From ec3f381f3e77f1a079e68202ce195ffd36eaa255 Mon Sep 17 00:00:00 2001 From: a829a5ef5d74929a9597c44c80f67c38 Date: Mon, 15 Mar 2021 20:54:35 +0000 Subject: [PATCH] no commit message --- module2/exo1/toy_notebook_en.ipynb | 48 +++++++++++++++++++++++++++++- 1 file changed, 47 insertions(+), 1 deletion(-) diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index b7968ed..75dbd28 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -6,7 +6,53 @@ "metadata": {}, "outputs": [], "source": [ - "module2/exo1/toy_notebook_en.ipynbWarning\n" + "# On the computation of $\\pi$\n", + "\n", + "## Asking the maths library\n", + "My computer tells me that $\\pi$ is *approximatively*\n", + "\n", + "from math import *\n", + "print(pi)\n", + "\n", + "## Buffon's needle\n", + "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__\n", + " \n", + "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)\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "## Using a surface fraction argument\n", + "A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:\n", + " \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", + "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')\n", + "\n", + "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:\n", + "\n", + "4*np.mean(accept)" ] } ], -- 2.18.1