141] 
A SECOND MEMOIR UPON QUANTICS. 
261 
and fiom this, by subtracting from each coefficient the coefficient which immediately 
precedes it, we form the table: 
0 ! 1 
1 1 
1 
0 
1 
1 
1 
0 
1 
1 
0 
1 
1 
1 
1 
1 
0 
1 
0 
1 
1 
1 
1 
2 
0 
1 
0 
The successive lines fix the number and character of the covariants of the degrees 
0, 1, 2, 3, &c. The line (0), if this were to be interpreted, would show that there is a 
single covariant of the degree 0 ; this covariant is of course merely the absolute con 
stant unity, and may be excluded. The line (1) shows that there is a single covariant 
of the degree 1, viz. a covariant of the order 3; this is the cubic itself, which I 
represent by U. The line (2) shows that there are two asyzygetic covariants of the 
degree 2, viz. one of the order 6, this is merely U 2 , and one of the order 2, this I 
represent by H. The line (3) shows that there are three asyzygetic covariants of the 
degree 3, viz. one of the order 9, this is U 3 ; one of the order 5, this is UH, and one of 
the order 3, this I represent by The line (4) shows that there are five asyzygetic 
covariants of the degree 4, viz. one of the order 12, this is U 4 ; one of the order 8, 
this is U 2 H ; one of the order 6, this is H 2 ; and one of the order 0, i.e. an invariant, 
this I represent by V. The line (5) shows that there are six asyzygetic covariants of 
the degree 5, viz. one of the order 15, this is U 5 ; one of the order 11, this is U 3 H ; 
one of the order 9, this is U 2i î> ; one of the order 7, this is UH 2 ; one of the order 5, 
this is H$> ; and one of the order 3, this is V U. The line (6) shows that there are 8 
asyzygetic covariants of the degree 6, viz. one of the order 18, this is U e ; one of the 
invariant or leading coefficient, we have the equations of connexion of these numerical coefficients. Thus, for 
the discriminant of a cubic, the terms of the next inferior weight are a-cd, ab 2 d, abc s , b 3 c, and operating on 
each of these separately with the symbol 
ind. Ô.- + 2 ind. c.7 + 3 ind. d.-, 
abc 
we find 
abed 
+ 6 a 2 d 2 
3 b 3 d 
+ 2 abed 
2 b 2 c 2 
+ 6 ac 3 
+ 3 abed 
+ 4 b 2 c 2 
+ 3 b 3 d 
and equating the horizontal lines to zero, and assuming a 2 d 2 = 1, we have a 2 d 2 — 1, abed—~ 6, ac 3 — 4, b 3 d — 4, 
b 2 c 2 = -3, or the value of the discriminant is that given in the text.