The surprise here is that every nontrivial symmetric $t$-block-design satisfies the $l^2/m$-cohesive property.
\begin{prop}
	Assume $1 < l < m-1$.  Every symmetric $t$-$(m,l,\lambda)$ block design, $(\INDEX{m},  \BLOCKSET)$, is $l^2/m$-cohesive.
\end{prop}
\begin{proof}
	By Theorem~\ref{th:sym.eq}, we only need to show that $\lambda \le l^2 /m$, which follows by applying the trivial necessary conditions from Proposition~\ref{block:triv} to obtain the elementary estimation,
	$$
	\lambda = \frac{r(l-1)}{m-1} = \frac{l(l-1)}{m-1} \le \frac{l^2}{m}.
	$$
\end{proof}

Thus, following the strategy, wherein we employ maximal MUBs, as outlined after the statement of Theorem~\ref{th_costruct_OGFF}, we seek symmetric block designs for which $m=b=q$, where $q$ is a prime power or a power of four, for the complex and real cases, respectively.

\begin{cor}\label{cor:symogffs}
	Assume $1 < l < m-1$.
	\begin{itemize}
		\item 	Every symmetric $t$-$(m,l,\lambda)$ block design for which $m$ is a prime power yields a tight $(n,l)$-OGFF for $\mathbb C^m$, where $n=m(m+1)$, according to the construction described in Theorem~\ref{th_costruct_OGFF}.
		\item 	Every symmetric $t$-$(m,l,\lambda)$ block design for which $m$ is a power of four yields a tight $(n,l)$-OGFF for $\mathbb R^m$, where $n=m(m/2 +1)$, according to the construction described in Theorem~\ref{th_costruct_OGFF}.
	\end{itemize}
\end{cor}

Examining the list of known families of symmetric designs in II.6.8 of \cite{MR2246267}, we identify numerous families of symmetric block designs for which Corollary~\ref{cor:symogffs} yields more tight OGFFs.

\subsubsection{Point-hyperplane symmetric designs; see Family 1 of~ \cite{MR2246267}}
Given a prime power $q$ and $t \ge 2$, a symmetric $1$-$\PARENTH{\frac{q^{t+1} -1}{q-1}, \frac{q^t-1}{q-1}, \frac{q^{t-1}-1}{q-1} }$ block design exists, known as a {\bf point-hyperplane design}.
\begin{itemize}
	\item
	By Corollary~\ref{cor:symogffs}, a tight $\PARENTH{m, \frac{q^t-1}{q-1}}$-OGFF for $\mathbb C^m$, comprised of $n:=\frac{(q^{t+1} -1)(q^{t+1} +q -2)}{(q-1)^2}$ projections,  exists whenever $m:=\frac{q^{t+1} -1}{q-1}$ is a prime power, leading to numerous families of examples of tight OGFFs.
	\item
	Unfortunately, the necessary conditions of these block designs seem too restrictive to admit real examples of OGFFs according to our approach.
\end{itemize}
Contingent upon the open question regarding the existence of an infinitude of Fermat or Mersenne primes -- a famous open problem in number theory~\cite{MR0327647} -- complex examples arising from this family may be infinite by taking $q=2$.

\subsubsection{Hadamard symmetric designs; see Family 2 of~ \cite{MR2246267}}
Given $t \ge 1$, a symmetric $1$-$\PARENTH{4t-1, 2t-1, t-1 }$ block design exists, known as a {\bf Hadamard design}.
\begin{itemize}
	\item
	By Corollary~\ref{cor:symogffs}, a tight $\PARENTH{m, 2t-1}$-OGFF for $\mathbb C^m$, comprised of $n:=4t(4t-1)$ projections,  exists whenever $m:= 4t-1$ is a prime power, leading to an infinite family of complex examples of tight OGFFs.
	\item
	Unfortunately,  $m:= 4t-1$ is never a power of four, so these designs yield no real examples.
\end{itemize}

\subsubsection{Menon symmetric designs; see Family 6 of~ \cite{MR2246267}}
The existence of a Hadamard matrix (see \cite{MR2246267} for details) of order $4t^2$, $t \in\mathbb N$, is equivalent to the existence of a symmetric $t$-$\PARENTH{4t^2, 2t^2-t, t^2-t}$ block design, called a {\bf Menon design}.  According to the well-known Hadamard conjecture~\cite{MR2246267}, such block designs exist for all values of $t$.  Besides the conjecture, a simple tensor construction~\cite{MR2246267}, among other constructions, assures their existence when $t=2^s$ for some $s \in \mathbb N$.
\begin{itemize}
	\item
	By Corollary~\ref{cor:symogffs}, when $t=2^s$ for some $s \in \mathbb N$, a tight $\PARENTH{4t^2, 2t^2-t}$-OGFF for $\mathbb C^m$ comprised of $n:=16t^4 + 4t^2$ projections exists,  leading to an infinite family of complex examples.

	\item
	Conveniently, these designs are well-suited for the real case.  Similarly, when $t=2^s$ for some $s \in \mathbb N$, where $s$ is even, then a tight $\PARENTH{4t^2, 2t^2-t}$-OGFF for $\mathbb R^m$ comprised of $n:=4t^2(2t^2 + 1)$ projections exists,  leading to an infinite family of tight OGFFs.
\end{itemize}

\subsubsection{Wallis symmetric designs; see Family 7 of~\cite{MR2246267}}
Given a prime power $q$ and $t \in \mathbb N$, a symmetric
$t$-$\PARENTH{m, l,\lambda  }$ block design exists, known as a {\bf Wallis design},
where
\begin{itemize}
	\item $m:=q^{t+1}\left(q^t + q^{t-1} + \dots + q^2 + q + 2 \right),$\\
	\item $l:= q^{t}(q^t + q^{t-1} + \dots + q^2 +  q + 1 )$, and \\
	\item $\lambda =  q^{t}(q^{t-1} + q^{t-2} + \dots + q^2 + q + 1 ).$
\end{itemize}
	  Taking $q=2$
	  and computing (inductively or directly) that
	  $$
	  2^{t+1}(2^t + 2^{t-1} + .... + 4 + 2 + 2) = 4^{t+1},
	  $$
 the Wallis family yields infinite families of both real and complex tight OGFFs.
\begin{itemize}
	\item
	By Corollary~\ref{cor:symogffs}, when $q=2$ and $t \in \mathbb N$, a tight $\PARENTH
	{n,l}$-OGFF for $\mathbb C^m$ comprised of $n:=4^{2t+2} + 4^{t+1}$ projections exists,  leading to an infinite family of complex examples.
	\item
	Once again, these designs are well-suited for the real case.  When $q=2$ and $t \in \mathbb N$, a tight $\PARENTH{n, l}$-OGFF for $\mathbb R^m$ comprised of $n:=4^{t+1}(\frac{4^{t+1}}{2}+1)$ projections exists,  leading to another infinite family of real examples.
\end{itemize}

\subsubsection{Wilson/Brouwer symmetric designs; see Family 1 of~11 \cite{MR2246267}}
Given an odd prime power $q$ and $t \in \mathbb N$, a symmetric
$1$-$\PARENTH{m, l,\lambda  }$ block design exists, which we refer to as a {\bf Wilson/Brouwer  design},
where
\begin{itemize}
	\item $m:=2\left(q^t + q^{t-1} + \dots + q^2 + q \right) +1,$\\
	\item $l:= q^{t},$ and \\
	\item $\lambda =  q^{t-1}\frac{q-1}{2}$.
\end{itemize}
Unfortunately, the addition of $1$ to an even number makes it impossible for $m$ to be a power of four here.  Still, for various parameters -- eg, $q=3$, $t=1$ yields $m=7$ -- this family produces various complex instances of tight OGFFs.

A comprehensive list of all tight OGFFs obtained from symmetric designs is beyond the scope of this work -- particularly due to outstanding open questions about their existence for various parameters~\cite{MR2246267}.  Having demonstrated families arising in this fashion, we move to tight OGFFs arising from affine designs.

\subsection{Affine Designs}
In the work where this recipe is presented, the authors combined real and complex even powers of maximal MUBs of Hadamard $3$-designs (see\cite{2016arXiv160704546B}), a special case of affine designs, to form maximal OGFFs according to Theorem~\ref{th_costruct_OGFF}.

\begin{defn}
A block design $\PARENTH{\INDEX{m}, \BLOCKSET}$  is {\bf resolvable} if  $\BLOCKSET$ partitions into subsets, called {\bf parallel classes}, such that
\begin{itemize}
\item the blocks with in each class are disjoint, and
\item for each parallel class, every element of $\INDEX{m}$ is contained in a block.
\end{itemize}
Moreover, if
the number of elements occurring in the intersection between blocks from different parallel classes is constant, then it is an {\bf affine design}.

\end{defn}

A simple result in~\cite{MR2246267} assures the existence of analogous block designs for the odd prime power case.
We restate these for our special case.

\begin{prop}\label{prop_affexist}[See II.7.10 \cite{MR2246267}]
	If $m=p^{t+1}$, $l=p^t$ and $\lambda=\frac{p^t-1}{p-1}$ for some prime, $p$, then a resolvable $1$-$(m,l,\lambda)$ block design exists and its remaining parameters are therefore  $r=\sum_{i=0}^t p^i$ and  $b=\sum_{i=1}^{t+1} p^i$.
\end{prop}

 According to Bose's condition, such designs are $l^2/m$-cohesive affine block designs.

 \begin{thm}[Bose's condition; see Theorem II.7.28 of \cite{MR2246267}]
 	Given any resolvable $1$-$(m,l,\lambda)$ block design, the number of blocks is bounded by the other parameters according to
 	  $$b \ge m +r-1,$$
 	  and this lower bound is achieved if and only if the design is an $l^2/m$-cohesive affine design.
 \end{thm}

 One may verify that the designs produced by Proposition~\ref{prop_affexist} achieve this lower bound.  Noting that such designs satisfy
 $$k_{\mathbb C^{m}} b= (p^{t+1} +1) \sum_{i=1}^{t+1} p^i > d_{\mathbb C^m}+1 = p^{2t+2},$$
 the conditions of Theorem~\ref{th_costruct_OGFF} are satisfied.

 \begin{cor}
 	For every prime $p$ and $t\in \mathbb N$, a tight $(n,l)$-OGFF, comprised of $n=(p^{t+1} +1) \sum_{i=1}^{t+1} p^i$ exists for $\mathbb C^m$, where $m=p^{t+1}$ and $l=p^t$.
 \end{cor}

Although these tight OGFFs are not maximal OGFFs, they are maximal with respect to the construction in Theorem~\ref{th_costruct_OGFF}, as each instance is produced by a maximal set of MUBs and a maximal affine design, according to Bose's condition.


TITLE: Rank of a hollow matrix
QUESTION [0 upvotes]: $A$ is a $d \times d$ square and symmetric matrix.
$A(i,i)=0$ for all $i$,
$A(i,j)>0$ and distinct for $i \neq j$.
Is $A$ full-rank, or invertible? This is true for $d=2,3,4$.

REPLY [1 votes]: A counterexample is given as follows:
\begin{pmatrix}
0 & 2 & 3 & 15 \\
2 & 0 & 5 & 9 \\
3 & 5 & 0 & 6 \\
15 & 9 & 6 & 0
\end{pmatrix}
This matrix has determinant zero.
Reference: Determinant of symmetric matrix with the main diagonal elements zero


TITLE: C/c: modding out an object
QUESTION [1 upvotes]: Given a category $\mathbf{C}$ and some object $c \in \mathbf{C}$, what's the meaning of $\mathbf{C}/c$?
I know that for a congruence relation $R$ on $\mathbf{C}$, the expression $\mathbf{C}/R$ denotes the quotient category of equivalence classes. But I don't know how to similarly "mod out" an object $c$.

REPLY [2 votes]: $\mathbf{C}/c$ denotes a slice category or over category of arrows pointing to $c$. On nLab it is defined like so:

The slice category or over category $\mathbf{C}/c$ of a category $\mathbf{C}$ over an object $c \in \mathbf{C}$ has

objects that are all arrows $f \in \mathbf{C}$ such that $cod(f) = c$, and

morphisms $g: X \to X' \in \mathbf{C}$ from $f:X \to c$ to $f': X' \to c$ such that $f' \circ g = f$.

$$
  C/c
  =
  \left\lbrace
    \array{
       X &&\stackrel{g}{\to}&& X'
       \\
       & {}_f \searrow && \swarrow_{f'}
       \\
       && c
    }
  \right\rbrace
$$

Slice categories ($\mathbf{C}/c$) don't seem to have a very obvious relation to quotient categories ($\mathbf{C}/R$); they just use the same syntax. The alternative syntax $\mathbf{C} \downarrow c$ avoids this potential confusion.
There is also the dual notion of a coslice category or under category of arrows pointing away from $c$, denoted $c/\mathbf{C}$ or $c \downarrow \mathbf{C}$.


TITLE: Reduce water pressure
QUESTION [0 upvotes]: I'm installing a drip irrigation system to my "garden" , some 6 plant pots, but not been successful so far due to the high water pressure.
Yes I know I can go buy a pressure regulator, but this will take out the fun.
So I'd like understand how using little of physics and a little ingenuity and of course little money, could I do simple water pressure reducer.

REPLY [1 votes]: the coolest pressure regulator I have seen is the simplest, and most fun: it is a rubber plug with a small hole through it, inserted into the piping of a drip system. High water pressure squeezes the plug, causing it to deform. This partially occludes the hole, increasing its flow resistance. low water pressure does not deform the plug so much, leaving the hole unobstructed, decreasing the flow resistance. Over a reasonable range of source pressures, this maintain flow rate almost constant! and it only costs a couple of dollars.


TITLE: Cumulative distribution function, integration problem
QUESTION [2 upvotes]: Given the continuous Probability density function $f(x)=\begin{cases} 2x-4, & 2\le x\le3  \\ 0 ,& \text{else}\end{cases}$
Find the cumulative distribution function $F(x)$.
The formula is $F(x)=\int _{ -\infty  }^{ x }{ f(x) } $

My Solution
The first case is when $2\le x\le3$  then $$\int_2^x {(2u-4)} \, du=[u^2-4u]_2^x=x^2-4x-4+8=x^2-4x+4$$
so $F(x)=x^2-4x+4 , \text{ for } 2\le x\le 3$

The Problem
Now I have the cases where $x<2 \text{ and } x>3$ , ($\int_{-\infty}^{x}0 \, dx$) but I am not sure how to do it. I would appreciate if someone could show me the solution of this two cases.

REPLY [1 votes]: A general technique to deal with such type of questions is to split up the integral at the points where the density $f$ has different specifications, i.e. at $y=2$ and $y=3$. If $y<2$, then
$$
F(y)=\int_{-\infty}^yf(x)\,\mathrm dx=\int_{-\infty}^y0\,\mathrm dx=0,
$$
and if $2\leq y\leq 3$, then
$$
\begin{align}
F(y)&=\int_{-\infty}^2f(x)\,\mathrm dx+\int_2^yf(x)\,\mathrm dx=\int_{-\infty}^20\,\mathrm dx+\int_2^y(2x-4)\,\mathrm dx\\
&=0+\left[x^2-4x\right]_2^y=y^2-4y+4.
\end{align}
$$
Lastly, if $y>3$ then
$$
\begin{align}
F(y)&=\int_{-\infty}^2 f(x)\,\mathrm dx+\int_2^3 f(x)\,\mathrm dx+\int_3^y f(x)\,\mathrm dx\\
&=\int_{-\infty}^2 0\,\mathrm dx+\int_2^3 (2x-4)\,\mathrm dx+\int_3^y 0\,\mathrm dx\\
&=0+1+0=1.
\end{align}
$$


TITLE: Understanding tension based on assumptions of pulley system
QUESTION [3 upvotes]: If we consider a simple pulley system with two masses hanging on each end of a MASSLESS and INEXTENSIBLE string around a MASSLESS and FRICTIONLESS pulley, how then can one reason that the tension at each end of the string must be the same?
My own reasoning:
MASSLESS ROPE means that for any segment of the rope with tension $T_1$ and $T_2$ we have that $\sum F = T_ 2 - T_1 = 0$ (since $m = 0$) and thus the tensions must be the same, on a non curved rope at least!
INEXTENSIBLE means that no energy can be stored in the string, however I fail to see how this is a neccesary condition (for equal tension)
MASSLESS PULLEY means that no rotational inertia exists, and thus no force can alter the tension of the string (?)
FRICTIONLESS PULLEY is hard for me to figure.
Needless to say, I feel quite at a loss conceptually!

REPLY [4 votes]: What we mean by a frictionless pulley is that the friction in the bearings of the pulley is negligible, and the pulley is free to rotate without any resistance.  We don't mean that the friction between the string and the pulley surface is negligible.  In fact, we assume that there is enough static friction between the string and pulley surface to prevent the string from slipping.  But, in this case, if you do a moment balance on the pulley, you must then conclude that the tensions in the string on either side are equal (even if the pulley has angular acceleration), since the moment of inertia of the pulley is assumed to be zero.


TITLE: finding operator norm $T_N$
QUESTION [2 upvotes]: How do find the operator norm of , $T_N\colon c_0\to \Bbb R$ given by
$T_N(y):=\sum_{j=1}^Nx_jy_j$ ,when $N$ is a integer .
TIA

REPLY [1 votes]: If $x=0$, $T_N=0$. So assume $x\neq 0$.
It is easily seen, by triangular inequality, that
$$
|T_N(y)|\leq \sum_{j=1}^N|x_j||y_j|\leq \|y\|_\infty \sum_{j=1}^N|x_j|=\|x\|_1\|y\|_\infty.
$$
Hence $\|T_N\|\leq \|x \|_1$.
Now consider the sequence $y$ such that $y_n=0$ whenever $x_n=0$ and also for $n>N$. And for the other values of $n$, take
$$
y_n=\frac{x_n}{|x_n|}.
$$
That's the sign function of $x$. What is $\|y\|_\infty$? What is $T_N(y)$? Conclude that the above inequality is an equality:
$$
\|T_N\|=\|x\|_1.
$$


TITLE: Matrix transposition and invertibility
QUESTION [1 upvotes]: The following question is a question in my midterm, I could not develop any simpler forms of $AA^T = kI$.

Let $A_{n\times n}$ verify the expression $AA^T=kI_{n\times n}$, $k\in R^{+}$.
a) Show that $A$ is invertible and find $A^{-1}$.
b) Calculate $A^{-1}(A^{-1})^T$.

REPLY [0 votes]: Since $AA^T=kI,$ we can have $A(\frac{1}{k}A^T)=I,$ so $A$ is invertible with $A^{-1} = \frac{1}{k}A^T.$
Then the rest is direct computation:
$$A^{-1}(A^{-1})^T=\frac{1}{k}A^T(\frac{1}{k}A^T)^T=\frac{1}{k^2}A^TA=\frac{1}{k^2}\cdot kI=\frac{1}{k}I.$$


TITLE: How do we find maxima and minima for a discrete-time function
QUESTION [1 upvotes]: To find maxima and minima of a continuous-time signal we can use differentiation. I want to know how do we find maxima and minima of a discrete-time signal?

REPLY [1 votes]: You can still do the differentiation but numerically. Then the argument which indicates ablolutely minimum value is what you are after.
Other way would be to sort you data in descending order, if it is finite, and take the first element.


TITLE: Mean-Width and Diam(K)
QUESTION [0 upvotes]: I believe the following is always true:
if $A$ is a Convex Hull of some points in $R^n$, then we have
$meanwidth(A)$$\ge$$C$$\cdot$$diam$$(A)$ ; (where the $diam$ stands for the maximum distance between 2 points in the same Convex Hull)
I need the above constant $C$ precisely, but I couldn't find any paper, text, book etc. containing the precise value of $C$. Any ideas?

REPLY [0 votes]: Hug, Daniel; Weil, Wolfgang, $\textit{A Course on Convex Geometry}$. Very nice book if you are interested in Convex Geometry; explains the above mentioned property.


TITLE: How to construct matrix from 4 sub matrices in Matlab?
QUESTION [1 upvotes]: I have an 8x8 matrix defined as
$T =  \begin{bmatrix}T_{UU} \quad T_{UF}\\T_{FU} \quad T_{FF}\end{bmatrix}$
I can define $T_{UU}$, $T_{UF}$, $T_{FU}$, and $T_{FF}$ as
TFU = T(5:8,1:4)
TFF = T(5:8,5:8)
TUU = T(1:4,1:4)
TUF = T(1:4,5:8)

Is there a nice way to build the following matrixes in Matlab or Octave?
$K1 =  \begin{bmatrix}0 \quad -T_{UF}\\-I \quad -T_{FF}\end{bmatrix}$
and
$K2 =  \begin{bmatrix}I \quad -T_{UU}\\0 \quad -T_{FU}\end{bmatrix}$
where I is an Identity matrix.

REPLY [1 votes]: Matlab is pretty good with block matrices like this. You can do
K1 = [zeros(4,4),-TUF; -eye(4), -TFF];


TITLE: Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$
QUESTION [1 upvotes]: Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of 3.

REPLY [0 votes]: A bit longer:

Assume $\tau(n)=\top,n\equiv 1\pmod{3}$
$$1+2^n+2^{2n}=7k\implies 2^{2n}=7k-2^n-1,\;k\in\mathbb Z$$
$\tau(n+1)=\top\implies\;n+1\equiv 2\pmod{3}$
$$1+2^{n+1}+2^{2(n+1)}=1+2\cdot 2^{n}+4\cdot2^{2n}=1+2\cdot2^n+4(7k-2^n-1)\\=28k-2\cdot2^n-3=28k-(2^{n+1}+3)$$
$$7\mid2^{n+1}+3\implies7\mid 2\cdot2^n-4=2(2^n-2)\iff 7\mid2^n-2=2(2^{n-1}-1)\\\iff 7\mid 2^{2n-1}-1$$
$$2^6\equiv 1\pmod{7}\implies 2^{6m}\equiv 1\pmod{7},\;m\in\mathbb N$$
$2n-1=6m\implies 3\mid 2n-1\implies 3\mid 2n+2\implies n\equiv 1\pmod{3}$


text:
\begin{document}
\title{\textbf{Ring of flows of one-dimensional differential equations}}
\author{Ronald Orozco L\'opez}

\newcommand{\Addresses}{{

\textit{E-mail address}, R.~Orozco: \texttt{rj.orozco@uniandes.edu.co}

}}

\maketitle

\begin{abstract}
In this article the ring of flows of autonomous differential equations of order one on integral domains is constructed. First we build the autonomous ring $\Opa(\hurw_{R}[[x]])$ and then its structure is studied. Next, we build the ring of formal exponential generating series of the ring
$\Opa(\hurw_ {R} [[x]]) $, where it is possible to find solutions of differential equations of order one when the vector field of the system can be decomposed in sum or product of functions.
\end{abstract}
{\bf Keywords:} Hurwitz ring, autonomous operator, autonomous ring, flows ring\\
{\bf Mathematics Subject Classification:} 47H99, 16W99, 34A34, 11B83

\section{Introducci\'on}
A very important problem in the theory of autonomous differential equations is to find exact solutions of these. In the one-dimensional case the solutions of $\phi^{\prime}=f(\phi)$
can be found by using the method of separation of variables. However, it is not always possible to find explicit solutions to these equations by this method. In this article we propose a different method that allows not only to find solutions to these equations but also to give a ring structure to the set of one-dimensional dynamic systems. This type of solutions are found if the vector field $f(x)$ is taken from the Hurwitz ring of power series $\hurw_{R}[[x]]$, where $R$ is a integral domain.

In article [5] we make a first approach to this type of solutions. In particular we find an analytical solution to the autonomous equation $\phi^{(k)}=f(\phi)$ of order $k$ in terms of autonomous polynomials. In this paper the equations will be limited to order one.

Autonomous polynomials are constructed using Bell polynomials. Bell polynomials are a very useful tool in mathematics to represent the $n$-th derivative of the composition of functions (see [3]). Indeed, let $f$ and $g$ be two analytic functions with representation
in power series $\sum a_{n}\frac{x^{n}}{n!}$ and $\sum b_{n}\frac{x^{n}}{n!}$, respectively, with $a_{n},b_{n}\in\C$. Then
\begin{equation}
f(g(x))= f(b_{0}) + \sum_{n=1}^{\infty}Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})\frac{x^{n}}{n!},
\end{equation}
where $Y_{n}$ is the $n$-th Bell polynomial. For example, it is well known that
\begin{eqnarray*}
Y_{1}(b_{1};a_{1})&=&a_{1}b_{1},\\
Y_{2}(b_{1},b_{2};a_{1},a_{2})&=&a_{1}b_{2}+a_{2}b_{1}^{2},\\
Y_{3}(b_{1},b_{2},b_{3};a_{1},a_{2},a_{3})&=&a_{1}b_{3}+a_{2}(3b_{1}b_{2})+a_{3}b_{1}^{3}.
\end{eqnarray*}

Bell's polynomials can be expressed explicitly by using the formula of Faá di Bruno ([3])
\begin{equation}\label{faa}
Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})=\sum_{k=1}^{n}B_{n,k}a_{k},
\end{equation}
where
\begin{equation}\label{partialBell}
B_{n,k}=\sum_{\vert p(n)\vert=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}},
\end{equation}
and the sum runs over all partitions $p(n)$ of $n$, that is, $n=j_{1}+2j_{2}+\cdots+nj_{n}$, $j_{h}$ denotes the number of parts of size $h$ y $\vert p(n)\vert=j_{1}+j_{2}+\cdots+j_{n}$ is the length
of the partition $p(n)$.

Suppose $ f $ with power series representation and let
\begin{equation}\label{autonoma}
\begin{cases}
u^{\prime} = f(u)\\
u(0)=x
\end{cases}
\end{equation}
be the autonomous differential equation with initial value problem.

In [5] autonomous polynomials were introduced, which are the coefficients of the analytical solution of the equation (\ref{autonoma}) expressed as a function of the initial value in the following way
\begin{small}
\begin{equation}\label{polinomio_autonomo}
\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}
\end{equation}
\end{small}

From (\ref{polinomio_autonomo}) we can see that each autonomous polynomial $A_{n}$ is defined on the sequence $(f^{(n)}(x))_{n\geq0}$ of derivatives of $f(x)\in C_{\infty}(\R)$, where $C_{\infty}(\R)$
is the set of all smoothness functions in $\R$. In this article, autonomous differential equations of order one with functions in $\hurw_{R}[[x]]$, the Hurwitz ring of exponential generating functions, are studied. The article is divided as follows. In the second section the ring $(\hurw_{R}[[x]],+,\cdot) $ and its structure are studied. In the third section we define the autonomous operator and construct the autonomous ring $(\Opa(\hurw_{R}[[x]]),\boxplus,\odot)$ that will serve as the basis for defining operations with flows of dynamical systems. Dynamic systems on rings are defined in section four. In this section it is proved that the flow of a dynamic system is a $R$-module acting on the phase space of the system. Finally, in section five the ring of generating functions of sequences in the autonomous ring is constructed. With this ring it is possible to define the one-dimensionals flows ring with which it will be possible to define flow operations. That is, we can decompose the flow of a dynamic system into simpler flows.

Throughout this paper $R$ will be a ring of characteristic zero. Also, $\N$ will be the monoid of non-negative integers and $\N_{1}$ will be the semigroup of positive integers.

\section{The Hurwitz ring $(\hurw_{R}[[x]],+,\cdot)$}

Throughout this paper let $(R,+,\cdot)$ denote an integral domain. Let $\hurw_{R}=\{(a_{n})_{n\in\N}: a_{n}\in R\}$ denote the set of sequences with elements in the ring $R$. Take $\textbf{a}=(a_{n})_{n\in\N}$ and $\textbf{b}=(b_{n})_{n\in\N}$ in $\hurw_{R}$. Define the sum in $\hurw_{R}$ as
$\textbf{a}+\textbf{b}=(a_{n}+b_{n})_{n\in\N}$. Let $\cdot$ denote the Hadamard product of $\textbf{a}$ and $\textbf{b}$ given by $\textbf{a}\cdot\textbf{b}=(a_{n}\cdot b_{n})_{n\in\N}$, that is, the product in $\hurw_{R}$ is a componentwise product. Then $(H_{R},+,\cdot)$ is a ring with unit element $\textbf{1}=(1,1,1,...)$. An element $\textbf{a}$ is invertible in $\hurw_{R}$ with respect to the product $\cdot$ if and only if $a_{n}\in R^{*}$ for all $n\geq0$.

Now let $*$ denote the Hurwitz product in $\hurw_{R}$ as
\begin{equation}
\textbf{a}\ast\textbf{b}=\left(\sum_{h=0}^{n}\binom{h}{n}a_{h}b_{n-h}\right)_{n\in\N}.
\end{equation}

Then $(\hurw_{R},+,\ast)$ is a ring with unit element $\textbf{e}=(1,0,0,...)$. An element
$\textbf{a}=(a_ {n})_{n\in\N}$ in $\hurw_{R}$ is invertible with respect to $\ast$ if and only if
$a_{0}\in R^{*}$. Let
$$\hurw_{R}^{*}=\{\textbf{a}^{-1}:\textbf{a}\in\hurw_{R}\}$$
denote the set of invertible elements in $(\hurw_{R},+,*)$, where $\textbf{a}^{-1}=\textbf{b}=(b_{n})_{n\in\N}$ with

\begin{equation}
\begin{cases}
b_{0}=a_{0}^{-1}\\
b_{n}=-a_{0}^{-1}\sum_{h=1}^{n}\binom{n}{h}a_{h}b_{n-h},\ n\geq1.
\end{cases}
\end{equation}

Now denote $\hurw_{R}[[x]]$ the set of formal power series of the form $\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}$ with coefficients in $R$. It is clear that
$(\hurw_{R}[[x]],+,\cdot)$ is a ring with sum and product of ordinary series, that is,

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

where $f(x)=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}$,
$g(x)=\sum_{n=0}^{\infty}b_{n}\frac{x^{n}}{n!}\in\hurw_{R}[[x]]$. The ring $\hurw_{R}[[x]]$ is
known as ring of Hurwitz of power series (ve\'ase [4]).

Let $\rho_{x}:(\hurw_{R},+,*)\rightarrow(\hurw_{R}[[x]],+,\cdot)$ be an isomorphism given by
$$\rho_{x}(\textbf{a})=\rho_{x}((a_{n})_{n\in\N})=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}=f(x).$$

If $\rho_{x}(\textbf{a})=f(x)$ y $\rho_{x}(\textbf{b})=g(x)$, then

$$\rho_{x}(\textbf{a}\ast\textbf{b})=\rho_{x}(\textbf{a})\cdot\rho_{x}(\textbf{b})$$
and
$$\rho_{x}(\textbf{a}^{-1})=\frac{1}{\rho_{x}(\textbf{a})}=(\rho_{x}(\textbf{a}))^{-1}$$

Let $\delta$ be a derivation on $\hurw_{R}$ define by

$$\delta((a_{n})_{n\in\N})=(a_{n+1})_{n\in\N}$$

with $\delta(\textbf{a}*\textbf{b})=\delta(\textbf{a})\ast\textbf{b}+\textbf{a}*\delta(\textbf{b})$
and let $\delta_{x}\equiv\frac{d}{dx}$ be a derivation on $\hurw_{R}[[x]]$ given by

$$\delta_{x}\left(\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}\right)=\sum_{n=0}^{\infty}a_{n}\frac{\delta_{x}x^{n}}{n!}=\sum_{n=0}^{\infty}a_{n+1}\frac{x^{n}}{n!}.$$

Then

\begin{equation}\label{relation_delta_mu}
\delta_{x}\rho_{x}(\textbf{a})=\rho_{x}(\delta\textbf{a}),
\end{equation}

that is, the following diagram commutes

\begin{equation}
\xymatrix{
 \hurw_{R} \ar[d]^{\rho_{x}} \ar[r]^{\delta} & \hurw_{R} \ar[d]^{\rho_{x}}\\
 \hurw_{R}[[x]] \ar[r]^{\delta_{x}} & \hurw_{R}[[x]]
}
\end{equation}

It is easy to see that $\rho_{x}(\ker\delta)=\ker\delta_{x}$, that is, the constant ring of
$\hurw_{R}[[x]]$ is the image by $\rho_{x}$ of the constant ring of $\hurw_{R}$.

\section{The Ring Autonomous $(\Opa(\hurw_{R}[[x]]),\boxplus,\odot)$}

In this section the autonomous ring is constructed. This is a ring of sequences of autonomous polynomials defined on the ring $\hurw_{R}[[x]] $. We first define the autonomous operator and study its properties. We then use this operator to construct the binary operations of the autonomous ring. We start with the following definition

\begin{definition}\label{def_opa}
Let $\hurw_{S}$ be the ring of sequences on the ring $S=\hurw_{R}[[x]]$. We define the
\textbf{autonomous operator} $\Opa$ as the linear map
$\Opa:\hurw_{R}[[x]]\rightarrow\hurw_{S}$ defined by
\begin{equation}
\Opa(f(x))=(A_{0}(f(x)),A_{1}(f(x)),A_{2}(f(x)),A_{3}(f(x)),...)
\end{equation}
where the $A_{n}$ are recursively defined by
\begin{eqnarray*}
A_{0}(f(x))&=&x,\\
A_{1}(f(x))&=&f(x),\\
A_{n+1}(f(x))&=&Y_{n}(A_{1}(f(x)),A_{2}(f(x))...,A_{n}(f(x));\delta^{1}f(x),...,\delta^{n}f(x)),
\end{eqnarray*}
$n\geq 1$.\\
The polynomials $A_{n}$ in the indeterminate $\delta^{0}f,\delta^{1}f,...,\delta^{n-1}f$ will be called \textbf{autonomous polynomials}.
\end{definition}

Some autonomous polynomials $A_{n}$ are
\begin{eqnarray*}
A_{2}(f(x)) &=& A_{1}(f(x))\delta^{1}f(x)=f(x)\delta f(x),\\
A_{3}(f(x)) &=& A_{2}(f(x))\delta f(x)+A_{1}^{2}(f(x))\delta^{2}f(x),\\
&=&f(x)(\delta f(x))^{2}+f^{2}(x)\delta^{2}f(x),\\
A_{4}(f(x)) &=& A_{3}(f(x))\delta^{1}f(x)+3A_{1}(f(x))A_{2}(f(x))\delta^{2}f(x)+A_{1}^{3}(f(x))\delta^{3}f(x),\\
&=&f(\delta f(x))^{3}+4f^{2}(x)\delta^{1}f(x)\delta^{2}f(x)+f^{3}(x)\delta^{3}f(x).\\
\end{eqnarray*}

In the following Lemma we will see how to construct the autonomous polynomial $A_{n}(f(x))$ from the polynomial $A_{n-1}(f(x))$

\begin{lemma}\label{lemma_delta_opa}
Take $f(x)\in\hurw_{R}[[x]]$. Then
\begin{equation}
A_{n+1}(f(x))=f(x)\delta(A_{n}(f(x))),
\end{equation}
con $n\geq1$.
\end{lemma}
\begin{proof}
By definition $A_{2}(f)=f\delta_{}(f)$. Now suppose by induction hypothesis that
$A_{n}(f)=f\delta(A_{n-1}(f))$. Put $A_{n}\equiv A_{n}(f)$. On the one hand, we have
\begin{eqnarray*}
A_{n+1}&=&\delta_{}fA_{n}+\delta_{}^{2}f\{A_{1}A_{n-1}+A_{2}A_{n-2}+\cdots\}+\cdots\\
&=&\delta_{}(f)f\delta(A_{n-1})+\delta_{}^{2}(f)\{ff\delta(A_{n-2})+f\delta(A_{1})A_{n-2}+\cdots\}+\cdots\\
&=&f[\delta_{}(f)\delta(A_{n-1})+\delta_{}^{2}(f)\{f\delta(A_{n-2})+\delta(A_{1})A_{n-2}+\cdots\}+\cdots]
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
f\delta(A_{n})&=&f\delta(\delta_{}(f)A_{n-1}+\delta_{}(f)\{A_{1}A_{n-2}+A_{2}A_{n-3}+\cdots\}+\cdots)\\
&=&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]\\
&=&f[\delta_{}(f)\delta(A_{n-1})+\delta_{}^{2}(f)\{f\delta(A_{n-2})+\delta(A_{1})A_{n-2}+\cdots\}+\cdots]
\end{eqnarray*}
Now by comparing the results we obtain $A_{n+1}=f\delta(A_{n})$.
\end{proof}

From the above we obtain the following corollary

\begin{corollary}\label{coro_delta_aut}
Take $f(x)\in\hurw_{R}[[x]]$. Then
\begin{equation}
A_{n}(f(x))=\overbrace{f(x)\delta(\cdots f(x)\delta(f(x)))}^{n}.
\end{equation}
\end{corollary}

Now we will define operations in $\Opa(\hurw_{R}[[x]])$ that allow to give it a ring structure.

\begin{definition}
We define the sum $\boxplus$ in $\Opa(\hurw_{R}[[x]])$ as follows
\begin{equation}
\Opa(f)\boxplus\Opa(g)=(B_{n}(f,g))_{n\in\N}
\end{equation}
where
\begin{eqnarray}
B_{0}(f,g)&=&x\\
B_{1}(f,g)&=&A_{1}(f)+A_{2}(g)\\
B_{n}(f,g)&=&A_{n}(f)+A_{n}(g)+H_{n}(f,g),\ \ \ n\geq2
\end{eqnarray}
where $H_{n}(f,g)$ is a polynomial in the variables $f,f^{\prime},...,f^{(n-1)}$ and
$ g,g^{\prime},...,g^{(n-1)}$ fulfilling
\begin{equation}\label{eqn_Hn}
H_{n+1}(f,g)=f\delta A_{n}(g)+g\delta A_{n}(f)+(f+g)\delta H_{n}(f,g)
\end{equation}
\end{definition}

\begin{lemma}\label{lemma_ope_+}
Take $f(x),g(x)\in\hurw_{R}[[x]]$. Then
\begin{equation}
\Opa(f(x))\boxplus\Opa(g(x))=\Opa(f(x)+g(x))
\end{equation}
\end{lemma}
\begin{proof}
Finding the autonomous polynomials for the sum $ f + g $ we obtain
\begin{equation}
A_{n}(f+g)=A_{n}(f)+A_{n}(g)+H_{n}(f,g).
\end{equation}
where $H_{n}(f,g)$ is a polynomial in the variables $f,f^{\prime},...,f^{(n-1)}$ and
$g,g^{\prime},...,g^{(n-1)}$. If $H_{n}(f,g)$ satisfies (\ref{eqn_Hn}) and if we use the Lemma \ref{lemma_delta_opa}, we have
\begin{eqnarray*}
A_{n+1}(f+g)&=&(f+g)\delta A_{n}(f+g)\\
&=&(f+g)\delta[A_{n}(f)+A_{n}(g)+H_{n}(f,g)]\\
&=&f\delta A_{n}(f)+g\delta A_{n}(g)+f\delta A_{n}(g)\\
&&+g\delta A_{n}(f)+(f+g)H_{n}(f,g)\\
&=&A_{n+1}(f)+A_{n+1}(g)+H_{n+1}(f,g)
\end{eqnarray*}
Then our affirmation is followed.
\end{proof}

We want to find the additive inverse of the element $\Opa(f)$ in $\Opa(\hurw_{R}[[x]])$. For this we have first the definition

\begin{definition}
Take $a\in R-\{0\}$. Let $\exp(a)$ denote the expansion in non-negative integer powers from $a$
\begin{equation}
\exp(a)=(1,a,a^{2},a^{3},...).
\end{equation}
For $a=0$ define
\begin{equation}
\exp(0)=(1,0,0,0,...).
\end{equation}
\end{definition}

The reason for denoting the expansion in non-negative integer powers from $a$ as $\exp(a)$ is because
$\rho_{x}\exp(a)=e^{ax}$ is the exponential function knowed. The following result tells us that the set of all expansions $\exp(a)$ form a commutative ring with unity.

\begin{theorem}
Let $\exp(R)$ denote the set of all expansions $\exp(a)$, $a\in R$. Then
$(\exp(R),\ast,\cdot)$ is a commutative ring with unity $\exp(1)$ and componentwise product.
\end{theorem}
\begin{proof}
It is easy to see that $\exp(a)\ast\exp(b)=\exp(a+b)$. Then the operation $\ast$ is closed in
$\exp(R)$. It is also clear that $(\exp(R),\ast)$ is a commutative group with unit $\exp(0)$ and
inverse elements of the form $\exp(-a)$. On the other hand, $\exp(a)\cdot\exp(b)=\exp(ab)$ and the
operation $\cdot$ is closed in $\exp(R)$. Then $(\exp(R),\cdot)$ is a commutative monoid with unity.
Finally
\begin{eqnarray*}
\exp(a)\cdot(\exp(b)\ast\exp(c))&=&(\exp(a)\cdot\exp(b))\ast(\exp(a)\cdot\exp(c))
\end{eqnarray*}
and $\exp(R)$ is a commutative ring.
\end{proof}

As we will notice below, the ring $\exp(R)$ acts on the set $\Opa(\hurw_{R}[[x]])$

\begin{proposition}\label{prop_mult_esc}
Take $f\in\hurw_{R}[[x]]$ and $a \in R$. Then
\begin{equation}
\exp(a)\cdot\Opa(f)=\Opa(af).
\end{equation}
where the product $\exp(a)\cdot\Opa(f)$ is componentwise.
\end{proposition}
\begin{proof}
Let $a\neq0$. We will prove that $A_{n}(af)=a^{n}A_{n}(f)$. The proof will be by induction. For $n=0$, it is trivial. It is true for $n=1$, since $A_{1}(af)=af=aA_{1}(f)$. Now suppose that our statement holds for every number smaller than $n$ and we will consider
the case $n+1$. We have
\begin{eqnarray*}
A_{n+1}(af)&=&\sum_{k=1}^{n}\sum_{\vert p(n)\vert=k}\dfrac{n!}{j_{1}!\cdots j_{n}!}\left[\frac{aA_{1}}{1!}\right]^{j_{1}}\cdots\left[\frac{a^{n} A_{n}}{n!}\right]^{j_{n}}a\delta^{k}f\\
&=&\sum_{k=1}^{n}\sum_{\vert p(n)\vert=k}\frac{n!a^{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\\
&=&a^{n+1}\sum_{k=1}^{n}\sum_{\vert p(n)\vert=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\\
&=&a^{n+1}A_{n+1}(f)
\end{eqnarray*}
Now set $a=0$. Then
$$\Opa(0)=\Opa(0f)=\exp(0)\Opa(f)$$
and our statement turns out to be true.
\end{proof}

With everything stated above, we are ready to show that $\Opa(\hurw_{R}[[x]])$ is a commutative
group with sum $\boxplus$ and inverse elements $\exp(-1)\cdot\Opa(f)$

\begin{theorem}
$(\Opa(\hurw_{R}[[x]]),\boxplus)$ is a commutative group.
\end{theorem}
\begin{proof}
From Lemma \ref{lemma_ope_+}, $\Opa(f)\boxplus\Opa(g)\in\Opa(\hurw_{R}[[x]])$. Now
take $0\in\hurw_{R}[[x]]$. So it's easy to see that $\Opa(0)=(x,0,0,...)=\textbf{0}_{\Opa}$ and by the Lemma above
\begin{eqnarray*}
\Opa(f)\boxplus\textbf{0}_{\Opa}&=&\Opa(f)\boxplus\Opa(0)\\
&=&\Opa(f+0)\\
&=&\Opa(f)
\end{eqnarray*}
then $\textbf{0}_{\Opa}$ is the neutral element in $(\Opa(\hurw_{R}[[x]]),\boxplus)$. By the Proposition above it is easy to show that $\exp(-1)\Opa(f)$ is the additive inverse of $\Opa(f)$.
Finally the associativity and commutativity of $(\Opa(\hurw_{R}[[x]]),\boxplus)$ follow from those of $(\hurw_{R}[[x]],+)$.
\end{proof}

From the above it follows that the map $\lambda:(\Opa(\hurw_{R}[[x]]),\boxplus)\rightarrow(\hurw_{R}[[x]],+)$ is a group isomorphism. Next we will define a product in $\Opa(\hurw_{R}[[x]])$

\begin{definition}
We define the product $\odot$ in $\Opa(\hurw_{R}[[x]])$ as follows
\begin{equation}
\Opa(f)\odot\Opa(g)=\left(C_{n}(f,g)\right)_{n\in\N}
\end{equation}
with
\begin{eqnarray}
C_{0}(f,g)&=&x\\
C_{n}(f,g)&=&\sum_{l=1}^{n}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+1}(g), \ \ n\geq1
\end{eqnarray}
where
\begin{equation}
\mathcal{A}_{l}(f)=\sum_{\vert p(n)\vert=l}\alpha_{\vert p(n)\vert} A_{j_{1}}(f)A_{j_{2}}(f)\cdots A_{j_{r}}(f)
\end{equation}
$p(n)=j_{1}+j_{2}+\cdots+j_{r}$ and the $\alpha_{\vert p(n)\vert}$ are suitable numbers.
\end{definition}

As noted, performing multiplications on $\Opa(\hurw_{R}[[x]])$ seems very hard with the definition above. However, the following lemma will allow us an easier way

\begin{lemma}\label{lemma_ope_prod}
$\Opa(f)\odot\Opa(g)=\Opa(fg)$ for all $f,g\in\hurw_{R}[[x]]$.
\end{lemma}
\begin{proof}
Set $A_{n}(f)\equiv A_{n}(f(x))$. Then
\begin{eqnarray*}
A_{1}(fg)&=&fg=A_{1}(f)A_{1}(g),\\
A_{2}(fg)&=&(ff^{\prime})(g^{2})+(f^{2})(gg^{\prime})=A_{2}(f)A_{1}^{2}(g)+A_{1}^{2}(f)A_{2}(g),\\
A_{3}(fg)&=&f^{3}(g(g^{\prime})^{2}+g^{2}g^{\prime\prime})+4f^{2}f^{\prime}g^{2}g^{\prime}+
(f(f^{\prime})^{2}+f^{2}f^{\prime\prime})g^{3}\\
&=&A_{1}^{3}(f)A_{3}(g)+4A_{1}(f)A_{2}(f)A_{1}(g)A_{2}(g)+A_{1}^{3}(g)A_{3}(f).
\end{eqnarray*}
Assume by induction that
\begin{equation}
A_{n}(fg)=\sum_{l=1}^{n}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+1}(g).
\end{equation}
By the Lemma \ref{lemma_delta_opa}, $A_{n+1}(fg)=fg\delta A_{n}(fg)$. So using the induction hypothesis
\begin{eqnarray*}
A_{n+1}(fg)&=&fg\delta A_{n}(fg)\\
&=&fg\delta\left(\sum_{l=1}^{n}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+1}(g)\right)\\
&=&fg\delta\left(\mathcal{A}_{1}(f)\mathcal{A}_{n}(g)+\sum_{l=2}^{n-1}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+1}(g)+\mathcal{A}_{n}(f)\mathcal{A}_{n}(1)\right)\\
&=&f\delta\mathcal{A}_{1}(f)g\mathcal{A}_{n}(g)+f\mathcal{A}_{1}(f)g\delta\mathcal{A}_{n}(g)\\
&&+\sum_{l=2}^{n-1}[f\delta\mathcal{A}_{l}(f)g\mathcal{A}_{n-l+1}(g)+f\mathcal{A}_{l}(f)g\delta\mathcal{A}_{n-l+1}(g)]\\
&&+f\delta\mathcal{A}_{n}(f)g\mathcal{A}_{1}(g)+f\mathcal{A}_{n}(f)g\delta\mathcal{A}_{1}(g)
\end{eqnarray*}
Again we will use the Lemma \ref{lemma_delta_opa} and also that $A_{1}(f)=f$, $A_{1}(g)=g$. We have
\begin{eqnarray*}
f\delta\mathcal{A}_{1}(f)&=&f\delta A_{n}(f)=A_{n+1}(f)\\
f\delta\mathcal{A}_{n}(f)&=&f\delta A_{1}^{n}(f)=nA_{1}^{n-1}(f)f\delta A_{1}(f)=nA_{1}^{n-1}(f)A_{2}(f)\\
f\mathcal{A}_{l}(f)&=&A_{1}(f)\mathcal{A}_{l}(f),\ \ \ 1\leq l\leq n
\end{eqnarray*}
and
\begin{eqnarray*}
f\delta\mathcal{A}_{l}(f)&=&f\sum_{\vert p(n)\vert=l}\alpha_{\vert p(n)\vert}\sum_{k=1}^{l}A_{j_{1}}(f)\cdots \delta A_{j_{k}}(f)\cdots A_{j_{l}}(f)\\
&=&\sum_{\vert p(n)\vert=l}\alpha_{\vert p(n)\vert}\sum_{k=1}^{l}A_{j_{1}}(f)\cdots f\delta A_{j_{k}}(f)\cdots A_{j_{l}}(f)\\
&=&\sum_{\vert p(n)\vert=l}\alpha_{\vert p(n)\vert}\sum_{k=1}^{l}A_{j_{1}}(f)\cdots A_{j_{k}+1}(f)\cdots A_{j_{l}}(f).
\end{eqnarray*}
Likewise, $g\delta\mathcal{A}_{n-l+1}(g)$ is obtained simply by replacing $f$ by $g$ and by noting
that $n-l+1$ takes the same values as $l$ when it ranging between $1$ and $n$. Further we note that $A_{1}(f)\mathcal{A}_{l}(f)$ and $f\delta\mathcal{A}_{l}(f)$ are contained in $\mathcal{A}_{l+1}(f)$ and that $A_{1}(g)\mathcal{A}_{n-l+1}(g)$ and $g\delta\mathcal{A}_{n-l+1}(g)$ are contained in
$\mathcal{A}_{n-l+2}(g)$ for $l$ ranging between $2$ and $n-1$. Finally we put all of the above together to obtain
\begin{eqnarray*}
A_{n+1}(fg)&=&A_{n+1}(f)A_{1}^{n+1}(g)+\sum_{l=2}^{n}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+2}(g)+
A_{n+1}(g)A_{1}^{n+1}(f)\\
&=&\mathcal{A}_{1}(f)\mathcal{A}_{n+1}(g)+\sum_{l=2}^{n}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+2}(g)+
\mathcal{A}_{1}(g)\mathcal{A}_{n+1}(l)\\
&=&\sum_{l=1}^{n+1}\mathcal{A}_{l}(f)\mathcal{A}_{n-l+2}(g)
\end{eqnarray*}
just as we wanted to show.
\end{proof}

Now we will show that $\Opa(\hurw_{R}[[x]])$ is a ring

\begin{theorem}
$(\Opa(\hurw_{R}[[x]]),\boxplus,\odot)$ is a commutative ring with neutral element
$\textbf{0}_{\Opa}=(x,0,0,...)$ and unity $\textbf{1}_{\Opa}=(x,1,0,0,...)$.
This ring will be called \textbf{autonomous ring}
\end{theorem}
\begin{proof}
It was already shown that $(\Opa(\hurw_{R}[[x]]),\boxplus)$ is a commutative group. By the Lemma \ref{lemma_ope_prod}, $\Opa(f(x))\odot\Opa(g(x))$ is contained in $\Opa(\hurw_{R}[[x]])$. Also $\textbf{1}_{\Opa}$ is a unit with respect to $\odot$ in
$\Opa(\hurw_{R}[[x]])$, since $\Opa(1)=\textbf{1}_{\Opa}$ and
\begin{eqnarray*}
\textbf{1}_{\Opa}\odot\Opa(f(x))&=&\Opa(1\cdot f(x))\\
&=&\Opa(f(x)).
\end{eqnarray*}
The associativity follows from the associativity of $\cdot$ in $\hurw_{R}[[x]]$. Finally
\begin{eqnarray*}
\Opa(f)\odot(\Opa(g)\boxplus\Opa(h))&=&\Opa(f)\odot\Opa(g+h)\\
&=&\Opa(f(g+h))\\
&=&\Opa(fg+fh))\\
&=&\Opa(fg)\boxplus\Opa(fh)\\
&=&[\Opa(f)\odot\Opa(g)]\boxplus[\Opa(f)\odot\Opa(h)]
\end{eqnarray*}
Then $(\Opa(\hurw_{R}[[x]]),\boxplus,\odot)$ is a ring with unity $\textbf{1}_{\Opa}$.
\end{proof}

From the above it follows that the map $\lambda$ is an isormorphism of the ring $(\hurw_{R}[[x]],+,\cdot)$ in the ring $(\Opa(\hurw_{R}[[x]]),\boxplus,\odot)$. With the following result we will show how to obtain the inverse elements in $\Opa(\hurw_{R}[[x]])$

\begin{theorem}
$(\Opa(\hurw_{R}[[x]]^{*}),\odot)$ is a commutative group.
\end{theorem}
\begin{proof}
We have that
\begin{eqnarray*}
\textbf{1}_{\Opa}&=&\Opa(1)\\
&=&\Opa(f(x)f^{-1}(x))\\
&=&\Opa(f(x))\odot\Opa(f^{-1}(x)).
\end{eqnarray*}
Then $\Opa(f(x))$ has an inverse with respect to $\odot$ and
$$\Opa(f(x))^{-1}=\Opa(f^{-1}(x)).$$
Thus $(\Opa(\hurw_{R}[[x]]*),\odot)$ is a group.
\end{proof}

Now it will be shown that $\Opa(\hurw_{R}[[x]])$ is an $\exp(R)$-algebra

\begin{theorem}\label{theo_exp}
$\Opa(\hurw_{R}[[x]])$ is an $\exp(R)$-algebra.
\end{theorem}
\begin{proof}
It is should show that
\begin{eqnarray}
\exp(a)[\Opa(f)\boxplus\Opa(g)]&=&\exp(a)\Opa(f)\boxplus\exp(a)\Opa(g)]\\
\exp(a)[\Opa(f)\odot\Opa(g)]&=&[\exp(a)\Opa(f)]\odot\Opa(g)=\Opa(f)\odot[\exp(a)\Opa(g)]
\end{eqnarray}
and this follows directly from $\Opa(a(f+g))$ and $\Opa(afg)$ along with Proposition \ref{prop_mult_esc}.
\end{proof}

In the following theorems we will discuss the structure of the autonomous ring.

\begin{theorem}\label{theo_delta_domi}
The ring $\Opa(\hurw_{R}[[x]])$ is an integral domain.
\end{theorem}
\begin{proof}
We know that $\hurw_{R}[[x]]$ is domain when $R$ is also domain. Now take $f$, $g$ from
$\hurw_{R}[[x]]$. If $\Opa(f)\odot\Opa(g)=\Opa(fg)=\textbf{0}_{\Opa}$, then $fg=0$. Then either
$f=0$ or $g=0$. Thus either $\Opa(f)=\textbf{0}_{\Opa}$ or $\Opa(g)=\textbf{0}_{\Opa}$ and
$\Opa(\hurw_{R}[[x]])$ is a domain as well.
\end{proof}

Let $I$ be an ideal in $R$. From [2] we use the following notation for ideals in $\hurw_{R}[[x]]$:

\begin{eqnarray}
I+\left\langle x\right\rangle&=&\left\{f(x)=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}:a_{0}\in I\right\}\\
H_{I}[[x]]&=&\left\{f(x)=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}:a_{n}\in I\right\}
\end{eqnarray}

Let $\chi:\Opa(\hurw_{R}[[x]])\rightarrow\hurw_{R}[[x]]$ be a homomorphism defined by
$\chi(\Opa(f(x)))=A_{1}[f(x)]=f(x)$. We have

\begin{proposition}\label{propo_iso}
Let $I$ be an ideal in $R$. Then
\begin{enumerate}
\item $\Opa(\hurw_{R}[[x]])/\chi^{-1}(I+\left\langle x\right\rangle)\simeq\hurw_{R}[[x]]/(I+\left\langle x\right\rangle)\simeq R/I$.
\item $\Opa(\hurw_{R}[[x]])/\Opa(\hurw_{I}[[x]])\simeq\Opa(\hurw_{R}[[x]]/\hurw_{I}[[x]])$.
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}
\item Let $J=I+\left\langle x\right\rangle$. Define the map $\psi:\Opa(\hurw_{R}[[x]])\rightarrow\hurw_{R}[[x]]/J$ by $\psi=\tau\circ\chi$, where $\tau:\hurw_{R}[[x]]\rightarrow\hurw_{R}[[x]]/J$
is the canonical map. Then $\psi$ is an overjective homomorphism with $Ker(\psi)=\phi^{-1}(J)$
and hence $\Opa(\hurw_{R}[[x]])/\chi^{-1}(J)\simeq\hurw_{R}[[x]]/J$. The remainder is a known result.
\item Since $\hurw_{R}[[x]]/\hurw_{I}[[x]]\simeq\hurw_{R/I}[[x]]$, then
$\Opa(\hurw_{R}[[x]]/\hurw_{I}[[x]])\simeq\Opa(\hurw_{R/I}[[x]])$. Define the map
$\sigma:\Opa(\hurw_{R}[[x]])\rightarrow\Opa(\hurw_{R/I}[[x]])$ by
$\sigma(\Opa(f))=\Opa(\overline{f})$, where
$\overline{f(x)}=\sum_{n=0}^{\infty}\overline{a}_{n}\frac{x^{n}}{n!}$. Then $\sigma$ is an
overjective homomorphism with $Ker(\sigma)=\Opa(\hurw_{I}[[x]])$. Then
$\Opa(\hurw_{R}[[x]])/\Opa(\hurw_{I}[[x]])\simeq\Opa(\hurw_{R/I}[[x]])$.
\end{enumerate}
\end{proof}

\begin{theorem}\label{theo_maxi}
If $M_{\alpha}$ is the set of maximal ideals of $\hurw_{R}[[x]]$, then
$\Opa(M_{\alpha})$ is the set of maximal ideals of $\Opa(\hurw_{R}[[x]])$.
\end{theorem}
\begin{proof}
On the one hand, if $M$ is a maximal ideal of $\hurw_{R}[[x]]$, then $\Opa(M)=\chi^{-1}(M)$
is a maximal ideal of $\Opa(\hurw_{R}[[x]])$. On the other hand, let $\Opa(M)$ denote the maximal ideal of $\Opa(\hurw_{R}[[x]])$, then $\chi(\Opa(M))$ is an ideal of
$\hurw_{R}[[x]]$. Take $\Opa(h)\in\Opa(\hurw_{R}[[x]])\setminus\Opa(M)$. Then
$(\Opa(M),\Opa(h))=\Opa(\hurw_{R}[[x]])$ and therefore there exists $\Opa(f)\in\Opa(M)$,
$\Opa(g)\in\Opa(\hurw_{R}[[x]])$ such that $\textbf{1}_{\Opa}=\Opa(g)\odot\Opa(h)\boxplus\Opa(f)=\Opa(gh+f)=\Opa(1)$. Thus $1=gh+f\in(\chi(\Opa(M)),h)$ and it follows that $\chi(\Opa(M))$ is a maximal ideal of $\hurw_{R}[[x]]$.
\end{proof}

\begin{theorem}\label{the_prime}
If $P+\left\langle x\right\rangle$ and $\hurw_{P}[[x]]$ are prime ideals in $\hurw_{R}[[x]]$, then
$\Opa(P+\left\langle x\right\rangle)$ and $\Opa(\hurw_{P}[[x]])$ are prime ideals in
$\Opa(\hurw_{R}[[x]])$.
\end{theorem}
\begin{proof}
If $\Opa(f)\odot\Opa(g)=\Opa(fg)\in\Opa(P+\left\langle x\right\rangle)$, then
$fg\in P+\left\langle x\right\rangle$ and therefore either $f\in P+\left\langle x\right\rangle$ or
$g\in P+\left\langle x\right\rangle$. Then either$\Opa(f)\in\Opa(P+\left\langle x\right\rangle)$
or $\Opa(g)\in\Opa(P+\left\langle x\right\rangle)$. Proof for the second part is the same
\end{proof}

The simplest autonomous ring arises when $R$ is a field. We denote this field by $\F$

\begin{theorem}\label{theo_ideal_princi}
Let $\F$ be a field. The ring $\Opa(\hurw_{\F}[[x]])$ is a domain of principal ideals being of the form
\begin{equation}
\left\langle\Opa(x^{n})\right\rangle=\Opa(x^{n})\odot\Opa(\hurw_{\F}[[x]])
\end{equation}
for all $n\geq1$ and with only ideal maximal
\begin{equation}
\left\langle\Opa(x)\right\rangle=\Opa(x)\odot\Opa(\hurw_{\F}[[x]])
\end{equation}
\end{theorem}
\begin{proof}
It is a known fact that if $\F$ is a field, then $(\hurw_{\F}[[x]],+,\cdot)$ is a domain
of prime ideals with ideals $\left\langle x^{n}\right\rangle$, $n\geq1$. Since $\lambda$  is an
is an isomorphism of $(\hurw_{\F}[[x]],+,\cdot)$ into $(\Opa(\hurw_{\F}[[x]]),\boxplus,\odot)$,
then it follows that $\left\langle\Opa(x^{n})\right\rangle$ is the set of ideals in
$\Opa(\hurw_{\F}[[x]])$ and likewise it is proved that $\left\langle\Opa(x)\right\rangle$ is its maximal ideal.
\end{proof}

\begin{theorem}\label{theo_campo}
Let $\F$ be a field. The set $\Opa[\hurw_{\F}((x))]=\Opa(\hurw_{\F}[[x]])/\left\langle\Opa(x)\right\rangle$ is a field.
\end{theorem}
\begin{proof}
Since $\left\langle\Opa(x)\right\rangle$ is maximal, it follows by the Proposition \ref{propo_iso}
that $\Opa[\hurw_{\F}((x))]$ is a field.
\end{proof}

We end this section by defining polynomials in $\Opa(\hurw_{R}[[x]])$. We begin by defining
the monomials
\begin{equation}
\Opa(x)_{\odot}^{k}=\Opa(x)\odot\Opa(x)\odot\cdots\odot\Opa(x)
\end{equation}
and then we have the
\begin{definition}
A polynomial of degree $n$ in $\Opa(\hurw_{R}[[x]])$ is one of the form
\begin{equation}
\bbox_{k=0}^{n}\exp(a_{k})\Opa(x)_{\odot}^{k}
\end{equation}
\end{definition}
Since
\begin{equation}
\bbox_{k=0}^{n}\exp(a_{k})\Opa(x)_{\odot}^{k}=\bbox_{k=0}^{n}\Opa(a_{k}x^{k})=\Opa\left(\sum_{k=0}^{n}a_{k}x^{k}\right),
\end{equation}
then a polynomial in $\Opa(\hurw_{R}[[x]])$ is the image for $\lambda$ of the polynomial
$\sum_{k=0}^{n}a_{k}x^{k}$. Then the ring $R[x]$ is mapped to $\Opa(R[x])$. If the polynomial $f(x)$
has factorization $f(x)=\prod_{i=1}^{n}(x-a_{i})$, then
\begin{eqnarray*}
\Opa(f(x))&=&\Opa\left(\prod_{i=1}^{n}(x-a_{i})\right)\\
&=&\bcast_{i=1}^{n}\Opa(x-a_{i})\\
&=&\bcast_{i=1}^{n}(x,x-a_{i},x-a_{i},...)\\
&=&\bcast_{i=1}^{n}[(x-a_{i},x-a_{i},x-a_{i},...)+(a_{i},0,0,...)]\\
&=&\bcast_{i=1}^{n}[(x-a_{i})\textbf{1}+(a_{i},0,0,...)]
\end{eqnarray*}

where $\textbf{1}=(1,1,1,...)$. Then the polynomial $\Opa(f(x))$ factors into
$\Opa(R[x])$.

Finally, let $\epsilon_{a}:R[x]\rightarrow R$ be the evaluation map given by
$\epsilon_{a}(f(x))=f(a)$. Now extend $\epsilon_{a}$ to $\Opa(R[x])$ by defining the map
$\overline{\epsilon_{a}}:\Opa(R[x])\rightarrow\hurw_{R}$ by
$\overline{\epsilon_{a}}(\Opa(f(x))=\Opa(\epsilon_{a}f(x))=\Opa(f(a))$. Denote the ring
$(Im(\overline{\epsilon_{a}}),\boxplus,\odot)$ the image ring of $\Opa(R[x])$ by the map
$\overline{\epsilon_{a}}$. Then $\overline{\epsilon_{a}}$ is a homorphism of rings
with kernel $Ker(\overline{\epsilon_{a}})=\left\langle\Opa(x-a)\right\rangle$. Thus
$\Opa(R[x])/\left\langle\Opa(x-a)\right\rangle\simeq Im(\overline{\epsilon_{a}})$.

\section{One-dimensional flow over domains}

A one-dimensional continuous dynamical system is a tuple $(S,X,\phi)$ in which $S\subseteq\R$ is
the set of times, $X\subseteq\R$ is the phase space and $\phi:S\times X\rightarrow X$ is
the system flow satisfying

\begin{enumerate}
\item $\phi(0,x)=x$
\item $\phi(t,\phi(s,x))=\phi(t+s,x)$
\end{enumerate}

When $S=R$ and putting $\phi_{t}(x)=\phi(t,x)$ we notice that $\phi_{t}$ is a group acting on
the phase space $X$. When we put $S=\Z$, the dynamical system is referred to as a
discrete dynamical system. In this section we define a dynamical system with $S=R$ a ring,
$X=\hurw_{T}[[t]]$ where $T$ is some ring with coefficients in $\hurw_{R}[[x]]$.

From [6] we know that the solution $\phi$ of (\ref{autonoma}) is given by
\begin{equation}
\phi(t,x)=x+\sum_{n=1}^{\infty}A_{n}(f(x))\dfrac{t^{n}}{n!}.
\end{equation}

This solution is a flow defined on $f(x)$. Then we can notice that the operator $\Opa$ maps $f(x)$ to the action group $\{\phi_{t}:t\in\R\}$.  With this in mind we have the following definition

\begin{definition}
We define a \textbf{dynamic system} over the ring $R$ as the tuple $(R,\hurw_{S}[[t]],\Phi)$ where
$R$ is the \textbf{set of times}, $\hurw_{S}[[t]]$ the \textbf{phase space} and $\Phi$ is the \textbf{flow}.
\begin{equation}\label{flujo_diferencial}
\Phi(t,x,f(x))=x+\sum_{n=1}^{\infty}A_{n}(f(x))\dfrac{t^{n}}{n!},
\end{equation}
that is, $\Phi$ is the map
$\Phi:R\times\hurw_{S}[[t]]\times\hurw_{R}[[x]]\rightarrow\hurw_{S}[[t]]$ where $S=\hurw_{R}[[x]]$.
\end{definition}

\begin{theorem}
$\Phi(t,x,f(x))$ is a flow.
\end{theorem}
\begin{proof}
When $t=0$ we have from (\ref{flujo_diferencial}) that $\Phi(0,x,f(x))=x$. We will now prove property 2) of a flow. On the one hand, the Taylor expansion of $\Phi(s+t,x,f(x))$ is $\sum_{n=0}^{\infty}\delta_{s}^{n}\Phi(s,x,f(x))\frac{t^{n}}{n!}$. On the other hand, making $\Phi_{s}=\Phi(s,x,f(x))$ we have
\begin{eqnarray*}
\Phi(t,\Phi(s,x,f(x)),f(x))&=&\Phi_{s}+\sum_{n=1}^{\infty}A_{n}(f(\Phi_{s}))\frac{t^{n}}{n!}\\
&=&\Phi_{s}+\sum_{n=1}^{\infty}\delta_{s}^{n}\Phi(s,x,f(x))\frac{t^{n}}{n!}\\
&=&\sum_{n=0}^{\infty}\delta_{s}^{n}\Phi(s,x,f(x))\frac{t^{n}}{n!}\\
&=&\Phi(s+t,x,f(x)).
\end{eqnarray*}
Then $\Phi(t,x,f(x))$ is a flow as stated.
\end{proof}

Clearly $\{\Phi_{t}:t\in R\}$, with $\Phi_{t}=\Phi(t,x,f(x))$, is a group acting on $\hurw_{S}[[t]]$. Let us define two derivatives on $\hurw_{S}[[t]]$, $\delta_{t}$ acting on the variable $t$ and $\delta_{x}=\delta$ acting on the variable $x$. We will use the Lemma \ref{lemma_delta_opa} to show a beautiful relation between $\delta_{t}Phi$ and $\delta_{x}\Phi$.

\begin{theorem}
Take $f(x)\in\hurw_{R}[[x]]$. Then
\begin{equation}
f(x)\delta_{x}\Phi(t,x,f(x))=\delta_{t}\Phi(t,x,f(x))=f(\Phi)
\end{equation}
\end{theorem}
\begin{proof}
From (\ref{flujo_diferencial}) and the Lemma \ref{lemma_delta_opa} we have
\begin{eqnarray*}
f(x)\delta_{x}\Phi(t,x,f(x))&=&f(x)\delta_{x}\left(x+\sum_{n=1}^{\infty}A_{n}(f(x))\dfrac{t^{n}}{n!}\right)\\
&=&f(x)+\sum_{n=1}^{\infty}f(x)\delta_{x}\left(A_{n}(f(x))\right)\dfrac{t^{n}}{n!}\\
&=&A_{1}([f(x)])+\sum_{n=1}^{\infty}A_{n+1}(f(x))\dfrac{t^{n}}{n!}\\
&=&\sum_{n=0}^{\infty}A_{n+1}(f(x))\dfrac{t^{n}}{n!}\\
&=&\delta_{t}\Phi(t,x,f(x)).
\end{eqnarray*}
\end{proof}

Next we will prove that the flows of $\delta_{t}\Phi=f(\Phi)$ and of $\delta_{t}Phi=af(\Phi)$,
with $a$in $R$, have matching trajectories. That is,

\begin{theorem}
For all $a$in $R$ it is fulfilled that $\Phi(t,x,af(x))=\Phi(at,x,f(x))$
\end{theorem}
\begin{proof}
From the Proposition \ref{prop_mult_esc} we know that $\Opa(af(x))=\exp(a)\Opa(f(x))$ for all
$a\in R$. For $a\neq0$ we have
\begin{eqnarray*}
\Phi(t,x,af(x))&=&x+\sum_{n=1}^{\infty}a^{n}A_{n}(f(x))\frac{t^{n}}{n!}\\
&=&x+\sum_{n=1}^{\infty}A_{n}(f(x))\frac{(at)^{n}}{n!}\\
&=&\Phi(at,x,f(x)).
\end{eqnarray*}
When $a=0$, $\Phi(t,x,0)=\Phi(0,x,f(x))=x$.
\end{proof}

The above theorem means that $\rho_{t}$ maps the class $\exp(R)\Opa(f)$ to the flow
$\Phi(t,x,f)$ for all $t\in R$. This is so because
$$\Phi(t,x,Rf(x))=\Phi(tR,x,f(x))=\Phi(R,x,f(x))$$
where $Rf(x)=\{af(x):a\in R\}$ is the set of scalar multiples of $f$.

When $I$ is a ideal in $R$, we have
$$\Phi(I,x,Rf(x))=\Phi(RI,x,f(x))=\Phi(I,x,f(x))$$

Now fix an $x$ in $R$ and $f\in\hurw_{R}[[x]]$. Define the homomorphism of groups
$\sigma:R\rightarrow\Phi_{R}(x,f)$ by $\sigma(a)=\Phi_{a}(x,f)$, where we make $\Phi_{t}(x,f)\equiv\Phi(t,x,f)$. We want to extend the homomorphism $\sigma$ to a homomorphism of $R$-modules by showing that $\Phi_{R}(x,f)$ has precisely $R$-module structure.
By the group property of $\Phi_{R}(x,f)$, for all $n\in\N$  we have
$$\Phi_{nt}(x,f)=\Phi_{t}(x,f)\circ\Phi_{t}(x,f)\circ\cdots\circ\Phi_{t}(x,f)$$
Then we compose the flow $\Phi_{t}(x,f)$ with itself $n$ times and we can define
$$n\star\Phi_{t}(x,f)=\Phi_{t}(x,f)\circ\Phi_{t}(x,f)\circ\cdots\circ\Phi_{t}(x,f).$$

In this way we can define the product $\star:R\times\Phi_{R}(x,f)\rightarrow\Phi_{R}(x,f)$ by
$$a\star\Phi_{t}(x,f)=\Phi_{at}(x,f)$$
for all $a\in R$ and it is very easy to observe that with the product $\star$ the subgroup $\Phi_{R}(x,f)$ adopts $R$-module structure and $\sigma$ becomes a homomorphism of $R$-module, since the ring $R$ is an $R$-module in itself. Now we want to understand under which conditions of the fixed element $x$ the $\sigma$-homomorphism becomes an isomorphism of $R$-module isomorphism.

Now fix a polynomial $f(x)\in R[x]$ and extend the map $\overline{\epsilon_{a}}$ to the flow $\Phi_{t}(x,f)$ by defining
\begin{equation}
\overline{\epsilon_{a}}\Phi_{t}(x,f)=\overline{\epsilon_{a}}\left(x+\sum_{n=1}^{\infty}A_{n}(f(x))\frac{t^{n}}{n!}\right)=a+\sum_{n=1}^{\infty}A_{n}(f(a))\frac{t^{n}}{n!}
\end{equation}

and let
\begin{equation}
\Gamma_{a}:=\{\overline{\epsilon_{a}}\Phi_{t}(x,f):t\in R\}
\end{equation}
be the orbit or trajectory of $a$.

Then $\Gamma_{0}$ is the exponential generating function of the sequence
$\overline{\epsilon_{0}}\Opa(f(x))$. If $\Gamma_{x_{0}}=\{x_{0}\}$, then $x_{0}$ is an equilibrium
point for $\Phi_{R}(x,f)$. Equilibrium points are obtained
when $A_{n}(f(x_{0}))=0$ for $n\geq1$, i.e., when $f(x_{0})=0$ for some $x_{0}\in R$. If $x_{0}$
is not an equilibrium point, then it will be called a regular point of $x_{0}$ regular point of
$\Phi_{R}(x,f)$.

Denote
\begin{eqnarray*}
\ker(\sigma)&=&\{t\in R:\sigma(t)=x\}\\
&=&\{t\in R:\Phi_{t}(x,f)=x\}
\end{eqnarray*}
the kernel of $\sigma$. If $\overline{x}$ is an equilibrium point, then
$\sigma:R$ and $\ker(\sigma)=R$. If $x$ is a regular point, $\ker(\sigma)=R$ and $\sigma$ is injective. Since $\sigma$ is overjective, then $\sigma$ is an isomorphism of $R$-modules. We will say that the module $\Phi_{R}(x,f)$ is a trivial module if $x$ is an equilibrium point. Otherwise it would be called a nontrivial module.

Let $I$ be an ideal of $R$ and $x$ be a regular point. Then $\sigma(I)=\Phi_{I}(x,f)$ is a submodule in $\Phi_{R}(x,f)$. Then we can establish the following correspondence
\begin{eqnarray}
\{\text{Ideals $I$ in }R\}\Leftrightarrow\{\text{Submodules }\Phi_{I} \text{ of }\Phi_{R}(x,f)\}\Leftrightarrow\{y^{\prime}=If(y)\}
\end{eqnarray}
where $\{y^{\prime}=If(y)\}$ denotes the set of all autonomous differential equations $y^{\prime}=af(y)$, with $a\in I$.

We end this section with the following results
\begin{theorem}
A nontrivial $R$-module $\Phi_{R}(x,f)$ is a torsion-free cyclic module.
\end{theorem}
\begin{proof}
Suppose $x$ is a regular point. To show that $\Phi_{R}(x,f)$ is a cyclic module, it is sufficient
to take a unit $u$ in $R$. Then $R\star\Phi_{u}(x,f)=\Phi_{uR}(x,f)=\Phi_{R}(x,f)$ and so
$\Phi_{R}(x,f)$ is cyclic. Now we will show that $\Phi_{R}(x,f)$ is torsion free.
On the one hand, there exists an ideal $I\subset R$ such that $\Phi_{R}(x,f)$ is isomorphic to $R/I$.
Since it was already shown that $R$ and $\Phi_{R}(x,f)$ are isomorphic, then it follows that $I$ is
the zero ideal. On the other hand, $\Phi_{R}(x,f)$ is a cyclic $R$-module is equivalent to saying
that the multiplication homomorphism $\tau_{s}:R\rightarrow\Phi_{R}(x,f)$,
$\tau_{s}(a)=a\star\Phi_{s}(x,f)$, is a homomorphism of overjective $R$-modules. Write
$I=\ker(\tau_{s})$. By the first isomorphism theorem for module, $\overline{\tau_{s}}$ is
an isomorphism from $R/I$ to $\Phi_{R}(x,f)$. Since $\overline{\tau_{s}}$ is the annihilator
$\ann(\Phi_{s}(x,f))$ of $\Phi_{s}(x,f)$, then $\ann(\Phi_{s}(x, f))=I=0$ for any
$\Phi_{s}(x,f)\in\Phi_{R}(x,f)$ and $0$ is the only torsion element in $\Phi_{R}(x,f)$.
\end{proof}

\section{Ring of one-dimensional flows}

Define the map $\rho_{t}:\Opa(\hurw_{R}[[x]])\rightarrow\hurw_{S}[[t]]$, where $S=\hurw_{R}[[x]]$, as $\rho_{t}\Opa(f(x))=x+\sum_{n=1}^{\infty}A_{n}(f(x))\frac{t^{n}}{n!}$. Then
\begin{eqnarray}
\Phi(t,x,f(x))&=&\rho_{t}\Opa(f(x)).
\end{eqnarray}

Then it is possible to extend the operations of the ring $\Opa(\hurw_{R}[[x]])$ to the set
$\rho_{t}\Opa(\hurw_{R}[[x]])$. We have
\begin{definition}
Take $f(x)$ and $g(x)$ from $\hurw_{R}[[x]]$. We define the sum $\boxplus$ and the product
$\odot$ in $\rho_{t}\Opa(\hurw_{R}[[x]])$ as follows
\begin{eqnarray}
\rho_{t}\Opa(f)\boxplus\rho_{t}\Opa(g)&=&\rho_{t}[\Opa(f)\boxplus\Opa(g)]\\
\rho_{t}\Opa(f)\odot\rho_{t}\Opa(g)&=&\rho_{t}[\Opa(f)\odot\Opa(g)]
\end{eqnarray}
\end{definition}

Then we have the following result
\begin{theorem}
The set $\rho_{t}\Opa(\hurw_{R}[[x]])$ with sum $\boxplus$ and product $\odot$ is a commutative ring with units $\rho_{t}\Opa(0)=x$ and $\rho_{t}\Opa(1)=x+t$. The ring $\rho_{t}\Opa(\hurw_{R}[[x]])$ will be called \textbf{ring of one-dimensional flows}.
\end{theorem}
\begin{proof}
By the definition above the set $\rho_{t}\Opa(\hurw_{R}[[x]])$ inherits the ring properties of
$\Opa(\hurw_{R}[[x]])$. On the other hand, $\rho_{t}\Opa(0)=\rho_{t}\textbf{0}_{\Opa}=x$ and $
\rho_{t}\Opa(1)=\rho_{t}\textbf{1}_{\Opa}=x+t$.
\end{proof}

Clearly $\rho_{t}$ is a ring isomorphism and therefore $\rho_{t}\Opa(\hurw_{R}[[x]])$ inherits all the algebraic properties of $\Opa(\hurw_{R}[[x]])$ from Section 3.
\begin{theorem}\label{theo_delta_domi_flujo}
$\rho_{t}\Opa(\hurw_{R}[[x]])$ is an integral domain.
\end{theorem}

Let $\chi:\rho_{t}\Opa(\hurw_{R}[[x]])\rightarrow\hurw_{R}[[x]]$ be an homomorphism defined by
$\chi(\rho_{t}\Opa(f(x)))=f(x)$. We have the proposition

\begin{proposition}\label{propo_iso_flujo}
Let $I$ be an ideal in $R$. Then
\begin{enumerate}
\item $\rho_{t}\Opa(\hurw_{R}[[x]])/\chi^{-1}(I+\left\langle x\right\rangle)\simeq\hurw_{R}[[x]]/(I+\left\langle x\right\rangle)\simeq R/I$.
\item $\rho_{t}\Opa(\hurw_{R}[[x]])/\rho_{t}\Opa(\hurw_{I}[[x]])\simeq\rho_{t}\Opa(\hurw_{R}[[x]]/\hurw_{I}[[x]])$.
\end{enumerate}
\end{proposition}

\begin{theorem}\label{theo_maxi_flujo}
If $M_{\alpha}$ is the set of maximal ideals of $\hurw_{R}[[x]]$, then $\rho_{t}\Opa(M_{\alpha})$
is the set of maximal ideals of $\rho_{t}\Opa(\hurw_{R}[[x]])$.
\end{theorem}

\begin{theorem}\label{the_prime_flujo}
If $P+\left\langle x\right\rangle$ and $\hurw_{P}[[x]]$ are prime ideals in
$\hurw_{R}[[x]]$, then $\rho_{t}\Opa(P+\left\langle x\right\rangle)$ and
$\rho_{t}\Opa(\hurw_{P}[[x]])$ are prime ideals in $\rho_{t}\Opa(\hurw_{R}[[x]])$.
\end{theorem}

\begin{theorem}\label{theo_campo_flujo}
Let $\F$ be a field. Then the set
$$\rho_{t}\Opa[\hurw_{F}((x))]=\rho_{t}\Opa(\hurw_{F}[[x]])/\left\langle\rho_{t}\Opa(x)\right\rangle$$ is a field.
\end{theorem}

The reason for constructing the ring $\rho_{t}\Opa(\hurw_{R}[[x]])$ is because it contains all the solutions to the autonomous differential equations $\delta_{t}\Phi=f(\Phi)$ for each function $f(x)$ in $\hurw_{R}[[x]]$. In this ring it will be possible to decompose the solutions
of an autonomous differential equation of order one into simpler solutions.\\
First suppose that $f=f_{1}+f_{2}+\cdots+f_{k}$. We wish to solve the differential equation $\delta_{t}\Phi=f(\Phi)$.  The flow of this equation becomes
\begin{eqnarray*}
\Phi(t,x,f(x))&=&\rho_{t}\Opa(f(x))\\
&=&\rho_{t}\Opa\left(\sum_{i=1}^{k}f_{i}(x)\right)\\
&=&\rho_{t}\left(\bbox_{i=1}^{k}\Opa(f_{i}(x))\right)\\
&=&\bbox_{i=1}^{k}\rho_{t}\Opa(f_{i}(x))\\
&=&\bbox_{i=1}^{k}\Phi(t,x,f_{i}(x)).
\end{eqnarray*}
Thus the flow of $\delta_{t}\Phi=f(\Phi)$ decomposes into summands where each summand is the flow of the equation $\delta_{t}\Phi=f_{i}(\Phi)$.

\begin{example}
Find the solution to the differential equation $y^{\prime}=e^{y}+sin(y)$, where $e^{ax}$ and $\sin(x)$ are in $\hurw_{\R}[[x]]$. We first calculate the flow of $y^{\prime}=e^{ay}$. We have
\begin{eqnarray*}
\rho_{t}\Opa(e^{ax})&=&\rho_{t}(x,e^{ax},ae^{2ax},2!a^{2}e^{3ax},...)\\
&=&x+\sum_{n=1}^{\infty}(n-1)!a^{n-1}e^{anx}\frac{t^{n}}{n!}\\
&=&x+\frac{1}{a}ln\left(\frac{1}{1-ate^{ax}}\right)
\end{eqnarray*}
In this way
\begin{equation}
\Phi(t,x,e^{ax})=x+\frac{1}{a}ln\left(\frac{1}{1-ate^{ax}}\right)
\end{equation}
is the flow we are looking for. Now we only have to use the previous result with $a=1,i,-i$. In this
way
\begin{small}
\begin{eqnarray*}
\Phi(t,x,e^{x}+\sin(x))&=&\Phi\left(t,x,e^{x}+\frac{1}{2i}e^{ix}-\frac{1}{2i}e^{-ix}\right)\\
&=&\Phi(t,x,e^{x})\boxplus\Phi(t,x,\frac{1}{2i}e^{ix})\boxplus\Phi(t,x,-\frac{1}{2i}e^{-ix})\\
&=&\left[x+ln\left(\frac{1}{1-te^{x}}\right)\right]\boxplus\left[x-(i)ln\left(\frac{2}{2-te^{ix}}\right)\right]\\
&&\boxplus\left[x+(i)ln\left(\frac{2}{2+te^{-ix}}\right)\right]
\end{eqnarray*}
\end{small}
is the flow of the equation $y^{\prime}=e^{y}+sin(y)$.
\end{example}

Now suppose that $f$ factorizes as $f=f_{1}f_{2}\cdots f_{k}$ in $\hurw_{R}[[x]]$ and let's find the solution to the equation $y^{\prime}=f(y)$. Then its flow will be
\begin{eqnarray*}
\Phi(t,x,f(x))&=&\bcast_{i=1}^{k}\rho_{t}\Opa(f_{i})\\
&=&\bcast_{i=1}^{k}\Phi(t,x,f_{i}(x)).
\end{eqnarray*}
where each $\Phi(t,x,f_{i}(x))$ is the flow of the equation
$y^{\prime}=f_{i}(y)$ associated with the dynamical system $(R,\hurw_{S}[[x]],\rho_{t}\Opa(f_{i}))$.

Now suppose $f(x)=\sum_{k=0}^{n}a_{k}x^{k}$ is a polynomial of degree $n$ in $R[x]$. We want to define a polynomial in the ring $\rho_{t}\Opa(R[x])$. Let us first define the monomials in $\rho_{t}\Opa(R[x])$ as
\begin{eqnarray}\label{eqn_pot_rho}
[\rho_{t}\Opa(x)]_{\odot}^{k}&=&\overbrace{\rho_{t}\Opa(x)\odot\cdots\odot\rho_{t}\Opa(x)}^{k}
\end{eqnarray}

and let's calculate its value
\begin{lemma}\label{lemma_potencia}
When $k=0$ we have that $[xe^{at}]_{\odot}^{0}=x+at$ and when $k=1$, clearly $[xe^{at}]_{\odot}^{1}=xe^{at}$. For $k\geq2$ we have
\begin{equation}
[xe^{at}]_{\odot}^{k}=\frac{x}{\sqrt[k-1]{1-a(k-1)x^{k-1}t}}
\end{equation}
\end{lemma}
\begin{proof}
Por (\ref{eqn_pot_rho}), $[xe^{at}]_{\odot}^{k}=\rho_{t}\Opa(ax^{k})$. Then
we need to find the flow of the equation $y^{\prime}=ay^{k}$, for $k\geq0$. When $k=0$,
the equation $y^{\prime}=a$ has solution $\Phi(t,x,a)=x+at$. When $k=1$, the equation
$y^{prime}=ay$ has solution $\Phi(t,x,ax)=xe^{at}$. For $k\geq2$ we will use the method of
separation of variables. Then the flow is
\begin{equation*}
\Phi(t,x,ax^{k})=\frac{x}{\sqrt[k-1]{1-a(k-1)x^{k-1}t}}.
\end{equation*}
\end{proof}

Then, by direct application of this Lemma, we have the following results
\begin{definition}
A polynomial of degree $n$ in $\rho_{t}\Opa(R[x])$ is of the form
\begin{equation}
\bbox_{k=0}^{n}[\rho_{t}\Opa(a_{k}x)]_{\odot}^{k}=(x+a_{0}t)\boxplus xe^{a_{1}t}\boxplus\bbox_{k=2}^{n}\frac{x}{\sqrt[k-1]{1-a_{k}(k-1)x^{k-1}t}}
\end{equation}
\end{definition}

Now let $\F$ be a field and let $f(x)$ be a polynomial of degree $n$ in $\F[x]\subset\hurw_{\F}[[x]]$. Further suppose $f(x)$ factorizes as $\prod_{i=1}^{n}(x-a_{i})$ in $\F[x]$. Then
\begin{eqnarray*}
\bbox_{k=0}^{n}[\rho_{t}\Opa(b_{k}x)]_{\odot}^{k}
&=&\rho_{t}\Opa\left(\prod_{i=1}^{n}(x-a_{i})\right)\\
&=&\rho_{t}\left(\bcast_{i=1}^{n}\Opa(x-a_{i})\right)\\
&=&\bcast_{i=1}^{n}\rho_{t}\Opa(x-a_{i})\\
&=&\bcast_{i=1}^{n}[(x-a_{i})e^{t}+a_{i}]
\end{eqnarray*}

\begin{theorem}
The irreducible polynomials in $\rho_{t}\Opa(\R[x])$ are of the form
\begin{equation}
(x-a)e^{t}+a
\end{equation}
\begin{equation}
\sqrt{d}\left(\frac{x-b/2+\sqrt{d}\tan(\sqrt{d}t)}{\sqrt{d}-(x-b/2)\tan(\sqrt{d}t)}\right)+\frac{b}{2}
\end{equation}
where $d=\frac{4c-b^{2}}{4}$.
\end{theorem}
\begin{proof}
We know that every irreducible polynomial in $\R[x]$ is of the form $x-a$ and $x^{2}-bx+c$. Then every polynomial in $\R[x]$ contains these factors. By solving the equations $y^{\prime}=y-a$ and $y^{\prime}=y^{2}-by+c$ we obtain the functions above.
\end{proof}

\begin{example}
Solve the equation $y^{\prime}=1-y+y^{2}-y^{3}$. Directly by the Lemma
\ref{lemma_potencia} la solución es
\begin{eqnarray*}
\Phi(t,x,f(x))&=&(x+t)\boxplus xe^{-t}\boxplus\left(\frac{x}{1-xt}\right)\boxplus\left(\frac{x}{\sqrt{1+2x^{2}t}}\right)
\end{eqnarray*}
By factoring $f(x)$ we obtain $f(x)=1-x+x^{2}-x^{3}=(1-x)(x^{2}+1)$. Then
\begin{eqnarray*}
\Phi(t,x,f(x))&=&\rho_{t}\Opa(1-x)\odot\rho_{t}\Opa(x^{2}+1)\\
&=&\rho_{t}\Opa(-(x-1))\odot\rho_{t}\Opa(x^{2}+1)\\
&=&[(x-1)e^{-t}-1]\odot\left(\frac{x+\tan(t)}{1-x\tan(t)}\right)
\end{eqnarray*}
would be the factored solution. A solution with the variable separation method leads to a function
that cannot be written explicitly. Then solve on the ring $\rho_{t}\Opa(\hurw_{\R}[[x]])$ is
the best way to go.
\end{example}

As something very easy to prove we have the following
\begin{theorem}
The irreducible polynomials in $\rho_{t}\Opa(\C[x])$ are of the form
\begin{equation}
\Phi(t,x,x-a)=(x-a)e^{t}+a
\end{equation}
\end{theorem}

Other important results that can be deduced from the above Lemma are as follows
\begin{theorem}
Let $\F$ be a field. Then the ideals in $\rho_{t}\Opa(\hurw_{\F}[[x]])$ are of the form
\begin{equation}
\left\langle\rho_{t}(\Opa(x))\right\rangle=xe^{t}\odot\rho_{t}\Opa(\hurw_{\F}[[x]])
\end{equation}
and
\begin{equation}
\left\langle\rho_{t}(\Opa(x))_{\odot}^{m}\right\rangle=\frac{x}{\sqrt[m-1]{1-(m-1)x^{m-1}t}}\odot\rho_{t}\Opa(\hurw_{\F}[[x]])
\end{equation}
for $m\geq2$, where $\left\langle xe^{t}\right\rangle$ is a maximal ideal.
\end{theorem}
\begin{proof}
It follows by taking into account that $\left\langle\Opa(x)_{\odot}^{m}\right\rangle$ are ideals in $\Opa(\hurw_{\F}[[x]])$.
\end{proof}

\begin{theorem}
Let $\F$ be a field. In the field $\rho_{t}\Opa[\hurw_{\F}((x))]$ we have the following identities
\begin{eqnarray}
(x+at)&\odot&\left(x+\frac{t}{a}\right)=x+t,\ \ \ a\neq0\\
xe^{t}&\odot&\sqrt{2t+x^{2}}=x+t\\
\frac{x}{\sqrt[m-1]{1-(m-1)x^{m-1}t}}&\odot&\sqrt[m+1]{(m+1)t+x^{m+1}}=x+t
\end{eqnarray}
for $m\geq2$.
\end{theorem}
\begin{proof}
We have to keep in mind that the flows $\{rho_{t}\Opa(f)$ and $\rho_{t}\Opa(g)$ are inverses in
the ring $\rho_{t}\Opa[\hurw_{\F}((x))]$ only if $fg=1$. Then the identities above arise from
solving the differential equations $y^{\prime}=a$, $y^{\prime}=a^{-1}$, $y^{\prime}=y^{m}$ and
$y^{\prime}=y^{-m}$ for $m\geq1$, respectively.
\end{proof}

We end this section with the following result that relates the operations $\circ$, $\boxplus$ and
$\odot$
\begin{theorem}
In the ring $\rho_{t}\Opa(\hurw_{R}[[x]])$ we have the following identities relating the operations $\circ$, $\boxplus$ and $\odot$.
\begin{enumerate}
\item $\Phi_{t}(x,f+g)\circ\Phi_{s}(x,f+g)=\Phi_{t+s}(x,f)\boxplus\Phi_{t+s}(x,g)$.
\item $\Phi_{t}(x,fg)\circ\Phi_{s}(x,fg)=\Phi_{t+s}(x,f)\odot\Phi_{t+s}(x,g)$
\end{enumerate}
\end{theorem}
\begin{proof}
The proof is straightforward.
\end{proof}

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\title{Cork twisting Schoenflies problem }
\author{Selman Akbulut}
\thanks{Partially supported by NSF grants DMS 0905917}
\keywords{}
\address{Department  of Mathematics, Michigan State University,  MI, 48824}
\email{akbulut@math.msu.edu }
\subjclass{58D27,  58A05, 57R65}
\date{\today}
\begin{abstract}
The stable Andrews-Curtis conjecture in combinatorial group theory is  the statement that every balanced presentation of the trivial group  can be simplified to the trivial form by elementary moves corresponding to ``handle-slides" together with ``stabilization" moves. Schoenflies conjecture is the statement that the complement of any smooth embedding $S^{3}\hookrightarrow S^{4}$  are pair of smooth balls. Here we suggest an approach to these problems by certain cork twisting operation on contractible manifolds, and demonstrate it on the example of the first  Cappell-Shaneson homotopy sphere.
\end{abstract}

\date{}
\maketitle

\setcounter{section}{-1}

\vspace{-.3in}

\section{Introduction}

Let $G(P)=\{ x_1,x_2,..,x_n \;|\; r_{1}(x_1,..,x_n),..,r_{n}(x_1,..x_n) \}$ be a balanced presentation $P$ of the trivial group. Here balanced means  the presentation has the same number of generators and relators. When there is no danger of confusion we will  abbreviate $r_{j}:=r_{j}(x_1,..,x_n)$. The presentation $P$ is called  {\it stably Andrews-Curtis trivial} (SAC-trivial in short) if by changing  relators by the following finite number of the steps, and their inverses, we obtain the trivial presentation:
\begin{itemize}

\item[(a)] $r_{i}\mapsto r_{i}r_{j} $ for some $j\neq i $.

\item[(b)] Add a new generator $x_{n+1}$ and a relation  $x_{n+1}\gamma$.
\item[(c)] $r_{i} \mapsto r_{i}^{-1}$ or
$r_{i} \mapsto \gamma r_{i}\gamma^{-1}$, where $\gamma$ represents any word in $G(P)$.
\end{itemize}

\vspace{.05in}

Fundamental group of any compact $2$-complex gives such a presentation, so any compact contractible $4$-manifold $W$, which is a {\it $2$-handlebody} (i.e.\;a handlebody  consisting of handles of index $\leq 2$) has such a presentation. Generators $\{x_{j} \}$ correspond to the $1$-handles, and the relations $\{r_{j}\}$ correspond to the $2$-handles. (a) corresponds to sliding $2$-handles over each other, and (b) corresponds to introducing (or taking away) a canceling pair of $1$ and $2$-handles. Call a pair of $2$-handlebodies SAC {\it equivalent} if they are related by these two steps.

\vspace{.05in}

Not much known about which presentations of the trivial group are SAC-trivial. in \cite{ak1} the following examples were proposed (n=0,1,..)

$$G(P_{n})=\{ x,y \;|\; xyx=yxy, \; x^{n+1}= y^{n}  \}$$

$G(P_{n})$ is the trivial group since the first relation gives $y = (yx)^{-1}x (yx)$, so $y^{n+1}=(yx)^{-1}x^{n+1} (yx)= (yx)^{-1}y^{n} (yx)=x^{-1}y^{n}x=x^{-1}x^{n+1}x=y^{n}$, hence $y=1$ and $x=1$. Gersten showed that $G(P_{2})$ is SAC-trivial (\cite{ge}, \cite{gst}), but it is not known whether the other $G(P_{n})$ are SAC-trivial. The case $G(P_{4})$ is particularly interesting, since it is the fundamental group  presentation of the $2$- handlebody of a Cappell-Shaneson homotopy $4$-ball $W_{0}= \Sigma_{0}-B^{4}$ constructed in \cite{ak1}, where $\Sigma_{0}$ is the $2$-fold covering space of the first Cappell-Shaneson exotic $\R\P^4$  defined in \cite{cs}.

\vspace{.05in}

The main reason topologists are interested SAC problem is its relation to the Schoenflies  conjecture, which says ``{\it The complement of a smoothly imbedded $S^{3}\hookrightarrow S^{4}$ is a disjoint union of two smooth $4$-balls}''. So far, in dimension $4$ only the topological version of this conjecture is known (\cite{b}, \cite{m}). If  the presentation of $\pi_{1}(W)$ a smooth contractible  $2$-handlebody $W^{4}$ is SAC-trivial then $W\times [0,1]=B^{5}$ (because in dimension $5$ canceling handles algebraically is equivalent canceling them geometrically), hence this gives an imbedding $W\hookrightarrow S^{4}$ via its double
$$D(W):=W\cup_{\partial} -W=\partial (W\times [0,1])=S^4$$ Then since $\partial W=S^{3}$ the topological Schoenflies theorem implies $W$ is homeomorphic to $B^{4}$. To apply this  to the smooth Poincare conjecture (PC), we first puncture a given smooth homotopy $4$-sphere $\Sigma$ to a homotopy $4$-ball $W=\Sigma -B^{4}$ and turn it to $2$-handlebody by canceling its $3$-handles (if we can), reducing it to a SAC problem. This is what is done for the first Cappell-Shaneson homotopy $4$-ball $W_{0}$ in \cite{ak1} (similar proof for the other ones), but there the associated SAC problem was bypassed by directly imbedding $W_{0}\hookrightarrow S^{4}$, hence reducing the smooth PC to Schoenflies problem. In particular if the complement of $W$ in $S^4$ is $B^4$ then $W$ itself must be $B^{4}$.
Figure~\ref{fig1} shows the $3$-handle free handlebody picture of $W_{0}$ which imbeds into $S^{4}$ (\cite{ak1} Figure 28).
\begin{figure}[ht]  \begin{center}
\includegraphics[width=.5\textwidth]{r1}
\caption{$W_{0}$}
\label{fig1}
\end{center}
\end{figure}

In the end it turned out that all Cappell-Shaneson homotopy balls $W_{0}, W_{1},..$ are diffeomorphic to $B^{4}$; without even appealing to the Schoenflies problem (\cite{g}, \cite{a1}). More specifically, the proofs proceed by first introducing canceling 2/3 handle pairs, then canceling all the handles ending up with $W_{i}=B^4$, $i=0,1,2..$.
Here we revisit the Schoenflies problem by analyzing  the approach of \cite{ak1} more closely, where another a  $2$-handlebody $W_{0}^{*}$ with $\partial W^{*}_{0}=S^3$ was constructed so that $$S^{4}=W_{0}\cup_{\partial} -W_{0}^{*}$$
We can reduce this Schoenflies problem to another Schoenflies problem which we can solve, i.e. by imbedding $W_{0}^{*}\hookrightarrow S^{4}$ with complement $B^{4}$
  $$S^{4}= W_{0}^{*}\cup_{\partial} B^4$$
which implies  $W_{0}^{*}=B^{4}$, and so $W_{0}=B^4$. Of course this last step is not new, it is just a case of proving some homotopy $4$-balls are standard by introducing a single canceling $2/3$ handle pair (\cite{g}, \cite{a1}). We  stated it this way to relate it to Schoenflies problem. Curiously the associated presentation of the fundamental group of $W_{0}^{*}$ is $G(P_{2})$, while $\pi_{1}(W_{0})$ is $G(P_{4})$. The hope is, associating $W$ another convenient ``{\it twin}" contractible manifold $W\mapsto W^{*}$ might help to resolve SAC triviality.

\section{Flexible contractible manifolds} A flexible contractible $4$-manifold is a smooth compact contractible $2$-handlebody, where its $2$-handles are represented by $0$-framed unknotted curves (i.e. after erasing circles with dots we get a $0$-framed unlink).

 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.53\textwidth]{r2}
\caption{A flexible contactible manifold and its twin}
\label{fig2}
\end{center}
\end{figure}

We call the $2$-handlebody $W^{*}$ obtained from $W$ by zero and dot exchanges of its handles ($S^{2}\times B^2 \leftrightarrow B^{3}\times S^{1}$ exchanges in  the interior of $W$) the {\it twin of W} . We call the operation $W\mapsto W^{*}$ {\it cork twisting the flexible manifold $W$}. Notice that this notion depends on the handles. Here we do not address the problem of how the twin of $W$ changes after handle slides of W. It is clear that  this  decomposes the $4$-sphere as $S^{4}= W\cup_{\partial} -W^{*}$, i.e. $W$ imbeds into $S^4$ with  complement $W^{*}$.

\begin{figure}[ht]  \begin{center}
\includegraphics[width=.34\textwidth]{r3}
\caption{A flexible contractible manifold and its twin}
\label{fig3}
\end{center}
\end{figure}

\section{The twin of $W_{0}$}

First  by applying the diffeomorphism
described in Section 2.3 of \cite{a3}, we identify the handlebody $W_{0}$ of Figure~\ref{fig1} with  Figure~\ref{fig4}. This diffeomorphism is a combination of introducing and canceling 1/2 handle pairs. Recall that the associated presentation of $\pi_{1}(W_{0})$ is $G(P_{4})$.

\begin{figure}[ht]  \begin{center}
\includegraphics[width=.5\textwidth]{r4}
\caption{$W_{0}$}
\label{fig4}
\end{center}
\end{figure}

\begin{Prop}
The twin $W_{0}^{*}$ of the $2$-handlebody of $W_{0}$ in Figure~\ref{fig4} is  given by Figure~\ref{fig5}, and the associated presentation of $\pi_{1}(W_{0}^{*})$ is  $G(P_{2})$.
\end{Prop}
\begin{figure}[ht]  \begin{center}
\includegraphics[width=.35 \textwidth]{r5}
\caption{$W_{0}^{*}$}
\label{fig5}
\end{center}
\end{figure}

\proof Clearly Figure~\ref{fig6} is the twin of $W_{0}$ in Figure~\ref{fig4}. Notice that the two circles with dots do not link each other, since the affect of blowing down one $+1$ is undone by blowing down another $-1$ (before putting dots on them),  of course during this process the framed circles representing the $2$-handles changed appropriately.

  \begin{figure}[ht]  \begin{center}
\includegraphics[width=.5 \textwidth]{r6}
\caption{$W_{0}^{*}$}
\label{fig6}
\end{center}
\end{figure}

By sliding $2$-handles over the $1$-handles (as indicated by dotted arrows) in Figure~\ref{fig6} we get Figure~\ref{fig7}. Then isotoping the two dotted circles away from each other and after rotating the figure $90^{0}$ we get Figure~\ref{fig5}.
 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.37 \textwidth]{r7}
\caption{$W_{0}^{*}$}
\label{fig7}
\end{center}
\end{figure}

Next we calculate the presentation of $\pi_{1}(W_{0}^{*})$  from Figure~\ref{fig5}:
\begin{itemize}
\item[(a)] $x^{2}yx^{-1}yx^{-1}y^{-1}=1$
\item[(b)] $y^{-2}x^{-1}yx^{-1}yx=1$
\end{itemize}
(a) $\implies x^{-1}yx y^{-1}=xyx^{-1}$, and (b) $\implies x^{-1}yx y^{-1}=y^{-1}xy$. Hence $xyx=yxy$. Also (a)
$\implies  x^{3}yx^{-1}y=xyx=yxy \implies x^{3}y=yx^{2}$. Hence
 $x^{3}=yx^{2}y^{-1}=(yxy^{-1})^{2}=(x^{-1}yx)^{2}=x^{-1}y^{2}x\implies x^{3}=y^{2}$.  Hence we get the presentation $G(P_{2})$. \qed

 \begin{Rm} As shown in \cite{ak1}, attaching pair of $2$-handles to $\partial W_{0}$ along the dotted circles of Figure~\ref{fig1} gives $\#_{2}(S^{2}\times B^{2})$, which can be capped by a pair of $3$-handles $\#_{2}(B^{3}\times S^{1})$. Reader can check that turning these handle pairs upside down gives the handlebody  $W_{0}^{*}$.
 \end{Rm}

 \begin{Prop} $W_{0}^{*}$ smoothly imbeds into $S^{4}$ with complement $B^{4}$, hence $W_{0}^{*} $ (therefore $W_{0}$) is diffeomorphic to $B^{4}$.
 \end{Prop}

 \proof By sliding the $2$-handles over the $1$-handles of Figure~\ref{fig5} (as indicated by the dotted arrows in the figure) we get Figure~\ref{fig8}. Again by applying the diffeomorphism of Section 2.3 of [A3] to this figure twice, we get Figure~\ref{fig9}, then after the  handle slides (indicated by dotted arrows) we get  Figure~\ref{fig10}, and then by an isotopy we get Figure~\ref{fig11}.

\begin{figure}[ht]
\centering
\begin{minipage}{.35\textwidth}
  \centering
  \includegraphics[width=.8\linewidth]{r8}
  \caption{$W_{0}^{*}$}
  \label{fig8}
\end{minipage}
 \begin{minipage}{.4\textwidth}

  \includegraphics[width=.85\linewidth]{r9}
  \caption{$W_{0}^{*}$}
  \label{fig9}
\end{minipage}
\end{figure}

\begin{figure}[ht]  \begin{center}
\includegraphics[width=.5\textwidth]{r10}
\caption{$W_{0}^{*}$}
\label{fig10}
\end{center}
\end{figure}

 We  will now modify Figure~\ref{fig11} by a sequence of handle slides, and  adding (and canceling) $1/2$ handle pairs to get a new $2$-handlebody presentation of $W_{0}^{*}$ which will have the required property. First by doing the handle slides indicated by the dotted arrows, we go from Figure~\ref{fig11} to Figures~\ref{fig12} and ~\ref{fig13}. Then by canceling a $1/2$ handle pair we obtain the second picture of Figure~\ref{fig13}, which is a ribbon $1$-handle, induced from $K\#-K$ where $K$ is the trefoil knot,  and a $2$-handle.

 \begin{figure}[ht]
\centering
\begin{minipage}{.4\textwidth}
  \centering
  \includegraphics[width=.7\linewidth]{r11}
  \caption{$W_{0}^{*}$}
  \label{fig11}
\end{minipage}
 \begin{minipage}{.45\textwidth}

  \includegraphics[width=.7\linewidth]{r12}
  \caption{$W_{0}^{*}$}
  \label{fig12}
\end{minipage}
\end{figure}
 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.65 \textwidth]{r13}
\caption{$W_{0}^{*}$}
\label{fig13}
\end{center}
\end{figure}

\noindent The first picture of Figure~\ref{fig14} is the short hand of this handlebody (dotted line indicates the ribbon move giving the ribbon $1$-handle in Figure~\ref{fig13}). The second picture of Figure~\ref{fig14} is drawn after this ribbon move. Then doing the indicated handle slide to Figure~\ref{fig14} we get the handlebody $C_{2}$, where $C_{n}$ is the handlebody described in Figure~\ref{fig15}.

 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.56 \textwidth]{r14}
\caption{$W_{0}^{*}$}
\label{fig14}
\end{center}
\end{figure}

 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.7 \textwidth]{r15}
\caption{$C_{n}$ and a useful isotopy}
\label{fig15}
\end{center}
\end{figure}

Notice  $C_{n}$ gives the presentation $G(P_{n})$ and is similar to $H_{n,1}$ of \cite{g}, but it differs in the way its $0$-framed $2$-handle links the $1$-handles (this fact provides us the useful isotopy of Figure~\ref{fig15}).  This difference is due to the fact that here we are getting a nonstandard ribbon which $K\#-K$ bounds.  Now the proof of the proposition follows from the following Lemma~\ref{lem3} whose proof is similar  to the one in \cite{g} for $H(n,1)$. \qed.

\begin{Lem}\label{lem3}
 $C_{n}$ smoothly imbeds into $S^{4}$ with complement $B^{4}$.
\end{Lem}

\proof Attach a $2$-handle to $C_{n}$ along the loop
$\gamma $ with $-1$ framing, as shown  in the first picture of Figure~\ref{fig15} (this framing corresponds to $0$-framing when viewed from  $S^{3}$). Denote this manifold by $C_{n} +\gamma^{-1}$. The steps of Figure~\ref{fig15} show the following equivalences by handle slides $$C_{n} +\gamma^{-1} \cong C_{n-1} +\gamma^{-1}...\cong C_{0}+\gamma^{-1}$$

 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.7 \textwidth]{r16}
\caption{$C_{n}+ \gamma^{-1} \cong C_{n-1}+ \gamma^{-1}$}
\label{fig16}
\end{center}
\end{figure}

\newpage

Furthermore, Figure~\ref{fig17} shows $C_{0} + \gamma^{-1} \cong S^{2}\times B^{2}$, hence we can cap $C_{n} + \gamma^{-1}$ with $S^{1}\times B^{3}$ to get $S^{4}$. Let $N$ be the handlebody consisting of $S^{1}\times B^{3}$  union the dual of the $2$-handle $\gamma^{-1}$. $N$ is a contractible manifold with boundary $S^{3}$, consisting of a single pair if  $1$- and $2$-handles, hence $N\cong B^{4}$. This fact follows from  Property-R theorem of \cite{ga}. \qed

 \begin{figure}[ht]  \begin{center}
\includegraphics[width=.57 \textwidth]{r17}
\caption{$C_{0}+ \gamma^{-1}\cong S^{2}\times B^{2}$}
\label{fig17}
\end{center}
\end{figure}

\begin{Rm} There is a certain dictionary  relating AC-triviality of a $2$-handlebody to its twin, which we didn't discussed here, opting directly dealing with Schoenflies problem. This is because the circle with dots can slide over each other just like $2$-handles slide over each other (e.g. Section 1.2 of \cite{a3}). We hope to address this in a future paper.
\end{Rm}

\section*{Acknowledgement}
I want to thank  Peter Teichner and Rob Kirby for stimulating my interest to the problems discussed here.


TITLE: Understanding a lemma for Implicit Function Theorem
QUESTION [0 upvotes]: The following lemma is a key for Implicit Function Theorem, given in Topology and Geometry by Bredon; but I am unable to see what the theorem is stating, what its conditions imply geometrically. I tried in small dimension case ($n=m=1$), but could not proceed.

Let $\xi\in\mathbb{R}^n$ and $\eta\in\mathbb{R}^m$ be given. Let $f:\mathbb{R}^n\times\mathbb{R}^m\rightarrow\mathbb{R}^m$ be $C^1$ and put $f=(f_1,\ldots, f_m)$. Assume that $f(\xi,\eta)=\eta$ and that all the following partial derivatives vanish at $(\xi,\eta)$:
$$
\frac{\partial f_i}{\partial y_j}(\xi,\eta)=0
$$
where $x_1,\ldots, x_n$ are coordinates in $\mathbb{R}^n$ and $y_1,\ldots,y_n$ are in $\mathbb{R}^m$. Then there exists real numbers $a>0$ and $b>0$ such that there exists a unique function $\phi:A\rightarrow B$ where $A=\{x\in\mathbb{R}^n \,:\, |x-\xi|\le a\}$ and $B=\{y\in\mathbb{R}^m \,:\, |y-\eta|\le b\}$, such that $$\phi(\xi)=\eta \hskip5mm \mbox{ and } \hskip5mm \phi(x)=f(x,\phi(x)) \hskip5mm \forall x\in A.$$
Moreover $\phi$ is continuous.

Can one explain what the lemma is saying?

The function $f$ is taken as $f(\xi,\eta)=\eta$; this looks like a projection map, but it is not overall projection map (since, only at point $(\xi,\eta)$, value is second component $\eta$). So, I did not get the importance of this assumption.
Further, the $m$ partial derivatives of $m$ components of $f$ are taken $0$, so in the $Df(\xi,\eta)$, the last $m\times m$ part is fully $0$; When one tries to put condition on an $m\times m$ matrix, the basic conditions come are whether it is singular or not; but he is putting more than singular condition. I did not get reason behind this.

REPLY [1 votes]: What the lemma is saying is that if $F(\xi, \eta) = \eta$ and $D_yF(\xi, \eta) = 0$, then the equation
$$F(x, y) = y$$
defines $y$ implicitly as a continuous function $y = \phi(x)$ of $x$ for $x$ near $\xi$.
Really the lemma is just the implicit function theorem applied to $G(x, y) = F(x, y) - y$.


TITLE: Calculate $\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k$?
QUESTION [2 upvotes]: Is there a way to calculate $\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k$?
If I would be able to calculate it I could find the solution at another task. So hopefully it is possible. I feel like I am missing out onto something here. Is ther a way to simplify the second part?

REPLY [0 votes]: Note that
$$\sum_{n=0}^\infty \color{blue}{\sum_{k=0}^n \binom nk (-r)^k}=\sum_{n=0}^\infty \color{blue}{(1-r)^k}
=\frac 1{1-(1-r)}=\frac 1r$$
Putting $r=\frac 13$ gives
$$\sum_{n=0}^\infty \sum_{k=0}^n \binom nk \left(-\frac 13\right)^k=\color{red}3$$