Determinants and Their Applications in Mathematical Physics

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).