Oh, yeah, `$` works too

module2/exo1/toy_notebook_en.ipynb

@@ -4,21 +4,15 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# On the computation of \\\\( \\pi \\\\)"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Asking the maths library"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "My computer tells me that \\\\(\\pi\\\\) is *approximatively*"
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -43,13 +37,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Buffon’s needle"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "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**"
    ]
   },
@@ -82,20 +70,14 @@
    "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 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\"\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 $[X^2 + Y^2 \\le 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
     "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 3,
    "metadata": {},
    "outputs": [
     {
@@ -114,12 +96,15 @@
    "source": [
     "%matplotlib inline\n",
     "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",
+    "\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",
@@ -130,13 +115,13 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to \\\\( \\pi \\\\) by counting how\n",
-    "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\n",
+    "many times, on average, $X^2 + Y^2$ is smaller than $1$:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 5,
+   "execution_count": 4,
    "metadata": {},
    "outputs": [
     {
@@ -145,7 +130,7 @@
        "3.112"
       ]
      },
-     "execution_count": 5,
+     "execution_count": 4,
      "metadata": {},
      "output_type": "execute_result"
     }