Solution Manual for An Introduction to Analysis, 4th Edition

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Solutions ManualAn Introductionto AnalysisFourth EditionWilliam R. WadeUniversity of Tennessee, Knoxville

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An Introduction to AnalysisTable of ContentsChapter 1: The Real Number System1.2Ordered field axioms……………………………………………..11.3The Completeness Axiom………………………………………..21.4Mathematical Induction…………………………………………..41.5Inverse Functions and Images……………………………………61.6Countable and uncountable sets………………………………….8Chapter 2: Sequences inR2.1Limits of Sequences……………………………………………..102.2Limit Theorems………………………………………………….112.3Bolzano-Weierstrass Theorem…………………………………..132.4Cauchy Sequences……………………………………………….152.5Limits Supremum and Infimum………………………………....16Chapter 3: Functions onR3.1Two-Sided Limits………………………………………………..193.2One-Sided Limits and Limits at Infinity…………………………203.3Continuity………………………………………………………..223.4Uniform Continuity……………………………………………...24Chapter 4: Differentiability onR4.1The Derivative…………………………………………………...274.2Differentiability Theorem………………………………………..284.3The Mean Value Theorem……………………………………….304.4Taylor’s Theorem and l’Hôpital’s Rule………………………....324.5Inverse Function Theorems……………………………………...34Chapter 5: Integrability onR5.1The Riemann Integral…………………………………………….375.2Riemann Sums……………………………………………………405.3The Fundamental Theorem of Calculus………………………….435.4Improper Riemann Integration…………………………………...465.5Functions of Bounded Variation…………………………………495.6Convex Functions………………………………………………..51

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Chapter 6: Infinite Series of Real Numbers6.1Introduction……………………………………………………….536.2Series with Nonnegative Terms…………………………………..556.3Absolute Convergence……………………………………………576.4Alternating Series………………………………………………...606.5Estimation of Series……………………………………………....626.6Additional Tests…………………………………………………..63Chapter 7: Infinite Series of Functions7.1Uniform Convergence of Sequences……………………………..657.2Uniform Convergence of Series………………………………….677.3Power Series……………………………………………………...697.4Analytic Functions……………………………………………….727.5Applications……………………………………………………...74Chapter 8: Euclidean Spaces8.1Algebraic Structure………………………………………………768.2Planes and Linear Transformations……………………………...778.3Topology ofRn…………………………………………………..798.4Interior, Closure, and Boundary…………………………………80Chapter 9: Convergence inRn9.1Limits of Sequences……………………………………………...829.2Heine-Borel Theorem…………………………………………….839.3Limits of Functions……………………………………………….849.4Continuous Functions…………………………………………….869.5Compact Sets……………………………………………………..879.6Applications………………………………………………………88Chapter 10: Metric Spaces10.1Introduction………………………………………………………..9010.2Limits of Functions………………………………………………..9110.3Interior, Closure, and Boundary…………………………………..9210.4Compact Sets……………………………………………………...9310.5Connected Sets……………………………………………………9410.6Continuous Functions……………………………………………..9610.7Stone-Weierstrass Theorem……………………………………….97

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Chapter 11: Differentiability onRn11.1Partial Derivatives and Partial Integrals……………………………9911.2The Definition of Differentiability…………………………………10211.3Derivatives, Differentials, and Tangent Planes…………………….10411.4The Chain Rule……………………………………………………..10711.5The Mean Value Theorem and Taylor’s Formula………………….10811.6The Inverse Function Theorem……………………………………..11111.7Optimization………………………………………………………...114Chapter 12: Integration onRn12.1Jordan Regions………………………………………………………11712.2Riemann Integration on Jordan Regions…………………………….11912.3Iterated Integrals……………………………………………………..12212.4Change of Variables…………………………………………………12512.5Partitions of Unity…………………………………………………...13012.6The Gamma Function and Volume………………………………….131Chapter 13: Fundamental Theorems of Vector Calculus13.1Curves………………………………………………………………..13513.2Oriented Curves……………………………………………………...13713.3Surfaces………………………………………………………………14013.4Oriented Surfaces…………………………………………………….14313.5Theorems of Green and Gauss……………………………………….14713.6Stokes’s Theorem…………………………………………………….150Chapter 14: Fourier Series14.1Introduction…………………………………………………………..15614.2Summability of Fourier Series……………………………………….15714.3Growth of Fourier Coefficients…………………………………...…15914.4Convergence of Fourier Series……………………………………....16014.5Uniqueness…………………………………………………………...163

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SOLUTIONS TO EXERCISESCHAPTER 11.2 Ordered field axioms.1.2.0.a) False. Leta= 2/3,b= 1,c=−2, andd=−1.b) False. Leta=−4,b=−1, andc= 2.c) True. Sincea≤bandb≤a+c,|a−b|=b−a≤a+c−a=c.d) True. Noa∈Rsatisfiesa < b−εfor allε >0, so the inequality is vacuously satisfied. If you want a moreconstructive proof, ifb≤0 thena < b−ε <0 + 0 = 0. Ifb >0, then forε=b,a < b−ε= 0.1.2.1.a) Ifa < bthena+c < b+cby the Additive Property. Ifa=bthena+c=b+csince + is a function.Thusa+c≤b+cholds for alla≤b.b) Ifc= 0 thenac= 0 =bcso we may supposec >0. Ifa < bthenac < bcby the Multiplicative Property. Ifa=bthenac=bcsince·is a function. Thusac≤bcholds for alla≤bandc≥0.1.2.2.a) Suppose 0≤a < band 0≤c < d. Multiplying the first inequality bycand the second byb, we have0≤ac≤bcandbc < bd. Hence by the Transitive Property,ac < bd.b) Suppose 0≤a < b. By (7), 0≤a2< b2. If√a≥ √bthena= (√a)2≥(√b)2=b, a contradiction.c) If 1/a≤1/b, then the Multiplicative Property impliesb=ab(1/a)≤ab(1/b) =a, a contradiction. If 1/b≤0thenb=b2(1/b)≤0 a contradiction.d) To show these statements may not hold whena <0, leta=−2,b=−1,c= 2 andd= 5. Thena < bandc < dbutac=−4 is not less thanbd=−5,a2= 4 is not less thanb2= 1, and 1/a=−1/2 is not less than1/b=−1.1.2.3.a) By definition,a+−a−=|a|+a2−(|a| −a2)= 2a2=aanda++a−=|a|+a2+(|a| −a2)= 2|a|2=|a|.b) By Definition 1.1, ifa≥0 thena+= (a+a)/2 =aand ifa <0 thena+= (−a+a)/2 = 0. Similarly,a−= 0ifa≥0 anda−=−aifa <0.1.2.4.a)|2x+ 1|<7 if and only if−7<2x+ 1<7 if and only if− −4< x <3.b)|2−x|<2 if and only if−2<2−x <2 if and only if−4<−x <0 if and only if 0< x <4.c)|x3−3x+1|< x3if and only if−x3< x3−3x+1< x3if and only if 3x−1>0 and 2x3−3x+1>0. The firstinequality is equivalent tox >1/3. Since 2x3−3x+ 1 = (x−1)(2x2+ 2x−1) implies thatx= 1,(−1± √3)/2,the second inequality is equivalent to (−1− √3)/2< x <(−1 +√3)/2 orx >1.Therefore, the solution is(1/3,(√3−1)/2)∪(1,∞).d) We cannot multiply by the denominatorx−1 unless we consider its sign.Case 1.x−1>0. Thenx < x−1 so 0<−1, i.e., this case is empty.Case 2.x−1<0. Then by the Second Multiplicative Property,x > x−1 so 0>−1, i.e., every number fromthis case works. Thus the solution is (−∞,1).e)Case 1. 4x2−1>0. Cross multiplying, we obtain 4x2<4x2−1, i.e., this case is empty.Case 2.4x2−1<0.Then by the Second Multiplicative Property, 4x2>4x2−1, i.e., 0>−1.Thus thesolution is (−1/2,1/2).1.2.5.a) Supposea >2.Thena−1>1 so 1<√a−1< a−1 by (6).Therefore, 2< b= 1 +√a−1<1 + (a−1) =a.b) Suppose 2< a <3. Then 0< a−2<1 so 0< a−2<√a−2<1 by (6). Therefore, 0< a <2 +√a−2 =b.c) Suppose 0< a <1. Then 0>−a >−1, so 0<1−a <1. Hence√1−ais real and by (6), 1−a <√1−a.Therefore,b= 1− √1−a <1−(1−a) =a.d) Suppose 3< a <5. Then 1< a−2<3 so 1<√a−2< a−2 by (6). Therefore, 3<2 +√a−2 =b < a.

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1.2.6.a+b−2√ab= (√a− √b)2≥0 for alla, b∈[0,∞).Thus 2√ab≤a+bandG(a, b)≤A(a, b).Onthe other hand, since 0≤a≤bwe haveA(a, b) = (a+b)/2≤2b/2 =bandG(a, b) =√ab≥ √a2=a. Finally,A(a, b) =G(a, b) if and only if 2√ab=a+bif and only if (√a− √b)2= 0 if and only if√a=√bif and only ifa=b.1.2.7.a) Since|x+ 2| ≤ |x|+ 2,|x| ≤2 implies|x2−4|=|x+ 2| |x−2| ≤4|x−2|.b) Since|x+ 3| ≤ |x|+ 3,|x| ≤1 implies|x2+ 2x−3|=|x+ 3| |x−1| ≤4|x−1|.c) Since|x−2| ≤ |x|+ 2,−3≤x≤2 implies|x2+x−6|=|x+ 3| |x−2| ≤6|x−2|.d) Since the minimum ofx2+x−1 on (−1,0) is−1.25,−1< x <0 implies|x3−2x+ 1|=|x2+x−1| |x−1|<5|x−1|/4.1.2.8.a) Since (1−n)/(1−n2) = 1/(1 +n), the inequality is equivalent to 1/(n+ 1)< .01 = 1/100. Since1 +n >0 for alln∈N, it follows thatn+ 1>100, i.e.,n >99.b) By factoring, we see that the inequality is equivalent to 1/(2n+ 1)<1/40, i.e., 2n+ 1>40. Thusn >39/2,i.e.,n≥20.c) The inequality is equivalent ton2+ 1>500. Thusn >√499≈22.33, i.e.,n≥23.1.2.9.a)mn−1+pq−1=mqq−1n−1+pq−1nn−1= (mq+pn)n−1q−1.Butn−1q−1nq= 1 and uniquenessof multiplicative inverses implies (nq)−1=n−1q−1.Therefore,mn−1+pq−1= (mq+pn)(nq)−1.Similarly,mn−1(pq−1) =mpn−1q−1=mp(nq)−1. By what we just proved and (2),mn+−mn=m−mn= 0n= 0.Therefore, by the uniqueness of additive inverses,−(m/n) = (−m)/n.Similarly, (m/n)(n/m) = (mn)/(mn) =mn(mn)−1= 1, so (m/n)−1=n/mby the uniqueness of multiplicative inverses.b) Any subset ofRwhich contains 0 and 1 will satisfy the Associative and Commutative Properties, theDistributive Law, and have an additive identity 0 and a multiplicative identity 1.By part a),Qsatisfies theClosure Properties, has additive inverses, and every nonzeroq∈Qhas a multiplicative inverse.Therefore,Qsatisfies Postulate 1.c) Ifr∈Q,x∈R\Qbutq:=r+x∈Q, thenx=q−r∈Q, a contradiction. Similarly, ifrx∈Qandr6= 0,thenx∈Q, a contradiction. However, the product of any irrational with 0 is a rational.d) By the First Multiplicative Property,mn−1< pq−1if and only ifmq=mn−1qn < pq−1nq=np.1.2.10.0≤(cb−ad)2=c2b2−2abcd+a2d2implies 2abcd≤c2b2+a2d2. Addinga2b2+c2d2to both sides,we conclude that (ab+cd)2≤(a2+c2)(b2+d2).1.2.11.LetP:=R+.a) Letx∈R. By the Trichotomy Property, eitherx >0,−x >0, orx= 0. ThusPsatisfies i). Ifx >0 andy >0, then by the Additive Property,x+y >0 and by the First Multiplicative Property,xy >0. ThusPsatisfiesii).b) To prove the Trichotomy Property, supposea, b∈R.By i), eithera−b∈P,b−a=−(a−b)∈P, ora−b= 0. Thus eithera > b,b > a, ora=b.To prove the Transitive Property, supposea < bandb < c. Thenb−a, c−b∈Pand it follows from ii) thatc−a=b−a+c−b∈P, i.e.,c > a.Sinceb−a= (b+c)−(a+c), it is clear that the Additive Property holds.Finally, supposea < b, i.e.,b−a∈P. Ifc >0 thenc∈Pand it follows from ii) thatbc−ac= (b−a)c∈P,i.e.,bc > ac. Ifc <0 then−c∈P, soac−bc= (b−a)(−c)∈P, i.e.,ac > bc.1.3 The Completeness Axiom.1.3.0.a) True. IfA∩B=∅, then sup(A∩B) :=−∞and there is nothing to prove. IfA∩B6=∅, then usethe Monotone Property.b) True. Ifx∈A, thenx≤supA. Sinceε >0, we haveεx≤εsupA, so the latter is an upper bound ofB. Itfollows that supB≤εsupA. On the other hand, ifx∈A, thenεx∈B, soεx≤supB, i.e., supB/εis an upperbound forA. It follows that supA≤supB/ε.c) True. Ifx∈Aandy∈B, thenx+y≤supA+ supB, so sup(A+B)≤supA+ supB. If this inequality isstrict, then sup(A+B)−supB <supA, and it follows from the Approximation Property that there is ana0∈Asuch that sup(A+B)−supB < a0. This implies that sup(A+B)−a0<supB, so by the Approximation Propertyagain, there is ab0∈Bsuch that sup(A+B)−a0< b0. We conclude that sup(A+B)< a0+b0, a contradiction.2

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d) False. LetA=B= [0,1]. ThenA−B= [−1,1] so sup(A−B) = 16= 0 = supA−supB.1.3.1.a) Sincex2+ 2x−3 = 0 impliesx= 1,−3, infE=−3, supE= 1.b) Sincex2−2x+ 3> x2impliesx <3/2, infE= 0, supE= 3/2.c) Sincep2/q2<5 impliesp/q <√5, infE= 0, supE=√5.d)Since 1 + (−1)n/n= 1−1/nwhennis odd and 1 + 1/nwhennis even, infE= 0 and supE= 3/2. e) Since1/n+ (−1)n= 1/n+ 1 whennis even and 1/n−1 whennis odd, infE=−1 and supE= 3/2.f) Since2−(−1)n/n2= 2−1/n2whennis even and 2 + 1/n2whennis odd, infE= 7/4 and supE= 3.1.3.2.Sincea−1/n < a+ 1/n, choosern∈Qsuch thata−1/n < rn< a+ 1/n, i.e.,|a−rn|<1/n.1.3.3.a < bimpliesa− √2< b− √2. Chooser∈Qsuch thata− √2< r < b− √2. Thena < r+√2< b.By Exercise 1.2.9c,r+√2 is irrational. Thus setξ=r+√2.1.3.4.Ifmis a lower bound ofEthen so is any˜m≤m. Ifmand˜mare both infima ofEthenm≤˜mand˜m≤m, i.e.,m=˜m.1.3.5.Suppose thatEis a bounded, nonempty subset ofZ. Since−Eis a bounded, nonempty subset ofZ, ithas a supremum by the Completeness Axiom, and that supremum belongs to−Eby Theorem 1.15. Hence by theReflection Principle, infE=−sup(−E)∈ −(−E) =E.1.3.6.a) Let≤ >0 andm= infE. Sincem+≤is not a lower bound ofE, there is ana∈Esuch thatm+≤ > a.Thusm+≤ > a≥mas required.b) By Theorem 1.14, there is ana∈Esuch that sup(−E)−≤ <−a≤sup(−E).Hence by the SecondMultiplicative Property and Theorem 1.20, infE+≤=−(sup(−E)−≤)> a >−sup(−E) = infE.1.3.7.a) Letxbe an upper bound ofEandx∈E. IfMis any upper bound ofEthenM≥x. Hence bydefinition,xis the supremum ofE.b) The correct statement is: Ifxis a lower bound ofEandx∈Ethenx= infE.Proof.−xis an upper bound of−Eand−x∈ −Eso−x= sup(−E). Thusx=−sup(−E) = infE.c) IfEis the set of pointsxnsuch thatxn= 1−1/nfor oddnandxn= 1/nfor evenn, then supE= 1,infE= 0, but neither 0 nor 1 belong toE.1.3.8.SinceA⊆E, any upper bound ofEis an upper bound ofA.SinceAis nonempty, it follows fromthe Completeness Axiom thatAhas a supremum.Similarly,Bhas a supremum.Moreover, by the MonotoneProperty, supA,supB≤supE.SetM:= max{supA,supB}and observe thatMis an upper bound of bothAandB.IfM <supE, thenthere is anx∈Esuch thatM < x≤supE. Butx∈Eimpliesx∈Aorx∈B. ThusMis not an upper boundfor one of the setsAorB, a contradiction.1.3.9.By induction, 2n> n. Hence by the Archimedean Principle, there is ann∈Nsuch that 2n>1/(b−a).LetE:={k∈N: 2nb≤k}. By the Archimedean Principle,Eis nonempty. Hence letm0be the least element inEand setq= (m0−1)/2n. Sinceb >0,m0≥1. Sincem0is least inE, it follows thatm0−1<2nb, i.e.,q < b.On the other hand,m0∈Eimplies 2nb≤m0, soa=b−(b−a)< m02n−12n=m0−12n=q.1.3.10.Since|xn| ≤M, the setEn={xn, xn+1,. . .}is bounded for eachn∈N. Thussn:= supEnexistsand is finite by the Completeness Axiom.Moreover, sinceEn+1⊆En, it follows from the Monotone Property,sn≥sn+1for eachn∈N. Thuss1≥s2≥. . . .By the Reflection Principle, it follows thatt1≤t2≤ · · ·.Or, if you prefer a more direct approach,σn:= sup{−xn,−xn+1,. . .}satisfiesσ1≥σ2≥. . . . Sincetn=−σnforn∈N, it follows from the Second Multiplicative Property thatt1≤t2≤. . . .1.3.11.LetE={n∈Z:n≤a}. Ifa≥0, then 0∈E. Ifa <0, then by the Archimedean Principle, thereis anm∈Nsuch thatm >−a, i.e.,n:=−m∈E. ThusEis nonempty. SinceEis bounded above (bya), itfollows from the Completeness Axiom and Theorem 1.15 thatn0= supEexists and belongs toE.Setk=n0+ 1.Sincek >supE,kcannot belong toE, i.e.,a < k.On the other hand, sincen0∈Eandb−a >1,k=n0+ 1≤a+ 1< a+ (b−a) =b.We conclude thata < k < b.3

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1.4 Mathematical Induction.1.4.0.a) False. Ifa=−b= 1 andn= 2, then (a+b)n= 0 is NOT greater thanb2= 1.b) False. Ifa=−3,b= 1, andn= 2, then (a+b)n= 4 is not less than or equal tobn= 1.c) True. Ifnis even, thenn−kandkare either both odd or both even. If they’re both odd, thenan−kbkis theproduct of two negative numbers, hence positive. If they’re both even, thenan−kbkis the product of two positivenumbers, hence positive. Thus by the Binomial Formula,(a+b)n=n∑k=0(nk)an−kbk=an+nan−1b+n∑k=2(nk)an−kbk=:an+nan−1b+C.SinceCis a sum of positive numbers, the promised inequality follows at once.d) True. By the Binomial Formula,12n=(1a+a−22a)n=n∑k=0(nk)1ak(a−2)n−k2n−kan−k=n∑k=0(nk)(a−2)n−kan2n−k.1.4.1.a) By hypothesis,x1>2. Supposexn>2. Then by Exercise 1.2.5a, 2< xn+1< xn. Thus by induction,2< xn+1< xnfor alln∈N.b) By hypothesis, 2< x1<3.Suppose 2< xn<3.Then by Exercise 1.2.5b, 0< xn< xn+1.Thus byinduction, 0< xn< xn+1for alln∈N.c) By hypothesis, 0< x1<1. Suppose 0< xn<1. Then by Exercise 1.2.5c, 0< xn+1< xn. Thus by inductionthis inequality holds for alln∈N.d) By hypothesis, 3< x1<5.Suppose 3< xn<5.Then by Exercise 1.2.5d, 3< xn+1< xn.Thus byinduction this inequality holds for alln∈N.1.4.2.a) 0 = (1−1)n=∑nk=0(nk)1n−k(−1)k=∑nk=0(nk)(−1)k.b) (a+b)n=an+nan−1b+· · ·+bn≥an+nan−1b.c) By b), (1 + 1/n)n≥1n+n1n−1(1/n) = 2.d) 2n= (1 + 1)n=∑nk=0(nk)so∑nk=1(nk)= 2n−1. On the other hand∑n−1k=02k= 2n−1 by induction.1.4.3.a) This inequality holds forn= 3. If it holds for somen≥3 then2(n+ 1) + 1 = 2n+ 1 + 2<2n+ 2<2n+ 2n= 2n+1.b) The inequality holds forn= 1. If it holds fornthenn+ 1<2n+ 1≤2n+n <2n+ 2n= 2n+1.c) Nown2≤2n+ 1 holds forn= 1,2, and 3. If it holds for somen≥3 then by a),(n+ 1)2=n2+ 2n+ 1<2n+ 2n= 2n+1<2n+1+ 1.d) We claim that 3n2+ 3n+ 1≤2·3nforn= 3,4,. . . . This inequality holds forn= 3. Suppose it holds forsomen. Then3(n+ 1)2+ 3(n+ 1) + 1 = 3n2+ 3n+ 1 + 6n+ 6≤2·3n+ 6(n+ 1).Similarly, induction can be used to establish 6(n+ 1)≤4·3nforn≥1. (It holds forn= 1, and if it holds fornthen 6(n+ 2) = 6(n+ 1) + 6≤4·3n+ 6<4·3n+ 8·3n= 4·3n+1.) Therefore,3(n+ 1)2+ 3(n+ 1) + 1≤2·3n+ 6(n+ 1)≤2·3n+ 4·3n= 2·3n+1.Thus the claim holds for alln≥3.Nown3≤3nholds by inspection forn= 1,2,3. Suppose it holds for somen≥3. Then(n+ 1)3=n3+ 3n2+ 3n+ 1≤3n+ 2·3n= 3n+1.4

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1.4.4.a) The formula holds forn= 1. If it holds fornthenn+1∑k=1k=n(n+ 1)2+n+ 1 = (n+ 1)(n2 + 1) = (n+ 1)(n+ 2)2.b) The formula holds forn= 1. If it holds fornthenn+1∑k=1k2=n(n+ 1)(2n+ 1)6+ (n+ 1)2=n+ 16(n(2n+ 1) + 6(n+ 1)) = (n+ 1)(n+ 2)(2n+ 3)6.c) The formula holds forn= 1. If it holds fornthenn+1∑k=1a−1ak= 1−1an+a−1an+1= 1−1an+1.d) The formula holds forn= 1. If it holds fornthenn+1∑k=1(2k−1)2=n(4n2−1)3+ (2n+ 1)2= 2n+ 13(2n2+ 5n+ 3)= 2n+ 13(2n+ 3)(n+ 1) = (n+ 1)(4n2+ 8n+ 3)3= (n+ 1)(4(n+ 1)2−1)3.1.4.5.0≤an< bnholds forn= 1. If it holds fornthen by (7), 0≤an+1< bn+1.By convention,n√b≥0. Ifn√a <n√bis false, thenn√a≥n√b≥0. Taking thenth power of this inequality, weobtaina= (n√a)n≥(n√b)n=b, a contradiction.1.4.6.The result is true forn= 1. Suppose it’s true for some odd number≥1, i.e., 22n−1+ 32n−1= 5`forsome`, n∈N. Then22n+1+ 32n+1= 4·22n−1+ 9·32n−1= 4·5`+ 5·32n−1is evidently divisible by 5. Thus the result is true by induction.1.4.7.We first prove that 2n! + 2≤(n+ 1)! forn= 2,3,. . . . It’s true forn= 2. Suppose that it’s true forsomen≥2. Then by the inductive hypothesis,2(n+ 1)! + 2 = 2(n+ 1)n! + 2 = 2n! + 2 + 2n·n!≤(n+ 1)! + 2n·n!.But 2< n+ 1 so we continue the inequality above by2(n+ 1)! + 2<(n+ 1)! +n·(n+ 1)! = (n+ 2)·(n+ 1)! = (n+ 2)!as required.To prove that 2n≤n! + 2, notice first that it’s true forn= 1. If it’s true for somen≥1, then by the inequalityalready proved,2n+1= 2·2n≤2(n! + 2) = 2n! + 2 + 2≤(n+ 1)! + 2as required.1.4.8.Ifn= 1 orn= 2, the result is trivial. Ifn≥3, then by the Binomial Formula,2n= (1 + 1)n=n∑k=0(nk)>(n3)=n(n−1)(n−2)6.1.4.9.a) Ifm=k2, then√m=kby definition.On the other hand, ifmis not a perfect square, then byRemark 1.28,√mis irrational. In particular, it cannot be rational.5

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b) If√n+ 3 +√n∈Qthenn+ 3 + 2√n+ 3√n+n= (√n+ 3 +√n)2∈Q. SinceQis closed under subtractionand division, it follows that√n2+ 3n∈Q. In particular,n2+ 3n=m2for somem∈N. Nown2+ 3nis a perfectsquare whenn= 1 but ifn >1 then(n+ 1)2=n2+ 2n+ 1< n2+ 2n+n=n2+ 3nn2+ 3nn2+ 3n=< n2+ 4n+ 4 = (n+ 2)2.Therefore, the original expression is rational if and only ifn= 1.c) By repeating the steps in b), we see that the original expression is rational if and only ifn(n+7) =n2+7n=m2for somem∈N. Ifn >9 then(n+ 3)2=n2+ 6n+ 9< n2+ 7nn2+ 7nn2+ 7n < n2+ 8n+ 16 = (n+ 4)2.Thus the original expression cannot be rational whenn >9. On the other hand, it is easy to check thatn2+ 7nisnot a perfect square forn= 1,2,. . .,8 but is a perfect square, namely 144 = 122, whenn= 9. Thus the originalexpression is rational if and only ifn= 9.1.4.10.The result holds forn= 0 sincec0−b0= 1 anda20+b20=c20.Suppose thatcn−1−bn−1= 1 anda2n−1+b2n−1=c2n−1hold for somen≥0. By definition,cn−bn=cn−1−bn−1= 1, so by induction, this differenceis always 1. Moreover, by the Binomial Formula, the inductive hypothesis, and what we just proved,a2n+b2n= (an−1+ 2)2+ (2an−1+bn−1+ 2)2=a2n−1+ 4an−1+ 4 + (2an−1+ 2)2+ 2bn−1(2an−1+ 2) +b2n−1=c2n−1+ 2(an−1+ 2) + (2an−1+ 2)2+ 2(cn−1−1)(2an−1+ 2)=c2n−1+ (2an−1+ 2)2+ 2cn−1(2an−1+ 2)= (2an−1+cn−1+ 2)2≡c2n.1.5 Inverse Functions and Images.1.5.0.a) False. Since (sinx)′= cosxis negative on [π/2,3π/2],fis 1–1 there, but the domain of arcsinxis[−π/2, π/2]. Thus here,f−1(x) = arcsin(π−x).b) True. By elementary set algebra and Theorem 1.37,(f−1(A)∩f−1(B))∪f−1(C) =f−1(A∩B)∪f−1(C)⊃f−1(A∩B)6=∅.c) False. IfX= [0,2],A= [0,1] andB={1}, thenB\A=∅but (A\B)c= [0,1)c= [1,2].d) False. Letf(x) =x+ 1 for−1≤x≤0 andf(x) = 2x−1 for 0< x≤1. Thenftakes [−1,1] onto [−1,1]andf(0) = 1, butf−1(f(0)) =f−1(1) ={0,1}.1.5.1.α)fis 1–1 sincef′(x) = 3>0 forx∈R. Ify= 3x−7 thenx= (y+7)/3. Thereforef−1(x) = (x+7)/3.By looking at the graph, we see thatf(E) =R.β)fis 1–1 sincef′(x) =−e1/x/x2>0 forx∈(0,∞). Ify=e1/xthen logy= 1/x, i.e.,x= 1/logy. Therefore,f−1(x) = 1/logx. By looking at the graph, we see thatf(E) = (1,∞).γ)fis 1–1 on (π/2,3π/2) becausef′(x) = sec2x >0 there. The inverse isf−1(x) = arctan(x−π). By lookingat the graph, we see thatf(E) = (−∞,∞).δ) Sincef′(x) = 2x+ 2<0 forx <−6,fis 1–1 on [−∞,−6]. Sincey=x2+ 2x−5 is a quadratic inx, wehavex= (−2±√4 + 4(5 +y))/2 =−1± √6 +y. Butxis negative on (−∞,−6], so we must use the negativesign. Hencef−1(x) =−1− √6 +x. By looking at the graph, we see thatf(E) = [19,∞).ε) By definition,f(x) =3x+ 2x≤0x+ 20< x≤23x−2x >2.Thusfis strictly increasing, hence 1–1, andf−1(x) =(x−2)/3x≤2x−22< x≤4(x+ 2)/3x >4,6

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Solution Manual for An Introduction to Analysis, 4th Edition - Page 12 preview image

i.e.,f−1(x) = (x+|x−2| − |x−4|)/3. By looking at the graph, we see thatf(E) = (−∞,∞).ζ) Sincef′(x) = (1−x2)/(x2+ 1)2is never zero on (−1,1),fis 1–1 on [−1,1].By the quadratic formula,y=f(x) impliesx= (1±√1−4y2)/2y. Sincex∈[−1,1] we must take the minus sign. Hencef−1(x) ={(1− √1−4x2)/2xx6= 00x= 0.By looking at the graph, we see thatf(E) = (−0.5,0.5).1.5.2.a)fdecreases andf(−1) = 5,f(2) =−4. Therefore,f(E) = (−4,5). Sincef(x) =−1 impliesx= 1andf(x) = 2 impliesx= 0, we also havef−1(E) = (0,1).b) The graph offis a parabola whose absolute minimum is 1 atx= 0 and whose maximum on (−1,2] is 5 atx= 2. Therefore,f(E) = [1,5]. Sinceftakes±1 to 2,f−1(E) = [−1,1].c) The graph offis a parabola whose absolute maximum is 1 atx= 1.Sincef(−2) =−8, it follows thatf(E) = [−8,1]. Since 2x−x2=−2 impliesx= 1± √3, we also havef−1(E) = [1− √3,1 +√3].d) The graph ofx2−2x+ 2 is a parabola whose minimum is 1 atx= 1.Since log increases on (0,∞),f(1) = log(1) = 0, andf(3) = log(5), it follows thatf(E) = [0,log(5)].Since 3 = log(x2+x+ 1) impliesx= 1± √e3−1, we also havef−1(E) = [1−√e3−1,1)∪(1,1 +√e3−1].e) Since cosxis periodic with maximum 1 and minimum−1,f(E) = [−1,1]. Since cosxis nonnegative when(4k−1)π/2≤x≤(4k+ 1)π/2 for somek∈Z, it follows thatf−1(E) =⋃k∈Z[(4k−1)π/2,(4k+ 1)π/2].1.5.3.a) The minimum ofx−2 on [0,1] is−2 and the maximum ofx+1 on [0,1] is 2. Thus∪x∈[0,1][x−2, x+1] =[−2,2].b) The maximum ofx−1 on [0,1] is 0 and the minimum ofx+ 1 on [0,1] is 1. Thus∩x∈(0,1][x−1, x+ 1] = (0,1].c) The minimum of 1/kfork∈Nis 0 and 0∈[−1/k,1/k] for allk∈N. Thus∩k∈N[−1/k,1/k] ={0}.d) The maximum of 1/kfork∈Nis 1. Thus∪k∈N[−1/k,0] = [−1,0].e) The maximum of 1/kfork∈Nis 1 and the minimum of−kfork∈Nis−∞. Thus∪k∈N[−k,1/k) = (−∞,1).f) The maximum of (k−1)/kand the minimum of (k+1)/kfork∈Nis 1. Thus∩k∈N[(k−1)/k,(k+1)/k) ={1}.1.5.4.Supposexbelongs to the left side of (16), i.e.,x∈Xandx /∈ ∩α∈AEα.By definition,x∈Xandx /∈Eαfor someα∈A. Therefore,x∈Ecαfor someα∈A, i.e.,xbelongs to the right side of (16). These stepsare reversible.1.5.5.a) By definition,x∈f−1(∪α∈AEα) if and only iff(x)∈Eαfor someα∈Aif and only ifx∈∪α∈Af−1(Eα).b) By definition,x∈f−1(∩α∈AEα) if and only iff(x)∈Eαfor allα∈Aif and only ifx∈ ∩α∈Af−1(Eα).c) To showf(f−1(E)) =E, letx∈E.SinceE⊆f(X), choosea∈Xsuch thatx=f(a).By definition,a∈f−1(E) sox≡f(a)∈f(f−1(E)).Conversely, ifx∈f(f−1(E)), thenx=f(a) for somea∈f−1(E).Bydefinition, this meansx=f(a) andf(a)∈E. In particular,x∈E.To showE⊆f−1(f(E)), letx∈E. Thenf(x)∈f(E), so by definition,x∈f−1(f(E)).1.5.6.a) LetC= [0,1] andB= [−1,0]. ThenC\B={0}andf(C) =f(B) = [0,1]. Thusf(C\B) ={0} 6=∅=f(C)\f(B).b) LetE= [0,1]. Thenf(E) = [0,1] sof−1(f(E)) = [−1,1]6= [0,1] =E.1.5.7.a) implies b). By definition,f(A\B)⊇f(A)\f(B) holds whetherfis 1–1 or not. To prove the reverseinequality, supposefis 1–1 andy∈f(A\B). Theny=f(a) for somea∈A\B. Sincefis 1–1,a=f−1({y}).Thusy6=f(b) for anyb∈B. In particular,y∈f(A)\f(B).b) implies c). By definition,A⊆f−1(f(A)) holds whetherfis 1–1 or not. Conversely, supposex∈f−1(f(A)).Thenf(x)∈f(A) sof(x) =f(a) for somea∈A. Ifx /∈A, then it follows from b) thatf(A) =f(A\ {x}) =f(A)\f({x}), i.e.,f(x)/∈f(A), a contradiction.c) implies d).By Theorem 1.37,f(A∩B)⊆f(A)∩f(B).Conversely, supposey∈f(A)∩f(B).Theny=f(a) =f(b) for somea∈Aandb∈B.Ify/∈f(A∩B) thena/∈Bandb/∈A.Consequently,f−1(f({a}))⊇ {a, b} ⊃ {a}, which contradicts c).d) implies a).Iffis not 1–1 then there exista, b∈Xsuch thata6=bandy:=f(a) =f(b).Hence by d),{y}=f({a})∩f({b}) =∅, a contradiction.7

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1.6 Countable and uncountable sets.1.6.0.a) False. The functionf(x) =xforx∈Nandf(x) = 1 forx∈R\NtakesRontoN, butRis not atmost countable.b) False. The setsAm:={kn:k∈Nand−2m≤k≤2m}are finite, hence at most countable. Since the dyadicrationals are the union of theAm’s asmranges overN, they must be at most countable by Theorem 1.42ii.c) True. IfBwere at most countable, then its subsetf(A) would be at most countable by Theorem 1.41, i.e.,there is a functiongwhich takesf(A) ontoN. Hence by Exercise 1.6.5a,g◦ftakesAontoN. It follows fromLemma 1.40 thatAis at most countable, a contradiction.d) False, beguiling as it seems!LetEn={0,1,. . .,9}and definefonE1×E2× · · ·by taking each point(x1, x2,. . . ) onto the number with decimal expansion 0.x1x2· · ·. Clearly (see the proof of Remark 1.39),ftakesEonto [0,1]. Since [0,1] is uncountable, it follows from 1.6.0c thatE1×E2× · · ·is uncountable.1.6.1.The function 2x−1 is 1–1 and takesNonto{1,3,5,. . .}. Thus this set is countable by definition.1.6.2.By two applications of Theorem 1.42i,Q×Qis countable, henceQ3:= (Q×Q)×Qis also countable.1.6.3.Letgbe a function that takesAontoB.IfAis at most countable, then by Lemma 1.40 there is afunctionfwhich takesNontoA. It follows (see Exercise 1.6.5a) thatg◦ftakesNontoB. Hence by Lemma1.40,Bis at most countable, a contradiction.1.6.4.By definition, there is ann∈Nand a 1–1 functionφwhich takesZ:={1,2,. . ., n}ontoA.Letψ(x) :=f(φ(x)) forx∈Z. Sincefandφare 1–1,ψ(x) =ψ(y) impliesφ(x) =φ(y) impliesx=y. Moreover,sincefandφare onto, givenb∈Bthere is ana∈Asuch thatf(a) =b, and anx∈Zsuch thatφ(x) =a, henceψ(x)≡f(φ(x)) =f(a) =b. Thusψis 1–1 fromZontoB. By definition, then,Bis finite.1.6.5.a) Repeat the proof in Exercise 1.6.4 without referring toNandZ.b) By the definition ofB0, it is clear thatftakesAontoB0. Supposef−1(x) =f−1(y) for somex, y∈B0.Sincefis 1–1 fromAontoB0, it follows from Theorem 1.30 thatx=f(f−1(x)) =f(f−1(y)) =y. Thusf−1is1–1 onB0.c) Iffis 1–1 (respectively, onto), then it follows from part a) thatg◦fis 1–1 (respectively, onto).Conversely, ifg◦fis 1–1 (respectively, onto), then by parts a) and b),f≡g−1◦g◦fis 1–1 (respectively, onto).1.6.6.a) We prove this result by induction onn.Supposen= 1. Sinceφ:{1} → {1}, it must satisfyφ(1) = 1. In particular, in this caseφis both 1–1 and ontoand there is nothing to prove.Suppose that the result holds for some integern≥1 and letφ:{1,2,. . ., n+ 1} → {1,2,. . ., n+ 1}.Setk0=φ(n+ 1) and defineψbyψ(`) ={`` < k0`−1` > k0.Theψis 1–1 from{1,2,. . ., k0−1, k0+ 1,. . ., n+ 1}onto{1,2,. . ., n}.Supposeφis 1–1 on{1,2,. . ., n+ 1}. Thenφis 1–1 on{1,2,. . ., n}, henceψ◦φis 1–1 from{1,2,. . ., n}into{1,2,. . ., n}. It follows from the inductive hypothesis thatψ◦φtakes{1,2,. . ., n}onto{1,2,. . ., n}. By Exercise1.6.5,φtakes{1,2,. . ., n}onto{1,2,. . ., k0−1, k0+ 1,. . ., n+ 1}. Sinceφ(n+ 1) =k0, we conclude thatφtakes{1,2,. . ., n+ 1}onto{1,2,. . ., n+ 1}.Conversely, ifφtakes{1,2,. . ., n+ 1}onto{1,2,. . ., n+ 1}, thenφtakes{1,2,. . ., n}onto{1,2,. . ., k0−1, k0+1,. . ., n+ 1}, soψ◦φtakes{1,2,. . ., n}onto{1,2,. . ., n}. It follows from the inductive hypothesis thatψ◦φis1–1 on{1,2,. . ., n}. Hence by Exercise 1.6.5 and construction,φis 1–1 on{1,2,. . ., n+ 1}.b) We may suppose thatEis nonempty. Hence by hypothesis, there is ann∈Nand a 1–1 functionφfromEonto{1,2,. . ., n}. Moreover, by Exercise 1.6.5b, the functionφ−1is 1–1 from{1,2,. . ., n}ontoE.Consider the functionφ−1◦f◦φ. Clearly, it takes{1,2,. . ., n}into{1,2,. . ., n}. Hence by part a),φ−1◦f◦φis 1–1 if and only if it is onto. In particular, it follows from Exercise 1.6.5c thatfis 1–1 if and only iffis onto.1.6.7.a) Letq=k/j. Ifk= 0 thennq= 1 is a root of the polynomialx−1. Ifk >0 thennqis a root of thepolynomialxj−nk. Ifk <0 thennqis a root of the polynomialn−kxj−1. Thusnqis algebraic.b) By Theorem 1.42, there are countably many polynomials with integer coefficients. Each polynomial of degreenhas at mostnroots. Hence the class of algebraic numbers of degreenis a countable union of finite sets, hencecountable.8

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Solution Manual for An Introduction to Analysis, 4th Edition - Page 14 preview image

c) Since any number is either algebraic or transcendental,Ris the union of the set of algebraic numbers andthe set of transcendental numbers. By b), the former set is countable. Therefore, the latter must be uncountableby the argument of Remark 1.43.9

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CHAPTER 22.1 Limits of Sequences.2.1.0.a) True. Ifxnconverges, then there is anM >0 such that|xn| ≤M. Choose by Archimedes anN∈Nsuch thatN > M/ε. Thenn≥Nimplies|xn/n| ≤M/n≤M/N < ε.b) False.xn=√ndoes not converge, butxn/n= 1/√n→0 asn→ ∞.c) False.xn= 1 converges andyn= (−1)nis bounded, butxnyn= (−1)ndoes not converge.d) False.xn= 1/nconverges to 0 andyn=n2>0, butxnyn=ndoes not converge.2.1.1.a) By the Archimedean Principle, givenε >0 there is anN∈Nsuch thatN >1/ε.Thusn≥Nimplies|(2−1/n)−2| ≡ |1/n| ≤1/N < ε.b) By the Archimedean Principle, givenε >0 there is anN∈Nsuch thatN > π2/ε2. Thusn≥Nimplies|1 +π/√n−1| ≡ |π/√n| ≤π/√N < ε.c) By the Archimedean Principle, givenε >0 there is anN∈Nsuch thatN >3/ε. Thusn≥Nimplies|3(1 + 1/n)−3| ≡ |3/n| ≤3/N < ε.d) By the Archimedean Principle, givenε >0 there is anN∈Nsuch thatN >1/√3ε. Thusn≥Nimplies|(2n2+ 1)/(3n2)−2/3| ≡ |1/(3n2)| ≤1/(3N2)< ε.2.1.2.a) By hypothesis, givenε >0 there is anN∈Nsuch thatn≥Nimplies|xn−1|< ε/2. Thusn≥Nimplies|1 + 2xn−3| ≡2|xn−1|< ε.b) By hypothesis, givenε >0 there is anN∈Nsuch thatn≥Nimpliesxn>1/2 and|xn−1|< ε/4. Inparticular, 1/xn<2. Thusn≥Nimplies|(πxn−2)/xn−(π−2)| ≡2|(xn−1)/xn|<4|xn−1|< ε.c) By hypothesis, givenε >0 there is anN∈Nsuch thatn≥Nimpliesxn>1/2 and|xn−1|< ε/(1 + 2e).Thusn≥Nand the triangle inequality imply|(x2n−e)/xn−(1−e)| ≡ |xn−1|∣∣∣∣1 +exn∣∣∣∣≤ |xn−1|(1 +e|xn|)<|xn−1|(1 + 2e)< ε.2.1.3.a) Ifnk= 2k, then 3−(−1)nk≡2 converges to 2; ifnk= 2k+ 1, then 3−(−1)nk≡4 converges to 4.b) Ifnk= 2k, then (−1)3nk+ 2≡(−1)6k+ 2 = 1 + 2 = 3 converges to 3; ifnk= 2k+ 1, then (−1)3nk+ 2≡(−1)6k+3+ 2 =−1 + 2 = 1 converges to 1.c) Ifnk= 2k, then (nk−(−1)nknk−1)/nk≡ −1/(2k) converges to 0; ifnk= 2k+1, then (nk−(−1)nknk−1)/nk≡(2nk−1)/nk= (4k+ 1)/(2k+ 1) converges to 2.2.1.4.Supposexnis bounded. By Definition 2.7, there are numbersMandmsuch thatm≤xn≤Mfor alln∈N. SetC:= max{1,|M|,|m|}. ThenC >0,M≤C, andm≥ −C. Therefore,−C≤xn≤C, i.e.,|xn|< Cfor alln∈N.Conversely, if|xn|< Cfor alln∈N, thenxnis bounded above byCand below by−C.2.1.5.IfC= 0, there is nothing to prove. Otherwise, givenε >0 use Definition 2.1 to choose anN∈Nsuchthatn≥Nimplies|bn| ≡bn< ε/|C|. Hence by hypothesis,n≥Nimplies|xn−a| ≤ |C|bn< ε.By definition,xn→aasn→ ∞.2.1.6.Ifxn=afor alln, then|xn−a|= 0 is less than any positiveεfor alln∈N. Thus, by definition,xn→aasn→ ∞.10

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2.1.7.a) Letabe the common limit point. Givenε >0, chooseN∈Nsuch thatn≥Nimplies|xn−a|and|yn−a|are both< ε/2. By the Triangle Inequality,n≥Nimplies|xn−yn| ≤ |xn−a|+|yn−a|< ε.By definition,xn−yn→0 asn→ ∞.b) Ifnconverges to somea, then givenε= 1/2, 1 =|(n+ 1)−n|<|(n+ 1)−a|+|n−a|<1 fornsufficientlylarge, a contradiction.c) Letxn=nandyn=n+ 1/n. Then|xn−yn|= 1/n→0 asn→ ∞, but neitherxnnorynconverges.2.1.8.By Theorem 2.6, ifxn→athenxnk→a.Conversely, ifxnk→afor every subsequence, then itconverges for the “subsequence”xn.2.2 Limit Theorems.2.2.0.a) False. Letxn=n2andyn=−nand note by Exercise 2.2.2a thatxn+yn→ ∞asn→ ∞.b) True. Letε >0. Ifxn→ −∞asn→ ∞, then chooseN∈Nsuch thatn≥Nimpliesxn<−1/ε. Thenxn<0 so|xn|=−xn>0.Multiplyxn<−1/εbyε/(−xn) which is positive.We obtain−ε <1/xn, i.e.,|1/xn|=−1/xn< ε.c) False. Letxn= (−1)n/n. Then 1/xn= (−1)nnhas no limit asn→ ∞.d) True.Since (2x−x)′= 2xlog 2−1>1 for allx≥2, i.e., 2x−xis increasing on [2,∞).In particular,2x−x≥22−2>0, i.e., 2x> xforx≥2. Thus, sincexn→ ∞asn→ ∞, we have 2xn> xnfornlarge, hence2−xn<1xn→0asn→ ∞.2.2.1.a)|xn| ≤1/n→0 asn→ ∞and we can apply the Squeeze Theorem.b) 2n/(n2+π) = (2/n)/(1 +π/n2)→0/(1 + 0) = 0 by Theorem 2.12.c) (√2n+ 1)/(n+√2) = ((√2/√n) + (1/n))/(1 + (√2/n))→0/(1 + 0) = 0 by Exercise 2.2.5 and Theorem2.12.d) An easy induction argument shows that 2n+ 1<2nforn= 3,4,. . . . We will use this to prove thatn2≤2nforn= 4,5,. . . . It’s surely true forn= 4. If it’s true for somen≥4, then the inductive hypothesis and the factthat 2n+ 1<2nimply(n+ 1)2=n2+ 2n+ 1≤2n+ 2n+ 1<2n+ 2n= 2n+1so the second inequality has been proved.Now the second inequality impliesn/2n<1/nforn≥4. Hence by the Squeeze Theorem,n/2n→0 asn→ ∞.2.2.2.a) LetM∈Rand choose by Archimedes anN∈Nsuch thatN >max{M,2}. Thenn≥Nimpliesn2−n=n(n−1)≥N(N−1)> M(2−1) =M.b) LetM∈Rand choose by Archimedes anN∈Nsuch thatN >−M/2. Notice thatn≥1 implies−3n≤ −3so 1−3n≤ −2. Thusn≥Nimpliesn−3n2=n(1−3n)≤ −2n≤ −2N < M.c) LetM∈Rand choose by Archimedes anN∈Nsuch thatN > M.Thenn≥Nimplies (n2+ 1)/n=n+ 1/n > N+ 0> M.d) LetM∈RsatisfyM≤0. Then 2 + sinθ≥2−1 = 1 impliesn2(2 + sin(n3+n+ 1))≥n2·1>0≥Mforalln∈N. On the other hand, ifM >0, then choose by Archimedes anN∈Nsuch thatN >√M. Thenn≥Nimpliesn2(2 + sin(n3+n+ 1))≥n2·1≥N2> M.2.2.3.a) Following Example 2.13,2 + 3n−4n21−2n+ 3n2= (2/n2) + (3/n)−4(1/n2)−(2/n) + 3→ −43asn→ ∞.b) Following Example 2.13,n3+n−22n3+n−2 = 1 + (1/n2)−(2/n3)2 + (1/n2)−(2/n3)→12asn→ ∞.11

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