96 
ON THE PARTITIONS OF A POLYGON. 
[915 
which I prove as follows. Writing U to denote the same function of u which X is 
of x, I start from the equation 
u = x + yU, 
which determines m as a function of the independent variables x, y. We have 
|a-,v)=v d £ ( i 
where the accent denotes differentiation in regard to u; hence 
or say 
Writing 
and therefore 
this equation may be written 
du _ jj du _u — x du 
dy dx y dx’ 
du .du 
y T y ={ ~ u - x) di- 
/ x =\udx, 
du 1 
Jx~ U ’ 
d 2 u x du x du 
y — =u 
du 
dxdy dx " dx U X dx ’ 
or, integrating with respect to x, we have 
or say 
that is, 
du x j 
y — u x = \u 2 — ux, 
2 du x 2 (u x — ^x 2 ) _(u — xf 
y dy 
r 
y 
2 d (u x — _(u — ocf 
dy V y )~ y 2 
6. But, from the equation 
we have 
u = x + yU, 
r , y 6 
u =x + yX + ^2( ZS )'+ 1>2> 3 
and thence 
% = K+ A. + o + iXa ( x ’)' + 
if for a moment X x is written for jxdx. And hence, from the relation obtained 
above, we have the required identity 
4y 
6y 
Î72 + T±3 + nfo + •— H+ifi < x> >' + ÏTO ( x ’r +