Determinants and Their Applications in Mathematical Physics

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,