Third attempt, cosmetic changes

module2/exo1/toy_document_orgmode_python_en.org

-#+TITLE:  On the computation of pi
-#+AUTHOR: Dorinel Bastide
-#+DATE:   Today's date
+#+TITLE: On the computation of pi
 #+LANGUAGE: en
-# #+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/readtheorg.css"/>
@@ -40,14 +37,13 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
 : 3.128911138923655
 
 * 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 ocde uses this approach:
-#+begin_src python :results output file :session :var matplot_lib_filename="C:/Users/Utilisateur/mooc-rr/module2/exo1/PictureRes.png" :exports results
+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 ocde uses this approach:
+
+
+#+begin_src python :results output file :var matplot_lib_filename="C:/Users/Utilisateur/mooc-rr/module2/exo1/PictureRes.png" :exports both :session *python*
 import matplotlib
 matplotlib.use('Agg')
 import matplotlib.pyplot as plt
-import numpy as np
 np.random.seed(seed=42)
 N = 1000
 x = np.random.uniform(size=N, low=0, high=1)
@@ -70,10 +66,9 @@ print(matplot_lib_filename)
 Type "help", "copyright", "credits" or "license" for more information.
 C:/Users/Utilisateur/mooc-rr/module2/exo1/PictureRes.png]]
 
-It is then straightforward to obtain a (not really good) approximation
-to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller
-than $1$:
-#+begin_src python :results output :exports both
+It is then straightforward to obtain a (not really good) approximation to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:
+
+#+begin_src python :results output :session *python* :exports both
 import numpy as np
 4*np.mean(accept)
 #+end_src