Exam (elaborations) TEST BANK FOR Vector Calculus, Linear Algebra and Differential Forms 2nd Edition By Hubbard J.H (Solution manual)

Solutions for Chapter 0 1 Solutions for Chapter 1 11 Review Solutions, Chapter 1 57 Solutions for Chapter 2 71 Review Solutions for Chapter 2 125 Solutions for Chapter 3 141 Review Solutions, Chapter 3 199 Solutions for Chapter 4 211 Review Solutions, Chapter 4 279 Solutions for Chapter 5 293 Review Solutions, Chapter 5 315 Solutions for Chapter 6 319 Review Solutions, Chapter 6 387 Solutions for Appendix A 401 iii Solutions for Chapter 0 0.2.1 (a) The negation of the statement is: There exists a prime number such that if you divide it by 4 you have a remainder of 1, but which is not the sum of two squares. Remark. By the way, this is false, and the original statement is true: 5 = 4+1; 13 = 9+4; 17 = 16 + 1; 29 = 25+4; : : : ; 97 = 81+16; : : : . If you divide a whole number (prime or not) by 4 and get a remainder of 3, then it is never the sum of two squares: 3 is not, 7 is not, 11 is not, etc. You may well be able to prove this; it isn't all that hard. But the original statement about primes is pretty tricky. (b) The negation of the statement is: There exist x 2 R and ² 0 such that for all ± 0 there exists y 2 R with jy ¡ xj ± and jy2 ¡ x2j ¸ ². Remark. In this case also, this is false, and the original statement was true: it is the de¯nition of the mapping x 7! x2 being continuous. (c) The negation of the statement is: There exists ² 0 such that for all ± 0, there exist x; y 2 R with jx ¡ yj ± and y2 ¡ x2j ¸ ². Remark. In this case, this is true, and the original statement is false. Indeed, if you take ² = 1 and any ± 0 and set x = 1 ± ; y= 1 ± + ± 2; then jy2 ¡ x2j = jy ¡ xjjy + xj = ± 2 µ 2 ± + ± 2 ¶ 1: The original statement says that the function x 7! x2 is uniformly continuous, and it isn't. 0.3.1 (a) (A ¤ B) ¤ (A ¤ B) (b) (A ¤ A) ¤ (B ¤ B) (c) A ¤ A 0.4.1 (a) No; many people have more than one aunt and some have none. (b) No, 1 0 is not de¯ned. (c) No. 0.4.2 None of the relations in Exercise 0.4.1 is a true function. (a) The aunt of," from people to people is unde¯ned for some people, and it takes on multiple values for others. Even if we considered the function "The aunt of" from people with one aunt to people, the function would be neither one to one nor onto: some aunts have more than one niece or nephew, and not all people are aunts. (b) f(x) = 1 x, from real numbers to real numbers is unde¯ned at 0. f(x) = 1 x, from real numbers except 0 to real numbers is a function. This function is one to one, but not onto (it never takes on the value 0). (c) The capital of," from countries to cities is not a function, because it takes on multiple values for some countries. The capital of," from countries with one capital city to cities is, of course, a 1 2 Solutions for Chapter 0 function. This function is one to one, because no two countries share the same capital, but it is not onto, because there are many cities which are not the capital of any country (New York is not a capital). 0.4.3 Here are some examples: (a) DNA of" from people (excluding clones and identical twins) to DNA patterns. (b) f(x) = x. 0.4.4 Here are some examples: (a) year of" from events to dates (b) sin : R ! [¡1; 1]. 0.4.5 Here are some examples: (a) Eldest daughter of" from fathers to children. (b) arctan(x) : R!R. 0.4.6 (a) You can make f(x) = x2 one to one by restricting its domain to positive reals. You cannot make it one to one by changing its range. (b) You can make the function not onto by restricting its domain (for example to: [¡1; 1]) or by changing its range, say to all R. 0.4.7 The following are well de¯ned: g ± f : A ! C, h ± k : C ! C, k ± g : B ! A, k ± h : A ! A, and f ± k : C ! B. The others are not, unless some of the sets are subsets of the others. For example, f ± g is not because the range of g is C, which is not the domain of f, unless C ½ A. 0.4.8 0.4.9 (a) f ¡ g(h(3)) ¢ = f ¡ g(¡1) ¢ = f(¡3) = 8 (b) f ¡ g(h(1)) ¢ = f ¡ g(¡2) ¢ = f(¡5) = 25 0.4.10 (a) Since f is onto, for every c 2 C there exists a b 2 B such that f(b) = c. Since g is onto, for every b 2 B there exists a 2 A such that g(a) = b. Therefore for every c 2 C there exists a 2 A such that (f ± g)(a) = c: (b) If f(g(a1)) = f(g(a2)), then g(a1) = g(a2) since f is one to one, hence a1 = a2, since g is one to one. 0.4.11 If the argument of the square root is nonnegative, the square root can be evaluated, so the open ¯rst and the third quadrants are in the natural domain. The x-axis is not (since y = 0 there), but the y-axis with the origin removed is in the natural domain, since x=y = 0 there. 0.4.12 0.4.13 The function is de¯ned for f x 2 R j ¡1 · x 0, or 0 x g. It is also de¯ned for the negative odd integers. Solutions for Chapter 0 3 0.5.1 Without loss of generality we may assume that the polynomial is of odd degree d and that the coe±cient of xd is 1. Write the polynomial xd + ad¡1xd¡1 + ¢ ¢ ¢+a0: Let A = ja0j + ¢ ¢ ¢+jad¡1j + 1. Then ¯¯ ad¡1Ad¡1 + ad¡2Ad¡2 + ¢ ¢ ¢+a0 ¯¯ · (A¡1)Ad¡1 and similarly ¯¯ ad¡1(¡A)d¡1 + ad¡2(¡A)d¡2 + ¢ ¢ ¢+a0 ¯¯ · (A¡1)Ad¡1: Therefore p(A) ¸ Ad ¡ (A ¡ 1)Ad¡1 0 and p(¡A) ¸ (¡A)d + (A¡1)Ad¡1 0: By the intermediate value theorem, there must exist c with jcj A such that p(c) = 0. 0.5.2 (a) The sequence of points xk = 1 ¼(2k + 1=2) converges to 0 as k!1, but f(xk) = 1 and does not converge to 0 = f(0). (b) We need to show that for any x1 x2 in R, and any number a between f(x1) = a1 and f(x2) = a2, there exists x 2 [x1; x2] such that f(x) = a. We only need to worry about the case where x1 · 0 and x2 ¸ 0, because otherwise f is continuous on [x1; x2]. If both are 0, there is nothing to prove, so we may assume that x2 0. Choose k 2 N such that 1=(2k¼) x2. Then the interval Ik = [1=(2(k + 1)¼);1=(2k¼)] is contained in [x1; x2], and for any a 2 [¡1; 1] there is a x 2 Ik (in fact two of them) such that f(x) = a. Since both a1; a2 2 [¡1; 1], this proves the result. 0.5.3 Suppose f : [a; b] ! [a; b] is continuous. Then the function g(x) = x¡f(x) satis¯es the hypotheses of the intermediate value theorem (Theorem 0.5.9): g(a) · 0 and g(b) ¸ 0. So there must exist x 2 [a; b] such that g(x) = 0, i.e., f(x) = x. 0.6.1 (a) There are in¯nitely many ways of solving this problem, and many seem just as natural as the one we propose. Begin by listing the integers between ¡1 and 1, then list the numbers between ¡2 and 2 that can be written with denominators · 2 and which haven't been listed before, then list the rationals between ¡3 and 3 that can be written with denominators · 3 and haven't already been listed, etc. This will eventually list all rationals. Here is the beginning of the list: ¡1; 0; 1;¡2;¡3 2;¡1 2; 1 2; 3 2; 2;¡3;¡8 3;¡5 2;¡7 3;¡5 3;¡4 3;¡2 3;¡1 3; 1 3; 2 3; : : : (b) Just as before, list ¯rst the ¯nite decimals in [¡1; 1] with at most one digit after the decimal point (there are 21 of them), then the ones in [¡2; 2] with at most two digits after the decimal, and which haven't been listed earlier (there are 380 of these), etc. 0.6.2 4 Solutions