Information Technology Reference
In-Depth Information
x
k
n
=
A
(
λ
)
B
(
λ
)
f
[
λ
]
k
=1
λ∈P
n,k
x
|
A
(
λ
)
B
(
λ
)
f
[
λ
]
=
λ
λ
|
∈
P
n
10
Proof of Entry 3 in Table 2
By definition,
G
(
F
(
z
)) =
k
g
[
k
]
k
!
F
(
z
)
k
.
≥
0
But by 1, since
F
(0) = 0,
z
n
n
!
F
(
z
)
k
=
α∈
C
n,k
f
[
α
]
Thus
h
[
n
]=
a
∈
C
n
1
|
a
|
!
f
[
a
]
g
[
|
a
|
]
=
π
∈
P
n
f
[
π
]
g
[
|
π
|
]
=
λ
∈
P
n
C
(
λ
)
f
g
[
λ
]
11
Proof of Entry 4 in Table 2
By Lagrange's theorem, if
F
(0) = 0, and
F
0
(
z
):=
F
(
z
)
/z
(whence
f
0
[
n
]=
f
[
n
+1]
/
(
n
+ 1)),
then
w
n
n
!
F
−
1
(
w
)=
z
n−
1
(
n
F
0
(
z
)
−n
−
1)!
C
(
λ
)
−
f
0
[
λ
]
=
λ∈P
n−
1
n
|
λ
|
=
λ∈P
n−
1
1)
|λ
+1
|
C
(
λ
+1)
f
[
λ
+1]
.
(
−
=
λ
1)
|λ|
C
(
λ
)
f
[
λ
]
(
−
∈
DP
n−
1
=
R
(
π
)
∈DP
n−
1
1)
|π|
f
[
π
]
(
−
