1916-17.] Operators applied to Solution of Equations. 19 
We have simply ax 2 — bx + c = 0 , and one obvious way of obtaining 
an operand or free term is to write 
c a „ 
X =7+7 X ". 
o b 
From the initial value x = T we derive a second, viz. 
b 
c a/ c \ 2 
x= i + i{b) ■ 
Now to convert the first term here into the second is the work of the 
operator D” 1 . D7 1 . D 7/ 2 , and cc solving the equation ” therefore is simply carry- 
ing out ad infinitum, or so long as the process yields results tangibly 
different from zero, the operation here initiated and inculcated. 
The other operand comes by writing 
be 1 . . . , b c a 
x = • - , leading to x = x — , 
a a x a a b 
and again the first term is converted into the second by the operator 
D" 1 . D7 1 . D & 2 . Here now we have b passing downwards and a passing 
upwards. 
The operations written full out are 
c a 
T ‘ 0 
0 
1 x 0 
or — 7 
b 
0 
That - should be evaluated as — 1 will cause no difficulty if, instead 
of the zeros, we substitute two infinitesimally small quantities h, neces- 
sarily of opposite signs. Thus the two operators D" 1 and D 6 may be 
supposed either to overshoot their mark, or to fall short of it by an 
infinitesimal distance. In the former case we shall have D” 1 . D b producing 
cut 1 b 
from — , and then Dfi . D 6 on this will give 
a 
h 
h 
-he c 
— • - or — - 
6 1 b 
In the latter case we have instead of in both cases therefore — 1 . 
— h h 
This step once taken, the way is clear. The operator acts now con- 
tinuously and without further novel feature, producing 
b + \J b~ — I cue 
b e ae 2 
X= a~b~ P 
n 2 e 3 
,a 3 c 4 
2 7 r - 5 + etc -> or 
2 a 
the second root, and if we confine ourselves to operands with integral 
indices, there is no third. 
