From 5f629081e5673f92cbc088d407ef9802d79a3273 Mon Sep 17 00:00:00 2001 From: a76a105ccce500acd74b5f6c5ef4f439 Date: Wed, 17 Nov 2021 16:26:39 +0000 Subject: [PATCH] Update file again after comparison --- module2/exo1/toy_notebook_en.ipynb | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 0995350..ad7d073 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -4,8 +4,8 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "# 1. On the computation of $\\pi$\n", - "## 1.1 Asking the maths library\n", + "# On the computation of $\\pi$\n", + "## Asking the maths library\n", "My computer tells me that $\\pi$ is *approximately*" ] }, @@ -64,8 +64,8 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "# 1.3 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:" + "# 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:" ] }, { @@ -95,7 +95,7 @@ "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", + "accept = (x*x+y*y) <= 1\n", "reject = np.logical_not(accept)\n", "\n", "fig, ax = plt.subplots(1)\n", @@ -108,7 +108,7 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "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:" + "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:" ] }, { -- 2.18.1