ME P. ALEXANDER ON THE EXPANSION OF FUNCTIONS. 



317 



and 



■ = h+ 



dx 



(16) 



An elementary solution of (4) is in this case 



A.-1 



2 



or 



G n = (a V# K ) * J p+ ^ (« Jg n ) 



A.-1 



(17) ! 



where J and K are Bessel's functions of the first and second orders. 



Supposing then that it is possible to expand <p (x) in terms of the first of 

 these and operating on (1) with the operator 



we have 



But 



O n = I dxX x G n 



(J>(x)x x G n dx = A /G G n x x dx + A l /G l G n x*dx+&c. 

 o «-/ o «-/o 



G m G„x k dx 



f r dGn p dGm I 



^ X^dx-^dx ) x=a 



(18) 



(19) 



9m 9n 



AH h+ ^} _ 



gm—g n 



from (5) and (16) 



= -ak\ Gn<r - Gm \ from (16) 



= if m is not =n 



= — if m = n 







} 



from (5) 



(20) 



By the method of vanishing fractions, 



f a Glx^dx =-ah\ G ^.G m \ 



'* ^ 9 m -9n > (»»=».'<*=«) 



(21) 



\-l 



X-l 



Or 



G„ = (a; ^J 2 J ^.(aj ^„) + B„(a; >Jg n ) 2 K *_i(a? ^/^„) 



p+ 



P+- 



