Seunghee Ye Ma 8: Week 2 Oct 6

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1 Week 2 Summary This week, we will learn about sequences and real numbers. We first define what we mean by a sequence and discuss several properties of sequences. Then, we will talk about what it means for a sequence to converge to a limit. In the second section, we will define real numbers. But first, we take a look at what it is about the reals that make them so special. The formal definition of real numbers will also give us a plethora of sequences to work with. Topics Page Sequences. First Definitions Convergence and Limit Arithmetic With Limits Real numbers 5 Sequences. First Definitions Let s begin by defining what we mean by a sequence of real numbers. Definition (Sequence) A sequence of real numbers is an infinite ordered set of real numbers {a, a 2,... } indexed by the natural numbers. A sequence of real numbers can be anything. For example, we could have the sequence a n n for all n; or also a n n for all n. Of course, the sequence could be constant as well e.g. a n for all n. However, it s important to note that a sequence does not have to have a nice pattern as the ones above. For example, we could consider the sequence {a n } where a n is the n-th decimal digit of π. When we talk about sequences, there are a few properties that Definition A sequence {a n } is said to be: bounded from below if there exists L R such that for all n, we have a n L. Then, L is called a lower bound for {a n } bounded from above if there exists U R such that for all n, we have a n U.Then, U is called an upper bound for {a n } bounded if it is bounded from above and from below monotonically increasing or nondecreasing if a n a n+ for all n N monotonically decreasing or nonincreasing if a n a n+ for all n N Example. (Fibonacci sequence). Some sequences are defined recursively. One of the most famous such sequences is the Fibonacci sequence which is defined as follows. a a 2, a n a n + a n 2 n 3 Page of 7

2 Is the Fibonacci sequence bounded from below/above or monotonically increasing/decreasing? Let s show that the Fibonacci sequence is bounded from below but not bounded from above; and it is monotonically increasing. Proof. Note that a a 2 > 0. Now, we show by induction that for all n N, a n > 0. Base case is true by the definition of the Fibonacci sequence. The induction step is equally as straightforward. Suppose a k > 0 for all k n. Now, a n+ a n + a n. By induction hypothesis, we know that a n, a n > 0 and thus, their sum must also be nonnegative. Therefore, a n+ > 0. We conclude that a n > 0 for all n N and thus, the sequence {a n } is bounded from below by 0. The second claim is that {a n } is not bounded from above. Let s proceed by contradiction. Suppose that there exists an upper bound U. We saw above that a n > 0 for all n. In fact, by exactly the same argument, we can show that is a lower bound for {a n }. The first few elements of the sequence are a a 2, a 3 2, a 4 3, a 4 5. In particular, a n+ a n + a n a n + for all n 5. Then, we see that a n+ a 2 + (n ) n for all n. Since U R we can find a natural number N such that rn > max(u, 5). However, since N 5, we have a N N > U and thus U is not an upper bound for {a n }. Hence, {a N } is not bounded from above. (Note that we also proved that the Fibonacci sequence is monotonically increasing!) Upper and lower bounds for sequences are very important since they give us a great deal of information about how the sequence behaves. It will also be important in deciding whether a sequence converges, as we will see later. Some values of upper/lower bounds are special: Definition (Least Upper Bound and Greatest Lower Bound) An upper bound U for the sequence {a n } is called the least upper bound if for any upper bound U, we have the inequality U U. A lower bound L for the sequence {a n } is called the greatest lower bound if for any lower bound L, we have the inequality L L. Example.2. Find the least upper bound and the greatest lower bound for the Fibonacci sequence. Proof. First, least upper bound is easy. {a n } is not bounded from above so there are no upper bounds. Hence, a least upper bound does not exist. The great lower bound is also not very hard to find. We saw above that 0 and are lower bounds for {a n }. Since > 0, will be our candidate for the greatest lower bound. Now suppose L is a lower bound. Then, we must have L a n for all n N. In particular, L a. But this precisely means that, which is a lower bound itself, satisfies the definition of the greatest lower bound. Hence, the greatest lower bound for {a n } is. Why are least upper bounds and greatest lower bounds so special? Are they even that special? To answer the second question, let s take a step back and think about why we want the natural numbers, N: we want to count things. Then, why do we need integers? We want to be able to subtract natural numbers. Subtraction is a very natural thing to do once we have the natural numbers so we expand our scope of numbers to the integers. Now that we can add and subtract, why talk about the rational numbers? Because we want to be able to divide integers. With our scope broadened to the rational numbers, we can add, subtract, multiply and divide. What more do we want? Why do we need the real numbers? Is it because some numbers such as 2, π, and e are irrational? That still doesn t justify all the other irrational numbers that we are adding to get the real numbers. The answer to this is the least upper bound property. Proposition.. Let A R be a subset of real numbers which is bounded from above. Then, A has a least upper bound Page 2 of 7

3 It turns out that the least upper bound property is very useful. So much so that it justifies talking about R instead of just Q. The following is an example of where the least upper bound property comes in handy. Proposition.2 (Archimedean property). Let x, y R and suppose x > 0. Then, we can find a natural number n N such that nx > y While this proposition might be familiar and even sound trivial to you, proof of it is not entirely trivial. Proof. Suppose we cannot find n N such that nx > y. Then, it means that for all n N, we nx y. In other words, the set {nx n N} is bounded from above by y. By the least upper bound property of R, we can find a least upper bound for this set, say z. Now, consider z x. Since z x < z and z is the least upper bound, we know that z x is not an upper bound for the set {nx n N}. In particular, we can find N N such that Nx > z x. However, this implies that Nx > z x Nx + x > z (N + )x > z Since (N + )x {nx n N}, the last inequality implies that z is not an upper bound for our set, which is a contradiction. Therefore, we conclude that we can find n N such that nx > y..2 Convergence and Limit Now, let s talk about convergence of sequences. We all have an intuitive idea of what it means for a sequence to converge. It should roughly be the sequence approaching a particular value as we go off to infinity. The following definition is the mathematical way of saying exactly what we think a convergent sequence should be. Definition (Limit and Convergence) Let {a n } be a sequence of real numbers. We say that the sequence converges if there exists a real number L R such that for all ε > 0, we can find N N such that for all n N, we have a n L < ε If such L exists, L is called the limit of the sequence {a n }. If this is your first time seeing a definition with ε, spend some time to take in what the definition is realy saying and convince yourself that it is exactly what we described in the paragraph above. We will introduce more definitions and theorems that will involve statement such as for all ε > 0 in the future so it will be worth your effort understanding this definition. Example.3. Show that the sequence {a n } defined by a n ( ) n n converges to 0 Proof. Let ε > 0 be a positive real number. Then, we want to find N N such that a n 0 a n < ε for all n N. Since ε > 0, ε > 0. Hence, we can find N N such that ε < N. Then, we have that N < ε. Now, for all n N we have: a n 0 a n n N < ε Therefore, the sequence {a n } converges to 0 as n approaches infinity. Example.4. Let a n n + n. Find wheter a n converges, and if it does, find its limit. Page 3 of 7

4 Solution. Let ε > 0. By plugging in a few large valuese of n, we guess that the limit of the sequence is 0. And note that n + n n + n () ( n + n + + n n) (2) n + + n < n + n n + + n (3) n + n (4) n (5) Hence, we want to find N such that for all n N, n is small (more precisely, less than ε). Since ε > 0 we know ε 2 > 0. By the Archimedean property, we can find N N such that N > ε 2. In particular, we have N < ε. Now, let n N. Then, we have a n 0 n + n 0 (6) n + n (7) < < ε n N (8) Hence, we conclude that the sequence a n converges to 0 as n approaches infinity..3 Arithmetic With Limits There are a few properties of limits that make working with them particularly nice. We will talk about these properties later this course but we preview a couple of them right here. Proposition.3. Let {a n } and {b n } be sequences which converge to L and M, respectively, as n approaches. Define a new sequence {c n } by letting c n a n + b n for all n. Then, {c n } approaches L + M as n approaches infinity. Proof. Let ε > 0. Since ε > 0, we know that ε 2 > 0. Applying the definition of convergence to {a n} and {b n } with their respective limit, we see that we can find:. N such that for all n N, we have a n L < ε 2 ; and 2. N 2 such that for all n N 2, we have b n M < ε 2 Now, let N max(n, N 2 ). The claim is that this particular N satisfies the condition in the definition of convergence. Let n N. Then, we have to show that c n (L + M) < ε Since c n a n + b n, by algebraic manipulation and triangle inequality we see that c n (L + M) (a n + b n ) (L + M) (a n L) + (b n M) a n L + b n M However, since N was chosen to be the maximum of N and N 2, we have that n N N and N 2. Page 4 of 7

5 Therefore, a n L < ε 2 and b n M < ε 2. Putting these together we see that c n (L + M) a n L + b n M < ε 2 + ε 2 ε Therefore, for any ε > 0, we can find N such that for all n N, c n (L + M) < ε. Thus, we conclude that {c n } approaches L + M as n approaches infinity. Proposition.4. Let {a n }, {b n } be sequences that converge to limits L and M respectively. Define a new sequence {c n } by c n a n b n. Then, the sequence {c n } converges to LM as n approaches infinity. The proof of Proposition.4 is similar to the proof of Proposition.3, only slightly more complicated. We will omit the proof of Proposition.4 but instead look at the following example. Example.5. Let {a n } be a sequence defined as follows: Show that {a n } does not converge to a limit. a, and a n+ + a 2 n for all n 2 Solution. Suppose {a n } does converge to a limit L. Then, by Proposition.4, we know that a 2 n must converge to L 2. Since a n+ + a 2 n, by squaring both sides we obtain a 2 n+ + a 2 n By taking the limit of both sides as n goes to infinity we get lim n a2 n+ lim ( + n a2 n) The left hand side is simply L 2 as we saw above. The right hand side, however, is lim ( + n a2 n) + lim n a2 n + L 2 Therefore, we get L 2 + L 2 which is a contradiction. Hence, the original sequence {a n } does not converge to a limit. 2 Real numbers There are lots of interesting sequences that arise naturally in mathematics but we now dive into studying how we can use sequences to define the real numbers. Definition (Real Numbers) A real number is an expression of the form x ±a a 2 a m.b b 2 where {b n } is a sequence of integers between 0 and 9, inclusive. Alternatively, we think of such an expression as x a + 0.b b 2 where a Z and {b n Z, 0 b n 9n N} (in other words, x is made up of its integral part before the decimal point and the part after the decimal part). Given a real number x as above, its truncation to n decimal places is the rational number m t n (x) a a m.b b n : a 0 m + + a m b 0 + b a i 0 m i + b j 0 j Note that given a real number x, we can associate to it a sequence rational numbers {t n (x)}. Suppose we have two real numbers x and y following above definition. We will say that x and y are equal as real numbers, and write x y, if the sequence {t n (x) t n (y)} converges to 0. i j Page 5 of 7

6 Proposition 2.. Let x and y be real numbers. Suppose x a a r.b b 2 and y a a s.b b 2. Then, x y if. a i a i for all i,..., r s; and 2. b i b i for all i; or there exists N such that b N b N, and b n 0, b n 9 for all n > N (or with roles of b i and b i reversed). In other words, real numbers x and y are equal if x and y are identically the same or if we can write x a a r a r.00, y a a r [a r ].99 or x a a r.b b N 000,, y a a r.b [b N ]999 Proof. Suppose x and y are real numbers with above properties. If x and y are identically the same, clearly we have t n (x) t n (y) for all n. Hence, {t n (x) t n (y)} is just the constant sequence {0} which clearly converges to 0. Now suppose x and y are not identically same but satisfy and 2 above. Note that since a i a i for all i,..., r s, we have that t 0 (x) t 0 (y). Moreover, since b i b i for i,..., N, we know that t i (x) t i (y) for all i,..., N Since t N (x) t N (x) + b N and t 0 N N (y) t N (y) + b N we get 0 N ( t N (x) t N (y) t N (x) + b ) ( ) N 0 N t N (y) + b N 0 N (9) ( bn b ) N (t N (x) t N (y)) + 0 N (0) 0 N () More over, since b i 0, b i 9 for all i > N, we have the following: ( bn+ b ) N+ t N+ (x) t N+ (y) (t N (x) t N (y)) + 0 N+ (2) (t N (x) t N (y)) 9 0 N+ (3) 0 N N+ 0 N+ 0 N+ (4) By induction, we can show that for all i > N, we have t i (x) t i (y) 0. Hence, we want to show that this i sequence converges to 0. Let ε > 0 be given. Then, we can find M large enough that 0 M > ε. Then, we have that ε <. 0 M Hence, for all i M we have t i (x) t i (y) 0 i 0 M < ε Therefore, {t n (x) t n (y)} converges to 0 as n approaches. Thus, x y as real numbers. Definition (Cauchy Sequences) Let {a n } be a sequence. We say that {a n } is a Cauchy sequence if for all ε > 0, we can find N N such that for all n, m N, we have a n a m < ε The essense of Cauchy sequences is that a sequence that is Cauchy is one whose terms get very close to each other as n goes off to infinity. This should strongly hint that the sequence should converge. Before we investigate the relationship between Cauchy sequences and convergent sequences, let s first look at a few examples of Cauchy sequences. Page 6 of 7

7 Example 2.. Let x be a real number represented as x a a m.b b 2. Show that {t n (x)} forms a Cauchy sequence. Solution. Let ε > 0. Then, we can find N N such that 0 < ε. Let N N N +. Now, take n, m N. Then, t n (x) t N (x) < and t 0 N m (x) t N (x) <. Hence, 0 N Thereore, {t n (x)} is a Cauchy sequence. t n (x) t m (x) t n (x) t m (x) + t N (x) t N (x) (5) (t n (x) t N (x)) (t m (x) t N (x)) (6) t n (x) t N (x) + t m (x) t N (x) (7) < 0 N + 0 N 2 0 N < < ε N 0 (8) Page 7 of 7