基本原理：冪函式可逼近任意函式。

上式中，N表示多項式階數，實際應用中一般取3或5；

假設N=5，則：

共有6個未知數，僅需6個點即可求解；

可表示為矩陣方程：

Y的維數為[R*1]，U的維數[R * 6]，K的維數[6 * 1]。

R> 6時，超定方程求解：

下面是使用C++實現的多項式擬合的程式，程式中使用opencv進行矩陣運算和影象顯示。程式分別運行了N=3,5,7,9時的情況，結果如下：

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

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

int main()
{
	//資料輸入
	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:多項式階次
	int n = 9;
	Mat mat_k = polyfit(in_point, n);


	//計算結果視覺化
	Mat out(150, 500, CV_8UC3,Scalar::all(0));

	//畫出擬合曲線
	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);
	}

	//畫出原始散點
	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次擬合", out);
	waitKey(0);
	
	return 0;
}

Mat polyfit(vector<Point>& in_point, int n)
{
	int size = in_point.size();
	//所求未知數個數
	int x_num = n + 1;
	//構造矩陣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;
	}

	//矩陣運算，獲得係數矩陣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;
}