5.1 Determinants Which Represent Particular Polynomials 171
b.ψn(x)=
1
n!
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
α 0 α 1 α 2 α 3 ··· αn− 1 αn
nx
n− 12 x
n− 23 x
.....................
1 nx
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
n+1
.
Both determinants are Hessenbergians (Section 4.6).
Proof of (a).Denote the determinant byHn+1, expand it by the two
elements in the last row, and repeat this operation on the determinants of
lower order which appear. The result is
Hn+1(x)=
n
∑
r=1
(
n
r
)
Hn+1−r(−x)
r
+(−1)
n
αn.
TheHn+1term can be absorbed into the sum, giving
(−1)
n
αn=
n
∑
r=0
(
n
r
)
Hn+1−r(−x)
r
.
This is an Appell polynomial whose inverse relation is
Hn+1(x)=
n
∑
r=0
(
n
r
)
(−1)
n−r
αn−rx
r
,
which is equivalent to the stated result.
Proof of (b).Denote the determinant byH
∗
n+1
and note that some
of its elements are functions ofn, so that the minor obtained by removing
its last row and column isnotequal toH
∗
nand hence there is no obvious
recurrence relation linkingH
∗
n+1,H
∗
n,H
∗
n− 1 , etc.
The determinantH
∗
n+1can be obtained by transformingHn+1by a series
of row operations which reduce some of its elements to zero. MultiplyRi
by (n+2−i), 2≤i≤n+ 1, and compensate for the unwanted factorn!by
dividing the determinant by that factor. Now perform the row operations
R
′
i=Ri−
(
i− 1
n+1−i
)
xRi+1
first with 2≤i≤n, which introduces (n−1) zero elements intoCn+1,
then with 2≤i≤n−1, which introduces (n−2) zero elements intoCn,
then with 2≤i≤n−2, etc., and, finally, withi= 2. The determinant
H
∗
n+1
appears.