Fibonacci Numbers and Binet Formula

(An Introduction to Number Theory)

By: (The Ladies) 2

Recurrence Sequence

• each further term of the sequence is defined as a function of the preceding terms (starting seed and rule)

• Fibonacci sequence(1,1,2,3,5,8,13,21,34,55...)

• Lucas sequence(2,1,3,4,7,11,18,29,47,76)

• take 3+8(1+2)+(3+5)(1+3)+(2+5)

(4)+(7) - can be shown to hold in general

Mathematical induction• Fibonacci (1,1,2,3,5,8,13...)

• 1,1+1,1+1+2,1+1+2+3,1+1+2+3+5,1+1+2+3+5+8...

• 1,2,4,7,12,20...

• +1 to each term

• 2,3,5,8,13,21...

• because 1+1+2+3+5+81+1+1+2+3+5+8(2+1)+2+3+5+8

(3+2)+3+5+8(5+3)+5+8(8+5)+8(13+8)

Fibonacci sequence patterns

Fibonacci (1,1,2,3,5,8,13...)neither arithmetic nor geometricso write it in a different way1/1=12/1=(1+1)/1 = (1+(1/1))3/2=(2+1)/2=1+(1/2)= 1+ 1/(1+(1/1))5/3=(3+2)/3=1+(2/3)=1+ 1/(1+ 1/(1+(1/1)))and so on (Golden ratio φ)

Golden Ratio φ

φ=1+1/φφ2=φ+1 quadratic equationφ=(1+sqrt(5))/2 (only the positive

answer)φ=1.618033989...

Golden Ratio and practical application

• most famous and controversial in history - human aesthetics

• Converting between km and miles• 1 mile= 1.6093 km• 13 km = 8 miles• Fibonacci (1,1,2,3,5,8,13,21...)

• OK, using Fibonacci numbers, how many miles are in 50 kilometers?? (show your work)

Binet Formula

• A formula to find a term in a Fibonacci numbers without generating previous terms

• Jacques Binet in 1843 - known to Euler and Bernoulli 100 years before

• Fibonacci numbers are actually a combo of two geometric progressions

• Recall φ2=φ+1 and τ2=τ+1 identities

• Use them to come up with a formula for the Fibonacci series

Binet Formula

• φ2=φ+1 and τ2=τ+1 identities

• φ2= φ+1• φ3=φ(φ2)=φ(φ+1)=(φ2)+φ=(φ+1)+φ= 2φ+1• φ4=φ(φ3)=φ(2φ+1)=(2φ2)+φ=2(φ+1)+φ= 3φ+2• φ5=φ(φ4)=φ(3φ+2)=(3φ2)+2φ=3(φ+1)+2φ= 5φ+3• φ6=φ(φ5)=φ(5φ+3)=(5φ2)+3φ=5(φ+1)+3φ= 8φ+5

• φ2=1φ+1 So, φn=Fnφ+Fn-1

• φ3=2φ+1 and • φ4=3φ+2 τn=Fnτ+Fn-1

• φ5=5φ+3• φ6=8φ+5

Binet Formula

φ^n=Fnφ+Fn-1

τ^n=Fnτ+Fn-1

φ^n - τ^n=Fnφ - Fnτ

Fn= (φ^n - τ^n) / (φ - τ)remember that φ = (1+sqrt(5))/2 and τ = (1+sqrt(5))/2therefore (φ - τ) = sqrt (5)Fn= (φ^n - τ^n) / sqrt(5) Fn=(φ^n/sqrt(5)) - (τ^n/sqrt(5)) (two geometric progressions)now for the Fibonacci term 1000 is F1000= (φ^(10000) - τ^(10000)) / sqrt(5) = 43466557686937456... (209 digits)

The Fibonacci Sequence in Nature

http://www.youtube.com/watch?v=ahXIMUkSXX0

Application : The Towers of Hanoi

The Rules:

From here: To here:

Without:1. Moving more than one disk at a time. 2. Placing a larger disk on top of a smaller

disk.

An Example:

1 2

3

5 76

4

The Goal:

To find the minimum number of moves necessary to complete the puzzle.

hn = # of moves required to transfer n disks.

Let us find a recurrence rule to predict hn.

What We Know:

h3 = 7 h5 = 31h7 = 123

h4 = 15 h6 = 63h8 = 247

hn = small disks + big disk + small disks

hn-1 + 1 + hn-1

2hn-1 + 1

Recursive Formula: hn= 2hn-1+ 1

Closed Formula:

3, 7, 15, 31, 63, ...

One less than a power of 2?

3 = 22 - 17 = 23 - 115 = 24 - 1

hn = 2n -1

Prediction : End of the World?

High on the mountaintops sat a monk who could foretell the end of the world. He had a Tower of Hanoi with 64 gleaming diamond disks and could move one a second. When he stopped the world would end.

How long do we have?

Prediction : Solution

Number of moves required 264 -1

So . . .

roughly 583,344,214,028 years.

Prime Numbers: How do we find them?

200 B.C. Eratosthenes invented the sieve.

Prime Numbers:

Prime Numbers:

Prime Numbers:

Prime Numbers:

Prime Numbers:

And it stops.

Why?

The number of tests is the# of primes < testing maximum

Proof by contradiction:1. A composite exists in 11-1002. Thus, it is not a multiple of a P < 103. Thus, both factors > 104. Therefore, the composite > 100

Prime Numbers: How many exist?

E = P1 * P2 * P3 * P4 ... Pn

now...

q = P1 * P2 * P3 * P4 * ... * Pn + 1

Following the Composite Theorems (must be factor of unique prime numbers), infinite prime numbers exist.

Where Aren't the Prime Numbers?

2*3 + 2 = composite 2*3 + 3 = composite

K = 2 * 3 * 4 * ... * (N+1) K+2 = 2 * 3 * 4 * ... * (N+1) +2 K+3 = 2 * 3 * 4 * ... * (N+1) + 3 K+(N+1) = 2 * 3 * 4 * ... * (N+1) + (N+1)

K+2, K+3, K+4, K+(N+1) --> all composite, there are runs infinitely long where there are no primes.