From 666eea8a3cdd61d3e1bfa0408fc6de5becb1645b Mon Sep 17 00:00:00 2001 From: 16fd6933414ad4ae0fb1412e7effdbc7 <16fd6933414ad4ae0fb1412e7effdbc7@app-learninglab.inria.fr> Date: Sun, 14 Jun 2020 14:36:17 +0000 Subject: [PATCH] Replace toy_document_fr.Rmd --- module2/exo1/toy_document_fr.Rmd | 497 ++++++++++++++++++++++++++++--- 1 file changed, 450 insertions(+), 47 deletions(-) diff --git a/module2/exo1/toy_document_fr.Rmd b/module2/exo1/toy_document_fr.Rmd index 0839e1d..f44eebc 100644 --- a/module2/exo1/toy_document_fr.Rmd +++ b/module2/exo1/toy_document_fr.Rmd @@ -1,47 +1,450 @@ ---- -title: "On the computation of pi" -author: "Tegegne" -date: "June 13, 2020 -output: html_document ---- - -```{r setup, include=FALSE} -knitr::opts_chunk$set(echo = TRUE) -``` - -## Asking the maths library -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__ - -```{r} -set.seed(42) -N = 100000 -x = runif(N) -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 $X\sim U(0,1)$ 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 -df = data.frame(X = runif(N), Y = runif(N)) -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 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) -``` \ No newline at end of file + + + + + + + + + + + + + + + +On the computation of pi + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + +
+

Asking the maths library

+

My computer tells me that \(\pi\) is approximatively

+
pi
+
## [1] 3.141593
+
+
+

Buffon’s needle

+

Applying the method of Buffon’s needle, we get the approximation

+
set.seed(42)    
+N = 100000  
+x = runif(N)    
+theta = pi/2*runif(N)   
+2/(mean(x+sin(theta)>1))
+
## [1] 3.14327
+
+
+

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 \(X\sim U(0,1)\) and \(Y\sim U(0,1)\), then \(P[X^2+Y^2\leq 1] = \pi/4\) (see “Monte Carlo method” on Wikipedia). The following code uses this approach:

+
set.seed(42)    
+N = 1000    
+df = data.frame(X = runif(N), Y = runif(N)) 
+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 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:

+
4*mean(df$Accept)
+
## [1] 3.156
+
+ + + + +
+ + + + + + + + + + + + + + + -- 2.18.1