152 4. Particular Determinants
The identity
xD
n
[x
n− 1
(1 +x)
n
]=nD
n− 1
[x
n
(1 +x)
n− 1
] (4.11.81)
can be proved by showing that both sides are equal to the polynomial
n!
n
∑
r=1
(
n
r
)(
n+r− 1
n
)
x
r
.
It follows by differentiating (4.11.79) that
(xQn)
′
=nPn. (4.11.82)
Hence,
{x(1 +x)E}
′
=(1+x)E+x{(1 +x)E}
′
=(1+x)E+KnxQ
2
n, (4.11.83)
{x(1 +x)E}
′′
=KnQ
2
n
+Kn(Q
2
n
+2xQnQ
′
n
)
=2KnQn(xQn)
′
=2nKnPnQn. (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 determinantE, has been found by Chalkley
using a method based on an earlier publication.
Exercises
1.Prove that
∣
∣
∣
∣
(1 +x)
m+1
+c− 1
m+1
∣
∣
∣
∣
n
=U=V, 0 ≤m≤ 2 n− 2.
2.Prove that
(1 +x)D
n
[x
n
(1 +x)
n− 1
]=nD
n− 1
[x
n− 1
(1 +x)
n
]
[(1 +x)Pn]
′
=nQn.
Hence, prove that
[X
2
(X
2
E)
′′
]
′
=4n
2
X(XE)
′
,
where
X=
√
x(1 +x).