OPERATIVE DIVISION 583 



greatest root, and the odd points ^50 — and 50/65 -f to the 

 least root. In the equation 



x* — i5a; 4 + 85^ — 225# 2 + 274a; — 120= o, 



of which the roots are 5, 4, 3, 2, 1, the critical points derived 

 from successive pairs of terms might appear to refer to each 

 root in succession. The point 85/15 — , however, is separated 

 from the root 4 by the root 5, and can therefore refer to no 

 root at all — as indicated by the qualifying condition in pro- 

 position (e). 



(4) Some more theorems on critical points will be given in 

 Notes II and III and Figures 8 and 9. Two of these are so 

 frequently useful for the analyses of equations that they 

 should be mentioned here. If all the roots of an equation 

 are positive and real, the critical points derived from 

 successive pairs of terms must be in diminishing order of 

 magnitude. In fact their successive ratios must be greater 

 than the ratios of the corresponding binomial coefficients — 

 which is indeed connected with Newton's Rule. For example, 

 x 4 — 2x 3 + 1 2X 2 — i ix+ 10= o, that is (x 2 — x -\- i)(x 2 — x-\- 10), 

 has no roots. But conversely the roots need not all be real even 

 if these conditions hold. If the equation consists of groups 

 of terms of alternate signs, the former rule, at least, applies 

 to the partial roots of successive pairs oj groups (Example P). 



VIII. (1) We are now able to return to the proper subject 

 of this paper, the inverts obtained by operative division. In 

 Sections III and IV we obtained a number of different inverts, 

 either by ascending or by descending division, by taking fx— o 

 in an ascending or in a descending series, and then dividing 

 both sides of the equation by different powers of x. We have 

 now to find in conclusion, (1) to which root each invert refers ; 

 (2) under what conditions it is convergent ; and (3) what is 

 its geometric interpretation. At the end of Section IV it 

 would have been difficult to answer these questions, but the 

 intervening sections will help us to make the attempt. 



Let the proposed equation be 



fx = p x n +p lX n - l + . . . + p q x n -« . . . + p n _,x+p n = o 



Suppose p always positive and let p q be one of the negative 

 coefficients. Then, dividing algebraically by p x n ~ q , we have 



** + P1/P0 • x*- 1 + . . . + ox . . . + A-i/A) • x~ n+q+1 + 



+ Pn/p0.X- n+q =-P q /P . 



