NCTM Annual Meeting St. Louis, April 2006 Intersections of Algebra and Counting Duane DeTemple...

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NCTM Annual MeetingSt. Louis, April 2006

Intersections of

Algebra and Counting

Duane DeTempleProfessor of Mathematics

Washington State UniversityPullman WA

The NCTM Algebra Standard

All students should “Understand patterns, relations, and functions.”

To meet the grades 9 – 12 expectations, students should “generalize patterns using explicitly defined and recursively defined functions.”

Preview of Coming Attractions

Multiplying Apples and Bananas: How to Count by Polynomial Multiplication

Counting Trains: How to Count by Obtaining Recurrence Relations

Solving Recurrence Relations: How to Find and Combine Geometric Sequences to Obtain Explicit Formulas

Multiplying Apples & Bananas

= A2 B

)( + = AB + A2 B

= B + B2 +AB + AB2 + A2B + A2B2

)(1+ + ( + )

Packing a Lunch

How many ways can up to 4 pieces of fruit be put into the lunch sack, where at least one banana is included?

Solving the Lunch Problem

B B2 AB AB2 A2B A2B2

= B + B2 +AB + AB2 + A2B + A2B2

)(1+ + ( + )

How many lunches include exactly 3 pieces of fruit, including at least one banana?

AB2 + A2B

XX2 + X2X = 2 X3

(1 + X + X2)(X + X2) = X + 2X2 + 2X3 + X4

What lunch packing problem is solved by

(1 + X + X2)(X + X2)(1 + X2)= X + 3X2 + 2X3 + 3X4 + 2X5 + X6

? A package of 2

cookies

What polynomial multiplication applies here?

Two packages

with 2 cookies

2 3 4 2 3 2 4(1 )( )(1 )x x x x x x x x x

0,1, …, 4 apples

1, 2, or 3 bananas

0, 2, or 4 cookies

How can polynomials be multiplied easily?

15 10 20

6 11 2 20

Example Synthetic Multiplication

15 10 20

6 11 2 20

Remark: Synthetic Multiplication on TI-

83Input "áP",áPInput "áQ",áQdim(áP)üMClrList áRdim(áQ)üdim(áR)For(J,1,M-1)áR+áP(J)*áQüáRaugment({0},áQ)üáQaugment(áR,{0})üáREndáR+áP(M)*áQüáRDisp áR

Example2 3 4 2 3 2 4

2 3 4 5

6 7 9 10 18 1

(1 )( )(1 )

2 4 5 7

7 7 25 4 +

x x x x x x x x x

x x x x x

x x x x x x

There are 5 ways to pack 8 items including up to 4 apples, at least 1 and up to 3 bananas, and up to 2 packages of cookies (2 cookies/package) :

A4B2C2, A3B1C4, A3B3C2, A2B2C4, A1B3C4

A Postage ProblemMore Problems Solved By Multiplying Polynomials

You discovered you have five 13¢, two 15¢, and three 20¢ stamps.

Can you put 39¢ (exactly) postage on a one-ounce letter?

How about 63¢ for a two-ounce letter?

Solving the Postage Problem

11 22 33 44 55

20 40 60

x x x x x

11 37 40

Making Change

The till has just 3 nickels, 4 dimes, and 2 quarters.

Can you give out 75¢ in change?

Solution to Change Problem

2 4 6 8

x x x x

15 211 3x x

Note: Use “nickels” (5 cents) as the unit.

Solutions of an Equation With Integer Unknowns

How many solutions are there of the equation

2 8a cb where

{0,1,2,3,4}

Answer: 5

Train Counting

Let a d-train have cars of lengths 1, 2, … , n in some order. How many trains, dn, have total length n?

There are d4 = 8 trains of length 4.

1+1+1+1

Seeing a Pattern

d3 = 4

d 2 = 2

d 1 = 1

stretch oldcaboose toget a cabooseof length > 1

add a unitlengthcaboose

Describing the d-train pattern

1 nn ndd d d-trains

of length n

Add 1-car caboose

to alldn-trains

Stretch the caboose of all dn-trains

Conclusion:The number of d-trains is

given by the doubling geometric sequence

Explicit formula: 2

Recursion formula: 2 ,

with initial condition 1

Counting f-trains

Let an f-train have cars of lengths 1 and 2 in some order. How many trains, fn , have total length n ?

There are f4 = 5 trains of length 4.

The Pattern of f-Trains

f 1 = 1

add acaboose oflength two

add a unitlengthcaboose

Describing the f-train pattern

2 1nn nff f

f-trains of length

Add 1-car caboose

to allfn+1-trains

Add a 2-car

caboose to all fn-trains

Conclusion:The number of f-trains is given by the Fibonacci

sequence!

1n nf F

1, 1, 2, 3, 5, 8, 13, …

Counting p-trainsA p-train has an engine of three types: A, B, or C, and has cars of lengths 2 or 3. A 2-car cannot be attached to engine C.

How many trains, pn, have cars of total length n?

p5 = 5

More Cases of p-trains

p1= 0p0= 3A-

p2= 2A-

B-p3= 3

p-train sequence:

3, 0, 2, 3, 2, 5, …

What’s the pattern?

p4= 2B-

A-p5= 5A-

The Pattern of the p-trains

The Recurrence for p-Trains

p-trains of

length n + 3

3 1nn npp p

Add a 2-car

caboose to all

pn+1-trains

Add a 3-car

caboose to all pn-trains

Foxtrot

Bill Amend, October 11, 2005

What should Jason say to score a touchdown?

3,0,2,3,2,5,5,7 Search

Greetings from The On-Line Encyclopedia of Integer Sequences!

Search: 3,0,2,3,2,5,5,7

Displaying 1-1 of 1 results found.

A001608Perrin sequence: a(n) = a(n-2) + a(n-3). +20

3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, 486, 644, 853, 1130,

http://www.research.att.com/~njas/sequences/Seis.html

An Amazing Property of the Perrin Sequence

n 0 1 2 3 4 5 6 7 8 9 10p(n) 3 0 2 3 2 5 5 7 10 12 17

n 11 12 13 14 15 16 17 18 19 20 21p(n) 22 29 39 51 68 90 119 158 209 277 367

Theorem: For all primes n, n divides p(n).Question: If n divides p(n), is n a prime?Answer: No. The smallest example is

n = 271441 = 5212 divides p(271441)

Solving Recurrences

Problem: How do you solve the Fibonacci RR?

2 1n n nx x x Idea: Look for solutions in the form of geometric sequences xn = xn

2 1n n nx x x

Divide out xn to get the quadratic equation

2 1x x Solve the quadratic to get the two roots

1 5 1 5and

2 2p q

Thus 2 21 and 1p p q q

Multiply equations by any constants a and b to get a general solution

n nnx ap bq

of the Fibonacci RR

2 1n n nx x x

What are good choices for the constants a and b?

The choice a = b = 1

1 5 1 51

2 22,1, 3, 4, 7,11,18, 29, 47,

n nnL p q

This is the Lucas sequence, named for Edouard Lucas (1842-1891).

The choice a = - b = 1/5

1 5 1 51

5 2 5 2 50,1,1, 2, 3, 5, 8,13, 21,

This is the Fibonacci sequence!

Problem: How do you solve the Perrin RR?

3 1n n nx x x

Use the same idea: Look for solutions in the form of geometric sequences xn = xn

3 1n n nx x x

3 1n n nx x x 3 1x x

by xn to get the cubic equation

Solve the cubic to get three roots u, v, and w and the solution

n n nau bv cw

Divide

where a, b, and c are any constants.

The choice a = b = c = 1 gives the solution

n n nnP u v w

We see that0 0 0

1 1 11

2 2 22

P u v w

x x x u x v x w

x u v w x

uv uw vw x uvw

Since u, v, and w are the roots of

3 1 0x x we have that

0, 1u v w uv uw vw

Equate coefficients of x2 and x1

u v w uv uw vw

We also have that

1 1 11 0P u v w

Therefore,

so2 2 2

2 2P u v w

Conclusion: the Perrin Sequence is given either by the recurrence relation

3, 0, 2n n nP P P

or explicitly by

n n nnP u v w

where u, v, and w are the roots of the cubic equation

3 1x x

For downloads of

This PowerPoint presentation

The paper From Fibonacci to Foxtrot: Investigating Recursion Relations with Geometric Sequences

TI-8X program to multiply polynomials

Go to: http://www.math.wsu.edu/math/faculty/detemple/

NCTM Annual Meeting St. Louis, April 2006 Intersections of Algebra and Counting Duane DeTemple...

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