426 
ON THE DOUBLE ^-FUNCTIONS. 
[697 
It may be remarked by way of verification that the equation remains unaltered 
on substituting for x, y, z, a, b, c, d their reciprocals and multiplying the whole by 
a 4 b 2 c 2 d 2 x 2 y 2 z 2 . 
I further remark that, writing a = 0, we have 
A = 0, B = 0, G = b 2 c 2 d 2 , D = - b 2 c 2 d 2 , E = 0, F= bed (be +bd + cd), 
G = — bed (6 + c + d), Ff=0, I = 2bed, J = b 2 + c 2 +d 2 — 2 (be + bd + cd); 
and writing also 
e = l, — S = (6 + c + c7), ry = bc + bd + cd, — /3 = bed, 
(whence 
a — x.b—x.c — x.d—x = /3x + <yx 2 + Ea? + ear 1 ), 
we have the formula 
/3 2 (x 2 + y 2 + z 2 — Zyz — 2 zx — 2 xy) 
— 4/3y xyz 
— 2/3S ary? (# + y + z) 
— 4/3e xyz (yz + zx + xy) 
+ (8 2 — 4ye) x 2 y 2 z 2 = 0, 
given p. 348 of my Elliptic Functions as a particular integral of the differential 
equation when the radical is V fix + yx 2 + Ex 3 4- ex 4 . 
Let the equation in (x, y, z) be called u = 0; u has been given in the form 
u = (Sz 2 — 2i&z + 21, and we thence have i = — 23 which, in virtue of the equation 
u— 0 itself, becomes ^ ^ = V23 2 - 21(1; we find easily 
23 2 — 21(5 = (a — b) 2 (a — cf (a — d) 2 {(adbA + ajdjbc) 2 - (be, ad) 2 (x — y) 2 }, 
or, attending to the relation 
this is 
or we have 
(adbjCj + ajdibc) 2 = (adh^ — a^bc) 2 + 4abcda 1 b 1 c 1 d 1 
= (be, ad) 2 (x — y) 2 + 4abcda 1 b 1 c 1 d 1 , 
23 2 — 21(5 = 4 (a —b) 2 (a — c) 2 (a — d) 2 abcdaJbjCjdj, 
Writing 
we have of course 
= (a— b) (a - c) (a — d)'Jabed Va^c^. 
a — z, b-z, c- z, d — z = a 2 , b 2 , c 2 , d 2 , 
the like formulae 
— (a — b) (a — c) (a — cl) V a^c^ V a 2 b 2 c 2 d 2 
— ( a — ^) { a — c ) ( a — d)*Jabed Va 2 b 2 c 2 d 2 ; 
ay