Update toy_document_en.Rmd

parent 2ecf8f5b
...@@ -13,14 +13,16 @@ knitr::opts_chunk$set(echo = TRUE) ...@@ -13,14 +13,16 @@ knitr::opts_chunk$set(echo = TRUE)
My computer tells me that $\pi$ is _approximatively_ My computer tells me that $\pi$ is _approximatively_
```pi ```{r}
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__
```set.seed(42) ```{r}
set.seed(42)
N = 100000 N = 100000
x = runif(N) x = runif(N)
theta = pi/2*runif(N) theta = pi/2*runif(N)
...@@ -29,7 +31,7 @@ theta = pi/2*runif(N) ...@@ -29,7 +31,7 @@ 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 thath if $\X ~ U (0,1)$ and $\Y ~ U (0,1)$, then $\P[X^2 + Y^2 <=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 $\le$ 1] = $\pi$/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
```set.seed(42) ```set.seed(42)
N = 1000 N = 1000
...@@ -38,7 +40,7 @@ df$Accept = (df$X**2 + df$Y**2 <=1) ...@@ -38,7 +40,7 @@ df$Accept = (df$X**2 + df$Y**2 <=1)
library(ggplot2) 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 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 : 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 :
```4*mean(df$Accept) ```4*mean(df$Accept)
``` ```
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