OPERATIVE DIVISION 407 



exactly at the root itself — which explains why Newton's method 

 gives the quickest approximation near the root. 



But any number of vicarious operations may be found 

 which will give the root more or less quickly. For example, 



F = *- (so that + F = o' 5 -f- 1 • 20 — c 1 o 3 ) gives results almost 

 10 



as quickly as Newton's method, and more easily. 



The reader will find the clearest example in the case of the 

 straight line a + bo. If b lies between zero and — 2, the 

 iteration of a + ( 1 + b)o gives convergent geometric pro- 

 gressions. If b = — 1, the root is obtained by the first iterant 

 Xi = [a]x . When b = — 2, the iterants oscillate between 

 x and a — x , [0 + /]" is a periodic function, and the iteration 

 " stagnates " — but the root is the mean of two successive 

 iterants. The iteration " fails" when b > or < — 2 ; but if 



in these cases we put F = — / and F = ^ respectively we 



obtain vicarious operations which certainly converge — so that 

 the root can always be obtained under this theorem. 



To find the roots of / in succession it generally suffices to 



iterate o 4- — /, where m is a numerical " modifier " — frequently 

 m 



an integer. To obtain the first, third, fifth . . . roots take that 



sign of + — / which makes the absolute term positive ; for the 

 nr 



second, fourth. . . roots take the other sign ; and for the negative 



roots put — for 0. The value of m should be sufficient to 



make ±_ —f conform to the conditions required by the theorem 

 m 



given above — since if F = ± - /, F' = ±—f ; but near the 



root m may be replaced by — /'. If, however, /' becomes nega- 

 tive infinity, other vicarious operations may be required. In 

 the next Part, I hope to examine the question how to select the 

 vicarious function in order to obtain the root with certainty 

 and rapidity, either arithmetically or algebraically. 



(3) We have now considered two methods for the solution 

 of certain equations — the Operative one, which gives an algebraic 

 series, and the Iterative one, which gives an arithmetic one. If 

 our theorems are sound, these should be equivalent. Take first 

 the simplest case, that of the general quadratic o = a — x + x % . 



