update toy_notebook_en

module2/exo1/toy_notebook_en.ipynb

@@ -4,9 +4,9 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# On the computation of π\n",
+    "# On the computation of $\\pi$\n",
     "## Asking the maths library\n",
-    "My computer tells me that π is approximatively"
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -31,7 +31,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Buffon’s needle"
+    "## Buffon's needle\n",
+    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
@@ -64,9 +65,8 @@
    "metadata": {},
    "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\n",
-    "fact that if X ∼ U(0, 1) and Y ∼ U(0, 1), then P[X2 + Y2 ≤ 1] = π/4 (see \"Monte Carlo method\"\n",
-    "on Wikipedia). The following code uses this approach:\n"
+    "\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"
    ]
   },
   {
@@ -102,6 +102,13 @@
     "ax.set_aspect('equal')"
    ]
   },
+  {
+   "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:"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 4,