for Numerical Analysis - Cengage

1 Student solutions manual and Study Guide Chapters 1 & 2 Preview for Prepared by Richard L. Burden Youngstown State University J. Douglas Faires Youngstown State University Australia Brazil Japan Korea Mexico Singapore Spain United Kingdom United States Numerical Analysis 9th EDITION Richard L. Burden Youngstown State University J. Douglas Faires Youngstown State University 2011 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below.

2 For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at Further permissions questions can be emailed to ISBN-13: 978-0-538-73563-6 ISBN-10: 0-538-73563-5 Brooks/Cole 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: Cengage Learning products are represented in Canada by Nelson Education, Ltd. For your course and learning solutions , visit Purchase any of our products at your local college store or at our preferred online store ContentsPrefa evMathemati alPreliminaries1 Exer alDifferentiationandIntegration61 Exer tMethodsforSolvingLinearSystems119 Exer hniquesinMatrixAlgebra151 Exer alSolutionsofNonlinearSystemsofEquations 199 Exer alSolutionstoPartialDifferentialEquation s233 Exer ePrefa eThisStudentSolutionsManualandStudyGuide forNumeri alAnalysis,NinthEdition.

3 ByBurdenandFaires ontainsrepresentativeexer isesthathavebeenworkedoutindetailforallt hete hniquesdis ularattentionwaspaidtoensurethattheexer isessolvedintheGuidearethoserequiringins ightintothetheoryandmethodsdis isesarealsointheba kofthebook, isestothetextthatinvolvetheuseofa omputeralgebrasystem(CAS).We hoseMapleasourstandardCAS,be ausetheirNumeri alAnalysispa ,anyofthe ommon omputeralgebrasystems,su hMathemati a,MATLAB,andthepubli domainsystem,Sage, anbeusedwithsatisfa enttea hingofthe oursewehavefoundthatstudentsunderstoodth e on eptsbetterwhentheyworkedthroughthealgori thmsstep-by-step,butleta omputeralgebrasystemdothetedious ti etoin ludestru turedalgorithmsinourNumeri alAnalysisbookforallthete hniquesdis anbe odedinanyappropriateprogramminglanguage, , faires/Numeri al- Analysis /youwill nd odeforallthealgorithmswrittenintheprogra mminglanguagesFORTRAN,Pas al,C, nd odeintheformofworksheetsforthe omputeralgebrasystems,Maple,MATLAB.

4 AndMathemati e ttheNumeri alAnalysispa kageandthenumerous ontainsadditionalinformationaboutthebook ,andwillbeupdatedregularlytore e tanymodi ,wewillpla ethereanyerrataweareawareof,aswellasresp onsestoquestionsfromusersofthebook on erninginterpretationsoftheexer isesandappropriateappli ationsofthete anbein orporatedintofutureeditionsofthebookorth esupplements,wewouldbemostgratefultore eiveyour anbemosteasily onta tedbyele troni , eMathemati alPreliminariesExer , (lnx)x= 0hasatleastonesolutionintheinterval[4,5] .SOLUTION:Itisnotpossibletoalgebrai allysolveforthesolutionx,butthisisnotreq uiredintheproblem, (x) =x (lnx)x=x exp(x(ln(lnx))).Sin efis ontinuouson[4,5]withf(4) (5) , (4,5)with0 =f(x) =x (lnx) .Findintervalsthat ontainasolutiontotheequationx3 2x2 4x+ 3 = :Letf(x) =x3 2x2 4x+ riti alpointsoffo urwhen0 =f (x) = 3x2 4x 4 = (3x+ 2)(x 2);thatis,whenx= 23andx= ano (x) = 0,be ausef(x) ef( 2) = 5andf 23 ;f(0) = 3andf(1) = 2;andf(2) = 5andf(4) = 19;solutionslieintheintervals[ 2, 2/3],[0,1],and[2,4].

5 X 1|f(x)|whenf(x) = (2 ex+ 2x) :Firstnotethatf (x) = ( ex+ 2)/3,sotheonly riti alpointoffo ursatx= ln 2,whi hliesintheinterval[0,1].Themaximumfor|f( x)|must onsequentlybemax{|f(0)|,|f(ln 2)|,|f(1)|}= max{1/3,(2 ln2)/3,(4 e)/3}= (2 ln 2) ' (x) =x3+ 2x+k rossesthex-axisexa tlyon e,regardlessofthevalueofthe :Forx <0,wehavef(x)<2x+k <0,providedthatx < ,forx >0,wehavef(x)>2x+k >0,providedthatx > ,thereexistsanumbercwithf(c) = (c) = 0andf(c ) = 0forsomec 6=c, ,thereexistsanumberpbetweencandc withf (p) = ,f (x) = 3x2+ 2> ontradi tiontothestatementthatf(c) = 0andf(c ) = 0forsomec 6= ethereisexa tlyonenumbercwithf(c) = ondTaylorpolynomialforf(x) =excosxaboutx0= 0, ( )toapproximatef( ), ndanupperboundfor|f( ) P2( )|,and omparethistothea |f(x) P2(x)|,forxin[0,1].

6 ApproximateZ10f(x)dxusingZ10P2(x) ( ).SOLUTION:Sin ef (x) =ex(cosx sinx), f (x) = 2ex(sinx),andf (x) = 2ex(sinx+ cosx),wehavef(0) = 1,f (0) = 1,andf (0) = (x) = 1 +xandR2(x) = 2e (sin + cos )3! ( ) = 1 + = |f( ) P2( )| max [ ] 2e (sin + cos )3!( )2 13( )2max [0, ]|e (sin + cos )|.Tomaximizethisquantityon[0, ], rstnotethatDxex(sinx+ cosx) = 2excosx >0,forallxin[0, ].Thisimpliesthatthemaximumandminimumval uesofex(sinx+ cosx)on[0, ]o urattheendpointsoftheinterval,ande0(sin 0 + cos 0) = 1< (sin + cos ) e|f( ) P2( )| 13( )3( ) (a)gives,forallx [0,1],|f(x) P2(x)| 13( )3e1(sin 1+cos 1) .Z10f(x)dx Z101 +x dx= x+x22 10= (b),Z10|R2(x)|dx Z1013e1(cos 1 + sin1)x3dx= eZ10excosx dx= ex2(cosx+ sinx) 10=e2(cos 1 + sin1) 12(1 + 0) ,thea tualerroris| | xtoapproximatesin 1.

7 SOLUTION:Firstweneedto onvertthedegreemeasureforthesinefun = radians,so1 = ef(x) = sinx,f (x) = cosx,f (x) = sinx,andf (x) = cosx,wehavef(0) = 0,f (0) = 1,andf (0) = xisgivenbyf(x) P2(x)andR2(x) = cos 3! |cos | 1,then sin 180 180 = R2 180 = cos 3! 180 3 10 (x) = laurinpolynomialP3(x). (4)(x)andboundtheerror|f(x) P3(x)|on[0,1]. nef(x)byf:=exp x2 sin x3 f:=e(1/2)xsin 13x Then ndthe rstthreetermsoftheTaylorserieswithg:=tay lor(f,x= 0,4)g:=13x+16x2+23648x3+O x4 Extra tthethirdMa laurinpolynomialwithp3 := onvert(g,polynom)p3 :=13x+16x2+ :=diff(f,x,x,x,x)f4 := 1191296e(1/2x)sin 13x +554e(1/2x)cos 13x Findthe :=diff(f4,x)f5 := 1992592e(1/2x)sin 13x +613888e(1/2x)cos 13x Seeifthefourthderivativehasany riti alpointsin[0,1].

8 P:=fsolve(f5 = 0,x, )p:=.6047389076 Theextremevaluesofthefourthderivativewil lo uratx= 0,1, :=evalf(subs(x=p,f4))c1 :=.09787176213c2 :=evalf(subs(x= 0,f4))c2 :=.09259259259c3 :=evalf(subs(x= 1,f4))c3 :=.09472344463 Themaximumabsolutevalueoff(4)(x)isc1andt heerrorisgivenbyerror:=c1/24error:=. |sinx| |x|. 0thefun tionf(x) =x sinxisnon-de reasing,whi himpliesthatsinx xwithequalityonlywhenx= tthatthesinefun tionisoddtorea hthe on :Firstobservethatforf(x) =x sinxwehavef (x) = 1 cosx 0,be ause 1 cosx ,thestatement learlyholdswhen|x| ,be ause|sinx| (x)isnon-de reasingforallvaluesofx,andinparti ularthatf(x)> f(0) = 0whenx > eforx 0,wehavex sinx,andwhen0 x ,wehave|sinx|= sinx x=|x|. < x <0,wehave x > esinxisanoddfun tion,thefa t(frompart(a))thatsin( x) ( x)impliesthat|sinx|= sinx x=|x|.

9 Asa onsequen e,forallrealnumbersxwehave|sinx| |x|. C[a,b],andthatx1andx2arein[a,b]. existsbetweenx1andx2withf( ) =f(x1) +f(x2)2=12f(x1) +12f(x2). existsbetweenx1andx2withf( ) =c1f(x1) +c2f(x2)c1+c2..Giveanexampletoshowthatth eresultinpart(b)doesnotne essarilyholdwhenc1andc2haveoppositesigns withc16= (f(x1) +f(x2))Mathemati alPreliminaries5istheaverageoff(x1)andf( x2), betweenx1andx2withf( ) =12(f(x1) +f(x2)) =12f(x1) +12f(x2). min{f(x1),f(x2)}andM= max{f(x1),f(x2)}.Thenm f(x1) Mandm f(x2) M,soc1m c1f(x1) c1 Mandc2m c2f(x2) (c1+c2)m c1f(x1) +c2f(x2) (c1+c2)Mandm c1f(x1) +c2f(x2)c1+c2 ,thereexistsanumber betweenx1andx2forwhi hf( ) =c1f(x1) +c2f(x2)c1+c2..Letf(x) =x2+ 1,x1= 0,x2= 1,c1= 2,andc2= (x)>0forallvaluesofx,butc1f(x1) +c2f(x2)c1+c2=2(1) 1(2)2 1= ,page282.

10 Findthelargestintervalinwhi hp mustlietoapproximate 2withrelativeerroratmost10 :Weneed p 2 2 10 4,so p 2 2 10 4;thatis, 2 10 4 p 2 2 10 mustbeintheinterval 2( ), 2( ) . to ompute1314 672e , :Usingthree-digitroundingarithmeti gives1314= ,67= ,ande= 67= = = e1314 672e orre ,sotheabsoluteandrelativeerrorstothreedi gitsare| |= | | 8,respe ise5(e)usingthree-digit hoppingarithmeti .SOLUTION:Usingthree-digit hoppingarithmeti gives1314= ,67= ,ande= 67= = = e1314 672e orre ,sotheabsoluteandrelativeerrorstothreedi gitsare| |= ,and| | ,respe ise5(e)were rstthreetermsoftheMa laurinseriesforthear tangentfun tiontoapproximate = 4 arctan12+ arctan13 , :LetP(x) =x 13x3+ 12 = 13 = ,so = 4 arctan12+ arctan13 ,respe tively,| | 10 3and| || | 10 (x) =ex e 0f(x).