250 
MR. A. CAYLEY’S INTRODUCTORY MEMOIR UPON QUANTICS. 
set (x,y..) and of course the coefficients of the quantic. It is in fact easy to show 
that any such derivative is a covariant according to the definition given in this Me- 
moir. But to do this some preliminary explanations are necessary. 
11. I consider any two operations P, Q, involving each or either of them differen- 
tiations in respect of variables contained in the other of them. It is required to 
investigate the effect of the operation P.Q, where the operation Q is to be in the first 
place performed upon some operand O, and the operation P is then to be performed 
on the operand QH. Suppose that P involves the differentiations B a , B 4 .. in respect 
of variables a, b.. contained in Q and CL, we must as usual in the operation P replace 
d a , .. by B a +d' a , .. where the unaccentuated symbols operate only upon Q, 
and the accentuated symbols operate only upon Q. Suppose that P is expanded in 
ascending powers of the symbols dl, d A .., viz. in the form P+Pi+P 2 -f &c., we have 
first to find the values of P^, P 2 Q, &c., by actually performing upon Q as operand 
the differentiations d' a , d' A .. The symbols PQ, PjQ, P 2 Q, &c. will then contain only 
the differentiations d a , d 6 .. which operate upon CL, and the meaning of the expression 
being once understood, we may write 
P.Q=PQ+P 1 Q + P 2 Q+ &C. 
In particular if P be a linear function of <3 A .., we have to replace P by PH-P 19 
where P, is the same function of d' A .. that P is of d a , d 6 .., and it is therefore clear 
that we have in this case 
P.Q=PQ-fP(Q), 
where on the right-hand side in the term PQ the differentiations B a , d A , .. are con- 
sidered as not in anywise affecting the symbol Q, while in the term P(Q) these differ- 
entiations, or what is the same thing, the operation P, is considered to be performed 
upon Q as operand. 
Again, if Q be a linear function of a, b, c.., then P 2 Q=0, P 3 Q = 0, &c., and there- 
fore P.Q=PQ-|-PiQ ; and I shall in this case also (and consequently whenever 
P 2 Q=0, P 3 Q=0, &c.) write 
P. Q=PQ + P(Q), 
where on the right-hand side in the term PQ the differentiations d a , d A ., are considered 
as not in anywise affecting the symbol Q, while the term P(Q) is in each case what 
has been in the first instance represented by P^. 
We have in like manner, if Q be a linear function of B a , d c .., or if P be a linear 
function of a, b, c..., 
Q. P=QP+Q(P) ; 
and from the two equations (since obviously PQ=QP) we derive 
P.Q-Q.P=P(Q)-Q(P), 
which is the form in which the equations are most frequently useful. 
