转载： https://blog.csdn.net/i_chaoren/article/details/79822574

基本原理：幂函数可逼近任意函数。

上式中，N表示多项式阶数，实际应用中一般取3或5；

假设N=5，则：

共有6个未知数，仅需6个点即可求解；

可表示为矩阵方程：

Y的维数为[R*1]，U的维数[R * 6]，K的维数[6 * 1]。

R> 6时，超定方程求解：

下面是使用C++实现的多项式拟合的程序，程序中使用opencv进行矩阵运算和图像显示。程序分别运行了N=3,5,7,9时的情况，结果如下：

#include <opencv2\opencv.hpp>
#include <iostream>
#include <vector>
using namespace cv;
using namespace std;

Mat polyfit(vector<Point>& in_point, int n);

int main()
{
//数据输入
Point in[19] = { Point(50,120),Point(74,110),Point(98,100),Point(122,100),Point(144,80)
,Point(168,80),Point(192,70),Point(214,50),Point(236,40),Point(262,20)
,Point(282,20),Point(306,30),Point(328,40),Point(356,50),Point(376,50)
,Point(400,50),Point(424,50),Point(446,40),Point(468,30) };

vector<Point> in_point(begin(in),end(in));

//n:多项式阶次
int n = 9;
Mat mat_k = polyfit(in_point, n);


//计算结果可视化
Mat out(150, 500, CV_8UC3,Scalar::all(0));

//画出拟合曲线
for (int i = in[0].x; i < in[size(in)-1].x; ++i)
{
Point2d ipt;
ipt.x = i;
ipt.y = 0;
for (int j = 0; j < n + 1; ++j)
{
ipt.y += mat_k.at<double>(j, 0)*pow(i,j);
}
circle(out, ipt, 1, Scalar(255, 255, 255), CV_FILLED, CV_AA);
}

//画出原始散点
for (int i = 0; i < size(in); ++i)
{
Point ipt = in[i];
circle(out, ipt, 3, Scalar(0, 0, 255), CV_FILLED, CV_AA);
}

imshow("9次拟合", out);
waitKey(0);

return 0;
}

Mat polyfit(vector<Point>& in_point, int n)
{
int size = in_point.size();
//所求未知数个数
int x_num = n + 1;
//构造矩阵U和Y
Mat mat_u(size, x_num, CV_64F);
Mat mat_y(size, 1, CV_64F);

for (int i = 0; i < mat_u.rows; ++i)
for (int j = 0; j < mat_u.cols; ++j)
{
mat_u.at<double>(i, j) = pow(in_point[i].x, j);
}

for (int i = 0; i < mat_y.rows; ++i)
{
mat_y.at<double>(i, 0) = in_point[i].y;
}

//矩阵运算，获得系数矩阵K
Mat mat_k(x_num, 1, CV_64F);
mat_k = (mat_u.t()*mat_u).inv()*mat_u.t()*mat_y;
cout << mat_k << endl;
return mat_k;
}