Solutions to Exercises

Chapter 1

1.1 If x = min were rational, then x 2 = m 2 /n2 would also be rational. The square of /2+V3 is 5+2V6.1f 5+2V6 = p/q E Q, then V6 = (p-5q)/lOq E Q, and this is not the case. 1.2 AnB = {1,3}, AU B = {1,2,3,4,5,7}, X\ A = {2,4,6,8}, X\ B = {5,6, 7,8}. X\(AnB) = (X\A)U(X\B) = {2,4,5,6, 7,8}, X\(AUB) = (X \ A) n (X \ B) = {6,8}. 1.3 437 = 19 x 23, 493 = 17 x 29. 1.4 a) True. Ifx = min E Q and x+y = p/q E Q, then y = (pn-qm)/qn E Q, a contradiction. b) False. Take x = 0. If we had insisted that x '" °this would have been true. c) False. Take x = ..;2, y = 1 - ..;2. d) False. Take x = y = ..;2. e) False. Take x =/2, y = V3. f) False. Again take x = ..;2, y = V3. 1.5 Take u = x + (l//2)(y - x). 1.6 Let N be a positive integer such that N > 1/(y - x). Then the difference between successive members of the sequence

... , -3/N, -2/N, -l/N, 0, l/N, 2/N, '" 238 Real Analysis

is less than y - x, and so x < q < y for at least one rational number q=M/N. 1. 7 True for n = 1. Let n ~ 1 and suppose true for n. Then 12+...+ (n + 1)2 = (1/6)n(n + 1)(2n + 1) + (n + I? = (1/6)(n + 1)(2n2 + n + 6n + 6) = (1/6)(n + 1)[(n + 1) + 1][2(n + 1) + 1]. 1.8 True for n = 1. Let n ~ 1 and suppose true for n. Then 13+...+(n + 1)3 = (1/4)n2(n+l)2+(n+l)3 = (1/4)(n+l)2(n2+4n+4) = (1/4)(n+l)2[(n+ 1) + IF. 1.9 True for n = 1. Let n ~ 1 and suppose true for n. Then a + (a + d) + ... + (a + nd) = na + !n(n - l)d + a + nd = (n + l)a + Hn(n - 1) + 2n]d = (n + l)a + Hn + 1)[(n + 1) - l]d. 1.10 True for n = 1. Let n ~ 1 and suppose true for n. Then a+ar +... +arn = [a(l-rn)l/(I-r)+arn = [a-arn+arn-arn+l]/(I-r) = [a(l-r n +ll/(I• r). 1.11 If n =1 then the right hand side is (x2 - 2x + 1)/(x - 1)2 = 1. So true for n = 1. Let n ~ 1 and suppose true for n. Then n n nxn+l - (n + l)x + 1 n 1 + 2x + ... + (n + l)x = (x _ 1)2 + (n + l)x nxn+l - (n + l)xn + 1 + (n + l)xn+2 - 2(n + l)xn+l + (n + l)xn =----'--~----'----:--'--~---'---'-----'-----=------(x - 1)2 (n + l)xn+2 - (n + 2) xn+l + 1 = (x - 1)2

1.12 (n + 1)4 < 4n4 if and only if (n + 1)/n < .;2, that is, if and only if n(.;2 - 1) > 1, that is, if and only if n (being an integer) is at least 3. 4n > n4 if n = 5. Let n ~ 5, and suppose true for n. Then 4n +l = 4.4n > 4n4 > (n + 1)4. 1.13 The formula is true for n = 1. Let n ~ 1 and suppose true for n. Then qn+l = 3qn -1 = (3/2)(3n+ 1) -1 = (1/2)(3n +l +3 - 2) = (1/2)(3n +1 + 1). 1.14 The formula is true for n = 0 and n = 1. Suppose that it is true for all k < n. Then an = 4[2n- 2(n+ 1) - 2n- 3n] = 2n- 1 [2n+2 -n] = 2n- 1(n+2). 1.15 a) x2 + 4x + 5 = (x + 2)2 + 1 ~ 1 > O. b) x2 + 5xy + 7y2 = (x + (5/2)y)2 + (3/4)y2 ~ O. c) a2 + b2 + e2 + (l/a2) + (l/b2) + (l/e2) - 6 = [a - (l/a)F + [a - (l/bW + [a - (l/eW ~ O. Equality occurs if and only if a = l/a, b = l/b, e = l/e, that is, if and only if a, b, e E {-I, I}. 1.16 2a ~ a+b ~ 2b, and so 2ab/2b ~ 2ab/(a+b) ~ 2ab/2a; that is, a ~ H ~ b. Also, G2 - H 2 = ab - (4a2b2/(a + b)2) = [ab(a - b)2l/(a + b)2 ~ O. Solutions to Exercises 239

1.17 Trick question! A = {(x,y) : (x + 1)2 + (y - 2)2 + 1 = O} = 0, B = {(x,y): (x+2y+l)2+ y2+2=0}=0.

1.18 Ix - al < 6 ¢=} [x - a < 6 and -(x - a) < 6] ¢=} a - 6 < x < a + 6.

1.19 From Exercise 1.10, a+ar+'" arn = a/(I-r) -arn+1 /(I-r) < a/(I-r). 1.20 Put a = 1 and r = IX2/Xll in the previous inequality, to obtain 1 + Ixdxli +.. ·IX2/Xlln-l < 1/(I-lx2/xll). The left hand side is not greater than n times its smallest term, so not greater than nlx2/xlln-l. Hence nIX2/Xlln-l < 1/(I-lx2/xll = IXll/(lxd -lx21).

1.21 (3x+2)/(x+l) <1 ¢=} (3x+2)/(x+l)-1<0 ¢=} (3x+2-x• 1)/(x + 1) < 0 ¢=} (2x + 1)/(x + 1) < 0 ¢=} x E (-1, -1/2]. 1.22 If x ~ y, then Ix - yl = y - x, so (I/2)(x + y + Ix - yl) = y = max {x,y}, (I/2)(x + y -Ix - yl) = x = min {x, y}. The case x ~ y is similar. 1.23 lab - cdl = Ib(a - c) + c(b - d)1 ~ Iblla - cl + Icllb - dl. So with the given information lab - cdl ~ Iblla - cl + Icllb - dl < K(€/2K) + L(€/2L) = €. 1.24 Since labl > k2, 1(I/a) - (l/b)1 = la - bl/labl ~ la - bl/k2.

Chapter 2

2.1 The following table gives the answer: 0.0001 100,000,000

2.2 If k = 0 then the sequence is (1,1,1, ...), with limit 1. If k > 0, then k k II/n - 01 < € whenever n > (I/€)l/k. Hence (I/n ) ~ O. If k = -l < 0, k l l then I/n = n' > M for any given M > 0, provided n > M / • (Note that, for any positive a in IR and any positive integer k, we can define al / k as sup {x E IR : x k < a}.) 2.3 n/(n2+ 1) < 0.0001 if and only if n2-10,000n +1> 0, that is, if and only if n > 5,000 + y'25,OOO,OOO - 1, that is, if and only if n > 9,999.

2 2.4 n + 2n ~ 9999 ¢=} n ~ -1 + VI + 9999 = 99.

2.5 an ~ -00 if and only if, for every K > 0 there exists a positive integer N with the property that an < - K for every n > N. 2.6 The formula is correct for n = 1 and n = 2. Suppose that it is true for l 2 all k < n. Then an = (1/2)[2 + 4(-(I/2)t- + 2 + 4(-(I/2)t- ] = 240 Real Analysis

(1/2)[4 - 8(-(1/2))n + 16(-(1/2)tl = 2 + 4(-(1/2)t. It is then clear that (an) --* 2. 2.7 Suppose that f3 > B. There exists N such that Ibn - f31 < f3 - B for all n > N. Thus f3 - bn < f3 - B and so bn > B for all n > N. This is a contradiction to the definition of B. 2.8 If lanl ~ A, then -A ~ an ~ A. So "bounded" implies "bounded above and below". Conversely, suppose that A ~ an ~ B. If 0 ~ A ~ B then lanl = an ~ B. If A ~ B ~ 0 then lanl = -an ~ -A. If A ~ 0 ~ B then lanl ~ max {IAI, IBI}. SO in all cases (an) is bounded.

2.9 We know that for every € > 0 there exists a positive integer N such that Ian - al < € for every n > N. It follows that there exists a positive integer N', namely N' = N - 1, such that Ibn - al < f for every n > N'. Thus (bn ) --* a.

2.10 Since (an) --* a, for every f > 0 there exists N such that Ian - al < f for all n > N. There exists an integer M with the property that bM = aK, with K ~ N. Then, for all m > M, bm = ak for some k > N, and so Ibm - al < f.

2.11 Suppose, for a contradiction, that L < O. Taking f = ILI/2 = -L/2, we know that there exists a positive integer N such that Ian - LI < ILI/2 for all n > N. This implies in particular that an < L/2 < 0 -a contradiction. 2.12 Suppose first that L > O. Since an is positive for all n, Iva;- JII = (Ian • LD/(~ + JI) < (l/JI)lan - LI· Choosing N so that Ian - LI < (JI)f for all n > N, we see that Iva;- JII < f for all n > N. So (va;) --* JI.

Suppose now that L = O. If, for a given f, we choose N so that lanl < €2 for every n > N, then Iva;I < €, and so (va;) --* O. 2.13 Since (an - xn) is a positive sequence with limit ~ - a, we deduce from Exercise 2.10 that ~ - a > O. Similarly f3 - ~ ~ O. 2.14 By Theorems 2.1 and 2.8,

lim max {an,bn} = max {a,f3} , lim min {an,bn} =min{a,f3}. n-too n-too Let an = (_l)n, bn = (_l)n+1. Then max {an, bn} = 1, min {an, bn} = -1. So (an) and (bn) both diverge, while (max {an, bn}) and (min {an, bn}) are both convergent. The final statement follows from the observation that

bn = max {an, bn} + min {an,bn} - an' Solutions to Exercises 241

l n 2.15 a / = l+hn, where hn > O. By the binomial theorem, a = (l+hn)n > 1+ nhn (since all the remaining terms are positive). Thus 0 < hn < (a -l)/n, l n and so, by the sandwich principle, (hn ) --t O. Hence (a / ) --t 1. If a = 1 the sequence is constant, with limit 1. If 0 < a < 1 then a = l/b, where b> 1. Since (b l/n) --t 1, it follows that al/n = l/bl/n --t 1 as n --t 00. 2.16 Certainly (2n + 3n)l/n > (3n)l/n = 3 for all n ~ 1. On the other hand, (2/3)n < 1 for all n ~ 1, and so (2 n + 3n)l/n = 3[1 + (2/3)np/n < 3(2l/n). Thus 3 < (2 n + 3n)l/n < 3(2l/n), and so, by the sandwich principle and the result of the previous exercise, ((2 n + 3n)l/n) --t 3. 2 2.17 Ian - 01 = la; - 031/(a; + anO + 0 ) = la; - 031/[(an + ~0)2 + ~02] ~ 3 2 la; - 0 1/(3/4)0 . So, for a given f > 0, choose N so that, for all n > N, 3 la~ - 0 1 < (3/4)02f . Then Ian - 01 < f, and so (an) --t o.

a) Yes. For a given f, choose N so that la~1 < f3 for all n > N. Then lanl < f. b) No. Consider an = (_l)n. 2.18 a) It is clear that an ~ 0 for all n. Certainly al < 1. Suppose inductively that an < 1. Then 1-an+! =1- (3an + l)/(an + 3) = 2(1- an)/(an + 3) > O. b) an+! - an = (3an + l)/(an + 3) - an = (1 - a;)/(an + 3) > O. c) Since the sequence is increasing, and bounded above by 1, it has a limit 0, which must be non-negative, and satisfies 0 = (30+1)/(0+3). Thus 0=1. 2.19 a) It is clear that ar > 3. Suppose inductively that a; > 3. Then a;+l • 3 = ... = (a; - 3)/(an + 2)2 > O. b) an - an+l = an - (2an + 3)/(an + 2) = (a; - 3)/(an + 2) > O. c) Since (an) is decreasing and bounded below by V3, it has a positive limit 0, satisfying 0 = (20 + 3)/(0 + 2). Thus 0 = V3. 2.20 If a limit 0 exists, then it is positive, and satisfies 02 = 2 + 20j thus 0= 1+ V3. The sequence begins (approximately) (2,2.45,2.62, ...), which suggests that it might be monotonic increasing. Were that to be the case, 1 + V3 would be the supremum. It is clear that al < 1 + V3, so suppose inductively that 0 < an < 1 + V3. Then a~+! = 2 + 2an < 2 +2(1 + V3) = 4+ 2V3 = (1 + V3) 2, and so an+! < 1+ V3. Thus, since it is clear that every 2 an is positive, 0 < an < 1 + V3 for all n. Since x - 2x - 2 takes negative values in the interval (1 - V3,1 + V3), a; - a;+! = a; - 2an - 2 < 0,

and so (an) is monotonic increasing. Hence limn --telO an exists, and equals 1 + V3. 242 Real Analysis

2.21 By the arithmetic-geometric inequality (1.26), we know that an ~ bn for all n ~ 2. Since an+l - an = t(an + bn) - an = t(bn - an) ~ 0, it follows that (an) is decreasing. Since bnH/bn = .,fanbn/bn = y'an/bn ~ 1, it follows that (bn) is increasing. Since an ~ bn ~ b2 and bn ~ an ~ a2, both sequences converge: say (an) -+ a, (b n ) -+ (3. But then a = (a + (3)/2, and so a = (3. 2.22 Let m > n. For the first sequence, lam - ani = 1(-l)m/m - (-l)n/nl ~ (l/m) + (l/n) ~ 2/n. So if, for a given f > 0, we choose NI ~ 2/f, then lam - ani < f for all m > n > NI. In the same way, for the second sequence n n 2 lam - ani ~ 2/2 -! ~ 1/2 - < f for all m > n > N2, where N2 is chosen so that 2N2 - 2 > l/f. 2.23 Let m, n and N be positive integers such that m > n > N. The given property implies that a2 =I al· Then lam - ani = l(anH - an) + (an+2 • an+d + ... + (am - am-I)1 ~ lanH - ani + lan+2 - anHI + ... + lam • am-II ~ la2 - all(kn- l + kn + ... + km- 2) < kn- I la2 - all/(l - k) < N k - I la2 - all/(l - k). If, for a given f > 0, we now choose N so that KN-I < f(l- k)/la2 - all, we find that lam - ani < f for all m > n > N. To see that k = 1 is not good enough, let an = .;n. Then lanH - ani = l/{../n + 1 +.Jii) < 1/(.;n + ..;n=I) = Ian - an-II, but (an) is not a Cauchy sequence.

2.24 Consider an arbitrary but fixed n > N, and let bm = lam - ani (m ~ n). By assumption, bm < f for all m, and so limm-too bm ~ f. That is, la - ani ~ f. 2.25 a) bnbn- l = (gn+I!gn)(gn/gn-l) = gn+I!gn-1 = (gn + 2gn-I)/gn-1 = bn - l + 2. b) Divide by bn- l to obtain bn = 1 + (2/bn- I). c) Observe first that gn > gn-l, and so bn- l > 1. Thus it follows from part a) that bnbn- l > 3. Hence IbnH - bnl = 11 + (2/bn) - 1 - (2/bn-I)1 = 21(bn - bn-I)/(bnbn-I)1 < (2/3)lbn - bn-d·

d) It follows from Exercise 2.23 that (bn ) is a Cauchy sequence, and so has a limit (3. From part a) it follows that (32 = (3 + 2, and so (3 = 2. 2.26 Ln+l - Ln = 1/(2n + 1) + 1/(2n + 2) - l/(n + 1) ~ 1/(2n + 2) + 1/(2n + 2) -l/(n+1) = 0, so (Ln) is monotonic increasing. Also 1/(2n) + 1/(2n) + ... + 1/(2n) < Ln < l/n + l/n + ... + l/n, and so 1/2 < Ln < 1 for all n. Hence (Ln ) -+ a, where 1/2 ~ a ~ 1. 2.27 For all n, 1/n(n+2) = (1/2)[(1/n)-(1/(n+2))]. Hence E;:'=l (l/n(n+2)) = (1/2)[(1- (1/3)) + (1/2) - (1/4)) + ... + (l/N) - (l/(N + 2)))]. Most Solutions to Exercises 243

terms cancel, and so 2::=1 (l/n(n + 2)) = (1/2)[1 + (1/2) - 1/(N + 1) • 1/(N + 2)] -t ~ as N -t 00. 2.28 I/n! -I/(n+ I)! = [(n+ 1) -I]/(n+ I)! = n/(n+ I)!, and so 2::=1(n/(n+ I)!) = ((1- (1/2!)) + ((1/2!) - (1/3!)) + ... + ((I/N!) - (1/(N + I)!)) = 1 - (I/(N + I)!) -t 1 as N -t 00. Observing that n/(n + I)! < (n + I)/(n + I)! = I/n! < (n - I)/n!, we see that 2:~=o(I/n!) = 1 + 2:~=1 (I/n!) ~ 1 + 2:~=1 (n/(n + I)!) = 2, and 2:~=o(l/n!) = 1 + 1 + 2:~=2(I/n!) :c:; 2 + 2:~=2((n - 1)/n!) = 2 + 2:~=1 (n/(n + I)!) = 3.

2.29 Let an = 1 + ~ + ... + :,. Then lan+1 - ani = l/n(n + 1) -t 0 as n -t 00, but (an) cannot be a Cauchy sequence, since it does not converge. 2.30 Let 2::=1 an = AN, 2::=1 bn = BN. Then 2::=1 (an+bn) = AN+BN, and so 2:~=1 (an + bn) =limN-+oo(AN + BN) = limN-+oo AN + limN-+oo BN = A + B. Also, 2::=1 (kan) = kAN -t kA as N -t 00. 2.31 a) False. The harmonic series is a counterexample. b) False. Take an = l/n and bn = l/n2. c) True. It follows from Theorem 2.28. d) True. Since (an) -t 0, we may suppose that 0 :c:; an < 1 beyond a certain point. Hence a; < an, and so 2:~=1 a; is convergent by the comparison test. e) False. Let a2k =l/k for k =0,1,2, ..., and otherwise let an = O. Then an+l + ... + a2n = I/k, where k is defined uniquely by the property k n+I :c:; 2 :c:; 2n. Since k -t 00 as n -t 00, we have that limn-+oo(an+l + ... +a2n) = O. On the other hand, 2:~=1 an = 2:~1 (l/k) is divergent. 2.32 Let 2::=1 an = AN, 2:::=1 bn = BN. Then AN :c:; BN, and so A = limN-+oo AN :c:; limN-+oo BN = B. 2.33 Since .;n+T./(n2 + 2) '" I/n3/2, the first series converges. So does the 4 3 second series, since (n +3) /(n - 1) '" 1/ n • The third series diverges, since 1 2 (n + I)/Jn3 + 2 '" I/n / . 2.34 In + 1 - y'ri = I/(Jn + 1 + y'ri) x I/y'ri, so the series is divergent. Alternatively, observe that 2::=1 (In + 1 - y'ri) = JN + 1 - 1 -t 00 as N -t 00. 2::=d(l/y'ri) - (l/Jn + 1] = 1 - (l/JN + 1) -t 1 as N -t 00, so the series is convergent. 2.35 For the first series an+l/an = [(n+ l)!nn]/[(n+ 1)n+1n !] = (n/(n+ l)r = 1/[1 + (l/n)]n -t I/e as n -t 00. (See Example 2.13.) Since I/e < 1, the 244 Real Analysis

series converges. For the second series an+I/an = [(n+ 1)3 n!]/[(n+ 1)!n3] = [(n + 1)/n]3[1/(n + 1)]-+ 0 as n -+ 00, and so the series is convergent. 2.36 Certainly L:~=l (an + bn) is convergent. Since max {an, bn} < an + bn, it follows from the comparison test that L:~=o max {an, bn} is convergent. 2.37 This follows from the previous exercise, since (anbn)1/2 ~ max {an, bn}. To show the converse false, let _ {1 if n is even b _ {1 if n is odd an - 0 if n is odd n - 0 if n is even.

Neither L:~=o an nor L:~=o bn converges, but (anbn)1/2 = 0 for every n. 2.38 Let _ {l/n if n is even b _ {l/n if n is odd an - 1/n2 if n is odd n - 1/n2 if n is even. Neither L:~=l an nor L::'obn is convergent, but min {an,bn} = 1/n2. 2.39 Suppose that limn-too(an+l/an) = L > 1. Choosing e > 0 so that R = L - e > 1, we obtain an integer N with the property that an+l/an > R for all n > N. It follows that an> aN+IRn-N-l for all n > N, and so, from the (N+1)th term onwards, we have a comparison with the divergent geometric series L:~=N+l aN+lRn-N-l. Hence the series L:~=l an is divergent. n n 2.40 an+I/an = (n + l)ka +l /nka = [(n + l)/n]ka -+ a as n -+ 00, and so the series is convergent. From Theorem 2.20 it follows that limn-too nkan = O. 2.41 The conditions of the Leibniz test are satisfied in both cases, and so the series are convergent. The second series is absolutely convergent, the first is not. 2.42 Let an = l/n if n is odd, and an = 1/n2 if n is even. Then (an) -+ 0 but is not monotonic decreasing. L:~=l (-l)n-lan is divergent, since the sum of the positive terms increases without limit, while the sum of the negative terms can never exceed the sum of the convergent series L:~=l (1/(2n)2). 2.43 a) 82n = 1+(1/2)+·· ·+(1/2n)-2[(1/2)+(1/4)+·· +(1/2n)] = H2n -Hn. b) T3n = [1 + (1/3) + (1/5) + ... + (1/(4n - 1))] - [(1/2) + (1/4) + ... + (1/2n)] = [H4n - (1/2)H2n ]- (1/2)Hn = [H4n - H2n ]+ (1/2)[H2n • Hn] = 84n + (1/2)82n . It follows that the rearranged series has sum 38/2. 2.44 Let Un be the sum of the series to n terms. Then U3n = [1 + (1/3) + ... + 1/(2n -1)]- [(1/2) + (1/4) + ...+ (1/4n)] = [H2n - (1/2)Hn]- (1/2)H2n = (1/2)(H2n - Hn) = (1/2)82n. It follows that the rearranged series has sum 8/2. Solutions to Exercises 245

Chapter 3

y

2

-2 -1 0 ------+-----x 2

-1

-2

(The dots serve to indicate that the value of the function at each integer n is n.) 3.2 a) y

2

----+------t------x-2 -1 0 2

b) y

2

--~----f'--~--_ X 2 246 Real Analysis

c) y

2

-2 -1 0 ------+----+-x 2

-2

3.3 a) dom(g 0 J) = {x E IR : 2x - 3 ~ O} = [3/2,00), and (g 0 J)(x) = -/2x- 3. 2 b) dom(g 0 J) = {x E IR : -/4 - x i= I} = [-2,2] \ {vf3, -vf3}, and (g 0 J)(x) = 1/[(4 - x2)3/2 -1].

c) Since f(x) is rational for every x, it follows that dom(g 0 J) = [0,00) and (g 0 J)(x) = 1 for every x in the domain. 3.4 a) (J/g)±h=(J±g·h)/g. b) (J/g)·h=(J·h)/g. c) By repeated use of the above observations, we can delay the use of division to the very end, and a single division is enough.

3.5 ((J + g) 0 h)(x) = (J + g)(h(x») = f(h(x») + g(h(x») = (J 0 h)(x) + (g 0 h)(x) = ((Joh)+(goh»)(x). Let g(x) = h(x) = x and let f(x) = x2. Then 2 2 2 2 (J 0 (g + h»)(x) = (2x)2 = 4x , ((J 0 g) + (J 0 h»)(x) = x + x = 2x . 3.6 Let q = fig, r = h/k, where f,g,h,k are polynomials. Then q + r = (J .k +9 .h) / (g .k), q. r = (J .h) / (g .k), q/ r = (J .k) / (g .h) are all rational functions. If q and r are rational functions, then, for each x, (q 0 r)(x) is obtained from r(x) by repeated applications of addition, multiplication and division. The result is a rational expression in x. 3.7 For example, 1 1 (g 0 h)(x) = g( -l/x) = -l/x = -x, (h 0 g)(x) = h(l/x) = -l/x = -x,

and so 9 0 h = hog = f. 3.8 Here are a few sample computations. Solutions to Exercises 247

(9 0 J)(x) =9 C~ X) =1 - 1/1 ~ x =1- (1- x) = x;

thus f 0 9 =9 0 f =i. 1 x (fop)(x) = f(l/x) = 1- (l/x) = x-I) (poJ)(x) =p(I/(I-x)) = I-x.

Thus fop =r, p 0 f = q. 1 1 (p 0 q)(x) p(1 - x) (q 0 p)(x) q(l/x) j = = -1-x-, = = 1 - -x

thus po q = f, q 0 p =g. The complete table is

0 i f 9 P q r i i f 9 P q r f f 9 i r p q 9 9 i f 9 r P p p q r i f 9 q q r P 9 i f r r p q f 9 i

3.9 This follows immediately from the identity cos 2 8 + sin2 8 = 1. 3.10 From the addition formula, cos 28 =cos2 8 - sin2 8 =cos2 8 - (1- cos2 9) = 2 cos2 8 - 1. Hence 2cos2 8 = 1 + cos 28, and the result follows. Similarly, cos 28 = (1-sin2 9) - sin2 9 =1-2sin2 8, and so sin2 8 = (1/2)(1- cos 28). 3.11 From the addition formulae, sin(9 +

2 2 2 2 2 3.13 (1 - t )/(1 + t ) = [cos 2 (0/2) - sin (8/2))/[cos (9/2) + sin (9/2)) = cos 9, 2 2 2 2t/(l + t ) = [2cos(8/2)sin(9/2))/[cos (9/2) + sin (9/2)] = sin8, tan 9 = 2 sin 0/ cos 8 = 2t/(1 - t ). 3.14 Suppose that f is increasing and bounded above. Then the set {/(x) : x E [a, co)}, being bounded above, has a supremum A. For all f > 0 there exists X such that f(X) > A - f, since otherwise A - f would be an upper bound for f. Since f is increasing, it follows that A - f < f(x) ~ A for all x> X. Thus limz-too f(x) = A. 248 Real Analysis

3.15 limz --+ oo I(-x) = L if and only if for all f. > 0 there exists M > 0 such that IJ(-x) - LI < f. for all x > M, that is (writing -x as y) if and only if for all f. > 0 there exists M > 0 such that I/(y) - LI < f. for all y < -M. 3.16 Let I, defined on [a, b], be increasing. Then I(a) :5 I(x) :5 I(b) for every x in [a, b], and so I is bounded both above and below. The result does not hold for an open interval: if I(x) = 1/(1 - x) (x E [0,1)), then I is increasing, but is not bounded above.

3.17 Since B ~ A, we have that {f(x) : x E B} ~ {f(x) : x E A}. It fol• lows that every upper bound of {f(x) : x E A} is also an upper bound of {f(x) : x E B}. In particular sUPA I is an upper bound of {f(x) : x E B}, and so supA I ~ supB I. A similar argument regarding lower bounds gives that infA I :5 infB I· 3.18 Let I, with domain [0,1], be given by I(x) = 1 if x is rational, and I(x) = -1 if x is irrational. Then limz --+ll/(x)1 = 1, but limz --+l/(x) does not exist. 3.19 This follows from Theorem 3.3 and the observation that max {f, g} = (1/2)(1 + 9 + II - gl) and min {f,g} = (1/2)(1 + 9 -II - gl). 3.20 The identity l-cosx = 2sin2(x/2) follows immediately from Exercise 3.10. Then 1- cos x _ ~ (Sin(x/2))2 ~ 0 x2 - 2 x/2 ~ 2 as x ~ .

3.21 This follows from Theorem 3.11, since max {f,g} = (1/2)(1 + 9 + II - gl) and min {f,g} = (1/2)(1 + 9 -II - gl). 3.22 (x-a)(b-x) ~ 0 for all x in [a, b]. Since both the functions x ~ (x-a)(b-x) and x ~ .;x are continuous, the function x ~ .../(x - a)(b - x) is contin• uous in [a,b]. The same applies to the function x ~ .../(x - a)/(b - x), except that the region of definition and of continuity is the interval [a, b). 3.23 The function x ~ cot x = cos x/sin x is continuous except when sin x = 0, that is, for x in IR \ {mr : n E Z}.

3.24 Let f. = I/(c)l/2. Then there exists b > 0 such that 11/(x)1 - I/(c)11 :5 I/(x) - l(c)1 < I/(c)1f2 for all x in (c - b, c+ b). It follows that I/(c)I/2 < I/(x)1 < 31/(c)1f2, and so certainly I(x) :I 0 for all x in (c - b,c + 0). 3.25 Let f. > 0 be given, and let b = f.. There is no harm in supposing that f. < 1. If 0 < x < b, then o {x < f. if xis rational I/(x) - I( )1 = x2 < f.2 < f. if x is irrational. Solutions to Exercises 249

Thus f is continuous at O. Similarly, for continuity at 1, define 8 = €/2. Also, observe that, for all x in (0,1), 1 - x2 = (1 - x)(l + x) < 2(1 - x). Then, if 11 - xl < 8, it follows that

f( )_ f(l)1 -{ 1 - x < 8 < € if x is rational I x-I_ x2 < 2(1 - x) < 28 = € if x is irrational.

2 Thus f is continuous at 1. Suppose finally that 0 < x < 1, so that x :f:. X. If x is rational, and if, for all n ~ 1, we define an = x + (V2In), then 2 limn--+ oo an = x, whereas limn--+ oo f(an) = limn --+ oo (an )2 = x :f:. f(x). If x is irrational, we define bn to be any chosen rational number in the interval (x,x + (lin)). Then limn--+oobn = x, whereas limn--+oof(bn) = limn --+ oo bn = x :f:. f(x). From Theorem 3.5, we deduce that, whether x is rational or irrational, f is not continuous at x. 3.26 Let g(x) = f(x) - x. Then 9 is continuous on [a, b]. Since a ~ f(x) ~ b for all x in [a, b], we have g(a) = f(a) - a ~ 0 and g(b) = f(b) - b ~ O. If either of these inequalities is an equality then the result is clear, with c = a or c = b. Otherwise, by the intermediate value theorem (Theorem 3.12) there exists c in (a, b) such that g(c) = O. 3.27 The function f - 9 is continuous, and (f - g)(O) < 0, (f - g)(l) > O. Hence, by the intermediate value theorem, there exists c in (a, b) such that (f - g)(c) = O. 3.28 Suppose that f is not monotonic. Then there exist a, b, c in lR. such that a < b < c and either (i) f(a) < f(b), f(b) > f(c), or (ii) f(a) > f(b), f(b) < f(c). Consider Case (i), and suppose first that f(a) ~ f(c) < f(b). Let d E (J(c) , (f(b)) ~ (J(a), f(b)). Then by the intermediate value theorem (Theorem 3.12) we obtain a contradiction, since there exist Cl in (a, b) and C2 in (b, c) such that f(Cl) = f(C2) = d. Similarly, if we suppose that f(c) ~ f(a) < f(b), then, for all din (J(a) , (f(b)) ~ (J(c),f(b)), there exist Cl in (a, b) and C2 in (b, c) such that f(Cl) = f(C2) = d. Case (ii) is dealt with in a similar way. 3.29 Consider the sequence

2") _ ( 2 4 8 ) ( X - x,x ,x ,x '00' ,

2 where x E (-1, 1). Then (x ") -+ 0, and so, by continuity at 0, the sequence 2n 2n (J(x )) -+ f(O). But (J(x )) is the constant sequence (J(x)) , with limit f(x), and so we conclude that f(x) = f(O) for all x in (-1,1). 3.30 Putting x = 0 in the functional equation, we find that f(O) = f(aO) = bf(O), where b > 1, and this gives a contradiction unless f(O) = O. Let 250 Real Analysis

If(x)1 ~ M for all x in [0,1]. Let f > 0 be given, and let n E N. Then f(anx) = bnf(x) for all x in [0, a-n], and so there exists N in N such that

N if n > N. Hence, defining 6 as a- , we see that If(x) - f(O)1 < f for all x in (-6,6) n dom f. Thus f is continuous at O. 3.31 For example, suppose that f and g are continuous at the point a. Then for every sequence (an) with limit a, we have that (J(an)) -+ f(a), (g(an)) -+ g(a). By Theorem 2.8, ((J +g)(an)) -+ (J +g)(a), and so, by Theorem 3.6, f + g is continuous at a.

3.32 Let c E dom f, and let (cn ) be a sequence contained in dom f such that (cn) -+ c. Then, using Theorem 3.5 twice, we deduce that (J(cn )) -+ f(c) and (g 0 1)(cn)) -+ (g 0 1)(c). Hence, by Theorem 3.6, g 0 f is continuous.

3.33 For each f > 0 there exists 61 > 0 such that If(x) - f(y)1 < f/2 for all x, y in [a, b] such that Ix - yl < 61 , and there exists 62 > 0 such that If(x) - f(y)1 < f/2 for all x, y in [b, c] such that Ix - yl < 61 • Let o= min {01,02}, and let x, y in [a,c] be such that Ix - yl < o. If both x and yare in [a, b], or if both x and y are in [b, c], then it is clear that If(x) - f(y)1 < f/2 < f. For the remaining case we may suppose without essential loss of generality that x < b < y. Then Ib - xl < 6 ~ 61 and so If(b) - f(x)1 < f/2. Similarly, Iy - bl < 6 ~ 02, and so If(y) - f(b)1 < f/2. Hence If(x) - f(y)1 = 1(J(x) - f(b)) + (J(b) - f(y))1 ~ If(b) - f(x)1 + If(y) - f(b)1 < f.

3.34 Let sin-1 x = 0, so that 0 E [0,71"/2] and sin 0 = x. Then cos ((71" /2) - 0) = 1 x, and (71"/2) - 0 E [0,71"]. It follows that cos- x = (71"/2) - o.

3.35 As an extreme case, consider the constant function Ck , defined by the rule that Ck(X) = k for all x. The image of the function is {k}, but there is no inverse function from {k} to lit 3.36 Since 8+2x-x2 = 9-(x-1)2, the function has maximum value 9, obtained 2 2 when x = 1. Now y = 8 + 2x - x ¢::::> x - 2x + (y - 8) = 0 ¢::::> x = 1 1 ± ,;g=y. There is an inverse function f- : (-00,9] -+ [1,00) given by f-l(y) = 1 +,;g=y. 3.37 The image of f is IR\ {O}, and this is the domain of the inverse function. The formula is obtained by observing that y = 1/(1 - x) ¢::::> x = 1 - (l/y). So f-l(y) = 1- (l/y) (y E IR \ {l}). Solutions to Exercises 251

Chapter 4

4.1 It is easy to see that f(x) -t 0 = f(O) as x -t 0, and so f is continuous at O. However,

f(x) - f(O) _ M_{ 1/-/X if x> 0 x - 0 - x - -l/M if x < 0, and so f is not differentiable at O. 4.2 a) Suppose that limx-tc[(f(x) - f(c))/(x - c)] = 1'(c). Then for every f > 0 there exists 6 > 0 such that ~(c) If(xl = - f'(C)! < f (9.16)

for all x in dam f \ {c} such that Ix - cl < 6. Then (9.16) holds for all x in dam f n (-00, c) such that Ix - cl < f, and it also holds for x in dam f n (c, 00) such that Ix - cl < f. Thus the left derivative and right derivative both exist, and both are equal to 1'(c). b) If x < 0 then x(x-1) > 0, and so (f(x) - f(O))/(x-O) = x(x-1)/x = x - 1 -t -1 as x -t 0-. If x > 0, and is sufficiently close to 0, then x(x -1) < 0, and so (f(x) - f(O))/(x -0) = -x(x-1)/x = -x+1 -t 1 as x -t 0+. Thus f{(x) = -1, f;(x) = 1.

4.3 Since limx-to(f(x) - f(O))/(x - 0) = xsin(l/x) -t 0 as x -t 0, we have that 1'(0) = O. If x I: 0 then, by ordinary calculus methods, 1'(x) = 2x sin(l/x) - cos(l/x), and this does not have a limit as x -t O. Thus l' is not continuous at O. 4.4 No: if f(x) = x and g(x) = -x, then max {f,g} and min {f,g} are not differentiable at O.

4.5 Let n = -m, where m is a positive integer. Then f(x) = l/xm , and so 1'(x) = (_1/x2m)mxm - 1 = (_m)x-m - 1 = nxn - 1 • 4.6 Note that Dx(tanx) = tan2 x + 1. By L'H6pital's rule,

· tan x - x l' tan2 x I' 2tanx 2tan3 x 11m = 1m -- = 1m ------+ x-tO x3 x-tO 3x2 x-tO 6x 2 = lim 2 + 8 tan x + 6 tan4 x = 1 x-tO 6 3 Again, by L'H6pital's rule, · sinx - xcosx l' xsinx l' sinx 1 11m = 1m--= 1m--=-. x-tO x3 x-tO 3x2 x-tO 3x 3 252 Real Analysis

4.7 a) If Q > 0, then If(x) - f(a)1 -t 0 as x -t a, and so f is continuous at a. If Q > 1, then l(f(x) - f(a))/(x - a)1 < Mix - al"-1, and so f'(a) = O. b) Let f(x) = Ixi- Then If(x) - f(O)1 = Ilxll = Ixl = Ix - 01, and so f, which is not differentiable at 0, nonetheless satisfies a HOlder condition in which M = 2 and Q = 1. 4.8 Let x E [a, b]. If x> e, then f(x) ~ f(e), and so (j(x) - f(e)]/(x - c) ~ O. Hence limz-tc+(j(x)- f(e))/(x-e) ~ O. Similarly limz-tc- f(x)- f(e))/(x• c) ~ O. Since f is differentiable, the two limits are both equal to f'(e), and so f'(e) = O. The proof for d is similar. 4.9 For every ein the domain, l(f(x)- f(e))/(x-e)1 < Ix-el, and so f'(e) = O. It follows that f is constant.

4.10 We show that (J(l/n)) is a Cauchy sequence. Let t > 0 be given, and choose an integer N such that liN < t. Let m, n > N. Then, for some e between 11m and lin, If(l/m) - f(l/n)1 = 1(l/m) - (1/n)llf'(e)1 < 1(l/m) - (1/n)1 < liN < t. 1 1 1 4.11 By Theorem 4.15 (cos- )'(y) = 1/[- sin(cos- y)]. Since 0 < cos- Y < 11', it follows that sin(cos-1 y), being positive, is equal to Jl- cos2 (cos-1 y) = 1 J1=Y2. Hence (cos- )'(y) = -IIJ1=Y2 for all y in (-1,1). 4.12 Since Dz(cos-1 x+sin-1 x) = 0, it follows that cos-1 x+sin-1 x is constant. Its value at 0 is cos-1 0 + sin-1 0 = 11'12 + 0 = 11'/2, and so this is its value throughout [-1,1]. 4.13 Since f'(x) = f(x) > 0 for all x, it follows that f is an increasing function. By Theorem 4.15, (f-1)'(X) = I/f'(J-1(x)) = Ilf(J-1(x)) = l/x. 4.14 We use L'Hopital's rule twice, differentiating with respect to h: thus, limh-to(llh2)[f(a + 2h) - 2f(a + h) + f(a)) = limh-to(1/2h)[2f'(a + 2h) • 2f'(a + h)] = limh-tO(1/2) [4f"(a + 2h) - 2f"(a + h)] = f"(a). 4.15 Denote x2cos x by h(x). Observe that COS(2k-1) x = (_I)k sin x, COS(2k) x = (_I)k cosx. Then, by Leibniz's Theorem, h2n- 1(X) = (_I)n-1 [((2n • 1)(2n - 2) - x2) sinx + 2(2n -1)xcosx]. 1 4.16 Since f'(x) = -msin(msin- x)IJl- x2 , if follows that Jl- x2f'(x) = -m sin(m sin-1 x). Differentiating again, we obtain (after some easy alge• bra) (1- x2)f"(x) - xj'(x) + m 2 f(x) = O. Hence the required result is true for n = O. Now suppose inductively that Solutions to Exercises 253

Then, differentiating, we have [(1- x2)j(n+2) (x) - 2xj(n+l) (X)] - (2n - 1)[xj(n+1) (x) + j(n) (X)] 2 2 + (m - n + 2n -l)j(n)(x) = 0,

and collecting terms gives us

Hence, by induction, the result holds for all n ~ O. Now put x = 0 to obtain j(n+2)(0) = (n 2 - m 2)j(n)(0) for all n ~ O. Since j(O) = 1 and 1'(0) = 0, we see that j(n)(o) = 0 for all odd n, while for 2 2 2 2 even n we have j(n) (0) = [(n - 2)2 - m ] ... [2 - m J[-m ]. 4.17 If j(x) = ao + a1X + ... + anxn, then j(r)(x) = r!ar + positive powers of x for all r < n, f(n)(x) = n!an, and f(r)(x) = 0 for all r > n. Thus

~ j(r)(o) = {ar ~f 0 ~ r ~ n r. 0 If r > n.

Chapter 5

5.1 In every subinterval of any dissection D, the supremum and infimum of the function is k, and so U(Ck , D) = .c(Ck , D) = k(b - a). If follows that I: Ck = k(b - a). 5.2 Let Dn = {O'~'~"'" n;;l, I}. Then, using Exercise 1.7, we see that 3 2 2 U(J, Dn) = (lJn) L:~1 (i2Jn2) = (lJn )(1 + 2 + ... + n2) = (1/6)[1 + 3 2 2 (1/n)][2+(1/n)] while .c(J, D n) = (1/n )(1 +2 +.. +(n-1)2) = (1/6)[1- (1/n)][2 - (l/n)]. Hence, for all n, I; j - J; f $ U(J, D n)- .cU, Dn) = l/n, 1 and so j is Riemann integrable. Since 10 f lies between the upper and lower sums, both of which have limit 1/3 as n -+ 00. we also deduce that I; j = 1/3.

5.3 Choose a dissection D = {xo, Xl, ... , xn } containing c. Then, recalling that M i and mi are (respectively) the supremum and the infimum of j in the open subinterval (Xi-l,Xi), we see that Mi = mi = 0 for all i, and so 1 UU, D) = .c(J, D) = O. Hence J;f - Jo j $ U(J, D) - .cU, D) = 0, and 1 so f E 'R[O, 1]. Since Jo j must lie between U(J, D) and U(J, D), we must have J; f = o. 254 Real Analysis

5.4 Suppose for convenience that C1 < C2 < ... < C1<, and look first at the case where C1 > a and C1< < b Let € > a be given. Define

8 = min { C1 - a, b - Ck, 4k(M€ _ m)} ,

where M = sUP[a,b) f, m = inf[a,b) f. Then let D be a partition in which the subinterval (Ci - 8, c; + 8) features, for i = 1,2, ... , k. In each of the intervals [a, Cl -8], [Cl +8,C2-8], ... , h-l +8,c1<-8], [Ck+8, b] the function is continuous, and so we can arrange for the contribution of those intervals to U(J, D) - £(J, D) to be less than €/2. The contribution of each of the intervals [Ci - 8, Ci + 8] cannot exceed 28(M - m) = €/ 2k, and so the total contribution of those intervals is at most €/2. Hence U(J, D) - .c(J, D) < (€/2)+(€/2) = €, and so f E n[a,b]. Small and straightforward adjustments in the argument are necessary to cope with the case where Cl = a and/or Ck = b.

5.5 Let € = f(c)/2. By continuity, there exists 8 > 0 such that If(x) - f(c)1 < € for all x in [a,b] such that Ix - el < fJ. It follows that f(e) - (J(c)/2) < f(x) < I(c) + (J(c)/2), and so in particular I(x) > l(c)/2 for all x in the interval [a, b) n (c - 8, C + fJ), an interval whose length is at least 8. Thus J: I ~ 8(J(c)/2) > O.

5.6 The function x I-t (J(X))2 is continuous and takes only non-negative values. By the previous exercise, J: (J(x) ) 2 dx = 0 implies that (J(x) )2 = 0 for all x in [a, b].

5.7 Let D be a dissection of [a, b) containing C and d, and let D1 = D n [a, c),

D2 = D n [e, d), D3 = D n [d, b]. (The situation simplifies if e = a or d = b.) Then U(J' D) - .c(J, D) = E:=1 (U(J, Di) - .c(J, Di)). If, for a given € > 0, we choose D so that U(J, D) - .c(J, D) < €, then certainly U(J, D2) - .c(J, D2) < €, and so I E nrc, d).

5.8 Let € > 0 be given. By uniform continuity, there exists 8 > 0 such that I/(x) - l(y)1 < €/2(b - a) for all x, y in [a, b) such that Ix - yl < 8. Choose n so that n > 1/8; thus each subinterval in D n has length less than 8. Then in each subinterval (Xi-I, Xi) there exists x such that Mi - I(x) < €/4(b - a) and there exists y such that f(y) - mi < €/4(b - a). It follows that Mi - mi = (Mi - I(x)) + (J(x) - I(Y)) + (J(y) - mi) < €/4(b - a) + I/(x) - l(y)1 + €/4(b - a) < €, since Ix - yl < 8. It follows that U(J, Dn )• .c(J, Dn ) = ((b - a) /n) E~=1 (Mi - mi) < € if n is sufficiently large. Now ~ ~ .c(J, Dn ) J: I U(J, Dn ), and so certainly IU(J, Dn ) - J: II < € and 1.c(J, Dn ) - J: II < € for sufficiently large n. It follows that the sequences (U(J, Dn )) and (.c(J, Dn )) both have limit J:f. Solutions to Exercises 255

5.9 Let D m be the dissection {0,1/22m,1/22m-l, ... ,1/2,1}. In all but the first subinterval, the function is constant, and so the supremum and infi• mum are equal. In the first subinterval the length is 1/22m , the supremum 2m 2m 1 is 1/2 and the infimum is _1/2 + . Hence U(J, Drn ) - .c(J, Dm ) = 2rn 2rn 2m 1 4m I (1/2 )[(1/2 ) + (1/2 + )] = 3/2 + , and this can be made less than any given € by taking m sufficiently large. Hence f E R[O, 1]. The value of the integral is the limit of U(J, Dm ), namely the sum of the geometric senes. 1 . 21 - 21 . 4" 1 + 11Th4" • 8 - .. '. e sum IS. 21/ (1 + 4"1) -_5' 2 1 2 1 l 5.10 Let f(x) = g(x) = x. Then fo f . 9 = f; x dx = h but Uo f) Uo g) = ~. 5.11 Properties a) and b) follow immediately from Theorem 5.15. It follows from Theorem 5.14 that (J, I) ~ 0, and from Exercise 5.6 that (J, I) = 0 only if f = O. The inner product (kf + g, kf + g) is non-negative for all k. That is, k2(J,1) + 2k(J, g) + (g,g) = 0 for all k. It follows that the discriminant is non-positive: 4(J,g)2_4(J, I)(g,g) ~ 0, and so U: f. g)2 ~ U: j2) U: g2). 5.12 Let f(x) = 0 for a ~ x ~ !(a + b), and f(x) = 1 for !(a + b) < x ~ b. Then f: f = !(b - a), but there is no c such that f(c) = !. 5.13 Since g(x) ~ 0, we have that mg(x) ~ f(x)g(x) ~ Mg(x) for all x in [a,b], where M = sUP[a,b] f, m = inf[a,b] f· Thus m f: 9 ~ f: f· 9 ~ M f: g, and so, since J: 9 > 0, m ~ U: f· g) / U: g) ~ M. By the intermediate value theorem, there exists c in [a, b] such that f(c) = U: f· g) / U: g). 2 x x 2 5.14 Since f(x) = !x J: g(t) dt - x Jo tg(t) dt + ! Jo t g(t) dt, we may deduce that f'(x) = !x2g(x) + x f: g(t) dt - x2g(x) - f: tg(t) dt + ~x2g(X) = x J: g(t) dt - foX tg(t) dt. Hence f"(x) = f: g(t) dt + xg(x) - xg(x) = f: g(t) dt, and fll/(x) = g(x). 2 3 5 4 5.15 f(x) = g(u) - g(v), where u = x , V = x , and g(u) = f; t /(1 + t ) dt. 5 4 5 4 2 ll 8 Hence f'(x) = [u /(1 + u )] • 2x - [v /(1 + v )] . 3x = 2x /(1 + x )• 3xI7 /(1 + x I2 ). 5.16 4 Jr/ tan2 xdx= [tanx-x]~/4=1-(71"/4); o 2 2 r/ 2 1 r/ 1 1 1r/2 1 2 J sin xdx="2Jo (1-CoS2x)dx="2[x-"2sin2xL =:i'lr . o 5.17 The argument takes no account of the indefiniteness of an indefinite inte• gral. 256 Real Analysis

f"/2 [],,/2 ,,/2 5.18 10 = Jo xdx = 1f2/8, h = X(-COSX) 0 - fo (-cosx)dx = 0 + ,,/2 []sin x 0 = 1. To obtain the reduction formula, integrate by parts, taking the factors as sin x and x sinn- 1 x: 1"/2 In =1 x sinn x dx = [(- cosx)x sinn- 1 X]~/2 0 1"/2 + 10 (cosx)[sinn-1x+(n-1)xsinn-2xcosx]dx

1"/2 1"/2 =0+1 sinn-1xcosxdx+(n-1)1 xsinn- 2x(1-sin2x)dx 0 0 1 ] ,,/2 = [ ~ sinn x 0 + (n - 1)In - 2 - (n - l)In 1 = - + (n - 1)In - 2 - (n - l)In . n Hence 1 n-1 In = 2' + --In- 2 • n n ~ ~h ~,and ~(i ~Io) 3 1f2 So 13 = + = 14 = 116 + + = i + 64 . 5.19

5 f1 x dx 1 f2 du 6 1 [ ]2 1 1 v"I+X6 = {; 1 ..;u (with u = 1 + x ) = 3 .;u 1 = 3(V2 - 1). 0 1

,,/2 sinxdx 3 -du 1 (3 )2 =1-2 (with u = 3 + cos x) = 12 . 1o + cos x 4 U 3 1"/4 11 1 10 cos2xV4 - sin2x = -2" 4 VUdu (with u = 4 - sin2x) = 3(8-3V3) .

5.20 Since sin(1f - x) = sinx, putting u = 1f - x gives I = fo" xf(sinx)dx = f~ (1f - u)f sin(1f - u)(-du) = 1f fo" f(sin u) du - fo1f uf(sin u) du 2 = 1f fo" f(sinx) dx - I, and the result follows. Since cos2 x = 1 - sin x, we can apply the result to the given integral I, making the substitution u = cos x, obtaining 1 I - ~ r sinxdx = ~ f- -du _ ~ [tan-1 u] 1 = 1f2 - 2 10 1 + cos2 X 2 11 1 + u2 - 2 -1 4' Solutions to Exercises 257

5.21 The result holds for n = 1: f(x) = f(a) + J: f'(t) dt. For the inductive step, note that integration by parts gives

Rn = 1 r (x _ t)n-1 f(n)(t) dt (n-l)!Ja x = 1 [_ (x - t)n f(n) (t)] t=x _ 1 l _ (x - t)n f(n+1) (t) dt (n - I)! n t=a (n - I)! a n = (x - a)n f(n) (a) + Rn+l n! and the result follows. 2 5.22 Since x/VX6 + 1 ~ 1/x as x -t 00, the first integral converges. Since (2x + 1)/(3x2 + 4y'X + 7) ;::: l/x as x -t 00, the second integral diverges. 5.23 Since Kx 1 (2K -1)x2+Kx-l {l/X if K"# 1/2 x2 + 1 - 2x + 1 = (x2 + 1)(2x + 1) ;::: l/x2 if K = 1/2, the integral converges if and only if K = 1/2. Denote the second integrand by F(x). Then 2 2 F x) _ (x + 1)2 - K (2x + 1) (- (x + I)V2x2 + l(x + 1 + K V2x2 + 1 x2(1 - 2K2) + 2x + (1 - K) =+~IO-;-( = -;"(x-+-=-1):-'-";"2:=X:;<'2 x-+--::l-+-'-K-=-=-..;r.:2=x2;;=+==='1 ~ {1/X if K "# ±1/V2 ~ l/x2 if K = ±1/V2. Thus the integral converges if and only if K = ±1/V2. 5.24 For 0 ~ x ~ 1, let g(x) = x(1-x), and let f(x) = g(x- LxJ) for x ~ O. Thus the graph of 9 repeats between any two positive integers, and fen) = 0 for n = 0,1,2, .... Trivially, E:'=l fen) is convergent, but, for N E N, 1 oo Ji' f = (N - 1) J0 x(1 - x) dx = (N - 1)/6, and so Jo f diverges. 5.25 a) Integration by parts gives Jt (sinx/x)dx = [- cosx/x]~­ Jt (cosx/x2) dx. Since J100 (cosx/x2)dx is (absolutely) convergent, and since cosK/K -t 0 as K -t 00, the integral is convergent. b) In the interval [2k1r,(2k + 1)11"], sinx ~ O. So J2(::+l)7r Isin x/xl dx = J;::+l)7r (sin x/x) dx ~ (1/(2k + 1)11") J;::+l)7< sin x dx = 2/(2k + 1)11". In the interval [(2k -1)1I",2k1r], sinx ~ O. So J(~~~l)7r Isinx/xidx = J(~~~l)7r (- sin x/x) dx ~ (1/2k1r) J(~~"-l)7< (- sin x)) dx = 2/2k1r. 258 Real Analysis

N1T 1T Hence f: Isin x/xl dx = f: Isin x/xl dx + f;: Isin x/xl dx + ... + f(~"%-':..l)1T Isin x/xl dx 2': (2/'rr) E;:'2 (l/r), and by the divergence of the harmonic series it follows that I is not absolutely convergent. 5.26 l/Jsinx '" l/yIX as x -t 0+. Hence the first integral converges. Since 2 sinx/x '" l/x as x -t 0+, the second integral diverges. Since, by L'Hopital's rule, (x - sinx)/x3 -t 1/6 as x -t 0, it follows that l/(x • sin x) x 1/x3 as x -t 0+. Hence the integral is divergent.

5.27 Since sin x/x -t 1 as x -t 0+, there is no problem regarding the lower limit. We have seen that the first integral converges, but is not absolutely convergent. The second integral is easily seen to be absolutely convergent.

2 rOO ~sin x [ 2 (1)] 00 roo ( 1) 10 dx = sin x -; 0 - 10 -; 2sinxcosxdx 00 sin 2x 00 sin u = 0+ --dx =1 --du, where u = 2x. 1o x 0 u

Chapter 6 2 6.1 L(2n ) = nlog2, and log2 = f1 (dx/x) > 1/2, since 1/2 = inf{l/x : 1 ~ x ~ 2}. 6.2 (l-u)(l+u) = l-u2 < 1 and so, ifn is positive, 1-u < l/(l+u) < 1. It follows that, for all positive x, foX (1 - u) du < f: (1/(1 + u)) du < foX du; that is, x - ~x2 < 10g(1 + x) < x. 6.3 Observe first that f(l) = O. Now, f'(x) = 1 - (l/x), and so f'(x) > 0 if o < x < 1, and f'(x) < 0 if x > 1. By the mean value theorem, f(x) = f(l) + (x - l)f'(c) = (x - l)f'(c), where c is between 1 and x. Since (x - l)f'(c) is positive for all x i: 1, we have logx < x - 1. Similarly, 2 g(l) = 0, and g'(x) = (l/x) - (1/x ), which is negative if x < 1 and positive if x> 1. Thus g(x) = (x - l)g'(c) (where c is between 1 and x) is positive for all x i: 1, and so 1 - (l/x) < logx. 6.4 From Taylor's Theorem, 10g(1 + x) = log 1 + x(log)'(l) + (x2 /2)(log)"(1) + (x3 /6)(log)"'(t), where 1 < t < x. That is, 10g(1 + x) = x + (x 2 /2) + 3 3 2 (x /6)(2/t ) > X + (x /2). Again, we have 10g(1 +x) = log 1+x(log)'(l) + (x2 /2)(log)N(1) + (x3 /6)(log)"'(1) + (x4/24)(log)(4)(t) = x + (x 2 /2) + 3 4 4 2 3 (x /3) - (x /24)(6/t ) < X + (x /2) + (x /3). Solutions to Exercises 259

6.5 Using integration by parts, we have m 1 £( )- xm+l(logx)n Jx + (I )n-l 1 d m,n - m + 1 - m + In ogx ~ X xm+1(logx)n n = m + 1 - m + 1£(m, n - 1) ,

and so (noting that £(1,0) = x 2 /2) we have 2 2 2 £(1,3) = 2(logx)3x - 2£(1,2)3 = 2(logx)3x - 23 [x2(logx)2 - £(1,1)]

2 2 =-x1 2(logx) 3 - -x3 2(logx) 2 + -3 [x-logx --x ] 2 4 2 2 2 1 = 4x2(2(logx)3 - 3(logx)2 + 310gx - 3).

6.6 By L'Hopital's rule,

· 10g(cosax) I' -asin ax/ cos ax a l' sinax I' cosbx 11m = 1m =-, 1m --, 1m-- x--to10g(cosbx) x--tO -bsinbx/cosbx b z--tosinbx z--tocosax 2 = ~ . lim a cos ax .1 = a . b z--tO bcos bx b2

6.7 By Taylor's Theorem, e-x = 1 - x + (x 2 /2)e-9z , where °< () < 1. Since both x2 and e-9z are positive for all x f:. 0, it follows that e-Z > 1 - x. Replacing x by -x gives eX > 1 + x, and this holds for all x f:. 0. Taking reciprocals gives eZ < 1/(1 - x), provided 1 - x is positive. 6.8 a) y E im cosh if and only if there exists x such that e2x - 2yeZ + 1 = O. This is a quadratic equation in eX, and so eX = y ± y'Y2=1. Since eZ must be positive, a suitable x exists only if y ~ 1. Similarly, y E im sinh 2z Z if and only if there exists x such that e - 2ye - 1 = 0, that is, if and only if eX = y ± Ji'+Y2. The appropriate x is log(y + Ji'+Y2), and exists for all y. b) These are all a matter of routine algebra. For example, sinh x cosh y + coshxsinhy = (1/4)(eX - e-X)(ell + e-II) + (eX + e-X)(ell - e-II)] = (1/4)(ex+II - e-x+II +eX-II _ e-(x+II) + eX+II +e-x+II - eX-II - e-(x+II] = (1/2)(ex+ II - e-(x+II») = sinh(x + y). c) sinh has a positive derivative throughout its domain, and so has a inverse function sinh-1 : IR ~ IR with positive derivative. Moreover

(sinh-1)'(x) = 1 = 1 = 1 1 2 cosh(sinh- x) VI + (sinh(sinh-1x))2 -./1 + x . 260 Real Analysis

In [0,00) the function cosh has a positive derivative, and so there is an inverse function cosh-1 : [000) -t [1,00), with positive derivative. Also, (cosh-1)'(x) =1/ sinh(cosh-1x) = I/Jx2 - 1. d) This amounts to solving the equations y = sinhx and y = coshx for x, something already done in part a). 6.9 Substitute u = logxj then du = dx/x, and so J(dx/x log x) = J(du/u) = logu = loglogx. Now loglogx -t 00 as x -t 00, and consequently the integral J200 (dx/ x log x) is divergent. By the integral test (Theorem 5.37) so is the series L:::'=3(I/nlogn). Since, for all integers k ~ 3,

kH ------,.----,-1 < l --dx < --1 (k + l)log(k + 1) - k xlogx - klogk it follows that n 1 n-l 1 L -1- ~ log log n -loglog3 ~ L -1-' r=4 r ogr r=3 r ogr from which it follows that 1 1 ~ ~ -1-n ogn t5n + log log 3 -31og 3' Since 1 dx - - in ~ 0, t5n t5n 1 = --- -- n logn n-l X log x

the decreasing sequence (t5n ) has a limit 15. Since K n ,..., loglogn, we require ee 5 ~ 2.85 x 1064 terms for the sum to exceed 5. 6.10 The same method can be used to compare the series

I ~ I with the integral I ~x I = log log log x . n=2f: n ogn og ogn j x ogx og ogx

The series diverges, but so slowly that after 10100 terms the sum is still less than 2.

Z a 1 a 1 6.11 Jt e-zxa- 1dx is convergent for all 0:, since e > Kx + and so e-zx - < a 1 I/Kx2 for all positive x. If 0: < 1 then the integral J; e-zx - dx is im• proper. Since e-zxa- 1 ,..., x a - 1 as x -t 0+, the integral converges if and only if 0: > O. To prove the functional equation, integrate by parts:

00 [_e-zxa-l]~ Z a 2 reo:) = -1 -e- (0:-I)x - dx=0+(0:-I)r(0:-1). Applying this repeatedly gives r(n) = (n - 1)!r(l) = (n - 1)1. Solutions to Exercises 261

6.12 We know that limy--+oo e-kyyo = O. Substituting log x for y gives k limx --+ oo x- (1ogx)O = O. In this latter limit, substituting l/t for x gives limt--+o+ tk (log t)° = O.

6.13 Since (l/n) logn -+ 0 as n -+ 00, it follows, by applying exp to both sides, that n 1/n -+ 1. 6.14 log{xlogx) = log x + log log x =1+ loglogx -+ 1 log x logx log x as x -+ 00. 6.15 By L'Hopital's rule, . aX - bX l' aX log a - bX10gb log a -10gb 11m = 1m = . x--+o eX - dx x--+o ex log e - dx log d log e - log d

6.16 Substituting u = st in the first integral, we have

by Exercise 6.11. Call the integrals C and S. Integrating by parts gives 00 1 -st 1 -st . d 1 as C =[ - -e cos at]00 --1 -e .- a sm at t = - - - , s 0 0 s s s and 00 1 1 a S= [--e-stsinat]00 - 1 --e-stacosatdt=-C, s 0 0 s s From the equations sC = 1 - as and sS = aC the required results follow. 6.17 The result is certainly true for n = 0, with Po (1 / x) = 1. If we suppose X2 inductively that f(n) (x) = Pn{1/x)e-1/ , then x2 f(nH) (x) = e-1/ (:3Pn{1/X) - :2P~{1/X)) = PnH(1/x)e-1/X2,

where PnH(l/x) = (2/x3)Pn{1/x) - (1/x2)P~(1/x) is a polynomial in l/x. If we now suppose inductively that f(n) (0) =0, then

f(nH) (0) = lim f(n)(x) - f(n){o) = lim ..!.P {1/x)e- l / X2 = o. x--+o X X--+OX n The Taylor-Maclaurin series of f is identically zero, and so cannot possibly converge to f. 262 Real Analysis

6.18 If we put x = n in the inequality eX > xnIn! we immediately obtain the desired result.

6.19 Comparing the lower sum, the integral from 1 to n and the upper sum gives log 1 + log 2 + ... + log(n - 1) ~ [x log x - xr ~ log 2 + log 3 + ... + log n . Hence log[(n-1)!] ~ nlogn-n+ 1 ~ log(n!), and so (n-1)! ~ nn/en-l ~ n!.

Chapter 7

7.1 a) Choose N so that llin - III < (./2 and IIgn - gil < (./2 for all n > N. Then, for all n > N, II Un + gn) - U + g) II ~ IIln - III + IIgn - gil < (..

b) Choose N 1 so that IIlnll < 11111 + 1 for all n > N 1. Choose N 2 so that IIln - III < (./2(lIgl1 + 1) and IIgn - gil < (./2(11111 + 1) for all n > N2 • Then, for all n > max {Nil N2 }, IIln' gn - I· gil = IIln(gn - g) + Un - J)gll ~ IIlnllllgn - gil + llin - IlIlIglI ~ (11111 + l)lIgn - gil + IIln - III (lIgll + 1) < (..

c) Since Un) -t I uniformly in [a,b], we may choose N 1 so that Iln(x)1 > 2 8/2 for all n > N1 and all x in (a,b]. Hence Iln(x)l(x)1 > 8 /2 for all 2 n > N1 and all x in [a, b]. Choose N2 so that IIln - III < 8 (./2 for all n > N2 • Then, for all n > max {Nil N2 } and for all x in [a, b], 1 I Iln(x) - l(x)1 IIn(x) - I(x) = Iln(x)l(x)/ < €. 7.2 Let In(x) = {l/n .ifx.is.rati~nal o If x IS irratIOnal. Then each In is discontinuous everywhere, but Un) tends uniformly to the zero function as n -t 00.

7.3 It is clear that In -t 0 pointwise. To show that the convergence is not uniform, we investigate the maximum value of In (x) in the interval [0,1]:

3 n 1 2 n 2 n 1 I~(x) =n x - - n (n + l)x =n x - (n - (n + l)x) , Solutions to Exercises 263

and so the maximum value, occurring when x =nl(n + 1), is

2 n 1 n+2 ( 1) n+I n . ()n: 1 . n + 1 = (n: l)n+l = n 1 - n + 1 (9.17) Thus IIfnll '" nle, and so certainly does not tend to zero. Now,

l 2 r 2 [xn+I xn+2 ] I n 10 fn(x) dx =n n + 1 - n + 2 °= (n + l)(n + 2) ,

which tends to 1 as n ~ 00. So the integral of the (pointwise) limit is J; 0 dx = 0, whereas the limit of the integral is 1. 7.4 Again it is clear that (fn) ~ 0 pointwise. Here we have IIfnll ~ lie, and so the convergence is not uniform. However, J; fn = nl(n + l)(n + 2) ~ 0 as n ~ 00. 7.5 Here IIfnll ~ 0 as n ~ 00, and so the convergence is uniform. On the other hand, f~(x) = nxn- I(l- x) - xn, and (f~) has pointwise limit g, where

g(x) = {O ~f 0~ x < 1 -11fx=1.

Since each f~ is continuous but 9 is not, the convergence cannot be uniform. 7.6 a) fn(O) = 0 for all n. If x> 0 then xl(x + n) ~ o. b) Since f~(x) = nl(x+n)2 > 0, fn is increasing in [0,00), and so IIfnll = sUP[O,b] Ifn(x)1 = bl(b+n). This tends to 0 as n tends to 00, and so the convergence is uniform. c) If we redefine Ilfnll as sup[O,oo) Ifn(x)l, we see that, for a fixed n, the function x f-t xl(x + n) increases steadily towards a limit 1 as x ~ 00. So IIfnll = 1 for all n, and so convergence to 0 is not uniform. 7.7 a) It is clear that fn(O) = 0 for all n, and that, for all x i= 0, nxl(l+nx) = xl[x + (lin)] ~ 1. b) For all x 2: b, Ifn(x) - 11 = 1/(1 + nx) ~ 1/(1 + nb). Sollfn - 111 = sUP[b,oo) Ifn(x) - 11 = 1/(1 + nb) ~ 0 as n ~ 00. Thus convergence is uniform in [b, 00). c) sup[O,oo) Ifn(x) - f(x)1 2: Ifn(lln) - f(lln)1 = 1/2. So the convergence in [0,00) is not uniform. 7.8 sUPR Ifn(x) - f(x)1 = lin ~ 0, and so (fn) ~ f uniformly in JIl On the 2 ~ other hand, SUPR I(In(X)) - (J(X))2 1 = SUPR 1(2xln)+(1In2)1 I(2n/n) + (1/n2 )1 ~ 2, and so the convergence is not uniform. 264 Real Analysis

7.9 For all x in [O,b], Inx2/(n3+x3 )j ~ na2/n3 = a2/n2. Since 2::::'=1(I/n2) is convergent, the given series is uniformly convergent in [0, b) by the Weier• strass test. 7.10 If x = 1, then xn/(xn + 1) = 1/2 for all n, and so the series diverges. If x> 1 then limn_H,o(xn/(xn + 1)) = 1, and so again the series diverges. If x = 0 then the series sums trivially to O. So suppose that 0 < x ~ a < 1. Then Ixn/(xn+ll < xn ~ an for all x in [0, a]. Since 2::::'=1 an is convergent, the series is uniformly convergent in [0, a] by the Weierstrass M-test. For all x in [0,00), l/n2(x+ 1)2 ~ l/n2, and so the given series is uniformly convergent in [0,00). l/(xn+ 1) does not tend to zero unless x > 1, and so the series is divergent for x in [0,1). So suppose that x ~ a > 1. Then 1/(xn + 1) ~ 1/(an + 1) < l/an, and so, by the Weierstrass M-test, the series is uniformly convergent in [a, 00). 7.11 Denote the sum of the series by 8(x); then clearly 8(0) = O. For x "lOwe have a convergent geometric series with first term x2 and common ratio 1/(1 + x2); its sum, after a bit of calculation, is 8(x) = 1 + x2. Since 8 is discontinuous at 0, the convergence in any interval containing 0 cannot be uniform. Consider the set J = (-00, -b]U [a, 00), where a, b> 0 and where we may assume without essential loss of generality that a ~ b. Then, for all x in J, x2/(1 + x2)n < 1/(1 + x2)n-1 ~ 1/(1 + a2)n-1, and so, by the Weierstrass test, the series is uniformly convergent in J. 7.12 For all x in [0,1], Ix/(n3/2 + n3/4x2)1 ~ 1/(n3/2 + n3/4x2) ~ l/n3/2. 3 2 Since 2::::'=1 l/n / is convergent, the given series is uniformly conver• gent in [0,1]. This technique will not work for the other series. We obtain the maximum value of x/(n3/4 + n3/2x2) by observing that its derivative, (n3/4 - n3/2x2)/(n3/4 + n3/2x2)2 is zero when x = n-3/ 8 • The maximum 9 8 3 4 3 2 2 9 8 value is l/n / • Thus, for all x in [0,1], Ix/(n / + n / x )1 ~ l/n / , and it follows by the Weierstrass M-test that the series is uniformly convergent in [0,1]. 7.13 For all x in [0,1], xn(1 - x)/n2 ~ l/n2, and so it is immediate by the Weierstrass M-test that the series is uniformly convergent in [0,1]. This simple argument will not work for the other series. From a simple calculus argument, the maximum value of xn(1 - x) is obtained when x = n/(n + 1). The value is [n/(n + 1)]n[I/(n + 1)] = [1 + (l/n)tn[I/(n + 1)]. Now [1 + (l/n)tn < 1 for all n ~ 1. Thus, for all n ~ 1 and for all x in [0,1], xn (l-x)/n ~ l/n(n+ 1). Hence the series is uniformly convergent in [0,1]. 7.14 For all x in [0,1], log(n + x) -logn ~ log(n + 1) -logn = 10g(1 + (l/n)). Solutions to Exercises 265

By the Mean Value Theorem, for all t > 0, log(1 + t) = log 1 + tic = tic, where 0< c < t. Certainly log(1 + (lin)) < lin, and so, for all x in [0,1], 2 (1 In) (log(n + x) - log x) ~ 1In • Hence the series is uniformly convergent in [0,1]. 7.15 (i) limn-too(anlan+d = 1/2, so R = 1/2. When x = 1/2 the se• ries is E~=o(1/(n + 1)) which diverges; when x = -1/2 the series is E~o( -1)n(l/(n + 1)), which converges. So the interval of convergence is [-1,1). (ii) limn-too lanlanHI = 1, so R = 1. The series diverges for x = -1 and converges for x = 1, so the interval of convergence is (-1, 1]. (iii) anlanH = n!(n + 2)nH I(n + 1)!(n + l)n = [1 + (1/(n + 1)]nH -t e n as n -t 00. So R = e. When x = e, the Stirling formula gives anx = n! en Inn::=:: nn+!e-nenInn = n!, and so the series diverges at ±e. The in• terval of convergence is (-e, e). (iv)a;;n = 2+ (lin) -t 2 as n -t 00. Hence n R = 2. When Ixl = 2, lanxnl = [2/(2 + (1/n))]n = [1 + (1/2n)t -t 1 2 e- / # °as n -t 00. So the interval of convergence is (-2,2). (v) anlan+l = (n + 2) log(n + 2)/(n + 3) log(n + 3) -t 1 as n -t 00. So R = 1. For x = 1 the series diverges (by the integral test); for x = -1 it con• verges by the Leibniz test. So the interval of convergence is [-1,1). (vi) anlan+l = (n!)2(2n+2)!/(2n)! ((n+ 1)!)2 = (2n+ 1)(2n+2)/(n+ 1)2 -t 4 as n -t 00. So R =4. If Ixl = 4, (n!)24n n2n+le-2n22n 1 2 lanxnl (2 ),::=:: 2! n / as n -t 00. = n . (2n) n+ e-2n = -1+ °

Hence the interval of convergence is (-4,4).

7.16 For ItI < 1, E~=o tn = 1/(1 - t). Differentiating term by term gives E~=l ntn- 1 = 1/(I-t)2. Hence E~=l ntn =t/(I-t)2. Finally, integrating term by term gives

~ n n+l t t dt t (1 1) d ~ n 1x (1 - t)2 (1 _ t)2 - 1 _ t t + =10 =10 1 ] z 1 = -1- + log(1 - t) = -1- + log(1 - x) - 1. [ - t o-x

7.17 sin 3x = sin 2x cosx+cos 2x sin x = 2sinx(l-sin2 x) +(1- 2sin2 x) sinx = 3 sin x - 4 sin3 x. Hence

00 x2nH 00 (3x)2nH 3 4sin x =3sinx - sin3x =3 ~)-lt (2 I)' -I)-It(2 )' . n=O n+. n=O n+l.

It follows that sin3 x = E~=o a2nHX2nH , where a2nH = (-I)n(3/4)(I• 32n )/(2n + I)!. 266 Real Analysis

7.18 Since 2::~=0 xr tends to 1/(l-x) as x -+ 00 uniformly in any closed interval contained in (-1, 1), and since 2::~=1 anxn tends to some function j (x) uniformly in any closed interval contained in (-R, R), it follows that, for all Ixl < min {I, R} (1 + x + x2+ ... + xn)(ao + alX + a2x2 + + anxn) = (ao + (ao + ar)x + (ao + al + a2)x2+ + (ao + al + ... + an)xn

tends to j(x)/(I-x) as n -+ 00. Thus (I/(I-x)) 2:::=0 anxn = 2::~0 snxn. For the last part, note that 10g(1 + x) = 2:::=0 anxn, where ao = 0 and an = (_I)n-l/n otherwise. 7.19 f'(x) = 1/";1 +x2, f"(x) = -x/(1 + x2)3/2, and so (1 + x2)f"(x) + x1'(x) = 0 for all x. Differentiating n times by Leibniz's Theorem gives (1 +x2)j(n+2) (x) +2nxj(n+l)(x) +n(n-l)j(n)(x) +xj(n+l)(x) +nj(n)(x) = o (n ~ 0), and putting x = 0 gives j(n+2)(o) = .:-n2j(n)(o). Since j(O) = 0 it follows that j(n)(o) = 0 for all even n. Since 1'(0) = 1, it follows that the coefficient of x2n+l in the Taylor-Maclaurin series is

qn=(_It(2n-I)2.....32. 12 = (_I)n (2n-I) 3.1 (2n + I)! (2n + 1)2.4 2n _ (_I)n (2n)! _ (_ )n (2n)! - (2n + 1)22 .42 •••. •(2n)2 - 1 (2n + 1)22n (n!)2 . Since qn/qn+l = (2n + 1)2/(2n + 2)(2n + 3) -+ 1 as n -+ 00, the radius of convergence is 1. Let Ixl = 1. Then, since (2n)2n+! e-2n 1 qn x (2n + 1)22nn2n+le-2n x n3/2 '

the interval of convergence is [-1,1].

Chapter 8

8.1 a) A(O) = 0 is clear. For A(-x) = -A(x), substitute u = -t in the integral. A is strictly increasing, since the integrand is positive. It is differentiable in (-1,1), since the integrand is continuous, and A'(x) = 1/";1 - x2 by the fundamental theorem. (This is actually a bit inaccurate, since the integral is improper - which is why I preferred to use the seemimgly more complicated parametric approach.) Solutions to l:.xercises 267

b) By Theorems 3.20 and 4.15, S = A-I : [-11"/2,11"/2] -+ [-1,1] exists. Since A(O) = 0 it follows that S(O) = O. Also, S'(x) = 1/[A'(A-1 (x))] = V1- (A-l(X))2 = V1- (S(x)( From this it fol• lows that S'(x) is differentiable in (-1,1), that S'(O) = 1 and that [S(X)]2 + [S'(xW = 1. Hence

l 2 S"(x) = (1/2)[1- (S(x))2r / (-2S(x)S'(x)) = -S(x).

Chapter 9

9.1 Use Stirling's formula: (l/n)(n!)l/n'" (1/n)(211"n)I/2n(n/e)-+ l/e. The Greek Alphabet

a A alpha /3 B beta "Y r gamma 6 L\ delta to E epsilon ( Z zeta TJ H eta 0, {) e theta

~ ~ xi £ I iota K K kappa A A lambda I-l M mu v N nu 0 0 omicron 11", tv II pi p P rho a E sigma T T tau v y upsilon ¢J,

[1] P. P. G. Dyke, An introduction to Laplace transforms and Fourier series. Springer, 2000. [2] D. A. R. Wallace, Groups, rings and fields. Springer, 1999. Index

Abel's test, 198 - below, 10 Abel's theorem, 208 - function, 74, 93 Abel, Nils Henrik (1802-1829), 198, 208 - sequence, 34 absolute magnitude, 18 bounded above absolutely convergent - function, 74 - integral, 155 - sequence, 34 - series, 58 bounded below additivity theorem, 224 - function, 74 analytic function, 118, 179 - sequence, 34 anti-differentiation, 119 Briggs, Henry (1561-1631), 170 antiderivative, 140 arbitrary constant, 143 Cartesian product, 63 arc-length function, 225 Cartesian product (of sets), 7 archimedean property, 9 Cauchy sequence, 42, 74 Archimedes of Syracuse (287-212 BC), 9 - is convergent, 43 arithmetic mean, 24 Cauchy's mean value theorem, 107 arithmetic progression, 24 Cauchy, Augustin-Louis (1789-1857), arithmetic-geometric inequality, 24 22, 107 associative law, 8 Cauchy-Schwarz inequality, 22, 221 axiom of Archimedes, 10 chain rule, 103 axiom of completeness, 10, 86 circular functions, 70, 217 - derivatives, 103 belongs to (E), 5 - sum formulae, 71, 218 binomial coefficient, 14 circular measure, 70 binomial series, 211 codomain (of a function), 63 - convergence at ±1, 211 commutative law, 8 binomial theorem, 14, 32 comparison test, 50 Bolzano, Bernhard Placidus Johann - "ultimate" version, 51 Nepomuk (1781-1848),86 - asymptotic version, 54 Boole, George (1815-1864), 7 - for integrals, 152, 153, 160 Boolean algebra, 7 compatibility Borel, Felix Edouard Justin Emile - with addition, 9 (1871-1956), 91 - with multiplication, 9 bounded complement (of a subset), 6 - above, 9 completing the square, 21 274 Real Analysis complex numbers, 5 Fibonacci sequence, 45 composition of functions, 67, 88 field, 8 conditionally convergent series, 59 function, 63 constant function, 67, 85 - as process, 63 continuity, 82 fundamental theorem of calculus, 140 - implies uniform continuity, 92 - sequential, 82, 90 Gamma function, 178 - uniform, 90 general principle continuous function, 128 - of convergence, 74 - at a point, 81 - of uniform convergence, 196, 197 - nowhere differentiable, 234 geometric mean, 24 - on its domain, 86 geometric progression, 24 convergent geometric series, 49 - integral, 151, 158 golden number, 46 - sequence, 30 graph (of a function), 64 - series, 48 Gregory's series, 209 cosecant, 72 Gregory, James (1638-1675), 209 cosine function, 70, 217 - derivative, 103 harmonic mean, 25 - is continuous, 88 harmonic series, 49, 171 cotangent, 72 - is divergent, 49 decreasing function, 74, 127 Heine, Heinrich Eduard (1821-1881), 91 defining property (of a set), 5 Heine-Borel theorem, 91 density (of rationals), 9 higher derivatives, 113 derivative, 99 Holder condition, 110 Descartes, Rene (1596-1650), 7 Holder, Otto Ludwig (1859-1937), 110 differentiable function, 99 differential coefficient, 99 identity function, 67, 85 differentiation, 99 image differentiation term by term, 193 - of a function, 64 Dirichlet, Johann Peter Gustav Lejeune - of a point, 63 (1805-1859), 84 improper integral discriminant, 22 - first kind, 150 disjoint sets, 6 - second kind, 158 dissection, 119 increasing function, 74, 127 distance (between functions), 182 indefinite integral, 143 distributive law, 8 inequalities, 18 divergent infimum, 11, 93 - sequence, 30 infinite integral, 150 domain (of a function), 63 infinitely differentiable function, 118, dummy symbol, 121 179 inner product, 221 e, 41, 167, 171 integers, 5 - is irrational, 178 integral part, 66 element (of a set), 5 integral test, 156 empty set (0), 6 integrand, 152 epsilon (f), 28 integration by parts, 144 equality of functions, 66 integration by substitution, 146 Euler's constant, 172 integration term by term, 192, 193 Euler, Leonard (1701-1783),172 interior, 99 expx, 170 intermediate value theorem, 86, 94 exponential function, 170 intersection (of sets), 6 - in growth of bacteria, 177 interval, 11 Index 275

- closed, 11 Napier, John (1550-1617), 170 - open, 11 natural logarithm, 170 inverse circular functions, 95 natural numbers, 5 - derivatives, 112 Newton, Isaac (1643-1727), 59, 99 inverse cosine, 95, 219 norm - derivative, 112 - of a function, 181 inverse function, 94, 110, 168 - of a vector, 221 inverse sine, 95, 219 null sequence, 31 - derivative, 112 inverse tangent, 96 open covering, 91 - derivative, 112 ordered field, 9 irrationality - of e, 178 - of..;2, 3 parametric equations, 220 isolated discontinuity, 130 Pascal triangle, 15 - identity, 15, 115 Kronecker, Leopold, 1823-1891, 10 Pascal, Blaise, 1623-1662, 15 1r, 219, 228 L'Hopital's rule, 108 - calculation, 195 L'Hopital, Guillaume Franc;ois Antoine, pointwise convergence, 183 Marquis de (1661-1704), 108 polynomial function, 67 Laplace transform, 179 - is continuous, 86 Laplace, Pierre Simon (1749-1827), 179 power series, 201 left limit, 78 principle of induction, 12 Leibniz test, 59 - second principle, 16 Leibniz's theorem, 114, 214 product rule, 101 Leibniz, Gottfried Wilhelm (1646-1716), 59,99 quadratic function, 21 length of a curve, 222 quotient rule, 102 limit (of a function) - as x -+ -00, 73, 80 radian, 70 - as x -+ a, 75 radius of convergence, 203 - as x -+ 00, 73 limit (of a sequence), 28 - nth root test, 205 lnx,170 - ratio test, 204 logx, 170 ratio test, 56 logarithm, 165 rational function, 67 - in calculation, 170 - is continuous, 86 lower bound, 10 rational numbers, 5 - greatest, 10 real numbers, 5 lower integral, 120 reciprocal (of a function), 67 lower sum, 119 reciprocal law, 8 rectifiable curve, 222 Maclaurin's theorem, 117 reduction formula, 146 Maclaurin, Colin (1698-1746), 117 refinement (of a dissection), 121 mean value theorem, 106 Riemann integrable function, 121 - Cauchy, 107 Riemann, Georg Friedrich Bernhard - integrals, 139 (1826-1866), 61, 119 modulus, 18 right limit, 78 monotonic decreasing Rolle's theorem, 105 - sequence, 37 Rolle, Michel (1652-1719), 105 monotonic function, 74, 127 monotonic increasing sandwich principle, 37 - sequence, 37 scalar product, 221 276 Real Analysis

Schwarz, Karl Hermann Amandus, Taylor-Maclaurin theorem, 117 1843-1921, 22 Theorem of Pythagoras, 3 secant, 72 transitive law, 8 sequence, 27 triangle inequality, 182 - as function, 64 trichotomy law, 9 series, 48 set, 5, 27 ultimately monotonic, 37 sine function, 70, 79, 217 uniform convergence, 183 - derivative, 103 - of a sequence, 183 - is continuous, 88 - of a series, 192 singleton sets, 5 - of power series, 201 square roots (calculation), 39 uniformly continuous function, 90, 128 Stirling's formula, 215, 230 union (of sets), 6 Stirling, James (1692-1770), 215 unit circle, 70 strictly decreasing function, 73, 95, 109 upper bound, 9 strictly increasing function, 73, 94, 109 - least upper bound, 9 strictly monotonic function, 74 upper integral, 120 subset, 5 upper sum, 119 - proper subset, 6 sum - of functions, 66 value (of a function), 63, 64 sum (of a series), 48 supremum, 11, 93 Waerden, Bartel Leendert van der (1903-1996), 234 tangent function, 72 Wallis's formula, 229, 233 - continuity, 88 Wallis, John (1616-1703), 229 Taylor's theorem, 116, 150 Weierstrass M-test, 197 Taylor, Brook (1685-1731), 116 Weierstrass, Karl Theodor Wilhelm Taylor-Maclaurin series, 118 (1815-1897), 197