By Jacob T Schwartz

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Let us now extend the concept of a feedback machine by equipping the processing unit with an internal memory unit Then the iteration of a two-step method Xn+1 = g(xn, Xn-1) can be implemented as follows. First note that to start the feedback machine two initial values xo and x1 are required. Preparation: Initialize the memory unit with xo and the input unit with x1. Iteration: Evaluate Xn+1 = g(xn, Xn-1) where Xn is in the input unit and Xn-1 is in the memory unit. Then update the memory unit with x n One-Step Machines With Two Variables Somehow it seems that feedback machines with memory should be more flexible in modeling different phenomena.

Such a test is not as straightforward as we might think, because in the course of the calculations the se- = The (3A +I)-Problem 1 40 The Backbone of Fractals quence may exceed the largest possible number which the computer is able to accurately represent. Thus, some variable precision routines must be programed in order to enlarge the range of numbers representable by a computer. The algorithm can easily be extended to negative integers. runs into CYCLE of length 2 Are there other cycles? Yes indeed: • -17, -50, -25, -74, -37, -110, -55, -164, -82, -41, -122, -61, -182, -91,-272,-136,-68,-34,-17, ...

Thus, using the definition in eqn. 2) again, we obtain an expression for the error en+l 2en . 3) Now zo > 0 and therefore eo > -1 and thus en > 0 for n = 1, 2, 3, ... But then Zn > y'O. for all n > 0. Finally, we can obtain estimates out of eqn. 3). If we drop the '2' in the denominator we obtain en+l en <2 and if we drop '2en' we obtain e2 en+l n < 2· The first inequality and the definition of en by eqn. 2) shows that Zt > Z2 > Z3 > ... ;a and that the limit is y'O.. The second inequality shows that if en < then en+l < w- 2n /2.