From d71f925e53deac9fdcaedec6f3b35e247f8ed2be Mon Sep 17 00:00:00 2001
From: 5ced80d094cc8bbd195a6caf0e47db49
 <5ced80d094cc8bbd195a6caf0e47db49@app-learninglab.inria.fr>
Date: Mon, 25 Jan 2021 18:23:35 +0000
Subject: [PATCH] Modificaciones menores

---
 module2/exo1/toy_notebook_en.ipynb | 9 +++++----
 1 file changed, 5 insertions(+), 4 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 9bb35ed..28a42fd 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -7,7 +7,7 @@
     "hidePrompt": false
    },
    "source": [
-    "## 1. On the computation of $\\pi$\n",
+    "## 1 On the computation of $\\pi$\n",
     "\n",
     "### 1.1 Asking the maths library\n",
     "\n",
@@ -30,7 +30,7 @@
     }
    ],
    "source": [
-    "from math import*\n",
+    "from math import *\n",
     "print(pi)"
    ]
   },
@@ -74,7 +74,7 @@
    "source": [
     "### 1.3 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 $P[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\"](https://en.wikipedia.org/wiki/Monte_Carlo_method) [on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
+    "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:"
    ]
   },
   {
@@ -105,6 +105,7 @@
     "y=np.random.uniform(size=N, low=0, high=1)\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",
@@ -115,7 +116,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting howmany 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