no commit message

module2/exo1/toy_notebook_en.ipynb

 {
  "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 2,
@@ -18,6 +37,16 @@
     "print(pi)"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "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__"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 4,
@@ -43,6 +72,16 @@
     "2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "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 $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 5,
@@ -79,6 +118,24 @@
     "ax.set_aspect('equal')"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "SyntaxError",
+     "evalue": "invalid syntax (<ipython-input-7-46b60a496a9b>, line 1)",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-7-46b60a496a9b>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m    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:\u001b[0m\n\u001b[0m                             ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
+     ]
+    }
+   ],
+   "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:"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 6,
@@ -98,13 +155,6 @@
    "source": [
     "4*np.mean(accept)"
    ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {