Math 410 Homework 6 Due Monday, October 26

Transcription

1 Math 40 Homework 6 Due Monday, October 26. Let c be any constant and assume that lim s n = s and lim t n = t. Prove that: a) lim c s n = c s We talked about these in class: We want to show that for all ɛ > 0 there exists an N such that c s n B(cs, ɛ) whenever n N. That is, show that c s c s n < ɛ whenever n N. Also, we can use the fact that lim s n = s, which means that for any ɛ > 0, Ñ such that s s n < ɛ whenever n Ñ. Let ɛ = ɛ c and N = Ñ the resulting cut off point from the convergence of {s n } for this choice of ɛ. Then for any n N, c s c s n = c s s n < c ɛ = ɛ which is what we wanted to show. c b) lim s nt n = st (there s a trick for this that we ll cover in class) We want to show that ɛ > 0 N st st s n t n < ɛ whenever n N and we can use the facts that s n and t n converge to s and t respectively. Namely, ɛ > 0, Ñ st s s n < ɛ whenever n Ñ and ɛ > 0, N st t t n < ɛ whenever n N. The trick from class is as follows: st s n t n = st st n + st n s n t n = s(t t n ) + t n (s s n ) (it doesn t matter if you add and subtract st n or s n t you will get to the same place in the end) so st s n t n s t t n + t n s s n by the triangle inequality and algebra. By a theorem from class, since {t n } converges, it is bounded, so there exists some M > 0 such that t n < M for all n. Now, let ɛ = ɛ ɛ and ɛ = M s and N = max{ñ, N}. Then for all n N, we get st s n t n s t t n + t n s s n < s t t n +M s s n < s ɛ+m ɛ = s ɛ s +M ɛ M = ɛ which is what we wanted to show! 2. Squeeze Theorem Show that if x n y n z n for all n N, and if lim x n = lim z n = l, then lim y n = l as well. We are assuming that lim x n = l, so for all ɛ > 0 there exists a Ñ such that d(x n, l) < ɛ whenever n Ñ. Similarly, for all ɛ > 0 there exists N such that d(z n, l) < ɛ whenever n N and we want to show that for all ɛ > 0 there exists a N such that d(y n, l) < ɛ whenever n N. The total inequality we come up with out of this situation is l ɛ < x n y n z n < l + ɛ as long as n is larger than both Ñ and N. Therefore, we only need to let ɛ = ɛ and ɛ = ɛ and N = max{ñ, N} so that l ɛ < x n y n z n < l + ɛ

2 whenever n N. In other words, d(y n, l) < ɛ whenever n N which is what we wanted to show. 3. Let {x n } and {y n } be given sequences and define {z n } to be the shuffled sequence {x, y, x 2, y 2,..., x n, y n,...}. Prove that {z n } converges if and only if x n and y n converge and lim x n = lim y n. I have a long and somewhat obvious proof below, but Meghan pointed out that if we use the version of convergence that {a n } n= converges to a if for all ɛ > 0, all but finitely many of the a n are inside B(a, ɛ), then this proof is really easy!! Observe: We assume that for all epsilon > 0, all but finitely many of the z n s are in B(z, ɛ). Given ɛ > 0, let ɛ = ɛ. Since each z n is either x i or y j for some i or j, only finitely many z n s outside B(z, ɛ) implies that all but finitely many of the x n s are inside B(z, ɛ) so x n x and similarly for the the sequence consisting of only y n s. The other direction is similar! Isn t that nice?!? Assume that {z n } converges to z. That is, for every ɛ > 0, there exists a Ñ such that d(z n, z) < ɛ whenever n Ñ. Show that {y n} converges to z as well. Let ɛ = ɛ and N = Ñ/2. Since z 2n = y n, the fact that d(z m, z) < ɛ = ɛ for every m Ñ = 2N, it certainly holds for m even with m = 2n Ñ = 2N, namely d(z m, z) = d(z 2n, z) = d(y n, z) < ɛ whenever n N which is what we wanted to show. Similarly, for {x n } let N = (Ñ + )/2. Since z 2n = x n, the fact that d(z m, z) < ɛ = ɛ for every m Ñ = 2N, implies that it certainly holds for m odd, so m = 2n Ñ = 2N, so d(z m, z) = d(z 2n, z) = d(x n, z) < ɛ whenever n N. Therefore, both {x n } and {y n } converge and lim x n = lim y n. Assume that there exists z such that for all ɛ > 0 and all ɛ > 0, there exist Ñ and N such that d(x n, z) < ɛ whenever n Ñ and d(y n, z) < ɛ whenever n N. Given ɛ > 0, choose ɛ = ɛ = ɛ and N = max{(ñ + )/2, N/2}. Then d(z 2n, z) = d(x n, z) < ɛ whenever m = 2n N and d(z 2n, z) = d(y n, z) < ɛ whenever m = 2n N, so d(z m, z) < ɛ whenever m N and the sequence {z n } converges. 4. a) Prove that the sequence defined by x = 3 and x n+ = 4 x n converges. We will want to use the monotone convergence theorem! Computing the first few elements of the sequence, we see that x = 3, x 2 = 4 3 =, x 3 = 4 = 3, x 4 = = 3 4, 3 so we are going to prove that the sequence is decreasing and bounded below by 0. Proof by induction: might need to do the bounded below by zero part st cause we ll probably need the 0 < a < b implies 0 < /b < /a equation. In fact, we ll prove by induction that 0 < x n 3 for all n N. 0 < x = 3 3 which starts our induction. Assume that 0 < x k 3. This implies that 0 > x k 3 so 4 > 4 x k 4 3 = > 0 and therefore 4 x k = x k+ > 4 > 0 which is an equation that is valid for k namely for x 2, x 3, x 4,... are between and 4 which proves our induction step. Combining this with the fact that x = 3 we see that 0 < x k 3 for all k N and therefore the sequence is bounded.

3 We will also prove that the sequence is decreasing by induction. x > x 2 which starts our induction. Assume that x k > x k and prove that x k > x k+. From the first induction, we know that 3 x k > x k > 0, so 3 x k < x k < 0 and 0 < = 4 3 < 4 x k < 4 x k < 4. Therefore 0 < 4 < 4 x k = x k+ < 4 x k = x k < which proves our induction. Therefore, our sequence is bounded and (strictly) decreasing so converges by the monotonic sequence theorem! b) Now that we know lim x n exists, explain why lim x n+ must also exist and equal the same value. Let z n = x n+ another sequence. By the exact same argument as above, z n is bounded and decreasing, so converges. Also, let x be the limit of the sequence x n. Therefore, for any ɛ > 0, there exists Ñ such that d(x, x n) < ɛ whenever n Ñ. Let ɛ = ɛ and N = Ñ. Then d(x, z n) = d(x, x n+ ) < ɛ whenever n + Ñ = N + aka whenever n N, so {z n } converges to x as well. c) Take the limit of each side of the recursive equation in a) like we did in class to explicitly compute lim xn. Therefore, we get x = lim x n+ = lim 4 x n = 4 lim x n = 4 x where the first equality is part b), the second is the definition of the sequence, the third is a combination of the limit theorems we had from class (limit of the sum is sum of limits, etc.) and the last equality is our assumption from part b). The total equation x = 4 x gives us x2 4x + = 0 which has roots x = 4 ± = 2 ± 3 Of these two roots, the larger one is outside the range of 4 < x n (the equation we found in part a) that applies to all n > ) so the final answer is lim x n = 2 3 which is just a bit more than 4. This does make sense because a final answer of /4 would only happen if the x n 0 which we know isn t actually the case! 5. Show that 2, 2 2, 2 2 2,... converges and find its limit. This is a lot like the previous problem in that we use induction to show that the sequence is increasing and bounded and therefore the monotone convergence theorem applies and the sequence converges. Then (skipping step b) above because it s like making curtains, doing it once was enough) we will use the recurrence relation x n+ = 2x n combined with the limit theorems to say that whatever the limit is it satisfies the relation x = lim x n+ = 2 lim x n = 2x and get two possible answers (here the equation is x 2 2x = 0 so the two answers are 0 and 2) one of which is clearly incorrect. Therefore, once we have proven that this sequence converges, we know that it converges to 2.

4 First show everything is bounded (and importantly, positive otherwise we can t apply the square root function to both sides of our inequalities!) Proof by induction. 0 < 2 < 2 which starts our induction. Assume that 0 < x k < 2. Multiply all sides by 2 to get 0 < 2x k < 4, now since everything is positive and y = x is an increasing function, we can take the square root of all sides and preserve the inequalities to get 0 < 2xk < 4, in other words 0 < x k = 2x k < 2 which is what we wanted to prove. Now prove that our sequence is increasing. This is also a proof by induction. 2 < 2 2 since the left hand side squared is 2 and the right hand side squared is 2 2 and < 2 which starts the induction. Assume that x k < x k which implies that 2x k < 2x k. Again, since everything is positive (st half of the proof) and y = x is an increasing function, this implies that x k = 2x k < 2x k = x k+ which completes our proof! 6. (Limit Superior) aka lim sup. Let {a n } be a bounded sequence in R. a) Prove that the sequence defined by y n = sup{a k k n} converges. Since the set {a n } is bounded, its subsets {a k k n} are also bounded and since R possesses the least upper bound property y n exists for all n. Now notice that the sets {a k k } {a k k 2} {a k k 3} and if we have A B, then sup A sup B so the sequence is decreasing. In addition, if b is a lower bound for {a n }, then b is also a lower bound for the sequence of y n s so we have a bounded decreasing sequence which converges by the monotone convergence theorem. b) The limit superior of {a n } (denoted lim sup a n ) is defined by lim sup a n = lim y n where y n is the sequence given above. Provide a reasonable definition for lim inf a n and briefly explain why it always exists for any bounded sequence. The definition of the limit inferior for a sequence would be the same as the limit superior but with inf instead of sup. Namely, let x n = inf{a k k n}, then the limit inferior would be lim inf a n = lim x n. This exists because the inf of a bounded set exists, inf of a subset is the inf of the superset so the lim inf sequence is increasing and bounded so converges just like in part a). c) Prove that lim inf a n lim sup a n for every bounded sequence and give an example of a sequence for which the inequality is strict. For every n, x n and y n are (respectively) the inf and the sup of the same set, so x n y n for all n. Let lim x n = x and lim y n = y and assume that x > y and let δ = x y. Then since {x n } converges to x, for all ɛ > 0 there exists Ñ such that x ɛ < x n < x + ɛ for all n Ñ. In particular, it is true for ɛ = δ/2. Similarly, there exists N such that y δ/2 < y n < y + δ/2} for all n Ñ. In total we have y δ/2 < y n < y + δ/2 = x δ/2 < x n < x + δ/2 for every n max{ñ, N} and hence y n < x n which is a contradiction to our above observation. An example where the inequality is strict is a n = ( ) n which has lim inf a n = and lim sup a n = and <. d) Show that lim inf a n = lim sup a n if and only if lim a n exists. In this case, all three are equal.

5 Assume that lim sup = lim inf = l. Let x n be the limit inferior sequence and y n be the limit superior sequence. Then, by definition, x N a n y N for any n N, and in particular, this is true for n = N. Therefore, by the Squeeze Theorem (problem ), since l = lim inf = lim x n = lim y n = lim sup, we can conclude that lim a n = l as well. Assume lim a n exists, so a n is bounded and therefore lim x n and lim y n also exist (part a)). Let lim x n = x, lim a n = a and lim y n = y. By part c) above, we also have x a y. Proof by contradiction. Assume that x < a and let δ = a x the distance between a and x. Now {a n } converges to a, so for every ɛ > 0, there exists a N such that d(a n, a) < ɛ whenever n N. In particular, this is true for ɛ = δ/2, so x = a δ < a δ/2 < a n < a + δ/2 for all n N. However, x N = inf{a n n N} and since a δ/2 < a n for all n N, x N their greatest lower bound must satisfy a δ/2 x N a n. However, since x n is an increasing sequence, so x n x for all n which contradicts the fact that we have just shown that x = a δ < a δ/2 x N. Therefore x can not be strictly less than a so must equal a. Similarly (the argument is identical, just replace inf with sup and use y a = η the fact that if we let ɛ = η/2, we get that a n < a + η/2 < a + η = y for all n N) we get that a = y and therefore x = y. e) Prove that if {b n } is another bounded sequence, then lim sup(a n + b n ) lim sup a n + lim sup b n. We don t have a pre-defined way to add sets which means that this is slightly subtle. The difference here between the combination lim sup(a n + b n ) and the individual limits lim sup a n and lim sup b n is that for the combination, we are taking the sup of a set that just consists of each a n +b n (the indices match) while taking the lim sup of the individual terms means it is the max of a n and the max of b n and then adding them which is at least the sum of the matched pairs but could be larger. Let lim sup(a n + b n ) = c, lim sup a n = a and lim sup b n = b, show that c a + b. Proof by contradiction: assume that a + b < c. Let {yc n }, {ya n } and {yb n } be the lim sup sequence for {a n + b n }, {a n } and {b n } respectively. Then by part a), the {y n } sequence is always decreasing, and by the proof of the monotone convergence theorem, bounded decreasing sequences converge to their infemium as a set, so c = inf{yc n }, a = inf{ya n } and b = inf{yc n }. By definition of inf, since a + b < c and c is the greatest lower bound, a + b is not a lower bound, so there exists N such that c a N + b N < a + b. Consider a N < a + b b N. Since b is the infemium of the b n s, b b N so we have a N < a ɛ for ɛ > 0 (here ɛ = b N b). In particular, this means that a N < a, but a is the infemium of the a N s which is a contradiction. 7. Give an example of each of the following, or argue that such a request is impossible. a) A Cauchy sequence that is not monotone. The alternating sequence a n = ( )n n is Cauchy but not monotone. b) A monotone sequence that is not Cauchy. The sequence a n = n is not Cauchy (since it s not bounded we can t use the monotone convergence theorem).