From 7da0e3f6d024bb72829f3189755973f7d057a8ef Mon Sep 17 00:00:00 2001 From: e3e97d29a12734436c722a96738e732b Date: Thu, 22 Oct 2020 10:38:58 +0000 Subject: [PATCH] Update toy_document_en.Rmd --- module2/exo1/toy_document_en.Rmd | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd index 0a35629..6a788fe 100644 --- a/module2/exo1/toy_document_en.Rmd +++ b/module2/exo1/toy_document_en.Rmd @@ -5,12 +5,12 @@ date: "25 juin 2018" output: html_document --- ## Asking the maths library -My computer tells me that π is approximatively +My computer tells me that $\pi$ is *approximatively* ```{r} pi ``` -## Buffon’s needle -Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation +## Buffon's needle +Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__ ```{r} set.seed(42) N = 100000 @@ -19,7 +19,7 @@ theta = pi/2*runif(N) 2/(mean(x+sin(theta)>1)) ``` ## Using a surface fraction argument -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[X2+Y2≤1]=π/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: ```{r} set.seed(42) N = 1000 @@ -28,7 +28,7 @@ df$Accept = (df$X**2 + df$Y**2 <=1) library(ggplot2) ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw() ``` -It is therefore straightforward to obtain a (not really good) approximation to π by counting how 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 : ```{r} 4*mean(df$Accept) ``` -- 2.18.1