title: "On the computation of pi"		
auteur : Meiling WU
output: html_notebook
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#+PROPERTY: header-args  :session  :exports both

* Asking the maths library	
My computer tells me that $\pi$ is /approximatively/

#+begin_src R :results output :session *R* :exports both
pi
#+end_src

#+RESULTS:	
: [1] 3.141593

* Buffon's needle
Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*

#+begin_src R :results output :session *R* :exports both	
set.seed(42)
N = 100000
x = runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)>1))
#+end_src

#+RESULTS:
: [1] 3.14327

* 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 [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach:

#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *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()
#+end_src

#+RESULTS:
[[file:figure_pi_mc1.png]]

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:
#+begin_src R :results output :session *R* :exports both
4*mean(df$Accept)
#+end_src


#+RESULTS:
: [1] 3.156