4.6 Hessenbergians 95
br=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=1
A
(n)
rr =nAn−^1.
c. Bn=
n
∑
r=0
(
n
r
)
Ar(−x)
n−r
.
Proof.
B 1 =−x+A 1 ,
B 2 =x
2
− 2 A 1 x+A 2 ,
B 3 =−x
3
+3A 1 x
2
− 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=0
an−rBr
r!
+(a 1 −x)Bn− 1 ,
B
′
n=−(n−1)!
n− 2
∑
r=1
an−rrBr− 1
r!
−(n−1)(a 1 −x)Bn− 2 −Bn− 1
=−(n−1)!
n− 2
∑
r=1
an−rBr− 1
(r−1)!
−(n−1)(a 1 −x)Bn− 2 −Bn− 1
=−(n−1)!
n− 3
∑
r=0
an− 1 −rBr
r!
−(n−1)(a 1 −x)Bn− 2 −Bn− 1
=−(n−1)!
n− 2
∑
r=0
bn− 1 −rBr
r!
−Bn− 1
=−(n−1)Bn− 1 −Bn− 1
=−nBn− 1 ,
which proves (a).