no commit message

module2/exo1/toy_notebook_en.ipynb

@@ -6,7 +6,53 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "module2/exo1/toy_notebook_en.ipynbWarning\n"
+    "# On the computation of $\\pi$\n",
+    "\n",
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*\n",
+    "\n",
+    "from math import *\n",
+    "print(pi)\n",
+    "\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__\n",
+    "                        \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",
+    "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
+    "2/(sum((x+np.sin(theta))>1)/N)\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## 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 $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:\n",
+    "    \n",
+    "%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",
+    "ax.set_aspect('equal')\n",
+    "\n",
+    "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:\n",
+    "\n",
+    "4*np.mean(accept)"
    ]
   }
  ],