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0a103cca4e060a3462e317be5772e4a4
mooc-rr
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55b68fb5
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55b68fb5
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Sep 18, 2024
by
0a103cca4e060a3462e317be5772e4a4
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toy_notebook_fr.ipynb
module2/exo1/toy_notebook_fr.ipynb
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module2/exo1/toy_notebook_fr.ipynb
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55b68fb5
{
"cells": [],
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**toy_notebook_fr**\n",
"\n",
"March 28, 2019\n",
"1 **A propos du calcul de** \\pi\n",
0a103cca4e060a3462e317be5772e4a4
@0a103cca4e060a3462e317be5772e4a4
·
Sep 18, 2024
Maintainer
how to write pi
how to write pi
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"1.1 **En demandant à la lib maths**\n",
"Mon ordinateur m'indique que \\pi vaut approximativement\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from math import *\n",
"print(pi)\n"
]
},
{
"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",
"execution_count": null,
"metadata": {},
"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": {},
"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 sinus se base sur le fait que si X"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
...
...
@@ -16,10 +69,9 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.
3
"
"version": "3.6.
4
"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
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