Problem Statement

Almkvist Berndt 1988 begins with an investigation of why the agm is such an efficient algorithm, and proves that it converges quadratically. This is an efficient method to calculate

π

{\displaystyle \pi }

. With the same notations used in Arithmetic-geometric mean, we can summarize the paper by writing:

π

4

a g m

( 1 , 1

/

2

)

2

1 −

∑

n

1

∞

2

n + 1

(

a

n

2

−

g

n

2

)

{\displaystyle \pi ={\frac {4;\mathrm {agm} (1,1/{\sqrt {2}})^{2}}{1-\sum \limits {n=1}^{\infty }2^{n+1}(a{n}^{2}-g_{n}^{2})}}}

This allows you to make the approximation, for any large   N:

π ≈

4

a

N

2

1 −

∑

k

1

N

2

k + 1

(

a

k

2

−

g

k

2

)

{\displaystyle \pi \approx {\frac {4;a_{N}^{2}}{1-\sum \limits {k=1}^{N}2^{k+1}(a{k}^{2}-g_{k}^{2})}}}

The purpose of this task is to demonstrate how to use this approximation in order to compute a large number of decimals of

π

{\displaystyle \pi }

.

Let's start with the solution:

import "./big" for BigRat

var digits = 500
var an = BigRat.one
var bn = BigRat.half.sqrt(digits)
var tn = BigRat.half.square
var pn = BigRat.one
while (pn <= digits) {
    var prevAn = an
    an = (bn + an) * BigRat.half
    bn = (bn * prevAn).sqrt(digits)
    prevAn = prevAn - an
    tn = tn - (prevAn.square * pn)
    pn = pn + pn
}
var pi = (an + bn).square / (tn * 4)
System.print(pi.toDecimal(digits, false))