《《数值分析》黄仿伦改编英文版课后习题答案》由会员分享，可在线阅读，更多相关《《数值分析》黄仿伦改编英文版课后习题答案（31页珍藏版）》请在金锄头文库上搜索。
1、Answers for ExercisesNumerical methods using MatlabChapter 1P10 2. Solution (a) produces an equation .)(xgx 0862 xxSolving it gives the roots and .2 x4 xSince and , thus, both and are fixed points of .2)2( g4)4( g2 P4 P)(xg(b) (d) The iterative rule using is.)(xg2 2144nnnppp The results for part (b)
2、-(d) with starting value and are listed in Table 9 . 10 p8 . 30 p1.Table 1kkpkAkRkkpkAkR01.90000.10000.050003.8000.20000.050011.79500.20500.102513.98000.02000.005021.56900.43100.215523.99980.00020.000131.04510.95490.477534.00000.00000.0000(e) Calculate values of at and .xxg 4)(2 x4 x, and .12)2( g10
3、)4( gSince is continuous, there exists a number such that)(xg 0 for all .1)( xg4,4 xThere also exists a number such that0 for all .1)( xg2,2 xTherefore, is an attractive fixed point. The sequence generated by 4 pwith starting value converges to . is a 2 2144nnnppp 8 . 30 p4 p2 prepelling fixed point
4、. The sequence generated by with 2 2144nnnppp starting value does not converge to .9 . 10 p2 pP11 4. Find the fixed point for : gives.)(xg)(xgx 2 pFind the derivative: .12)( xxgEvaluate and : , .)2( g)2(g 3)2( g5)2( gBoth and gives . There is no reason to find the solution(s) 2 p2 p1)( pgusing the f
5、ixed-point iteration.P11 6. Proof )()()(010112ppgpgpgpp )()()( 0101ppKppg P21 4. False position method: Assume thatcontains the root. The equation of the secand ,nnbaline through and is . It intersects )(,(nnafa)(,(nnbfb)()()()(n nnnn nbxabafbfbfy x-axise at (Eq. 1.36, p18)()()(nnnnn nnafbfabbfbc ,
6、1981. 0)6 . 1()(,4907. 0)4 . 2()(00 fbffaf;8301. 1)()()(00000 00 afbfabbfbcSince , then .0095. 0)(0 cf8301. 1, 4 . 2,11 baSimilarly, we have, 1.84093- 1 c1.84093- , 4 . 2,22 ba, 1.84139- 2 c1.84139- , 4 . 2,33 ba-1.841403 c10. Bisection method: Assume that contains the root. Then .,nnba2nn nbac (a)
7、, then .1587. 1)4(, 4;1425. 0)3(, 300 fbfa5 . 30 cSince , then .03746. 0)5 . 3()(0 fcf5 . 3, 3,11 baSimilarly, we can obtain . The results are listed in Table 3.L,321cccTable 3 nanf(an)bnf(bn)cnf(cn)03-0.1425 41.15783.50.374613-0.14253.50.37463.250.108823-0.14253.250.10883.125-0.016633.125-0.01663.2
8、50.10883.18750.045943.125-0.01663.18750.04593.15630.014753.125-0.04793.15630.01473.1406-9.4265e-004The values of tan(x) at midpoints are going to zero while the sequence converges (b) Since , , there exist a root in .05574. 1)1tan( 01425. 0)3tan( )3, 1(The results using Bisection method are listed i
9、n Table 4.Table 4 nanf(an)bnf(bn)cnf(cn)011.55743-0.14252-2.1850111.55742-2.18501.514.101421.514.10142-2.18501.75-5.520431.514.10141.75-5.52041.625-18.430941.514.10141.625-18.43091.5625120.532551.5625120.53251.625-18.43091.5938-43.5584Although the sequence converges, the values of tan (x) at midpoin
10、ts are not going to zero. P36 2. has two zeros . ()3)(2 xxxf2131 x3028. 2,3028. 121 xxThe first derivative of is.3)(2 xxxf12)( xxfThe Newton-Raphson iterative function is .123 )()()(2 xx xfxfxxgThe Newton-Raphson formula is , . 12321 nn npppL, 2, 1, 0 nThe results are listed in Table 5 with starting
11、 value p0=1.6 and p0=0.0 respectively.Table 5p0p1p2p3p41.62.52732.31522.30282.30280.0-3.0000-1.9615-1.1472-0.0066Obviously, the sequence generated by the starting value p0=0.0 does not converge.11. Use Newton-raphson method to solve .0)(3 AxxfThe derivative of is .)(xf23)(xxf .3232 )()()(223xAxxAx x
12、fxfxxg Newton-Raphoson formula is , .322 11 nnnpAp pL, 2, 1 nSince is a zero of and ,3Ap Axxf 3)(10332)(33 ApxApgThe sequence generated by the recursive formula will converge to 322 11 nnnpAp pfor any starting value , where .3Ap ,33 0 AAp0 Answers for ExercisesNumerical methods using MatlabChapter 2
13、P44 2. Solution The 4th equation yields . 24 xSubstituting to the 3rd equation gives .24 x53 xSubstituting both and to the 2nd equation produces .24 x53 x32 xis obtained by sustituting all , and to the 1st equation.21 x32 x53 x24 xThe value of the determinant of the coefficient matrix is .115573115 D4. Proof (a) Calculating the product of the two given upper-triangular matrices gives. 3333332323222222331323121311221212111111332322131211332322131211000000000bababababababababababbbbbbaaaaaa BAIt is also an upper-triangular matrix.(b) Let and where and when .NNijaA
