diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 5bca4a19c56ffc2b6f2a55c6665449807045e9c6..669b41709a42c1d016568307b9fdb888c578fdf0 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -2,20 +2,20 @@
"cells": [
{
"cell_type": "markdown",
- "id": "93d7cca7",
+ "id": "fa916478",
"metadata": {},
"source": [
- "# 1 On the computation of π\n",
+ "# On the computation of $\\pi$\n",
"\n",
- "## 1.1 Asking the maths library\n",
+ "## Asking the maths library\n",
"\n",
- "My computer tells me that π is approximativel"
+ "My computer tells me that $\\pi$ is *approximatively*"
]
},
{
"cell_type": "code",
- "execution_count": 2,
- "id": "d0bc647a",
+ "execution_count": 1,
+ "id": "08c5c563",
"metadata": {},
"outputs": [
{
@@ -33,18 +33,17 @@
},
{
"cell_type": "markdown",
- "id": "99049b47",
+ "id": "8c9199c1",
"metadata": {},
"source": [
- "## 1.2 Buffon’s needle\n",
- "\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__"
]
},
{
"cell_type": "code",
- "execution_count": 3,
- "id": "d9073c83",
+ "execution_count": 2,
+ "id": "abff8880",
"metadata": {},
"outputs": [
{
@@ -53,7 +52,7 @@
"3.128911138923655"
]
},
- "execution_count": 3,
+ "execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
@@ -69,18 +68,17 @@
},
{
"cell_type": "markdown",
- "id": "2b0f149b",
+ "id": "4bd4c67b",
"metadata": {},
"source": [
- "## 1.3 Using a surface fraction argument\n",
- "\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 ∼ 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:\n"
+ "## 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:"
]
},
{
"cell_type": "code",
- "execution_count": 4,
- "id": "6b5b21e8",
+ "execution_count": 3,
+ "id": "5413190d",
"metadata": {},
"outputs": [
{
@@ -97,9 +95,9 @@
}
],
"source": [
- "%matplotlib inline\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",
@@ -107,6 +105,7 @@
"\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",
@@ -115,16 +114,16 @@
},
{
"cell_type": "markdown",
- "id": "ca88818e",
+ "id": "c1e500ca",
"metadata": {},
"source": [
- "It is then straightforward to obtain a (not really good) approximation to π by counting how 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:"
]
},
{
"cell_type": "code",
- "execution_count": 5,
- "id": "fe6e57c1",
+ "execution_count": 4,
+ "id": "fc8a2a3e",
"metadata": {},
"outputs": [
{
@@ -133,7 +132,7 @@
"3.112"
]
},
- "execution_count": 5,
+ "execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}