Page 167 - Determinants and Their Applications in Mathematical Physics

152 4. Particular Determinants

The identity
xD [x n−1 (1 + x) ]= nD n−1 [x (1 + x) n−1 ] (4.11.81)
n
n
n
can be proved by showing that both sides are equal to the polynomial
n
n n + r − 1
n! x .
r
r n
r=1
It follows by differentiating (4.11.79) that

(xQ n ) = nP n . (4.11.82)
Hence,

{x(1 + x)E} =(1 + x)E + x{(1 + x)E}
2
=(1 + x)E + K n xQ , (4.11.83)
n
2
2

{x(1 + x)E} = K n Q + K n (Q +2xQ n Q )
n n n
=2K n Q n (xQ n )
=2nK n P n Q n . (4.11.84)
The theorem follows from (4.11.80).
A polynomial solution to the differential equation in Theorem 4.47, and
therefore the expansion of the determinant E, has been found by Chalkley
using a method based on an earlier publication.

Exercises

1. Prove that
m+1
(1 + x)
+ c − 1
= U = V, 0 ≤ m ≤ 2n − 2.
m +1

n
2. Prove that
(1 + x)D [x (1 + x) n−1 ]= nD n−1 [x n−1 (1 + x) ]
n
n
n

[(1 + x)P n ] = nQ n .
Hence, prove that
2
2
2

[X (X E) ] =4n X(XE) ,
where
+
X = x(1 + x) .