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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# **A propos du calcul de $\\pi$**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## **En demandant à la lib maths**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3.141592653589793\n"
     ]
    }
   ],
   "source": [
    "from math import pi\n",
    "print (pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## **En utilisant la méthode des aiguilles de Buffon**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.128911138923655"
      ]
     },
     "execution_count": 10,
     "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 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] = \\frac{pi}{4}"
   ]
  }
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