56 
Proceedings of the Royal Society of Edinburgh. [Sess. 
and the common roots a, /3, y, it follows that 
a m a n A (a) 
a m a” aA(a) 
a m a n B(a) 
a m a n aB(a) 
a m a n a 2 B(a) 
p m p* AOS) 
/3~ 0* pm 
/T /3 n B (P) 
/3 m / 3 n (3B((3) 
/ 3 m (3 n /3 2 B(/3) 
T 7“ A(y) 
5 
7” 7" yMy) 
5 
7“ 7“ b <7> 
i m T 7 B(y) 
5 
T y n 7 2 B(y) 
are all equal to 0 ; so that, if we temporarily write 
(m , n , p) for the alternant | a m /3 n y p | , 
we have 
(mn5)a 0 + (mni)^ + (mn?>)a. 2 + (mn2)a s + (mn\)a b + (mnO)a b — 0 
(mw6)a 0 + (innbya-^ + (mni)a 2 + ( mn3)a s + (mn2)a 4 + (mnl)a 5 = 0 
(mn4:)b 0 + (mu 3)b 1 + (mn2)b 2 + (mnl)b^ + (mn0)b 4 = 0 
(mn5)b 0 + (mni)b l + ( mn3)b. 2 + ( mn2)b 3 + (mn\)b 4 = 0 
(mn6)b 0 + (mn5)b 1 + (mni)b 2 + (mn3)b s + (mn2)b 4 = 0. 
Here, however, by taking any two of the numbers 0, 1, 2, 3, 4, 5, 6 as 
values for m, n two of the alternants will disappear, and we shall be able 
to eliminate the five others, the final and complete result thus being in 
Cayley’s notation 
% 
a 2 
00 
a l 
a b 
a o 
a i 
a B 
a 4 
“5 
\ 
h 
b 2 
h 
*4 
h 
h 
h 
For example, putting m, n equal to 0, 1, then equal to 0, 2, and finally 
equal to 1, 2, we should have the particular three results which in accord- 
ance with our usage under Trudi we might write 
(a 0 , . . 
■ 
(*„ ■ • 
• ■ > ^t)s ! 
All this, however, is considerably modified from Mansion’s exposition. 
Gunther, S. (1879). 
[Eine Relation zwischen Potenzen und Determinanten. Zeitschrift f. 
Math. u. Phys., xxiv. pp. 244-248.] 
The subject here is simply the evaluation of the bigradient which is the 
discriminant of (x m+2 — l)/(x— 1), the result being 
(m + 2 ) m . 
For example, when m is 2, 
1111 . 
. 1111 
12 3 .. 
. 12 3 . 
. 1 2 3 
4 2 . 
