1. From Counting to Pascal: A Journey through Number Theory, Geometry, and Calculus NYS Master Teachers March 9, 2015 Dave Brown – dabrown@ithaca.edu Slides available at http://faculty.ithaca.edu/dabrown/docs/masterteachers/

2. 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

3. 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 2

4. 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

5. Dividing Rectangles 1.Constructive approach Systematically build all divisions of a given length 2.Recursive approach Use divisions of shorter rectangles to build longer one

6. Dividing Rectangles 1.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 9 2 2-block 5-block3-block 1 3 4 5 6 7 8 9 2 2-block 4-block

7. Dividing Rectangles 1.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 9 2

8. Dividing Rectangles 2.Recursive approach Use divisions of shorter rectangles to build longer one Every division has a final (rightmost) block of length 1 OR of length greater than 1

9. Dividing Rectangles 2.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?

10. Dividing Rectangles 2.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!!

11. Dividing Rectangles 2.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

12. Dividing Rectangles 2.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-blocks to the end Lengthen the end-block by 1

13. Dividing Rectangles 2.Recursive approach Use divisions of shorter rectangles to build longer one How would you build all rectangles of length 6?

14. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions

15. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 1

16. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 12

17. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 124

18. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 1248

19. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 124816

20. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 1248162 n-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 2 9 decisions about dividing to make; hence, 2 9 divisions How does this generalize to rectangles of length n?

21. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 1248162 n-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?

22. Number of Divisions of Length n How many divisions of a rectangle of length n are possible? Rectangle Length 12345n # of different divisions 1248162 n-1 2.Recursive approach – idea of building up Reverse this recursive count! #(length 1) = 1 = 2 0 #(length 2) = 2*#(length 1) = 2 = 2 1 #(length 3) = 2*#(length 2) = 4 = 2 2 #(length 4) = 2*#(length 3) = 8 = 2 3 Continue via induction #(length n) = 2*#(length n-1) = 2*2 n-2 = 2 n-1

23. 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 23

24. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 Rectangle of length 3 Rectangle of length 4 Rectangle of length 5 Rectangle of length 6

25. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 Rectangle of length 4 Rectangle of length 5 Rectangle of length 6

26. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 Rectangle of length 5 Rectangle of length 6

27. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 Rectangle of length 6

28. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6

29. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

30. Pascal’s Triangle!

31. Pascal’s Triangle Named after Blaise Pascal (1623-1662) but known much earlier.

32. 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) = 1 Pas(n,k) = Pas(n-1,k-1) + Pas(n-1,k) How do we relate this to the block divisions?

33. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

34. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 Pas(0,0)00000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

35. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 Pas(0,0)00000 Rectangle of length 2 Pas(1,0)Pas(1,1)0000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

36. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 Pas(0,0)00000 Rectangle of length 2 Pas(1,0)Pas(1,1)0000 Rectangle of length 3 Pas(2,0)Pas(2,1)Pas(2,2)000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

37. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 Pas(0,0)00000 Rectangle of length 2 Pas(1,0)Pas(1,1)0000 Rectangle of length 3 Pas(2,0)Pas(2,1)Pas(2,2)000 Rectangle of length 4 Pas(3,0)Pas(3,1)Pas(3,2)Pas(3,3)00 Rectangle of length 5 146410 Rectangle of length 6 1510 51 The # of ways to divide a rectangle of length n into k blocks is Pas(n-1,k-1)

38. 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

39. 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

40. 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

41. Paths to Work 1 4 6 1 4 1 1 1 11 1 1 1 1 1 1 1 1 1 1 2 3 3 5 10 5

42. 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?

43. 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?

44. 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 5 C 2 = 5!/(2!3!)

45. Counting Paths # paths to (3,2) is 5 C 2 By symmetry of paths, same as # paths to (2,3) # paths to (2,3) is 5 C 3 Confirms 5 C 2 = 5 C 3 But, # paths is also Pas(5,2) # of paths to (n,k) is (n+k) C k = Pas(n+k,k) Entries of Pascal’s Triangle are the binomial coefficients! Do we already know this? (a+b) n+k

46. Summing Up First, how long until Iva Jean can retire? # of paths to (8,8) is Pas(16,8) = 16 C 8 16 C 8 = 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

47. Pascal Patterns + + + + + + 1 2 4 8 Where have we seen this?

48. Pascal Patterns - - + - + - 0 0 0 An interesting pattern, but any relationship to either of our problems?

49. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51

50. A Finer Counting # of blocks in Rectangle 123456 Rectangle of length 1 100000 Rectangle of length 2 110000 Rectangle of length 3 121000 Rectangle of length 4 133100 Rectangle of length 5 146410 Rectangle of length 6 1510 51 +++ - -- The # of divisions into even # of blocks equals the # of divisions into odd# of blocks!

51. 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)=0

52. Pascal Patterns

54. 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)

55. 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!

56. 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)

57. 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 Formula

58. 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)

59. 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 = 5 3

60. Golden Ratio A Rose by any other name... Golden Section Golden Mean Divine Proportion Divine Section Golden Number Part II – Geometric and Algebraic Patterns

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

62. 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

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

65. Solve it!

66. Algebraic Implications The reciprocal of phi is one less than phi.

67. Algebraic Implications What number is one more than phi?

68. Algebraic Implications Let’s play with the recursive nature: Can we do it again? Have far can this go?

69. Algebraic Implications Continued Fraction Expansion

70. Algebraic Implications Convergents of the continued fraction What do you notice?

71. Algebraic Implications Will Fibonacci return?

72. Algebraic Implications Try these! Compute the numbers How do these relate to the Golden Ratio?

73. Algebraic Implications Prove that the continued square root (infinite surd) also holds:

74. Trigonometric Implications How would you get this one?! When you see the number 5, what shape do you think of?

75. Trigonometric Implications Compute the lengths of the segments L and M.

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

77. Trigonometric Implications HINT: Consider the center triangle, ABC. Also, consider the smaller (similar) triangle, BCD that is indicated. BC D A α=36 β/2 β=72 1 1 1 L-1

78. Trigonometric Implications BC D A α=36 β/2 β=72 1 ° ° 1 1 L-1 ABC and BCD are similar triangles. So,

79. Trigonometric Implications

80. Now, how do we get : π/5 cos(π/5) = (|L|/2)/1

81. Golden Rectangle – Building the Golden Ratio Start with a square; side length 1

82. Golden Rectangle – Building the Golden Ratio Bisect the square.

83. Golden Rectangle – Building the Golden Ratio Draw the diagonal. How long is the diagonal?

84. Golden Rectangle – Building the Golden Ratio Swing the diagonal down along the base. How long is the extended base?

85. Golden Rectangle – Building the Golden Ratio What is the ratio of the length to width of the rectangle?

86. Golden Rectangle – Who Cares?

87. Golden Rectangle – Who Cares? 13.380 8.280 Ratio=1.6159 0.1% difference From Golden Ratio

88. Golden Rectangle – Another Construction 11 2 3 5 Rectangle Ratios: 1:1 2:1 3:2 5:3 8:5 Where did we see these? Convergents of continued fraction

89. Golden Spiral

92. Back to Fibonacci We saw that the ratios of consecutive Fibonacci numbers approach the golden ratio. Why?!

93. Back to Fibonacci This is a statement about limits. Proof?

94. Back to Fibonacci

97. So, the Golden Ratio is the limit of Fibonacci ratios. Can the Golden Ratio tell us anything about the Fibonacci numbers??

100. Binet’s Formula

101. Plotting

103. Pascal’s Triangle

104. 112358

105. Pascal’s Triangle & the Number e? 1 1 2 9 96 162,000 2500 26,471,025

106. Pascal’s Triangle & the Number e? 1; 1; 2; 9; 96; 2500; 162,000; 26,471,025

107. Pascal’s Triangle & the Number e? 1; 1; 2; 9; 96; 2500; 162,000; 26,471,025 Do you see the patterns?

108. Pascal’s Triangle & the Number e? 1; 1; 2; 9; 96; 2500; 162,000; 26,471,025

109. 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 2

110. Pi in Pascal’s Triangle Similarly, we can add consecutive triangular numbers Tetrahedral numbers

111. Pi in Pascal’s Triangle