178 
FURTHER INVESTIGATIONS ON THE DOUBLE ^-FUNCTIONS. 
[663 
Reverting to the before-mentioned comparison of the first and second columns of 
Table III., four of the equations are 
f 
X 
y> y' 
X, X 
y> y' 
{<=! = 
\d], 
that is, Vc [c] = 
Vrf [rf], 
{d} = 
that is, Vrf [rf] = 
Vc [c], 
{*} = 
- {a6|, 
that is, Ve [e] = 
- V«6 [ab], 
{/} = 
- {erf}, 
that is, V/[/] = 
— V erf [erf] ; 
viz. the four terms on the left-hand side are not absolutely equal, but only proportional, 
to those on the right-hand side. Substituting for Vc, Vrf, etc., their values, and in 
troducing on the right-hand side the factor 
the equations become 
V ac .be .ce. cf .ad. bel. de. df 
XX yy 
[c] = ac .be. ce . ef [rf], 
[rf] = ad .bd.de . df[c], 
[e] = — ce.de [ab], 
[/]=- cf.df[cd]. 
The functions on the left-hand satisfy the identity 
def[c] - efc [d] +fcd [e] - ede [/] = 0, 
nr, as this may also be written, 
def[c\ - cef\d\ + cdf[e] - ede [/] = 0. 
Hence substituting the right-hand values, the whole equation divides by ce.de.cf.df-, 
omitting this factor, it becomes 
ef. ac . be [d] — ef .ad. bd [c] — cd {[«,&] — [erf]} = 0, 
where the variables are y, y : it is to be shown that this is in fact an identity, and (as 
it is thus immaterial what the variables are) I change them into x, x'. 
We have 
ac . be [rf] — ad .bd [c] = (a — c) (b — c) (rf — x) (rf — x) 
— (a — d) (b — rf) (c — x) (c — x) 
— (c — rf) 1, x + x\ xx 
1, a + b, ab 
1, c -1- rf, cd 
= erf \xx abed], 
suppose.