solution2

module2/exo1/toy_notebook_en.ipynb

@@ -4,14 +4,14 @@
    "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"
+    "## Asking the maths library"
    ]
   },
   {
@@ -43,19 +43,19 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.2 Buffalo's needle"
+    "## Buffalo's needle"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
+    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 5,
    "metadata": {},
    "outputs": [
     {
@@ -64,13 +64,13 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 2,
+     "execution_count": 5,
      "metadata": {},
      "output_type": "execute_result"
     }
    ],
    "source": [
-    "In [2]: import numpy as np\n",
+    "import numpy as np\n",
     "np.random.seed(seed=42)\n",
     "N = 10000\n",
     "x = np.random.uniform(size=N, low=0, high=1)\n",
@@ -89,12 +89,12 @@
    "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\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). 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": 3,
+   "execution_count": 6,
    "metadata": {},
    "outputs": [
     {
@@ -113,29 +113,31 @@
    "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",
-    "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",
-    "ax.set_aspect('equal')\n"
+    "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:"
+    "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,
+   "execution_count": 7,
    "metadata": {},
    "outputs": [
     {
@@ -144,7 +146,7 @@
        "3.112"
       ]
      },
-     "execution_count": 4,
+     "execution_count": 7,
      "metadata": {},
      "output_type": "execute_result"
     }