# Fibonacci Numbers

(Difference between revisions)
 Revision as of 15:22, 30 June 2010 (edit)← Previous diff Revision as of 15:29, 30 June 2010 (edit) (undo)Next diff → Line 102: Line 102: [[image:identity1.gif|thumb|Image 11|none|600px]] [[image:identity1.gif|thumb|Image 11|none|600px]] - To see the proof, click below. + {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage=Click here to hide proof. - + |HiddenText=:$F_1=F_3-F_2$ - {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage= + - :$F_1=F_3-F_2$ + :$F_2=F_4-F_3$ :$F_2=F_4-F_3$ :$F_3=F_5-F_4$ :$F_3=F_5-F_4$ Line 115: Line 113: Adding up all the equations, we get : Adding up all the equations, we get : - :$F_1+F_2+\dots+F_n=F_{n+2}-F_2$.|HiddenText=Click here to hide proof}} + :$F_1+F_2+\dots+F_n=F_{n+2}-F_2$.}} ====Sum of Fibonacci numbers with odd indices==== ====Sum of Fibonacci numbers with odd indices==== Line 130: Line 128: [[image:fidentity2.gif|thumb|Image 12|none|800px]] [[image:fidentity2.gif|thumb|Image 12|none|800px]] - To see the proof, click below. + {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage=Click here to hide proof. - + |HiddenText= - {{HideShow| + - + :$F_1=F_2$ :$F_1=F_2$ :$F_3=F_4-F_2$ :$F_3=F_4-F_2$ Line 163: Line 159: To see the proof, click below. To see the proof, click below. - {{HideShow| + + {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage=Click here to hide proof. + |HiddenText= Subtracting {{EquationNote|Eq. (2)}}, the sum of Fibonacci numbers with odd indices, from {{EquationNote|Eq. (1)}}, the sum of the first $n$ Fibonacci numbers, we get the identity of the sum of Fibonacci numbers with even indices. }} Subtracting {{EquationNote|Eq. (2)}}, the sum of Fibonacci numbers with odd indices, from {{EquationNote|Eq. (1)}}, the sum of the first $n$ Fibonacci numbers, we get the identity of the sum of Fibonacci numbers with even indices. }} Line 173: Line 171: :$\sum_{i=1}^n {F_i}^2=F_n F_{n+1}$ :$\sum_{i=1}^n {F_i}^2=F_n F_{n+1}$ - This identity can be proved by studying the area of the rectangles in Image 14. To see the proof, click below. + This identity can be proved by studying the area of the rectangles in Image 14. - {{HideShow| + {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage=Click here to hide proof. + |HiddenText= The rectangle is called a Fibonacci rectangle, which is further described in [[Fibonacci_Numbers#Fibonacci_Numbers_in_Nature| Fibonacci Numbers in Nature]]. The numbers inside each square indicate the length of one side of the square. Notice that the lengths of the squares are all Fibonacci numbers. The rectangle is called a Fibonacci rectangle, which is further described in [[Fibonacci_Numbers#Fibonacci_Numbers_in_Nature| Fibonacci Numbers in Nature]]. The numbers inside each square indicate the length of one side of the square. Notice that the lengths of the squares are all Fibonacci numbers. Line 229: Line 228: To see the proof for this special case, click below. To see the proof for this special case, click below. - {{HideShow| + {{HideShowThis|ShowMessage=Click here to see proof.|HideMessage=Click here to hide proof. + |HiddenText= Assume that $F_n$ and $F_{n+1}$ have some integer $k$ as their common divisor. Then, both $Assume that [itex] F_n$ and $F_{n+1}$ have some integer $k$ as their common divisor. Then, both [itex]

## Revision as of 15:29, 30 June 2010

Fibonacci Spiral

The spiral curve of the Nautilus sea shell follows the pattern of the spiral drawn in a Fibonacci rectangle, a collection of squares with sides that have the length of Fibonacci numbers.

# Basic Description

The Fibonacci sequence is the sequence $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots,$ where the first two numbers are 1s and every later number is the sum of the two previous numbers. So, given two $1$'s as the first two terms, the next terms of the sequence follows as : $1+1=2, 1+2=3, 2+3=5, 3+5=8, \dots$
Image 1

The Fibonacci numbers can be discovered in nature, such as the spiral of the Nautilus sea shell, the petals of the flowers, the seed head of a sunflower, and many other parts of the nature. The seeds at the head of the sunflower, for instance, are arranged so that one can find a collection of spirals in both clockwise and counterclockwise ways. The number of spirals differs depending on whether one counts in a clockwise or a counterclockwise way because different patterns of spirals are formed depending on the counting direction, as shown by Image 1. The two numbers of spirals are always consecutive numbers in the Fibonacci sequence.

Nature prefers this way of arranging the seeds because it seems to allow the seeds to be uniformly distributed. For more information about Fibonacci patterns in nature, see Fibonacci Numbers in Nature

## Origin

The Fibonacci sequence was studied by Leonardo of Pisa, or Fibonacci (1770-1240). In his work Liber Abacci, he introduced a problem involving the growth of the rabbit population. The assumptions were

• there is one pair of baby rabbits placed in an enclosed place on the first day of January
• this pair will grow for one month before reproducing and produce a new pair of baby rabbits on the first day of March
• each new pair will mature for one month and produce a new pair of rabbits on the first day of their third month
• the rabbits never die, so after they mature, the rabbits produce a new pair of baby rabbits every month.

The problem was to find out how many pairs of rabbits there will be after one year.

Image 2

On January 1st, there is only 1 pair. On February 1st, the baby rabbits matured to be grown up rabbits, but they have not reproduced, so there will only be the original pair present.

Now look at any later month. June is a good example. As you can see in Image 2, all 5 pairs of rabbits that were alive in May continue to be alive in June. Furthermore, all 3 pairs of rabbits that were also alive on April 1st, which all became or were adult rabbit pairs on May 1st, reproduce, creating 3 new pairs of rabbits born in June.

This means that on June 1st, there are 8 pairs of rabbits. This is equal to the 5 pairs from May 1st plus the 3 new pairs, which is the number of pairs from April 1st. This same reasoning can be applied to any month, March or later, so the number of rabbits pairs at a certain point is the same as the sum of the number of rabbit pairs in the two previous months.

This is exactly the rule that defines the Fibonacci sequence. As you can see in the image, the population by month begins: 1, 1, 2, 3, 5, 8, ..., which is the same as the beginning of the Fibonacci sequence. The population continues to match the Fibonacci sequence no matter how many months out you go.

An interesting fact is that this problem of rabbit population was not intended to explain the Fibonacci numbers. This problem was originally intended to introduce the Hindu-Arabic numerals to Western Europe, where people were still using Roman numerals, and to help people practice addition. It was coincidence that the number of rabbits followed a certain pattern which people later named as the Fibonacci sequence.

# Fibonacci Numbers in Nature

### Leaf Arrangement

Fibonacci numbers appear in the arrangement of leaves in certain plants. Take a plant, locate the lowest leaf and number that leaf as 0. Number the leaves by order of creation starting from 0, as shown in Image 3. Then, count the number of leaves you encounter until you reach the next leaf that is directly above and pointing in the same direction as the lowest leaf, which is the leaf with number 8 in this image. The number of leaves you pass, in this case, 8, will be a Fibonacci number.

Image 3

Moreover, the number of rotations you make around the stem until you reach that leaf will also be a Fibonacci number. You make rotations up the stem by following ascending order of the leaf's number. In the image, if you follow the red arrows, the number of rotations you make until you reach 8 will be 5, which is a Fibonacci number.

In Image 4, the leaf that is pointing in the same direction as the lowest leaf 0 is the leaf number 13. The number of leaves in between these two leaves is 13, which is a Fibonacci number. Moreover, going up the stem in a clockwise direction, such that we follow leaves 0, 1, 2, ..., 13, we make 8 rotations, and going up the stem in a counterclockwise direction, we make 5 rotations. The number of clockwise rotations and the number of counterclockwise rotations are always consecutive Fibonacci numbers.

Image 4

### Spirals

Image 5

Fibonacci numbers can be seen in nature through spirals, such as the spirals in sea shells, snails, or the galaxy. Fibonacci numbers appear in spirals through Fibonacci rectangles as shown in Image 5.

Image 6

We can build Fibonacci rectangles first by drawing two squares with length 1 next to each other. Then, we draw a new square with length 2 that is touching the sides of the original two squares. We draw another square with length 3 that is touching one unit square and the latest square with length 2. We can build Fibonacci rectangles by continuing to draw new squares that have the same length as the sum of the length of the latest two squares.

After building Fibonacci rectangles, we can draw a spiral in the squares, each square containing a quarter of a circle. Such spirals are called the Fibonacci spirals, and they can be seen in sea shells, snails, the spirals of the galaxy, and other parts of nature, as shown in Image 6 and Image 7.

Image 7

### Ancestry of Bees

Fibonacci numbers also appear when studying the ancestry of bees. Bees reproduce according to the following rules:

• male bees hatch from an unfertilized egg, and have only a mother and no father,
• female bees hatch from a fertilized egg, and require both a mother and a father.

The table below starts with a male bee, and tracks the ancestors of the male bee. Only one female was needed to produce the male bee. This female bee, on the other hand, must have had both a mother and a father to be hatched; thus, the third row of the bee family tree has one male and a female.

For each male and female, such pattern repeats. When we count the number of bees for each generation, we get a Fibonacci sequence as we go up the generations, similar to the way we got Fibonacci numbers in the rabbit population problem.

Image 8

# A More Mathematical Explanation

## Symbolic Definition of Fibonacci Sequence

The Fibonacci sequence is the sequence UNIQ5b47be4b40 [...]

## Symbolic Definition of Fibonacci Sequence

The Fibonacci sequence is the sequence $F_1, F_2, F_3, \ldots, F_n, \ldots$ where

$F_n = F_{n-1} + F_{n-2} \quad \hbox{ for } n>2$,

and

$F_1 = 1,\ F_2 = 1$.

The Fibonacci sequence is recursively defined because each term is defined in terms of its two immediately preceding terms.

## Identities and Properties

### Idenitities

There are some interesting identities, including formula for the sum of first $n$ Fibonacci numbers, the sum of Fibonacci numbers with odd indices and sum of Fibonacci number with even indices. Note that all the identities and properties in this section can be proven in a more rigorous way through mathematical induction.

#### Sum of first $n$ Fibonacci numbers

The sum of first$n$ Fibonacci numbers is one less than the value of the ${(n+2)}^{\rm th}$ Fibonacci number:

Eq. (1)        $F_1+F_2+\dots+F_n=F_{n+2}-F_2=F_{n+2}-1$

For example, the sum of first $5$ Fibonacci number is :

$F_1+F_2+F_3+F_4+F_5= 1 + 1 + 2 + 3 +5=F_7-1=12$

The example is demonstrated below. The total length of red bars that each correspond to $F_1, F_2, F_3, F_4, F_5$ is one unit less than the length of $F_7$.

Image 11
$F_1=F_3-F_2$
$F_2=F_4-F_3$
$F_3=F_5-F_4$
$\dots$
$F_{n-1}=F_{n+1}-F_n$
$F_n=F_{n+2}-F_{n+1}$

Adding up all the equations, we get :

$F_1+F_2+\dots+F_n=F_{n+2}-F_2$.

#### Sum of Fibonacci numbers with odd indices

The sum of first $n$ Fibonacci numbers with odd indices is equal to the ${(2n)}^{\rm th}$ Fibonacci number:

Eq. (2)        $F_1+F_3+F_5+\dots+F_{2n-1}=F_{2n}$

For instance, the sum of first $4$ Fibonacci numbers with odd indices is:

$F_1+F_3+F_5+F_7=1+2+5+13=21=F_8$

This example is shown below.

Image 12
$F_1=F_2$
$F_3=F_4-F_2$
$F_5=F_6-F_4$
$\dots$
$F_{2n-1}=F_{2n}-F_{2n-2}$

Adding all the equations, we get :

$F_1+F_3+F_5+\dots+F_{2n-1}=(F_2-F_2)+(F_4-F_4)+(F_6-F_6)+\dots+(F_{2n-2}-F_{2n-2})+F_{2n}$
$=F_{2n}$
Except for $F_{2n}$, all the terms on the right side of the equation disappear because each term is canceled out by another term that has the opposite sign and the same magnitude.

#### Sum of Fibonacci numbers with even indices

The sum of first $n$ Fibonacci numbers with even indices is one less than the ${(2n+1)}^{\rm th}$ Fibonacci number:

$F_2+F_4+\dots+F_{2n}=F_{2n+1}-1$

For example, the sum of first $4$ Fibonacci numbers with even indices is :

$F_2+F_4+F_6= 1+3+8=F_7-1=13-1=12$

This example is shown below.

Image 13

To see the proof, click below.

Subtracting Eq. (2), the sum of Fibonacci numbers with odd indices, from Eq. (1), the sum of the first $n$ Fibonacci numbers, we get the identity of the sum of Fibonacci numbers with even indices.

#### Sum of the squares of Fibonacci numbers

The sum of the squares of the first $n$ Fibonacci numbers is the product of the $n^{\rm th}$ and the ${(n+1)}^{\rm th}$ Fibonacci numbers.

Image 14
$\sum_{i=1}^n {F_i}^2=F_n F_{n+1}$

This identity can be proved by studying the area of the rectangles in Image 14.

The rectangle is called a Fibonacci rectangle, which is further described in Fibonacci Numbers in Nature. The numbers inside each square indicate the length of one side of the square. Notice that the lengths of the squares are all Fibonacci numbers.

Any rectangle in the picture is composed of squares with lengths that are Fibonacci numbers. In fact, any rectangle is composed of every square with side lengths $F_1$ through $F_n$. Moreover, the dimension of this rectangle is $F_n$ by $F_{n+1}$.

We can prove this identity by computing the area of the rectangle in two different ways. The first way of finding the area is to add the area of each squares. That is, the area of the rectangle will be :

${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2$.

Another way of computing the area is by multiplying the width by the height. Using this method, the area will be :

$F_n F_{n+1}$.

Because we are computing the area of the same rectangle, the two methods should give the same results. Thus,

${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2=F_n F_{n+1}$.

For example, for the red rectangle, the width is $5$ and the height is $8$. Since $5$ is the $5^{\rm th}$ Fibonacci number and $8$ is the $6^{\rm th}$ Fibonacci number, let

$n=5$.

The area of the rectangle is :

$1^2+1^2+2^2+3^2+5^2={F_1}^2+{F_2}^2+{F_3}^2+{F_4}^2+{F_5}^2=\sum_{i=1}^5 {F_i}^2=40$,

or

$5 * 8 = F_5 F_{5+1} = F_5 F_6 = 40$.

Thus,

$\sum_{i=1}^5 {F_i}^2=F_5 F_{5+1}$.

### Properties

#### Greatest Common Divisor

The greatest common divisor of two Fibonacci numbers is the Fibonacci number whose index is the greatest common divisor of the indices of the original two Fibonacci numbers. In other words,

$\gcd(F_n,F_m) = F_{\gcd(n,m)}$.

For instance,

$\gcd(F_9,F_6)=\gcd(34,8)=2=F_3=F_{\gcd(9,6)}$.

In a special case where $F_n$ and $F_m$ are consecutive Fibonacci numbers, this property says that

$\gcd(F_n, F_{n+1})=F_{\gcd(n,n+1)}=F_1=1$.

That is, $F_n$ and $F_{n+1}$are always relatively prime.

To see the proof for this special case, click below.

Assume that $F_n$ and $F_{n+1}$ have some integer $k$ as their common divisor. Then, both $F_{n+1}$ and $F_n$ are each multiples of $k$:

Eq. (3)        $F_{n+1}=ka$
Eq. (4)        $F_n=kb$

Subtracting Eq. (4) from Eq. (3), we get :

$F_{n-1}=k(a-b)$,
which means that if two consecutive Fibonacci numbers, $F_n$ and $F_{n+1}$, have $k$ as their common divisor, then the previous Fibonacci number, $F_{n-1}$ must also be a multiple of $k$. In that case, $F_{n-1}$ and $F_n$, which are also two consecutive Fibonacci numbers, will have $k$ as a common divisor. Then, it follows that $F_{n-2}$ must also be a multiple of $k$. Repeating the subtraction of consecutive Fibonacci numbers, we can conclude that the very first Fibonacci number, $F_1 = 1$ must also be a multiple of $k$. So $k=1$, and the only common divisor between two consecutive Fibonacci numbers is 1. Thus, two consecutive Fibonacci numbers are relatively prime.

#### Finite Difference of Fibonacci Numbers

One of the interesting properties of Fibonacci numbers is that the sequence of differences between consecutive Fibonacci numbers also forms a Fibonacci sequence, as shown in the table below. For more information about the difference table, click Difference Tables.

Because the first sequence of differences of the Fibonacci sequence also includes a Fibonacci sequence, the second difference also includes a Fibonacci sequence. The Fibonacci sequence is thus reproduced in every sequence of differences.

We can see that the sequence of differences is composed of Fibonacci numbers by looking at the definition of Fibonacci numbers :

$F_n = F_{n-1} + F_{n-2}$.

The difference between two consecutive Fibonacci numbers is :

$F_n - F_{n-1} = F_{n-2}$.
Thus, the difference between two consecutive Fibonacci numbers, $F_n$ and $F_{n-1}$, is equal to the value of the previous Fibonacci number, $F_{n-2}$.

## Golden Ratio

Image 9

The golden ratio appears in paintings, architecture, and in various forms of nature. Two numbers are said to be in the golden ratio if the ratio of the smaller number to the larger number is equal to the ratio of the larger number to the sum of the two numbers. In Image 9, the width of A and B are in the golden ratio if$a : b = (a+b) : a$.

The golden ratio is represented by the Greek lowercase phi ,$\varphi$, and the exact value is

$\varphi=\frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,$

This value can be found from the definition of the golden ratio. To see an algebraic derivation of the exact value of the golden ratio, go to Golden Ratio : An Algebraic Representation.

An interesting fact about golden ratio is that the ratio of two consecutive Fibonacci numbers approaches the golden ratio as the numbers get larger, as shown by the table below.

$\frac{F_{n+1}}{F_n}$ $\frac{1}{1}$=1 $\frac{2}{1}$=2 $\frac{3}{2}$=1.5 $\frac{5}{3}$=1.66667 $\frac{8}{5}$=1.6 $\frac{13}{8}$=1.625 $\frac{21}{13}$=1.61538 $\frac{34}{21}$=1.61904 $\frac{55}{34}$=1.61765 $\frac{89}{55}$=1.61818

Lets assume that the ratio of two consecutive Fibonacci numbers have a limit and verify that this limit is, in fact, the golden ratio. Let $r_n$ denote the ratio of two consecutive Fibonacci numbers, that is,

$r_n=\frac{F_{n+1}}{F_n}$.

Then,

$r_{n-1}=\frac{F_n}{F_{n-1}}$.

$r_n$ and $r_{n-1}$ are related by :

$r_n=\frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}=1+\frac{F_{n-1}}{F_n}=1+\cfrac{1}{{F_n}/{F_{n-1}}}=1+\frac{1}{r_{n-1}}$.

Assuming that the ratio $r_n$ has a limit, let $r$ be that limit:

$\lim_{n \to \infty} r_n=\lim_{n \to \infty}\frac{F_{n+1}}{F_n}=r$.

Then,

$\lim_{n \to \infty} r_n = \lim_{n \to \infty} r_{n-1} = r$.

Taking the limit of $r_n=1+\frac{1}{r_{n-1}}$ we get :

$r=1+\frac{1}{r}$

Multiplying both sides by $r$, we get

Eq. (5)        ${r}^2=r+1$

which can be written as:

$r^2 - r - 1 = 0$.

Applying the quadratic formula , we get $r = \frac{1 \pm \sqrt{5}} {2}$.

Because the ratio has to be a positive value,

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

which is the golden ratio. Thus, if $r_n$ has a limit, then this limit is the golden ration. That is, as we go farther out in the sequence, the ratio of two consecutive Fibonacci numbers approaches the golden ratio. In fact, it can be proved that $r_n$ does have a limit; one way is to use Binet's formula in the next section. For a different proof using infinite continued fraction go to Continued Fraction Representation and Fibonacci Sequences

Image 10: Vitruvian man

Many people find the golden ratio in various parts of nature, art, architecture, and even music. However, there are some people who criticize this viewpoint. They claim that many mathematicians are wishfully trying to make the connection between the golden ratio and other parts of the world even though there is no real connection.

One example of the golden ratio that mathematicians found in nature is the human body. According to many, an ideal human body have proportions that show the golden ratio, such as:

• distance between the foot and navel : distance between the navel and the head
• distance between the finger tip and the elbow : distance between the wrist and the elbow
• distance between the shoulder line and top of the head : length of the head.

Leonardo da Vinci's drawing Vitruvian man shown in Image 10 emphasizes the proportion of human body. This drawing shows the proportions of an ideal human body that was studied by a Roman architect Vitruvius in his book De Architectura. In the drawing, a man is simultaneously inscribed in a circle and a square. The ratio of the square side to the radius of the circle in the drawing reflects the golden ratio, although the drawing deviates from the real value of the golden ratio by 1.7 percent. The proportions of the body of the man is also known to show the golden ratio.

Although people later found the golden ratio in the painting, there is no evidence whether Leonardo da Vinci was trying to use the golden ratio in his painting or not. For more information about the golden ratio, go to Golden Ratio

## Binet's Formula for Fibonacci Numbers

Binet's Formula gives a formula for the $n^{\rm th}$ Fibonacci number as :

$F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$,

where $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (5), that is,

$\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$.

Here is one way of verifying Binet's formula through mathematical induction, but it gives no clue about how to discover the formula. Let

$F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$

as defined above. We want to verify Binet's formula by showing that the definition of Fibonacci numbers holds true even when we use Binet's formula. First, we will show through inductive step that:

$F_n=F_{n-1}+F_{n-2}\quad\hbox{ for } n>2$

and then we will show the base case that:

$F_1=1,\quad F_2=1$.

First, according to Binet's fromula,

$F_{n-1}+F_{n-2} = \frac{{\varphi}^{n-1}-{\bar{\varphi}}^{n-1}}{\sqrt5}+ \frac{{\varphi}^{n-2}-{\bar{\varphi}}^{n-2}}{\sqrt5}$
$=\frac{({\varphi}^{n-1}+{\varphi}^{n-2})-({\bar{\varphi}}^{n-1}+{\bar{\varphi}}^{n-2})}{\sqrt5}$
$=\frac{({\varphi}+1){\varphi}^{n-2}-(\bar{\varphi}+1){\bar{\varphi}}^{n-2}}{\sqrt5}$.

Because $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (1), the above equation becomes :

$F_{n-1}+F_{n-2}=\frac{{{\varphi}^2}{\varphi}^{n-2}-{{\bar{\varphi}}^2}{\bar{\varphi}}^{n-2}}{\sqrt5}$
$=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$
$=F_n$, as desired.

Now, because $\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$,

$F_1=\frac{\varphi-\bar{\varphi}}{\sqrt5}=\frac{1}{\sqrt5}\left (\frac{1 + \sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right)=\frac{1}{\sqrt5} {\sqrt5} = 1$
$F_2=\frac{{\varphi}^2-{\bar{\varphi}}^2}{\sqrt5}=\frac{(\varphi+\bar{\varphi})(\varphi-\bar{\varphi})}{\sqrt5}=\frac{1*{\sqrt5}}{\sqrt5}=1$.
Binet's formula thus is a correct formula of Fibonacci numbers.

## Fibonacci Numbers and Fractals

### Fibonacci Numbers and the Mandelbrot Set

Image 15

The Mandelbrot set is a set of points in which the boundary forms a fractal. It is a set of all complex numbers $c$ for which the sequence

Eq. (6)        :$z_{n+1}=(z_n)^2+c \quad \hbox{ for } n=0,1,2,\dots$

does not go to infinity, starting with $z_0=0$.

For instance, $c=0$ is included in the Mandelbrot set because

$z_1=(z_0)^2+c=0^2+0 = 0$
$z_2=(z_0)^2+c=0^2+0=0$
$\dots$
$z_n=0^2+0=0$ for any $n$.

Thus, the sequence defined by $c=0$ is bounded and $0$ is included in the Mandelbrot set.

On the other hand, when we test$c=1$,

Image 16
$z_1=(z_0)^2+c=1$
$z_2=(z_1)^2+c=2$
$z_3=(z_2)^2+c=5$
$\dots$

The terms of this sequence will increase to infinity. Thus, $c=1$ is not included in the Mandelbrot set.

Image 17

People have been drawn to study the Mandelbrot set because of its aesthetic beauty. It is surprising to many people how a simple formula like Eq. (6) can generate a complex structure of the Mandelbrot set. The Fibonacci sequence is related to the Mandelbrot set through the period of the main cardioid and some large primary bulbs. For each bulb, there are many antennas, and the largest antenna is called the main antenna. The number of spokes in the main antenna is the period of the bulb.

The period of the main cardioid is considered to be 1. In Image 17, the main antenna has five spokes, including the one connecting the primary bulb and the junction point of the antenna. The period of this bulb is five.

Now, we will consider the period of the largest primary bulbs that are attached to the main cardioid and are in between two larger bulbs. In Image 18, the largest bulb between the bulb of period 1 and the bulb of period 2 is the bulb of period 3, and this bulb was found by looking for the largest bulb on the periphery of the main cardioid. The largest bulb between the bulb of period 2 and period 3 is the bulb of period 5, and the one between bulb of period 3 and period 5 is the bulb of period 8. The sequence generated in this way proceeds as 1, 2, 3, 5, 8, 13, ..., following the pattern of Fibonacci sequence.

Image 18

# References

## Things to add(possible ideas for future)

• Fibonacci numbers and Pascal's triangle
• A helper page for recursively defined sequence
• A section describing the Fibonacci numbers with negative subscripts.

• A derivation of the exact value of the golden ratio. The derivation is redundant with the information in the golden ratio page.

[[Category:]]