586 
PROFESSOR SYLVESTER ON THE REAL 
It must be remembered that we know, from the form of the criteria-series to the 
derivatives in respect to either x or y (indifferently), that the equation must have some 
imaginary roots ; and the question therefore lies between its having two or four. If the 
discriminant is negative, the former will be the case, if positive, the latter. 1 shall show 
that in each equation the discriminant is positive. 
Let s, t represent in general the quartic invariants, then we have to show that s 3 — 2 7 i? 
is positive. 
In case (a), s=l + 4e 2 +3e 4 
=( 1 + 0(1 + 30 
t= 
—e 
1 
=e 2 —e 4 — e i —e 2 —e 6 —e 2 
= - e 2 _2e 4 -e 6 
= - 01 + 0 2 , 
so that 
s 3 -27f=(l-0 3 {(l + 30 3 - 2 Ml+0}=(l+0 3 (l + 9 0> 
and is positive. 
In case ( b ), 
s=(l — 4e 2 + 30= (1 -0(1 - 3 0 
and 
t= 
lee 2 
e e 2 e 
e 2 e 1 
=e 2 +e 4 +e 4 — e 2 — e 6 — e 2 
= -e 2 +2e 4 -e 6 =-e 2 (l-e 2 ) 2 , 
s 3 — 2 7^ 2 = (1 — 0 3 (( 1 — 30 3 — 27^ 4 (1 — O) 
=(i-0 3 (i-90- 
The above can only be negative when e 2 lies between 1 and ^ ; but in the case supposed 
e>l. Hence the discriminant is positive, and the roots are all imaginary ( 12 ). Thus, 
then, the theorem is established for n= 4, as well as for the cases where the criteria 
are zero (as will have been observed in the course of the demonstration), as for those 
where they are plus or minus ; and it should be observed that the demonstration proceeds 
upon our being able to show that the quartic, in the case where it resists reduction to 
the case of the cubic, viz. where the criteria are negative at the two extremes and positive 
in the middle, may by real linear transformations be changed into a form where either 
the middle criterion is zero and the two extremes negative, or the two extremes zero, 
and the middle one positive. 
( 12 ) The reader conversant only with ordinary algebra may easily verify this result. For writing 
y x 
the equation becomes z 2 +4ez + 6e 2 — 2=0, and this will have its roots impossible unless 4e 2 ^~6e 2 —2, or 2e 2 —2 
negative, which it cannot be, since e 2 >l, and consequently cc : y has all its roots impossible. Moreover 
the same conclusion would (as before shown) hold good unless (? lay between 1 and -J- ; for on making z=2, 
the function above written in z becomes 2+8e-f 6<r, or 2(l+e)(l-f-3<?) ; and making z=— 2, it becomes 
2 — 8e+6e 2 , or 2(1— e)(l — 3e), which two quantities evidently have both positive signs unless e lies between 
1 and 1, or between — 1 and — -i; so that the first and third Sturmian functions are (except on that supposition) 
respectively positive and negative for z=2, and also for z=— 2, showing that no root of z can lie between 
2 and —2, and consequently that all the roots of oc : y remain impossible. 
