Okay? Let's move on. Well, so far, everything’s been very intuitive. Let's get a little bit more formal now. And we've got a sequence a n, n from 1 to infinity. The intuitive explanation or the intuitive description I gave of this. That the sequence or the members of the sequence tend to a limit a as n tends to infinity. That sort of corresponds fairly roughly to the fact that the absolute value of an minus a becomes arbitrarily close to 0. Boy, my writing really is a problem, isn't it? Close. Now let me give you the formal definition. And the formal definition involves being precise about what that means. A n tends to a as n tends to infinity, if and only if for all real numbers epsilon greater than 0, there is a natural number n such that for all m greater than or equal to n, absolute value of am minus a is less than epsilon. And now perhaps for the first time. You see why we spent so much efforts understanding quantifiers, and in particular, about the order in which quantifiers are paired. This definition is absolutely crucial in real analysis. It's absolutely crucial to our definition and our understanding of the real numbers. And that means it's absolutely crucial to calculus. And hence it's absolutely crucial to physics, science, engineering, technology, etc, etc, etc. This is the real stuff, folks. Okay, let's try and understand how that captures this intuition. The geometric intuition we have I think is fairly clear, that you're going along a sequence and the numbers get closer and closer and closer to some fixed number. How does this somewhat complicated looking expression capture that in a precise formal way? Well, let's peel away this part for a minute. By the way, it's traditional in mathematics to use epsilon in this context. And at the back of your mind, you should think that epsilon is not just positive, but it's very small and positive, like a tenth or a hundredth or a millionth or a zillionth or whatever. As we'll see when I see when I go through this, it's when epsilon becomes small but remains positive that this thing kicks into power. Okay, so let's forget that a bit, let's look at the other part. Exists an n such that for all m greater than or equal to n, absolute value a m minus a is less than epsilon. And I've avoided mentioning explicitly the set N now. Notice that in the first place, I didn't mention that m was also a natural number, because the context makes that clear. n is a natural number, and we're talking about all m's greater than or equal to n. And when we discussed quantifiers earlier on, I pointed out that mathematicians do this kind of simplification all the time. So it makes it easy to read, that's all. Okay, what does this part mean? So we've been given an epsilon, we assume, we're given some positive number epsilon. And at the back of your mind, think of it as a small positive number. What this means is that from some point onwards, so from some point n onwards. All the members of the sequence are within a distance epsilon of a. So all the numbers in the sequence {an} n equals 1 to infinity are within. A distance of epsilon from a. You still with me? Now let's see what this thing means when we have the whole expression. This says that for any epsilon greater than 0, this thing holds. So for any epsilon 0, there is a point such that from that point onwards, all the numbers are within a distance epsilon from a. As I mentioned a moment ago, the intuition here is that we can take epsilon greater than 0 as small as we want. So let me draw a picture now. We've got some number a, and there's a sequence bouncing around here somewhere. Maybe it's coming in from the left, from the right, or maybe it's bouncing around. But there were these numbers, a1, a2, scattered along this line here. Now, we're given an epsilon greater than zero, so let's suppose it's down here. So here's, let's say, here's a minus epsilon. And here is a plus epsilon. Okay, I've just got a distance epsilon to the left and a distance epsilon to the right. And the argument I'm about to give will hold for any epsilon, at least that's what this says. So I'm given an epsilon and I'll look at this interval. And what the rest of this formula says is that from some point on, all the elements of the sequence are in here. If I take a smaller epsilon, let's do it, say, here and here. Let's call it epsilon prime, so that would be a minus epsilon prime. And here would be a plus epsilon prime. So I'll take a smaller epsilon, still positive, so the formula will hold for the epsilon prime. There will be some point from which all the elements of the sequence beyond that point are within this region. And then I could take an even smaller one and an even smaller one. Notice that the n depends on the epsilon. Remember that example about the American Melanoma Foundation? Wasn't a problem for them getting the quantifiers the wrong way around. If we got the quantifiers the wrong way around here, we'd be in big trouble, this wouldn't work. The point is, given an epsilon, you could find an n. From each epsilon, you may have to go further out in the sequence until you comes to a point where all the elements are within that distance. But what this says is that you could always by going sufficiently follow out the sequence. Reach a stage where all of the numbers in the sequence from then onwards are within a given distance of a. And you can do that for smaller, and smaller, and smaller a. And that's capturing this intuition. And it does it beautifully and elegantly and with enormous power. Okay, let me give you a couple of examples. First example, let's look at the sequence one over n, n goes from one to infinity. Now we know that one over n turns to zero as n turns to infinity. Let's prove this vigorously, in terms of the definition of limits that I just gave. Okay. Some of it prove that fact. What after sure is that for all epsilon greater than zero, all the other numbers that are greater than zero, there's an end in the natural numbers. So it's that for all natural numbers m greater than or equal to n. Absolute value 1 over m minus 0 is less than epsilon. Well, let's simplify that. For all real numbers epsilon greater than 0, there is an n, and let me just drop that explicit mention of the natural numbers, for all m greater than equal to n. That just says absolute value of 1 over m, less than epsilon, which actually it just means 1 over m. Less than epsilon, because m is a positive integer, okay? So I have to verify this, in order to prove that. Okay so how do I verify that? Well at epsilon greater than 0 be given Another way of saying it would be to say less epsilon greater than 0 would be arbitrary. Different way of saying the same thing essentially, at least in this circumstance. What I'll need to do, Is find an n such that, For all m greater or equal to n. 1 over m is less than epsilon. Well. Pick any n Such that n is bigger than one over epsilon. If epsilon is a very small number, this will be a very large number, so I might have to pick a larger. Now, intuitively, you know that since the natural numbers go on forever as it were, they get bigger and bigger, that's always possible. And actually uses a principle of mathematics called the Archimedean Property. So named after Archimedes. See Assignment 10.2 for a little bit more about that. So pick an n big enough, so that n is greater than 1 over epsilon. Then, If m greater than or equal to n, one over m is less than or equal to one over n, is less than epsilon, and we're done. Notice that the choice of n depended on epsilon. Given an epsilon, I picked my n in terms of the epsilon. Please note American Melanoma Foundation. At least when you apply quantify as in mathematics, it's a big deal if you get in the wrong order. The n depends upon the epsilon. The small the epsilon is, the bigger the n it has to be intuitively for this sequence. The smaller the tolerance you impose on the number being close to zero, the further out in the sequence you have to go, the bigger the n is before you in that tolerance. So quantify order matters. The choice of n depended on epsilon. Different epsilon, different n. Okay, let me give you one more example. And we look at this sequence, n over n plus one or n equals one to infinity, that's the sequence a half, two-thirds, three-quarters, four-fifths, and so on. And let me prove in terms of the definition. That's n over n plus one turns to one as n turns to infinity. I mean true to believe obviously does, these numbers get closer and closer and closer to one. But to prove that, what I have to show is the following. For all epsilon greater than 0, there is a natural number n, such that for all natural numbers m greater than or equal to n absolute value of m over m plus 1 minus 1 is less than epsilon. Okay, I'm going to prove it the same was as before. I'm going to assume I'm giving it epsilon, then I'm going to find an n dependent on epsilon to make this thing true. Let epsilon greater than zero be given. We need to find an n such that for all m greater or equal to n, M over m + 1- 1 in absolute value is less than epsilon. Well I'm going to pick n so large that n is bigger than one over epsilon. That's actually the same choice as in the previous example. It just works out that way, because I'm using very simple examples. In more complicated examples The choice is perhaps not quite so simple as straight forward as this. Okay, but let see why that won't works in this case. Then for any m greater or equal to n, I've got the following, m over m + 1 minus 1, an absolute value, equals minus 1 over m + 1, an absolute value. Just work that out as a single fraction, which is just 1 over m + 1. Which is less than 1 over m. Which is less than or equal to 1 over n. because m is greater or equal to n which is less than epsilon. And we're done. Okay, you should now be in a position to attempt the questions on assignment ten point two. Bear in mind that the two examples I've given were particularly simple ones. Where the choice of the n was very simple. You won't always be so lucky, you might have to work a little bit harder to find the n. But my focus is here, as it's been through out the course, wasn't so much on the details on the procedures. And if you graduated from high school, I assume you can do mathematical procedures, you can do the algebra. The focus is on the reasoning, in particular in the context of sequences. The key thing is how given an epsilon you have to choose an n that depends upon it. It's this quantifier switch that's really crucial. For all epsilon there is an n. If you go on to further studies in mathematics, in particular real analysis, you're going to see variance of these kind of definition. Not just for little bits of sequences but for the definition of continuity. And then when you come to look at the forward definitions of integration and so forth. So you're going to come across this kind of a definition a lot. Coming up with this formulation was one of the jewels in the crown of late 19th century mathematics. Absolutely brilliant piece of work, and it gave advice to pretty well all of modern mathematics. At least all of modern mathematics has to do with real numbers. Okay, well good luck with assignment ten point two. Well, you made it. All you have to do now is complete the final assignment and you should be well set for taking college level mathematics courses. Or for simply reasoning more effectively in your job or in your everyday life. Since, you got this far I'm sure you'll complete these last steps. So let me offer you my congratulations now. These massive online courses are still hugely experimental so simply sticking with us and finishing the course is an accomplishment all by itself. Future generations of students will surely have it easier in some ways. What they won't find easy, however, is the material itself. Learning facts is relatively easy. Learning to think a whole new way is extremely difficult. However well you think you've mastered the material, you should feel pleased with completing the entire process under your own motivation. As a life long teacher much of my motivation and reward has been getting to know each generation of students and working with them, helping them overcome difficulties, and then seeing them succeed. With a MOOK that regular close contacts absence and I miss it. On the other hand, being able to reach tens of thousands of students as opposed to 25 brings its own reward. The next two weeks are devoted to the test flight process. See the website for details
