MR P. ALEXANDER ON THE EXPANSION OF FUNCTIONS. 315 



The solution of this in all its generality has not yet been obtained, but in 

 most of the particular cases which have been solved the method seems to be to 

 operate on (1) with an operator 0„ such that 



O„.G m =0, (2) 



except m = n. 



And, therefore, 



O n . <j>(x) = A„O n . G n (x) 



A _ On.<p(x) .... (3) 



" On. Unix) 



Following this lead, I have found an operator of this nature in the case where 

 G , Gi, G 2 , &c, are elementary solutions of the equation, 



(S+g)G = 0, (4) 



when g has the values g , g u g 2 , &c, derived from the condition 



*- G U = ° (5) 



where S and cr are operations which may have the forms 



d \ , v / d \2 



*= x »+ x .(^) +x 44y+ & (^ 



where X , X 1} X 2 , &c, and P , Pi, P 2 , &c, may be either constants or functions 

 of x. 



The operator I have discovered for the solution of this problem is — 



o n= xs+</„n, =a (8) 



The proof is as follows : — 



But from (4), 







8 - G m = -9m- G m 





. * 2 -G m = -9j& m = 9j-& m 





* 3 -G m = <CSG m =-g m 3 G m 



(10) 



