126 
MR. E. P. CULVEKWELL ON DISCRIMINATION OE MAXIMA AND 
multiplied by positive powers of a. Similarly, if (c) = 0, p = 2+ 2p 2 , and from 
(a) 2 p 2 — 0>p { = > 0, but from (d) 2 p } — 3 p 2 = >0, and, as these are incompatible for 
positive values of yq and p 2) d^&yjdxf dx 2 z cannot appear in the quadratic function, 
on whose sign the problem depends. For, if we make any assumption which gives to 
it a finite value, we give, to other quantities in the integral, values involving negative 
powers of (3. If /3 be infinitely small, this would imply infinite values for the A varia¬ 
tions. But the values p =8, yq = 1, p 2 = f» will satisfy (d) — 0, and the rest either 
= or >0. Hence, in the case in question, the sign of the variation depends on 
that of 
If the highest fluxions had been 
Aq d h!\ dhy l dhy g 
dxy dx x 7 dx o dxp dxrp dx£ 
the conditions would be, if we write p 1 = 1 all through, p — 8 = > 0 (a ); 
p — 7—p 2 —>0(b); p — 5 — 2p z = > 0 (c); p — 3 p 2 — > 0 (d). Putting p = 8, 
p 2 = 1, («) and (6) = 0, and (c) and (h) > 0 : hence c£ 8 A yjdxy and d s A yjdxy dx 2 will 
remain in the condition. Putting p = 9, p 2 •= 2, (6) (and (c))=0, while (a) and 
(h) > 0 : hence d 7 A yjdx-f dx 2 remains. Again, putting p 2 = 5 and p) = 15, we get 
(c/) = 0, while (c) = 0 and (a) and ( b) > 0, and hence h 3 Ayjdx^ appears. Hence in 
this case all the fluxions remain. 
To complete the proof, we must show that it is possible to assign values for the 
variations A y x , A y 2 , &c., which shall satisfy the limiting conditions as well as those 
given above. In order to show what kind of assumption must be made, it is 
necessary to remark that, if z be any quantity such that z = 0, and dz/dx 1 = 0 all 
over any surface </> (aq, X 2 ,. . . x m ) = 0, then will every other fluxion, as dz/dx r , = 0 
over that surface. For, as z = 0, 
0 — dz— ----- dx x + ~~ dx 2 + . . . dx m 
dx 1 dx 2 dx m 
whenever dx x , dx 2 , . . . dx m satisfy the equation = 0. The only relation among 
dx x , dx 2 , . . . dx m imposed by this limitation is 
~ dxj -f ~ dx 0 4- &c. = 0. 
Ct0C-y CtXc J 
Hence, comparing coefficients, 
dz/dx-y dz/dx 2 
d^/dxy d(j)/dx 2 C ‘ ’ 
