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
= AB
= 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
each
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?
2
2
2 3
2 3
3 2 4
2 5
6 4 8
15 10 20
6 11 2 20
x x
x
x x
x x x
x x x
Example Synthetic Multiplication
3 2 4
2 5
6 4 8
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
15 30
20 40 60
(1 )
1
1
x x x x x
x x
x x x
11 37 40
70 1
6
45
3
1
2
x x x
x
x x
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 3
2 4 6 8
5 10
1
(1 )
1
x x x
x x 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}
1,2,3
0,1c
b
a
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
1+1+2
1+2+1
2+1+1
2+2
3+1
1+3
4
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
+ 1
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
1
1
1
Explicit formula: 2
Recursion formula: 2 ,
with initial condition 1
nn
n n
d
d d
d
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
f3
= 3
f2
= 2
f 1 = 1
add acaboose oflength two
add a unitlengthcaboose
Describing the f-train pattern
2 1nn nff f
f-trains of length
n + 2
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
A- B-
C-
B-
A-
More Cases of p-trains
p1= 0p0= 3A-
B-
C-
p2= 2A-
B-p3= 3
C-
B-
A-
p-train sequence:
3, 0, 2, 3, 2, 5, …
What’s the pattern?
p4= 2B-
A-p5= 5A-
B-
C-
B-
A-
The Pattern of the p-trains
A-
B-
C-
B-
A-A-
B-
C-
B-
A-
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
22
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
0 00
1 11
1 1 2
1 5 1 51
2 22,1, 3, 4, 7,11,18, 29, 47,
n nnL p q
L p q
L p q
This is the Lucas sequence, named for Edouard Lucas (1842-1891).
The choice a = - b = 1/5
0 0
0
1 1
1
5
1 10
5 5
1 5 1 51
5 2 5 2 50,1,1, 2, 3, 5, 8,13, 21,
n n
n
p qF
p qF
p qF
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
0
1 1 11
2 2 22
1 1
?
n
1
a
3
?
d
P u v w
P u v w
P u v w
3
3 2
1 ( )
( )
( )
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
2
2 2 2
2 2 2
0
2
2 1
u v w
u v w uv uw vw
u v w
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 1
0 1 2
,
3, 0, 2n n nP P P
P 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/
Top Related