4.11 Hankelians 4 141
The only element which remains in columnnis a 1 in position (n+1,n).
Hence,
En+1=−
∣ ∣ ∣ ∣ ∣ ∣ ∣
h 11 h 12 ··· h 1 ,n− 1 vn 1 /n
h 21 h 22 ··· h 2 ,n− 1 vn 2 /(n+1)
.....................................................
(hn 1 −t)(hn 2 −t) ··· (hn,n− 1 −t) vnn/(2n−1)
∣ ∣ ∣ ∣ ∣ ∣ ∣ n
.
(4.11.20)
It is seen from (4.11.3) (withi=n) that the sum of the elements in the
last column is unity and it is seen from the lemma that the sum of the
elements in columnjisx
2 j− 1
,1≤j≤n−1. Hence, after performing the
row operation
R
′
n=
n
∑
i=1
Ri, (4.11.21)
the result is
En+1=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
h 11 h 12 ··· h 1 ,n− 1 vn 1 /n
h 21 h 22 ··· h 2 ,n− 1 vn 2 /(n+1)
................................................
hn− 1 , 1 hn− 1 , 2 ··· hn− 1 ,n− 1 vn,n− 1 /(2n−2)
xx
3
··· x
2 n− 3
1
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n
. (4.11.22)
The final set of column operations is
C
′
j=Cj−x
2 j− 1
Cn, 1 ≤j≤n− 1 , (4.11.23)
which removes thex’s from the last row. The result can then be expressed
in the form
En+1=−
∣
∣h
(n)
∗
ij
∣
∣
n− 1
, (4.11.24)
where, referring to (4.11.5),
h
(n)
∗
ij
=h
(n)
ij
−
vnix
2 j− 1
n+i− 1
=vnix
2 j− 1
(
1
i+j− 1
−
1
i+n− 1
)
+δijt
=
(
vni
i+n− 1
)(
(n−j)x
2 j− 1
i+j− 1
)
+δijt
=−
(
vn− 1 ,i
n−i
)(
(n−j)x
2 j− 1
i+j− 1
)
+δijt
=−
(
n−j
n−i
)(
vn− 1 ,ix
2 j− 1
i+j− 1
−δijt
)
=−
(
n−j
n−i
)
̄h
(n−1)
ij
,
∣
∣h
(n)
∗
ij
∣
∣
n− 1
=−
∣
∣− ̄h
(n−1)
ij
∣
∣
n− 1