newton.cpp

计算数值分析中Newton插值算法的演示算法

#include <stdio.h>
#include <math.h>

int fac(int n)
{
	int i,m=1;
	for(i=1;i<=n;i++)
	{
		m*=i;
	}
	return m;
}

void main()
{
	double x[5]=
	{
		1.566,1.567,1.568,1.569,1.570
	};
	double tanx[5]=
	{
		208.49128,263.41125,357.61106,556.69098,1255.76559
	};
	double sinx[5]=
	{
		0.9999885,0.9999928,0.9999961,0.9999984,0.9999997
	};
	double cosx[5]=
	{
		0.0047963,0.0037963,0.0027963,0.0017963,0.0007963
	};
	double chazhi[9][5]={0};
	double chazhi2[9][5]={0};
	double chazhi3[9][5]={0};
	
	int i,j;

	double h=0.001;
	double xi=1.5695;
	double t=(xi-x[0])/h;
	
	double result,result2,result3;
	double tem;

	for(i=0;i<5;i++)
	{
		chazhi[2*i][0]=sinx[i];
		chazhi2[2*i][0]=cosx[i];
		chazhi3[2*i][0]=tanx[i];
	}

	for(i=1;i<5;i++)
	{
		for(j=i;j<5;j++)
		{
			chazhi[i+2*(j-i)][i]=chazhi[i+2*(j-i)+1][i-1]-chazhi[i+2*(j-i)-1][i-1];
			chazhi2[i+2*(j-i)][i]=chazhi2[i+2*(j-i)+1][i-1]-chazhi2[i+2*(j-i)-1][i-1];
			chazhi3[i+2*(j-i)][i]=chazhi3[i+2*(j-i)+1][i-1]-chazhi3[i+2*(j-i)-1][i-1];
		}
	}

	

	result=sinx[0];
	result2=cosx[0];
	result3=tanx[0];
	for(i=1;i<5;i++)
	{
		tem=1;
		for(j=0;j<i;j++)
		{
			tem*=(t-j);
		}
		result+=(chazhi[i][i]/fac(i)*tem);
	}
	for(i=1;i<5;i++)
	{
		tem=1;
		for(j=0;j<i;j++)
		{
			tem*=(t-j);
		}
		result2+=(chazhi2[i][i]/fac(i)*tem);
	}
	for(i=1;i<5;i++)
	{
		tem=1;
		for(j=0;j<i;j++)
		{
			tem*=(t-j);
		}
		result3+=(chazhi3[i][i]/fac(i)*tem);
	}
	printf("sinx=%f,cosx=%f,sinx/cosx=%f,直接差值tanx=%f\n",result,result2,result/result2,result3);
}