Suppose we solve Exercise 8.2 and use the estimate y0 of the square root of x to start two Heron iterations, y1 = 1 2 (y0 + x/y0 ) and y2 = y1 − 1 2 (y1 − x/y1). What is the relative error of y2?...

Suppose we solve Exercise 8.2 and use the estimate y0 of the square root of x to start two Heron iterations, y1 = 1 2 (y0 + x/y0 ) and y2 = y1 − 1 2 (y1 − x/y1). What is the relative error of y2? (This is the single precision algorithm for IBM mainframes; its accuracy is advertised as 2−25.9.)