Using basic arithmetics for calculating Pi with arbitary precision

I am looking for a formula/algorithm to calculate PI~3.14 in a given precision.

The formula/algorithm must have only very basic arithmetic as

• +: Addition
• -: Subtraction
• *: Multiplication
• /: Divison

because I want to implement these operations in C++ and want to keep the implementation as simple as possible (no bignum library is allowed).

I have found that this formula for calculating Pi is pretty simple:

Pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...  = sum( (-1)^(k+1)/(2*k-1) , k=1..inf )

(note that (-1)^(k+1) can be implemented easily by above operators).

But the problem about this formula is the inability to specify the number of digits to calculate. In other words, there is no direct way to determine whe开发者_高级运维n to stop the calculation.

Maybe a workaround to this problem is calculating the difference between n-1th and nth calculated term and considering it as the current error.

Anyway, I am looking for a formula/algorithm that have these properties and also converges faster to Pi

Codepad link:

#include <iostream>
#include <cmath>
int main()
{
    double p16 = 1, pi = 0, precision = 10;

    for(int k=0; k<=precision; k++)
    {
        pi += 1.0/p16 * (4.0/(8*k + 1) - 2.0/(8*k + 4) - 1.0/(8*k + 5) - 1.0/(8*k+6));
        p16 *= 16;
    }
    std::cout<<std::setprecision(80)<<pi<<'\n'<<M_PI;
}

Output:

3.141592653589793115997963468544185161590576171875
3.141592653589793115997963468544185161590576171875

This is actually the Bailey-Borwein-Plouffe formula, also taken from the link from wikipedia.

In your original (slowly converging) example, the error term can be computed because this is an alternating series; see http://en.wikipedia.org/wiki/Alternating_series#Approximating_Sums

Essentially, the next uncomputed term is a bound on the error.

You can just do the Taylor envelope of the arctan(1) and then you will get pi/4 just summing all the rest part. The taylor envelope of arctan(1)

http://en.wikipedia.org/wiki/Taylor_series

also you can use the euler formula with z=1 and then multiply the result by 4.

http://upload.wikimedia.org/math/2/7/9/279bed5a2ea3b80a71f5b22078090168.png