164 4. Particular Determinants
=
∞
∑
r=0
(p)re
(r+p)t
r!
,
where
(p)r=p(p+ 1)(p+2)···(p+r−1) (4.12.52)
and denote the corresponding determinant byE
(p)
n :
E
(p)
n =
∣
∣
ψ
(p)
m
∣
∣
n
, 0 ≤m≤ 2 n− 2 ,
where
ψ
(p)
m
=D
m
f
=
∞
∑
r=0
(p)r(r+p)
m
e
(r+p)t
r!
. (4.12.53)
Theorem 4.59.
E
(p)
n =
e
n(2p+n−1)t/ 2
(1−e
t
)
n(p+n−1)
n− 1
∏
r=1
r!(p)r.
Proof. Put
gr=
αre
t
(1−e
t
)
2
,αrconstant,
and note that, from (4.12.48),
g 1 =D
2
(logf)
=
pe
t
(1−et)^2
,
so thatα 1 =pand
loggr= logαr+t−2 log(1−e
t
),
D
2
(loggr)=
2 e
t
(1−e
t
)
2
. (4.12.54)
Substituting these functions into the differential–difference equation, it is
found that
αn=nα 1 +2
n− 1
∑
r=1
(n−r)
=n(p+n−1). (4.12.55)
Hence,
gn=
n(p+n−1)e
t
(1−e
t
)
2
,
gn−r=
(n−r)(p+n−r−1)e
t
(1−e
t
)
2