数值分析/计算方法

Lagrange插值多项式

实验要求和提示

 实验代码（C·无画图）

#define N 13
#include<iostream>
using namespace std;
int main()
{
    double x[N]={0,10,20,30,40,50,60,70,80,90,100,110,120};
    double y[N]={5,1,7.5,3,4.5,8.8,15.5,6.5,-5,-10,-2,4.5,7};
    double l; 
    double X=65;
    double Y=0;
    int i,k;
    for(k=0;k<N;k++)
    {
        l=1;
        for(i=0;i<N;i++)  
            if(i!=k)
                l=l*((X-x[i])/(x[k]-x[i]));
        
        Y=Y+y[k]*l;
    }
    cout<<Y;
}

埃尔米特插值

实验要求和提示

 实验代码 （C·无画图）

#define N 10
#include<iostream>

using namespace std;
int main()
{
    double x[N]={0.10,0.20,0.30,0.40,0.50,0.60,0.70,0.80,0.90,1.00};
    double y[N]={0.904837,0.818731,0.740818,0.670320,0.606531,0.548812,0.496585,0.449329,0.406570,0.367879};
    double m[N]={-0.904837,-0.818731,-0.740818,-0.670320,-0.606531,-0.548812,-0.496585,-0.449329,-0.406570,-0.367879};
    double l;
    double ldao[N]={0};   //ldao(x)为l(x)的导数
    double X=0.55;
    double Y=0;
    int j,k,i;
    for(j=0;j<N;j++)
    {
     l=1;
        for(k=0;k<N;k++)
            if(k!=j)
            {
                ldao[j]=ldao[j]+(1/(x[j]-x[k]));
                l=l*((X-x[k])/(x[j]-x[k]));
            }
        Y=Y+y[j]*((1-2*(X-x[j])*ldao[j])*l*l)+m[j]*((X-x[j])*l*l);
    }
    cout<<Y;
    return 0;
}

最小二乘法

实验要求和提示

 实验代码（Matlab·有画图）

x = [0 10 20 30 40 50 60 70 80 90];
y = [68 67.1 66.4 65.6 64.6 61.8 61.0 60.8 60.4 60];
xifsum = 0;
yisum = 0;
xisum = 0;
xycsum = 0;
for i = 1:10
xifsum = x(i) * x(i) + xifsum;
yisum = y(i) + yisum;
xisum = x(i) + xisum;
xycsum = x(i) * y(i) + xycsum;
end
m = 9;
b = (xifsum*yisum - xisum*xycsum) / ((m+1)*xifsum - xisum*xisum);
a = ((m+1)*xycsum - xisum*yisum) / ((m+1)*xifsum - xisum*xisum);
X = 55;
S = a * X + b;
disp('X的值为55时，表达式的值为');
disp(S);
X = 0 : 1 : 90;
yy = a * X + b;
figure
plot(X, yy, 'k')
for j = 1:10
text(x(j), y(j), '*', 'color', 'r');
end
text(55, S, '*', 'color', 'b');

实验演示

复化求积公式

 实验代码（C）

#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
double function(double x);
double T(double a,double b,double h,int N);
double Fh(double a,double b,double h,int N);
void getIntegralValue(double a,double b,int N);//复化梯形求积
int main()
{
    double a,b ;//产生的节点数
    int N=1;
    cout<<"请输入左右区间范围：";
    cin>>a;
    cin>>b;
    //getIntegralValue(a,b,N);
    double t,TN,Tn,Hn=0,h=b-a;
    t=T(a,b,h,N);
    do
    {
        Hn=Fh(a,b,h,N);
        Tn=t;
        TN=0.5*(T(a,b,h,N)+Hn);
        N=2*N;  h=h/2;
        t=TN;
    }while(fabs(TN-Tn)>=0.000001);
    cout<<"Tn="<<setprecision(8)<<Tn<<endl;
    cout<<"等分数为："<<N/2<<endl;
    return 0;
}

double function(double x)
{
    return exp(x)*cos(x);
}
double T(double a,double b,double h,int N)
{
    double f=0;
    for(int k=1;k<=N-1;k++)
        f+=function(a+k*h);
    return (h/2)*(function(a)+2*f+function(b));
}

double Fh(double a,double b,double h,int N)
{
    double Hn=0;
    for(int i=1;i<=N;i++)
        {
            h=(b-a)/N;
            Hn+=function(a+(h*(2*i-1)/2));
        }
    return Hn=h*Hn;
}

标签：10,xisum,插值,double,求积,int,xifsum,复化,yisum
来源： https://www.cnblogs.com/WScoconut/p/16387484.html