{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "<p style=\"text-align:center\";>__toy\\_notebook\\_fr__</p>\n",
    "\n",
    "<p style=\"text-align:center\";>March 28, 2019</p>\n",
    "\n",
    "#  propos du calcul de $\\pi$\n",
    "\n",
    "## En demandant à la lib maths\n",
    "\n",
    "Mon ordinateur m’indique que $\\pi$ vaut _approximativement_"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "outputs": [],
   "source": [
    "from math import *\n",
    "print(pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "## En utilisant la méthode des aiguilles de Buffon"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "Mais calculé avec la méthode des ![aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_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,
    "hideOutput": true,
    "hidePrompt": false
   },
   "source": [
    "## Avec un argument \"fréquentiel\" de surface"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "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\n",
    "\n",
    "\n",
    "X\u0018U(0, 1)etY\u0018U(0, 1)alorsP[X2+Y2\u00141]=p/4 (voirméthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :"
   ]
  }
 ],
 "metadata": {
  "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
}