From d70ed8e867221007700b0be53a968974cbf1cb94 Mon Sep 17 00:00:00 2001
From: 8a31633c40889a7a89c831544b7d92e5
 <8a31633c40889a7a89c831544b7d92e5@app-learninglab.inria.fr>
Date: Sun, 14 Aug 2022 13:03:11 +0000
Subject: [PATCH] Update toy_document_fr.Rmd

---
 module2/exo1/toy_document_fr.Rmd | 6 ++----
 1 file changed, 2 insertions(+), 4 deletions(-)

diff --git a/module2/exo1/toy_document_fr.Rmd b/module2/exo1/toy_document_fr.Rmd
index f965d17..6abff95 100644
--- a/module2/exo1/toy_document_fr.Rmd
+++ b/module2/exo1/toy_document_fr.Rmd
@@ -26,8 +26,7 @@ theta = pi/2*runif(N)
 2/(mean(x+sin(theta)>1))
 ```
 
-## Using Fraction arguement 
-
+## Using Fraction arguement
 A method that is easier to understand and does not make use of the sin function is based on the fact that if  $X\sim U(0,1)$ et $Y\sim U(0,1)$ then $P[X^2+Y^2\leq 1] = \pi/4$ [Mont carlo method](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:
 
 ```{r}
@@ -39,8 +38,7 @@ library(ggplot2)
 ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
 ```
 
-It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, X2+Y2 is smaller than 1 :
-
+It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average,  $X^2 + Y^2$  is smaller than 1 :
 
 ```{r}
 4*mean(df$Accept)
-- 
2.18.1