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p
μ H , p
κ α α (α) = κ
α μ α
(1.313)
p
1
v α ) 1
p μ α κ α λ α
p
1
t
t
v H , p
κ α α (α) =
v α + (
1
(
1
λ α )
0 κ
(
1
λ α )
α
t
=
(1.314)
=
+
then, when n
p
1, we have
p
p +
1
μ H , p + 1
κ α α (α) = κκ
α μ α = κ
μ α
(1.315)
α
p
1
v α ) 1
p μ α κ α λ α
p
1
t
t
v H , p + 1
κ α α (α) =
v α + (
1
(
1
λ α )
0 κ
(
1
λ α )
α
t
=
v α ) 1
p
1
p + 1
α
+ λ α
κ
μ α
v α (
1
(
1
λ α )
p
1
p
1
t
t
+ μ α κ α λ α
0 κ
(
1
λ α )
α
t
=
α ) 1
(
p
p
=
v
α + (
1
v
1
λ α )
λ α (
1
λ α )
p
1
p
1
t
t
+
1
+ κ
p
α
μ α κ α λ α
0 κ
(
1
λ α )
α
t
=
p
t
1
1
p
+
p
t
=
v
α + (
1
v
α )
(
1
λ α )
μ α κ α λ α
0 κ
(
1
λ α )
α
t
=
(1.316)
=
+
and thus, (5) holds for n
p
1. Therefore, (5) holds for all n .
(6) For k
=
1, we have
J 1
κ α α (α) = α + κ α π α α
v
α )
1
1
=
μ α + (
1
μ α )
(
1
κ α )
1 1
t
t = 0 λ
1
1
t
1
α
v
α κ α
(
1
κ α )
v
(1.317)
α
α
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