Two Heads Are Better Than None Or, Me and the Fibonaccisskennedy/KSTwoHeads.pdf · The Problem The Game: Flip a fair coin repeatedly until two consecutive heads appear, stop. The

Two Heads Are Better Than NoneOr, Me and the Fibonaccis

Steve Kennedy(with help from Matt Stafford)

Carleton College andMAA Books

Problems are the lifeblood of mathematics.

— David Hilbert

Indiana MAA Spring Section Meeting 2014

The Problem

The Game:

Flip a fair coin repeatedly until two consecutiveheads appear, stop.

The Problem:

I What is the probability that the game will ever end?(Intuitively, this is big, but consider HTHTHTHT. . ., orTTTTTT. . ..)

The Problem

The Game:


The Problem:

I What is the probability that the game will ever end?(Intuitively, this is big, but consider HTHTHTHT. . ., orTTTTTT. . ..)

The Problem

The Game:


The Problem(s):

I What is the probability that the game will ever end?(Intuitively, this is big, but consider HTHTHTHT..., orTTTTTT...)

I What is the probability that the game ends after exactly nflips?

Counting Arguments

n = 2HH THHT TT

Each event equally likely, so P2 = 1/4.

Counting Arguments

n = 2HH THHT TT

Each event is equally likely, so P2 = 1/4.

n=3

HHH THHHHT THTHTH TTHHTT TTT


Counting Arguments: Wait just one minute!

n = 2HH THHT TT


n=3



Non–Counting Arguments

Yes, neither HHH nor HHT can actually happen. So, theprobability of seeing THH is really 1/6. Right?


Neither HHH nor HHT can actually happen. So, the probability ofseeing THH is really 1/6?

Well, not exactly. The probability of seeing THH is 1/6 times theprobability that the game didn’t end in exactly two flips.


Neither HHH nor HHT can actually happen. So, the probability ofseeing THH is really 1/6?


So, actually, the probability of seeing THH is:

1

6× 6

8.


Yes, neither HHH nor HHT can actually happen. So, theprobability of seeing THH is really 1/6.



1

6× 6

8=

1

8.

This had to work out.


Yes, neither HHH nor HHT can actually happen. So, theprobability of seeing THH is really 1/6.



1

6× 6

8=

1

8.

This had to work out—because the coin doesn’t know the rules ofthe game.

Counting Arguments

n = 2HH THHT TT


n=3



n=4

HHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTT


Counting Arguments

n=5

HHHHH HTHHH THHHH TTHHHHHHHT HTHHT THHHT TTHHTHHHTH HTHTH THHTH TTHTHHHHTT HTHTT THHTT TTHTTHHTHH HTTHH THTHH TTTHHHHTHT HTTHT THTHT TTTHTHHTTH HTTTH THTTH TTTTHHHTTT HTTTT THTTT TTTTT

So, P5 = 3/32.

n=6 HTHTHH, HTTTHH, THTTHH, TTHTHH, TTTTHH

So, P6 = 5/64.

n=7 Homework

Counting Arguments

n=5


So, P5 = 3/32.


So, P6 = 5/64.

n=7 Homework

Counting Arguments

n=5


So, P5 = 3/32.


So, P6 = 5/64.

n=7

Homework

Counting Arguments

n=5


So, P5 = 3/32.


So, P6 = 5/64.

n=7 Homework

Search for Pattern

14,

18,

216,

332,

564, . . .

Pn = xn2n , What’s xn?

xn = 1, 1, 2, 3, 5, 8, 13, 21, . . .

Search for Pattern

14,

18,

216,

332,

564, . . .


xn = 1, 1, 2, 3, 5, 8, 13, 21, . . .

Search for Pattern

14,

18,

216,

332,

564, . . .


xn = 1, 1, 2, 3, 5, 8, 13, 21, . . .

An Unnecessary(?) Aside

The Fibonacci Numbers — 1202

The Rules for Rabbit Reproduction

1. 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 nmonths later?











































Month: 0 1 2 3 4 5Mature Pairs 0Immature Pairs 1

Total Pairs 1



Month: 0 1 2 3 4 5Mature Pairs 0 1Immature Pairs 1

Total Pairs 1



Month: 0 1 2 3 4 5Mature Pairs 0 1Immature Pairs 1 0

Total Pairs 1 1



Month: 0 1 2 3 4 5Mature Pairs 0 1 1Immature Pairs 1 0 1

Total Pairs 1 1 2



Month: 0 1 2 3 4 5Mature Pairs 0 1 1 2Immature Pairs 1 0 1 1

Total Pairs 1 1 2 3



Month: 0 1 2 3 4 5Mature Pairs 0 1 1 2 3 5Immature Pairs 1 0 1 1 2 3

Total Pairs 1 1 2 3 5 8


The Fibonacci Numbers — 1202Month: 0 1 2 3 4 5Mature Pairs 0 1 1 2 3 5Immature Pairs 1 0 1 1 2 3

Total Pairs 1 1 2 3 5 8

Thus, Fn+1 = Fn + Fn−1.

An Answer

TheoremThe sequence xn is the Fibonacci sequence.

Example

n=6T HTT HTT

An Answer


Example

n=6THTTHH HTHTHHTTHTHH HTTTHHTTTTHH

So, x6 ≤ x5 + x4.

An Answer


Example


So, x6 ≤ x5 + x4.

Conversely*HTTHH **HTHH*THTHH **TTHH*TTTHH

An Answer


Example


So, x6 ≤ x5 + x4.

ConverselyTHTTHH HTHTHHTTHTHH HTTTHHTTTTHH

So, x6 ≥ x5 + x4.

More Questions

I What is the probability that the game ends after two or threeflips?

1

4+

1

8=

3

8

I What is the probability that the game ends in four or fewerflips?

1

4+

1

8+

2

16=

1

2

I What is the probability that the game ends in twenty or fewerflips?

20∑n=2

Pn ≈ .983

More Questions


1

4+

1

8=

3

8


1

4+

1

8+

2

16=

1

2


20∑n=2

Pn ≈ .983

More Questions


1

4+

1

8=

3

8


1

4+

1

8+

2

16=

1

2


20∑n=2

Pn ≈ .983

More Questions


1

4+

1

8=

3

8


1

4+

1

8+

2

16=

1

2


20∑n=2

Pn ≈ .983

More Questions


1

4+

1

8=

3

8


1

4+

1

8+

2

16=

1

2


20∑n=2

Pn ≈ .983

More Questions


1

4+

1

8=

3

8


1

4+

1

8+

2

16=

1

2


20∑n=2

Pn ≈ .983

Original Question

What is the probability that the game ever ends?

limn→∞

n∑k=2

Pn =∞∑k=0

Fk

2k+2

Does this series converge?

Original Question


limn→∞

n∑k=2

Pn =∞∑k=0

Fk

2k+2


Original Question


limn→∞

n∑k=2

Pn =∞∑k=0

Fk

2k+2


Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

2

1

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

8

1

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

16

1

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

32

1

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

Convergence

1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . . = S

1

4+

1

8+

1

16+

1

32+

1

64+

1

128+ . . . =

1

21

16+

1

32+

1

64+

1

128+ . . . =

1

81

32+

1

64+

1

128+ . . . =

1

161

64+

1

128+ . . . =

1

321

64+

1

128+ . . . =

1

321

128+ . . . =

1

64. . .

So, S = 12 + 1

8 + 116 + 2

32 + 364 + . . ..

The Sum

Recall,

S =1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . .

We just decided,

S =1

2+

1

8+

1

16+

2

32+

3

64+

5

128+ . . .

So,

S =1

2+

1

2

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .

S =1

2+

1

2S

S = 1

The Sum

Recall,

S =1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . .

We just decided,

S =1

2+

1

8+

1

16+

2

32+

3

64+

5

128+ . . .

So,

S =1

2+

1

2

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .

S =1

2+

1

2S

S = 1

The Sum

Recall,

S =1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . .

We just decided,

S =1

2+

1

8+

1

16+

2

32+

3

64+

5

128+ . . .

So,

S =1

2+

1

2

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .

S =1

2+

1

2S

S = 1

The Sum

Recall,

S =1

4+

1

8+

2

16+

3

32+

5

64+

8

128+ . . .

We just decided,

S =1

2+

1

8+

1

16+

2

32+

3

64+

5

128+ . . .

So,

S =1

2+

1

2

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .

S =1

2+

1

2S

S = 1

The Sum

1

4+

1

8+

2

16+

3

32+

5

64+ . . . = 1

∞∑n=2

Fn−2

2n= 1

The Sum

∞∑n=2

Fn−2

2n= 1

!!!

The Sum

∞∑n=2

Fn−2

2n= 1!!!

The Miracle

∑∞n=2

Fn−22n = 1!!!

THIS IS A MIRACLE!!!

The Miracle

∑∞n=2

Fn−22n = 1!!!

THIS IS A MIRACLE!!!

The Miracle

DefinitionA miracle is the simultaneous occurrence of two or morezero-probability events.

I The series converged. (And I could prove it!)

I We could find the value to which it converged.

I That value was rational.

I That value was the simplest possible rational.

The Miracle


I The series converged.

(And I could prove it!)




The Miracle






The Miracle






The Miracle






The Miracle






The Miracle

The Very Good Reason

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

x g(x) =∞∑k=0

Fkxk+1 = x + x2 + 2x3 + 3x4 + . . .

x2g(x) =∞∑k=0

Fkxk+2 = x2 + x3 + 2x4 + . . .

So,g(x)− xg(x)− x2g(x) = 1

g(x) =1

1− x − x2


g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

x g(x) =∞∑k=0

Fkxk+1 = x + x2 + 2x3 + 3x4 + . . .

x2g(x) =∞∑k=0

Fkxk+2 = x2 + x3 + 2x4 + . . .

So,g(x)− xg(x)− x2g(x) = 1

g(x) =1

1− x − x2


g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

x g(x) =∞∑k=0

Fkxk+1 = x + x2 + 2x3 + 3x4 + . . .

x2g(x) =∞∑k=0

Fkxk+2 = x2 + x3 + 2x4 + . . .

So,g(x)− xg(x)− x2g(x) = 1

g(x) =1

1− x − x2


g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

x g(x) =∞∑k=0

Fkxk+1 = x + x2 + 2x3 + 3x4 + . . .

x2g(x) =∞∑k=0

Fkxk+2 = x2 + x3 + 2x4 + . . .

So,g(x)− xg(x)− x2g(x) = 1

g(x) =1

1− x − x2


g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

x g(x) =∞∑k=0

Fkxk+1 = x + x2 + 2x3 + 3x4 + . . .

x2g(x) =∞∑k=0

Fkxk+2 = x2 + x3 + 2x4 + . . .

So,g(x)− xg(x)− x2g(x) = 1

g(x) =1

1− x − x2


g(x) =1

1− x − x2

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

g

(1

2

)=∞∑k=0

Fk

(1

2

)k

= 1+1

2+2

(1

2

)2

+3

(1

2

)3

+5

(1

2

)4

+. . .

= 1 +1

2+

2

4+

3

8+

5

16+ . . .

= 4

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .


g(x) =1

1− x − x2

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

g

(1

2

)=∞∑k=0

Fk

(1

2

)k

= 1+1

2+2

(1

2

)2

+3

(1

2

)3

+5

(1

2

)4

+. . .

= 1 +1

2+

2

4+

3

8+

5

16+ . . .

= 4

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .


g(x) =1

1− x − x2

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

g

(1

2

)=∞∑k=0

Fk

(1

2

)k

= 1+1

2+2

(1

2

)2

+3

(1

2

)3

+5

(1

2

)4

+. . .

= 1 +1

2+

2

4+

3

8+

5

16+ . . .

= 4

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .


g(x) =1

1− x − x2

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

g

(1

2

)=∞∑k=0

Fk

(1

2

)k

= 1+1

2+2

(1

2

)2

+3

(1

2

)3

+5

(1

2

)4

+. . .

= 1 +1

2+

2

4+

3

8+

5

16+ . . .

= 4

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .


g(x) =1

1− x − x2

g(x) =∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

g

(1

2

)=∞∑k=0

Fk

(1

2

)k

= 1+1

2+2

(1

2

)2

+3

(1

2

)3

+5

(1

2

)4

+. . .

= 1 +1

2+

2

4+

3

8+

5

16+ . . .

= 4

(1

4+

1

8+

2

16+

3

32+

5

64+ . . .

Technicalities

∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .

This is the Taylor series about 0 for 11−x−x2 .

The interval of convergence is(−1

λ ,1λ

), where λ = 1+

√5

2 , i.e. thegolden mean.

(NB The radius of convergence, 1λ , is approximately .618, so 1

2 iscomfortably inside.)

Technicalities

∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .



λ ,1λ

), where λ = 1+

√5




Technicalities

∞∑k=0

Fkxk = 1 + x + 2x2 + 3x3 + 5x4 + . . .



λ ,1λ

), where λ = 1+

√5




Two Great Tastes That Taste Great Together

The function, g(x) = 11−x−x2 , is called the generating function for

the Fibonacci numbers.

Turning the PageRewrite g(x) using partial fractions:

1

1− x − x2=

A

1− λx+

B

1 + (λ− 1)x

Solve for A and B:

A =λ√5

B =λ− 1√

5

By the known formula for geometric series:

A

1− λx=∞∑k=0

A (λx)k

Similarly,

B

1 + (λ− 1)x=∞∑k=0

B(−1)k ((λ− 1)x)k


1

1− x − x2=

A

1− λx+

B

1 + (λ− 1)x

Solve for A and B:

A =λ√5

B =λ− 1√

5


A

1− λx=∞∑k=0

A (λx)k

Similarly,

B

1 + (λ− 1)x=∞∑k=0

B(−1)k ((λ− 1)x)k


1

1− x − x2=

A

1− λx+

B

1 + (λ− 1)x

Solve for A and B:

A =λ√5

B =λ− 1√

5


A

1− λx=∞∑k=0

A (λx)k

Similarly,

B

1 + (λ− 1)x=∞∑k=0

B(−1)k ((λ− 1)x)k


1

1− x − x2=

A

1− λx+

B

1 + (λ− 1)x

Solve for A and B:

A =λ√5

B =λ− 1√

5


A

1− λx=∞∑k=0

A (λx)k

Similarly,

B

1 + (λ− 1)x=∞∑k=0

B(−1)k ((λ− 1)x)k

Never in a Million Years

1

1− x − x2=∞∑k=0

A (λx)k +∞∑k=0

B(−1)k ((λ− 1)x)k

Recalling the values of A and B,

∞∑k=0

Fkxk =∞∑k=0

[λk+1

√5

+(−1)k(λ− 1)k+1

√5

]xk

Power series expansions are unique! So,

Fk =λk+1

√5

+(−1)k(λ− 1)k+1

√5


1

1− x − x2=∞∑k=0

A (λx)k +∞∑k=0

B(−1)k ((λ− 1)x)k


∞∑k=0

Fkxk =∞∑k=0

[λk+1

√5

+(−1)k(λ− 1)k+1

√5

]xk


Fk =λk+1

√5

+(−1)k(λ− 1)k+1

√5


1

1− x − x2=∞∑k=0

A (λx)k +∞∑k=0

B(−1)k ((λ− 1)x)k


∞∑k=0

Fkxk =∞∑k=0

[λk+1

√5

+(−1)k(λ− 1)k+1

√5

]xk


Fk =λk+1

√5

+(−1)k(λ− 1)k+1

√5

Be wise, generalize!

Heads Numerators Generatingand Recursion Function

One 1, 1, 1, 1, . . . 11−x

an = an−1Two 1, 1, 2, 3, . . . 1

1−x−x2an = an−1 + an−2

Three 1, 1, 2, 4, 7, . . . 11−x−x2−x3

an = an−1 + an−2 + an−3Four 1, 1, 2, 4, 8, 15, . . . 1

1−x−x2−x3−x4an = an−1 + an−2 + an−3 + an−4

And so on . . .

“Suppose you had a three-sided coin?”

Heads Numerators Generatingand Recursion Function

One 1, 2, 4, 8, . . . 11−2x

an = 2an−1Two 1, 2, 6, 16, 48, . . . 1

1−2x−2x2an = 2(an−1 + an−2)

Three 1, 2, 6, 18, 52, 152, . . . 11−2x−2x2−2x3

an = 2(an−1 + an−2 + an−3)Four 1, 2, 6, 18, 54, 160, . . . 1

1−2x−2x2−2x3−2x4an = 2(an−1 + an−2 + an−3 + an−4)

Homework: Do the n consecutive heads from an m-sided coinproblem.

Fibonacci Fractions

Do the long division problem:

10, 000

9, 899=

1.010203050813213455 . . .

What just happened?

g

(1

100

)= 1 +

1

100+

2

(100)2+

3

(100)3+ . . .

= 1 + .01 + .0002 + .000003 + . . .

g

(1

1000

)=

1, 000, 000

998, 999

Fibonacci Fractions


10, 000

9, 899= 1.010203050813213455 . . .

What just happened?

g

(1

100

)= 1 +

1

100+

2

(100)2+

3

(100)3+ . . .

= 1 + .01 + .0002 + .000003 + . . .

g

(1

1000

)=

1, 000, 000

998, 999

Fibonacci Fractions


10, 000

9, 899= 1.010203050813213455 . . .

What just happened?

g

(1

100

)= 1 +

1

100+

2

(100)2+

3

(100)3+ . . .

= 1 + .01 + .0002 + .000003 + . . .

g

(1

1000

)=

1, 000, 000

998, 999

Fibonacci Fractions


10, 000

9, 899= 1.010203050813213455 . . .

What just happened?

g

(1

100

)= 1 +

1

100+

2

(100)2+

3

(100)3+ . . .

= 1 + .01 + .0002 + .000003 + . . .

g

(1

1000

)=

1, 000, 000

998, 999

Multiplying Weirdness

Recall

Fk =1√5

[λk+1 + (−1)k(λ− 1)k+1

]

Here’s a picture of those fractional parts:

Pictured is the fractional part of 1√5λk , k = 1, 2, . . . , 15.

Multiplying Weirdness

Recall

Fk =1√5

[λk+1 + (−1)k(λ− 1)k+1

]Here’s a picture of those fractional parts:

Pictured is the fractional part of 1√5λk , k = 1, 2, . . . , 15.

Your Real Homework

The picture for the fractional part of 1.5n for n = 10 . . . 80. (Thisis what we expect to see.)

Your Real Homework

Conjecture

There are not very many real numbers, γ, that have the propertythat there exists a constant C so that the sequence consisting ofthe fractional parts of Cγn, n = 1, 2, . . . has only finitely manylimit points.

Not very many = measure zero= first category= nowhere dense

Your Real Homework

Conjecture

There are not very many real numbers, γ, that have the propertythat there exists a constant C so that the sequence consisting ofthe fractional parts of Cγn, n = 1, 2, . . . has only finitely manylimit points.

Not very many = measure zero= first category= nowhere dense

Your Real Homework: Hints

The number 1 +√

2 = p is even better than the golden mean.

p1 ≈ 2.41421

p2 ≈ 5.82842

p3 ≈ 14.07106

. . .

p8 ≈ 1153.99913

p9 ≈ 2786.00035

p10 ≈ 6725.99985

Your Real Homework: Hints

p1 = 1 +√

2

p2 = 3 + 2√

2

p3 = 7 + 5√

2

. . .

p8 = 577 + 408√

2

p9 = 1393 + 985√

2

p10 = 3363 + 2378√

2

A Final Word

Thanks to the local organizers: Adam Coffman, Rob Merkovskyand Marc Lipman.

Thanks to IUPU–Fort Waye.

Thank you for your kind attention.

A Final Word




A Final Word




Documents

Two Heads Are Better Than None Or, Me and the Fibonaccisskennedy/KSTwoHeads.pdf · The Problem The Game: Flip a fair coin repeatedly until two consecutive heads appear, stop. The