4.13 Hankelians 6 165
Substituting this formula into (4.12.50) withAn→E
(p)
n andE
(1)
n =f,
E
(p)
n
=
(
e
t
1 −e
t
)p n− 1
∏
r=1
[
(n−r)(p+n−r−1)
e
t
(1−e
t
)
2
]r
, (4.12.57)
which yields the stated formula.
Note that the substitutionx=e
t
yields
ψ
(1)
m =ψm,
E
(1)
n =En,
so thatψ
(p)
m may be regarded as a further generalization of the geometric
seriesψmandE
(p)
n is a generalization of Lawden’s determinantEn.
Exercise.If
f=
{
sec
p
x
cosec
p
x
,
prove that
An=
{
sec
n(p+n−1)
x
cosec
n(p+n−1)
n− 1
∏
r=1
r!(p)r.
4.13 Hankelians 6
4.13.1 Two Matrix Identities and Their Corollaries....
Define three matricesM,K, andNof ordernas follows:
M=[αij]n (symmetric),
K=[2
i+j− 1
ki+j− 2 ]n (Hankel),
N=[βij]n (lower triangular),
(4.13.1)
where
αij=
{
(−1)
j− 1
ui−j+ui+j− 2 ,j≤i
(−1)
i− 1
uj−i+ui+j+2,j≥i;
(4.13.2)
ur=
N
∑
j=1
ajfr(xj),ajarbitrary; (4.13.3)
fr(x)=
1
2
{
(x+
√
1+x
2
)
r
+(x−
√
1+x
2
)
r
}
; (4.13.4)
kr=
N
∑
j=1
ajx
r
j
; (4.13.5)
βij=0,j>iori+jodd,