nctm annual meeting st. louis, april 2006 intersections of algebra and counting duane detemple...
TRANSCRIPT
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NCTM Annual MeetingSt. Louis, April 2006
Intersections of
Algebra and Counting
Duane DeTempleProfessor of Mathematics
Washington State UniversityPullman WA
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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.”
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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
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Multiplying Apples & Bananas
= AB
= A2 B
)( + = AB + A2 B
= B + B2 +AB + AB2 + A2B + A2B2
)(1+ + ( + )
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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?
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Solving the Lunch Problem
B B2 AB AB2 A2B A2B2
= B + B2 +AB + AB2 + A2B + A2B2
)(1+ + ( + )
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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
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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
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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
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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
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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
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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
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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?
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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
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Making Change
The till has just 3 nickels, 4 dimes, and 2 quarters.
Can you give out 75¢ in change?
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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.
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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
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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
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Seeing a Pattern
d3 = 4
d 2 = 2
d 1 = 1
stretch oldcaboose toget a cabooseof length > 1
add a unitlengthcaboose
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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
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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
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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.
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The Pattern of f-Trains
f3
= 3
f2
= 2
f 1 = 1
add acaboose oflength two
add a unitlengthcaboose
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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
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Conclusion:The number of f-trains is given by the Fibonacci
sequence!
1n nf F
1, 1, 2, 3, 5, 8, 13, …
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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-
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More Cases of p-trains
p1= 0p0= 3A-
B-
C-
p2= 2A-
B-p3= 3
C-
B-
A-
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p-train sequence:
3, 0, 2, 3, 2, 5, …
What’s the pattern?
p4= 2B-
A-p5= 5A-
B-
C-
B-
A-
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The Pattern of the p-trains
A-
B-
C-
B-
A-A-
B-
C-
B-
A-
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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
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Foxtrot
Bill Amend, October 11, 2005
What should Jason say to score a touchdown?
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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
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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)
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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
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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
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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?
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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).
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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!
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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
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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.
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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
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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
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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
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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
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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/