From 0699f9b4d2560e2f2a86acd019388b3e7c76af22 Mon Sep 17 00:00:00 2001
From: 8057a7ae58e587eb7d45460ceced4594
 <8057a7ae58e587eb7d45460ceced4594@app-learninglab.inria.fr>
Date: Thu, 23 Sep 2021 17:51:37 +0000
Subject: [PATCH] minor changes

---
 module2/exo1/toy_notebook_en.ipynb | 13 ++-----------
 1 file changed, 2 insertions(+), 11 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 90c97c5..9be2a9e 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*"
    ]
   },
@@ -33,13 +31,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Buffon's needle"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
+    "## Buffon's needle\n",
     "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
    ]
   },
@@ -73,7 +65,6 @@
    "metadata": {},
    "source": [
     "## Using a surface fraction argument\n",
-    "\n",
     "A method that is easier to understand and does not make use of thesin function is based on thefact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$, then $P[X^2 + Y^2 \\leq 1] = \\frac{\\pi}{4}$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
    ]
   },
@@ -116,7 +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, $X^2+Y^2$ 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