Delete toy_document_en.Rmd

module2/exo1/toy_document_en.Rmd

-# On the computation of pi
-
-TEGEGNe
-
-11 June 2020
-
-## Asking the maths library
-
-My computer tells me that π is approximatively
-
----
-pi
-
-
-```
-## [1] 3.141593
-```
-
-## Buffon’s needle
-
-Applying the method of (https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), 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∼U(0,1)**) and (**Y∼U(0,1),**) then P(**[X2+Y2≤1]=π/4 **)(see “Mhttps://en.wikipedia.org/wiki/Monte_Carlo_method") 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$Y**2 <=1)
-library(ggplot2)
-ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
-```
-
-It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, (**X2**)+(**Y2**) is smaller than
-1 :
-
-```
-4*mean(df$Accept)
-```
-
-```
-## [1] 3.156
-```
-
-