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On the computation of pi

Arnaud Legrand

25 juin 2018

Asking the maths library

My computer tells me that π is approximatively

{r}
pi

Buffon's needle

Applying the method of Buffon's needle, 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∼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:

{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

{r}
4*mean(df$Accept)