210 Prof. F. Y. Edgeworth on the 



And a compound of symmetrical elements must itself be 

 symmetrical. "We have, therefore, 



1 (^ f\ 



z =V^<r y ™) K> 



To the five equations which have been stated there is to be 

 added the condition that the integral of zdxdy between 

 extreme limits =1. 



To solve this system : substitute in (3) and (4) the values 



dz dz 



of -pr and -j — given in (1) and (2) respectively. We have, 



then, 



, dz . d 2 z 

 z+ VTy +m ^ =0 W 



Integrating (6) with respect to #, and (7) with respect to 

 y, we have 



».+ *-|=*(y), (8) 



dz 



^+™^=f(>); ( 9 ) 



where <f> and yjr are arbitrary functions. 

 Both these functions reduce to zero; as may thus be proved: — 



dz 

 From (5) it appears that when #=0, -7- also =0, whatever 



the value of y. If, then, we put # = 0, the left side of equation 

 (8) vanishes for all values of y. Therefore the right side of 

 the equation vanishes for all values of y. Therefore <j>(y) is 

 identical with zero. By parity yjr(y) is null. 

 By equation (8) thus reduced we have 



z=®(y)xe u , (10) 



z=V(x)e 2m (11) 



Identifying the right-hand members of (10) and (11) we have 



Z=Ce 2fc 2m y (12) 



where C is a constant : which is found to be - — -7= from the 



2-7T \km 



