v3

module2/exo1/toy_notebook_en.ipynb

@@ -4,7 +4,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1 On the computation of π\n",
+    "## 1 On the computation of $\\pi$\n",
     "\n",
     "### 1.1 Asking the maths library\n",
     "\n",
@@ -13,7 +13,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 14,
+   "execution_count": 20,
    "metadata": {},
    "outputs": [
     {
@@ -40,7 +40,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 15,
+   "execution_count": 21,
    "metadata": {},
    "outputs": [
     {
@@ -49,7 +49,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 15,
+     "execution_count": 21,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -69,14 +69,12 @@
    "source": [
     "### 1.3 Using a surface fraction argument\n",
     "\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\" on Wikipedia).  \n",
-    "The following code uses this approach:"
+    "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": 18,
+   "execution_count": 22,
    "metadata": {},
    "outputs": [
     {
@@ -116,7 +114,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 19,
+   "execution_count": 23,
    "metadata": {},
    "outputs": [
     {
@@ -125,7 +123,7 @@
        "3.112"
       ]
      },
-     "execution_count": 19,
+     "execution_count": 23,
      "metadata": {},
      "output_type": "execute_result"
     }