Update

module2/exo1/toy_notebook_fr.ipynb

@@ -9,7 +9,92 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# **1 A propos du calcul de $\\pi$**"
+    "# **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}"
    ]
   }
  ],