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Counting III: Pascal’s Triangle, Polynomials, and Vector ProgramsPowerPoint Presentation

Counting III: Pascal’s Triangle, Polynomials, and Vector Programs

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Counting III: Pascal’s Triangle, Polynomials, and Vector Programs

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Counting III: Pascal’s Triangle, Polynomials, and Vector Programs

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- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

- X1

- X2

- +

- +

- X3

Counting III: Pascal’s Triangle, Polynomials, and Vector Programs

Xn+1 - 1

1 + X1 + X2 + X 3 + … + Xn-1 + Xn =

X- 1

1

1 + X1 + X2 + X 3 + … + Xn + ….. =

1 - X

1

1 + X1 + X2 + X 3 + … + Xn + ….. =

1 - X

- (X-1) ( 1 + X1 + X2 + X 3 + … + Xn + … )
- = X1 + X2 + X 3 + … + Xn + Xn+1 + ….
- - 1 - X1 - X2 - X 3 - … - Xn-1 – Xn - Xn+1 - …
- = 1

1

1 + X1 + X2 + X 3 + … + Xn + ….. =

1 - X

- 1 + x + …..
- ____________________
- 1-x | 1
- -[1 – x]
- _____
- x
- -[x – x2 ]
- ______
- x2
- -....

1

1 + aX1 + a2X2 + a3X 3 + … + anXn + ….. =

1 - aX

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1

(1 – aX)(1-bX)

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1 + c1X1 + .. + ck Xk + …

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1 + c1X1 + .. + ck Xk + …

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1 + c1X1 + .. + ck Xk + …

an+1 – bn+1

a- b

a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0 =

- (a-b) (a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0)
- = a1bn +… ai+1bn-i… + anb1 + an+1b0
- - a0bn+1 – a1bn… ai+1bn-i… - an-1b2 - anb1
- = - bn+1 + an+1
- = an+1 – bn+1

an+1 – bn+1

a- b

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1 + c1X1 + .. + ck Xk + …

an+1 – bn+1

a- b

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1

=

(1 – aX)(1-bX)

n=0..1

Xn

=

n=0..1

or

Xn

(n+1)an

when a=b

Previously, we saw that

Polynomials Count!

What is the coefficient of BA3N2 in the expansion of

(B + A + N)6?

The number of ways to rearrange the letters in the word BANANA.

1

1

1

1

X

X

X

X

1

X

1

X

1

X

X2

X2

X3

X

X

X2

X

1

Combine like terms to get 1 + 3X + 3X2 + X3

Binomial Coefficients

binomial expression

“Product form” or

“Generating form”

“Additive form” or

“Expanded form”

“Closed form” or

“Generating form”

“Power series” (“Taylor series”) expansion

Let x=1.

The number of

subsets of an

n-element set

By playing these two representations against

each other we obtain a new representation of

a previous insight:

By varying x, we can discover new identities

Let x= -1.

Equivalently,

The number of even-sized subsets of an n element set is the same as the number of odd-sized subsets.

Let x= -1.

Equivalently,

We could discover new identities by substituting in different numbers for X. One cool idea is to try complex roots of unity, however, the lecture is going in another direction.

Proofs that work by manipulating algebraic forms are called “algebraic” arguments.Proofs that build a 1-1 onto correspondence are called “combinatorial” arguments.

Let On be the set of binary strings of length n with an odd number of ones.

Let En be the set of binary strings of length n with an even number of ones.

We gave an algebraic proof that

On = En

Let On be the set of binary strings of length n with an odd number of ones.

Let En be the set of binary strings of length n with an even number of ones.

A combinatorial proof must construct a one-to-one correspondence betweenOn and En

Let fn be the function that takes an

n-bit string and flips all its bits.

...but do even n work? In f6 we have

110011 001100

101010 010101

Uh oh. Complementing maps evens to evens!

fn is clearly a one-to-one and onto function

for odd n. E.g. in f7 we have

0010011 1101100

1001101 0110010

Let fn be the function that takes an

n-bit string and flips only the first bit.

For example,

0010011 1010011

1001101 0001101

110011 010011

101010 001010

The binomial coefficients have so many representations that many fundamental mathematical identities emerge…

1

(1+X)0 =

1 + 1X

(1+X)1 =

1 + 2X + 1X2

(1+X)2 =

1 + 3X + 3X2 + 1X3

(1+X)3 =

1 + 4X + 6X2 + 4X3 + 1X4

(1+X)4 =

1

(1+X)0 =

1 + 1X

(1+X)1 =

1 + 2X + 1X2

(1+X)2 =

1 + 3X + 3X2 + 1X3

(1+X)3 =

1 + 4X + 6X2 + 4X3 + 1X4

(1+X)4 =

1

(1+X)0 =

1 + 1X

(1+X)1 =

1 + 2X + 1X2

(1+X)2 =

1 + 3X + 3X2 + 1X3

(1+X)3 =

1 + 4X + 6X2 + 4X3 + 1X4

(1+X)4 =

1

(1+X)0 =

1 + 1X

(1+X)1 =

1 + 2X + 1X2

(1+X)2 =

1 + 3X + 3X2 + 1X3

(1+X)3 =

1 + 4X + 6X2 + 4X3 + 1X4

(1+X)4 =

“Pascal’s Triangle”

Al-Karaji, Baghdad 953-1029

Chu Shin-Chieh 1303The Precious Mirror of the Four Elements

. . . Known in Europe by 1529

Blaise Pascal 1654

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

“It is extraordinary

how fertile in

properties the

triangle is.

Everyone can

try his

hand.”

Summing The Rows

1

1 + 1

1 + 2 + 1

1 + 3 + 3 + 1

1 + 4 + 6 + 4 + 1

1 + 5 + 10 + 10 + 5 + 1

1 + 6 + 15 + 20 + 15 + 6 + 1

=1

=2

=4

=8

=16

=32

=64

6+20+6

=

1+15+15+1

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Summing on 1st Avenue

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Summing on kth Avenue

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

=2

=3

=5

=8

=13

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

2

2

2

2

2

2

2

Al-Karaji Squares

1

1 1

1 2 +2*1

1 3 +2*3 1

1 4 +2*6 4 1

1 5 +2*10 10 5 1

1 6 +2*15 20 15 6 1

=0

=1

=4

=9

=16

=25

=36

All these properties can be proved inductively and algebraically. We will give combinatorial proofs using the Manhattan block walking representation of binomial coefficients.

A

B

0

0

j’th Street

k’th Avenue

1

1

2

2

3

3

4

4

Level n

0

2

4

3

1

0

k’th Avenue

………

1

………….

2

……………………

3

…………………………

4

…………………………………

Level n

0

2

4

3

1

0

k’th Avenue

………

1

………….

2

……………………

3

…………………………

4

…………………………………

. . . . .

k’th Avenue

1

. . . . .

level n

1

1

. . . . .

2

1

1

. . . . .

3

3

1

1

. . . . .

4

1

6

1

4

1

1

5

5

10

10

6

6

15

15

20

. . . . .

k’th Avenue

1

. . . . .

level n

1

1

. . . . .

2

1

1

. . . . .

3

3

1

1

. . . . .

4

1

6

1

4

1

1

5

5

10

10

6

6

15

15

20

. . . . .

k’th Avenue

1

. . . . .

level n

1

1

. . . . .

2

1

1

. . . . .

3

3

1

1

. . . . .

4

1

6

1

4

1

1

5

5

10

10

6

6

15

15

20

. . . . .

k’th Avenue

1

. . . . .

level n

1

1

. . . . .

2

1

1

. . . . .

3

3

1

1

. . . . .

4

1

6

1

4

1

1

5

5

10

10

6

6

15

15

20

By convention:

(empty product = 1)

. . . . .

k’th Avenue

1

. . . . .

level n

1

1

. . . . .

2

1

1

. . . . .

3

3

1

1

. . . . .

4

1

6

1

4

1

1

5

5

10

10

6

6

15

15

20

Application (Al-Karaji):

- Let’s define a (parallel) programming language called VECTOR that operates on possibly infinite vectors of numbers. Each variable V! can be thought of as:
- < * , * , * , * , *, *, . . . . . . . . . >
- 0 1 2 3 4 5 . . . . . . . . .

- Let k stand for a scalar constant
- <k> will stand for the vector <k,0,0,0,…>
- <0> = <0,0,0,0,….>
- <1> = <1,0,0,0,…>
- V! +T!means to add the vectors position-wise.
- <4,2,3,…> + <5,1,1,….> = <9,3,4,…>

- RIGHT(V!) means to shift every number in V! one position to the right and to place a 0 in position 0.
- RIGHT( <1,2,3, …> ) = <0,1,2,3,. …>

- Stare
- V! = <6,0,0,0,..>
- V! = <42,6,0,0,..>
- V! = <2,42,6,0,..> V!= <13,2,42,6,.>

- Example:
- V! := <6>;
- V! := RIGHT(V!)+ <42>;
- V! := RIGHT(V!)+ <2>;
- V! := RIGHT(V!)+ <13>;

V! = < 13, 2, 42, 6, 0, 0, 0, . . . >

- Stare
- V! = <1,0,0,0,..>
- V! = <1,1,0,0,..>
- V! = <1,2,1,0,..> V! = <1,3,3,1,.>

- Example:
- V! := <1>;
- Loop n times:
- V! := V! +RIGHT(V!);

V! = nth row of Pascal’s triangle.

Vector programs can be implemented by polynomials!

- X1

- X2

- +

- +

- X3

- The vector V! = < a0, a1, a2, . . . > will be represented by the polynomial:

- The vector V! = < a0, a1, a2, . . . > will be represented by the formal power series:

<0>is represented by 0

<k>is represented by k

V! + T! is represented by (PV + PT)

RIGHT(V! ) is represented by (PV X)

- Example:
- V! := <1>;
- Loop n times:
- V! := V! +RIGHT(V!);

PV := 1;

PV := PV + PV X;

V! = nth row of Pascal’s triangle.

- Example:
- V! := <1>;
- Loop n times:
- V! := V! +RIGHT(V!);

PV := 1;

PV := PV (1+ X);

V! = nth row of Pascal’s triangle.

- Example:
- V! := <1>;
- Loop n times:
- V! := V! +RIGHT(V!);

PV = (1+ X)n

V! = nth row of Pascal’s triangle.

What is the coefficient of Xk in the expansion of:

( 1 + X + X2 + X3 + X4 + . . . . )n ?

Each path in the choice tree for the cross terms has n choices of exponent e1, e2, . . . , en ¸ 0. Each exponent can be any natural number.

Coefficient of Xk is the number of non-negative solutions to: e1 + e2 + . . . + en = k

What is the coefficient of Xk in the expansion of:

( 1 + X + X2 + X3 + X4 + . . . . )n ?

( 1 + X + X2 + X3 + X4 + . . . . )n =

What is the coefficient of Xk in the expansion of:

(a0 + a1X + a2X2 + a3X3 + …) ( 1 + X + X2 + X3 + . . . )

= (a0 + a1X + a2X2 + a3X3 + …) / (1 – X) ?

a0 + a1 + a2 + .. + ak

(a0 + a1X + a2X2 + a3X3 + …) / (1 – X)

=

=

PREFIXSUM(a0 + a1X + a2X2 + a3X3 + …)

Let’s add an instruction called PREFIXSUM to our VECTOR language.

W! := PREFIXSUM(V!)

means that the ith position of W contains the sum of all the numbers in V from positions 0 to i.

0

k’th Avenue

1

2

3

4

- V! := 1! ;
- Loop k times: V! := PREFIXSUM(V!) ;

Can you see how PREFIXSUM can be represented by a familiar polynomial expression?

W! := PREFIXSUM(V!)

is represented by

PW = PV / (1-X)

= PV (1+X+X2+X3+ ….. )

- Zero_Ave := PREFIXSUM(<1>);
- First_Ave := PREFIXSUM(Zero_Ave);
- Second_Ave :=PREFIXSUM(First_Ave);
- Output:= First_Ave + 2*RIGHT(Second_Ave)

OUTPUT! = <1, 4, 9, 25, 36, 49, …. >

- Zero_Ave = 1/(1-X);
- First_Ave = 1/(1-X)2;
- Second_Ave = 1/(1-X)3;
- Output = 1/(1-X)2 + 2X/(1-X)3

(1-X)/(1-X)3 + 2X/(1-X)3

= (1+X)/(1-X)3

- Zero_Ave := PREFIXSUM(<1>);
- First_Ave := PREFIXSUM(Zero_Ave);
- Second_Ave :=PREFIXSUM(First_Ave);
- Output:= RIGHT(Second_Ave) + Second_Ave

Second_Ave = <1, 3, 6, 10, 15,.

RIGHT(Second_Ave) = <0, 1, 3, 6, 10,.

Output = <1, 4, 9, 16, 25

(1+X)/(1-X)3

outputs <1, 4, 9, ..>

X(1+X)/(1-X)3

outputs <0, 1, 4, 9, ..>

The kth entry is k2

X(1+X)/(1-X)3 = k2Xk

What does X(1+X)/(1-X)4 do?

X(1+X)/(1-X)4 expands to :

Sk Xk

where Sk is the sum of the first k squares

Aha! Thus, if there is an alternative interpretation of the kth coefficient of X(1+X)/(1-X)4we would have a new way to get a formula for the sum of the first k squares.

Using pirates and gold we found that:

THUS:

Expressions of the form

Finite Polynomial / Finite Polynomial

are called Rational Polynomial Expressions.

Clearly, these expressions have some deeper interpretation as a programming language.

What about this one?

X/(1-X-X2)

The action of dividing one polynomial by another can simulate a program to recursively compute Fibonacci numbers.

- Example:
- INPUT I!; /* not allowed to alter I! */
- V! := I! + 1;
- Loop n times:
- V! := V! +RIGHT(V!) + I! ;
- OUTPUT V!

- Let P be a vector program that takes input.
- A vector relation is any statement of the form:
- V! = P( V! )
- If there is a unique V! satisfying the relation, then V! is said to be defined by the relation V! = P( V! ).

Vector Recurrence Relation Definition:

F! = RIGHT( F! + <1> ) + RIGHT( RIGHT( F! ) )

Recurrence Relation Definition:

F! = RIGHT( F! + <1> ) + RIGHT( RIGHT( F! ) )

F! = a0, a1 , a2, a3, a4, . . .

RIGHT(F! +<1>) = 0, a0 + 1, a1, a2, a3,

RIGHT( RIGHT( F! ) )

= 0, 0 , a0, a1, a2, a3, .

F! = RIGHT( F! + 1 ) + RIGHT( RIGHT( F! ) )

F = a0 + a1 X + a2 X2 + a3 X3 +

RIGHT(F + 1) = (F+1)X

RIGHT( RIGHT( F ) )

= FX2

F = (F + 1) X + F X2

F = a0 + a1 X + a2 X2 + a3 X3 + . . .

RIGHT(F + 1) = (F+1)X

RIGHT( RIGHT( F ) )

= F X2

F = F X + X + F X2

Solve for F.

F - FX - FX2 = X

F(1-X-X2) = XF = X/(1-X-X2)

What is the Power Series Expansion of x / (1-x-x2) ?

Since the bottom is quadratic we can factor it.

X / (1-X-X2) =

X/(1- X)(1 – (-)-1X)

where =

“The Golden Ratio”

X

(1 – X)(1- (--1X)

?

n=0..∞

Xn

=

an+1 – bn+1

a- b

(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1

=

(1 – aX)(1-bX)

n=0..1

Xn

=

n+1 – (--1)n+1

5

1

(1 – X)(1- (--1X)

n=0.. ∞

Xn

=

n+1 – (--1)n+1

5

X

(1 – X)(1- (--1X)

n=0.. ∞

Xn+1

=

The ith Fibonacci number is:

- Applied Combinatorics, by Alan Tucker
- Generatingfunctionology, Herbert Wilf