binomial generating functions

7/28/2019 Binomial Generating Functions

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Binomial Coefficients and Generating FunctionsITT9130 Konkreetne Matemaatika

Chapter Five

Basic Identities

Basic Practice

Tricks of the Trade

Generating Functions

Hypergeometric Functions

Hypergeometric Transformations

Partial Hypergeometric Sums


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Contents

1 Binomial coefficients

2 Generating FunctionsOperations on Generating FunctionsBuilding Generating Functions that Count

Identities in Pascals Triangle


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Next section





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Binomial theorem

Theorem 1

(a+b)n =n

k=0

n

k

akbnk,

for any integer n 0.Proof. Expanding (a + b)n = (a + b)(a + b) (a + b) yields the sum of the

2n products of the form e1e2 en, where each ei is a or b. Theseterms are composed by selecting from each factor (a + b) either a orb. For example, if we select a k times, then we must choose b n ktimes. So, we can rearrange the sum as

(a + b)n =n

k=0

Ckakbnk,

where the coefficient Ck is the number how many possibilities are toselect k elements (k factors (a + b)) from a set of n elements (fromthe production of n factors (a + b)(a + b) (a + b)).That is why the coefficient C

kis called "(from) n choose k" and

denoted byn

k

. Q.E.D.


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Binomial coefficients and combinations

Theorem 2

The number of k-subsets of an n-set isn

k

=

n!

k!(n k)!

Proof. At first, determine the number of k-element sequences: there are nchoices for the first element of the sequence; for each, there are n 1choices for the second; and so on, until there are n k+ 1 choicesfor the k-th. This gives n(n 1)...(n k+ 1) = nk choices in all. Andsince each k-element subset has exactly k! different orderings, thisnumber of sequences counts each subset exactly k! times. To getour answer, divide by k!:

n

k

=

nk

k!=

n!

k!(n k)!

Q.E.D.

Some other notations used for the "n choose k" in literature:

Cnk, C(n, k), nCk,

n

Ck.


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Binomial coefficients and combinations

Theorem 2

The number of k-subsets of an n-set isn

k

=

n!

k!(n k)!

Proof. At first, determine the number of k-element sequences: there are nchoices for the first element of the sequence; for each, there are n 1choices for the second; and so on, until there are n k+ 1 choicesfor the k-th. This gives n(n 1)...(n k+ 1) = nk choices in all. Andsince each k-element subset has exactly k! different orderings, thisnumber of sequences counts each subset exactly k! times. To getour answer, divide by k!:

n

k

=

nk

k!=

n!

k!(n k)!

Q.E.D.

Some other notations used for the "n choose k" in literature:

Cnk, C(n, k), nCk,

n

Ck.


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Properties of Binomial Coefficients

1 nk=0

nk

= 2n (number of all subsets ofn-set);

2 nk=0(1)

k

nk

= 0 (number of subsets with odd cardinality is

equel to the number of sets with even cardinality).

Testus. Take a = b= 1 in the binomial theorem:

n

k=0

n

k

=

n

k=0

n

k

1k1nk = (1 + 1)n = 2n

Take a = 1 ja b= 1:

n

k=0

(1)kn

k

=

n

k=0

n

k

(1)k1nk = (1 + 1)n = 0

Q.E.D.


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Another Property

3 nk = nnk ;Proof. Direct conclusion from the Theorem 2:

n

k =n!

k!(nk)!=

n

nkQ.E.D.


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Yet Another Proerty

4 Ifn,k> 0, then

n1k1

+

n1k

=

nk

.

Proof.

Note that 1

n k+

1

k=

k+nk

k(nk)=

n

k(nk).

Multiplying this by (n 1)! and dividing by (k 1)!(n k 1)!, we get

(n 1)!

(k 1)!(n k)(n k 1)!+

(n 1)!

k(k 1)!(n k 1)!=

n(n 1)!

k(k 1)!(n k)(n k 1)!

This can be rewritten after some simplifying transformations as:

(n 1)!

(k 1)!(n k)!+

(n 1)!

k!(n k 1)!=

n!

k!(n k)!

Q.E.D.


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Pascals Triangle

n

n0

n1

n2

n3

n4

n5

n6

0 1

1 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 1

6 1 6 15 20 15 6 1 Blaise Pascal(16231662)


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Pascals Triangle

11

11

111 6

56 15

4

3

4 610 1020 15

21

2 13 1

4 15 16 1

Blaise Pascal

(16231662)

Pascal Triangle is symmetric with respect to the vertical linethrough its apex.

Every number is the sum of the two numbers immediately above it.


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Pascals Triangle

11

11

111 6

56 15

4

3

4 610 1020 15

21

2 13 1

4 15 16 1

Blaise Pascal

(16231662)

Pascal Triangle is symmetric with respect to the vertical linethrough its apex.

Every number is the sum of the two numbers immediately above it.


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Next section





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Generating Functions

DefinitionThe generating function for the infinite sequence g0,g1,g2, . . . isthe power series

G(x) = g0 +g1x+g2x2 +g3x

3 + =

n=0gnx

n

Ssome simple examples

0, 0, 0, 0, . . . 0 + 0x+ 0x2 + 0x3 + = 0

1, 0, 0, 0, . . . 1 + 0x+ 0x2 + 0x3 + = 1

2, 3, 1, 0, . . . 2 + 3x+ 1x2 + 0x3 + = 2 + 3x+ 1x2


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More examples (1)

1, 1, 1, 1, . . . 1 +x+x2 +x3 + = 11x

S = 1 + x+ x2 + x3 +

xS = x+x2 +x3 +

Subtract the equations:

(1 x)S= 1 ehk S=1

1 x

NB! This formula converges only for 1 < x< 1.

Actually, we dont worry about convergence issues.


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More examples (1)

1, 1, 1, 1, . . . 1 +x+x2 +x3 + = 11x

S = 1 + x+ x2 + x3 +

xS = x+x2 +x3 +

Subtract the equations:

(1 x)S= 1 ehk S=1

1 x

NB! This formula converges only for 1 < x< 1.

Actually, we dont worry about convergence issues.


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More examples (2)

a,ab,ab2,ab3, . . . a+abx+ab2x2 +ab3x3 + = a1bx

Like in the previous example:

S = a+abx+ab2x2 +ab3x3 +

bxS = abx+ab2x2 +ab3x3 +

Subtract and get:

(1 bx)S= a ehk S= a1 bx


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More examples (3)

Taking in the last example a = 0, 5 and b= 1 yields

0, 5 + 0, 5x+ 0, 5x2 + 0, 5x3 + =0, 5

1 x(1)

Taking a = 0, 5 and b= 1, gives

0, 5 0, 5x+ 0, 5x2 0, 5x3 + =0, 5

1 + x(2)

Adding equations (1) and (2), we get the generating function of the

sequence 1, 0, 1, 0, 1, 0, . . .:

1 +x2 +x4 +x6 + =0, 5

1 x+

0, 5

1 +x=

1

(1 x)(1 +x)=

1

1 x2


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More examples (3)


0, 5 + 0, 5x+ 0, 5x2 + 0, 5x3 + =0, 5

1 x(1)


0, 5 0, 5x+ 0, 5x2 0, 5x3 + =0, 5

1 + x(2)


sequence 1, 0, 1, 0, 1, 0, . . .:

1 +x2 +x4 +x6 + =0, 5

1 x+

0, 5

1 +x=

1

(1 x)(1 +x)=

1

1 x2


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More examples (3)


0, 5 + 0, 5x+ 0, 5x2 + 0, 5x3 + =0, 5

1 x(1)


0, 5 0, 5x+ 0, 5x2 0, 5x3 + =0, 5

1 + x(2)


sequence 1, 0, 1, 0, 1, 0, . . .:

1 +x2 +x4 +x6 + =0, 5

1 x+

0, 5

1 +x=

1

(1 x)(1 +x)=

1

1 x2


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Next subsection





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Operations on Generating Functions

1. Scaling

Iff0, f1, f2, . . . F(x),

then

cf0,cf1,cf2, . . . cF(x),for any c R.

Proof.

cf0,cf1,cf2, . . . cf0 + cf1x+ cf2x2

+ == c(f0 + f1x+ f2x

2 + ) =

= cF(x)

Q.E.D.


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Operations on Generating Functions (2)

2. AdditionIf f0, f1, f2, . . . F(x) and g0,g1,g2, . . . G(x), then

f0 +g0, f1 +g1, f2 +g2, . . . F(x) +G(x).

Proof.

f0 +g0, f1 +g1, f2 +g2, . . .

n=0

(fn +gn)xn =

= n=0

fnxn+

n=0

gnxn =

= F(x) +G(x)

Q.E.D.


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O G F ( )


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4. Differentiation

If f0, f1, f2, . . . F(x), then

f1, 2f2, 3f3, . . . F(x).

Proof.

f1, 2f2, 3f3, . . . f1 + 2f2x+ 3f3x2 + . . . =

=

d

dx(f0

+f1x

+f2x

2

+f3x

3

+ ) =

=d

dxF(x)

Q.E.D.

O i G i F i (4)


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4. Differentiation

If f0, f1, f2, . . . F(x), then

f1, 2f2, 3f3, . . . F(x).

Example

1, 1, 1, . . . 11x

1, 2, 3, . . . ddx

1

1x =1

(1x)2

O i G i F i (5)


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5. IntegrationIf f0, f1, f2, . . . F(x), then

0, f0,1

2f1,

1

3f2,

1

4f3, . . .

x0

F(z)dz.

Proof.

0, f0,1

2f1,

1

3f2,

1

4f3, . . . f0x+

1

2f1x

2 +1

3f2x

3 +1

4f3x

4 + . . . =

= x0

(f0 + f1z+ f2z2 + f3z3 + )dz=

=

x0

F(z)dz

Q.E.D.

O ti G ti F ti (5)


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5. Integration

If f0, f1, f2, . . . F(x), then

0,f0

,

1

2f1

,

1

3f2

,

1

4f3

, . . . x

0

F(z

)dz

.

Example

1, 1, 1, 1 . . . 1

1x

0, 1, 12

, 13

, . . . x0

1

1zdz= ln(1 x) = ln1

(1x)

O ti G ti F ti (6)


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6. Convolution (product)

If f0, f1, f2, . . . F(x) and g0, g1, g2, . . . G(x), then

h0, h1, h2, . . . F(x) G(x),

where hn = f0gn + f1gn1 + f2gn2 + + fng0.

O e ations on Gene atin F nctions (6)


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h0, h1, h2, . . . F(x) G(x),


Proof.

F(x) G(x) = (f0 + f1x+ f2x2 + . . .)(g0 + g1x+ g2x

2 + . . .)

= f0g0 + (f0g1 + f1g0)x+ (f0g2 + f1g1 + f2g0)x2 + . . .

Q.E.D.



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h0, h1, h2, . . . F(x) G(x),


Proof.

F(x) G(x) = (f0 + f1x+ f2x2 + . . .)(g0 + g1x+ g2x

2 + . . .)

= f0g0 + (f0g1 + f1g0)x+ (f0g2 + f1g1 + f2g0)x2 + . . .

Q.E.D.Notice that all terms involving the same power of xlie on a /sloped diagonal:

g0x0 g1x

1 g2x2 g3x

3 . . .a0x

0 f0g0x0 f0g1x

1 f0g2x2 f0g3x

3 . . .f1x

1 f1g0x1 f1g1x

2 f1g2x3 . . .

f2x2 f2g0x

2 f2x3 . . .

f3x3 f3g0x

3 . . .... . . .



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h0, h1, h2, . . . F(x) G(x),


Example

1, 1, 1, 1 . . . 0, 1,1

2,

1

3, . . . =

= 10, 10 + 11, 10 + 11 + 11

2

, 10 + 11 + 11

2

+ 11

3

,

= 0, 1, 1 +1

2, 1 +

1

2+

1

3,

= 0, H1, H2, H3,

Hence

k0 Hkxk

=

1

(1 x) ln

1

(1 x) .



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Example

1, 1, 1, 1, . . . 1

1 x

1, 2, 3, 4, . . . ddx

11 x

= 1(1 x)2

0, 1, 2, 3, . . . x1

(1 x)2=

x

(1 x)2

1, 4, 9, 16, . . . ddx

x(1 x)2 = 1 +

x(1 x)3

0, 1, 4, 9, . . . x1 + x

(1 x)3=

x(1 +x)

(1 x)3

Counting with Generating Functions


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Choosing k-subset from n-set

Binomial theorem yields:n0

,

n

1

,

n

2

, . . . ,

n

n

, 0, 0, 0, . . .

n0+n1x+n2x2 . . . ,nnxn = (1 + x)n

Thus, the coefficient ofxk

in (1 + x)n

is the number of ways tochoose k distinct items from a set of size n.

For example, the coefficient of x2 is , the number of ways to choose2 items from a set with n elements.

Similarly, the coefficient of xn+1 is the number of ways to choosen+ 1 items from a n-set, which is zero.



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Choosing k-subset from n-set

Binomial theorem yields:n0

,

n

1

,

n

2

, . . . ,

n

n

, 0, 0, 0, . . .

n0+n1x+n2x2 . . . ,nnxn = (1 + x)n

Thus, the coefficient ofxk

in (1 + x)n

is the number of ways tochoose k distinct items from a set of size n.

For example, the coefficient of x2 is , the number of ways to choose2 items from a set with n elements.

Similarly, the coefficient of xn+1 is the number of ways to choosen+ 1 items from a n-set, which is zero.

Next subsection


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Next subsection




Building Generating Functions that Count


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The generating function for the number of ways to choose n elements from thisset A is the function A(x) thats expansion into power series has coefficient

ai = 1 of xi iff i can be selected into the subset, otherwise ai = 0

.

Examples of GF selecting items from a set A:

If any natural number can be selected:

A(x) = 1 + x+ x2

+ x3

+ =

1

1 x



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.



A(x) = 1 + x+ x2 + x3 + =1

1 x

If any even number can be selected:

A(x) = 1 + x2 + x4 + x6 + =1

1 x2



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u g G g u C u

The generating function for the number of ways to choose n elements from thisset A is the function A(x) thats expansion into power series has coefficientai = 1 of x

i iff i can be selected into the subset, otherwise ai = 0

.



A(x) = 1 + x+ x2 + x3 + =1

1 x

If any even number can be selected:

A(x) = 1 + x2 + x4 + x6 + =1

1 x2

If any positive even number can be selected:

A(x) = x2

+ x4

+ x6

+ =x

1 x2



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g g



.



A(x) = 1 + x+ x2 + x3 + =1

1 x

If any number multiple of 5 can be selected:

A(x) = 1 + x5 + x10 + x15 + =1

1 x5



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g g

The generating function for the number of ways to choose n elements from thisset A is the function A(x) thats expansion into power series has coefficientai = 1 of x

i iff i can be selected into the subset, otherwise ai = 0

.



A(x) = 1 + x+ x2 + x3 + =1

1 x

If at most four can be selected:

A(x) = 1 + x+ x2 + x3 + x4 =1 x5

1 x

If zero or one number can be selected:

A(x) = 1 + x

Counting elements of two sets


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g

Convolution Rule

Let A(x) be the generating function for selecting items from set A, andlet B(x), be the generating function for selecting items from set B.

IfA and B are disjoint, then the generating function for selecting items from

the union AB is the product A(x) B(x).

Proof. To count the number of ways to select n items from AB we have to select jitems from A and n j items from B, where j {0, 1, 2, . . . ,n}.Summing over all the possible values of j gives a total of

a0bn + a1bn1 + a2bn2 + + anb0

ways to select n items from AB. This is precisely the coefficient of xn in the seriesfor A(x) B(x) Q.E.D.

How many positive integer solutions does the equationx + x n have?


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x1+ x2 = n have?

We accept any natural number can be solution for x1, i.e generating function for

selection a value for this variable is A(x) = 1 + x+ x2

+ x3

+ =

1

1x;same for variable x2;all pairs of numbers for (x1, x2) are selected by the function H(x) = A(x) A(x);numer of solutions is the coefficient of xn.

H(x) = (1 + x+ x2 + x3 + )(1 + x+ x2 + x3 + ) =

= 1 (1 + x+ x2 + x3 + ) + x(1 + x+ x2 + x3 + ) +

+x2(1 + x+ x2 + x3 + ) + x3(1 + x+ x2 + x3 + ) + =

= 1 + 2x+ 3x2 + 4x3 + + (n + 1)xn + =1

(1 x)2

Indeed, this equation has n + 1 solutions:

0 + n = n

1 + (n 1) = n

2 + (n 2) = n

n + 0 = n

How many positive integer solutions does the equationx + x n have?


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x1+ x2 = n have?


selection a value for this variable is A(x) = 1 + x+ x2

+ x3

+ =

1


H(x) = (1 + x+ x2 + x3 + )(1 + x+ x2 + x3 + ) =

= 1 (1 + x+ x2 + x3 + ) + x(1 + x+ x2 + x3 + ) +

+x2(1 + x+ x2 + x3 + ) + x3(1 + x+ x2 + x3 + ) + =

= 1 + 2x+ 3x2 + 4x3 + + (n + 1)xn + =1

(1 x)2


0 + n = n

1 + (n 1) = n

2 + (n 2) = n

n + 0 = n

How many positive integer solutions does the equationx1+ x2 = n have?


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x1+ x2 = n have?


selection a value for this variable isA

(x

) = 1 +x

+x2

+x3

+ =

1


H(x) = (1 + x+ x2 + x3 + )(1 + x+ x2 + x3 + ) =

= 1 (1 + x+ x2 + x3 + ) + x(1 + x+ x2 + x3 + ) +

+x2(1 + x+ x2 + x3 + ) + x3(1 + x+ x2 + x3 + ) + =

= 1 + 2x+ 3x2 + 4x3 + + (n + 1)xn + =1

(1 x)2


0 + n = n

1 + (n 1) = n

2 + (n 2) = n

n + 0 = n

Number of solutions of the equation x1+x2+ + xk= n


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Theorem

The number of ways to distribute n identical objectsinto k bins is

n+k1

n

.

Proof.

The number of ways to distribute n objects equals to the number of solutions of

x1 + x2 + + xk = n that is coefficient of xn of the generating function

G(x) = 1/(1 x)k = (1 x)k.

For recollection: Maclaurin series (a Taylor seriesexpansion of a function about 0):

f(x) = f(0) + f(0)x+f(0)

2!x2 +

f(0)

3!x3 + +

f(n)(0)

n!xn +

.

Number of solutions of the equation x1+x2+ + xk= n


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Theorem

The number of ways to distribute n identical objectsinto k bins is

n+k1

n

.

Proof.

The number of ways to distribute n objects equals to the number of solutions of

x1 + x2 + + xk = n that is coefficient of xn of the generating function

G(x) = 1/(1 x)k = (1 x)k.

For recollection: Maclaurin series (a Taylor seriesexpansion of a function about 0):

f(x) = f(0) + f(0)x+f(0)

2!x2 +

f(0)

3!x3 + +

f(n)(0)

n!xn +

.

Equation x1+x2+ +xk= n (2)


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Continuation of the proof ...

lets differentiate G(x) = (1 x)k:

G(x) = k(1 x)(k+1)

G(x) = k(k+ 1)(1 x)(k+2)

G(x) = k(k+ 1)(k+ 2)(1 x)(k+3)

G(n)(x) = k(k+ 1) (k+ n 1)(1 x)(k+n)

Coefficient of xn can be evaluated as:

G(n)(0)

n!=

k(k+ 1) (k+ n 1)

n!=

= (k+ n 1)!(k 1)!n!

=

=

n + k 1

n

Q.E.D.

Distribute n objects into k bins so that there is at least oneobject in each bin


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object in each bin

Theorem

The number of positive solutions of the equationx1 +x2 + + xk = n is

n1k1

.

Idea of the proof. Possible number of objects in a single bin (xi > 0)could be generated by the function

C(x) = x+ x2 +x3 + = x(1 + x+ x2 + ) =x

1 x

Similarly to the previous theorem, the number distributions is thecoefficient ofxn of the generating function

H(X) = Ck(x) =xk

(1 x)k

Q.E.D.

Example: 100 Euro


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How many ways 100 Euro can be changed using smaller banknotes?

Generating functions for selecting banknotes of 5, 10, 20 or 50 Euros:

A(x) = x0 + x5 + x10 + x15 + =1

1 x5

B(x) = x0 + x10 + x20 + x30 + = 11 x10

C(x) = x0 + x20 + x40 + x60 + =1

1 x20

D(x) = x0 + x50 + x100 + x150 + =1

1 x50

Generating function for the task

P(x) = A(x)B(x)C(x)D(x) =1

(1 x5)(1 x10)(1 x20)(1 x50)

Example: 100 Euro


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A(x) = x0 + x5 + x10 + x15 + =1

1 x5

B(x) = x0 + x10 + x20 + x30 + = 11 x10

C(x) = x0 + x20 + x40 + x60 + =1

1 x20

D(x) = x0 + x50 + x100 + x150 + =1

1 x50


P(x) = A(x)B(x)C(x)D(x) =1

(1 x5)(1 x10)(1 x20)(1 x50)

Example: 100 Euro


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A(x) = x0 + x5 + x10 + x15 + =1

1 x5

B(x) = x0 + x10 + x20 + x30 + = 11 x10

C(x) = x0 + x20 + x40 + x60 + =1

1 x20

D(x) = x0 + x50 + x100 + x150 + =1

1 x50


P(x) = A(x)B(x)C(x)D(x) =1

(1 x5)(1 x10)(1 x20)(1 x50)

Example: 100 Euro (2)


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1. Observation:

(1 x5)(1 + x5 + + x45 + 2x50 + 2x55 + + 2x95 + 3x100 + 3x105 + + 3x145 + 4x150 + ) =

1 + x5 + + x45 + 2x50 + 2x55 + + 2x95 + 3x100 + 3x105 + + 3x145 + 4x150

x5

x45

x50

2x55

2x95

2x100

3x105

3x145

3x150

4x155

=

= 1 + x50 + x100 + x150 + x200 + =1

1 x50

Thus:

F(x) = A(x)D(x) =1

(1 x5)(1 x50) = k0 k

10+ 1x5k = k0 fkx5k



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1. Observation:

(1 x5)(1 + x5 + + x45 + 2x50 + 2x55 + + 2x95 + 3x100 + 3x105 + + 3x145 + 4x150 + ) =

1 + x5 + + x45 + 2x50 + 2x55 + + 2x95 + 3x100 + 3x105 + + 3x145 + 4x150

x5 x45 x50 2x55 2x95 2x100 3x105 3x145 3x150 4x155 =

= 1 + x50 + x100 + x150 + x200 + =1

1 x50

Thus:

F(x) = A(x)D(x) =1

(1 x5)(1 x50)=

k0

k

10

+ 1

x5k =

k0

fkx5k

2. Similarly:

G(x) = B(x)C(x) =1

(1 x10)(1 x20)=

k0

k

2

+ 1

x10k =

k0

gkx10k



55/63

Convolution:P(x) = F(x)G(x) =

k0

cnx5n

The coefficient of x100 equals to

c20 = f0g10 + f2g9 + f4g8 + + f20g0

=10

k=0

f2kg10k

=10

k=0

2k

10

+ 1

10 k

2

+ 1

=

10

k=0

k+ 55

12 k2

= 1(6 + 5 + 5 + 4 + 4) + 2(3 + 3 + 2 + 2 + 1) + 3 1 =

= 24 + 22 + 3 = 49

Next subsection


56/63


2 Generating FunctionsOperations on Generating FunctionsBuilding Generating Functions that CountIdentities in Pascals Triangle

Vandermondes identity


57/63

Binomial theorem provides the generating function:

(1 + x)m = k0

m

k

xk and (1 + x)n =

k0

n

k

xk

Convolution yields

k0

m + n

k

xk = (1 + x)m+n

= (1 + x)m (1 + x)n

=

k0

mk

xk

k0

nk

xk

= k0

k

j=0

m

j

n

kj

xk

Vandermondes identity (2)


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Equating coefficients of zk on both sides of this equation gives:

Vandermondes convolution:

k

j=0

m

j

n

kj

=

m + n

k

Convolution yields

1

1

1

1

1

1

1

1

1

1 9

8

7

6

5

4

3

2

1

1

3

6

10

15

21

28

36 84

56

35

20

10

4

1

1

5

15

35

70

126 126

56

21

6

1

1

7

28

84 36

8

1

1

9 1

9

2

=

3

0

6

2

+

3

1

6

1

+

3

2

6

0



59/63

Equating coefficients of zk on both sides of this equation gives:

Vandermondes convolution:

k

j=0

m

j

n

kj

=

m + n

k

Convolution yields

1

1

1

1

1

1

1

1

1

1 9

8

7

6

5

4

3

2

1

1

3

6

10

15

21

28

36 84

56

35

20

10

4

1

1

5

15

35

70

126 126

56

21

6

1

1

7

28

84 36

8

1

1

9 1

9

2

=

3

0

6

2

+

3

1

6

1

+

3

2

6

0



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Special case m = n = k yields:

2n

n

=

n

j=0

n

j

n

n j

=

j0

n

j

2

Example: 402

+412

+ 422

+432

+ 442

= 841

1

1

1

11

1

1

1 8

7

6

54

3

2

1

1

3

610

15

21

28 56

35

20

104

1

15

15

35

70 56

21

6

1

1

7

28 8

1

1

1

1

1

1

1 16

9

4

1

1

9

36 16

1

1

1 + 16 + 36 + 16 + 1 = 70



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Special case m = n = k yields:

2n

n

=

n

j=0

n

j

n

n j

=

j0

n

j

2

Example: 402

+412

+ 422

+432

+ 442

= 841

1

1

1

11

1

1

1 8

7

6

54

3

2

1

1

3

610

15

21

28 56

35

20

104

1

15

15

35

70 56

21

6

1

1

7

28 8

1

1

1

1

1

1

1 16

9

4

1

1

9

36 16

1

1

1 + 16 + 36 + 16 + 1 = 70

Sequence

m0

, 0,

m1

, 0,m2

, 0,

m3

, 0,m4

, 0, . . . ,


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Lets take sequences

m

0

,

m

1

,

m

2

, . . . ,

m

n

, . . . F(X) = (1 + x)m

and

m0, m1,m2, m3, . . . , (1)nmn, . . . G(X) = (1 x)mThen the convolution corresponds to the function (1 x)m(1 + x)m = (1 x2)m thatgives the identity of binomial coefficients:

n

j=0

m

j

m

n j

(1)j = (1)n/2

m

n/2

[n is even] .

Some other useful sequences


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For any n 0:

1(1 x)n+1

= k0

n 1k

(x)k =

k0

n + kn

xk

Here the identity negative coefficients

nk

= (1)k

nk1

k

is used.

For any n 0, using right-shift :

xn

(1 x)n+1 = k0knxk

Exponential series

ex = 1 + x+1

2!x2 +

1

3!x3 + =

k0

1

k!xk

Logarithmic series

ln (1 + x) = x1

2x2 +

1

3x3 =

k1

(1)k+1

kxk

1

ln(1 + x)= x+

1

2x2 +

1

3x3 + =

k1

1

kxk

binomial generating functions

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