section 1.1 à 1.3

parent 43aabe0a
{ {
"cells": [ "cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# À propos du calcul de $\\pi$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## En demandant à la lib maths\n",
"\n",
"Mon ordinateur m’indique que $\\pi$ vaut *approximativement*"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [
"from math import *\n",
"print(pi)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## En utilisant la méthode des aiguilles de Buffon\n",
"\n",
"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": 2,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"3.128911138923655"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"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": [
"## 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 $X \\cdot U(0, 1)$ et $Y \\cdot U(0, 1)$ alors $P[X2+Y2\u00141]=p/4$ (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 :"
]
},
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": null, "execution_count": null,
......
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