SOME FIBONACCI SURPRISES - PowerPoint Presentation

Slide1

SOME FIBONACCI SURPRISES

The Power of Visualization

James

Tanton

MAA Mathematician-at-Large

t

anton.math@gmail.com

Curriculum Inspirations:

www.maa.org/ci

Mathematical Stuff:

www.jamestanton.com

Mathematical Courses:

www.gdaymath.comSlide2
Slide3
Slide4

Visualization in the curriculum

* “Visual” or “Visualization” appears 34 times in the ninety-three pages of the U.S. Common Cores State Standards

- 22 times in reference to grade 2-6 students using visual models for fractions

- 1 time in grade 2 re comparing shapes

- 5 times re representing data in statistics and modeling

- 4 times re graphing functions and interpreting features of graphs

- 2 times in geometry re visualizing relationships between two- and three-dimensional objects.

* Alberta curriculum: Recognised HS core mathematical process: [V] Visualization “involves thinking in pictures and images, and the ability to perceive, transform and recreate different aspects of the

world” (Armstrong, 1993, p. 10). The use of visualization in the study of mathematics provides students with opportunities to understand mathematical concepts and make connections among them. Slide5

The Fibonacci numbers…

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

… arise in a myriad of contexts. Slide6

For example, count the number of sequences of

Os and Xs of a given length, avoiding two consecutive Xs. Slide7

Go the other extreme: Insist that

Xs come in pairs!Slide8

Count the number of ordered partitions of a given integer that avoid 1:Slide9

Or count the number of ordered partitions that use two different types of 1! Slide10

Count the number of ways to arrange non-nested parentheses around

a string of objects: Slide11

Count the number of ways to stack (two-dimensional) cannon balls so that each row is contiguous. Slide12

OR …

The language of ABEEBA uses only three letters of the alphabet: A, B, and E. * No word begins with an E. * No word has the letter E immediately following an A.

* All other combinations of letters are words.Slide13

Actually, words that begin with an E are allowed. They are swear words.

Let’s count the swear words in the language of ABEEBA.Slide14

And so on!

There is one visual model that explains all these examples - and so much more!Slide15

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

The Fibonacci numbers

arise from a classic

honeycomb path-counting

puzzle: Slide16
Slide17

We see that the number of paths from the left cell to the

N

th cell of the honeycomb is the Nth Fibonacci number F .

NSlide18

Actually … the number of paths between any two cells

N cells apart is F .

N

LET’S NOW HAVE SOME FUN!Slide19

Here’s a path between two cells on the top row.

The path “touches down” one groups of dots in the bottom row.

This matches placing non-nested parentheses around those dots. Slide20

Each step in a path either follows a diagonal step or skips over

two diagonal steps. We see a partition of a number into 1s and 2s.

N

diagonals

corresponds to a path to the

N+1

th

cell.Slide21

Indeed every second Fibonacci number.

2 =

1

+ 1 = 1 +

1

=

1

+

1

= 1 + 1 5

partitions

3 = 2 + 1 = 2 +

1

= 1 + 2 = 1

+ 2

= 1+1+1 =

1

+1+1 = 1+

1

+1 = 1+1+

1

=

1

+

1

+1 =

1

+1+

1

= 1+

1

+

1

=

1

+

1

+

1

13

partitions

1 =

1

2

partitions

Count the ordered partitions of a number with two different types of 1.Slide22

Each diagonal step in a path is a “break” between dots.

There are an odd number of dots between

breaks.

1

+

1

+

3

+

1

+

5

+

1 = 12

EXAMPLE

: 5 = 3+1+1 = 1+3+1 = 3+1+1

= 1+1+1+1+1

There are five odd partitions of five.Slide23

M

issed circles define the path.

Focus on the dots that a path misses.

The first dot is never circled, the last dot is never circled,

no two consecutive dots are ever circled.

Ignoring end dots, draw

Os

on the dots hit and

Xs

on the dots missed.

Get sequence of

Os

and

Xs

along the zigzag avoiding two consecutive

Xs

.

EXAMPLE

: {1,2,3} has

{ }, {1}, {2}, {3}, {1,3}

as five such subsets.

OXOOXOXOOXSlide24

No two consecutive dots circled

 No section

just 1 segment long.

Each dot a path misses breaks the zigzag line of segments:

Draw extra line segments at the beginning and end of the zigzag.

So we have a one-free partition of the number of the zig-zag steps.

EXAMPLE

: 5 = 2+3 = 3+2

There are three one-free partitions of five.Slide25

The language of ABEEBA.

Consider paths between cells on the top row. Slide26

Cannonball stacks:

Each stack gives a sequence of diagonal and horizontal steps…

… which gives a path between two cells on the top row. Slide27

Another approach to cannonballs:

There are three ways to make a stack with an extra row:

* add a ball to the left of a previously made stack

* add a ball to the right of a previously made stack

* place a previously made stack on top of a next row

But there is double counting. Slide28

But we saw today

We have the identity: Slide29

(Inspired by

a conversation with Sam

Vandervelde)

PRODUCTS OF PARTITIONS

ALWAYS FIBONACCI?

Take all partitions on

N

, multiply terms, and add.Slide30

Consider paths

that end on a lower

cell. Consider all possible locations of

the UP

steps.

Single

DOWN

in each section.

The sum of all such products counts all paths.

Answer must be

.

F

24

A partition of 12 with terms multiplied together.

12 dots on top row

24 dots in all

There

2x4x1x2x3 ways to place the

DOWNs

.Slide31

Consider every second Fibonacci number:

1, 2, 5, 13, 34, 89, …

SUM ALWAYS ONE LESS THAN A FIBONACCI NUMBER?

TRIANGULAR SUMSSlide32

Consider paths that end on a particular top dot,

2

N

+ 1

.

(N+

1 dots on top row,

N

dots on bottom row.)

1 path

N

paths touch just 1 lower dot

This accounts for all paths:

.

(N-

k

)

places for span of

k

dots.

F paths using those dots.

2k+1

2k+1

(

N-k

) x F paths touch a span of

k

dots.Slide33

FIBONACCI IDENTITIES

Take your

favourite

Fibonacci identity and

try to prove

via paths.

EXAMPLE

: Slide34

PROOF

: Slide35

Here is my

favourite

identity:

I’ve always wondered …

Is there a formula for the quotient?

EXAMPLE

: 12 is divisible by 2, 3, 4 and 6, and

F = 144 is divisible by F = 1, F = 2, F = 3 and F = 8.

12

2

3

6

4Slide36

What if there is a remainder?

So in full generality …Slide37

PROOF:Slide38
Slide39
Slide40

There are a myriad of Fibonacci identities that perhaps can be proved via path walking. (Care to try?)Slide41

CHALLENGE FOR TODAY: Slide42

Weeks of fun to be had all with the

POWER OF A PICTURE!

THANK YOU!Slide43

SOME FIBONACCI SURPRISES

The Power of Visualization

James

Tanton

MAA Mathematician-at-Large

t

anton.math@gmail.com

Curriculum Inspirations:

www.maa.org/ci

Mathematical Stuff:

www.jamestanton.com

Mathematical Courses:

www.gdaymath.com