And what kinds of mathematical things can we observe?

Well, one is that the limit of

incident terms in the Fibonacci sequence approaches.

Phi, this golden ratio, that is the limit as n goes to infinity of Fn plus

1 over Fn equals 1 plus root 5 over 2.

How do we prove that? Begin with recursion

relation Fn plus 2 equals Fn plus 1 plus Fn, and divide all terms by Fn plus 1.

This simplifies the middle expression to a 1.

Call this limit of incident terms L and then compute

the limit of everything that we have. Well, on the left,

you'll see Fn plus 2 over Fn plus 1. That, of course, is going to limit

to the same thing as Fn plus 1 over Fn. That is L.

On the right, we have something that will limit to 1 over L.

And so we obtain the expression L equals 1 plus 1 over L.

By multiplying through,

we see a familiar quadratic expression that factors

with root v and c, where c is 1

minus root 5 over 2. Now that, again, is a negative number.

It's equal to 1 minus phi.

In fact, we know that, that is not the limit.

And so the limit is phi, this golden ratio.

Now, this still doesn't really feel like calculus.

Even though we're supposed to be doing discrete calculus at this point.

Let me tell you what calculus is good for. Let's say you are asked to compute

an exact expression for the Fibonacci number or maybe an approximation.

Let's say, what is the one thousandth Fibonacci number or the one millionth.

Well, of course, you could get there by adding, adding, adding many times over.

But there's a more intelligent way, using calculus.

The solution that we'll

present to the few lectures is that F sub n is exactly equal to 1 over

root 5 times quantity phi to the n minus psi to the n.

Where phi and psi are as computed before, 1 plus

root 5 over 2 and 1 minus root 5 over 2.

Now since

the latter is negative, when we take a high

power of that, it's going to be something that is negligible,

very close to 0. And so we could approximate the nth

Fibonacci number by phi to the n over the square root of 5.

That is a very convenient result.

It's not hard at all to do that computation.

This is one of things that we'll see discrete calculus is good for.

It's going to take a little bit of time to get to the point where we can do this.

We have to build up some of the rest of

our techniques involving discrete versions of derivatives and differential equations.

We'll move on to discrete versions of integrals, stay tuned.

And keep

your eyes open for interesting sequences.