ilie calculus of functions. 21 g 
f{x,y); and since taking the second functions is a direct ope- 
ration, it is evident that the original function i[/ [x, y) will 
produce the same result, whether we take the second func- 
tion relative to x or relative to y ; therefore it must be simi- 
larly composed of x and y ; that is to say, it must be symme- 
trical relative to x and y : but we have before shown that no 
symmetrical function can satisfy the equation, consequently 
the equation is contradictory. 
This train of reasoning I offer with considerable hesitation, 
well aware of the extreme difficulty of reasoning correctly on 
a subject so very general, and which, from its novelty, the 
mind has not been sufficiently habituated to consider, so as to 
rely with confidence on any lengthened process of reasoning. 
I thought it, however, right to mention this proof, that those 
who may seek for particular cases, might first enquire whether 
the equation be possible. 
Problem XXI. 
Given the equation 
xtyl^(x,y) z=y^>'(x,y) 
Substituting <p f(q>x, <py) for (x,y) in this equation we have 
x<p 1 f*>*((px 9 cpy)=y <p'f 2 ’ 1 (cpx, <py) 
putting f x for x and < p y for y it becomes 
f X . cp'f '’ 2 (x,y) = fy - Vf 2>1 { x >y) 
This equation will be satisfied if we could find such a form 
for f, that the two following equations might be fulfilled. 
f'' 2 (x,y)=y and f»fx,y) = x 
for in that case it would become 
fx y = f y . f x 
which is identical. 
MDCccxyi, G g 
