diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 78c8123715d8176873bec5e57e51a428b705494c..97c2ee9a031fce317ee36d55d2a5c61e0780fd99 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -6,6 +6,7 @@ "source": [ "# On the computation of $\\pi$\n", "## Asking the maths library\n", + "\n", "My computer tells me that $\\pi$ is *approximatively*" ] }, @@ -32,7 +33,7 @@ "metadata": {}, "source": [ "## Buffon’s needle\n", - "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__" ] }, { @@ -65,12 +66,12 @@ "metadata": {}, "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* ∼ *U*(0, 1) and *Y* ∼ *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:" + "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": 5, "metadata": {}, "outputs": [ { @@ -89,12 +90,15 @@ "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", + "\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", @@ -105,8 +109,7 @@ "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:" + "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:" ] }, {