Page 180 - Determinants and Their Applications in Mathematical Physics

4.13 Hankelians 6 165
(p) (1)
Substituting this formula into (4.12.50) with A n → E n and E n = f,

p n−1 r
e t e t
(p)
E = (n − r)(p + n − r − 1) , (4.12.57)
1 − e t (1 − e )
n t 2
r=1
which yields the stated formula.
Note that the substitution x = e yields
t
ψ (1) = ψ m ,
m
E (1) = E n ,
n
(p)
so that ψ m may be regarded as a further generalization of the geometric
(p)
series ψ m and E n is a generalization of Lawden’s determinant E n .
Exercise. If
sec x
/
p
f = ,
cosec x
p
prove that
n−1

sec n(p+n−1) x
A n = r!(p) r .
cosec n(p+n−1)
r=1
4.13 Hankelians 6
4.13.1 Two Matrix Identities and Their Corollaries
Define three matrices M, K, and N of order n as follows:

(symmetric),
M =[α ij ] n
K =[2 i+j−1 k i+j−2 ] n (Hankel), (4.13.1)
(lower triangular),
N =[β ij ] n
where
j−1
(−1) u i−j + u i+j−2 ,j ≤ i
α ij = i−1 (4.13.2)
(−1) u j−i + u i+j+2 , j ≥ i;
N

u r = a j f r (x j ), a j arbitrary; (4.13.3)
j=1
/ 0
+ +
1
f r (x)= (x + 1+ x ) +(x − 1+ x ) ; (4.13.4)
2 r
2 r
2
N

k r = a j x ; (4.13.5)
r
j
j=1
β ij =0, j > i or i + j odd,