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4.12 Hankelians 5 157
n
i−1
n−1
=2 (uv)
i=2
=2 n−1 (uv) 1+2+3+···+n−1
2 2 1 n(n−1)
−(n−1)
=2 (x − 1) 2 .
which completes the proof.
Exercises
1. Prove that
|H m (x)| n =(−2) n(n−1)/2 1! 2! 3! ··· (n − 1)!,
0≤m≤2n−2
where H m (x) is the Hermite polynomial.
2. If
P n−1
A n = P n ,
P n+1
P n
prove that
2
n(n +1)A =2(P ) . (Beckenbach et al.)
n n
4.12.2 The Generalized Geometric Series and Eulerian
Polynomials
Notes on the generalized geometric series φ m (x), the closely related function
ψ m (x), the Eulerian polynomial A n (x), and Lawden’s polynomial S n (x) are
given in Appendix A.6.
∞
ψ m (x)= r x ,
m r
r=1
xψ (x)= ψ m+1 (x), (4.12.13)
m
S m (x)=(1 − x) m+1 ψ m , m ≥ 0, (4.12.14)
A m (x)= S m (x), m > 0,
A 0 =1, S 0 = x. (4.12.15)
Theorem (Lawden).
λ n x n(n+1)/2
E n = |ψ i+j−2 | n = ,
(1 − x) n 2
λ n n! x n(n+1)/2
F n = |ψ i+j−1 | n = ,
(1 − x) n(n+1)
2 n(n+1)/2
λ n (n!) x (1 − x n+1 )
G n = |ψ i+j | n = ,
(1 − x) (n+1) 2