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Solutions to Exercises

Chapter 1

1.1 If x = min were rational, then x2 = m 2 /n2 would also be rational. Thesquare of /2+V3 is 5+2V6.1f 5+2V6 = p/q E Q, then V6 = (p-5q)/lOq EQ, and this is not the case.

1.2 An B = {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. If x = 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 beentrue.

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 differencebetween 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 numberq=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- rn +ll/(Ir).

1.11 If n =1 then the right hand side is (x2 - 2x + 1)/(x - 1)2 =1. So true forn = 1. Let n ~ 1 and suppose true for n. Then

nxn+l - (n + l)xn + 11 + 2x + ... + (n + l)xn = (x _ 1)2 + (n + l)xn

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 ifn(.;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. Thenqn+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 allk < 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 ifa = 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 greaterthan n times its smallest term, so not greater than nlx2/xlln-l. HencenIX2/Xlln-l < 1/(I-lx2/xll = IXll/(lxd -lx21).

1.21 (3x+2)/(x+l) <1 ¢=} (3x+2)/(x+l)-1<0 ¢=} (3x+2-x1)/(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 giveninformation 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:

0.0001100,000,000

2.2 If k = 0 then the sequence is (1,1,1, ...), with limit 1. If k > 0, thenII/nk - 01 < € whenever n > (I/€)l/k. Hence (I/nk

) ~ O. If k = -l < 0,then I/nk = n' > M for any given M > 0, provided n > M l

/l

• (Note that,for any positive a in IR and any positive integer k, we can define al / k assup {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 onlyif n > 5,000 + y'25,OOO,OOO - 1, that is, if and only if n > 9,999.

2.4 n2 + 2n ~ 9999 ¢=} n ~ -1 + VI + 9999 = 99.

2.5 an ~ -00 if and only if, for every K > 0 there exists a positive integer Nwith 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 forall k < n. Then an = (1/2)[2 + 4(-(I/2)t-

l + 2 + 4(-(I/2)t-2

] =

240 Real Analysis

(1/2)[4 - 8(-(1/2))n + 16(-(1/2)tl = 2 + 4(-(1/2)t. It is then clearthat (an) --* 2.

2.7 Suppose that f3 > B. There exists N such that Ibn - f31 < f3 - B for alln > N. Thus f3 - bn < f3 - B and so bn > B for all n > N. This is acontradiction to the definition of B.

2.8 If lanl ~ A, then -A ~ an ~ A. So "bounded" implies "bounded aboveand below". Conversely, suppose that A ~ an ~ B. If 0 ~ A ~ B thenlanl = an ~ B. If A ~ B ~ 0 then lanl = -an ~ -A. If A ~ 0 ~ B thenlanl ~ 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 thatIan - al < € for every n > N. It follows that there exists a positive integerN', 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 forall 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 soIbm - al < f.

2.11 Suppose, for a contradiction, that L < O. Taking f = ILI/2 = -L/2, weknow that there exists a positive integer N such that Ian - LI < ILI/2 forall 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)ffor 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 fromExercise 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}) areboth convergent. The final statement follows from the observation that

bn = max {an, bn} + min {an,bn} - an'


2.15 al /n = 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,and so, by the sandwich principle, (hn ) --t O. Hence (a l / n ) --t 1. If a = 1the sequence is constant, with limit 1. If 0 < a < 1 then a = l/b, whereb> 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 (2n + 3n)l/n = 3[1 + (2/3)np/n < 3(2l /n).Thus 3 < (2n + 3n)l/n < 3(2l /n), and so, by the sandwich principle andthe result of the previous exercise, ((2n + 3n)l/n) --t 3.

2.17 Ian - 01 = la; - 031/(a; + anO + 0 2) = la; - 031/[(an + ~0)2 + ~02] ~

la; - 0 3 1/(3/4)02. So, for a given f > 0, choose N so that, for all n > N,la~ - 0 3

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. Thenlanl < f.

b) No. Consider an = (_l)n.

2.18 a) It is clear that an ~ 0 for all n. Certainly al < 1. Suppose inductivelythat 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 limit0, which must be non-negative, and satisfies 0 = (30+1)/(0+3). Thus0=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 positivelimit 0, satisfying 0 = (20 + 3)/(0 + 2). Thus 0 = V3.

2.20 If a limit 0 exists, then it is positive, and satisfies 0 2 = 2 + 20j thus0= 1+ V3. The sequence begins (approximately) (2,2.45,2.62, ...), whichsuggests 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 supposeinductively 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 everyan is positive, 0 < an < 1 + V3 for all n. Since x2 - 2x - 2 takes negativevalues 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 equals1 + V3.

242 Real Analysis

2.21 By the arithmetic-geometric inequality (1.26), we know that an ~ bn forall n ~ 2. Since an+l - an = t(an + bn) - an = t(bn - an) ~ 0, it followsthat (an) is decreasing. Since bnH/bn = .,fanbn/bn = y'an/bn ~ 1, itfollows that (bn) is increasing. Since an ~ bn ~ b2 and bn ~ an ~ a2, bothsequences converge: say (an) -+ a, (bn ) -+ (3. But then a = (a + (3)/2, andso 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, thenlam - ani < f for all m > n > N I . In the same way, for the second sequencelam - ani ~ 2/2n-! ~ 1/2n- 2 < f for all m > n > N2, where N2 is chosenso that 2N2 - 2 > l/f.

2.23 Let m, n and N be positive integers such that m > n > N. The givenproperty 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) <kN - I la2 - all/(l - k). If, for a given f > 0, we now choose N so thatKN-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 aCauchy 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 parta) 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 sohas 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


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, andso 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 acertain point. Hence a; < an, and so 2:~=1 a; is convergent by thecomparison test.

e) False. Let a2k = l/k for k =0,1,2, ..., and otherwise let an =O. Thenan+l + ... + a2n = I/k, where k is defined uniquely by the propertyn+I :c:; 2k :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 thesecond series, since (n +3) / (n4 - 1) '" 1/ n3 • The third series diverges, since(n + I)/Jn3 + 2 '" I/n1

/2

.

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 asN -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)3n!]/[(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, itfollows from the comparison test that L:~=omax {an, bn} is convergent.

2.37 This follows from the previous exercise, since (anbn)1/2 ~ max {an, bn}. Toshow the converse false, let

_ {1 if n is even b _ {1 if n is oddan - 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 oddan - 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 forall 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 geometricseries L:~=N+l aN+lRn-N-l. Hence the series L:~=l an is divergent.

2.40 an+I/an = (n + l)kan+l /nkan = [(n + l)/n]ka -+ a as n -+ 00, and so theseries 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 theseries are convergent. The second series is absolutely convergent, the firstis not.

2.42 Let an = l/n if n is odd, and an = 1/n2 if n is even. Then (an) -+ 0 butis not monotonic decreasing. L:~=l (-l)n-l an is divergent, since the sumof the positive terms increases without limit, while the sum of the negativeterms 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 sum38/2.

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 sum8/2.


Chapter 3

y

2

-2 -1 0------+-----x2

-1

-2

245

(The dots serve to indicate that the value of the function at each integer nis n.)

3.2 a)

y

2

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

2

b)y

2

--~----f'--~--_ X2

246

c)y

2

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

2

-2

Real Analysis

3.3 a) dom(g 0 J) = {x E IR : 2x - 3 ~ O} = [3/2,00), and (g 0 J)(x) =-/2x- 3.

b) dom(g 0 J) = {x E IR : -/4 - x2 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 ofdivision 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(J 0 (g + h»)(x) = (2x)2 = 4x2, ((J 0 g) + (J 0 h»)(x) = x2 + x2 = 2x2.

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 rationalfunctions. If q and r are rational functions, then, for each x, (q 0 r)(x) isobtained from r(x) by repeated applications of addition, multiplication anddivision. 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.


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

thus f 0 9 =9 0 f =i.

247

3.12

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) = -1-, (q 0 p)(x) =q(l/x) = 1 - - j

-x x

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

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

9 9 i f 9 r Pp p q r i f 9q q r P 9 i fr r p q f 9 i

3.9 This follows immediately from the identity cos2 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 + <p) + sin(8 - <p) =2sin 9 cos <p. Now letx =8 + <p, y =8 - <p. Then 8 = (x + y)/2, <p = (x - y)/2, and the resultfollows. Similarly cos(8 + <p) + cos(8 - <p) = 2cos 8 cos ¢, and the resultfollows, with x and y defined as before.

( )sin(x+y) sinxcosx+cosxsiny

tan x +y = = .cos(x + y) cos x cos y - sin x sin y

Dividing the numerator and denominator by cos x cos y gives the result.

3.13 (1 - t2 )/(1 + t2 ) = [cos2 (0/2) - sin2 (8/2))/[cos2 (9/2) + sin2 (9/2)) = cos 9,2t/(l + t2 ) = [2cos(8/2)sin(9/2))/[cos2 (9/2) + sin2 (9/2)] = sin8, tan 9 =sin 0/ cos 8 = 2t/(1 - t2 ).

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 > 0there exists X such that f(X) > A - f, since otherwise A - f would be anupper bound for f. Since f is increasing, it follows that A - f < f(x) ~ Afor 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 thatIJ(-x) - LI < f. for all x > M, that is (writing -x as y) if and only if forall 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 everyx in [a, b], and so I is bounded both above and below. The result doesnot hold for an open interval: if I(x) = 1/(1 - x) (x E [0,1)), then I isincreasing, but is not bounded above.

3.17 Since B ~ A, we have that {f(x) : x E B} ~ {f(x) : x E A}. It follows 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 givesthat 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 notexist.

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 ~ 0x2 - 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 continuous 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 :5I/(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 rationalI/(x) - I( )1 = x2 < f.2 < f. if x is irrational.


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

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

Thus f is continuous at 1. Suppose finally that 0 < x < 1, so that x 2 :f:. X.

If x is rational, and if, for all n ~ 1, we define an = x + (V2In), thenlimn--+oo an = x, whereas limn--+ oo f(an) = limn --+ oo (an )2 = x2 :f:. f(x).If x is irrational, we define bn to be any chosen rational number in theinterval (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 isrational 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 forall x in [a, b], we have g(a) = f(a) - a ~ 0 and g(b) = f(b) - b ~ O. If eitherof these inequalities is an equality then the result is clear, with c = a orc = b. Otherwise, by the intermediate value theorem (Theorem 3.12) thereexists 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 thata < 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 valuetheorem (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 supposethat 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' ,

where x E (-1, 1). Then (x2") -+ 0, and so, by continuity at 0, the sequence(J(x2n )) -+ f(O). But (J(x2n )) is the constant sequence (J(x)) , with limitf(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. Thenf(anx) = bnf(x) for all x in [0, a-n], and so there exists N in N such that

if n > N. Hence, defining 6 as a-N , we see that If(x) - f(O)1 < f for all xin (-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 forevery 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 forall x, y in [a, b] such that Ix - yl < 61 , and there exists 62 > 0 suchthat If(x) - f(y)1 < f/2 for all x, y in [b, c] such that Ix - yl < 61• Leto= min {01,02}, and let x, y in [a,c] be such that Ix - yl < o. If bothx and yare in [a, b], or if both x and y are in [b, c], then it is clear thatIf(x) - f(y)1 < f/2 < f. For the remaining case we may suppose withoutessential loss of generality that x < b < y. Then Ib - xl < 6 ~ 61 and soIf(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) =x, and (71"/2) - 0 E [0,71"]. It follows that cos- 1 x = (71"/2) - o.

3.35 As an extreme case, consider the constant function Ck , defined by the rulethat Ck(X) = k for all x. The image of the function is {k}, but there is noinverse function from {k} to lit

3.36 Since 8+2x-x2 =9-(x-1)2, the function has maximum value 9, obtainedwhen x = 1. Now y = 8 + 2x - x2

¢::::> x2- 2x + (y - 8) = 0 ¢::::> x =

1 ± ,;g=y. There is an inverse function f- 1 : (-00,9] -+ [1,00) given byf-l(y) = 1 +,;g=y.

3.37 The image of f is IR\ {O}, and this is the domain of the inverse function. Theformula is obtained by observing that y = 1/(1 - x) ¢::::> x = 1 - (l/y).So f-l(y) = 1- (l/y) (y E IR \ {l}).


Chapter 4

251

(9.16)

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

f(x) - f(O) _ M _{ 1/-/X if x> 0x - 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 everyf > 0 there exists 6 > 0 such that

If(xl =~(c) - f'(C)! < f

for all x in dam f \ {c} such that Ix - cl < 6. Then (9.16) holds for allx in dam f n (-00, c) such that Ix - cl < f, and it also holds for x indam f n (c, 00) such that Ix - cl < f. Thus the left derivative and rightderivative 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, thenx(x -1) < 0, and so (f(x) - f(O))/(x -0) = -x(x-1)/x = -x+ 1 -t 1as 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 havethat 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' isnot continuous at O.

4.4 No: if f(x) = x and g(x) = -x, then max {f,g} and min {f,g} are notdifferentiable at O.

4.5 Let n = -m, where m is a positive integer. Then f(x) = l/xm , and so1'(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,

1· tan x - x l' tan2 x I' 2tanx + 2tan3 x1m = 1m -- = 1m ------

x-tO x3 x-tO 3x2 x-tO 6x

= lim 2 + 8 tan2

x + 6 tan4

x = 1x-tO 6 3

Again, by L'H6pital's rule,

1· sinx - xcosx l' xsinx l' sinx 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 conditionin 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))/(xc) ~ O. Since f is differentiable, the two limits are both equal to f'(e), andso 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, andchoose an integer N such that liN < t. Let m, n > N. Then, for somee 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.

4.11 By Theorem 4.15 (cos-1)'(y) = 1/[- sin(cos-1 y)]. Since 0 < cos-1 Y < 11',

it follows that sin(cos-1 y), being positive, is equal to Jl - cos2 (cos- 1 y) =J1=Y2. Hence (cos-1)' (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 valuethroughout [-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].

4.16 Since f'(x) = -msin(msin-1 x)IJl- x2 , if follows that Jl- x2f'(x) =-m sin(m sin-1 x). Differentiating again, we obtain (after some easy algebra)

(1- x2)f"(x) - xj'(x) + m 2 f(x) = O.

Hence the required result is true for n = O. Now suppose inductively that


Then, differentiating, we have

[(1- x2)j(n+2) (x) - 2xj(n+l) (X)] - (2n - 1)[xj(n+1) (x) + j(n) (X)]

+ (m2 - n2+ 2n -l)j(n)(x) = 0,

and collecting terms gives us

253

Hence, by induction, the result holds for all n ~ O.

Now put x = 0 to obtain j(n+2)(0) = (n2 - m 2)j(n)(0) for all n ~ O. Sincej(O) = 1 and 1'(0) = 0, we see that j(n)(o) = 0 for all odd n, while foreven n we have j(n) (0) = [(n - 2)2 - m2] ... [22 - m2J[-m2].

4.17 If j(x) = ao + a1X + ... + anxn, then j(r)(x) = r!ar + positive powers ofx 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 ~ nr. 0 If r > n.

Chapter 5

5.1 In every subinterval of any dissection D, the supremum and infimum ofthe function is k, and so U(Ck , D) = .c(Ck , D) = k(b - a). If follows thatI:Ck = k(b - a).

5.2 Let Dn = {O'~'~"'" n;;l, I}. Then, using Exercise 1.7, we see thatU(J, Dn) = (lJn) L:~1 (i2Jn2) = (lJn3 )(12 + 22 + ... + n2) = (1/6)[1 +(1/n)][2+(1/n)] while .c(J, D n) = (1/n3)(12+22+.. +(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,

and so j is Riemann integrable. Since 101 f lies between the upper andlower sums, both of which have limit 1/3 as n -+ 00. we also deduce thatI; j = 1/3.

5.3 Choose a dissection D = {xo, Xl, ... , xn } containing c. Then, recalling thatM i and mi are (respectively) the supremum and the infimum of j in theopen subinterval (Xi-l,Xi), we see that Mi = mi = 0 for all i, and so

UU, D) = .c(J, D) = O. Hence J;f - Jo1j $ U(J, D) - .cU, D) = 0, and

so f E 'R[O, 1]. Since Jo1 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 casewhere 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 whichthe subinterval (Ci - 8, c; + 8) features, for i = 1,2, ... , k. In each of theintervals [a, Cl -8], [Cl +8,C2-8], ... ,h-l +8,c1<-8], [Ck+8, b] the functionis continuous, and so we can arrange for the contribution of those intervalsto U(J, D) - £(J, D) to be less than €/2. The contribution of each of theintervals [Ci - 8, Ci + 8] cannot exceed 28(M - m) = € / 2k, and so the totalcontribution 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 adjustmentsin the argument are necessary to cope with the case where Cl = a and/orCk = 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 theinterval [a, b) n (c - 8, C + fJ), an interval whose length is at least 8. ThusJ: 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 forall 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 ord = b.) Then U(J' D) - .c(J, D) = E:=1 (U(J, Di) - .c(J, Di)). If, for agiven € > 0, we choose D so that U(J, D) - .c(J, D) < €, then certainlyU(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 thatI/(x) - l(y)1 < €/2(b - a) for all x, y in [a, b) such that Ix - yl < 8. Choosen 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 followsthat 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.


5.9 Let D m be the dissection {0,1/22m,1/22m-l, ... ,1/2,1}. In all but thefirst subinterval, the function is constant, and so the supremum and infimum are equal. In the first subinterval the length is 1/22m , the supremumis 1/22m and the infimum is _1/22m+1. Hence U(J, Drn ) - .c(J, Dm ) =(1/22rn )[(1/22rn ) + (1/22m+1)] = 3/24m+I , and this can be made less thanany given € by taking m sufficiently large. Hence f E R[O, 1]. The valueof the integral is the limit of U(J, Dm ), namely the sum of the geometric

. 1 1 1 1 11Th . 1/ (1 1) _ 2senes . 2 - 2 . 4" + 4" • 8 - .. '. e sum IS 2 + 4" - 5'

5.10 Let f(x) = g(x) = x. Then fo1

f . 9 = f; x2dx = h but Uo1

f) Uolg) = ~.

5.11 Properties a) and b) follow immediately from Theorem 5.15. It followsfrom Theorem 5.14 that (J, I) ~ 0, and from Exercise 5.6 that (J, I) = 0only if f = O. The inner product (kf + g, kf + g) is non-negative forall k. That is, k2(J,1) + 2k(J, g) + (g,g) = 0 for all k. It follows that thediscriminant 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).

5.14 Since f(x) = !x2J: g(t) dt - x Jox

tg(t) dt + ! Jox

t2g(t) dt, we may deducethat 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).

5.15 f(x) = g(u) - g(v), where u = x2, V = x3 , and g(u) = f; t5 /(1 + t4 ) dt.Hence f'(x) = [u5 /(1 + u4 )] • 2x - [v5 /(1 + v4 )] . 3x2 = 2xll /(1 + x8 )

3xI7 /(1 + x I2 ).

5.16 r/4

Jo

tan2 xdx= [tanx-x]~/4=1-(71"/4);

r/2 1 r/2 1 1 1r/2 1Jo

sin2

xdx="2Jo (1-CoS2x)dx="2[x-"2sin2xL =:i'lr2.

5.17 The argument takes no account of the indefiniteness of an indefinite integral.

256 Real Analysis

Hence

f" /2 [],,/2 ,,/25.18 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"/2In =1

0x sinn x dx = [(- cosx)x sinn- 1X]~/2

1"/2+ 10 (cosx)[sinn- 1x+(n-1)xsinn-2xcosx]dx

1"/2 1"/2=0+ 1

0sinn-1xcosxdx+(n-1) 1

0xsinn- 2x(1-sin2x)dx

[1 ] ,,/2

= ~ sinn x 0 + (n - 1)In - 2 - (n - l)In

1= - + (n - 1)In - 2 - (n - l)In .n

1 n-1In = 2' +-- In- 2 •

n n

So 13 = ~ + ~h = ~,and 14 = 116 + ~(i + ~Io) = i + 634

1f2 .

5.19

f1 x5dx 1 f2 du 1 [ ]2 11

0v"I+X6 = {; 1

1..;u (with u = 1 + x

6) = 3 .;u 1 = 3(V2 - 1).

1,,/2 sinxdx 13 -du 1(3 )2 = -2 (with u = 3 + cos x) = 12 .

o + cos x 4 U

1"/4 113 110 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= 1f fo" f(sinx) dx - I, and the result follows. Since cos2 x = 1 - sin2 x,we can apply the result to the given integral I, making the substitution

u =cos x, obtaining

I - ~ r sinxdx = ~ f-1

-du _ ~ [tan-1 u] 1 = 1f2- 2 10 1 + cos2 X 2 11 1 + u2 - 2 -1 4'


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

Rn = 1 r (x _ t)n-1 f(n)(t) dt(n-l)!Ja

= 1 [_ (x - t)n f(n) (t)] t=x _ 1 l x_ (x - t)n f(n+1) (t) dt

(n - I)! n t=a (n - I)! a n

= (x - a)n f(n) (a) + Rn+ln!

and the result follows.

5.22 Since x/VX6 + 1 ~ 1/x2 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"# 1/2x2 + 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

F x) _ (x + 1)2 - K 2(2x2+ 1)

( - (x + I)V2x2 + l(x + 1 + K V2x2 + 1

x2(1 - 2K2) + 2x + (1 - K)= -;"(x-+-=-1):-'-";"2:=X:;<'2=+~IO-;-(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) = 0for n = 0,1,2, .... Trivially, E:'=l fen) is convergent, but, for N E N,

Ji' f = (N - 1) J01x(1 - x) dx = (N - 1)/6, and so Jo

oof 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

Hence f: N1T Isin x/xl dx = f: 1T 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 theharmonic series it follows that I is not absolutely convergent.

5.26 l/Jsinx '" l/yIX as x -t 0+. Hence the first integral converges. Sincesinx/x2

'" l/x as x -t 0+, the second integral diverges. Since, byL'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 lowerlimit. We have seen that the first integral converges, but is not absolutelyconvergent. The second integral is easily seen to be absolutely convergent.

rOO sin2

x [ (1)] 00 roo ( 1)10 ~ dx = sin2 x -; 0 - 10 -; 2sinxcosxdx

100 sin 2x 100 sin u= 0+ --dx = --du, where u = 2x.o x 0 u

Chapter 6

6.1 L(2n ) = nlog2, and log2 = f12(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. Itfollows 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 ifo< 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,g(l) = 0, and g'(x) = (l/x) - (1/x2), which is negative if x < 1 andpositive if x> 1. Thus g(x) = (x - l)g'(c) (where c is between 1 and x) ispositive 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 + (x2 /2) +(x3 /6)(2/t3 ) > X + (x2/2). 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 + (x2 /2) +(x3 /3) - (x4 /24)(6/t4

) < X + (x2 /2) + (x3 /3).


6.5 Using integration by parts, we have

£( ) - xm+l(logx)n Jxm+

1(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) = x2 /2) we have

259

x2

3 x2

3 [x2

]£(1,3) = 2(logx)3 - 2£(1,2) = 2(logx)3 - 2 2(logx)2 - £(1,1)

1 2 3 3 2 2 3 [x2 x2]=-x (logx) - -x (logx) + - -logx - -

2 4 2 2 21= 4x2(2(logx)3 - 3(logx)2 + 310gx - 3).

6.6 By L'Hopital's rule,

1· 10g(cosax) I' -a sin ax/ cos ax a l' sinax I' cosbx1m = 1m =-, 1m --, 1m--

x--to10g(cosbx) x--tO -bsinbx/cosbx b z--tosinbx z--tocosax

= ~ . lim a cos ax .1 = a2

.

b z--tO bcos bx b2

6.7 By Taylor's Theorem, e- x = 1 - x + (x2 /2)e-9z , where °< () < 1. Sinceboth 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. Takingreciprocals 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 sinhif and only if there exists x such that e2z - 2yeZ

- 1 = 0, that is, if andonly if eX = y ± Ji'+Y2. The appropriate x is log(y + Ji'+Y2), andexists 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 ainverse function sinh-1 : IR ~ IR with positive derivative. Moreover

(sinh-1)'(x) = 1 = 1 = 1cosh(sinh-

1x) VI + (sinh(sinh-1x))2 -./1 + x2 .

260 Real Analysis

Since

In [0,00) the function cosh has a positive derivative, and so there isan 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 forx, 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 theintegral J2

00(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,

1 l kHdx 1------,.----,- < -- < --

(k + l)log(k + 1) - k xlogx - klogk

it follows that

n 1 n-l 1L -1- ~ log log n -loglog3 ~ L -1-'r=4 r ogr r=3 r ogr

from which it follows that

1 1-1- ~ t5n + log log 3 ~ -313'n ogn og

1 in dxt5n - t5n - 1 =-- - -- ~ 0,n logn n-l X log x

the decreasing sequence (t5n ) has a limit 15. Since K n ,..., loglogn, we requireee

5~ 2.85 x 1064 terms for the sum to exceed 5.

6.10 The same method can be used to compare the series

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

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

6.11 Jt e-zxa- 1dx is convergent for all 0:, since eZ > Kxa+1 and so e-zxa- 1 <I/Kx2 for all positive x. If 0: < 1 then the integral J; e-zxa- 1dx is improper. Since e-zxa- 1 ,..., x a - 1 as x -t 0+, the integral converges if andonly if 0: > O.

To prove the functional equation, integrate by parts:

reo:) = [_e-zxa-l]~ -100

-e-Z (0:-I)xa -2dx=0+(0:-I)r(0:-1).

Applying this repeatedly gives r(n) = (n - 1)!r(l) = (n - 1)1.


6.12 We know that limy--+oo e-kyyo = O. Substituting log x for y giveslimx--+oo x-k(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.

6.14log{xlogx) = log x + log log x =1+ loglogx -+ 1

log x logx log xas x -+ 00.

6.15 By L'Hopital's rule,

1. aX - bX l' aX log a - bX10gb log a -10gb1m = 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

C [ 1 -st ]00 1001 -st . d 1 as= - -e cos at - - -e . - a sm at t = - - - ,

s 0 0 s s s

and

[ 1 ]00 1001 aS= --e-stsinat - --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 supposeinductively that f(n) (x) = Pn{1/x)e-1/ X2 , then

f(nH) (x) = e-1/x2

(: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 ..!.Pn{1/x)e- l / X2 = o.x--+o X X--+OX

The Taylor-Maclaurin series of f is identically zero, and so cannot possiblyconverge to f.

262 Real Analysis

6.18 If we put x = n in the inequality eX > xnIn! we immediately obtain thedesired 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 N1 so that IIlnll < 11111 + 1 for all n > N 1. Choose N2 so thatIIln - 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 N1 so that Iln(x)1 >8/2 for all n > N1 and all x in (a,b]. Hence Iln(x)l(x)1 > 82 /2 for alln > N1 and all x in [a, b]. Choose N2 so that IIln - III < 82(./2 for alln > N2 • Then, for all n > max {Nil N2 } and for all x in [a, b],

I1 I Iln(x) - l(x)1

In(x) - I(x) = Iln(x)l(x)/ < €.

7.2 Let

In(x) = {l/n .ifx.is.rati~nalo If x IS irratIOnal.

Then each In is discontinuous everywhere, but Un) tends uniformly to thezero function as n -t 00.

7.3 It is clear that In -t 0 pointwise. To show that the convergence is notuniform, we investigate the maximum value of In (x) in the interval [0,1]:

I~(x) =n3 xn- 1- n2 (n + l)xn =n2 xn- 1 (n - (n + l)x) ,


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

( )n 1 n+2 ( 1)n+I

n2

. n: 1 . n + 1 = (n: l)n+l = n 1 - n + 1

Thus IIfnll '" nle, and so certainly does not tend to zero. Now,

263

(9.17)

rl [xn+I xn+2 ] I n2

10 fn(x) dx =n2

n + 1 - n + 2 °= (n + l)(n + 2) ,

which tends to 1 as n ~ 00. So the integral of the (pointwise) limit isJ; 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, andso the convergence is not uniform. However, J; fn = nl(n + l)(n + 2) ~ 0as n ~ 00.

7.5 Here IIfnll ~ 0 as n ~ 00, and so the convergence is uniform. On the otherhand, 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 theconvergence is uniform.

c) If we redefine Ilfnll as sup[O,oo) Ifn(x)l, we see that, for a fixed n, thefunction 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 isuniform in [b, 00).

c) sup[O,oo) Ifn(x) - f(x)1 2: Ifn(lln) - f(lln)1 = 1/2. So the convergencein [0,00) is not uniform.

7.8 sUPR Ifn(x) - f(x)1 = lin ~ 0, and so (fn) ~ f uniformly in JIl On the

other hand, SUPR I(In(X))2

- (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) isconvergent, the given series is uniformly convergent in [0, b) by the Weierstrass test.

7.10 If x = 1, then xn/(xn + 1) = 1/2 for all n, and so the series diverges. Ifx> 1 then limn_H,o(xn/(xn + 1)) = 1, and so again the series diverges. Ifx = 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 uniformlyconvergent in [0,00).

l/(xn+ 1) does not tend to zero unless x > 1, and so the series is divergentfor 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 convergentin [a, 00).

7.11 Denote the sum of the series by 8(x); then clearly 8(0) = O. For x "lOwehave a convergent geometric series with first term x2 and common ratio1/(1 + x2); its sum, after a bit of calculation, is 8(x) = 1 + x2. Since 8 isdiscontinuous at 0, the convergence in any interval containing 0 cannot beuniform. Consider the set J = (-00, -b]U [a, 00), where a, b> 0 and wherewe may assume without essential loss of generality that a ~ b. Then, forall x in J, x2/(1 + x2)n < 1/(1 + x2)n-1 ~ 1/(1 + a2)n-1, and so, by theWeierstrass 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.Since 2::::'=1 l/n3

/2 is convergent, the given series is uniformly conver

gent in [0,1]. This technique will not work for the other series. We obtainthe 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 maximumvalue is l/n9

/8

• Thus, for all x in [0,1], Ix/(n3/4 + n3/2x2)1 ~ l/n9/ 8 , and

it follows by the Weierstrass M-test that the series is uniformly convergentin [0,1].

7.13 For all x in [0,1], xn(1 - x)/n2 ~ l/n2, and so it is immediate by theWeierstrass M-test that the series is uniformly convergent in [0,1]. Thissimple argument will not work for the other series. From a simple calculusargument, 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)).


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],(1 In) (log(n + x) - log x) ~ 1In2 • Hence the series is uniformly convergentin [0,1].

7.15 (i) limn-too(anlan+d = 1/2, so R = 1/2. When x = 1/2 the series is E~=o(1/(n + 1)) which diverges; when x = -1/2 the series isE~o( -1)n(l/(n + 1)), which converges. So the interval of convergenceis [-1,1). (ii) limn-too lanlanHI = 1, so R = 1. The series diverges forx = -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 eas n -t 00. So R = e. When x = e, the Stirling formula gives anxn =n! en Inn::=:: nn+! e-nenInn = n!, and so the series diverges at ±e. The interval of convergence is (-e, e). (iv)a;;n = 2+ (lin) -t 2 as n -t 00. HenceR = 2. When Ixl = 2, lanxnl = [2/(2 + (1/n))]n = [1 + (1/2n)tn -t

e-1/ 2 # °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 converges 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 4as n -t 00. So R =4. If Ixl =4,

(n!)24n n2n+le-2n22nlanxnl = (2 ),::=:: 2! =n

1/

2 -1+ °as n -t 00.n . (2n) n+ e-2n

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

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

~ n n+l t t dt t (1 1) d~ n + 1x =10 (1 - t)2 =10 (1 _ t)2 - 1 _ t t

[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 x 2nH 00 (3x)2nH4sin

3x =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)(I32n )/(2n + I)!.

266 Real Analysis

7.18 Since 2::~=0 xr tends to 1/(l-x) as x -+ 00 uniformly in any closed intervalcontained 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, forall 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 andan = (_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). Sincej(O) = 0 it follows that j(n)(o) = 0 for all even n. Since 1'(0) = 1, itfollows 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 ofconvergence is 1. Let Ixl = 1. Then, since

(2n)2n+! e-2n 1qn 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 inthe integral. A is strictly increasing, since the integrand is positive.It is differentiable in (-1,1), since the integrand is continuous, andA'(x) = 1/";1 - x2 by the fundamental theorem. (This is actually abit inaccurate, since the integral is improper - which is why I preferredto 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 follows that S'(x) is differentiable in (-1,1), that S'(O) = 1 and that[S(X)]2 + [S'(xW = 1. Hence

S"(x) =(1/2)[1- (S(x))2rl/2 (-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 gamma6 L\ deltato E epsilon( Z zeta

TJ H eta0, {) e theta

~ ~ xi£ I iotaK K kappaA A lambda

I-l M muv N nu0 0 omicron11", tv II pip P rhoa E sigmaT T tauv y upsilon¢J, <p ~ phi

1/J 1[/ psi

X X chiw {} omega

Bibliography

[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.

Abel's test, 198Abel's theorem, 208Abel, Nils Henrik (1802-1829), 198, 208absolute magnitude, 18absolutely convergent- integral, 155- series, 58additivity theorem, 224analytic function, 118, 179anti-differentiation, 119antiderivative, 140arbitrary constant, 143arc-length function, 225archimedean property, 9Archimedes of Syracuse (287-212 BC), 9arithmetic mean, 24arithmetic progression, 24arithmetic-geometric inequality, 24associative law, 8axiom of Archimedes, 10axiom of completeness, 10, 86

belongs to (E), 5binomial coefficient, 14binomial series, 211- convergence at ±1, 211binomial theorem, 14, 32Bolzano, Bernhard Placidus Johann

Nepomuk (1781-1848),86Boole, George (1815-1864), 7Boolean algebra, 7Borel, Felix Edouard Justin Emile

(1871-1956), 91bounded- above, 9

Index

- below, 10- function, 74, 93- sequence, 34bounded above- function, 74- sequence, 34bounded below- function, 74- sequence, 34Briggs, Henry (1561-1631), 170

Cartesian product, 63Cartesian product (of sets), 7Cauchy sequence, 42, 74- is convergent, 43Cauchy's mean value theorem, 107Cauchy, Augustin-Louis (1789-1857),

22, 107Cauchy-Schwarz inequality, 22, 221chain rule, 103circular functions, 70, 217- derivatives, 103- sum formulae, 71, 218circular measure, 70codomain (of a function), 63commutative law, 8comparison test, 50- "ultimate" version, 51- asymptotic version, 54- for integrals, 152, 153, 160compatibility- with addition, 9- with multiplication, 9complement (of a subset), 6completing the square, 21

274

complex numbers, 5composition of functions, 67, 88conditionally convergent series, 59constant function, 67, 85continuity, 82- implies uniform continuity, 92- sequential, 82, 90- uniform, 90continuous function, 128- at a point, 81- nowhere differentiable, 234- on its domain, 86convergent- integral, 151, 158- sequence, 30- series, 48cosecant, 72cosine function, 70, 217- derivative, 103- is continuous, 88cotangent, 72

decreasing function, 74, 127defining property (of a set), 5density (of rationals), 9derivative, 99Descartes, Rene (1596-1650), 7differentiable function, 99differential coefficient, 99differentiation, 99differentiation term by term, 193Dirichlet, Johann Peter Gustav Lejeune

(1805-1859), 84discriminant, 22disjoint sets, 6dissection, 119distance (between functions), 182distributive law, 8divergent- sequence, 30domain (of a function), 63dummy symbol, 121

e, 41, 167, 171- is irrational, 178element (of a set), 5empty set (0), 6epsilon (f), 28equality of functions, 66Euler's constant, 172Euler, Leonard (1701-1783),172expx, 170exponential function, 170- in growth of bacteria, 177

Real Analysis

Fibonacci sequence, 45field, 8function, 63- as process, 63fundamental theorem of calculus, 140

Gamma function, 178general principle- of convergence, 74- of uniform convergence, 196, 197geometric mean, 24geometric progression, 24geometric series, 49golden number, 46graph (of a function), 64Gregory's series, 209Gregory, James (1638-1675), 209

harmonic mean, 25harmonic series, 49, 171- is divergent, 49Heine, Heinrich Eduard (1821-1881), 91Heine-Borel theorem, 91higher derivatives, 113Holder condition, 110Holder, Otto Ludwig (1859-1937), 110

identity function, 67, 85image- of a function, 64- of a point, 63improper integral- first kind, 150- second kind, 158increasing function, 74, 127indefinite integral, 143inequalities, 18infimum, 11, 93infinite integral, 150infinitely differentiable function, 118,

179inner product, 221integers, 5integral part, 66integral test, 156integrand, 152integration by parts, 144integration by substitution, 146integration term by term, 192, 193interior, 99intermediate value theorem, 86, 94intersection (of sets), 6interval, 11

Index

- closed, 11- open, 11inverse circular functions, 95- derivatives, 112inverse cosine, 95, 219- derivative, 112inverse function, 94, 110, 168inverse sine, 95, 219- derivative, 112inverse tangent, 96- derivative, 112irrationality- of e, 178- of..;2, 3isolated discontinuity, 130

Kronecker, Leopold, 1823-1891, 10

L'Hopital's rule, 108L'Hopital, Guillaume Franc;ois Antoine,

Marquis de (1661-1704), 108Laplace transform, 179Laplace, Pierre Simon (1749-1827), 179left limit, 78Leibniz test, 59Leibniz's theorem, 114, 214Leibniz, Gottfried Wilhelm (1646-1716),

59,99length of a curve, 222limit (of a function)- as x -+ -00, 73, 80- as x -+ a, 75- as x -+ 00, 73limit (of a sequence), 28lnx,170logx, 170logarithm, 165- in calculation, 170lower bound, 10- greatest, 10lower integral, 120lower sum, 119

Maclaurin's theorem, 117Maclaurin, Colin (1698-1746), 117mean value theorem, 106- Cauchy, 107- integrals, 139modulus, 18monotonic decreasing- sequence, 37monotonic function, 74, 127monotonic increasing- sequence, 37

Napier, John (1550-1617), 170natural logarithm, 170natural numbers, 5Newton, Isaac (1643-1727), 59, 99norm- of a function, 181- of a vector, 221null sequence, 31

open covering, 91ordered field, 9

parametric equations, 220Pascal triangle, 15- identity, 15, 115Pascal, Blaise, 1623-1662, 151r, 219, 228- calculation, 195pointwise convergence, 183polynomial function, 67- is continuous, 86power series, 201principle of induction, 12- second principle, 16product rule, 101

quadratic function, 21quotient rule, 102

radian, 70radius of convergence, 203- nth root test, 205- ratio test, 204ratio test, 56rational function, 67- is continuous, 86rational numbers, 5real numbers, 5reciprocal (of a function), 67reciprocal law, 8rectifiable curve, 222reduction formula, 146refinement (of a dissection), 121Riemann integrable function, 121Riemann, Georg Friedrich Bernhard

(1826-1866), 61, 119right limit, 78Rolle's theorem, 105Rolle, Michel (1652-1719), 105

sandwich principle, 37scalar product, 221

275

276

Schwarz, Karl Hermann Amandus,1843-1921, 22

secant, 72sequence, 27- as function, 64series, 48set, 5, 27sine function, 70, 79, 217- derivative, 103- is continuous, 88singleton sets, 5square roots (calculation), 39Stirling's formula, 215, 230Stirling, James (1692-1770), 215strictly decreasing function, 73, 95, 109strictly increasing function, 73, 94, 109strictly monotonic function, 74subset, 5- proper subset, 6sum- of functions, 66sum (of a series), 48supremum, 11, 93

tangent function, 72- continuity, 88Taylor's theorem, 116, 150Taylor, Brook (1685-1731), 116Taylor-Maclaurin series, 118

Real Analysis

Taylor-Maclaurin theorem, 117Theorem of Pythagoras, 3transitive law, 8triangle inequality, 182trichotomy law, 9

ultimately monotonic, 37uniform convergence, 183- of a sequence, 183- of a series, 192- of power series, 201uniformly continuous function, 90, 128union (of sets), 6unit circle, 70upper bound, 9- least upper bound, 9upper integral, 120upper sum, 119

value (of a function), 63, 64

Waerden, Bartel Leendert van der(1903-1996), 234

Wallis's formula, 229, 233Wallis, John (1616-1703), 229Weierstrass M-test, 197Weierstrass, Karl Theodor Wilhelm

(1815-1897), 197

Solutions to Exercises - Home - Springer978-1-4471-0341... · 2017-08-23 · Solutions to Exercises...

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