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    "# 1 À propos du calcul de $\\pi$\n",
    "\n",
    "## 1.1 En demandant à la lib maths\n",
    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*\n",
    "\n",
    "\n",
    "In \\[1]:\n",
    ">from math import *\n",
    ">print (pi)\n",
    "    \n",
    "3.141592653589793\n",
    "  \n",
    "\n",
    "## 1.2 En utilisant la méthode des aiguilles de Buffon\n",
    "Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on \n",
    "obtiendrait comme **approximation** :\n",
    "\n",
    "\n",
    "In \\[2]:\n",
    ">import numpy as 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]: 3.1289111389236548\n",
    "\n",
    "\n",
    "## 1.3 Avec un argument \"fréquentiel\" de surface\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 $P[X^2+Y^2\\leq1] = \\pi/4$ (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",
    "\n",
    "\n",
    "In \\[3]:\n",
    "\n",
    ">%matplotlib inline\n",
    ">import matplotlib.pyplot as plt\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",
    ">accept = (x*x+y*y) <=1\n",
    ">reject = np.logical_not(accept)\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",
    "\n",
    "\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",
    "\n",
    "\n",
    "In \\[4]: 4*np.mean(accept)\n",
    "\n",
    "Out\\[4]: 3.1120000000000001\n",
    "\n",
    "\n"
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