/**
拉格朗日插值计算
拉格朗日插值公式:
n n
Pn(x(i))= ∑〔 ∏( x-x(j))/(x(k)-x(j)) 〕y(k)
k=0 j=0,
j≠k
属性:插值计算法
n
精度(局部截断误差):| f(x) - Pn(x) | = [f(ε)] / (n+1)! ∏ ( x - x(k) )
(注:其中[f(ε)]为f(ε)第n+1次求导的表达式) k=0
《数值计算方法与算法》第二版 - 科学出版社 P19
代码维护:2007.04.18 pengkuny
**/
float Lagrange(float px[], float py[], int n, float x)
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{//px,py:插值点(Xi,Yi) n:插值点个数 x:待计算的函数点
float y = 0;
for(int k=0; k<n; k++)//k控制Lagrange基函数序列
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{
float tmp = 1;//tmp表示Lagrange基函数
for(int i=0;i<n;i++)
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{
if(i!=k)
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{
tmp = tmp * (x-px[i])/(px[k]-px[i]); //key step
}
}
y = y + py[k]*tmp;
}
return y;
}
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int main()
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{
float x;//插值
float px[10];//已知(x0,y0),(x1,y1)
float py[10];
int n;//输入已知插值组数
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cout<<"输入插值组数:"<<endl;
cin>>n;
cout<<"输入"<<n<<"组已知插值数(X,Y)"<<endl;
for(int i=0; i<n; i++)
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{
cin>>px[i]>>py[i];
}
cout<<"输入插值:"<<endl;
cin>>x;
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cout<<"Lagrange插值结果:"<<Lagrange(px, py, n, x)<<endl;
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system("pause");
return 0;
}
posted on 2007-04-20 10:54
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