Skip to content
Projects
Groups
Snippets
Help
Loading...
Help
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
M
mooc-rr
Project
Project
Details
Activity
Releases
Cycle Analytics
Repository
Repository
Files
Commits
Branches
Tags
Contributors
Graph
Compare
Charts
Issues
0
Issues
0
List
Board
Labels
Milestones
Merge Requests
0
Merge Requests
0
CI / CD
CI / CD
Pipelines
Jobs
Schedules
Charts
Wiki
Wiki
Snippets
Snippets
Members
Members
Collapse sidebar
Close sidebar
Activity
Graph
Charts
Create a new issue
Jobs
Commits
Issue Boards
Open sidebar
5c13e2907b128387adfbf972d1b506b3
mooc-rr
Commits
78ccd1a8
Commit
78ccd1a8
authored
Dec 26, 2024
by
5c13e2907b128387adfbf972d1b506b3
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
Add new file
parent
9665f34e
Changes
1
Hide whitespace changes
Inline
Side-by-side
Showing
1 changed file
with
32 additions
and
0 deletions
+32
-0
no name
module2/exo1/no name
+32
-0
No files found.
module2/exo1/no name
0 → 100644
View file @
78ccd1a8
# À propos du calcul de pi
_Arnaud Legrand_
_25 juin 2018_
## En demandant à la lib maths
Mon ordinateur m’indique que $\pi$ vaut _approximativement_
```{r}
pi
```
##En utilisant la méthode des aiguilles de Buffon
Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme _approximation_ :
```{r}
set.seed(42)
N = 100000
x = runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)>1))
```
## 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∼U(0,1)
et Y∼U(0,1)
alors P[X2+Y2≤1]=π/4
(voir méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait:
\ No newline at end of file
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment