Determinants and Their Applications in Mathematical Physics

210 5. Further Determinant Theory

b.Q 2 n=

1

Bn

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

c 1 c 2 c 3 ··· cn+1

c 2 c 3 c 4 ··· cn+2

...................................

cn cn+1 cn+2 ··· c 2 n

ψ 0 x

n

ψ 1 x

n− 1

ψ 2 x

n− 2

··· ψn

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

n+1

.

Proof. From the second equation in (5.5.24) in the previous section and

referring to Theorem 5.11a,

q 2 n− 1 ,r=

r

∑

t=0

cr−tp 2 n− 1 ,t, 0 ≤r≤n− 1

=

1

An

r

∑

t=0

cr−tA

(n+1)

n+1,n+1−t

.

Hence, from the second equation in (5.5.21) withn→n−1 and applying

Lemma (a) withur→x
r
andvs→A

(n+1)

n+1,s

,

AnQ 2 n− 1 =

n− 1

∑

r=0

x

r

r

∑

t=0

cr−tA

(n+1)

n+1,n+1−t

=

n

∑

j=1

A

(n+1)

n+1,j+1

j− 1

∑

r=0

crx

n+r−j

=

n

∑

j=1

x

n−j

A

(n+1)

n+1,j+1

j− 1

∑

r=0

crx

r

=

n

∑

j=1

ψj− 1 x

n−j

A

(n+1)

n+1,j+1

.

This sum represents a determinant of order (n+ 1) whose firstnrows are

identical with the firstnrows of the determinant in part (a) of the theorem

and whose last row is

[

0 ψ 0 x

n− 1

ψ 1 x

n− 2

ψ 2 x

n− 3

···ψn− 1

]

n+1

.

The proof of part (a) is completed by performing the row operation

R

′

n+1=Rn+1+x

n

R 1.

The proof of part (b) of the theorem applies Lemma (b) and gives the

required result directly, that is, without the necessity of performing a

row operation. From (5.5.27) in the previous section and referring to

Theorem 5.11b,

q 2 n,r=

r

∑

t=0

cr−tp 2 n,t, 0 ≤r≤n