Page 179 - Determinants and Their Applications in Mathematical Physics

164 4. Particular Determinants
∞
(p) r e (r+p)t
= ,
r!
r=0
where
(p) r = p(p + 1)(p +2) ··· (p + r − 1) (4.12.52)
(p)
and denote the corresponding determinant by E n :

E (p) = ψ (p) , 0 ≤ m ≤ 2n − 2,

n m
n
where
ψ (p) = D f
m
m
∞
m (r+p)t
(p) r (r + p) e
= . (4.12.53)
r!
r=0
Theorem 4.59.
e n(2p+n−1)t/2 n−1

(p)
E = r!(p) r .
(1 − e )
n t n(p+n−1)
r=1
Proof. Put
α r e t
g r = , α r constant,
t 2
(1 − e )
and note that, from (4.12.48),
2
g 1 = D (log f)
pe t
= ,
t 2
(1 − e )
so that α 1 = p and
log g r = log α r + t − 2 log(1 − e ),
t
2e t
2
D (log g r )= . (4.12.54)
t 2
(1 − e )
Substituting these functions into the differential–difference equation, it is
found that
n−1

α n = nα 1 +2 (n − r)
r=1
= n(p + n − 1). (4.12.55)
Hence,
n(p + n − 1)e t
g n = ,
t 2
(1 − e )
(n − r)(p + n − r − 1)e t
g n−r = . (4.12.56)
t 2
(1 − e )