From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus NYS Master Teachers March 9, 2015 Dave Brown – [email protected] Slides.

From Counting to Pascal: A Journey through Number Theory, Geometry,

and Calculus

NYS Master TeachersMarch 9, 2015

Dave Brown – [email protected] available at http://faculty.ithaca.edu/dabrown/docs/masterteachers/

Goals

• Use counting (combinatorics) to generate patterns

• Link counting patterns to algebra, geometry, trigonometry, & calculus

• Explore patterns in counting structures• Draw connections among various branches of

math• Learn some discrete mathematics

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 12 2

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 twice

Dividing Rectangles

1. Constructive approachSystematically build all divisions of a given length

2. Recursive approachUse divisions of shorter rectangles to build longer one

Dividing Rectangles1. Constructive approach

Systematically 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 92

2-block 5-block3-block

1 3 4 5 6 7 8 92

2-block 4-block4-block

Dividing Rectangles1. Constructive approach

Systematically 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 92

Dividing Rectangles

2. 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 1

Dividing Rectangles2. Recursive approach

Use divisions of shorter rectangles to build longer one• Remove all final blocks of length 1• Do you get all rectangles of length 4?


Use 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!!


Use 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 1


Use 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-blocksto the end

Lengthen the end-blockby 1


Use divisions of shorter rectangles to build longer one

How would you build all rectanglesof length 6?

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


Rectangle Length

1 2 3 4 5 n


1


Rectangle Length

1 2 3 4 5 n


1 2


Rectangle Length

1 2 3 4 5 n


1 2 4


Rectangle Length

1 2 3 4 5 n


1 2 4 8


Rectangle Length

1 2 3 4 5 n


1 2 4 8 16


Rectangle Length

1 2 3 4 5 n


1 2 4 8 16 2n-1

• How do we prove this?1. 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 divisions• How does this generalize to rectangles of length n?


Rectangle Length

1 2 3 4 5 n


1 2 4 8 16 2n-1

• How do we prove this?2. 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 continues• How does it help get to the formula?


Rectangle Length

1 2 3 4 5 n


1 2 4 8 16 2n-1

2. Recursive approach – idea of building up• Reverse 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 = 23

• Continue via induction• #(length n) = 2*#(length n-1) = 2*2n-2 = 2n-1

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 12 2 3

A Finer Counting# of blocks in Rectangle

1 2 3 4 5 6

Rectangle of length 1

1 0 0 0 0 0

Rectangle of length 2Rectangle of length 3Rectangle of length 4Rectangle of length 5Rectangle of length 6


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0

Rectangle of length 3Rectangle of length 4Rectangle of length 5Rectangle of length 6


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0

Rectangle of length 4Rectangle of length 5Rectangle of length 6


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0

Rectangle of length 5Rectangle of length 6


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0



1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1

Pascal’s Triangle!

Pascal’s TriangleNamed after Blaise Pascal (1623-1662) but known much earlier.

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?


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1


1 2 3 4 5 6


Pas(0,0) 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1


1 2 3 4 5 6


Pas(0,0) 0 0 0 0 0


Pas(1,0) Pas(1,1) 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1


1 2 3 4 5 6


Pas(0,0) 0 0 0 0 0


Pas(1,0) Pas(1,1) 0 0 0 0


Pas(2,0) Pas(2,1) Pas(2,2) 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1


1 2 3 4 5 6


Pas(0,0) 0 0 0 0 0


Pas(1,0) Pas(1,1) 0 0 0 0


Pas(2,0) Pas(2,1) Pas(2,2) 0 0 0


Pas(3,0) Pas(3,1) Pas(3,2) Pas(3,3) 0 0


1 4 6 4 1 0


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)

Going to Work• Iva Jean lives 16 blocks from work• Each day, she will take a different path• After exhausting all paths, Iva Jean can retire from

her latest job• Question: When does Iva Jean get to retire?

work

home

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 paths

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 paths

Paths to Work

1

4

6

1

4

1 1111 1 1 1

1

1

1

1

1

1

1

2

3

3

5

10

10

5

Paths to Work - Observations

• If the office is at (3,2), then is how many blocks from home?

• (3,2) is 5 blocks from home• The number of paths to (3,2) is: 10• In 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?

Closer look at (3,2)

• Let E=East and N=North• Think of paths as combos of E’s and N’s• How 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, NEEEN• How do we more easily compute this?

Closer look at (3,2)

• Every path to (3,2) has length 5• Every path to (3,2) is a word consisting of 5

letters• Not just any word of 5 letters• 5 letter words with exactly 3 E’s and 2 N’s• Think about placing the 2N’s into the 5 spots for

letters (3 E’s, 2 N’s)• This is 5C2 = 5!/(2!3!)

Counting Paths

• # paths to (3,2) is 5C2

• By symmetry of paths, same as # paths to (2,3)• # paths to (2,3) is 5C3

• Confirms 5C2 = 5C3

• 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+k

Summing Up

• First, how long until Iva Jean can retire?• # of paths to (8,8) is Pas(16,8) = 16C8

• 16C8 = 16!/(8!8!) = 12,870 days ≈ 35 years• Counting of block divisions of rectangles &

Counting of paths both lead to same structure• Pascal’s Triangle

Pascal Patterns

++

++ + +

1

24

8

Where have we seen th

is?

Pascal Patterns

--

+- + -

00

0An interesting pattern,but any relationship toeither of our problems?


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1


1 2 3 4 5 6


1 0 0 0 0 0


1 1 0 0 0 0


1 2 1 0 0 0


1 3 3 1 0 0


1 4 6 4 1 0


1 5 10 10 5 1

+ + +-- -

The # of divisions into even # of blocks equals the # of divisions into odd# of blocks!

Pascal Patterns

--

+- + -

0

00

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)kPas(n,k)+ ... +(-1)nPas(n,n)=0

Pascal Patterns

Pascal Patterns

Pascal PatternsPas(4,2) = 6 = 3+2+1

= Pas(3,1)+Pas(2,1)+Pas(1,1)

= Pas(3,2)+Pas(2,1)+Pas(1,0)

Pascal PatternsPas(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!

Pascal PatternsPas(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)

Pascal PatternsPas(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 Formula

Pascal Patterns10*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)

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 sixentries 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 = 53

Golden RatioA Rose by any other name...• Golden Section• Golden Mean• Divine Proportion• Divine Section• Golden Number

Part II – Geometric and Algebraic Patterns

Golden Ratio• Studied for thousands of years• Mathematicians• Artists• Biologists• Physicists• Musicians

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 is

What is the Golden Ratio?This implies that the golden ratio is a fixed constant, like π. Can we compute it?

Solve it!

Algebraic Implications

The reciprocal of phi is one less than phi.


What number is one more than phi?


Let’s play with the recursive nature:

Can we do it again?

Have far can this go?


Continued Fraction Expansion


Convergents of the continued fraction

What do you notice?


Will Fibonacci return?


Try these! Compute the numbers

How do these relate to the Golden Ratio?


Prove that the continued square root (infinite surd) also holds:

Trigonometric Implications

How would you get this one?!

When you see the number 5, what shape do you think of?


Compute the lengths of the segments L and M.


HINT:Consider the center triangle, ABC.Also, consider the smaller (similar) triangle, BCD that is indicated.You figure it out!

A

B C

D


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-1


B C

D

A α=36

β/2

β=72

1

°

°

1

1

L-1

ABC and BCD are similar triangles.

So,



Now, how do we get:

π/5

cos(π/5) = (|L|/2)/1

Golden Rectangle – Building the Golden Ratio

Start with a square; side length 1


Bisect the square.


Draw the diagonal.

How long is the diagonal?


Swing the diagonaldown along the base.

How long is the extendedbase?


What is the ratio of the length to width of the rectangle?

Golden Rectangle – Who Cares?

Golden Rectangle – Who Cares?

13.380

8.280

Ratio=1.6159

0.1% differenceFrom Golden Ratio

Golden Rectangle – Another Construction

1 1

23

5

Rectangle Ratios:1:12:13:25:38:5

Where did we see these?

Convergents of continued fraction

Golden Spiral

Golden Spiral

Golden Spiral

Back to Fibonacci

We saw that the ratios of consecutive Fibonacci numbersapproach the golden ratio.

Why?!

Back to Fibonacci

This is a statement about limits.

Proof?

Back to Fibonacci

Back to Fibonacci

Back to Fibonacci

Back to Fibonacci

So, the Golden Ratio is the limit of Fibonacci ratios.

Can the Golden Ratio tell us anything about theFibonacci numbers??

Binet’s Formula

Plotting

Plotting

Pascal’s Triangle

Pascal’s Triangle

1 1 2 3 5 8

Pascal’s Triangle & the Number e?

1

1

2

9

96

162,000

2500

26,471,025


1; 1; 2; 9; 96; 2500; 162,000; 26,471,025


1; 1; 2; 9; 96; 2500; 162,000; 26,471,025

Do you see the patterns?


1; 1; 2; 9; 96; 2500; 162,000; 26,471,025

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+1C2


• Similarly, we can add consecutive triangular numbers• Tetrahedral numbers


From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus NYS Master Teachers March 9, 2015 Dave Brown – [email protected] Slides.

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