Minor changes

parent 68d94d17
......@@ -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)
......
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