Commit b1095d7c authored by ZackOSX's avatar ZackOSX

syncrho 3

parent a948527b
#+TITLE: À propos du calcul de $\pi$
#+AUTHOR: Zack Zox
# +DATE: La date du jour
#+OPTIONS: toc:1
#+LANGUAGE: fr
# +PROPERTY: header-args :eval never-export
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# +PROPERTY: header-args :session both
* En demandant à la lib maths
Mon ordinateur m'indique que $\pi$ vaut /approximativement/
#+begin_src R :results output :exports both
#+begin_src R :results output :session *R :exports both
pi
#+end_src
......@@ -40,27 +34,27 @@ theta = pi/2*runif(N)
#+end_src
#+RESULTS:
:
: [1] 3.14327
* Avec un argument "fréquentiel" de surface
Sinon, une méthode plus simple à comprendre et ne faisant pas
intervenir d'appel à la fonction sinus se base sur le fait que si
$X \sim U(0,1)$ et $Y \sim U(0,1)$ alors $P[X^2 + Y^2 ≤ 1] = \pi/4$ (voir
$X\sim U(0,1)$ et $Y\sim U(0,1)$ alors $P[X^2+Y^2≤ 1] = \pi/4$ (voir
[[https://fr.wikipedia.org/wiki/Méthode_de_Monte-Carlo#Détermination_de_la_valeur_de_π][méthode de Monte Carlo sur Wikipedia]]). Le code suivant illustre ce fait :
#+begin_src R :results output graphics :file (org-babel-temp-file "figure" ".png") :exports both :width 600 :height 400 :session *R*
#+begin_src R :results output graphics :file (org-babel-temp-file "figure" ".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
Il est alors aisé d'obtenir une approximation (pas terrible) de $\pi$
en comptant combien de fois, en moyenne, \$X^2+Y^2 est inférieur à 1 :
#+begin_src R :results output :session *R* :exports both
4*mean(df$Accept)
#+end_src
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