AND ON NEWTON'S METHOD OF CO-ORDINATED EXPONENTS. 623 



Begin from any point B, and taking the rule for the signs given above, shew the numerators 

 and denominators in the vertical and horizontal projections of BC, CD, DE, EF, FG, GH. 

 We are at liberty to add the points p, q, and r, which affect no dimension, but only the 

 values of u. The essential terms of the equation, for satisfaction of these lower conditions, 

 are the terms whose exponent points are B, p, C, q, D, r, E, F, G, H : but in estimating the 

 terms, the origin should be removed to A, to avoid branches of the form x = 0, y = 0. We 

 may add terms to the equation from any or all the points on or within Bb, Hh, produced 

 upwards ad infinitum, without affecting the character of any root. We may now ask, in what 

 manner are we limited as to the assignment of superior conditions. Write the desired dimen- 

 sions, with properly modified denominators, also in descending order. Then the sums of the 

 numerators, with the signs, must be equal in the two sets, and the sums of the denominators, 

 taken positively, must be equal ; and further, there must be ascent from the last of the 

 inferiors to the first of the superiors, and also from the last of the superiors to the first of the 

 inferiors. Thus 1, 2, 3, 3, 2, 1 solutions of the superior dimensions 3, i, 0, - ^, - 1, - 3 

 will do ; for, writing down the two sets together 



220-1-3-2 

 2 4 2 1 2 1 



3 10-1-2-3 

 12 3 3 2 "T~ 



3 _ 2 2 — 3 



2+2+0-1-3-2 = 3 + 1+0-1-2-3, 2 + 4+2 + 1+2 + 1 = 1+2 + 3 + 3+2+1, -> , -> 



1 12 1 • 



Proceeding as before, the essential terms are derived from H, I, K, L, M, N; to which 

 s, t, u, may be added. We may further add any terms derived from points inside the 

 polygon ; and the equation, and the fullest equation, which satisfies the conditions, is obtained 

 by supplying any numerical coefficients we please which are finite both ways to the. under- 

 marked terms (xf, &c.) of the following, and finite or zero coefficients to the rest : 



jr/fH + (*/ + ...+*,») y n + (*» + ... + x')y ] ° + (x + ... + * 10 ) y 9 



+ f> ( ° + ... + a?' ) if + (x° + ... + a; 10 ) y~ + (x° + ... + x ™) y 6 + (x + ... + a? 9 ) y 5 



+ (.r, +...+ < v 9 )y* + {or +...+ * 9 )y 3 + (xf + ...+ a? 8 ) y 1 + (a? 3 +...+ x;)y + x< = 0. 



By inverting the polygon we obtain what we may properly call the inverted curve, in 

 which the initial and terminal characters are interchanged : that is, from <p (x, y) = 0, we get 

 (h (x _1 , y' 1 ) = ; and by properly shifting the origin, we clear the second equation of 

 fractions. This may all be done at once, by shifting the origin to a, and changing the order 

 of signs on both axes. By taking Z for an origin, and reading ZA positively, we obtain 

 (p (x, y' 1 ) = 0, in which all the dimensions simply change sign. By taking z for an origin, and 

 reading zA positively, we interchange initial and terminal characters, and also make all dimen- 

 sions change sign ; as in <p (x~ l , y) = 0. Hence, by throwing one of two given singular points 

 to an infinite distance, it might be settled whether the two can coexist in an algebraic curve. 



By looking at the polygon from the axis of m, we see that if y have /3 root-functions of x 

 of the degree a : /3, x has a root-functions of y of the degree /3 : a. This seems to follow 

 very easily from an obvious mode of algebraical proof, which, however, contains either much 

 assumption on the inversion of expansions, or some prolixity of operation. 



Vol. IX. Paet IV. 80 



