648 PROFESSOR CHRYSTAL ON THE HESSIAN. 



Ex. 5. U=X 3 W 5 +X2l g + u 10 



Y~x i v 1 +x 3 v 3 +v 7 

 K—xv^U — u 5 V =x 2 u 9 +u 12 

 L=ow 5 K — u d JJ EEa:w 17 -}- w 19 

 M.=zxu a ~L — w 17 KE=w 29 



llx29 = MK = KL+^K +UgK 



= KL+aw 12 +« 9 w 12 

 = KU + Ku g +xio li + u 9 u Vi 

 = K U 4- xu 12 + 2u g u u 

 = JJY + u^J+ &c. 

 = tJV+w 6 +u 5 u s + 2u g u 12 +xu li 

 UV = 319-5 -40-216 -12 = 46. 



3. The process just exemplified is tedious to apply and capricious in its 

 action/ and affords, besides, no indication of generality. It clearly contains 

 redundant steps, for in the three examples (1) (4) (5) the same final result, viz., 

 U V = 46, is obtained by extremely different developments. Yet it is obvious, 

 a priori, that the same final result ought to be arrived at in all these cases, 

 since the additional terms which appear in U and V in examples (4) and (5) 

 are such that they do not affect the forms of U and V at the point * = 0j/ = 0, 

 which alone can be supposed to affect TXV. 



It is at once suggested, therefore, that the problem will be simplified by 

 substituting for U and V the approximations to their branches at the origin 

 determined by the rule of Newton and Cramer. In this way we can in general 

 reduce the problem to a series of others, of which the following is a type. 



To determine the number of intersections of 



U=x m -?/ n = (1) and V=as?-y=0 .... (2) 



at the point x — y — 0. 



Since imaginary branches must be considered as well as real branches, 

 it may be well to give a rigorous proof of the solution in this simple case. 



If a x a 2 .... a„ be the n roots of + 1, the n values of y given by y" = x m are 



m to 



n n n 



a-1% , ct 2 £ .... a n % . 



The eliminant of the two equations (1) and (2) with respect to y is therefore 



{^-(a^yjj^-W-T} {•-(<J=") F }-0, 



* T ){* B -a[\ \x ' -al\ \x " -a.J-0, 



Now, if g be the G.C.M. of n and v, and n = gn', v=gv', then the series 



a\, <%, <, simply consists of the roots of x n ' - 1 = repeated g times. Hence 



the equation last written reduces to 



xf mv '(x^ n '- mv '—iy=0 , that is (x !in '-of" , y j = .... (3) , 



