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
)