A. Ritchie-Scott 
407 
Now representing the left side of the above equations by means of operators 
such as 
01 (z) = x^.z; <^.,{z) = X., .z, ... (f>, {z) = x,. z 
.(29), 
.(30). 
we have 
Further 
</>! = - ^■^ = - 1 : 01 (a'o) = - v,, : 0., {x,) = - ?•„ ; 0, (a',) = - r,, = - 1 
and generally 0, (a;^..) = 0^.. (^;,) = _ (31 ). 
Since these operators are weighted sums of differential coefficients they are 
distributive, commutative and iterative. 
Hence 0r(^) = 0i (01 (4) 
= 0, (X) . Z - X, . 01 (Z) 
= — ?-ii^ + X^'^Z 
0iH4=0,(0r(^)) 
= 01 (- + Xi'z) 
= - rn 01 (z) ■+ 2x, .(f>,{x).z + x,^(P(z) y (32). 
= - rii^^i^ — 27\iXiZ + x{'z 
= — 3i\iXj^z + x{'z 
etc.. 
0102 (2) = 01 {x.,z) 
= 01 (a;,) . z + it.v,0i (z) 
= — 1\.,Z + X„XiZ 
In this manner any function (f)/'(f)J' ... (z) may be expanded into a series of 
terms each of which involves the ?''s, the x's and z. That is we may write 
Sr,,. /•.,",■■' . . . x/^ x./-^ ...z = 0/' 0,A' ... (2) (33). 
Now if both sides be integrated with respect to all the variables we shall have 
a sum of mixed moments on the left side and on the right side the integral 
r.h, 
0/' 0./^ . . . 0,/« (z) dxi dx.y . . . dx, 
.(34), 
on the reduction of which the solution depends. 
Consider the numerator of the index of e 
SAss*'/ + 2'S<AstXsXt =f{a.\X2 ... Xn) 
Write oei = yi + ruhi, 
«2 = y„+ri,Ai, 
.(35). 
Xs = ys + rjti, 
26—2 
