Optimal order and efficiency for iterations with two evaluations

The problem is to calculate a simple zero of a non-linear function f. We consider rational iterations without memory which use two evaluations of f or its derivatives. It is shown that the optimal order is 2. This settles a conjecture of Kung and Traub that an iteration using n evaluations without memory is of order at most 2 "^, for the case n = 2. Furthermore we show that any rational two-evaluation iteration of optimal order must use either two evaluations of f or one evaluation of f and one of f . From this result we completely settle the question of the optimal efficiency, in our efficiency measure, for any two-evaluation iteration without memory. Depending on the relative cost of evaluating f and f , the optimal efficiency is achieved by either Newton iteration or the iteration ^ defined by * ( f ) ( x ) = X f(x+f(x)1-f(x) •