150 
Proceedings of Royal Society of Edinburgh. [sess. 
Thus, selecting out of C (r, s) terms a number quite sufficient to 
indicate how all the other terms are formed, 
A r (r, s) = Y. cqa 2 . . . a r V a r+1 a r+2 . . . a r+s - Y. oqoq... a r _ x a r+ ff a r a r+2 .... a r+s + . . . 
+ V . cqa 2 . . . a r _ 2 a r a r+1 V a r _ x a r+2 .... a r+s -...+(- 1 ) rs V . a s+1 a s+2 . . . a r+s Y cqa 
Changing, in each quaternion, the second Y into S, we get the 
series for B(r, s ) : changing, in each quaternion, the first Y only into 
S, we get the series for Z(r , s ) : changing throughout Y into S, we 
get the series for S(r, s). 
In the investigation we have to use series formed in exactly the 
same way, but with changed vectors (as, for instance, cq being 
omitted). It will be found, practically, quite sufficient to indicate 
what is meant by writing S', S", or S^, &c., instead of S., &c. I 
have found it convenient that £ shall represent Ycqcq, and x the 
quaternion a 3 a 4 .... a r+s , i.e. } the product of all the quaternions 
except the first two. 
Of course S and Z are always scalars, and Y and B are always 
vectors. 
B. Statement of Theorem. 
S(2r, 2 s) = C(r, s) Scqoqa? ; 
Z(2r, 2s) = 0 ; 
Y(2r, 2s) = C(r,s-l)Y.a 1 a 2 £; 
( 1 ) 
B(2r, 2s) = C(r - 1, s) Y. a 4 a 2 x • 
S(2r+1, 2s- 1) = 0; 
Z(2r + 1, 2s - 1) = 0 ; 
Y(2r + l„2s - 1) = 2C(r, s - 1)Y. a 4 a 2 x ; 
B(2r +1, 2s - 1) = 0 ; 
S(2r, 2s + 1) = C(r, s - ^Soqa^a? ; 
Z(2?’, 2s + 1) = 3C (r - 1, s)Sa 1 a 2 a? ; 
Y(2r, 2s + 1) = C(r, s)Y ai a 2 x ; 
B(2r, 2s+ 1) = 0 ; 
S(2r + 1, 2s) = C(r - 1, s^cqcq# ; 
Z(2r + 1, 2s) = 3C(r, s - ^Scqa^ ; 
Y(2r+1, 2s) = 0; 
B(2r + 1, 2s) = C(r,s)Ya 1 a 2 a?. 
1 
(2) 
( 3 ) 
• • ( 4 ) 
