MR. R. C. ROWE ON ABEL’S THEOREM. 
725 
Also take 
%) = V~ ( 1 +px+qx 2 ), 
so that removing the factor x=0 (see p. 702), we get by elimination of y between 
X and 6 
E = (cf—c z )x ?J +2 pqxr-\- ( p 2 +2^+c 2 -f-1 )x + 2p = 0. 
It is clear that, in general, no linear relation connects the coefficients of this 
equation, so that F 0 (x)=l ; and the formula reduces to 
t 
X -dx — 
y) 
x\y) 
© 
u 
1 T" TldC? 1 
v a -f foe 2 
V 
log By 
log {y — (1+pT+qx 2 )} 
n 
ci -T for" 
y 
log 
{y— {l+px + qx 2 )} 
where, as usual, l= x / — 1. 
b 
r / a — 
Therefore nl, XcLc= — —-£ -- log 
J 2i y & 
1 
vp 
\/n 
b 
a - 
\\/' n < _ P 
2t y 
log; 
o 
i + 
pi 
P n 
i 1 -f TlP 
Now the last term in general vanishes. 
_ n Q } a + bz 2 r log {y — (l+px + qx 2 )} 
For 
v 1 Qg {y - (i +pz +wC ^ log (—gs 8 ) r 
y ~ y 
log 1 
px +1 — y 
qx 2 
and 
while Ci in the term 
X 
and this vanishes. 
£ 
log 1 4 
2-=0 
y 
px+l —y 
qx? 
is % 
A 
yy 
Therefore the first term of the descending expansion of % involves x 3 , while that ol 
a + bx 2 
1 + nP 
wherefore 
2 begins with x° ; 
Ci a + bx 2 ^ log {y-(l +px + gx 2 } _ 0 # 
x 1 “f" TlX" 
y 
n —0 ] 
, a + bx 2 
* There is an exceptional case if ^ _j_ q j , for then the expansion of pqypb begins with x 2 ; and the 
Ci is not necessarily zero. 
