316 MR P. ALEXANDER ON THE EXPANSION OF FUNCTIONS. 



= (0n-9 m )' X <r& m + ° G <n 



+ *<*« ■ ■ (11) 



But from (5), 

 Hence (11) gives — 



Hence from (3), 



Ab _ O»-0(s) 

 0„.G„ 



i 

 = if to is not = ™ ^ (12) 



0., I 



By the method of vanishing fractions, 



(m = n, x = a) 



\- Or. 



G m ) 







£<<r.GU)l 



Hence (13) becomes 



(13) 



A " = il <*■«»> L <14) 



which is the required solution. 



As new results, especially when very general, are liable to suspicion, I 

 proceed to test this by a particular example whose solution can be otherwise 

 found. 



Let 



H£f-m- t± ^ 2k ™ 



* In the proof of (11) it has been assumed that g m is less than g„. The same may be proved for 

 f/ m greater than y„ by expanding {h + g n )~ l in the reverse order. 



