《数值分析》黄仿伦改编英文版课后习题答案推荐.docVIP

下载本文档

674
1
约2万字
约 31页
2018-03-15 发布于贵州
举报
版权申诉

《数值分析》黄仿伦改编英文版课后习题答案推荐.doc

1、有哪些信誉好的足球投注网站（book118）网站文档一经付费（服务费），不意味着购买了该文档的版权，仅供个人/单位学习、研究之用，不得用于商业用途，未经授权，严禁复制、发行、汇编、翻译或者网络传播等，侵权必究。。
2、本站所有内容均由合作方或网友上传，本站不对文档的完整性、权威性及其观点立场正确性做任何保证或承诺！文档内容仅供研究参考，付费前请自行鉴别。如您付费，意味着您自己接受本站规则且自行承担风险，本站不退款、不进行额外附加服务；查看《如何避免下载的几个坑》。如果您已付费下载过本站文档，您可以点击这里二次下载。
3、如文档侵犯商业秘密、侵犯著作权、侵犯人身权等，请点击“版权申诉”（推荐），也可以打举报电话：400-050-0827(电话支持时间：9:00-18:30)。
4、该文档为VIP文档，如果想要下载，成为VIP会员后，下载免费。
5、成为VIP后，下载本文档将扣除1次下载权益。下载后，不支持退款、换文档。如有疑问请联系我们。
6、成为VIP后，您将拥有八大权益，权益包括：VIP文档下载权益、阅读免打扰、文档格式转换、高级专利检索、专属身份标志、高级客服、多端互通、版权登记。
7、VIP文档为合作方或网友上传，每下载1次，网站将根据用户上传文档的质量评分、类型等，对文档贡献者给予高额补贴、流量扶持。如果你也想贡献VIP文档。上传文档

《数值分析》黄仿伦改编英文版课后习题答案推荐

Answers for Exercises—Numerical methods using Matlab Chapter 1 P10 2. Solution (a) produces an equation . Solving it gives the roots and . Since and , thus, both and are fixed points of . (b) –(d) The iterative rule using is. The results for part (b)-(d) with starting value and are listed in Table 1. Table 1 0 1.9000 0.1000 0.0500 0 3.800 0.2000 0.0500 1 1.7950 0.2050 0.1025 1 3.9800 0.0200 0.0050 2 1.5690 0.4310 0.2155 2 3.9998 0.0002 0.0001 3 1.0451 0.9549 0.4775 3 4.0000 0.0000 0.0000 (e) Calculate values of at and . , and . Since is continuous, there exists a number such that for all . There also exists a number such that for all . Therefore, is an attractive fixed point. The sequence generated by with starting value converges to . is a repelling fixed point. The sequence generated by with starting value does not converge to . P11 4. Find the fixed point for : gives. Find the derivative: . Evaluate and : , . Both and gives . There is no reason to find the solution(s) using the fixed-point iteration. P11 6. Proof P21 4. False position method: Assume thatcontains the root. The equation of the secand line through and is . It intersects x-axise at (Eq. 1.36, p18) , ; Since , then . Similarly, we have , , 10. Bisection method: Assume that contains the root. Then . (a) , then . Since , then . Similarly, we can obtain . The results are listed in Table 3. Table 3 n an f(an) bn f(bn) cn f(cn) 0 3 -0.1425 4 1.1578 3.5 0.3746 1 3 -0.1425 3.5 0.3746 3.25 0.1088 2 3 -0.1425 3.25 0.1088 3.125 -0.0166 3 3.125 -0.0166 3.25 0.1088 3.1875 0.0459 4 3.125 -0.0166 3.1875 0.0459 3.1563 0.0147 5 3.125 -0.0479 3.1563 0.0147 3.1406 -9.4265e-004 The values of tan(x) at midpoints are going to zero while the sequence converges (b) Since , , there exist a root in . The results using Bisection method are listed in Table