From b20d1229a5c280f7c8e44281f065447df1068c96 Mon Sep 17 00:00:00 2001 From: 16fd6933414ad4ae0fb1412e7effdbc7 <16fd6933414ad4ae0fb1412e7effdbc7@app-learninglab.inria.fr> Date: Sun, 14 Jun 2020 15:26:01 +0000 Subject: [PATCH] Replace toy_document_fr.Rmd --- module2/exo1/toy_document_fr.Rmd | 486 +++---------------------------- 1 file changed, 37 insertions(+), 449 deletions(-) diff --git a/module2/exo1/toy_document_fr.Rmd b/module2/exo1/toy_document_fr.Rmd index 22ba455..cb6727f 100644 --- a/module2/exo1/toy_document_fr.Rmd +++ b/module2/exo1/toy_document_fr.Rmd @@ -1,449 +1,37 @@ - - - - - - - - - - - - - - - -On the computation of pi - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
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Asking the maths library

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

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pi
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## [1] 3.141593
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Buffon’s needle

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

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

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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:

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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()
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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:

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4*mean(df$Accept)
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## [1] 3.156
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- - - - - - - - - - - - - - - +--- +title: "On the computation of pi" +author: "Tegegne" +date: "14 june 2020" +output: html_document +--- +```{r setup, include=FALSE} +knitr::opts_chunk$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) +``` -- 2.18.1