Leibniz formula for Pi with a given accuracy - kotlin

Question

I was asked to calculate the Pi number using the Leibniz formula for Pi with a given accuracy (eps).
The formula looks like this:
Initially, I wrote the following code:
fun main() {
val eps = 0.005
var n = 2
var r = row(n) // current row
var r0 = row(n-1)
var s = r0 + r
while (Math.abs(r) > eps) {
n++
r = row(n)
s += r
}
println(r.toString() + " <-- Leibniz(" + n.toString() + ")")
println(Math.abs(s*4).toString() + " <-- our evaluation with eps")
println(Math.PI.toString() + " <-- real Pi")
println((Math.abs(s*4)) in (Math.PI-eps..Math.PI+eps))
}
fun row(n: Int) = ((Math.pow(-1.0, n.toDouble()))/(2*n-1))
Then I found out that it doesn't work correctly, because
println((Math.abs(s*4)) in (Math.PI-eps..Math.PI+eps)) printed false.
I went deeper, made a debug, and realised that if went with
while (Math.abs(r) > eps/2)
over
while (Math.abs(r) > eps) everything works fine.
Could someone please provide any explanation on what I did wrong or why I have to divide eps by 2 if that is correct.
Thanks.

Answer 1

Each term r_i in that series is summed up to PI with a factor of 4 because sum(r_0, .., r_n) = PI/4. So of course, when you stop at the first r_i <= eps that only means that sum(r_0, ..., r_(i-1)) has an accuray of eps, ie it is somewhere in between [PI/4 - eps/2, PI/4 + eps/2]. But PI it self is 4*sum thus the accuracy is of course 4*eps ie the approximation lies somewhere inbetween [PI-2*eps ,PI+2*eps]
For your value of eps = 0.005:
The first r_100 = 0.00497512... is the first r <= eps
sum(r0, ..., r_99) = 0.782829, so PI at that point would be approximated as 3.1315929
EDIT
Also you are actually calculating -PI because are flipping the sign of each term in the series. So what you call r0 in your code (it should rather be called r1 because it's the result of row(1)) is -1 instead of +1

Answer 2

When you check Math.abs(r) > eps you're looking at the size of the n-th element of the series.
The distance of your current approximation from PI is the sum of all the terms in the series after that one.
As far as I know the relationship between the size of the n-th element of a convergent series and how good of an approximation you have depends on the specific series you are summing.