diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 0bbbe371b01e359e381e43239412d77bf53fb1fb..f414b1af310a21da1b835efa8de3074f12fc3fd6 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -1,5 +1,85 @@ { - "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 } - diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb index 0bbbe371b01e359e381e43239412d77bf53fb1fb..18d275c354c0e515ce4213614aa196632d45f734 100644 --- a/module2/exo1/toy_notebook_fr.ipynb +++ b/module2/exo1/toy_notebook_fr.ipynb @@ -1,5 +1,96 @@ { - "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 } -