From 2db144d6141d827fa394d04e7e4ef6771c7089ee Mon Sep 17 00:00:00 2001
From: couchot <jf.couchot@gmail.com>
Date: Thu, 21 Nov 2013 09:22:21 +0100
Subject: [PATCH] j

---
 complexite.tex           |   6 +-
 equations.tex            | 168 ++++++++++++++++++++-------------------
 interpol.tex             |   6 +-
 partiels/13mesi/main.log |  32 ++++----
 partiels/13mesi/main.pdf | Bin 113686 -> 113815 bytes
 partiels/13mesi/main.tex |  20 +++--
 partiels/main.tex        | 117 ---------------------------
 tps/chap2/diff_div.py    |  18 +++--
 tps/chap3/jjm.py~        |  23 ------
 tps/chap3/methodes.py    |  86 ++++++++------------
 tps/chap3/methodes.pyc   | Bin 3502 -> 4964 bytes
 tps/chap3/methodes.py~   | 128 -----------------------------
 tps/chap3/ordre.py       |  13 +--
 13 files changed, 171 insertions(+), 446 deletions(-)
 delete mode 100644 partiels/main.tex
 delete mode 100644 tps/chap3/jjm.py~
 delete mode 100644 tps/chap3/methodes.py~

diff --git a/complexite.tex b/complexite.tex
index 4fdf00d..6d66125 100644
--- a/complexite.tex
+++ b/complexite.tex
@@ -36,7 +36,7 @@ $5n^3\leq 10.\dfrac{n^3}{2}$.
 constante $c$ telle que 
 $\dfrac{n^3}{2} \leq c.(n^2 +17n + 5)$ soit encore 
 $\dfrac{n^3}{2(n^2 +17n + 5)} \leq c$ et donc  
-$\dfrac{n^3}{2} \leq c$ ce qui est impossible. 
+$\dfrac{n}{2} \leq c$ ce qui est impossible. 
 \end{itemize}
 
 
@@ -200,7 +200,7 @@ On considère tout d'abord l'algorithme suivant:
  \KwData{$n$: degré, $(a_i)_{0 \le i \le n}$: coefficients, $t$: 
    réel en lequel on évalue}
  \KwResult{$\textit{val}$: réel tel que $\textit{val}= p(t)$}
- $a' = a$ \; \nllabel{laff1} 
+ $a' = a_n$ \; \nllabel{laff1} 
  \For{$i=n-1$ \KwTo $0$}{\nllabel{laford} 
     $a' = a_i + t \times a'$ \; \nllabel{lafori}
  \nllabel{laforf}}
@@ -280,7 +280,7 @@ n & T_1(n) & T_2(n) \\
 \begin{enumerate}
 \item En déduire quel sera le temps nécessaire au traitement de données 
   de taille $n'=10^4$.
-\item Même question avec $n'=200\lambda$ où $\lambda$ réel de $[1, +\infty[$.
+\item Même question avec $n'=200\lambda$ où $\lambda$ réel de $[0, +\infty[$.
 \item Déterminer la taille maximale des données que peuvent traiter 
   $A_1$ et $A_2$ si le temps disponible est $t=10^5$.
 \end{enumerate}
diff --git a/equations.tex b/equations.tex
index 8fec68d..bd10efc 100644
--- a/equations.tex
+++ b/equations.tex
@@ -1,6 +1,5 @@
 \section{Motivation}
-Bien qu’il puisse être posé simplement (trouver tous les $x$ tel que $P(x) =0$),
-résoudre algébriquement des équations
+Résoudre algébriquement des équations
 est un problème difficile.
 Depuis l'antiquité, l’homme a cherché des algorithmes donnant les
 valeurs des racines d'un polynôme 
@@ -12,18 +11,17 @@ la forme
 $ 
 x^3 + px+q = 0
 $ avec $p$ et $q$ non nuls. Calculons le discriminant 
-$\Delta = \frac{4}{27}p^3+ p^2$ et discutons de son signe:
+$\Delta = \frac{4}{27}p^3+ q^2$ et discutons de son signe:
 \begin{itemize}
 \item si $\Delta$ est nul, l'équation possède
   deux solutions réelles, une simple et une double :
-
-$$
+$
 \begin{cases}
   x_0=2\sqrt[3]{\frac{-q}2}=\frac{3q}p
   \\
   x_1=x_2=-\sqrt[3]{\frac{-q}2}=\frac{-3q}{2p}
 \end{cases}
-$$
+$
 
 
 \item Si $\Delta$ est positif, l'équation possède une solution réelle
@@ -31,21 +29,21 @@ $$
 $u = \sqrt[3]{\frac{-q + \sqrt{\Delta}}2}$ et $v = \sqrt[3]{\frac{-q - \sqrt{\Delta}}2}$.
 La seule solution réelle est alors $x_0=u+v$. 
 Il existe également deux solutions complexes conjuguées l'une de l'autre:
-$$
+$
 \begin{cases}
 x_1= j u +\bar{j} v \\ 
 x_2= j^2u +\overline{j^2}\end{cases}
-\qquad\textrm{ où }\qquad j=-\frac12+ i \frac{\sqrt3}2=e^{i\frac{2\pi}3}
-$$
-
-\item si $\Delta$ est négatif,  l'équation possède trois solutions réelles:
-$$
-x_k = 2 \sqrt{\frac{-p}{3}} \cos{\left(\frac13\arccos{\left(\frac{-q}{2}\sqrt{\frac{27}{-p^3}}\right)}+ \frac{2k\pi}{3}\right)}\qquad\mbox{ avec }\qquad k\in\{0,1,2\}.
-$$
+\textrm{ où } j=-\frac12+ i \frac{\sqrt3}2=e^{i\frac{2\pi}3}
+$
 
+\item sinon,  l'équation a trois solutions réelles:
+$
+x_k = 2 \sqrt{\frac{-p}{3}} \cos{\left(\frac13\arccos{\left(\frac{-q}{2}\sqrt{\frac{27}{-p^3}}\right)}+ \frac{2k\pi}{3}\right)}\textrm{ avec } k\in\{0,1,2\}.
+$
 
 \end{itemize}
-Donnée à titre d'exemple, ce travail montre que rapidement on obtient 
+%Donnée à titre d'exemple, 
+Ce travail montre que rapidement on obtient 
 des algorithmes compliqués.
 De plus, Abel a montrée en 1824 qu'il n'est pas toujours possible 
 d'exprimer les racines de l'équation générale de
@@ -83,7 +81,7 @@ le résultat est établi, c.-à-d. $f(a_{n+1})f(b_{n+1})\leq 0$;
 \item sinon, $f(a_n)f(x_n)> 0$. Ainsi $f(a_{n+1})f(b_{n+1})$ a le même 
 signe que $f(a_{n+1})f(b_{n+1})f(a_n)f(x_n)$ c.-à-d. le signe de  
 $f(x_{n})f(b_{n})f(a_n)f(x_n)$ soit encore le signe de  $f(a_n)f(b_n)$.
-Ce dernier est positif d'après l'hypothèse de récurence. 
+Ce dernier est négatif d'après l'hypothèse de récurence. 
 \end{itemize}
 
 A chaque itération, l'intervalle $[a_n,b_n]$ est découpé 
@@ -221,9 +219,9 @@ Dans cet exercice, on suppose en plus que:
 \begin{enumerate}
 \item Montrer que l'on a $x_{n+1} \neq x_{n}$.
 \item Montrer que (\ref{eq:num:class}) est équivalente à 
-\begin{equation}
+$
 x_{n+1} = x_{n} - \frac{f(x_n)}{q_n} \label{eq:num:gen}
-\end{equation}
+$
 \item Dans la représentation graphique donnée figure~\ref{fig:graphe1}:
   \begin{enumerate}
   \item donner l'équation de la droite tracée;
@@ -234,14 +232,14 @@ x_{n+1} = x_{n} - \frac{f(x_n)}{q_n} \label{eq:num:gen}
 \end{enumerate}
 
 \item Dans la méthode de la corde, $q_n$ est constante et égale à 
-\begin{equation}
-q_{n} = \frac{f(b) - f(a)}{b - a} \label{eq:num:corde}
-\end{equation}
+%\begin{equation}
+$q_{n} = \frac{f(b) - f(a)}{b - a} \label{eq:num:corde}
+$%\end{equation}
 
 \item Dans la méthode de Lagrange on a 
-\begin{equation}
-q_{n} = \frac{f(x_{n}) - f(x_{n-1})}{x_n - x_{n-1}} \label{eq:num:lagrange}
-\end{equation}
+%\begin{equation}
+$q_{n} = \frac{f(x_{n}) - f(x_{n-1})}{x_n - x_{n-1}} \label{eq:num:lagrange}
+$%\end{equation}
 
 \item Dans la méthode de Newton on a $q_n = f'(x_n)$
 \end{enumerate}
@@ -282,6 +280,8 @@ solution $\alpha$ de cet intervalle.
 
 
 \begin{Prop}[Vitesse de convergence]
+Si la suite des itérés obtenus par dichotomie converge,
+alors la convergence est linéaire.
 Si la suite des itérés de Lagrange converge, alors la convergence est 
 d'ordre $(1+\sqrt{5})/2$.
 Si la suite des itérés de Newton converge, 
@@ -310,6 +310,67 @@ fixe $g_i(x) =x$ avec
 \end{itemize}
 
 
+
+\begin{TP}[Comparaison d'approches]
+
+\begin{enumerate}
+\item Écrire la fonction 
+\verb+[n,X] = iteration_dichotomie(a,b,m,epsilon,f)+ où
+\begin{itemize}
+\item \verb+a+, \verb+b+ sont les bornes de l'intervalle, \verb+m+
+  est le nombre maximal 
+  d'itérations, \texttt{epsilon} est la précision souhaitée 
+  (voir équation (\ref{eq:erreur:epsilon})) et \verb+f+ la fonction à itérer;
+\item \verb+n+ est le nombre d'itérations réalisées pour que 
+\verb+f(+$\verb+x+_{\verb+n+}$\verb+)+=0 ou que 
+$|\verb+x+_{\verb+n+}- \verb+x+_{\verb+n-1+}| \leq \verb+epsilon+$, \verb+n+ étant inférieur à \verb+m+ et \verb+X+ est 
+  le vecteur contenant les 
+  valeurs $\verb+x+_{\verb+0+},\ldots,\verb+x+_{\verb+n+}$.
+\end{itemize}
+\item Écrire la fonction 
+\verb+[n,X] = iteration_corde(a,b,+$\verb+x+_{\verb+0+}$\verb+,m,epsilon,f)+ où
+\begin{itemize}
+\item $\verb+x+_{\verb+0+}$ est le premier terme de la suite;
+\end{itemize}
+\item Écrire la fonction 
+\verb+[n,X] = iteration_Newton(+$\verb+x+_{\verb+0+}$\verb+,m,epsilon,f)+.
+\end{enumerate}
+\end{TP}
+
+
+\begin{TP}
+L'objectif du TP est de mesurer l'ordre de grandeur de la convergence d'une 
+méthode.
+On suppose que la suite $(x_n)$ converge vers $l$ avec l'ordre $p\geq 1$. 
+On note $e_n= l -x_n$. On a $|e_{n+1}| \approx c |e_n|^p$ et donc 
+$\ln(|e_{n+1}|) \approx p \ln(|e_n|) + \ln(c)$.
+En posant $y = \ln(|e_{n+1}|)$, $x = \ln(|e_n|)$ et $k = \ln(c)$ on a 
+$y = px + k$ soit l'équation d'une droite de pente $p$.
+Pour estimer $p$, on peut donc tracer l'ensemble de points 
+$(\ln(|e_{n}|),\ln(|e_{n+1}|))$, contruire la droite de regression linéaire
+et prendre son coefficient directeur.
+
+\begin{enumerate}
+\item Construire la méthode 
+\verb+p=ordre_convergence(X,l)+ telle que 
+\begin{itemize}
+\item \texttt{X} est le vecteur contenant les valeurs des itérés $\verb+x+_{\verb+0+}$, \ldots, $\verb+x+_{\verb+n+}$ et 
+  \verb+l+ est la limite présumée de la suite;
+\item cette fonction exploite la fonction 
+\verb+scipy.stats.linregress(x, y=None)+;
+\item \verb+p+ est l'ordre de convergence calculé numériquement.
+\end{itemize}
+
+\item Tester les méthodes du TP précédent 
+  (dichotomie, corde, Lagrange, Newton) pour la fonction 
+  $f$ définie par $f(x)=\cos(x)-x$ sur $[0,\pi/2]$.
+  Calculer l'ordre à l'aide de la fonction développée à la première question.
+\item Comparer avec la fonction \verb+scipy.optimize.newton+.
+\end{enumerate}
+\end{TP}
+
+
+
 \subsection{Algorithme du point fixe}
 L'algorithme du point fixe donné ci dessous (Algorithme~\ref{algo:pf}) est 
 une version constructive de la suite $(x_n)$ définie par $x_0$ et pour tout $n \in \N$, 
@@ -358,7 +419,7 @@ Pour chacune des équations suivantes, étudier la convergence de la suite des
 itérés du point fixe pour un $x_0$ choisi dans l'intervalle proposé.
 \begin{enumerate}
 \item fonction $g_3$ de Ferrari sur $[2;4]$;
-\item fonction $g_2$ de Ferrari sur $[3;\textrm{3,1}]$;
+\item fonction $g_2$ de Ferrari sur $[3;\textrm{3,1}]$.
 \end{enumerate}
 \end{Exo}
 
@@ -376,62 +437,5 @@ proposition~\ref{th:csconv:pf}  sont suffisantes, pas nécessaires.
 \end{enumerate}
 \end{Exo}
 
-\begin{TP}
-Tout le code suivant est à faire en python.
-\begin{enumerate}
-\item Écrire la fonction 
-\verb+[n,X] = iteration_dichotomie(a,b,m,epsilon,f)+ où
-\begin{itemize}
-\item \verb+a+, \verb+b+ sont les bornes de l'intervalle, \verb+m+
-  est le nombre maximal 
-  d'itérations, \texttt{epsilon} est la précision souhaitée 
-  (voir équation (\ref{eq:erreur:epsilon})) et \verb+f+ la fonction à itérer;
-\item \verb+n+ est le nombre d'itérations réalisées pour que 
-\verb+f(+$\verb+x+_{\verb+n+}$\verb+)+=0 ou que 
-$|\verb+x+_{\verb+n+}- \verb+x+_{\verb+n-1+}| \leq \verb+epsilon+$, \verb+n+ étant inférieur à \verb+m+ et \verb+X+ est 
-  le vecteur contenant les 
-  valeurs $\verb+x+_{\verb+0+},\ldots,\verb+x+_{\verb+n+}$.
-\end{itemize}
-\item Écrire la fonction 
-\verb+[n,X] = iteration_corde(a,b,+$\verb+x+_{\verb+0+}$\verb+,m,epsilon,f)+ où
-\begin{itemize}
-\item $\verb+x+_{\verb+0+}$ est le premier terme de la suite;
-\end{itemize}
-\item Écrire la fonction 
-\verb+[n,X] = iteration_Newton(+$\verb+x+_{\verb+0+}$\verb+,m,epsilon,f)+.
-\end{enumerate}
-\end{TP}
-
-
-\begin{TP}
-L'objectif du TP est de mesurer l'ordre de grandeur de la convergence d'une 
-méthode.
-On suppose que la suite $(x_n)$ converge vers $l$ avec l'ordre $p\geq 1$. 
-On note $e_n= l -x_n$. On a $|e_{n+1}| \approx c |e_n|^p$ et donc 
-$\ln(|e_{n+1}|) \approx p \ln(|e_n|) + \ln(c)$.
-En posant $y = \ln(|e_{n+1}|)$, $x = \ln(|e_n|)$ et $k = \ln(c)$ on a 
-$y = px + k$ soit l'équation d'une droite de pente $p$.
-Pour estimer $p$, on peut donc tracer l'ensemble de points 
-$(\ln(|e_{n}|),\ln(|e_{n+1}|))$, contruire la droite de regression linéaire
-et prendre son coefficient directeur.
-
-\begin{enumerate}
-\item Construire la méthode 
-\verb+p=ordre_convergence(X,l)+ telle que 
-\begin{itemize}
-\item \texttt{X} est le vecteur contenant les valeurs des itérés $\verb+x+_{\verb+0+}$, \ldots, $\verb+x+_{\verb+n+}$ et 
-  \verb+l+ est la limite présumée de la suite;
-\item cette fonction exploite la fonction 
-\verb+scipy.stats.linregress(x, y=None)+;
-\item \verb+p+ est l'ordre de convergence calculé numériquement.
-\end{itemize}
-
-\item Tester les méthodes du TP précédent 
-  (dichotomie, corde, Lagrange, Newton) pour la fonction 
-  $f$ définie par $f(x)=\cos(x)-x$ sur $[0,\pi/2]$.
-  Calculer l'ordre à l'aide de la fonction développée à la première question.
-\item Comparer avec la fonction \verb+scipy.optimize.newton+.
-\end{enumerate}
-\end{TP}
 
 
diff --git a/interpol.tex b/interpol.tex
index b5eef1f..a3d0cda 100644
--- a/interpol.tex
+++ b/interpol.tex
@@ -74,11 +74,11 @@ $$p_i(x) = p_{i-1}(x) + d_i  \left[  \prod_{j=0}^{i-1} (x - x_j)\right].$$
 
 Ainsi pour obtenir $p$ sur $x_0$, $x_1$, \ldots, $x_n$, il suffit:
 \begin{itemize}
-\item de calculer $d_0$, pour définir $p_0$ qui interpole $p$ sur $x_0$,
-\item de calculer $d_1$, pour définir $p_1$  qui interpole $p$ sur $x_0$ 
+\item de calculer $d_0$, pour définir $p_0$ qui interpole $f$ sur $x_0$,
+\item de calculer $d_1$, pour définir $p_1$  qui interpole $f$ sur $x_0$ 
   et $x_1$,
 \item \ldots
-\item de calculer $d_n$, pour définir $p_n$  qui interpole $p$ sur 
+\item de calculer $d_n$, pour définir $p_n$  qui interpole $f$ sur 
   $x_0$, $x_1$, \ldots, $x_n$.
 \end{itemize}
 
diff --git a/partiels/13mesi/main.log b/partiels/13mesi/main.log
index 5f4531e..401fce1 100644
--- a/partiels/13mesi/main.log
+++ b/partiels/13mesi/main.log
@@ -1,4 +1,4 @@
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+This is pdfTeX, Version 3.1415926-2.5-1.40.14 (TeX Live 2013/Debian) (format=pdflatex 2013.11.3)  4 NOV 2013 10:10
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diff --git a/partiels/13mesi/main.pdf b/partiels/13mesi/main.pdf
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diff --git a/partiels/13mesi/main.tex b/partiels/13mesi/main.tex
index e8c2577..5ed0b08 100644
--- a/partiels/13mesi/main.tex
+++ b/partiels/13mesi/main.tex
@@ -1,4 +1,4 @@
-\documentclass[10pt,a4paper,french]{article}
+\documentclass[11pt,a4paper,french]{article}
 \usepackage[francais]{babel}
 \usepackage[utf8]{inputenc}
 \usepackage{a4}
@@ -29,7 +29,7 @@
   Novembre 2013 (durée 45 mn).  J.-F.  Couchot,}
 
 \maketitle
-\vspace{-5em}
+\vspace{-3em}
 % \begin{tabular}{ll}
 % Nom:& ........................................\\
 % Prénom:& ........................................\\
@@ -82,31 +82,35 @@ Vos réponses seront données directement ci-dessous.
 \item (2pts) Montrer que l'équation de la droite $(D_n)$ est  
 $y - f(b_n) = \frac{f(b_n)-f(a_n)}{b_n-a_n} (x-b_n)$. 
 
-\vspace{4cm}
+\vspace{3cm}
 
 \item (2pts) Montrer que le nombre $x_n$ est donné par l'équation
 $x_n = a_n - \frac{a_n-b_n}{f(a_n)-f(b_n)} f(a_n)$.
  
-\vspace{4cm}
+\vspace{3cm}
 
 
 \item (2pts) En moyenne, l'ordre de cette méthode est 1,618. 
   Comparer cet ordre avec celui des autres méthodes du cours. 
-\vspace{4cm}
+\vspace{3cm}
 
 \item (5pts)  Quelle partie de cette méthode est commune avec la 
   méthode par dichotomie? Est-elle toujours plus efficace? 
-  Comparer les approches par exemple 
-  sur l'intervalle $[-1,1]$ avec fonction $f$ définie sur $\mathds{R}$ par
+  Comparer les deux approches par exemple 
+  sur l'intervalle $[-1,1]$ avec la fonction $f$ définie sur $\mathds{R}$ par
   $f(x)= 2x^3-4x^2+3x$. 
 \vspace{6cm}
 
+\newpage
+\vspace{5cm}
 \item  (4pts) Quelle partie de cette méthode est commune avec la méthode de Lagrange? 
 Est-ce la même méthode? Si ce n'est pas le cas, Expliquer ce qui diffère.
 \vspace{4cm}
 
 
-\item (5pts) Donner le code d'un programme qui implanterait cette méthode.
+\item (5pts) Donner le code d'un programme qui implanterait cette méthode, 
+  et ce dans le langage de votre choix.
+  
 
 \vspace{4cm}
 
diff --git a/partiels/main.tex b/partiels/main.tex
deleted file mode 100644
index 57c391d..0000000
--- a/partiels/main.tex
+++ /dev/null
@@ -1,117 +0,0 @@
-\documentclass[10pt,a4paper,french]{article}
-\usepackage[francais]{babel}
-\usepackage[utf8]{inputenc}
-\usepackage{a4}
-\usepackage{amsmath}
-\usepackage{amsfonts}
-\usepackage{amssymb}
-\usepackage{framed}
-\usepackage{dsfont}
-\usepackage[amsmath,thmmarks,thref,framed]{ntheorem}
-\usepackage[dvips]{graphics}
-\usepackage{epsfig}
-\usepackage{calc}
-\usepackage{tabls}
-\usepackage{slashbox}
-\usepackage{times}
-\usepackage{multicol}
-\usepackage{tabularx}
-\usepackage{textcomp}
-
-\usepackage{pst-all}
-
-\usepackage[a4paper]{geometry}
-
-
-\date{}
-\geometry{hmargin=1cm, tmargin=1cm,bmargin=1.5cm}
-\begin{document}
-\title{UE MESI, Master IMR 2ème année.\\
-  Novembre 2012 (durée 1h).  J.-F.  Couchot,}
-
-\maketitle
-\vspace{-5em}
-\begin{tabular}{ll}
-Nom:& ........................................\\
-Prénom:& ........................................\\
-\end{tabular}
-
-
-On s'intéresse à résoudre une équation de la forme $f(x)=0$ par la 
-méthode de Müller. Dans cette méthode, on considère que l'on 
-a le triplet de points $(x_{n-2},x_{n-1},x_{n})$. Pour calculer 
-$x_{n+1}$, on fait comme suit:
-\begin{enumerate}
-\item \label{itm:1} on approche $f(x)$ par un polynôme $P(x)$ aux points
-  $(x_{n-2},x_{n-1},x_{n})$,
-\item\label{itm:2} on résout l'équation $P(x)=0$. La racine la plus proche de $x_n$ est $x_{n+1}$;
-\item on recommence avec le triplet $(x_{n-1},x_{n},x_{n+1})$\ldots
-\end{enumerate}
-
-
-
-
-
-
-Vos réponses seront données directement ci-dessous.
-
-\begin{enumerate}
-
-\item En utilisant une base de Lagrange, montrer que le polynôme $P(x)$ 
-  obtenu à l'étape 1. de la première itération est défini par 
-$$
-P(x) =  \dfrac{(x-x_{n-1})(x-x_{n-2})f(x_n)}{(x_n-x_{n-1})(x_n-x_{n-2})}  +
-\dfrac{(x-x_{n})(x-x_{n-2})f(x_{n-1})}{(x_{n-1}-x_{n})(x_{n-1}-x_{n-2})} +
-\dfrac{(x-x_{n})(x-x_{n-1})f(x_{n-2})}{(x_{n-2}-x_{n})(x_{n-2}-x_{n-1})} 
-$$   
-
-\vspace{4cm}
-
-\item Montrer que le polynôme de la question précédente est de degré 2. 
-  Est-ce cohérent avec le fait qu'on veuille approximer $f$ en trois points? 
-\vspace{4cm}
-
-\item Montrer que le polynôme de la première question peut s'écrire 
-  sous la forme $P(x) = a_n x^2 + b_n x + c $ où 
-\begin{eqnarray*}
-a_n & = & \dfrac{f(x_n)}{(x_n-x_{n-1})(x_n-x_{n-2})} +
-\dfrac{f(x_{n-1})}{(x_{n-1}-x_{n})(x_{n-1}-x_{n-2})} +
-\dfrac{f(x_{n-2})}{(x_{n-2}-x_{n})(x_{n-2}-x_{n-1})} \\
-b_n & = &-\dfrac{f(x_n)(x_{n-1}+x_{n-2})}{(x_n-x_{n-1})(x_n-x_{n-2})} -
-\dfrac{f(x_{n-1})(x_{n}+x_{n-2})}{(x_{n-1}-x_{n})(x_{n-1}-x_{n-2})} -
-\dfrac{f(x_{n-2})(x_{n}+x_{n-1})}{(x_{n-2}-x_{n})(x_{n-2}-x_{n-1})} \\
-c_n & = & \dfrac{f(x_n)x_{n-1}x_{n-2}}{(x_n-x_{n-1})(x_n-x_{n-2})} +
-\dfrac{f(x_{n-1})x_{n}x_{n-2}}{(x_{n-1}-x_{n})(x_{n-1}-x_{n-2})} +
-\dfrac{f(x_{n-2})x_{n}x_{n-1}}{(x_{n-2}-x_{n})(x_{n-2}-x_{n-1})} 
-\end{eqnarray*}
-\vspace{8cm}
-
-
-\item Exprimer les deux racines $x'_{n}$ et $x''_{n}$ du polynôme précédent
-en fonctions de $a_n$, $b_n$ et  $c_n$ lorsqu'on itère dans les réels.
-\vspace{3cm}
-
-\item Comment est alors défini $x_{n+1}$?
-\vspace{3cm}
-
-\item On pourrait montrer que l'ordre de la convergence est 1,84. Comparer cette vitesse de convergence avec celle de Newton et celle de Lagrange.
-\vspace{3cm}
-
-
-\item Donner le code Python de la fonction 
-  $\verb+[n,X] = iteration_muller(x+_{\verb+0+},\verb+x+_{\verb+1+},\verb+x+_{\verb+2+}\verb+,m,epsilon,f)+$ où
-\begin{itemize}
-\item $\verb+x+_{\verb+0+}$, $\verb+x+_{\verb+1+}$ et $\verb+x+_{\verb+2+}$ sont les trois premières valeurs des itérés, \verb+m+
-  est le nombre maximal 
-  d'itérations, \texttt{epsilon} est la précision souhaitée 
-  et \verb+f+ la fonction à itérer;
-\item \verb+n+ est le nombre d'itérations réalisées pour que 
-\verb+f(+$\verb+x+_{\verb+n+}$\verb+)+=0 ou que 
-$|\verb+x+_{\verb+n+}- \verb+x+_{\verb+n-1+}| \leq \verb+epsilon+$, \verb+n+ étant inférieur à \verb+m+ et \verb+X+ est 
-  le vecteur contenant les 
-  valeurs $\verb+x+_{\verb+0+},\ldots,\verb+x+_{\verb+n+}$.
-\end{itemize}
-\end{enumerate}
-
-
-\end{document}
\ No newline at end of file
diff --git a/tps/chap2/diff_div.py b/tps/chap2/diff_div.py
index 8ac41e3..a2c7573 100644
--- a/tps/chap2/diff_div.py
+++ b/tps/chap2/diff_div.py
@@ -4,12 +4,12 @@ from math import *
 
 
 def coeffs(X,Y):
-    n = len(X)-1
-    diff_div=[[0 for _ in xrange(n+1)] for _ in xrange(n+1)]
-    for i in xrange(n+1):
+    n = len(X)
+    diff_div=np.zeros((n,n))
+    for i in range(n):
         diff_div[i][0] = Y[i]
-    for j in xrange(1,n+1):
-        for i in xrange(n-j+1):
+    for j in range(1,n):
+        for i in range(n-j):
             r = float(diff_div[i+1][j-1]-diff_div[i][j-1])/(X[i+j]-X[i])
             diff_div[i][j] = r
     print diff_div
@@ -34,6 +34,7 @@ def construit_et_eval_pol(X,Y,x):
 
 def construit_pol(X,Y):
     d = coeffs(X,Y)
+    print "les coeffs sont",d
     n = len(X)-1
     return lambda x : \
         sum([d[i]*prod([x - X[j] for j in xrange(i)]) for i in xrange(n+1)])
@@ -41,17 +42,18 @@ def construit_pol(X,Y):
 
 
 
-"""
+
 
 print "tp 2.1....."
 X = [10,25,60]
 XX= [15,40,100]
 Y = [2.3, 8, 24.6]
 p = construit_pol(X,Y)
+
 Yp =[p(x) for x in XX]
 Ypp = [construit_et_eval_pol(X,Y,x) for x in XX]
 
-print Yp, Ypp
+print "Yp", Yp#, Ypp
 
 
 x=np.linspace(10,100,100)
@@ -106,4 +108,4 @@ plt.ylabel('exp')
 plt.xlabel("x")
 plt.show()
 
-
+"""
diff --git a/tps/chap3/jjm.py~ b/tps/chap3/jjm.py~
deleted file mode 100644
index da9ef8b..0000000
--- a/tps/chap3/jjm.py~
+++ /dev/null
@@ -1,23 +0,0 @@
-def iteration_lagrange(x0, q0, m, epsilon, f):
-    def maj_test(xn,xnm1):
-        return f(xn)!=0 and abs(xn-xnm1)>epsilon
-
-    n=0
-    test=f(x0)!=0
-    X=[x0]
-    xnm1=x0
-    xn=float(x0-f(x0))/q0
-    qn=q0
-    print(xn,q0)
-    while n<=m and test:
-        qn=float((f(xn)-f(xnm1)))/(xn-xnm1)
-        xnm1=xn
-        xn=xnm1-f(xnm1)/qn
-
-        X+=[xn]
-        test=maj_test(xn,xnm1)
-        n+=1
-    return (n,X)
-
-////////////////////////////////////////////////////////////////////////////
-print(iteration_lagrange(pi/4,1,200,0.0000001,f))
diff --git a/tps/chap3/methodes.py b/tps/chap3/methodes.py
index 8ae3b71..9c9424e 100644
--- a/tps/chap3/methodes.py
+++ b/tps/chap3/methodes.py
@@ -16,79 +16,61 @@ def fp(x):
 
 
 def  iteration_dichotomie(a,b,m,epsilon,f):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    xnm1 = a 
-    xn= a
+    def maj_test(xn,xnm1,n,m):
+        return f(xn) == 0 or abs(xnm1-xn) <= epsilon or n >= m    
+    xn= float(a)
+    xnm1 =float(b) 
     X=[]
-    n = 1
-    an= a
-    bn=b
-    test = True
-    while n <= m and test:
+    n = 0
+    an= float(a)
+    bn=float(b)
+    test = maj_test(xn,xnm1,n,m)
+    while not test:
         xnm1 = xn
         xn=float(an+bn)/2
-        test = maj_test(xn,xnm1)
         X +=[xn]
         if f(an)*f(xn)<=0 : 
             bn=xn
         else :
             an=xn
         n +=1
+        test = maj_test(xn,xnm1,n,m)
     return (n,X)
 
 def  iteration_newton(x0,m,epsilon,f,fp):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    n=0;
-    test= f(x0) != 0
-    xn=x0
-    X=[x0]
-    while n < m and test:
+    def maj_test(xn,xnm1,epsilon,n,m,f):
+        return f(xn) == 0 or abs(xnm1-xn) <= epsilon or n >= m
+    n = 0
+    xn = float(x0)
+    X = [x0]
+    test = maj_test(xn,xn+2*epsilon,epsilon,n,m,f)
+    while not test:
         qn=fp(xn)
         xnm1=xn
         xn= xn-f(xn)/qn
         X += [xn]
         n=n+1
-        test= maj_test(xn,xnm1)
-
-#f(x) !=0 and n<m  and abs(x-xm1)>epsilon
+        test= maj_test(xn,xnm1,epsilon,n,m,f)
     return (n,X)
 
 
 def  iteration_corde(a,b,x0,m,epsilon,f):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    n=0;
+    def maj_test(xn,xnm1,n,m,f):
+        return f(xn)== 0 or abs(xnm1-xn) <= epsilon or n >= m
+    n=0
     q=float(f(b)-f(a))/(b-a)
-    test= f(x0) != 0
-    xn=x0
-    X=[x0]
-    while n < m and test:
+    xnm1 = float(b)
+    xn=float(x0)
+    test= maj_test(xn,xnm1,n,m,f)
+    X=[float(x0)]
+    while not test:
         xnm1=xn
-        xn= xn-f(xn)/q
+        xn= xn-float(f(xn))/q
         X += [xn]
         n=n+1
-        test= maj_test(xn,xnm1)
-
-#f(x) !=0 and n<m  and abs(x-xm1)>epsilon
+        test= maj_test(xn,xnm1,n,m,f)
     return (n,X)
 
-"""def  iteration_newton(x0,m,epsilon,f,fp):
-    n=0;
-    delta=float(1)/fp(x0)
-    test= f(x0) != 0
-    x=x0
-    X=[x0]
-    while(test):
-        xm1=x
-        x= x-delta*f(x)
-        delta=float(1)/fp(x)
-        X += [x]
-        n=n+1
-        test= not (f(x)==0 or n>=m  or abs(x-xm1)<=epsilon)
-    return (n,X)
-"""
 
 def  iteration_lagrange(x0,x1,m,epsilon,f):
     n=0;
@@ -138,24 +120,24 @@ def  iteration_muller(x0,x1,x2,m,epsilon,f):
 
 def main():
     print "TP 3.1 ............ dichotomie"
-    print iteration_dichotomie(0,pi/2,200,0.00000001,f)
+    print iteration_dichotomie(0,pi/2,45,1E-9,f)
 
-    
     print "TP 3.1 ............ corde"
-    print iteration_corde(0,pi/2,0,200,0.00000001,f)
-
+    print iteration_corde(0,pi/2,0,200,1E-9,f)
 
     print "TP 3.1 ............ newton"
-    print iteration_newton(0,200,0.00000001,f,fp)
+    print iteration_newton(0,200,1E-9,f,fp)
 
 
+
+"""
     print "TP 3.1 ............ lagrange"
     print iteration_lagrange(0,pi/2,200,0.00000001,f)
 
     print "TP 3.1 ............ muller"
     print iteration_muller(0,pi/4,pi/2,200,0.00000001,f)
     
-
+"""
 if __name__ == '__main__':
     main()
 
diff --git a/tps/chap3/methodes.pyc b/tps/chap3/methodes.pyc
index 29f025b357c103ef7ae073440747650c45c18c7f..92edf69e31dd458a3155546e98878358085851e2 100644
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diff --git a/tps/chap3/methodes.py~ b/tps/chap3/methodes.py~
deleted file mode 100644
index c7080e2..0000000
--- a/tps/chap3/methodes.py~
+++ /dev/null
@@ -1,128 +0,0 @@
-import matplotlib.pyplot as plt
-import numpy as np
-from math import *
-
-
-
-
-
-def f(x):
-    return cos(x)-x
-
-def fp(x):
-    return -sin(x)-1
-
-
-
-
-def  iteration_dichotomie(a,b,m,epsilon,f):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    xnm1 = a 
-    xn= a
-    X=[]
-    n = 1
-    an= a
-    bn=b
-    test = True
-    while n <= m and test:
-        xnm1 = xn
-        xn=float(an+bn)/2
-        test = maj_test(xn,xnm1)
-        X +=[xn]
-        if f(an)*f(xn)<=0 : 
-            bn=xn
-        else :
-            an=xn
-        n +=1
-    return (n,X)
-
-def  iteration_newton(x0,m,epsilon,f,fp):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    n=0;
-    test= f(x0) != 0
-    xn=x0
-    X=[x0]
-    while n < m and test:
-        qn=fp(xn)
-        xnm1=xn
-        xn= xn-f(xn)/qn
-        X += [xn]
-        n=n+1
-        test= maj_test(xn,xnm1)
-
-#f(x) !=0 and n<m  and abs(x-xm1)>epsilon
-    return (n,X)
-
-
-def  iteration_corde(a,b,x0,m,epsilon,f):
-    def maj_test(xn,xnm1):
-        return f(xn) != 0 and abs(xnm1-xn) > epsilon  
-    n=0;
-    q=float(f(b)-f(a))/(b-a)
-    test= f(x0) != 0
-    xn=x0
-    X=[x0]
-    while n < m and test:
-        xnm1=xn
-        xn= xn-f(xn)/q
-        X += [xn]
-        n=n+1
-        test= maj_test(xn,xnm1)
-
-#f(x) !=0 and n<m  and abs(x-xm1)>epsilon
-    return (n,X)
-
-"""def  iteration_newton(x0,m,epsilon,f,fp):
-    n=0;
-    delta=float(1)/fp(x0)
-    test= f(x0) != 0
-    x=x0
-    X=[x0]
-    while(test):
-        xm1=x
-        x= x-delta*f(x)
-        delta=float(1)/fp(x)
-        X += [x]
-        n=n+1
-        test= not (f(x)==0 or n>=m  or abs(x-xm1)<=epsilon)
-    return (n,X)
-"""
-
-def  iteration_lagrange(x0,x1,m,epsilon,f):
-    n=0;
-    delta=float(x1-x0)/(f(x1)-f(x0))
-    test= f(x0) != 0
-    x=x1
-    X=[x1]
-    while(test):
-        xm1=x
-        x= x-delta*f(x)
-        delta=float(x-xm1)/(f(x)-f(xm1))
-        X += [x]
-        n=n+1
-        test= not (f(x)==0 or n>=m  or abs(x-xm1)<=epsilon)
-    return (n,X)
-
-
-def main():
-    print "TP 3.1 ............ dichotomie"
-    print iteration_dichotomie(0,pi/2,200,0.00000001,f)
-
-    
-    print "TP 3.1 ............ corde"
-    print iteration_corde(0,pi/2,0,200,0.00000001,f)
-
-
-    print "TP 3.1 ............ newton"
-    print iteration_newton(0,200,0.00000001,f,fp)
-
-
-    print "TP 3.1 ............ lagrange"
-    print iteration_lagrange(0,pi/2,200,0.00000001,f)
-    
-
-if __name__ == '__main__':
-    main()
-
diff --git a/tps/chap3/ordre.py b/tps/chap3/ordre.py
index 95517ff..00283a8 100644
--- a/tps/chap3/ordre.py
+++ b/tps/chap3/ordre.py
@@ -22,16 +22,17 @@ def ordre_convergence(X,l=None) :
 
 def main():
 
-    print "TP 3.2 ............ ordre convergence corde"
-    print ordre_convergence(iteration_corde(0,pi/2,0,200,0.00000001,f)[1])
+    print "TP 4.2 ............ ordre convergence corde"
+    print ordre_convergence(iteration_corde(0,pi/2,0,200,1E-9,f)[1])
 
 
-    print "TP 3.2 ............ ordre convergence newton"
-    print ordre_convergence(iteration_newton(0,200,0.00000001,f,fp)[1])
+    print "TP 4.2 ............ ordre convergence lagrange"
+    print ordre_convergence(iteration_lagrange(0,pi/2,200,1E-9,f)[1])
 
 
-    print "TP 3.1 ............ ordre convergence lagrange"
-    print ordre_convergence(iteration_lagrange(0,pi/2,200,0.00000001,f)[1])
+    print "TP 4.2 ............ ordre convergence newton"
+    print ordre_convergence(iteration_newton(0,200,1E-9,f,fp)[1])
+
 
 
 if __name__ == '__main__':
-- 
2.39.5