brouillon

module2/exo1/toy_notebook_fr.ipynb

 {
  "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "March 28, 2019"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
@@ -7,6 +14,13 @@
     "# Premier test jupyter"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# À propos du calcul de π $\\pi$"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 1,
@@ -24,6 +38,128 @@
     "import os\n",
     "print('hello world')"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## En demandant à la lib maths\n",
+    "Mon ordinateur m’indique que π vaut approximativement"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.141592653589793\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# En utilisant la méthode des aiguilles de Buffon\n",
+    "Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation** :\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.128911138923655"
+      ]
+     },
+     "execution_count": 5,
+     "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": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "ut[2]: 3.1289111389236548\n",
+    "1.3 Avec un argument \"fréquentiel\" de surface\n",
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction\n",
+    "sinus se base sur le fait que si X ∼ U(0, 1) et Y ∼ U(0, 1) alors P[X\n",
+    "2 + Y\n",
+    "2 ≤ 1] = π/4 (voir\n",
+    "méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :\n",
+    "In [3]: %matplotlib inline\n",
+    "import matplotlib.pyplot as plt\n",
+    "np.random.seed(seed=42)\n",
+    "N = 1000\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "y = np.random.uniform(size=N, low=0, high=1)\n",
+    "1\n",
+    "accept = (x*x+y*y) <= 1\n",
+    "reject = np.logical_not(accept)\n",
+    "fig, ax = plt.subplots(1)\n",
+    "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
+    "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
+    "ax.set_aspect('equal')\n",
+    "Il est alors aisé d’obtenir une approximation (pas terrible) de π en comptant combien de fois,\n",
+    "en moyenne, X\n",
+    "2 + Y\n",
+    "2\n",
+    "est inférieur à 1 :\n",
+    "In [4]: 4*np.mean(accept)\n",
+    "Out[4]: 3.1120000000000001\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {