4.1 Alternants 59Wn=(−1)n(n+1)/ 2
XnYnn
∏i=1(xi+ 1)(yi−1).Removingf(x 1 ),f(x 2 ),...,f(xn), from the firstnrows inVnandWn,and expanding each determinant by the last row and column, deducethat∣
∣
∣
∣1 −xiyjxi−yj∣
∣
∣
∣
n=
1
2
∣
∣
∣
∣
1
xi−yj∣
∣
∣
∣
n{
n
∏i=1(xi+ 1)(yi−1)+
n
∏i=1(xi−1)(yi+1)}
4.1.6 A Determinant Related to a Vandermondian
LetPr(x) be a polynomial defined as
Pr(x)=r
∑s=1asrxs− 1
,r≥ 1.Note that the coefficient isasr, not the usualars.
LetXn=|xi− 1
j
|n.Theorem.
|Pi(xj)|n=(a 11 a 22 ···ann)Xn.Proof. Define an upper triangular determinantUnas follows:
Un=|aij|n,aij=0, i>j,=a 11 a 22 ···ann. (4.1.7)Some of the cofactors ofUiare given by
U
(i)
ij=
{
0 ,j>i,Ui− 1 ,j=i,U 0 =1.Those cofactors for whichj<iare not required in the analysis which
follows. Hence,|U
(i)
ij
|nis also upper triangular and|U
(i)
ij
|n={
U
(1)
11 U(2)
22 ···U(n)
nn,U(1)
11 =1,
U 1 U 2 ···Un− 1.(4.1.8)
Applying the formula for the product of two determinants in Section 1.4,
|U
(j)
ij
|n|Pi(xj)|n=|qij|n, (4.1.9)