Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 10)

424 
ON THE DOUBLE ^-FUNCTIONS. 
[697 
Here the product on the left-hand side is 
= (a — c) {bid Vabcjdi — bd x Va^cd} + (b — d) {- a x c VabCidj + ac x Va^cd}, 
viz. this is 
= Vabcxdj {(a — c) bjd — (b — cl) a x c} — Va^cd {(a — c) bdj — (b — d) ac x }, 
and in this last expression the two terms in { } are at once shown to be each 
= (be, acl); whence the identity in question. 
Comparing in like manner the first expressions for 
spectively, we have 
re- 
(b — d) (be, ad)- - = (a - b) (a — c) (b — d) {adbjCj + ajdjbc + 2 Vabcdajb^dj}, 
CL Z 
(d —a) (be, ad) 2 ^ = 
— (a — b) {(a — c) 2 bdbxdx + (b — d) 2 aca^ +2 (a — c)(b — d) VabcdaJbjCidx}, 
whence, adding, the radical on the right-hand side disappears; the whole equation 
divides by — (a — b), and omitting this factor, the relation to be verified is 
(be, ad) 2 = (a — c) 2 bdbjdj + (b - d) 2 aca^ — (a - c) (b - d) (adbjCj + a^bc); 
the right-hand side is here 
= [(a — c) b x d — (b — d) a x c} {(a — c) bdj — (b — d) acj}, 
and each of the two factors being = (be, ad), the identity is verified. It thus appears 
that the twelve equations are in fact equivalent to a single equation in x, y, z. 
Writing in the several formulae x = a, b, c, d successively, they become 
x = 
a, 
x = b, 
x — e, 
x — d, 
a — z 
a x 
c — a 
bi 
b — a Ci 
a — b . a — c 
di 
d — z 
di ’ 
cl-b 
' Ci 
d — c ' bi ’ 
d — b . d —c‘ 
aC 
b-z_ 
b x 
c - b 
a x 
b — a .b — c d x 
a — b 
Ci 
d — z 
dC 
d — a 
' Ci 
d — a .d — c ’ bj ’ 
d — c ’ 
a/ 
c — z 
Ci 
c — 
a . c —b 
di 
b — c a! 
a — c 
bi 
cl — z 
di’ 
d- 
a. cl — b 
' Cj 
d — a ’ bi ’ 
d-b* 
a,’ 
viz. for 
x — a, the 
relation 
is 0 = 
V> 
but in the other 
three cases 
respectively 
relation 
is a linear 
one, 
z = 
«3/ + /3 
7?/ + 8 ' 
we have 
Rationalising the first equation for /\]^, 
(be, ad) 2 (a — z) = (a — b) (a — c) (d — z) {adb^j + ajdjbc -1- 2 Vabcda 1 b 1 c 1 d 1 ], 
and thence 
{(be, ad) 2 (a — z) — (a — b)(a — c) (d — z) (adh^ + a^bc)} 2 
= (a —b) 2 (a — c) 2 (d — z) 2 .4abcda 1 b 1 c 1 d 1 .
	        
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