sujet 2

parent c9d479bc
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"cell_type": "code",
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"source": [
"# À propos du calcul de $\\pi$"
]
},
{
"cell_type": "code",
"execution_count": 8,
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"outputs": [],
"source": [
"## En demandant à la lib maths"
]
},
{
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"source": [
"Mon ordinateur m'indique que $\\pi$ vaut *approximativement*"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [
"from math import *\n",
"print(pi)"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [],
"source": [
"## En utilisant la méthode des aiguilles de Buffon"
]
},
{
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"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": 11,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"3.128911138923655"
]
},
"execution_count": 11,
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"output_type": "execute_result"
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"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": "code",
"execution_count": 12,
"metadata": {},
"outputs": [],
"source": [
"## Avec un argument \"fréquentiel\" de surface"
]
},
{
"cell_type": "markdown",
"metadata": {},
"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 se $X \\sim U(0,1)$ et $Y \\sim U(0,1)$ alors $ P[X^2+Y^2\\leq 1] = \\pi/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 :"
]
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{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
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"display_name": "Python 3",
......@@ -16,10 +124,9 @@
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