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

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.