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00457cfdbb94b4c89e4b2ff8b27dc825
mooc-rr
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28ea4a3b
Commit
28ea4a3b
authored
Apr 01, 2020
by
Clément Courageux
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À propos du calcul de
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<div
id=
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>
<h1
class=
"title"
>
À propos du calcul de
π
</h1>
<div
id=
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>
<h2>
Table des matières
</h2>
<div
id=
"text-table-of-contents"
>
<ul>
<li><a
href=
"#org3de6e6c"
>
1. En demandant à la lib maths
</a></li>
<li><a
href=
"#org833fcbb"
>
2. En utilisant la méthode des aiguilles de Buffon
</a></li>
<li><a
href=
"#org9658f1f"
>
3. Avec un argument "fréquentiel" de surface
</a></li>
</ul>
</div>
</div>
<div
id=
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class=
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>
<h2
id=
"org3de6e6c"
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class=
"section-number-2"
>
1
</span>
En demandant à la lib maths
</h2>
<div
class=
"outline-text-2"
id=
"text-1"
>
<p>
Mon ordinateur m'indique que
π
vaut
<i>
approximativement
</i>
:
</p>
<div
class=
"org-src-container"
>
<pre
class=
"src src-python"
><span
style=
"color: #0000FF;"
>
from
</span>
math
<span
style=
"color: #0000FF;"
>
import
</span>
*
pi
</pre>
</div>
<pre
class=
"example"
>
3.141592653589793
</pre>
</div>
</div>
<div
id=
"outline-container-org833fcbb"
class=
"outline-2"
>
<h2
id=
"org833fcbb"
><span
class=
"section-number-2"
>
2
</span>
En utilisant la méthode des aiguilles de Buffon
</h2>
<div
class=
"outline-text-2"
id=
"text-2"
>
<p>
Mais calculé avec la
<b>
méthode
</b>
des
<a
href=
"https://fr.wikipedia.org/wiki/Aiguille_de_Buffon"
>
aiguilles de Buffon
</a>
, on obtiendrait comme
<b>
approximation
</b>
:
</p>
<div
class=
"org-src-container"
>
<pre
class=
"src src-python"
><span
style=
"color: #0000FF;"
>
import
</span>
numpy
<span
style=
"color: #0000FF;"
>
as
</span>
np
np.random.seed(seed=42)
<span
style=
"color: #BA36A5;"
>
N
</span>
= 10000
<span
style=
"color: #BA36A5;"
>
x
</span>
= np.random.uniform(size=N, low=0, high=1)
<span
style=
"color: #BA36A5;"
>
theta
</span>
= np.random.uniform(size=N, low=0, high=pi/2)
2/(
<span
style=
"color: #006FE0;"
>
sum
</span>
((x+np.sin(theta))
>
1)/N)
</pre>
</div>
<pre
class=
"example"
>
3.128911138923655
</pre>
</div>
</div>
<div
id=
"outline-container-org9658f1f"
class=
"outline-2"
>
<h2
id=
"org9658f1f"
><span
class=
"section-number-2"
>
3
</span>
Avec un argument "fréquentiel" de surface
</h2>
<div
class=
"outline-text-2"
id=
"text-3"
>
<p>
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 \(P[X^2+Y^2 \leq1] = \pi/4\) (voir
<a
href=
"https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80"
>
méthode
yde Monte Carlo
</a>
sur Wikipedia). Le code suivant illustre ce fait :
</p>
<div
class=
"org-src-container"
>
<pre
class=
"src src-python"
><span
style=
"color: #0000FF;"
>
import
</span>
matplotlib.pyplot
<span
style=
"color: #0000FF;"
>
as
</span>
plt
np.random.seed(seed=42)
<span
style=
"color: #BA36A5;"
>
N
</span>
= 1000
<span
style=
"color: #BA36A5;"
>
x
</span>
= np.random.uniform(size=N, low=0, high=1)
<span
style=
"color: #BA36A5;"
>
y
</span>
= np.random.uniform(size=N, low=0, high=1)
<span
style=
"color: #BA36A5;"
>
accept
</span>
= (x*x+y*y)
<
= 1
<span
style=
"color: #BA36A5;"
>
reject
</span>
= np.logical_not(accept)
<span
style=
"color: #BA36A5;"
>
fig
</span>
,
<span
style=
"color: #BA36A5;"
>
ax
</span>
= plt.subplots(1)
ax.scatter(x[accept], y[accept], c=
<span
style=
"color: #008000;"
>
'b'
</span>
, alpha=0.2, edgecolor=
<span
style=
"color: #D0372D;"
>
None
</span>
)
ax.scatter(x[reject], y[reject], c=
<span
style=
"color: #008000;"
>
'r'
</span>
, alpha=0.2, edgecolor=
<span
style=
"color: #D0372D;"
>
None
</span>
)
ax.set_aspect(
<span
style=
"color: #008000;"
>
'equal'
</span>
)
plt.savefig(matplot_lib_filename)
matplot_lib_filename
</pre>
</div>
<p>
<img
src=
"fige.png"
alt=
"fige.png"
/>
Il est alors aisé d'obtenir une approximation (pas terrible) de
π
en comptant
combien de fois, en moyenne, \(X^2+Y^2\) est inférieur à 1 :
</p>
<div
class=
"org-src-container"
>
<pre
class=
"src src-python"
>
4*np.mean(accept)
</pre>
</div>
<pre
class=
"example"
>
3.112
</pre>
</div>
</div>
</div>
<div
id=
"postamble"
class=
"status"
>
<p
class=
"date"
>
Date: 01/04/2020
</p>
<p
class=
"author"
>
Auteur: Clément
</p>
<p
class=
"date"
>
Created: 2020-04-01 Wed 10:17
</p>
<p
class=
"validation"
><a
href=
"http://validator.w3.org/check?uri=referer"
>
Validate
</a></p>
</div>
</body>
</html>
module2/exo1/toy_document_orgmode_python_fr.org
View file @
28ea4a3b
#+TITLE:
Votre titre
#+TITLE:
À propos du calcul de \pi
#+AUTHOR:
Votre nom
#+AUTHOR:
Clément
#+DATE:
La date du jour
#+DATE:
01/04/2020
#+LANGUAGE: fr
#+LANGUAGE: fr
# #+PROPERTY: header-args :eval never-export
# #+PROPERTY: header-args :eval never-export
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
#
#
+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/readtheorg.css"/>
#
#
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#+HTML_HEAD: <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
#+HTML_HEAD: <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
#+HTML_HEAD: <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
#+HTML_HEAD: <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
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#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
* Quelques explications
* En demandant à la lib maths
Mon ordinateur m'indique que \pi vaut /approximativement/ :
#+begin_src python :session :exports both
from math import *
pi
#+end_src
#+RESULTS:
: 3.141592653589793
* En utilisant la méthode des aiguilles de Buffon
Mais calculé avec la *méthode* des [[https://fr.wikipedia.org/wiki/Aiguille_de_Buffon][aiguilles de Buffon]], on obtiendrait comme
*approximation* :
#+begin_src python :session :exports both
import numpy as np
np.random.seed(seed=42)
N = 10000
x = np.random.uniform(size=N, low=0, high=1)
theta = np.random.uniform(size=N, low=0, high=pi/2)
2/(sum((x+np.sin(theta))>1)/N)
#+end_src
#+RESULTS:
: 3.128911138923655
* 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 \(P[X^2+Y^2 \leq1] = \pi/4\) (voir
[[https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80][méthode
yde Monte Carlo]] sur Wikipedia). Le code suivant illustre ce fait :
#+begin_src python :exports both :session :var matplot_lib_filename="fige.png" :results file
import matplotlib.pyplot as plt
np.random.seed(seed=42)
N = 1000
x = np.random.uniform(size=N, low=0, high=1)
y = np.random.uniform(size=N, low=0, high=1)
accept = (x*x+y*y) <= 1
reject = np.logical_not(accept)
fig, ax = plt.subplots(1)
ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)
ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)
ax.set_aspect('equal')
plt.savefig(matplot_lib_filename)
matplot_lib_filename
#+end_src
#+RESULTS:
[[file:fige.png]]
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 python :results output :session :exports both
4*np.mean(accept)
#+end_src
#+RESULTS:
: 3.112
* Quelques explications :noexport:
Ceci est un document org-mode avec quelques exemples de code
Ceci est un document org-mode avec quelques exemples de code
python. Une fois ouvert dans emacs, ce document peut aisément être
python. Une fois ouvert dans emacs, ce document peut aisément être
...
...
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