Fourth

module2/exo1/toy_notebook_en.ipynb

@@ -4,15 +4,15 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# 1. On the computation of $\\pi$"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.1 Asking the maths library\n",
-    "My computer tells me that $\\pi$ is _approximately_"
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximately*"
    ]
   },
   {
@@ -37,7 +37,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.2 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__"
    ]
   },
@@ -70,8 +70,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.3 Using a surface fraction argument\n",
-    "\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 $X$ $\\sim$ $U(0, 1)$ and $Y$ $\\sim$ $U(0,1)$, then $P[X^2 + Y^2 ≤ 1]$ = $\\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
@@ -115,7 +114,7 @@
    "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:"
+    "many times, on average, $X^2 + Y^2$ is smaller than 1:\n"
    ]
   },
   {