4.11 Hankelians 4 145where
C
′
n=λn−^1 ,n−^1 Cn+n− 1
∑j=1λn− 1 ,j− 1 Cj=
n− 1
∑j=1λn− 1 ,j− 1[
νj− 1 νj···νn+j− 3 νn+j− 2]T
n=2
−(4n−5)[
00 ··· 01
]T
n. (4.11.42)
Hence,
An=2−4(n−1)+1
An− 1An− 1 =2−4(n−2)+1
An− 2.......................................A 2 =2
−4(1)+1
A 1 , (A 1 =ν 0 =1).(4.11.43)
The theorem follows by equating the product of the left-hand sides to the
product of the right-hand sides.
It is now required to evaluate the cofactors ofAn.Theorem 4.49.
a.A(n)
nj=2
−(n−1)(2n−3)
λn− 1 ,j− 1 ,b.A
(n)
n 1 =2−(n−1)(2n−3)
,c. Anj
n =22(n−1)
λn− 1 ,j− 1.Proof. Thenequations in (4.11.40) can be expressed in matrix form as
follows:
AnLn=C′
n, (4.11.44)where
Ln=[
λn 0 λn 1 ···λn,n− 2 λn,n− 1]
Tn. (4.11.45)
Hence,
Ln=A− 1
nC′
n=A
− 1
n[
A
(n)
ji]
nC
′
n=2
(n−1)(2n−1)−2(n−1)[
An 1 An 2 ···An,n− 1 Ann]T
n, (4.11.46)
which yields part (a) of the theorem. Parts (b) and (c) then follow
easily.
Theorem 4.50.
A
(n)
ij =2−n(2n−3)[
2
2 i− 3
λi− 1 ,j− 1 +n− 1
∑r=i+1λr− 1 ,i− 1 λr− 1 ,j− 1]
,j≤i<n− 1.