Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 127

=V 1 −(r−1)

∑

j

K

j 1

h+r+j− 1

=V 1 −(r−1)δr 1 , 1 ≤r≤n.

The second term is zero. The result follows.

The proof of (4.10.18) whens=rfollows from the identity

1

(h+r+j−1)(h+s+j−1)

=

1

s−r

(

1

h+r+j− 1

−

1

h+s+j− 1

)

and (4.10.15). Whens=r, the proof follows from (4.10.8) and (4.10.16):

Vr

∑

s

Vs

(h+r+s−1)

2

=

∑

s

K

rs

h+r+s− 1

=1.

To prove (4.10.19), apply (4.10.4) and (4.10.16):

∑

j

jK

1 j

h+r+j− 1

=

∑

j

(

1 −

h+r− 1

h+r+j− 1

)

K

1 j

=V 1 −hδr 1 −(r−1)δr 1 , 1 ≤r≤n.

The third term is zero. The result follows.

Equation (4.10.20) follows from (4.10.4) and the double-sum identity (C)

(Section 3.4) withfr=randgs=s+h−1, and (4.10.21) follows from

the identity (4.10.9) in the form

jK

1 j

=V 1 Vj−hK

1 j

by summing overjand applying (4.10.4) and (4.10.20).

4.10.2 Three Formulas of the Rodrigues Type.......

Let

Rn(x)=

n

∑

j=1

K

1 j

x

j− 1

=

1

Kn

∣ ∣ ∣ ∣ ∣ ∣ ∣

1 xx

2

··· x

n− 1

k 21 k 22 k 23 ··· k 2 n

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

kn 1 kn 2 kn 3 ··· knn

∣ ∣ ∣ ∣ ∣ ∣ ∣ n

.

Theorem 4.37.

Rn(x)=

(h+n)!

(n−1)!

2

h!x

h+1

D

n− 1

[x

h+n

(1−x)

n− 1

].

Proof. Referring to (4.10.9), (4.10.5), and (4.10.6),

D

n− 1

[

x

h+n

(1−x)

n− 1

]

=

n− 1

∑

i=0

(−1)

i

(

n− 1

i

)

D

n− 1

(x

h+n+i

)