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    "# À propos du calcul de $\\pi$\n",
    "\n",
    "## 1.1 En demandant à la lib maths\n",
    "\n",
    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*\n",
    "\n",
    "\n",
    "In [1]:\n",
    "1 from math import *\n",
    "2 print (pi)\n",
    "    \n",
    "3.141592653589793\n",
    "  \n",
    "\n",
    "## 1.2 En utilisant la méthode des aiguilles de Buffon\n",
    "\n",
    "Mais calculé avec la **méthode** des [aiguilles de Buffon]\n",
    "(https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on \n",
    "obtiendrait comme **approximation** :\n",
    "\n",
    "In [2]:\n",
    "1 import numpy as np\n",
    "2 np.random.seed(seed=42)\n",
    "3 N = 10000\n",
    "4 x = np.random.uniform(size=N, low=0, high=1)\n",
    "5 theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
    "6 2/(sum((x+np.sin(theta))>1)/N)\n",
    "    \n",
    "3.1289111389236548\n",
    "\n",
    "1 ##  Avec un argument \"fréquentiel\" de surface\n",
    "2 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 $P[X^2+Y^2\\leq\n",
    "1] = \\pi/4$ (voir [méthode de Monte Carlo sur Wikipedia]\n",
    "(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",
    "In [3]:\n",
    "\n",
    "1%matplotlib inline\n",
    "2 iport matplotlib.pyplot as plt\n",
    "3\n",
    "4 np.random.dees(seed=42)\n",
    "5 N = 1000\n",
    "6 x = np.random.uniform(size=N, low=0, high=1)\n",
    "7 y = np.random.uniform(size=N, low=0, high=1)\n",
    "8\n",
    "9 accept = (x*x+y*y) <=1\n",
    "10 reject = np.logical_not'accept)\n",
    "11\n",
    "12 fig, ax = plt.subplots(1)\n",
    "13 ax.scatter(x[accept], y [accept], c='b', alpha=0.2, edgecolor=None)\n",
    "14 ax.scatter(x[reject], y [reject], c='r', alpha=0.2, edgecolor=None)\n",
    "15 ax.set_aspect('equal')\n",
    "\n"
   ]
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