A Journey through Number Theory Geometry and Calculus NYS Master Teachers March 9 2015 Dave Brown dabrownithacaedu Slides available at http facultyithacaedu dabrown docs ID: 618381
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Slide1
From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus
NYS Master TeachersMarch 9, 2015
Dave Brown –
dabrown@ithaca.edu
Slides available at
http
://
faculty.ithaca.edu
/
dabrown
/docs/
masterteachers
/Slide2
Goals
Use counting (combinatorics) to generate patternsLink counting patterns to algebra, geometry, trigonometry, & calculusExplore patterns in counting structuresDraw connections among various branches of math
Learn some discrete mathematicsSlide3
Dividing Rectangles
Start with a simple subdividing game.Divide rectangles into shorter rectangles and count.
For example, we can divide a length 3 rectangle
How many divisions of a length 3 rectangle are possible?
Explore subdivisions using Activity 1.
These two are considered different
1
1
1
1
1
2
2Slide4
Dividing Rectangles
What strategies can we use to list ALL possible divisions?
Important: What is your process?Two main ideas in counting:
Make sure we have counted everything
Make sure nothing has been counted twiceSlide5
Dividing Rectangles
Constructive approachSystematically build all divisions of a given length
Recursive approachUse divisions of shorter rectangles to build longer oneSlide6
Dividing Rectangles
Constructive approachSystematically build all divisions of a given length
Suppose you want to find all divisions of rectangle of length 10
If we want all three block divisions, we can use 2 separators in any of 9 spots.
1
3
4
5
6
7
8
9
2
2-block
5-block
3
-block
1
3
4
5
6
7
8
9
2
2-block
4
-block
4-blockSlide7
Dividing Rectangles
Constructive approachSystematically build all divisions of a given length
Suppose you want to find all divisions of rectangle of length 10
If we want all 5 block divisions, how many separators do we use?
4 separators will give us all divisions into 5 blocks
1
3
4
5
6
7
8
9
2Slide8
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
Every division has a final (rightmost) block of length 1 OR of length greater than 1Slide9
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
Remove all final blocks of length 1
Do you get all rectangles of length 4?Slide10
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
What about the other 8 length 5 rectangles?
Can we do anything to final blocks to get all rectangles of length 4?
SHRINK!!Slide11
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
We get all 8 length 4 rectangles by either removing the final 1-blocks, Or
We
get all 8 length 4 rectangles
by shrinking the final blocks bigger than 1Slide12
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
Reverse the logic.
How do you get all of the length 5 rectangles from two copies of the length 4 rectangles?
Add 1-blocks
to the end
Lengthen the end-block
by 1Slide13
Dividing Rectangles
Recursive approachUse divisions of shorter rectangles to build longer one
How would you
build all rectangles
of length 6?Slide14
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisionsSlide15
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1Slide16
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2Slide17
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
4Slide18
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
48Slide19
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
48
16Slide20
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
48
162n-1
How do we prove this?Constructive approach – separators
For the length 10 rectangle, in how many places could we use separators?9 locations (can use 0-9 separators)In each location, you can either use a separator or not
So, there are 29 decisions about dividing to make; hence, 29
divisionsHow does this generalize to rectangles of length n?Slide21
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
48
162n-1
How do we prove this?Recursive approach – idea of building up
For the length 5 rectangles, how many ended with 1-blocks vs not?What is the relationship btw # of length 5 rectangles and # of length 4?
#(length 5) = 2*#(length 4)Similarly, #(length 4) = 2
*#(length 3)This pattern continuesHow does it help get to the formula?Slide22
Number of Divisions of Length n
How many divisions of a rectangle of length n are possible?
Rectangle Length
1
2
3
4
5
n
# of different divisions
1
2
48
162n-1
Recursive approach – idea of building upReverse this recursive count!
#(length 1) = 1 = 20#(length 2) = 2*#(length 1) =
2 = 21 #(length 3) = 2*#(length 2) = 4 = 22
#(length 4) = 2*#(length 3) = 8 = 2
3Continue via induction
#(length n) = 2*#(length n-1) = 2*2n-2
= 2n-1 Slide23
A Finer Counting
Consider the number of blocks in each division.For example, in how many ways can a length 3 rectangle be broken
into 2 blocks?
Explore block counts in Activity 2.
1
1
1
1
1
2
2
3Slide24
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2
Rectangle of length 3
Rectangle of length 4
Rectangle of length 5
Rectangle of length 6Slide25
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 3
Rectangle of length 4
Rectangle of length 5
Rectangle of length 6Slide26
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 4
Rectangle of length 5
Rectangle of length 6Slide27
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
Rectangle of length 6Slide28
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6Slide29
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6
1
5
10
10
5
1Slide30
Pascal’s Triangle!Slide31
Pascal’s Triangle
Named after Blaise Pascal (1623-1662) but known much earlier.Slide32
Pascal’s Triangle
Row 0
Row 1
Row 2
Row 3
Pas(0,0) = 1
Pas(1,0) = Pas(1,1) = 1
Pas(2,0) = Pas(2,2) = 1,
Pas(2,1) = 1 + 1 =
Pas(1,0)
+
Pas(1,1
)Pas(3,0) = Pas(3,3)
= 1, Pas(3,1) = Pas(2,0) + Pas(2,1) Pas(
3,2) = Pas(2,1) + Pas(2,2)
Pas(n,0) = Pas(n,n) = 1Pas(n,k) = Pas(n-1,k-1) + Pas(n-1,k)
How do we relate this to the block divisions?Slide33
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6
1
5
10
10
5
1Slide34
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
Pas(0,0)
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6
1
5
10
10
5
1Slide35
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
Pas(0,0)
0
0
00
0
Rectangle of length 2Pas(1,0)Pas(1,1)
0000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6
1
5
10
10
5
1Slide36
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
Pas(0,0)
0
0
00
0
Rectangle of length 2Pas(1,0)Pas(1,1)
0000
Rectangle of length 3Pas(2,0)
Pas(2,1)Pas(2,2)0
00
Rectangle of length 41
33
10
0
Rectangle of length 51
464
10
Rectangle of length 6
1
5
10
10
5
1Slide37
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
Pas(0,0)
0
0
00
0
Rectangle of length 2Pas(1,0)Pas(1,1)
0000
Rectangle of length 3Pas(2,0)
Pas(2,1)Pas(2,2)0
00
Rectangle of length 4Pas(3,0)
Pas(3,1)Pas(3,2)
Pas(3,3)0
0
Rectangle of length 51
464
10
Rectangle of length 6
1
5
10
10
5
1
The # of ways to divide a rectangle of length n into k blocks is Pas(n-1,k-1)Slide38
Going to Work
Iva Jean lives 16 blocks from workEach day, she will take a different pathAfter exhausting all paths, Iva Jean can retire from her latest jobQuestion: When does Iva Jean get to retire?
work
homeSlide39
Going to Work
Being a practical person, Iva Jean only moves systematically toward her goal.Moves only up (North) or to the right (East)Activity 3 – counting pathsSlide40
Going to Work
How many paths to an office 4 blocks away?(4,0)
(3,1)
(2,2)
(1,3)
(0,4)
Only 1 path
4
paths
Only 1 path
6 paths
4
pathsSlide41
Paths to Work
1
4
6
1
4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3
3
5
10
10
5Slide42
Paths to Work - Observations
If the office is at (3,2), then is how many blocks from home?(3,2) is 5 blocks from homeThe number of paths to (3,2) is: 10In terms of Pascal’s Triangle: The number of paths to (3,2) is:
Pas(5,2)Notice that this is the same as Pas(5,3)Why is this all true? Slide43
Closer look at (3,2)
Let E=East and N=NorthThink of paths as combos of E’s and N’sHow many E’s and how many N’s in a path to (3,2)?3 E’s and 2 N’s(# of paths) = (# of words with
3 E’s and 2 N’s)EEENN, EENNE, ENNEE, NNEEE, EENEN, ENENE, NENEE, ENEEN, NEENE, NEEENHow do we more easily compute this?Slide44
Closer look at (3,2)
Every path to (3,2) has length 5Every path to (3,2) is a word consisting of 5 lettersNot just any word of 5 letters5 letter words with exactly 3 E’s and 2 N’sThink about placing the 2N’s into the 5 spots for letters (3 E’s, 2 N’s)
This is 5C2 = 5!/(2!3!)Slide45
Counting Paths
# paths to (3,2) is 5C2By symmetry of paths, same as # paths to (2,3)# paths to (2,3) is
5C3Confirms 5C2
=
5
C
3
But, # paths is also Pas(5,2)
# of paths to (
n,k) is (n+k
)Ck = Pas(n+k,k)Entries of Pascal’s Triangle are the binomial coefficients!
Do we already know this?
(a+b)n+kSlide46
Summing Up
First, how long until Iva Jean can retire?# of paths to (8,8) is Pas(16,8) = 16C816C
8 = 16!/(8!8!) = 12,870 days ≈ 35 yearsCounting of block divisions of rectangles & Counting of paths both lead to same structurePascal’s TriangleSlide47
Pascal Patterns
+
+
+
+
+
+
1
2
4
8
Where have we seen this? Slide48
Pascal Patterns
-
-
+
-
+
-
0
0
0
An interesting pattern,
b
ut any relationship to
e
ither of our problems?
Slide49
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
00
0
Rectangle of length 2110
000
Rectangle of length 31
210
00
Rectangle of length 41
33
100
Rectangle of length 5
14
64
10
Rectangle of length 6
1
5
10
10
5
1Slide50
A Finer Counting
# of blocks in Rectangle
12
3
4
5
6
Rectangle of length 1
1
0
0
0
00
Rectangle of length 21
100
00
Rectangle of length 312
10
00
Rectangle of length 4
13
31
00
Rectangle of length 5
14
6
4
1
0
Rectangle of length 6
1
5
10
10
5
1
+
+
+
-
-
-
The # of divisions into even # of blocks equals the # of divisions into odd# of blocks! Slide51
Pascal Patterns
-
-
+
-
+
-
0
0
0
0
How do we express this pattern in terms of the triangle entries?
That is, use Pas(
n,k
).
Pas(n,0)-Pas(n,1)+Pas(n,2)- ... +(-1)
k
Pas
(
n,k
)+ ... +(-1)
n
Pas
(
n,n)=0Slide52
Pascal PatternsSlide53
Pascal PatternsSlide54
Pascal Patterns
Pas(4,2) = 6 = 3+2+1
= Pas(3,1)+Pas(2,1)+Pas(1,1)
= Pas(3,2)+Pas(2,1)+Pas(1,0) Slide55
Pascal Patterns
Pas(8,3) = 56 = 35+15+5+1
= Pas(7,3)+Pas(6,2)+Pas(5,1)+Pas(4,0)
= Pas(7,4)+Pas(6,4)+Pas(5,4)+Pas(4,4) = Pas(8,5)
Notice, in either formula, need to be
decreasing something!Slide56
Pascal Patterns
Pas(10,6) = 210 = 126+56+21+6+1
=
Pas(9,5)+Pas(8,5)+Pas(7,5)+Pas(6,5)+Pas(5,5)
=
Pas(9,4)+Pas(8,3)+Pas(7,2)+Pas(6,1)+Pas(5,0)
= Pas(10,4) Slide57
Pascal Patterns
Pas(
n,k
) = Pas(n-1,k-1)+Pas(n-2,k-1)+Pas(n-3,k-1)+...+Pas(k-1,k-1)
=
Pas(n-1
,n-k)
+Pas(n-2
,n-k
-1)+Pas(n-2
,n-k-2)+...+Pas(n-k-1,0) = Pas(n,n
-k) Hockey Stick Pattern FormulaSlide58
Pascal Patterns
10*6*35 = 2100 = 5*20*21
Pas(5,3)*Pas(6,5)*Pas(7,4) =
Pas(
5,4)
*Pas(
6,3)
*Pas(
7,5)
7*56*36 = 14112 = 21*8*84
Pas(7,1)*Pas(8,3)*Pas(9,2) =
Pas
(7,2)
*Pas
(8,1)
*Pas(9,3) Slide59
Pascal Patterns
Pas(
n,
k
)*Pas(n+1,k+2)*Pas(n+2,k+1) =
Pas
(n,k+1)
*Pas
(n+1,k)
*Pas
(n+2,k+2)
Star of David Pattern
Also tells us that the product of all six
entries around Pas(n+1,k+1) is a
Perfect Square!
And, if you take the pattern to the edge,
you get Perfect Cubes! 1*10*6 + 4*1*15 + 5 = 125 = 53Slide60
Golden Ratio
A Rose by any other name...Golden SectionGolden Mean
Divine ProportionDivine Section
Golden Number
Part II – Geometric and Algebraic PatternsSlide61
Golden Ratio
Studied for thousands of yearsMathematiciansArtists
BiologistsPhysicists
MusiciansSlide62
What is the Golden
Ratio?Two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger quantity.
The common ratio isSlide63
What is the Golden
Ratio?This implies that the golden ratio is a fixed constant, like π. Can we compute it?Slide64Slide65
Solve it!Slide66
Algebraic Implications
The reciprocal of phi is one less than phi.Slide67
Algebraic Implications
What number is one more than phi?Slide68
Algebraic Implications
Let’s play with the recursive nature:
Can we do it again?
Have far can this go?Slide69
Algebraic Implications
Continued Fraction ExpansionSlide70
Algebraic Implications
Convergents
of the continued fraction
What do you notice?Slide71
Algebraic Implications
Will Fibonacci return?Slide72
Algebraic Implications
Try these! Compute the numbers
How do these relate to
the Golden Ratio?Slide73
Algebraic Implications
Prove that the continued square root (infinite surd) also holds:Slide74
Trigonometric Implications
How would you get this one?!
When you see the number 5,
what shape do you think of?Slide75
Trigonometric Implications
Compute the lengths of the segments
L
and
M
.Slide76
Trigonometric Implications
HINT:
Consider the center triangle, ABC.Also, consider the smaller (similar) triangle, BCD that is indicated.
You figure it out!
A
B
C
DSlide77
Trigonometric Implications
HINT:
Consider the center triangle, ABC.
Also, consider the smaller (similar) triangle, BCD that is indicated.
B
C
D
A
α=36
β/2
β=72
1
1
1
L-1Slide78
Trigonometric Implications
B
C
D
A
α=36
β/2
β=72
1
°
°
1
1
L-1
ABC and BCD are similar triangles.
So,Slide79
Trigonometric ImplicationsSlide80
Trigonometric Implications
Now, how do we get
:
π/5
cos
(
π/
5) = (|L|/2)/1Slide81
Golden Rectangle –
Building the Golden Ratio
Start with a square;
side length 1Slide82
Golden Rectangle –
Building the Golden Ratio
Bisect the square.Slide83
Golden Rectangle –
Building the Golden Ratio
Draw the diagonal.
How long is the diagonal?Slide84
Golden Rectangle –
Building the Golden Ratio
Swing the diagonaldown along the base.
How long is the extended
base?Slide85
Golden Rectangle –
Building the Golden Ratio
What is the ratio of the length to width of the rectangle?Slide86
Golden Rectangle –
Who Cares?Slide87
Golden Rectangle –
Who Cares?
13.380
8.280
Ratio=1.6159
0.1% difference
From Golden RatioSlide88
Golden Rectangle – Another Construction
1
1
2
3
5
Rectangle Ratios:
1:1
2:1
3:2
5:3
8:5
Where did we see these?
Convergents
of
continued fractionSlide89
Golden SpiralSlide90
Golden SpiralSlide91
Golden SpiralSlide92
Back to Fibonacci
We saw that the ratios of consecutive Fibonacci numbers
approach the golden ratio.
Why?!Slide93
Back to Fibonacci
This is a statement about limits.
Proof?Slide94
Back to FibonacciSlide95
Back to FibonacciSlide96
Back to FibonacciSlide97
Back to Fibonacci
So, the Golden Ratio is the limit of Fibonacci ratios.
Can the Golden Ratio tell us anything about the
Fibonacci numbers??Slide98Slide99Slide100
Binet’s
FormulaSlide101
PlottingSlide102
PlottingSlide103
Pascal’s TriangleSlide104
Pascal’s Triangle
1
1
2
3
5
8Slide105
Pascal’s Triangle & the Number e?
1
1
2
9
96
162,000
2500
26,471,025Slide106
Pascal’s Triangle & the Number e?
1; 1; 2; 9; 96; 2500; 162,000; 26,471,025Slide107
Pascal’s Triangle & the Number e?
1; 1; 2; 9; 96; 2500; 162,000; 26,471,025
Do you see the patterns?Slide108
Pascal’s Triangle & the Number e?
1; 1; 2; 9; 96; 2500; 162,000; 26,471,025Slide109
Pi in Pascal’s Triangle
What is special about these numbers?
Triangular numbers
Sum of consecutive counting numbers
1+2+…+n = n(n+1)/2 =
n+1
C
2Slide110
Pi in Pascal’s Triangle
Similarly, we can add consecutive triangular numbers
Tetrahedral numbersSlide111
Pi in Pascal’s Triangle