21121058 Fibonacci Numbers

7/27/2019 21121058 Fibonacci Numbers

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FibonacciFibonacciNumbersNumbers

By: Sara Miller

Advisor: Dr. Mihai Caragiu


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AbstractAbstractv We will investigate various ways of

proving identities involving FibonacciNumbers, such as, induction, linear

algebra (matrices), and combinatorics(0-1 sequences).

vWe will also look at one educationalactivity.


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SummarySummary

v 1. Introductionv 2. Mathematical Induction

v 3. Linear Algebra (matrices)v 4. Combinatorics (0-1 sequences)

v 5. Educational Activity


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1. Introduction1. Introduction

vLeonardo PisanoO Born around 1175 in Pisa, Italy

O His nickname was Fibonacci

O He traveled extensively with his father

O Wrote book: Liber Abbaciin 1220

Presented a numerical series now referredto as the Fibonacci numbers


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Fibonacci NumbersFibonacci Numbers

v 1, 1, 2 , 3, 5, 8, 13, 21, 34, 55,

89, 144, 233,


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Rabbit ProblemRabbit Problemv

A certain man put a pair of rabbits in aplace surrounded by a wall. How manypairs of rabbits can be produced from

that pair in a year if it is supposed thatevery month each pair begets a new pair

from which the second month on becomesproductive? (Liber Abbaci, chapter 12, p.283-4)


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Chart of Rabbit ProblemChart of Rabbit ProblemMonth Adult Pairs Baby Pairs Total Pairs

January 1 0 1

February 1 1 2

March 2 1 3

April 3 2 5

May 5 3 8June 8 5 13

July 13 8 21

August 21 13 34

September 34 21 55

October 55 34 89

November 89 55 144

December 144 89 233

January 233 144 377


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The Fibonacci LinearThe Fibonacci LinearRecurrenceRecurrence

1 11 2

f and f= =

1 2f f fn n n

= +

Initial Conditions:

Recurrence relation:


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1 2 3 4 5 6 7

1 1 2 3 5 8 13fn

n


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2. Mathematical Induction2. Mathematical Induction

2.1 Sum of Squares of Fibonacci Numbers:

2 2 2...1 2 1 , 1+ + + = + f f f f fn n n n

I will initially carry out the proof of this identityby induction. Then I will provide a visual proof.


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Proof by induction:

(a)For 1n = the formula takes the form

2 21 (1)(1) thus 1 11 1 2

f f f

= = =

Thus the first part of the induction(basis step) is finished.


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(b) Assuming the formula holds true for n, we willprove it for 1n+ .

Therefore,

2 2 2...

1 2 1

f f f f fn nn

+ + + =

+

And we will prove

2 2 2 2...1 2 1 1 2

f f f f f fn n n n+ + + + =+ + +


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Indeed,2 2 2 2

...1 2 1f f f fn n+ + + +

+

2

1 1

f f f

n n n

= +

+ +

By inductive hypothesis

1 1

f f fnn n

= ++ +

1 2

f fn n

=+ +

By1 2

f f fnn n+ =

+ +

Thus the induction is complete, and I have proved that

the formula holds for all n.

QED.


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Visual ProofVisual Proof

2 2 2...

1 2 1f f f f fn n n

+ + + =+


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... , 151 3 2 1 2

+ + + + =

f f f f f nn n

2.2 Sum of Odd Fibonacci Numbers:


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... 1 , 00 2 4 2 2 1

+ + + + =

+

f f f f f n

n n

2.3 Sum of Even Fibonacci Numbers:


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2... , 1

1 2 2 3 3 4 2 1 2 2

f f f f f f f f f n

n n n

+ + + + =

Proof:

(a)For 1n = the formula takes the form 2 21 1 =1 thus 1 1

1 2 2

f f f

= =

Thus the first part of the induction (basis step)

is finished.

2.4 Sum of Products of Consecutive Fibonacci

Numbers:


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(b) Assuming the formula holds true for n, we will proveit for

1n+.

For the inductive step, we will assume

2...1 2 2 3 3 4 2 1 2 2

f f f f f f f f fn n n

+ + + + =

And we will prove

2...1 2 2 3 3 4 2 1 2 2 2 1 2 1 2 2 2 2f f f f f f f f f f f f fn n n n n n n

+ + + + + + = + + + +


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Indeed,

...1 2 2 3 3 4 2 1 2 2 2 1 2 1 2 2

f f f f f f f f f f f fn n n n n n

+ + + + + + + + +

2 2 2 2 1 2 1 2 2

f f f f fn n n n n

= + ++ + +

By inductive Hypothesis

2 2 2 1 2 1 2 2

f f f f fn n n n n

= + ++ + +


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2 2 2 1 2 1 2 2f f f f f

n n n n n

= + ++ + +

2 2 2 2 1 2 2f f f f

n n n n= +

+ + +

2 2 2 2 1f f f

n n n

= ++ +

2 2 2

fn= + By

2 2 1 2 2f f fn n n+ =+ +

Thus the induction is complete, and I have provedthat the formula holds for all n.

QED


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3. Linear Algebra (Matrices)3. Linear Algebra (Matrices)

3.1 Cassinis Identity:

2 11 1

nf f fnn n = +

First we will introduce the transition matrix1 1

1 0T

=

It is easy to see that this matrix satisfies

1 for all 1

1

f fnn T nff nn

+ =


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It is easy to prove (by induction) that

1

1

f fnnnT

f fn n

+=

for any n=1,2,3,


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Lets take the determinants of both sides:

2det 1 1nT f f f nn n = +

But,

det det 1n nnT T

= =

Therefore,

2 11 1

nf f fnn n

= +

Thus we have proven the Cassinis Identity.

QED


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4. Combinatorics4. Combinatorics

4.1 0 - 1 Sequences

One important fact is that the number of 0 1 sequences

of length n without consecutive 1s is for every 12f nn + .

Lets prove this!


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First denote by An the number of 0-1 sequences of

length n without consecutive 1s.

Here is an example of a string of length 8 without

consecutive 1s:

0 1 0 0 1 0 1 0


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For 1n = , we have a single , so we have two possibilities a 0

or a 1,

Thus 2 , 21 1 2 3

A f f= = =+

therefore2

A fn n=

+

holds true for 1n= .

For 2n= , we have two , so we have three possibilities 00,

01, 10

Thus 3 , 3

2 2 2 4

A f f= = =+

and

2

A fnn

=+

holds true

for 2n = .

Thus the first part of the induction (basis step) is finished.


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I shall prove the induction step in the following way:

Assume is true for all .2A f k nk k=


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0 1 0 0 1 0 1 0

1A

n cellsn Ending with a 0

0 1 0 0 1 0 1 0

2 cellsn

0 1 0 0 1 0 0 1 2

An

Ending with a 1

Therefore,1 2 1 2 2 2 1 2

A A A f f f f fn nn n n n n n= + = + = + =

+ + + +QED


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We consider the identity 2 21 2 1

f f fn n n+ =

+ +

1n 1n 1n 1n 2n 2n

0

0 1 0

central cell central cell is 0 central cell is 1

2 1 1 2 1n n

+ = 21 1 1f f f

n n n =+ + + 2f f fn n n =

Therefore, 2 22 1 1f f fnn n= ++ + .

4.2 An Example of Combinatorial Proof:


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Many activities can be used in the classroom to

generate and investigate Fibonacci sequences. One isto have students place 1 and 2 cent stamps across the

top of a postcard (facing with correct side up) in

different arrangements to make up certain postageamounts. The number of different arrangements will

be a Fibonacci number.

5. Educational Activity:


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21121058 Fibonacci Numbers