modified

module2/exo1/toy_notebook_fr.ipynb

@@ -4,30 +4,94 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# Amalia exercise\n"
+    "# 1. A propos de calcul du $\\uppi$\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "1. A propos de calcul du \\uppi\n"
+    "## 1.1 En demandant à lib maths\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "1.1 En demandant à lib maths \n",
+    " Mon ordinateur m’indique que $\\uppi$ vaut *approximativement*\n",
     "\n"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 1,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "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 :"
+   ]
   }
  ],
  "metadata": {