From ac3e98940dd4aa916674c2fa98025f37ae084fa7 Mon Sep 17 00:00:00 2001
From: 5ced80d094cc8bbd195a6caf0e47db49
 <5ced80d094cc8bbd195a6caf0e47db49@app-learninglab.inria.fr>
Date: Mon, 25 Jan 2021 18:36:23 +0000
Subject: [PATCH] =?UTF-8?q?Correcciones=20luego=20de=20comparaci=C3=B3n?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 module2/exo1/toy_notebook_en.ipynb | 50 ++++++++++++++----------------
 1 file changed, 23 insertions(+), 27 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 28a42fd..6012e44 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -7,10 +7,14 @@
     "hidePrompt": false
    },
    "source": [
-    "## 1 On the computation of $\\pi$\n",
-    "\n",
-    "### 1.1 Asking the maths library\n",
-    "\n",
+    "## On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Asking the maths library\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
@@ -38,14 +42,13 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### 1.2 Buffon's needle\n",
-    "\n",
+    "### 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": 2,
+   "execution_count": 5,
    "metadata": {},
    "outputs": [
     {
@@ -54,7 +57,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 2,
+     "execution_count": 5,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -62,9 +65,9 @@
    "source": [
     "import numpy as np\n",
     "np.random.seed(seed=42)\n",
-    "N=10000\n",
-    "x=np.random.uniform(size=N, low=0, high=1)\n",
-    "theta=np.random.uniform(size=N, low=0, high=pi/2)\n",
+    "N = 10000\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
     "2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
@@ -72,8 +75,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### 1.3 Using a surface fraction argument\n",
-    "\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 $P[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
@@ -100,13 +102,14 @@
     "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",
-    "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 = 1000\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "y = np.random.uniform(size=N, low=0, high=1)\n",
     "\n",
-    "fig, ax=plt.subplots(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",
     "ax.set_aspect('equal')"
@@ -116,7 +119,7 @@
    "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:"
    ]
   },
   {
@@ -138,13 +141,6 @@
    "source": [
     "4*np.mean(accept)"
    ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {
-- 
2.18.1