数值分析作业 - 非线性求解方程

数值分析

Problem 1

分析收敛阶:

二分法每步将误差缩小 , 线性收敛, 收敛阶为 ,
割线法收敛阶为
不动点迭代 , 收敛阶为 ,
不动点迭代 , 收敛阶为 ,
牛顿法收敛阶为

先比较收敛阶大小, 收敛阶为时比较的大小, 可以判断收敛速度:

Problem 2

则牛顿法对应的不动点迭代阶收敛, 误差满足

Problem 3

局部收敛

不收敛

Problem 4

则是三重根, 牛顿法不二次收敛,

Problem 5

则牛顿法局部二次收敛, 有

估计

由的结论

估计

Problem 6

Problem 7

二分法

precision = 10;
binarySearch[fun_, var_, l0_, r0_, delta_] :=
  Module[{f, a, b, fa, fb, l, r, \[Rho],
    n}, (f[x_] = N[fun /. var -> x, precision];
    \[Rho] = 1/2;
    l[1] = l0;
    r[1] = r0;
    n = 1;
    While[r[n] - l[n] > delta,
     mid = (l[n] + r[n])/2;
     If[f[mid] > 0, (
       l[n + 1] = l[n];
       r[n + 1] = mid;
       ), (
       l[n + 1] = mid;
       r[n + 1] = r[n];
       )];
     n++;];
    Print[
     TableForm[
      Table[NumberForm[#, precision] & /@ {l[i], r[i],
         f[(l[i] + r[i])/2]}, {i, 1, n - 1}],
      TableHeadings -> {Range[n - 1], {"l", "r", "f(mid)"}}]];
    Return[{l[n], r[n]}];)];
binarySearch[E^x + x - 7, x, 0., 3., 1.*^-8]

不动点迭代

构造一种可行的不动点迭代形式

1	NestList[Log[7 - #] &, 0., 20] // InputForm

{0., 1.9459101490553132, 1.6201977869925166, 1.6826516101163946,
 1.6709747561453339, 1.673168340375046, 1.6727566260886084,
 1.6728339137527832, 1.6728194056442924, 1.6728221290601764,
 1.6728216178297926, 1.6728217137962444, 1.672821695781745,
 1.6728216991633662, 1.6728216985285795, 1.6728216986477396,
 1.6728216986253712, 1.67282169862957, 1.672821698628782,
 1.67282169862893, 1.6728216986289022}

牛顿法

1	NestList[# - (E^# + # - 7)/(E^# + 1) &, 0., 20] // InputForm

{0., 3., 2.2371293658878337, 1.7930473102816238, 1.6787764832017877,
 1.6728366106108064, 1.6728216987225175, 1.6728216986289066,
 1.6728216986289064, 1.6728216986289066, 1.6728216986289064,
 1.6728216986289066, 1.6728216986289064, 1.6728216986289066,
 1.6728216986289064, 1.6728216986289066, 1.6728216986289064,
 1.6728216986289066, 1.6728216986289064, 1.6728216986289066,
 1.6728216986289064}

试位法

tryPosition[fun_, var_, l0_, r0_, it_] := Module[{f, l, r, c, n}, (
    f[x_] = N[fun /. var -> x, precision];
    l[1] = l0;
    r[1] = r0;
    n = 1;
    Do[
     c[n] = (r[n] f[l[n]] - l[n] f[r[n]])/(f[l[n]] - f[r[n]]);
     If[f[l[n]] f[c[n]] > 0, (
       l[n + 1] = l[n];
       r[n + 1] = c[n];
       ), (
       l[n + 1] = c[n];
       r[n + 1] = r[n];
       )];
     n++;, it];
    Print[
     TableForm[
      Table[NumberForm[#, precision] & /@ {l[i], r[i], c[i],
         f[c[i]]}, {i, 1, n - 1}],
      TableHeadings -> {Range[n - 1], {"a", "b", "c", "f(c)"}}]];
    Return[{l[n], r[n]}];)];
tryPosition[E^x + x - 7, x, 0., 3., 20]