Update toy_document_en.Rmd

parent 369c4335
......@@ -5,12 +5,12 @@ date: "25 juin 2018"
output: html_document
---
## Asking the maths library
My computer tells me that π is approximatively
My computer tells me that $\pi$ is *approximatively*
```{r}
pi
```
## Buffons needle
Applying the method of [Buffon’s needle](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)
N = 100000
......@@ -19,7 +19,7 @@ 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” 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}
set.seed(42)
N = 1000
......@@ -28,7 +28,7 @@ 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, 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)
```
......
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