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.