save progress

parent 4b5d7dc7
...@@ -13,13 +13,64 @@ ...@@ -13,13 +13,64 @@
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": null, "execution_count": 1,
"metadata": {}, "metadata": {},
"outputs": [], "outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [ "source": [
"from math import *\n", "from math import *\n",
"print(pi)" "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": 4,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"3.128911138923655"
]
},
"execution_count": 4,
"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 appel à la fonction sinus se base sur le fait que si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\le 1] =\\pi / 4$"
]
} }
], ],
"metadata": { "metadata": {
......
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