665] 
A MEMOIR OX THE DOUBLE ^-FUNCTIONS. 
207 
The formula thus becomes 
П 2 
Д-A = aaj 
M + (~!jf — ) (Зет + aj Эм) 2 + ( — 
2 U 1 _U 1 / 
0 s 9 
¡j- ^ (Зет + a du) 3 
- |^ 2 (abcdjejfj + ajbjCjdef) - (Vde) 3 j (du) 3 
— ~ (dvr + aj Эм) 2 — (Зет + a du) 3 
+ |^(abcd iei fi + a 1 b 1 c 1 def) — (Vde) 2 ] (Зет) 2 , 
“ ' ' "''J 
viz. substituting for M its value, the term in (\/de) 3 (du) 3 disappears, and the formula is 
4 
П 
, AA = aa 2 
%' (du) 3 + 233'ЭмЭет + (£' (Зет) 2 + ^ j (Зет + а. Эм.) 2 
+ ~ + a du) 3 - ^ (abcdjejfj + ajbjCjdef) (ЭгО 2 
(Зет + aj du) 3 — a ^ 1 (Зет + а Эг^) 2 
+ (abcd^fi + ajbjCjdef) — (Vde) 3 \ (Зет) 2 : 
say for shortness this is 
4 
П 2 
AA 
Я 77 я 77 1 /— 
= aa, X —(Зет + а, Эм) 2 —^ (Зет + а Эм) 2 + ^ (abcd^h + ajbjCjdef) - (fde) 3 (Зет) 2 . 
Second step of the reduction. 
In the reductions which follow, we make as many terms as may be to contain 
the factor аа 1; so as to simplify as much as possible the portion not containing this 
factor. 
We have Зет + du = (Зет + 0 Эм) + а Эм, and consequently 
(Зет + aj Эм) 2 = (Зет + 9 du) 3 + a P, 
where P = 2 du Зет + (a + 20) (du) 3 : similarly Зет + а Эм = (Зет - 0 du) + ^ Эм, and therefore 
(Зет + a du) 3 = (Зет — 9 du) 3 + a l P 1 , 
where Р г = 2 Эм Зет + (a 2 - 20) (du) 3 : the values may also be written 
P = 2 Эм Зет + (2a : — а) (Зм) 2 , P 1 = ’2 du Зет + (2a — a!) (du)-.