[bsteinbach]

Not only similar, it's identical to Newton Raphson in ideal arithmetic!

The nth order product approximation for the reciprocal of d = 1 - x is y_n = product_{i=1}{n} (1 + x^{2^(i-1)}) = sum_{i=0}^{2^n - 1} x^i = (1 - x^(2 n)) / (1 - x).

The Newton-Raphson iteration for the reciprocal of d is z_{n+1} = z_n * (2 - (1-x) * z_n).

Induction with the base case z_0 = 1 = y_0 shows the sequences are equal by inserting y_n for z_n in the Newton Raphson iteration: z_{n+1} = (1 - x^(2 n)) * (2 - (1 - x^(2 n))) / (1 - x) = ( 1 - x^(4 n) ) / ( 1 - x) = y_{n+1}.

So I guess you could call it an explicit product formula for the Newton Raphson reciprocal.

[perihelions]

This 100% checks out! I didn't recognize this at all.

- "...in ideal arithmetic!"

Comparing the two, Newton-Raphson seems rather better conditioned. I guess it's the (1.0 + x^(2^i)) part that holds back the explicit-product version (adding a large float with a small one).