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<div id="content" class="content">
<h1 class="title">À propos du calcul de \(\pi\)</h1>
<div id="table-of-contents" role="doc-toc">
<h2>Table des matières</h2>
<div id="text-table-of-contents" role="doc-toc">
<ul>
<li><a href="#org4e24512">1. En demandant à la lib maths</a></li>
<li><a href="#orgef227a1">2. En utilisant la méthode des aiguilles de Buffon</a></li>
<li><a href="#org37354cb">3. Avec un argument "fréquentiel" de surface</a></li>
</ul>
</div>
</div>
<div id="outline-container-org4e24512" class="outline-2">
<h2 id="org4e24512"><span 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 \(\pi\) vaut <i>approximativement</i>:
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #b6a0ff;">from</span> math <span style="color: #b6a0ff;">import</span> *
<span style="color: #f78fe7;">print</span>(pi)
</pre>
</div>
<pre class="example">
3.141592653589793
</pre>
</div>
</div>
<div id="outline-container-orgef227a1" class="outline-2">
<h2 id="orgef227a1"><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 méthode 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: #b6a0ff;">import</span> numpy <span style="color: #b6a0ff;">as</span> np
np.random.seed(seed=42)
<span style="color: #00d3d0;">N</span> = 10000
<span style="color: #00d3d0;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #00d3d0;">theta</span> = np.random.uniform(size=N, low=0, high=pi/2)
<span style="color: #f78fe7;">print</span>(2/(<span style="color: #f78fe7;">sum</span>((x+np.sin(theta))&gt;1)/N))
</pre>
</div>
<pre class="example">
3.128911138923655
</pre>
</div>
</div>
<div id="outline-container-org37354cb" class="outline-2">
<h2 id="org37354cb"><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
\(X ~ U(0, 1)\) et \(Y ~ U(0, 1)\) alors \(P[X^2 + Y^2 \le 1] = \pi/4\)
(voir
<a href="https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80">méthode de Monte Carlo sur Wikipedia</a>).
Le code suivant illustre ce fait :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #b6a0ff;">import</span> matplotlib.pyplot <span style="color: #b6a0ff;">as</span> plt
np.random.seed(seed=42)
<span style="color: #00d3d0;">N</span> = 1000
<span style="color: #00d3d0;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #00d3d0;">y</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #00d3d0;">accept</span> = (x*x+y*y) &lt;= 1
<span style="color: #00d3d0;">reject</span> = np.logical_not(accept)
<span style="color: #00d3d0;">fig</span>, <span style="color: #00d3d0;">ax</span> = plt.subplots(1)
ax.scatter(x[accept], y[accept], c=<span style="color: #79a8ff;">'b'</span>, alpha=0.2, edgecolor=<span style="color: #00bcff;">None</span>)
ax.scatter(x[reject], y[reject], c=<span style="color: #79a8ff;">'r'</span>, alpha=0.2, edgecolor=<span style="color: #00bcff;">None</span>)
ax.set_aspect(<span style="color: #79a8ff;">'equal'</span>)
plt.savefig(matplot_lib_filename)
matplot_lib_filename
</pre>
</div>
<div id="orge303c2e" class="figure">
<p><img src="file:///tmp/babel-ZDryvm/figureyRFOxR.png" alt="figureyRFOxR.png" />
</p>
</div>
<p>
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 :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #f78fe7;">print</span>(4*np.mean(accept))
</pre>
</div>
<pre class="example">
3.112
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="date">Date: 12/12/2024</p>
<p class="author">Auteur: Matteo Chancerel</p>
<p class="date">Created: 2024-12-12 Thu 11:27</p>
<p class="validation"><a href="https://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
</html>
module2/exo1/toy_document_orgmode_python_fr.org
View file @ 67f9d0e6
#+TITLE: À propos du calcul de $\pi$
#+AUTHOR: Matteo Chancerel
#+DATE: 25/09/2024
#+DATE: 12/12/2024
#+LANGUAGE: fr
# #+PROPERTY: header-args :eval never-export
# À propos du calcul de $\pi$
## Table des matières
## 1 En demandant à la lib maths
* En demandant à la lib maths
Mon ordinateur m'indique que $\pi$ vaut /approximativement/:
......@@ -19,7 +18,7 @@ print(pi)
#+RESULTS:
: 3.141592653589793
## 2 En utilisant la méthode des aiguilles de Buffon
* 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* :
......@@ -35,8 +34,7 @@ print(2/(sum((x+np.sin(theta))>1)/N))
#+RESULTS:
: 3.128911138923655
## 3 Avec un argument "fréquentiel" de surface
* 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
......@@ -62,11 +60,11 @@ ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)
ax.set_aspect('equal')
plt.savefig(matplot_lib_filename)
print(matplot_lib_filename)
matplot_lib_filename
#+end_src
#+RESULTS:
[[file:]]
[[file:/tmp/babel-ZDryvm/figure8q72fF.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 :
......@@ -76,3 +74,4 @@ print(4*np.mean(accept))
#+end_src
#+RESULTS:
: 3.112
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