Numerical approximations
Sometimes an analytical solution is hard to derive or is simply not available. One way to get around this is to approximate a solution by doing what we already know.
Approximating pi
Believe it or not, mathematics used to be an experimental science. Before the the concept of infinity or calculus was a thing, people had no way of dealing with curves. But curves are everywhere! To estimate the circumference of a circle, the Greek polymath Archimedes started by fitting a hexagon (a shape with straight lines!) inside a circle. Since he knew the distance between the corners of the polygon to the center of the circle is equal to the radius (r), he can apply Pythagorean’s theorem to get the distance between those corners. Summing up those distances would then give an approximation of the circle’s circumference. But the approximation of a hexagon is crude. Instead, Archimedes continued adding more edges to the polygon, and with each added edge, the approximation gets closer and closer to the actual value.
Instead of drawing polygons and solving the Pythagorean’s equation by hand(!), we can now have computers do it for us!
Activity 1: Approximating \(\pi\)
Suppose we have a unit circle enclosed within a unit square, like so:
It is said that the value of \(\pi\) is equal to 4 times the ratio between the circle’s area and the square’s area:
\[ \pi = 4 * \frac{Area_{circle}}{Area_{square}} \]
Obviously, anyone who’s taken elementary geometry would tell you “but we need \(\pi\) to get the area of the circle!” Instead of using equations, we can fill up the space (that make up the shapes) with quantities that we already know and use that to approximate area. Like estimating the volume of a vase by adding water from a graduated cylinder. Instead of water, we can fill up that space with points of data on a computer. For example, we can use the seq() function to fill up the space between 0 and 1, and the amount of points within that range gives us some quantitative information about the space. Since the circle is smaller than the square, some points will lie within the circle while some are outside. The number of points that lie within and outside of the circle therefore gives us an approximation of the ratio between the area of the circle and the square. The more points we use, the closer we are to the actual value. Just as we can fill our vase with discrete and coarse objects like marbles, pebbles, or coins, instead of water. Coarser objects simply reduces the resolution of our approximation.
res <- 100 #resolution; the higher the better the approximation
points <- seq(from = 0, to = 1, length.out = res)
xy_points <- expand.grid(x = points, y = points) #gives a pairwise combination of each value
#Plot with points
circle_square_p +
geom_point(data = xy_points, aes(x = x, y = y), size = 0.5, alpha = 0.3)+
lims(x = c(0, 1), y = c(0,1))
To check whether a point lies within the circle, we can use equation of the circle:
\[ (x - x_o)^2 + (y - y_o)^2 = r^2 \]
\[ (x - 0.5)^2 + (y - 0.5)^2 = 0.5^2 \]
\[ \sqrt((x−0.5)^2+(y−0.5)^2) = 0.5 \]
Any point that gives a value greater than \(r\) (or 0.5) would lie outside of the circle. To implement this, we simply substitute the x and y with our “data” (apply the LHS of the equation), and check whether each result is < than 0.5.
approx_pi_fun <- function(n = 100){
#Number of points used to approximate
points <- seq(from = 0, to = 1, length.out = n)
#Expand it to span 2 dimenisons (x and y)
xy_points <- expand.grid(x = points, y = points)
#Check whether points are within circle
area_with_points <- xy_points%>%
mutate(distance_to_center = sqrt((x-0.5)^2 + (y-0.5)^2))%>% #circle equation
mutate(in_circle = ifelse(distance_to_center < 0.5, yes = 1, no = 0))
#check condition
#Number of points inside square
area_square <- n*n #all points should be within square
area_circle <- sum(area_with_points$in_circle) #sum up all the 1s
#Approximate pi
approx_pi <- 4*(area_circle/area_square)
return(approx_pi)
}
approx_pi_fun(n = 100)## [1] 3.0672Task 1a: Use the approx_pi_fun() function and apply it to varying levels of sample sizes (n) and see how the approximation of \(\pi\) changes
Task 1b: Instead of using seq() can you think of a stochastic way of approximating pi? Imagine throwing darts and adding up the darts that lie within and outside of the circle.
Hint: draw from a probability distribution but think about which one specifically
# #I. take random shots with "equal probablity" within the unit square
#
# # random_darts <- data.frame(x = some distribution b/w 0 and 1,
# # y = some distribution b/w 0 and 1)
#
# #Check where darts lie
# pi_exp2 <- random_darts%>%
# mutate(distance_to_center = sqrt((x-0.5)^2 + (y-0.5)^2))%>% #circle equation
# mutate(in_circle = ifelse(distance_to_center < 0.5, yes = 1, no = 0))
#
# #Approximate value of pi
# # No. of points inside circle
# # No. of points inside square
# # ratio * 4
