"Mon ordinateur m'indique que $\\pi$ vaut *approximativement*"
]
},
{
...
...
@@ -57,14 +58,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"### 1.2 Buffon’s needle"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation"
"## 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__ :"
]
},
{
...
...
@@ -96,16 +91,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"### 1.3 Using a surface fraction argument"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"A method that is easier to understand and does not make use of the sin function is based on the\n",
"fact that if $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X^2 + Y^2 ≤ 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
"## 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 $X\\sim U(0,1)$ 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 :"
]
},
{
...
...
@@ -127,12 +114,14 @@
}
],
"source": [
"%matplotlib inline\n",
"%matplotlib inline\n",
"import matplotlib.pyplot as plt\n",
"\n",
"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",
"\n",
"accept = (x*x+y*y) <= 1\n",
"reject = np.logical_not(accept)\n",
"\n",
...
...
@@ -146,8 +135,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how\n",
"many times, on average, $X^2 + Y^2$ is smaller than 1:"
"Il est alors aisé d'obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :"
