From 37d32bfad42aa46a6284a6ff47cdded71fde5946 Mon Sep 17 00:00:00 2001
From: 5ced80d094cc8bbd195a6caf0e47db49
 <5ced80d094cc8bbd195a6caf0e47db49@app-learninglab.inria.fr>
Date: Mon, 25 Jan 2021 18:42:31 +0000
Subject: [PATCH] =?UTF-8?q?Correcci=C3=B3n=20luego=20de=20comparaci=C3=B3n?=
 =?UTF-8?q?=202?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 module2/exo1/toy_notebook_en.ipynb | 20 ++++++++++----------
 1 file changed, 10 insertions(+), 10 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 6012e44..3ca66cf 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -7,14 +7,14 @@
     "hidePrompt": false
    },
    "source": [
-    "## On the computation of $\\pi$"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### Asking the maths library\n",
+    "## Asking the maths library\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
@@ -42,13 +42,13 @@
    "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**"
+    "## 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": 5,
+   "execution_count": 2,
    "metadata": {},
    "outputs": [
     {
@@ -57,7 +57,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 5,
+     "execution_count": 2,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -75,8 +75,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### 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 $P[X^2 + Y^2 \\le 1] = \\pi/4$ (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:"
    ]
   },
   {
@@ -98,7 +98,7 @@
     }
    ],
    "source": [
-    "%matplotlib inline\n",
+    "%matplotlib inline  \n",
     "import matplotlib.pyplot as plt\n",
     "\n",
     "np.random.seed(seed=42)\n",
@@ -106,7 +106,7 @@
     "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",
-- 
2.18.1