MR. R. C. ROWE ON ABEL’S THEOREM. 
745 
Now, to write k +1 for k is to change the right hand side of this inequality by 
y k — (n—k—r)y t + (n—k—r—l)y^ +1 ; i.e., by (n—k—r—l)(y, c+1 —y,). 
This is negative if k<n — r — 1 
vanishes if k—n — r —1 
is positive if k>n — r —1. 
So there is a minimum value when k—n — r —1, and we must therefore have 
Pr< —1+2/1 + 2/2+ • • • -\- y n - r - l - 
Let n — r — l=k a + (3 (and lie between k a and k a+1 ), 
then P r < —. . . J r n a m a J r l3 mo+l 
A^ad- 1 
Therefore 
£ ni'rn,+/3 
\ i =1 
— I. 
(B) 
If this is to be true whatever r is, it must hold when we put a=0 ; 
wherefore P r ^Ef/3 —\ — 1 ^E(/3cr 1 )— 1 
where r is one of the numbers n —1, n — 2, . . . n — /q and f3 is Jess than k x : for 
/3=n — r — 1 . 
Now P,. cannot be negative, therefore the smallest value assignable to /3 is the least 
which makes 
E(£oq)=l ; i.e., is (p'= )E^ + 1 . 
We must then have y n ~^- 1 as the highest term in j\{x, y). 
Tliis condition, necessary—and, as we see without difficulty, sufficient also ; for the 
values assigned by equation (B) to P r are clearly positive when a is greater than zero— 
can always be satisfied unless j3'=n. 
1 1 
This can only happen in two cases, viz. : when a— - 7 or cr=- In these two cases 
J ri n — 1 n 
it can be easily shown that a single integral of the given form can be expressed by 
MDCCCLXXXI. 5 D 
