no commit message

module2/exo1/toy_notebook_en.ipynb

 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# toy_notebook_fr\n",
+    "\n",
+    "## March 28, 2019\n",
+    "\n",
+    "## 1 À propos du calcul de p\n",
+    "\n",
+    "## 1.1 En demandant à la lib maths\n",
+    "\n",
+    "Mon ordinateur m’indique que _p_ vautapproximativement\n",
+    "\n",
+    "In [ 1 ]: frommathimport *\n",
+    "print(pi)\n",
+    "\n",
+    "3.\n",
+    "\n",
+    "## 1.2 En utilisant la méthode des aiguilles de Buffon\n",
+    "\n",
+    "Mais calculé avec la **méthode** desaiguilles de Buffon, on obtiendrait comme **approximation** :\n",
+    "\n",
+    "In [ 2 ]: import numpyas np\n",
+    "np.random.seed(seed= 42 )\n",
+    "N= 10000\n",
+    "x= np.random.uniform(size=N, low= 0 , high= 1 )\n",
+    "theta =np.random.uniform(size=N, low= 0 , high=pi/ 2 )\n",
+    "2 /(sum((x+np.sin(theta))> 1 )/N)\n",
+    "\n",
+    "Out[ 2 ]: 3.\n",
+    "\n",
+    "## 1.3 Avec un argument \"fréquentiel\" de surface\n",
+    "\n",
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction\n",
+    "sinus se base sur le fait que siX \u0018U(0, 1)etY \u0018U(0, 1)alorsP[X^2 +Y^2 \u0014 1 ]= _p_ /4 (voir\n",
+    "méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :\n",
+    "\n",
+    "In [ 3 ]: %matplotlib inline\n",
+    "import matplotlib.pyplotas plt\n",
+    "\n",
+    "```\n",
+    "np.random.seed(seed= 42 )\n",
+    "N= 1000\n",
+    "x= np.random.uniform(size=N, low= 0 , high= 1 )\n",
+    "y= np.random.uniform(size=N, low= 0 , high= 1 )\n",
+    "```\n",
+    "### 1\n",
+    "\n",
+    "\n",
+    "```\n",
+    "accept =(x*x+y*y) <= 1\n",
+    "reject =np.logical_not(accept)\n",
+    "```\n",
+    "```\n",
+    "fig, ax =plt.subplots( 1 )\n",
+    "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
+    "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
+    "ax.set_aspect('equal')\n",
+    "```\n",
+    "Il est alors aisé d’obtenir une approximation (pas terrible) de _p_ en comptant combien de fois,\n",
+    "en moyenne,X^2 +Y^2 est inférieur à 1 :\n",
+    "\n",
+    "In [ 4 ]: 4 *np.mean(accept)\n",
+    "\n",
+    "Out[ 4 ]: 3.\n",
+    "\n",
+    "### 2\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +96,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-

module2/exo1/toy_notebook_fr.ipynb

 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "1\n",
+    "# À propos du calcul de $\\pi$\n",
+    "\n",
+    "1\n",
+    "## En demandant à la lib maths\n",
+    "2\n",
+    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*\n",
+    "In [1]:\n",
+    "\n",
+    "1\n",
+    "from math import *\n",
+    "2\n",
+    "print(pi)\n",
+    "3.141592653589793\n",
+    "\n",
+    "1\n",
+    "## En utilisant la méthode des aiguilles de Buffon\n",
+    "2\n",
+    "Mais calculé avec la __méthode__ des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme __approximation__ :\n",
+    "3\n",
+    "​\n",
+    "In [2]:\n",
+    "\n",
+    "1\n",
+    "import numpy as np\n",
+    "2\n",
+    "np.random.seed(seed=42)\n",
+    "3\n",
+    "N = 10000\n",
+    "4\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "5\n",
+    "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
+    "6\n",
+    "2/(sum((x+np.sin(theta))>1)/N)\n",
+    "3.1289111389236548\n",
+    "\n",
+    "1\n",
+    "## Avec un argument \"fréquentiel\" de surface\n",
+    "2\n",
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d'appel à la fonction sinus se base sur le fait que  si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\leq 1] = \\pi/4$ (voir [méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :\n",
+    "In [3]:\n",
+    "\n",
+    "1\n",
+    "%matplotlib inline  \n",
+    "2\n",
+    "import matplotlib.pyplot as plt\n",
+    "3\n",
+    "​\n",
+    "4\n",
+    "np.random.seed(seed=42)\n",
+    "5\n",
+    "N = 1000\n",
+    "6\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "7\n",
+    "y = np.random.uniform(size=N, low=0, high=1)\n",
+    "8\n",
+    "​\n",
+    "9\n",
+    "accept = (x*x+y*y) <= 1\n",
+    "10\n",
+    "reject = np.logical_not(accept)\n",
+    "11\n",
+    "​\n",
+    "12\n",
+    "fig, ax = plt.subplots(1)\n",
+    "13\n",
+    "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
+    "14\n",
+    "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
+    "15\n",
+    "ax.set_aspect('equal')\n",
+    "\n",
+    "\n",
+    "1\n",
+    "Il est alors aisé d'obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :\n",
+    "In [4]:\n",
+    "\n",
+    "1\n",
+    "4*np.mean(accept)\n",
+    "3.1120000000000001\n"
+   ]
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +107,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-