From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus NYS Master Teachers...
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Transcript of From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus NYS Master Teachers...
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?
Dividing Rectangles2. Recursive approach
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!!
Dividing Rectangles2. Recursive approach
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
Dividing Rectangles2. Recursive approach
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
Dividing Rectangles2. Recursive approach
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
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
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
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 4
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 4 8
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 4 8 16
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 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?
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 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?
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 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
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 2
1 1 0 0 0 0
Rectangle of length 3Rectangle of length 4Rectangle of length 5Rectangle of length 6
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4Rectangle of length 5Rectangle of length 6
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5Rectangle of length 6
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
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?
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
A Finer Counting# of blocks in Rectangle
1 2 3 4 5 6
Rectangle of length 1
Pas(0,0) 0 0 0 0 0
Rectangle of length 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
A Finer Counting# of blocks in Rectangle
1 2 3 4 5 6
Rectangle of length 1
Pas(0,0) 0 0 0 0 0
Rectangle of length 2
Pas(1,0) Pas(1,1) 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
A Finer Counting# of blocks in Rectangle
1 2 3 4 5 6
Rectangle of length 1
Pas(0,0) 0 0 0 0 0
Rectangle of length 2
Pas(1,0) Pas(1,1) 0 0 0 0
Rectangle of length 3
Pas(2,0) Pas(2,1) Pas(2,2) 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
A Finer Counting# of blocks in Rectangle
1 2 3 4 5 6
Rectangle of length 1
Pas(0,0) 0 0 0 0 0
Rectangle of length 2
Pas(1,0) Pas(1,1) 0 0 0 0
Rectangle of length 3
Pas(2,0) Pas(2,1) Pas(2,2) 0 0 0
Rectangle of length 4
Pas(3,0) Pas(3,1) Pas(3,2) Pas(3,3) 0 0
Rectangle of length 5
1 4 6 4 1 0
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)
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 - 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
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 6 4 1 0
Rectangle of length 6
1 5 10 10 5 1
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 2
1 1 0 0 0 0
Rectangle of length 3
1 2 1 0 0 0
Rectangle of length 4
1 3 3 1 0 0
Rectangle of length 5
1 4 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!
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 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?
Algebraic Implications
Let’s play with the recursive nature:
Can we do it again?
Have far can this go?
Trigonometric Implications
How would you get this one?!
When you see the number 5, what shape do you think of?
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
D
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-1
Trigonometric Implications
B C
D
A α=36
β/2
β=72
1
°
°
1
1
L-1
ABC and BCD are similar triangles.
So,
Golden Rectangle – Building the Golden Ratio
Swing the diagonaldown along the base.
How long is the extendedbase?
Golden Rectangle – Building the Golden Ratio
What is the ratio of the length to width of the rectangle?
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
Back to Fibonacci
We saw that the ratios of consecutive Fibonacci numbersapproach the golden ratio.
Why?!
Back to Fibonacci
So, the Golden Ratio is the limit of Fibonacci ratios.
Can the Golden Ratio tell us anything about theFibonacci numbers??
Pascal’s Triangle & the Number e?
1; 1; 2; 9; 96; 2500; 162,000; 26,471,025
Do you see the patterns?
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