test 1

parent b44d43a5
{
"cells": [],
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# toy_notebook_fr"
]
},
{
"cell_type": "markdown",
"metadata": {
"hideOutput": true
},
"source": [
"March 28, 2019"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## A propos du calcul de *π*"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### En demandant à la lib maths\n",
"Mon ordinateur m'indique que *π* vaut *approximativement*"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"ename": "SyntaxError",
"evalue": "invalid syntax (<ipython-input-4-4ac309e0cc7a>, line 1)",
"output_type": "error",
"traceback": [
"\u001b[0;36m File \u001b[0;32m\"<ipython-input-4-4ac309e0cc7a>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m In[1] : from $\\math$ import *\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
]
}
],
"source": [
"In[1] : from $\\math$ import *\n",
" 'print(pi)'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##### En utilisant la méthode des aiguilles de Buffon\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": 2,
"metadata": {},
"outputs": [
{
"ename": "SyntaxError",
"evalue": "invalid syntax (<ipython-input-2-e0eb810442a1>, line 1)",
"output_type": "error",
"traceback": [
"\u001b[0;36m File \u001b[0;32m\"<ipython-input-2-e0eb810442a1>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m In[2]:import numpy as np\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
]
}
],
"source": [
"In[2]:import numpy as np\n",
"np.random.seed(see=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",
"Sinon une méthode plus simple à comprendre et ne faisant par intervenir d'appel à la fonction sinus se base sur le fait si X ~U(0,1) alors P[X²+Y²<=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 : "
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"in[3] %matplotlib inline\n",
"import matplotlib.pyplot as plt"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"np.random.seed(seed=42)\n",
"N=1000\n",
"x=np.random.uniform(size=N, low=0, high=1)\n",
"y=np.random.uniform(size=N, low=0, high=1)\n",
"accept=(x*x+y*y)<=1\n",
"reject=np.logical_not(accept)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"fig, ax=plt.subplots(1)\n",
"ax.scatter(x[accept], y[accept], c='b', alpha=0,2, edgecolor=None)\n",
"ax.scatter(x[reject], y[reject], c='r', alpha=0,2, edgecolor=None)\n",
"ax.set_aspect('equal')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Il est alors aisé d'obtenir une approximation (pas terrible) de *π* en comptant combien de fois, en moyenne, X²+Y² est inférieur à 1 : "
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"In [4] 4*np.mean(accept)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
......@@ -16,10 +165,9 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.3"
"version": "3.6.4"
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"nbformat": 4,
"nbformat_minor": 2
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