Welcome to Calculus.
I'm Professor Greist and we're about to begin Lecture 45, Bonus Material. In our main lesson, we considered sequences as the discreet version of
functions. Ending with a particular sequence due to
Fibonacci and its relation to the golden ratio, that's what we're
going to explore here. The Fibonacci sequence is a sequence,
we'll denote it F sub n and it is defined through the recursion relation F sub n plus 2 equals F
sub n plus 1 plus F sub n.
And so, beginning with the first two terms, 0 and 1. One computes the remaining terms by
summing up the previous two. Now, people have found the Fibonacci
sequence hiding in all sorts of natural objects. That's not really our concern in this
class. We're going to stick to the math. 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.