Lecture 32: Fibonacci Numbers

## The Fibonacci Numbers

1 1 2 3 5 8 13 21 34 …

Virtually everyone in the world of Mathematics has either heard of or encountered the Fibonacci Numbers in their life. The Fibonacci Series is a simple recursive formula such that Fn+2 = Fn+1 + Fn. Depending on which circle you are in, the series will start with F0 = 0 or F1 = 1. Regardless, the same conclusions can be drawn. This has no effect on any of the identities or theorems.

## Origins of the Fibonacci Numbers

#### Pingala

Although this series is named for Fibonacci, the series existed long before. The first documented occurrence of the series was in the 4th Century BC by an Indian writer by the name of Pingala. Pingala studied grammar. In Sanskrit, vowels are either long or short. Short vowels are one syllable and long vowels are two syllables long. The combination of the vowel sounds are called metres. Pingala was interested in the question, "how many variations exist that are a given length?" In the following chart, s will stand for a short syllable, and L will stand for a long syllable. One can see that this is the beginning of the Fibonacci Numbers.

 Syllables Variations 0 1 1 1 s 2 2 ss L 3 3 sss sL Ls 4 5 ssss ssL sLs Lss LL 5 8 sssss sssL ssLs sLss Lsss sLL LsL LLs

#### Fibonacci

Although almost everyone in the world of Mathematics knows of him as Fibonacci, his real name is Leonardo de Pisa. He was born in the 12th Century AD, almost 1600 years after Pingala did his work. Fibonacci concentrated his work in the Hindu-Arabic numerical system, and in the 13th Century, he wrote a book titled Liber Abaci. Translated, this means the Book of Calculation. This book advocated the use of the numbers 0 through 9 and the use of decimal places. Throughout this book, Fibonacci posed Mathematical questions and showed his solutions to the problems. One of these problems was concerned with the theoretical population of rabbits. Fibonacci made four assumptions about the rabbits.

1. Rabbits don't die
2. Every month, every mature pair of rabbits produce exactly on pair of new rabbits
3. It takes one month for a pair of new rabbits to mature
4. The rabbit population in month one is one pair of new rabbits

Assuming N means a new pair of rabbits and M means a mature pair of rabbits, Fibonacci obtained the following results:

 Month Pairs of Rabbits 0 0 1 1 N 2 1 M 3 2 MN 4 3 MNM 5 5 MNMMN 6 8 MNMMNMNM

## The Golden Ratio

The Golden Ratio is a term that has come up in Mathematics a lot. To the Greeks, the Golden Ratio was a way to make rectangles as beautiful as possible. To the Fibonacci Numbers, the Golden Ratio is the limit as n goes to infinity of $\frac{F_{n+1}}{F_{n}}$. The Golden Ratio is one of the solutions to the equation $\frac{x}{1} = \frac{1}{x-1}$.

$\varphi = \frac{1 + \sqrt{5}}{2}$

$1-\varphi = \frac{1 - \sqrt{5}}{2} = \frac{1-5}{2(1+\sqrt{5})}=-\frac{2}{1+\sqrt{5}}=-\frac{1}{\varphi}$

$\varphi^2=\frac{1+2\sqrt{5}+5}{4}=\frac{3+\sqrt{5}}{2}=1+\frac{1 + \sqrt{5}}{2}=1+\varphi$

$(1-\varphi)^2=1-2\varphi+\varphi^2=1-2\varphi+1+\varphi=2-\varphi$

$F_n=\frac{\varphi^n-(1-\varphi)^n}{\sqrt{5}}$

$F_0=\frac{1-1}{\sqrt{5}}=0\quad F_1=\frac{\varphi-(1-\varphi)}{\sqrt{5}}=\frac{2*\varphi-1}{\sqrt{5}}=1$

$F_n=F_{n-2}+F_{n-1}=\frac{\varphi^{n-2}-(1-\varphi)^{n-2}}{\sqrt{5}}+\frac{\varphi^{n-1}-(1-\varphi)^{n-1}}{\sqrt{5}}=\frac{\varphi^{n-2}(1+\varphi)-(1-\varphi)^{n-2}(1+1-\varphi)}{\sqrt{5}}$

$=\frac{\varphi^n-(1-\varphi)^n}{\sqrt{5}}$

## Fibonacci Identities

#### Identity Number One

$\sum_{i=0}^n F_i=F_{n+2}-1$

$n=0\quad F_0 = F_2-1=0$

$n=1\quad F_0+F_1=F_3-1=1$

$n=n+1\quad \sum_{i=1}^{n+1} F_i=\sum_{i=1}^n F_i+F_{n+1}=F_{n+2}-1+F_{n+1}=F_{n+3}-1$

#### Identity Number Two

$\sum_{i=0}^n i F_i=nF_{n+2}-F_{n+3}+2$

$n=0\quad 0 F_0 = 0 F_2-F_3+2=-2+2=0$

$n=1\quad 0 F_0+F_1=F_2-F_3+2=1-2+2=1$

$n=n+1\quad \sum_{i=0}^{n+1} i F_i=\sum_{i=0}^n i F_i +(n+1)F_{n+1}$

$=nF_{n+2}-F_{n+3}+2+(n+1)F_{n+1}$

$=nF_{n+3}-F_{n+3}+F_{n+1}+2=(n+1)F_{n+3}-2F_{n+3}+F_{n+1}+2$

$=(n+1)F_{n+3}-F_{n+3}-F_{n+2}-F_{n+1}-+F_{n+1}+2=(n+1)F_{n+3}-F_{n+4}+2$

#### Identity Number Three

$\sum_{i=0}^n {F_i}^2=F_nF_{n+1}$

$n=0\quad F_0^2 = F_0F_1=0*1=0$

$n=1\quad F_0^2+F_1^2=F_1F_2=1*1=1$

$n=n+1\quad \sum_{i=0}^{n+1} {F_i}^2=\sum_{i=0}^n {F_i}^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2$

$=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$

#### Other Identities

Calculation of large Fibonacci numbers can take a long time, but there are identities which speed up the process.

F3n = 5 Fn3 + 3 (-1)n Fn

Example:

If n = 3

then Fn = 2

5 (23) + 3 (-1)3 (2) =

5 (8) + 3 (-1) (2) =

5 (8) - 3 (2) =

40 - 6 = 34

F3(3) = 34 indeed F9 = 34

F4n = 4 Fn Fn+1 (Fn+12 + 2 Fn2) - 3 Fn2 (Fn2 + 2 Fn+12)

Example:

If n =2

then Fn = 1

and Fn+1 = 2

4 (1) (2) (22 + 2 (12)) - 3 (12) (12 + 2 (22)) =

4 (2) (4 + 2) - 3 (1) (1 + 2 (4)) =

8 (6) - 3 (1 + 8) =

8 (6) - 3 (9) =

48 - 27 = 21

F4(2) = 21 indeed F8 = 21

## Fibonacci Numbers Related to Other Topics

#### Right Triangles

The Fibonacci Numbers have multiple ties to Pythagorean Triples. First of all, after n = 5, every other Fibonacci Number is the hypotenuse of a Pythagorean Triple. Also, with any four consecutive Fibonacci numbers, a Pythagorean Triple can be formed.

$a=F_nF_{n+3}\quad b=2F_{n+1}F_{n+2}\quad c=F_{n+1}^2+F_{n+2}^2$

$n=3\quad a=16\quad b=30\quad c=34\quad 16^2+30^2=34^2$

$a^2+b^2$

$=F_{n}^2F_{n+3}^2+4F_{n+1}^2F_{n+2}^2 =(F_{n+2}-F_{n+1})^2 (F_{n+2}+F_{n+1})^2+4F_{n+1}^2F_{n+2}^2$

$=(F_{n+2}^2-F_{n+1}^2)^2+4F_{n+1}^2F_{n+2}^2= F_{n+2}^4-2F_{n+1}^2F_{n+2}^2+F_{n+1}^4+4F_{n+1}^2F_{n+2}^2$

$=F_{n+2}^4+2F_{n+1}^2F_{n+2}^2+F_{n+1}^4=(F_{n+2}^2+F_{n+1}^2)^2=c^2$

#### Relatively Prime

Two consecutive Fibonacci Numbers are relatively prime. Meaning that gcd(Fn+1, Fn) = 1

###### Proof using Euclid's Algorithm

We know Fn+1 = d (Fn) + r where 0<= r < Fn

However we know:

Fn+1 = Fn + Fn-1 and Fn = Fn-1 + Fn-2

so that means that we know d = 1 and r = Fn-1. So,

Fn+1 = Fn + Fn-1

Fn = d2 Fn-1 + r where 0<= r < Fn-1

However, we can apply the same formula and find:

Fn = Fn-1 + Fn-2

We can now continue doing this.

Fn+1 = Fn + Fn-1

Fn = Fn-1 + Fn-2

Fn-1 = Fn-2 + Fn-3

.

.

.

F5 = F4 + F3

F4 = F3 + F2

However, this states that 3 = 1 (2) + 1

So, we just showed gcd(Fn+1, Fn) = 1

###### Proof using Contradiction

Assume there exists a d>1 such that d|Fn+1 and d|Fn

Well, we then know that d divides and linear integral combination of the two.

So, d should divide Fn+1 - Fn

Notice that Fn+1 = Fn + Fn-1, which means Fn+1 - Fn = Fn-1

So, d|Fn-1

We can then continue to apply this process until we get to d|F1

However, F1 = 1 and d>1 so d can't divide 1. So, we have a contradiction.

Our assumption must be wrong, and so there does not exist a d>1 that divides both Fn+1 and Fn.

In other words, gcd(Fn+1, Fn) = 1

#### Legendre

Although the Fibonacci numbers appear to be the random answer to a theoretical question, they in fact have a relationship to many other parts of Mathematics. For example, the Fibonacci numbers relate to the Legendre symbol. There are a few equations that hold. For any prime p:

1. Fp $\equiv (\frac{p}{5})$ mod( p )

2. If n = p - ($\frac{p}{5}$) then Fn $\equiv 0$ mod( p )

3. if p $\equiv 1 mod( 4 )$ and n = $(\frac{p \pm 1}{2})$ then 5 Fn2$\equiv \frac{5(\frac{p}{5}) \pm 5}{2}$ mod( p )

4. if $p \equiv 3 mod( 4 )$ and n =$(\frac{p \pm 1}{2})$ then 5 Fn2$\equiv \frac{5(\frac{p}{5}) \mp 3}{2}$ mod( p )

If p = 3 then

$(\frac{3}{5})$ = -1

Fp = 2

F3 - (-1) = F4 = 3

F1 = 1

F2 = 1

1. 2 $\equiv -1$ mod( 3 )

2. 3 $\equiv 0$ mod( 3 )

4a. 5 (12) = 5 $\equiv \frac{5 (-1) - 3}{2}$ mod( 3 )

5 $\equiv -1$ mod( 3 )

4b. 5 (12) = 5 $\equiv \frac{5 (-1) + 3}{2}$ mod( 3 )

5 $\equiv -1$ mod( 3 )

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