My computer tells me that $\pi$ is *approximatively*
```{r}
```{r}
pi
pi
```
```
##Buffon’s needle
## Buffon's needle
Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**
Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
```{r}
```{r}
set.seed(42)
set.seed(42)
N = 100000
N = 100000
...
@@ -20,8 +24,8 @@ x = runif(N)
...
@@ -20,8 +24,8 @@ x = runif(N)
theta = pi/2*runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)>1))
2/(mean(x+sin(theta)>1))
```
```
##Using a surface fraction argument
##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[X^2+Y^2≤1]=π/4 **(see [“Monte Carlo method” on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
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:
It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, ***X*^2+*Y*^2** is smaller than 1 :
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: