@@ -18,34 +18,34 @@ My computer tells me that $\pi$ is /approximatively/
...
@@ -18,34 +18,34 @@ My computer tells me that $\pi$ is /approximatively/
pi
pi
#+end_src
#+end_src
#+RESULTS:
#+RESULTS:
: [1] 3.141593
: [1] 3.141593
* Buffon's needle
* Buffon's needle
Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*
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
#+begin_src R :results output :session *R* :exports both
set.seed(42)
set.seed(42)
N = 100000
N = 100000
x = runif(N)
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))
#+end_src
#+end_src
#+RESULTS:
#+RESULTS:
: [1] 3.14327
: [1] 3.14327
* 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 $X\sim U(0,1)$ 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:
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*
#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R*