ASSIGNMENT 12

by Soo Jin Lee

4. Generate a Fibonnaci sequence in the first column using f(0) = 1, f(1) = 1,

f(n) = f(n-1) + f(n-2)

.

a. Construct the ratio of each pair of adjacent terms in the Fibonnaci sequence. What happens as n increases? What about the ratio of every second term? etc.

b. Explore sequences where f(0) and f(1) are some arbitrary integers other than 1. If f(0)=1 and f(1) = 3, then your sequence is a Lucas Sequence. All such sequences, however, have the same limit of the ratio of successive terms.

By using spreadsheet I can build Fibonnaci sequence.

(in the first column using f(0)=1, f(1)=1 , generate when f(n)=f(n-1) + f(n-2))

A. From the above picture, I can investigate some interesting points.

As the number of 'n' increases, we can see that the ratio of the terms converges to the same number.

First we can see the ratio of every other terms converges to "1.61803399" as the "n" increases.

fixing x= 1.61803399 ----------------ratio of the adjacent terms

x^2 - x - 1 = 0

x^2 = x + 1 = 2.61803399 ---------------ratio of every second term

x^3 = x(x+1) = x^2 + x = x + 1 +x = 2x +1 =4.236076798 ------------------------------ratio of every third term

x^4 = x(2x + 1) = 2x^2 +x = 2( x + 1) + x = 2x + x + 2 =3x +2 = 6.85410197 ------ratio of every fourth term

x^5 = x(3x + 2) = 3x^2 + 2x = 3(x +1) +2x =5x +3 = 11. 0901699 ----------------------ratio of every fifth term

x^6 = x(5x+3)=5x^2 + 3x = 5(x+1)+3x= 8x+5= 17.9442719 -----------------------------ratio of every sixth term

.

by continuing, we can get the ratio of every "nth" term now.

The ratio we have got is called the "GOLDEN RATIO".

B. When f(0)=1 and f(1)=3,

The Lucas numbers are a slight variation of Fibonacci numbers.

These numbers follow the same recursion: f(n) = f(n-1) + f(n-2)

but having different initial conditions: f(0)=1, f(1)=3 leading to the sequence as the picture of the first column and this sequence ic called "Lucas Sequence"

By investigating the Lucas sequence, I can find the similar things happened as I did at the Fibonacci sequence.

The ratio of the first term converges to "1.61803399" from the 22nd row,

fixing x=1.61803399

x^2-x-1=0

x^2=x+1=2.61803399------------ratio of every second term

x^3 = x(x+1) = x^2 + x = x + 1 +x = 2x +1 =4.23606798----------------------------ratio of every third term

x^4 = x(2x + 1) = 2x^2 +x = 2( x + 1) + x = 2x + x + 2 =3x +2 = 6.85410197 ------ratio of every fourth term

x^5 = x(3x + 2) = 3x^2 + 2x = 3(x +1) +2x =5x +3 = 11. 0901699 ----------------------ratio of every fifth term

By exploring Fibonacci and Lucas Sequence,

I can predict that I can also get some nice golden ratio when I fix f(0)=2, f(1)=3

by the picture above, you can see the graph we have got here is similar to the other two's.

So this proves that all such sequences have the same limit of the ratio of successive terms.