Divisor Reciprocation

f(x) = 1/x - B so f(x)=0 when x = 1/B

f'(x) = -1/x^2

So xi+1 = xi - ((1/xi) - B)/(-1/x^2))

xi+1 = xi (2 - Bxi )

But must form Bxi product every time (waste), so multiply both sides by B and let ai = Bxi :

xi+1 = xi (2 - ai )

ai+1 = ai (2 - ai )

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