Solution Manual For Numerical Methods Using Matlab, 4th Edition

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SolutionManualforNumericalMethods:UsingMATLABJohnH.MathewsandKurtisD.FinkAugust2002

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Contents1Preliminaries5itdfibReview(oflGaleulnstIHITHTATRETRHTHIEECARTARITITEERSE231avyNeversijiHTAGTTAHREECRCTTIRIHIRRICEdt3iierrorEAmalyeistildFT—HIEIRATGERTHOUECHRHACATIEEATof2TheSolutionofNonlinearEquationsf(z)=0152.91irveracianfarSolvingaiiioe1NARRECAPREACHARCHECITRHG2.2BracketingMethodsforLocatingaRoot..............182.3InitialApproximationandConvergenceCriteria..........202.4Newton-RaphsonandSecantMethods...............212.5Aitken’sProcessandSteffensen’sandMuller'sMethods.....273TheSolutionofLinearSystemsAX=B333.1IntroductiontoVectorsandMatrices................33BallPropertiesofVectors|andMatrices(lIIHHIRMAHITERIITREAAIRITNTZ3.3Upper-TriangularLinearSystems..................373.4GaussianEliminationandPivoting.................38disiiltranenlar|kactorization(i(TIHUIIARIRTAIATIETATIRTTIETETRSG3.6IterativeMethodsforLinearSystems................40SiidiifiTveration|forNonlinearSystems!EIHIHHEEIHWITHEECITIHHTEGS4InterpolationandPolynomialApproximation514.1TaylorSeriesandCalculationofFunctions.............51dillIntraduction|tolInterpolation([LHLHIIHHATHERIRTIRIGIRTRIHIETETNGddlNasrange[ApprosimationiilHARTIRIRCIIHITREARNIERENRTITsoddlNewtonBolynovayals[RHLHEAMHRHRGVETREIIRGIRIGRIITITHARITIGdoiChebysheviFolynomialsi(Gptionat)iILIMHTARIHIITIEGIEGITITITHGGSdibilliFadeAppraxamationsfltiditEITHERGS5CurveFitting73ia:iliffIeast.SquaresilefillyHRHGRIEEATIHEEIERTIRIes2curyetineRTTIHEATERIHETAUHSIEHINTNERATIONHeoeffInterpolationbylSphnekmetionsITHATITHETNCNEATIEREIEITITTNGq3

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4CONTENTS5.4FourierSeriesandTrigonometricFrotynoratalsutRGHIATERTIEREIEEEerowlifiisezerGurves(HLERIEIHGREEMIEHHITITTTTIIIHIAIIRTRIIETTNS6NumericalDifferentiation93631ADproximatintheDerivativeMTIITITHITIHENRTIRTHTRITITITHITGS6.2NumericalDifferentiationFormulas.................077NumericalIntegration101[{:dllilintroduction|tolQuadrature(itEHTIRATITIRICLETVATABIAICRRARICEOATG17.2CompositeTrapezoidalandSimpson'sRule............1057.3RecursiveRulesandRombergIntegration.............114EeCREbfallGaussTreocndraiTntegration|litfHIHEMEARITGITITTIEITITIEITCHS8NumericalOptimization121a:dliliMminmizationiofafimetionltHTHRHEATARIETAEEEIEIEEACROSS)82Nelder-MeadandPowell'sMethods.................128.3Gradient/andNewtonsMethods([HHTTHHICITARGRIRRTETTIITRog9SolutionofDifferentialEquations1359.1IntroductiontoDifferentialEquations...............1350:2lllEnle=Nethod|iLMMRIMIACIHEARERIIERCRENERTIETITRIEATSS3Hens!Method](LHCHTRCRERRTARENIEAERENEEAtdiTaviorSeriesiMcthodMLILFTMITTTRERIFIIARIRICARRINUARINColliiRumeeKattalncthodstWTLURIHITRRNHITREITIIRAINCOREFAalibliiBredictor.(orrectoniMethodsflITARMRRFRAHITATIRHRCHAIICIIING9urlsystems!ofInferentialiauations|IITHTHTMTENTIRITIRITITORITAATII9:8BoundaryValueProblems|MTHTITATITHITEITIMTIETTRIEITTIHITGGo.9kmiteDifferenceMethod!ftHIHIARITGIATRIREATIIEICINIRIRRNRSG10SolutionofPartialDifferentialEquations15504hByperboliclEquationsHAGCHGTLHTETBETEITITITIRTTT621araboliclEquations(MHHEARTRIRHIREIEITIEEITTIEIRRe03likllipticikquationsi|HsuECEHREARIEITEEETTITRRTIEITINTHL6011EigenvaluesandEigenvectors16311.1HomogeneousSystems:TheEigenvalueProblem.........163C2owerihettiodiREHEATING6si3Jacobisivethad]ititiiuiipmsgurgigRmtiRigiraiiiirmitGeo11.4EigenvaluesforSymmetricMatrices................170

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Chapter1...Preliminaries1.1ReviewofCalculusL(8)L=limyoofol=2limpoo€n=lipo(2—faut)=2=%=0(b)limy,oo2mptonl—21liceen=(3—ect)=4-1=02.(a)limyoosin(zn)=sin(limy_,oozn)=sin(2)(b)limyooIn(x2)=In(limpoo22)=In(4)3.(a)Sincefiscontinuouson[1,0];solveinatebbsHLkoiRGJH24/7ACT)¢c=1-vZel-1,0](b)Sincefiscontinuouson[6,8];solveViT-55-2=3Z?-5x—11=05£/Faminboll5e=EBs4.(a)f'(z)=22—3=0,thusthecriticalpointsare¢=+1.Thusmin{f(~1),f(1),£(2)}=min{5,—1,~1}=~1andmaz{f(-1),f(1),/(2)}=maz{5,-1,~1}=5(b)f(z)=~2cos(x)sin(x)—cos(z)=—cos(z)(2sin(z)+1)=0,thusthecriticalpointsare¢=7,77/6,117/6.Thus5+StudyXY

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6CHAPTER1.PRELIMINARIESmin{f(0),f(r),f(77/6),f(117/6),f(2m)}=min{1,1,5/4,5/4,1}=1andmaz{f(0),f(r),f(7x/6),f(117/6),f(2r)}=max(1,1,5/4,5/4,1}=5/45.(a)f(z)=42°—82=d(x?—2)=0,thusc=0,42€[2,2](b)f(x)=cos(z)+2co8(2z)=cos(z)+2(2co8*(x)—1)=4cos?(z)+cos(z)—2=0T=(-1£33)/8¢=cos'((-1£/33)/8),2r—cos™I((—1=v/33)/8)6.(a)f(x)=ghzandL200)~1SolvingLo=§yields=1.(b)f(x)=(2?+22)/(x+1)?andLO=LO—1Solvingf(z)=(#2+22)/(z+1)>=Lyieldsc=—1+27.ThegivenfunctionsatisfiesthehypothesesoftheGeneralizedRolle’sThe-orem.Sincef(0)=f(1)=f(3)=0,thereexistaa¢€(0,3)suchthatJ"(€)=0.Solve6c—8=0tofind¢=4/3.8.(a)[Pzetdr=e"—eof=e?+1(b)J2,#5dz=$1n(2?+1)[',=0(Theintegrandisanoddfunction)9.(a)4fit*cos(t)dt=22cos(z)Ob)£7edt=oo?(322)=3o2en®10.(8)gpg[26e%dz=22°14=52.Solving62%=52yields¢=LapropeMJ(6)&Jy"?zeos(z)dz=(wsin(e)+cos(x))3=—(1+2).Useacalculatortoapproximatethesolution(s):cos(z)=—(1+=a12.16506,4.43558€(0,37/2].na)2p=2(b)or=3(©)Comy=30(4—aby)=3limicoo5(2—2p)=3(d)inw=C=i1)=ShinoYi(ty—me)=§—StudyXY

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1.2.BINARYNUMBERSKf12.(a)“gl-D+k@-1P-Je-12+A=—1)+1(b)dz?+3z+1(c)got—$2?+113.TheTaylorpolynomialofdegreen=4expandedaboutzo=0forflx)=sin(z)isP(x).14.(a)P(3)=-24EfUGFLEE15.Theaverageareaisgivenby;71;Iride=(23=i16.AnypolynomialP(x)satisfiesthehypothesesofRolle’sTheoremontheinterval[a,b].ThusP'()hasatleastn—1realrootsintheinterval[a,b],P(x)hasatleastn—2realrootsintheinterval[a,b],...,andP("=1hasatleast7—(n—1)=1rcalrootintheinterval[a,b].17.Iff,fandf”aredefinedontheinterval[a,b],thenfiscontinuousontheinterval[a,b]andfisdifferentiableontheinterval(a,b).ByTheorem1.6(MeanValueTheorem)thereexistsnumbers¢;€(a,c)and¢;€(¢,b)suchthat:1)—I(a),10)=fe)(ey)=om)=1)=LTDngpep)=LO-But,sincef(a)=f(b)=0itfollowsthatf/(c;)=f(c)/(c—a)andf'(e2)=f(c)/(c—b).Giventhatf’andf”aredefinedintheinterval[a,b],itfollowsthatf”alsosatisfiesthehypothesesofTheorem1.6.Thusthereexistsanumberd€(a,b)suchthat:fayl@-re)8-82 jev-ocy—cyca-e(2—e)e=b)(c—a)~~sincef(c)>0.1.2BinaryNumbers1.Answerswilldependonspecificplatform.2.(a)21(b)56(c)254(d)5193.(a)0.75(b)0.65625(c)0.6640625(d)0.855468754.(a)14140625(b)314160156255.(a)v2—1.4140625=0.00015109...(b)7~3.1416015625=—0.000008908...+StudyXY

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8CHAPTER1.PRELIMINARIES6.(2)23=10111,2=2011)+1b=1nifAveyIR,UHATR|5=22)+1b=02=201)+0b3=0DUETYsHERSEES(b)87=10101114,87=2(43)+1bo=1silionyapgl21=2010)+1by=110=2(5)4+0by3=05=2AD+1B=12=201)+0bs=0EE"EDME(c)378=101111010,(d)2388=10010101010007.(a)00111ee (b)0.1101uo (c)0.101114(d)010010114,8.(a)0.00011,(b)§=0.102000=0.01402R=%d=0=INT(3)Fi=3=FRAC(3)2F,=ida=1=-INT(3)F=3=FRAC(3)Wr=Fdy=0=INT(3)Fo=§=FRAC(H)23=3di=1=INT(})F=1=FRrAC()(¢)4=0.d1d2d3140=0.001102R=%di=0=INT()Fy=?=FRAC(3)2F=Hdy=0=INT(2)P=dmFRAC(3)2Wy=gdy=1=INT(Z)Fy=g=FRAC(3)2F=3di=0=INT(3)F,=2=FRAC()9.(a)5=0.0001100iw0=0.0001Trap—00001100410=0.0000000TT00z4,.,==0.00625I~StudyXY

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1.2.BINARYNUMBERS9(b)#=0.0010010r0=0.00Tsuo—0.010010,=0.00000000100T0-=00022321428...10.InTheorem1.14let¢=§and7=1,thenELUMJAI8764"512EE11.InTheorem1.14let¢=3/16andr=1/16,thenELE)3J1611256iffanaeinifamin=12.1=5.Assume(H*er=.ThenHA[ilarGB-665*5-(#)(%)ph+1MifinotealTherefore,bytheprincipleofmathematicalinduction,2=Vcanberepre-sentedasadecimalnumberthathasNdigits.13.(a)i~0.10114x271=0.1011x27!silo:onsWi2d5l0:011010b<|27z0.1000111px20%~0.1001rwox2°=0.100140x2°5=0.101Luox272=0.001011,x2°px0.101110x2°Thus(§+1)+&~0.1100¢w0(b)%20.1100nex27%=0.001101x2713~0101lpex27!=0.101100x2-1pes0.171001x2-7+StudyXY

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10CHAPTER1.PRELIMINARIESi#&0.1110pex271=0.011100;x2°=~0.1101peXx272=0.001101,x2°pi0.101001100x20Thus(&+3)+£~0.101040(c)%~0.1011x272=0.01011x2715=0.1110x27%=0.001110x2-1STITTspec50.10010X271=0.100140x2717=~0.10014Xx272=0.010014,x271FoDITOTImEdomThus(%+2)+4~0.111040x271(d)5~0.1011pex2°=0.1011000x2°3~0.1110mex27%=0.0001110x2°%20.1101x2°=0.1101004,0x2°7~0.1011x272=0.001001,x2°TESSmnrnmeThus(F5+5)+4~0.11114,14.(2)10=101¢hree(b)23=212uree(c)421=1201214pree(d)1784=211002¢nree15.(a)3=0.Lnree(b)3=0Tohree(€)15=0.0022¢hree[IEQE16.(a)(a)10=20se(b)(b)35=12040,(c)(c)721=103440(d)(d)734=1041444,17.(a)3=0T3fi0e(b)£=0Zp.(c)75=0.02510(d)$2=0.1104,

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1.3.ERRORANALYSIS111.3ErrorAnalysisI(a)©—&=0.00008182,22%=(.0000300998.....,4-significantdigitsb)y—§=350,42=0.0355871.....,2-significant,digitsv(¢)z=2=0.000008,===0.117647,O-significantdigitsalRienegHypoil(r+54shay+stm)=itmmtsem=Goad?”~0.2553074428=p3.(a)p1+p2=1.41440.09125=1.505pip2=(2.1414)(0.09125)=0.1290(b)p1+p2=31.415+0.027182=31.442P1p2=(31.415)(0.27182)=0.853924.(a)OT0711885222010710678110ik0.0000C767108=0.707103Theerrorin-volveslossofsignificance.(b)0.09317218025-0.6931478036J0.00009499909=0.4999938Theerrorin-volveslossofsignificance.5.(a)In(2)lu0)For(c)cos(2x)(d)cos(z/2)6.(a)bdHECKplEEfpJHEELE=20.12-3(7.398)+8.16—1=20.12-22.194+8.16—1=5.09rileRRRfolRUEdrEBSAME=(~0.2800)(2.72)+3)(2.72)~1=(-0.7616+3)(2.72)—1=(2.2384)(2.72)1=6.088-1=5.088R(2.72)=(272-1)EGUh=5.088+StudyXY

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12CHAPTER1.PRELIMINARIES(b)P(0.975)=(((0.975)°—3(0.975)%)+3(0.975))—1=(0.9268—3(0.9506))+2.925)—1=(0.9268—2.852)+2.925)+1=(~1.925+2.925)~1=1-1=0Q(0.975)=((0.975—3)(0.975)+3)(0.975)—1=((-2.025)(0.975)+3)(0.975)—1=(~1.9774+3)(0.975)—1=(1.026)(0.975)—1=1-1=0R(0.975)=(0.975—1)%=(-0.025)=—0.00001562To(a)F+§+++ohog0.498(0)Ag+ais+a+a++in04998.(a)Thepropagationoferroris€,+¢,+€,.(b)p_itejothaalllatrisallaiiaiedHence,if1<|g]<|p|,thenthereisapossibilityofmagnificationoftheoriginalerror.(e)paro=(Ptep)d+e)F+e)=PUT+preg+acy+Ply+Fepey+Gener+Peger+peer=dF+(Beg+dep+Pier)H(Pepeq+Geper+Pegi)+epcqerDependingontheabsolutevaluesofp,,and#,thereisapossibilityofmagnificationoftheoriginalerrorsep,€4,and€,.9.2toos(h)=2+h+841HORY)(25)cosh)=1+h+5+540h)I~StudyXY

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1.3.ERRORANALYSIS1310.ef4sin(h)=1+2%+4+O(h?)etsin(h)=h+h+ELOR)AnintermediatecomputationwasBaillAalllad3SALcorceAthtsrg+=gr)=hthHETEr{meme11.Iiilalllsllcon)£5)=14hKBBE4OK)cos(h)sin(h)=h-2-42+O(hT)AnintermediatecomutationwasLallbSl2ril1i2RsTrITEDITA12.x==b/FdaEe=)famiimiBa(VFac)=taeThecaseforx7ishandledinasimilarmanner.13.(a)==—0.001000,x2=—1000(b)1=—0.00100,7=—10000(¢)1=—0.000010,z5=—100000(d)z;=—0.000001,22=~1000000+StudyXY

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14CHAPTER1.PRELIMINARIES+Studyxy

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Chapter2..TheSolutionofNonlinear.Equationsf(z)=02.1IterationforSolvingr=g(x)1.(a)Clearly,g(z)€C[0,1].Since¢'(z)=—/2<0ontheinterval[0,1],thefunctiong(z)isstrictlydecreasingontheinterval[0,1].Ifgisstrictlydecreasingon[0,1],theng(0)=1andg(1)=0implythat9([0,1])=[0,1]€[0,1].Thus,byTheorem2.2,thefunctiong(x)hasafixedpointontheinterval[0,1].Inaddition:|f'(z)}=|—#/2|=z/2<1/2<1ontheinterval[0,1].Thus,byTheorem2.2,thefunctiong(x)hasauniquefixedpointontheinterval[0,1].(b)Clearly,g(x)€C[0,1].Sinceg/(x)=—In(2)2™<0ontheinterval[0,1],thefunctiong(x)isstrictlydecreasingontheinterval[0,1].Tfgisstrictlydecreasingon[0,1],then(0)=1andg(1)=1/2implythatg([0,1])=[1/2,1]C[0,1].Thus,byTheorem2.2thefunctiong(x)hasafixedpointontheinterval[0,1].Inaddition:|¢’(2)|=|~1n(2)272|=In(2)272<In(2)<In(e)=1ontheinterval[0,1].Thus,byTheorem2.2,thefunctiong(x)hasanuniquefixedpointontheinterval[0,1].(c)Clearlyg(x)iscontinuouson[0.5,5.2]andg([0.5,5.2])€[0.5,5.2].But,9([0.5,2])C[0.5,2].Thus,thehypothesesofthefirstpartofTheorem2.2aresatisfiedandghasafixedpointin[0.5,2].While(1,1)istheuniquefixedpointin[0.5,2],[f/(1)]=1#£1,thusthehypotheseeinpart(4)ofTheorem2.2cannotbesatisfied.15+StudyXY

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