update

module2/exo1/toy_notebook_en.ipynb

@@ -4,7 +4,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# On the computation of $\\pi$ \n",
+    "# On the computation of $\\pi$\n",
+    "\n",
     "## Asking the maths library\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
@@ -31,7 +32,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Buffon's needle ##\n",
+    "## Buffon's needle\n",
     "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the _approximation_"
    ]
   },
@@ -66,10 +67,9 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Using a surface fraction argument ##\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  \n",
-    "fact that if X ~ *U*(0,1) and Y ~ *U*(0,1), then *P*[X<sup>2</sup> + Y<sup>2</sup> $\\le$ 1] = $\\pi$/4 (see [\"Monte Carlo method\"  \n",
-    "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach."
+    "fact that if X ~ *U*(0,1) and Y ~ *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:"
    ]
   },
   {
@@ -100,9 +100,10 @@
     "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",
-    "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",
@@ -113,8 +114,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "&nbsp;&nbsp;&nbsp;&nbsp; It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how  \n",
-    "many times, on average, X<sup>2</sup> + Y<sup>2</sup> is smaller than 1:"
+    "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:"
    ]
   },
   {