分別用1~~4次Lagrange，Newton插值計算函數(shù)在0.9處的函數(shù)值

分別用1~~4次Lagrange，Newton插值計算函數(shù)在0.9處的函數(shù)值

Lagrange

• 插值基函數(shù)

  
  
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• Lagrange插值多項式
  
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  [圖片來源]: https://blog.csdn.net/u013871100/article/details/41249285

Newton

• 均差

  
  
  image3.png
  
  
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• Newton插值多項式
  
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  [圖片來源]:https://blog.csdn.net/Gy_1096440940/article/details/78445088

#include <iostream>
using namespace std;

//求yi的值
float Evaluation(float x)
{
    return 1/(1+25*x*x);
}

//Lagrange
float Lagrange(float x,int n,float a[],float b[])
{
    float t,y=0;
    int j,k;
    for (k = 0;k <= n;k++)
    {
        t = 1;
        for (j = 0;j <= n;j++)
        {
            if (j != k)
                t = ((x - a[j])/(a[k]-a[j]))*t;
        }
        y = t * b[k]+y;
    }
    return y;
}

// Newton均差
void AverageDeviation(int n,float ad[][100],float a[],float b[])
{
    int i,j;
    for (i = 0;i <= n; i++)
    {
        ad[i][0] = b[i];
        for (j = 1;j <= n; j++)
        {
            if(j>i)
                ad[i][j] = 0;
            else
            {
                ad[i][j] = (ad[i-1][j-1]-ad[i][j-1])/(a[i-j]-a[i]);
            }
        }
    }
    
}

//Newton
float Newton(int n,float x,float a[],float b[],float ad[][100])
{
    int i,j;
    float y=ad[0][0];
    float p;
    for (i=1;i<=n;i++)
    {
        p=ad[i][i];
        for(j=0;j<=i-1;j++)
            p *= (x-a[j]);
        y += p;
    }
    return y;

}

int main()
{
    float x,a[100],b[100];
    float ad[100][100];
    int i,j,k,n;
    cout << "輸入插值次數(shù)n："<<endl;
    cin >> n;
    cout << "輸入你要估算對象x?："<<endl;
    cin >> x;
    for (i=0;i<=n;i++)
    {
        cout<< "輸入x"<<i<<"的值：";
        cin >> a[i];
        b[i] = Evaluation(a[i]);
        cout<< "輸入y"<<i<<"的值："<<b[i]<<endl;
    }
    cout << "Lagrange得y = "<< Lagrange(x,n,a,b)<<endl;
    // 調(diào)用函數(shù)求Newton均差
    AverageDeviation(n,ad,a,b);
    for(i=0;i<=n;i++)
    {
        cout<<"第"<<i<<"階均差為："<<ad[i][i]<<endl;
    }
    cout << "Newton得y = "<< Newton(n,x,a,b,ad)<<endl;
    return 0;
}