9

module2/exo1/toy_notebook_fr.ipynb

@@ -10,8 +10,7 @@
     "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*\n",
     "\n",
     "\n",
-    "In \\[1]:\n",
-    ">from math import *\n",
+    "In \\[1]:>from math import *\n",
     ">print (pi)\n",
     "    \n",
     "3.141592653589793\n",
@@ -23,14 +22,15 @@
     "\n",
     "\n",
     "In \\[2]:\n",
-    "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)\n",
+    ">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)\n",
     "    \n",
-    "3.1289111389236548\n",
+    "Out\\[2]: 3.1289111389236548\n",
+    "\n",
     "\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 sinus se base sur le fait que  si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\leq1] = \\pi/4$ (voir [méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :\n",
@@ -38,29 +38,31 @@
     "\n",
     "In \\[3]:\n",
     "\n",
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
+    ">%matplotlib inline\n",
+    ">import matplotlib.pyplot as plt\n",
+    "\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",
+    "\n",
+    ">accept = (x*x+y*y) <=1\n",
+    ">reject = np.logical_not(accept)\n",
+    "\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",
     "\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",
     "\n",
-    "accept = (x*x+y*y) <=1\n",
-    "reject = np.logical_not(accept)\n",
     "\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 $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :\n",
     "\n",
-    "Il est alors aisé d'obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :\n",
     "\n",
+    "In \\[4]: 4*np.mean(accept)\n",
     "\n",
-    "In \\[4]:\n",
+    "Out\\[4]: 3.1120000000000001\n",
     "\n",
-    "4*np.mean(accept)\n",
-    "3.1120000000000001\n",
     "\n"
    ]
   },