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...@@ -18,14 +18,15 @@ ...@@ -18,14 +18,15 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## 1 On the computation of $\\pi$" "# À propos du calcul de $\\pi$"
] ]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"### 1.1 Asking the maths library" "## En demandant à la lib maths\n",
"Mon ordinateur m'indique que $\\pi$ vaut *approximativement*"
] ]
}, },
{ {
...@@ -57,14 +58,8 @@ ...@@ -57,14 +58,8 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"### 1.2 Buffon’s needle" "## 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": "markdown",
"metadata": {},
"source": [
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation"
] ]
}, },
{ {
...@@ -96,16 +91,8 @@ ...@@ -96,16 +91,8 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"### 1.3 Using a surface fraction argument" "## 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 :"
},
{
"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:"
] ]
}, },
{ {
...@@ -127,12 +114,14 @@ ...@@ -127,12 +114,14 @@
} }
], ],
"source": [ "source": [
"%matplotlib inline\n", "%matplotlib inline \n",
"import matplotlib.pyplot as plt\n", "import matplotlib.pyplot as plt\n",
"\n",
"np.random.seed(seed=42)\n", "np.random.seed(seed=42)\n",
"N = 1000\n", "N = 1000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n", "x = np.random.uniform(size=N, low=0, high=1)\n",
"y = 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", "accept = (x*x+y*y) <= 1\n",
"reject = np.logical_not(accept)\n", "reject = np.logical_not(accept)\n",
"\n", "\n",
...@@ -146,8 +135,7 @@ ...@@ -146,8 +135,7 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how\n", "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 :"
"many times, on average, $X^2 + Y^2$ is smaller than 1:"
] ]
}, },
{ {
......
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