Stoneham's series

[Albert Chan]

(06-08-2024 05:02 AM)Johnh Wrote:  I put it into Free42, similar code that I posted above, with a 'View' line added to monitor the convergence towards Pi. I let it run 30 minutes, by which time it had completed 84 million loops ... stopped it it was at 3.14159266291 and counting down maybe one digit every few seconds in the 11th decimal place, slowing down exponentially

Convergence is slow, similar to Basel problem, error ≈ 1/n
see thread Evaluation of ζ(2) by the definition (sort of) [HP-42S & HP-71B]

(06-07-2024 08:54 PM)Albert Chan Wrote:  Or, we do product backwards, for more accuracy.

Code:

function q(n) -- 1 - product((1-1/b^2), b = n-(n+1)%2 .. 3 step -2)
    local a = 0
    for b=n-(n+1)%2,3,-2 do a = a + (1-a)/(b*b) end
    return a
end

With b going big, (1-a) does not change much.
If we assume (1-a) = 1, product turned into summation.

a = 1/3^2 + 1/5^2 + 1/7^2 + ... + 1/n^2 = (pi^2/8 - 1) if n→∞

Because this only sum odd square reciprocal, error ≈ .5/n

We are using (1-a)*4 for pi estimate, so we expected error ≈ -2/n
Of course (1-a) ≈ pi/4 ≠1 --> my guess is error around (-pi/2)/n

lua> n = 84e6*2+1
lua> (1-q(n)) * 4
3.1415926629397712
lua> _ + (-pi/2)/n
3.141592653589793