My computer tells me that $ \ pi $ is * approximately *
`` `` r cars}
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
## Asking the maths library
My computer tells me that $\pi$ is *approximatively*
```{r}
pi
`` ''
## Using the Buffon needle method
But calculated with the __method__ of [Buffon needles] (https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), we would obtain as __approximation__:
`` `{r}
set.seed (42)
N = 100,000
x = runif (N)
theta = pi / 2 * runif (N)
2 / (mean (x + sin (theta)> 1))
`` ''
## With a surface "frequency" argument
Otherwise, a simpler method to understand and not involving a call to the sine 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://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte- Carlo # D% C3% A9termination_de_la_valeur_de_% CF% 80)). The following code illustrates this fact:
`` `{r}
set.seed (42)
```
## Buffon's needle
Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), 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 $X\sim U(0,1)$ 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}
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 easy to obtain an approximation (not great) of $ \ pi $ by counting how many times, on average, $ X ^ 2 + Y ^ 2 $ is less than 1:
