Second trial

module2/exo1/toy_notebook_en.ipynb

@@ -4,27 +4,20 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# A propos de pi"
+    "# 1. On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1. On the computation of $\\pi$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### 1.1 Asking the maths library\n",
+    "## 1.1 Asking the maths library\n",
     "My computer tells me that $\\pi$ is approximately"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 10,
+   "execution_count": 14,
    "metadata": {},
    "outputs": [
     {
@@ -44,13 +37,13 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### 1.2 Buffon’s needle\n",
+    "## 1.2 Buffon’s needle\n",
     "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": 11,
+   "execution_count": 15,
    "metadata": {},
    "outputs": [
     {
@@ -59,7 +52,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 11,
+     "execution_count": 15,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -73,9 +66,19 @@
     "2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "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 $X$ $\\sim$ $U(0, 1)$ 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:"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 12,
+   "execution_count": 16,
    "metadata": {},
    "outputs": [
     {
@@ -107,6 +110,34 @@
     "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\n",
+    "many times, on average, $X^2 + Y^2$ is smaller than 1:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.112"
+      ]
+     },
+     "execution_count": 17,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "4*np.mean(accept)"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,