A.7 Symmetric Polynomials 327Also,
gnj(xi)=0,j=i. (A.7.5)Examples
σ(3)
2
=x 1 x 2 +x 1 x 3 +x 2 x 3 ,σ(n)
r 0
=1, 1 ≤r≤n,σ(3)
21
=x 1 +x 3 ,σ(3)
22
=x 1 x 3 ,σ(4)
31
=x 1 +x 2 +x 4 ,σ(4)
32
=x 1 x 2 +x 1 x 4 +x 2 x 4 ,σ(4)
33 =x^1 x^2 x^4.Lemma.
σ(n)
rs=
s
∑p=0σ(n)
p
(−xr)s−p
.Proof. Since
gr(x)=−1
xr(
1 −
xxr)− 1
f(x)=−
f(x)xr∞
∑q=0(
xxr)q,
it follows that
n− 1
∑s=0(−1)
s+1
σ(n)
rsxn− 1 −s
=n
∑p=0∞
∑q=0(−1)
p
σ(n)
p xn−p+qxq+1
r.
Equating coefficients ofx
n− 1 −s
,
(−1)
s+1
σ(n)
rs=
n
∑p=s+1(−1)
p
σ(n)
p
xs−p
r.
Hence
(−1)
s+1
σ(n)
rs+
s
∑p=0(−1)
p
σ(n)
p
xs−p
r=
n
∑p=0(−1)
p
σ(n)
p
xs−p
r=xs−n
r f(xr)=0.The lemma follows.
Symmetric polynomials appear in Section 4.1.2 on Vandermondians.