From 4523f1173798057bb88653e7070bdb15bcc8b809 Mon Sep 17 00:00:00 2001
From: 973469d19e8da3800b283676fe06a1b5
 <973469d19e8da3800b283676fe06a1b5@app-learninglab.inria.fr>
Date: Fri, 20 Jun 2025 14:42:34 +0000
Subject: [PATCH] entrega 1

---
 module2/exo1/toy_notebook_en.ipynb | 131 ++++++++++++++++++++++++++++-
 1 file changed, 128 insertions(+), 3 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 0bbbe37..20dd65c 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,6 +1,132 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "# On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "## Asking the maths library"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "My computer tells me that $\\pi$ *is approximatively*"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.141592653589793\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Buffon’s needle"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "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,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.128911138923655"
+      ]
+     },
+     "execution_count": 2,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "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",
+    "2/(sum((x+np.sin(theta))>1)/N)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##  Using a surface fraction argument"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "A method that is easier to understand and does not make use of the sin function is based on the\n",
+    "fact that if $$X ∼ \\mathcal{U}(0, 1)$$ and $$\\mathcal{Y} ∼ \\mathcal{U}(0, 1)$$, then $$\\mathcal{P}[X^2 + Y^2 ≤ 1] = π/4$$ (see \"[Monte Carlo method]\"(https://en.wikipedia.org/wiki/Monte_Carlo_method)\n",
+    "on Wikipedia). The following code uses this approach:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    " %matplotlib inline\n",
+    "import matplotlib.pyplot as plt\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)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
  "metadata": {
+  "hide_code_all_hidden": false,
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -16,10 +142,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-
-- 
2.18.1