23/2 DESIGN FOR A BRAIN 



computed to any degree of accuracy by a numerical method. The 

 proof given in Chapter 8, though verbal, is adequate to establish 

 the elementary properties of the system. A rigorous statement 

 and proof would add little of real value. 



23/2. How many trials will be necessary, on the average, for a 

 terminal field to be found ? If an ultrastable system has a 

 probability p that a new field of the main variables will be stable, 

 and if the fields' probabilities are independent, then the number 

 of fields occurring (including the terminal) will be, on the average, 



i/ P . 



For at the first field, a proportion p will be terminal, and 

 q (= 1 — p) will not. Of the latter, at the second field, the pro- 

 portion p will be terminal and q not ; so the total proportion stable 

 at the second field will be pq, and the number still unstable q 2 . 

 Similarly the proportion becoming terminal at the w-th field will 

 be pq u ~ A . So the average number of trials made will be 

 p -f 2pq + 3pq 2 + . . . + upq u ~ x + . . . _ 1 

 V + V9. + Vf + • • • + Vt~ x + • • • ~ V 



23/3. In an ultrastable system, a field may be terminal and yet 

 show little resemblance to the ' normal ' equilibrium which is 

 necessary if the system is to show, after each of a variety of dis- 

 placements, a return to the resting state. A field, for instance, 

 might have a resting state at which only a single line of behaviour 

 terminated : if the representative point were on that line the field 

 would be terminal ; but hardly any displacement would be 

 followed by a return to the resting state. 



It can, however, be shown that if a proportion of the fields 

 evoked by the step-function changes are of this or similar type, 

 then the terminal fields will contain them in smaller proportion. 

 For, given a field and a closed critical surface, let k ± be the pro- 

 portion of lines of behaviour crossing the boundary which are 

 stable. Thus in Figure 8/7/1, in I k x = 0, in II h t = 0, in III 

 /c 1 = i approximately, and in IV k ± = 1. To count the lines, 

 the boundary surface could be divided into portions of equal 

 area, small enough so that stable and unstable lines do not pass 

 through the same area. Then if we assume that in any field the 

 representative point is equally likely to start at any of the small 

 areas, a field's chance of being terminal is proportional to k v 



238 



