• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The Fibonacci numbers and the golden ratio

Extracts from this document...

Introduction

Sofie Bronée 1v.         Fibonacci project.         11/04/08

The Fibonacci Numbers and the Golden Ratio

_____________________________________

image00.jpg

The Fibonacci Numbers        

The Fibonacci numbers are sequence of numbers. They are named after the Italian mathematician Leonardo of Pisa, known as Fibonacci. He published a book called “Liber Abaci”, and he was the first person to publish a book in Western Europe that used the Indian numerals 9, 8, 7, 6, 5, 4, 3, 2, 1. Fibonacci was perhaps the greatest mathematician of his time. But he is most famous for the numbers which has his name.

These are the first ten terms in the Fibonacci sequence.

U1

U2

U3

U4

U5

U6

U7

U8

U9

U10

1

1

2

3

5

8

13

21

34

55

To get the next term in the sequence the two previous terms must be added. This can be written as:  

Un = Un-1 + Un-2

A number in the sequence is the sum of its two predecessors.

The Golden Ratio

To find the Golden Ratio, a term has to be divided with the previous term

(i.e. . image01.png.. etc).

...read more.

Middle

0


image11.png

By investigating the Fibonacci numbers, I can make a conjecture, that the ratio of two consecutive terms gets closer, as they increase, to the Golden Ratio. By changing the start numbers I proved that it doesn’t make a difference which numbers we use.

To prove my conjecture I’ve changed the equation and solved it. Afterwards I found the discriminant (5), which leads me to find the roots. I plugged in the discriminant to the quadratic equation for finding the roots. The results for the positive root is, 1.618 and the negative root is, -0.618. This proves my conjecture. Look below to see my calculations and explanations.

I divided both sides of this equation Un = Un-2 + Un-1, by Un-1.

image13.png = 1 +  image14.png = Why we do it?

image15.png

image16.png= 1 + image17.png

Here, image16.png is one of the numbers in the column above, it gets closer and closer to 1.618, which I will call x. On the other side of the equation, image17.png is the reciprocal

of one of the numbers in the same column, so it gets closer to image02.png.

...read more.

Conclusion

The Greeks used the golden ratio in the design and construction of the Greek temple (Parthenon).

image06.jpg

image07.jpg

n

Un

Un+1/Un

1

1

1

2

1

2

3

2

1,5

4

3

1,666667

5

5

1,6

6

8

1,625

7

13

1,615385

To prove that the Greeks really used the golden ratio in the design, I’ve measured the sides. The first square is 1cm by 1cm; the next is 1cm by 2 cm and so on. Look in the table below to see the ratio between each side.

Appendix

1)

I changed to two “start numbers” to 1 and 7, and by making a table and a graph I proved that it doesn’t matter which numbers we use to get the Golden Ratio.

image08.png

To show that we would eventually get the same result, no matter what “start numbers” were, I plugged in -9 and 5. To prove that with even a negative number, we would reach the Golden Ratio.

image09.png

I also tried to plug in decimal numbers to see what happened. I plugged in 2,6 and -1,9.

image10.png

2)

The inverted Golden Ratio can be found by dividing Un-1 by Un-2. As you can see below, theinverted ratio is reached much faster, than the opposite.

image12.png

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Stellar numbers

    The same method utilized in the triangle general statement will be utilized here: The reason for adding F(n)+F(n) is because of the geometric construction of the stellar shape as seen below there are 12 triangles which can be drawn in any stage of the stellar shape with 6 vertices.

  2. Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the ...

    To continue on this trend, I set S2 as equal to and S3 as equal to . Eventually, when I continued this trend all the way up until S10, the equation that was determined is N(N+1). Essentially, the idea behind this attempt at the general statement can be backed by

  1. Maths Project. Statistical Analysis of GCSE results at my secondary school summer 2010 ...

    12 We 9 40 f 40 11 We 10 34 m 34 10 We 9 58 f 58 9 Wh 10 46 m 46 8 Wh 9 46 f 46 7 Wi 11 52 f 52 6 Wi 10 40 m 40 5 Wi 10 28 m 28 4 Wo

  2. Stellar Numbers. In this task geometric shapes which lead to special numbers ...

    Consider stellar (star) shapes with p vertices, leading to p-stellar numbers. The first four representations for a star with six vertices are shown in the four stages S1-S4. The 6-stellar number at each stage is the total number of dots in the diagram. 3. Find the number of dots (i.e.

  1. Lacsaps fractions are an arrangement of numbers that are symmetrically repeating based on a ...

    But the formula does seems to work for the 2nd term of the 2nd element but not the rest. I will let the difference be x, hence the equation (for the difference between n and r) should be Table 9 the value difference between the numerators (N)

  2. Stellar numbers. This internal assessment has been written to embrace one of the ...

    wheren?N.y represents the total number of dots and n represents the term of the triangular number. Table 2 n y Sum of y 1 1 1 2 3 1+2 3 6 1+2+3 4 10 1+2+3+4 5 15 1+2+3+4+5 6 21 1+2+3+4+5+6 7 28 1+2+3+4+5+6+7 8 36 1+2+3+4+5+6+7+8 However, this is not the only way such a statement can be generated.

  1. Stellar Numbers. In this folio task, we are going to determine difference geometric shapes, ...

    Sn is the number of dots that the diagram contains. When n=1, S1=1 n=2, S2=1+12×(2-1) =1+1×12 n=3, S3=1+12×(2-1)+12×(3-1)=1+3×12 n=4, S4=1+12×(2-1)+12×(3-1)+12×(4-1)=1+6×12 n=5, S5=1+12×(2-1)+12×(3-1)+12×(4-1)+12×(5-1)=1+10×12 n=6, S6=1+12×(2-1)+12×(3-1)+12×(4-1)+12×(5-1)+12×(6-1)=1+15×12 Therefore, we know that the number of dots in each stage above is, Sn= 1+12(X) We are now considering the triangular numbers at the first page,

  2. Math Portfolio Stellar Numbers. This task is an investigation of geometric numbers, the ...

    The series is arithmetic because the difference remains constant. The number of dots in the nth square number can be expressed as: 1+3+5+7+....+( The following formula can be used to find the sum of the series: Sn= Sn = Sn = Sn = Sn= n2 Where n Therefore the relationship

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work