AND IMAGTNABY BOOTS OB EQUATIONS. 
625 
their resultant in respect to u, v is obviously 
(rsty(r-s)(s—t)(t- 
■r)n 
( e ) Again, if we write 
rst(gu + <rv + 7w ) = £ 
rst(w— r)(<r+r— f)w+ (r— f)(v+ g — <r)v + (g — cr)(g + <r— r)w)— 7J> 
u-\-v+w = 0 , 
and from these equations deduce the values of u, v,w, and substitute them in rid -(- sv' a -f tw s , we shall obtain 
M. Hermite’s “ forme-type ” expressed in terms of the parameters of the reduced form, and every coefficient 
therein will be invariantive. 
The resultant of the equations above written (on making £=0, £=0) will appear in the denominator of each 
such coefficient. Hence it appears, from M. Hermite’s expressions (Camb. and Dubl. Math. Journal, vol. ix. 
p. 193), where J 3 will be seen to enter into the denominator of A, B, C, C', B', A', that this resultant to a factor 
pres is his J 3 . Its value may easily be calculated, and will be found to be 
j errfy + cr + t) 3 — 4(j + cr + r)fy s- + f v + err) + 9^> or = JK + 9L. 
Accordingly as L (to use Dr. Salmon’s convenient elliptical expression) is the condition of the failure of my 
general reduced form, so is 9L+ JK the condition of the failure of M. Heemite’s “forme-type.” As particular 
cases of this last failure, we may suppose J=0, L=0, or H=0, L=0. In the former case the reduced form is 
ax s +5ecc*y, of which the simplest quadratic and cubic covariants are respectively aex 2 ; ae 2 y 2 x. Thus to find 
L, the first linear covariant, we have to operate upon ae 2 y\v with ael 
d\ 2 
—J , which gives a?e 3 x; and to find L a , we 
cl' 
have to operate on (aex r ) 2 with ae 2 ^— j — , or, if we please (according to M. Heemite’s method), with ^a?e 3 -^-^ 
on aex 2 , showing that L 2 vanishes, but Lj continues to subsist. When, secondly, K=0, L=0, the reduced form 
is ax’ + ey 5 , and the canonizant disappears entirely, so that the first, and consequently also the second, linear 
covariants, each of them becomes a null. 
( 37 ) By aid of the reduced forms of the invariants J, K, L, I given in the text, it is easy to prove that every 
other invariant, say £2 of a quintic, is a rational integral function of these four. In what follows, let a paren- 
thesis enclosing the symbol of any invariant signify its value when any two of the quantities u, v, w in the 
reduced form ru 5 + sv° -f tvd ; [u+v+w= 0] are taken as the independent variables. We have then 
(J)=f z +cr 2 +T 2 — 2g<r— 2gt— 2crr, (K)=f(rrfy + cr+r), (L)=fW, (I)=pWfy — cr)(<r— r)(r— g), 
g, cr, r meaning st, tr, st. 
The degree of £2 must be of the degree 4m or 4m +2. 1. Let it be of the form 4m. Then, since the in- 
terchange of any two of the variables u, v, w must leave (£1) unaltered, (£1) will be unaltered by the interchange 
of any two of the letters r, s, t, and is consequently a symmetric function of g, cr, r, the roots of the equation 
Hence 
fl3 _(K) 62+ 0 MJ)(L)_ (L i 
(L)3 ( L ) 
^ ' (L) 2m 
F denoting a rational integral function-form of the quantities it affects. 
£ 2 = 
F(J, K, L) 
L 2 ™ 
Consequently 
Hence since £2 cannot become infinite when L=0, which merely implies that the general form reduces to 
{a, 0, 0, 0, e, C^x, y^, 
£2=$(J, K, L), a rational integral function of J, K, L. 
2. If the degree £2 is of the form 4m +2, (£2) will be a function of r, s, t, which changes its sign when u and v 
