Question

Estimating a square root.

Answer 1

The way that cheap calculators do it is with Newton's approximation.

x=√n

x^2=n

x^2-n=0

Newton's approximation is:

x-f(x)/(dy/dx)

x-(x^2-n)/(2x) which if you include the improved approximation

(2x^2-x^2+n)/(2x)

x↔(x^2+n)/(2x)

It converges very quickly to the exact answer (if there is one) with relatively few iterations...

For example:  √5 ?  take an initial guess of 2 and you get:

(x^2+5)/(2x)

2, 2.25, 2.2361, 2.2360, 2.236067977499

Answer 2

I'm sorry but I'm not into square roots yet :(