essai 3

parent d150cbcc
{ {
"cells": [ "cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"1 À propos du calcul de π\n",
"1.1 En demandant à la lib maths\n",
"Mon ordinateur m’indique que π vaut approximativement"
]
},
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 2, "execution_count": null,
"metadata": {}, "metadata": {},
"outputs": [ "outputs": [],
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [ "source": [
"In [1]: from math import *\n", "In [1]: from math import *\n",
"print(pi)" "print(pi)"
] ]
}, },
{
"cell_type": "markdown",
"metadata": {},
"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", "cell_type": "code",
"execution_count": 3, "execution_count": 3,
...@@ -33,10 +42,21 @@ ...@@ -33,10 +42,21 @@
"Out[2]: 3.1289111389236548" "Out[2]: 3.1289111389236548"
] ]
}, },
{
"cell_type": "markdown",
"metadata": {},
"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\n",
"sinus se base sur le fait que si X ∼ U(0, 1) et Y ∼ U(0, 1) alors P[X2 + Y2 ≤ 1] = π/4 (voir méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :"
]
},
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 4, "execution_count": 4,
"metadata": {}, "metadata": {
"scrolled": true
},
"outputs": [ "outputs": [
{ {
"data": { "data": {
...@@ -68,21 +88,22 @@ ...@@ -68,21 +88,22 @@
] ]
}, },
{ {
"cell_type": "code", "cell_type": "markdown",
"execution_count": 5,
"metadata": {}, "metadata": {},
"outputs": [],
"source": [ "source": [
"In [4]: 4*np.mean(accept)\n", "Il est alors aisé d’obtenir une approximation (pas terrible) de π en comptant combien de fois,\n",
"Out[4]: 3.1120000000000001" "en moyenne, X2 + Y2 est inférieur à 1 :"
] ]
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": null, "execution_count": 5,
"metadata": {}, "metadata": {},
"outputs": [], "outputs": [],
"source": [] "source": [
"In [4]: 4*np.mean(accept)\n",
"Out[4]: 3.1120000000000001"
]
} }
], ],
"metadata": { "metadata": {
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment