168 Proceedings of Royal Society of Edinburgh. [sess. 
and therefore 
(SH 
0 
for it is impossible that the functional determinant A could he 0 , 
as the functions involved in it are by hypothesis mutually indepen- 
dent. The vanishing of however implies that x is not in- 
volved in /: and this, if we bear in mind the meaning of the 
brackets, further implies that / is a function of only f x , / 2 , . . . ,/„ 
— that is to say, that f,f l ,f 2i . . . , f n are not mutually indepen- 
dent. 
There now only remains to establish the proposition for the 
case where the determinant is of the second order, — that is to 
say, 
where 
Vffi 
dx dx. 
m = o 
^ dx l dx 
Here the function f 1 must involve both variables, or only one of 
them, x say, or be a constant. If it be not a constant, x x is 
expressible in terms of f x or at most in terms of f x and a?, and thus, 
by substitution, / is expressible in terms of the same. In this way 
we have 
0/ -l 
fbf\ 
+ 
Wjdx 
a* ' 
\dx) 
</ = 
(V\V . 1 
dx Y 
w a* J 
and therefore, by elimination of ( J 
y 
8/i 
3/\§/i _ V % 
dx 
dx x dx dx 1 
= 0. 
M.¥i 
dx x dx 
Now it cannot be that the second factor on the left vanishes, 
because f x is known to involve x x ; consequently 
9/ 
dx 
= 0 . 
From this it follows that of the two variables f x and x, supposed 
to be involved in /, the second is awanting, and that therefore/ is 
expressible in terms of f x alone. Our result thus is that either f x 
is a constant or that / and f x are dependent. 
