How to resolve the algorithm Arithmetic-geometric mean/Calculate Pi step by step in the Pascal programming language

Published on 12 May 2024 09:40 PM
#Pascal

How to resolve the algorithm Arithmetic-geometric mean/Calculate Pi step by step in the Pascal programming language

Table of Contents

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:

Step by Step solution about How to resolve the algorithm Arithmetic-geometric mean/Calculate Pi step by step in the Pascal programming language

Source code in the pascal programming language

program AgmForPi;
{$mode objfpc}{$h+}{$b-}{$warn 5091 off}
uses
  SysUtils, Math, GMP;

const
  MIN_DIGITS = 32;
  MAX_DIGITS = 1000000;

var
  Digits: Cardinal = 256;

procedure ReadInput;
var
  UserDigits: Cardinal;
begin
  if (ParamCount > 0) and TryStrToDWord(ParamStr(1), UserDigits) then
    Digits := Min(MAX_DIGITS, Max(UserDigits, MIN_DIGITS));
  f_set_default_prec(Ceil((Digits + 1)/LOG_10_2));
end;

function Sqrt(a: MpFloat): MpFloat;
begin
  Result := f_sqrt(a);
end;

function Sqr(a: MpFloat): MpFloat;
begin
  Result := a * a;
end;

function PiDigits: string;
var
  a0, b0, an, bn, tn: MpFloat;
  n: Cardinal;
begin
  n := 1;
  an := 1;
  bn := Sqrt(MpFloat(0.5));
  tn := 0.25;
  while n < Digits do begin
    a0 := an;
    b0 := bn;
    an := (a0 + b0)/2;
    bn := Sqrt(a0 * b0);
    tn := tn - Sqr(an - a0) * n;
    n := n + n;
  end;
  Result := Sqr(an + bn)/(tn * 4);
  SetLength(Result, Succ(Digits));
end;

begin
  ReadInput;
  WriteLn(PiDigits);
end.

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