From 1b848681db6aca68ff17e56e96f4db947d563c0a Mon Sep 17 00:00:00 2001
From: 16fd6933414ad4ae0fb1412e7effdbc7
 <16fd6933414ad4ae0fb1412e7effdbc7@app-learninglab.inria.fr>
Date: Sat, 13 Jun 2020 13:40:56 +0000
Subject: [PATCH] Update toy_document_fr.Rmd

---
 module2/exo1/toy_document_fr.Rmd | 72 ++++++++++++++++++--------------
 1 file changed, 41 insertions(+), 31 deletions(-)

diff --git a/module2/exo1/toy_document_fr.Rmd b/module2/exo1/toy_document_fr.Rmd
index 0427748..0839e1d 100644
--- a/module2/exo1/toy_document_fr.Rmd
+++ b/module2/exo1/toy_document_fr.Rmd
@@ -1,37 +1,47 @@
+---
 title: "On the computation of pi"	
 author: "Tegegne"	
 date: "June 13, 2020  
-
-output: html_document	
+output: html_document
 ---	
-`` `{r setup, include = FALSE}	
-knitr :: opts_chunk $ set (echo = TRUE)	
-`` ''	
-## By asking the math lib	
-My computer tells me that $ \ pi $ is * approximately *	
-`` `` r cars}	
+
+```{r setup, include=FALSE}	
+knitr::opts_chunk$set(echo = TRUE)	
+```	
+
+## Asking the maths library	
+My computer tells me that $\pi$ is *approximatively*	
+
+```{r}	
 pi	
-`` ''	
-## Using the Buffon needle method	
-But calculated with the __method__ of [Buffon needles] (https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), we would obtain as __approximation__:	
-`` `{r}	
-set.seed (42)	
-N = 100,000	
-x = runif (N)	
-theta = pi / 2 * runif (N)	
-2 / (mean (x + sin (theta)> 1))	
-`` ''	
-## With a surface "frequency" argument	
-Otherwise, a simpler method to understand and not involving a call to the sine 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 [Monte Carlo method on Wikipedia] (https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte- Carlo # D% C3% A9termination_de_la_valeur_de_% CF% 80)). The following code illustrates this fact:	
-`` `{r}	
-set.seed (42)	
+```	
+
+## Buffon's needle	
+Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__	
+
+```{r}	
+set.seed(42)	
+N = 100000	
+x = runif(N)	
+theta = pi/2*runif(N)	
+2/(mean(x+sin(theta)>1))	
+```	
+
+## 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 ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:	
+
+```{r}	
+set.seed(42)	
 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 ()	
-`` ''	
-It is then easy to obtain an approximation (not great) of $ \ pi $ by counting how many times, on average, $ X ^ 2 + Y ^ 2 $ is less than 1:	
-`` `{r}	
-4 * mean (df $ Accept)	
-`` ''
\ No newline at end of file
+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()	
+
+```	
+
+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:	
+
+```{r}	
+4*mean(df$Accept)	
+```
\ No newline at end of file
-- 
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