Exo 1 Draft

module2/exo1/test.ipynb

+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1] 13\n"
+     ]
+    }
+   ],
+   "source": [
+    "xa = 12\n",
+    "b = xa + 1\n",
+    "print(b)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1] 12\n",
+      "[1] 13\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(xa)\n",
+    "print(b)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "R",
+   "language": "R",
+   "name": "ir"
+  },
+  "language_info": {
+   "codemirror_mode": "r",
+   "file_extension": ".r",
+   "mimetype": "text/x-r-source",
+   "name": "R",
+   "pygments_lexer": "r",
+   "version": "3.4.1"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

module2/exo1/toy_notebook_fr.ipynb

 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "toy_notebook_fr"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "March 28, 2019"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "#A propos du calcul de pi"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "hideCode": false
+   },
+   "outputs": [],
+   "source": [
+    "## En demandant à la lib maths"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "Mon ordi m'indique que pi vaut approximativement"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "In [1]: from math import *\n",
+    "    print(pi)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "##En utilisant la méthode des aiguilles de Buffon"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "Mais calculé avec la méthode des aiguilles de Buffon, on obtiendrait comme approximation:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "In [2]: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": [
+    "## Avec un argument \"fréquentiel\" de surface"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d'apel à la fonction sinus se base sur le fait que si X ~ U(0,1) et Y ~ U(0,1) alors P[X² + Y² < 1] = pi/4 (voir méthode de Monte-Carlo sur Wikipedia). Le code suivant illustre ce fait :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "IndentationError",
+     "evalue": "unexpected indent (<ipython-input-1-c47c06b12806>, line 2)",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-1-c47c06b12806>\"\u001b[0;36m, line \u001b[0;32m2\u001b[0m\n\u001b[0;31m    import matplotlib.pyplot as plt\u001b[0m\n\u001b[0m    ^\u001b[0m\n\u001b[0;31mIndentationError\u001b[0m\u001b[0;31m:\u001b[0m unexpected indent\n"
+     ]
+    }
+   ],
+   "source": [
+    "In [3]: %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",
+    "    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')"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "Il est alors aisé d'obtenir une approximation (pas terrible) de pi en comptant combien de fois, en moyenne, X² + Y² est inférieur à 1:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "In [4] : 4*np.mean(accept)"
+   ]
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +175,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-