187 



in §1. This gives 



f(x+l)ufx+l) — r(x)f(x)u(x) = s(x). 



Since f(x+l) — r(x)f(x)=0 we can divide by r(x)f(x) and we have 



s(x) 

 u(x+l) — u(\) = 



r(x)f(x) 



X s(x+t) 

 u(xj = p(x) — S 



t = o r(x+t)f(x+t) 



where p(x) is an arbitrary periodic function of period 1. Xow 



f(x+t) = rfx+t— l)f(x+t^l) = = r(x+t— l)r(x+t— 2) 



Making this substitution in the preceding equation we have 



yr. s(x+t) 



u(x)=p(x)- ^ 



r(x)f(x) 



t = o r(x+t) r(x+t — 1) r(x)f(x) 



If we choose p(x)=0 we have 



oc s(x+t) 



F(x)=f(x)u(x)=-S 



t = o r(x+t) r(x+t — 1) . . . . r(x) 

 F(x) is then a solution satisfying I and II provided that 



s(x+n) 



S(x) = uo(x)+u,(x)+u.,(x) + un(x)=- 



r(x+n) r(x+n — 1) . . . r(x) 



is analj'tic. 8(x) is analytic jirovided that it converges uniformly in any 



closed region T lying in the strip defined by the relation A<R(x)<A+l. 



In the region T in the strip under consideration the following ratio of the 

 (t+l)th term to the t-th term holds for everj- value of x in that region. 



t + l 



1 



s(x+t) 



r(x+t) s(x+t — 1) I 



I 1 f n — m n — m — 1 ] 



(5) =\—\t — fk — n— m R(x)]t + I- 



|a[ J 



( ! l,(x) M 



U+- + ^+ ... M, 

 i t t== J I 



where k has the same meaning as in §1 and 1 = g — n. When n = m (5) becomes 



1^+1 I 1 If — k+1 11 ]\ 



(6) 1 = — -|^ l-\ h . . . • terms in — , — , etc.;- 1 . 



u. I |a| I I t t^ t'^ II 



In considering the value of this ratio we shall need to examine the fol- 

 lowing cases: 



