Information Technology Reference
In-Depth Information
g [3] = x
1
f [3] + 6 x
2
f [2] f [1] + 6 x
3
f [1] 3
g [4] = x
1
f [4] + 8 x
2
f [3] f [1] + 6 x
2
f [2] 2 +36 x
3
f [2] f [1] 2 +24 x
4
f [1] 4
That concludes the illustrative example. In Table 2, which follows, we list a few
of many possible variations on the theme. For the record, entry 1 is Leibniz' rule,
entry 5 is Faa di Bruno, and entries 6 and 7 are essentialy due to Lagrange.
Table 2. Some (
H
(
z
)
↔ h
[
n
]) pairs
H ( z )
h A [ n ]
h B [ n ]
1 .
1 ( z )
···
F k ( z )
f 1 [ α 1 ]
···
f k [ α k ]
A (
c
) f 1 [ c 1 ]
···
f k [ c k ]
α C n,k
c
C n,k
x
|
f [ α ]
x
|
A ( λ ) B ( λ ) f [ λ ]
2 . a,d
F ( z ) x
α
|
λ
|
α
C n
λ
P n
3 . b,c
G ( F ( z ))
f
g [
π
]
C ( λ ) f
g [ λ ]
π P n
λ
P n
H ( z )= G ( F ( z )) F ( z )
( 1) | π | f [ π ]
( 1) |λ| C ( λ ) f [ λ ]
4 . b
F 1 ( z )
R (
π
)
DP n− 1
λ DP n− 1
F ( z )=
1
G ( F ( z ))
1) | π | f
1) |λ| C ( λ ) f
5 . b,c
G ( F 1 ( z ))
(
g [
π
]
(
g [ λ ]
λ
DV n− 1
R ( π ) DV n− 1
6 . ( H, z )=0 (See[4])
7 . b H = G ( w ) ,z = F ( w ) (see no. 5)
8 . ( z )= G ( F 1 ( z ) ,...F q ( z )) (requires q -chromatic partitions)
........................................................................................
( a ) F (0) = 1, ( b ) F (0) = 0 ,F (0) = 1, ( c ) f g [ α ] = f [ α ] g [ |α| ], ( d ) x ∈ R
c
an integer composition
λ
an integer partition
C n
a set of integer compositions
P n
a set of integer partitions
α
a set composition
π
a set partition
C n
a set of set compositions
P n
a set of set partitions
n
sum of parts
k
number of parts
j
number of null parts
null parts ok
7
The Lagrange Inversion Formula
The following is required for some of our proofs.
Lagrange Inversion Formula [2] [6]. Suppose F (0) = 0, and F 0 ( z )= F ( z ) /z .
Then
z n
n !
F 1 ( z )= z n− 1
( n
F 0 ( z ) −n .
1)!
 
Search WWH ::




Custom Search