1887.] 
Dr T. Muir on Simple Alternants. 
443 
duads 
\ 0, y + 8 + 1 
j 7 + S 
< 2, y + 8 - 1 
L ^ 7 + 1 
with the triads 
0, /5 + y + 8 — s f 1, 2+5 
1, /l + y + 8 — S , 2 + 5 
have to be shown to be 1 more than those of the same duads 
with the triads j + y + ^ s + 1, 1+5 
* 1, /? + y + 8 - 5 , 2 + 5 . 
The second triad in the two cases being the same may be neglected ; 
and then we have only to note that the middle elements of the 
remaining triads are the same, and that the final elements 2+5 
and 1+5 are, for the purpose in view, as good as if they were the 
same, because the lesser of them is greater than y + 8. 
Returning now to the third theorem, and putting q = 2 + 5 , and 
r 4 + 5 we have 
0, 1,2 + 5, 4 + 5) = 1 b' + ] + ct(0, 2, 2 + 5 , 3 + 5 ) — ct(l, 2, 2 + 5 , 2 + 5 ). 
But by the preceding theorem (v.) 
ct(0, 2, 2 + 5 , 3 + 5 ) = \%a s b s c + ] + abed . a(0, 1,5, s + 2) . 
Hence 
a(0, 1,2 + 5,4 + 5 ) = p& s-1 & s + ] + [3« s & s c + ] + aM.a(0, 1,5,5 + 2) (vi.) 
— a theorem which enables us to write down the full expansion 
of a(0, 1, 2 + 5, 4 + 5) in terms of symmetric functions, because 
«(0, 1, 5 , 5 + 2) is the same function of 5 that a(0, 1, 2 + 5 , 4 + 5 ) is of 
5+2. 
Further, the fifth theorem, with the help of the expansion just 
obtained, gives us the like expansion for a(0, 2, 2 + 5, 3 + 5 ) . 
All the theorems assist materially in lightening the labour of 
tabulating the expansions of the a functions. The following are 
the tables for the functions of order 4 and degrees 1 to 9. 
Degree 1. 
a(0124) = %a. 
cc(0125 ) = ta‘ 1 + tab. 
(0134)= %ab. 
Degree 2. 
