Update format fix

module2/exo1/toy_document_orgmode_python_en.org

-#+TITLE:  On the computation of pi
+#+TITLE: On the computation of pi
 #+LANGUAGE: en
 
 #+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
@@ -10,7 +10,7 @@
 
 #+PROPERTY: header-args  :session  :exports both
 
-* Asking the maths library
+* Asking the math library
 My computer tells me that $\pi$ is /approximatively/
 
 #+begin_src python :results value :session *python* :exports both
@@ -18,28 +18,22 @@ from math import *
 pi
 #+end_src
 
-#+RESULTS:
-: 3.141592653589793
-
 * Buffon's needle
 Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*
 
 #+begin_src python :results value :session *python* :exports both
 import numpy as np
 np.random.seed(seed=42)
-N = 1000
+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:
-: 3.144654088050314
-
 * 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 $X\simU(0,1)$ and $Y\simU(0,1)$, then $P[X^2+Y^2 \le1]=\pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach:
+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 $P[X^2+Y^2\le 1]=\pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach:
 
-#+begin_src python :results output :session *python* :var matplot_lib_filename="figure_pi_mc2.png" :exports both
+#+begin_src python :results output file :var matplot_lib_filename="figure_pi_mc2.png" :exports both :session *python*
 import matplotlib.pyplot as plt
 
 np.random.seed(seed=42)
@@ -59,14 +53,8 @@ plt.savefig(matplot_lib_filename)
 print(matplot_lib_filename)
 #+end_src
 
-#+RESULTS:
-: figure_pi_mc2.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 value :session *python* :exports both
+#+begin_src python :results output :session *python* :exports both
 4*np.mean(accept)
 #+end_src
-
-#+RESULTS:
-: 3.112