4.6 Hessenbergians 95br=ar,r> 1.Bn(0) =An,B
(n)
ij(0) =A
(n)
ij. (4.6.11)
Theorem 4.23.
a.B′
n
=−nBn− 1.b.
n
∑r=1A
(n)
rr =nAn−^1.c. Bn=n
∑r=0(
nr)
Ar(−x)n−r
.Proof.
B 1 =−x+A 1 ,B 2 =x2
− 2 A 1 x+A 2 ,B 3 =−x3
+3A 1 x2
− 3 A 2 x+A 3 , (4.6.12)etc., which are Appell polynomials (Appendix A.4) so that (a) is valid for
small values ofn. Assume that
B
′
r=−rBr−^1 ,^2 ≤r≤n−^1 ,and apply the method of induction.
From (4.6.10),Bn=(n−1)!n− 2
∑r=0an−rBrr!+(a 1 −x)Bn− 1 ,B
′
n=−(n−1)!n− 2
∑r=1an−rrBr− 1r!−(n−1)(a 1 −x)Bn− 2 −Bn− 1=−(n−1)!n− 2
∑r=1an−rBr− 1(r−1)!−(n−1)(a 1 −x)Bn− 2 −Bn− 1=−(n−1)!n− 3
∑r=0an− 1 −rBrr!−(n−1)(a 1 −x)Bn− 2 −Bn− 1=−(n−1)!n− 2
∑r=0bn− 1 −rBrr!−Bn− 1=−(n−1)Bn− 1 −Bn− 1=−nBn− 1 ,which proves (a).