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6115f23799e79a0082756425c9fe750d
mooc-rr
Commits
d6dc05cc
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d6dc05cc
authored
Nov 07, 2023
by
6115f23799e79a0082756425c9fe750d
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toy_notebook_fr.ipynb
module2/exo1/toy_notebook_fr.ipynb
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module2/exo1/toy_notebook_fr.ipynb
View file @
d6dc05cc
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@@ -4,8 +4,9 @@
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@@ -4,8 +4,9 @@
"cell_type": "markdown",
"cell_type": "markdown",
"metadata": {},
"metadata": {},
"source": [
"source": [
" \n",
" __1.À propos du calcul de π__ \n",
"__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\n"
]
]
},
},
{
{
...
@@ -27,11 +28,12 @@
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@@ -27,11 +28,12 @@
]
]
},
},
{
{
"cell_type": "code",
"cell_type": "markdown",
"execution_count": null,
"metadata": {},
"metadata": {},
"outputs": [],
"source": [
"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",
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@@ -59,11 +61,16 @@
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@@ -59,11 +61,16 @@
]
]
},
},
{
{
"cell_type": "code",
"cell_type": "markdown",
"execution_count": null,
"metadata": {},
"metadata": {},
"outputs": [],
"source": [
"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[X\n",
"2 + Y\n",
"2 ≤ 1] = π/4 (voir\n",
"méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :"
]
},
},
{
{
"cell_type": "code",
"cell_type": "code",
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@@ -87,11 +94,15 @@
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@@ -87,11 +94,15 @@
]
]
},
},
{
{
"cell_type": "code",
"cell_type": "markdown",
"execution_count": null,
"metadata": {},
"metadata": {},
"outputs": [],
"source": [
"source": []
"Il est alors aisé d’obtenir une approximation (pas terrible) de π en comptant combien de fois,\n",
"en moyenne, X\n",
"2 + Y\n",
"2\n",
"est inférieur à 1 "
]
},
},
{
{
"cell_type": "code",
"cell_type": "code",
...
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