Assignment 12

Investigation 3

Exploring the Fibonacci Sequence

First we will generate a Fibonacci sequence in the first column of our spreadsheet using f(0) = 1, f(1) = 1, and f(n) = f(n-1) + f(n-2)

 n Value Fibonnaci Value 0 1 1 1 2 2 3 3 4 5 5 8 6 13 7 21 8 34 9 55 10 89 11 144 12 233 13 377 14 610 15 987 16 1597 17 2584 18 4181 19 6765 20 10946

Now if we create a third column with the ratio of each pair of adjacent terms in the Fibonacci sequence.

 n Value Fibonnaci Value Ratio 0 1 1 1 1 1 2 2 2 3 3 1.5 4 5 1.66666666666667 5 8 1.6 6 13 1.625 7 21 1.61538461538462 8 34 1.61904761904762 9 55 1.61764705882353 10 89 1.61818181818182 11 144 1.61797752808989 12 233 1.61805555555556 13 377 1.61802575107296 14 610 1.61803713527851 15 987 1.61803278688525 16 1597 1.61803444782168 17 2584 1.61803381340013 18 4181 1.61803405572755 19 6765 1.61803396316671 20 10946 1.6180339985218

It is obvious that the ratios are approaching 1.618 or something close to 1.618. Now lets do the same thing with the ration of every second term.

 n Value Fibonnaci Value Adjacent Ratio Second Ratio 0 1 1 1 1 1 2 2 2 2 3 3 1.5 3 4 5 1.66666666666667 2.5 5 8 1.6 2.66666666666667 6 13 1.625 2.6 7 21 1.61538461538462 2.625 8 34 1.61904761904762 2.61538461538462 9 55 1.61764705882353 2.61904761904762 10 89 1.61818181818182 2.61764705882353 11 144 1.61797752808989 2.61818181818182 12 233 1.61805555555556 2.61797752808989 13 377 1.61802575107296 2.61805555555556 14 610 1.61803713527851 2.61802575107296 15 987 1.61803278688525 2.61803713527851 16 1597 1.61803444782168 2.61803278688525 17 2584 1.61803381340013 2.61803444782168 18 4181 1.61803405572755 2.61803381340013 19 6765 1.61803396316671 2.61803405572755 20 10946 1.6180339985218 2.61803396316671 21 17711 1.61803398501736 2.6180339985218 22 28657 1.6180339901756 2.61803398501736 23 46368 1.61803398820532 2.6180339901756 24 75025 1.6180339889579 2.61803398820533 25 121393 1.61803398867044 2.6180339889579

What would happen to our ratios if we used two different values for f(0) and f(1). For example, like the Lucas sequence let f(0)=1 and f(1)=3.

 n Value Fibonnaci Value Adjacent Ratio Second Ratio 0 1 0.333333333333333 1 3 3 2 4 1.33333333333333 4 3 7 1.75 2.33333333333333 4 11 1.57142857142857 2.75 5 18 1.63636363636364 2.57142857142857 6 29 1.61111111111111 2.63636363636364 7 47 1.62068965517241 2.61111111111111 8 76 1.61702127659574 2.62068965517241 9 123 1.61842105263158 2.61702127659574 10 199 1.61788617886179 2.61842105263158 11 322 1.61809045226131 2.61788617886179 12 521 1.61801242236025 2.61809045226131 13 843 1.61804222648752 2.61801242236025 14 1364 1.61803084223013 2.61804222648752 15 2207 1.61803519061584 2.61803084223013 16 3571 1.6180335296783 2.61803519061584 17 5778 1.61803416409969 2.6180335296783 18 9349 1.61803392177224 2.61803416409969 19 15127 1.61803401433308 2.61803392177224 20 24476 1.61803397897799 2.61803401433308 21 39603 1.61803399248243 2.61803397897799 22 64079 1.61803398732419 2.61803399248243 23 103682 1.61803398929446 2.61803398732419 24 167761 1.61803398854189 2.61803398929446 25 271443 1.61803398882935 2.61803398854189

We can see that our ratios still approach the same values as the original Fibonacci sequence.