牛顿迭代法求解多元高阶方程组

#include<iostream>
#include<cmath>
#define N 2                     // 非线性方程组中方程个数、未知量个数 



#define Epsilon 0.0001             // 差向量1范数的上限
#define Max     100               //最大迭代次数
#define k1       -1.22785792e-001 
#define k2       1.37477946e-002 
using namespace  std;
const int N2=2*N;
int X = 0;
int Y = 1;

int test(float x, float y);

int main()
{
void ff(float xx[N],float yy[N]);            //计算向量函数的因变量向量yy[N]
void ffjacobian(float xx[N],float yy[N][N]);//计算雅克比矩阵yy[N][N]
void inv_jacobian(float yy[N][N],float inv[N][N]);     //计算雅克比矩阵的逆矩阵inv
void newdundiedai(float x0[N], float inv[N][N],float y0[N],float x1[N]);   //由近似解向量 x0 计算近似解向量 x1

float x0[N]={2.0,0.25}, y0[N], jacobian[N][N], invjacobian[N][N], x1[N], errornorm;
int i,j,iter=0;

                                   //如果取消对x0的初始化，撤销下面两行的注释符， 就可以由键盘向x0读入初始近似解向量
//for( i=0;i<N;i++)
 //  cin>>x0[i];

cout<<"初始近似解向量："<<endl;
	for (i=0;i<N;i++)
cout<<x0[i]<<"  ";
    cout<<endl;
	cout<<endl;

do
{
	iter=iter+1;
	cout<<"第 "<<iter<<" 次迭代开始"<<endl;    //计算向量函数的因变量向量 y0

ff(x0,y0);                                    //计算雅克比矩阵 jacobian

ffjacobian(x0,jacobian);                        //计算雅克比矩阵的逆矩阵 invjacobian

inv_jacobian(jacobian,invjacobian);               //由近似解向量 x0 计算近似解向量 x1
 
newdundiedai(x0, invjacobian,y0,x1);               //计算差向量的1范数errornorm
	errornorm=0;
	for (i=0;i<N;i++)
		errornorm=errornorm+fabs(x1[i]-x0[i]);
	if (errornorm<Epsilon) break;

	for (i=0;i<N;i++)
		x0[i]=x1[i];

} while (iter<Max);
   test( x1[0],  x1[1]);
return 0;
}

void ff(float xx[N],float yy[N])              //调用函数
{
	float x,y;
	int i;	
    x=xx[0];
	y=xx[1];
	float r = x * x + y * y;
	yy[0]=(1 + k1 * r + k2 * r * r) * x - X;                       
	yy[1]=(1 + k1 * r + k2 * r * r) * y - Y;                   //计算初值位置的值

    cout<<"向量函数的因变量向量是： "<<endl;
	for( i=0;i<N;i++)
      cout<<yy[i]<<"  ";
    cout<<endl;
	cout<<endl;

}

void ffjacobian(float xx[N],float yy[N][N])
{
  float x,y;
  int i,j;

	x=xx[0];
	y=xx[1];
	float r = x * x + y * y;
	//jacobian have n*n element                   //计算函数雅克比的值
	yy[0][0]=1 + k1 * r + k2 * r * r + 2 * k1 * x * x + 4 * k2 * x * x * r ;
	yy[0][1]=2 * k1 * x * y + 4 * k2 * x * y * r;
	yy[1][0]=2 * k1 * x * y + 4 * k2 * x * y * r;
	yy[1][1]=1 + k1 * r + k2 * r * r + 2 * k1 * y * y + 4 * k2 * y * y * r;

   cout<<"雅克比矩阵是： "<<endl;
	for( i=0;i<N;i++)
   {for(j=0;j<N;j++)
		 cout<<yy[i][j]<<"  ";
	   cout<<endl;
   }
	cout<<endl;
}

void inv_jacobian(float yy[N][N],float inv[N][N])
{float aug[N][N2],L;
 int i,j,k;

 cout<<"开始计算雅克比矩阵的逆矩阵 ："<<endl;
 for (i=0;i<N;i++)
	{  for(j=0;j<N;j++)
		 aug[i][j]=yy[i][j];
	   for(j=N;j<N2;j++)
		if(j==i+N) aug[i][j]=1;
		else  aug[i][j]=0;
	}

 
 for (i=0;i<N;i++)
	{  for(j=0;j<N2;j++)
		 cout<<aug[i][j]<<"  ";
       cout<<endl;
	}
cout<<endl;

 
for (i=0;i<N;i++)
	{
	  for (k=i+1;k<N;k++)
	  {L=-aug[k][i]/aug[i][i];
		for(j=i;j<N2;j++)
         aug[k][j]=aug[k][j]+L*aug[i][j];
	  }
	}


 for (i=0;i<N;i++)
	{  for(j=0;j<N2;j++)
		 cout<<aug[i][j]<<"  ";
       cout<<endl;
	}
cout<<endl; 
 
 
 for (i=N-1;i>0;i--)
	{  
	 for (k=i-1;k>=0;k--)
	  {L=-aug[k][i]/aug[i][i];
		for(j=N2-1;j>=0;j--)
         aug[k][j]=aug[k][j]+L*aug[i][j];
	  }
	}




for (i=0;i<N;i++)
	{  for(j=0;j<N2;j++)
		 cout<<aug[i][j]<<"  ";
       cout<<endl;
	}
cout<<endl;


for (i=N-1;i>=0;i--)
   for(j=N2-1;j>=0;j--)
		aug[i][j]=aug[i][j]/aug[i][i];


for (i=0;i<N;i++)
	{  for(j=0;j<N2;j++)
		 cout<<aug[i][j]<<"  ";
       cout<<endl; 
	   for(j=N;j<N2;j++)
	   inv[i][j-N]=aug[i][j];
	}
cout<<endl;


cout<<"雅克比矩阵的逆矩阵： "<<endl;
for (i=0;i<N;i++)
	{  for(j=0;j<N;j++)
		 cout<<inv[i][j]<<"  ";
       cout<<endl;
	}
cout<<endl;

}

void newdundiedai(float x0[N], float inv[N][N],float y0[N],float x1[N])
{
	int i,j;
	float sum=0;
	
	for(i=0;i<N;i++)
	{ sum=0;
	  for(j=0;j<N;j++)
	    sum=sum+inv[i][j]*y0[j];
		x1[i]=x0[i]-sum;
    }

    cout<<"近似解向量："<<endl;
	for (i=0;i<N;i++)
	 cout<<x1[i]<<"  ";
    cout<<endl;cout<<endl;


}

int test(float x, float y)
{
	float ep = 0.01;
	float r = x * x + y * y;
	if ( ((1 + k1 * r + k2 * r * r) * x - X) < ep && ((1 + k1 * r + k2 * r * r) * y - Y)<ep)
      printf("ok!\n");
	else
		printf("error!\n");
	return 0;


}