Update toy_document_en.Rmd

parent 7da0e3f6
...@@ -4,13 +4,20 @@ author: "Arnaud Legrand" ...@@ -4,13 +4,20 @@ author: "Arnaud Legrand"
date: "25 juin 2018" date: "25 juin 2018"
output: html_document output: html_document
--- ---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
## Asking the maths library ## Asking the maths library
My computer tells me that $\pi$ is *approximatively* My computer tells me that $\pi$ is *approximatively*
```{r} ```{r}
pi pi
``` ```
## Buffon's needle ## Buffon's needle
Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__ Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
```{r} ```{r}
set.seed(42) set.seed(42)
N = 100000 N = 100000
...@@ -20,6 +27,7 @@ theta = pi/2*runif(N) ...@@ -20,6 +27,7 @@ theta = pi/2*runif(N)
``` ```
## Using a surface fraction argument ## 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: 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} ```{r}
set.seed(42) set.seed(42)
N = 1000 N = 1000
...@@ -29,6 +37,7 @@ library(ggplot2) ...@@ -29,6 +37,7 @@ 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()
``` ```
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 :
```{r} ```{r}
4*mean(df$Accept) 4*mean(df$Accept)
``` ```
......
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