4.11 Hankelians 4 141The 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 /nh 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=1Ri, (4.11.21)the result is
En+1=∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
h 11 h 12 ··· h 1 ,n− 1 vn 1 /nh 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−x2 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−
vnix2 j− 1n+i− 1=vnix2 j− 1(
1
i+j− 1−
1
i+n− 1)
+δijt=
(
vnii+n− 1)(
(n−j)x2 j− 1i+j− 1)
+δijt=−
(
vn− 1 ,in−i)(
(n−j)x
2 j− 1i+j− 1)
+δijt=−
(
n−jn−i)(
vn− 1 ,ix2 j− 1i+j− 1−δijt)
=−
(
n−jn−i)
̄h
(n−1)
ij,
∣
∣h
(n)
∗ij∣
∣
n− 1=−
∣
∣− ̄h
(n−1)
ij∣
∣
n− 1