From 0188de7d873cd459a3374ebdba687245e2ff6fec Mon Sep 17 00:00:00 2001
From: de005631c87798887a112e0e26e250dc
 <de005631c87798887a112e0e26e250dc@app-learninglab.inria.fr>
Date: Wed, 1 Apr 2020 15:45:34 +0000
Subject: [PATCH] first commit

---
 module2/exo1/toy_notebook_fr.ipynb | 129 ++++++++++++++++++++++++++++-
 1 file changed, 127 insertions(+), 2 deletions(-)

diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0bbbe37..9fd116e 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -1,6 +1,132 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "# Toy_notebook_fr"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "\\date{March 28, 2019}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 1 A propos du calcul de $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 1.1 En demandant à la lib maths\n",
+    "\n",
+    "Mon ordinateur m’indique que $\\pi$ vaut -approximativement-"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 1.2 En utilisant la méthode des aiguilles de Buffon\n",
+    "\n",
+    "Mais calculé avec la **méthode** des <span style=\"color: blue;\">aiguilles de Buffon</span>, on obtiendrait comme **approximation** :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "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": [
+    "### 1.3 Avec un argument \"fréquentiel\" de surface\n",
+    "\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 $X∼U(0,1)$ et $Y∼U(0,1)$ alors $P[X^2+Y^2 ≤ 1]=\\pi/4$ (voir <span style=\"color: blue;\">méthode de Monte Carlo sur Wikipedia</span>). Le code suivant illustre ce fait :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "%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",
+    "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')"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Il est alors aisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois,\n",
+    "en moyenne, $X^2 + Y^2$ est inférieur à $1$ :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "4*np.mean(accept)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
  "metadata": {
+  "hide_code_all_hidden": false,
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -22,4 +148,3 @@
  "nbformat": 4,
  "nbformat_minor": 2
 }
-
-- 
2.18.1