188 A MEMOIR ON THE DOUBLE ^-FUNCTIONS. [665 
Proceeding next to find the value of A A, = A d 2 A — (dA) 2 , =A 2 d 2 logA, it is to 
be remarked that we have in general 
A PQ = P 2 AQ + Q 2 AP, 
and therefore also AP 2 = 2P 2 AP, and consequently A \/P = -A AP. Hence starting from 
A = O Va, we have 
AA = aAO -f H 2 - Aa, 
where Aa = — a d 2 x — (dx) 2 . I assume that we have Afi = il 2 ilf = \ fl 2 S (du) 2 , where S 
denotes a function of x which is to be determined: the equation thus becomes 
A A = ^il 2 {a S (du) 2 — 2d 2 x — 2 (9a;) 2 } ; 
we have (dx) 2 = abed (du) 2 , and thence, differentiating and omitting on each side the 
factor dx, we obtain 
2d 2 x — — (abc + abd + acd + bed), 
and the equation becomes 
A A = £fl 2 {a (S + be + bd + cd) — bed} (du) 2 , 
which is to be simplified by assuming a proper value for S; in order that the same 
simplification may apply to the formulae for AB, etc., it is necessary that S be 
symmetrical in regard to a, b, c, d. 
Writing for the moment b', c, d' to denote b — a, c — a, d — a respectively, we have 
b', c', d" = b — a, c — a, d — a, and thence 
b'c’d' — bed — a (be + bd + cd) + a 2 (b + c + d) — a 3 , 
and consequently 
a (be + bd + cd) — bed = — b'cd' + a 2 (b 4- c + d — a) : 
hence, in the expression of A A, the factor which multiplies ^£l 2 (du) 2 is 
a {/S + a (b + c + d — a)} — b'c'd', 
viz. the expression added to S is 
(a — x) (b + c + d — a — 2x), 
= a(b+c + d — a) — x (a + b + c + d) + 2a; 2 . 
Hence assuming 
S — — 2x 2 -(- x (ci b c d) k, 
k being a constant symmetrical in regard to a, b, c, d, which may be at once taken 
to be = a 2 + b 2 + c 2 + d 2 — ab — ac — ad— be — bd — cd] then writing also 
A = b 2 + c 2 + d 2 — be — bd — cd, ¡jl = — b'c'd' — a — b. a — c. a — d, 
the expression a {S + a (b + c + d — a)} — b'c'd' becomes = aA + /j, ; and the sought for 
equation thus is 
AA = A d 2 A — (dA) 2 = ¿fl 2 (aX, + n) (du) 2 ,