SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. ERNESTYEUNG FundScience! &HelpErnestfinishhisPhysicsResearch! : quantumsuper-A-polynomials-athesisbyErnestYeung http://igg.me/at/ernestyalumni2014 Facebook: ernestyalumni gmail: ernestyalumni google: ernestyalumni linkedin: ernestyalumni tumblr: ernestyalumni twitter: ernestyalumni weibo: ernestyalumni youtube: ernestyalumni indiegogo: ernestyalumni ErnestYeungissupportedbyMr. andMrs. C.W.Yeung,Prof. RobertA.Rosenstone,MichaelDrown,ArvidKingl,Mr. andMrs. ValerieCheng,andtheFoundationforPolishSciences,WarsawUniversity. SOLUTIONSTO VOLUME1One-VariableCalculus,withanIntroductiontoLinearAlgebra I2.5Exercises-Introductiontosettheory,Notationsfordesignatingsets,Subsets,Unions,intersections,complements. Exercise10. Distributivelaws LetX =A∩(B∪C),Y =(A∩B)∪(A∩C) Supposex∈X x∈Aandx∈(B∪C)=⇒x∈AandxisinatleastBorinC thenxisinatleasteither(A∩B)or(A∩C) x∈Y,X ⊆Y Supposey ∈Y yisatleastineither(A∩B)orA∩C theny ∈AandeitherinBorC y ∈X,Y ⊆X X =Y LetX =A∪(B∩C),Y =(A∪B)∩(A∪C) Supposex∈X thenxisatleasteitherinAorin(B∩C) ifx∈A,x∈Y ifx∈(B∩C),x∈Y x∈Y,X ⊆Y Supposey ∈Y thenyisatleastinAorinBandyisatleastinAorinC ify ∈A,theny ∈X ify ∈A∩Bory ∈A∪C,y ∈X (variouscarvingsoutofA,simply) ify ∈(B∩C),y ∈X y ∈X,Y ⊆X X =Y 1 Exercise11. Ifx∈A∪A,thenxisatleastinAorinA. Thenx∈A. SoA∪A⊆A. OfcourseA⊆A∪A. Ifx∈A∩A,thenxisinAandinA. Thenx∈A. SoA∩A⊆A. OfcourseA⊆A∩A. Exercise12. Letx∈A. y ∈A∪BifyisatleastinAorinB. xisinAsox∈A∪B. =⇒A⊆A∪B. Suppose∃b∈Bandb∈/ A. b∈A∪Bbutb∈/ A. soA⊆A∪B. Exercise13. Letx ∈ A∪∅, thenxisatleastinAorin∅. Ifx ∈ ∅, thenxisanullelement(notanelementatall). Then actualelementsmustbeinA. =⇒A∪∅⊆A. Letx∈A. Thenx∈A∪∅. A⊆A∪∅. =⇒A=A∪∅. Exercise14. Fromdistributivity,A∪(A∩B)=(A∪A)∩(A∪B)=A∩(A∪B). Ifx∈A∩(A∪B),x∈Aandx∈A∪B,i.e. x∈AandxisatleastinAorinB. =⇒ x is in A and is in B or is not in B. Then x ∈ A. =⇒ A ∩ (A ∪ B) ⊆ A. Of course, A ⊆ A ∩ (A ∪ B). =⇒A∩(A∪B)=A∪(A∩B)=A. Exercise15. ∀a ∈ A,a ∈ C and ∀b ∈ B,b ∈ C. Consider x ∈ A∪B. x isat leastin A orin B. In eithercase, x ∈ C. =⇒A∪B ⊆C. Exercise16. ifC ⊆AandC ⊆B, thenC ⊆A∩B ∀c∈C,c∈Aandc∈B x∈A∩B,x∈Aandx∈B.Then∀c∈C,c∈A∩B. C ⊆A∩B Exercise17. (1) ifA⊂BandB ⊂C then ∀a∈A,a∈B.∀b∈B,b∈C. thensincea∈B,a∈C,∃c∈C suchthatc∈/ B. ∀a∈A,a∈Bsoa(cid:54)=c∀a.=⇒A⊂C (2) IfA⊆B,B ⊆C,A⊆C since,∀a∈A,a∈B,∀b∈B,b∈C.Thensincea∈B,a∈C. A⊆C (3) A⊂BandB ⊆C. B ⊂C orB =C. A⊂Bonly. ThenA⊂C. (4) Yes,since∀a∈A,a∈B. (5) No,sincex(cid:54)=A(setsaselementsaredifferentfromelements) Exercise18. A−(B∩C)=(A−B)∪(A−C) Supposex∈A−(B∩C) thenx∈Aandx∈/ B∩C =⇒x∈/ B∩C thenxisnotinevenatleastoneBorC =⇒x∈(A−B)∪(A−C) Supposex∈(A−B)∪(A−C) thenxisatleastin(A−B)orin(A−C)=⇒xisatleastinAandnotinBorinAandnotinC thenconsiderwhenoneofthecasesistrueandwhenbothcasesaretrue =⇒x∈A−(B∩C) Exercise19. (cid:91) Supposex∈B− A A∈F (cid:91) thenx∈B,x∈/ A A∈F (cid:91) x∈/ A=⇒x∈/ A,∀A∈F A∈F (cid:92) since∀A∈F,x∈B,x∈/ A, thenx∈ (B−A) A∈F 2 (cid:92) Supposex∈ (B−A) A∈F thenx∈B−A andx∈B−A and ... 1 2 then∀A∈F,x∈B,x∈/ A thenx∈/ evenatleastoneA∈F (cid:91) =⇒x∈B− A A∈F (cid:92) Supposex∈B− A A∈F (cid:92) thenx∈/ A A∈F thenatmostx∈Afor∀A∈F butone thenxisatleastinoneB−A (cid:91) =⇒x∈ (B−A) A∈F (cid:91) Supposex∈ (B−A) A∈F thenxisatleastinoneB−A thenforA∈F,x∈Bandx∈/ A Consider∀A∈F (cid:92) =⇒ thenx∈B− A A∈F Exercise20. (1) (ii)iscorrect. Supposex∈(A−B)−C thenx∈A−B,x∈/ C thenx∈Aandx∈/ Bandx∈/ C x∈/ Bandx∈/ C =⇒x∈/ evenatleastBorC x∈A−(B∪C) Supposex∈A−(B∪C) thenx∈A,x∈/ (B∪C) thenx∈Aandx∈/ Bandx∈/ C =⇒x∈(A−B)−C Toshowthat(i)issometimeswrong, Supposey ∈A−(B−C) y ∈Aandy ∈/ B−C y ∈/ B−C theny ∈/ Bory ∈C ory ∈/ C (wheredoesthisleadto?) Considerdirectly, Supposex∈(A−B)∪C thenxisatleastinA−BorinC thenxisatleastinAand∈/ BorinC Supposex=c∈C andc∈/ A 3 (2) IfC ⊆A, A−(B−C)=(A−B)∪C I3.3Exercises-Thefieldaxioms. Thegoalseemstobetoabstracttheseso-calledrealnumbersintojustx’sandy’sthat arepurelybuiltupontheseaxioms. Exercise1. Thm. I.5. a(b−c)=ab−ac. Lety =ab−ac;x=a(b−c) Want: x=y ac+y =ab(byThm. I.2,possibilityofsubtraction) NotethatbyThm. I.3,a(b−c)=a(b+(−c))=ab+a(−c)(bydistributivityaxiom) ac+x=ac+ab+a(−c)=a(c+(−c))+ab=a(0+b)=ab ButthereexistsexactlyoneyorxbyThm. I.2. x=y. Thm. I.6. 0·a=a·0=0. 0(a)=a(0)(bycommutativityaxiom) Givenb∈Rand0∈R,∃exactlyone −bs.t. b−a=0 0(a)=(b+(−b))a=ab−ab=0(byThm. I.5. andThm. I.2) Thm. I.7. ab=ac ByAxiom4,∃y ∈Rs.t. ay =1 sinceproductsareuniquelydetermined,yab=yac=⇒(ya)b=(ya)c=⇒1(b)=1(c) =⇒b=c Thm. I.8. PossibilityofDivision. Givena,b,a(cid:54)=0,chooseysuchthatay =1. Letx=yb. ax=ayb=1(b)=b Therefore, there exists at least one x such that ax = b. But by Thm. I.7, there exists only one x (since if az −b, and so x=z). Thm. I.9. Ifa(cid:54)=0,thenb/a=b(a−1). b Letx= forax=b a y =a−1foray =1 Want: x=by Nowb(1)=b, soax=b=b(ay)=a(by) =⇒x=by(byThm. I.7) Thm. I.10. Ifa(cid:54)=0,then(a−1)−1 =a. Now ab = 1 for b = a−1. But since b ∈ R and b (cid:54)= 0 (otherwise 1 = 0, contradiction), then using Thm. I.8 on b, ab=b(a)=1; a=b−1. Thm.I.11. Ifab=0,a=0orb=0. ab=0=a(0)=⇒b=0orab=ba=b(0)=⇒a=0. (weusedThm. I.7,cancellationlawformultiplication) Thm. I.12. Want: x=yifx=(−a)bandy =−(ab). ab+y =0 ab+x=ab+(−a)b=b(a+(−a))=b(a−a)=b(0)=0 0isunique,soab+y =ab+ximpliesx=y(byThm. I.1) Thm. I.13. Want: x+y =z,ifa=bx,c=dy,(ad+bc)=(bd)z. (bd)(x+y)=bdx+bdy =ad+bc=(bd)z Sousingb,d(cid:54)=0,whichisgiven,andThm. I.7,thenx+y =z. 4 Thm. I.14. Want: xy =zforbx=a,dy =c,ac=(bd)z. (bd)(xy)=(bx)(dy)=ac=(bd)z b,d(cid:54)=0,sobyThm. I.7,xy =z. Thm.I.15. Want: x=yz,ifbx=a,dy =c,(bc)z =ad (bc)z =b(dy)z =d(byz)=da d(cid:54)=0sobyThm. I.7,byz =a,byz =abx b(cid:54)=0sobyThm. I.7,yz =x Exercise2. Consider0+z =0. ByThm. I.2,thereexistsexactlyonez,z =−0. ByAxiom4,z =0. 0=−0. Exercise3. Consider1(z)z(1)=1. Thenz =1−1. ButbyAxiom4,thereexistsdistinct1suchthatz(1)=1,soz =1. Exercise4. Supposethereexistsxsuchthat0x=1,but0x=0and0and1aredistinct,so zerohasnoreciprocal. Exercise5. a+(−a)=0,0+0=0. Then a+(−a)+b+(−b)=(a+b)+(−a)+(−b)=0 −(a+b)=−a+(−b)=−a−b Exercise6. a+(−a)=0,b+(−b)=0,so a+(−a)+b+(−b)=a+(−b)+(−a)+b=(a−b)+(−a)+b=0+0=0 −(a−b)=−a+b. Exercise7. (a−b)+(b−c)=a+(−b)+b+(−c)=a+(b+(−b))+(−c)=a−c Exercise8. (ab)x=1 (ab)−1 =x a(bx)=1 a−1 =bx b(ax)=1 b−1−ax a−1b−1 =(abx)x=1(x)=(ab)−1 Exercise9. Want: x=y =z,if a z = a=zt b+t=0 −b (−a) y = by =u a+u=0 b (cid:16)a(cid:17) (cid:16)a(cid:17) x=− +x=v+x=0 vb=a b b a+(−a)=vb+by =b(v+y)=0 ifb(cid:54)=0,v+y =0, butv+x=0 byThm. I.1,x=y b+t=0, thenz(b+t)=zb+zt=zb+a=z(0)=0 a+zb=0=⇒−a=zb=by sinceb(cid:54)=0,z =ysox=y =z Exercise10. Sinceb,d(cid:54)=0,Let ad−bc z = (bd)z =ad−bc bypreviousexerciseorThm. I.8,thepossibilityofdivision bd a x= bx=a b −c t= dt=−c(ByThm. I.3,weknowthatb−a=b+(−a)) d 5 dbx+bdt=(bd)(x+y)=ad−bc=(bd)z b,d(cid:54)=0, sox+y =z I3.5Exercises-Theorderaxioms. Theorem1(I.18). Ifa0thenac0,thenac0 Theorem4(I.21). 1>0 Theorem5(I.22). Ifabc. Theorem6(I.23). Ifa−b. Inparticular,ifa<0,then−a>0. Theorem7(I.24). Ifab>0,thenbothaandbarepositiveorbotharenegative. Theorem8(I.25). Ifa0 a(−c)0.Thenac>bc(bydefinitionof>) (2) a0,thenifb>0,ab>0(b)=0. Ifb<0,ab<0(b)=0. Soifa>0,thenb>0. Ifa<0,thenifb>0,ab<0(b)=0. Ifb<0,ab>0(b)=0. Soifa<0,thenb<0. (4) a0, andbyThm. I.21,1>0 x2+1>0+0=0→x2+1(cid:54)=0 =⇒(cid:64)x∈Rsuchthatx2+1=0 Exercise3. a<0,b<0,a+b<0+0=0(ByThm. I.25) Exercise4. Considerax=1. ax=1>0.ByThm. I.24,a,xarebothpositiveora,xarebothnegative Exercise5. Definex,ysuchthatax=1,by =1. Wewantx>ywhenb>a. xb−ax=xb−1>0=⇒bx>1=by b>0sox>y Exercise6. 6 Ifa=bandb=c, thena=c Ifa=bandb0. Exercise9. Supposea≥x. Thena−x≥0. Ifa∈Rso∃y ∈R,suchthata−y =0. Considery+1∈R(byclosureunderaddition). a−(y+1)=a−y−1=0−1<0Contradictionthata≥y+1 Exercise10. Ifx=0, done. x Ifx>0,xisapositiverealnumber. Leth= . 2 x =⇒ >xContradiction. 2 I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line, Upperboundofaset,maximumelement,leastupperbound(supremum),Theleast-upper-boundaxiom(completeness axiom),TheArchimedeanpropertyofthereal-numbersystem,Fundamentalpropertiesofthesuprenumandinfimum. WeuseThmI.30,theArchimedeanpropertyofrealnumbers,alot. Theorem9(I.30). Ifx>0andifyisanarbitraryrealnumber,thereexistsapositiveintegernsuchthatnx>y. We will use the least upper-bound axiom (completeness axiom) alot for continuity and differentiation theorems later. Apostol states it as an axiom; in real analysis, the existence of a sup for nonempty, bounded sets can be shown with an algorithmtozoomintoasupwithmonotonicallyincreasingandmonotonicallydecreasingsequenceof“guesses”andshowing itsdifferenceisaCauchysequence. Axiom1(Leastupper-boundaxiom). EverynonemptysetS ofrealnumberswhichisboundedabovehasasuprenum;that is,there’sarealnumberBs.t. B =supS. Exercise1. 0h>0,n∈Z+,harbitrary y−x>h/n=⇒y >x+h/n>x soletz =x+h/nDone. Exercise2. x∈Rso∃n∈Z+suchthatn>x(Thm. I.29). Setofnegativeintegersisunboundedbelowbecause If∀m∈Z−,−x>−m,then−xisanupperboundonZ+. ContradictionofThm. I.29. =⇒∃m∈Zsuchthatm0sofor1,∃n∈Z+suchthatnx>1,x> 1. n Exercise4. x is an arbitrary real number. By Thm. I.29 and well-ordering principle, there exists a smallest n+1 positive integersuchthatx < n+1(considerthesetofallm+1 > xandsobywell-orderingprinciple,theremustbeasmallest elementofthisspecificsetofpositiveintegers). Ifx=nforsomepositiveintegern,done. Otherwise,notethatifxx. x>n. Exercise5. Ifx=n,done. Otherwise,considerallm>x. Bywell-orderingprinciple,thereexistsasmallestelementnsuch 7 thatn>x. Ifx+1n. Exercise6. y−x>0. n(y−x)>h, harbitrary,n∈Z+ y >x+h/n=z >x Sincehwasarbitrary,thereareinfinitelymanynumbersinbetweenx,y. Exercise7. x= a ∈Q,y ∈/ Q. b a±by x±y = b (cid:18) (cid:19) a−mb Ifa±bywasaninteger,saym,theny =± whichisrational. Contradiction. b ay ay xy = = b 1 b n Ifaywasaninteger,ay =n,y = ,butyisirrational. =⇒xyisirrational. a x y yisnotaninteger Exercise8. Proofbycounterexamples. Wewantthatthesumorproductof2irrationalnumbersisnotalwaysirrational. Ify isirrational,y+1isirrational,otherwise,ify+1∈Q, y ∈Qbyclosureunderaddition. =⇒y+1−y =1 Likewise,y1 =1. y Exercise9. y−x>0=⇒n(y−x)>k,n∈Z+,karbitrary. Choosektobeirrational. Thenk/nirrational. k k y > +x>x.Letz =x+ ,zirrational. n n Exercise10. (1) Supposen=2m andn+1=2m . 1 2 1 2m +1=2m 2(m −m )=1 m −m = .Butm −m canonlybeaninteger. 1 2 1 2 1 2 2 1 2 (2) By the well-ordering principle, if x ∈ Z+ is neither even and odd, consider the set of all x. There must exist a smallestelementx ofthisset. Butsincex ∈Z+,thentheremustexistan p2 sinceifq ,q >0, q q1 q2 1 2 np q q p 1 2 > 1 2 np q >q p q q q q 1 2 1 2 1 2 1 2 nexistssince(p q ),(q p )∈R. 1 2 1 2 Thesetofrationalnumbersdoesnotsatisfytheleast-upper-boundproperty. ConsideranonemptysetofrationalnumbersS boundedabovesothat∀x= r ∈S,xb ,butalsob >b . 1 2 2 1 I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, The well-orderingprinciple. Considerthese2proofs. N +N +···+N =N2 N N−1 (cid:88) (cid:88) (N −1)+(N −2)+···+(N −(N −1))+(N −N)=N2− j = j j=1 j=1 N N (cid:88) (cid:88) N(N +1) N2+N =2 j =⇒ j = 2 j=1 j=1 Aninterestingpropertyisthat n n (cid:88) (cid:88) S = j = (n+m−j) j=m j=m Sothat N N m N (cid:88) (cid:88) (cid:88) (cid:88) m(m+1) N(N +1) j = j+ j = j+ = 2 2 j=1 j=m j=1 j=m N (cid:88) N(N +1)−m(m+1) (N −m)(N +m+1) j = = 2 2 j=m Anotherwaytoshowthisisthefollowing. S = 1+ 2+ ···+ (N −2)+ (N −1)+ N butS = N+ N −1+ ···+ 3+ 2+ 1 N(N +1) 2S =(N +1)N S = 2 9 Telescopingserieswillletyouget(cid:80)N j2andotherpowersofj. j=1 N (cid:88) N(N +1) (2j−1)=2 −N =N2 2 j=1 N N N (cid:18) (cid:19) (cid:88) (cid:88) (cid:88) N(N +1) (j2−(j−1)2)= (j2−(j2−2j+1))= (2j−1)=2 −N =N2 2 j=1 j=1 j=1 N N N (cid:88) (cid:88) (cid:88) (j3−(j−1)3)=N3 = (j3−(j3−3j2+3j−1))= (3j2−3j+1) j=1 j=1 j=1 (cid:88)N N(N +1) 2N3+2N −3N2−3N N(N +1)(2N +1) (cid:88)N =⇒3 j2 =−3 +N =N3 =⇒ = = j2 2 2 6 j=1 j=1 N N N (cid:88) (cid:88) (cid:88) j4−(j−1)4 =N4 = j4−(j4−4j3+6j2−4j+1)= 4j3−6j2+4j−1= j=1 j=1 j=1 N (cid:88) N(N +1)(2N +1) N(N +1) =4 j3−6 +4 −N =N4 6 2 j=1 N (cid:88) 1 1 =⇒ j3 = (N4+N(N +1)(2N +1)−2N(N +1)+N)= (N4+(2N)N(N +1)−N(N +1)+N) 4 4 j=1 1 1 1(N(N +1))2 = (N4+2N3+2N2−N2−N +N)= N2(N2+2N +1)= 4 4 4 2 Exercise1. Inductionproof. N+1 n 1(1+1) (cid:88) (cid:88) n(n+1) n(n+1)+2(n+1) (n+2)(n+1) j = j+n+1= +n+1= = 2 2 2 2 j=1 j=1 Exercise6. (1) 1 1 8k+8 (2k+3)2 A(k+1)=A(k)+k+1= (2k+1)2+k+1= (4k2+4k+1)+ = 8 8 8 8 (2) Then=1caseisn’ttrue. (3) (n+1)n n2+n n2+n+ 1 1+2+···+n= = < 4 2 2 2 (cid:18)2n+1(cid:19)2 1 (n+1/2)2 n2+n+1/4 and = = 2 2 2 2 Exercise7. (1+x)2 >1+2x+2x2 1+2x+x2 >1+2x+2x2 0>x2 =⇒ Impossible (1+x)3 =1+3x+3x2+x3 >1+3x+3x2 =⇒x3 >0 Bywell-orderingprinciple,wecouldarguethatn=3mustbethesmallestnumbersuchthat(1+x)n >1+2x+2x2. Or wecouldfind,explicitly n (cid:18) (cid:19) n (cid:18) (cid:19) (cid:88) n n(n−1) (cid:88) n (1+x)n = xj =1+nx+ x2+ xj j 2 j j=0 j=3 10

SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. Fund Science!

Similar SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. Fund Science!