147] 
OF THE ROOTS OF AN EQUATION. 
423 
And in general it is easy to see that the left-hand side of M. Borchardt’s 
formula is equal to 
[0] + [1] (1) (1)' + [2] (2) (2)' + [P] (P) (l»y + Ac., 
where (1), (2), (P), «fee. are the symmetric functions of the roots (a, ¡3, ... k), (1)', (2)', 
(l 2 )', &c. are the corresponding symmetric functions of (x, y,...u), and [0], [1], [2], [P], 
&c. are mere numerical coefficients; viz. [0] is equal to 1.2.3...n, and [1], [2], [P], &c. 
are such that the product of one of these factors into the number of terms in the 
corresponding symmetric function (1), (2), (P), &c. may be equal to 1.2.3... ft. The 
right-hand side of M. Borchardt’s formula is therefore, as in the particular case, the 
generating function of the symmetric functions of the roots (a, /3, ... k), and if a 
convenient expression of such right-hand side could be obtained, we might by means 
of it express all the symmetric functions of the roots in terms of the coefficients. 
Tables relating to the Symmetric Functions of the Roots of an Equation. 
The outside line of letters contains the combinations (powers and products) of the 
coefficients, the coefficients being all with the positive sign, and the coefficient of the 
highest power being unity; thus in the case of a cubic equation the equation is 
x 3 + bx 2 + cx + d = {x — a) (x — f3)(x — 7) = 0. 
The outside line of numbers is obtained from that of letters merely by writing 1, 2, 3... 
for b, c, cl..., and may be considered simply as a different notation for the combinations. 
The outside column contains the different symmetric functions in the notation above 
explained, viz. (1) denotes Sa, (2) denotes Sa 2 , (P) denotes 2a/3, and so on. The Tables 
(a) are to be read according to the columns; thus Table 11(a) means 6 2 = 1 (2) + 2 (l) 2 , 
c = (P). The Tables (b) are to be read according to the lines ; thus Table II (6) 
means (2) = - 2c + lb 3 , (P) = + lc. 
I (a). 
II (a). 
Ill (a). 
1 
2 
P 
3 
12 
P 
II 
b 
II 
c 
P 
II 
d 
be 
b 3 
(1) 
- 1 
(2) 
+ 1 
(3) 
-1 
(l 2 ) 
+ 1 
+ 2 
(21) 
- 1 
- 3 
(P) 
- 1 
- 3 
- 6 
1(6). 
II (6). 
III (6). 
1 
2 
P 
3 
12 
p 
= 
b 
c 
P 
= 
d 
be 
b 3 
(1) 
-1 
(2) 
_ 2 
+ 1 
(3) 
- 3 
+ 3 
-1 
(P) 
+ 1 
(21) 
+ 3 
- 1 
(P) 
-1