chapter-math.tex

\chapter{An Example of Mathematical Writing}
\index{An Example of Mathematical Writing%
@\emph{An Example of Mathematical Writing}}%

\section{Generalized Fatou's Lemma}
\index{Generalized Fatou's Lemma%
@\emph{Generalized Fatou's Lemma}}%

Here we show an application of the following lemma:

\begin{lem}[Generalized Fatou's Lemma] \label{l:fatou}

Let $A$ be a Dedekind ring and $F$ a rational series 
in $A[[X]]$, i.e., $F = p/q$ for some 
$p, q \in A[X]$. Then there exist two polynomials 
$P, Q \in A[X]$ such that $F = P/Q$, 
where $P$ and $Q$ are relatively prime and 
$Q(0) = 1$.

\end{lem}

\proof
See \cite{bertin:psn}, p.~15, theorem~1.3.
\endproof

\begin{thm} \label{l:req}
Let $\{c_n\}_{n=-\infty}^{\infty}$ a set of 
elements from $K$ such that $c_n \in k'$ for every 
$n \geq n_0$, and verifying the following recurrence 
relation of order M:
\begin{equation}
c_n\ =\ r_1\,c_{n-1} + r_2\,c_{n-2} + \dots + r_M\,c_{n-M}
\end{equation}
for every $n \in \mathbb Z$, where $r_1,r_2,\dots,r_M$ are in 
$K$, $r_M \neq 0$. 
Then:

\item{(i)} The coefficients $r_1,r_2,\dots,r_M$ are in 
$k'$, and for every $n \in \mathbb Z$, $c_n \in k'$.

\item{(ii)} If $c_n \in \mathcal O_{k',v}$ 
for every $n \geq n_0$, then the coefficients 
$r_1,r_2,\dots,r_M$ are all in 
$\mathcal O_{k',v}$.

\end{thm}


\proof 

\item{(i)} Let $C_n$ and $R$ be the matrices:

\begin{equation}
C_n\ =
\ \left(
\begin{array}{llll}
              c_n & c_{n+1} & \hdots & c_{n+M-1} \\
              c_{n+1} & c_{n+2} & \hdots  & c_{n+M} \\
              \vdots & \vdots & \ddots & \vdots \\
              c_{n+M-1} & c_{n+M} & \hdots & c_{n+2M-2}
\end{array}
\right)
\end{equation}
and
\begin{equation}
R\ =
\ \left(
\begin{array}{lllll}
              0 & 1 & 0 & \hdots & 0 \\
              0 & 0 & 1 & \hdots & 0 \\
             \vdots & \vdots & \vdots & \ddots & \vdots \\
              0 & 0 & 0 & \hdots & 1 \\
              r_M & r_{M-1} & r_{M-2} & \hdots & r_1 
\end{array}
\right)
\end{equation} 

We have that $C_{n+1} = R\,C_n$. Since the recurrence 
relation is of order M, $C_n$ is non singular. 
On the other hand, $R = C_{n+1}\,C_{n}^{-1}$. Since the 
elements of $C_n$ are in $k'$ for $n \geq n_0$, the entries 
of $R$, and those of $R^{-1}$, will be in $k'$. Since 
$C_{n-1} = R^{-1}\,C_n$, we get that the entries of 
$C_n$ will be in $k'$ also for $n < n_0$. 

\item{(ii)} For each $t \geq n_0$ define the formal 
power series 

\begin{equation}
F_t(X)\ =\ \sum_{n=0}^{\infty} c_{t+n}\,X^n
\end{equation}
which is in $\mathcal O_{k',v}[[X]]$. 
We have $F_t(X) = p_t(X)/q(X)$, 
where $p_t(X),q(X) \in k'[X]$ are the following:
\begin{equation}
p_t(X)\ =\ \sum_{j=0}^{M-1} \Bigl( c_{t+j} - 
                    \sum_{i=1}^{j} r_i\,c_{t+j-i} \Bigr)\,X^j
\end{equation}
\begin{equation}
q(X)\ =\ 1 - r_1\,X - r_2\,X^2 - \dots - r_M\,X^M
\end{equation}
This can be checked by multiplying $F_t(X)$ by $q_t(X)$ 
and using the recurrence relation, which gives 
$F_t(X)\,q(X) = p_t(X)$ (see \cite{poorten:sp}). 

Now we will prove that $p_t(X)$ and $q(X)$ are relatively 
prime. To do so, we will see that they cannot have any 
common root (in $\overline {k'}$). In fact, assume 
that $\alpha$ is a common root of $p_{t_0}(X)$ and $q(X)$ 
for some $t_0 \geq n_0$, i.e.: 
$p_{t_0}(\alpha) = q(\alpha) = 0$. 
Since $q(0)=1$, then $\alpha \neq 0$. Now we have:
\begin{equation}
X\,F_{t_0+1}(X) = F_{t_0}(X) - c_{t_0}
\end{equation}
so:
\begin{multline}
X\,p_{t_0+1}(X) = X\,q(X)\,F_{t_0+1}(X) \\
= q(X)\,(F_{t_0}(X) - c_{t_0}) = p_{t_0}(X) - c_{t_0}\,q(X)
\end{multline}
Hence $p_{t_0+1}(\alpha) = 0$, which means that $\alpha$ is 
also a root of $p_{t_0+1}(X)$. By induction we get that 
$p_t(\alpha) = 0$ for every $t \geq t_0$. Grouping the 
terms of $p_t(X)$ with respect to $c_t,c_{t+1},\dots,c_{t+M-1}$, 
we get:
\begin{equation}
p_t(X) = \sum_{j=0}^{M-1} a_j(X)\,c_{t+j}
\end{equation}
where 
\begin{equation}
a_j(X) = X^j\,\Bigl( 1 - \sum_{i=1}^{M-j-1} r_i\,X^i \Bigr)
\end{equation}
Note that $a_0(X),a_1(X),\dots,a_{M-1}(X)$ do not depend on t. 
On the other hand $p_t(\alpha)=0$ implies
\begin{equation}
\label{e:coldep}
\sum_{j=0}^{M-1} a_j(\alpha)\,c_{t+j} = 0
\end{equation}
for every $t \geq t_0$. Note that $a_{M-1}(\alpha)=\alpha^{M-1}
\neq 0$, so $a_0(\alpha),a_1(\alpha),\dots,a_{M-1}(\alpha)$ 
are not all zero, and (\ref{e:coldep}) means that the columns 
of the matrix $C_{t_0}$ are linearly dependent, so 
$\det C_{t_0}=0$, which contradicts the fact that $C_{t_0}$ 
is non singular. Hence, the hypothesis that $p_t(X)$ and 
$q(X)$ have a common root has to be false. This proves that 
$p_t(X)$ and $q(X)$ are relatively prime. 

By (generalized Fatou's) lemma~\ref{l:fatou}, 
and taking into account that 
$\mathcal O_{k',v}$ is a Dedekind ring, 
we get that there exist two relatively prime 
polynomials $P_t(X)$ and $Q_t(X)$ in 
$\mathcal O_{k',v}[X]$ such that 
$F_t(X) = P_t(X)/Q_t(X)$ and $Q_t(0)=1$. Hence: 
$p_t(X)\,Q_t(X) = q(X)\,P_t(X)$. By unique factorization 
of polynomials in $k'[X]$, there is a $u \in k'$ such that 
$P_t(X) = u\,p_t(X)$ and $Q_t(X) = u\,q_t(X)$. Since 
$Q_t(0)=q(0)=1$, we get that $u=1$, so 
$P_t(X) = p_t(X)$ and $Q_t(X) = q(X)$. 
Hence, the coefficients of $q(X)$ are in 
$\mathcal O_{k',v}$. 

\endproof


\section{Other Examples of Mathematical Writing}

\subsection{An Example of a Commutative Diagram}
\index{An Example of a Commutative Diagram%
@{An Example of a Commutative Diagram}}%

The following is an example of a commutative diagram.
\index{commutative diagram}%
It requires the \texttt{amscd} package.
\index{amscd package@{\texttt{amscd} package}}

\begin{equation*}
\newcommand{\End}{\operatorname{End}}
\begin{CD}
S^{{\mathcal{W}}_\Lambda}\otimes T   @>j>>   T\\
@VVV                                    @VV{\End P}V\\
(S\otimes T)/I                  @=      (Z\otimes T)/J
\end{CD}
\end{equation*}

That diagram has been made with the following commands:

\begin{verbatim}
\newcommand{\End}{\operatorname{End}}
\begin{CD}
S^{{\mathcal{W}}_\Lambda}\otimes T   @>j>>   T\\
@VVV                                    @VV{\End P}V\\
(S\otimes T)/I                  @=      (Z\otimes T)/J
\end{CD}
\end{verbatim}

\subsection{Using AMS Fonts}
\index{Using AMS Fonts@{Using AMS Fonts}}

To use AMS fonts it is necessary to choose from an assortment 
of \LaTeX{} packages. For instance the command 
\cn{usepackage\{amsfonts\}} calls in the \emph{amsfonts} package, 
which provides blackboard bold letters (e.g. $\mathbb{R}$) and 
some math symbols. A superset of that package is 
\emph{amssymb}. Other packages are \emph{eufrak} 
for Frankfurt letters (e.g. $\mathfrak{R}$)
and \emph{eucal} for Euler script 
(e.g. $\mathcal{R}$). 
Consult the \LaTeX{} documentation about this subject 
for additional information.