Name: Anonymous 2012-01-01 10:53
Can anone here explain to me how to find the computational complexity of Newtons Method for roots?
Eva demonstrates the program for Alyssa:
(new-if (= 2 3) 0 5)
5
(new-if (= 1 1) 0 5)
0
Delighted, Alyssa uses new-if to rewrite the square-root program:
(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x)
x)))
What happens when Alyssa attempts to use this to compute square roots? Explain.