94 4. Particular Determinants
Also, adjusting the dummy variable inS 1 and referring to (4.6.4) with
n→n−1,
S 1 =
n− 1
∑i=0(−1)
i
(i+1+a)φiψn− 1 −i=
n− 1
∑i=1(−1)
i
iφiψn− 1 −i+(1+a)n− 1
∑i=0(−1)
i
φiψn− 1 −i=−S 3.
Hence,ψ
′
n=(a+2−n)Fψn−^1 , which is equivalent to the stated result.Note that ifφ
′
m=(m−1)φm−^1 , thenψ′
m=−(m−1)ψm−^1. 4.6.3 A Hessenberg–Appell Characteristic Polynomial
Let
An=|aij|n,where
aij={
aj−i+1,j≥i,−j, j=i−1,0 , otherwise.In some detail,
An=∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣
∣
∣
∣
∣
a 1 a 2 a 3 a 4 ··· an− 1 an− 1 a 1 a 2 a 3 ··· an− 2 an− 1− 2 a 1 a 2 ··· ··· ···− 3 a 1 ··· ··· ······ ··· ···a 1 a 2−(n−1) a 1∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ n. (4.6.7)
Applying the recurrence relation in Theorem 4.20,
An=(n−1)!n− 1
∑r=0an−rArr!,n≥ 1 ,A 0 =1. (4.6.8)LetBn(x) denote the characteristic polynomial of the matrixAn:
Bn=∣
∣
An−xI∣
∣
. (4.6.9)
This determinant satisfies the recurrence relation
Bn=(n−1)!n− 1
∑r=0bn−rBrr!,n≥ 1 ,B 0 =1, (4.6.10)where
b 1 =a 1 −x,