approximating square roots using first-order Taylor series

To estimate the square root of a number using only simple arithmetic, the first-order Taylor series of the square root function provides a convenient method. As any calculus student knows, the first-order Taylor expansion around x² is given by

sqrt(x² + a)  ~  x + a / 2x

In practice, this can provide a good, quick estimate of a number's square root, which we illustrate by calculating the square root of 40. Obviously the nearest integer square is 36, so in the equation above we can put x = 6 and a = 40-36 =4. This gives sqrt(40) as approximately 6+4/12 = 6.3333..., whereas the true value is around 6.324 (note that this method always yields a slight over-estimate, since we're approximating a concave-down curve by its tangent at x²).

Of course, if the number in question is slightly below the nearest integer square, we simply use a negative value of a in the formula above; for example, to find the square root of 250 we note that 16² is 256, so a good approximation is 16 - 6/32 = 15.8125, as compared to the true value of 15.81138...