{ "cells": [ { "cell_type": "markdown", "metadata": { "hideCode": false, "hidePrompt": false }, "source": [ "**toy_notebook_fr**\n", "\n", "March 28, 2019\n", "1 **A propos du calcul de** \\pi\n", "1.1 **En demandant à la lib maths**\n", "Mon ordinateur m'indique que \\pi vaut approximativement\n" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "hideCode": false, "hidePrompt": false }, "outputs": [], "source": [ "from math import *\n", "print(pi)\n" ] }, { "cell_type": "markdown", "metadata": { "hideCode": false, "hidePrompt": false }, "source": [ "**1.2 En utilisant la méthode des aiguilles de Buffon**\n", "Mais calculé avec la **méthode** des aiguilles de Buffon, on obtiendrait comme **approximation :**" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "hideCode": false, "hidePrompt": false }, "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": { "hideCode": false, "hidePrompt": false }, "source": [ "**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" ] } ], "metadata": { "celltoolbar": "Tags", "hide_code_all_hidden": false, "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.4" } }, "nbformat": 4, "nbformat_minor": 2 }