From 00d16c82bc2926b2507ee671e00d3de53ea5168b Mon Sep 17 00:00:00 2001 From: ae44f8007c63991902a8055ff9736871 Date: Mon, 13 May 2024 11:33:33 +0000 Subject: [PATCH] Update toy_document_en.Rmd --- module2/exo1/toy_document_en.Rmd | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd index 48a544d..a96387b 100644 --- a/module2/exo1/toy_document_en.Rmd +++ b/module2/exo1/toy_document_en.Rmd @@ -11,14 +11,14 @@ knitr::opts_chunk$set(echo = TRUE) ``` ## Asking the maths library -My computer tells me that π is *approximatively* +My computer tells me that $\pi$ is *approximatively* -```{r cars} +```{r} pi ``` -##Buffon's needle -Applying the method of [Buffon's neddle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the *approximation* +## 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) @@ -28,8 +28,8 @@ 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∼U(0,1) and Y∼U(0,1), then P[X2+Y2≤1]=π/4 (see [“Monte Carlo method”](https://en.wikipedia.org/wiki/Monte_Carlo_method) on Wikipedia). The following code uses this approach: +## 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) @@ -40,7 +40,7 @@ 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 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 -- 2.18.1