Page 133 - Determinants and Their Applications in Mathematical Physics

118 4. Particular Determinants

5. If

A n = |φ m | n , 0 ≤ m ≤ 2n − 2,
F n = |φ m | n , 1 ≤ m ≤ 2n − 1,
G n = |φ m | n , 2 ≤ m ≤ 2n,

where φ m is an Appell polynomial, apply Exercise 3a in which the
cofactors are scaled to prove that
(n) (n) (n)
D(A )= − iA + jA i,j+1
ij i+1,j
in which the cofactors are unscaled. Hence, prove that
(n+1)
a. D (F n )=(−1) n+r r!A , 0 ≤ r ≤ n;
r
r+1,n+1
b. D (F n )= n!A n ;
n
c. F n is a polynomial of degree n;
(n+1)
d. D (G n )=(−1) r! A pq , 0 ≤ r ≤ 2n;
!
r
r
p+q=r+2
e. D (G n )=(2n)!A n ;
2n
f. G n is a polynomial of degree 2n.
6. Let B n denote the determinant of order (n + 1) obtained by bordering
A n (0) by the row
2 3 n−1
R = 1 − xx − x ··· (−x) •
n+1
at the bottom and the column R on the right. Prove that
T
2n−2
(n)
B n = − (−x) r A (0).
pq
r=0 p+q=r+2
Hence, by applying a formula in the previous exercise and then the
Maclaurin expansion formula, prove that
B n = −G n−1 .

7. Prove that
(−1) r! (i + r − s − 1)!(j + s − 1)!
r
r
D (A ij )= A i+r−s,j+s .
r
(i − 1)!(j − 1)! s!(r − s)!
s=0
8. Apply the double-sum relation (A 1 ) in Section 4.8.7 to prove that G n
satisfies the differential equation
2n−1 m+1
m
(−1) φ m D (G n )
=0.
m!
m=0