C语言实现最小二乘法

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在此之前我们再回顾下上一篇文章中的系数求解公式。二次拟合曲线的方程为:

系数行列式为:

另外:

所求系数为:

使用C实现最小二乘法为:

#include#include #include

#define DATA_NUM (6)#define DOUBLE_PRECISION (1e-15)

double x[DATA_NUM]={0,2,4,6,8,10};double y[DATA_NUM]={0,6,25,42,70,110};

//y = a*x^2 + b*x + cbool LeastSquares(double *x, double *y, unsigned int

data_num, double *a, double *b, double *c) { double

sumx=0,sumx2=0,sumx3=0,sumx4=0,sumy=0,sumxy=0,sumx2y=0; double D=0;

if(!data_num) return false;

for(int i=0;i

sumx2+=pow (x[i],2); sumxy+=x[i]*y[i]; sumx3+=pow(x[i],3);

sumx2y+=pow(x[i],2)*y[i]; sumx4+=pow(x[i],4); }

D = sumx2*sumx2*sumx2 + sumx*sumx*sumx4 + data_num*sumx3*sumx3 -

data_num*sumx2*sumx4 - 2*sumx*sumx2*sumx3;

if (fabs(D) < DOUBLE_PRECISION) { return false; } *a =

(sumy*(sumx2*sumx2-sumx*sumx3) + sumxy*(data_num*sumx3-sumx*sumx2) +

sumx2y*(sumx*sumx-data_num*sumx2))/D; *b =

(sumy*(sumx*sumx4-sumx2*sumx3) + sumxy*(sumx2*sumx2-data_num*sumx4) +

sumx2y*(data_num*sumx3-sumx*sumx2))/D; *c =

(sumy*(sumx3*sumx3-sumx2*sumx4) + sumxy*(sumx*sumx4-sumx2*sumx3) +

sumx2y*(sumx2*sumx2-sumx*sumx3))/D; return true;}

int main() { double a,b,c; LeastSquares(x, y, DATA_NUM, &a, &b, &c); printf("a=%9.8f,\nb=%9.8f,\nc=%9.8f\n",a,b,c); printf

("y=%9.6fx*x+%9.6fx+%9.6f",a,b,c); return 0;}

上述在求解行列式值的时候使用的是代数余子式方法求解。即用某一个列与代数余子式相乘。