数值分析上级报告2

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1、姓名：许名珞学号:120169第二章一、题目(1)给定初值兀。及容许误羌编制牛顿法解方程f(x)=0的通用程序。(2)给定方程/(X)二坯-兀=0,易知其有三个根Q*=-竝2*=0,琼=73a)由牛顿方法的局部收敛性可知存在5〉0,当X。w(-》,+/)吋，Newton迭代序列收敛于根X2*。试确定尽可能犬的5。b)试取若干初始值,观察当00,-1),(-1,-»),(-/,+/),(&1),(1,XO)时Newton序列的收敛性以及收敛于哪一个根。(3)通过本上机题，你明片了什么？二、程序1.fx.mfunctionFx=fx(x)Fx=xA3/3-x;2.dfx.mfunctionFx

2、=dfx(x)Fx=xA2-l;3.Newton.mclearef^lOA-6;k=0;xO=input('PleaseinputinitialvalueX0:‘);disp('kfprintfroXkJ;%fxO);flag=l;whileflag==l&&k<=10A3xl=xO-fx(xO)/dfx(xO);ifabs(xl-xO)

3、0;flagl=l;whileflagl==l&&m<=10人3x1=xO-fx(xO)/dfx(xO);ifabs(xl-x0)<10A-6flag1=0;endm=m+1;x0=x1;endifflag1=11

4、abs(x0)>=10A-6flag=0;endendfprintfifThemaximundeltaisdelta);三、运行结果1.运行search.m文件结果为：Themaximumdeltais0.774597即得最大的5为0.774597,Newton迭代序列收敛于根兀；=0的最大区间为(-0.774597,0.774597)o2.运行Newton.m文件在区间(一0

5、0,-1),(一1,-5),(-氏+5),(6,1),(1,+00)上各输入若干个数，计算结果如下:1)在区间(-a),-l)上取-1000,・100,・10,・5,・3,-1.5PleaseinputinitialvalueXO:-10000kXk0-10000.0000001・6666.6667332・4444.4445893・2962.9632094-1975.3090315-1316.8730256-877.9158567-585.2779978-390.1864709-260.12602210-173.41991111-115.61711812・77.08384513・51.39

6、788014-34.27822915-22.87161816-15.27694917-10.22845918-6.88478019-4.68877220-3.27480721-2.40771422-1.93975023-1.76125924-1.73276225-1.73205126-1.732051PleaseinputinitialvalueX0:-100kXk0-100.0000001・66.6733342・44.4588913・29.6542634-19.7920165-13.2284476-8.8696517-5.9892318・4.1073249-2.91075510-2.20

7、018911-1.84868712-1.74223513-1.73213914-1.73205115-1.732051PleaseinputinitialvalueX0:-10kXk0123456789-10.000000-6.734007-4.590570-3.212840-2.371653-1.922981-1.757175-1.732580-1.732051-1.732051PleaseinputinitialvalueX0:-5kXk0123456785.0000003.4722222.5241801.9960681.7766181.7336741.7320531.7320511.

8、732051PleaseinputinitialvalueX0:-3kXk3.0000002.2500001.8692313-1.7458104・1.7322125-1.7320511.732051PleaseinputinitialvalueX0:-1.5kXk0123451.5000001.8000001.7357141.7320621.7320511.732051结果显示，以上初值迭代序列均收敛于-1.732051