diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd index 3225870d8acfe7033f0ba830fd9dc7d1a6e7daa1..e3238f5831c63a230b736f819ea58923266c09e6 100644 --- a/module2/exo1/toy_document_en.Rmd +++ b/module2/exo1/toy_document_en.Rmd @@ -1,51 +1,59 @@ --- -title: "Your title" -author: "Your name" -date: "Today's date" +title: "On the computation of pi" +author: "Jayashri Govindan" +date: "23 Octobre 2023" output: html_document -editor_options: - markdown: - wrap: 72 --- -On the computation of pi +```{r setup, include=FALSE} +knitr::opts_chunk$set(echo = TRUE) +``` -Arnaud Legrand +## Asking the maths library -25 juin 2018 +My computer tells me that $\pi$ is *approximatively* -Asking the maths library +# On the computation of pi + +*Arnaud Legrand* + +*25 juin 2018* + +## Asking the maths library My computer tells me that π is approximatively -{r} +```{r} pi +``` -Buffon's needle +## Buffon's needle -Applying the method of Buffon's needle, we get the approximation +Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation -{r} +```{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 +## 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). 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} +```{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 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 therefore 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} +```{r} 4*mean(df$Accept) - +```