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)
f(0) = 0
f(1) = 1
It's easy to show that we can also write
f(n) = (an – bn) / (a – b)
a = (1 + √5) / 2
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) = (ax – bx) / (a – b)
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)):
If we graph f(x) for -5 ≤ x ≤ 0, it looks like this:
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:
What about derivatives of f(x)?
It's fairly easy to see that
f(n)(x) = (αn ax - βn bx) / (a – b)
αn = (log a)n
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:
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.