4.12 Hankelians 5 157
=2
n− 1
n
∏
i=2
(uv)
i− 1
=2
n− 1
(uv)
1+2+3+···+n− 1
=2
−(n−1)
2
(x
2
−1)
1
2
n(n−1)
.
which completes the proof.
Exercises
1.Prove that
|Hm(x)|n=(−2)
n(n−1)/ 2
1! 2! 3!···(n−1)!,
0 ≤m≤ 2 n− 2
whereHm(x) is the Hermite polynomial.
2.If
An=
∣
∣
∣
∣
Pn− 1 Pn
Pn Pn+1
∣
∣
∣
∣
,
prove that
n(n+1)A
′′
n=2(P
′
n)
2
. (Beckenbach et al.)
4.12.2 The Generalized Geometric Series and Eulerian
Polynomials
Notes on the generalized geometric seriesφm(x), the closely related function
ψm(x), the Eulerian polynomialAn(x), and Lawden’s polynomialSn(x) are
given in Appendix A.6.
ψm(x)=
∞
∑
r=1
r
m
x
r
,
xψ
′
m(x)=ψm+1(x), (4.12.13)
Sm(x)=(1−x)
m+1
ψm,m≥ 0 , (4.12.14)
Am(x)=Sm(x),m> 0 ,
A 0 =1,S 0 =x. (4.12.15)
Theorem (Lawden).
En=|ψi+j− 2 |n=
λnx
n(n+1)/ 2
(1−x)
n^2
,
Fn=|ψi+j− 1 |n=
λnn!x
n(n+1)/ 2
(1−x)
n(n+1)
,
Gn=|ψi+j|n=
λn(n!)
2
x
n(n+1)/ 2
(1−x
n+1
)
(1−x)
(n+1)^2