Introduction to Mathematical Thinking: Real Analysis 3

Introduction to Mathematical Thinking. Real Analysis 3 We can construct the sequence an, n from 1 to infinity. We can describe the behavior of a sequence as its members approach a limit from above by saying that the sequence tends to the limit as n approaches infinity. Roughly speaking, the absolute value of an minus a becomes arbitrarily close to 0 when the absolute value of a approaches infinity. Now let us define the term formally. And the formal definition involves a precise statement of what an arbitrarily close to 0 means. An tends to a as n tends to infinity, if and only if for all real numbers epsilon greater than 0, for every number m greater than or equal to n, there exists a natural number n such that the absolute value of am minus a is less than epsilon. Now you see why we spent so much time on quantifiers and being careful about the order in which to use them. The formal definition of a limit is crucial to the study of real analysis. It's crucial to our understanding of the real numbers, and hence it's crucial to calculus. Hence, it's crucial to physics, science, engineering, technology and so on. Let's try to understand how a formal definition captures this intuition: an-a being arbitrarily close to 0. Our geometric intuition is fairly clear: we see a sequence of numbers getting closer and closer to some fixed number. Formal definition is a complicated-looking expression that can be precisely stated. Let's set aside, for the moment, the part (∀ε>0). In mathematics, the letter ε is traditionally used in this (∀ε>0) context. It is important to remember that epsilon is not just positive but very small, like a tenth or a hundredth, or a millionth, or a zillionth. It is when ε becomes small but remains positive that the limit of (∃n∊N)(∀m≥n)[|am-a|<ε) is useful. Let's put that aside for the moment and look at the other part. For every n greater than or equal to 1, there exists a natural number m such that the absolute value of am minus a is less than epsilon. We have not mentioned explicitly the set N. We didn't mention that m was also a natural number in the first place because the context makes it clear. n is a natural number, and we'll be considering all m's greater than or equal to n. We mentioned earlier how mathematicians commonly use quantifiers to simplify expressions. Thus, it is easy to read. Okay, what does this part (∃n)(∀m≥n)[|am-a|<ε] mean? We'll assume for this problem that we're given some positive number epsilon. And keep it in mind as a small positive number. This means that after some point n, all members of the sequence will be within distance epsilon of a. All of the numbers in the sequence {an} n = 1 to infinity are within a distance of epsilon from a. Let's see what (∀ε>0)(∃n∊N)(∀m≥n)[|am-a|<ε) means when we have the whole expression. This states that for any epsilon greater than 0, the following is true. For any epsilon 0, there is a point at which all the numbers are within a distance epsilon of a. The intuition is that as long as we take ε small enough, we can approximate epsilon with any value we choose. Look at the interval in the picture below. We have a number, a, and a sequence is moving around in here. There were these numbers, a1, a2, scattered along this line. Now, we're given an epsilon greater than zero, so we'll assume it's below zero. Let's say that, here on the left of the interval is a minus epsilon, and here on the right of the interval is a plus epsilon. Let us assume that we have an epsilon of distance to the left and another epsilon of distance to the right. Our argument will hold for any such epsilon. Let us consider the epsilon interval for a moment. The rest of the formula states that all elements of the sequence are contained within this interval. Suppose we take a smaller epsilon. We can call this a minus epsilon prime (on the left) and this a plus epsilon prime (on the right). So we take a smaller positive value for the epsilon, which will ensure that the formula will still work when we use it with epsilon prime. There will be a point, however small, from which all elements in the sequence are within the region between a - ε and a + ε. We can take an even smaller interval and an even smaller one. Note that the n depends on the value of epsilon. If we reversed the quantifiers here, our results would not be valid. Given an epsilon, you could find an n. From each epsilon, you must proceed one-by-one until you reach a point at which all elements of the sequence fall within that distance. But what this (∀ε>0)(∃n∊N)(∀m≥n)[|am-a|<ε) says, is that by always following out the sequence sufficiently far, you could reach a stage where all of the numbers in the sequence were within a given distance of a. And so, you can continue this process for a smaller and smaller value of a. That is captured by the (|an-a| being arbitrary close to 0) intuition. It does so beautifully, elegantly and with enormous power. Let's get to the examples. First, let’s consider the sequence {1/n} as n goes from one to infinity. Now we know that {1/n} tends to zero as n tends to infinity. Let's prove this rigorously, in terms of the definition given above. Some of these facts are proven; in fact, for all epsilon greater than zero, all integers greater than zero, there is a number in the set of natural numbers. Hence, for all natural numbers m greater than or equal to n. The absolute value of 1 over m – 0 is less than epsilon. Let's simplify that. For all epsilon greater than 0, there exists an n such that m is greater than or equal to n. The expression 1/m ± ε simply refers to 1/(m + ε) because m is a positive integer. In order to prove the result, we must check whether the limit of one divided by n as n gets larger is zero. We have shown that as n approaches infinity, one divided by n approaches zero. How can we verify that? When epsilon is greater than zero, another way to say it would be that epsilon must be less than or equal to zero. Essentially, in this context the two phrases mean the same thing. We will need to find a value of n for which the inequality (∀ε>0)(∃n)(∀m≥n)[1/m<ε], for all m greater or equal to n, is valid. The ratio 1 over m is less than epsilon. For any n greater than 1/ε, let n be defined as the number equal to n.If epsilon is a very small number, this will be a very large number, and so it might be prudent to choose a larger. Now, intuitively, you know that the natural numbers go on forever, so they grow larger and larger. It actually uses a principle of mathematics known as the Archimedean Property. So choose a big enough value of n so that n is greater than 1 over epsilon. If m is greater than or equal to n, then one over m is less than or equal to one over n, which is less than epsilon and completes the proof. We chose n based on epsilon. Given an epsilon, we picked our n in terms of the epsilon. In mathematics, getting the order of quantify and what it applies to wrong can have serious repercussions. The sequence converges to the limit of n, which depends upon the epsilon. The smaller the epsilon is, the bigger n must be in order for it to converge to a finite value. The smaller the tolerance you impose on the number being close to zero, the farther along in the sequence of numbers you have to go; in other words, the larger n will be in that tolerance. Order matters. The choice of n depends on epsilon. Different epsilons, different n. Look at one more example. We can generate a sequence n over n plus one, n equals one to infinity, by starting with 1/2, 2/3, 3/4 and continuing in that pattern. Let us prove in terms of the definition, that as n approaches infinity, n/n+1 converges to 1. For all values of epsilon greater than 0, there is a natural number n such that for any m greater than or equal to n, the 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. Given a positive number epsilon, we seek to find an integer n such that for all integers m greater than or equal to n, m over m + 1- 1 is less than epsilon. We are going to pick a number, n, so large that n is greater than 1 over epsilon. The choice is the same as the one in the previous example. The examples we are using are simple, so the results are predictable. In more complicated examples, the choice is not so simple. However, let's consider why this won't work in this case. If m is greater than or equal to n, then m/m + 1 – 1 = [absolute value of (m/m + 1)], which is less than or equal to [absolute value of (m/m + 1)]/[absolute value of (m + 1)], which is less than or equal to epsilon. Remember that the examples we saw were very simple ones. Where the n was chosen very simply. You will not always be so lucky. Sometimes you will have to work harder to find the n. But focus is here, as it has been throughout this course, not on the details or procedures but on the big picture and how you can use it. The focus is on reasoning about infinite sequences. The key idea is that, given some epsilon, you can choose an n that depends on it. This quantifier switch is really crucial. The formulation (∀ε>0)(∃n∈N) was a brilliant and important achievement in late 19th century mathematics. All of modern mathematics—at least, all of it dealing with real numbers—owes a debt to this work.