Math 414, Fall 2016, Test I Dr. Holmes September 23, 2016 The test begins at 10:30 am and ends officially at 11:45 am: what will actually happen at 11:45 is that I will give a five minute warning. The test is divided into four sections, each containing two questions. You should do one question from each section first, then go back and do the second question in each section. The reason for this is that the question you do better on in each section (meaning you, individually) will count significantly more than the other question in the section. If a section happens to contain a third question, you only need to do two of the three questions in that section. The sections are of equal weight. No list of definitions or theorems is provided: I believe that everything you need is of such a basic nature that you can be expected to know it. You are permitted nonetheless to quietly request a definition or statement of a theorem which I have not provided during the exam. I do not guarantee that I will provide such things: some things I believe you should know if you have been paying attention. But you are allowed to ask. You may use your writing instrument, your test paper, and a calculator without graphing or symbolic computation capabilities. Nothing else is permitted. Your cell phone must be turned off and inaccessible to you. 1
1 Section I: Inequality Reasoning 1. Prove the Triangle Inequality a + b a + b from basic properties of inequalities and absolute value. You may if you wish use the approach which considers what happens when you square both sides of the inequality. You are allowed to assume the definition of absolute value, the multiplication property of absolute values, the ability to add inequalities, and (for the approach using squares) the theorem that if a and b are positive real numbers, a < b iff a 2 < b 2 (and similarly for the other inequality relations). Certainly a a a and b b b. So a b a + b a + b so ( a + b ) a + b a + b (this line really is needed to make it clear what property is being applied to get the next line) so a + b a + b (using the fact that r x r always implies x r). Alternatively, ab ab = a b so a + b 2 = a 2 + 2ab + b 2 a 2 + 2 a b + b 2 = ( a + b ) 2, so a + b a + b. 2
2. Prove the transitive property of approximation using algebra and properties of inequalities and absolute value (you may use the triangle inequality). This property is the assertion a ɛ b and b ɛ c implies a ɛ+ɛ c. Why do I put transitive in quotes? Assume that a ɛ b and b ɛ c. By definition of the approximation relation, a b ɛ and b c < ɛ, Thus a c = (a b) + (b c) (triangle inequality) a b + b c < ɛ+ɛ (by addition of inequalities). So a ɛ+ɛ c as desired. 3
2 Section II: Explicit Limit Proofs I will be grading quite a lot on carefully written proofs. I do not want to see just calculations: they must be supported with adequate English. Notice that there are three questions in this section, of which you only need to do two. If you do all three, your best work will count. If you do very well on all three, you might earn some additional credit. 1. Show that lim { n + 1 n} = 0, n directly from the definition of limit. This is an example from the text which we did in class, and should be done exactly as I did it in class. In the text, page 37. 4
2. Show that lim {ln(n)} =, n directly from the definition of infinite limits. Of course, this means you have to know the definition of infinite limits. You are allowed to use the fact that the logarithmic and exponential functions are inverses and the fact that both are increasing functions: this is an example we did in class, and you should do it the same way that it was done in class. In the text, page 41. 5
3. Show that lim {1 +... + n rn } = 1 1 r if r < 1. I m going to deliberately leave you floating on what you are allowed to use here: carefully cite any facts about limits that you use in your argument. Proposition 4.2, page 52. I would have expected a correct answer to make use of (and cite) the fact that r n 0 as n if r < 1 (the partially correct answers generally did). Error term analysis is a good approach, as in the proof in the book. 6
3 Section III 1. Prove the Squeeze Theorem. This asserts that if {a n }, {b n }, {c n } are sequences with a n b n c n for every n and then lim n {b n } = L. lim {a n} = lim {c n } = L, n n You do not need anything for this except the definition of limit and algebra and basic properties of absolute values. Hint: you really want to unpack assertions about absolute values into chains of inequalities in this proof: calculations with absolute values are not involved. p. 64 in the book, proof of Theorem 5.2. 7
2. Prove the following: If for any ɛ > 0, a ɛ b, then a = b. I m unhappy with the phrasing here: any has unfortunate usage patterns in English. I should have said if for all. This can be best proved by contrapositive. Suppose that a b. Our aim is to prove ( ɛ > 0 : a ɛ b). We prove this by assuming ( ɛ > 0 : a ɛ b) and reasoning to a contradiction. Since a b, a b > 0, so a a b b, so a b < a b, which is absurd, completing the argument. 8
4 Section IV Notice that there are three problems in this section, of which you only need to do two. If you do all three your best work will count; if you do very well on all three you might receive a small amount of additional credit. 1. Prove the Nested Intervals Theorem. You may assume usual algebraic knowledge, the completeness property = the monotone convergence theorem (it is an axiom for us), known definitions and the addition or subtraction property of limits. The Nested Intervals Theorem says that given sequences {a n } and {b n } with the following properties: (a) a n a n+1 b n+1 b n for each n (b) lim n {b n a n } = 0 we can draw the following conclusion: there is exactly one real number L which belongs to the closed interval [a n, b n ] for every n and which is the limit of both sequences {a n } and {b n }. By (a), it is immediate that {a n } is increasing and {b n } is decreasing. For any n, a n b n b 1, by (a) and the fact that {b n } is decreasing. For any n, a 1 a n b n by (a) and the fact that {a n } is increasing. So we have shown that {a n } is increasing and bounded above by b 1, so lim{a n } = L exists. and we have shown that {b n } is decreasing and bounded below by a 1, so lim{a n } = L exists. Now L M = lim{a n } lim{b n } = lim{a n b n } = 0, so L = M. You were given on the board the theorem that the limit of an increasing sequence is greater than or equal to all of its terms and the limit of a decreasing sequence is less than or equal to all of its terms, so we have for any n that a n L = M b n, so L belongs to each closed interval [a n, b n ]. Suppose some other number N belonged to every interval [a n, b n ]. We will show that L ɛ N for any ɛ, so L = N by an earlier problem on this test. Fix ɛ. For large enough n, the length of [a n, b n ], which 9
is 1 2 n (b n a n ), will be less than ɛ. Since both L and N belong to an interval of length < ɛ, L ɛ N as required, and since ɛ is arbitrarily chosen, we must have L = N. 10
2. Define the sequence {a n } recursively, defining a 0 = 0 and a n+1 = 1+an 2. Prove by induction from the definition of the sequence (no fair writing its closed non-recursive formula) that this sequence is monotone and bounded. I m going to use a student s answer (better than mine). First, prove that a n < 1 for every n. Clearly a 0 = 0 < 1. Suppose a k < 1. Then a k = a k+a k < 1+a k < 1+1 = 1, so a 2 2 2 k+1 = 1+a k < 1, 2 completing the proof of the claim by induction. Then prove that the sequence is monotone: actually this has already been shown. We repeat from above: for any k, a k = a k+a k < 1+a k = 2 2 a k+1. 11
3. Prove the Product Property of Limits: if lim n {a n } = L and lim n {b n } = M then lim n {a n b n } = LM. This is to be proved from the definition of limit: the author s approach using error term analysis is recommended. (so you are allowed to use Theorem 4.1: if e n is defined as a n L, then lim n {a n } = L if and only if lim n {e n } = 0.) Look at the proof of the Product Theorem on page 62 in the book. It is interesting to note that you can complete this proof by using error analysis and appealing directly to the sum and constant multiple properties of limits, as some of you did on the test. I didn t give full credit for this, because I asked for a proof directly from the definition of limit, but perhaps this is how I should actually do it in lecture. 12