141] 
A SECOND MEMOIR UPON QUANTICS. 
255 
and I represent by YA, Y 2 A, Y 3 A, «fee. the results obtained by successive operations 
with F upon the function A. The function Y S A will be of the degree 0 and of the 
weight ^(m6 -fi) + s. And it is clear that in the series of terms Y A, Y 2 A, Y Z A, &e., 
we must at last come to a term which is equal to zero. In fact, since m is the 
greatest weight of any coefficient, the weight of F s is at most equal to md, and therefore 
if \ (md ~/jl)+s> md, or s > \(m0 + ¿¿), we must have Y S A = 0. 
Now writing for greater simplicity XY instead of X. Y, and so in similar cases, we 
have, as regards Y S A, 
Hence 
and consequently 
Similarly 
and therefore 
And again, 
and therefore 
XY - YX = — 2s. 
(XY- YX)A = ¡¿A, 
XYA=YXA + f iA= t iA. 
(XY- YX) Y A =(/jl — 2) Y A, 
XY 2 A = YXYA + (fi-2) YA 
= fiYA + (fi-2) YA = 2(n-l)YA. 
(XY-YX)Y 2 A=(n-‘i)Y 2 A, 
XY S A = YXY 2 A + (fji - 4) Y-A 
= 20a-l)F 2 A+0a-4)F 2 A = 30*-2) YA, 
or generally 
+ Y S A. 
Hence putting s = ya+l, /¿ + 2, &c., we have 
XY^A = 0, 
X Y* +2 A = - (ii + 2) 1. Y^A, 
XY» +3 A = - (fi + 3) 2. Y» +2 A, 
«fee., 
equations which show that 
Y* +1 A = 0; 
for unless this be so, i.e. if Y^A 4= 0, then from the second equation X1 4= 0, and 
therefore Y^A + 0, from the third equation XP‘ +3 4 0, and therefore 4 0, and so 
on ad infinitum, i.e. we must have F U+1 H = 0.