DUNKEL. — LINEAR DIFFERENTIAL EQUATIONS. 365 



We will now show that the n solutions (62) are linearly independent. 

 Suppose they were not and that there were n constants C Ky \ not all zero 

 such that : 



It will be convenient to suppose that our notation is such that: 



(66) B ri SXr 2 < . ..<Rr,„. 



Consider first those equations of (65) for which Rr k = R r x . We 

 have: 



(67) limit 2 2 C , a x ~ Tk <■' ) = ° Zr k = R ri 



1=0 =* *=* (1= 1, 2, . . . e k ). 



Now let ^ = 1 in (67), and consider the limit of each term for any 

 given value, within the range indicated, for k. For each term in which 

 Rr K > R r k = Rr 1 the limit is zero by II. We have left, now, only 

 the terms : 



(68) C K ,x x ~* k *k,l Rr K = Rr k = R ri . 



According to I. no logarithms appear explicitly in (68), and we can 

 write : 



Now by (63) the limit of all such terms is zero except in the one case 

 k =■ k and A = 1, and for this term we have : 



(70) limit G kil x~ rk z k k \ = limit C M &{ = C k>1 . 



x=0 x—Q 



So in the case of I = 1 the limit (67) turns out to be C kl when 

 we evaluate the limit term by term. Now this is impossible unless 

 C kl = ; and so we must have, writing now « in place of k : 



(71) C*,i = ° Rr K = Rr x . 



Now consider in the same way the cases of (67) in which 1=2; and 

 choose any one of the values of k indicated. Here again by II. the limit 

 of each term is zero when R r K > R r k ; and we have left the terms : 



(72) 0., **"**« Rr K = Rr k = R ri 



X> 2. 



