Update toy_document_en.Rmd

parent f7733a41
...@@ -11,14 +11,14 @@ knitr::opts_chunk$set(echo = TRUE) ...@@ -11,14 +11,14 @@ knitr::opts_chunk$set(echo = TRUE)
``` ```
## Asking the maths library ## Asking the maths library
My computer tells me that π is *approximatively* My computer tells me that $\pi$ is *approximatively*
```{r cars} ```{r}
pi pi
``` ```
##Buffon's needle ## Buffon's needle
Applying the method of [Buffon's neddle](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)
...@@ -28,8 +28,8 @@ theta = pi/2*runif(N) ...@@ -28,8 +28,8 @@ theta = pi/2*runif(N)
2/(mean(x+sin(theta)>1)) 2/(mean(x+sin(theta)>1))
``` ```
##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∼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: 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)
...@@ -40,7 +40,7 @@ library(ggplot2) ...@@ -40,7 +40,7 @@ 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 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} ```{r}
4*mean(df$Accept) 4*mean(df$Accept)
``` ```
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