Two Heads Are Better Than None
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1 Two Heads Are Better Than None Or, Me and the Fibonaccis Steve Kennedy (with help from Matt Stafford) Carleton College and MAA Books Problems are the lifeblood of mathematics. David Hilbert Indiana MAA Spring Section Meeting 204
2 The Problem The Game: Flip a fair coin repeatedly until two consecutive heads appear, stop.
3 The Problem The Game: Flip a fair coin repeatedly until two consecutive heads appear, stop. The Problem: What is the probability that the game will ever end? (Intuitively, this is big, but consider HTHTHTHT..., or TTTTTT....)
4 The Problem The Game: Flip a fair coin repeatedly until two consecutive heads appear, stop. The Problem(s): What is the probability that the game will ever end? (Intuitively, this is big, but consider HTHTHTHT..., or TTTTTT...) What is the probability that the game ends after exactly n flips?
5 Counting Arguments HH TH n = 2 HT TT Each event equally likely, so P 2 = /4.
6 Counting Arguments HH TH n = 2 HT TT Each event is equally likely, so P 2 = /4. HHH THH HHT THT n=3 HTH TTH HTT TTT Each event is equally likely, so P 3 = /8.
7 Counting Arguments: Wait just one minute! HH TH n = 2 HT TT Each event is equally likely, so P 2 = /4. HHH THH HHT THT n=3 HTH TTH HTT TTT Each event is equally likely, so P 3 = /8.
8 Non Counting Arguments Yes, neither HHH nor HHT can actually happen. So, the probability of seeing THH is really /6. Right?
9 Non Counting Arguments Neither HHH nor HHT can actually happen. So, the probability of seeing THH is really /6? Well, not exactly. The probability of seeing THH is /6 times the probability that the game didn t end in exactly two flips.
10 Non Counting Arguments Neither HHH nor HHT can actually happen. So, the probability of seeing THH is really /6? Well, not exactly. The probability of seeing THH is /6 times the probability that the game didn t end in exactly two flips. So, actually, the probability of seeing THH is:
11 Non Counting Arguments Yes, neither HHH nor HHT can actually happen. So, the probability of seeing THH is really /6. Well, not exactly. The probability of seeing THH is /6 times the probability that the game didn t end in exactly two flips. So, actually, the probability of seeing THH is: = 8. This had to work out.
12 Non Counting Arguments Yes, neither HHH nor HHT can actually happen. So, the probability of seeing THH is really /6. Well, not exactly. The probability of seeing THH is /6 times the probability that the game didn t end in exactly two flips. So, actually, the probability of seeing THH is: = 8. This had to work out because the coin doesn t know the rules of the game.
13 Counting Arguments HH TH n = 2 HT TT Each event is equally likely, so P 2 = /4. HHH THH HHT THT n=3 HTH TTH HTT TTT Each event is equally likely, so P 3 = /8. HHHH HTHH THHH TTHH HHHT HTHT THHT TTHT n=4 HHTH HTTH THTH TTTH HHTT HTTT THTT TTTT Each event is equally likely, so P 4 = 2/6.
14 Counting Arguments n=5 HHHHH HTHHH THHHH TTHHH HHHHT HTHHT THHHT TTHHT HHHTH HTHTH THHTH TTHTH HHHTT HTHTT THHTT TTHTT HHTHH HTTHH THTHH TTTHH HHTHT HTTHT THTHT TTTHT HHTTH HTTTH THTTH TTTTH HHTTT HTTTT THTTT TTTTT So, P 5 = 3/32.
15 Counting Arguments n=5 HHHHH HTHHH THHHH TTHHH HHHHT HTHHT THHHT TTHHT HHHTH HTHTH THHTH TTHTH HHHTT HTHTT THHTT TTHTT HHTHH HTTHH THTHH TTTHH HHTHT HTTHT THTHT TTTHT HHTTH HTTTH THTTH TTTTH HHTTT HTTTT THTTT TTTTT So, P 5 = 3/32. n=6 HTHTHH, HTTTHH, THTTHH, TTHTHH, TTTTHH So, P 6 = 5/64.
16 Counting Arguments n=5 HHHHH HTHHH THHHH TTHHH HHHHT HTHHT THHHT TTHHT HHHTH HTHTH THHTH TTHTH HHHTT HTHTT THHTT TTHTT HHTHH HTTHH THTHH TTTHH HHTHT HTTHT THTHT TTTHT HHTTH HTTTH THTTH TTTTH HHTTT HTTTT THTTT TTTTT So, P 5 = 3/32. n=6 HTHTHH, HTTTHH, THTTHH, TTHTHH, TTTTHH So, P 6 = 5/64. n=7
17 Counting Arguments n=5 HHHHH HTHHH THHHH TTHHH HHHHT HTHHT THHHT TTHHT HHHTH HTHTH THHTH TTHTH HHHTT HTHTT THHTT TTHTT HHTHH HTTHH THTHH TTTHH HHTHT HTTHT THTHT TTTHT HHTTH HTTTH THTTH TTTTH HHTTT HTTTT THTTT TTTTT So, P 5 = 3/32. n=6 HTHTHH, HTTTHH, THTTHH, TTHTHH, TTTTHH So, P 6 = 5/64. n=7 Homework
18 Search for Pattern 4, 8, 2 6, 3 32, 5 64,...
19 Search for Pattern 4, 8, 2 6, 3 32, 5 64,... P n = x n 2 n, What s x n?
20 Search for Pattern 4, 8, 2 6, 3 32, 5 64,... P n = x n 2 n, What s x n? x n =,, 2, 3, 5, 8, 3, 2,...
21 An Unnecessary(?) Aside The Fibonacci Numbers 202
22 An Unnecessary(?) Aside The Fibonacci Numbers 202 The Rules for Rabbit Reproduction. Gestation period is one month.
23 An Unnecessary(?) Aside The Fibonacci Numbers 202 The Rules for Rabbit Reproduction. Gestation period is one month. 2. Rabbits born in male/female pairs.
24 An Unnecessary(?) Aside The Fibonacci Numbers 202 The Rules for Rabbit Reproduction. Gestation period is one month. 2. Rabbits born in male/female pairs. 3. Rabbits reach sexual maturity in one month.
25 An Unnecessary(?) Aside The Fibonacci Numbers 202 The Rules for Rabbit Reproduction. Gestation period is one month. 2. Rabbits born in male/female pairs. 3. Rabbits reach sexual maturity in one month. 4. Rabbits are always pregnant. (And never die.)
26 An Unnecessary(?) Aside The Fibonacci Numbers 202 The Rules for Rabbit Reproduction. Gestation period is one month. 2. Rabbits born in male/female pairs. 3. Rabbits reach sexual maturity in one month. 4. Rabbits are always pregnant. (And never die.) Start with one newborn pair, how many pairs will you have n months later?
27 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs 0 Immature Pairs Total Pairs
28 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs 0 Immature Pairs Total Pairs
29 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs 0 Immature Pairs 0 Total Pairs
30 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs 0 Immature Pairs 0 Total Pairs 2
31 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs 0 2 Immature Pairs 0 Total Pairs 2 3
32 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs Immature Pairs Total Pairs
33 An Unnecessary(?) Aside The Fibonacci Numbers 202 Month: Mature Pairs Immature Pairs Total Pairs Thus, F n+ = F n + F n.
34 An Answer Theorem The sequence x n is the Fibonacci sequence. Example n=6 T T T HT HT
35 An Answer Theorem The sequence x n is the Fibonacci sequence. Example THTTHH n=6 TTHTHH TTTTHH So, x 6 x 5 + x 4. HTHTHH HTTTHH
36 An Answer Theorem The sequence x n is the Fibonacci sequence. Example THTTHH n=6 TTHTHH TTTTHH So, x 6 x 5 + x 4. HTHTHH HTTTHH Conversely *HTTHH *THTHH *TTTHH **HTHH **TTHH
37 An Answer Theorem The sequence x n is the Fibonacci sequence. Example THTTHH n=6 TTHTHH TTTTHH So, x 6 x 5 + x 4. HTHTHH HTTTHH THTTHH Conversely TTHTHH TTTTHH So, x 6 x 5 + x 4. HTHTHH HTTTHH
38 More Questions What is the probability that the game ends after two or three flips?
39 More Questions What is the probability that the game ends after two or three flips? = 3 8
40 More Questions What is the probability that the game ends after two or three flips? = 3 8 What is the probability that the game ends in four or fewer flips?
41 More Questions What is the probability that the game ends after two or three flips? = 3 8 What is the probability that the game ends in four or fewer flips? = 2
42 More Questions What is the probability that the game ends after two or three flips? = 3 8 What is the probability that the game ends in four or fewer flips? = 2 What is the probability that the game ends in twenty or fewer flips?
43 More Questions What is the probability that the game ends after two or three flips? = 3 8 What is the probability that the game ends in four or fewer flips? = 2 What is the probability that the game ends in twenty or fewer flips? 20 P n.983 n=2
44 Original Question What is the probability that the game ever ends?
45 Original Question What is the probability that the game ever ends? lim n k=2 n P n = k=0 F k 2 k+2
46 Original Question What is the probability that the game ever ends? lim n k=2 Does this series converge? n P n = k=0 F k 2 k+2
47 Convergence = S
48 Convergence = S = 2
49 Convergence = S = = 8
50 Convergence = S = = = 6
51 Convergence = S = = = = = 32
52 Convergence = S = = = = = = 64...
53 Convergence = S = = = = = = So, S =
54 The Sum Recall, S =
55 The Sum Recall, S = We just decided, S =
56 The Sum Recall, S = We just decided, So, S = S = ( S = S
57 The Sum Recall, S = We just decided, So, S = S = ( S = S S =
58 The Sum = n=2 F n 2 2 n =
59 The Sum n=2 F n 2 2 n =
60 The Sum n=2 F n 2 2 n =!!!
61 The Miracle n=2 F n 2 2 n =!!!
62 The Miracle n=2 F n 2 2 n =!!! THIS IS A MIRACLE!!!
63 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events.
64 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events. The series converged.
65 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events. The series converged. (And I could prove it!)
66 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events. The series converged. (And I could prove it!) We could find the value to which it converged.
67 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events. The series converged. (And I could prove it!) We could find the value to which it converged. That value was rational.
68 The Miracle Definition A miracle is the simultaneous occurrence of two or more zero-probability events. The series converged. (And I could prove it!) We could find the value to which it converged. That value was rational. That value was the simplest possible rational.
69 The Miracle
70 The Very Good Reason g(x) = F k x k = + x + 2x 2 + 3x 3 + 5x k=0
71 The Very Good Reason g(x) = F k x k = + x + 2x 2 + 3x 3 + 5x x g(x) = k=0 F k x k+ = x + x 2 + 2x 3 + 3x k=0
72 The Very Good Reason g(x) = F k x k = + x + 2x 2 + 3x 3 + 5x x g(x) = x 2 g(x) = k=0 F k x k+ = x + x 2 + 2x 3 + 3x k=0 F k x k+2 = x 2 + x 3 + 2x k=0
73 The Very Good Reason g(x) = F k x k = + x + 2x 2 + 3x 3 + 5x x g(x) = x 2 g(x) = k=0 F k x k+ = x + x 2 + 2x 3 + 3x k=0 F k x k+2 = x 2 + x 3 + 2x k=0 So, g(x) xg(x) x 2 g(x) =
74 The Very Good Reason g(x) = F k x k = + x + 2x 2 + 3x 3 + 5x x g(x) = x 2 g(x) = k=0 F k x k+ = x + x 2 + 2x 3 + 3x k=0 F k x k+2 = x 2 + x 3 + 2x k=0 So, g(x) xg(x) x 2 g(x) = g(x) = x x 2
75 The Very Good Reason g(x) = x x 2
76 The Very Good Reason g(x) = g(x) = x x 2 F k x k = + x + 2x 2 + 3x 3 + 5x k=0
77 The Very Good Reason g ( ) = 2 g(x) = g(x) = x x 2 F k x k = + x + 2x 2 + 3x 3 + 5x k=0 ( F k 2 k=0 ) k = ( ) ( ) ( )
78 The Very Good Reason g ( ) = 2 g(x) = g(x) = x x 2 F k x k = + x + 2x 2 + 3x 3 + 5x k=0 ( F k 2 k=0 ) k = ( ) ( ) = ( )
79 The Very Good Reason g ( ) = 2 g(x) = g(x) = x x 2 F k x k = + x + 2x 2 + 3x 3 + 5x k=0 ( F k 2 k=0 ) k = ( ) ( ) = = 4 ( ( )
80 Technicalities F k x k = + x + 2x 2 + 3x 3 + 5x k=0
81 Technicalities F k x k = + x + 2x 2 + 3x 3 + 5x k=0 This is the Taylor series about 0 for The interval of convergence is ( golden mean. λ, λ x x 2. ), where λ = + 5 2, i.e. the
82 Technicalities F k x k = + x + 2x 2 + 3x 3 + 5x k=0 This is the Taylor series about 0 for The interval of convergence is ( golden mean. λ, λ x x 2. ), where λ = + 5 2, i.e. the (NB The radius of convergence, λ, is approximately.68, so 2 is comfortably inside.)
83 Two Great Tastes That Taste Great Together The function, g(x) = x x 2, is called the generating function for the Fibonacci numbers.
84 Turning the Page Rewrite g(x) using partial fractions: x x 2 = A λx + B + (λ )x
85 Turning the Page Rewrite g(x) using partial fractions: Solve for A and B: x x 2 = A λx + B + (λ )x A = λ 5 B = λ 5
86 Turning the Page Rewrite g(x) using partial fractions: Solve for A and B: x x 2 = A λx + B + (λ )x A = λ 5 B = λ 5 By the known formula for geometric series: A λx = A (λx) k k=0
87 Turning the Page Rewrite g(x) using partial fractions: Solve for A and B: x x 2 = A λx + B + (λ )x A = λ 5 B = λ 5 By the known formula for geometric series: Similarly, A λx = A (λx) k k=0 B + (λ )x = B( ) k ((λ )x) k k=0
88 Never in a Million Years x x 2 = A (λx) k + B( ) k ((λ )x) k k=0 k=0
89 Never in a Million Years x x 2 = A (λx) k + B( ) k ((λ )x) k k=0 Recalling the values of A and B, F k x k = k=0 k=0 k=0 [ λ k+ 5 + ( )k (λ ) k+ 5 ] x k
90 Never in a Million Years x x 2 = A (λx) k + B( ) k ((λ )x) k k=0 Recalling the values of A and B, F k x k = k=0 k=0 k=0 Power series expansions are unique! So, [ λ k+ 5 + ( )k (λ ) k+ 5 ] x k F k = λk+ 5 + ( )k (λ ) k+ 5
91 Be wise, generalize! Heads Numerators Generating and Recursion Function One,,,,... x a n = a n Two,, 2, 3,... x x 2 a n = a n + a n 2 Three,, 2, 4, 7,... x x 2 x 3 a n = a n + a n 2 + a n 3 Four,, 2, 4, 8, 5,... x x 2 x 3 x 4 a n = a n + a n 2 + a n 3 + a n 4 And so on...
92 Suppose you had a three-sided coin? Heads Numerators Generating and Recursion Function One, 2, 4, 8,... 2x a n = 2a n Two, 2, 6, 6, 48,... 2x 2x 2 a n = 2(a n + a n 2 ) Three, 2, 6, 8, 52, 52,... 2x 2x 2 2x 3 a n = 2(a n + a n 2 + a n 3 ) Four, 2, 6, 8, 54, 60,... 2x 2x 2 2x 3 2x 4 a n = 2(a n + a n 2 + a n 3 + a n 4 ) Homework: Do the n consecutive heads from an m-sided coin problem.
93 Fibonacci Fractions Do the long division problem: 0, 000 9, 899 =
94 Fibonacci Fractions Do the long division problem: 0, 000 9, 899 =
95 Fibonacci Fractions Do the long division problem: 0, 000 9, 899 = What just happened? ( ) g = (00) (00) =
96 Fibonacci Fractions Do the long division problem: 0, 000 9, 899 = What just happened? ( ) g = (00) (00) = g ( ), 000, 000 = , 999
97 Multiplying Weirdness Recall F k = 5 [ λ k+ + ( ) k (λ ) k+]
98 Multiplying Weirdness Recall F k = 5 [ λ k+ + ( ) k (λ ) k+] Here s a picture of those fractional parts: Pictured is the fractional part of 5 λ k, k =, 2,..., 5.
99 Your Real Homework The picture for the fractional part of.5 n for n = (This is what we expect to see.)
100 Your Real Homework Conjecture There are not very many real numbers, γ, that have the property that there exists a constant C so that the sequence consisting of the fractional parts of Cγ n, n =, 2,... has only finitely many limit points.
101 Your Real Homework Conjecture There are not very many real numbers, γ, that have the property that there exists a constant C so that the sequence consisting of the fractional parts of Cγ n, n =, 2,... has only finitely many limit points. Not very many = measure zero = first category = nowhere dense
102 Your Real Homework: Hints The number + 2 = p is even better than the golden mean. p p p p p p
103 Your Real Homework: Hints p = + 2 p 2 = p 3 = p 8 = p 9 = p 0 =
104 A Final Word Thanks to the local organizers: Adam Coffman, Rob Merkovsky and Marc Lipman.
105 A Final Word Thanks to the local organizers: Adam Coffman, Rob Merkovsky and Marc Lipman. Thanks to IUPU Fort Waye.
106 A Final Word Thanks to the local organizers: Adam Coffman, Rob Merkovsky and Marc Lipman. Thanks to IUPU Fort Waye. Thank you for your kind attention.
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