1916-17.] Operators applied to Solution of Equations. 
To apply it, e.g ., to the solution of the quintic, see § III (/), 
29 
Aar 3 + Bad — Cad + Ead + Fa? + G = 0 
E 
belonging to the operand g, we put y — 1, f(x) — x, so that f'(x) = 1, so that 
we have 
where 
E JL/ x 
X = Q + <K X )’ 
, / x _F 1 G 1 B A q 
9(X) — p, 9 " + • -9 + ^ar" + -,ar 
C a? 0 ad C C 
I d 
We have, therefore, for _ ^ (#) | 
1 c i 
2! da* 
or 
F 2 1 G 2 1 B 2 4 , A 2 6 » 0 FG J , 0 FB , 0 FA „ , 0 GB 0 I 
+ & + + + * + *-& a + 2 -&* + 2 rS + * ^ x + 
GA , ,BA , 
Q-2 X+ J G 2 r 
F" 1 96 “ 1 , qB 2 3 5 o-*- - - ^ , n , 
~ Q2 J Q2 ^C 2 “ 6 TV ^ 4 +n 2 + -p2^ + (,+ r«2 + 0 7 ^» 
,A 2 5 qFG 1 FB , 0 FA 
C 2 ^ 3 C 2 a: 4 + C 2 + " C 2 
GA BA 
C 2 G- 
E 
which, with x = ^, the operand or initial value, gives the third line in the 
development previously obtained from the operators, as 
E , F C G C 2 BE 2 AE 3 
— and — . • — i — • 1 1 
C C E^C E- C 3 C 4 
are the first and second lines. 
The reader can easily verify that the trend of the two expansions is 
the same, a term in Lagrange’s development being a line, in the develop- 
ment from the operators. 
If <p(z) contain n — 1 terms consisting of positive or negative powers 
of 0 , the number of terms in its square, third power, etc., is also ^FL, 
„_ 1 H 3 , etc. See § III (/). 
If we split Lagrange’s <j>(z) into \fr{z) -\- f{z), his theorem still holds. We 
may now, starting from the case of the quadratic, build up a proof for 
the general case, by Induction, make f(z) include z +3 , z +4 , etc., in succession. 
Lagrange showed further that the equation Ax n + Bx + C = 0 always 
yielded its least root from his application of his own Theorem. This was 
because he always took for operand 
C 
B 
the operand for the direct solution 
which always leads to the numerically smallest root. 
