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QuestionA small exercise Pin
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AnswerRe: A small exercise Pin
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AnswerRe: A small exercise Pin
BobJanova10-Nov-11 1:30
BobJanova10-Nov-11 1:30 
Well, this is a quadratic equation, with roots at 0 and 1 and therefore a midpoint at ½.

y = (x - ½)² - ¼ (expand and compare with the initial constant which was 0)
(x - ½)² = y + ¼
x - ½ = sqrt(y + ¼)
x = sqrt(y + ¼) + ½

Since you're already using a square root this won't be much more expensive than what you're doing now (maybe less, not sure of the cost of ceil) and it's accurate.

x(x - 1) isn't that close to x², it differs by x and therefore the relative error only goes down as 1/x. That's a very poor approximation and I don't see why you'd use it if it's also an expensive one with a sqrt in it.
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