Last updated Wed Mar 11 17:39 EDT 2020
Due-date: Wednesday, 3/18/20. Homework will be collected by email; send me a pdf file no later than 1:55 p.m.
Students who handwrite their homework may send me a scan—not a photograph—but before sending, please make sure that your scan has not cut off any of your writing, or reduced any margins.You are required to do all of the problems below. You will not be required to hand them all in. I will announce later which ones you do have to hand in on the due-date. Don't make the mistake of thinking that I'm collecting only the problems I think are important.
The "due date" above is the date that your written-up problems should be handed in, but don't wait to get started on the assignment. You should always get started on problems as soon as we cover the relevant material in class.
A. Rosenlicht problems (some from Chapter VI, some from Chapter VII)
- Rosenlicht Chap. VI/ 13, 15, 18, 20, 22, 23, 24ab, 25ab, 26. Of these, hand in only 13, 15, 20, 22, 23, 26.
Some notes/comments:
- Note on #13. To do this problem 13 (which is tricky enough to warrant my giving you a hint below), you'll need to use the fact that \(\lim_{n\to\infty} c^{1/n}=1\) for every real number \(c>0\). You may assume this fact for this problem. It follows from another problem you'll be handing in (Exercise 6.26(a) in the Notes on Integration), but you have enough tools now to prove it by simpler means.
Hint for #13: By definition, a sequence \( (A_n)_{n=1}^\infty\) in \({\bf R}\) converges to \(M\in {\bf R}\) if and only if given any \(\epsilon > 0\), we have \(A_n\in (M-\epsilon, M+\epsilon)\) for all \(n\) sufficiently large. In problem 13, for the relevant sequence \(A_n\) and number \(M\), and a given \(\epsilon>0\), the way that you show \(A_n > M-\epsilon\) for all \(n\) sufficiently large will be quite different from the (easier) way that you show \(A_n < M+\epsilon\) for all \(n\) sufficiently large. Continuity of \(f\) is critical to the "\(A_n > M-\epsilon\)" argument, but irrelevant to the "\(A_n < M+\epsilon\)" argument.- Note on #18: No trig functions or their inverses are allowed. The point of the problem is to show directly that the two integrals are equal to each other, not that they are equal because they both yield \(\tan^{-1} x\).
- Note on #20: This problem is not easy! (It is definitely the hardest one on this assignment. That would be true even if I removed all hints for other problems.) If you have what you think is a quick proof, you are probably overlooking something, making an implicit assumption, etc.
- Note on #22: In this problem, "between 0 and 1" means strictly between 0 and 1. Hint for #22: \([1,n] = [1,2]\cup [2, 3] \cup \dots \cup [n-1,n].\)
- Hint for #23: Divide one function by another.
- Note on #25: Do this problem without using l'Hôpital's Rule; that's why Rosnelicht gave the pre-(a) part of the problem. (The method I showed you in class to do part (c) is essentially equivalent to the pre-(a) part of the problem, though this may not be obvious at first glance.)
As mentioned in class: a good way to think of (and remember) what #25 is showing is that "logs are weaker than powers, and powers are weaker than exponentials." Equivalently: "exponentials are stronger than powers, and powers are stronger than logs." In any battle between functions of two of these types—where "battle" means an (initially) indeterminate limit of type "\(\frac{\infty}{\infty}\)" or "\(0\times\infty\)"—the stronger function wins. For example, in part (b), a limit of type "\(0\times\infty\)", the factor \(x^{\alpha}\) wants the limit to be \(0\), while the factor \(\log x\) wants the limit to be \(-\infty\). "Puny log!" scoffs the power-function \(x^\alpha\), as the Hulk to Loki. "You are helpless to stop me from making the limit \(0\)!" As \(x\to\infty\), exponential functions \(x\mapsto e^{\epsilon x}\) (with \(\epsilon>0\)) similarly laugh at power-functions \(x\mapsto x^\alpha\) , no matter how large \(\alpha\) is (say, \(\alpha=1,000,000,000\)) or how small \(\epsilon\) is (say, \(\epsilon=10^{-100}\)). "You tortoise!" says the exponential function to the power function. "In a race to infinity, you eat my dust!"
Die-hard fans of l'Hôpital's Rule should do the following exercise:(a) Use l'Hôpital's Rule alone to compute \(\lim_{x\to\infty}\frac{(x^3+e^{x/2})^{\sqrt{2}}}{(x^2+e^{x\sqrt{2}})^{1/2}}\). You are not permitted to divide numerator and denominator by anything, or to use ideas like "This quantity is negligible compared to that one" that amount to multiplying and dividing something by an appropriate quantity; with those operations allowed, you might as well not be using l'Hôpital's Rule. No stopping till you get an answer or you die, whichever comes first. (b) In your next life (or this one, if you disobeyed the last instruction in part (a)), ask yourself the question, "Which is better: knowing the algebraic and growth properties of common functions, or knowing l'Hôpital's Rule?"
- Rosenlicht Chap. VII/ 3–5. Of these, hand in only 3. In these exercises, remember that for a sequence of functions, Rosenlicht's "converges" is my "converges pointwise" (and similarly for "convergent").
B. Click here for a non-book, non-notes problem. C. In the Notes on Integration:
- Read the portions of Section 6.9 not covered in class (various proofs, proof-parts, and Remarks), up through and including Remark 6.97. Optional, not required: Read the material on the Fundamental Theorem of Calculus for vector-valued functions. (This starts on the middle of p. 57 and goes through the end of Section 6.9 on p. 59.)
- In Section 6.10, read Remarks 6.115–6.117, the first paragraph of Remark 1.21 (the remainder of Remark 1.21 is optional reading), and Remark 1.22.
- Do exercises 6.9, 6.12–6.16, 6.25–6.27, 6.29. (Also, in Proposition 6.110, derive all the properties of "exp" that were not derived in class.) Of these, hand in only 6.26ab. If you work on the "just or fun" exercise 6.28, and think you've pretty much nailed it, and would like to hand something in, let me now and I'll think about what I'd accept for some number of extra-credit points TBD.
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