Is this an efficient way to compute the value of Pi, where the limit of j \$\propto\$ accuracy?

PI = 0;

for (var j = 1; j < 100; j+=2) PI += (4/j)*((j+1)%4?1:-1);

The formula I used to write this is:

$$\pi \equiv \frac{4}{1}-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\cdots$$

  • \$\begingroup\$ Could you please add a description of the mathematics behind your algorithm? That should help us to understand the merits of your approach, over, say something from here \$\endgroup\$ Dec 27, 2016 at 8:34
  • \$\begingroup\$ I've edited the question, adding the formula used. \$\endgroup\$
    – Tobi
    Dec 27, 2016 at 18:51

1 Answer 1


Your code will work eventually, but this series converges very very slowly. In fact, as you can read on Wolfram's web site:

... this sum converges so slowly that 300 terms are not sufficient to calculate pi correctly to two decimal places!

Here's a better way to do it which implements this function:

$$ \pi = \sum_{k=0}^\infty \frac{4(-1)^k}{2k+1} \bigg(\frac{1}{2^{2k+1}} + \frac{1}{3^{2k+1}}\bigg)$$

  for(k=0;k<100;k++) {
    z = 2*k+1;
    n *= -1;
    PI += n/z*(Math.pow(2, -z) + Math.pow(3, -z));

  • \$\begingroup\$ I know this is code-review, but is that equation the fastest converging one, or are the limitless / ever-accurate equations. \$\endgroup\$
    – Tobi
    Dec 27, 2016 at 21:47
  • \$\begingroup\$ There are equations that converge faster, but I chose this one that was close in computational complexity to the one you originally used. See Approximations of Pi on Wikipedia. \$\endgroup\$
    – Edward
    Dec 27, 2016 at 22:43

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