华科计算方法上机作业代码.doc

计算方法上机实验试题1. (25分)计算积分， n=0,1,2,20若用下列两种算法(A) (B) 试依据积分In的性质及数值结果说明何种算法更合理。i0=1.8232000e-001;disp(i0);for n=1:30 i1=double(-5.0*i0+1.0/n); disp(n); disp(i1); i0=i1;end0.1823 1 0.0884 2 0.0580 3 0.0433 4 0.0333 5 0.0333 6 2.4261e-13 7 0.1429 8 -0.5893 9 3.0575 10 -15.1877 11 76.0294 12 -380.0637 13 1.9004e+03 14 -9.5019e+03 15 4.7510e+04 16 -2.3755e+05 17 1.1877e+06 18 -5.9387e+06 19 2.9693e+07 20 -1.4847e+08 21 7.4234e+08 22 -3.7117e+09 23 1.8558e+10 24 -9.2792e+10 25 4.6396e+11 26 -2.3198e+12 27 1.1599e+13 28 -5.7995e+13 29 2.8998e+14 30 -1.4499e+15i30=5.9140e-003;disp(i30);for n=30:-1:1 i1=double(-0.2*i30+1.0/(5*n); disp(n); disp(i1); i30=i1;end 0.0059 30 0.0055 29 0.0058 28 0.0060 27 0.0062 26 0.0065 25 0.0067 24 0.0070 23 0.0073 22 0.0076 21 0.0080 20 0.0084 19 0.0088 18 0.0093 17 0.0099 16 0.0105 15 0.0112 14 0.0120 13 0.0130 12 0.0141 11 0.0154 10 0.0169 9 0.0188 8 0.0212 7 0.0243 6 0.0285 5 0.0343 4 0.0431 3 0.0580 2 0.0884 1 0.18232. (25分)求解方程f(x)=0有如下牛顿迭代公式， n1，x0给定(1) 编制上述算法的通用程序，并以（为预定精度）作为终止迭代的准则。(2) 利用上述算法求解方程 这里给定x0=/4，且预定精度=10-10。fun=inline(cos(x)-x)fun = Inline function: fun(x) = cos(x)-x dfun=inline(-sin(x)-1)dfun = Inline function: dfun(x) = -sin(x)-1 x=agui_newton(fun,dfun,pi/4,1e-10) 0.7395 0.7391 0.7391 0.7391x = 0.73913. (25分) 已知，(1) 利用插值节点x0=1.00，x1=1.02，x2=1.04，x3=1.06，构造三次Lagrange插值公式，由此计算f(1.03)的近似值，并给出其实际误差；function f = Language(x,y,x0)syms t;if(length(x) = length(y) n = length(x); else disp(x和y的维数不相等！); return;end %检错f = 0.0;for(i = 1:n) l = y(i); for(j = 1:i-1) l = l*(t-x(j)/(x(i)-x(j); end; for(j = i+1:n) l = l*(t-x(j)/(x(i)-x(j); %计算拉格朗日基函数 end; f = f + l; %计算拉格朗日插值函数 simplify(f); %化简 if(i=n) if(nargin = 3) f = subs(f,t,x0); %计算插值点的函数值 else f = collect(f); %将插值多项式展开 f = vpa(f,6); %将插值多项式的系数化成6位精度的小数 end endendfunction f=fun(x)f=exp(x)*(3*x-exp(x);end x0=1.00 1.02 1.04 1.06x0 =1.0000 1.0200 1.0400 1.0600y0=fun(1.00) fun(1.02) fun(1.04) fun(1.06)y0 = 0.7658 0.7954 0.8227 0.8475 f=Language(x0,y0,1.03)f =116635870560615609/144115188075855872 116635870560615609/144115188075855872ans = 0.8093误差：e=abs(fun(1.03)-f)e =45008364041/144115188075855872 45008364041/144115188075855872ans = 3.1231e-07(2) 利用插值节点x0=1，x1=1.05构造三次Hermite插值公式，由此计算f(1.03)的近似值，并给出其实际误差。 function yi=Hermite(x,y,yd,xi)%x全部的插值节点；%y插值节点处的函数值；%yd插值节点处的导数值；%如果此处缺省，则用均差代替导数；%端点用向前、向后均差，中间点用中心均差；%xi为标量，自变量x；%yi为xi处的函数值；%如果没有给出y的导数值，则用均差代替导数.if isempty(yd)=1 yd=gradient(y,x);endn=length(x);m1=length(y);m2=length(yd);%输入x,y和y的导数个数必须相同if n=m1|n=m2|m1=m2 error(The length of X,Y and Yd must be equal);endp=zeros(1,n);q=zeros(1,n);yi=0;for k=1:n t=ones(1,n);z=zeros(1,n); for j=1:n if j=k %插值节点必须互异 if abs(x(k)-x(j) diff(exp(x)*(3*x-exp(x)ans =exp(x)*(3*x - exp(x) - exp(x)*(exp(x) - 3) x=1.0; exp(x)*(3*x - exp(x) - exp(x)*(exp(x) - 3)ans = 1.5316 x=1.05; exp(x)*(3*x - exp(x) - exp(x)*(exp(x) - 3)ans = 1.2422 y=fun(1.0) fun(1.05)y