{
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   "cell_type": "markdown",
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   "source": [
    "# 1. A propos de calcul du $\\uppi$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.1 En demandant à lib maths\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " Mon ordinateur m’indique que $\\uppi$ vaut *approximativement*\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3.141592653589793\n"
     ]
    }
   ],
   "source": [
    "from math import *\n",
    "print(pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.2 En utilisant la méthode des aiguilles de Buffon"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mais calculé avec la **méthode** des aiguilles de Buffon, on obtiendrait comme **approximation** :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.128911138923655"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "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)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Avec un argument \"fréquentiel\" de surface"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction sinussebasesurlefaitquesi $X \\sim U(0,1)$ et $Y \\sim U(0,1)$ alors $P[X\\^2+Y\\^2 $\\leq$ 1]$= $\\uppi \\frac 4$  (voir méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :"
   ]
  }
 ],
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