-\begin{figure}[t]
-\begin{center}
- \subfigure[ICs on the interval
-$(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad
- \subfigure[ICs on the interval
-$(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad
-\end{center}
-\caption{General aspect of the chaotic iterations.}
-\label{fig:ICs3}
-\end{figure}
-
-
-We have written a Python program to represent the chaotic iterations with the
-vectorial negation on the real line $\mathds{R}$. Various representations of
-these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}.
-It can be remarked that the function $g$ is a piecewise linear function: it is
-linear on each interval having the form $\left[ \dfrac{n}{10},
-\dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its
-slope is equal to 10. Let us justify these claims:
-
-\begin{proposition}
-\label{Prop:derivabilite des ICs}
-Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on
-$\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{
-\dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$.
-
-Furthermore, on each interval of the form $\left[ \dfrac{n}{10},
-\dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$,
-$g$ is a linear function, having a slope equal to 10: $\forall x \notin I,
-g'(x)=10$.
-\end{proposition}
-
-
-\begin{proof}
-Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket
-0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral
-prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$
-and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all
-the images $g(x)$ of these points $x$:
-\begin{itemize}
-\item Have the same integral part, which is $e$, except probably the bit number
-$s^0$. In other words, this integer has approximately the same binary
-decomposition than $e$, the sole exception being the digit $s^0$ (this number is
-then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$,
-\emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$).
-\item A shift to the left has been applied to the decimal part $y$, losing by
-doing so the common first digit $s^0$. In other words, $y$ has been mapped into
-$10\times y - s^0$.
-\end{itemize}
-To sum up, the action of $g$ on the points of $I$ is as follows: first, make a
-multiplication by 10, and second, add the same constant to each term, which is
-$\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$.
-\end{proof}
-
-\begin{remark}
-Finally, chaotic iterations are elements of the large family of functions that
-are both chaotic and piecewise linear (like the tent map).
-\end{remark}
-
-
-
-\subsection{Comparison of the two metrics on $\big[ 0, 2^\mathsf{N} \big[$}
-
-The two propositions bellow allow to compare our two distances on $\big[ 0,
-2^\mathsf{N} \big[$:
-
-\begin{proposition}
-Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0,
-2^\mathsf{N} \big[, D~\right)$ is not continuous.
-\end{proposition}
-
-\begin{proof}
-The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is
-such that:
-\begin{itemize}
-\item $\Delta (x^n,2) \to 0.$
-\item But $D(x^n,2) \geqslant 1$, then $D(x^n,2)$ does not converge to 0.
-\end{itemize}
-
-The sequential characterization of the continuity concludes the demonstration.
-\end{proof}
-
-
-
-A contrario:
-
-\begin{proposition}
-Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0,
-2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction.
-\end{proposition}
-
-\begin{proof}
-If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given
-threshold, because $D_e$ only returns integers. So, after this threshold, the
-integral parts of all the $x^n$ are equal to the integral part of $x$.
-
-Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k
-\in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This
-means that for all $k$, an index $N_k$ can be found such that, $\forall n
-\geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the
-digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the
-result.
-\end{proof}
-
-The conclusion of these propositions is that the proposed metric is more precise
-than the Euclidian distance, that is:
-
-\begin{corollary}
-$D$ is finer than the Euclidian distance $\Delta$.
-\end{corollary}
-
-This corollary can be reformulated as follows:
-
-\begin{itemize}
-\item The topology produced by $\Delta$ is a subset of the topology produced by
-$D$.
-\item $D$ has more open sets than $\Delta$.
-\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than
-to converge with the one inherited by $\Delta$, which is denoted here by
-$\tau_\Delta$.
-\end{itemize}
-
-
-\subsection{Chaos of the chaotic iterations on $\mathds{R}$}
-\label{chpt:Chaos des itérations chaotiques sur R}
-
-
-
-\subsubsection{Chaos according to Devaney}
-
-We have recalled previously that the chaotic iterations $\left(\Go,
-\mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We
-can deduce that they are chaotic on $\mathds{R}$ too, when considering the order
-topology, because:
-\begin{itemize}
-\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10}
-\big[_D\right)$ are semiconjugate by $\varphi$,
-\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic
-according to Devaney, because the semiconjugacy preserve this character.
-\item But the topology generated by $D$ is finer than the topology generated by
-the Euclidian distance $\Delta$ -- which is the order topology.
-\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the
-chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order
-topology on $\mathds{R}$.
-\end{itemize}
-
-This result can be formulated as follows.
-
-\begin{theorem}
-\label{th:IC et topologie de l'ordre}
-The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the
-Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the
-order topology.
-\end{theorem}
-
-Indeed this result is weaker than the theorem establishing the chaos for the
-finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre}
-still remains important. Indeed, we have studied in our previous works a set
-different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$),
-in order to be as close as possible from the computer: the properties of
-disorder proved theoretically will then be preserved when computing. However, we
-could wonder whether this change does not lead to a disorder of a lower quality.
-In other words, have we replaced a situation of a good disorder lost when
-computing, to another situation of a disorder preserved but of bad quality.
-Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary.
-
