Visualizing the Fibonacci sequence

Here are few interesting things I came across recently when I was thinking about the Fibonacci sequence. In particular, it involves visualizing the Fibonacci sequence.

The Fibonacci sequence is defined by the recurrence

f(n) = f(n – 1) + f(n – 2)

where

f(0) = 0

and

f(1) = 1

It's easy to show that we can also write

f(n) = (anbn) / (ab

where

a = (1 + √5) / 2

and

b = (1 - √5) / 2

It's also easy to see the behavior of f(n) for negative values of n: we have that

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

What happens if we think of f as a function of a real variable instead of an integer variable?

In this case we have the function

f(x) = (axbx) / (ab)

This turns out to be a complex-valued function that happens to have a real part of zero at integer values. If we graph f(x) for 0 ≤ x ≤ 5, it looks like this (the complex value z = x + i y is plotted as (x, y)): 

Image001

If we graph f(x) for -5 ≤ x ≤ 0, it looks like this:

Image002
 

In that graph we can easily see how the fact f(-n) = (-1)n+1 f(n) is reflected.

To see what the graph for both positive and negative values of x look like together, here's the graph of f(x) for -4 ≤ x ≤ 5:

 

Image003 

What about derivatives of f(x)? 

It's fairly easy to see that 

f(n)(x) = (αn ax - βn bx) / (ab

where

αn = (log a)n

and

βn = (log b)n

so that the graph of f(n) looks somewhat similar to the graph of f.

Here's how the graphs of f, f' and f'' look, for example:

Image002 
If we look at how the differences between Fibonacci numbers behave, like

Δf(n) = f(n + 1) – f(n) = f(n – 1)

we might expect to see this reflected in f', but there doesn't appear to be a nice, clean relationship like

f'(x) = f(x – 1)

I'm not even sure how close you can get in this particular case.

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