φ and Fibonacci
by Emily Kennedy

Let's move on to a seemingly unrelated idea--the Fibonacci sequence.
Check out this site to see a little more about its origins.
The Fibonacci sequence is defined as follows:

F₀ = 1,         F₁ = 1,     and     F_n = F_n-2 + F_n-1 for n ≥ 2.

That is, the first and second (or "zeroth" and first) terms of the sequence are both 1, and each of the other terms is defined to be the sum of the two previous terms. So the first few terms of our sequence are

F₀ = 1
F₁ = 1
F₂ = 1 + 1 = 2
F₃ = 1 + 2 = 3
F₄ = 2 + 3 = 5
F₅ = 3 + 5 = 8
F₆ = 5 + 8 = 13
etc.

Spreadsheets are a great way to investigate recursively-defined functions and sequences like this one, because one can define a certain cell in terms of other cells. Try creating a spreadsheet with the first 50 Fibonacci numbers in the leftmost column.

You can type 1 into cells A1 and A2, then type =A1+A2 into cell A3, and copy and paste cell A3 into cells A4, A5, A6, ..., all the way to cell A50. See how easy that was?

Here's what mine looks like. You can click on a cell and then look at the function bar at the top of the screen to see the formula used to calculate that cell.

Now, in the second column, use the formula capability again to calculate the ratio of consecutive Fibonacci numbers (with the later number on top; that is, make it , not ). Make sure your cells show their value to several decimal places. Calculate the ratio many times for different pairs of consecutive Fibonacci numbers, until the ratio starts to "level off." Notice anything? Here is my file.

The ratio of consecutive Fibonacci numbers
seems to be approaching the Golden Ratio!
Why is this? Let's find out!

We have shown that .

Dividing both sides by , we have

We can replace the second in this equation with
(they are equal, after all!)

So

Again, we can replace the second in this equation with :

Repeating this process "forever," we get the continued fraction representation of :

Let's use this fact to get better and better approximations of .
(Of course, we already know the exact value of : , but soon you'll see why these approximations will help us.)

We can compute just part of this continued fraction, and as we compute longer and longer pieces, we'll get better and better approximations.

Let , , , , etc.

Since these are increasingly long "pieces" of the continued fraction representation of , we must have that
s_n approaches as n approaches ∞.

Also note that

, , , etc.

More generally,

Let's calculate some values of our s_i's:

etc.

Conjectures?

It looks like for all n ≥ 1.
Let's prove this, using proof by induction (see this page for a more detailed description of proof by induction--see, in particular, the ladder analogy on the first page).

Base case: n = 1

Induction hypothesis:
Assume that for some k ≥ 2, we have that

Induction step:
We need to show that .

By hypothesis, we have ,
and we have seen that
So

Thus, for all n ≥ 1, we have

And we have shown that s_n approaches as n approaches ∞,
so approaches as n approaches ∞.

Q.E.D.

Now go back to your spreadsheet and add a few more columns:
in the third column, calculate the ratio of every third Fibonacci number (); in the fourth column, every fourth Fibonacci number (); etc. Make as many columns as you like. Here is my file.

Do you see any patterns? Look for functions of .

It looks like the third column approaches ², the fourth column approaches ³, the fifth column approaches ⁴, etc.

Wow! Why is this? Let's find out!

We showed above that the ratio of consecutive Fibonacci numbers approaches as n approaches ∞.
Let's use this fact to prove our observations about the third and fourth columns:

Third column:

since F_n+2 and F_n+1 are consecutive,
and F_n+1 and F_n are consecutive
(so each of these ratios approaches as n approaches ∞)

Fourth column:

since F_n+3 and F_n+2 are consecutive,
F_n+2 and F_n+1 are consecutive,
and F_n+1 and F_n are consecutive
(so each of these ratios approaches as n approaches ∞)

Now let's generalize this idea to the k^th column
(the ratio of every k^th Fibonacci number):

There are k-1 factors in this product,
each of which approaches as n approaches ∞,
so the product approaches ^k-1.

We have shown that the values in the k^th column (the ratio of every k^th Fibonacci number) approaches ^k-1. Does this match your conjecture?

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