From 1ed0c9cb7b5d8f5087373ee0dbd24310e517db83 Mon Sep 17 00:00:00 2001
From: 50b5daa9521febd9b1bedafb5c2ad791
 <50b5daa9521febd9b1bedafb5c2ad791@app-learninglab.inria.fr>
Date: Tue, 13 Oct 2020 14:35:24 +0000
Subject: [PATCH] commit

---
 module2/exo1/toy_notebook_en.ipynb | 7 +------
 1 file changed, 1 insertion(+), 6 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index e20592e..56e3d42 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -5,9 +5,7 @@
    "metadata": {},
    "source": [
     "# On the computation of $\\pi$\n",
-    "\n",
     "## Asking the maths library\n",
-    "\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
@@ -34,7 +32,6 @@
    "metadata": {},
    "source": [
     "## Buffon’s needle\n",
-    "\n",
     "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
    ]
   },
@@ -68,7 +65,6 @@
    "metadata": {},
    "source": [
     "## 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\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
    ]
   },
@@ -111,8 +107,7 @@
    "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<sup>2</sup> + Y<sup>2</sup>\n",
-    "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<sup>2</sup> + Y<sup>2</sup> is smaller than 1:"
    ]
   },
   {
-- 
2.18.1