no commit message

module2/exo1/toy_notebook_en.ipynb

@@ -6,15 +6,17 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "## toy_notebook_en\n",
-    "\n",
-    "March 28, 2019 \n",
-    "\n",
-    "# On the computation of π\n",
-    "\n",
+    "# On the computation of $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
     "## Asking the maths library\n",
-    "\n",
-    "My computer tells me that π is approximatively"
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -37,26 +39,27 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 7,
+   "execution_count": 12,
    "metadata": {},
    "outputs": [
     {
      "ename": "SyntaxError",
-     "evalue": "invalid syntax (<ipython-input-7-8cfaff9333ae>, line 2)",
+     "evalue": "invalid syntax (<ipython-input-12-78e089221e89>, line 2)",
      "output_type": "error",
      "traceback": [
-      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-7-8cfaff9333ae>\"\u001b[0;36m, line \u001b[0;32m2\u001b[0m\n\u001b[0;31m    Applying the method of Buffon’s needle, we get the approximation\u001b[0m\n\u001b[0m               ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
+      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-12-78e089221e89>\"\u001b[0;36m, line \u001b[0;32m2\u001b[0m\n\u001b[0;31m    Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__\u001b[0m\n\u001b[0m               ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
      ]
     }
    ],
    "source": [
-    "## Buffon’s needle\n",
-    "Applying the method of Buffon’s needle, we get the approximation\n"
+    "## Buffon's needle\n",
+    "Applying the method of [Buffon's needle](https://en.wikipedia.org\n",
+    "/wiki/Buffon%27s_needle_problem), we get the __approximation__\n"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 5,
+   "execution_count": 13,
    "metadata": {},
    "outputs": [
     {
@@ -65,7 +68,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 5,
+     "execution_count": 13,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -80,7 +83,6 @@
     "x = np.random.uniform(size=N, low=0, high=1)\n",
     "\n",
     "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
-    "\n",
     "2/(sum((x+np.sin(theta))>1)/N)\n"
    ]
   },
@@ -91,9 +93,11 @@
    "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\n",
-    "fact that if X ∼ U(0, 1) and Y ∼ U(0, 1), then P[X2 + Y2 ≤ 1] = π/4 (see \"Monte Carlo method\"\n",
-    "on Wikipedia). The following code uses this approach:"
+    "A method that is easier to understand and does not make use of the $\\sin$\n",
+    "function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, \n",
+    "then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia]\n",
+    "(https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this\n",
+    "approach:"
    ]
   },
   {
@@ -136,8 +140,8 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to π by counting how\n",
-    "many times, on average, X2 + Y2 is smaller than 1:"
+    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ \n",
+    "by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:"
    ]
   },
   {
@@ -160,6 +164,13 @@
     " 4*np.mean(accept)"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
   {
    "cell_type": "code",
    "execution_count": null,