module 2 exercise 1 - fixes

module2/exo1/toy_document_orgmode_python_en.org

@@ -12,7 +12,7 @@
 #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
 
 * Asking the math library
-My computer tells me that \pi is approximatively
+My computer tells me that $\pi$ is /approximatively/
 #+begin_src python :results value :session *python* :exports both
 from math import *
 pi
@@ -21,8 +21,9 @@ pi
 #+RESULTS:
 : 3.141592653589793
 
-* Buffon's needle
- Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][Buffon's]] needle, we get the approximation 
+* Buffon's needle 
+Applying the method of
+[[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][Buffon's needle]], we get the *approximation*
 
 #+begin_src python :results value :session *python* :exports both
 import numpy as np
@@ -44,9 +45,7 @@ then $P[X^2+Y^2 \le 1]=\pi/4$ (see
 method" on Wikipedia]]). The following code uses this approach:
 
 
-
-
-#+begin_src python :results output :session *python* :var matplot_lib_filename="figure.png" :exports  code
+#+begin_src python :results output file :session *python* :var matplot_lib_filename="figure.png" :exports both
 import matplotlib.pyplot as plt
 
 np.random.seed(seed=42)
@@ -67,18 +66,21 @@ print(matplot_lib_filename)
 #+end_src
 
 #+RESULTS:
-: figure.png
+[[file:figure.png]]
+
+
+
 
-[[./figure.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
+
+