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X 1. X 2. +. +. X 3. Counting III: Pascal’s Triangle, Polynomials, and Vector Programs. X n+1 - 1. 1 + X 1 + X 2 + X 3 + … + X n-1 + X n =. X - 1. The Geometric Series. 1. 1 + X 1 + X 2 + X 3 + … + X n + ….. =. 1 - X. The Infinite Geometric Series. 1.

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Counting iii pascal s triangle polynomials and vector programs

  • X2

  • +

  • +

  • X3

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


The geometric series

Xn+1 - 1

1 + X1 + X2 + X 3 + … + Xn-1 + Xn =

X- 1

The Geometric Series


The infinite geometric series

1

1 + X1 + X2 + X 3 + … + Xn + ….. =

1 - X

The Infinite Geometric Series


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

  • -....


Geometric series linear form
Geometric Series (Linear Form)

1

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

1 - aX


Geometric series quadratic form
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 c n
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 n a 0 b n a 1 b n 1 a i b n i a n 1 b 1 a n b 0
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 b then c n n 1 a n a 0 b n a 1 b n 1 a i b n i a n 1 b 1 a n b 0
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


If a b then c n a 0 b n a 1 b n 1 a i b n i a n 1 b 1 a n b 0

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 + …


Geometric series quadratic form1

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.


Choice tree for terms of 1 x 3

1 polynomial.

1

1

1

X

X

X

X

1

X

1

X

1

X

Choice tree for terms of (1+X)3

X2

X2

X3

X

X

X2

X

1

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


The binomial formula
The Binomial Formula polynomial.

Binomial Coefficients

binomial expression



One polynomial two representations
One polynomial, polynomial.two representations

“Product form” or

“Generating form”

“Additive form” or

“Expanded form”


Power series representation
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:


By varying x, we can discover new identities polynomial.

Let x= -1.

Equivalently,


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
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
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 n
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 formula2
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 k th row are the coefficients of 1 x k
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 th row of pascal s triangle
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
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
How many shortest routes from A to B? algebraically. We will give combinatorial proofs using the

A

B


Manhattan

0 algebraically. We will give combinatorial proofs using the

0

j’th Street

k’th Avenue

1

1

2

2

3

3

4

4

Manhattan


Manhattan1

Level n algebraically. We will give combinatorial proofs using the

0

2

4

3

1

0

k’th Avenue

………

1

………….

2

……………………

3

…………………………

4

…………………………………

Manhattan


Manhattan2

Level 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
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 programs1
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 programs2
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 programs3
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 programs4
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
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
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 a 0 a 1 a 2
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 programs5
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 programs6
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 programs7
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.


What does this program output

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
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 program1
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 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 k in p v x 2 x 1 x 4 is the sum of the first k squares
Coefficient of X kk in PV = (X2+X)(1-X)-4 is the sum of the first k squares:


Vector programs polynomials closed form expression
Vector programs -> Polynomials k-> Closed form expression


Expressions of the form k

Finite Polynomial / Finite Polynomial

are called Rational Polynomial Expressions.

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


What about this one? k

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
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
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! ).


Fibonacci numbers

Vector Recurrence Relation Definition: simulate a program to recursively compute Fibonacci numbers.

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

Fibonacci Numbers

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

X

(1 – X)(1- (--1X)

?

n=0..∞

Xn

=


Geometric series quadratic form2

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

=


Geometric series quadratic form3

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

=


Power series expansion of f

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

5

Power Series Expansion of F

X

(1 – X)(1- (--1X)

n=0.. ∞

Xn+1

=


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


References
References simulate a program to recursively compute Fibonacci numbers.

  • Applied Combinatorics, by Alan Tucker

  • Generatingfunctionology, Herbert Wilf


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