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Interview Question for Software Engineer at Facebook:
Nov 7, 2011
Compute the square root of a number down to a certain precision. ie sqrt(num, precision) returns a number that is in-between sqrt(num) - precision and sqrt(num) + precision.
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1 of 1 people found this helpful
by Newton Rapheson:
while (x * x - num).abs > precision
x -= (x * x - num) / (2 * x)
end
More generally, x -= (f(x) / f'(x)), and the derivative of (x * x - num) is 2x.