From 8211c0c4e4017d24b0569022388c08d770c20d1f Mon Sep 17 00:00:00 2001 From: 616531ae6e3981887d590c5b8da5026e <616531ae6e3981887d590c5b8da5026e@app-learninglab.inria.fr> Date: Wed, 11 Jan 2023 08:36:17 +0000 Subject: [PATCH] update --- module2/exo1/toy_notebook_en.ipynb | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 3d4081e..78e09b3 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -4,7 +4,8 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "# On the computation of $\\pi$ \n", + "# On the computation of $\\pi$\n", + "\n", "## Asking the maths library\n", "My computer tells me that $\\pi$ is *approximatively*" ] @@ -31,7 +32,7 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Buffon's needle ##\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_" ] }, @@ -66,10 +67,9 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Using a surface fraction argument ##\n", + "## 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 $\\le$ 1] = $\\pi$/4 (see [\"Monte Carlo method\" \n", - "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach." + "fact that if X ~ *U*(0,1) and Y ~ *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:" ] }, { @@ -100,9 +100,10 @@ "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", @@ -113,8 +114,7 @@ "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, X2 + Y2 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:" ] }, { -- 2.18.1