the calculus of functions. 191 
only of one value, let z equal the function, then we have the 
equation z = <px. If from this we endeavour to discover the 
value of x in terms of z, the operation is an inverse one and 
x admits of one or more values according to the nature of the 
operations denoted by <p. This number may even be infinite; 
if <p denotes an equation of the n th degree, there are n values 
of x in terms of %. It may then, be enquired whether in using 
the substitution employed in the latter part of this Problem, any 
of these (perhaps infinite number) may be taken, or whether 
only certain particular values should be used ? without attend- 
ing to this circumstance, our conclusions may become erro- 
neous : all these different values will satisfy the equation 
<pfxz —x, but only those must be used which also satisfy the 
equation (pep x — x : thus if z— <px = a — x 1 we shall have 
x = cp'z~ + V a—z if we employ the upper sign we have 
cp^ 1 x = + s/ a — [a — x*) = + V x* = + x 
If we use the lower one 
<p 1 <p.r ™ — - V a — ■ (a — x 2 ) = — Vx* = — x 
the upper sign must therefore be taken, because in the latter 
—1 
part of the Problem we suppose cp q>x = x and cp <py=y. 
This remark, which is of some importance, extends to the 
conclusions in my former Paper and to the whole of the 
subsequent enquiries. 
The equation ( 1 ) might be considered as similar to the 
original one, and the same transformation might be performed 
on this, and thus we might continue to deduce new conditions. 
In the first part we found that the equation tyx = rf/etx always 
admitted of an easy solution when u n x = x and by continuing 
the substitutions already pointed out, we should arrive at 
