538 J. D. HAMILTON DICKSON ON RELATIONS BETWEEN FUNCTIONS 
In reducing the second factor of one of the terms under a >,—take an 
example from equation (2),—the process was as follows :— 
1. Subtract the first column from the second,—thus, all terms in the second 
column which contain neither a nor 8 are removed, and what remains 
contains a—£ as a factor. 
2. Now divide the second column by a—8. It is thus reduced one degree 
lower than the first, and does not contain 8, but only terms which when 
multiplied by 8 and subtracted from the first column leave those original 
terms of the first column which do not contain £. 
Hence such a factor as that considered is always (1) divisible by a—8; and 
(2) after the above two operations contains neither a nor B. 
This process is general, and gives for an equation of the mth degree 
a* f(m) =v?.n—m.m.(s? —7t) = a?. 24 (a— B)?(Q2in-1 — Qn-2Qm)} —-- (6). 
If the roots of the equation are all real, the functions f/(m) are all positive. 
For; the expression Q?,,_;—Q,,-2Q,, contains only real roots, therefore (as it 
may be written, changing the notation) 
NM— NN mm 
P?,.-1—Pn-2P, is w fortiori positive if Le pele EST) 
| Soe pier 
is positive ; but this is, d@ wn facteur pres f(m—1) for an equation of the (7 —2)th 
degree with all its roots real. Thus, finally Q?,,_1:—-Q,-2Q, is positive if f(1), 
or f(r) is positive in an equation of the rth degree, 7 being one of the numbers 
n,n—2,n—4,... a according as 7 is ae But in such a case f(1) and 
J(r) are positive. Hence f(m) is positive always, when the roots are real. 
Hence NewtTon’s quadratic elements are always positive when the roots of 
the equation are real. 
This supplies the desideratum mentioned by SyLvEsTER, “Proc. Lond. Math. 
Soc.” vol. i. p. 13, note. 
II. The results thus obtained may be condensed as follows :— 
We may write 
hegre ey 1 ; i : 1 ; 1 | 
bed B+ty +8 , aty +8 , a+ B+8 , a+ Bty 
By+Bd+y5, ayt+ad+y5 , aB+ad5+ 85 , aB+ay+ By 
| Bys ; ayd ‘ ad ; aBy 
={(a—8) | 1 1 |} =2 {(a—B)? | 
B+yt8d , atyt+6 | 
By+Bd+y5 , ay+ad+y6 8 ,yts 
Byd ? ays | Oo , 8 
