Update toy_document_orgmode_python_en.org

module2/exo1/toy_document_orgmode_python_en.org

-#+TITLE:  Mr.
+
+
+#+TITLE:  On the computation of pi
 #+AUTHOR: Bikash Adhikari
 #+DATE:   23/03/2020
 #+LANGUAGE: en
-# #+PROPERTY: header-args :eval never-export
 
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+ #+PROPERTY: header-args :session  :export both
 
-* Table of Contents
- * [[1. Asking the math library]]
- * [[2. *Buffon's needle]]
- * [[3.  Using a surface fraction argument]]
 
-
-* 1. Asking the math library
-My computer tells me that \pi is /approximately/
-#+begin_src python :results output :exports both
+*  Asking the math library
+My computer tells me that $\pi$ is /approximately/
+#+begin_src python :results value :session *python* :export both 
 from math import *
 pi
 #+end_src
 
-* 2. *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 output :exports both
+
+
+* * 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* :export both
 import numpy as np
 np.random.seed(seed=42)
-N=10000
-x=np.random.uniform(size=N, low =0, high=pi/2)
-theta=np.random.uniform(size=N, low=0, high = pi/2)
+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
 
 
-
-* 3.  Using a surface fraction argument
+* 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 code use this approach:
+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 code use this approach:
 
-#+begin_src python :results output :exports both
+#+begin_src python :results value :session *python* :export both 
 import matplotlib.pyplot as plt
 
 np.random.seed(seed=42)
@@ -68,7 +61,6 @@ print(matplot_lib_filename)
 It is then straightforward to obtain a (not really good) approximation
 of \pi by counting how many time, on average, $X^2$ + $^{}Y^2$ is smaller
 than 1:
-#+begin_src python :results output :exports both
+#+begin_src python :results value :session *python* :export both 
 4*np.mean(accept)
 #+end_src
-