Download Exponential Generating Functions COS 341 Discrete Mathematics 1

COS 341 Discrete Mathematics
Exponential Generating
Functions
1
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and practice problems
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•
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Tutoring:
Seniors: See Dean Richard Williams (408 West College, 8-5520)
Juniors: See Dean Frank Ordiway (404 West College, 8-1998)
Sophomores: See Director of Studies in your home college
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Ordinary Generating Functions
(a0 , a1 , a2 ,…) : sequence of real numbers
Ordinary
Generating Function of this sequence is
∧
∞
i
the power series a( x) = ∑ i=0 ai ⋅ x
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Exponential Generating Functions
(a0 , a1 , a2 ,…) : sequence of real numbers
Exponential Generating function of this
sequence is the power series
ai i
a ( x ) = ∑ i =0 ⋅ x
i!
∞
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Exponential generating function
examples
What is the generating function for the sequence (1,1,1,1,…) ?
∞
1 i
x x 2 x3
x = 1+ + + +
∑
1! 2! 3!
i =0 i !
= ex
What is the generating function for the sequence (1, 2, 4,8,…) ?
∞
2i i
2 x (2 x) 2 (2 x)3
+
+
x = 1+ +
∑
1!
2!
3!
i =0 i !
= e2 x
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Operations on exponential generating
functions
• Addition
(a0 + b0 , a1 + b1 ,…) has generating function a ( x) + b( x)
• Multiplication by fixed real number
(αa0 , αa1 ,…) has generating function αa ( x)
• Shifting the sequence to the right
(0,… 0 , a0 , a1 ,…) has generating function x n a ( x)
n×
• Shifting to the left
k −1
(ak , ak +1 ,…) has generating function
a( x) − ∑ i=0 ai ⋅ xi
xn
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• Substituting αx for x
(a0 , αa1 , α 2 a2 …) has generating function a (αx)
• Substitute xn for x
(a0 , 0,… 0, a1 , 0,… 0, a2 …) has generating function a ( x n )
n−1×
n−1×
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• Differentiation
d
(a1 , 2a2 ,3a3 …) has generating function
a ( x) (or a '( x))
dx
• Integration
x
(0,a0 , 12 a1 , 13 a2 …) has generating function
∫
f (t )dt
0
• Multiplication of generating functions
(
∞
∑ n=0 an ⋅ x
n
)(
∞
) (
∑ n=0 bn ⋅ x =
n
∞
n
c
⋅
x
∑ n=0 n
)
cn = ∑ k =0 ak ⋅ bn−k
n
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Differentiation
a ( x) is the exponential generating function for (a0 , a1 , a2 ,…)
∞
ai i
a( x) = ∑ ⋅ x
i =0 i !
∞
∞
ai
ai
d
i−1
⋅ xi−1
a ( x) = ∑ ⋅ i ⋅ x = ∑
dx
i =0 i !
i =1 (i −1)!
d
a ( x) is the exponential generating function for (a1 , a2 , a3 ,…)
dx
Differentiation is equivalent to shifting the sequence to the left
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Integration
a ( x) is the exponential generating function for (a0 , a1 , a2 ,…)
∞
ai i
a( x) = ∑ ⋅ x
i =0 i !
x
∫
x
0
x
i +1
∞
∞
a
x
ai
ai
i
i
=∑
⋅ x i+1
a (t )dt = ∑ ⋅ ∫ t dt = ∑ ⋅
i =0 i ! (i + 1)
i =0 i ! 0
i =0 (i + 1)!
∞
∫ a(t )dt
is the exponential generating function for (0, a0 , a1 , a2 ,…)
0
Integration is equivalent to shifting the sequence to the right
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Multiplication
Ordinary
(
∞
∑ n=0 an ⋅ x
n
)(
∞
) (
∑ n=0 bn ⋅ x =
n
∞
n
c
x
⋅
∑ n=0 n
)
cn = ∑ k =0 ak ⋅ bn−k
n
Exponential
 ∞ an n 
  ∞ cn n 
∞ bn
n

∑ n=0 ⋅ x ∑ n=0 ⋅ x  = ∑ n=0 ⋅ x 

n ! 
n!  
n! 
n
cn
ak bn−k
= ∑ k =0 ⋅
n!
k ! (n − k )!
n  
n
n!

cn = ∑
ak ⋅ bn−k = ∑   ak ⋅ bn−k
 
k =0 k !( n − k )!
k =0  k 
n
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Implications of product rule
C ( x) = A( x) B( x)
Ordinary
cn = ∑ k =0 ak ⋅ bn−k
n
Useful for counting with
indistinguishable objects
Exponential
n
cn = ∑   ak ⋅ bn−k
 
k =0  k 
n
Useful for counting with
ordered objects
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Interpretation of Multiplication: Product Rule
Given arrangements of type A and type B, define
arrangements of type C for n labeled objects as follows:
Divide the group of n labeled objects into two groups,
the First group and the Second group;
arrange the First group by an arrangement of type A
and the Second group by an arrangement of type B.
an : number of arrangements of type A for n objects
bn : number of arrangements of type B for n objects
cn : number of arrangements of type C for n objects
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Interpretation of Multiplication: Product Rule
an : number of arrangements of type A for n people
bn : number of arrangements of type B for n people
cn : number of arrangements of type C for n people
n
cn = ∑   ak ⋅ bn−k
 
k =0  k 
n
an : exponential generating function A( x)
bn : exponential generating function B( x)
cn : exponential generating function C ( x)
C ( x) = A( x) B( x)
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A( x) : exponential generating function for arrangements of type A
a0 = 0 : no empty group allowed
Define arrangements of type D for n labeled objects as follows:
Divide the group of n labeled objects into k groups,
the First group, Second group,… , kth group (k = 0,1, 2,… )
arrange each group by an arrangement of type A .
D( x) : exponential generating function for arrangements of type D
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Dk ( x) : exponential generating function for arrangements of type D
with exactly k groups
Dk ( x) = A( x) k
∞
D( x) = ∑ Dk ( x)
k =0
∞
= ∑ A( x) k
k =0
1
=
1− A( x)
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A( x) : exponential generating function for arrangements of type A
a0 = 0 : no empty group allowed
Define arrangements of type E for n labeled objects as follows:
Divide the group of n labeled objects into k groups,
and arrange each group by an arrangement of type A
(the groups are not numbered) .
E ( x) : exponential generating function for arrangements of type E
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Ek ( x) : exponential generating function for arrangements of type E
with exactly k groups
A( x) k
Ek ( x) =
k!
∞
E ( x) = ∑ Ek ( x)
k =0
∞
=∑
k =0
A( x) k
k!
= e A( x )
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Example
How many ways can n people be arranged into pairs,
(the pairs are not numbered) ?
A( x) : exponential generating function for a single pair
a2 = 1, ai = 0 for i ≠ 2
x2
A( x) =
2
E ( x) : exponential generating function for arranging
n people into pairs
E ( x) = e
n/2
2

e
x
x n term = n x n =  
 2 
n!
1
(n / 2)!
x2
2
n!
en = n / 2
2 (n / 2)!
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Derangements (or Hatcheck lady revisited)
d n : number of permutations on n objects without a fixed point
D ( x) : exponential generating function for number of derangements
A permutation on [n] can be constructed by picking
a subset K of [n], constructing a derangement of K and
fixing the elements of [n] - K .
Every permutation of [n] arises exactly once this way.
∞
n! n
1
EGF for all permutations = ∑ x =
1− x
n=0 n !
∞
1 n
x
EGF for permutations with all elements fixed = ∑ x = e
n=0 n !
1
= D( x) ⋅ e x
1− x
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Derangements
1
= D( x) ⋅ e x
1− x
−x 1
D( x) = e
1− x
n  ∞
∞

x
n
n

∑ x 
= ∑ (−1)
 n=0
n !  n=0 
k
n

dn
(
−
1
)

= coefficient of x n = ∑
 k =0 k ! 
n!
 n (−1) k 

d n = n ! ∑
 k =0 k ! 
Different proof in Matousek 10.2, problem 17
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Example
How many sequences of n letters can be formed from
A, B, and C such that the number of A's is odd and the
number of B's is odd ?
x
xn
e
− e−x
EGF for A's = ∑
=
2
n odd n !
e x − e−x
EGF for B's =
2
∞
xn
EGF for C's = ∑ = e x
n=0 n !
− x 2
e −e

required EGF = 

2
x
 e x =

e3 x − 2e x + e−x
4
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Example
How many sequences of n letters can be formed from
A, B, and C such that the number of A's is odd and the
number of B's is odd ?
required EGF =
e3 x − 2e x + e−x
4
n
n

1
3
(
1)
2
−
+
−

coefficient of x n = 
n !
4

3n − 2 + (−1) n
required number =
4
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