426 Proceedings of Royal Society of Edinburgh. [sess. 
thereby takes the form 0, and 0, f 2 , f 3 , . . . , f n be introduced into 
the expression for f 1 which thereby takes the form 0 15 then 
V + jL . A . . . . Al.— s? +*?$ m *Qi A d/» . 
^ — dx dx Y dx n ^ ~ dx dx 1 dx 2 dx n 
In other words, instead of stipulating that 0, / 2 , / 3 , . . . , f n 
be introduced into f 1 he merely stipulates that /, / 2 , / 3 , . . . , f n 
be introduced. His proof is thus defective. 
(8) The nature of the oversight is possibly made clearer by 
observing what 
Y + A.A . . . % 
^ dx dx 1 dx n 
becomes, when, in addition to substituting for 
V, A 
dx dx , 
v_ 
dx. 
as was done in the first case, we also substitute for 
9/i ; : §/i_ . 
dx dx 1 dx n 
Even the first column of the altered Jacobian cannot now be 
simplified to the same extent as before, because part of the simpli- 
fication consisted in subtracting a multiple of the second column in 
its unaltered form. In fact the result instead of being 
y ± d A . d h ¥x . . 
" “ dx dXj dx 2 dx n 
is 
00 
00 
d$i 
.3*1 
¥_ 
¥2 
d A 
. A 
dx 
Vi 
dx 
dx 
3/ 
dx 
dx 
dx 
dx 
00 
00 , 
90i 
, 3< £i 
¥_x 
d A . 
. . A 
dX] 
3/i 
0^ 
03 ^ 
¥ 
dx 1 
dx Y 
dx 1 
dxi 
00 
dx n 
00 
9/i‘ 
dx n 
90i 
9a? n 
+ d <h. 
¥ 
df 
dx n 
s 
dx n 
A . 
dx n 
. .A 
dx n 
(9) As an example let us take the case where 
u 1 = x(y + z), u 2 = y(z + x), u^ = z(x + y), 
and where therefore 
J(u 1 ,u 2 ,u s ) = 
y + z 
y 
z 
X X 
+ x y 
z x + y 
— 4=xyz, 
