My computer tells me that $\pi$ is _approximatively_
My computer tells me that $\pi$ is _approximatively_
```pi
```{r}
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__
```set.seed(42)
```{r}
set.seed(42)
N = 100000
N = 100000
x = runif(N)
x = runif(N)
theta = pi/2*runif(N)
theta = pi/2*runif(N)
...
@@ -29,7 +31,7 @@ theta = pi/2*runif(N)
...
@@ -29,7 +31,7 @@ theta = pi/2*runif(N)
## 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 thath if $\X ~ U (0,1)$ and $\Y ~ U (0,1)$, then $\P[X^2 + Y^2 <=1] = $\pi$/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 $\le$ 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 $\pi$ 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 :