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\begin{document}
\newtheorem{lemma}{Lemme}[section]
\newtheorem{notations}{Notations}[section]
\newtheorem{notation}{Notation}[section]
\newtheorem{theorem}{Théorème}[section]
\newtheorem{remark}[theorem]{Remarque}
\newtheorem{proposition}{Proposition}[section]
\newtheorem{definition}{Définition}[section]
\newtheorem{definitions}{Définitions}[section]
\newtheorem{remarque}{Remarque}[section]
\newtheorem{remarques}{Remarques}[section]
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\newtheorem{corollary}{Corollaire}[section]
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 \mainmatter
\parskip=8pt
\chapter*{Introduction}
\pagestyle{fancy}\lhead{\textbf{\textit{Introduction} }}\rhead{}
\addcontentsline{toc}{chapter}{Introduction}








 \chapter{ Equations aux différences }
\pagestyle{fancy}\lhead{\textbf{\textit{Equations aux
diff\'{e}rences } }}\rhead{}

\section{Première équation }

\subsection{titre 2 }

\begin{definition}
La suite de Fibonacci est la suite $\left\{ F_{n}\right\} _{n\geq
0}$ telle que $F_{0}=0,\,F_{1}=1$ et
\begin{equation}
F_{n+2}=F_{n+1}+F_{n},\label{eq102}
\end{equation}
pour tout $n\geq 0$.
\end{definition}


\begin{description}
  \item[i)] \emph{\textbf{L'identité de Cassini}}: Pour $n>0$, on a
\begin{equation}
F_{n-1}F_{n+1}-F_{n}^{2}=(-1)^{n}.
\end{equation}

  \item[ii)]  \emph{\textbf{L'identité d'Ocagne}}: Pour $n, r \in\mathbb{N}$, on a
  \begin{equation}
F_{n+r}F_{n+1}-F_{n+r+1}F_{n}=(-1)^{n}F_{r}.
\end{equation}
  \item [iii)] \emph{\textbf{ L'identité de Johnson}}: Pour $k,l,m,n$ et $r \in\mathbb{N}$ tels que $k+l=m+n$,
   \begin{equation}
F_{k}F_{l}-F_{m}F_{n}=(-1)^{r}(F_{k-r}F_{l-r}-F_{m-r}F_{n-r}).
\end{equation}
\end{description}

\begin{lemma}\cite{rab}
yyyyyyyyyyy\label{lem1}
\end{lemma}

\begin{theorem}
Soit $\left\{ x_{n}\right\} _{n\geq -k}$ une solution de
\eqref{eq601}. Alors pour $n=0,1,\ldots,$
\begin{equation*}
x_{(k+1)n+i}=\frac{W_{n+1}+W_{n}x_{i-(k+1)}}{W_{n+2}+W_{n+1}x_{i-(k+1)}}q,\;i=1,2,...,k+1.
\end{equation*}\label{th601}
\end{theorem}


\begin{proof}
 On a
$$x_{1}=\frac{q}{p+x_{-k}},$$
donc le résultat est vérifié pour  $n=0$.  Supposons que  $n>0$ et
que notre proposition est vraie pour $n-1$. C'est à dire,
\begin{equation*}
x_{(k+1)n-(k+1)+i}=\frac{W_{n}+W_{n-1}x_{i-(k+1)}}{W_{n+1}+W_{n}x_{i-(k+1)}}q.
\end{equation*}
Maintenant, il découle de  Eq.\eqref{eq601} que
\begin{eqnarray*}
x_{(k+1)n+i} &=&\dfrac{q}{p+x_{(k+1)-(1+k)+i}}\\
&=&\dfrac{q}{p+\dfrac{W_{n}+W_{n-1}x_{i-(k+1)}}{W_{n+1}+W_{n}x_{i-(k+1)}}}\\
&=&\dfrac{q}{\dfrac{pW_{n+1}+qW_{n}+(pW_{n}+qW_{n-1})x_{i-(k+1)}}{W_{n+1}+W_{n}x_{i-(k+1)}}}.\\
\end{eqnarray*}
Donc, on a
\begin{equation*}
x_{(k+1)n+i}=\frac{W_{n+1}+W_{n}x_{i-(k+1)}}{W_{n+2}+W_{n+1}x_{i-(k+1)}}q.
\end{equation*}
\end{proof}

\begin{corollary}
tttttttttttttttttttttttt
\end{corollary}

\begin{eqnarray*}
\lim_{n\rightarrow\infty}x_{(k+1)n+i}&=&\lim_{n\rightarrow\infty}\frac{W_{n+1}+W_{n}x_{i-(k+1)}}{W_{n+2}+W_{n+1}x_{i-(k+1)}}q\\
 &=&\lim_{n\rightarrow\infty}\frac{q+\dfrac{q}{\Phi_{+}}x_{i-(k+1)}}{\Phi_{+}+x_{i-(k+1)}}\\
 &=&\frac{-p+\sqrt{p^{2}+4q}}{2}.
\end{eqnarray*}

\begin{enumerate}
  \item yyyyyyyyyy gggggggggggg
  \item rrrrrrrrrrrrrrrrf  hn
  \item tttt dd dddd
\end{enumerate}
 \chapter{ Titre chapire 2 }
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\chapter{ Titre chapire 3}
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\begin{thebibliography}{99}\pagestyle{fancy}\lhead{\textbf{\textit{Bibliographie} }}\rhead{}



\bibitem {2}S. Elaydi, \emph{An introduction
to difference equation}, Departement of mathematics, Trinity
University San Antonio, Texas 78212 USA, 2005.






\bibitem {8}N. Touafek, Y.\ Halim, \emph{Global attractivity of a rational
difference equation}, Mathematical Sciences Letters 2, No.3, (2013),
161-165.


\end{thebibliography}





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