part1

module2/exo1/toy_notebook_en.ipynb

@@ -4,6 +4,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
+<<<<<<< HEAD
     "# On the computation of $\\pi$"
    ]
   },
@@ -13,6 +14,11 @@
    "source": [
     "## Asking the maths library\n",
     "My computer tells me that $\\pi$ is *approximatively*"
+=======
+    "#  À propos du calcul de $\\pi$\n",
+    "##  En demandant à la lib maths\n",
+    "Mon ordinateur m’indique que $\\pi$ vaut *approximativement*"
+>>>>>>> ff5bf62e6a54b420ff906287a58051154df436c2
    ]
   },
   {
@@ -37,8 +43,13 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
+<<<<<<< HEAD
     "## Buffon's needle\n",
     "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__\n"
+=======
+    "## En utilisant la méthode des aiguilles de Buffon\n",
+    "Mais calculé avec la **méthode** [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation** :\n"
+>>>>>>> ff5bf62e6a54b420ff906287a58051154df436c2
    ]
   },
   {
@@ -70,8 +81,15 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
+<<<<<<< HEAD
     "## 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:\n"
+=======
+    "## Avec un argument \"fréquentiel\" de surface\n",
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction\n",
+    "sinus se base sur le fait que si $X \\sim U(0,1)$ et $Y \\sim U(0, 1)$ alors $P[X^2 + Y^2 ≤ 1] = \\pi/4$\n",
+    "(voir [méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :\n"
+>>>>>>> ff5bf62e6a54b420ff906287a58051154df436c2
    ]
   },
   {
@@ -114,7 +132,12 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
+<<<<<<< HEAD
     "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:"
+=======
+    "Il est alors aisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois,\n",
+    "en moyenne, $X^2 + Y^2$ est inférieur à 1 :"
+>>>>>>> ff5bf62e6a54b420ff906287a58051154df436c2
    ]
   },
   {

module2/exo1/toy_notebook_fr.ipynb

@@ -23,3 +23,6 @@
  "nbformat_minor": 2
 }
 
+#  À propos du calcul de $\pi$
+##  En demandant à la lib maths
+Mon ordinateur m’indique que $\pi$ vaut *approximativement*

module2/exo2/exercice.ipynb

 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "a = np.array([14.0, 7.6, 11.2, 12.8, 12.5, 9.9, 14.9, 9.4, 16.9, 10.2, 14.9, 18.1, 7.3, 9.8, 10.9,12.2, 9.9, 2.9, 2.8, 15.4, 15.7, 9.7, 13.1, 13.2, 12.3, 11.7, 16.0, 12.4, 17.9, 12.2, 16.2, 18.7, 8.9, 11.9, 12.1, 14.6, 12.1, 4.7, 3.9, 16.9, 16.8, 11.3, 14.4, 15.7, 14.0, 13.6, 18.0, 13.6, 19.9, 13.7, 17.0, 20.5, 9.9, 12.5, 13.2, 16.1, 13.5, 6.3, 6.4, 17.6, 19.1, 12.8, 15.5, 16.3, 15.2, 14.6, 19.1, 14.4, 21.4, 15.1, 19.6, 21.7, 11.3, 15.0, 14.3, 16.8, 14.0, 6.8, 8.2, 19.9, 20.4, 14.6, 16.4, 18.7, 16.8, 15.8, 20.4, 15.8, 22.4, 16.2, 20.3, 23.4, 12.1, 15.5, 15.4, 18.4, 15.7, 10.2, 8.9, 21.0])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "std = np.std(a, axis=None, dtype=None, out=None, ddof=1)\n",
+    "mini = min(a)\n",
+    "maxi = max(a)\n",
+    "med = np.median(a)\n",
+    "mean = np.mean(a)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "4.334094455301447"
+      ]
+     },
+     "execution_count": 15,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "std"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "2.8"
+      ]
+     },
+     "execution_count": 16,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "mini"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "23.4"
+      ]
+     },
+     "execution_count": 17,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "maxi"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "14.5"
+      ]
+     },
+     "execution_count": 18,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "med"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "14.113000000000001"
+      ]
+     },
+     "execution_count": 21,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "mean"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
  "metadata": {
   "kernelspec": {
    "display_name": "Python 3",
@@ -16,10 +155,9 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "version": "3.6.4"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-