Information Technology Reference
In-Depth Information
More generally,
z
n
n
!
G
(
F
−
1
(
z
)) =
z
n−
1
(
n
G
(
z
)
F
0
(
z
)
−n
.
−
1)!
8 Proof of Entry 1 in Table 2
Let
F
j
(
z
):=
j≥
0
f
j
[
n
j
]
z
n
j
n
j
!
,for
j
=1
....,k
.Then
k
k
f
j
[
n
j
]
z
n
j
n
j
!
F
j
(
z
)=
j
=1
j
=1
n
j
≥
0
=
n
1
,...,n
k
∈
N
f
k
[
n
k
]
z
n
1
+
···
+
n
k
n
1
!
f
1
[
n
1
]
···
(Distributive Law.)
···
n
k
!
z
n
n
!
k
F
j
(
z
)=
c
A
(
c
)
f
1
[
c
1
]
···
f
k
[
c
k
]
j
=1
∈
C
n,k
=
α∈
C
n,k
f
1
[
α
1
]
···f
k
[
α
k
]
.
9
Proof of Entry 2 in Table 2
Since
F
(0) = 1, we may write
F
(
z
)=1+
G
(
z
)with
G
(0) = 0 and use the
binomial theorem:
F
(
z
)
x
=(1+
G
(
z
))
x
=
j≥
0
x
j
G
(
z
)
j
But by One, since
g
[0] = 0,
z
n
n
!
G
(
z
)
j
=
α∈
C
n,j
g
[
α
]=
α∈
C
n,j
f
[
α
]
so
x
j
h
[
n
]=
j≥
0
f
[
α
]
α∈
C
n,j
x
|
f
[
α
]
=
α∈
C
n
α
|
A
(
c
)
x
|
f
[
c
]
=
c∈C
n
c
|
