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Solution Manual For Numerical Methods Using Matlab, 4th Edition

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    Solution Manual for Numerical Methods: Using MATLAB John H. Mathews and Kurtis D. Fink August 2002

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    Contents 1 Preliminaries 5 itd fib Review (ofl Galeulnst I HITHTATRETRHTHIEECARTARITITE ERS E231 avy Nevers ij iHTAGTTAHREECRCTTIRIHIR RICE dt3iierror EAmalyeistil dF T—HIEIRATGERTHOUECHRHACATIE EAT of 2 The Solution of Nonlinear Equations f(z) = 0 15 2.91 ir vera cian far Solving aii ioe 1 NAR RECA PREACH ARCHECITRHG 2.2 Bracketing Methods for Locatinga Root . . . . .......... 18 2.3 Initial Approximation and Convergence Criteria. . . . ...... 20 2.4 Newton-Raphson and Secant Methods . . . . ........... 21 2.5 Aitken’s Process and Steffensen’s and Muller's Methods . . . . . 27 3 The Solution of Linear Systems AX = B 33 3.1 Introduction to Vectors and Matrices . . . ............. 33 Ball Properties of Vectors|and Matrices (lI IHHIRMAHITERIITREAAIRITNTZ 3.3 Upper-Triangular Linear Systems . . . . .............. 37 3.4 Gaussian Elimination and Pivoting . . . .............. 38 disiiltranenlar|kactorization (i (TIHUIIARIRTAIATIETATIRTTIETETRSG 3.6 Iterative Methods for Linear Systems . . . .. ........... 40 SiidiifiTveration|for Nonlinear Systems! EI HIHHEEIHWITHEECITIHHTE GS 4 Interpolation and Polynomial Approximation 51 4.1 Taylor Series and Calculation of Functions . . . . .. ....... 51 dill Intraduction|tol Interpolation ([LHLHIIHHATHERIRTIRIGIRTRIHIETETNG ddl Nasrange [Approsimationiil HARTIRIRCIIHITREARNIERENRTIT so ddl Newton Bolynovayals [RHLHEAMHRHRGVETREIIRGIRIGRIITITHARITIG doi ChebysheviFolynomialsi(Gptionat)i IL IMHTARIHIITIEGIEGITITITHG GS dibilliFade Appraxamationsfltidit EITHER GS 5 Curve Fitting 73 ia:iliff Ieast. Squares ile filly HRHGRIEEATIHEEIERTIRIes 2c urye tine RTTIHEATERIHETAUHSIEHINTN ERATION He oe ff Interpolation byl Sphne kmetions ITHATITHETNCNEATIEREIEITITTN Gq 3

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    4 CONTENTS 5.4 Fourier Series and Trigonometric Frotynoratals ut RGHIATERTIERE IEEE er owlifiisezer Gurves(HLERIEIHGREEMIEHHITITTTTIIIHIAIIRTRIIETTNS 6 Numerical Differentiation 93 631A Dproxim atin the Derivative MTIITITHITIHENRTIRTHTRITITITHIT GS 6.2 Numerical Differentiation Formulas . . . . ............. 07 7 Numerical Integration 101 [{:dllilintroduction| tol Quadrature (it EHTIRATITIRICLETVATABIAICRRARICEOAT G1 7.2 Composite Trapezoidal and Simpson's Rule . . .......... 105 7.3 Recursive Rules and Romberg Integration . . . .......... 114 Ee CR Eb fall Gauss TreocndraiTntegration|litfHIHEMEARITGITITTIEITITIEI TCHS 8 Numerical Optimization 121 a:dliliMminmizationi ofa fimetion lt HTHRHEATARIETAEEEIEIEEACRO SS) 82 Nelder-Mead and Powell's Methods . . . ..............12 8.3 Gradient/and Newtons Methods ([HHTTHHICITARGRIRRTETTIITR og 9 Solution of Differential Equations 135 9.1 Introduction to Differential Equations . .............. 135 0:2lllEnle = Nethod|iLMMRIMIACIHEARERIIERCRENERTIETITRI EAT SS 3H ens! Method] (LHC HTRCRERRTARENIEAE RENEE A tdi Tavior SeriesiMcthod MLILFTMITTTRERIFIIARIRICARRINU ARINC olliiRumee Ka ttalncthodst WTLURIHITRRNHITREITIIRAINC ORE FA al ib liiBredictor. (orrecton iMethodsflITARMRRFRAHITATIRHRCHAIICIIING 9 urls ystems! of In ferentialiauations|IITHTHTMTENTIRITIRITITORITAATII 9:8 Boundary Value Problems| MTHTITATITHITEITIMTIETTRIEITTIHIT GG o.9kmite Difference Method! ft HIHIARITGIATRIREATIIEICINIRIR RNR SG 10 Solution of Partial Differential Equations 155 04h Byperbolicl Equations HAGCHGTLHTETBETEITITITIR TTT 6 21 arabolicl Equations (MHHEARTRIRHIREIEITIEEITTI EIR Re 03 likllipticikquationsi| Hsu ECEHREARIEITEEETTITRRTIEITINTHL 60 11 Eigenvalues and Eigenvectors 163 11.1 Homogeneous Systems: The Eigenvalue Problem . ........ 163 C2 ower ihettiod i REHEATING 6s i3 Jacobi sivethad]ititiiuiipmsgurgigRmtiRigiraiiiirmitG eo 11.4 Eigenvalues for Symmetric Matrices . ............... 170

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    Chapter 1 . . . Preliminaries 1.1 Review of Calculus L (8) L=limy oo fol =2 limp oo €n = lipo (2 — faut) =2=%=0 (b) limy, oo 2mptonl — 2 1 lice en = (3 — ect) = 4-1 =0 2. (a) limy oo sin(zn) = sin(limy_,oo zn) = sin(2) (b) limyoo In(x2) = In(limp oo 22) = In(4) 3. (a) Since fis continuous on [1,0]; solve inate bbs HL koi RG JH 24/7 ACT) ¢c = 1-vZel-1,0] (b) Since f is continuous on [6,8]; solve ViT-55-2 = 3 Z?-5x—11 = 0 5£/F amin boll 5 e = EBs 4. (a) f'(z) = 22 — 3 =0, thus the critical points are ¢ = +1. Thus min{ f(~1), f(1), £(2)} = min{5,—1,~1} = ~1 and maz{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, thus the critical points are ¢ = 7,77 /6,117/6. Thus 5 + StudyXY

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    6 CHAPTER 1. PRELIMINARIES min{f(0), f(r), f(77/6), f(117/6), f(2m)} = min{1,1,5/4,5/4,1} = 1 and maz{f(0), f(r), f(7x/6), f(117/6), f(2r)} = max(1,1,5/4,5/4,1} = 5/4 5. (a) f(z) = 42° — 82 = d(x? — 2) = 0, thus c = 0,42 € [2,2] (b) f(x) = cos(z) + 2co8(2z) = cos(z) + 2(2co8*(x) — 1) = 4cos?(z) + cos(z) — 2 =0 T = (-1£33)/8 ¢ = cos'((-1£/33)/8),2r — cos™I((—1 = v/33)/8) 6. (a) f(x) = ghz and L200) ~ 1 Solving Lo = § yields = 1. (b) f(x) = (2? + 22)/(x + 1)? and LO=LO — 1 Solving f(z) = (#2 +22) /(z + 1)> = L yields c = —1 + 2 7. The given function satisfies the hypotheses of the Generalized Rolle’s The- orem. Since f(0) = f(1) = f(3) = 0, there exista a ¢ € (0, 3) such that J"(€) = 0. Solve 6c — 8 = 0 to find ¢ = 4/3. 8. (a) [Pzetdr =e" — eof =e? +1 (b) J2, #5 dz = $1n(2? + 1)[', = 0 (The integrand is an odd function) 9. (a) 4 fi t* cos(t)dt = 22 cos(z) Ob) £7 edt = oo? (322) = 3o2en® 10. (8) gpg [2 6e%dz = 22°14 = 52. Solving 62% = 52 yields ¢ = La prope MJ (6) & Jy"? zeos(z)dz = (wsin(e) + cos(x))3 = —(1 + 2). Use a calculator to approximate the solution(s): cos(z) = —(1 + = a1 2.16506, 4.43558 € (0, 37/2]. na) 2p=2 (b) or =3 (©) Co my = 30 (4 — aby) = 3limicoo 5 (2 — 2p) = 3 (d) in w= C= i 1) = Shino Yi (ty — me) = § — StudyXY

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    1.2. BINARY NUMBERS Kf 12. (a) “gl -D + k@-1P- Je - 12+ A= —1) +1 (b) dz? +3z +1 (c) got — $2? +1 13. The Taylor polynomial of degree n = 4 expanded about zo = 0 for flx)= sin(z) is P(x). 14. (a) P(3)=-24 Ef UG FLEE 15. The average area is given by; 71; I ride = (23 =i 16. Any polynomial P(x) satisfies the hypotheses of Rolle’s Theorem on the interval [a,b]. Thus P'() has at least n— 1 real roots in the interval [a,b], P(x) has at least n — 2 real roots in the interval [a,b], ..., and P("=1 has at least 7 — (n — 1) = 1 rcal root in the interval [a,b]. 17. If f, f and f” are defined on the interval [a,b], then f is continuous on the interval [a,b] and f is differentiable on the interval (a,b). By Theorem 1.6 (Mean Value Theorem) there exists numbers ¢; € (a,c) and ¢; € (¢,b) such that: 1) — I(a) , 10) = fe) (ey) = om) = 1) = LTD ng pep) = LO - But, since f(a) = f(b) = 0 it follows that f/(c;) = f(c)/(c — a) and f'(e2) = f(c)/(c—b). Given that f’ and f” are defined in the interval [a,b], it follows that f” also satisfies the hypotheses of Theorem 1.6. Thus there exists a number d € (a,b) such that: fay l@-re) 8-82 jev-o cy —cy ca-e (2 —e)e=b)(c—a) ~~ since f(c) > 0. 1.2 Binary Numbers 1. Answers will depend on specific platform. 2. (a) 21 (b) 56 (c) 254 (d) 519 3. (a) 0.75 (b) 0.65625 (c) 0.6640625 (d) 0.85546875 4. (a) 14140625 (b) 31416015625 5. (a) v2 — 1.4140625 = 0.00015109. .. (b) 7 ~ 3.1416015625 = —0.000008908. . . + StudyXY

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    8 CHAPTER 1. PRELIMINARIES 6. (2) 23 = 10111, 2 = 2011) +1 b=1 nifA vey IR, UH ATR | 5 = 22) +1 b=0 2 = 201)+0 b3=0 DUET Ys HER SEES (b) 87 = 10101114, 87 = 2(43)+1 bo=1 sili ony ap gl 21 = 2010) +1 by=1 10 = 2(5)4+0 by3=0 5 = 2AD+1 B=1 2 = 201)+0 bs=0 EE" ED ME (c) 378 = 101111010, (d) 2388 = 1001010101000 7. (a) 00111ee (b) 0.1101uo (c) 0.101114 (d) 010010114, 8. (a) 0.00011, (b) § = 0.102000 = 0.0140 2R=% d=0= INT(3) Fi= 3 = FRAC(3) 2F, = i da =1=-INT(3) F=3= FRAC(3) Wr=F dy=0= INT(3) Fo=§= FRAC(H) 23=3 di=1=INT(}) F=1=FRrAC() (¢) 4 = 0.d1d2d3140 = 0.00110 2R=% di=0= INT() Fy = ? = FRAC(3) 2F = H dy=0= INT(2) P= dm FRAC(3) 2Wy=g dy=1= INT(Z) Fy=g= FRAC(3) 2F=3 di=0=INT(3) F,=2=FRAC() 9. (a) 5 = 0.0001100iw0 = 0.0001 Trap — 00001100410 = 0.0000000TT00z4,., = = 0.00625 I~ StudyXY

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    1.2. BINARY NUMBERS 9 (b) #=0.0010010r0 = 0.00Tsuo — 0.010010, = 0.00000000100T 0 - = 00022321428... 10. In Theorem 1.14 let ¢ = § and 7 = 1, then ELUM JAI 87 64" 512 EE 11. In Theorem 1.14 let ¢ = 3/16 and r = 1/16, then ELE) 3 J 1611256 iffanae i nif a min = 12. 1 = 5. Assume (H* er =. Then HA [ilar GB - 66 5* 5 - (#)(%) ph+1 Mifinoteal Therefore, by the principle of mathematical induction, 2=V can be repre- sented as a decimal number that has N digits. 13. (a) i ~ 0.10114 x 271 = 0.1011 x 27! silo: onsWi2d 5 l0:011010b<| 27 z 0.100011 1p x 20 % ~ 0.1001rwo x 2° = 0.100140 x 2° 5 = 0.101Luo x272 = 0.001011, x 2° px 0.101110 x 2° Thus (§ + 1) + & ~ 0.1100¢w0 (b) % 2 0.1100ne x 27% = 0.001101 x 271 3 ~ 0101lpe x27! = 0.101100 x 2-1 pes 0.171001 x 2-7 + StudyXY

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    10 CHAPTER 1. PRELIMINARIES i # & 0.1110pe x 271 = 0.011100; x 2° =~ 0.1101pe Xx 272 = 0.001101, x 2° pi 0.101001 100 x 20 Thus (& +3) + £ ~ 0.101040 (c) % ~ 0.1011 x 272 = 0.01011 x 271 5 = 0.1110 x 27% = 0.001110 x 2-1 STITT spec 5 0.10010 X 271 = 0.100140 x 271 7 =~ 0.10014 Xx 272 = 0.010014, x 271 Fo DITOTIm Ed om Thus (% + 2) + 4 ~ 0.111040 x 271 (d) 5 ~ 0.1011pe x 2° = 0.1011000 x 2° 3 ~ 0.1110me x 27% = 0.0001110 x 2° % 2 0.1101 x 2° = 0.1101004,0 x 2° 7 ~ 0.1011 x 272 = 0.001001, x 2° TESS mn rn me Thus (F5 +5) + 4 ~ 0.11114, 14. (2) 10 = 101¢hree (b) 23 =212uree (c) 421 = 120121 4pree (d) 1784 = 211002¢nree 15. (a) 3 =0.Lnree (b) 3 =0Tohree (€) 15 = 0.0022¢hree [IE QE 16. (a) (a) 10 = 20 se (b) (b) 35 = 12040, (c) (c) 721 = 103440 (d) (d) 734 = 10414 44, 17. (a) 3 =0T3fi0e (b) £=0Zp. (c) 75 = 0.02510 (d) $2 = 0.1104,

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    1.3. ERROR ANALYSIS 11 1.3 Error Analysis I (a) © —& = 0.00008182, 22% = (.0000300998....., 4- significant digits b) y— § = 350,42 = 0.0355871 ....., 2-significant, digits v (¢) z= 2=0.000008, == = 0.117647, O-significant digits al Rien eg Hy po il (r+ 54 shay + stm) = itm mt sem = Goad?” ~ 0.2553074428 = p 3. (a) p1+p2 = 1.4144 0.09125 = 1.505 pip2 = (2.1414)(0.09125) = 0.1290 (b) p1 + p2 = 31.415 + 0.027182 = 31.442 P1p2 = (31.415)(0.27182) = 0.85392 4. (a) OT0711885222 010710678110 ik 0.0000C767108 = 0.707103 The error in- volves loss of significance. (b) 0.09317218025-0.6931478036 J 0.00009499909 = 0.4999938 The error in- volves loss of significance. 5. (a) In(2) lu 0) For (c) cos(2x) (d) cos(z/2) 6. (a) bd HECK pl EE fp JHE ELE = 20.12 -3(7.398) + 8.16 — 1 = 20.12 -22.194+8.16—1 = 5.09 rile RRR fol RUE dr EB SAME = (~0.2800)(2.72) + 3)(2.72) ~ 1 = (-0.7616+3)(2.72) — 1 = (2.2384)(2.72) 1 = 6.088-1 = 5.088 R(2.72) = (272-1) EG Uh = 5.088 + StudyXY

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    12 CHAPTER 1. 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=0 Q(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=0 R(0.975) = (0.975 —1)% = (-0.025) = —0.00001562 To (a) F+§+ ++ oh og 0.498 (0) Ag +ais +a +a ++ in0499 8. (a) The propagation of error is €, + ¢, + €,. (b) p_ite j otha alll atrisalla iia ied Hence, if 1 < |g] < |p|, then there is a possibility of magnification of the original error. (e) paro= (Ptep)d+e)F +e) = PUT + preg + acy + Ply + Fepey + Gener + Peger + peer = dF + (Beg + dep + Pier) H(Pepeq + Geper + Pegi) + epcqer Depending on the absolute values of p, , and #, there is a possibility of magnification of the original errors ep, €4, and €,. 9. 2 toos(h) = 2+h+ 8 41H ORY) (25) cosh) = 1+h+5 +5 40h) I~ StudyXY

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    1.3. ERROR ANALYSIS 13 10. ef 4sin(h) = 1+2%+ 4 + O(h?) etsin(h) = h+h +E LOR) An intermediate computation was Baill Aalllad 3 SAL cor ce Atht srg += gr) =hth HET Er {meme 11. I i il alll sll con) £5) = 14h KB BE 4 OK) cos(h)sin(h) = h- 2-42 + O(hT) An intermediate comutation was Lal lb Sl 2ril1i2Rs Tr I TE DI TA 12. x = =b/Fda Ee =) fami imi Ba( VF ac) = tae The case for x7 is handled in a similar manner. 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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    14 CHAPTER 1. PRELIMINARIES + Studyxy

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    Chapter 2 . . The Solution of Nonlinear . Equations f(z) =0 2.1 Iteration for Solving r = g(x) 1. (a) Clearly, g(z) € C[0,1]. Since ¢'(z) = —/2 < 0 on the interval [0,1], the function g(z) is strictly decreasing on the interval [0,1]. If g is strictly decreasing on [0, 1], then g(0) = 1 and g(1) = 0 imply that 9([0,1]) = [0,1] € [0,1]. Thus, by Theorem 2.2, the function g(x) has a fixed point on the interval [0,1]. In addition: |f'(z)} =| — #/2| = z/2 < 1/2 < 1 on the interval [0,1]. Thus, by Theorem 2.2, the function g(x) has a unique fixed point on the interval [0,1]. (b) Clearly, g(x) € C[0,1]. Since g/(x) = —In(2)2™ < 0 on the interval [0,1], the function g(x) is strictly decreasing on the interval [0,1]. Tf gis strictly decreasing on [0,1], then (0) = 1 and g(1) = 1/2 imply that g([0,1]) = [1/2,1] C [0,1]. Thus, by Theorem 2.2 the function g(x) has a fixed point on the interval [0,1]. In addition: |¢’(2)| = | ~1n(2)272| = In(2)27 2 < In(2) < In(e) = 1 on the interval [0,1]. Thus, by Theorem 2.2, the function g(x) has an unique fixed point on the interval [0,1]. (c) Clearly g(x) is continuous on [0.5,5.2] and g([0.5,5.2]) € [0.5,5.2]. But, 9([0.5,2]) C [0.5,2]. Thus, the hypotheses of the first part of Theorem 2.2 are satisfied and g has a fixed point in [0.5,2]. While (1,1) is the unique fixed point in [0.5,2], [f/(1)] = 1 #£ 1, thus the hypothesee in part (4) of Theorem 2.2 cannot be satisfied. 15 + StudyXY

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