226 
AN INTRODUCTORY MEMOIR UPON QUANTICS. 
[139 
these sets by the original set (x, y,...), we have a covariant involving only the original 
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 
Memoir. 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 il, and the operation P is then to be per 
formed on the operand QCl. Suppose that P involves the differentiations d a , d b) ... in 
respect of variables a, b, ... contained in Q and fi, we must as usual in the operation 
P replace d a , d b) ... by d a + d' a> d b + d' b , ... where the unaccentuated symbols operate 
only upon il, and the accentuated symbols operate only upon Q. Suppose that P is 
expanded in ascending powers of the symbols d' a , d' b , ..., viz. in the form P + P 1 + P 2 + &c., 
we have first to find the values of P^Q, P 2 Q, &c., by actually performing upon Q as 
operand the differentiations d' a> d' b .... The symbols PQ, P 1 Q, P 2 Q, &c. will then contain 
only the differentiations d a , d b ,... which operate upCn il, and the meaning of the ex 
pression being once understood, we may write 
P. Q = PQ + PiQ + P 2 Q + &c. 
In particular if P be a linear function of d a , d b , ..., we have to replace P by P + P u 
where P x is the same function of d' a , d' b , ... that P is of d a , d b) ..., and it is therefore 
clear that we have in this case 
P.Q = PQ + P(Q), 
where on the right-hand side in the term PQ the differentiations d a , d b) ... are con 
sidered as not in anywise affecting the symbol Q, while in the term P (Q) these 
differentiations, or what is the same thing, the operation P, is considered to be per 
formed 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 
therefore P. Q = PQ + P X Q; and I shall in this case also (and consequently whenever 
P 2 Q — ^y 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 b ,... are con 
sidered 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 X Q. 
We have in like manner, if Q be a linear function of 3 a , d b , 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.