SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL

SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. ERNEST YEUNG Fund Science! & Help Ernest finish his Physics Research! : quantum super-A-polynomials - a thesis by Ernest Yeung ... I 2.5 Exercises - Introduction to set theory, Notations for designating sets, Subsets, Unions, intersections, complements. Exercise 10.


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Exercise11.Ifx2A[A,thenxisatleastinAorinA.Thenx2A.SoA[AA.OfcourseAA[A.Ifx2A\A,thenxisinAandinA.Thenx2A.SoA\AA.OfcourseAA\A.Exercise12.Letx2A.y2A[BifyisatleastinAorinB.xisinAsox2A[B.=)AA[B.Suppose9b2Bandb=2A.b2A[Bbutb=2A.soAA[B.Exercise13.Letx2A[;,thenxisatleastinAorin;.Ifx2;,thenxisanullelement(notanelementatall).ThenactualelementsmustbeinA.=)A[;A.Letx2A.Thenx2A[;.AA[;.=)A=A[;.Exercise14.Fromdistributivity,A[(A\B)=(A[A)\(A[B)=A\(A[B).Ifx2A\(A[B),x2Aandx2A[B,i.e.x2AandxisatleastinAorinB.=)xisinAandisinBorisnotinB.Thenx2A.=)A\(A[B)A.Ofcourse,AA\(A[B).=)A\(A[B)=A[(A\B)=A.Exercise15.8a2A;a2Cand8b2B;b2C.Considerx2A[B.xisatleastinAorinB.Ineithercase,x2C.=)A[BC.Exercise16.ifCAandCB;thenCA\B8c2C;c2Aandc2Bx2A\B;x2Aandx2B:Then8c2C;c2A\B:CA\BExercise17.(1)ifABandBCthen8a2A;a2B:8b2B;b2C:thensincea2B;a2C;9c2Csuchthatc=2B:8a2A;a2Bsoa6=c8a:=)AC(2)IfAB;BC;ACsince,8a2A;a2B;8b2B;b2C:Thensincea2B;a2C.AC(3)ABandBC.BCorB=C.ABonly.ThenAC.(4)Yes,since8a2A;a2B.(5)No,sincex6=A(setsaselementsaredifferentfromelements)Exercise18.A�(B\C)=(A�B)[(A�C)Supposex2A�(B\C)thenx2Aandx=2B\C=)x=2B\CthenxisnotinevenatleastoneBorC=)x2(A�B)[(A�C)Supposex2(A�B)[(A�C)thenxisatleastin(A�B)orin(A�C)=)xisatleastinAandnotinBorinAandnotinCthenconsiderwhenoneofthecasesistrueandwhenbothcasesaretrue=)x2A�(B\C)Exercise19.Supposex2B�[A2FAthenx2B;x=2[A2FAx=2[A2FA=)x=2A;8A2Fsince8A2F;x2B;x=2A;thenx2\A2F(B�A)2
(2)IfCA;A�(B�C)=(A�B)[CI3.3Exercises-Theeldaxioms.Thegoalseemstobetoabstracttheseso-calledrealnumbersintojustx'sandy'sthatarepurelybuiltupontheseaxioms.Exercise1.Thm.I.5.a(b�c)=ab�ac.Lety=ab�ac;x=a(b�c)Want:x=yac+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)=abButthereexistsexactlyoneyorxbyThm.I.2.x=y.Thm.I.6.0a=a0=0.0(a)=a(0)(bycommutativityaxiom)Givenb2Rand02R;9exactlyone�bs.t.b�a=00(a)=(b+(�b))a=ab�ab=0(byThm.I.5.andThm.I.2)Thm.I.7.ab=acByAxiom4;9y2Rs.t.ay=1sinceproductsareuniquelydetermined,yab=yac=)(ya)b=(ya)c=)1(b)=1(c)=)b=cThm.I.8.PossibilityofDivision.Givena;b;a6=0,chooseysuchthatay=1.Letx=yb.ax=ayb=1(b)=bTherefore,thereexistsatleastonexsuchthatax=b.ButbyThm.I.7,thereexistsonlyonex(sinceifaz�b,andsox=z).Thm.I.9.Ifa6=0,thenb=a=b(a�1).Letx=b
aforax=by=a�1foray=1Want:x=byNowb(1)=b;soax=b=b(ay)=a(by)=)x=by(byThm.I.7)Thm.I.10.Ifa6=0,then(a�1)�1=a.Nowab=1forb=a�1.Butsinceb2Randb6=0(otherwise1=0,contradiction),thenusingThm.I.8onb,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=0ab+x=ab+(�a)b=b(a+(�a))=b(a�a)=b(0)=00isunique,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)zSousingb;d6=0,whichisgiven,andThm.I.7,thenx+y=z.4
dbx+bdt=(bd)(x+y)=ad�bc=(bd)zb;d6=0;sox+y=zI3.5Exercises-Theorderaxioms.Theorem1(I.18).Ifabandc&#x-277;0thenacbcTheorem2(I.19).Ifabandc&#x-277;0,thenacbcTheorem3(I.20).Ifa6=0,thena2&#x-277;0Theorem4(I.21).1&#x-277;0Theorem5(I.22).Ifabandc0,thenac&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;֕ ;� Td;&#x [00;bc.Theorem6(I.23).Ifaband�a&#x-277;�b.Inparticular,ifa0,then�a&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;U 0;&#x Td ;&#x[000;0.Theorem7(I.24).Ifab&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;U 0;&#x Td ;&#x[000;0,thenbothaandbarepositiveorbotharenegative.Theorem8(I.25).Ifacandbd,thena+bc+d.Exercise1.(1)ByThm.I.19,�c&#x-278;0a(�c)b(�c)!�ac�bc�bc�(�ac)=ac�bc&#x]TJ/;༔ ; .96;& T; 22;&#x.861;&#x 0 T; [0;0:Thenac&#x]TJ/;༔ ; .96;& T; 22;&#x.861;&#x 0 T; [0;bc(bydenitionof&#x]TJ/;༔ ; .96;& T; 22;&#x.861;&#x 0 T; [0;)(2)ab!a+0b+0!a+b+(�b)b+a+(�a)!(a+b)�b(a+b)+(�a)ByThm.I.18(a+b)+�(a+b)+(�b)(a+b)�(a+b)+(�a)�b�a(3)Ifa=0orb=0,ab=0,but00Ifa&#x]TJ/;༔ ; .96;& T; 17;&#x.559;&#x 0 T; [0;0,thenifb&#x]TJ/;༔ ; .96;& T; 17;&#x.559;&#x 0 T; [0;0,ab&#x]TJ/;༔ ; .96;& T; 17;&#x.559;&#x 0 T; [0;0(b)=0.Ifb0;ab0(b)=0.Soifa&#x]TJ/;ø 9;&#x.962; Tf;&#x 27.;ɓ ;� Td;&#x [00;0,thenb&#x]TJ/;ø 9;&#x.962; Tf;&#x 27.;ɓ ;� Td;&#x [00;0.Ifa0,thenifb&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;Չ ;� Td;&#x [00;0;ab0(b)=0.Ifb0;ab&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;0(b)=0.Soifa0,thenb0.(4)acsoa+bc+b=b+cbdsob+cd+cByTransitiveLaw;a+bd+cExercise2.Ifx=0;x2=0.0+1=16=0.Sox6=0.Ifx6=0;x2&#x-278;0;andbyThm.I.21;1&#x-278;0x2+1&#x-278;0+0=0!x2+16=0=)@x2Rsuchthatx2+1=0Exercise3.a0;b0;a+b0+0=0(ByThm.I.25)Exercise4.Considerax=1.ax=1&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;0:ByThm.I.24;a;xarebothpositiveora;xarebothnegativeExercise5.Denex;ysuchthatax=1;by=1.Wewantx&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;ywhenb&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;a.xb�ax=xb�1&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;0=)bx&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;1=byb&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;0sox&#x]TJ/;ø 9;&#x.962; Tf;&#x 17.;ՙ ;� Td;&#x [00;yExercise6.6
thatn�x.Ifx+1n,thenxn�1,contradictingthefactthatnisthesmallestelementsuchthatxn.Thusx+1&#x-278;n.Exercise6.y�x&#x-278;0.n(y�x)&#x-278;h;harbitrary;n2Z+y&#x-278;x+h=n=z&#x-278;xSincehwasarbitrary,thereareinnitelymanynumbersinbetweenx;y.Exercise7.x=a
b2Q;y=2Q.xy=aby
bIfabywasaninteger,saym,theny=a�mb
bwhichisrational.Contradiction.xy=a
by
1=ay
bIfaywasaninteger,ay=n,y=n
a,butyisirrational.=)xyisirrational.x
yyisnotanintegerExercise8.Proofbycounterexamples.Wewantthatthesumorproductof2irrationalnumbersisnotalwaysirrational.Ifyisirrational,y+1isirrational,otherwise,ify+12Q;y2Qbyclosureunderaddition.=)y+1�y=1Likewise,y1
y=1.Exercise9.y�x�0=)n(y�x)�k;n2Z+;karbitrary.Choosektobeirrational.Thenk=nirrational.y�k
n+x�x:Letz=x+k
n;zirrational:Exercise10.(1)Supposen=2m1andn+1=2m2.2m1+1=2m22(m1�m2)=1m1�m2=1
2:Butm1�m2canonlybeaninteger.(2)Bythewell-orderingprinciple,ifx2Z+isneitherevenandodd,considerthesetofallx.Theremustexistasmallestelementx0ofthisset.Butsincex02Z+,thentheremustexistanxsuchthatn+1=x0.nisevenoroddsinceitdoesn'tbelongintheaboveset.Sox0mustbeoddoreven.Contradiction.(3)(2m1)(2m2)=2(2m1m2)even2m1+2m2=2(m1+m2)even(2m1+1)+(2m2+1)=2(m1+m2+1)=)sumoftwooddnumbersiseven(n1+1)(n2+1)=n1n2+n1+n+2+1=2(2m1m2)2(2m1m2)�(n1+n2)�1odd,theproductoftwooddnumbersn1;n2isodd(4)Ifn2even,niseven,sinceforn=2m,(2m)2=4m2=2(2m2)iseven.a2=2b2.2(b2)even.a2even,soaeven.Ifaevena=2n:a2=4n2Ifbodd,b2odd.bhasnofactorsof2b26=4n2Thusbiseven.8
TelescopingserieswillletyougetPNj=1j2andotherpowersofj.NXj=1(2j�1)=2N(N+1)
2�N=N2NXj=1(j2�(j�1)2)=NXj=1(j2�(j2�2j+1))=NXj=1(2j�1)=2N(N+1)
2�N=N2NXj=1(j3�(j�1)3)=N3=NXj=1(j3�(j3�3j2+3j�1))=NXj=1(3j2�3j+1)=)3NXj=1j2=�3N(N+1)
2+N=N3=)2N3+2N�3N2�3N
2=N(N+1)(2N+1)
6=NXj=1j2NXj=1j4�(j�1)4=N4=NXj=1j4�(j4�4j3+6j2�4j+1)=NXj=14j3�6j2+4j�1==4NXj=1j3�6N(N+1)(2N+1)
6+4N(N+1)
2�N=N4=)NXj=1j3=1
4(N4+N(N+1)(2N+1)�2N(N+1)+N)=1
4(N4+(2N)N(N+1)�N(N+1)+N)=1
4(N4+2N3+2N2�N2�N+N)=1
4N2(N2+2N+1)=1
4(N(N+1))2
2Exercise1.Inductionproof.1(1+1)
2N+1Xj=1j=nXj=1j+n+1=n(n+1)
2+n+1=n(n+1)+2(n+1)
2=(n+2)(n+1)
2Exercise6.(1)A(k+1)=A(k)+k+1=1
8(2k+1)2+k+1=1
8(4k2+4k+1)+8k+8
8=(2k+3)2
8(2)Then=1caseisn'ttrue.(3)1+2++n=(n+1)n
2=n2+n
2n2+n+1
4
2and2n+1
221
2=(n+1=2)2
2=n2+n+1=4
2Exercise7.(1+x)2�1+2x+2x21+2x+x2�1+2x+2x20�x2=)Impossible(1+x)3=1+3x+3x2+x3�1+3x+3x2=)x3�0Bywell-orderingprinciple,wecouldarguethatn=3mustbethesmallestnumbersuchthat(1+x)n�1+2x+2x2.Orwecouldnd,explicitly(1+x)n=nXj=0njxj=1+nx+n(n�1)
2x2+nXj=3njxj10
(1)am+am+1++am+n(2)n=11
2=1
1�1
2=1
2n+12(n+1)Xk=n+21
k=2nXm=1(�1)m+1
m�1
n+1+1
2n+1+1
2n+2=2nXm=1(�1)m+1
m+�1
2(n+1)+(�1)2n+1+1
(2n+1)=2(n+1)Xm=1(�1)m+1
mExercise13.n=12(p
2�1)12since1
2�p
2�1ncase(p
n+1�p
n)(p
n+1+p
n)=n+1�n=1p
n+1+p
n
2p
n=1
2(r
1+1
n+1)n+1case(p
n+2�p
n+1)(p
n+2+p
n+1)=n+2�(n+1)=1p
n+2+p
n+1
2p
n+1=1+q
1+1
n+1
2�1Sothen,usingthetelescopingproperty,n�1Xn=12(p
n+1�p
n)=2(p
m�1)mXn=11
p
nmXn=12(p
n�p
n�1)=2(p
m�1)2p
m�1I4.9Exercises-Absolutevaluesandthetriangleinequality.Exercise1.(1)jxj=0iffx=0Ifx=0,x=0;�x=�0=0.Ifjxj=0;x=0;�x=0.(2)j�xj=(�xif�x0xif�x0=(xifx0�xifx0(3)jx�yj=jy�xjbypreviousexerciseand(�1)(x�y)=y�x(bydistributivity)(4)jxj2=((x)2ifx0(�x)2ifx0=x2(5)p
x2=(xifx0�xifx0=jxj(6)Wewanttoshowthatjxyj=jxjjyjjxyj=(xyifxy0�xyifxy0=(xyifx;y0orx;y0�xyifx;�y0or�x;y0jxjjyj=(xjyjifx0�xjyjifx0=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:xyifx;y0�xyifx;�y0�xyif�x;y0xyif�x;�y0(7)Bypreviousexercise,since x
y =jxy�1j=jxjjy�1j 1
y =(1
yif1
y0�1
yif1
y01
jyj=(1
yif1
y0�1
yif1
y012
(1)1+1
nn=nXj=0nk1
nj=nXk=0n!
(n�k)!k!1
nkk�1Yr=01�r
n=k�1Yr=0n�r
n=1
nkn!
(n�k)!nXk=11
k!k�1Yr=01�r
n=1
nkn!
(n�k)!(2)(1+1
n)n=1+nXk=1 1
k!k�1Yr=0(1�r
n)!1+nXk=11
k!1+nXk=11
2k=1+1
2��1
2n+1
1
2=1+(1�1
2n)3Therstinequalityobtainedfromthefactthatif0x1;xnx1.Thesecondinequalitycamefromthepreviousexercise,that1
k!1
2k.(1+1
n)n=nXk=0nk1
nk=1+1
n+n�1Xk=1nk1
nk=1+1
n+n�1Xk=2nk1
nk+n
11
n��2Exercise13.(1)S=p�1Xk=0b
ak=1��b
ap
1�b
ap�1Xk=0bkap�1�k=ap�11��b
ap
1�b
a=bp�ap
b�a(2)(3)Givennp(n+1)p+1�np+1
p+1(n+1)pWewantn�1Xk=1kpnp+1
p+1nXk=1kpn=21p2p+1
p+11p+2pp=1122=2=2:21+2=3p�218=31+4=5I4.10Miscellaneousexercisesinvolvinginduction.Exercise13.(1)(2)(3)Letn=2.2�1Xk=1kp=1p=1;np+1
p+1=2p+1
p+12Xk=1kp=1+2pWhatmakesthisexercisehardisthatwehavetouseinductiononpitself.Letp=1.121+1
1+2=21+21=314
Nowassumepthcase.Testthep+1case.2p+2
p+2=2(p+1)
p+22p+1
p+1�1sincep+22p+2=2(p+1)forp2Z+Fortheright-handinequality,wewillusethefactjustproven,that2p�(p)&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0andpthcaserewritteninthismanner(1+2p)&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;2p+1
p+1=)(1+2p)(p+1)�2p+1So(p+2)(1+2p+1)=(p+2)+((p+1)+1)2p(2)=(p+2)+2(p+1)2p+2p(2)��(p+2)+2(2p+1�(p+1))+2p(2)=�p+2p+2+2p+1�2p+2Sothen=2caseistrueforallp2Z+.Assumenthcaseistrue.Wenowprovethen+1case.nXk=1kp=n�1Xk=1kp+npnp+1
p+1+npnp+1
p+1+(n+1)p+1�np+1
p+1=(n+1)p+1
p+1n+1Xk=1kp=nXk=1kp+(n+1)p�np+1
p+1+(n+1)p+1�np+1
p+1=(n+1)p+1
p+1Wehadusedtheinequalityproveninpartb,np(n+1)p+1�np+1
p+1(n+1)p.Exercise14.UseinductiontoproveageneralformofBernoulli'sinequality.1+a1=1+a1(1+a1)(1+a2)=1+a2+a1+a1a21+a1+a+2Testthen+1case(1+a1)(1+a2):::(1+an+1)(1+a1+a2++an)(1+an+1)==1+a1+a2++an+an+1+an+1(a1+a2+:::an)1+a1+a2++an+an+1Notethatthelaststepdependeduponthegivenfactthatallthenumberswereofthesamesign.Fora1=a2==an=x,thenwehave(1+x)n1+nx.(1+x)n=nXj=0njxj=1+nxSincexandnarearbitrary,wecancomparetermsofxj's.Thenx=0.Exercise15.2!
22=1
23!
33=2
91.Sowe'veshownthen=2;n=3cases.Assumethenthcase,thatn!
nn�1
2k,wherekisthegreatestintegern
2.(n+1)!
(n+1)n+1(n+1)nn�1
2k
(n+1)n+1=n
n+1n1
2k=1�1
n+1n1
2k1
21
2k=1
2k+1whereinthesecondtothelaststep,wehadmadethisimportantobservation:kn
2=)k+1
2n+1
2=)1
n+11
2k+11
2Exercise16.15
1.7Exercises-Theconceptofareaasasetfunction.Wewillusethefollowingaxioms:AssumeaclassMofmeasurablesets(i.e.setsthatcanbeassignedanarea),setfunctiona,a:M!R.Axiom2(Nonnegativeproperty).(1)8S2M;a(S)0Axiom3(Additiveproperty).IfS;T2M,thenS[T;S\T2Mand(2)a(S[T)=a(S)+a(T)�a(S\T)Axiom4(Differenceproperty).IfS;T2M;STthenT�S2Mand(3)a(T�S)=a(T)�a(S)Axiom5(Invarianceundercongruence).IfS2M;T=S,thenT2M,a(T)=a(S)Axiom6(Choiceofscale).8rectangleR2M,ifRhasedgelengthsh;kthena(R)=hkAxiom7(Exhaustionproperty).LetQsuchthat(4)SQTIf9onlyonecsuchthata(S)ca(T);8S;TsuchthattheysatisfyEqn.(??)thenQmeasurableanda(Q)=cExercise1.(1)Weneedtosaythatweconsideralinesegmentorapointtobeaspecialcaseofarectangleallowinghork(orboth)tobezero.LetTl=flinesegmentcontainingx0g;Q=fx0g.ForQ,only;QByAxiom3,letT=S.a(T�S)=a(;)=a(T)�a(T)=0;QTl=)a(;)a(Q)a(Tl)=)0a(Q)0=)a(Q)=0(2)[email protected][j=1Qj1A=NXj=1a(Qj)ifQj'sdisjoint.LetQj=fxjg.Sincea(Qj)=0.Bypreviouspart,aSNj=1Qj=0Exercise2.LetA;Bberectangles.ByAxiom5,A;Baremeasurable.ByAxiom2,A\Bmeasurable.a(A\B)=p
a2+b2d+ab�(1
2ab+p
a2+b2d)=1
2abExercise3.Provethateverytrapezoidandeveryparallelogramismeasurableandderivetheusualformulasfortheirareas.Atrapezoidissimplyarectanglewitharighttriangleattachedtoeachendofit.Tr=R+T1+T2.T1;T2arerighttrianglesandsobythepreviousproblem,T1;T2aremeasurable.ThenTrismeasurablebytheAdditivepropertyaxiom(notethatthetrianglesandtherectangledon'toverlap).17
1.11Exercises-Intervalsandordinatesets,Partitionsandstepfunctions,Sumandproductofstepfunction.Exercise4.(1)[x+n]=yx+n;y2Z;y�nx[x]+n=z+nx+nIfy�nz;thenyz+nx+n:thenywouldn'tbethegreatestintegerlessthanx+n=)y=z+n(2)=y2x�[x]=�y2�x�y2�1x�xy1=[�x]=�y2�1=�[x]�1;(andy1=�y2�1since�y2&#x-277;�x)Ifxisaninteger�[x]=[�x](3)Letx=q1+r1;y=q2+r2;0r1;r21.=[q1+q2+r1+r2]=(q1+q2q1+q2+1ifr1+r21[x]+[y]=q1+q2[x]+[y]+1=q1+q2+1(4)Ifxisaninteger;[2x]=2x=[x]+[x+1
2]=[x]+[x]=2x[x]+[x+1
2]=q+(qifr1
22q+1ifr�1
2[2x]=[2(q+r)]=[2q+2r]=(2qifr1
22q+1ifr�1
2(5)[x]+[x+1
3]+[x+2
3]=q+(qifr2
3q+1ifr�2
3+(qifr1
3q+1ifr�1
3=8�&#x]TJ ;� -1;.93; Td;&#x [00;:3qifr1
33q+1if1
3r2
33q+2ifr�2
3[3x]=[3(q+r)]=[3q+3r]=8�&#x]TJ ;� -1;.93; Td;&#x [00;:3qifr1
33q+1if1
3r2
33q+2ifr�2
3Exercise5.Directproof.[nx]=[n(q+r)]=8�&#x]TJ ;� -1;.93; Td;&#x [00;:nqifr1
nnq+1if1
nr2
nnq+n�1ifr�n�1
nExercise6.a(R)=hk=IR+1
2BR�1bXn=a[f(n)]=[f(a)]+[f(a+1)]++[f(b)][f(n)]=gf(n);g2Z,sothatiff(n)isaninteger,g=f(n),andiff(n)isnotaninteger,gisthelargestintegersuchthatgf(n),sothatalllatticepointsincludedandlessthangareincluded.Exercise7.19
Exercise3.[x]=yxso�y�x.�y�1�x,otherwiseif�y�1�x,y+1xandsoywouldn'tbethelargestintegerx.=)[x]+[�x]=y�y�1=�1OruseExercise4(c),pp.64.Zba([x]+[�x])dx=Zba[x�x]dx=Zba(�1)dx=a�bExercise4.(1)n2Z+;Rn0[t]dt=Pn�1t=0t=(n�1)(n�1+1)
2=(n�1)n
2(2)Exercise5.(1)R20[t2]dt=R21[t2]dt=1(p
2�1)+2(p
3�p
2)+3(2�p
3)=5�p
2�p
3(2)R3�3[t2]dt=R30[t2]dt+R0�3[t2]dt=R30[t2]dt+�R03[t2]dt=2R30[t2]dtZ32[t2]dt=4(p
5�2)+5(p
6�p
5)+6(p
7�p
6)+7(p
8�p
7)+8(3�p
8)16�p
5�p
6�p
7�p
8Z20[t2]dt+Z32[t2]dt=21�3p
2�p
3�p
5�p
6�p
7�p
8=)Z3�3[t2]dt=42�2(3p
2+p
3+p
5+p
6+p
7)Exercise6.(1)Rn0[t]2dt=Rn1[t]2dt=Pn�1j=1j2=(n�1)n(2n�1)
6(2)Rx0[t]2dt=P[x�1]j=1j2+q2rwherex=q+r;q2Z+;0r1.Zx0[t]2dt=q(q�1)(2q�1)
6+q2r=2(x�1)=2(q+r�1)=)q(q�1)(2q�1)+6q2r=12q+12r�12=)x=1;x=5=2Exercise7.(1)Z90[p
t]dt=Z91[p
t]dt=3(1)+5(2)=13Z106[p
t]dt=3(1)+5(2)+7(3)=34=(4)(3)(17)
6AssumeZn20[p
t]dt=n(n�1)(4n+1)
6Z(n+1)20[p
t]dt=Zn20[p
t]dt+Z(n+1)2n2[p
t]dt=n(n�1)(4n+1)
6+n((n+1)2�n2)==(n2�n)(4n+1)+6n(2n+1)
6=4n3+n2�4n2�n+12n2+6n
6=4n3+9n2+5n
6indeed;(n+1)(n)(4(n+1)+1)
6=(n2+n)(4n+5)
6=4n3+5n2+4n2+5n
621
ThenRbasRbat.Exercise12.(1)Rbas+Rcbs=Pn1k=1sk(x2k�x2k�1)+Pn2k=n1sk(x2k�x2k�1)=Pn3k=1sk(x2k�x2k�1)=Rcas(2)Rba(s+t)=Pn3k=1(s+t)k(x2k�x2k�1)=Pn3k=1(sk+tk)(x2k�x2k�1)=Pn3k=1sk(x2k�x2k�1)+Pn3k=1tk(x2k�x2k�1)sinceP3=fxkgisanerpartitionthanthepartitionfors;P1;t;P2,thenconsidersk(y2j�y2j�1)=sk((x2k+1�x2k)+(x2k�x2k�1)),son3Xk=1sk(x2k�x�k�12)+n3Xk=1tk(x2k�x�k�12)=n1Xj=1sj(x2j�x�j�12)+n2Xj=1tj(x2j�x�j�12)==Zbas+Zbat(3)Rbacs=Pnk=1csk(x2k�x2k�1)=cPnk=1sk(x2k�x2k�1)=cRbas(4)Rb+ca+cs(x)dx=Pnk=1sk(x2k�x2k�1)wheres(x)=skifxk�1xxkx(y+c)=skifxk�1y+cxk=)xk�1�cyxk�c=)yk�1yykwhereP0=fykgisapartitionon[a;b]Zbas(y+c)dy=nXk=1sk(y2k�y2k�1)==nXk=1sk((xk�c)2�(xk�1�c)2)=nXk=1sk(x2k�2xkc+c2�(x2k�1�2xk�1c+c2))==nXk=1sk(x2k�x2k�1�2c(xk�xk�1))6=nXk=1sk(x2k�x2k�1)(5)Sincex2k�x2k�1&#x-278;0;Rbasdx=Pnk=1sk(x2k�x2k�1)Pnk=1tk(x2k�x2k�1)=RbatdxNotethatwehadshownpreviouslythattheintegraldoesn'tchangeundernerpartition.Exercise13.Zbas(x)dxnXk=1sk(xk�xk�1);Zbat(x)dx=n2Xk=1tk(yk�yk�1)P=fx0;x1;:::;xng;Q=fy0;y1;:::;yngNotethatx0=y0=a;xn=yn2=b.ConsiderPSQ=R.Rconsistsofn3elements,(sincen3n+n2someelementsofPandQmaybethesame.Risanotherpartitionon[a;b](bypartitiondenition)sincexk;yk2Randsincerealnumbersobeytransitivity,fxk;ykgcanbearrangedsuchthataz1z2zn3�2bwherezk=xkoryk.(s+t)(x)=s(x)+t(x)=sj+tkifxj�1xxj;yj�1xyjIfxj�17yj�1;letzl�1=yj�1;xj�1andIfxj7yj;letzl=xj;yjLetsj=sl;tk=tl(s+t)(x)=s(x)+t(x)=sl+tl;ifzl�1xzlZba(s(x)+t(x))dx=Zba((s+t)(x))dx=n3Xl=1(sl+t)l)(zl�zl�1)=n3Xl=1sl(zl�zl�1)+n3Xl=1tl(zl�zl�1)Ingeneral,itwasshown(ApostolI,pp.66)thatanynerpartitiondoesn'tchangetheintegralRisanerpartition.Son3Xl=1sl(zl�zl�1)+nXl=1tl(zl�zl�1)=nXk=1sk(xk�xk�1)+n2Xk=1tk(yk�yk�1)=Zbas(x)dx+Zbat(x)dx23
Z21(x�1)(3x�1)dx=Z21(3x2�4x+1)dx=(x3�2x2+x) 21=2Z11=3(1�x)(3x�1)dx=�(x3�2x2+x) 11=3=4
27Z1=30(x�1)(3x�1)dx=4
27Sothenalanswerfortheintegralis62=27.Exercise17.R30(2x�5)3dx=8R30(x�5
2)3dx=8R3�5=2�5=2x3dx=81
4x4 1=2�5=2=39
2Exercise18.R3�3(x2�3)3dx=R30(x2�3)3+Rx�3(x2�3)3=R30(x2�3)2+�R30(x2�3)3=02.4Exercises-Introduction,Theareaofaregionbetweentwographsexpressedasanintegral,Workedexamples.Exercise15.f=x2;g=cx3;c�0For0x1
c,cx1(sincec&#x]TJ/;ø 9;&#x.962; Tf;&#x 23.;ʉ ;� Td;&#x [00;0).Socx3x2(sincex2&#x-278;0).Zf�g=Zx2�cx3=1
3x3�c
4x4 1=c0=1
12c3Zf�g=2
3=1
12c3;
c=1
2p
2
Exercise16.f=x(1�x);g=ax.Zf�g=Z1�a0x�x2�ax=(1�a)1
2x2�1
3x3 1�a0=(1�a)31
6=9=2=)
a=�2
Exercise17.=2R1�1p
1�x2dx(1)Z3�3p
9�x2dx=3Z3�3r
1�x
32=3(3)Z1�1p
1�x2=
9
2
NowZkbkafx
kdx=kZbafdx(2)Z20r
1�1
4x2dx=2Z10p
1�x2dx=2
4=

2
(3)R2�2(x�3)p
4�x2dxZ2�2xp
4�x2dx=(�1)Z�22�xp
4�x2=)2Z2�2xp
4�x2=0�3Z2�22r
1�x
22dx=(�6)(2)Z1�1p
1�x2=
�6
Exercise18.Consideracircleofradius1andatwelve-sideddodecagoninscribedinit.Dividethedodecagonbyisoscelestrianglepieslices.Theinterioranglethatisthevertexangleofthesetrianglesis360=12=30degrees.Thenthelengthofthebottomsideofeachtriangleisgivenbythelawofcosines:c2=1+1�2(1)(1)cos30=2 1�p
3
2!=)c=p
2s
1�p
3
225
Asinx+Bcosx=B(cosx+A
Bsinx)=B
cos (cosxcos +sin sinx)=Ccosx+ whereA
B=tan , =� ,C=B
cos .Exercise12.sinx=cosx=p
1�cos2x=)cosx=1=p
2=)x=
4Try5=4.sin5=4=cos3=4=�sin=4=�1=p
2.cos5=4=�sin3=4=�cos=4=�1=p
2.Sosin5=4=cos5=4.x=5=4mustbetheotherroot.So==4+n(byperiodicityofsineandcosine).Exercise13.sinx�cosx=1=p
1�cos2x=1+cosx=)1�cos2x=1+2cosx+cos2x=)0=2cosx(1+cosx)cosx=�1;x==2+2nExercise14.cosx�y+cosx+y=cosxcosy+sinxsiny+cosxcosy�sinxsiny=2cosxcosycosx�y�cosx+y=sinxcosy�sinycosx+sinxcosy+sinycosx=2sinxcosysinx�y+sinx+y=sinxcosy�sinycosx+sinxcosy+sinycosx=2sinxcosyExercise15.sinx+h�sinx
h=sin(x+h=2)cosh=2+cos(x+h=2)sinh=2�sin(x+h)cosh=2�cosx+h=2sinh=2
h=sinh=2
h=2cos(x+h=2)cosx+h�cosx
h=cos(x+h=2)cosh=2�sin(x+h=2)sinh=2�(cos(x+h=2)cosh=2+sin(x+h=2)sinh=2)
h=�sinh=2
h=2sin(x+h=2)Exercise16.(1)sin2x=2sinxcosxifsin2x=2sinxandx6=0;x6=n;cosx=1butx6=n=)
x=2n
(2)cosx+y=cosxcosy�sinxsiny=cosx+cosy.cosxcosy�cosx�cosy=sinyp
1�cos2xLettingA=cosx;B=cosy;A2B2+A2+B2�2A2B�2AB2+2AB=1�A2�B2+A2B2A2+B2�A2B�AB2+AB=1=2B2(1�A)+B(A�A2)+A2�1=2=0B=A(1�A)p
A2(1�A)2�4(1�A)(A2�1=2)
1�A=A1
p
1�A(A2(1�A)�4(A2�1=2))1=2==A1
p
1�A(�3A2�A3+2)1=2Notethat�1B1,butforjAj1.Solvefortherootsof�3A2�A3+2,A0=�1;�1+p
3;�1�p
3.Sosupposecosx=9=10.Thenthereisnorealnumberforysuchthatcosywouldberealandsatisfytheaboveequation.(3)sinx+y=sinxcosy+sinycosx=sinx+siny=)siny(1�cosx)+siny+�cosxsiny=0;=)y=2nCheckingourresult,wendthatsin(2n+y)=sin2n+siny(1)28
Zx0cos(a+bt)dt=Zx0(cosacosbt�sinasinbt)dt=cosa
bsinbt�sina(�cosbt=b) x0==cosa
bsinbx+sina
b(cosbx�1)=1
bsina+bx�sina=bZx0sin(a+bt)dt=Zx0(sinacosbt+sinbtcosa)dt=sina
bsinbt�cosa
bcosbt x0==1
b(cosbx+a+cosa)Exercise29.(1)Zx0sin3tdt=Zx03sint�sin3t
4dt=�3
4cost+cos3t=12 x0=�3=4(cosx�1)+cos3x�1
12==1
3�3
4cosx+1
12(cos2xcosx�sin2xsinx)=2=3�1=3cosx(2+sin2x)(2)Zx0cos3tdt=Zx01
4(cos3t+3cost)dt=1
4sin3t
3+3
4sint x0==1
12(sin2xcosx+sinxcos2x)+3
4sinx=1
12(2sinxcosx+sinx(2cos2x�1))==sinxcos2x+2sinx
3Exercise30.Nowusingthedenitionofaperiodicfunction,f(x)=f(x+p);f(x+(n+1)p)=f(x+np+p)=f(x+np)=f(x)andknowingthatwecouldwriteanyrealnumberinthefollowingform,a=np+r;0p;r2R;n2ZthenZa+paf(x)dx=Zr+prf(x+np)dx=Zr+prf(x)dx=Zprf+Zr+ppf(x)dx==Zprf+Zr0f(x�p)dx=Zprf+Zr0f=Zp0fExercise31.(1)Z20sinnxdx=1
nZ2n0sinxdx=1
n(�cosx) 2n0=�1
n(1�1)=0Z20cosnxdx=1
nZ2n0cosxdx=1
nsinx 2n0=0(2)Z20sinnxcosmxdx=Z201
2(sin(n+m)x+sin(n�m)x)dx=0+0=0Z20sinnxsinmxdx=Z201
2(cos(n�m)x+cos(n+m)x)dx=0+0=0Z20cosnxcosmxdx=Z201
2(cos(n�m)x+cos(n+m)x)dx=0+0=0WhileZ20sin2nxdx=Z201�cos2nx
2dx=Z20cos2nxdx=Z201+cos2nx
2dx=30
Exercise9.1
=2�0Rsin2x=1
(x�sin2x=2) 0=1
2Exercise10.1
�0Rcos2x=1
2Exercise11.(1)1
a�0Rx2=a2=3=c2=)c=a=p
3(2)1
a�0Rxn=1
a1
n+1xn+1 a0=an
n+1=cn=)c=a
(n+1)1=nExercise12.A=Zwf=ZwZwx2=kZxZx3=1
4x4=k1
2x2;k=1
2;w=xZx4=1
5x5=k1
3x3;k=3
5;w=x2Zx5=1
6x6=k1
4x4;k=2
3;w=x3Exercise13.A(f+g)=1
b�aZf+g=1
b�aZf+1
b�aZg=A(f)+A(g)A(cf)=1
b�aZcf=c1
b�aZfA(f)=1
b�aZf1
b�aZg=A(g)Exercise14.A(c1f+c2g)=Rw(c1f+c2g)
Rw=c1Rwf
Rw+c2Rwg
Rw=c1A(f)+c2A(g)fgw�0(nonnegative);=)wfwgExercise15.Aba(f)=1
b�aZbaf=1
b�a Zcaf+Zbcf!=c�a
b�a Rcaf
c�a!+b�a�(c�a)
b�aRbaf
b�cacb0c�a
b�a1Lett=c�a
b�a=)Aba(f)=tAca(f)+(1�t)Abc(f)Aba(f)=Rbawf
Rbaw=Rcaw
RbawRcawf
Rcaw+ Rbaw�Rcaw
Rbaw!Rbcwf
Rbcw0Rcaw
Rbaw1sincewisanonnegativefunction.Lett=Rcaw
Rbaw=)Aba(f)=tAca(f)+(1�t)Abc(f)Exercise16.Recallthatxcm=Rx
Rorrcm=Rrdm
M.32
Let=cxncZL0xndx=1
n+1Ln+1c=M=)c=(n+1)M
Ln+1cZL0xxndx=c1
n+2Ln+2=n+1
n+2ML=3ML
4xcm=Rx
M=3L
4=)Zx=n+1
n+2=3
4=)n=2
=3M
L3x2
Exercise23.(1)1
=2�0Z3sin2t=6
(2)1
=2�0Z9sin22t=9=2=)vrms=3p
2=2Exercise24.T=2(justlookatthefunctionsthemselves)1
2Z20160sint2sin(t�=6)=80p
32.19Exercises-Theintegralasafunctionoftheupperlimit.Indeniteintegrals.Exercise1.Rx0(1+t+t2)dt=x+1
2x2+1
3x3Exercise2.2y+2y2+8y3=3Exercise3.2x+2x2+8x3=3�(�1+1=2+�1=3)=2(x+x2+4x3=3)+5=6Exercise4.R1�x1(1�2t+3t2)dt=(t�t2+t3) 1�x1=�2x+2x2�x3Exercise5.Rx�2t4+t2=1
5t5+1
3t3 x�2=x5
5+x3
3+40
3Exercise6.Rx2xt4+2t2+1=t5
5+2
3t3+t x2x=1
5(x10�x5)+2
3(x6�x3)+x2�xExercise7.�2
3t3=2+t x1=2
3(x3=2�1)+(x�1)Exercise8.�2
3t3=2+4
5t5=4 x2x=2
3(x3�x3=2)+4
5(x5=2�x5=4)Exercise9.sintjxi=sinxExercise10.�t
2+sint x20=x2
2+sinx2Exercise11.�1
2t+cost x2x=x2�x
2+cosx2�cosxExercise12.�1
3u3+�1
3cos3u x0=x3
3+�1
3(cos3x�1)Exercise13.1
3v3+cos3v
�3 x2x=x6�x3
3+�1
3(cos3x2�cos3x)Exercise14.R1�cos2x
2+x=�1
2x�sin2x
4+1
2x2 y0=
y
2�sin2y
4+y2
2
34
Exercise19.g(2n)=R20f(t)dtConsiderZ1�1f(t)dt=Z10f(t)dt+Z0�1f(t)dt=Z10f(t)dt+1
�1Z01f(�1t)dt==Z10f+Z01f(t)dt=0ConsiderthatR31f(t)dt=R1�1f(t+2)dt=R1�1f(t)dt=0.Then,byinduction,Z2n+11f=Z2n�11f+Z2n+12n�1f(t)dt=0+Z1�1f(t+2n)dt=Z1�1f(t)dt=0(1)g(2n)=Z10f+Z2n�11f+Z2n2n�1f=Z10f+Z0�1f(t)dt=Z10f+�Z01f(�t)dt=Z10f+Z01f=0(2)g(�x)=Z�x0f=�Zx0f(�t)dt=Zx0f(t)dt=g(x)g(x+2)=Zx+20f(t)dt=Z20f+Zx+22f=Zx0f(t+2)dt=Zx0f(t)dt=g(x)Exercise20.(1)gisoddsinceg(�x)=Z�x0f(t)dt=�Zx0f(�t)dt=�Zx0f(t)dt=�g(x)Nowg(x+2)=Zx+20f=Z20f+Zx+22f=g(2)+Zx0f(t+2)dt=g(2)+Zx0f(t)dt=g(2)+g(x)=)g(x+2)�g(x)=g(2)(2)g(2)=Z20f=Z21f+Z10f=Z21f+A=Z0�1f(t+2)dt+A=Z0�1f(t)dt+A=�Z01f(�t)dt+A=2Ag(5)�g(3)=g(2)g(3)=g(2)+Z32f(t)dt=2A+Z10f(t+2)dt=2A+A=3A=)g(5)=3A+2A=5A(3)Thekeyobservationistoseethatgmustrepeatitselfbyachangeof2intheargument.Tomakeg(1)=g(3)=g(5),they'redifferent,unlessA=0!Exercise21.Fromthegiven,wecanderiveg(x)=f(x+5);f(x)=Zx0g(t)dt=)f(5)=Z50g(t)dt=g(0)=7(1)ThekeyinsightIuncoveredwas,whenstuck,oneofthethingsyoucando,istothinkgeometricallyanddrawapicture.g(�x)=f(�x+5)=g(x)=�f(x�5)=)�g(x)=f(x�5)36
limx!0sin�x+a
2+x�a
2�sin�x+a
2��x�a
2
x�a==limx!0 sinx+a
2cosx�a
2+sinx�a
2cosx+a
2��sinx+a
2cosx�a
2�sinx�a
2cosx+a
2
x�a!==limx!a2sinx�a
2cosx+a
2
x�a=cosaExercise20.limx!02sin2x=2
4(x=2)2=1
2limx!0sinx=2
x=2=1
2Exercise21.limx!01�p
1�x2
x21+p
1�x2
1+p
1�x2=limx!01�(1�x2)
x2(1+p
1�x2)=
1
2
Exercise22.b;caregiven.sinc=ac+b;a=sinc�b
c;c6=0.ifc=0,thenb=0;a2R.Exercise23.b;caregiven.2cosc=ac2+b;a=2cosc�b
c2;c6=0.Ifc=0,thenb=2;a2R.Exercise24.tangentiscontinuousforx=2(2n+1)=2cotangentiscontinuousforx=22nExercise25.limx!0f(x)=1.Nof(0)cannotbedened.Exercise26.(1)jsinx�0j=jsinxjjxj.Choose=foragiven.Then8&#x]TJ/;༔ ; .96;& T; 10;&#x.516;&#x 0 T; [0;0;9&#x]TJ/;༔ ; .96;& T; 10;&#x.516;&#x 0 T; [0;0suchthatjsinx�0jwhenjxj.(2)jcosx�1j=j�2sin2x=2j=2jsinx=2j22jx
2j2=jxj2
22=2=Ifwehadchosen0=p
2foragiven.jx�0j=p
2.(3)jsinx(cosh�1)+cosxsinhjjsinxjjcosh�1j+jcoshjjsinhj
2+
2=jcosx+h�cosxj=jcosxcosh�sinxsinh�cosxj=jcosx(cosh�1)�sinxsinhjjcosxjjcosh�1j+jsinxjjsinhj
2+
2=since8�091;2�0suchthatjcosh�1j0;jsinhjwheneverjhjmin(1;2)Choose3suchthatifjhj3;jcosh�1j
2;jsinhj
2Exercise27.f(x)�A=sin1
x�A.Letx=1
n.jf(x)�Aj=jsinn�Aj�jjsinnj�jAjj�j1�jAjjConsiderjx�0j=jxj=1
n(n).Consider0=j1�jAjj
2.Thensupposea(n)jx�0jbutjf(x)�Aj�0.Thus,contradiction.Exercise28.Considerx1
n;n2Z+;n�M(n)(nisagivenconstant)f(x)=1
x=[n]=n;form�M(n);x=1
mf(x)�M(n)so8�0,wecannotnd=1
nsuchthatjf(x)�Ajforx.Sof(x)!1asx!0+.38
(3) Zbaf(x)dx�(b�a)f(c) = Zba(f(x)�f(c))dx Zbajf(x)�f(c)jdxZbajx�cjdx==Zca(c�x)dx+Zbc(x�c)dx=c(c�a)�1
2(c�a)(c+a)+1
2(b�c)(b+c)�c(b�c)==1
2((c�a)2+(b�c)2)Drawagureforclear,geometricreasoning.Considerasquareoflength(b�a)anda45�45righttriangleinside.Fromthegure,it'sobviousthatrighttrianglesofc�alengthand(b�c)lengthliewithinthe(b�a)righttriangle.Comparethetrapezoidofc�a;b�abaseswiththeb�arighttriangle.1
2(b�c)(b�a+c�a)=1
2(b�c)(b�c+2(c�a))�1
2(b�c)2Indeed,thetrapezoidandc�arighttriangleequalstheb�atrapezoidsince1
2(b�c)(b�a+c�a)+1
2(c�a)2=1
2(b2�c2�2ab+2ac+c2�2ca+a2)=1
2(b�a)2=)1
2(b�a)2�1
2(b�c)2+1
2(c�a)2sothen Zbaf(x)dx�(b�a)f(c) (b�a)2
23.11Exercises-Bolzano'stheoremforcontinuousfunctions,Theintermediate-valuetheoremforcontinuousfunc-tions.Thesetheoremsformthefoundationforcontinuityandwillbevaluablefordifferentiationlater.Theorem10(Bolzano'sTheorem).Letfbecont.at8x2[a;b].Assumef(a);f(b)haveoppositesigns.Then9atleastonec2(a;b)s.t.f(c)=0.Proof.Letf(a)0;f(b)&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0.Want:Fineonevaluec2(a;b)s.t.f(c)=0Strategy:ndthelargestc.LetS=fallx2[a;b]s.t.f(x)0g.Sisnonemptysincef(a)0.SisboundedsinceallS[a;b].=)Shasasuprenum.Letc=supS.Iff(c)&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0;9(c�;c+)s.t.f&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0c�isanupperboundonSbutcisaleastupperboundonS.Contradiction.Iff(c)0;9(c�;c+)s.t.f0c+isanupperboundonSbutcisanupperboundonS.Contradiction.Theorem11(Sign-preservingPropertyofContinuousfunctions).Letfbecont.atcandsupposethatf(c)6=0.then9(c�;c+)s.t.fbeon(c�;c+)hasthesamesignasf(c).Proof.Supposef(c)&#x]TJ/;ø 9;&#x.962; Tf;&#x 19.;ȳ ;� Td;&#x [00;0.8&#x]TJ/;ø 9;&#x.962; Tf;&#x 19.;ȳ ;� Td;&#x [00;0;9&#x]TJ/;ø 9;&#x.962; Tf;&#x 19.;ȳ ;� Td;&#x [00;0s.t.f(c)�f(x)f(c)+ifc�xc+(bycontinuity).Choosefor=f(c)
2.Thenf(c)
2f(x)3f(c)
28x2(c�;c+)Thenfhasthesamesignasf(c).40
f=1
1+x6g=1�x2+x4Za01�x2+x4=x�1
3x3+1
5x5 a0=a�a3
3+a5
5m=1
1+a6M=11
1+a6a�a3
3+a5
5Za01
1+x2dxa�a3
3+a5
5Soifa=1
10,(a�a3=3+a5=5)=a�0:333:::a3+0:2a5=0:099669Exercise4.(b)iswrong,sinceithadchoseng=sint,butgneededtobenonnegative.Exercise5.Atworst,wecouldhaveutilizedthefundamentaltheoremofcalculus.Zsint2dt=Z1
2t(2tsint2)dt=1
2c(�cost2) p
(n+1)p
n==�1
2c((�1)n+1�(�1)n)=
1
c(�1)n
Exercise6.Rba(f)(1)=f(c)Rba1=f(c)(b�a).Thenf(c)=Rbaf
b�a=0forsomec2[a;b]byMean-valuetheoremforintegrals.Exercise7.fnonnegative.Considerfatapointofcontinuityc,andsupposef(c)�0.Then1
2f(c)�0.jf(x)�f(c)j=)f(c)�f(x)f(c)+Let=1
2f(c)9�0for=1
2f(c)Zc+c�f(x)dx�1
2f(c)(2)=f(c)�0ButRbaf(x)dx=0andfisnonnegative.f(c)=0.Exercise8.mZgZfgMZg=)mZg0MZg8gm0MforZg=1butalso�m0�M=)m0M0forZg=�1Sobecauseofthiscontradiction,m=M=0.Byintermediatevaluetheorem,f=0;8x2[a;b].4.6Exercises-Historicalintroduction,Aprobleminvolvingvelocity,Thederivativeofafunction,Examplesofderiva-tives,Thealgebraofderivatives.Exercise1.f0=1�2x,f0(0)=1;f0(1=2)=0;f0(1)=�1;f0(10)=�19Exercise2.f0=x2+x�2(1)f0=0;x=1;�2(2)f0(x)=�2;x=0;�1(3)f0=10;x=�4;3Exercise3.f0=2x+3Exercise4.f0=4x3+cosxExercise5.f0=4x3sinx+x4cosxExercise6.f0=�1
(x+1)2Exercise7.f0=�1
(x2+1)2(2x)+5x4cosx+x5(�sinx)Exercise8.f0=x�1�(x)
(x�1)2=�1
(x�1)243
g=f1f2g0=f01f2+f1f02g0
g=f01
f1+f02
f2g=f1f2:::fnfn+1g0=(f1f2:::fn)0fn+1+(f1f2:::fn)f0n+1;g0
g=(f1f2:::fn)0
f1f2:::fn+f0n+1
fn+1=f01
f1+f02
f2++f0n
fn+f0n+1
fn+1Exercise25.(tanx)0=cosx
sinx0=cos2x�(�sinx)sinx
cos2x=sec2x(cotx)0=cosx
sinx0=�sinxsinx�cosxcosx
sin2x=�csc2x(secx)0=�1
cos2x(�sinx)=tanxsecx(cscx)0=�1
sin2xcosx=�cotxcscxExercise35.f0=(2ax+b)(sinx+cosx)�(cosx�sinx)(ax2+bx+c)
(sinx+cosx)2==(2ax+b)(sinx+cosx)�(cosx�sinx)(ax2+bx+c)
(sinx+cosx)2Exercise36.f0=asinx+(ax+b)cosx+ccosx+(cx+d)(�sinx)=axcosx+(b+c)cosx+(a�d)sinx�cxsinxSothena=1;d=1;b=d;c=0.Exercise37.g0=(2ax+b)sinx+(ax2+bx+c)cosx+(2dx+e)cosx+(dx2+ex+f)(�sinx)==ax2cosx�dx2sinx+(2a�e)xsinx+(b+2d)xcosx+(b�f)sinx+(c+e)cosxg=x2sinx.Sod=�1;b=2;f=2;a=0;e=0;c=0.Exercise38.1+x+x2++xn=xn+1�1
x�1(1)(1+x+x2++xn)0=1+2x++nxn�1=(n+1)xn(x�1)�(1)(xn+1�1)
(x�1)2=(n+1)(xn+1�xn)�xn+1+1
(x�1)2=nxn+1�(n1)xn+1
(x�1)2x(1+2x++nxn�1)=x+2x2++nxn=nxn+2�(n+1)xn+1+x
(x�1)2(2)(x+2x2++nxn)0=(1+22x1++n2xn�1)==(n(n+2)xn+1�(n+1)2xn+1)(x�1)2�2(x�1)(nxn+2�(n+1)xn+1+x)
(x�1)4x+22x2++n2xn=(n(n+2)xn+2�(n+1)2xn+1+x)(x�1)�2(nxn+3�(n+1)xn+2+x2)
(x�1)3==n2xn+3+(�2n2�2n+1)xn+2+(n+1)2xn+1�x2�x
(x�1)345
Thenrecallthisneattrigonometricfact:cos3x=cos2xcosx�sin2xsinx=4cos3x�3cosx=)cos3x=3
4cosx�cos3x
4=0Particularlyforthisproblem,wehavecos3x=1.Sox=0;2=3;4=3.cosx=1;�1
2.Pluggingcosx!xbackintowhatwehavefora,a=�1,whichwealreadyhaveinthepreviouspart,anda=1
2.So
f1
2=3
8
y(x)=1
4x+1
4
Exercise9.f(x)=(x2ifxcax+bifx�cf0(x)=(2xifxcaifx�c
a=2c;b=�c2
Exercise10.f(x)=(1
jxjifjxj�ca+bx2ifjxjcNotethatc0sincejxjc,forthesecondcondition.f0(x)=8�&#x]TJ ;� -1;.93; Td;&#x [00;:�1
x2ifx�c1
x2ifxc2bxifjxjcSo
b=�1
2c3;a=3
2c
.Exercise11.f0=(cosxifxcaifa�cExercise12.f(x)=1�p
2
1+p
2=1�A
1+AA=p
xA0=a=1
2x�1=2=1
2A;A00=�1
4x�3=2=�1
4A3f0=�A0(1+A)�A0(1�A)
(1+A)2=�2A0
(1+A)2=�1
p
x(1+p
x)2f00=1
(A(1+A2))2(A0(1+A)2+A(2)(1+A)A0)=3p
x+1
2x3=2(1+p
x)3f000=1
2�1
A2A0(A2(1+A)3)�(2AA0(1+A)3+3A2(1+A)2A0)(3+1
A)
(A2(1+A)3)2=�3
41
A+4+5A
A4(1+A4)=�3
4(1+4p
x+5x)
p
x(x+p
x)4Exercise13.47
D(f�g)=limh!0(f(x+h)�g(x+h))2�(f(x)�g(x))2
h==limh!0(F�G)2�(f�g)2
h==Df+Dg+�limh!02FG�2fg
h=Df+Dg�g
fDf+f
gDgD(fg)=limh!0((fg)(x+h))2�((fg)(x))2
h==limh!0(f2(x+h))(g2(x+h))�f2(x)g2(x+h)+(g2(x+h)�g2(x))f2(x)
h==g2Df+f2DgD(f=g)=limh!0f2(x+h)
g2(x+h)�f2(x)
g2(x)
h=limh!0f2(x+h)�f2(x)
g2(x+h)+f2(x)
g2(x+h)�f2(x)
g2(x)
h==Df
g2+f2
g4(�Dg)wheng(x)6=0(2)(3)4.12Exercises-Thechainrulefordifferentiatingcompositefunctions,Applicationsofthechainrule.Relatedratesandimplicitdifferentiation.Exercise1.�2sin2x�2cosxExercise2.x
p
1+x2Exercise3.�2xcosx2+2x(x2�2)sinx2+2sinx3+6x3cosx3Exercise4.f0=cos(cos2x)(�2cosxsinx)cos(sin2x)+sin(cos2x)sin(sin2x)(2sinxcosx)==�sin2x(cos(cos2x))Exercise5.f0=nsinn�1xcosxcosnx+�nsinnxsinnxExercise6.f0=cos(sin(sinx))(cos(sinx))(cosx)Exercise7.f0=2sinxcosxsinx2�2xcosx2sin2x
sin2x2=sin2xsinx2�2xsin2xcosx2
sin2x2Exercise8.f0=1
2sec2x
2+1
2csc2x
2Exercise9.f0=2sec2xtanx+�2csc2xcotxExercise10.f0=p
1+x2+x2
p
1+x2=1+2x2
p
1+x2Exercise11.f0=4
(4�x2)3=2Exercise12.f0=1
31+x3
1�x3�2=33x2(2)
(1�x3)2=2x2
(1�x3)21+x3
1�x3�2=3Exercise13.Thisexerciseisimportant.Itshowsaneatintegrationtrick.49
Exercise20.V=s3;s=s(t)dV
dt=3s2ds
dt.s=5cm75cm3=secs=10cm300cm3=secs=xcm3x2cm3=secExercise21.l=p
x2+h2dl
dt=1
lxdx
dtdx
dt=l
xdl
dt=10mi
�p
102�82(�4mi=sec)=20
3mi
sec3600sec
1hrExercise22.l2=x2+s22ldl
dt=2xdx
dtdl
dt=x
ldx
dtdl
dtx=s
2=20p
5dl
dt(x=s)=50p
2Exercise23.dl
dt=x
ldx
dt=3
512=
36
5mi=hr
Exercise24.Giventhepreliminaryinformationr
h=2
5= ;V=1
3r2h=1
3 2h3(1)V=r2
h2(h2y�hy2+1
3y3)dV
dt=r2
h2(h2�2hy+y2)dy
dtdy
dt=h2
r21
h2�2hy+y2dV
dt=102
421
102�2(10)5+255=5
4(2)dV
dt= 2h2dh
dt;dh
dt=1
 2h2dV
dt=5
4Exercise25. =r
h=3
2dV
dt= 2h2dh
dtc�19
4(22)4=36=)
c=36+1
Exercise26.Theconstraintequation,usingPythagoreantheoremonthegeometryofabottomhemisphere,isr2=R2�(R�h)2=2Rh�h2Sothenrdr
dt=(R�h)dh
dtV=Zr2dh=)dV
dh=r2=(2Rh�h2)=)dV
dh=(2(10(5)�25))=5051
6x+8yy0=0=)y0=�3x
4y3+4(y02+yy00)=0y00=�3
4�y021
y=�9
4y3Exercise33.sinxy+xcos2xy(y+xy0)+4x=0y0x2cosxy+xycosxy+sinxy+4x=0Exercise34.y=x4.yn=xm.yn=xm;y0nyn�1=mxm�1;y0=mxm�1
nyn�1=m
nxm�1
xm(1�1=n)=y0=m
nxm=n�14.15Exercises-Applicationsofdifferentiationtoextremevaluesoffunctions,Themean-valuetheoremforderivatives.Let'srecapwhatwasshowninthepasttwosections:Theorem13(Theorem4.3).LetfbedenedonI.Assumefhasarel.extremaatanint.pt.c2I.If9f0(c);f0(c)=0;theconverseisnottrue.Proof.Q(x)=f(x)�f(c)
x�cifx6=c;Q(c)=f0(c)9f0(c),soQ(x)!Q(c)asx!csoQiscontinuousatc.IfQ(c)�0,f(x)�f(c)
x�c�0.Forx�c?0,f(x)?f(c),thuscontradictingtherel.maxorrel.min.(noneighborhoodaboutcexistsforone!)IfQ(c)0,f(x)�f(c)
x�c0.Forx�c?0,f(x)7f(c),thuscontradictingtherel.maxorrel.min.(noneighborhoodaboutcexistsforone!)Converseisnottrue:e.g.saddlepoints.Theorem14(Rolle'sTheorem).Letfbecont.on[a;b],9f0(x)8x2(a;b)andletf(a)=f(b)then9atleastonec2(a;b),suchthatf0(c)=0.Proof.Supposef0(x)6=08x2(a;b).Byextremevaluetheorem,9abs.max(min)M;msomewhereon[a;b].M;monendpointsa;b(Thm4.3).F(a)=f(b),som=M.fconstanton[a;b].Contradictf0(x)6=0Theorem15(Mean-valuetheoremforDerivatives).Assumefiscont.everywhereon[a;b],9f0(x)8x2(a;b).9atleastonec2(a;b)suchthat(6)f(b)�f(a)=f0(c)(b�a)Proof.h(x)=f(x)(b�a)�x(f(b)�f(a))h(a)=f(a)b�f(a)a�af(b)+af(a)h(b)=f(b)(b�a)�b(f(b)�f(a))=bf(a)�af(b)=h(a)=)9c2(a;b);suchthath0(c)=0=f0(c)(b�a)�(f(b)�f(a))Theorem16(Cauchy'sMean-ValueFormula).Letf;gcont.on[a;b],9f0;g08x2(a;b)Then9c2(a;b).x(7)f0(c)(g(b)�g(a))=g0(c)(f(b)�f(a))(notehowit'ssymmetrical)53
Forinstance,fora=0;b=2,thenf(b)�f(a)
b�a=�1=2,soc=1=2orc=p
2Exercise4.f(1)=1�12=3=0=f(�1)=1�((�1)2)3=0f0=�2
3x�1=36=0forjxj1Thisispossiblesincefisnotdifferentiableatx=0.Exercise5.x2=xsinx+cosx.g=xS+C�x2.g0=S+xC�S�2x=xC�2x=x(C�2).SincejCj1then(C�2)isnegativeforallx.Thenforx?0,g070.Sinceg(0)=1andforx!1,g!1,thenwecouldconcludethatgmustbecomezerobetween0and1and�1and0.Exercise6.f(b)�f(a)
b�a=f0(c)b=x+hb�a=ha=xxx+hx+h=)f(x+h)�f(x)=hf0(x+h)(1)f(x)=x2;f0=2x.(x+h)2�x2=2xh+h2=h(2(x+h))2x+h
2�x=h=)
=1
2
sothenlimh!0=1
2(2)f(x)=x3,f0=3x2.(x+h)3�x3=3x2h+3xh2+h3=h3(x+h)2=) r
3x2+3xh+h2
3�x!=h==r
3x2+3xh+h2
3h2�x
h=q
x2+xh+h2
3�x
[email protected]
x2+xh+h2
3+x
q
x2+xh+h2
3+x1A==x+h
3
x+q
x2+hx+h2
3=)limh!0
=1
2
Noticethetrickofmultiplyingbytheconjugateontopandbottomtogetawaytoevaluatethelimit.Exercise7.f(x)=(x�a1)(x�a2):::(x�ar)g(x).(1)a1a2ar.Sincef(a1)=f(a2)=0.f0(c)=0forc12(a1;a2).Considerthatf(a2)=f(a3)=0aswellasf0(c2)=0forc2(a2;a3).Indeed,sincef(aj)=f(aj+1)=0,f0(c)=0forc2(aj;aj+1).Thus,9r�1zero's.f(k)hasr�kzerosin[a;b].f(k)=(x�a1)(x�a2):::(x�ar�k)gk(x)Sincef(a1)=f(a2)=0,f(k+1)(c1)=0forc12(a1;a2).f(k)(aj)=f(k)(aj+1)=0;f(k+1)(cj)=0forcj2(aj;aj+1)=)f(k)(x)hasatleastr�kzerosin[a;b]Wehadshowntheabovebyinduction.(2)Wecanconcludethatthere'satmostr+kzerosforf(sincef(k)hasexactlyrzeros,theintervalscontainingtherzerosaredenite).55
(3)f00=(�2)(x�1)2(x�3)2�(x�2)(2(x�1)(x�3)2+2(x�3)(x�1)2)
(x�1)4(x�3)4==(6)x2�4x+13
3
(x�1)3(x�3)3x2�4x+13
3�0since144�4(�3)(�13)=144+12(�13)0sof00&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0ifx&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;3;x1f000if1x3(4)Seesketch.Exercise9.f(x)=x=(1+x2)(1)f0=(1+x2)�x(2x)
(1+x2)2=1�x2
(1+x2)2f0(x)=0whenx=1(2)f0?0whenjxj71(3)f00=�2x(1+x2)2�2(1+x2)(2x)(1�x2)
(1+x2)4=2x(x2�3)
(1+x2)3f00�0whenx�p
3f000when0xp
3f00�0when�p
3x0f000whenx�p
3(4)Seesketch.Exercise10.f(x)=(x2�4)=(x2�9)(1)f0=2x(x2�9)�(x2�4)(2x)
(x2�9)2=�10x
(x2�9)2f0(0)=0(2)f0?0whenx70,x6=3(3)f00=(�10)(x2�9)2�2(x2�9)(2x)x
(x2�9)2=(30)(x2+3)
(x2�9)3f00?0whenjxj?3(4)Seethesketch.Exercise11.f(x)=sin2x(1)f0=sin2xSothenf0=0whenx=
2n(2)f0�0when0x
2nx
2+nf00when
2+nx(n+1)(3)f00=2cos2xf00&#x-278;0when�
4+nx
4+nf000when
4+nx3
4+n(4)Seesketch.58
f=x+yf0=1+y0=0=1+�x
y=0=)
y=x
f00=y00=�1�y02
yfory�0,f000sothatfismax.wheny=xNotethat2x+2yy0=0y0=�x
yx+yy0=01+y02+yy00=0=)yy00=�1�y02Exercise6.l2=(L�x)2+x2=L2�2Lx+2x2=AdA
dx=�2L+4x=0=)x=L
2d2A
dx2=4�0=)Aminimizedl(x=L
2)=Lp
2
2Exercise7.(x+p
L2�x2)2=AA0=2(x+p
L2�x2)(1+�x
p
L2�x2)=0whenL2�x2=x2orx=L
p
2sothenthesideofthecircumscribingandarea-maximizedsquareisL
p
2+r
L2�L2
2=2L
p
2Exercise8.A=(2x)(2p
R2�x2)=4xp
R2�x2A0=4(p
R2�x2+�x2
p
R2�x2)=4R2�2x2
p
R2�x2=)x=R
p
2sinceA0?0whenx7R
p
2;soAismaximizedatx=R
p
22x=2R
p
2;2p
R2�x2=2R
p
2sothentherectanglethathasmaximumsizeisasquare.Exercise9.Provethatamongallrectanglesofagivenarea,thesquarehasthesmallestcircumscribedcircle.A0=(2x)(2p
r2�x2)=4xp
r2�x2(xtheareatobeA0)�A0
4x2=r2�x2=)x4�x2r2+A20
16=0=)0=2xr2+x22rdr
dx�4x3dr
dx=0(forextrema)=)x=r
p
2andp
r2�x2=r
p
2Wecouldarguethatwehadfoundaminimumbecauseatthe“innity”boundaries,thecircumscribingcirclewouldbeinnitelylarge.Exercise10.GivenasphereofradiusR,ndtheradiusrandaltitudehoftherightcircularcylinderwiththelargestlateral60
Consideringthegeometricorphysicalconstraints,sincelimV!1V=limh!1V=0,sothenr=2p
2R
3mustmaximizeV.Exercise15.Findtherectangleoflargestareathatcanbeinscribedinasemicircle,thelowerbasebeingonthediameter.A=p
R2�x2xA0=p
R2�x2+�x2
p
R2�x2=0=)
x=R
p
2;h=R
p
2
Exercise16.Findthetrapezoidoflargestareathatcanbeinscribedinasemicircle,thelowerbasebeingonthediameter.A=1
2h(2p
R2�h2+2R)dA
dh=p
R2�h2+R+h�h
p
R2�h2dA
dh=0=)
h=p
3R
2
=)A=5p
3R2
8p
R2�h2=2r
R2�3
4R2=2R
2=
R
Exercise17.Anopenboxismadefromarectangularpieceofmaterialbyremovingequalsquaresateachcornerandturningupthesides.Findthedimensionsoftheboxoflargestvolumethatcanbemadeinthismannerifthematerialhassides(a)10and10;(b)12and18(1)(x�2r)(Y�2r)r=(xy�2rx�2ry+4r2)r=xyr�2r2x�2r2y+4r3=VdV
dr=xy�4rx�4ry+12r2=0=)r=4(x+y)p
16(x+y)2�4(12)xy
2(12)=(x+y)p
x2+y2�xy
6d2V
dr2=�4x�4y+24r=�4(x+y)+24rWecanpluginourexpressionforrintothesecondderivativeofV,thevolumeofthebox,tondoutthatwewanttopickthe“negative”rootfromr,inordertomaximizetheboxvolume.Thenforx=10;y=10,wehaver=5
3,sothattheboxdimensionsare5
320
320
3.(2)12and18=)5�p
72+2p
78+2p
7Exercise18.Ifaandbarethelegsofarighttrianglewhosehypotenuseis1,ndthelargestvalueof2a+b.L=2a+b=2a+p
1�a2L0=2+�a
p
1�a2L0=0=)a
22=1�a2=)a=2
p
5L00=(�1) p
1�a2��a
p
1�a2a
1�a2!=(�1)1
(1�a2)3=20(soa=2
p
5maximizesL)Exercise19.2+x2
600gallonsperhour.l0=300mix=constantspeed.l0
x=timespent.K=gascost=0:30.62
So,isolatingl,thelengthofthecrease,andthentakingthederivative,l=w0
2sin( )cos2 =w0
2csc( )sec2( )dl
d =w0
2��cot csc sec2 +csc 2sec sec tan ==w0
2�C
S1
S1
C2+1
S21
C2S
C=w0
2�1
S2C+2
C3dl
d =0����!
sin =1
p
3
ortan =1
p
2where istheangleofthecrease.Thecorrespondingminimumlengthofthecreasewillbel=w0
21
1
p
32
3=
9p
3
2
Exercise22.(1)ConsiderthecenterofthecircleO,theapexoftheisoscelestrianglethatmakesanangle2 ,A,andoneofitsothervertices,B.DrawalinesegmentfromOtoBandsimplyconsiderthetwotrianglesmakinguponehalfoftheisoscelestriangle.Findalltheangles.AngleAOBis�2 bythegeometryori.e.inspectionofthegure.Thecomplementofthatangleis2 .Beforehand,wecangetthelengthoftheisoscelestrianglelegfromthelawofcosines.cos(�2 )=�cos(2 )s2=R2+R2�2R2cos(�2 )+2R2(1+cos(2 ))=2R2(2cos2 )=4R2cos2 s=2Rcos Theconstraintequationis(8)P=4Rcos +2Rsin(2 )SothenP0=4R(�sin )+4Rcos(2 )=0=)cos2 =sin sin =�1
2q
1
4�4(1)(�1
2)
2(1)=1
2�0=)
P=3p
3R
P=3p
3isamaxbecauseLookatthe“boundaryconditions”imposedonPbythephysical-geometry. =0,triangleiscompletelyattened, =,triangle“completelydisappears.”(2)IhadoriginallythoughttoReusetheconstraintequation,Eqn.(??).Thisiswrong!Thinkabouttheproblemdirectlyandforwhatitactuallyis;lesswishfulthinking.ConsideraxedperimeterLandimagineLtobeastringthatcanbestretchedintoanisocelestriangle.A“trivial”isocelestriangleisacollapsedtrianglewithtwosidesoflengthL=2only.ThentheradiusofthediskneedstobeL=4.Considerageneralisoscelestrianglewith2 asthevertexangleandisoscelessidesofh.Theperimeterforthistriangle,P,isthenP=2h+2hsin =2h(1+sin )=)h=P
2(1+sin )h
2=Rcos 64
(2)M(t)=(t�a
aift�a
a�b�t
bi.e.t�b+a
ab�2b�t
bifb�t
b�t�a
adM
dt=(1
aift�2ab
a+b�1
biftab
a+b2SincedM
dt?0whent?2ab
a+b,Misminimizedfor
t=2ab
a+b
4.23Exercises-PartialDerivatives.Exercise8.f(x;y)=x
p
x2+y2.fx=1
p
x2+y2+�x2
(x2+y2)3=2=y2
(x2+y2)3=2fy=�xy
(x2+y2)3=2fxx=�3y2x
(x2+y2)5=2fyy=(�x)x2�2y2
(x2+y2)5=2fxy=(�y) (x2+y2)3=2�x3
2(x2+y2)1=2(2x)
(x2+y2)3!==(�y)�2x2+y2
(x2+y2)5=2fyx=2y
(x2+y2)3=2+�3y2y
(x2+y2)5=2=(�y)(�2x2+y2)
(x2+y2)5=2Exercise9.(1)z=(x�2y)2zx=2(x�2y)=2p
zzy=2(x�2y)(�2)=�4p
zx(2z)�4zy=(x�2y)2p
z=2z(2)z=(x4+y4)1=2zx=1
z2x3zy=2y3
zx(2z)�4zy=(x�2y)2p
z=2zExercise10.f=xy
(x2+y2)2;fx=y
(x2+y2)2+�4x2y
(x2+y2)3=y3�3x2y
(x2+y2)3Sofxx=�6xy(x2+y2)3�3(x2+y2)2(2x)(y3�3x2y)
(x2+y2)6==12xy(x2�y2)
(x2+y2)4Bylabelsymmetry,fxx+fyy=12xy(x2�y2)
(x2+y2)4+12yx(y2�x2)
(x2+y2)4=067
Zx0f(t)dt=A(x)�A(0)A0(x)=f(x)=)2x+sin2x+2xcos2x+�sin2xf0=2+4cos2x+�4xsin2x�2cos2x=2+2cos2x�4xsin2x
f
4=
2f0
4=2�
Exercise15.Rxcf(t)dt=cosx�1
2f(x)=�sinxc=�
6.Exercise16.Supposef(x)=sinx�1andc=0.Zx0tsint�t=�tcost+sint�1
2t2 x0=sinx�xcosx�1
2x2Soc=0.Exercise17.Forf(x)=�x2f(x)+2x15+2x17(foundbytakingthederivativeofRx0f=R1xt2f+x16
8+x18
9+C,)Supposethatf=2x15.=)x16
8=�x18
9+�1
9+x16
8+x18
9+C=)C=1
9Exercise18.Byplugginginx=0intothedenedf(x),f(x)=3+Rx01+sint
2+t2dt,wegetforp(x)=a+bx+cx2,a=3Continuingon,f0=1+sinx
2+x2;f0(0)=1
2=bf00=(cosx)(2+x2)�2x(1+sinx)
(2+x2)2f00(0)=1
2+2c;c=1
4Exercise19.f(x)=1
2Zx0(x�t)2g(t)dt=1
2Zx0(x2�2xt+t2)g(t)dt==1
2x2Zx0g�2xZx0tg+Zx0t2gf0=xZx0g+x2
2g(x)�Zx0tg�x(xg(x))+1
2x2g(x)==xZx0g�Zx0tgf00=Zx0g+xg�xg=Zx0g
f00(1)=2
f000=g
f000(1)=5
Exercise20.(1)(Rx0(1+t2)�3dt)0=(1+x2)�3(2)(Rx20(1+t2)�3dt)0=(1+x4)�3(2x)=2x
(1+x4)3(3)(Rx2x3(1+t2)�3dt)0=(1+x4)�3(2x)�(1+x6)�3(3x2)=2x
(1+x4)3�3x2
(1+x6)369
(1)df
dt=1+2sintcost
1+t2=v(t)a(t)=2(cos(2t))(1+t2)�2t(sin(2t))
(1+t2)2a(t=2)=a(t=1)=4
4=(2)v(t=1)1
2(3)v(t)=(t�1)+1
2;t�1(4)f(t)�f(1)=Zt1v(t)dt=Zt1(t�1)+1
2=t2
2�t+1
2t t1=t2
2+�t+t
2+
2�1
2Exercise26.(1)f00(x)�08xf0(0)=1;f0(1)=0Z10f00(t)dt=f0(1)�f0(0)=0�10Thus,it'simpossible,sincef00(x)&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0,soR10f00(t)dt&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0(2)Z103�
2sinx
2dx=3x+cosx
2 10=3�1=2f(x)=3x2
2+2
sinx
2+C(3)f00(0)�08xf0(0)=1;f(x)1008x�0Zbaf00(t)dt=f0(b)�f0(a);Zkcf0(t)dt=f(k)�f(c)Zb0f00=f0(b)�f0(0)=f0(b)�1?0ifb?0Zkc(f0(b)�1)db=f(k)�f(c)�(k�c)�0ifk�c�0f(k)�f(c)�k�cf(k)�f(0)�k�0f(100)�f(0)�100f(x)100isuntrueforallx�0(4)f00(x)=ex�0f0(x)=ex;f0(0)=1f(x)=ex8x0;ex1Exercise27.f00(t)6.b�a=1
2.f0(0)=0Zbaf00=f0(b)�f0(a)6(b�a)=3sinceb�a=1
2Za0f00=f0(a)�f0(0)=f0(a)6(a�0)=6aIfa=1
2;f0(1=2)3Thenbyintermediatevaluetheorem,withfbeingcontinuousandf0(0)=0;f0(1=2)3,f0musttakeonthevalueof3somewherebetween0and3.Thusthereisaninterval[a;b]oflength1=2wheref03.71
(8x3+27)5=33
51
24=1
40(8x3+27)5=3Exercise18.3
2(sinx�cosx)2=3Exercise19.Zxdx
p
1+x2+(1+x2)3=2=Z1
2du
p
u+u3=2==1
2Zdu
p
up
1+u1=2=2(1+u1=2)1=2==2(1+p
1+x2)1=2+CExercise20.Z(x2�2x+1)1=5dx
1�x=Z�(x�1)2=5
x�1dx=�Z(x�1)�3=5dx=�5=2(x�1)2=5Exercise21.Thm.1.18.invarianceundertranslation.Rbaf(x)dx=Rb+ca+cf(x�c)dx.Thm.1.19.expansionorcontractionoftheintervalofintegration.Zbaf(x)dx=1
kZkbkafx
kdxyx+cdy=dxZbaf(x)dx=Zb+ca+cf(y�c)dyy=kxdy=kdxZbaf(x)dx=1
kZkbkafy
kdyExercise22.Fx
a;1=Zx=a0up
(u2+12)qduu=t
adu=dt
aFx
a;1=1
aZx0(t=a)pdt
�t
a2+12q==a�p�1+2qZx0tp
(t2+12)qdt=a�p�1+2qF(x;a)Exercise23.Z1xdt
1+t2=F(1)�F(x)Zx1dt
1+t2=F(x)�F(1)Z1=x1dt
1+t2=F1
x�F(1)u=1
tdu=�1
t2dt;�1
u2du=dtZx1dt
1+t2=Z1=x1�du
u2�1+1
u2==�Z1=x1du
u2+1=Z11=xdt
t2+1Exercise24.Z10xm(1�x)ndx=�Z01(1�u)m(un)du=Z10(1�x)mxndxusingu=1�xx=1�uExercise25.73
(1)Zsin2x=�1
2sinxcosx+1
2Z1=�1
2sinxcosx+1
2xZ=20sin2xdx=
4(2)R=20sin4x=�1
4sin3xcosx =20+3
4R=20sin2x=3
16(3)R=20sin6x=5
6R=20sin4x=
5
32
Exercise10.(1)Zsin3xdx=�1
3sin2xcosx+2
3Zsinx=�1
6sin2xcosx�2
3cosx=�3
4cosx+1
12cos3xsince�3
4cosx+1
12cos3x=�3
4cosx+1
12(cosxcos2x�sin2xsinx)==�3
4cosx+1
12(cosx(1�2sin2x)+�2sin2xcosx)=
�2
3cosx�1
3sin2xcosx
(2)Zsin4xdx=�1
4sin3xcosx+3
4Zsin2x=�1
4sin3xcosx+3
4(x
2�sin2x
4)=�1
4sin3xcosx+3x
8�3sin2x
16Now1
32sin4x=1
32(2sin2xcos2x)=1
8(sinxcosx(1�2sin2x)=sin2x
16�1
4sin3xcosx=)�1
4sin3xcosx+3x
8�3sin2x
16=3x
8�1
4sin2x+1
32sin4x(3)Zsin5xdx=Zsin4xsinxdx=�cosxsin4x+Zcos2x4sin3x==�cosxsin4x+4(Zsin3x�Zsin5x)=�cosxsin4x+4Zsin3x�4Zsin5x5Zsin5dx=�cosxsin4x+4Zsin3x5Zsin5dx=�cosx(1�cos2x)2+4(�3
4cosx+1
12cos3x)=�cosx(1�2cos2x+cos4x)+�3cosx+1
3cos3x=�cosx+2cos3x�cos5x�3cosx+1
3(cosxcos2x�sinxsin2x)==�4cosx+2cos3x�cos5x+1
3(4cos3x�3cosx)=�5cosx+10cos3x
3�cos5xZsin5xdx=�cosx+2cos3x
3�1
5cos5xMysolutiontothelastpartofthisexerciseconictswithwhat'sstatedinthebook.Exercise11.(1)Zxsin2xdx=(Zsin2x)x�Z(sin2t)=x2
2�xsin2x
4�x2
4+cos2x
8==x2
8�xsin2x
4�cos2x
8wehadusedZsin2x=x
2�sin2x
475
(1)Z(a2�x2)ndx=x(a2�x2)n�Zn(a2�x2)n�1(�2x)xdx=x(a2�x2)n+2nZx2(a2�x2)n�1dxZx2(a2�x2)n�1dx=Z((x2�a2)+a2)(a2�x2)n�1dx=Z�(a2�x2)n+a2(a2�x2)n�1dx=)Z(a2�x2)ndx=x(a2�x2)n
2n+1+2a2n
2n+1Z(a2�x2)n�1dx(2)Z(a2�x2)dx=x(a2�x2)
3+2a2
3x=�x3
3+a2xZ(a2�x2)5=2dx=x(a2�x2)5=2
6+a25
6Z(a2�x2)3=2dxZ(a2�x2)3=2dx=x(a2�x2)3=2
4+3a2
4Z(a2�x2)1=2dxZ(a2�x2)1=2=aZr
1�x
a2dx=a2Zcos2d==a2Z1+cos2
2=a2
2+sin2
4=a2 arcsinx
a+1
2x
ar
1�x
a2!sin=x
acosd=dx
aZa0p
a2�x2=a2
2�0=a2
2Za0(a2�x2)3=2dx=3a2
4a2
2=3a4
8
Za0(a2�x2)5=2dx=5a2
63a4
8=
5
16a6
Exercise16.In(x)=Rx0tn(t2+a2)�1=2dt(1)In(x)=(t2+a2)1=2tn�1�Z(n�1)tn�2(t2+a2)1=2=tn�1(t2+a2)1=2�(n�1)Ztn�2(t2+a2)
(t2+a2)1=2(n)In=xn�1(x2+a2)1=2�a2(n�1)Ztn�2
(t2+a2)1=2=xn�1p
x2+a2�(n�1)a2In�2(2)n=5;x=2;a=p
5.I1(2)=Z20x(x2+5)�1=2dx=(x2+5)1=2 20=3�p
55I5(2)=Z20t5(t2+5)�1=2dt=25�1(4+5)1=2�5(5�1)I3(2)=48�20I3(2)3I3(2)=22p
4+5�5(3�1)I1(2)=12�10(3�p
5)I5(2)=1
5(48�20(�6+10p
5
3))=
168
5�40p
5
3
Exercise17.Zt3(c+t3)�1=2dt=t2(c+t3)1=2
3�Z2(c+t3)1=2
3Z3�1t3(4+t3)�1=2dt=t2(4+t3)1=2
3 3�1�2
3Z3�1(4+t3)1=2=
2p
31+2p
3
3�2
3(11:35)
Exercise18.77
Ifj=n,gn(0)=n!Exercise2.P(x)=5Xj=0ajxjP0(x)=5Xj=1jajxj�1P00(x)=5Xj=2j(j�1)ajxj�2P(0)=1=a0P0(0)=0=a1P00(0)=0=2(1)a2a1=a2=0=)P(x)=a5x5+a4x4+a3x3+1P0(x)=5a5x4+4a4x3+3a3x2P00(x)=20a5x3+12a4x2+6a3xP(1)=a5+a4+a3+1=2P0(1)=5a5+4a4+3a3=0P00(1)=20a5+12a4+6a3=0SolvefortheundeterminedcoefcientsbyGauss-Jordaneliminationprocess24543201261113524a5a4a335=24001352454320126111 00135=241001100 �1510635=)a5=6a4=�15a3=10
P(x)=6x5�15x4+10x3+1
Exercise3.Iff(x)=cosxandg(x)=sinx,Provethatf(n)=cos(x+n
2)andg(n)(x)=sin(x+n
2)f(n)(x)=cos(x+n
2)=(sinx(�1)j+1ifn=2j+1cosx(�1)jifn=2jg(n)(x)=sin(x+n
2)=(cosx(�1)jifn=2j+1sinx(�1)jifn=2jf(x)=cosxf0(x)=�sinxf00(x)=�cosxf000(x)=sinxf0000(x)=cosxf(2j)(x)=cosx(�1)jf(2(j+1))(x)=(cosx(�1)j)0=cosx(�1)j+1f(2j+1)(x)=sinx(�1)j+1f(2j+3)(x)=(sinx(�1)j+1)=sinx(�1)j+2g(x)=sinxg0(x)=cosxg00(x)=�sinxg000(x)=�cosxg0000(x)=sinxg(2j)(x)=sinx(�1)jg(2(j+1))(x)=(sinx(�1)j)00=sinx(�1)j+1g(2j+1)(x)=cosx(�1)jg(2j+3)(x)=(cosx(�1)j+1)Exercise4.79
Exercise18.Rsin(x�1)1=4dxu=(x�1)1=4du=1
4(x�1)�3=4dx=1
41
u3dx=)Zsin(x�1)1=4dx=Z(sinu)4u3du=4Zu3sinuduZu3sinudu=�u3cosu+3u2sinu+6ucosu+06sinu=�(x�1)3=4cos(x�1)1=4+3(x�1)1=2sin(x�1)1=4+6(x�1)1=4cos(x�1)1=4�6sin(x�1)1=4sin(x�1)1=4dx==
�4(x�1)3=4cos(x�1)1=4+24(x�1)1=4cos(x�1)1=4+12(x�1)1=4sin(x�1)1=4�24sin(x�1)1=4
Exercise19.Rxsinx2cosx2dx=(1=4)sin2x2+CExercise20.Rp
1+3cos2xsin2xdx=Rp
1+3cos2x2sinxcosxdx=Ru1=2du
�3=2u3=2
�9=
�2
9(1+3cos2x)3=2
,wherewehadusedthissubstitution:u=1+3cos2xdu=�6cosxsinxdxdu
�3=2cosxsinxdxExercise21.R20375x5(x2+1)�4dxu=x2+1du=2xdx(u�1)2=x4=)Z20375
2du(u�1)2u�4=375
2Z51du(u2�2u+1)
u4=375
2Z51du1
u2�2
u3+1
u4==375
2�1
u+1
u2+1
�3u3 51=
64=26
Exercise22.R10(ax+b)(x2+3x+2)�2dx=3
2Since�1
x2+3x+2 10=�1
6+1
2=2
3,thenifa=9=2,b=27
2,we'llobtain3=2Exercise23.In=R10(1�x2)ndxIn=Z10(1�x2)ndx=
x(1�x2)n 10�Z10xn(1�x2)n�1(�2x)dx
==2Z10x2n(1�x2)n�1dx=2nZ10((x2�1)+1)(1�x2)n�1dx==2nIn�1�2nIn=)In=2n
2n+1In�1I2=Z10(1�x2)2dx=Z10dx(1�2x2+x4)=x�2x3
3+1
5x5 10=
8
15
I3=6
78
15
16
35
I4=8
9I3=
128
315
I5=10
11I4=
256
693
Exercise24.F(m;n)=Rx0tm(1+t)ndt;m�0;n�082
Exercise29.Zxnp
(ax+b)dx=2xn(ax+b)3=2
3a�Znxn�12(ax+b)3=2
3a=2xn(ax+b)3=2
3a�2n
3aZxn�1(ax+b)p
ax+b==2xn(ax+b)p
ax+b
3a�2n
3Zxnp
ax+b�2nb
3aZxn�1p
ax+b
Zxnp
ax+bdx=2
(2n+3)axn(ax+b)3=2�nbZxn�1p
ax+b+C
n6=�3
2Exercise30.Zxm
p
a+bxdx=2xm(a+bx)1=2
b�Zmxm�12(a+bx)1=2
b=2
bxm(a+bx)1=2�2m
bZxm�1(a+bx)
p
a+bx=2
bxm(a+bx)1=2�2mZxm
p
a+bx�2ma
bZxm�1
p
a+bx
Zxm
p
a+bxdx=1
2m+12
bxm(a+bx)1=2�2ma
b(2m+1)Zxm�1
p
a+bx
Exercise31.Zdx
xnp
ax+b=2p
ax+b
axn+Zn2p
ax+b
axn+1r
ax+b
ax+b=2p
ax+b
axn+2n
aZax+b
xn+1p
ax+b==2p
ax+b
axn+2nZ1
xnp
ax+b+2nb
aZ1
xn+1p
ax+b=)(1�2n)Zdx
xnp
ax+b�2p
ax+b
axn=Zb�2n
a
xn+1p
ax+b
Z1
xnp
ax+b=�p
ax+b
(n�1)bxn�1�(2n�3)a
(2n�2)bZ1
xn�1p
ax+b
Exercise32.IderivedtheformulasforthisandExercise33bydoingthefollowingtrick.
(Cm+1S1�n)0=(m+1)Cm(�S2�n)+Cm+2(1�n)S�n=�(m+1)CmS2�n+(1�n)CmS�n(1�S2)==�(m+1+1�n)CmS2�n+(1�n)CmS�n=�(m�n+2)CmS2�n+(1�n)CmS�n
=)ZCmS�n=�(Cm+1S1�n)
n�1�(m�n+2)
n�1ZCmS2�nExercise33.(Cm�1S1�n)0=(m�1)Cm�2(�S2�n)+(1�n)CmS�n=�(m�1)Cm�2(S�n)(1�C2)+(1�n)CmS�n==�(m�1)Cm�2S�n+(m�1)CmS�n+(1�n)CmS�nCm�1S1�n=�(m�1)ZCm�2S�n+(m�n)ZCmS�nm�1
m�nZCm�2S�n+Cm�1S1�n
m�n=ZCmS�nExercise34.(1)P0(x)�3P(x)=4�5x+3x2P=nXj=0ajxjP0=nXj=1ajjxj�1=n�1Xj=0aj+1(j+1)xj=)n�1Xj=0(aj+1(j+1)�3aj)xj=4�5x+3x284
(3)Therst,second,anduptofthcasehasalreadybeenproven.Assumethenthcase,thatPn(0)=Pn(1).ZP0n+1=Pn+1(1)�Pn+1(0)(bythesecondfundamentaltheoremofcalculus)P0n+1=(n+1)PnZ10(n+1)Pn(t)=0(bythegivenpropertiesofBernoullipolynomials)=)Pn+1(1)=Pn+1(0)(4)Pn(x+1)�=Pn(x)=nxn�1istrueforn=1;2,byquickinspection(anddoingsomealgebramentally).P0n+1=(n+1)Pn=)ZP0n+1=(n+1)ZPnPn+1(x+1)�Pn+1(x)=Pn+1(a1)+(n+1)Zx+1a1Pn(t)�(Pn+1(a2)+(n+1)Zxa2Pn(t))a1=1;a2=0;soPn+1(1)�Pn+1(0)=0(frompreviousproblems)=)Pn+1(x+1)�Pn+1(x)=(n+1)(Zx+11Pn(t)�Zx0Pn(t))==(n+1)Zx0Pn(t+1)�Pn(t)=(n+1)Zx0ntn�1=(n+1)xn(5)Zk0Pn=Zk0P0n+1
n+1=Pn+1(k)�Pn+1(0)
n+1Pn(x+1)�Pn(x)=nxn�1Pn(x+1)�Pn(x)
n=xn�1=)k�1Xx=1Pn+1(x+1)�Pn+1(x)
n+1=k�1Xx=1xn=k�1Xr=1Pn+1(r+1)�Pn+1(r)
n+1==k�1Xr=1rn=Pn+1(k)�Pn+1(0)
n+1(telescopingseriesandPn+1(1)=Pn+1(0))(6)Thispartwasfairlytricky.Ahorriblecluewasthatthispartwillrelydirectlyonthelastpart(becauseofthewaythisquestionisasked),whichgaveusPx�1j=1jn=Rx0Pn(t)dt=Pn+1(x)�Pn+1(0)
n+1.Useinduction.Itcanbeeasilyveried,pluggingin,thatPn(1�x)=(�1)nPn(x)istrueforn=0:::5.Assumethenthcaseistrue.u=1�tdu=�dtZx0Pn(t)dt=Z1�x1�Pn(1�u)du=�Z1�x1Pn(u)(�1)ndu(sincePn(1�x)=(�1)nPn(x),assumednthcaseistrue)=(�1)n+1Z1�x1Pn(t)dt==(�1)n+1Z1�x0Pn(t)dt(sinceZ10Pn=0)=)=(�1)n+1Pn+1(1�x)�Pn+1(0)
n+1=Pn+1(x)�Pn+1(0)
n+1=)Pn+1(1�x)=(�1)n+1Pn+1(x)Inthesecondtolastandlaststep,wehadused(�1)n+1Pn+1(0)=Pn+1(0).Forn+1even,thisisdenitelytrue.Ifn+1wasodd,Doingsomealgebrafortherstvecases,wecanshowthatP2j�1(0)=0forj=2;3.Assumethejthcaseis86
(2)log(1+x)=1+log(1�x)ln(1+x)=1+ln(1�x)=ln(e)+ln(1�x)lne(1�x)=)1+x=e�ex=)
x=e�1
1+e
(3)2logx=xlog2lnx2+ln2�x=0=ln1=lnx22�x=)
x=2
(4)log(p
x+p
x+1)=1p
x+1=e�p
x=)x+1=e2�2p
xe+x2p
xe=e2�1=)
x=e2�1
2e2
Exercise3.f=lnx
xf0=1�lnx
x2f00=�2
x3�1
xx2�2xlnx
x4=2lnx�3
x3f00whenx&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;ef0&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0when0xeforx3&#x-277;0,2lnx�3&#x-277;0=)f00(x)0(concave),when0xe3=2;f00(x)&#x-277;0(convex),whenx&#x-277;e3=2Exercise4.f(x)=log(1+x2)f0=2x
1+x2Exercise5.f(x)=logp
1+x2f0=x
1+x2Exercise6.f(x)=logp
4�x2f0=1
2�2x
4�x2=�x
4�x2Exercise7.f(x)=log(logx)f0=1
lnx1
x=1
xlnxExercise8.f(x)=logx2logxf0=(2logx+loglogx)0=2
x+1
xlogxExercise9.f(x)=1
4logx2�1
x2+1f0=1
4�logx2�1�logx2+10=1
42x
x2�1�2x
x2+1=x1
x4�1Exercise10.f(x)=(x+p
1+x2)nlnf=nln(x+p
1+x2)f0
f=n1
x+p
1+x21+x
p
1+x2=n
p
1+x2f0=(x+p
1+x2)nn
p
1+x2Exercise11.f(x)=p
x+1�log(1+p
x+1)88
Zxnlogax=Zxnloga+Zxnlogx=xn+1
n+1loga+ZxnlogxZxnlnx=xn+1
n+1lnx�Zxn+1
n+11
x=xn+1
n+1logx�xn+1
(n+1)2=)Zxnlogax=xn+1
n+1loga+xn+1
n+1logx�xn+1
(n+1)2Exercise23.Rx2log2xdxZx2log2x=1
3x3ln2x�Zx2
32lnx=x3ln2x
3�2
3x3lnx
3�x3
9=
x3ln2x
3�2x3lnx
9+2x3
27
Exercise24.Rdx
xlogxZdx
xlnx=ln(lnx)+CExercise25.R1�e�21log(1�t)
1�tdtZ1�e20ln(1�t)
1�tdt=�1
2(ln(1�t))2 1�e�20=
�2
Exercise26.Rlogjxj
xp
1+logjxjdxZlogx
xp
1+logx=Z(2(1+logx)1=2)0logx=2(1+logx)1=2logx�Z2(1+logx)1=2
x==
2logx(1+logx)1=2�4
3(1+logx)3=2
Exercise27.DeriveZxmlognxdx=xm+1
m+1lnnx�n
m+1Zxmlnn�1xByinspection,wejustneededintegrationbyparts.Zx3ln3x=x4
4ln3x�3
4Zx3ln2x=x4
4ln3x�3
4x4
4ln2x�2
4Zx3lnx=x4ln3x
4�3x4ln2x
16+3x4lnx
32�3x4
128Exercise28.Givenx�0,f(x)=x�1�lnx;g(x)=lnx�1+1
x(1)f0=1�1
xg0=1
x�1
x2=1
xf0xg0=f0sotheniff0�0;g0�0;f00;g00Forf000x1f0&#x]TJ/;ø 9;&#x.962; Tf;&#x 29.;Ҕ ;� Td;&#x [00;0x&#x]TJ/;ø 9;&#x.962; Tf;&#x 29.;Ҕ ;� Td;&#x [00;1f0(1)=g0(1)=0f(1)=g(1)=0x�1�lnx&#x]TJ/;ø 9;&#x.962; Tf;&#x 29.;Ҕ ;� Td;&#x [00;0sincef(1)=0isarel.min.0lnx�1+1
xsinceg(0)isarel.min.(2)Seesketch.Exercise29.limx!0log(1+x)
x=1(1)L(x)=Rx11
tdt;L0(x)=1
x;L0(1)=1(2)Usethistheorem.90
Zxyxf(t)dt=B(xy)�B(x)d
dxZxyxf(t)dt=d
dx(B(xy)�B(x))=dB(xy)
d(xy)y�dB(x)
dx=f(xy)y�f(x)=0=)f(xy)=f(x)
yA(x)=Zx1f(t)dt=Zx1f((2)t
2)dt=Zx1f(2)
(t=2)dt=
4lnx
Exercise35.GivenRxy1f(t)dt=yRx1f(t)dt+xRy1f(t)dt,andlettingFbetheantiderivativeoff,F(xy)�F(1)=y(F(x)�F(1))+x(F(y)�F(1))d=dx���!f(xy)(y)=y(f(x))+Zy0f(t)dtd=dx���!f0(xy)y2=y(f0(x))x=1��!f0(y)=(f0(1))
yR�!f(y)=klny+Cy=1��!f(1)=0+C=3Rxy1(klnt+3)=yRx1(klnt+3)+xRy1(klnt+3)k((xy)lnxy�(xy)+1)+3(xy�1)=y(k((x)lnx�x+1)+3(x�1))+x(k(ylny�y+1)+3(y�1))=)k�3=ky�3y�kxy+kx+3xy�3xy(3�k)+xy(k�3)+k�3+(3�k)x=0=)k=3=)f(x)=3lnx+3Exercise36.6.11Exercises-Polynomialapproximationstothelogarithm.Exercise1.Theorem20(Theorem6.5).If0x1andifm1,ln1+x
1�x=2(x+x3
3++x2m�1
2m�1)+Rm(x)wherex2m+1
2m+1Rm(x)2�x
1�xx2m+1
2m+1Rm(x)=E2m(x)�E2m(�x)whereE2m(x)istheerrortermforlog1�xm=5,x=1
3ln4=3
2=3=ln2'2 1
3+�1
33
3+�1
35
5+�1
37
7+�1
39
9!'0:693146047Theerrorform=5,x=1
3is�1
311
11R5(x)5
2�1
311
11Exercise2.ln1+x
1�x=ln3
2=ln3�ln292
Exercise8.cosxesinxExercise9.�2cosxsinxecos2xExercise10.1
xelogxExercise11.exeexExercise12.eeex(exeex)Exercise13.Rxexdx=xex�exExercise14.Rxe�xdx=�xe�x+�e�xExercise15.Rx2ex=x2ex�2xex+2exExercise16.Rx2e�2xdx=x2e�2x
�2+xe�2x
�2+�e�2x
4Exercise17.Rep
x=ep
x(2p
x)�R1
p
xep
x=ep
x(2p
x)�2ep
x=
2(p
xep
x�ep
x)
Exercise18.Rx3e�x2.x2e�x2=�2x3e�x2+2xe�x2e�x2=�2xe�x21
�2(x2e�x2+e�x2)0=1
�22xe�x2+�2x3e�x2+�2xe�x2=
x2e�x2+e�x2
�2
Exercise19.ex=b+Rbaetdt=b+ex�ea,ea=b.Exercise20.A=ZeaxcosbxdxB=ZeaxsinbxdxA=eax
acosbx�Z�bsinbxeax
a=eax
acosbx+b
aB=)aA+bB=eaxcosbx+CB=eax
asinbx�Zeax
abcosbx=)aB+bA=eaxsinbx+CA=1
a2+b2(aeaxcosbx+beaxsinbx)B=�beaxcosbx+aeaxsinbx
a2+b2Exercise21.lnf=xlnx;f0
f=lnx+1;f0=xx(lnx+1)Exercise22.lnf0
f=1
1+x+1
1+ex2(2xex2);f0=1+ex2+2(1+x)xex2Exercise23.f=ex�e�x
ex+e�x.94
lnf1=cosxlnsinxf01
f1=�sinxlnsinx+cos2x
sinxf01=�sinxcosx(lnsinx)2+cos3xlnsinx
sinxlnf2=sinxlncosxf02
f2=cosxlncosx+�sin2x
cosxf02=sinxcosx(lncosx)2�sin3xlncosx
cosx=)f=sinxcosx(�(lnsinx)2+(lncosx)2)+cos3xlnsinx
sinx+�sin3xlncosx
cosxExercise32.lnf=1
xlnxf0
f=�1
x2lnx+1
x2=)f0=�x1=x�2lnx+x1=x�2Exercise33.lnf=2lnx+1
3ln(3�x)�ln(1�x)+�2
3ln(3+x)f0
f=2
x+�1
3(3�x)��1
1�x�2
31
3+xf0=2x(3�x)1=3
(1�x)(3+x)2=3+�1
3x2(3�x)�2=3
(1�x)(3+x)2=3+x2(3�x)1=3
(1�x)2(3+x)2=3�2
3x2(3�x)1=3
(1�x)(3+x)5=3=x(18�12x+4
3x2+2
3x3)
(1�x)2(3+x)5=3(3�x)2=3Exercise34.lnf=Pni=1biln(x�ai)f0
f=nXi=1bi
x�ai=)
f0=nXj=1bj
x�ajnYi=1(x�ai)bi
Exercise35.(1)Showthatf0=rxr�1forf=xrholdsforarbitraryrealr.xr=erlnx(erlnx)0=erlnxr
x=rxr�1(2)Forx0,byinspectionofxr=erlogx,thenifxr�0,thentheequalitywouldremainvalid.Sothenxr=jxrj=jxjrandsolnjf(x)j=rlnjxjf0(x)
f(x)=r1
x=)f0(x)=rxr�1Exercise36.Usethedenitionax=exloga(1)logax=xlogaTakingtheexponentialisawell-denedinversefunctiontologsotakingthelogofbothsidesofthedenition,wegetlogax=xloga(2)(ab)x=axbx(ab)x=exlogab=ex(loga+logb)=axbx(3)axay=ax+yax+y=e(x+y)loga=exlogaeyloga=axay(4)(ax)y=(ay)x=axy(ax)y=exyloga=(ay)x=axy96
(1)f=ex�1�xf0=ex�1?0ifx?0f(0)=e0�1�0=0(ex�1+xforx�0e�x�1�xforx0(2)Zx0et=ex�1&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;ॷ ;� Td;&#x [00;x+1
2x2=)ex�1+x+1
2x2�e�x��1+x�x2
2=)e�x1�x+x2
2(3)Zx0et=ex�1�x+1
2x2+1
32x3=)ex�1+x+1
2x2+1
32x3�e�x��1+x�x2
2+x3
32=)e�x1�x+x2
2�x3
32(4)Supposethenthcaseistrue.ex�nXj=0xj
j!e�x=(�P2m+1j=0xj
j!P2mj=0xj
j!ex�1+nXj=0xj+1
(j+1)!=1+n+1Xj=1xj
j!=n+1Xj=0xj
j!�e�x+1(�P2m+1j=0xj+1
(j+1)!P2mj=0xj+1
(j+1)!=e�x(P2m+2j=0xj
j!�P2m+1j=0xj
j!Exercise42.UsingtheresultfromExercise41,1+x
nn=nXj=0nj1n�jx
nj=nXj=0n!
(n�j)!j!xj
nj=nXj=0n(n�1):::(n�j+1)
j!xj
njnXj=0xj
j!exIfyoumakethiscleverobservation,thesecondinequalityiseasytoderive.x&#x-278;0x
n�0e�x
n�1�x
n=)e�x=nn�1�x
nne�x�(1�x=n)n=)ex(1�x=n)�nExercise43.f(x;y)[email protected][email protected]=xylnx6.19Exercises-Thehyperbolicfunctions.Exercise7.2sinhxcoshx=2ex�e�x
2ex+e�x
2=1
2(e2x�e�x)=sinh2xExercise8.cosh2x+sinh2x=ex+e�x
22+ex�e�x
22=1
4(e2x+2+e�2x+e2x�2+e�2x)=cosh2xExercise9.coshx+sinhx=ex+e�x
2+ex�e�x
2=exExercise10.coshx�sinhx=ex+e�x
2�ex�e�x
2=e�xExercise11.Useinduction.98
(coshx+sinhx)2=cosh2x+2sinhxcoshx+sinh2x=cosh2x+sinh2x(coshx+sinhx)n+1=(coshx+sinhx)(coshnx+sinhnx)==coshnxcoshx+coshnxsinhx+sinhnxcoshx+sinhxsinhnx==enx+e�nx
2ex+e�x
2+enx+e�nx
2ex�e�x
2++enx�e�nx
2ex+e�x
2+enx�e�nx
2ex�e�x
2=cosh(n+1)x+sinh(n+1)xExercise12.cosh2x=cosh2x+sinh2x=1+2sinh2x6.22Exercises-Derivativesofinversefunctions,Inversesofthetrigonometricfunctions.Exercise1.(cosx)0=�sinx=�p
1�cos2xDarccosx=1
�p
1�x2�1x1Exercise2.(tanx)0=sec2x=sin2x+cos2x
cos2x=tan2x+1Darctanx=1
1+x2Exercise3.(cotx)0=�csc2x=�(sin2x+cos2x)
sin2x=�(1+cot2x)=)arccotx=�1
1+x2Exercise4.(secy)0=tanysecy=p
sec2y�1secy;jsecyj�18y2RIfwechoosetorestrictysuchthat0y,then(secy)0�0.Thenwemustmakesecy!jsecyj.Darcsecx=1
jxjp
x2�1Exercise5.(cscy)0=�cotxcscx=�cscy(p
csc2y�1)Letysuchthat�
2y
2(cscy)0Darccscx=1
�jxjp
x2�1Exercise6.(xarccotx)0=arccotx�x
1+x21
2ln(1+x2)0=1x
(1+x2)Zarccotx=xarccotx+1
2ln(1+x2)+CExercise7.(xarcsecx)0=arcsecx+x
jxjp
x2�1x
jxjlnjx+p
x2�1j0=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:1+x
p
x2�1
jx+p
x2�1jx�1��1+x
p
x2�1
jx+p
x2�1jx�1=x
jxjp
x2�1=)Zarcsecxdx=xarcsecx�x
jxjlogjx+p
x2�1j+C99
Exercise12.f0=1
q
1��x
221
2Exercise13.f0=�1
r
1�1�x
p
22�1
p
2=1
p
1+2x�x2Exercise14.f=arccos1
x.f0=�1
q
1��1
x2�1
x2=1
p
x2�1jxjExercise15.f(x)=arcsin(sinx)=1
p
1�sin2xcosx=cosx
jcosxjExercise16.1
2sqrtx�1
x+11
2p
x=p
x
2(x+1)Exercise17.1
1+x2+x2
1+x6Exercise18.1
r
1�q
1�x2
1+x22�2x(2)
(1+x2)2=p
1+x2
p
1+x2�(1�x2)�4x
(1+x2)2=�4
(1+x2)3=2p
2Exercise19.f=arctantan2x1
1+tan4x�2tanxsec2x=2tanxsec2x
1+tan4xExercise20.f0=1
1+(x+p
1+x2)2(1+x
p
1+x2)=p
1+x2+x
1+(x+p
1+x2)2Exercise21.f0=1
q
1�(sin2x+cos2x�2sinxcosx)(cosx+sinx)=cosx+sinx
p
2sinxcosxExercise22.f0=(arccosp
1�x2)0=1
�p
1�(1�x2)=�1
jxjExercise23.f0=1
1+1+x
1�x22
(1�x)2=2
(1�x)2+(1+x)2=2
(1�2x+x2+1+2x+x2)=1
1+x2Exercise24.f=(arccos(x2))�2f0=�2(arccosx2)�3�1
p
1�x2(2x)=4x(arccosx2)�3
p
1�x4Exercise25.101
Exercise34.x2arctanx
20=xarctanx+x2
21
1+x2=xarctanx+1
21�1
1+x21
2(x�arctanx)0=1
21�1
1+x2Zxarctanx=xarctanx+�1
2(x�arctanx)Exercise35.x3
3arccosx0=x2arccosx+x3
3�1
p
1�x2(x2p
1�x2)0=2xp
1�x2+�x3
p
1�x2((1�x2)3=2)0=3
2(�2x)(1�x2)1=2=�3x(1�x2)1=2
Zx2arccosx=x3
3arccosx�1
3x2p
1�x2�9
2(1�x2)3=2
Exercise36.x2(arctanx)2
20=x(arctanx)2+x21
1+x2arctanx=x(arctanx)2+1�1
1+x2arctanx(arctanx)2
20=arctanx
1+x2(xarctanx)0=arctanx+x
1+x2Zx(arctanx)2dx=x2(arctanx)2
2�xarctanx�ln(1+x2)
2+(arctanx)2
2Exercise37.(arctanp
x)0=1
1+x1
2p
x(xarctanp
x)0=arctanp
x+p
x
2(1+x)=arctanp
x+1
21
p
x�1
p
x(1+x)(xarctanp
x+arctanp
x+�x1=2)0=arctanp
x+0
Zarctanp
x=xarctanp
x+arctanp
x�x1=2
Exercise38.Fromthepreviousexercise,Zarctanp
x
p
x(1+x)dx=(arctanp
x)2Exercise39.Letx=sinuZp
1�x2dx=Zcos2udu=u
2+sin2u
4=arcsinx
2+xp
1�x2
4Exercise40.103
Zp
x�ap
b�xdx=Zp
bx�ab�x2+axdx==Zs
�x�a+b
2x�a+b
2+a2+b2
4�2ab
4==Zs
a�b
22�x�a+b
22=a�b
2Zvuut
1� x��a+b
2
�a�b
2!2dx==a�b
22Zp
1�u2==
a�b
22arcsin2x�(a+b)
a�b
2+2x�(a+b)
2(a�b)2p
(a�b)2�(2x�(a+b))2
Since,recall,arcsinx
2+1
2xp
1�x20=1
21
p
1�x2+p
1�x2
2+1
4x(�2x)
p
1�x2=p
1�x2Exercise47.Wow!
Zdx
p
(x�a)(b�x)
x�a=(b�a)sin2udx=(b�a)(2)sinucosudub�x=(a�b)sin2u+b�a=
(b�a)(cos2u)
Zdx
p
(x�a)(b�x)=Z(b�a)(2)sinucosudu
p
b�acosup
b�asinu=2u=
2arcsinr
x�a
b�a
6.25Exercises-Integrationbypartialfractions,Integralswhichcanbetransformedintointegralsofrationalfunc-tions.Exercise1.R2x+3
(x�2)(x+5)=R1
x�2+1
x+5=ln(x�2)+ln(x+5)Exercise2.Rxdx
(x+1)(x+2)(x+3)A
x+1+B
x+2+C
x+3=A(x2+5x+6)+B(x2+4x+3)+C(x2+3x+2)=)241115436323524ABC35=2401035=)24111543632 01035=24101001 �1=22�3=235A=�1=2;B=2;C=�3=2=)
�1
2ln(x+1)+2ln(x+2)+�3
2ln(x+3)
Exercise3.Rx
(x�2)(x�1)=R2
x�2+�1
x�1=2lnx�2�ln(x�1)Exercise4.Rx4+2x�6
x3+x2�2xdx105
A(x4+2x2+1)+x(Bx+C)(x2+1)+Dx2+Ex
x(x2+1)2=A(x4+2x2+1)+Bx4+Cx3+Bx2+Cx+Dx2+Ex
x(x2+1)2=)A=1;B=�1;D=�1;C=0;E=0Z1
x+�x
x2+1+�x
(x2+1)2=
lnx+�lnjx2+1j
2+(x2+1)�1
2
Exercise10.Rdx
(x+1)(x+2)2(x+3)3Exercise11.Rx
(x+1)2dxx
(x+1)2=A
x+1+B
(x+1)2=)x=A(x+1)+BA=1;B=�1Z1
x+1+�1
(x+1)2dx=lnx+1+1
x+1+CExercise12.Rdx
x(x2�1)=Rdx
x(x�1)(x+1)=RA
x+B
x�1C
x+1A(x2�1)+Bx(x+1)+Cx(x�1)=Ax2�A+Bx2+Bx+Cx2�Cx=)A=�1;B=1
2=CZ�1
x+1=2
x�1+1=2
x+1=�lnx+1
2lnjx�1j+1
2lnjx+1jExercise13.Rx2dx
x2+x�6=Rx2dx
(x+3)(x�2)Theeasiestwaytoapproachthisproblemistonoticethatthisisanimproperfractionandtodolongdivision:x2
x2+x�6=1+6�x
x2+x�6A
x+3+B
x�2=)6=A(x�2)+B(x+3)A=�6
5;B=6
5�x=(A+B)x�2A+3B2A=3BA=3B
2=)B=�2=5A�3=5=)Z1+�6=5
x+3+6=5
x�2+�3=5
x+3+�2=5
x�2=
�9
5lnjx+3j+4
5lnjx�2j+x+C
Exercise14.Rx+2
(x�2)2=Rx�2+4
(x�2)2=lnjx�2j+R4
(x�2)2=
lnjx�2j+�4(x�2)�1
Exercise15.Rdx
(x�2)2(x2�4x+5)Considerthedenominatorwithitsx2�4x+5.Usually,wewouldtryapartialfractionformsuchasA
x�2+B
(x�2)2+Cx+D
x2�4x+5,butthealgebrawillgetmessy.Instead,ithelpstobecleverhere.1
(x�2)2(x2�4x+4+1)=1
(x�2)2((x�2)2+1)=1
(x�2)2�1
(x�2)2+1=)Zdx
(x�2)2(x2�4x+5)=Z1
(x�2)2�1
(x�2)2+1=
�(x�2)�1�arctan(x�2)+C
Exercise16.R(x�3)dx
x3+3x2+2x=R(x�3)dx
x(x+2)(x+1)Z(x�3)dx
x(x+2)(x+1)=Z1
(x+2)(x+1)+�3Z1
x(x+2)(x+1)1
(x+2)(x+1)=�1
x+2+1
x+11
x(x+2)(x+1)=A
x+B
x+2+C
x+1107
Workingoutthealgebraforthepartialfractionsmethod,weobtain1
x3(x�2)=�1=2
x3+�1=4
x2+�1=8
x+1=8
x�2Sothen
Zdx
x3(x�2)=1
4x2+1
4x+�1
8lnx+1
8lnjx�2j+C
Exercise21.Z1�x3
x(x2+1)=�Zx3�1
x(x2+1)=�Zx2
x2+1+Z1
x(x2+1)==�Z1�1
x2+1+Z1
x+�x
x2+1==
�x+arctanx+lnx�lnjx2+1j+C
Exercise22.Zdx
x4�1=Z1
x2+11
x2�1=Z1=2
x2�1�1=2
x2+1==Z1
21=2
x�1�1=2
x+1�1=2
x2+1==
1
4ln(x�1)�1
4ln(x+1)�1
2arctanx+C
Exercise23.
Zdx
x4+1
Ihadtorelyoncomplexnumbers.Noticethatwithcomplexnumbers,youcansplituppolynomialpowersumsx4+1=(x2+i)(x2�i)=(x+ie(i
4))(x�ie(i
4))(x+e(i
4))(x�e(i
4))==(x+e(i3
4))(x�e(i3
4))(x+e(i
4))(x�e(i
4))A
(x+e(i3
4))+B
(x�e(i3
4))+C
(x+e(i
4))+D
(x�e(i
4))=1
x4+1A(x2�i)(x�e(i3
4))+B(x2�i)(x+e(i3
4))+C(x2+i)(x�e(i
4))+D(x2+i)(x+e(i
4))=1dothealgebra��������!x3:A+B+C+D=0x2:�Ae(�3
4)+Be(i3
4)�Ce(i
4)+De(i
4)=0x1:�iA�iB+iC+iD=0x0:�e(i
4)A+Be(i
4)+C(�e(i3
4))+D(e(i3
4))=1=)26641111�e(i3
4)e(i3
4)�e(i
4)e(i
4)�i�iii�e(i
4)e(i
4)�e(i3
4)e(i3
4)37752664ABCD3775=266400013775TodothecomplexalgebraforthedesiredGaussianeliminationprocedure,Itreatedthecomplexnumbersasvectorsandaddedthemandrotatedthemwhenmultiplied.109
Exercise27.Rdx
1+acosx(0a1)Again,usingthehalf-anglesubstitution,
u=tanx
2du=sec2x
2
2dx
,1
aZdx
1
a+cosx=1
aZdx
1
a+C2�S2=1
aZsec2x
2dx
1
asec2x
2+1�T2==1
aZsec2x
2dx
1
a+1+T2(1
a�1)=1
aZ2du
1
a+1+u2�1
a�1=2
1+aZdu
1+uq
1�a
1+a2=2
1+aarctanq
1�a
1+au
q
1�a
1+a==)
2
p
1�a2arctan r
1�a
1+atanx
2!
Exercise28.Rdx
1+acosxHalf-anglesubstitution.Zdx
1+acosx=Zdx
1+a(C2�S2)=Zsec2x
2dx
sec2x
2+a(1�T2)=u=tan=2=Tdu=sec2=21
2d=)=Z2du
1+T2+a(1�T2)=2Zdu
(1�a)T2+(1+a)=2
1�aZdu
u2�a+1
a�1==2
1�[email protected]
u�q
a+1
a�1�1
u+q
a+1
a�11A1
2q
a+1
a�1==r
a�1
a+11
1�a ln(u�r
a+1
a�1)�ln(u+r
a+1
a�1)!==
�1
p
a2�[email protected]@tanx
2�q
a+1
a�1
tanx
2+q
a+1
a�11A1A
Exercise29.Rsin2x
1+sin2xdxZs2
1+s2dx=Zs2+1�1
1+s2dx=x+�Zdx
1+sin2xZdx
1+sin2x=Zdx
1+�1�cos2x
2=Z2dx
3�cos2x=2
3Zdx
1�cos2x
3=2
3Zdx
1��c2�s2
3=2
3Zsec2xdx
sec2x��1�T2
3=u=tanxdu=sec2xdx=)=2
3Zdu
1+u2��1�u2
3=2
3Zdu
2
3+4
3u2=Zdu
1+(p
2u)2==1
p
2arctanp
2tanx=)
Zsin2x
1+sin2xdx=x�1
p
2arctan(p
2tanx)
Itseemslikeforhere,whendealingwithsquaresoftrig.functions,“stepup”todoubleangle.Exercise30.Rdx
a2sin2x+b2cos2x(ab6=0)Takenote,weneednotchangetheangletohalf-angleordouble-angle.111
Z(x)0p
3�x2dx=xp
3�x2�Zx(�x)
p
3�x2=xp
3�x2�Z�x2+3�3
p
3�x2=xp
3�x2�Zp
3�x2+3Z1
p
3�x2=)2Zp
3�x2=xp
3�x2+p
3Z1
r
1�x
p
32=)
Zp
3�x2=x
2p
3�x2+3
2arcsinx
p
3
Exercise34.R1
p
3�x2dx=�(3�x2)1=2+C.arccosx
p
30=1
p
3�1
q
1�x2
3=�1
p
3�x2xp
3�x20=p
3�x2+�x2
p
3�x2
xp
3�x2
2+�3
2arccosx
p
3
Exercise35.Rp
3�x2
xdx=Rq
3
x2�1dx.p
3
x=secp
3cos=xdx=�sinp
3Zp
sec2�1(�sin)p
3=Ztansin(�p
3)=�p
3Z(sec�cos)==�p
3lnjsec+tanj+p
3sin==
�p
3ln p
3
x+r
3
x2�1 +p
3r
1�x2
3
Exercise36.Rq
1+1
xdx xr
1+1
x!0=r
1+1
x+x
2q
1+1
x�1
x=r
1+1
x+�1=2
p
x2+xlnx+1
2+p
x2+x0=1
x+1
2+p
x2+x1+x+1
2
p
x2+x=1
p
x2+x=)Z1+1
xdx=xr
1+1
x+1
2lnx+1
2+p
x2+xExercise37.(xp
x2+6)0=p
x2+5+x2
p
x2+5(ln(x+p
x2+b))0=1
x+p
x2+b1+x
p
x2+b=1
p
x2+bZp
x2+5=1
2xp
x2+5+5ln(x+p
x2+5)Exercise38.113
Otherwise,rememberthatforrationalexpressionsinvolvingsinglepowersofsinandcos,wecanmakeau=tan=2substitution.u=tanx
22du=sec2x
2dxC=cosx=2;S=sinx=2Zsinx
2+cosxdx=Z2SCdx
2+C2�S2=Z2T
2sec2x=2+1�T22du
sec2x=2==4Zudu
(1+u2)(3+u2)=2ZT
T2+1�T
T2+3=21
2lnT2+1�1
2lnT2+3=lnT2+1
T2+3=ln2
4+2cosxwheretan2x
2=sin2x
2
cos2x
2=1�cosx
1+cosxExercise3.Zex
xdx=ex�Zex(x�1)
x2xdx=ex�Zex(x�1)
xdx:::Noway.Exercise4.R=20ln(ecosx)dx=�cosxj3=20=
1
.Exercise5.(1)f=p
4x+2x(x+1)(x+2)lnf=1
2ln4(x+2)
x(x+1)(x+2)=1
2(ln(4x+2)�lnx�ln(x+1)�ln(x+2))f0
f=1
24
4x+2�1
x�1
x+1�1
x+2=)f0=1
2s
4x+2
x(x+1)(x+2)4
4x+2�1
x�1
x+1�1
x+2f0(1)=�7
12(2)Z414x+2
x(x+2)(x+1)dx=2Z41(2x+1)
x(x+2)(x+1)dx==21
2lnx+�3
2lnjx+2j+lnjx+1j 41=
ln25
8
sincewecanndtheantiderivativethroughpartialfractions:A
x+B
x+2+C
x+1=2x+1
x(x+2)(x+1)A(x2+3x+2)+B(x2+x)+C(x2+2x)=2x+1241113122003524ABC35=2402135=)24010001100 �3=211=235Exercise6.115
(2)Sinceforf(x)=ex,wedenedexsuchthatf0=f,if(ex+g)0=ex+g0=ex+g=)ex+g=Cex=)g=(C�1)exbutf0(0)=1so
g=ex
Exercise9.(1)g(2x)=2exg(x)g(3x)=exg(2x)+e2xg(x)=ex2exg+e2xg=3e2xg(2)Assumeg(nx)=ne(n�1)xgg((n+1)x)=exg(nx)+e(n+1)xg(x)=nenxg(x)+enxg=(n+1)enxg(3)Fromg(x+y)=eyg(x)+exg(y),g(0)=g(0)+g(0)=)g(0)=0g(x+h)�g(x)
h=ehg(x)+exg(h)�g(x)
h=g(x)eh�1
h+exg(h)
hf0(0)=2=limh!0g(x+h)�g(x)
h=limh!0g(h)
h(4)g0(x)=g(x)+2ex
C=2
Exercise10.8x2R;f(x+a)=bf(x);f(x+2a)=bf(x+a)=b2f(x)f(x+(n+1)a)=f(x+na+a)=bf(x+na)=bn+1f(x)f(x+na)=bnf(x)
f(x)=bx=ag(x)wheregisperiodicina
Exercise11.(ln(fg))0=f0
f+g0
g=)(fg)0=f0g+fg0lnf
g0=f0
f�g0
g=)f0g�g0f
g2=f
g0Exercise12.A=R10et
t+1dt(1)Zaa�1e�t
t�a�1dtu=t�a=)Z0�1e�t�a
t�1dt=�Z01et�a
�t�1dt=�e�aZ10et
t+1=
�e�aA
(2)R10tet2
t2+1dt=R101
2dueu
u+1=
1
2A
(3)R10et
(t+1)2dt=�et
(t+1) 10�R10�et
t+1=
�e1
2+1+A
(4)R10etln(1+t)dt=etln(1+t)�Ret
1+t=
eln2�A
Exercise13.117
(3)f(t)=e�jtjF(x)=Zx0f(t)dt=Zx0e�jtjdt==(Rx0e�tdt=e�tj0x=1�e�xifx0Rx0etdt=etjx0=ex�1ifx0(4)f(t)=max.of1andt2F(x)=Zx0f(t)dt=8&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;ॷ ;� Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:Rx01dt=xifjxj11+Rx1t2dtifx&#x]TJ ;� -1;.93; Td;&#x [00;1�Rx0fifx�1==8&#x]TJ/;༔ ; .96;& T; 18;&#x.978;&#x 0 T; [0;&#x]TJ ;� -1;.93; Td;&#x [00;:xifjxj11+1
3t3 x1=x3
3+2
3ifx�1�R�1xt2+�R0�11=1
3t3 x�1�1=x3
3+2
3ifx�1Exercise17.Rf2=Ra0f2=a2+a.x2+x
0=r
2x+1
forZa02x+1
=�x2+x a0=a2+aExercise18.f(x)=e�2x.(1)A(t)=Rt0e�2xdx=e�2x
�2 t0=e�2t�1
�2(2)V(t)=Rt0e�4xdx=
�4(e�4t�1)(3)y=e�2x=)lny
�2=xW(t)=Z1e�2tlny
�22dy=
4(y(lny)2�(2(ylny�y))) 1e�2t==
4�2��e�2t4t2�(2(e�2t(�2t)�e�2t))==
2�te�2t�
2e�2twheretheantiderivativeusedwas(y(lny)2)0=(lny)2+2lny(4)
4(1�e�4t)
1�e�2t
2=
2e�4t�1
t
e�2t�1
t=

whereweusedthelimitlimx!0ecx�1
x=cExercise19.sinhc=3
4(1)ec=ex+p
e2x+1ec�e�c
2=ex+p
e2x+1�1
ex+p
e2x+1
2==e2x+2exp
e2x+1+e2x+1�1
2(ex+p
e2x+1)=ex=3
4
x=ln3�2ln2
119
(2)ec�e�c
2=ex�p
e2x�1�1
ex�p
e2x�1
2==e2x�2exp
e2x�1+e2x�1�1
2(ex�p
e2x�1)=ex
+�1
ex�p
e2x�1=ex�1
ec=3
4=)
x=ln5�2ln2
Exercise20.(1)True.ln(2log5)=ln5ln2=(ln2)ln5.(2)log35=log25
log23Thisisatruefact.log35
log23=log25
(log23)2=log25=)1=log23False(3)Useinductionn=11�1=22p
1n=21+1
p
22p
2n+1casen+1Xk=1k�1=2=nXk=1k�1=2+1
p
(n+1)2p
n p
(n+1)
p
(n+1)!+1
p
(n+1)Now(n+1
2)2=n2+n+1
2�n2+n,certainly.Sothenn+1
2�p
n2+n=)n+1�p
n2+n=)n+1Xk=1k�1=22p
n+1(4)f=(coshx�sinhx�1)=ex+e�x�ex+e�x
2�1=e�x�10forx&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0False.Exercise21.For0x
2,(sinx)0=cosx�0for0x
2(x�sinx)0=1�cosx0for0x
2(x�sinx)(x=0)=0=)sinxxExercise22.1
t1
tx+1
tif0xtx+1Zx+1x1
tdt=ln(x+1)�lnx;Zx+1xx+1
t=
1
x
Solnx+1
x1
xExercise23.120
a1;b1;a;b.Matchinguptermbytermthecoefcientsforsinandcosseparately,Aa2�abB=aa1Ab2+Bab=b1b
A=aa1+bb1
a2+b2
�Aab+Bb2=�a1bAab+Ba2=ab1
B=ab1�a1b
a2+b2
Soifnotbotha;b=0;Za1sinx+b1cosx
asinx+bcosx=ZA(asinx+bcosx)+B(acosx�bsinx)
asinx+bcosx==
Ax+Blnjasinx+bcosxj+C
Exercise27.(1)f0(x2)=1
xdf
du=u�1=2f(x2)=2x�1(2)f0(sin2x)=1�sin2xf0(u)=1�uf=u�1
2u2+C=)
f(x)=x�x2
2+1
2
(3)f0(sinX)=(1�sin2x)f(u)=u�1
3u3+C
f(x)=x�x3
3+1
3
(4)f0(lnx)=(1forx1xforx�1=(1for0x1elnxx&#x-278;1
f(y)=(yfory0ey�1fory&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;Ԧ ;� Td;&#x [00;0
Exercise28.(1)Li(x)=Zx2dt
lntifx2Li(x)=x
lnx�21
ln2�Zx2�1
(lnx)2dt=x
lnx+Zx2dt
(lnx)2�2
ln2(2)Li(x)=x
lnx�2
lnx�2
(ln2)2+x
(lnx)2�Zxa�2
(lnt)3dtLi(x)=x
lnx+n�1Xk=1k!x
lnk+1x+n!x
lnn+1x�Zx2�(n+1)dt
lnn+2tLi(x)=x
lnx+nXk=1k!x
lnk+1x+(n+1)!Zx2dt
ln(n+1)+1tC2=�2
lnx+�2Xj=22(j�1)!
(ln2)jCn=�21
ln2�nXj=22(j�1)!
(ln2)jCn+1=�21
ln2�n+1Xj=22(j�1)!
(ln2)j122
Theorem24.Pnisapolynomialofdegreen1.Letf;gbe2functionswithderivativesofordernat0.(13)f(x)=Pn(x)+xng(x)whereg(x)7!0asx7!0.ThenPn=Tn(f;x=0).Exercise3.Tnf(x)=nXj=0f(j)(a)
j!(x�a)jax=exlna(ax)0=(ax)lna(ax)(n+1)=(ax(lna)n)0=ax(lna)n+1Tn(ax)=nXj=0(lna)j
j!xjExercise4.1
1+x0=�1
(1+x)2;1
1+x00=(�1)22
(1+x)31
1+x(n+1)=(�1)nn!
(1+x)n+10=(�1)n+1(n+1)!
(1+x)n+2Tn1
1+x=nXj=0(�1)jxjExercise5.
UseTheorem7.4.
.Theorem7.4saysforf(x)=Pn(x)+xng(x),Pn(x)istheTaylorpolynomial.1
1�[email protected]=0(x2)j1A+(x2)n+1
1�[email protected]=0x2j1A+xnxn+2
1�x2x
1�[email protected]=0x2j+11A+(x2n+3)
1�x2T2n+1x
1�x2=nXj=0x2j+1Exercise6.(ln(1+x))0=1
1+xTn1
1+x=nXj=0(�x)jTn(ln1+x)=nXj=0(�1)jxj+1
j+1=nXj=1(�1)j+1xj
jExercise7. logr
1+x
1�x!0=r
1�x
1+xr
1�x
1+x1
(1�x)2=1
(1+x)(1�x)=1
1�x2Z1
1�x2=logr
1+x
1�xsoZnXj=0x2j=nXj=0x2j+1
2j+1=T2n+1 lnr
1+x
1�x!Exercise8.Tn1
2�x=Tn1=2
1�x=2=1
2Tn 1
1��1
2x!=1
[email protected]=0(1
2x)j1A=nXj=0xj
2j+1Exercise9.Wecanshowthisintwoways.124
Exercise4.(1)x2=sinx=x�x3
6=)x3
6+x2�x=x
6x��3+p
15x�(�3�p
15)
x=p
15�3
(2)E4(r;0)=1
4!Zr0(r�t)4costdt�0sinr�r2=0+E4(r)r5
5!r
5!=3
5(2)(5)(4)(3)(2)1=3
4
(5)(4)(2)51
200Exercise5.arctanr�r2=r�r3
3�r2+E4(r;0)=0+E4(r;0)E4(r;0)=1
4!Zr0(x�t)4f(5)(t)dtM(r5)
5!r5
5!=0:0655367
100E4(r;0)Ej(r;0);j&#x-278;4the5thdegreetermisf(5)(0)
5!r5=24
5!r5�0sor2�arctanr=�E4(r;0)0Exercise6.Applylongdivisiononthefractionintheintegrand.Z101+x30
1+x60dx=Z101+x30�x60
1+x60dx=1+Z10x301�x30
1+x60dx==1+c1
31x31 10=1+c
31Exercise7.R1=201
1+x4dx.1
1+x4=1Xj=0(�x4)j=n�1Xj=0(�x4)j+En=1+�x4+x8:::16
17x4n+1
(n+1)!En(x;0)1
(n+1)!(x4)n+1=)Z1=20En=�1
24n+5
(n+1)!1
4n+5Z1=201
1+x4'1
2+�1
51
250:4938520:493750:493858Exercise8.(1)0x1
2sinx=x�x3
3!+E4(x)jE4(x)jMjxj5
5!=sin1
3jxj5
5!1�1
25
5!126
x(cosx)=((x�1)+1)(cosx)=(x�1)cos1+(�sin1)(x�1)2�cos1(x�1)3
2+cos1++(�sin1)(x�1)�cos1(x�1)2
2+sin1(x�1)3
3!==cos1+(cos1�sin1)(x�1)+�sin1�cos1
2(x�1)2+sin1�3cos1
3!(x�1)3+o(x�1)3Exercise3.Justtreattheargumentofsinx�x2justlikeuwithu!0.sin(x�x2)=(x�x2)�(x�x2)3
3!+(x�x2)5
5!+o(x�x2)5==(x�x2)�1
6�x3�3x4+3x5�x6+1
120�x5�5x6+10x7�10x8+5x9+x10==(x�x2)�1
6x3�1
2x4+61
120x5�25
120x6Exercise4.logx=log(1+(x�1))=(x�1)�(x�1)2
2+(x�1)3
3=)
a=0;b=1;c=�1
2
Exercise5.cosx=1�1
2x2+o(x3)asx!01�cosx=1
2x2+o(x3)1�cosx
x2=1
2+o(x3)
x2
since1�cosx
x2=1
2+o(x);1�cosx
x2!1
2asx!0
cosx=1�1
2x2+x4
4!+o(x5)=)cos2x=1�2x2+2
3x4+o(x5)1�cos2x�2x2
x4=�2
3x4�o(x5)
x4=�2
3�o(x)!�2
3asx!0Exercise6.limx!0sinax
sinbx=limx!0ax�(ax)3
3!+o(x4)
bx+o(x2)=a
bExercise7.limx!0sin2x
cos2xsin3x=limx!0(2x)+(2x)3
3!+o(x4)
1�(2x)2
2!+(2x)4
4!+o(x5)(3x)�(3x)3
3!+o(x4)=
2
3
Exercise8.limx!0sinx�x
x3=
�1
6
Exercise9.limx!0ln1+x
e2x�1=limx!0x�o(x)
2x+o(x)=1
2Exercise10.Don'tdothetrig.identity.limx!01�cos2x
xtanx=limx!01�1�x2
2!+o(x2)2
x�x+x3
6!+o(x3)=limx!01�(1+x2+o(x2))
x2+o(x2)=1Exercise11.128
Exercise23.limx!1x1
1�x=limx!1e1
1�xlnx=explimx!1lnx
1�x=explimx!1(x�1)+o(x�1)2
1�x=
e�1
Exercise24.limx!0(x+e2x)1=x=explimx!01
xln(x+e2x)limx!0ln(x+e2x)
x=limx!0ln(1+x+e2x�1)
x==limx!0x+e2x�1+o(x2)
x=limx!03x+o(x2)
x=3=)limx!0(x+e2x)1=x=e3Exercise25.limx!0(1+x)1=x�e
x=limx!0e1
xln(1+x)�e
x=limx!0ex+�x2
2+o(x2)
x�e
x==limx!0e1�x
2+o(x)�e
x=limx!0e(1+�x
2+o(x))�e
x=�e
2Exercise26.limx!0(1+x)1=x
e1=x=limx!0exp1
xln(1+x)�11=x=limx!0ex�x2
2+o(x3)�x
x2=e�1=2Exercise27.(arcsinx)0=1
p
1�x2(arcsinx)00=x
(1�x2)3=2(arcsinx)000=1
(1�x2)3=2+3x2
(1�x2)5=2explimx!01
x2lnarcsinx
x=explimx!01
x2ln1+arcsinx
x�1==explimx!01
x2ln1+x+x3=6+o(x4)�x
x=explimx!01
x2ln1+x2
6+o(x3)==explimx!01
x2x2+o(x3)
6=
e1=6
Exercise28.limx!01
x�1
ex�1=limx!0ex�1�x
x(ex�1)=limx!0x2
2+o(x3)
x2+o(x3)=
1
2
Exercise29.limx!11
logx�1
x�1=limx!1(x�1)�logx
(x�1)logx==limx!1(x�1)�((x�1)�(x�1)2
2+o(x�1)3)
(x�1)((x�1)+o(x�1)2)=
1
2
Exercise30.limx!0eax�ex�x
x2=limx!01+ax+(ax)2
2+o(x3)�1�x�x2
2�x
x2ifa=2,thelimitis
2
Exercise31.130
7.13Exercises-L'Hopital'srulefortheindeterminateform0=0.Exercise1.limx!23x2+2x�16
x2�x�2=limx!2(3x+8)(x�2)
(x�2)(x+1)=
14
3
Exercise2.limx!3x2�4x+3
2x2�13x+21=limx!3(x�3)(x�1)
(2x�1)(x�3)=
�2
Exercise3.limx!0sinhx�sinx
x3=0
0=limx!0coshx�cosx
3x2=0
0=limx!0sinhx+sinx
6x=limx!0coshx+cosx
6=
1
3
Exercise4.limx!0(2�x)ex�x�2
x3=limx!0�ex+(2�x)�1
3x2=limx!0(2�x)(1+x+x2=2+x3=6+o(x3))
x3=
�1
6
Exercise5.limx!0log(cosax)
log(cosbx)=limx!0�cosbxsinaxa
�cosaxsinbxb=limx!0cosaxa2
sinbxb2=a2
b2Exercise6.Whenitdoubt,Taylorexpand.limx!0+x�sinx
(xsinx)3=2=limx!0+1�cosx
3
2(xsinx)1=2(sinx+xcosx)=2
3limx!0+1�(1�x2
2+x4
24+o(x4))
e1
2ln(xsinx)(x�x3
6+o(x3)+x�x3
2+o(x3))==2
[email protected]
2�x4
24+o(x4)
q
x(x+x3
6+o(x3))(2x�2
3x3+o(x3))1A==2
[email protected]
2�x2
24+o(x2)
q
1+x2
6+o(x2)(2+�2
3x2+o(x2))1A=2
3limx!0+1=2
2=
1
6
Noticeinthethirdstephowingeneralwedealwithpowers,(xsinx)1=2,istoconvertitintoexponentialform,e1
2ln(xsinx),butitwasn'tnecessary.Exercise7.DoL'Hopital'srst.limx!a+p
x�p
a+p
x�a
p
x2�a2=limx!a+1
2p
x+1
2p
x�a
x
p
x2�a2=1
2limx!a+p
x2�a2
x3=2+p
x+a
x==1
2limx!a+p
x2�a2+x1=2p
x+a
x3=2=1
2p
2a
a3=2=p
2
2p
aExercise8.DoL'Hopital'satthesecondstep.limx!1+exp(xlnx)�x
1�x+lnx=limx!1+expxlnx(lnx+1)�1
�1+1
x=limx!1+exp(xlnx)(lnx+1)2+1
xexp(xlnx)
�1=x2==limx!1+x2exp(xlnx)(lnx+1)2+xexp(xlnx)=1+1=
2
Exercise9.KeepdoingL'Hopital's.132
Exercise16.(1)angleABCisx
2;lengthBCistanx
22tanx
2cosx
2=2sinx
2isthebaselengthofABCtanx
2sinx
2istheheightoftriangleABC=)T(x)=tanx
2sin2x
2=1�cos2x
2
cosx
2sinx
2=tanx
2�1
2sinx(2)S(x)=x
2((1))�1
2cosx
2(2sinx
2)=x
2�sin2
2(3)UseL'Hopital'stheorem.T(x)
S(x)=tanx
2�1
2sinx
x�sinx
2d
dx��!1
2sec2x
2�cosx
2
1�cosx
2d
dx��!sec2x
2tanx
2+sinx
sinxd
dx��!sec2x
2tan2x
2+1
2sec4x
2+cosx
cosx!x!0���!3
2Exercise17.UseL'Hopital'srule.I(t)=E
R(1�e�Rt
L)limR!0I(t)=limR!0E(�1)(e�Rt=L)��t
L
1=
Et
L
Exercise18.c�k!0sincec!kk�c=uk�u=cf(t)=A(sinkt�sinct)
c2�k2=A(sin(kt)�sin(k�u)t)
�u(2k�u)==A(cos(k�u)t)(t)
�(2k�u)+u!�Atcoskt
2k7.17Exercises-Thesymbols+1and�1.ExtensionofL'Hopital'srule;Innitelimits;Thebehavioroflogxandexforlargex.Exercise15.UseL'Hopital'satthesecondtolaststep.limx!1�1(lnx)(ln(1�x))=limx!1�1ln(1�x)
1
lnx=limx!1�11
1�x(�1)
�1
(lnx)2�1
x=limx!1�1x
1�x(lnx)2=limx!1�12lnx�1
x
(�1)=
0
Exercise16.PersistinusingL'Hopital'sandtryingallpossibilitiessystematically.limx!0+xxx�1=limx!0+e(xx�1)lnx=limx!0+e(exlnx�1)lnx=explimx!0+lnx
(exlnx�1)�1==explimx!0+1=x
(�1)(exlnx�1)�2(exlnx(lnx+1))=exp�limx!0+e2xlnx�2exlnx+1
exlnx(xlnx+x)==exp�limx!0+exlnx(lnx+1)+e�xlnx(�lnx�1)
(lnx+1+1)=exp�limx!0+exlnx�e�xlnx
1+1
lnx+1=
1
Exercise17.134
limx!0+(xxx�1)=limx!0+exxlnx�1=limx!0+eexlnxlnx�1=elimx!0+exlnxlnx�1==eelimx!0+xlnxlimx!0+lnx�1=0�1=�1Weusedlimx!0+x logx=08 �0sincet=1
x;x logx=�logt
t !0ast!1.Exercise18.limx!0�esinxln(1�2x)=explimx!0�ln(1�exln2)
1=sinx=exp limx!0�1
1�exln2��ln2exln2
�1
sin2xcosx!==explimx!0�(sin2x)ln2exln2
(1�exln2)cosx=exp(ln2)limx!0�2sinxcosx
�ln2exln2=
1
Exercise19.limx!0+e1
lnxlnx=
e
Exercise20.Attheend,L'Hopital'scouldbeusedtoverifythatindeedsinxlnsinx!0asx!0.limx!0+esinxlncotx=elimx!0+sinx(lncosx�lnsinx)=elimx!0+�sinxlnsinx=
1
Exercise21.RewritetanintosinandcosanduseL'Hopital's.limx!
4(tanx)tan2x=limx!
4etan2xlntanx=explimx!
41
cos2x(lnsinx�lncosx)==explimx!
41
sinxcosx�1
cosx(�sinx)
�2sin2xcos2x=explimx!
41
�sin22x=
e�1
Exercise22.Exercise23.UseL'Hopital'stheorem,takingderivativesoftopandbottom.limx!0+expe
1+lnxlnx=expelimx!0lnx
1+lnx=expelimx!01=x
1=x=eeExercise24.RewritetanintosinandcosandtakeoutsinsincewecoulddothelimitbeforedoingL'Hopital's.limx!1(2�x)tan(x=2)=limx!1etanx
2ln(2�x)=elimx!1sinx=2ln(2�x)
cosx=2=explimx!1(�1)
2�x

2sinx=2=
exp�2

Exercise26.limx!+1expxlnx+c
x�c=explimx!+1ln1+c=x
1�c=x
1=x=explimx!+11
(1+c=x
1�c=x)(x�c)�(x+c)
(x�c)2
�1
x2==exp(2c)=4=)
c=ln2
Exercise27.(1+x)c=exp(cln(1+x))=exp(c(x�o(x)))=1+c(x�o(x))+o(x�o(x))=1+cx+o(x)x21+1
x21=2�x2=x21+1
x21=2�x2Letx2=1
t.Sot!0asx!+1.=)(1+t)1=2�1
t=1+1
2t�1+o(t)
t=
1
2
135
7.17Exercises-Thesymbols+1and�1.ExtensionofL'Hopital'srule;Innitelimits;Thebehavioroflogxandexforlargex.Exercise1.limx!0e�1=x2
x1000=limu!1e�uu500=limu!1u500
eu=0u=1
x2x2=1
ux1000=1
u500WherewehadusedTheorem7.11,whicharetwoveryusefullimitsforlogandexp.Theorem26.Ifa;b�0,limx!+1(logx)b
xa=0(17)limx!+1xb
eax=0(18)Proof.Trick-usethedenitionofthelogarithmasanintegral.Ifc�0;t1;tc1=)tc�1t�1.0lnx=Zx11
tdtZx1tc�1dt=1
c(xc�1)xc
c=)0(lnx)b
xaxcb�a
cbChoosec=a
2b;xcb�a
cb=x�a=2
cb!0asx!1then(lnx)b
xa!0asx!0Forexp,Lett=ex.lnt=x.xb
eax=(lnt)b
ta!0ast!1asx!1.Exercise2.limx!0sin1
x
arctan1
x=limx!01
x�o�1
x
1
x�o�1
x=
1
Exercise3.UseL'Hopital's.limx!
2tan3x
tanx=limx!
2cosx
cos3x=limx!
2�sinx
�3sin3x=
�1
3
Exercise4.UseL'Hopital's.limx!1ln(a+bex)
p
a+bx2=limx!11
a+bex(bex)
bx
p
a+bx2=limx!11
ae�x+b
1
p
a
x2+b=1
p
bExercise5.Makethesubstitutionx=1
t.limx!1x4cos1
x�1+1
2x2=limt!11
t4cost�1+t2
2=limt!1t4=4!+o(t4)
t4=1
4!=1
120Exercise6.limx!lnjsinxj
lnjsin2xj=limx!1
sinxcosx
2
sin2xcos2x=�1
2limx!sin2x
sinx=�1
2limx!2cos2x
cosx=
1
Exercise7.limx!1
2�ln(1�2x)
tanx=limx!1
2�1
1�2x(�2)
(sec2x)=limx!1
2��2(cos2x)
(1�2x)==�2
limx!1
2�2cosx�sinx
�2=1137
A(x)=Zx0P(t)dt=Zx0tantdt=�lnjcostjjx0=�lncosxy=e�A(x)ZxaQ(t)eA(t)dt+b=cosx Zbasin2te�lncostdt+2!==cosxZxa2sintdt+2=�2cos2x+4cosxExercise4.y0+xy=x3.y=0;x=0.A(x)=1
2x2y=e�1
2x2Zx0t3et2
2dt+0=e�x2
2t2et2
2�2et2
2 x0==x2�2+2e�x2
2Exercise5.y0+y=e2t.y=1;t=0.A=xy(x)=e�xZx0e2tetdt+1=e�xe3t
3 x0+1=
e2x
3+2
3e�x
Exercise6.y0sinx+ycosx=1;(0;).=)y0+cotxy=cscxA(x)=Zxacottdt=lnsinx
sinay(x)=e�ln(sinx
sina)Z(csct)eln(sint
sina)dt+b=sina
sinxx�a
sina+bindeedy=x�a
sinx+bsina
sinxx!0fory=x�a
sinx+bsina
sinx=x+bsina�a
sinx
bsina=a
forx!0
a�bsina=
forx!Exercise7.x(x+1)y0+y=x(x+1)2e�x2=)y0+1
x(x+1)y=(x+1)e�x2A(x)=Zxa1
t�1
t+1dt=lnx
a�lnx+1
a+1eA(x)=(a+1)x
a(x+1)y=a(x+1)
(a+1)xZxa(t+1)e�t2(a+1)t
a(t+1)dt+b=(x+1)
xZxate�t2dt+ba(x+1)
(a+1)x==x+1
2xe�a2�e�x2+a(x+1)b
(a+1)xIt'seasytoseethatthelastequationabovegoesto0asx!�1.y=(e�a2�e�x2)(1=2)(1+1
x)+ab
a+1(1+1
x)=limx!0y=1
x e�a2�e�x2
2+ab
a+1!
a=0
139
Sothenf0+x�1
xf=�1
xA(x)=ZP(t)dt=Zt�1
tdt=1
2x2�lnx�1
2a2�lnaeA(x)=ex2�a2
2a
x;e�A(x)=x
aea2�x2
2
f(x)=x
aea2�x2
2Zxa�1
t2aet2�a2
2dt+b
Exercise13.v=ykv0=kyk�1y0v0+kPv=kQ=kyk�1y0+kPyk=kQwherek=1�n=)y0y�n+Py1�n=Q=)
y0+Py=Qyn
Exercise14.y0�4y=2exy1=2n=1
2k=1�1
2=1
2;v=y1=2;v0+1
2(�4)v=1
2(2ex)=v0�2v=exA(x)=ZxaP(t)dt=�2(x�a)=�2xv=e2xZete�2tdt+b=e2x��e�x+b=
be2x�ex
=)y=(1+p
2)2e4x�2(1+p
2)e3x+e2xy(x)=b2�2b+1=2b=1+p
2Exercise15.y0�y=�y2(x2+x+1);n=2k=1�n=�1;v=ykv=y�1.v0+kPv=kQv0+(�1)(�1)v=(�1)(�(x2+x+1))=v0+v=x2+x+1A(x)=ZP(t)dt=Zx01dt=xv=e�xZ(t2+t+1)etdt+b=e�x�t2et�tet+2et x0+be�x=(x2�x+2)�(2e�x)+be�xy=1
x2�x+2�2e�xExercise16.v0+�1
xv=2x2v=elnxZ2x2e�lnxdx+C=x(x2+C)=x3+CxThensincev=yk;k=1�n,y=(x3+Cx)2=x2(x2+C)2;x6=0
y=x2(x2�1)2
Check:y0=2x(x2�1)2+x3(4)(x2�1)2x2+x4(4)(x2�1)�2x2(x2�1)2=4x3Exercise17.xy0+y=y2x2logxon(0;+1)withy=1
2whenx=1
2;x6=0.141
Ifuisaknownsolution,y=u+1
visalsoasolutionifvsatisesarst-orderODE.(u+1=v)0=u0+�1
v2v0y0+Py+Qy2=R=)u0+�v0
v2+P(u+1
v)+Q(u2+2u
v+1
v2)=R=)
v0�Pv=Q(2uv+1)
Exercise20.y0+y+y2=2;y=1;�2.(1)If�2b1,y0+y+y2=2P=1;Q=1;R=2v0+(�P�2Qu)v=Qy=u+1
vu=1v0+(�1�2(1)(1))v=v0�3v=Q=1v=e3xZ1e�3tdt+b=e3xe�3t
�3 xa+b=be3x�1
3�1�e3x�3ay=1+3
3be3x�(1�e3x�3a)y(0)=1+3
3b�(1�e�3a)=)b=1+1
bb2�b�1=0b=1p
5
2
y=1+3
3be3x�(1�e3x)b=1�p
5
2
(2)u=�2v0+(�1�2(1)(�2))v=v0+3v=1v=e�3xZxae3tdt+b=e�3xe3x�e3a
3+be�3x=1�e3a�3x
3+be�3xy=�2+3
1�e3a�3x+3be�3x;y(0)=�2+3
1�e3a+3ba=0��!y(0)=�2+3
3b=)b=�1p
2
b1orb�2;y=�2+3
1�e�3x+3be�3xb=�1p
2
8.7Exercises-Somephysicalproblemsleadingtorst-orderlineardifferentialequations.Exercise3.(1)y0=� y(t).y(T)=y0e� T=y0
n.n=e TsotherelationshipbetweenTandndoesn'tdependupony0.
1
klne=T
.143
(4)M=M(t)=M0� t =1
10y=ektZ�kMe�ktdt+b=ektZt0�k(M0� u)e�kudu+b==�kektZt0(M0e�ku� ueku)du+bekt==�ke M0e�ku
�k t0� ue�ku
�k+�e�ku
k2 t0!+bekt==�kektM0
k(1�e�kt)� te�kt
�k�e�kt
k2+1
k2+bekt=y(t)=�M0ekt+M0� t� =k+ ekt=k+bekt=(�M0+ =k+b)ekt+(M0� t� =k)=
y(t)=(140+3
(ln3�ln7))e�kt+(60�t
10�3
ln3=7)
Exercise8.y0(t)=�k(y�M0);y0+ky=kM0.y=e�ktZtftikM0ekudu+b=e�ktf�M0(ektf�ekti)+b=M0(1�e�k(tf�ti))+be�ktf=)y(tf)�M0=�M0e�k(tf�ti)+be�ktf65�M0=�M0e�k(5)+75e�k(5)=(75�M0)e�5k=)ln65�M0
75�M0=�5k;k=1
5ln75�M0
65�M060�M0=�M0e�k(5)+65e�k(5)=(65�M0)e�5k=)ln60�M0
65�M0=�5k=)65�M0
75�M0=60�M0
65�M0
M0=55
Exercise9.Lety(t)=absoluteamountofsalt.Waterisleavingaccordingtow(t)=w0+(3�2)t=w0+t.Saltleaving=2gal
min:y(t)salt
w(t)Sotheny0=�2y
w0+t=)lny=�2(ln(w0+t)�lnw0)=�2lnw0+t
w0istheequationofmotiongivenbytheproblem.y(t)=Celnw0+t
w0�2=Cw0+t
w0�2
y(t)=50100
100+t2
y(t=60min:)=50100
1602=5025
64=625
32'19:53Exercise10.Letybethedissolvedsalt(totalamountof)atttime.The(total)amountofwateratanygiventimeinthetankisw=w0+t.Thereisdissolvedsaltinmixturethatisleavingthetankatanyminute.Thereisalsosaltfromundissolvedsaltinthetankthatis“cominginto”thedissolvedsalt,addingtotheamountofdissolvedsaltinthemixture.Thusy0(t)=(�2)y
w+ y
w�3; =�1gal
3minWeobtained easilybyconsideringonlythedissolvingpartandhowitdissolves1poundofsaltperminuteifthesaltconcentration,y
wwaszero,i.e.waterisfresh.145
dx
dt=kx(M�x)=kMx�kx2;=)dx
kMx�kx2=dt=1=kdx
x(M�x)=dt=)kdt=1
x+1
M�x1
Mdx=lnx+�ln(M�x)
MMk(t�ti)=lnx
M�x;eMk(t�ti)(M�x)=xx(t)=MeMk(t�ti)
1+eMk(t�ti)=
M
1+e�Mk(t�ti)
Exercise14.Notethatwearegiventhreeequallyspacedtimes.M=x2+x2e� (t2�t0);M�x2
x2=e� (t2�t0)M�x2
x2x1
M�x1=e� t2+ t0+ t1� t0=e� (t2�t1)M�x3
x3x2
M�x2=e� (t3�t2)=M�x2
x2x1
M�x1(M�x3)(M�x1)x22=x1x3(M�x2)2=x1x3(M2�2Mx2+x22)=x22(M2�M(x1+x3)+x1x3)(x22�x1x3)M2=M(x22(x1+x3)+�2x2x1x3)=(�x1(x3�x2)+x3(x2�x1))x2=)
M=x2(x3(x2�x1)�x1(x3�x2))
x22�x1x3
Exercise15.dx
dt=k(t)Mx�k(t)x2dx
Mx�x2=k(t)dt=)MZttik(u)du=lnx
M�xx
M�x=eMRttik(u)dux=MeMRttik(u)du
1+eMRttik(u)du=
M
1+e�MRttik(u)du
Exercise16.(1)M=2392(23�3:9)�3:9(92�23)
232�3:9(92)=201(2)M=122150(122�92)�92(150�122)
(122)2�92(150)=122150(30)�92(28)
(122)2�92(150)=216(3)Reject.8.14Exercises-Linearequationsofsecondorderwithconstantcoefcients,Existenceofsolutionsoftheequationy00+by=0,Reductionofthegeneralequationtothespecialcasey00+by=0,Uniquenesstheoremfortheequationy00+by=0,Completesolutionoftheequationy00+by=0,Completesolutionoftheequationy00+ay0+by=0.Exercise1.y00�4y=0y=c1e2x+c2e�2x.Exercise2.y00+4y=0y=c1cos(2x)+c2sin(2x).UseTheorem8.7.Theorem27.Letd=a2�4bbethediscrimnantofy00+ay0+by=0.Then8solutionson(�1;1)hastheform(22)y=e�ax=2(c1u1(x)+c2u2(x))where(1)Ifa=0,thenu1(x)=1andu2(x)=x(2)Ifd�0,thenu1(x)=ekxandu2(x)=e�kx,wherek=p
d
2(3)Ifd0,thenu1(x)=coskxandu2(x)=sinkx;wherek=1
2p
�d147
Exercise15.y00�4y0+29y=0d=16�4(1)(29)=�100=)u=e2x(c1sin5x+c2cos5x).v:y00+4y0+13y=0d=10�4(13)=�36=)v=e�2x(b1sin3x+b2cos3x)v(0)=b2u(0)=1(0+c2)=c2=0u=e2xc1sin5xu0=2e2xc1sin5x+e2xc15cos5xu0
2=1=2ec1(1)c1=1
2eu0(0)=1
2e5u0(0)=v0(0)=)b1=5
6ev=e�2xb1sin3xv0(0)=3b1
u=1
2ee2xsin5x
v=e�2x5
6esin3x
Exercise16.y00�3y0�4y=0u9�4(1)(�4)=25u=e3x
2(c1e5x
2+c2e�5x
2)y00+4y0�5y=0v16�4(1)(�5)=36v=e�2x(b1e3x+b2e�3x)u(0)=c1+c2=0v(0)=b1+b2=0=)u=2e3x
2c1(sinh5x
2)v=2b1e�2x(sinh(3x))u0=c13
2e3x
2(e5x
2�e�5x
2)+2e3x
2c15
2cosh5x
2u0(0)=5c1v0=�4b1e�2xsinh(3x)+6b1e�2xcosh(3x)v0(0)=6b1c1=6b1
5Exercise17.y00+ky=0d=�4(k)p
d
2=p
�4k
2=p
�kAssumek�0y=c1sinp
kx+c2cosp
kxy(0)=c2=0y(1)=c1sinp
k1=0=)p
k=nk0;�k=&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;ރ ;� Td;&#x [00;0y=c1ep
x+c2e�p
x;y(0)=c1+c2=0y=c1sinhp
xy=c1sinhp
1=0c1=0soifk0,therearenonontrivialsolutionssatisfyingfk(0)=fk(1)=0Exercise18.y00+k2y=0d=�4k20p
d
2=2k
2=k�0149
(2)u1=e2xu01=2e2xu001=4e2xu2=xe2xu02=e2x+2xe2xu002=2e2x+2e2x+4xe2x=4e2x+4xe2x=4e2x(1+x)u002�4u02+4u2=0=)
y00�4y0+4y=0
(3)u1(x)=e�x=2cosx;u01=�1
2e�x=2cosx+�e�x=2sinxu001=1
4e�x=2cosx+e�x=2sinx+�e�x=2cosx=�3
4e�x=2cosx+e�x=2sinxu2(x)=e�x=2sinxu02=�1
2e�x=2sinx+e�x=2cosxu002=1
4e�x=2sinx+�e�x=2cosx�e�x=2sinx==�3
4e�x=2sinx�e�x=2cosxu002+u02+5
4u2=0
y00+y0+5
4y=0
(4)u1(x)=sin(2x+1);u2(x)=sin(2x+2)u01=2cos(2x+1)u001=�4sin(2x+1)u02=2cos(2x+2)u002=�4sin(2x+2)
y00+4y=0
(5)u1=coshxu01=sinhxu001=coshx
y00�y=0
Exercise21.w=u1u02�u2u01.(1)w=08x2openintervalI,u2
u10=u02u1�u01u2
u21=0=)u2
u1=cIfu2
u1isnotconstant,thenw(0)6=0foratleastonecinI(otherwise,it'dbeconstant).(2)w0=u1u002�u2u001Exercise22.(1)w0+aw=u1u002�u2u001+a(u1u02�u2u01)=u1(�bu2)+�u2(bu1)=0w(x)=w(0)e�axifw(0)6=0,thenw(x)6=08x.(2)u16=0Ifw(0)=0;w(x)=08x,sou2
u1constant.Ifu2
u1constant,w(0)=0sincefromthepreviouspart.Exercise23.RecallthepropertiesoftheWronskian.(1)IfW(x)=v1(x)v02�v2v01=v1v02�v2v01=08x2I,thenv2
v1constantonI(2)W0=v1v002�v2v001(3)W0+aW=0ifv1;v2aresolutionstoy00+ay0+by=0W(x)=W(0)e�axSoifW(0)6=0;W(x)6=08x151
Consideraddingtogetherthesolutionsandthesolution'sderivativesintosomefunctionf.Bythelinearityofthedifferentialequation,weknowthatfisalsoasolutionsinceitisalinearsuperpositionofsolutions.y(x)=Av1(x)+Bv2(x)f(0)=Av1(x)+Bv2(0)y0(x)=Av01(x)+Bv02(x)f0(0)=Av01(0)+Bv02(0)v1(0)v2(0)v01(0)v02(0)AB=f(0)f0(0)=)1
W(0)v02(0)�v2(0)�v01(0)v1(0)f(0)f0(0)=ABsinceW(0)6=0;thisdivisionisallowedaboveso
y(x)=v02(0)f(0)�v2(0)f0(0)
W(0)v1(x)+v1(0)f0(0)�v01(0)f(0)
W(0)v2(x)
f(0);f0(0)areinitialconditionsfory.f(0);f0(0)arearbitrary.ButsinceW(0)6=0,W(0)=v1(0)v02(0)�v2(0)v01(0),wecandothingslikef(0)=v02(0)f(0)�v2(0)f0(0)
v1(0)v02(0)�v2(0)v01(0)v1(0)+v2(0)f0(0)�v01(0)f(0)
W(0)v2(0)8.17Exercises-Nonhomogeneouslinearequationsofsecondorderwithconstantcoefcients,Specialmethodsfordeterminingaparticularsolutionofthenonhomogeneousequationy00+ay0+by=R.Exercise1.y00�y=xhomogeneoussolutionc1ex+c2e�x.yp=�xy=c1ex+c2e�x�xExercise2.y00�y0=x2Forthehomogeneoussolutiony00�y0=0d=(�1)2�4(1)(0)=1yh=ex
2�c1ex
2+c2e�x
2=c2ex+c1yp=Ax3+Bx2+Cx+Dy0p=3Ax2+2Bx+Cy00p=6Ax+2B
y=c1+c2ex+�1
3x3+x2
Exercise3.y00+y0=x2+2xe�x
2�c1e�x
2+c2ex
2=c1e�x+c2P=Ax3+Bx2+Cx+D;P0=3Ax2+2Bx+CP00=6Ax+2B3Ax2+2Bx+C+6Ax+2B=x2+2xA=1
3B=0C=0
y=c1e�x+c2+1
3x3
Exercise4.y00�2y0+3y=x3u=ex(c1sinp
2x+c2cosp
2x)3(Ax3+Bx2+Cx+D)2(3Ax2+2Bx+C)(6Ax+2B)A=1
3B=2
3C=8
9D=16
27y=C1exsinp
2x+C2excosp
2x+1
3x3+2
3x2+8
9x+16
27Exercise5.y00�5y0+4y=x2�2x+1yh=e5x
2e3x
2+e�3x
2=c1e4x+c2ex152
y00+4y=e�2xyh=c1sin2x+c2cos2xw=sin2x�sin2x(2)�2cos2xcos2x=2t1=�Zcos2xe�2xdx
2=�1
2Ze�2xcos2xdxt2=Zsin2xe�2x
2=1
2Ze�2xsin2xdx(e�2xcos2x)0=�2e�2xcos2x+�2e�2xsin2x(e�2xsin2x)0=�2e�2xsin2x+2e�2xcos2xe�2xsin2x�e�2xcos2x
40=e�2xcos2xe�2xsin2x+e�2xcos2x
�40=e�2xsin2xt1=e�2xcos2x�e�2xsin2x
8t2=e�2xsin2x+e�2xcos2x
�8yp=e�2xsin2xcos2x�e�2xsin22x
8+e�2xsin2xcos2x+e�2xcos22x
�8=
e�2x
8
y=c1sin2x+c2cos2x+e�2x
8
Exercise9.y00+y0�2y=exd2=1�(4)(1)(�2)=9yh=e�1x
2e3x
2+e�3x
2=ex+e�2x(xex)0=ex+xex+(xex)00=+(2ex+xex)=)3ex+2xex
y=c1ex+c2e�2x+1
3xe2
Exercise10.y00+y0�2y=e2x.yh=e�x
2c1e3x
2+c2e�3x
2=c1ex+c2e�2x.W(x)=v1v02�v2v01=ex(�2)e�2x�e�2xex=�3e�xt1=�Zv2R
Wt2=Zv1R
Wt1=�Ze�2xe2x
�3e�x=1
3ext2=Zexe2x
�3e�x=�1
12e4xy1=t1v1+t2v2=1
3exex+�1
12e4xe�2x=1
4e2xy=c1ex+c2e�2x+1
4e2xExercise11.y00+y0�2y=ex+e2x.ConsidersolutionstoExercise9,10.L(ya)=ex;L(yb)=e2x;L(ya+yb)=ex+e2x=)
y=c1ex+c2e�2x+1
3xex+1
4e2x
Exercise12.y00�2y0+y=x+2xex.d=4�4(1)=0.Recallthedenitiontobelearnedforthissectionofexercises:Theorem29.Letd=a2�4bbethediscriminantofy00+ay0+by=0.Theneverysolutionofthisequationon(�1;1)hastheform(24)y=e�ax=2(c1u1(x)+c2u2(x))(1)Ifd=0thenu1=1;u2=x(2)Ifd�0,u1=ekx;u2=e�kx;k=p
d
2(3)Ifd0,u1=coskx;u2=sinkx;k=p
�d
2154
Discriminant:�1p
12�4(�2)
2=�2;1=)yh=c1ex+c2e�2x=)W=ex(�2)e�2x�e�2xex=�3e�xt1=�Ze�2tet
1+et
�3e�t=1
3Z1
1+et=�1
3ln(1+e�x)t2=Zetet
1+et
�3e�t=1
�3Ze3t
1+etu=et���!�1
3Zu2du
1+u==�1
3Zu+�u
u+1=�1
3Zu+�1+1
u+1=�1
3(1
2u2�u+lnu+1)=�1
6e2x+ex
3+�1
3ln(ex+1)
y1=�1
3exln(1+e�x)+�1
6+e�x
3�e�2x
3ln(ex+1)+c1ex+c2e�2x
Exercise17.y0+6y0+9y=f(x);wheref(x)=1for1x2.f(x)=0forallotherx.d=36�4(1)(9)=0yh=e�3x(c1+c2x)=c1e�3x+c2xe�3xW(x)=e�3x(e�3x�3xe�3x)�(xe�3x)(�3e�3x)=e�6xt1(x)=8��&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:a1xRx1�te3tdt=�te3t
3+e3t
9 x1=�3xe3x+e3x
9+2
9e3a12xR21�te3tdt=�6e6+e6
9+2e3
9=�5e6
9+2e3
91ax2Rxa�te3tdt=�te3t
3+e3t9 xa=�xe3x
3+e3x
9+ae3a
3�e3a
9t2(x)=Ze�3tf(t)
e�6t=Ze3tf(t)=Zxae3t=1
3�e3x�e3ay1=e�3x�3xe3x+e3x
9+C+x
3
y=c1e�3x+c2xe�3x+1
9when1x2;otherwisey=yh
Exercise18.Startfromy00�k2y=R(x).SupposeL(yp)=y00p�k2yp=R(x).yh=c1sinh(kx)+c2cosh(kx);L(yh)=0SoconsiderL(yp+yh)=R(x);yp+yh=y1isanotherparticularsolution.ThekeytothisproblemistoapplytheintegrationdirectlyontheODEitself,nottogotheotherwayaroundbydifferenti-atingthesupposedparticularsolution.Rx0dtsinh(k(x�t))������������!Zx0dtd2y
dt2(t)sinh(k(x�t))�k2Zx0dty(t)sinh(k(x�t))=Zx0dtR(t)sinh(k(x�t))dtZx0y00sinh()=�y0(0)sinh(kx)�Zy0cosh()(�k)==�y0(0)sinh(kx)+k(y(x)�y(0)cosh(kx)+kZy(t)sinh())==�y0(0)sinh(kx)+ky(x)�ky(0)cosh(kx)+k2Zy(t)sinh()=)y(x)�y0(0)sinh(kx)
k�y(0)cosh(kx)=1
kZx0dtR(t)sinh(k(x�t))dtNownotethatL(yh)=0,soapplyingRx0dtsinh(k(x�t))resultsin0still.156
ItcouldbeshownwithsomecomputationthatthisparticularsolutionsatisestheODEwithouthavingtoaddorsubtractpartsofahomogeneoussolution.Exercise20.y00+y=sinxyh=c1sinx+c2cosx=)W(x)=�s2�c2=�1t1=�Zcs
�1=�cos2x
4;t2=Zss
�1=�Z1�cos2x
2=�x
2�sin2x
4yp=�sinxcos2x
4+sin2x�2x
4cosx=sinxcos2x+sin3x�2xcosx
4
y=c1sinx+c2cosx+sinxcos2x+sin3x�2xcosx
4
Exercise21.y00+y=cosxyh=c1sinx+c2cosx=c1S+c2CW(x)=�1t1=Z�CC
�1=Z1+cos2x
2=x+sin2x
2
2;t2=ZSC
�1=cos2x
4yp=xsinx
2+sin2xsinx
4+cosxcos2x
4=)
y=xsinx
2+sin2xsinx
4+cosxcos2x
4+c1sinx+c2cosx
Exercise22.y00+4y=3xcosxyh=c1sin2x+c2cos2xW(x)=�sin22x(2)+�cos22x(2)=�2t1=Z�cos(2x)(3xcosx)
�2=3
2Zxcosxcos(2x)=3
2Zxc(1�2s2)=3
2Zxc�3Zxcs2==3
2(xs+c)�3Zxc3
30=3
2(xs+c)�(xs3�Zs3)=3
2(xs+c)�xs3+Zs(1�c2)==3
2(xs+c)�xs3+�c+1
3c3=3
2xs+c
2�xs3+1
3c3t2=Zsin(2x)(3xcosx)
�2=�3Zxsc2=Zx(c3)0=xc3�Zc3=xc3�Zc(1�s2)==xc3�s+1
3s3yp=3
2xs+c
2�xs3+1
3c3(2sc)+(xc3�s+s3
3)(1�2s2)=(lotsofalgebra)==
xc2+2
3s
=)
y=c1sin2x+c2cos2x+xsinx�2
3cosx
Remember,persistenceiskeytoworkthroughthealgebra,quickly.Exercise23.y00+4y=3xsinx.Fromtheworkabove,wecouldguessatthesolution.(xs)0=s+xc(xs)00=2c+�xs(xs)00+4(xs)=2c�xs+4xs=2c+3xs(c)00+4c=3c=)�2
3c00+4�2
3c=�2c=)
yh=xs�2
3c
y=xsinx�2
3cosx+c1sin2x+c2cos2x
158
Exercise5.y=Ccos(x+ )y0=Ccos(x0+ )y0=�Csin(x+ )=v0v20+y20=C2sin2(x+ )+C2cos2(x0+ )=C2=)
C=q
v20+y20
Exercise6.y=Ccos(kx+ )y(0)=Ccos( )=1y00(0)=�k2Ccos( )=�12y0=�kCsin(kx+ )y0(0)=�kCsin( )=2y00(0)
y(0)=�k2=�12
1=)k=2p
3y0(0)
y(0)=�kCsin( )
Ccos( )=2
1=2=�ktan( )=)
=�
6
Exercise7.k=2
3y=�Csin(kx)
y=�Csin2x
3
;C�0Exercise8.Let'srstsolvethehomogenousequation.y00+y=0yh=C1sinx+C2cosxW(x)=�S2�C2=1t1=Zx0�cost(1)
�1=sinxt2=Zx0(sint)(1)
�1=(cosx�1)for0x2otherwise,forx�2;t1=0;t2=0y1=sin2x+cos2x+(1�cosx)y0(x)=C1cosx+sinxy(0)=0=c2y0(0)=c1=1y=sinx+(1�cosx)=)
I(t)=sint+(1�cost)
0t2Exercise9.(1)Considerlarget.ThenI(t)=F(t)+Asin(!t+ )!Asin(!t+ )I=AS(!t+ )=A(S(!t)C( )+C(!t)S( ))I0=!AC(!t+ )=!A(C(!t)C( )�S(!t)S( ))I00=�!2AS(!t+ )=�!2A(S(!t)C( )+C(!t)S( ))I00+RI0+I=I00+I0+I==A((�!2C( )+�!S( )+C( ))S(!t)+(�!2S( )+!C( )+S( ))C(!t))=S(!t)=)tan( )=�!
1�!2Withtan =�!
1�!2andthetrigidentitiest2+1=sec2,1
C2=sec2,andS2+C2=1,wecangetC=1�!2
p
1�!2+!4S=�!
p
1�!2+!4NotethatthesignofSisxedbytan.A(�!2C( )+�!S( )+C( ))S(!t)=S(!t)=)A=1
(1�!2)C( )�!S( )=D
(1�!2)2�!(�!)=)
A=1
p
!4�!2+1
160
Exercise12.MvR=(M�m)(vR(t+h))+0M(vR(t+h)�vR(t))=(m)vR(t+h)Mv0R=k
gvR=)v0R
vR=k
g(M0�kt
g)=k
M0g(1�kt
M0g)lnvR=(k=w)ln(1�kt
w)�w
k=�ln(1�kt
w)vR=v0
1�kt
w=)x(t)=v0�w
kln(1�kt
w)=�v0w
kln(1�kt
w)8.22Exercises-Remarksconcerningnonlineardifferentialequations,Integralcurvesanddirectionelds.Exercise1.2x+3y=C=)y0=�2
3Exercise2.y=Ce�2x=)y0=�2yExercise3.x2�y2=c=)yy0=x=)y0=x
y;y6=0Exercise4.xy=c=)y0=�y
x;x6=0Exercise5.y2=cx=)y2
x=c=)y0=y
2xx6=0Exercise6.x2+y2+2Cy=1x2
y+y�1
y=�2C2xy�y0x2
y2+y0+1
y2y0=0y0=�2xy
1+y2�x2Exercise7.y=C(x�1)exy
(x�1)ex=Cy0(x�1)ex�(ex+(x�1)ex)y
(x�1)2e2x=0
y0=xy
x�1
Exercise8.y4(x+2)=C(x�2)y4(x+2)
x�2=C(4y3y0(x+2)+y4)(x�2)�y4(x+2)
(x�2)2=0=)4y3y0(x+2)+y4=y4(x+2)
x�2
y0=y
(x�2)(x+2)
Exercise9.y=ccosx=)y0=�tanxyExercise10.arctany+arcsinx=C1
1+y2y0+1
p
1�x2=0=)
y0=�(1+y2)
p
1�x2
Exercise11.Allcirclesthroughthepoints(1;0)and(�1;0).Startwiththecircleequation:(x�A)2+(y�B)2=R2162
Exercise7.(1�x2)1=2y0+1+y2=0arctany=arccosx+CExercise8.xy(1+x2)y0�(1+y2)=01
2ln(1+y2)=Z1
x�x
1+x2+C=lnx�1
2lnj1+x2jy2=kx
p
1+x22Exercise9.(x2�4)y0=ylny=�1
2arctanhx
2sinceZ1
x2�4dx=1
4Zdx
�x
22�1==1
2Zdu
u2�1(whereu=x
2)(tanh(u))0=cosh2u�sinh2u
cosh2u=1�tanh2u=)
y=kexp(�1
2arctanhx
2)
Exercise10.xyy0=1+x2+y2+x2y21
2ln(1+y2)=lnx+1
2x2+C=)
y2=kx2ex2�1
Exercise11.yy0=ex+2ysinxye�2y
�2�e�2y
4=exsinx�excosx
2+C(2y+1)e�2y=�2ex(sinx�cosx)+CExercise12.xdx+ydy=xy(xdy�ydx)y(1�x2)dy=x(�y2�1)dxydy
1+y2=xdx
x2�1=)1
2lnj1+y2j=1
2lnjx2�1j+C1+y2=(x2�1)K=)
y2=K(x2�1)�1
Exercise13.f(x)=2+Rx1f(t)dtf0(x)=f(x)=)f(x)=Cex=)Cex=2+Cex�Ce1C=2
e
f(x)=2
eex
Exercise14.f(x)f0(x)=5xf(0)=1f(x)2=5x2+C=)f(x)=p
5x2+C
f(x)=p
5x2+1
Exercise15.f0(x)+2xef(x)=0f(0)=0164
y0=x2+2y2
xy=x
y+2y
xv=y
x=)v+xv0=1
v+2v=)v0
1
v+v=1
x1
2lnj1+v2j=lnx+C=)
y2=(Cx2�1)x2
Exercise5.(2y2�x2)y0+3xy=0if2y26=x2;y0=3xy
x2�2y2y=vxy0=v0x+v=)y0=v0x+v=3vx2
x2�2v2x2=3v
1�2v2=)1�2v2
2v(1+v2)v0=1
x1
21
v+�3v
1+v2v0=1
x=)1
2lnv+�3
2ln(1+v2)=lnx+C=)v
(1+v2)3=Cx2=y=x
x2+y2
x23
yx3=C(x2+y2)3
However,y0=3xy
x2�2y2v=y
xy0=v0x+v=)v0x+v=3x2v
x2�2v2x2=3v
1�2v2v0x=3v
1�2v2�(v�2v3)
1�2v2=2(v+v3)
1�2v2=)1
v+�3v
1+v2v0=2
x=)lnv+�3
2lnj1+v2j=2lnx+Cv
(1+v2)3=2=Cx2=)y2=x4
(x2+y2)3=Cx4=)
y2=C(x2+y2)3
Exercise6.xy0=y�p
x2+y2=)y0=y
x=r
1+y
x2v=y
xvx=yv0x+v=v�p
1+v2=)v0x=�p
1+v2�v0
p
1+v2=1
x=)ln(v+p
1+v2)=lnx+Csince(ln(v+p
1+v2))0=1
v+p
1+v21+v
p
1+v2=1
p
1+v2v+p
1+v2=Cx=)1+v2=C2x2�2vCx+v2=)v=Cx
2�1
2Cx=)
y=Cx2
2�1
2C
Exercise7.x2y0+xy+2y2=0166
x+yy0+Cy0=0=)y0(y+C)=�xy0=�x
y+C=�x
y+1�x2�y2
2y=�2xy
y2�x2+1orthogonalcurves�����������!y0=y2�x2+1
2xy=1
2xy+1
21
x�xy�1RecognizethatthisisaRicattiequationandweknowhowtosolvethem.y0+�1
2xy=y�1�x
2+1
2xn=�1k=1�n=1�(�1)=2v=yk=y2v0+2�1
2xv=2
21
x�xA(x)=ZxaP(t)dt=Zxa�1
t=lna
xZxaQeA=Zxa1
t�ta
t=�a
x+1�a(x�a)
y2=v=�1+x
a�x(x�a)+bx
a
Exercise4.y2=Cx.y2
x=Cd=dx���!2yy0x�y2
x2=0y0=y
2x=)y0=1
��y
2x=�2x
y=)
y2+2x2=C
Exercise5.x2y=C.2xy+x2y0=0y0=�2y
x=)y0=x
2y1
2y2=x2
4+C
2y2�x2=C
Exercise6.y=Ce�2xe2xy=C=)2e2xy+e2xy0=0y0=�2yinvert���!y0=1
2y=)
y2=x+C
Exercise7.x2�y2=C2x�2yy0=0y0=x
y=)y0=�y
x=)lny=�lnx+C=)
y=C
x
Exercise8.ysecx=Cy0secx+ytanxsecx=0y0=�ytanx(invert)����!y0=1
ytanx
1
2y2=lnjsinxj+C
168
Exercise12.y=x
2k�1
2kxExercise13.y=f(x).nZx0f(t)dt=xf(x)�Zx0f(t)dt;(n+1)Zx0f(t)dt=xy=)(n1)y=y+xy0=)ny=xy0n
x=y0
ynlnx=lny
y=Cxn
ofy=Cx1=nExercise14.nZx0f2(t)dt=Zx0((y(x))2�(f(t))2)dt(n+1)Zx0f2(t)dt=xy2(x)=xy2;(n+1)f2(x)=y2+2xyy0y0=ny
2x=)lny=n
2lnx+C
y=Cxn=2
ofy=Cx1=2nExercise15.Zx0f2(t)dt=x2f(x)f2(x)=2xf+x2f0=)f0=f2�2xf
x2Thelefthandsideofthelastexpressionshownishomogeneous.Thusdothey=vxsubstitution.v0x+v=v2x2�2x2v
x2=v2�2vv0x
v2�3v=1=)v0
(v2�3v
)=1
x=1
31
v�3
�1
vv0ln(v�3
)�lnv=3lnx+Clnv�3=
v=3lnx+Cvx�3
xCvx4y�3
x=Cyx3=)
y=3x=
1+x3
2
Exercise16.A=Ra0f;B=R1afA�B=Ra0f+Ra1f=2f(a)+3a+bd=dx���!2f(a)=2f0(a)+3=)1=f0(a)
f(a)�3
2Sothena+C=ln(y�3
2);f(a)=Cea+3
2f(1)=0=Ce1+3
2;=)C=�3
2e
f(x)=�3
2ex�1+3
2
170
Adh
0�h1=2=dt=(A= 0)dh
1�
0h1=2ln1�ah1=20=1
1�ah1=2�a
21
h1=2(wherea=
0)(h1=2)0=1
2h1=21�ah1=2
1�ah1=2=�1
2�a
2h1=2
h1=2�1�ah1=2=)Z(A= 0)dh
1�
0h1=2=�(A= 0)2 20
2ln1�
0h1=2+2 0h1=2
 hfhi==T=)exp� 0
At�2 0
h1=2f�h1=2i2
2 20=1�
0h1=2f
1�
0h1=2it!1���!hf= 20
2=(100in3=s)2
c2(2)(32ft=s2)(5=3in2)2(12in=1ft)=(25=24)2Exercise20.V0=1
3R20H0V(h)=V0�1
3hhR0
H02=V0�1
3R20
H20h3=V0� h3mg(H0�h)=1
2mv2f;p
2g(H0�h)=vf(energyconservation)cA0vf=cA0s
2gH01�h
H0= r
1�h
H0dV
dt=� p
1�h=H=�3 h2dh
dt=)dh
dt=
3 h2r
1�h
HZh2=H20
q
1�h
H0= =H20
3 T==H0Zu2du
p
1�u=H0Z(1�y)2(�dy)
p
y=�H0Z1�2y+y2
p
y==�H02y1=2�2
3y3=2+2
5y5=2 hfhi==�H0 21�h
H01=2�4
31�h
H03=2+2
51�h
H05=2! hfhi==cA0p
2gH0=H20
31
3R20
H20TForhi=0;hf=H,H0(2(1)�4=3(1)+2=5)=H0(16=15)=cA0p
2gH0T=(R20);T=16
15p
H0R20
cA0p
2g=2
9R20p
H0
A0Exercise21.m2x�m+(1�x)=0=)(m2�1)x+1�m=0;=)m=1Exercise24.Givenfs.t.2f0(x)=f�1
xifx�0,f(1)=2andx2y00+axy0+by=0172
(3)Sincef0(x)2
3foreachx10
3,thenf!1forx!1(otherwise,fwouldhavetodecreasesomewhere,whichwouldcontradictthegivenfactaboutf).RewritetheDEfory0tobey0=2y2+x
3y2+5=)(3y2+5)y0=2y2+x=)(3+5
y2)y0=2+x
y2Considery0=2y2+x
3y2+5=2+x
y2
3+5
y2specically,x
y2.Nowymust,attheveryleast,havesomelinearincreasebecausewehadalreadyshownthaty02
3.Soy2wouldgotoinnityfasterthanlinearx.Thuslimx!1y0=2
3.Sothen(3+5
y2)y0x!1����!(3+0)2
3=2=2+x
y2.
0=x
y2
(4)y0=2+x
y2
3+5
y2x!1����!2
3.=)y=2
3xory
x=2
3.Exercise31.Givenafunctionfwhichsatisesthedifferentialequationxf00(x)+3x(f0(x))2=1�e�x(1)c6=0foranextrenum.cf00(c)+3c(f0(c))2=cf00(c)=1�e�c=)f00(c)=1�e�c
c�0(2)Cleverly,considerthelimit.xf00(x)+3x(f0(x))2=1�e�x=)f00(x)+3(f0(x))2=1�e�x
xx!0���!f00(0)+0=1Soacriticalpointatx=0wouldbeaminimum.(3)We'llhaveto“cheat”alittleandusetheideaofpowerseriesearlyonhere.f00+3(f0)2=1�e�x
xsuggeststhatweconsidertheTaylorseriesofe�x.1�e�x
x=�P1j=1(�x)j
j!
x=1Xj=0(�1)jxj
(j+1)!Thisfurthersuggeststhatfitselfhasapowerseriesrepresentationbecauseitsrstandsecondorderderivativesaresimplyacombinationofinnitelymanytermscontainingpowersofx.Thensupposef=P1j=0ajxj.f00+3(f0)2=1�e�x
x=)f0=1Xj=0(j+1)aj+1xjf00=1Xj=0(j+2)(j+1)aj+2xj=)1Xj=0(j+2)(j+1)aj+2xj+31Xj=01Xk=0(j+1)(k+1)aj+1ak+1xj+k=1Xj=0(�1)jxj
(j+1)!Iff(0)=0a0=0Iff0(0)=0a1=0Thenf=a2x2+P1j=3ajxj.Considerthex0termsintheDE.(f0)2doesn'tcontribute,becausef0'sleadingordertermisx1.Sothen2(1)a2+0=1=)
a2=1
2
i.e.f=1
2x2+1Xj=3ajxj
A=1
2
inorderforf(x)Ax2174
Z20einxe�inx=2=Z20(cosnx+isinnx)(cosnx�isinnx)==Z20cos2nx+sin2nxZ20einxeinx=0=Z20cos2nx�sin2nx+i(2cosnxsinnx)=)Z20cos2nx�sin2nx=0Z20cosnxsinnx=0Summingthetworesultsabove,weobtain2=Z202cos2nx=)Z20cos2nx=Thenalso,Z20sin2nx=Exercise8.z=rei=rei(+2m);m2Zz1=n=r1=nei(=n+2m=n)m=0;1;:::n�1=)z1=n=Rei m=z1mTherootsarespacedequallybyanangle2=ni=ei=2+i2n=)i1=3=ei=6;ei5=6;ei3=2i1=4=ei=8;e5i=8;e9i=8;e13i=8�i=e�i=2+i2n=)(�i)1=4=e�i=8;e3i=8;e7i=8;e11i=8Exercise9.eiueiv=ei(u+v)=cosu+v+isinu+v==(cosu+isinu)(cosv+isinv)=cosucosv�sinusinv+i(cosvsinu+cosusinv)=)sinu+v=cosvsinu+cosusinv=)cosu+v=cosucosv�sinusinvsin2z+cos2z=eiz�e�iz
2i2+eiz+e�iz
22==�(e2iz+e�2iz�2)+(e2iz+2+e�2iz)
4=1cosiy=eiiy+e�iiy
2==e�y+ey
2=coshysiniy=eiiy�e�iiy
2i==e�y�ey
2i=isinhyeiz=ei(x+iy)=eixe�y=(cosx+isinx)e�ye�iz=e�i(x+iy)=e�ixey=(cosx�isinx)eyThusitisclear,bymentallyaddingandsubtractingtheaboveresultsthat=)cosz=cosxcoshy�isinxsinhysinz=icosxsinhy+sinxcoshyExercise10.(1)Log(�1)=ilog(i)=ln1+i
2=i
2(2)Log(z1z2)=Log(jz1jjz2jei(1+2))=lnjz1jjz2j+i(1+2+2n)=Logz1+Logz2+i2n177
Exercise4.f(n)=1
5+3
5n�2
5n2!limn!1f(n)=1
5Exercise5.f(x)=x
2x=x
exp(xln2)!0sincelimx!1x
(ex) .Exercise6.f(n)=1+(�1)n=0of1.Thus,choosing1=j1�jLjj
2;jf(n)�Ljjjf(n)j�jLjj=j1�jLjj�1foranyn=2mExercise7.f(n)=1+(�1)n
n.Suppose=3
N.Soforn�N;1
N�1
n,nN=N()=3=.jf(n)j= 1+(�1)n
n 2
n3
n3
N=Exercise8.f(n)=(�1)n
n+1+(�1)n
2jf(n)�Lj= (�1)n
n+1+(�1)n
2�L  L�(�1)n
n � 1+(�1)n
2  jLj� (�1)n
n � 1+(�1)n
2 = jLj�1
n = 1+(�1)n
2  jLj�1
n �1
2 Thus,consider jLj�1
n �1
2 � 1
N�jLj �1 =0forn�NExercise9.f(x)=exp�1
xln2;limx!1f(x)=0.Exercise10.jf(n)�Lj=jn(�1)n�Ljjjn(�1)nj�jLjj=jjnj�jLjj=jnj�jLj�N�jLjThus,forn�N,N()=+jLj,sothenjf(n)�Lj�.Exercise11.f(n)=n2=3sinn!
n+1.jf(n)j= n2=3sin(n!)
n+1 = sin(n!)
n1=3+n�2=3  1
n1=3 Thus,forn�N,N()=1
3,jf(n)j.Exercise12.Converges,sincef(n)�1
3=3n+1+3(�2)n�3n+1�(�2)n+1
3(en+1+(�2)n+1)=(�2)n(3+2)
3(3n+1+(�2)n+1)==5
3(�2)n
3n+1+(�2)n+1 f(n)�1
3 =5
3 (�2)n
3n+1+(�2)n+1  �(�2)n+1
3n+1+(�2)n+1 == �1
3
�2n+1+1 = 1
1+3
�2n+1 1
�3
2n+11
�3
2nForn�N,consider=�2
3�N,i.e.N=�ln
ln2=3=N().Thus179
N()=1
=1;10;100;1000;10000.Exercise25.janj=1
nN()=1;10;100;1000;10000.Exercise26.janj= 1
n! 1
expnlnn1
expnForn�N,1
expN=,sothatN=ln1=.N()=1;2;4;6;9Exercise27.an=2n
n2+1;janj= 2
n2+1=n  2
n2 .N()=p
2
p
=1;4;14;44;141Exercise28.janj= 9
10 n=�9
10n=enln9=10N()=ln
ln�9
10=�ln1=
ln(9=10)==1;21;43;65;87Exercise30.If8�0,9N2Z+suchthatn�N,janj.janj2janj2ja2nj2Sofor81&#x-278;0,1=2and9N=N()=N(1),sothatja2nj1.Exercise31.jan+bn�(A+B)j=jjan�Ajjbn�Bjjjan�Aj+jbn�Bj+=28&#x-278;0;9NA;NB2Z+;jan�Ajifn&#x-278;NA;jbn�Bjifn&#x-278;NBConsidermax(NA;NB)=NA+Bjan+bn�(A+B)j281&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0;1=2;then9NA+B=NA+B(1)2Z+suchthatj(an+bn)�(A+B)j1ifn&#x-278;NA+Bjcan�cAj=cjan�Ajc81&#x-278;0;1=c;then9NcA=N()=N(1)2Z+suchthatjcan�cAj1forn&#x-278;N(1)Exercise32.Givenlimn!1an=A,limn!1(an�A)(an+A)=limn!1(an�A)limn!1(an+A)=0(2A)=02anbn=(an+bn)2�a2n�b2n2limn!1anbn=limn!1(an+bn)2�limn!1a2n�limn!1b2n=limn!1(an+bn)2�A2�B2=2AB=)limn!1anbn=ABExercise33.� n= ( �1)( �2):::( �n+1)
n!181
(3)limn!11
nPnk=11
1+(k
n)2=R101
1+x2dx=arctanxj10=

4
(4)limn!1Pnk=11
p
n2+k2=limn!11
nPnk=11
q
1+(k
n)2=R101
p
1+x2dx=ln(x+p
1+x2) 10=ln(1+p
2)(5)limn!1Pnk=11
nsink
n=limn!11
nPnk=1sink
n=R10sinxdx=��cosx
 10=�(�1�1)
=
2

(6)limn!1Pnk=11
nsin2k
n=Rn0sin2x=
1
2
10.9Exercises-Inniteseries,Thelinearitypropertyofconvergentseries,Telescopingseries,Thegeometricseries.Exercise1.P1n=11
(2n�1)(2n+1)=P1n=11=2
2n�1�1=2
2n+1=1
2Exercise2.P1n=12
3n�1=2P1n=01
3n=21
1�1=3=
3
Exercise3.P1n=21
n2�1=P1n=21=2
n�1�1=2
n+1=P1n=21=2
n�1�1=2
n+�1=2
n+1�1=2
n=1
2+1
4=3
4Exercise4.P1n=12n+3n
6n=P1n=1�1
3n+P1n=1�1
2n=1=3
1�1=3+1=2
1�1=2=1
2+1=3
2Exercise5.P1n=1p
n+1�p
n
p
n2+n=P1n=11
p
n�1
p
n+1=1Exercise6.1Xn=1n
(n+1)(n+2)(n+3)=1Xn=13=2
(n+2)(n+3)+�1=2
(n+1)(n+2)=1Xn=11
(n+2)(n+3)+1
2�1
6==1Xn=11
1+2�1
n+3�1
12=1
3�1
12=
1
4
Exercise7.P1n=12n+1
n2(n+1)2=P1n=11
n2�1
(n+1)2=1Exercise8.P1n=12n+n2+n
2n+1n(n+1)=1=P1n=11
2n(n+1)+1
2n+1=1
2P1n=11
n�1
n+1+1=4
1�1=2=1
2+1
2=1Exercise9.P1n=1(�1)n�1(2n+1)
n(n+1)=P1n=1(�1)n�11
n+1
n+1.(�1)n�11
n+1
n+1+(�1)n1
n+1+1
n+2=(�1)n�1
n+1
n+21Xj=1�1
2j�1+1
2j+1=1
21Xj=1�1
(j�1=2)+1
(j+1=2)=1
2�1
1=2=�1Exercise10.1Xn=2log((1+1
n)n(1+n))
(lognn)(log(n+1)n+1)=1Xn=2log��n+1
nn(1+n)
log(n+1)n+1lognn==1Xn=2log(n+1)n+1�lognn
log(n+1)n+1lognn=1Xn=21
lognn�1
log(n+1)n+1==1
2log2=log2p
esinceif1
2log2=y,theny=log2p
2.Exercise11.P1n=1nxn=xd
dxP1n=1xn=xx
1�x0=x
(1�x)2.Exercise12.183
1Xn=1nk+1xn=xd
dx1Xn=1nkxn=xd
dxPn(x)
(1�x)k+1=xP0k(x)(1�x)k+1+(k+1)(1�x)kPk(x)
(1�x)2k+2==xP0k(x)(1�x)+(k+1)Pk(x)
(1�x)k+2=(k+1)xPk(x)+x(1�x)P0k(x)
(1�x)k+2((k+1)Pk(x)+(1�x)P0k(x))xhasxasitslowestdegreetermfromxP0x(x)and(k+1)xk+1+�kxk+1=xk+1highestdegreetermisobtainedfrom(k+1)Pk(x)+�xP0k(x).Exercise21.P1n=0�n+kkxn=1
(1�x)k+1=dk
dxkP1n=0xn+k
k!.1Xn=0n+k+1k+1xn=1Xn=0dk+1
dxk+1xn+k+1
(k+1)!=1
(k+1)d
dx1Xn=0dk
dxkx(n+1)+k
k!==1
k+1d
dx1Xk=1dk
dxkxn+k
k!=1
k+1d
dx1Xn=0dk
dxkxn+k
k!�dk
dxkxk
k!==1
k+1d
dx1
(1�x)k+1�1=1
(1�x)k+2Exercise22.(1)P1n=2n�1
n!=P1n=21
(n�1)!�P1n=21
n!=P1n=11
n!�P1n=21
n!=
1
.(2)P1n=2n
n!+P1n=21
n!=P1n=21
(n�1)!+P1n=01
n!�1�1=P1n=11
n!+P1n=01
n!�2=P1n=02
n!�3=
2e�3
.(3)1Xn=2(n�1)(n+1)
n!=1Xn=2n2
n!+1Xn=2�1
n!=1Xn=2n
(n�1)!+1Xn=2�1
n!==1Xn=1n+1
n!+1Xn=2�1
n!=1Xn=11
(n�1)!+1=1Xn=01
n!+1=
e+1
Exercise23.(1)xd
dx�xd
dxP1n=1xn
n!=xd
dxxP1n=1nxn�1
n!=xd
dxP1n=1nxn
n!=P1n=1n2xn
n!=xd
dx�xd
dxex=x2ex+xex(2)xd
dxP1n=1n2xn
n!=P1n=1n3xn
n!=xd
dx�(x2+x)ex=x�(2x+1)ex+(x2+x)ex=(x3+3x2+x)exx=1
k=5
Exercise24.(1)P1n=2(�1)n=P1n=2(�1)n(n�(n�1)).Identical.(2)P1n=2(1�1)=P1n=2(�1)n.Notidentical.(3)Notidentical.P1n=2(�1)nvs.(P1n=2(�1+1))+1.(4)Identical.P1n=0�1
2n=1+P1n=1�1
2n�1��1
2n=P1n=1�1
2n(2�1)=P1n=0�1
2nExercise25.(1)1+x2+x4++x2n+=1
1�x2ifjxj1=)1+0+x2+0+x4+=1
1�x2ifjxj1(2)Thm.10.2.P1n=1( an+ bn)= P1n=1an+ P1n=1bn.Sothen1Xj=0xj�1Xj=0xj+(�x)j
2=1Xj=0xj�(�x)j
2=1Xj=0x2j+1=1
1�x�1
1�x2=x
1�x2(3)1Xj=0(x2)j+1Xj=0xj=1Xj=0(xj�x2j)=x
1�x2185
Byintegraltest,P1j=1j+1
2jconverges.Exercise4.P1j=1j2
2j.Zn1x2
2xdx=Zn1x2
exln2=x2e�xln2
�ln2+2xe�xln2
�(�ln2)2+2e�xln2
(�ln2)3 n1limn!1Zn1x2
2xdx=e�ln21
ln2+2
(ln2)2+2
(ln2)3=1
21
ln2+2
(ln2)2+2
(ln2)3Byintegraltest,P1j=1j2
2jconverges.Exercise5.1Xj=1jsinjxj
j2=1Xj=1aj1Xj=11
j2P1j=11
j2convergessincelimn!1Zn11
xsdx=limn!1x�s+1
�s+1 n1=limn!11
1�s1
ns�1�1=1
s�1ifs�1P1j=1jsinjxj
j2convergesbycomparisontestandintegraltest.Exercise6.1Xj=12+(�1)j
2j=1Xj=11
22j�1+3
22j=2(1=4)
1�1=4+3(1=4)
1�1=4=
4
3
Exercise7.P1j=1j!
(j+2)!.aj=j!
(j+2)!=aj=1
(j+1)(j+2)1
j2=bjSincePbjconverges,Pajconverges,bycomparisontest.Exercise8.P1j=2logj
jp
j+1=P1j=2ajP1j=2logj
j3=2Zn2logx
x3=2dx=Zn2(�2x�1=2)0logx=(�2x�1=2logx) n2�Zn2�2x�1=2
xdx==(�2)logn
n1=2�log2
21=2+�4x�1=2 n2limn!1Zn2logx
x3=2dx=21=2log2+4
p
2SoPajconvergesbycomparisontest.Exercise9.P1j=11
p
j(j+1)=P1j=1aj.Letbj=1
j.limj!1aj
bj=limj!1j
p
j(j+1)=limj!11
p
1+1=j=1Bylimitcomparisontest,sincePbjdiverges,Pajdiverges.Exercise10.P1j=11+p
j
(j+1)3�1=P1j=1ajbj=1
j5=2limj!1aj
bj=limj!11+p
j
(j+1)3�1j5=2=limj!1j3+j5=2
(j+1)3�1=limj!11+1=j1=2
(1+1=j)3�1
j3Bylimitcomparisontest,sincePbjconverges,Pajconverges.Exercise11.P1j=21
(logj)s=Paj187
(e�p
x)0=e�p
x�1
2p
x(p
xe�p
x)0=�e�p
x
2+1
2p
xe�p
x(p
xe�p
x+e�p
x)0=e�p
x
2Zn+1ne�p
xdx=2(p
x+1)e�p
x n+1n=2p
x
ep
x+1
ep
x n+1n==2p
n+1
ep
n+1+1
ep
n+1�p
n
ep
n�1
ep
n21Xj=1p
j+1
ep
j+1�p
j
ep
j+1
ep
j+1�1
ep
j=
�1
e
Notetheuseoftelescopingsuminthelaststep.Theseriesconverges.Exercise19.Rn1f(x)dx=Rn1logxdx=(xlnx�x)jn1=nlnn�n+1.n�1Xk=1lnkZn1lnxnXk=2ln(k)=)n�1Xk=1lnknlnn�n+1nXk=2ln(k)exp n�1Xk=1lnk!=(n�1)!nne�n+1exp nXk=2lnk!=(n)!nne�n+1=)e1=n
e(n!)1=n
ne1=nn1=n
e10.16Exercises-Theroottestandtheratiotestforseriesofnonnegativeterms.Exercise1.P1j=1(j!)2
(2j)!((j+1)!)2
(2j+2)(2j)!
(j!)2=(j+1)2
(2j+2)(2j+1)=j2+2j+1
4j2+6j+2j!1���!1
4Convergesbyratiotest.Exercise2.P1j=1(j!)2
2j2.((j+1)!)2
2(j+1)22j2
(j!)2=(j+1)22j2
2j2+2j+1=j2+2j+1
2ejln2j!1���!0Convergesbyratiotest.Exercise3.P1j=12jj!
jj2j+1(j+1)!
(j+1)j+1jj
2jj!=2(j+1)
(j+1)1
1+1=jjj!1���!2
e1Convergesbyratiotest.Exercise4.P1j=13jj!
jj3j+1(j+1)!
(j+1)j+1jj
3jj!=31
(1+1=j)jj!1���!3
e�1Divergesbyratiotest.Exercise5.P1j=1j!
3j.(j+1)!
3j+13j
j!=j+1
3Divergesbyratiotest.Exercise6.P1j=1j!
22j189
j3(p
2+(�1)j)j
3j!1=j=j3=j(p
2+(�1)j)
3=e3
jlnj(p
2+(�1)j)
3j!1���!(p
2+(�1)j)
31Convergesbyroottest.Exercise14.P1j=1rjjsinjxj.If0r1.1Xj=1rjjsinjxj1Xj=1rjsobycomparisontest,1Xj=1rjjsinjxjconvergesfor0r1Ifr1,limj!1rjjsinjxj6=0so1Xj=1rjjsinjxjdiverges,unlessjx=jExercise15.(1)cj=bj�bj+1aj+1
aj�08jN.Thentheremustbeapositivenumberrthat'sinbetweencjand0.ajbj�aj+1bj+1rajrnXj=NajnXj=N(ajbj�aj+1bj+1)=aNbN�an+1bn+1aNbN=)nXj=NajaNbN
r(2)cn0bj�bj+1aj+1
aj0ajbjbj+1aj+1=)bj
bj+1aj+1
ajX1
bjdiverges,solimj!1bj
bj+11byratiotestj!1���!1bj
bj+1aj+1
ajSobyratiotest,Pajdiverges.Exercise16.bn+1=n;bn=n�1.cn=n�1�nan+1
anr=)an+1
an1�1
n�r
n.UsingExercise15,Panconverges.P1
bndivergessinceP1
bnisaharmonicseriesofs=1.n�1�nan+1
an0=)1�1
nan+1
anExercise17.ForsomeN1;s�1;M�0,andgiventhatan+1
an=1�A
n+f(n)
ns=1� A�f(n)
ns�1
n!ConsiderA�f(n)
ns�1.Sincejf(n)jM,f(n)isnite,soconsiderslargerthan1andngoingtoinnitysothatf(n)
ns�1!0.UsingExercise16,forPajtoconverge,A�f(n)
ns�1=1+rwherer�0,forallnN,whereNissomepositivenumber.Letr=M
Ns�1sothatA=1+r+f(n)
ns�1�1191
Exercise3.P1j=1(�1)j�1
jsIfs�1,thentheseriesabsolutelyconverges.limj!11
js=0ifs�0.Convergesconditionallyfor0s1.Otherwise,ifs0theseriesdivergesabsolutely.Exercise4.P1j=1(�1)j135:::(2j�1)
246:::(2j)3.aj+1
aj=135(2j+1)
246:::(2j+2)246:::(2j)
135:::(2j�1)=2j+1
2j+2Absolutelyconverges.Exercise5.P1j=1(�1)j(j�1)=2
2jconvergessincelimj!1(�1)j(j�1)=2
2j=0;P1j=11
2j=1=2
1�1=2=
1
.Absolutelyconverges.Exercise6.P1j=1(�1)j2j+100
3j+1j.expjln2j+100
3j+1=expjln2
3+298
9j+3expjln2
3+1
2
3298
9j+3==expjln2
3+146
3+1=j0limj!1expjln2j+100
3j+1limj!1expjln2
3+146
3+1=j=0=)limj!1expjln2j+100
3j+1=0Sothealternatingseriesconverges.2j+100
3j+12j+100
3j2:5j
3j=5
6(forj200)=)2j+100
3j+1j5
6jforj200Sotheseriesabsolutelyconvergesbycomparisonwithageometricseries.Exercise7.P1j=2(�1)j
p
j+(�1)j.limj!11
p
j+(�1)jdoesn'texistsince???Toshowdivergence,weusuallythinkofeithertakingthegeneraltermandndingthelimit(andifitgoestoanonzeroconstant,thenitdiverges),orweuseratio,root,comparisontestonthegeneralterm.Sincethisisanalternatingseries,I'veobservedthatthegeneraltermisasumoftwoadjacentterms,oneevenandoneodd.(�1)j
p
j+(�1)j(�1)2j
p
2j+(�1)2j+(�1)2j+1
p
2j+1+(�1)2j+1=1
p
2j+1+�1
p
2j+1+1=p
2j+1�1�(p
2j+1)
(p
2j+1)(p
2j+1�1)==p
2j+1�p
2j�2
(p
2j+1)(p
2j+1�1)=p
2jq
1+1
2j�p
2j�2
(p
2j+1)(p
2jq
1+1
2j�1)=forjlarge������!'p
2j1+1
4j�p
2j�2
(p
2j+1)(p
2j1+1
4j�1)=�2
j 1�1
4p
2j
2�1
2j+1
2p
2j3=2!Everyterm,sinceweconsideredanyj,willcontain�2.Sowefactoritout.Then1
j 1�1
4p
2j
2�1
2j+1
2p
2jj3=2!�1
j 1�1
4p
2j
4�1
p
2j!=1
4j193
1�xsin1
x0=�sin1
x�xcos1
x�1
x2==�sin1
x+1
xcos1
x=�xsin1
x+cos1
x
xsin1
x=1Xj=0�1
x2j+1(�1)j
(2j+1)!�xsin1
x+cos1
x
x==�1+(1
x)3(+1)
3!+P1j=2(1
x)2j+1(�1)j
(2j+1)!+1��1
x2=2+P1j=2(1
x)2j(�1)j
(2j)!
x0forxlargeenoughP1j=1(�1)j1�jsin1
jconvergessinceaj=1�jsin1
jismonotonicallydecreasingsequencewithlimit0.1�jsin1
j=1�j1Xk=01
j2k+1(�1)k
(2k+1)!=1�[email protected]
j+1Xk=11
j2k+1(�1)k
(2k+1)!1CA==1�[email protected]+1Xk=11
j2k(�1)k
(2k+1)!1CA=1Xk=11
j2k(�1)k+1
(2k+1)!1
6j2Theseriesconvergesabsolutelysincethetermitselfisaseriesthatisdominatedby1
6j2,sothatbycomparisontest,theseriesmustconverge.Exercise18.P1j=1(�1)j1�cos1
j.(cos1
x)0=�sin1
x�1
x2=�1
x2sin1
x0P1j=1(�1)j(1�cos1
j)convergessinceaj=(1�cos1
j)ismonotonicallydecreasingto0(1�cos1
j)=1�1Xk=0(1=j)2k(�1)k
(2k)!=1Xk=1(1=j)2k(�1)k+1
(2k)!1
2j2Sotheseriesconvergesabsolutely,bycomparisontestwithP1
j2whichconverges.Exercise19.P1j=1(�1)jarctan1
2j+1.(arctan1
2j+1)0=1
1+1
2j+12�1
(2j+1)2(2)=�2
(2j+1)2+10P1j=1(�1)jarctan1
2j+1converges,sinceaj=arctan1
2j+1ismonotonicallydecreasingto01
1+x2=(arctanx)0=1Xj=0(�x2)j=1Xj=0(�1)jx2j=)arctanx=1Xj=0(�1)jx2j+1
2j+1arctan1
2j+1=1Xk=0(�1)k1
2j+12k+1
(2k+1)=1
2j+1+(�1) 1
(2j+1)3
3!+1Xk=2(�1)k1
2j+12k+1
2k+1��1
2j+1+(�1)1
3(2j+1)3=3(4j2+4j+1)+(�1)
3(2j+1)3=12j2+12j+2
3(2j+1)3�2j
(2j+1)2�2
91
jforj�2SobycomparisontesttoP1
j,Parctan1
2j+1divergesabsolutely.Theseriesisconditionallyconvergent.Exercise20.P1j=1(�1)j�
2�arctanlogj196
Xn=1nnzn = 1Xj=1(jz)j =X�elnjzjnXj=1(elnjz)j==elnjz�(elnjz)n
1�elnjz�!1Soz=0Exercise34.P1j=1(�1)jz3j
j!=P1j=1(�z3)j
j!=e�z3.
C
.Exercise35.P1j=0zj
3j=P1j=0�1
3jzjPzjbeconvergentorPnj=1zjbounded.
jzj3
andjzj=3ifz6=3Exercise36.P1j=1zj
jj
fzg=C
since z
j 1forjN&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;jzjExercise37.P1j=1(�1)j
z+jByLeibniz'sRule,since1
z+jj!1���!0,thentheseriesconverges.However,zcannotbeequaltoanynegativeintegersinceonetermintheserieswillthenblowup.Exercise38.P1j=1zj
p
jlog2j+1
j.zj
p
jlog2+1
j=zjlog2+1
j
p
jSincelimj!1log(2+1
j)
p
j=0sothatlog(2+1
j)
p
jisamonotonicallyconvergentsequence.ThenbyDirchlet'stest,Pnj=1zjmustbebounded.jzj�1,andjzj=1ifz6=1.Exercise39.P1j=11+1
5j+1j2jzj17j=P1j=11+1
5j+1j2(jzj17)j 1+1
5j+1jjzj17!jlimj!11+1
5j+1jlimj!11+1=5
jj=e1=5e1=5jzj171=)
jzje�1=85
Exercise40.P1j=0(z�1)j
(j+2)! 1Xj=0(z�1)j
(j+2)! 1Xj=0j(z�1)jj
(j+2)!1Xj=0j(z�1)jj+2
(j+2)!=
ejz�1j�1�jz�1j
1!
Theseriesconverges8z.Exercise41.1Xj=1(�1)j(z�1)j
j=1Xj=1(1�z)j
j=log(1�(1�z))=
logz
Sotheseriesconverges8zexceptforz=0.198
1
1+jzj2111+jzj20jzj28zexceptz=0Exercise47.P1j=1(�1)j2jsin2jx
jUseDirichlet'sTest.1
jisamonotonicallydecreasingsequenceconvergingtozero.Consider(�2)jsin2jx.Theconditionforconvergenceisj(�2sin2x)j1forjNforsomeN=)x2(�
4+n;
4+n);n2ZExercise49.Pajconverges.Paj1
ajdiverges.ThensincePajisaconvergentseries(byAbel'stest),1
ajisadivergentsequence.ThenP1
ajisdivergent(sincelimj!11
ajdoesn'texist).Exercise50.Pjajjconverges.Pjajjconverges,thenPajconverges.jajj2=a2j.jajjconverges,thenlimj!1jajj=0limj!1jajj2=0limj!1jajj2
a2j=1Bylimitcomparisontest,Pa2jconverges.Counterexample:P1
j2converges,butP1
jdiverges.Exercise51.GivenPaj;aj0.Pajconverges.Xp
aj1
(j)plimj!1aj=0limj!1p
aj=limj!1aj1=2=0Zn�1Xj=0xj=n�1Xj=0xj+1
j+1=nXj=1xj
j=Z1�xn
1�xAcounterexamplewouldbep
aj
jp=q
aj
j.Exercise52.(1)Pajconvergesabsolutely,thenifPjajjconverges,Pa2jconverges.a2j
1+a2j=1+�1
1+a2ja2j
1+a2ja2jsinceXa2jconverges;Xa2j
1+a2jconverges(2)Pajconvergesabsolutely,limj!1jajj=0P aj
1+aj =Pjajj1
j1+ajj.ByAbel'stest,sincelimj!11
j1+ajj=1
j1+limj!1ajj=1showsthat1
j1+ajj0isamonotonicallyconvergentsequenceByAbel'stest,Paj
1+ajisconvergent.200
Exercise10.R1dx
x(logx)s11.7Exercises-Pointwiseconvergenceofsequencesoffunctions,Uniformconvergenceofsequencesoffunctions,Uniformconvergenceofsequencesoffunctions,Uniformconvergenceandcontinuity,Uniformconvergenceandinte-gration,Asufcientconditionforuniformconvergence,Powerseries.Circleofconvergence.Exercise1.P1j=0zj
2j=P1j=0�z
2jUsingthecomparisontest, z
2 jtj; z
2 1=)jzj2Supposejzj=2,z6=2P1j=0�z
2j=P1j=0(e2i)jNowPnj=0(e2i)j1
sin+1,sincenXj=1(e2i)j=e2i�e2i(n+1)
1�e2i=ein�e�inein+i
�e�i+ei=sinnein+i
sin1
sinSoP1j=0�z
2jconvergesforjzj=24,z6=2Exercise2.P1j=0zj
(j+1)2jUseratiotest.aj+1
aj=zj+1
(j+2)2j+1(j+1)2j
zj=z
2(j+1)
(j+2)=z
2(1+1=j)
(1+2=j)j!1���!z
2Ifjzj2,P1j=0ajconverges,ifjzj&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;2,Pajdiverges.Ifjzj=2,Pzj
(j+1)2j=P(e2i)j1
j+1Now1
j+1isamonotonicallydecreasingsequenceofrealterms.P(e2i)jisaboundedseries.ByDirichlet'stest,Pajconvergesifjzj=2,z6=2Exercise3.P1j=0(z+3)j
(j+1)2jUseratiotest:aj+1
aj=(z+3)j+1
(j+2)2j+1(j+1)2j
(z+3)j=(z+3)
2j+1
j+2=(z+3)
21+1=j
1+2=jj!1���!z+3
2UsingTheorem11.7,Theorem34(Existenceofacircleofconvergence).AssumeXajzjconvergesforatleastz16=0divergesforatleastonez26=09r�0;suchthatXajzjabsolutelyconvergesforjzjrdivergesforjzj&#x-278;rWecanpluginrealnumberstosatisfytheconditionjz+3j
21forconvergence.Pajconvergesforjz+3j2;divergesforjz+3j&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;ԗ ;� Td;&#x [00;2.Considerjz+3j=2;z6=1Paj=P(e2i)j1
j+1.Since1
j+1isamonotonicallydecreasingsequenceofrealnumbersandP(e2i)jisaboundedseries,byDirichlet'stest,Pajconvergesforjz+3j=2;z6=�1.Exercise4.P1j=1(�1)j22jz2j
2j=�P1j=1(�1)j�1(2z)2j
(2j).Lookatwhatthetermslooklike.ConsiderusingLeibniz'sRule,Theorem10.14.Theorem35(Leibniz'srule).Ifajisamonotonicdecreasingsequenceandlimj!1aj=0,thenP1j=1(�1)j�1ajconverges.202
SincePn�1j=1f(j)Rn1f(x)dxn�1Xj=1lnjZn1lnx=nlnn�n+1nXj=2lnj(n�1)!nne�nnn!n!
nnne�nlimn!1n!
nnlimn!1ne�n=0limn!1n!
nnlimn!1n2
en=0=)
limn!1n!
nn=0
Sothensincen!
nnisamonotonicallydecreasingconvergentsequenceofrealterms;ifPzisaboundedseries,thenbyDirichlet'stest,Pn!zn
nnisconvergent.jzj1;jzj=1ifz6=1Trytheratiotest,becauseit'sclearfromtheresultsoftheratiotestwhereconvergenceanddivergencebeginsandends.(n+1)!jzjn+1
(n+1)n+1nn
n!jzjn=n
n+1njzj==1
1+1=nnjzjn!1����!1
ejzjConvergesfor
jzje
.Nowtryplugginginacomplexiede:ej
jj
j!=j!ejei2j
(j)jjje�jjejei2j
jjei2j!1Sotheseriesdivergesforjzj=e.Exercise7.P1j=0(�1)j(z+1)j
j2+1Bytheratiotest,jz+1jj+1
j2+2j+2j2+1
jz+1jj=jz+1j1+1=j2
1+2=j+2=j2j!1���!jz+1jTheseriesabsolutelyconvergesforjz+1j1.Forz=0,P(�1)j
j2+1convergessince1
j2+1j!1���!0Forz=�2,P(�1)j(�1)j
j2+1=P1
j2+1and1
j2+11
j2,butP1
j2isconvergent(byintegraltest).Sotheseriesconvergesforz=�2.ByDirichlet'stest,theseriesconvergesaswell,ifwetreataj=(�1)j(z+1)jandbj=1
j2+1tobethemonotonicallydecreasingsequence.=)
jz+1j1
forconvergencexExercise8.P1j=0aj2zj,0a1Usetheroottest.(aj2zj)1=j=ajzj!1���!0Sotheseriesconverges8z2CExercise9.P1j=1(j!)2
(2j)!zj204
Exercise13.P1j=0(sinaj)zja�0jsin(aj)zjjjzjjBycomparisontest,P1j=0(sinaj)zjconverges,sinceP1j=0jzjjconvergesabsolutely,forjzj1.Notethatifa=,theseriesiszero.P1j=0(sinaj)!1fora=2so
r=1indeed:
Exercise14.P1j=0(sinhaj)zj=P1j=0eaj�e�aj
2zj=1
2P1j=0(eaz)j�P1j=0�z
eaj;a�0Ifjzj1
ea,thenPsinhajzjconverges.Sothentheradiusofconvergenceisr=1
eaExercise15.P1j=1zj
aj+bj.Assumeabzj
bj(1+�a
bj)(ratiotest)zj+1
bj+1(1+�a
bj+1)[email protected]1+�a
bj
zj1A=z
b 1+�a
bj
1+�a
bj+1!j!1���!z
bSothenjzj7bconverges(diverges)byratiotest.Ifa=b,1Xj=1zj
2aj=1
21Xj=1z
ajBycomparisontestwithxj,ifjzj7jaj,theseriesconverges(diverges).Exercise16.P1j=1aj
j+bj
j2zjUseratiotestoneachofthesums,separately.(ajzj)j+1
j+1j
(ajzj)j=ajzj
1+1
jj!1���!ajzj=)jzj1
athentheseriesconverges(bjzj)j+1
(j+1)2j2
(bjzj)j=bjzj 1
1+1
j!2j!1���!bjzj=)jzj1
b
Soifa?b,thenr=a;(b)
Exercise17.R10fn(x)=R10nxe�nx2=e�nx2
�2 10=e�n�1
�2n!1����!1
2However,limn!1nxe�nx2=0Thisexampleshowsthattheoperationsofintegrationandlimitcannotalwaysbeinterchanged.Weneeduniformconver-gence.Exercise18.fn(a)=sinnx
nlimn!1sinnx
n=0f(x)=limn!1fn(x)f0n(x)=ncosnx
n=)limn!1f0n(0)=1Thisexampleshowsthatdifferentiationandlimitcannotalwaysbeinterchanged.Exercise19.P1j=1sinjx
j2=f(x)206
Forf(x)=Pnj=0f(j)(0)
j!xj+En(x)=)En(x)�x
rn+1En(r)f(r)=nXj=0f(j)(0)
j!rj+En(r)En(r)sincef(j)(0)08jSothen0En(x)�x
rn+1f(r)n!1andf(t)willbesomenon-innitevalue,soEn(x)n!1����!0.Exercise1.P1j=0(�1)jx2jConsiderabsoluteconvergence.limj!1(x2)j=0Ifx21Ifjxj=1,thenconsiderx2(2j)�x2(2j+1)=x4j(1�x)1Xj=0(1�x)x4j=(1�x)1Xj=0(x4)j
Indeed1Xj=0(�1)jx2jconvergesforjxj1
Exercise2.P1j=0xj
3j+1=1
3P1j=0�x
3jTheseriesconvergesfor
jxj3
Exercise3.Pj=01jxjZx01Xj=0jtj�1=1Xj=0xjSotheseriesconvergesforjxj1.Notethatwehadusedtheintegrabilityofpowerseries.Exercise4.P1j=0(�1)jjxj.jxjjejlnxlimj!1jejlnx=(1ifx&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;10if0x1Ifx=1,(2j)x2j�(2j+1)x2j+1=x2j(2j+�(2j+1)x)=
�1
X(�1)=1SoP(�1)jjxjconvergesonlyfor0x1;jxj1.Exercise5.P1j=0(�2)jj+2
j+1xj=P1j=0(�1)jj+2
j+1(2x)jlimj!1j+2
j+1(2x)j=limj!1(2x)jifj2xj1Sowhenjxj1
2,theseriesconverges.2j+2
2j+1�2j+3
2j+2=(2j+2)(2j+2)�(2j+3)(2j+1)
(2j+1)(2j+2)4j2+8j+4�(4j2+5j+3)
4j2+6j+2=3j+1
4j2+6j+2=(3j+1)=2
2j2+3j+1=(3j+1)=2
(2j+1)(j+1)3j+1
4j2+6j+23j+1
4j2+4
33(j+1=3)
4(j2+1=3)3
41
j+1
12j2Thus,itdiverges,bycomparisontestwith1
jforx=1
2.Theorem38.Letfberepresentedbyf(x)=P1j=0aj(x�a)jinthe(a�r;a+r)intervalofconvergence(1)P1j=1jaj(x�a)j�1alsohasradiusofconvergencer.208
Exercise15.e�x2=P1j=0(�1)jx2j
j!x2j+2
(j+1)!j!
x2j=x2
j+1j!1���!0Exercise16.sin3x=sin2xcosx+sinxcos2x=3sinx�4sin3x.sin3x=3sinx�sin3x
4=3
[email protected]=0x2j+1(�1)j
(2j+1)!�1Xj=0(3x)2j+1(�1)j
(2j+1)!1A==3
[email protected]=0(�1)jx2j+1(1�32j+1)
(2j+1)!1A=3
[email protected]=0(�1)j+1(32j+1�1)x2j+1
(2j+1)!1AExercise17.logq
1+x
1�x=1
2(log(1+x)�log(1�x))=1
2Pj=1(+x)j
j(�1)j�1�Pj=1(xj)(�1)j
jln(1+x)=1Xj=0(�x)j+1
j+1(�1)=1Xj=1(�x)j
j(�1)�ln(1�x)=1Xj=0xj+1
j+1=1Xj=1xj
j=)�1Xj=1((�1)j+1)xj
j=)1X
x2j+1
2j+1
x2j+3
2j+32j+1
x2j+1j!1���!x2jx2j1;converges,withradiusofconvergenceof1Exercise18.3x
1+x�2x2=1
1�x�1
1+2x=P1j=0xj
j�P1j=0(�2x)j
j=P1j=0xj
j(1�(�2)j)xj+1j(1�(�2)j+1)j
j+1j
jxj(1�(�2)j)j=x 1
(�2)j+2
1
(�2)j�1 j!1���!2x1jxj1
2Forx=1
2x2j
2j(1�(�2)2j)+x2j+1
2j+1(1�(�2)2j+1)=x2j((2j+1)(1�22j)+x(1+22j+1)(2j))
(2j)(2j+1)�1
22j(2j+1�(2j+1)22j+2jx+22j+2jx)
(2j)(2j+1)j!1���!�(2j+1)+4j
(2j)(2j+1)=2j�1
(2j)(2j+1)!0So3x
1+x�2x2convergesforjxj1
2Exercise19.12�5x
6�5x�x2=P1j=01+(�1)j
6jxj(jxj1).12�5x
6�5x�x2=5x�12
(x+6)(x�1)=1
1�x+6
6+x=1Xj=0xj+1Xj=0�x
6j=1Xj=0xj(1+�1
6j)jxj1sinceforx=1,limj!1(1+��1
6j)=1.Exercise20.1
x2+x+1=2
p
3P1j=0sin2(j+1)
3xj(jxj1)x3�1=(x�1)(x2+x+1)1
x2+x+1=x�1
x3�1=1�x
1�x31
1�x3=1Xj=0(x3)j1�x
1�x3=1Xj=0�(x3)j�x3j+1Byinduction,itcouldbeobservedthatx3j;�x3j+1;0appearinsequencesof3terms.210
Useratiotest:(�2(j+1))!
(�2j�2�k)!(j+1)!j!(�2j�k)!
(�2j)!x�2(j+1)
x�2j=(�2j�k)(�2j�k�1)
(j+1)(�2j)(�2j�1)x�2j!1���!0Thus,byratiotest,everyorderofderivativeexistsforeveryxonthereallinesincetheseriesrepresentingthederivativeconvergesforeveryx.(2)f(x)=P1j=0�x�2j
j!.Therearenononzerotermsofpositivepower,i.e.noxj;j1.=)f(j)(0)=08j111.16Exercises-Powerseriesanddifferentialequations,binomialseries.Exercise1.For(1�x2)y00�2xy0+6y=0,y=1Xj=0ajxjy0=1Xj=1jajxj�1y00=1Xj=2j(j�1)ajxj�2=1Xj=0(j+2)(j+1)aj+2xjf(0)=1=)a0=1f0(0)=0=)a1=02(1)a2+3(2)a3x+�2(1)a1x+6a0+6a1x+1Xj=2((j+2)(j+1)aj+2�j(j�1)aj�2jaj+6aj)xj==2a2+6+6a3x+1Xj=2((j+2)(j+1)aj+2+aj(j+3)(j�2))xj=)a2=�3a3=0
aj+2=(j+3)(j�2)
(j+2)(j+1)aj
Forj=2;a4=0,sothenaj+2=0forj=2;4;:::.Likewise,sincea3=0,thenaj+2=0forj=3;5;:::.=)f(x)=1�3x2Exercise2.Usingtheworkfromabove,thenfor(1�x2)y00�2xy0+12y=0f(0)=0=)a0=0f0(0)=2=)a1=22(1)a2+3(2)a3x+�2(1)a1x+12a0+12a1x+1Xj=2((j+2)(j+1)aj+2�j(j�1)aj�2jaj+12aj)xj==2a2+6a3x+�4x+0+24x+1Xj=2((j+2)(j+1)aj+2�aj(j+4)(j�3))xj=)a2=0a3=�10=3
aj+2=(j+4)(j�3)
(j+2)(j+1)aj
Forj=3;a5=0,sothenaj+2=0forj=3;5;:::.Likewise,sincea2=0,thenaj+2=0forj=2;4;:::.=)f(x)=�10=3x3+2Exercise3.f(x)=P1j=0x4j
(4j)!;d4y
dx4=y212
f0=1Xj=2xj�1
(j�1)!=1Xj=1xj
j!=x+yxj+1
(j+1)!j!
xj=x
jj!1���!0Sotheseriesconverges8x2Rbyratiotest.Exercise8.f(x)=1Xj=0(�1)j(kx)2j
(2j)!f0=1Xj=1(�1)j(kx)2j�1k
(2j�1)!=1Xj=0(�1)j+1(kx)2j+1k
(2j+1)!f00=1Xj=1(�1)j+1(kx)2j
(2j)!k2f00�k2f=0(kx)2j+1
(2j+1)!(2j)!
(kx)2j=kx
2k+1j!1���!0byratiotest,fconverges8x2R.Exercise9.f00=1Xj=1(3x)2j�1
(2j�1)!9=1Xj=09(3x)2j+1
(2j+1)!9(f�x)=9(x+1Xj=0(3x)2j+1
(2j+1)!�x)(3x)2j+3
(2j+3)!(2j+1)!
(3x)2j+1=9x2
(2j+3)(2j+2)j!1���!0(byratiotest,fconverges8x2R)Exercise10.J0(x)=P1j=0(�1)jx2j
(j!)222jJ1(x)=P1j=0(�1)jx2j+1
j!(j+1)!22j+1.(1)x2j+2
((j+1)!)222j+2(j!)222j
x2j=x2
(j+1)24j!1���!0byratiotest,fconverges8x2Rx2j+3
(j+2)!22j+3j!22j+1
x2j+1=x2
(j+2)(j+1)4j!1���!0byratiotest,fconverges8x2R(2)J00(x)=1Xj=1(�1)jx2j�1
(j�1)!(j!)22j�1=1Xj=0(�1)j+1x2j+1
j!(j+1)!22j+1=�J1(x)(3)j0(x)=xJ0(x)=1Xj=0(�1)jx2j+1
(j!)222jj1(x)=xJ1(x)=1Xj=0(�1)jx2j+2
j!(j+1)!22j+1j01=1Xj=0(�1)jx2j+1
(j!)222j=)j0=j01Exercise11.x2y00+xy0+(x2�n2)y=0.214
y0(0)[email protected]=3ajxj1A2++2a0a1x+2a0a2x2+2a01Xj=3ajxj++2a1a2x3+2a11Xj=3ajxj+1+2a21Xj=3ajxj+2y0=a1+2a2x+1Xj=3jajxj�1
a1=1
sincey0(0)=1Considertherstfewtermsofx2+y2a20+2a0a1x+a21x2+2a0a2x2+x2=a1+2a2x+3a3x2=)a1=1=a20=)
a0=1
2a2=2a0a1=)
a2=1
3a3=a21+2a0a2+1=4=)
a3=4
3
Exercise13.y0=1+xy2withy=0whenx=0=)a0=0y=1Xj=1ajxjy=1Xj=1jajxj�1=1Xj=0(j+1)[email protected]=4ajxj1A2++2a1a2x3+2a1a3x4+2a11Xj=4ajxj+1++2a2a3x5+2a21Xj=4ajxj+2+2a31Xj=4ajxj+3
a1=1
x:2a2=0=)a2=0x2:3a3=0=)a3=0x3:4a4=12=)
a4=1
4
x4:5a5=0=)a5=0x5:6a6=0=)a6=0x6:7a7=2a1a4+2a2a3=)
a7=1
14
x7:8a8=0+2a2a4=0=)a8=0x8:9a9=0=)a9=0x9:10a10=1
42+2(1)1
14=)
a10=23
1120
Exercise14.y0=x+y2y=0whenx=0=)a0=0216
Plugging0asagoodguessbackintotheODE,cos0=1=y(0)(2)=)y(0)=1
2Withthisinitialcondition,weget
f(x)=sinx
x+cosx�1
x2
ifx6=0Sof(0)=1
2andf()=�2
2Exercise20.(1)(1�x)�1=2=1Xj=0�1=2j(�x)j==1+1
2x+��1
2��3
2
2x2+��1
2��3
2��5
2
3!x3+��1
2��3
2��5
2��7
2
4!x4++��1
2��3
2��5
2��7
2��9
2
5!x5+==1+1
2x+3
8x2+5
8x3+35
128x5+63
256x5+:::(2)Tomakethenotationclear,(1�x)�1=2=P1j=0��1=2j(�x)j=P1j=0bjxj=P1j=0ajNow� j+1
� j= ( �1):::( �(j+1)+1)
(j+1)!j!
( �1):::( �j+1)=( �j)
(j+1)Sofor =�1
2,aj+1
aj=�1=2+j
j+1(�x)xUsingthis,wefurtherndthatbj+1bj1
50bj+2bj+11
50bj1
502Forx=1
50.Sobyinduction,bn+jbn�1
50jrn=1Xj=1an+j1Xj=1an1
50j=an1=50
1�1=50=an
49
rnan
49
(3)Notethat(1�x)�1=2=�1�1
50�1=2=�49
50�1=2=5p
2
77
51�1
50�1=2=1+1
100+3
21
2(50)2+5
21
2(50)3+35
81
2(50)4+63
81
2(50)5
p
2'1:4142135624
Exercise21.(1)1732
1000�1�176
3000000�1=2=1732
1000�3000000
29998241=2Obviously,(3000000)1=2=1000p
3sothatwehave1732(3=2999824)1=2.Withlongmultiplication,wecouldshoweasilythat17321732=2999824(it'shardertodivide).Sothen
1732
10001�176
3000000�1=2=p
3
(2)219
ed(X;L)=kX�Fk=ej(X�F)N�dj=ejrcos�dj=ercos�ed=r�ed
1�ecos=r=ed
ecos�1Soforr=2
2cos�1;e=2;d=1.Exercise8.r=4
1+2cose=2;d=2.Exercise9.r=4
1+cose=1d=4.Exercise10.3x+4y=25=)3
5x+4
5y=5.N=�3
5;4
5.L=fx=P+tAg;NX=NP.Tondthedistancefromthefocus,attheorigin,tothedirectrix,dN=P+tA;dNN=d=NPSoforthisproblem,d=5.r=kX�Fk=ed(X;L)=ej(X�F)N�dj=ejXN�dj= 3
5rcos+4
5rsin�5 r=1
25�3
5rcos�4
5rsinr=5=2
1+3
10cos+4
10sinExercise11.e=1;4x+3y=254
5x+3
5y=5;N=�4
5;3
5.d=5.kX�Fk=ed(X;L)=ej(X�F)N�dj=e 4
5rcos+3
5rsin�5 r=5�r4
5cos+3
5sinr=5
1+4
5cos+3
5sinExercise12.e=2,hyperbola,sothere's2branches.1
p
2x+1
p
2y=1
p
2L=fx=P+tAgXN=NPdN=P+tA;dNN=d=NP=1
p
2Notethatthesignofdheretellsyouwhatsidethefocus,attheorigin,lieson.kX�Fk=ed(X;L)=kXk=ej(X�F)N�dj=e(d�1
p
2rcos�1
p
2rsin)r=2=p
2
1+2
p
2cos+2
p
2sinButfortherightsidebranch,kX�Fk=ed(X;L)=kXk=ej(X�F)N�dj=�e(d�1
p
2rcos�1
p
2rsin)r=�2=p
2
1�2
p
2cos�2
p
2sinExercise13.e=1parabola.221
q
1�b2
a2=q
1�16
25=3
5=e;jFj=ae=53
5=3.Foci:(�1;�1);(�1;�5).Vertices:(�1;3);(�1;�7);(3;�2);(�5;�2).Exercise7.F=ae=3
4.a=1;e=3
4.b2=a2(1�e2);b2=1�1
4.
x2+4y2=1
.Exercise8.2a=4.a2=4.2b=3.b2=9=4.=)(x+3)2
4+(y�4)2
4=1Exercise9.(x+3)2
9=4+(y�4)2
4=1.Exercise10.2a=6;a=3.(x+4)2
9+(y�2)2
1=1.Exercise11.2a=10;a=5.jFj=ae=5e=4e=4=5.b2=a2(1�e2)=25�1�16
25=9.Exercise12.(x�2)2
a2+(y�1)2
b2=1;a=4from(6;1).b=2from(2;3).=)(x�2)2
42+(y�1)2
4=1Exercise13.b2=a2(1�e2).x2
100�y2
64=1;b2=100(1�e2)=�64.1+64
100=e2.Center(0;0).e=2p
41
10=p
41
5.Vertices;(10;0).F=ae=2p
41.Foci:(2p
41;0).x2
100=y2
64+1x;y!1�����!;y=4
5xExercise14.y2
100�x2
64=1;Center(0;0),a2=100;b2=�64.b2=a2(1�e2).e=p
41
5.Vertices(0;10).F=ae=(0;2p
41).x2
64+1=y2
100x;y!1�����!5
4x=y.Exercise15.(x+3)2
4�(y�3)2=1.Center(�3;3).e=q
1�b2
a2=q
1��1
4=p
5
2.Foci:ae=2p
5
2=p
5.(�3+p
5;3);(�3�p
5;3).Vertices:(�3;4);(�3;2);(1;3);(�7;3).(x+3)2
4=1+(y�3)2x;y!1�����!
(x+3)
2=y�3
Exercise16.x2
144=9�y2
144=16=1=x2
16�y2
9.e=q
1��9
16=5
4.Center(0;0).jFj=ae=5.Foci:(5;0);(�5;0).Vertices(4;0).Exercise17.20=5y2�4x2.Center(0;0).jFj=ae=2�3
2=3.Foci:(0;3).1=y2
4�x2
5.e=q
1��5
4=3
2.Vertices:(0;2)Exercise18.(x�1)2
4�(y+2)2
9=1.Center(1;�2).e=q
1��9
4=p
13
2;jFj=2p
13
2=p
13.Foci:(1+p
13;�2);(1�p
13;�2).Vertices:(5;�2);(�3;�2).Exercise19.F=ae=2(2)=4.x2
4+y2
�12=1.y2
12+1=x2
4x;y!1�����!y=p
3x.223
Exercise33.x2=8y.Exercise34.(y�3)=�8(x+4)2.Exercise35.c=5
45(x�7
4)=(y+1)2Exercise36.y=ax2+bx+c(0;1)!c=1(1;0)!0=a+b+1(2;0)!0=4a+2b+1a=1
2=)y=1
2x2�3
2x+1Exercise37.4c(x�1)=(y�3)2.4c(�2)=(�4)2=16.c=�2.�8(x�1)=(y�3)2.Exercise38.kX�Fk=ed(X;L)=j(X�F)N�djL=f(x;y)j2x+y=10;2
p
5x+y
p
5=10
p
5g.d=N=xLdNN=d=xLN=10
p
5.F=0=)kXk2=jXN�dj2=�2
p
5x+�y
p
5+10
p
52=x2+y25x2+5y2=(�2x�y+10)2=4x2+y2+100+4xy�40x�20y=)x2+4y2�4xy+40x+20y�100=013.25Miscellaneousexercisesonconicsections.Exercise1.y2
b2=1�x2
a2y2=b2�bx
a2=b21�x
a2y=2Za�abr
1�x
a2dx=2Z1�1abp
1�x2dx=(ab)areaofacircleofradius1Exercise2.(1)Withoutlossofgenerality,letthemajoraxisbe2ainthex-axis.y=bq
1��x
a2V=Za�ab21�x2
a2dx=b2aZ1�1(1�x2)dx=4
3(1)3b2a(2)Ifrotatedabouttheminoraxis,suppose,withoutlossofgenerality,2aistheminoraxis(justnotethatx2
a2+b2
a2=1havex;y;a;basdummylabels).=)V=4
3(1)3b2a,where2aistheminoraxis,2bisthemajoraxis.Exercise3.x2
(3=A)+y2
(3=B)=1By2=3�Ax2=)y2=3
B�Ax2
B;y=q
3
B�Ax2
B.Sotheareainsidethisellipseis2r
1
BZp
3=A�p
3=Ap
3�Ax2dx=2r
3
BZp
3=A�p
3=As
1�x2
�3
AFortheotherellipseequation,x2
3=(A+B)+y2
3=(A�B)=1.y2=3
A�B1�x2
(3=(A+B));y=q
3
A�Bq
1�x2
(3=(A+B)).Thus,theareainsidethisellipseis2r
3
A�BZp
3
A+B�p
3
A+Bvuuut
1�[email protected]
q
3
A+B1A2Equatingthetwoareasaftermakinganappropriatescalechange,2r
3
Br
3
AZ1�1p
1�x2dx=2r
3
A�Br
3
A+BZ1�1p
1�x2dx225
Sobysquaringbothsidesofthevectorequation,x2+y2=x2
2+xy+y2
2+1
2+x+yx2
2+y2
2�xy�x�y=1
2x2+y2�2xy�2x�2y=1Exercise9.Center(1=2;2)becauseweequatetheasymptotestoseewheretheyintersect:y=2x+1=�2x+3.(y�2)2
a2�(x�1=2)2
a2=4=1(0;0)���!4
a2�1
a2=3
a2=1(y�2)2
3�(x�1=2)2
3=4=1Exercise10.px2+(p+2)y2=p2+2p.x2
p+2+y2
p=1.(1)Sincep+2�p,thefocimustlieonthexaxis.a2=p+2;b2=a2(1�e2)=p=(p+2)(1�e2).e=q
2
p+2F=ae=p
2.(p
2;0).(2)F=ae=p
2=a(p
3)=)a=q
2
3;b2=2
3(1�3)=�4
3.x2
2=3�y2
4=3=1Exercise11.e=1foranellipse.kX�Fk=jXN�aj=a�XNk�X�Fk=kX+Fk=j�XN�aj=a+XNkX�Fk+kX+Fk=2aExercise12.kX�Fk=ej(X�F)N�dj=e(d�(X�F)N)kX+Fk=ed(X;L)=ej(X�F)N+dj=e(�d�(X�F)N)kX�Fk�kX+Fk=2edX!�Xsofortheotherbranch,kX+Fk�kX�Fk=2edExercise13.(1)(tx)2
a2+(by)2
b2=1�b
t2=a2(1�e2)
t2=�a
t2(1�e2)(2)b21=a21(1�e2)b22=a22(1�e2).1�b21
a21=1�b22
a22;b21
a21=b22
a22x21
a21+y2
b21=1=b2
b1x2
a22+b2
b1y2
(b2y)2(3)(tx)2
a2(ty)2
b2=1�b
t2=�a2(e2�1)
t2=��a
t2(e2�1).b21=a21(e2�1)b22=a22(e2�1)b21
a21+1=e2b22
a22+1=e2b21
a21=b22
a22x
a12y
b12=[email protected]b2
b1x
a21A2 b2
b1y
b2!2=1227
Considerasymptotesingeneral.kX�Fk=ed(X;L).kX�Fk
d(X;L)=e=kX�Fk
jXN�(FN�d)j=kX�Fk
(X�F)N+djForkX�Fk!1,kX�Fk�d.Tokeepratioofe,X�FmustbeultimatelydirectedbyNbyaratioofe.=)e=kX�Fk
kX�Fkcos=1
cose.g.ConsiderN=~ex.x2
a2�y2
b2=1=)y=b
ax=p
e2�1x.Fromthevectorequation,(X�F)N=(x�c;y)N=p
(x�c)2+y2cos=x�cp
(x�c)2+y2
x�c=1
cos=e;(x�c)2+y2
(x�c)2=e2;y2
(x�c)2=e2�1=)y=p
e2�1xForourproblem,considertheconicsectionapproachingtheasymptote.Thentheconicsectionwilllookmorelikethoselinearasymptotes.p
(x�2)2+(y�3)2=x+yy�1
2= (x+1
2)���������!s
x+1
2�5
22+ x+1
2�5
22=x+ x+1
2+1
2=)s
(1+ 2)x+1
22�5(1+ )x+1
2+25
2x!1����!p
1+ 2
(1+ )=) =0Theasymptotesarey=1
2andx=�1
2.Inthesecondpart,eachquadrantmustbechecked.Sofar,IonlyhavethatquadrantIIislled:pointsinquadrantIIIandquadrantIVcannotsatisfythegivencondition.Toseethis,considerquadrantII.kx�Ak=�x+y=p
2(x;y)�1
p
2;1
p
2ForquadrantII,N=�1
p
2;1
p
2.Bydiagram,(X�F)N�0andXN�0.AN=1
p
2�AN=�1
p
2d=1
p
2j(X�F)N+dj=(X�F)N+dTheequationfortheaxisoftheconicsectionisy=�(x�5).Bytakingtheasymptoticlimitlikeabove,wecanshowthat =0again.WeonlysketchthepartofthehyperbolainquadrantII.Bysimilarprocedure,IfoundthatquadrantIII,IVcannotsatisfythecondition.Exercise19.kX�Fk=d(X;L)=j(X�F)N+djx2+y2=(XN+d1)2=y2+2yd1+d21F=0���!x2=2yd1+d21y01=x
d1kX�Fk=j(X�F)N�d2j=d2�(X�F)NF=0���!kXk=d2�yx2+y2=d22�2d2y+y2=)x2=d2�2d2yy02=�x
d2229
vertexisequidistantfromthefocusandthedirectrix.Thus,weneedtoshowthatd(X;L)isequaltothedistancefromthecirclecenterCtothebottompointofQ.LetNbeaunitnormalvectorpointingfromthelinetowardsthefocus,placingthefocusinthepositivehalf-plane.LetCbethecenterofanarbitrarycircleinthefamilyandr1itsradius.LetX1bethepointoftangencybetweencircleQandcircleC.Wewantk(Q+r0N)�Ck=kX2�Ck.ThetangencyconditionbetweencircleQandCmeansthat(X1�C)=� (X1�Q); �0 =r1
r0Q�r0N�C=Q�X1�r0N�C+X1takethemagnitude����������!kQ�X1k2+kX1�Ck2+r20+2(Q�X1)(X1�C)+2(X1�Q)r0N+2(C�X)r0Nr20+r21+r20+2 r20+2r0(1+ )(x1�Q)N2r20+r21+2r1r0+2(r1+r0)(X1�Q)NIhadthoughtthekeyistousethelawofcosinestoevaluate(X1�Q)N=1
(C�X1)N.Lengthl=d(X;L)=d(C;L).Butthatjustgetsusbacktothesameplace.Ihadfoundthesolutionbyacleverconstruction.Buttocometothatconclusionitrequiredmetobe“unstuck”-ifsomethingdoesn'twork,moveontothenext-don'ttrytomakesomethingworkandgoincircles.Andpersistenceiskeybecausetherecanbemanyfalseeurekas.Again,consideraparticularcirclewithitscenterC2rightbelowthegivenQcirclethatjustmakesC2tangentwiththegivenlineL2.ThedirectrixisnotgoingtobeL2butL1,alinetranslatedbelowL2,lineoftangency,byr0,sothatkQ�C2k=r2+r0=d(C2;L1).Itisacleverarticialconstruction.Let'sshowthisforanycircleCofradiusr1inthefamily.TangenttothecircleQcondition:X1�C= (Q�X1).SothenkQ�Ck=r1+r0TangenttothelineL2=B2+tA2:(X2�C)A2=0kX2�Ck=r1ConsiderL1,alinetranslatedbyr0fromL2awayfromQ.IfL2=B2+tA2,L1=B2�r0N+tA2.SinceX2�C=r1(�N)thenX2�r0N�C=(r1+r0)(�N)willpointfromCtoL1,because(X2�r0N)=((B2+tA2)�r0N)2L1.=)kQ�Ck=r1+r0=kX2�r0N�Ck=d(C;L1).Exercise23.Withoutlossofgenerality,usey2=4cx.Thelatusrectumintersecttheparabolaat(c;+2c);(c;�2c).Thus4c=lengthoflatusrectum=2d=2(distancefromfocustodirectrix).y=2p
cxy0=p
c=p
xy0(c)=1Tangentlines:y=(x+c).intersection������!+(x+c)=�(x+c)x=�c(atthedirectrix)Exercise24.Centerofcircleisgiventobe0.Collinearwithcenterandcenternotbetweenthem:P= Q; �0kPkkQk=r20= kQk2231
F0=2e2tA+�2e�2tBF00=4e2tA+4e�2tB=4(F)Exercise15.G0=F0F0+FF00=FF00Exercise16.G=F(F0F00)G0=F0(F0F00)+F(F00F00+F0F000)=F(F0F000)Exercise17.Iflimt!pF(t)=A,8jthcomponent,8r

n�0;9j�0suchthatjFj(t)�Ajjr

nifjt�pjjConsiderminj=1;:::nj=0nXj=1jFj(t)�Ajj2nXj=1r

n2=wheneverjt�pj0=)limt!pkF(t)�Ak=0Iflimt!pkF(t)�Ak=0,8&#x-278;0;9&#x-278;0suchthatq
Pnj=1(Fj(t)�Aj)2ifjt�pj.=)Pnj=1(Fj(t)�Aj)2&#x-278;Pnj=1(Fj(t)�Aj)2&#x-278;(Fk(t)�Ak)2&#x-278;0=)&#x-278;jFk(t)�Akjifjt�pj.Exercise18.IfFisdifferentiableonI,thenF0=nXj=1f0j~ejf0j=limh!01
h(fj(t+h)�fj(t))F0=nXj=1limh!01
h(fj(t+h)�fj(t))=limh!01
hnXj=1(fj(t+h)�fj(t))ej=limh!01
h(F(t+h)�F(t))IfF0(t)=limhto01
h(F(t+h)�F(t))=limh!01
hPnj=1(fj(t+h)�fj(t))ej==Pnj=1limh!01
h(fj(t+h)�fj(t))ej=Pnj=1f0j(t)ejSoF0isdifferentiable.Exercise19.F0(t)=0;8j=1:::n,f0j(t)=0.Byone-dimensionalzero-derivativetheorem,fj(t)=cjconstant.ThusF(t)=Pnj=1cj~ej=ConanopenintervalI.Exercise20.1
6t3A+1
2t2B+Ct+DExercise21.Y0(x)+p(x)Y(x)=Q(x).Then8j=1;:::;ny0j(x)+p(x)yj(x)=Qj(x)Sincep;QarecontinuousonI,andgiventhisinitialvalueconditionyk(a)=bk,yj(x)=e�Rxap(t)dtbj+ZxaQj(t)eRtap(u)dudt=)nXj=1jj(x)=Y(x)=e�RxapB+ZxaQeRtapdtExercise22.233
(b).v=dt
dt~er+rd
dt~e+logsect~ez=cost~er+r~e+tansec
sec~ezvz=tan;v2=cos2t+r2+tan2=sec2cos=tan
sec=sin=r=sint
=arccos(sin)
Exercise6.A=ZR2()d=Z201
2e2cd=e2c
4c 20=e4c�1
4cExercise7.Z01
2sin4d=1
2Z0sin2(1�cos2)d=1
2Z01�cos2
2�1�cos4
2(4)d=3=16Exercise15.Placetargetatthecenter(withoutlossofgenerality).Thestrategyistobreakupvintothepolarcoordinateunitvectors.r=r~erv=dr
dt~er+rd
dt~edr
dt=vr=vcos(� )=�vcos rd
dt=vsin vsin
�vcos =�tan =rd
dt
dr
dt=rd
dr;1
rdr
d=�tan r=e�tan Exercise17.Arstorderdifferentialequationoftheformy0=f(x;y)ishomogeneousiff(tx;ty)=f(x;y).Thenf(rcos;rsin)=f(cos;sin)=f()Wendthatdy
d=dr
dsin+rcosdx
d=dr
dcos�rsinThusdy
dx=dr
dsin+rcos
dr
dcos�rsin=f()Exercise18.v=!krv=dr
dt~er+rd
dt~ev~er=0;sodt
dt=0;!kr=rd
dt~e=!r~e=rd
dtj!krj2=!2r2=r2d
dt2
!= d
dt ;!�0
Exercise19.(a)235

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