Understanding Real Analysis, Second Edition [2nd Ed] (Instructor Solution Manual, Solutions) [2 ed.] 9781138033016, 1138033014

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Exercises

1.1 1.

Numbers 101: The Very Basics (a) The claim makes sense and is true. √ (b) The claim makes no sense; 8 isn’t a subset. (c) The claim makes sense and is true. (d) The claim makes sense but is false; consider a = 0 and b =

√

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(e) The claim makes sense and is true.

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(f) The claim makes sense but is false: consider a = 0. √ (a) The claim is false; let a = 2. (b) The expression Q2 doesn’t make sense. (c) The claim is true. Since a2 > 0, some large n will work. (d) The claim is true; see Theorem ??. √ / Q. (e) If a ∈ Q and a 6= 0, then a 2 ∈

3. The number 1/a is an integer only if a = ±1. The number 1/a is rational for all nonzero integers a. The equation 1/a = a holds only if a = ±1. 4.

(a) 1 ∈ S1 but −1 ∈ / S1 (b) 2 ∈ S2 but 1/2 ∈ / S2 √ √ √ / S3 (c) 2 ∈ S3 but 1/ 2 = 2/2 ∈

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(a) 1 ∈ S1 but −1 ∈ / S1 (b) 2 ∈ S2 but 1/2 ∈ / S2 √ √ √ / S3 (c) 2 ∈ S3 but 1/ 2 = 2/2 ∈ (d) π ∈ S4 but π 2 ∈ / S4

6. Theorem ?? is useful. Because Q is closed under addition, multiplication, and division (if denominators aren’t zero), expressions like those in (a) and (b) are rational—unless, for (b), p = q = 0. Expressions involving square √roots are different. If p = q = 1, for instance, then p p2 + q 2 = 2 is irrational;pthe same expression is rational if p = 3pand q = 4. Thepquantity p2 + 2pq + q 2 is always p rational, since 2 2 2 p + 2pq + q = (p + q) = ±(p+q). (Note that (p + q)2 = p+q may be false.) 7. All of xy, x+y, x−y and x/y can be either rational or irrational. Examples are easy to find.

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8. The quantities in (a), (b), (c), (d) and (f) are all irrational; proofs are by contradiction. (E.g., if x/r were rational, then we could multiply by √ the rational number r. Then the product x is also rational, a contradiction.) r can go either way. √ 9. Assume toward contradiction that 3 = a/b for integers a and b, where a/b is in reduced form. Then squaring both sides gives 3b2 = a2 . This implies (essentially as in the proof of Theorem ??) that 3 divides both a and b, which contradicts the assumption that a/b is in reduced form. 10.

(a) Say x2 ∈ / Q. If x ∈ Q, then (by Theorem ??) x2 is rational, too, which contradicts our assumption. √ √ (b) Another proof√by contradiction. If x = 2 + 3 is rational,√then x2 = 5 + 2 6 is rational, too. This implies, in turn, that 6 is rational, which is absurd. √ √ √ (c) Yet another proof√by contradiction. Let’s write x = 2 + 3 + 5, √ √ then we have x− 5 = 2+ 3, and suppose x is rational. Squaring both sides of the last equation and simplifying gives √ √ x2 − 2x 5 = 2 6, which is progress, since only two square roots remain. Squaring again gives √ x4 − 4x3 5 + 20x2 = 24, which is even better, as only one square root is left. The last equation implies that √ x4 + 20x2 − 24 5= . 4x3 Because x is rational, so is the right-hand side above, and thus so is the left. This absurdity completes the proof.

11. Parts (i) and (ii) follow from the fact that 1 < a/b < 2. For part (iii), note that (2b − a)2 4b2 − 4ab + a2 a02 = = , b02 (a − b)2 a2 − 2ab + b2 and substituting a2 = 2b2 shows that the last fraction is 2. √ This all shows that if 2 = a/b holds for √any positive integers a and b, then we can find a new fraction a0 /b0 with 2 = a0 /b0 and b0 < b, which is absurd. 12.

(a) Z2 is not closed under addition: 1 + 1 = 2 ∈ / Z2 . (b) Z2 satisfies all the requirements in Theorem ??.

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13. Matrix addition in M2×2 is commutative, but multiplication is not; examples are easy to find. Every matrix A in M2×2 has an additive inverse −A, but multiplicative inverses exist only for some nonzero matrices (those with nonzero determinant); again, examples are easy to find. Distributivity does indeed hold in M2×2 . 14. If a and b are rational, then √ a−b 2 1 √ = 2 . a + 2b2 a+b 2 This shows that elements of F have multiplicative inverses in F . The rest is easier. 15. Since ln ln ln n tends to infinity, it must exceed two for large n. The wellordering property guarantees that a smallest such n0 exists. (Using a calculator we can see that n0 has about 703 decimal digits.) 16.

(a) The answer is no. For instance, the set {1, 1/2, 1/3, . . . } is a subset of Q, but has no least element. (b) The set R = { 1, 10, 100, 1000, . . . } does have the well-ordering property; every nonempty subset includes a smallest power of 10. (c) The set T = { −3, −2, −1, . . . , 41, 42} (like all finite sets of real numbers) does have the well-ordering property, since every nonempty subset of T is also finite, and hence has a least element. (d) If we trade “least” for “greatest” in the well-ordering property, the result no longer holds for N, since N itself has no greatest element. The property does hold for the finite set T , and also for Z \ N = {. . . , −3, −2, −1, 0}.

1.2 1.

Sets 101: Getting Started (a) D ⊂ I; D ∈ C. (b) B = {m ∈ A | m has 31 days}. (c) A × D is the set of ordered pairs (January, 2), (February, 2), . . . , (December, 2), (January, 3), (February, 3), . . . , (December, 3). There are 24 such pairs.

2.

(d) A \ B = {February, April, June, September, November}; B \ A = ∅; A ∩ C = {November}; B ∩ A = B; D ∩ I = D; D ∪ I = I. √ (a) S = {0, −1}; T is the √ interval of numbers between (−1 − 21)/2 ≈ −2.791 and (−1 + 21)/2 ≈ 1.791.

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(b) Decide whether each of the following statements is true or false, and explain: S ⊂ N is false because −1 ∈ / N; S ⊂ T is true; T ∩ Q 6= ∅ is true, since 0 ∈ T ∩ Q; −2.8 ∈ Q \ T is true.  (c) The quadratic formula shows that U = x ∈ R | x2 + x < 0 = (−1, 0). 3.

(a) R \ A = (−∞, 1) ∪ (3, ∞) (c) R \ A = (−∞, 1]) ∪ [2, 3] ∪ [4, ∞) (e) R \ A = {0}

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(a) R \ [a, b] = (−∞, a) ∪ (b, ∞) (b) I = (−∞, ∞) has empty complement; I = (−∞, 17) has closed complement [17, ∞); I = (0, 17) has complement (−∞, 0]∪[17, ∞). (c) R \ Z = (0, 1) ∪ (−1, 0) ∪ (1, 2) ∪ (−2, −1) ∪ . . . .

5. To say that a is in R \ (R \ A) means that a is not in R \ A; this means, in turn that a ∈ A. 6.

(a) Claim (i) is false. As one example, take A = R and B = ∅. Then R \ (A ∪ B) = ∅ but (R \ A) ∪ (R \ B) = R. (b) If A = B, then A ∪ B = A ∩ B = A, and both claims are clearly true. (c) To prove that (ii) holds, suppose x ∈ R \ (A ∪ B). Thus x ∈ / A ∪ B, so x ∈ / A and x ∈ / B; in other words, x ∈ R \ A and x ∈ R \ B. This is just another way of saying that x ∈ (R \ A) ∩ (R \ B). The “vice versa” implication is similar.

7. We know R\A1 = (−∞, 1]∪[3, ∞) and R\A2 = (−∞, 2]∪[5, ∞). Also, R \ (A1 ∩ A2 ) = (−∞, 2] ∪ [3, ∞) and R \ (A1 ∪ A2 ) = (−∞, 1] ∪ [5, ∞).

It’s easy to see that, as claimed, R \ (A1 ∩ A2 ) = (−∞, 2] ∪ [3, ∞) = (−∞, 1] ∪ [3, ∞) ∪ (−∞, 2] ∪ [5, ∞) = (R \ A1 ) ∪ (R \ A2 ). Similarly, R \ (A1 ∪A2 ) = (−∞, 1]∪[5, ∞) = ((−∞, 1] ∪ [3, ∞))∩((−∞, 2] ∪ [5, ∞) = (R \ A1 ) ∪ (R \ A2 ) 8. Here we have A1 ∪ A2 = (0, 1) ∪ (2, ∞) and A1 ∩ A2 = ∅. Thus R \ A1 = (−∞, 0] ∪ [1, ∞) and R \ A2 = (−∞, 2]. This implies that (R \ A1 ) ∪ (R \ A2 ) = (−∞, ∞) and (R \ A1 ) ∩ (R \ A2 ) = (−∞, 0) ∪ [1, 2]. Consistent with De Morgan, R\(A1 ∩A2 ) = (−∞, inf ty); R\(A1 ∪A2 ) = (−∞, 0] ∪ [1, 2]. 9. If x ∈ T 0 then x ∈ / T . Since T ⊃ S, we have x ∈ / S, which means x ∈ S 0 , as desired.

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10. Many possibilities exist for (a), (b), and (c). For (d) we could use I = (0, ∞) and J = (1, ∞); note that here, as in all possibilities for (d), one interval is contained in the other. 11.

(a) I = (−42, 0) and J = (0, ∞) work. (b) I = (−42, 0) and J = [0, ∞) work. (c) The given conditions (draw a picture) mean that a < c < 0 < b < d, so I ∪ J = (a, d) which is indeed an open interval.

12. Suppose a ∈ I ∪ J, b ∈ I ∪ J, and a < x < b. We’re done if we show that x ∈ I ∪ J. This is trivial if x = c, so we assume x 6= c. Now if both a ∈ I and b ∈ I, then x ∈ I by Definition ??, and we’re done. Similarly, we’re done if both a ∈ J and b ∈ J. So let’s assume a ∈ I and b ∈ J. If x < c, then we have a < x < c with a and c in I; by Definition ??, x ∈ I, too. Similarly, if x > c, then we have c < x < b and so x ∈ J. We’re done. 13. It’s easy for I and R \ I to be intervals. For instance, if I = (−∞, 0), then R\I = [0, ∞) is another interval. I and R\I cannot both be bounded intervals; two bounded intervals can’t “add up” to the unbounded set (−∞, ∞). 14. No. Any finite set I of numbers contains a smallest number, say a, and a second smallest, say b. If I were an interval, it would also have to contain the average, (a + b)/2, which lies (illegally) between a and b. 15. No. Suppose a and √ b are rational numbers in I, with a < b. Consider c = a + (b − a)/ 2. Note that c ∈ R \ Q and that a < c < b. If I were an interval, we’d have c ∈ I, which is impossible. 16.

(a) (1, 2)∪(3, ∞) is the union of two open intervals, and hence also open. The complement, (−∞, 2] ∪ [2, 3], is therefore closed. (b) R \ {a} = (−∞, a) ∪ (a, ∞). (c) (−∞, a) is itself an open interval, and therefore open. The complement of (−∞, a] is the open interval (a, ∞), so (−∞, a] is closed. (d) If I = (0, 1) were closed, then R \ I would be open. This is false, because 1 ∈ R \ I, but no open interval containing 1 is contained in R \ I.

17.

(a) The complement of {1, 2, 3} consists of four open intervals. (b) R \ Z is the union of all open intervals of the form (n, n + 1), where n ∈ Z. (c) If Q were open, we could find for each rational q an open interval I with q ∈ I ⊆. But I ⊆ Q is impossible.

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(d) If (0, 1] were open there would be an open interval I with 1 ∈ I ⊆ (0, 1]. That’s impossible—every such I includes points “to the right” of x = 1. A similar argument with 0 ∈ R \ (0, 1] shows that R \ (0, 1] isn’t open, so (0, 1] isn’t closed. 18.

(a) If S = {1, 2, 3} then P (S) = { ∅, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3} }, a set with 8 members. (b) Every subset of S is clearly a subset of T ; that’s our claim. (c) Each subset A of N10 corresponds to two subsets of N11 : A and A ∪ {11}.

19.

(a) There are 10 × 9 × 8 = 720 ways to choose 3 different elements in order. There are six ways to reorder each such choice, so the answer is that S had 720/6 = 120 elements. (b) T has 10 × 10 × 10 = 1000 elements. (c) S10 has 10! = 10 × 9 × 8 × · · · × 1 = 3628800 elements. (d) The sets N10 , S, T , S10 have no elements in common.

20. S has n elements. Each element of S is formed by dropping one element from S. 21. We have S ∈ A42 if and only if N100 \ S ∈ A58 , so there is a one-toone correspondence between A42 and A58 , which therefore have the same number of elements. 22.

(a) Three vertical lines. (b) Three horizontal lines. (c) The set of integer “lattice points” above the x-axis. (d) The parabola y = x2 . (e) A vertical sine curve. (f) The empty set.

23.

(a) The picture is a diagonal stripe from upper left to lower right. (b) The black squares can be described by the set {(x, y) | x + y is even}. (c) The element (2, 3, black) corresponds to a black square at position (2, 3). The set G × {black, white} represents all possible ways to choose a square and color it. (d) A picture is, in effect, a subset of squares to be colored black. Thus P (G) corresponds to the full set of possible pictures.

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1.3

Sets 102: The Idea of a Function

1. There are many possibilities; following are some. (a) Let A be the set of all humans who have ever lived, and B the set of all women who have ever lived. The function is not injective, because siblings have the same mother. The function is not surjective, either, since some women are not mothers. (b) Let A be the set of all mothers of sons, and B the set of all male humans. Then F IRST B ORN S ON : A → B is one-to-one but not onto. (c) Let A be the set of all humans and B the set of all colors. Then E YE C OLOR : A → B is neither one-to-one nor onto, since several people have blue eyes, and nobody has silver eyes. (d) Let A be the set of all US citizens and B = {January 1, January 2, . . . , December 31}. Then B IRTHDAY : A → B is onto (every day is someone’s birthday) but not one-to-one (several people have the same birthday). 2. There are many possibilities; following are some. (a) Let A be the set of humans and B = {A, B, AB, 0}. (There are 4 blood types.) B LOOD T YPE is onto but not one-to-one. (b) Let A be the set of all books published in English and B = N. With these definitions N UMBERO F PAGES is neither one-to-one nor onto. (c) Let A = R = B. C UBE ROOT is one-to-one and onto. (d) Let A = {0, 1, 2, . . . } and B = N. Then FACTORIAL is not one-toone (since 0! = 1! = 1) and is not onto. 3.

(a) The natural domain is all of R. (b) The natural domain is all of R except for points x at which cos x = 0. (c) The natural domain is (−∞, −1] ∪ [1, ∞).

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(a) The natural domain is the set of x for which 1 − ex ≥ 0; this is the set (−∞, 0]. (b) The natural domain is the set of x for which x2 + πx + 1 ≥ 0. With the quadratic formula we solve x2 + πx + 1 = 0; approximate values are x = −2.78 and x = −0.36. Thus the domain is (approximately) (−∞, −2.78] ∪ [−0.36, ∞). (c) The natural domain is the set of x for which ex > 0; that’s all of R.

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(a) f (x) = x2 + 1 works (b) f (x) = (x − 2)2 = x2 − 4x + 4 works

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(c) f (x) = (x + 5)2 = x2 + 10x + 25 works 6.

(a) f (x) = (x − 42)2 works (b) f (x) = 2x − 1 works; note that f is actually linear (c) f (x) = x2 + x − 1 works (d) f (x) = 3 − x2 works

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(a) Horizontal lines intersects the graph at most once. (b) Every horizontal line y = a, where 0 ≤ a ≤ 1, intersects the graph. (c) f (x) = 1 − x, f (x) = 1 − x2 , f (x) = 1 − x42 all work. So do many others. √ (d) Here f (1/ 2) = f (1/2) = 1/2, so f is not one-to-one. But f is onto, since every irrational number in [0, 1] has an irrational square root, also in [0, 1].

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(a) f is neither one-to-one nor onto; look at its graph (b) g is both one-to-one and onto; look at the graph for x ≥ 0 (c) h is both one-to-one and onto; look at its graph (d) k is one-to-one (everybody has her/his own SSN) but not onto

9. It’s reasonable to use f : (0, ∞) → (−∞, ∞) and g : (−∞, ∞) → (0, ∞). Graphs of f and g are mirror images across the line y = x. We have f (g(x)) = x and g(f (x)) = x for x in the appropriate domains. 10.

(a) G is a set of 12 ordered pairs; each of the form (month, n), where n is the number of letters in the month. (b) The graph G contains both (March, 5) and (April, 5). (c) The graph G contains no element of the form (Month, 2).

11.

(a) The domain is the set A = {a, e, i, o, u}; a good codomain is B = {1, 2, 3, 4, 5}. (b) f is one-to-one because no two elements of G have the same second coordinate. f is onto because every element of B is the second coordinate of an element of G. (c) The graph of f −1 is the set {(1, a), (2, e), (3, i), (4, o), (5, u)}. (d) We might call these functions VOWEL N UMBER and N UMBERO F VOWL, or something similar.

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(a) The domain is the set of states that begin with M. The range is the finite set {3, 4, 6, 10, 11, 12, 17}. The codomain can be any set that contains the range, such as N or even R.

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(b) f is not injective; f (Maryland) = f (Minnesota). (c) f is surjective only if the range and codomain are the same set. (d) A better name than f for this function might be E LECTORALVOTES. 13.

(a) One approach is to look at the graph. Or note that L(x) = L(y) means that 3x + 2 = 3y + 2, and this can happen only if x = y. Thus L is one-to-one. Also, for any b ∈ R we can set a = (b − 2)/3, and see that f (a) = b. Thus L is onto.

14.

(a) L is one-to-one since, if L(x1 ) = ax1 + b = ax2 + b = L(x2 ), then basic algebra gives ax1 + b = ax2 + b =⇒ ax1 = ax2 =⇒ x1 = x2 , (note that a 6= 0 is essential!) as the definition of injectivity requires. L is onto because if y0 is any real number, then we can set x0 = (y0 − b)/a, and we find f (x0 ) = ax0 + b = a

y0 − b + b = y0 . a

(again, a 6= 0 matters) as the definition of surjectivity requires. (b) If we define M (x) = (x − b)/a, then we have, for any x ∈ R, L( M (x) ) = aM (x) + b = a

x−b + b = x, a

as desired. A similar calculation shows that M (L(x)) = x for all x ∈ R. Thus, M = L−1 , as hoped for. 15.

(a) f is one-to-one and onto for all odd positive integers, and not for even positive integers. (b) Here f is one-to-one and onto for all integers, positive or negative.

16.

(a) Claim: If f and g are one-to-one, then so is g ◦ f . Proof: Suppose a1 ∈ A and a2 ∈ A, and that g(f (a1 )) = g(f (a2 )). We’re done if we show that a1 = a2 . Well, because g is one-to-one, we know g(f (a1 )) = g(f (a2 )) =⇒ f (a1 ) = f (a2 ), and since f is one-to-one, we have f (a1 ) = f (a2 ) =⇒ a1 = a2 , as desired.

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(b) Claim: If g ◦ f is onto, then so is g. Proof: We need to show that for any c ∈ C there is some b ∈ B with g(b) = c. By hypothesis, g ◦ f is onto, so there is a ∈ A with g(f (a)) = c. If we now set b = f (a), then g(b) = g(f (a)) = c, as desired. (c) For f (x) = ex + 1 and g(x) = x2 , both with domain R and codomain R, then g is not one-to-one, since, for example, g(−2) = 4 = g(2). Nevertheless, g ◦ f is one-to-one because if g ◦ f (x1 ) = (ex1 + 1)2 = (ex2 + 1)2 = g ◦ f (x2 ), then (taking positive square roots of both sides and manipulating symbolically) we have ex1 + 1 = ex2 + 1 =⇒ ex1 = ex2 =⇒ x1 = x2 , as desired. Done. 17. ` is both one-to-one and onto; make a table of some early values to see why. This might seem weird since N might seem “smaller” than Z. In fact (as we’ll see) they’re the same “size” in the sense at hand. 18.

(a) We have f (g(x)) = g(f (x)) = x for all x in the respective domains. (b) h and g are not inverse functions because, for instance, g(h(−4)) = 4 6= −4.

19. Use domain [−π, π] and codomain [−1, 1] for the sine function; and codomain [−π, π] and domain [−1, 1] for the arcsine function. Then the arcsine becomes an honest inverse for the sine. 20. All parts are easy. For instance, the relation is transitive because if both x and y share any blood type, and so do y and z, then x and z have the same type. 21.

(a) That M OD 5 is reflexive and symmetric is obvious. To see it’s transitive, suppose xM OD 5y and yM OD 5z. This means that 5 divides both x − y and y − z. But then 5 also divides the sum (x − y) + (y − z) = x − z, which means xM OD 5z. (b) The equivalence class [0] is the set {. . . , −10, −5, 0, 5, 10, . . . } of multiples of 5. The equivalence class [1] is the set {. . . , −9, −4, 1, 6, 11, . . . } of numbers one more than a multiple of 5. Similarly, [2] = {. . . , −8, −3, 2, 7, 12, . . . } and so on.

22.

(a) S AME L AST NAME is reflexive, symmetric, and transitive, but not a function, since a person x may share last names with many persons y.

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(b) A NDERSON is not reflexive, but is symmetric and transitive. A NDERSON is not a function on P . (c) S IBLING is symmetric, but not reflexive, not transitive (think about half-brothers and -sisters), and not a function. (d) M AYOR is not reflexive and not symmetric. It is transitive—if xM AYORy and yM AYORz, then y = z, so xM AYORz. M AYOR is a function—a “constant” function. (e) O LDEST N EIGHBOR is not reflexive, symmetric, or transitive, but it is a function.

1.4

Proofs 101: Proofs and Proof-Writing

1. Statement P is false, R is true, and Q is nonsense. R is the negation of P. 2. Statement S is true (check n = 1/4), T is false, and U is true. None are negations of each other. 3.

(a) If a function f is increasing for all x, then f 0 (x) ≥ 0 for all x. (b) If a function f has a maximum at x = a, then either f 0 (a) = 0 or f 0 (a) does not exist. (c) If f 00 (x) > 0 for all x, then the graph of f is concave up. P (d) If a series an converges, then lim an = 0. (This is known as the nth term test.)

4.

(a) If x2 ∈ / Q, then x ∈ / Q. √ 2 √ / Q but 2 ∈ Q. (b) 2 ∈ √ (c) ∀n ∈ N n ∈ R. (d) −1 ≤ sin x ≤ 1 for all x ∈ R.

5.

(a) The statement is false; no integer is greater than all real numbers. (b) The statement is true; x = 0.1 works, for instance.

6.

(a) The statement is true. For given x > 0, let y = 1/x. (b) The statement is false. For given x > 0 choose y =

7.

p 2/x.

(a) Given: Raining implies not sunny (true). Converse: Not sunny implies raining (false). Contrapositive: Sunny implies not raining (true). (b) Given: Raining implies clouds (true). Converse: Cloudy implies raining (false). Contrapositive: Not cloudy implies not raining (true).

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(c) Given: Raining implies not sunny and cloudy (true). Converse: Not sunny and cloudy implies raining (false). Contrapositive: Sunny or not cloudy implies not raining (true). (d) Given: Cloudy implies not sunny (true). Converse: Not sunny implies cloudy (false . . . it could be night). Contrapositive: Sunny implies not cloudy (true). 8.

(a) given: if x > 3, then 9 < x2 < 16. (false; look at x = −10.) converse: if 9 < x2 < 16 then x > 3. (false; look at x = −3.5.) contrapositive: if x2 ≥ 16 or x2 ≤ 9, then x ≤ 3. (false. look at x = −10.) (b) given: if x < 4, then 9 < x2 < 16. (false. look at x = −10.) converse: if 9 < x2 < 16 then x < 4. (true.) contrapositive: if x2 ≥ 16 or x2 ≤ 9, then x ≥ 4. (false; look at x = −10.)

9.

(c) given: if 3 < x < 4, then 9 < x2 < 16. (true.) converse: if 9 < x2 < 16 then 3 < x < 4. (false; try x = −3.5.) contrapositive: if x2 ≥ 16 or x2 ≤ 9, then x < 3 or x > 4. (true.) √ / N. This is true. R: Negation: ∃n ∈ N such that n2 + 6 ∈ √ S: Negation: ∀ ∈ N, n2 + 6 ∈ / N. This is true. T: Negation: ∃x ∈ [0, 1] such that cos(x) < 0.6. This is true; look at x = 1, for instance. U: Negation: Some positive even integer n with 4 < n < 24 is not the sum of two odd primes. This is false; one just checks.

10.

(a) Converse: If a + b is rational, then a and b are both rational. (False.) Contrapositive: If a + b is irrational, then at least one of a and b is irrational, too. (True.) (b) Converse: If 1/a is irrational, then a is irrational, too. (True.) Contrapositive: If 1/a is rational, then a is rational. (True.) (c) Converse: If ab is irrational, then both factors a and b are rational. (False.) Contrapositive: If ab is rational, then at least one of a and b is rational. (False.) P (d) Converse: If limn→∞ an = 0, then an converges. (False; look at the series 1 + 1/2 + 1/3 + . . . .) Contrapositive: If limn→∞ an 6= 0, P then an diverges. (True; this is known as the nth term test.)

11.

(a) Negation: All of a, b, and c are negative. (b) Negation: f (x) > 3 for some x ∈ [2, 7]. (c) Negation: sin n is rational for some positive integer n.

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(d) Negation: ∀x ∈ R we have x2 6= −1. (e) Negation: ∃x ∈ R such that ∀y ∈ R we have xy 6= 1. 12.

P : All dogs have fleas. Negation: Some dog has no fleas. Statement P is (probably) false. Q: Several dogs have fleas. Negation: No dog has fleas. Statement Q is true. R: My dog, Spike, is flea-free. Negation: Spike has fleas. The truth of statement R depends on Spike. S: 3 < 5. Negation: 3 ≥ 5. Statement S is true. √ √ T : √n − 1 + √n + 1 is irrational for every positive integer n. Negation: n − 1 + n + 1 is rational for some positive integer n. Statement T is true. √ √ U : √n − 1 + √ n + 1 is rational for every positive integer n. Negation: n − 1 + n + 1 is irrational for some positive integer n. Statement U is false. V : cos(x) < 1.001 for all x > 0. Negation: cos(x) ≥ 1.001 for some x > 0. Statement V is false. W : cos(x) < 0 for all x ∈ [2, 3]. Negation: cos(x) ≥ 0 for some x ∈ [2, 3]. Statement W is true. X: Every even integer n > 2 is the sum of two (not necessarily distinct) primes. Negation: Some even integer n > 2 is not the sum of two primes. Statement X is Goldbach’s conjecture—an important unsolved mathematical problem. Y : Every set of ten distinct numbers has a largest and a smallest element. Negation: Some set of ten distinct numbers has either no largest or no smallest element. Statement Y is true. Z: Every subset of the interval [0, 1] contains a largest element. Negation: Some subset of [0, 1] contains no largest element. Statement Z is false: (0, 1), for example, has no largest element.

13.

(a) Statement: If x > 3 then x2 > 9. Converse: If x2 > 9 then x > 3. Contrapositive: If x2 ≤ 9 then x ≤ 3. These are true, false, and true, respectively. (b) Statement: If x > 3 then x3 − 4x2 + 3x > 0. Converse: If x3 − 4x2 + 3x > 0 then x > 3. Contrapositive: If x3 − 4x2 + 3x ≤ 0 then x ≤ 3. The statements are true, false, and true, respectively. (It may be useful to plot the function.)

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(c) Statement: If a > 0 and b > 0 then |a + b| = |a| + |b|. Converse: If |a + b| = |a| + |b|, then a > 0 and b > 0. Contrapositive: If |a + b| = 6 |a| + |b|, then a ≤ 0 or b ≤ 0. The statements are true, false, and true, respectively. 14. The contrapositive of P =⇒ Q is Q =⇒ P . Taking the contrapositive again gives P =⇒ Q back again.

1.5 1.

Types of Proof (a) True; the simplest proof is indirect. √ (b) False; x = − 2 is a counterexample. (c) True; the simplest proof is indirect. (d) False; p = 11 is a counterexample.

2.

(e) False; x = 0 and y = 1/2 is a counterexample. √ / Q, then x ∈ Q. This is false; consider (a) The converse: if 2 + x ∈ √ x = 3. √ (b) The converse: if 2 + x ∈ / Q, then x ∈ / Q. This is false; consider x = 3. (c) The converse: if at least one of x and √ y is irrational, √ then x + y is irrational. This is false; consider x = 2 and y = − 2.

3. The claim is obvious for n = 1. Given F = {x1 , x2 , . . . , xn }, consider F 0 = {x1 , x2 , . . . , xn−1 }. By the inductive hypothesis F 0 has a largest member, say xn−1 . The largest member of F is then xn if xn > xn−1 , and xn−1 otherwise. 4.

(a) There’s a one-to-one correspondence between subsets and their complements; in other words, subsets come in pairs. (b) The one-to-one correspondence between subsets and their complements pairs “odd” subsets with “even” subsets, so we have the same number of each. (c) The inductive step: If S has n elements, then removing one of them, say s0 , gives a set S 0 with n−1 elements. By the inductive hypothesis the claim holds for S 0 . New subsets of S are produced by including the one new element s0 . But adding one more element turns odd subsets into even, and vice versa. The result follows.

5.

(a) If n = 1, then both sides have value 1; if n = 10, then both sides are 100.

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(b) The base case, P (1), holds as indicated in (a). For the inductive step we assume P (n) and try to prove P (n + 1), which is the claim that 1 + 2 + · · · + (n + 1) = (n + 1)(n + 2)/2. We start by adding (n + 1) to both sides of P (n), and then do algebra: n(n + 1) + (n + 1) 2 n  (n + 1)(n + 2) = (n + 1) +1 = . 2 2

1 + 2 + 3 + · · · + n + (n + 1) =

This shows P (n + 1), and completes the proof. n n n X X X n(n + 1) (2n − 1) = 2 n− 1=2 (c) Calculate with sums: − 2 k=1

k=1

k=1

n = n2 . 6. Checking some early cases suggests an answer: 1 1 1 n 1 + + + ··· + = . 1·2 2·3 3·4 n · (n + 1) n+1 Let’s prove this by induction. Note first that P (1) is clearly true, since both sides have value 1/2. For the inductive step, we assume P (n) and try to show P (n + 1). Adding 1/((n + 1)(n + 2)) to both sides of P (n), and manipulating the result algebraically does the trick. 7. Prove this by induction. The base case n = 1 is clear. For the inductive step, suppose that (1 + x)k ≥ 1 + kx for a particular k. Multiplying both sides by (1 + x) and doing a little algebra shows that (1 + x)k+1 ≥ 1 + (k + 1)x, as desired. 8. The base case P (1) is easy, so we assume P (k) and show P (k + 1). In P (k + 1), the left side is (1 + 2 + 3 + · · · + k + (k + 1))2 ; we’ll show that it equals 13 + 23 + 33 + · · · + (n + 1)3 . Along the way we’ll use the result of Example ??. Here goes (some details left to you): (1 + 2+ · · · + k + (k + 1))2 = ( (1 + 2 + 3 + · · · + k) + (k + 1) )2 = (1 + 2 + 3 + · · · + k)2 + 2(1 + 2 + 3 + · · · + k)(k + 1) + (k + 1)2 k(k + 1) (k + 1) + (k + 1)2 2 = 13 + 23 + · · · + k 3 + (k + 1)3 , = 13 + 23 + · · · + k 3 + 2

as desired.

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9.

(a) Rewrite f1 f2 f3 as (f1 f2 )f3 and use the ordinary product rule twice. (b) The formula is (f1 f2 f3 · · · fn )0 = f10 f2 f3 · · · fn + f1 f20 f3 · · · fn + f1 f2 f30 · · · fn + · · · + f1 f2 f3 · · · fn0 . We’ve proved the first few cases. For the inductive step, we write f1 f2 f3 · · · fn+1 = (f1 f2 f3 · · · fn )fn+1 and apply the ordinary product rule.

10.

(a) We can simply check by direct calculation (technology helps) that 5n > n! for n = 1, 2, 3, . . . , 11. (b) The proof is by induction. The base case n = 12 holds by a direct calculation. For the inductive step, suppose 5k < k! for a positive integer k ≥ 12. Then clearly 5 < k + 1, and so 5 · 5k < (k + 1) · k!; equivalently, 5k+1 < (k + 1)!, which completes the inductive step. Done.

11.

(a) Checking early cases suggests the answer n(n + 1)(n + 2)/3; this is readily proved by induction. (b) Notice that n X j=1

j(j + 1) =

n X j=1

2

j +

n X

j=

j=1

n(n + 1)(2n + 1) n(n + 1) + . 6 2

The result now follows from basic algebra. 12. let P (n) denote the claim for any given positive integer n. Then P (1) says 1 + r = (r2 − 1)/(r − 1), which is so by factoring the numerator on the right. For the inductive step we assume P (n) and show P (n + 1) (convince yourself of each step): 1 + r + r2 + r3 + · · · + rn + rn+1 = (1 + r + r2 + · · · + rn ) + rn+1 associative law

=

r

n+1

−1 + rn+1 r−1 inductive hypothesis

n+1

− 1) r−1 + rn+1 algebra r−1 r−1 rn+2 − 1 = . more algebra r−1 =

r

This shows that P (n + 1) holds, and completes the proof.

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13. Note first that P (n) is obviously true for the base case n = 4; just check both sides. To complete the inductive proof, we assume P (n) (for n ≥ 4) and show P (n + 1). By P (n), we have 2n < n!. We know also that 2 < (n + 1). Multiplying these inequalities gives 2n < n! =⇒ 2n+1 < n!(n + 1) = (n + 1)!. This shows that P (n + 1) holds, and we’re done. 14. The base case is our assumption (i). The inductive step follows easily from the product rule. 15. The base case is an assumption. (We’ll prove it later in the course.) The inductive step follows readily from the product rule. 17. Among small amounts, it’s clear we can pay out $3, $5, $6, $8, $9. In fact, we can pay all numbers greater than $9, too. Here’s an informal inductive proof. Note first that the base case $10 is clearly possible. Suppose we can pay out all amounts from $10 to $k. To show $k + 1 is also possible, note that k + 1 − 3 = k − 2 ≥ 8, so $k − 2 is payable. Adding one more $3 bill brings the total to $k. 18. The inductive step fails in moving from n = 1 to n = 2.

1.6

Sets 103: Finite and Infinite Sets; Cardinality

1. Here’s a sketch: The base case is given. For the inductive step, write F1 ∪ F2 ∪ · · · ∪ Fn = (F1 ∪ F2 ∪ · · · ∪ Fn−1 ) ∪ Fn and use the inductive hypothesis. 2.

(a) f (a) = 2a works (b) f (a) = a − 4 works (c) f (a) = 1 − a works (d) f (a) = ea works

3.

(a) f (x) = 7x − 2 is one-to-one: If f (a) = f (b), then 7a − 2 = 7b − 2, which implies a = b. g(x) = x/(1 + x) is also one-to-one: if g(a) = g(b), then a/(1 + a) = b/(1 + b). A little algebra (do it—keep in mind that a and b are positive) shows that a = b. (b) For any b ∈ R we can set a = (b + 2)/7 and see that f (a) = b. If b ∈ (−2, 5) then b + 2 ∈ (0, 7) and (b + 2)/7 ∈ (0, 1) as desired.

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(c) We know that both f and g are one-to-one and onto; it follows (from one of our theorems) that f ◦ g is one-to-one and onto. This means (among other things) that all of the sets under consideration here have the same cardinality. 4. We can use a linear function passing through the points (a, c) and (b, d). Such a line has slope m = (d − c)/(b − a) and so can be written in the form f (x) = m(x − a) + c. Note that every linear function with nonzero slope is one-to-one; in this case, m > 0. Also, for any c < y < d, we can set x = (y − c)/m + a (note that x ∈ (a, b)) and check that f (x) = y. Thus f is onto. 5. f can be one-to-one but g cannot. The latter holds by the pigeonhole principle, since g tries, in effect, to put 43 pigeons into 42 holes. 6. We prove the contrapositive. Let R be the range of f ; note that we can write f : S → R. If is not onto, then R has fewer members than S. By the pigeonhole principle, f is not one-to-one, as desired. 7.

(a) The function f : N → N given by f (n) = n + 1 works. (b) The function g : R → R given by g(x) = arctan x works. (c) Define h : R → R by h(x) = x if x ∈ / N and h(x) = x + 1 if x ∈ N.

8. By hypothesis, there are bijective functions f : A → B, and g : B → C. The composition g ◦ f : A → C is also bijective (why?). 9. Proof by contradiction. If the hypotheses hold and A \ B is finite, then A = (A \ B) ∪ B is the union of two finite sets, and hence finite itself. This is absurd. 10.

(a) We can write f (n) = n/2 for even n; f (n) = (1 − n)/2 for odd n. (b) We can write g(z) = 2z if z ≥ 0, and g(z) = −2z + 1 if z < 0. (c) Both compositions f ◦ g and g ◦ f give the identity function. For g ◦ f , for instance, we check that if n is even, then g(f (n)) = g(n/2) = n, and similarly for odd inputs.

11.

(a) Members of Q3 correspond to (some of the) points (p, q) with integer coordinates inside the circle p2 + q 2 = 3—there are nine such points (p, q). Since we’re interested in fractions p/q, we may as well require q > 0. This reduces the eligible points to just three: (−1, 1), (0, 1), and (1, 1), all of which give different reduced fractions p/q. Note that Q4 is the same set as Q3 . For Q10 , similar reasoning finds 11 points (p, q) with q > 0 and p2 + q 2 < 10. These 11 points correspond to 7 different rational numbers: ±2, ±1, ±0.5, and 0.

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(b) If p/q ∈ Qi , then clearly both |p| < i and |q| < i. There are at most 2i−1 choices for each of p and q, and so no more that (2i−1)2 choices for both p and q. (In fact, the number of choices is less, but that’s OK for present purposes.) An even rougher (but simpler-looking) upper bound is that Qi has fewer than 4i2 elements. (c) All parts are straightforward implications of the definitions. 12. The hint suggests what to do. 13. As one possible bijection f : [0, ∞) → (0, ∞) we can define f (x) = x + 1 if x ∈ Z and f (x) = x otherwise. 14. Examples and exercises given here show that all intervals of the forms (a, b), [a, b], (0, ∞), [0, ∞) have the same cardinality as (0, 1). Intervals of the form [a, ∞), (a, ∞), (−∞, a], (−∞, a), (a, b], [a, b), and (−∞, ∞) are readily put in one-to-one correspondence with one of (0, 1), [0, 1], (0, ∞), or [0, ∞), or a close relative. For example, f (x) = arctan(x) shows that (−∞, ∞) and (−π/2, π/2) have the same cardinality. 15. The set B is countably infinite. The set Bn of n-character books is clearly finite for each n, and B is the union of the (countably many) Bn . 16.

(a) A polynomial of degree n in Z[x] is essentially an (n + 1)-tuple of integers (don’t forget the constant term), i.e., an element of the (n+1)fold product Z × Z × . . . Z. This product is countable, by Proposition ??, page ??. The set Z[x] consists of polynomials of all possible degrees, and is therefore a countable union of countable sets. (b) If r = a/b, with a and b integers, then set p(x) = bx − a. It is easy to see that p(r) = 0. (c) We showed above that Z[x] is countable. Each p in Z[x] has only finitely many roots (no more than the degree of p) so the set of algebraic numbers is the union of countably many finite sets. (d) Let A be the set of algebraic numbers, so R \ A is the set of transcendental numbers. Since A is countable and R is uncountable, R \ A must be uncountable.

17. Different approaches are possible. In the case discussed, it’s fun to see that, in binary digit notation, h(0.1stuff) = 0.01stuff, h(0.01stuff) = 0.1stuff, h(0.001stuff) = 0.0001stuff, h(0.0001stuff) = 0.001stuff, etc.

1.7 1.

Numbers 102: Absolute Values (a) From elementary calculus we can see that |sin(x) + cos(x)| ≤

√

2.

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(b) The RTI says that |sin(x) − cos(x)| ≥ || sin(x)| − | cos(x)|| ≥ 0. We can’t improve on this, since |sin(x) − cos(x)| = 0 when x = π/4. (c) Yes. A famous trigonometric identity says that sin(x)2 + cos(x)2 = 1 for all x. 2.

(a) It is enough to prove the RTI under the assumption that |x| ≥ |y|. (If |x| < |y| we can just reverse the roles of x and y.) In other words, we need to show that |x − y| ≥ |x| − |y|. But this is equivalent to |x − y| + |y| ≥ |x|, which is just a form of the ordinary triangle inequality. (b) The distance from x to y is at least as great as the distance from |x| to |y|. In other words, taking absolute values of numbers either shrinks distances or leaves them alone. (c) The RTI is an equation except when x and y have opposite signs.

3. For the first inequality, use the ordinary triangle inequality and the fact that |x − y| = |x + (−y)|. The second inequality is half of the RTI. 4.

(a) x ∈ (−9, 2). (b) x ∈ (−∞, −9) ∪ (2, ∞).  √ √  (c) x ∈ − 2, 2 .

5.

(a) x ∈ (1, ∞)  √  √  √ √ (b) x ∈ − 4.07, − 3.93 ∪ 3.93, 4.07 (c) x ∈ (−3, 1)

6.

(a) Multiply both sides of |x − 3| < 0.07 by 5. (b) By the triangle inequality, |x + y − 8| ≤ |x − 3| + |y − 5| < 0.11. (c) Multiply the inequalities 2.93 < x < 3.07 and 4.96 < y < 5.04.

7.

(a) Subtracting 4 from all parts of 4.96 < y < 5.04 gives 0.96 < y − 4 < 1.04; note also that y − 4 = |y − 4|. (b) Note that 2.93 < x < 3.07 and 4.96 < y < 5.04. Thus |y − x| < 5.04 − 2.93 = 2.11 = K. (c) Since 2.93 < x < 3.07 and 4.96 < y < 5.04, we have |y − x| > 4.96 − 3.07 = 1.89 = L.

8. Note that the given conditions means that x ∈ (0.97, 1.03) and y is outside [−5, 5]. (a) |y − x| > 5 − 1.03 = 3.97 = L.

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(b) |y − x| can take all values greater than 3.97. More formally, we have 5 < y < ∞ and −1.03 < −x < −0.97. Adding these gives 3.97 < y − x = |y − x| < ∞. 9. The base case (n = 2) is the ordinary TI. For the inductive step, let’s assume that the TI holds for n summands. To show it holds for n + 1 summands we calculate just as in the problem, except that we use n + 1 summands instead of three. |x1 + · · · + xn + xn+1 | = |(x1 + · · · + xn ) + xn+1 | ≤ |x1 + · · · + xn | + |xn+1 | ≤ |x1 | + · · · + |xn | + |xn+1 | .

10. This view of things helps explain the triangle inequality name: the length of one side of a triangle is not greater than the sum of the other two sides. 11.

(a) Look separately at the cases x ≥ y (so |x − y| = x − y) and x < y (so |x − y| = y − x). x + y − |x − y| . 2 f (x) + g(x) − |f (x) − g(x)| . (c) We can use h(x) = 2 (d) A good interval is something like [−20, 2].

(b) min{x, y} =

12. Use the triangle inequality: |x − y| = |x − 7 + 7 − y| ≤ |x − 7|+|7 − y| = |x − 7| + |y − 7| <  +  = 2. 13.

(a) |x − 1| < 0.5 implies 1 − x < 0.5, or x > 0.5. (b) Use the triangle inequality: |c| = |c − x + x| ≤ |c − x| + |x| < |c| |c| |c| 2 + |x|. This implies |c| < 2 + |x|, or |x| > 2 . R1

f (x) dx is a limit of approximating sums, R like f (x1 )∆x1 + 1 f (x2 )∆x1 + · · · + f (xn )∆xn . Thus 0 f (x) dx is approximated by R1 |f (x1 )∆x1 + f (x2 )∆x1 + · · · + f (xn )∆xn |, and the integral 0 |f (x)| dx by |f (x1 )∆x1 | + |f (x2 )∆x1 | + · · · + |f (xn )∆xn |. The usual triangle inequality applies to these sums.

14. An integral

1.8

0

Bounds

1. Following are sketches.

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(a) Say m ≤ t ≤ M for all t ∈ T . If s ∈ S, then s ∈ T , too, so m ≤ s ≤ M . Thus S is bounded. (c) If |S| is bounded, then |s| < K for all s ∈ S and some K > 0. But then −K < s < K for all s ∈ S, so S is bounded. The converse is similar. 2. Fact ??, page ?? says almost exactly what’s claimed here. 3.

(a) I = [1, 3) works (b) I = (−∞, 3) works (c) I = (1, 3) works (d) No; if max(S) exists then max(S) = sup(S).

4.

(a) M = 5 works. (b) The number M = max{|a|, |b|} works. One way to see this is to notice that b ≤ |b| ≤ M , and a ≥ −|a| ≥ −M . Thus, for all s ∈ S we have −M ≤ a ≤ s ≤ b ≤ M , as desired. (c) If |s| ≤ 42 for all s ∈ S, then clearly for −42 ≤ s ≤ 42, so −42 and 42 are upper and lower bounds. (d) If S is bounded, then S has a lower bound a and an upper bound b. Set M = max{|a|, |b|}. As shown above, |s| ≤ M for all s ∈ S. For the converse, suppose |s| ≤ M for all s ∈ S. Then, as shown above, −M ≤ s ≤ M for all s ∈ S, so S is bounded above and below.

5.

(a) We have 1 = e0 ≤ f (x) = ex ≤ e10 and −1 ≤ g(x) = sin(x) ≤ 1 for x ∈ A. We know from calculus that (i) f (x) = ex is bounded below by 0 but is unbounded above on R, and (ii) g(x) = sin(x) is bounded below by −1 and above by 1 on all of R. (b) For f ◦ g: When x ∈ [0, 10] we have 0.37 ≈ e−1 ≤ esin(x) ≤ e1 ≈ 2.72. For g ◦ f : When x ∈ [0, 10] we have −1 ≤ sin(ex ) ≤ 1.

6.

(a) The graph of f is an upward-opening parabola. (b) A = [−12, 10] works, since f (−12) = f (10) = 120. (c) A = R works; f (x) = (x − 1)2 − 1 ≥ −1 for all x. (d) A = [−2, 0] works, since f (−2) = f (0) = 0. (e) f (x) takes at most 1234 different values for x ∈ A. One of these values is largest and another smallest.

7. Since A is bounded we know that −M ≤ x ≤ M for some M > 0 and all x ∈ A. But then also −3M + 5 ≤ 3x + 5 = f (x) ≤ 3M + 5 for all x ∈ A, as desired.

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8.

(a) Adding the inequalities gives −5 ≤ f (x) + g(x) ≤ 5 for all x. (b) Since |f (x)| ≤ 2 and |g(x)| ≤ 4, we have |f (x)g(x)| ≤ 8 for all x. (c) Upper and lower bounds for f ◦ g and g ◦ f are the same as those for f and g, respectively.

9.

(a) inf(S) = min(S) = 2; sup(S) = max(S) = 3. (b) inf(S) = min(S) = 2; sup(S) does not exist, as there are infinitely many primes. (c) Here S = {2}, so inf(S) = min(S) = 2 = sup(S) = max(S) (d) inf(S) = min(S) = 41; sup(S) = max(S) = 43. √ √ (e) inf(S) =√− 2; sup(S) = 2. Maximum and minimum values don’t exist, as 2 ∈ / S. √ √ (f) inf(S) = min(S) = − 2; sup(S) = max(S) = 2.

10.

(a) The set is bounded below by 1 but unbounded above, since the series in question diverges. (b) The set is bounded below by 1 and above by 2, since the series in question converges. (c) The set is bounded below by 0 and above by a small number like 0.4—look at the graph of an appropriate function. (d) The set is bounded below by (say) 0.8 but unbounded above—look at the graph of an appropriate function. (e) The set is unbounded above and below. (f) A cubic polynomial can have three roots at most, so the set is clearly bounded. In this case there’s only one root, around 12.4.

11.

(a) The shortest English words have one letter, so 1 is a lower bound for f on EW. It’s hard to identify the longest English word, but 1000 letters is probably a safe upper bound. (b) Words in this problem vary in length from 1 to 10 letters.

12. English sentences vary in length from one word (Stop!) to many, many words, but 1000 words is probably a safe upper bound. On the set SITP, values of g vary from 5 words to about 30 words, depending on how you count words in equations. 13.

(a) All subsets S ⊆ R have this property (b) The given property means that S is bounded. (c) For all s ∈ S, |s| ≥ 1.

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14.

(a) The maximum value is f (0) = 1; the minimum is f (±10) = 1/101. (b) The graph is a parabola with vertex at x = −1/3, so the minimum value is g(−1/3) = −22/3. The maximum occurs at the right endpoint; g(10) = 313. (c) Note that h(x) = 1 + x, so the function is largest at the right endpoint (h(10) = 11) and smallest at the left (h(−10) = −9).

15.

(a) The maximum value is f (0) = 1; the minimum is f (±K) = 1/(K + 1). (b) The graph is a parabola with vertex at x = −1/3, so the maximum occurs at the r ight endpoint; g(K) = 3K 2 +2K−7. If K ≥ 1/3, then −2/3 ∈ [−K, K], and so the minimum value is g(−1/3) = −22/3. If K < 1/3 then the minimum occurs at x = −K, and the value is g(−K) = 3K 2 − 2K − 7. (c) Note that h(x) = 1 + x, so the function is largest at the right endpoint (h(K) = 1 + K) and smallest at the left (h(−K) = 1 − K).

16. If A = 0 then f (x) = B, a constant, so B itself can be an upper and lower bound. If A > 0, then −K ≤ x ≤ K implies −KA + B ≤ x ≤ KA + B, so lower and upper bounds are −KA + B and KA + B, respectively. If A < 0, then KA + B ≤ x ≤ −KA + B, so lower and upper bounds are KA + B and −KA + B, respectively. 17.

(a) S = (1, ∞) is one possibility; there are many. (b) Let S be bounded away from zero. Then there exists δ > 0 so |s| > δ 1 1 1 for all s ∈ S. Thus, if t ∈ T we have |t| = = < , so T is s |s| δ bounded. The proof of the converse is similar. (c) If S is bounded away from 0 then—by definition—S has no points in common with the interval (−δ, δ), which is of the desired form. If, conversely, S has nothing in common with (a, b), where a < 0 < b, then we can take δ to be the smaller of −a and b.

18.

(a) The interval (0, 0.9) is bounded away from 1 but not from 0. (b) There is no number a such that Q is bounded away from a, because every interval contains rational numbers. (c) One possibility is Z \ {42}.

19.

(a) If a ∈ / I, then either a < 0 or a > 1. In the former case, we can take δ = −a/2; in the latter, δ = (a − 1)/2 works. (b) The points 2 and 3 are not in J, but J is not bounded away from either 2 or 3.

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(c) N does indeed have the property in question: if a ∈ / N. then we can take δ to be the distance to the nearest integer. 20. Suppose S were bounded away from β. Then there is δ > 0 such that β − s > δ for all s ∈ S. But this means β − δ > s for all s ∈ S, and so β − δ is an upper bound for S. This means β is not the least upper bound, a contradiction.

1.9 1.

2.

Numbers 103: Completeness 1 1 1 1 , , , ..., , .... 508 509 510 507 + n 1 1 1 1 1 1 1 1 (b) One possibility is the list + , + , + , . . . , + , .... 3 7 3 8 3 9 3 7+n (c) One possibility is to let h = b − a, and use the list a + h/2, a + h/3, a + h/4, . . . , a + h/n, . . . . “Successive averaging” is another possibility. (a) One possibility is the list

(a) One possibility is the list

1 1 1 507+π , 507+2π , 507+3π ,

....

h h (b) One possibility is to let h = b−a, and use the list a+ , a+ , . . . . π 2π 3. For given  > 0, consider the positive number 1/. Since N is unbounded, there exists an integer n with n > 1/. But then 0 < 1/n < , as desired. 4.

(a) The claim is true if 0 < b ≤ a; we can just let n = 2 (or even n = 1 if b < a). (b) The Archimedean claim is clearly false if a = 0 and a < b. It is also false if, say, a = −1 and b = 1.

5.

(a) The statement is false if a = 0 and b = 1. (b) The statement is true. If 0 < a < b then the the Archimedean principle applies. If a < b < 0 and n = −1, then na > b. The case a < 0 < b is left to you. (c) Clearly b + 1 > b, so we can choose n ∈ R with na = b + 1; i.e., n = (b + 1)/a. Nothing Archimedean needed.

6.

(a) As we’ve shown, the interval (0, y) includes a rational and an irrational. They lie in (x, y), too. (b) If a rational r and an irrational z lie in (−y, −x), then the rational −r and the irrational −z lie in (x, y).

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7.

(a) The italicized statement is true; this is essentially the well-ordering principle. (b) The italicized statement is now false. The set S of rationals less than π has supremum π, but π ∈ / Q.

8. Let −S be as in the hint. Because S is bounded below, −S must be bounded above, and so −S has a supremum, say β. We can then set α = −β and check that α is the desired infimum of S. Details left to you. 9.

(a) In = (3 − 1/n, 3 + 1/n) works. (b) In = [−3 − 1/n, 42 + 1/n] works. (c) In = (−3 − 1/n, 42 + 1/n) works.

10. Let In = (a + 1/n, b + 1/n) for n = 1, 2, . . . . 11. The hint explains the problem: If (a, b) is contained in all the Ii , then [a, b] is contained in all these intervals, too. 12.

(a) Here’s the special case with a42 and b73 : From the hypotheses, a42 ≤ a73 . But I73 = [a73 , b73 ], so a42 ≤ a73 ≤ b73 . The general case is similar. (b) The previous part shows that bn is an upper bound for all the am ; thus α = sup A ≤ bn . (c) We’ve shown everything except α ≤ β. But the previous part says α ≤ bn for all n, α is a lower bound for the bi . Since β is the greatest lower bound, we have α ≤ β.

13.

(a) The set S is certainly bounded above; by completeness there’s a least upper bound, β. (b) Note that (β+h)2 = β 2 +2βh+h2 < β 2 +2βh+h = β 2 +(2β+1)h. Now use property (ii) of h. (c) The preceding calculation shows β +h—a number larger than β—has square less than 2. This contradicts the supremum property of β. (d) Note that (β − k)2 = β 2 − 2βk + k 2 < β 2 − 2βk. Substitute for k and simplify.

14. The proof for any positive number a is almost identical to that for a = 2. At the appropriate point, choose h so (i) 0 < h < 1 and (ii) h ≤ (a − β 2 )/(2β + 1). And similarly for k. p√ √ p√ √ 4 15. The point is simply that 4 a = a, 8 a = a, and so on. 16. Imitate the proof in either Problem ?? or Problem ??. Note that (β + h)3 = β 3 + 3β 2 h + 3βh2 + h3 < β 3 + 3β 2 h + 3βh + h = β 3 + h(3β 2 + 3β + 1).

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17. We have, for instance,

√ 6

a=

p√ 3

a; other parts are similar.

18. Let S = {0.d1 , 0.d1 d2 , 0.d1 d2 d3 , 0.d1 d2 d3 d4 , . . . }. The set S is bounded above (by 1, for instance) and therefore has a (unique) supremum β—i.e., a real number somewhere in interval [0, 1].

2.1

Sequences and Convergence

1. For the three sequences given: 

2.

1.000

0.100 √

10 ≈ 3.16

0.010 10

0.001 √

1000 ≈ 31.6

(a)

N

1

(b)

N

1

100

104

106

(c)

N

1

e9 ≈ 8103

e99

e999

(a) If a given N works for  = 0.001, then the same N works for larger . We could use this N in all positions in the row. (b) Any N works in this case; we could use N = 0 in all positions, for example. (c) The number of digits in a number changes after 9, 99, 999, etc. So the second-row entries would be 9, 109 − 1, 1099 − 1, and 10999 − 1.

3. Suppose toward contradiction that L > b. Set  = b−L and choose N as in the definition of convergence. Then (explain why!) we must have xn > b for n > N , which is absurd. A similar argument shows that L ≥ a. 4. This is false. As a counterexample, find a sequence that increases to 17. 5.

(a) Set  = 0.1 and choose a corresponding N as in the definition. All xn with n > N lie in the desired interval. (b) Set  = 0.001 and choose a corresponding N as in the definition. Then all xn with n > N lie in (4.999, 5.001), and therefore exceed 4.999. (c) If we set  = 1 and choose a corresponding N , then all xn with n > N lie in (4, 6). Thus it is possible that xn > 6 only for members of the finite set {x1 , x2 , . . . xN }.

6.

(a) Yes. The divergent sequence 3, 0, 3, 0, 3, 0, 3, 0, . . . has infinitely many terms with value 3.

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(b) No. If, say, {xn } = 3 for all n > N , then this value of N works in the definition of convergence for any  > 0. 7.

(a) Yes, it’s possible. We might have xn = 4 − 1/n, for instance. (b) Yes, it’s possible. Our series could have the pattern 4.1, 3.9, 4.01, 3.99, 4.001, 3.999, . . . .

8.

(a) Yes. We could have xn = 4 for n ≤ 1234 and xn = L thereafter. (b) No. There would be trouble if we set  = |L − 4|.

9.

(a) Algebraic manipulation gives 2n 2 10 − = |an − L| = 3n + 5 3 9n + 15 and 10 0, the value N = 10/−15 works in the definition. 9 A formal proof resembles that in Example ??. (b) The proof is like that for (a), except that now we have |bn − L| = Thus N =

899900 0.

(c) We’ll show cn → 0. Observe first that |cn − L| = and

2 0 be given; set N = works, since if n > N then |cn − L| =

2 3 .

This N

2n 2 2 < < = . 2 3n + 5 3n 3N

10. Formal proofs resemble those in the preceding problem. Following are sketches. (a) The limit is 2/3. For given  > 0 we can use N =

15+10/ . 9

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(b) The limit is L = 2/3. Basic algebra gives |bn − L| =

10 − 2 sin n 8 ≤ . 9n + 3 sin n + 15 9n

(Convince yourself of the last inequality.) This implies that, for given  > 0 we can use N = 8/(9). (c) As in the preceding problem, cn → 0. Indeed, |cn − 0| =

2n 2 < , 3n2 + 5 3n

and so, for given  > 0 we can use N = 2/(3). 11. Here’s a sketch. Let  > 0 be given, and set 0 = /17. Since 0 > 0 we can choose N so that n > N implies |xn − 1| < 0 . This same N works in the definition of convergence of {17an } to 17. 12. This is a special case of Theorem ??, page ??. 13. Suppose {an } is decreasing. Since {an } is bounded, this set has an infimum; call it L. To see L is also the limit, let  > 0 be given. Because L is the infimum, L +  is not a lower bound for {an }, and so there is some aN with L +  > aN . This N “works”: If n > N , then an ≤ aN (because {an } is decreasing) and so L ≤ aN < L + , as desired. 14. Suppose {an } converges to L. Then for  = 1 we can choose N such that n > N implies |an − L| < 1. Consider the sets S1 = {an | n > N } and S2 = {an | n ≤ N }. Our choice of N implies that S1 is bounded (above by L + 1 and below by L − 1). The set S2 is a finite set, and therefore also bounded. Thus {an } is the union of two bounded sets, and hence also bounded. 15. Suppose xn → 5. To show yn → 0, let  > 0 be given. Since xn → 5 we can choose N so that n > N =⇒ |xn − 5| < . This is just another way of saying that |yn | < , and so yn → 0 as desired. The converse is almost identical. 16. The sequence {yn } converges to 212. To see why, notice that |yn − 212| = |5xn + 2 − 212| = 5 |xn − 42| , and so

 . 5 For a given  > 0, therefore, we can set 0 = /5, and choose N that works for the sequence {xn } and 0 . The same N does what’s needed for {yn }. |yn − 212| < 

⇐⇒

|xn − 42|
0 we have |1/n−0| ≥  only when n ≤ 1/. There are only finitely many such n. (b) Let  > 0 be given, and consider the set F = {n | |xn − L| ≥ }. By hypothesis, F is finite. If F is nonempty, then F has a largest element, say N ; this N works in the definition of convergence. If F is empty, then N = 0 works. (c) A sequence {xn } does not converge to zero if, for some  > 0, we have |xn | ≥  for infinitely many n. (d) A sequence {xn } does not converge to zero if, for some  > 0, there is no N such that |xn | <  whenever n > N .

18. If β has decimal expansion 0.d1 d2 d3 d4 d5 . . . , then the increasing rational sequence 0.d1 , 0.d1 d2 , 0.d1 d2 d3 , 0.d1 d2 d3 d4 , . . . converges to β. 19. Every constant sequence has such a table. 20. A sequence has such a table if and only if the sequence is constant from the sixth term on: x6 = x7 = x8 = . . . . (Such a sequence is called eventually constant.) 21.

(a) true (b) true; 0 is a lower bound (c) false; n = 43 is a counterexample (d) true; N = 10000 works (e) true; N = 1 works, for instance (f) false (g) true; for a given  we can choose N = 1/ (or the next integer if 1/ isn’t an integer)

22.

(a) true (b) true (c) false; n = 43 is a counterexample (d) true; N = 10000 works (e) true; N = 20000 works (f) false; this would contradict convergence (g) true; N = 1/ (or the next integer) works

23.

(a) true (b) true

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(c) false; use technology to find a counterexample, like n = 45 (d) true; N = 10000 works (e) true; N = 20000 works . . . use the triangle inequality (f) false; see (d) (g) true; N = 1/ (or the next integer) works 24. Suppose that {xn } converges to L = 1000. Let  = 1 and choose N as in the definition of convergence. Then we have xn ∈ (999, 1001) for all n > N , which means that 1001 is an upper bound for these xn . The remaining xn form a finite set, which is automatically bounded above. This contradicts the assumption of unboundedness. 25. The limits L are 0, 0, 0, 3, 42, respectively. Possible tables:

26.

10−10



1.000

0.100

0.010

(a)

N

1

(b)

N

1

(c)

N

1

100 √ 10 √ 10

10000 106 √ √ 100 1000 √ √ 100 1000

(d)

N

6

60

600

6000

6 · 1010

(e)

N

41

41

41

41

41

0.001

1020 105 105

(a) Yes. In effect, increasing the N -entry is always OK. (b) Yes and yes.

27.

(a) We can check explicitly that 1.511 ≈ 86.5 < 89 = f11 , and 1.512 ≈ 129.7 < 144 = f12 . For the inductive step, note that fk+1 > 1.5k−1 + 1.5k = 1.5k−1 · 2.5 > 1.5k−1 · 1.52 = 1.5k+1 , as desired. (Note that the proof worked because 1 + 1.5 > 1.52 . A similar result can be shown to hold if 1.5 is replaced by any number b with 1 + b > b2 .) (b) Using technology one can check that n ≥ 72 works.

28.

(a) The first few terms are 1/1, 2/1, 3/2, 5/3, 8/5; notice the Fibonacci numbers as numerators and denominators. (b) The claim is obviously true for n = 1. The inductive step is easy, too. Clearly, gn > 1 for all n, and so gn+1 = 1 + 1/gn < 1 + 1/1 = 2.

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(c) Easy algebra is enough. 29. Implications are as follows: . (a) The statement says that an → π. (b) The statement says that for some n, an = π for n > N . (c) The statement says that an = π for n > 1. In particular, (c) =⇒ (b) =⇒ (a).

2.2

Working with Sequences

1. Here is a sketch. For any constant c 6= 0, we have |can − ca| = |c| |an − a| <  ⇐⇒ |an − a|
0, we can choose N so that |an − a| < /|c| for n > N . (Why can we do this?) This N “works” for the given  and the original sequence {can }. (Assemble the pieces into an efficient proof.) 2.

(a) This is basic algebra; use a common denominator. (b) The directions give the main idea. (c) For (i), we can take δ to be the smaller of |b|/2 and the minimum mentioned in the problem. Part (ii) follows because |bn | ≥ δ for all n.

3. Both equivalences follows directly from the  − N definition. The second also follows from (a) of Proposition ??, page ??. 4.

(a) The lemma says √ that there exist both rational and irrational sequences that tend to 2, and that these can be chosen to be monotone. √ (b) One possibility is pn = 2 + 1/n for all n. √ (c) One possibility comes from the infinite decimal approximation 2 = 1.414213562 . . . ; we can use r1 = 1, r2 = 1.4, r3 = 1.414, etc. (We need a little fix when a zero-digit comes along.) Another possibility, a bit less concrete, is to use the fact that every interval contains rational numbers.

5. Imitate the (partial) proof given for Lemma ??. 6. We need sequences of irrationals decreasing to 0 and increasing to 1. These are easy to find. 7.

(a) We can define {bn } by bn = 1 − n1 .

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(b) We can define {bn } by bn = 1 for all n. (c) Because β − n1 is not an upper bound for C, there must exist bn as desired. (Note the strict inequality.) (d) Apply the squeeze principle to β −

1 n

< bn ≤ β.

8. This is trivial if β ∈ S; if so, we can use the constant sequence sn = β for all n. So we assume in what follow that β ∈ / S. Choose s1 ∈ S with β − 1 ≤ s1 < β; this is possible since β − 1 is not an upper bound for S. Now choose s2 with s2 ≥ s1 and s2 ≥ β − 12 ; this is possible since neither s1 nor β − 12 is an upper bound for S. Similarly, we can choose s3 with s3 ≥ s2 and s3 ≥ β − 13 , and so on. The sequence {sn } define in this way does what’s desired. 9. The n = 1 (base) case is easy: s2 = 1.5 < 2 = s1 . The inductive step is to show that if sn − sn−1 < 0, then sn+1 − sn < 0, too. Doing so involves careful but straightforward algebra. 10. For every positive integer n we have the estimate sn = Clearly,

√

√ 1 1 1 1 1 n + √ + · · · + √ > √ + · · · + √ = √ = n. 1 n n n n 2

n → ∞, and so sn → ∞, too.

11. It is clear that {hn } is increasing, and the inequality h2n ≥ hn + 12 implies that {hn } is unbounded. Since h1 = 1, the inequality means that h2 ≥ 1.5, h4 ≥ 2.0, h8 ≥ 2.5, h21 000 > 501, etc. 12.

(a) Clearly, Tn < Sn for all n, and both sequences appear to converge. (b) It is natural (and correct) to guess Tn = straightforward. (c) We can use algebraic facts: Tn =

n n+1

=

n n+1 .

The inductive proof is

1 1+1/n

→

1 1+0 .

1 1 1 ≥ 2 = for all n, and so Sn ≤ n2 + n n + n2 2n2 2Tn < 2 for all n ≥ 1. Thus {Tn } is bounded above, and clearly increasing, so {Tn } converges.

(d) Observe that

13.

(a) Let M > 0 be given. Since xn → ∞, we can choose N so xn > M whenever n > N . But then also −xn < −M when n > N . Thus −xn → −∞. (b) Suppose xn → ∞. Let  > 0 be given; we need to find N so n > N implies x1n = x1n < . Well, if we set M = 1/ then (since xn → ∞) we can choose N so n > N implies xn > M = 1/. But this implies 1/xn < , as desired.

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(c) Suppose xn < 0 for all n. Then xn → −∞ if and only if 1/xn → 0. This follows from parts (a) and (b): xn → −∞ ⇐⇒ −xn → ∞ ⇐⇒ − x1n → 0 ⇐⇒ x1n → 0. √ √ 14. The inequality an − 3 < |an − 3| implies that √ √ − |an − 3| < an − 3 < |an − 3| . Since the left and right quantities tend to zero, so must the middle. √ √ Proving the inequality an − 3 < |an − 3| involves multiplication and √ √ division by the “conjugate” expression an + 3 . 15.

(a) The limit is 2/5. (b) The sequence diverges to ∞. (c) The sequence converges to 1/2. (d) The sequence diverges to ∞. One strategy is to observe that n2 −4 n+2

n2 +arctan n n+2

>

= n − 2.

16. From its definition we see that {bn } is “obviously” monotone non-decreasing; think it through. Thus {bn } converges if but only if {bn } is bounded. Finally, convince yourself that {bn } is bounded if but only if {an } is bounded. 17.

(a) One possibility is to let an = (−1)n and bn = (−1)n+1 for all n. (b) We’ll show that cn → max{a, b}. One approach is to use the curious fact (which appears in a later section) that for any numbers an and bn , cn = max{an , bn } =

an + bn + |an − bn | . 2

Invoking Theorem 2.5 (on algebra with convergent sequences) and Proposition 2.7.b (on how absolute values play nicely with limits) we see that cn =

a + b + |a − b| an + bn + |an − bn | converges to c = = max{a, b}, 2 2

as desired. Here is another approach. We assume WLOG that a ≤ b and show that cn → b. Suppose first that a = b. Then for given  > 0 we can choose Na and Nb such that n > Na implies |an − b| < , and n > Nb implies

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|bn − b| < . Let N = max{ Na , Nb }. If n > N , then we have both |an − b| <  and |bn − b| < . Now recall that, for all n, either cn = an or cn = bn . Thus |cn − b| <  for all n > N , and we’re done. The other possibility is that a < b. Set  = b−a 2 and choose N (as above) so that both |an − a| <  and |bn − b| <  hold whenever n > N . This means (draw a picture if this isn’t clear) that an < bn for all n > N . In other words cn = bn for all n > N , and so surely {cn } and {bn } converge to the same limit, b. 18.

(a) Yes: xn > 0 for n > 25, so N = 25 works. (b) No: The sequence oscillates in sign after n = 10000. (c) The sequence {xn } is quasi-positive. Since a > 0 we can let  = a, and choose N as in the definition of convergence. Then for n > N we have |xn − a| < a which implies xn > 0.

19.

(a) We have x1 = 1, x2 = 2, x3 = 3/2, x4 = 5/3, x5 = 8/5, x6 = 13/8, x7 = 21/13, x8 = 34/21. Note the connection to the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, . . . . The sequence is not monotone, but has the pattern x1 < x2 > x3 < x4 > x5 < . . . . 1 . Since {xn } converges to L we can take (b) We know xn+1 = 1 + xn 2 limits on both sides to get L = 1 + 1/L. Solving √ for L gives L − L − 1 = 0. The quadratic formula says L = (1 ± 5)/2; from context it’s clear we want the positive root L ≈ 1.61803.

20.

(a) The first few values are x2 = 7/4, x3 = 97/56, x4 = 18817/10864. Decimal values (rounded) are 1.75, 1.73214, 1.73205. √ (b) The function x + 3/x has a minimum at 3. One can also do this algebraically, without calculus. (c) Use algebra: When x > 0, we have xn+1 < xn ⇐⇒ xn + 3/xn < 2xn ⇐⇒ x2 > 3. (d) Our sequence is bounded and monotone, so it converges. (e) L = (L + 3/L)/2 follows from our√algebra-with-limits theorem. Solving this equation for L gives L = 3.

21.

(a) Check first that since x1 = 3, we have 1 < xn < 4 for all n. Also, xn+1 = (x2n + 4)/5 < xn ⇐⇒ x2n − 5xn + 4 < 0, and it’s easy to see that this holds when 1 < xn < 4. (b) The sequence {xn } converges because it’s monotone and bounded. (c) Because the sequence converges, we must have L = (L2 + 4)/5, so L = 1 or L = 4. In context it’s clear that L = 1.

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22. The sequence pseudodiverges to infinity. For given M and N , let n be the first even integer greater than both M and N . This n satisfies the desired condition. 23. We’ll show |xn /yn | → 0. (Why is this enough?) First choose M > 0 such that |xn | < M for all n. (Why can we do this?) Now choose  > 0. Since yn → ∞ we can find N so that n > N implies |yn | > M  . (Why can we do this?) This N works for the sequence {|x /y |}: If n > N , then n n xn M M < yn yn < M/ = .

2.3 1.

Subsequences (a) The sequence 1, 2, 3, 1, 2, 3, . . . works. (b) The sequence 1, −1, 1/2, −1/2, 1/3, −1/3, . . . works. (c) The sequence 0, 1, 0, 2, 0, 3, . . . works. (d) The sequence 1, 1, 1/2, 2, 1/3, 3, . . . works. (e) The sequence 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, . . . works.

2.

(a) The converse says (falsely) that if {xn } has a convergent subsequence, then {xn } is bounded. Counterexamples are easy to find. The contrapositive says that if {xn } has no convergent subsequence, then {xn } is unbounded. (b) The BWT says that {yn } has a convergent subsequence. The BWT also applies to {zn }, but isn’t really needed, since it is easy to see that zn → 0. Thus, every subsequence of {zn } converges to zero. (c) If {xn } is bounded, then the BWT guarantees that a convergent subsequence exists. If {xn } is unbounded, then Lemma ?? applies. Both can occur, as they do in the sequence 1, 0, 2, 0, 3, 0, dots.

3.

(a) Given a sequence that lists the rationals, we can just form the subsequence of nonnegative rationals. (b) Look at any two successive terms, say p1 and p2 . Between these two rational numbers like other rationals. If the sequence were monotone, these other rationals would lie between p1 and p2 in the sequence. (c) We can choose n1 so that pn1 = 1. Then we choose n2 with n2 > n1 and pn2 ≥ 2. Continuing this process completes the proof; details are left to the reader.

4. Both parts follow from unraveling the –N definitions.

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5. In general it’s possible for a subsequence to behave differently from its parent. The point here is that this particular kind of subsequence—the “tail” of a given sequence—behaves in a more special way. Here’s the idea: Let  > 0 be given, and suppose that some N1 , say N1 = 1000, works for the subsequence. Note that the 1001th term of the subsequence is x5242 , and we know that from then on, all terms of the subsequence are within  of L. But this is just another way of saying that all terms past x5242 in the original sequence are within  of L; that is N = 5242 works for the given  in the parent sequence. In general, if N1 works for the subsequence, then N = 4242 + N1 works in the parent sequence. Clean this up a little to get an impeccable proof. 6. (As with many facts about subsequences, this one depends on the fact that for all k, nk ≥ k.) Let M > 0 be given. Since xn → ∞, there’s an N such that n > N implies xn > M . The same N works for {xnk }, since if k > N , then nk ≥ k > N , so xnk > M , as desired. 7. Statements (a), (c), (e), and (f) are all equivalent to each other. Statements (b) and (d) are also equivalent. 8.

2k , which converges to one as 2n + 1 2k − 1 k → ∞. Substituting n = 2k − 1 gives x2k−1 = − , which 2k converges to −1. With two different subsequence limits, {xn } must diverge.

(a) Substituting n = 2k gives x2k =

(b) Consider any convergent subsequence xn1 , xn2 , xn3 , xn4 , . . . , with limit L. Since this sequence has infinitely many terms, it must include either infinitely many even-indexed or infinitely many odd-indexed terms from the original sequence {xn }. If there are, say, infinitely many even-indexed terms, then they form a new subsequence, which obviously tends to one. But every subsequence of a convergent sequence tends to the same limit, so we must have L = 1. 9.

(a) The sequence 0, 3, 0, 3, 0, 3, . . . is one example. (b) The statement xn → 3 means that, for every  > 0, all but finitely many xn are within  of 3. Negating this condition means that, for some  > 0, infinitely many xn are at least away from 3. These {xn } can be taken in order to give the desired subsequence.

10. Since {xn } diverges, x0 is not the limit. Negating the definition of convergence to x0 means that, for some  > 0, no N works in the definition of convergence. We can use this to construct a sequence.

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Because N = 1 doesn’t work in the definition, there is some term—call it xn1 —for which |xn1 − x0 | ≥ . Now consider N = n1 . But this N doesn’t work either, so there must be some term—call it xn2 —for which n2 ≥ n1 and |xn2 − x0 | ≥ . Continuing this process gives the desired subsequence: n1 < n2 < n3 < . . . and |xnk − x0 | ≥  for all k. 11.

(a) For M > 0, choose N such that xn > M whenever n > N . The same N works for the subsequence {xnk }, since if k > N , then nk ≥ k > N , and so xnk > M , as desired. (b) The contrapositive of Theorem ??(a) says that if some subsequence {xnk } does not converge to L, then the original sequence {xn } doesn’t converge to L either. This implies Theorem ??(c), because if L is any number, then at least one subsequence fails to converge to L, and so L can’t be the limit.

12. If xn → x0 , then we know that every subsequence, and hence every monotone subsequence, must also converge to x0 . Now we show that if every monotone subsequence converges to x0 , then xn → x0 , too. Suppose, toward contradiction, that xn does not converge to x0 . Then (see earlier problems) for some  > 0 there is a subsequence {xnk } with |xnk − x0 | ≥  for all k. By Proposition ??, there is a monotone subsequence {yl } of {xnk }. Now |xnk − x0 | ≥  holds for all the xnk , and so it also holds for all the yl , too. Thus yl → x0 is impossible, which contradicts our assumption. 13. The sequence {x11 , x101 , x1001 , . . . } does the job. For this sequence we have nk = 10k + 1. 14. We’ll choose n1 , n2 , n3 , . . . in order. To choose n1 , let  = 0.1. By the definition of convergence there exists N1 such that n > N1 implies |xn − L| < 0.1, so we can let n1 be any such n, such as N1 + 1. To choose n2 , let  = 0.01 and choose N2 such that n > N2 implies |xn − L| < 0.01. Now we can choose any n2 with n2 > N2 and n1 < n2 . Similarly, we can choose n3 with n1 < n2 < n3 and |xn3 − L| < 0.001. Continuing this process gives the desired subsequence. 15. Notice first that zn = bn/2 for even n and zn = a(n+1)/2 for odd n. In particular, both sequences {an } and {bn } are subsequences of {zn }, so if zn → L we must have an → L and bn → L, too. For the converse, we suppose that both an → L and bn → L. Let  > 0 be given. We can choose N1 and N2 such that n > N1 =⇒ |an − L| < , and n > N2 =⇒ |bn − L| < . Now let N = 2 max{N1 , N2 }. This N “works” for {zn }, because if n > N and n is even, then n/2 > N ≥ N2 ,

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so |zn − L| = |bn/2 − L| < . Similarly, if n > N and n is odd, then (n + 1)/2 > N ≥ N1 , so |zn − L| = |a(n+1)/2 − L| < .

2.4 1.

Cauchy Sequences (a) For given  > 0 we can choose N = 2/. This N works because if 1 1   m, n > N then |xn − xm | ≤ |xn | + |xm | = + < + = . n m 2 2 (b) The sequence is not Cauchy. If we choose, say  = 0.001, then no N works, since if N is any positive number, no matter how big, then with m = N + 1 and n = N + 2 we have |yn − ym | = 2/1234 > 0.001. (c) The sequence is Cauchy. Note that if n > m, we have |zn − zm | = n m n−m n−m n 1 n + 1 m + 1 = (n + 1)(m + 1) < nm < nm = m . It follows that for  > 0 we can choose N = 1/. sin n sin m (d) The sequence is Cauchy. If n > m, we have |wn − wm | = 2 − ≤ n + 1 m2 + 1 p sin n sin m 2 2 2 n2 + 1 + m2 + 1 ≤ m2 + 1 < m2 , and m2 <  if m > 2/. p Thus, N = 2/ works in the definition.

2. For given  > 0 choose N1 such that |xm − xn | < /2 when n > m > N1 , and choose N2 such that |ym − yn | < /2 when n > m > N2 . Then N = max{N1 , N2 } does what’s needed. 3. For  > 0 choose N that works for  in the sense of Definition ??. This same N works in the definition of convergence to zero. To see why, let m > N be given. Then choose any n of the form n = 10k , with n > m (n = 10m is one possibility). Then we have n > m > N , and so |xm − xn | = |xm − 0| < , as desired. 4. Set  = 1 (any smaller  works, too) and choose N as in the Cauchy definition. This N works in the problem, since if n > m > N then |xn − xm | < 1, which can occur for integers xn and xm only if xn = xm . 5.

(a) Any sequence of rationals tending to an irrational will do. (b) Yes. The only Cauchy sequences of integers are eventually constant. (c) [0, 1] is complete because if {xn } is any Cauchy sequence in [0, 1], then {xn } tends to some limit L. Because 0 ≤ xn ≤ 1 for all n, the limit L is in [0, 1], too. (0, 1] is not complete because {1/n} is a Cauchy sequence in (0, 1], but its limit lies outside (0, 1].

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6.

(a) For any m and n, |yn − ym | = |5xn + 2 − 5xm − 2| = 5 |xn − xm | .  . For a 5 given  > 0, therefore, we can set 0 = /5, and choose N that works for the sequence {xn } and 0 . The same N does what’s needed for {yn }. It follows that |yn − ym | < 

⇐⇒

|xn − xm |
0 be given and set 0 = /42. Because {an } is Cauchy sequence there is some N such that |an − am | < 0 whenever n > N . This N works for the sequence {xn }; details left to you. 8.

(a) The sequence is monotone only if the coin falls heads (or tails) forever. (b) If the first 6 coin tosses show HT HT HT , then a6 = 21/32, and it follows that all further an , and therefore the limit, must lie in the interval [20/32, 22/32]. 1 1 1 1 (c) For every n, |an | ≤ 1 + + + + · · · + n−1 < 2. 2 4 8 2

9. Use the fact that two numbers with the same first n decimal places can differ by no more than 10−n . 10.

(a) By Theorem ?? the result is equivalent to the (known) fact that the product of convergent sequences is convergent. (b) Suppose |xn | ≤ M1 for all n and |yn | ≤ M2 for all n. Now choose N1 so  |xn − xm | < when n > m > N1 , 2M2 and choose N2 such that |yn − ym |
m > N2 .

Now N = max{N1 , N2 } does what’s needed, since if n > m > N we have |xn yn − xm ym | ≤ |xn yn − xn ym | + |xn ym − xm ym | ≤ |xn | |yn − ym | + |ym | |xn − xm | ≤ M1 |yn − ym | + M2 |xn − xm | < M1

 + M2 2M1 2

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1 1 1 1 11. The key point is that m+1 + m+2 + · · · + n < m , regardless of n. 2 2 2 2 This fact can be parlayed into a proof. 12.

(a) One possibility: 2, 1, 3, 2, 4, 3, 5, 4, . . . n n a o an o n converges, then the sequence 1717 (b) This is impossible: If 1717 1717 converges, too. (c) We could set an = n and bn = 17 − n. (d) We could set an = 1/n and bn = 1/n2 . (e) We could set an = n. (f) This is impossible: Every Cauchy sequence is bounded; the same is true of subsequences.

2.5 1.

Series 101: Basic Ideas (a) Sn = 0 for all n, and so the series converges to 0. (b) Sn = 42n for all n; this diverges (to ∞).

2.

(c) The partial sum sequence {Sn } has the form −1, 0, −1, 0, . . . ; this diverges. n(n + 1) (d) Sn = for all n; this diverges (to ∞). 2 1 − rn (e) The series is geometric, with r = 0.99, so Sn = = 100 − 1−r n 100 · 0.99 , and Sn → 100. n (f) One shows by induction that Sn = . Thus Sn → 1. n+1 P (a) The series an converges to 4 because the sequence An = 4 + 1/n of partial sums converges to 4. P (b) Because an converges we must have an → 0; the nth term test says so. P (c) The series An diverges by the nth term test: An → 4 6= 0. (d) We know A1 = a1 = 5 and A2 = a1 + a2 = 4.5; thus a2 = −0.5.

3.

(e) We know that a1 = 5. For n > 1 we have an = An − An−1 = 1 1 1 4+ −4− =− 2 . n n−1 n −n n (a) Since H2n > for all n and the right-had sequence diverges to in2 finity, we must have H2n → ∞, too. Since {Hn } has a divergent subsequence, {Hn } must diverge, too.

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(b) Suppose toward contradiction that Hn → H for some finite number H. Since {H2n } is a subsequence, we’d have H2n → H, too. But 1 we also have H2n > + Hn ; taking limits of both sides gives H ≥ 2 1 + H, which is absurd. 2 4.

(a)

∞ X k=1

∞

X 1 1 converges by comparison to the geometric series . k + 2k 2k k=1

∞ X

∞ X k 1 (b) diverges by comparison to the divergent series . 2k 2 − 1 2k k=1 k=1 (Convince yourself that the comparison works.) ∞ X k (c) converges by comparison to the convergent geometric series 3k k=1 ∞ X 2k . 3k k=1

(d)

∞ X k2 + 2 1 k2 + 2 diverges by the nth term test: lim = 6= 0. 2 k→∞ 3k 2 + 4 3k + 4 3

k=1

5.

(a) The series converges absolutely by comparison to

∞ X 1 . k2

k=1

(b) The series converges absolutely by comparison to

∞ X 1 . k2

k=1

(c) The series converges absolutely by comparison to the geometric series ∞ X 1 . 3k k=1

(d) The series diverges by the nth term test—all terms exceed 1/3. 6.

1 , which clearly tends to the limit 1. n+1 1 1 1 1 (b) Here we get Sn = 1+ − − , so Sn → 1+ as n → ∞. 2 n+1 n+2 2 1 1 1 1 (c) The pattern suggests that Sn = 1 + + . . . −− − − 2 10 n+1 n+2 1 ··· − . As n → ∞ the last 10 terms all tend to zero, so the n + 10 1 1 limit is S = 1 + + . . . ≈ 2.929. 2 10 (a) Sn = 1 −

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P P P 7. Let ak and bk be series, and let ck be the “sum series,” defined by ck = ak + bk for all k. Let An , Bn , and Cn denote the partial sums for these series. The key point is that—thanks to the commutative law for addition of finitely many numbers—Cn = An + Bn for all n. It follows that if An → A and Bn → B, then we must also have Cn → A + B, which is what we wanted to prove. 8.

(a) If r = 1 the series is just a + a + a + . . . , and so Sn = na, which clearly diverges. If r = −1 the series is just a − a + a − . . . , and so the successive Sn have the form a, 0, a, 0, . . . ; this, too, diverges.  (b) i. The sequence rn+1 is a subsequence of {rn }, so both converge to the same place, L. ii. Let xn = rn . We know that xn → L; one of our theorems says rxn → rL. iii. The preceding parts say that L = rL; if r 6= 0 we must have L = 0. iv. If 0 ≤ r < 1, then rn → 0. (Hint: Use a theorem about monotone sequences.) If 0 ≤ r < 1, rn+1 ≤ rn for all n, so the sequence is monotone decreasing and bounded below. Thus it converges, and the limit is zero by the preceding part. v. If −1 < r ≤ 0, then |r| < 1, and so |r|n → 0 by the preceding part. Now we have −|r|n ≤ rn ≤ |r|n , and so the middle limit is squeezed to zero. vi. If |r| ≥ 1 then clearly |rn | ≥ 1 for all n. In particular, 0 certainly isn’t the limit. Thus the sequence diverges.

9.

(a) It’s enough to show (i) if n > 1000 then Sn ≥ S1000 ; (ii) if n > 1000 then Sn ≤ S1001 . Claim (i) amounts to observing that every string of 1 1 1 1 the form − + −· · ·± adds up to a positive 1001 1002 1003 1000 + n result; group summands in pairs to see why. Claim (ii) holds because 1 1 1 1 every string of the form − + − + ··· ± 1002 1003 1004 1000 + n has negative sum; again, group summands in pairs to see why. (b) For given  > 0 we can take any integer N ≥ 1/. Then |SN − SN +1 | = 1 < . N +1 (c) For given  > 0 choose N as in the preceding part. Then for n > m > N we have both Sn and Sm between SN and SN +1 , and thus within  of each other.

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10.

11.

P P P (a) Suppose dk /10 converges. By Theorem ??, 10 dk /10 = dk converges, too, which contradicts our hypotheses. P (b) The product series ck dk may converge or diverge. If, say, dk = 1 for all k and ck = 1/k 2 , then the product series converges. But if dk = k for all k and ck = 1/2k , then the product series diverges. P (c) If A is constant then Ack converges by Theorem ??. (d) It is possible that dk ≤ ck for all k. This happens if, say, dk = −1 and ck = 0 for all k. P P (a) Let An and Bn be the partial sums for ak and bk . The hypothesis boils down to the fact that for n > 42 we have Bn = An − A42 + B42 . Thus, the sequences {An } and {Bn } differ (for large n) by an additive constant, and so both converge or both diverge. (b) We know (see the preceding part) that Bn = An −A42 +B42 . Because 4 X 2 bk − ak = k , we must have B42 − A42 = 2k 2 = 25585. Thus, k=1

Bn = An + 25585, and since An → 100 we must have Bn → 25685. P 12. Since ak converges we must have ak → 0, so there must exist K such 2 that 0 < aP k < 1 for k > K. Hence ak < ak for k > PK. It follows that 2 the series ak converges by comparison to the series ak . (Note that the inequality a2k < ak might not hold for k ≤ K. Do you see why this doesn’t matter?) 13. A similar problem appears in Section 1.5.

2.6

1.

Series 102: Testing for Convergence and Estimating Limits (a)

(b)

(c)

(d)

∞ X k=1 ∞ X k=1 ∞ X k=1 ∞ X k=1

∞

X 1 1 converges by comparison limit comparison to . 2k − 1 2k k=1 ∞ X

k diverges; one approach is limit comparison to 2k 2 − 1

k=1

1 . k

k converges by the ratio test. 3k k2 + 2 diverges; use the nth term test. 3k 2 − 2

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2.

(a)

∞ X (−1)k converges by the alternating series test; it converges ab3k 2 + 1 k=1 ∞ X 1 . solutely, too, by comparison or limit comparison to the p-series k2 k=1

(b)

∞ X k=1

2

k +k converges by limit comparison with k4

∞ X k=1

1 . k2

∞ ∞ X X sin(k) 1 (c) . converges absolutely by limit comparison to 3k − 1 3k k=1

3.

k=1

(a) The series diverges by the ratio test. (b) The series converges by the ratio test. (c) The series converges by the ratio test; a little algebra is needed.

4.

(a) The series is geometric; it converges for 0 ≤ a < 1 and diverges for a ≥ 1. (b) The series converges for all a ≥ 0; a is just a multiplicative constant. P k (c) The series can be written in geometric form r , r = a/1.3. Thus the series converges for 0 ≤ a < 1.3 and diverges for a ≥ 1.3. (d) The series converges for 0 ≤ a ≤ 0.5 and diverges otherwise; see the preceding part. (e) The series converges for all a > 0.

5.

(a) The series converges absolutely for −1 < x < 1; it’s geometric. (b) The series converges absolutely for −1 < x < 1; use the ratio test when x 6= 0. (c) The series converges absolutely for −1 < x < 1; use the ratio test when x 6= 0. converges conditionally when x = −1.) (d) The series converges absolutely for all x; use the ratio test when x 6= 0.

6.

(a)

∞ X 100k k=1

(b)

k!

converges by the ratio test.

∞ X sin k converges absolutely. The absolute value series con1.0001k k=1 ∞ X 1 . verges by comparison to the geometric series 1.0001k k=1

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(c)

∞ X k=1 ∞ X k=1

(d)

∞ X k=1

7.

k2 + k + 1 diverges; one approach is limit comparison to k3 + k2 + k + 1 1 . k (−1)k converges by the alternating series test. ln ln(k + 2)

X X X 1 X X1 √ then a2k diverges. If then (a) If ak = ak = k k X a2k converges. a2 (b) Use the limit comparison test: Here k = ak , and this tends to zero ak X by the kth term test. Hence a2k converges.

P 8. Suppose toward contradiction that ak converges. Since ak /bk → ∞ thenP (see the hint) bk /ak → 0, and the limit comparison test then implies that bk converges, a contradiction. 9. If

P

bk converges to B, then

P

ak converges to B −

K X

bk +

k=1

K X

ak . This

k=1

is similar to a problem in the preceding section, where K = 42. k 10. One can show easily (e.g., by induction) P that k! > 2 for k > 3. Thus, k 1/k! P 3, and hence 1/k! converges by comparison to 1/2 .

1 1 1 1 1 1 11. Notice that the sum Sn = √ + √ +· · ·+ √ > √ + √ +· · ·+ √ = n n n n 1 2 √ √ n √ = n. Since n → ∞, we must have Sn → ∞, too. n 13.

(a) The series converges by comparison to

∞ X k=1

1 2k−1

, which converges to

2. (b) We have R10 =

∞ X k=11

1 2k−1 + 1


0 be given and set δ = 1. This δ works for both limits; give details. 3.

(a) lim (2x + 3) = 5. Note that |f (x) − 5| = 2, |x − 1|. It follows that x→1

for given  > 0 we can set δ = /2.

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x2 − 1 = lim (x − 1) = −2. x→−1 x + 1 x→−1 (For given  > 0 we can set δ = .)

(b) Factoring the numerator gives lim

4.

x2 + x − 2 lim (x + 2) = 3. (For x→1 x − 1 x→1

(a) Factoring the numerator gives lim given  > 0 we can set δ = .)

2 + sin x ≤ (b) The limit is zero. To see why, note that |f (x) − 0| = x 3 − cos x 3 |x| . (The factor 3/2 comes from considering the possible sizes of 2 numerator and denominator.) It follows that for given  > 0 we can set δ = 2/3. 5.

(a) For f (x) = x2 , we have lim f (x) = lim x2 = 422 = f (42), so x→42 x→42 the condition does hold at a = 42. (b) For f (x) = x2 and any input a, we have lim f (x) = lim x2 = a2 = x→a

x→a

f (a), so the condition holds at every a. (c) Since the function f (x) = can’t possibly hold.

x2 −4 x−2

is not defined at a = 2, the condition

x2 − 4 32 − 4 = = f (3), as desired. x→3 x − 2 3−1

(d) We have lim f (x) = lim x→3

6.

(a) Note that a = 0 is√an endpoint of the domain. Also, we know lim+ f (x) = lim+ x = 0 = f (0), so the condition holds in the x→0

x→0

one-sided sense. (b) We have lim f (x) = lim |x| = |a| = f (a) for all a. x→a

x→a

(c) Theorem ??, Page ??, assures that lim f (x) = lim (1 + πx + ex2 + x→a

x→a

πex3 ) = 1 + πa + ea2 + πea3 = f (a), so the condition holds for all a. 7. Let {xn } be any sequence with xn → a and xn 6= a for all n. Then {f (xn )}, {g(xn )}, and {h(xn )} are all sequences, and the hypotheses imply that (i) f (xn ) ≤ g(xn ) ≤ h(xn ) for all n; (ii) f (xn ) → L and h(xn ) → L. Now the sequence version of the squeezing theorem implies that g(xn ) → L, too. Since this applies to any sequence {xn } of the given type, Lemma ?? implies that lim g(x) = L, as desired. x→a

8. Suppose that lim f (x) = L, that xn → a, and that xn 6= a for all n. To x→a

see that f (xn ) → L, let  > 0 be given. Because lim f (x) = L, we can x→a

choose δ > 0 so that |f (x) − L| <  whenever 0 < |x − L| < δ. Because

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xn → a and δ > 0, we can choose N so that |xn − a| < δ whenever n > N. This N works to show f (xn ) → L, since if n > N then we have |xn − a| < δ and so so |f (xn ) − L| < , as desired. 9.

(a) We say lim f (x) = L if for all  > 0 there exists M < 0 such that x→−∞

|f (x) − L| <  whenever x < M . (b) We say lim− g(x) = L if for all  > 0 there exists δ > 0 such that x→0

|g(x) − L| <  whenever −δ < x < 0. (c) Here’s the idea of a proof that lim f (x) = L implies that lim− f (1/x) = x→−∞

x→0

L. The converse is proved similarly. Let  > 0 be given. We want to show that |f (1/x) − L| <  for x in some interval (δ, 0). By hypothesis, there is some M < 0 so that |f (t) − L| <  when t < M . Writing t = 1/x, this means that |f (1/x) − L| <  when 1/x < M < 0, or, equivalently, when 0 > x > 1/M . (The inequality algebra is a bit tricky because both M and x are negative.) This implies that we can take −δ = 1/M , or δ = −1/M . 10.

(a) We’ll use the inequality√mentioned in the problem. Thus, let  > 0 be given, and set δ =  a. This δ works, since if |x − a| < δ, then √ √ √ − a| δ f (x) − a = x − a ≤ |x√ < √ = . a a (b) The proof that lim+ f (x) = f (0) = 0 is easier. For given  > 0 we x→0

can use δ = 2 . (c) To show that the squeezing inequality holds for all x > 0, use the “conjugate trick”: p p p ( 1 + 2/x − 1)( 1 + 2/x + 1) 2/x p 1 + 2/x−1 = =p < 2/x. 1 + 2/x + 1 1 + 2/x + 1 11.

(a) We can use δ ≈ 0.0024 (or less). Note δ ≈ /4. (b) We can use δ ≈ 0.00005 (or less). Note δ ≈ /200. (c) With  = 0.01, the value δ = 0.001 does not quite work at a = 10. (The value δ = 0.0005 does work.)

12. First let a = 0. Let  > 0 be given. Set δ = 1. (Any other positive δ would work, too.) This δ works, since if 0 < |x − 0| < δ, then in particular x 6= 0, and so |f (x) − 0| = 0 < , as desired. Now suppose a 6= 0. Then, for given  > 0 we can set δ = |a|. This δ works, since if 0 < |x − a| < δ then (as above) x 6= 0, and so |f (x) − 0| = 0 < , as desired.

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13. First suppose a ∈ Z0. Let  > 0 be given. Set δ = 1. This δ works, since if 0 < |x − 0| < δ, then x ∈ / Z, and so |f (x) − 0| = 0 < , as desired. If a ∈ / Z, then we can take δ to be the (positive!) distance from a to the nearest integer. 14. In this case limx→a h(x) = 0 unless a = 0. If a = 0 the limit lim h(x) x→0 does not exist. 15. The key idea is that, because S is a finite set, there is a smallest distance, say d, between any two points in S. Therefore, to show that limx→a j(x) = 0 when a ∈ S, we can use δ = d works for any  > 0. To show that limx→a j(x) = 0 when a ∈ / S, we can choose any  > 0 and let δ be the smallest distance from a to any point of S. Such a δ exists because S is finite. 16. The squeezing inequality itself is readily shown. The rest is easy: Since the left-hand and right-hand quantities tend to zero, the middle quantity must do so, too. 17.

(a) The point is that all three quantities in the claimed inequality are even functions of θ—i.e., they have the same value for θ and for −θ. (b) Draw the right triangles indicated, and note that θ is the area of the pie-shaped wedge with angle θ at the origin.

3.2

Continuous Functions

1. For given  > 0, δ =  works. 2. For given  > 0, any choice of δ, say δ = 1, works. 3. Give details as needed. (a) δ = 0.001/345 works. (b) δ = /345 works. (c) δ = /345 works regardless of the value of a. 4. Following are sketches; give details. (a) δ = 0.16 works because for x ∈ (3 − .16, 3 + .16) we have f (x) ∈ (8.07, 9.99), as required. (b) δ = 0.17 does not work because for 3.065 ∈ (3 − .17, 3 + .17) but f (x) = 10.02 ∈ / (8, 10).

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(c) δ = 0.16 does not work, since 10.159 ∈ (10−δ, 10+δ) √ but f (10.159) ≈ 103.21 ∈ / (99, 101) = (100 − , 100 + ). Since 101 ≈ 10.05 and √ 99 ≈ 9.95, we could use δ = 0.049. (d) δ = 0.0049 works. (e) δ = 0.00049 works. √ (f) Since 10 2 ≈ 1.072, it follows that δ = 0.071 works. 5.

(a) The idea: If f + g is continuous, then (f + g) − f = g is continuous, too. Give details. (b) There are many possiblities.

6.

(a) The problem, of course, is the jump at x = 2. Formally, let  = 1. Let’s show that no δ > 0 works for this . To that end, choose any δ > 0, and any number h with 0 < h < δ. Set x = 2 + h. Then we have |x − 2| < δ but |f (x) − f (2)| = 4 − 3 = 1 = . Thus this (and every) positive δ fails. (b) The problem here is that f (2) = 2 but f (x) ≈ 3 when x > 2 and x ≈ 2. More formally, let  = 1/2. To see that no δ > 0 works for this choice of , choose any δ > 0. Next, choose any small positive number h with h < 1/2 and h < δ, and set x = 2 + h. Note also f (x) > 2. Then |x − 2| < δ but |f (x) − f (2)| = |f (x) − 2| ≥ f (x) − 2 = 5 − 2 − h − 2 = 1 − h > 1/2 > δ.  ; note that δ > 0. |A| If |x − c| < δ, then |f (x) − f (c)| = |Ax + B − (Ac + B)| = |A||x − c| < |A|δ = , as desired.

7. Suppose first that A 6= 0. Let  > 0 be given. Set δ =

If A = 0 then f is a constant function. In this case any number delta > 0 works for a given  > 0. 8. Let  > 0 and set δ = . This works in the definition: If |x − a| < δ = , then |f (x) − f (a)| = | |x| − |a| | ≤ |x − a| < . (Note that we used the reverse triangle inequality.) 9.

(a) Let  > 0. Since f is continuous at 42 we can choose δ > 0 so |x − 42| < δ =⇒ |f (x) − f (42)| < . The same δ works for g, because |g(x) − g(42)| < |f (x) − f (42)|. (b) For given  > 0 the same δ works for f and for h. (c) For given  > 0 the same δ works for f and for k.

10. The function cos(x) sin(x) is continuous; see the relevant fact on page ??. applies.

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11. The composite function cos(sin(x)) is continuous; see the relevant fact on page ??. 12.

(a) One approach is to note that lim+ sign(x) = 1 but lim− sign(x) = x→0

−11. This implies that the ordinary limit can’t exist.

x→0

(b) If a 6= 0 we have sign(x) = sign(a) for x in some interval about a, and so it follows that lim sign(x) = sign(a). x→a

13. By the continuity hypothesis, f (0) = limn→∞ f (1/n). Since all terms of the sequence f (1/n) are positive, the limit can’t be negative. (One can argue this more formally using general properties of sequences. Or see Problem ??, page ??.) 14. Note that limx→3 f (x) = 5. Let  = 0.001. If we choose any δ > 0 that works for this , we’re done. 15. Note that limx→3 f (x) = f (3). Let  = f (3) − 5. If we choose any δ > 0 that works for this , we’re done. ( x if 0 ≤ x ≤ 1 16. The function f has formula f (x) = . Thus, −2x + 3 if 1 ≤ x ≤ 2 lim+ f (x) = 0 = f (0), and so f is continuous at the left endpoint. x→0

(Note the one-sided limit.) At x = 1 we have both lim f (x) = 1 = x→1−

lim+ f (x) = f (1), so f is continuous at x = 1, too.

x→1

17.

(a) The discontinuity of f (x) = 1/x at a = 0 is not removable; lim f (x) x→0 does not exist. (b) The discontinuity is removed if we set f (−2) = −4 = lim f (x). x→−2

(c) The discontinuity is removed if we set f (0) = 0 = lim f (x). x→0

18.

(a) No big deal; note the sharp points. (b) The point is that |f (x) − g(x)| is either f (x) − g(x) or g(x) − f (x), depending on which of f (x) and g(x) is larger. (c) By (b), both max{f, g} and min{f, g} are built up from sums, differences, and the absolute value—all continuous ingredients.

19. We need to show that f (c) = 0 if c is any irrational number. Recall (why?) that there is a sequence {rn } of rationals such that rn → c; note that f (rn ) = 0 for all n. Continuity of f at c requires that f (rn ) → f (c). In other words, f (c) = lim f (rn ) = lim 0 = 0. n→∞

n→∞

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20.

(a) Note that g(0) = 0; we need to show limx→0 g(x) = g(0) = 0. To do this, first choose M > 0 so that | f (x) | ≤ M for all x ∈ (−1, 1). Then we have 0 ≤ |g(x)| = |x| |f (x)| ≤ M |x| . Since the right-most quantity tends to zero as x → 0, so must g(x). (b) The function I(x) = x is clearly continuous at x = a (and everywhere else, for that matter). If f is continuous at x = a, then so (by Proposition ??) is the product function g(x) = f (x)I(x). For the converse, suppose that g is continuous at x = a, for a 6= 0. Then f (x) = g(x)/I(x) is the quotient of two functions, each continuous at a, with nonzero denominator at x = a. By Proposition ??, f is continuous at x = a.

3.3 1.

Why Continuity Matters: Value Theorems (a) There are many possibilities; one has the tent-shaped graph through (0, 0), (1, 1) and (2, 0). (b) If f (0) = f (1), f (2) = f (1), or f (0) = f (2) we’re done. We might as well assume, then, that either f (0) < f (2) < f (1) or f (2) < f (0) < f (1). In the former case, by the IVT, there exists c with 0 < c < 1 and f (c) = f (2), so indeed f is not one-to-one. The remaining case is similar.

2.

(a) p(x) = x(x2 + 1)2 is one possibility (b) p(x) = x2 (x − 1)(x − 2) is one possibility (c) p(x) = (x2 + 1)21 is one possibility (d) p(x) = x2 (x − 1)(x − 2) · · · (x − 40) is one possibility

3.

(a) First write p(x) = xn + an−1 xn−1 + · · · + a1 x + a0  an−1 an−2 a1 a0  = xn 1 + + 2 + · · · + n−1 + n . x x x x Now as x → ∞ the quantity in parentheses tends to 1 and so the product tends to ∞. (b) Let M = 0. Then there exists N so that f (x) > M = 0 whenever x > N . Thus if b is any number with b > N , we have f (b) > 0.

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(c) Numerical experiments reveal that q(−1) < 0, q(0) = 1, q(1) = 0, q(1.5) < 0, and q(2) > 0. This means, by the IVT, that there must be a root between −1 and 0 and another between 1.5 and 2. Factoring gives q(x) = (x − 1)(x2 + x + 1)(x2 − x − 1), from which we can find the real roots exactly. 4.

(a) The EVT says nothing about maximum and minimum values in this case. (b) We explained above that every odd-degree polynomial tends to ±∞ as x → ±∞. (c) Let q(x) have even degree. Then either (i) q(x) → ∞ as x → ±∞, or (ii) q(x) → −∞ as x → ±∞. Assuming case (i), let’s show q achieves has a minimum. Let q(0) = a. Then (i) implies that we can find N > 0 with f (x) > a for x > N and we can find n < 0 with f (x) > a for x < n. Now look at the interval I = [n, N ]. The EVT guarantees f achieves a minimum value f (c), so that f (x) ≥ f (c) for all x ∈ I. By construction that minimum value must apply to x∈ / I, too, since for such x we have f (x) > a ≥ f (c). Thus f (c) is the desired minimum. A similar argument shows that f achieves a maximum in case (ii). (d) An even-degree polynomial can achieve both a maximum and a minimum value on R only if the polynomial is constant.

5.

(a) The minimum value of p on [0, 3] is p(0) = 0. The EVT guarantees that p must assume such a minimum value. (b) The minimum value of q on (0, 3] is q(3) = 0. The EVT doesn’t apply here, since (0, 3] is not a closed interval. (c) Factoring gives p(a) − 4 = a(3 − a)2 − 4 = (a − 4)(a − 1)2 . This shows that p(1) = 4 and p(a) − 4 ≤ 0 when 0 ≤ a ≤ 3. In other words, p(1) is the maximum value.

6. Apply Proposition ?? to the odd-degree polynomial Q(x) = P (x) − π. 7.

(a) f (x) = |2x − 1| is one possibility sin(2πx) + 1 is one possibility. (b) f (x) = 2 (c) No such function exists. A continuous function on [0, 1] must achieve a minimum value.

8. The graph of f is an upward-opening parabola with vertex at x = 1. Thus f (x) takes the minimum value f (1) = −1 at x = 1. No maximum value is achieved on (0, 3). But note that f (−1) = 3 is the supremum of the set {f (x) | x ∈ (0, 3)}.

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9

(a) f (x) = −(x − 0.5)2 is one possibility. (b) No such function g can exist. Note that we have f (a) = e < 3 < π = f (b) for some a and b in [0, 1]. Thus 3 is an “intermediate value” and so must be attained for some c ∈ [0, 1]. (c) p(x) = −3 + x2 is one possibility. (d) No such function q can exist. If q has odd degree then q is unbounded above and below. If q has even degree then q may be bounded below, but in this case q achieves a minimum value rather than just approaching a minimum. (e) f (x) = ex − 3 is one possibility.

10. The function f is not onto because, for instance, f (x) = 43 is impossible. To see f is not one-to-one, choose any numbers a and b with a < π < c. If f (a) = f (π), f (b) = f (π), or f (a) = f (b) then clearly f is not one-toone. Thus we have either f (a) < f (b) < f (π) or f (b) < f (a) < f (π). In the first case, by the IVT, we must have f (c) = f (b) for some c with a < c < π. The other case is similar. 11. Suppose f attains a maximum value at some c ∈ (a, b). If f (c) = f (a) or f (c) = f (b) then we’re done already, so assume that f (c) > f (a) and f (c) > f (b). Let v be any number less than f (c) but greater than both f (a) and f (b). (There are infinitely many possibilities.) By the IVT, f (d) = v for some d ∈ (a, c), and f (e) = v for some e ∈ (c, b). Thus f (d) = f (e), so f is not one-to-one. 12. Suppose toward contradiction that f is not strictly monotone. Then there exists points a, b, and c in I with a < b < c but either (i) f (a) < f (b) and f (b) > f (c); or or (ii) f (a) > f (b) and f (b) < f (c). Both (i) and (ii) lead to contradictions. In case (i), let v be any number such that both f (a) < v < f (b) and f (b) > v > f (c). By the IVT we have f (d) = v for some d between a and b, and also f (e) = v for some e between b and c. This contradicts the assumption that f is one-to-one. 13. Suppose toward contradiction f (2) ≤ f (1). If f (2) = f (1) or f (2) = f (0), then f is not one-to-one, a contradiction. Thus we have either f (0) < f (2) < f (1) or f (2) < f (0) < f (1). In the former case, by the intermediate value theorem there exists c with 0 < c < 1 and f (c) = 2, which again contradicts the one-to-one property. The remaining case, f (2) < f (0) < f (1) is similar. Alternatively, just note that Problem ?? says that f is strictly monotone, and therefore strictly increasing.

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14.

(a) One approach is to modify the proof given for the PoPI. Another approach is to define g(x) = −f (x), which is also continuous at x = c. Now f (x) > K is equivalent to g(x) < −K. Applying the (alreadyknown) PoPILTV to g gives the desired result. (b) We have g(0) = 0 < 1, but there is no two-sided input interval (−δ, δ) on which g(x) < 1. Thus g fails the PoPIGTV. On the other hand, if we have g(0) = 0 > K, then g(x) > K for all x, so g satisfies the PoPILTV. (c) The function h satisfies the PoPILTV but fails the PoPIGTV (consider K = 0.5, for instance). (d) The “only if” part has already been shown. For the “if” part, let  > 0 be given. Then f (c) < f (c) + , and so the PoPIGTV gives δ1 > 0 such that f (x) < f (c) +  for x ∈ (c − δ1 , c + δ1 ). By the PoPILTV there is δ2 > 0 such that f (x) > f (c) −  for x ∈ (c − δ2 , c + δ2 ). Setting δ to be the smaller of δ1 and δ2 completes the proof.

15. If f is not constant then the range of f is an interval. But every interval contains both rational and irrational numbers. 16. (a) With δ = 1 we can use M = 432 , since |f (x)| = x2 ≤ 432 when x ∈ (41, 43). (Other values of δ and M are possible, too.) (b) Let  = 1. Since f is continuous at c there exists δ > 0 such that x ∈ (c − δ, c + δ) implies f (x) ∈ (f (c) − 1, f (c) + 1). In particular, for these x we have |f (x)| ≤ |f (c)| + 1, so we can take M = |f (c)| + 1. ( 1 if x ∈ Q . This example works be(c) One possibility is f (x) = 0 if x ∈ /Q cause f is not continuous anywhere. (d) One possibility is the function defined by f (x) = 1/x if x 6= 0 and f (0) = 42.

3.4 1

Uniform Continuity (a) For given  > 0 we can use any positive δ, such as δ = 1. (b) For any s and t, we have |g(s) − g(t)| = |2s + 7 − 2t − 7| = 2 |s − t|. This implies that for any given  > 0 the value δ = /2 works.

2.

(a) For any  > 0, the value δ = 1 works. (Would larger values of δ, say δ = 700, also work?) (b) If A = 0 the function is constant. For A 6= 0 imitate the proof in Example ??.

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3.

4.

(a) Let  > 0. The value δ =  “works” because if |x − y| < δ then |f (x) − f (y)| ≤ |x − y| < δ = . 1 1 . Work with a common (b) We have |f (x) − f (y)| = 2 − x + 1 y2 + 1 denominator inside the absolute value to complete the proof. (a) Note that |h(s) − h(t)| = s2 + 7 − t2 − 7 = s2 − t2 = |s + t| |s − t|. Now for s and t in [−100, 42] we have |s + t| ≤ 200, so |h(s) − h(t)| ≤ 200 |s − t|. This implies that for any given  > 0 the value δ = /200 works. (b) Let  = 1 and let δ > 0 be any positive number. Let’s see that δ, no matter how small, does not work. To do so, choose any integer n with 1/n < δ, and consider the numbers s = 1/n and t = 1/(n + 1). Then |s − t| < δ but |l(s) − l(t)| = 1. Thus, δ doesn’t work.

5. Show first that for x and y in [1/4, 10], |f (x) − f (y)| = 5

|x − y| ≤ 90|x− xy

y|. Then, for given  > 0 we can set δ = /90. 6.

(a) Given  > 0, set 0 = /137. Since f is uniformly continuous there is a corresponding δ > 0 that works for f and 0 —i.e., if x and y are in I and |x − y| < δ, then |f (x) − f (y)| < /137. It’s not hard to see that this same δ works for g. (b) For given  > 0 there is a corresponding δ > 0 that works for f . This same δ works for h, since if x and y are in I and |x − y| < δ, then | h(x) − h(y) | = | |f (x)| − |f (y)| | ≤ | f (x) − f (y) | ≤ δ.

7. In part (c), values of δ that work for f at c = 1, c = 10, c = 100, and c = 1000 are δ ≈ 0.414, δ ≈ 0.05, δ ≈ 0.005, and δ ≈ 0.0005, respectively. The fact that these δ’s decrease to zero means they can’t be chosen “uniformly” in R. On the other hand, δ = 0.0005 works for  = 1 on the entire interval [1, 1000]. The difference for g is that δ = 0.1 works for all of c = 1, c = 10, c = 100, and c = 1000, and would work for all c. This is what uniform continuity on R means. 8. Let  > 0 be given. Since f is uniformly continuous on J, there exists δ > 0 so that if s and t are in J and |s − t| < δ, then |f (s) − f (t)| < . The same condition holds if s and t are in I, since then s and t are also in J. In other words, for given  > 0, any δ that works for J automatically works for I, too. 9.

(a) Start by squaring both sides of the desired inequality.

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(b) For given  > 0 it works to set δ = 2 .√If 0 < x − y < δ then √ √ √ |f (x) − f (y)| = x − y ≤ x − y < 2 = . 10. First find a constant K such that for all x and y in R, |f (x) − f (y)| ≤ K|x − y|. 2 11. (a) Note that if a and b are in [−3, 42), then a2 + ab + b2 ≤ 33 · 423 = 5292. The hint implies, therefore, that |f (s) − f (t)| = s − t = 5292 |s − t|. This implies that for any  > 0 the value δ = /5292 works in the definition of uniform convergence on [−3, 42). (b) The function f (x) = x3 is continuous on the closed and bounded interval [−3, 42], and is therefore also uniformly continuous on the same interval, by Theorem ??. Since [−3, 42) ⊂ [−3, 42], f is also uniformly continuous on the smaller interval. 12. Let a ∈ I; we’ll show f is continuous at a. Set  > 0 and choose δ > 0 as in the definition of uniform continuity. This δ clearly works for  in the sense of ordinary continuity, since if x ∈ I and |x − a| < δ, then |f (x) − f (a)| < , as desired. 13.

(a) For given  > 0. Choose δ1 and δ2 such that, for x, y ∈ I,  ; 2  =⇒ |g(x) − g(y)| < . 2

|x − y| < δ1 =⇒ |f (x) − f (y)| < |x − y| < δ2

Now show that δ = min(δ1 , δ2 ) works for f + g. (b) One of many possibilities is f (x) = x = g(x) on I = R. 14.

(a) We know from earlier work that the product f g is continuous on I; because I is also closed and bounded, f g must also be uniformly continuous, by Theorem ??. (b) Suppose |f (x)| ≤ Mf and |g(x)| ≤ Mg for x ∈ I. The first step is a flashy calculation: |f (x)g(x) − f (y)g(y)| = |f (x)g(x) − f (y)g(x) + f (y)g(x) − f (y)g(y)| adding and subtracting

≤ |f (x)g(x) − f (y)g(x)| + |f (y)g(x) − f (y)g(y)| triangle inequality

= |g(x)| |f (x) − f (y)| + |f (y)| |g(x) − g(y)| ≤ Mg |f (x) − f (y)| + Mf |g(x) − g(y)| using the bounds

From this point we can mimic the proof in (a).

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15.

(a) For given  > 0 the number δ = /K works in the definition. (b) If A = 0 then L is constant, and K = 0 works. Otherwise, for given  > 0 the number δ = /|A| works. (c) The ratio in question is bounded by K. √ g(x) − g(0) √ = x = √1 blows up (d) For g(x) = x, the ratio x−0 x x + as x → 0 . By the preceding part, g is not Lipschitz continuous on [0, 1].

16.

(a) Mimic Example ??. √ √ x − y (b) For x, y ∈ [1, ∞), we have |f (x) − f (y)| = x − y = √ √ ≤ x + y |x − y|. This implies that for any given  > 0 the value δ =  works. 1 1 1 1 , , , . . . is Cauchy, but applying h gives the (c) The sequence , π 2π 3π 4π sequence 1, −1, 1, −1, . . . , which is not Cauchy. (d) If we set k(0) = 0, then k becomes continuous on the closed interval [0, 1], and so Theorem ?? applies.

17.

(a) The function f must be constant (and hence uniformly continuous) on R. (b) The function f need only be bounded (and not necessarily uniformly continuous) on R. (c) The function f must be constant (and hence uniformly continuous) on R. (d) For some δ, values of f can’t rise or fall value by more than 1 on intervals of length δ. Such a function could have jumps, and so need not even be continuous.

18.

(a) Proposition ?? guarantees that the output sequence is Cauchy, and therefore convergent to some number, which we call f (0). (b) The number n0 can be chosen because the sequence {f (1/n)} converges to f (0). The remaining key step is the triangle inequality:     1   1 |f (x) − f (0)| ≤ f (x) − f + f − f (0) < + . n0 n0 2 2

3.5

Topology of the Reals

1. By De Morgan, the intersection of two closed sets is the complement of the union of two open sets.

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(a) U1 ∪ U2 ∪ U3 ∪ . . . is the (gigantic) union of all the intervals that make up the various Ui . (b) Since Ai is closed for each i, we can write Ai = R \ Ui , where Ui is open. By De Morgan, we have R\(A1 ∩ A2 ∩ A3 ∩ . . . ) = (R \ A1 )∪(R \ A2 )∪· · · = U1 ∪U2 ∪. . . . We showed in (a) that the right side is open; thus, A1 ∩ A2 ∩ . . . is closed, as desired.

3.

(a) Let Ui = (−1 − 1/i, 1 + 1/i). (b) Let Ai = [−1 + 1/i, 1 − 1/i].

5. 1.

(a) The value  = 0.0042 works. (b) The value  = p works. (c) Suppose U is open and p ∈ U . Since U is a union of open intervals, we must have p ∈ (a, b) ⊆ U for some open interval (a, b). Since a < p < b we can take  to be the smaller of p − a and b − p.

6.

(a)

7. Let  > 0 be given, and consider the open interval U = (a − , a + ). Our assumption means that there is a largest N for which aN ∈ / U . This N works in the usual definition of sequence convergence. 9.

(a) All points in (0, 1) are interior to [0, 1]. (b) Every point p in U has N (p) ⊂ U for some  > 0. (c) A finite set contains no open intervals. (d) No points of Q are interior, because Q contains no intervals.

11.

(a) x ∈ f −1 ( R \ S ) ⇐⇒ f (x) ∈ / S ⇐⇒ x ∈ R \ S (b) Suppose f : RtoR is continuous, and A ⊆ R is closed. Then U = R \ A is open, so f −1 (U ) is open. By the previous part, f −1 (A) = R \ f −1 (U ), which is closed. The converse is similar.

13. 12.

(a) As we know, every interval (a, b) contains both rationals and irrationals. (b) The set S is not dense in R. The interval (0, .0005), for instance, has empty intersection with S.

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14. A sketch is given in the text; only minor details are needed. 15. Proofs are routine. Show that every element of the right-hand set is an element of the left-hand set, and vice versa. 16.

(a) f −1 ( (0, 1) ) = ∅, which is open. (b) Let U = (−.5, .5). Then U is open but f −1 (U ) = {1}, which is not open. (c) Let U = (−.5, .5). Then U is open but f −1 (U ) = {1}, which is not open. (d) A = {1} works.

17.

(a) (0, 1) = B0.5 (0.5) (b) R = B1 (0) ∪ B2 (0) ∪ B3 (0) ∪ . . . (c) R \ Z = B0.5 (0.5) ∪ B0.5 (−0.5) ∪ B0.5 (1.5) ∪ B0.5 (−1.5) . . . (d) (0, ∞) = B1 (1) ∪ B2 (2) ∪ B3 (3) ∪ . . .

18.

(a) All parts of Definition ?? are easily checked, including the triangle inequality. (b) B1 (p) = {p}; B2 (p) = R. (c) Every point is open, and so every set (a union of points) is open.

19.

(a) All parts of Definition ?? are easily checked, including the triangle inequality. (b) B1 (p) = (p − 1, p + 1); B2 (p) = R. (c) This topology has the same open sets as with the usual metric.

3.6 1.

Compactness (a) Each xi is in some Ui , so it takes at most n such Ui to cover F . (b) We can cover R with open sets Ui = (−i, i) for i ∈ N. A finite subcollection covers only a bounded subset of R. (c) It takes zero open sets to cover ∅; that’s certainly finite!

2. Cover Z with Ui of the form (k − 0.1, k + 0.1) for integers k ∈ Z. No finite subcover suffices. 3.

(a) Cover each term 1/k with a small open interval that contains no other term of the sequence. (b) The set {an } is compact in this case, because it contains its own limit.

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(c) Many sequences are possible. One is 0, 1, 0, 1, . . . .

5. Find a linear (therefore certainly continuous) function f : R → R that maps 1. [0, 1] onto [a, b]. 6.

(a) x0 = 1 is the closest point. (b) Both points in K = {π − 1, π + 1} are closest to p = π. Three closest points are impossible. (c) There is no closest point to p = π in [0, 1). That’s OK because [0, 1) is not compact. (d) 3 is the closest point in Z to p = π. There is no closest rational number to π (or to any irrational, for that matter).

7. The Ui are open because they are inverse images of the open intervals (−i, i). The Ui cover all of R, and so all of K. Compactness of K means that we need only finitely many Ui to cover K. This amounts to boundedness of f on K. 8. The point is that every closed and bounded interval [a, b] is also compact. 9. Proposition ?? says that the image of a compact set is compact. 10. Yes, it’s possible. f (x) = sin(100x) is one example.

4.1

Defining the Derivative

1. The difference quotients for g 0 (a) and for f 0 (a) are identical, so their limits are equal. 2. The difference quotient at x = a is itself constant, with value B. Thus L0 (a) = B for all a. 3.

(a) Draw your own picture. (b) The difference quotient calculations are straightforward. (c) The graphs of f and g are reflections of each other in the line y = x. The same is true of the respective tangent lines. In particular, the two tangent lines have reciprocal slopes.

4.

(a) f 0 (1) = 0 (b) g 0 (1) = 3 (c) h0 (1) = 42

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(d) k 0 (1) = −

1 16

(e) m0 (1) = 0 + 3 + 42 − 5.

1 16

(a) Calculation with the difference quotient gives f 0 (0) = 0. (b) Calculation with the difference quotient gives f 0 (c) =

6.

(c2

2c . + 5)2

(a) Calculate carefully with the difference quotient; use a common denominator. (b) Part (a) is the base step for induction; the product rule drives the inductive step.

7.

f (x) − f (0) is strictly positive for all x, and x−0 0 so f (0) must be nonnegative.

(a) The difference quotient

(b) We have −1 ≤ 8.

f (x) − f (0) ≤ 1 for all x, and so −1 ≤ f 0 (0) ≤ 1. x−0

f (x) − f (0) is nonnegative for x > 0, so the x−0 right-hand limit is nonnegative. Since the two-side limit exists, it too must be nonnegative. f (x) − f (0) (b) We have f 0 (0) = lim− . Since the difference quotient is x−0 x→0 nonpositive, so is its limit. (a) The difference quotient

(c) If f has a minimum value at x = 0, then f 0 (0) = 0. f (x) f (x)g(x) = lim g(x) lim g(x) = x→0 x→0 x x 0 g(0)f (0) = 0. Both limits exist by hypothesis.

9. Definition ?? gives (f g)0 (0) = lim

x→0

10. Definition ?? gives (f g)0 (0) = lim x→0 6.

f (x)g(x) f (x) = lim g(x) = g(0)f 0 (0) = x→0 x x

11. The limit-of-difference-quotient calculation for g 0 (2) is like that for f , except for a factor of 4. Note that the g-graph is formed by vertical stretching of the f -graph. 12.

(a) The g-graph comes from shifting the f -graph 4 units left and 5 units up. Slopes don’t change in the process. (b) Replacing x with y − 4 turns the difference quotient for g 0 (−2) into the difference quotient for f 0 (2).

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13.

(a) It’s easy to show that both lim f (x) = 0 and lim f (x) = 0; thus x→0−

x→0+

lim f (x) = 0 = f (0), which means that f is continuous at 0.

x→0

(b) Yes, f is differentiable at x = 0. Again, we have lim− x→0

f (x) − f (0) = x−0

f (x) − f (0) lim = 0. x−0 x→0+ 14.

(a) The easy one-sided limits lim− f (x) = 1 = lim+ f (x) imply lim f (x) = x→1

x→1

x→1

1 = f (1), and so f is continuous at x = 1. The limit that defines f 0 (1) does not exist because the corresponding one-sided limits are unequal. (b) Set f (x) = 2x − 1 for x ≥ 1. Then f 0 (1) = 2. f (x) . This limit exists, and x→0 x f (x) is zero, by squeezing: |f (x)| ≤ x2 implies − |x| ≤ ≤ |x|. Since the x right and left quantities tend to zero, so must the middle.

15. Note first that f (0) = 0. Thus, f 0 (0) = lim

16.

(a) Since g(x) = 0, we need to show that g(x) → 0 when x → 0. Boundedness of f means that −M ≤ f (x) ≤ M for some M > 0. Therefore, when −1 < x < 1, we have −M |x| ≤ g(x) ≤ M |x|. By the squeeze principle we have limx→0 g(x) = 0. (b) If f is the “sign function” (see Example ??, page ??); then g(x) = |x|. Other functions f are possible, too. (c) The difference quotient limit that defines h0 (0) is the limit treated in (a).

17. We need to show that lim x sin(1/x) = 0. Exactly this limit is discussed x→0

in Example ??, page ??. g(x) cos x = lim x = x→0 3 + cos x x 0. For this we can either use a squeezing argument or the fact that cos x is continuous at x = 0.

18. Note that g(x) = 0, so we need to show that lim

x→0

19.

(a) This is essentially the Principle of Persistent Inequalities. Assume toward contradiction that f (0) < 42. Then, says the PoPI, for some δ > 0 we have f (x) < 42 for x ∈ (−δ, δ), contradicting our assumption. h(x) (b) Clearly h(0) = 0. Now h0 (0) = lim = lim g(x) = 42. (The x→0 x→0 x last equality holds because g is continuous.)

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20. 21.

(a) Claim (iv) of Lemma ?? says that since f 0 (0) = 100 > 0, we must have f (x) < 0 on (−δ, 0) and f (x) > 0 on (0, δ) for suitable small δ. (b) δ = π/100 ≈ 0.03 works in this case.

4.2

Calculating Derivatives


1. One can show that xn −an = (x−a) xn−1 + axn−2 + · · · + an−1 ; there are n summands. Taking the limit as x → a gives nan − 1 as desired. 2.

0

0

(a) By the ordinary product rule, (f gh) = (f (gh)) = f 0 (gh)+f (gh)0 = f 0 gh + f ( g 0 h + gh0 ) = f 0 gh + f g 0 h + f gh0 —a nice pattern. (b) Iterating the usual chain rule gives ( f ◦ g ◦ h )0 (a) = f 0 ( (g ◦ h)(a) ) · (g ◦h)0 (a) = f 0 ( g(h(a) )·g 0 (h(a))·h0 (a). Here’s an even more compact version, omitting the “argument” a: ( f ◦g◦h )0 = ( f 0 ◦ (g ◦ h) ) · ( g 0 ◦ h ) · h0 (c) Finding the second derivative involves both the chain (twice) and the 0 00 product rules: ( f ◦ g ) (a) = ( f 0 (g(a)) · g 0 (a)) = ( f 0 (g(a)) )0 · g 0 (a) + f 0 (g(a)) · g 00 (a) = f 00 (g(a)) · g 0 (a) · g 0 (a) + f 0 (g(a)) · g 00 (a). 0

(d) Using first the chain and then the product rule gives ( f ◦ (gh) ) (a) = f 0 (gh(a)) · (gh)0 (a) = f 0 (gh(a)) · ( gh0 (a) + g 0 h(a) ), or, alternatively, f 0 (g(a)h(a)) · g(a)h0 (a) + f 0 (g(a)h(a)) · g 0 (a)h(a). Leaving out the arguments give a more cryptic form: ( f ◦ (gh) )0 = f 0 ◦ (gh) · gh0 + ( f 0 ◦ (gh) ) · g 0 h. (e) If f (x) = Ax+B and g is any function, then (f ◦g)0 (a) = Ag(x)+B. To find the derivative (it’s Ag 0 (x)) we can use either the chain rule or the linear combination rule. h(x) − h(a) . We can evaluate this x−a using the definition of h from f and g and algebraic properties of the limit:

3. The derivative h0 (a) is defined by lim

x→a

lim

x→a

3f (x) − 42g(x) − 3f (a) + 42g(a) h(x) − h(a) = lim x→a x−a x−a f (x) − f (a) g(x) − g(a) = 3 lim − 42 lim = 3f 0 (a) − 42g 0 (a). x→a x→a x−a x−a

4. See the preceding solution, but work with coefficients A and B rather than 3 and −42. 5.

(a) The chain rule says that if h(x) = 1/g(x), then h0 (a) = −g 0 (a)/g(a)2 .

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(b) Applying the product rule to f (x)/g(x) = f (x) · h(x) gives the derivative f 0 (a) · h(a) + f (a)h0 (a) = f 0 (a)/g(a) − f (a)g 0 (a)/g(a)2 , which is (after a bit of algebra) seen to be the desired expression. 6. One approach is to recall that (i) cos x = sin(x + π/2); and (ii) − sin x = cos(x + π/2). Differentiating both sides of (i) and then using (ii) gives (cos x)0 = cos(x+π/2) = − sin x. The remaining trigonometric functions are quotients or reciprocals of sines and cosines. 7. All parts are straightforward—but perhaps messy—calculations. 8.

(a) The two vertical asymptotes are at places (x = −1 and x = 0) where the derivative fails to exist. (b) The number 1 is a root of both the numerator and the denominator. Cancelling x − 1 from both numerator and denominator and simplifying algebraically gives the new formula q(x) = 3x4 + 3x2 + 3 + 2 2 − —which shows more clearly how the asymptotes arise. x+1 x (c) From the second formula we get q(1) = 8. The new formula is obviously differentiable at x = 1 since all its parts are.

9. Clearly, g(a) = f (a)2 = 0. Also the chain rule gives g 0 (a) = 2f (a)f 0 (a) = 0. 10.

(a) That (f ◦ g)0 (n) = 0 for all integers n follows immediately from the chain rule. (b) The product rule gives h0 (x) = − sin (πx) · π · g(x) + cos (πx) · g 0 (x); this is clearly zero for each integer n.

11.

(a) Notice first that f is continuous at x = 17, by Theorem ??, page ??. We also know, or can easily prove, that h(x) = |x| is continuous everywhere. Since g(x) is the composition h(f (x)), we know that g(x) must also be continuous. (b) If f (x) = x − 17, then g(x) = |x − 17|, which is clearly not differentiable at x = 17. (c) Note that g(x)2 is really the same thing as f (x)2 , which is the product of two differentiable functions, and therefore differentiable.

12. Parts (a) is a simple calculation. In (b) the function vanishes to order 3. For (c), use the product rule for various derivatives up to 5, and note that each summand has a factor that vanishes at x = a.

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4.3

The Mean Value Theorem

1. If a and b are any two roots of f , then Rolle’s theorem implies that f 0 (c) = 0 for some c between a and b. Our claim follows. If f (x) = 2+sin x, then f has no real roots, but f 0 (x) = cos x has infinitely many roots. 2.

(a) Note that p(x) = x3 + x = x(x2 + 1). The first factor is zero at x = 0 and the second is always positive. (b) If p had two roots a and b then, by Rolle’s theorem, there would be some c with p0 (c) = 0. But p0 (x) = 3x2 +1, which is always positive. (c) If q(x) = x13 + x11 + x9 + x7 + x5 + x3 + x, then clearly q(0) = 0, so there’s at least one root. If there were two roots then, as in the preceding part, there must also a root of q 0 . This is impossible, since q 0 (x) > 0 for all x. (d) Every polynomial of odd degree has at least one root. The argument in the preceding part shows that there can’t be two roots.

3.

(a) Every odd-degree polynomial has at least one root; we’ve shown this using the intermediate value theorem. (b) There are many possibilities. The functions f1 (x) = x3 (with a = b = 0), f2 (x) = x3 − 3x + 2 (with a = −3 and b = 2), and f3 (x) = x3 − 3x + 1 (with a = −3 and b = 1) have one, two, and three roots, respectively. (c) Rolle’s theorem implies that between any two distinct roots of f lies a root of f 0 . If f had at least 4 roots, f 0 (a quadratic polynomial) would have at least 3 roots, which is impossible. (d) We know f has at least one root. If a > 0 then f 0 (x) = 3x2 + a, so f 0 (x) > 0 for all x. Thus f 0 has no roots, and so f can have at most one root.

4. The claim is clearly true for n = 1. Assume it’s true for polynomials of degree n − 1, and suppose p has degree n. If p had n + 1 roots then, by Rolle’s theorem, there would be at least one root of p0 between any two successive roots of p. Thus p0 would have at least n roots, which violates the induction hypothesis. 5.

(a) Rolle’s theorem gives x1 between a and c and x2 between c and b. (b) Rolle’s theorem applied to f 0 on the interval [x1 , x2 ] gives the desired x0 .

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(c) If f : [a, c] → R has continuous derivatives f 0 , f 00 , . . . , f (42) on [a, c], and there are inputs x1 , x2 , . . . , x43 with f (x1 ) = f (x2 ) = · · · = f (x43 ), then the desired x0 exists. 6. Consider g(x) = f (x) − x. Then g(0) = g(1) = g(2). By the preceding problem, g 00 (x0 ) = 0 for some x0 ∈ (0, 2). Since g 00 (x) = f 00 (x), we’re done. 7. Rolle’s theorem implies that f 0 has a root in each interval (0, a), (a, 2a), (2a, 3a), . . . . Since f 0 is also periodic with period a, (f 0 )0 = f 00 also has roots in each of these intervals, and similarly for higher-order derivatives. 8.

f (b) − f (a) = f 0 (c) > M for all a and b. b−a If b > a we can multiply both sides by (b − a) to get the desired inequality.

(a) By the MVT we have

(b) Applying the previous part to f 0 , we have f 0 (x) − f 0 (0) = f 0 (x) > 2x when x > 0. Now consider g(x) = f (x) − x2 . Then we have g 0 (x) = f 0 (x) − 2x > 0 when x > 0. Applying the previous part to g gives g(x) > 0 when x > 0, which is what we wanted to show. 9.

(a) Every number c in (1, 7) works. (b) The only solution is c = 4. (c) In this case the MVT equation f 0 (c) =

f (b) − f (a) reduces to 100c99 = b−a

1 ≈ 0.955. 1001/99 f (b) − f (a) (d) In this case the equation f 0 (c) = reduces to 6c + 5 = b−a 3a + 3b + 5, which gives c = (a + b)/2, the midpoint of (a, b). 1, which gives c =

(e) If q(x) = Ax2 +Bx+C and [a, b] is any interval, then the mean value equation holds if and only if c = (a + b)/2. To see why, note that in f (b) − f (a) this case the equation f 0 (c) = reduces to B + 2Ac = b−a B + A(a + b). 10.

(a) No value of c works; f is not differentiable at 0. (b) The value c = 5 works. (c) No value of c works; f is not differentiable at 3.

11. If a car has the same velocity at times a and b, then there must be some intermediate time c at which the acceleration is zero.

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12. Here’s one possibility: Define f : [0, 1] → R by f (0) = 1 and f (x) = 0 for x 6= 0. 13.

(a) Consider h(x) = f (x) − g(x). By hypothesis, h0 (x) = 0 for all x ∈ (a, b). By Proposition ??, h is constant, as desired. (b) Let h(x) = f (x) − g(x) − 5x. Then h0 (x) = 0 for all x ∈ (a, b), and so h(x) = C for some constant C. It follows that f (x) = g(x) + 5x + C.

14.

(a) The C appears because (i) the derivative of the right side is independent of C; (ii) every antiderivative differs from sin x by a constant. (The MVT is involved in (ii).) ( ln |x| + 1 if x > 0 (b) The function f (x) = has derivative 1/x for all ln |x| + 2 if x < 0 x 6= 0, but is not of the form ln |x| + C. This can happen because the functions in question are not defined at x = 0.

15. If we write m = follows. 16.

f (b) − f (a) , then L(x) = m(x − a) + f (a), and the rest b−a

(a) This is straightforward. (b) The proof of (i) can be essentially rerun with strict inequalities. (c) If f were not monotone, then we’d have f (a) = f (b) for some a and b in I. Rolle’s theorem now says there is c ∈ I with f 0 (c) = 0, which contradicts the hypothesis.

17.

(a) This should be straightforward. (b) The proof of (i) can be essentially rerun with strict inequalities. (c) If f were not monotone, then we’d have f (a) = f (b) for some a and b in I. Rolle’s theorem now says there is c ∈ I with f 0 (c) = 0, which contradicts the hypothesis.

18. Suppose f (x) intersects the graph at both x = a and x = b. Then f (a) = a f (b) − f (a) and f (b) = b, so = 1. By the MVT there exists c with b−a 0 f (c) = 1, which contradicts the hypothesis. 19. Suppose f (b) ≥ b for some b > 0. Then the MVT implies that f 0 (c) = f (b) − f (0) f (b) = ≥ 1 for some c between 0 and b; this contradicts our b−0 b assumption.

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20.

(a) Let  > 0 be given; set δ = . If s and t are in D and |s − t| < δ, then |f (s) − f (t)| ≤ |s − t| < δ = . Thus f is uniformly continuous on D. (b) Let x and y be any numbers. By the MVT, we have (it’s OK to take absolute values) |f (x) − f (y)| = |f 0 (c)| |x − y| ≤ |x − y|. (c) The function f (x) = sin x is a contraction, since |f 0 (x)| = |cos x| ≤ 1. The function g(x) is not a contraction. One reason is that g(1) − g(0) = e − 1 ≈ 1.72 > 1 − 0.

21.

(a) The function f is continuous, and so must attain a minimum on the interval [s, t] by the extreme value theorem. The minimum can’t be at either endpoint because f 0 (s) < 0 and f 0 (t) > 0. (b) The function g(x) = f (x) − vx satisifies the conditions in (a), so there is some c between s and t with 0 = g 0 (c) = f 0 (c) − v. Thus f 0 (c) = v as desired.

22. We have f 0 (0) = 0; use a squeezing argument. For other values of x we can use the derivative rules, and simplify, to find that f 0 (x) = 2x sin(1/x) − cos(1/x). This has no limit as x → 0. 23.

(a) All parts follow by plugging t = 0 into f and p2 . (b) That g(0) = 0 = g(1) is an easy calculation. The t1 in question exists by Rolle’s theorem applied to g. (c) That g 0 (0) = 0 = g 0 (t1 ) is an easy calculation. The t2 in question exists by Rolle’s theorem applied to g 0 . (d) As in the previous parts, the claimed t3 exists by Rolle’s theorem applied to g 00 .

24. Very careful calculation is all that’s needed. Notice that h(0) = f (a) and h(1) = f (b).

4.4 1.

Sequences of Functions (a) For any real number x and  > 0, let N = 1/. If n > N then 1 < . Thus fn → f pointwise on R. |fn (x) − f (x)| = fn (x) = n Because the same inequality holds for all x, the convergence is also uniform. (b) For any fixed real number x we have fn (x) = nx → 0 as n → ∞, so fn → f pointwise on R. The convergence is not uniform on R. For  = 1, for instance, no N works: for every n we have |fn (x)−f (x)| = |x/n| > 1 whenever |x| > n.

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(c) For any fixed real number x we have fn (x) = sinn x → 0 as n → ∞. Thus fn → f pointwise on R. The convergence is also uniform; the proof is similar to that in (a). 2.

(a) The convergence is pointwise but not uniform. (b) The convergence is now uniform, because for n > 1000 we have fn (x) = 0 for all x ∈ [−1000, 1000].

3.

(a) Fix a ∈ R and let an = a + 1/n. Note that an → a and that f (an ) = fn (a) for all n. Since f is continuous at a, fn (a) = f (an ) → f (a), as desired. (c) Let  > 0 be given. Choose δ > 0 that works for  in the definition of uniform continuity, and choose an integer N with 1/N < δ. This N works for  in the definition of uniform convergence: if n > N then—for all real x—we have |fn (x) − f (x)| = |f (x + 1/n) − f (x)| < . (The last inequality holds because |(x + 1/n) − x| < δ.)

4. The result is just a restatement of the analogous fact for sequences of numbers. 5. The result is just a restatement of the analogous fact for Cauchy sequences of numbers. 6. Both parts are straightforward exercises in the definitions. In each case, /2 plays a role. 7.

(a) The statement is true. A convergent sequence of integers must eventually be constant. √ (b) The statement is false. The sequence { 2/n} is irrational, but converges to 0. (c) The statement is true by a “squeezing” property of limits.

8.

(a) We need to show that lim+ x→0

fn (x) fn (x) = lim− = 0. For the rightx x x→0

hand limit we have lim+

x→0

x1+1/n = lim+ x1/n = 0, x x→0

as desired, and the left-hand limit is similar.

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(b) For 0 ≤ x ≤ 1 we have |f (x) − fn (x)| = x − x1+1/n . With g(x) = x − x1+1/n , elementary calculus methods (set an appropriate derivative equal to zero) show that g(x) < n1 for 0 ≤ x ≤ 1. In other words, |f (x) − fn (x)| < n1 holds if 0 ≤ x ≤ 1; the same inequality holds for −1 ≤ x ≤ 0 because both f and fn are even functions. 9.

(a) The claim follows from analogous properties of number sequences. 2x 1 + 2 . The right(b) Here is the key identity: (x + 1/n)2 − x2 = n n hand side is large for large |x|.

10. Fix any point x in I. Since fn → f pointwise on I we know fn (x) → f (x). Since the sine function is continuous at f (x), we must have sin (fn (x)) → sin (f (x)). In other words, gn → g pointwise on I. 11. Let  > 0 be given. Since h is uniformly continuous on R we can choose δ > 0 such that |h(s) − h(t)| <  whenever |s − t| < δ. Because fn → f uniformly on I we can choose N such that |fn (x) − f (x)| < δ holds for all x ∈ I whenever n > N . By our choice of δ, this guarantees that |h (fn (x)) − h (f (x))| <  for all x ∈ I, as desired. 12.

(a) N = 3 works here, because f4 (x) approximates ex to within 0.01 for −1 ≤ x ≤ 1. (b) N = 3 works here, because f4 (x) approximates ex to within 1 for −2 ≤ x ≤ 2.

13.

(a) The ratio calculation is slightly messy, but it works. (b) We get f 00 = −f . This combined with f (0) = 0 and f 0 (0) = 1 says that f (x) = sin(x). (c) Differentiate f term by term to find a power series for f 0 . What famous function is this? How do you know? (d) The graphs should resemble the sine and cosine functions, respectively.

14.

|x|n+1 /(n + 1) n+1 = lim |x| = |x|, so the series conn→∞ n→∞ |x|n /n n verges when |x| < 1.

(a) lim

(b) Differentiating g term by term gives the geometric series, which converges to 1/(1 − x) for −1 < x < 1. (c) g(x) = − ln(1 − x) because g 0 (x) = 1/(1 − x) and g(0) = 0. (d) f5 (1/2) = 661/960 ≈ − ln(1 − 1/2).

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5.1

The Riemann Integral: Definition and Examples

1.

(a) Calculation gives RS(f, P, S) = 0 · 0.8 + 1 · 0.2 + 2 · 0.7 + 3 · 0.3 + 4 · 0.9 + 9 · 0.1 = 7.0; kPk is 0.9, the largest width. (b) To make RS(f, P, S) as small as possible, choose left endpoint samples: S = {0, 0.8, 1.0, 1.7, 2, 2.9}. This gives RS(f, P, S) = 0·0.8+ 0.82 · 0.2 + 1 · 0.7 + 1.72 · 0.3 + 4 · 0.9 + 2.92 · 0.1 = 6.136. (c) To make RS(f, P, S) as bad as possible, choose right endpoint samples: S = {0.8, 1.0, 1.7, 2, 2.9, 3}. This gives RS(f, P, S) = 0.82 · 0.8 + 12 · 0.2 + 1.72 · 0.7 + 22 · 0.3 + 2.92 · 0.9 + 32 · 0.1 = 12.404, an error of 3.04. √ (d) The sample set S = { 3} gives RS(f, P, S) = 3 · 3 = 9.

2.

(a) (b) (c) (d) (e)

Sampling at midpoints gives RS(f, P, S) = 8. Samples S = {0, 2, 3, 4} give RS(f, P, S) = 9. If P = {0, 4} and S = {0} then RS(f, P, S) = 0. For P = {0, 4} find a sample set S so that RS(f, P, S) = 8. No two-piece partition P gives RS(f, P, S) = 0; the second summand has to be positive. (f) Choose all partition points except the last one very near zero.

3. The fact that f (x) ≥ 3 for all x ∈ [0, 2] implies that RS(f, P, S) ≥ 6 for every partition P and sample points S. This implies, as in the proof of Proposition ??, that the integral itself cannot be less than 6. 4. Mimic Example ??, page ??. 7. Notice first that the inequality 0 ≤ f (x) ≤ g(x) implies that 0 ≤ RS(f, P, S) ≤ RS(g, P, S) holds for all suitable partitions P and sample sets S. To show f is integrable on [a, b], let  > 0 be given. Since g is integrable, there is some δ > 0 that works for g (and the value I = 0) in Definition ??. The preceding inequality shows that the same δ > 0 works for f . Rb 8. The converse says that if a f ≥ 0, then f (x) ≥ 0 for x ∈ [a, b]. This is false, and it is easy to give counterexamples. 9. Let  > 0 be given. Since I1 satisfies the definition, can choose δ1 > 0 that works for I1 and . Since I2 satisfies the definition, we can choose δ2 > 0 that works for I2 and . Let δ be the smaller of δ1 and δ2 , and calculate any Riemann sum RS(f, P, S) for kPk < δ. Then both I1 and I2 are within  of RS(f, P, S), and so within 2 of each other. Since this holds for all positive , we must have I1 = I2 .

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10. For given  > 0, use Definition ?? to choose a δ that works for /2. Then any two partitions P1 and P2 with norm less than δ have values within /2 of I, and hence within  of each other. 11.

(a) The condition holds by uniform continuity of f on [a, b]. (b) Note that n X |RS(f, P, S) − RS(f, P, T )| = (f (si ) − f (ti )) ∆xi i=1

≤

n X

|f (si ) − f (ti )| ∆xi
0. Now f (x) ≥ m for all x ∈ [a, b], and so Corollary ?? gives Z

b

Z

a

b

m = m (b − a) > 0,

f≥ a

as claimed. (b) Part (a) implies that f (c) ≤ 0 must hold for some c ∈ [a, b]. For similar reasons, f (d) ≥ 0 must hold for some d ∈ [a, b]. If either f (c) = 0 or f (d) = 0, we’re done; otherwise, the intermediate value theorem says that f (e) = 0 for some e between c and d. R1 (c) Let f (x) = 1 for x ≥ 0 and f (x) = 1 for x < 0. Then −1 f = 0, but f (c) 6= 0 for all c. Rb Rb Rb 7. The claim amounts to saying that − a |f | ≤ a f ≤ a |f |. This follows from Corollary ??, and from the fact that − |f (x)| ≤ f (x) ≤ |f (x)| for all x ∈ [a, b]. 8. The integral is zero, since f (x) = 0 for all but finitely many x in [−a, a]. The values sin n are irrelevant. (See Example ??, page ??.) 9.

(a) The function f is integrable on every interval [a, b] because it is built up from integrable functions as allowed by Theorem ??, page ??. R1 (b) A look at the graph shows −1 f = −1. Rn (c) A look at the graph shows −n f = −n.

10. Both parts are elementary calculus-style calculations; the point is that the theorems justify the calculations. 11. All parts are basic elementary calculus calculations. 12. Just fill in details. 13. Suppose f (x) ≤ g(x) holds except at some finite list x1 , x2 , . . . xn of points in [a, b]. If we redefine f at these points (we could set f (xi ) = g(xi ), for instance, without changing the integral) then f (x) ≤ g(x) holds throughout [a, b], and so Corollary ?? gives the desired result.

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5.3

Integrability

1. All of the concocted functions are continuous. 2. A regular partition with 1000 subdivisions works. 3. Here is one possibility: Set h = /100 and use P = {0, h, 1 − h, 1 + h, . . . , 10 − h, 10}. 4. The function f defined by f (x) = 1 if x ∈ Q and f (x) = 0 otherwise works; we get box sum one for every partition P of [0, 1]. R2 7. Given any box sum for 0 f with value less than , add (if necessary) one more point at x = 1 to get another box sum with value less than . This R1 same box sum works for 0 f ; just ignore boxes to the right of x = 1. 8. The function f (x) = cos x is monotone, and hence integrable, on intervals [0, π], [π, 2π], . . . . Theorem ?? lets us glue these intervals together. 10. Given  > 0, choose an integer N > 0 so that 1/N < . Suppose n > m > N . Then In and Im are Riemann sums for partitions Pn and Pm such that kPn k < δn and kPm k < δm . Because we chose the δk to be decreasing, we have both δn ≤ δN and δm ≤ δN . Thus both Pn and Pm have norm less than δN , and so |In − Im | < 1/N < , as desired. 11. Let mi and Mi be the inf and sup of f on [xi−1 , xi ]. For each i, it is clear that mi ≤ f (si ) ≤ Mi , and hence that mi ∆xi ≤ f (si )∆xi ≤ Mi ∆xi . Summing over i gives the desired result. 13. The function fbig is constant on each partition subinterval, and is therefore integrable on [a, b] by Theorem ??, page ??. By construction, moreover, Rb the integral a fbig has the same value as the upper sum US (f, P). It is also clear that fbig (x) ≥ f (x) for all x ∈ [a, b], and so (by Corollary ??, page ??) we have Z b Z b US (f, P) = fbig ≥ f = I, a

a

The proof that LS (f, P) underestimates I is almost identical. 14. The point is to find, for any given  > 0, any partition P of [0, 1] with box sum less than . To do so, start at the left, with x0 = 0 and x1 = /2; this box element contributes /2. Since only finitely many of the 1/n lie to the right of /2, one can choose the remaining partition points so that each remaining “bad point” lies in an interval small enough so that the total length of these intervals is less than /2. Complete this sketch to give a careful proof.

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5.4

Some Fundamental Theorems

1. The first claim follows immediately from the mean value theorem for integrals. Finding examples for the second part is easy. 2. Apply the mean value theorem for integrals to h(x) = f (x) − g(x). 3. The new average value is 5 · 3 + 7 = 22; the average value behaves “linearly.” ≈ 1.886; av ≈ 0.942; c ≈ 0.889. = 2; av = 1; c = 1. = 8/3; av = 4/3; c ≈ 1.155. ≈ 2.046 × 1011 ; av = 1.023 × 1011 ; c ≈ 1.829.

4.

(a) (b) (c) (d)

I I I I

5.

(a) (b) (c) (d)

b = 9/4. b = 2. √ b = 3 ≈ 1.732. b ≈ 1.094.

6. Since H 0 (x) = h(x) ≥ 0, H must be nondecreasing. Also, H(a) = H(b) = 0, so we must have H(x) = 0 for all x. Thus H 0 (x) = h(x) = 0, too. 7. Let M be the maximum value of f . Then h(x) = M − f (x) ≥ 0 for all Rb x ∈ [a, b]. The hypotheses imply that a h = 0; hence h(x) = 0 for all x ∈ [a, b]. The result need not hold if f is not continuous. Consider the function f given by f (0) = 1 and f (x) = 0 on [0, 1]. Rb 8. The hypothesis means that F (b) = 0 f = 3b for all x. Thus F 0 (b) = f (b) = 3 for all b. Rb 9. The hypothesis means that F (b) = 0 f = b2 /2 for all x. Thus f (b) = F 0 (b) = b all b. 10. Let f (x) = Cx + D, and do the calculation. 11. The interval [0, 8/3] is one possibility. Rb 12. By the fundamental theorem we can write a f = F (b) − F (a), where F is an antiderivative of f . Applying the ordinary mean value theorem to F gives, for some c in (a, b), Z b f = F (b) − F (a) = F 0 (c)(b − a) = f (c)(b − a), a

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which is the mean value theorem for integrals. 13.

(a) Note that hn (0) = 0 for all n, so clearly hn (0) → h(0) = 0. If x > 0, then hn (x) = 0 for all x with x < 1/n, so again hn (x) → h(x) = 0. Show that the sequence {hn } converges pointwise to h on [0, 1]. By R1 R1 contrast, 0 hn = 1 for all n, while 0 h = 0. (b) The sequence {hn } converges uniformly to h on [0.1, 1], because R1 hn (x) = 0 for all x ∈ [0.1, 1] whenever n > 10. Similarly, 0.1 hn = 0 for all n > 10, so the integrals converge as well.

15.

(a) The uniform limit function f is continuous on [0, 1] and hence is also integrable there. (b) For given  > 0, choose N such that |fn (x) − f (x)| <  for all n > N . Now Z 1 Z 1 Z 1 Z 1 Z 1 ≤ f − = (f − f ) |f − f | < f  = , n n n 0

0

0

0

0

which completes the proof.

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