187 



in §1. This gives 



f(x+l)u(x+l) — r(x)f(x)u(x) = s(x). 



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



.s(x) 



u(x+l) — u(x)= • 



r(x)f(x) 

 or S 



oo s(x+t) 

 u(x) = p(x)-S 



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



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



f(x + t) = rrx+t— l)f(x+t— 1) = = r(x+t.— l)r(x+t— 2) . . 



Making this substitution in the ])rece(ling equation we have 



-J-. s(x+t) 



u(x) = p(x) 



r(x)f(x). 



t = o r(x+t) r(x+t — 1) 

 If we choose p(x)=0 we have 



F(x)=f(x)u(x)^ 



. . r(x)f(x) 



.s(x+t) 



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)+Ui(x)+Uo(x) + . . . . , un(x) = 



r(x+n) r(x+n — 1) . . . r(x) 

 is analytic. S(x) is analytic provided 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 every value of x in that region. 



't + 1 



s(x+t) 



.(5) 



1 f n — m 



-u - 



al 



I |r(x+t) s(x+t-l) 



n — m — 1 



[k — n— m R(x)]t + • - 



f 1 l,(x) 



U+- + — + 



I t t= 



]\ 



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



(6) 



-k+1 11 1 



[-•••• terms in — , — , etc.f 



t t- t3 J 



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

 lowing cases: 



