title: "On the computation of pi" title: "On the computation of pi"
author: "Lijuan Ren"
author: "Arnaud Legrand" author: "Lijuan Ren"
date: "20 Sept 2021"
date: "25 juin 2018" date: "20 Sept 2021"
output: html_document
output: html_document output: html_document
---
--- ---
```{r setup, include=FALSE} ## Asking the maths library
## Asking the maths library
knitr::opts_chunk$set(echo = TRUE) pi
pi
``` ## [1] 3.141593
## [1] 3.141593
## Asking the maths library
My computer tells me that $\pi$ is *approximatively*
```{r}
## Buffon’s needle
pi
set.seed(42)
```
N = 100000
## Buffon's needle ## Buffon’s needle
x = runif(N)
Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
theta = pi/2*runif(N)
```{r}
2/(mean(x+sin(theta)>1))
set.seed(42) set.seed(42)
N = 100000 N = 100000
x = runif(N) x = runif(N)
## Using a surface fraction argument
theta = pi/2*runif(N) theta = pi/2*runif(N)
set.seed(42)
2/(mean(x+sin(theta)>1)) 2/(mean(x+sin(theta)>1))
N = 1000
```
df = data.frame(X = runif(N), Y = runif(N))
## Using a surface fraction argument ## Using a surface fraction argument
df$Accept = (df$X**2 + df$Y**2 <=1)
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: