From c9e1817035355917f51e990a24f03001de87e7be Mon Sep 17 00:00:00 2001 From: ee86fa7a3582ca7f1b880e3ab22059ce Date: Wed, 9 Jun 2021 12:38:15 +0000 Subject: [PATCH] change of math --- module2/exo1/toy_notebook_en.ipynb | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 9b92b9b..5cbade1 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -8,7 +8,7 @@ "\n", "March 28, 2019\n", "\n", - "# 1 On the computation of π\n", + "# 1 On the computation of $\\pi$\n", "\n", "## 1.1 Asking the maths library\n", "\n", @@ -73,8 +73,8 @@ "## 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 ∼ U ( 0, 1 ) and Y ∼ U ( 0, 1 ) , then P [ X 2 + Y 2 ≤ 1 ] = π/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 $X\\sim U(0,1)$ and $Y\\sim U(0,1)$ then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see (\"[Monte Carlo method\"\n", + "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:" ] }, { @@ -115,8 +115,8 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "It is then straightforward to obtain a (not really good) approximation to π 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\n", + "many times, on average, $X^2 + Y^2$ is smaller than 1:" ] }, { -- 2.18.1