diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index bae231ee293ba5a32f95504444dd5123e66034ba..b72b8e52b8f1d34b469d6cd51720858627755c1c 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -1,32 +1,12 @@ { "cells": [ - { - "cell_type": "code", - "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "ename": "SyntaxError", - "evalue": "invalid syntax (, line 1)", - "output_type": "error", - "traceback": [ - "\u001b[0;36m File \u001b[0;32m\"\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m 1 On the computation of π\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n" - ] - } - ], - "source": [ - "1 On the computation of π\n", - "1.1 Asking the maths library\n", - "My computer tells me that π is approximatively" - ] - }, { "cell_type": "markdown", "metadata": {}, "source": [ - "# 1 On the computation of π\n", - "## 1.1 Asking the maths library\n", - "My computer tells me that π is approximatively" + "# On the computation of $\\pi$\n", + "## Asking the maths library\n", + "My computer tells me that $\\pi$ is *approximatively*" ] }, { @@ -51,8 +31,8 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## 1.2 Buffon’s needle\n", - "Applying the method of Buffon’s needle, we get the approximation" + "## Buffon's needle\n", + "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__" ] }, { @@ -84,9 +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\n", - "fact that if X ∼ U(0, 1) and Y ∼ U(0, 1), then P[X2 + Y2 ≤ 1] = π/4 (see \"Monte Carlo method\" on Wikipedia). 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:" ] }, { @@ -126,8 +105,7 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "It is then straightforward to obtain a (not really good) approximation to π by counting how\n", - "many times, on average, X2 + Y2 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:" ] }, {