From 5611c5b8b8d918e8f2d3c930b0850359f6f88e2d Mon Sep 17 00:00:00 2001 From: feee8ecec645c53ceeda57e948ea51be Date: Fri, 23 Jul 2021 22:17:56 +0000 Subject: [PATCH] Minor changes --- module2/exo1/toy_document_en.Rmd | 12 ++++++++---- 1 file changed, 8 insertions(+), 4 deletions(-) diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd index 42a848f..d69a0ce 100644 --- a/module2/exo1/toy_document_en.Rmd +++ b/module2/exo1/toy_document_en.Rmd @@ -5,9 +5,13 @@ date: "23/7/2021" output: html_document --- -## Asking the math library +```{r setup, include=FALSE} +knitr::opts_chunk$set(echo = TRUE) +``` + +## Asking the math library -My computer me that $\pi$ is *approximatively* +My computer tells me that $\pi$ is *approximatively* ```{r} pi @@ -27,7 +31,7 @@ theta = pi/2*runif(N) ## 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 $P[X^2+Y^2 \le 1]=\pi/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code use 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 $X \sim U(0,1)$ and $Y \sim U(0,1)$, then $P[X^2+Y^2 \le 1]=\pi/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code use this approach: ```{r} set.seed(42) @@ -37,7 +41,7 @@ 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 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 : +It is therefore 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) -- 2.18.1