046 PROFESSOR CHRYSTAL ON THE HESSIAN. 



The direct calculation of the reduction is therefore a general problem, whose 

 interest is quite equal to its difficulty. With a view to clear the way for a 

 general solution (if such be attainable) I have worked out a number of cases, 

 some quite special, others of a more general character, and propose to com- 

 municate the solutions to the Society in the following paper. 



2. In its ultimate stage the problem reduces to the following : — 



To determine the number UV of the intersections of two algebraical curves 

 U = 0, V = 0, which coincide with a common point which is multiple on one or 

 both. 



Let us suppose that the common point is a multiple point of order k on U 

 and of order k on V. 



So long as no one of the k tangents of U coincides with any one of the 

 k tangents of V, there is no difficulty; the number of intersections absorbed at 

 the common point is kK. 



But let us suppose that I of the k tangents and X of the k tangents coincide 

 with x = 0, then we have 



U^XlU^i +U k+ i+ 



V = x K u K _ k + u K+1 + 



or, what is still worse, that x — is a multiple inflexional or undulatory tangent, 

 so that 



TJ=Z ! Uik-i +X m U k+ i_ m + X n U k+2 -n + 



V=a^v K - K +a^v K+ i^ +x v v K + 2 -v + . . . • 



and the jDroblem becomes one of some difficulty. 



In many cases the solution may be obtained by the following process : — 



Ex. 1. 



Let us suppose 



V=a3%i + v 7 . 

 Let K^-xvjXJ — u 5 Y=xv 1 u 10 — v, h v 1 =u n , 

 say, where u n does not contain *asa factor. Then, since K passes through all 



the intersections of U and V, taking VU to denote the number of those inter- 

 sections which coincide with x = y = we have, since 



K = | gives « 6 V = and.-. ^ = 0}' 



UK =tJV +Uw 6 

 i.e., Uw 12 = U V + Um 5 , 

 whence UV =TJw 18 — U« 6 . 



