数值分析pdf.doc

1、证明1 ? x ? sin x ? 0 在[0,1]内有一个根，使用二分法求误差不大于 ?10?4 根要迭代 多少次并输出每一步的迭代解和迭代误差。 a = 0;b = 1; fprintf(n || a || b || c \n) for k=1:14 c = (a+b)/2; fa = 1-a-sin(a); fb = 1-b-sin(b); fc = 1-c-sin(c); fprintf(%d || %f || %f || %f \n,k,a,b,c); if abs(fc)(1/2)*10^(-4) r=c; sprintf(the root is: %d , r); elseif fa*fc0 b=c; elseif fb*fc0 a=c; end end root = (a+b)/2 运行结果如下： n || a || b || c 1 || 0.000000 || 1.000000 || 0.500000 2 || 0.500000 || 1.000000 || 0.750000 3 || 0.500000 || 0.750000 || 0.625000 4 || 0.500000 || 0.625000 || 0.562500 5 || 0.500000 || 0.562500 || 0.531250 6 || 0.500000 || 0.531250 || 0.515625 7 || 0.500000 || 0.515625 || 0.507813 8 || 0.507813 || 0.515625 || 0.511719 9 || 0.507813 || 0.511719 || 0.509766 10 || 0.509766 || 0.511719 || 0.510742 11 || 0.510742 || 0.511719 || 0.511230 12 || 0.510742 || 0.511230 || 0.510986 13 || 0.510742 || 0.511230 || 0.510986 14 || 0.510742 || 0.511230 || 0.510986 root = 0.5110 二 求解方程的根，要求，分别用简单迭代法，迭代法的加速方法：，以及艾特金方法求解，要求误差应满足｜｜ 解：简单迭代法： 程序如下： x0 = 0.5; epsilon = 10^(-5); e = 1; fprintf(k || xk ||error \n); k = 1; while eepsilon x1 = exp(-x0); e = abs(x1-x0); x0=x1; k=k+1; fprintf(%d || %f ||%f\n,k,x1,e); end x0 k || xk ||error 2 || 0.606531 ||0.106531 3 || 0.545239 ||0.061291 4 || 0.579703 ||0.034464 5 || 0.560065 ||0.019638 6 || 0.571172 ||0.011108 7 || 0.564863 ||0.006309 8 || 0.568438 ||0.003575 9 || 0.566409 ||0.002029 10 || 0.567560 ||0.001150 11 || 0.566907 ||0.000652 12 || 0.567277 ||0.000370 13 || 0.567067 ||0.000210 14 || 0.567186 ||0.000119 15 || 0.567119 ||0.000067 16 || 0.567157 ||0.000038 17 || 0.567135 ||0.000022 18 || 0.567148 ||0.000012 19 || 0.567141 ||0.000007 x0