Do a writeup of either problem 4 or 7 from Jim Wilson's Problem Set #12.
Generate a Fibonacci 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 Fibonacci sequence.
What happens as n increases?
What about the ratio of every second term, etc.
B) Explore sequence 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.
The first place to start this problem is to go ahead and create the Fibonacci sequence:
As we can see, the numbers get really large, really fast.
Now, what were to happen if we would look at the ratio of each pair of adjacent terms, that is f(n)/f(n-1)? What about the ratio of every second term, or even every third term?
Well, lets look at the sequences and see what happens:
Well, if we look at the ratio of adjacent terms, you can notice that as n increases, the value reaches the limit of 1.618033989. What is the significance of this number, well it is known to be the golden ratio. This decimal can be represented as the fraction: [1 + sqrt(5)]/2.
Notice the second and third ratios, they also approach a limit, but theirs is different then the Golden Ratio. What happens when we look at the ratio of the second ratios to the first ratios, that is the second ratio / the first ratio. The limit of this ratio turns out to be the Golden Ratio again. Why is that?
Well, if we look at this algebraically:
2nd ratio/1st ratio = [f(n)/f(n-2)]/[f(n)/f(n-1)] = [f(n)/f(n-2)]*[f(n-1)/f(n)] = f(n-1)/f(n-2).
Recall that f(n-1)/f(n-2) is the ratio of the 2 previous terms, which we have already seen to reach a limit of the Golden Ratio, so it makes sense that this 2nd to 1st ratio reaches the same limit.
This will also be true when we look at the ratio of the third ratios to the second ratios:
Can you see why this relationship should hold true?
What would happen now if we were to look at sequences described in part B.
Well let's start by looking at the results:
Notice that the limits of the ratios of this sequence are the same as the limits of our previous example. That also means that the limits of the ratios will be the same as well. Will this relationship hold true for other examples?
Let's look at one more sequence and see what happens. This time we will look at the Lucas sequence, where f(0)=1 and f(1)=3:
Again we are faced with the same results.
So, in conclusion, it seems that whenever we define a sequence with f(n)=f(n-1) +f(n-2), and consequently look at the ratios of previous terms, we will result in getting a similar limit, The Golden Ratio.
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