NCTM Annual Meeting
St. Louis, April 2006
Intersections
of
Algebra and Counting
Duane DeTemple
Professor of Mathematics
Washington State University
Pullman 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
(
+
)
(1+
+
)(
=B+
2
B
+AB +
2
AB
= AB +
2
A
B
+
+
2
AB
)
+
2
2
AB
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
2
B
AB
2
AB
2
AB
2
2
AB
(1+ +
)(
+
)
= B + B2 +AB + AB2 + A2B + A2B2
How many lunches include
exactly 3 pieces of fruit,
including at least one banana?
2
AB
2
XX
+
2
AB
2
XX
3
X
+
=2
2
2
(1 + X + X )(X + X )
2
3
4
= X + 2X + 2X + X
What lunch packing
problem is solved by
2
2
2
(1 + X + X )(X + X )(1 + X )
2
3
4
5
6
= X + 3X + 2X + 3X + 2X + X
?
A package
of 2
cookies
What polynomial
multiplication applies here?
Two
packages
with 2
cookies
each
(1  x  x  x  x )( x  x  x )(1  x  x )
2
0,1, …, 4
apples
3
4
1, 2, or 3
bananas
2
3
0, 2, or 4
cookies
2
4
How can polynomials be
multiplied easily?
Example
3  2x  4x
2  5x
Synthetic Multiplication
3 2 4
2
5
2
6  4x  8x
2
6
15 x  10 x  20 x
3
6  11x  2 x  20 x
3
2
2
6
4 8
15  10
20
11  2
20
Remark:
Synthetic Multiplication on TI-83
Input "áP",áP
Input "áQ",áQ
dim(áP)üM
ClrList áR
dim(áQ)üdim(áR)
For(J,1,M-1)
áR+áP(J)*áQüáR
augment({0},áQ)üáQ
augment(áR,{0})üáR
End
áR+áP(M)*áQüáR
Disp áR
Example
(1  x  x  x  x )( x  x  x )(1  x  x )
2
3
4
2
 x2 x 4 x 5 x 7 x
2
3
4
3
2
4
5
7 x  7 x  5 x  4 x  2 x + x
6
7
8
9
10
11
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
More Problems Solved By Multiplying Polynomials
A Postage Problem
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 twoounce letter?
Solving the Postage Problem
(1  x  x  x  x  x )
11
22
 1  x  x
15
33
40
60
 1 x 
11

2 x 
70
55

30
 1  x  x  x
20
44

x x
37
x 
63
x
145
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
1  x  x
2
x
3

Note:
Use
“nickels”
(5 cents)
as the
unit.
(1  x  x  x  x )
2
4
 1  x  x
5
 1
10
6
8

 3x 
15
x
21
Solutions of an Equation
With Integer Unknowns
How many solutions are there of the equation
a  b  2c  8
where
a  {0,1, 2,3, 4}
b  1, 2,3
c  0,1
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?
1+1+1+1
1+1+2
1+2+1
2+1+1
2+2
3+1
There are d4 = 8
trains of length 4.
1+3
4
Seeing a Pattern
d1 = 1
d2 = 2
d3 = 4
add a unit
length
caboose
stretch old
caboose to
get a caboose
of length > 1
Describing the d-train
pattern
dn1  dn  dn
d-trains
of length Add 1-car
caboose
n+1
to all
dn-trains
Stretch the
caboose of
all dn-trains
Conclusion:
The number of d-trains is
given by the doubling
geometric sequence
Explicit formula: d n  2
n 1
Recursion formula: d n 1  2d n ,
with initial condition d1  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
f2 = 2
f3 = 3
add a
caboose of
length two
add a unit
length
caboose
Describing the f-train
pattern
f n2  f n1  f n
f-trains of
length
n+2
Add 1-car
caboose
to all
fn+1-trains
Add a 2car
caboose to
all fn-trains
Conclusion:
The number of f-trains is
given by the Fibonacci
sequence!
f n  Fn 1
1, 1, 2, 3, 5, 8, 13, …
Counting p-trains
A 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
BA
-BC
-
More Cases of p-trains
ABC-
AB-

p0= 3
p2= 2
ABC-
p1= 0
p3= 3
AB-
ABp4= 2 AB-
p5= 5
C-
p-train sequence:
3, 0, 2, 3, 2, 5, …
What’s the pattern?
The Pattern of the p-trains
ABABC-
ABABC-
The Recurrence for p-Trains
pn3  pn1  pn
Add a
2-car
p-trains
of length caboose
to all
n+3
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?
http://www.research.att.com/~njas/sequences/Seis.html
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.
A001608
Perrin sequence:
a(n) = a(n-2) + a(n-3).
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,
An Amazing Property of
the Perrin Sequence
n
p(n)
0
3
1
0
2
2
3
3
4
2
5
5
6
5
7
8 9 10
7 10 12 17
n 11 12 13 14 15 16 17 18 19 20 21
p(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?
xn 2  xn1  xn
Idea: Look for solutions in the form
of geometric sequences xn = xn
x
n 2
x
n 1
x
n
Divide out xn to get the quadratic
equation
x  x 1
2
Solve the quadratic to get the two roots
1 5
p
2
and
1 5
q
2
Thus
p  p 1
2
and
q  q 1
2
Multiply equations by any constants a
and b to get a general solution
xn  ap  bq
n
n
of the Fibonacci RR
xn 2  xn1  xn
What are good choices for the
constants a and b?
The choice a = b = 1
Ln  p  q
n
n
L0  p  q  1  1  2
0
0
1 5 1 5
L1  p  q 

1
2
2
2, 1, 3, 4, 7, 11, 18, 29, 47,
1
1
This is the Lucas sequence, named
for Edouard Lucas (1842-1891).
The choice a = - b = 1/5
p q
Fn 
5
0
0
p  q 11
F0 

0
5
5
n
n
p  q 1 5 1 5
F1 


1
5
2 5
2 5
0, 1, 1, 2, 3, 5, 8, 13, 21,
1
1
This is the Fibonacci sequence!
Problem: How do you solve
the Perrin RR?
xn3  xn1  xn
Use the same idea: Look for solutions in
the form of geometric sequences xn = xn
x
n 3
x
n 1
x
n
Divide
x
n 3
x
n 1
x
n
by xn to get the cubic equation
x  x 1
3
Solve the cubic to get three roots
u, v, and w and the solution
au  bv  cw
n
n
n
where a, b, and c are any constants.
The choice a = b = c = 1 gives the solution
Pn  u  v  w
n
n
n
We see that
P0  u  v  w  1  1  1  3
0
0
0
P1  u  v  w  ?
1
1
1
and
P2  u  v  w  ?
2
2
2
Since u, v, and w are the roots of
x  x 1  0
3
we have that
x  x  1  ( x  u )  x  v  x  w 
3
 x  (u  v  w) x
(uv  uw  vw) x  uvw
3
2
Equate coefficients of x2 and x1
u  v  w  0, uv  uw  vw  1
Therefore,
P1  u  v  w  0
1
1
1
We also have that
0   u  v  w
2
 u  v  w  2  uv  uw  vw 
2
2
2
 u  v  w  2  1
2
so
2
2
P2  u  v  w  2
2
2
2
Conclusion: the Perrin Sequence is given
either by the recurrence relation
Pn 3  Pn 1  Pn ,
P0  3, P1  0, P2  2
or explicitly by
Pn  u  v  w
n
n
n
where u, v, and w are the roots of the
cubic equation
x  x 1
3
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/