Fix formatting exo1

module2/exo1/toy_document_orgmode_R_en.org

@@ -8,7 +8,7 @@
 #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
 #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
 
-#+PROPERTY: header-args :eval never-export
+#+PROPERTY: header-args :session :exports both
 
 * Asking the maths library
 My computer tells me that $\pi$ is /approximatively/
@@ -23,7 +23,7 @@ pi
 * 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 R :results output :session *R* :exports results
+#+begin_src R :results output :session *R* :exports both
 set.seed(42)
 N = 100000
 x = runif(N)
@@ -35,7 +35,7 @@ theta = pi/2*runif(N)
 : [1] 3.14327
 
 * 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 $X\simU(0,1)$ and $Y\simU(0,1)$, 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 uses this approach:
 
 #+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R* 
 set.seed(42)
@@ -43,13 +43,13 @@ N = 1000
 df = data.frame(X = runif(N), Y = runif(N))
 df$Accept = (df$X**2 + df$Y**2 <=1)
 library(ggplot2)
-ggplot(df, aes(x=X, y=Y, color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
+ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
 #+end_src
 
 #+RESULTS:
 [[file:figure_pi_mc1.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:
+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 R :results output :session *R* :exports both
 4*mean(df$Accept)
 #+end_src