# On the computation of pi

## Table of Contents

- 1. Asking the maths library
- 2. Buffon's needle
- 3. Using a surface fraction argument

## 1 Asking the maths library

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

    pi

    [1] 3.141593

## 2 Buffon's needle

Applying the method of Buffon's needle, 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

## 3 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). 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 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:

    4*mean(df$Accept)

    [1] 3.156

Author: Konrad Hinsen

Created: 2019-03-28 Thu 11:13

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