Loading in 5 sec....

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

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

- 68 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Counting III: Pascal’s Triangle, Polynomials, and Vector Programs' - ishmael-burks

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

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

Presentation Transcript

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 + X1 + X2 + X 3 + … + Xn + ….. =

1 - X

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

Geometric Series (Quadratic Form)

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

1

(1 – aX)(1-bX)

Suppose we multiply this out to get a single, infinite polynomial.What is an expression for Cn?

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

1 + c1X1 + .. + ck Xk + …

c polynomial.n =a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0

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

1 + c1X1 + .. + ck Xk + …

If a polynomial.= b then cn = (n+1)(an)a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0

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

1 + c1X1 + .. + ck Xk + …

a polynomial.n+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

a polynomial.n+1 – bn+1

a- b

if a b then cn =a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1 + c1X1 + .. + ck Xk + …

a polynomial.n+1 – bn+1

a- b

Geometric Series (Quadratic Form)(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 polynomial.

Polynomials Count!

What is the coefficient of BA polynomial.3N2 in the expansion of

(B + A + N)6?

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

1 polynomial.

1

1

1

X

X

X

X

1

X

1

X

1

X

Choice tree for terms of (1+X)3X2

X2

X3

X

X

X2

X

1

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

The Binomial Formula polynomial.

One polynomial, polynomial.two representations

“Product form” or

“Generating form”

“Additive form” or

“Expanded form”

Power Series Representation polynomial.

“Closed form” or

“Generating form”

“Power series” (“Taylor series”) expansion

Let polynomial.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:

The number of even-sized subsets of an polynomial.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 “algebraic” arguments.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

A Combinatorial Proof “algebraic” 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.

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

An attempt at a correspondence “algebraic” arguments.

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

A correspondence that works for all “algebraic” arguments.n

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…

The Binomial Formula 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 =

Pascal’s Triangle: many fundamental mathematical identities emerge…kth row are the coefficients of (1+X)k

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 =

k many fundamental mathematical identities emerge…th Row Of Pascal’s Triangle:

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 =

Inductive definition of kth entry of nth row: many fundamental mathematical identities emerge…Pascal(n,0) = Pacal (n,n) = 1; Pascal(n,k) = Pascal(n-1,k-1) + Pascal(n,k)

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” many fundamental mathematical identities emerge…

Al-Karaji, Baghdad 953-1029

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

. . . Known in Europe by 1529

Blaise Pascal 1654

Pascal’s Triangle many fundamental mathematical identities emerge…

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 many fundamental mathematical identities emerge…

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 many fundamental mathematical identities emerge…

=

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 1 many fundamental mathematical identities emerge…st 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 k many fundamental mathematical identities emerge…th 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 many fundamental mathematical identities emerge…

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 many fundamental mathematical identities emerge…

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 many fundamental mathematical identities emerge…

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.

How many shortest routes from A to B? algebraically. We will give combinatorial proofs using the

A

B

0 algebraically. We will give combinatorial proofs using the

0

j’th Street

k’th Avenue

1

1

2

2

3

3

4

4

ManhattanLevel n algebraically. We will give combinatorial proofs using the

0

2

4

3

1

0

k’th Avenue

………

1

………….

2

……………………

3

…………………………

4

…………………………………

ManhattanLevel n algebraically. We will give combinatorial proofs using the

0

2

4

3

1

0

k’th Avenue

………

1

………….

2

……………………

3

…………………………

4

…………………………………

Manhattan. . . . . algebraically. We will give combinatorial proofs using the

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

. . . . . algebraically. We will give combinatorial proofs using the

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

. . . . . algebraically. We will give combinatorial proofs using the

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

. . . . . algebraically. We will give combinatorial proofs using the

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: algebraically. We will give combinatorial proofs using the

(empty product = 1)

. . . . . algebraically. We will give combinatorial proofs using the

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): algebraically. We will give combinatorial proofs using the

Vector Programs algebraically. We will give combinatorial proofs using the

- 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 . . . . . . . . .

Vector Programs algebraically. We will give combinatorial proofs using the

- 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,…>

Vector Programs algebraically. We will give combinatorial proofs using the

- 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,. …>

Vector Programs algebraically. We will give combinatorial proofs using the

- 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, . . . >

Vector Programs algebraically. We will give combinatorial proofs using the

- 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! algebraically. We will give combinatorial proofs using the

- X1

- X2

- +

- +

- X3

Programs -----> Polynomials algebraically. We will give combinatorial proofs using the

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

Formal Power Series algebraically. We will give combinatorial proofs using the

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

V algebraically. We will give combinatorial proofs using the ! = < a0, a1, a2, . . . >

<0>is represented by 0

<k>is represented by k

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

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

Vector Programs algebraically. We will give combinatorial proofs using the

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

PV := 1;

PV := PV + PV X;

V! = nth row of Pascal’s triangle.

Vector Programs algebraically. We will give combinatorial proofs using the

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

PV := 1;

PV := PV (1+ X);

V! = nth row of Pascal’s triangle.

Vector Programs algebraically. We will give combinatorial proofs using the

- 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 X algebraically. We will give combinatorial proofs using the k 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 X algebraically. We will give combinatorial proofs using the k in the expansion of:

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

( 1 + X + X algebraically. We will give combinatorial proofs using the 2 + X3 + X4 + . . . . )n =

What is the coefficient of X algebraically. We will give combinatorial proofs using the k in the expansion of:

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

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

a0 + a1 + a2 + .. + ak

(a algebraically. We will give combinatorial proofs using the 0 + a1X + a2X2 + a3X3 + …) / (1 – X)

=

=

PREFIXSUM(a0 + a1X + a2X2 + a3X3 + …)

Let’s add an instruction called algebraically. We will give combinatorial proofs using the 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 algebraically. We will give combinatorial proofs using the

k’th Avenue

1

2

3

4

What does this program output?- V! := 1! ;
- Loop k times: V! := PREFIXSUM(V!) ;

Can you see how algebraically. We will give combinatorial proofs using the PREFIXSUM can be represented by a familiar polynomial expression?

W algebraically. We will give combinatorial proofs using the ! := PREFIXSUM(V!)

is represented by

PW = PV / (1-X)

= PV (1+X+X2+X3+ ….. )

Al-Karaji Program algebraically. We will give combinatorial proofs using the

- 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, …. >

Al-Karaji Program algebraically. We will give combinatorial proofs using the

- 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

(1+X)/(1-X) algebraically. We will give combinatorial proofs using the 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) algebraically. We will give combinatorial proofs using the 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) algebraically. We will give combinatorial proofs using the 3 = k2Xk

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

X(1+X)/(1-X) algebraically. We will give combinatorial proofs using the 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 kpirates and gold we found that:

THUS:

Coefficient of X kk in PV = (X2+X)(1-X)-4 is the sum of the first k squares:

Vector programs -> Polynomials k-> Closed form expression

Finite Polynomial / Finite Polynomial

are called Rational Polynomial Expressions.

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

X/(1-X-X2)

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

Vector Program I/O 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!

Vector Recurrence Relations simulate a program to recursively compute Fibonacci numbers.

- 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: simulate a program to recursively compute Fibonacci numbers.

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

Fibonacci NumbersRecurrence Relation Definition:

F simulate a program to recursively compute Fibonacci numbers.! = 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 simulate a program to recursively compute Fibonacci numbers.! = 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 X simulate a program to recursively compute Fibonacci numbers.2

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

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

RIGHT( RIGHT( F ) )

= F X2

F = F X + X + F X simulate a program to recursively compute Fibonacci numbers.2

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-x simulate a program to recursively compute Fibonacci numbers.2) ?

Since the bottom is quadratic we can factor it. simulate a program to recursively compute Fibonacci numbers.

X / (1-X-X2) =

X/(1- X)(1 – (-)-1X)

where =

“The Golden Ratio”

Linear factors on the bottom simulate a program to recursively compute Fibonacci numbers.

X

(1 – X)(1- (--1X)

?

n=0..∞

Xn

=

a simulate a program to recursively compute Fibonacci numbers.n+1 – bn+1

a- b

Geometric Series (Quadratic Form)(1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) =

1

=

(1 – aX)(1-bX)

n=0..1

Xn

=

simulate a program to recursively compute Fibonacci numbers.n+1 – (--1)n+1

5

Geometric Series (Quadratic Form)1

(1 – X)(1- (--1X)

n=0.. ∞

Xn

=

simulate a program to recursively compute Fibonacci numbers.n+1 – (--1)n+1

5

Power Series Expansion of FX

(1 – X)(1- (--1X)

n=0.. ∞

Xn+1

=

The i simulate a program to recursively compute Fibonacci numbers.th Fibonacci number is:

References simulate a program to recursively compute Fibonacci numbers.

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

Download Presentation

Connecting to Server..