华科计算方法上机作业代码

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1、《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);forn=1:30i1=double(-5.0*i0+1.0/n);disp(n);disp(i1);i0=i1;end0.182310.088420.058030.043340.033350.033362.4261e-1370.14298-0.589393.057510-15.18771176.029412-380.0637

2、131.9004e+0314-9.5019e+03154.7510e+0416-2.3755e+05171.1877e+0618-5.9387e+06192.9693e+0720-1.4847e+08217.4234e+0822-3.7117e+09231.8558e+1024-9.2792e+10254.6396e+1126-2.3198e+12271.1599e+1328-5.7995e+13292.8998e+1430-1.4499e+15i30=5.9140e-003;disp(i30);forn=30:-1:1i1=

3、double(-0.2*i30+1.0/(5*n));disp(n);disp(i1);i30=i1;end0.0059300.0055290.0058280.0060270.0062260.0065250.0067240.0070230.0073220.0076210.0080200.0084190.0088180.0093170.0099160.0105150.0112140.0120130.0130120.0141110.0154100.016990.018880.021270.024360.028550.034340.

4、043130.058020.088410.18231.(25分)求解方程f(x)=0有如下牛顿迭代公式，n≥1，x0给定(1)编制上述算法的通用程序，并以（ε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=π/4，且预定精度ε=10-10。fun=inline('cos(x)-x')fun=Inlinefunction:fun(x)=cos(x)-x>>dfun=inline('-sin(x)-1')dfun=Inlinefunction:dfun(x)=-sin(x)-1>>x=agui_newton

5、(fun,dfun,pi/4,1e-10)0.73950.73910.73910.7391x=0.73911.(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；functionf=Language(x,y,x0)symst;if(length(x)==length(y))n=length(x);elsedisp('x和y的维数不相等！');return;end%检错f=0.0;for(i=1:n)

6、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elsef=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endendendfunctionf=fun(x)f=exp(

7、x)*(3*x-exp(x));endx0=[1.001.021.041.06]x0=1.00001.02001.04001.0600y0=[fun(1.00)fun(1.02)fun(1.04)fun(1.06)]y0=0.76580.79540.82270.8475>>f=Language(x0,y0,1.03)f=116635870560615609/144115188075855872>>116635870560615609/144115188075855872ans=0.8093误差：e=abs(fun(1.03)-

8、f)e=45008364041/144115188075855872>>45008364041/144115188075855872ans=3.1231e-07(1)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