24 
Proceedings of the Royal Society of Edinburgh. [Sess. 
the appearance of the counteracting zero in the denominator, then, the 
intervening elements are zero. But we shall first write down the solution 
itself. 
_C 
B 
HsXr 
'E N 
x = 
-¥ E 
cAc 
+ 1 JAB 
_2(W/C\ 3 
- 
B/ VB 
2 
B\ 2 /E\ 3 
C 
+ 0 + 
CM 
B, 
A\/E\ 3 
14 
BWE\ 5 
C 
O/VC / 
CAcXS 
-0 + 10^ 
A\ 3 /E\/C 
B / VB/VB 
-14 A 
A\ 4 /C\ 5 
AV 2Y5Y+o 
- 2 
0 /A\ 3 /EY 
B/liJ +35 
B / VB 
J( yc\ 
B ) VB/VB/ 
MYIY£Y_42^ 
etc., etc. 
Here, when the operator Dp . Dp . Dp acts on the element (g)(g) 
in the second horizontal line, the result is zero. But we must keep record 
of the operations, which are as follows : — 
MxAx?x(lxi 
C’ 2 V - 1 B 
or zero. But with another action by the same operator there arises 
(OK - 1)( - 2 )( - 3 ) \ E 3 „ B° _ AE 3 
C 4 ~ 2-3 ( - 1)(0) O' ' 
If the development be carried further, the same mode of calculating the 
coefficients must be observed. A little reflection will explain the appear- 
ance of these zero elements here. The two former roots had C and B 
for the denominators of their every element respectively. This root 
has both C and B. 
In this root occurs, with changed sign, every element occurring in 
either of the preceding roots, with the solitary exception of the element 
which belongs to the second root, and is the sum of all three. 
The 
function of the second operator here, therefore, is to take every element 
belonging to the first root and convert it gradually into an element of the 
second root. This it does by making the element pass through the value 
zero. 
It ought to be added that a zero never appears in the denominator of 
