将代码和实际理论结合起来才能更好的理解理论上是怎么实现的，参考用高博十四讲的理论加实践亲手试一下，感觉公式和代码才能结合起来。不能做到创新，至少做到了解和理解
曲线拟合问题：
考虑这样一条曲线：$y = \exp (a{x^2} + bx + c) + w$，其中a,b,c为曲线的参数，w为高斯噪声，满足$w = (0,{\sigma ^2})$，假设有N个关于x,y的观测数据点，想根据这些数据点求出曲线的参数，（误差噪声）最小二乘
\[\mathop {\min }\limits_{a,b,c} \frac{1}{2}\sum\limits_{i = 1}^N {{{\left\| {{y_i} - \exp (ax_i^2 + b{x_i} + c)} \right\|}^2}} \]
我们首先明确我们要估计的变量是a,b,c这三个系数，思路是先根据模型生成x,y的真值，然后在真值中加入高斯分布的噪声。随后使用高斯牛顿法从带噪声的数据拟合参数模型。定义误差为：${e_i} = {y_i} - \exp (ax_i^2 + b{x_i} + c)$，求取雅克比矩阵：
\[{J_i} = \left[ {\begin{array}{*{20}{c}}
{\frac{{\partial {e_i}}}{{\partial a}} = - x_i^2\exp (ax_i^2 + b{x_i} + c)}\\
{\frac{{\partial {e_i}}}{{\partial b}} = - {x_i}\exp (ax_i^2 + b{x_i} + c)}\\
{\frac{{\partial {e_i}}}{{\partial c}} = - \exp (ax_i^2 + b{x_i} + c)}
\end{array}} \right]\]
高斯牛顿法的增量方程为：
\[\left( {\sum\limits_{i = 1}^{100} {{J_i}{{({\sigma ^2})}^{ - 1}}{J_i}^T} } \right)\Delta {x_k} = \sum\limits_{i = 1}^{100} { - {J_i}{{({\sigma ^2})}^{ - 1}}{e_i}} \]
1 #include <iostream> 2 #include <chrono> 3 #include <opencv2/opencv.hpp> 4 #include <Eigen/Core> 5 #include <Eigen/Dense> 6 using namespace cv; 7 using namespace std; 8 using namespace Eigen; 9 10 int main(int argc, char **argv) { 11 double ar = 1.0, br = 2.0, cr = 1.0; // 真实参数值 12 double ae = 2.0, be = -1.0, ce = 5.0; // 估计参数值,赋给一个初值，然后在这个初值的基础上进行变化量的迭代 13 int N = 100; // 数据点 14 double w_sigma = 1.0; // 噪声Sigma值 15 double inv_sigma = 1.0 / w_sigma; // 后面噪声1/（Sigma^2)值会用 16 cv::RNG rng; // OpenCV随机数产生器 17 18 vector<double> x_data, y_data; // 用于拟合的观测数据 19 for (int i = 0; i < N; i++) { 20 double x = i / 100.0; 21 x_data.push_back(x); 22 y_data.push_back(exp(ar * x * x + br * x + cr) + rng.gaussian(w_sigma * w_sigma)); 23 } 24 25 // 开始Gauss-Newton迭代 26 int iterations = 100; // 迭代次数 27 double cost = 0, lastCost = 0; // 本次迭代的cost和上一次迭代的cost最小二乘值，通过此值衡量迭代是否到位，进行终止 28 29 chrono::steady_clock::time_point t1 = chrono::steady_clock::now();//计算迭代时间用 30 for (int iter = 0; iter < iterations; iter++) { 31 32 Matrix3d H = Matrix3d::Zero(); // Hessian = J^T W^{-1} J in Gauss-Newton 33 Vector3d b = Vector3d::Zero(); // bias 34 cost = 0; 35 36 for (int i = 0; i < N; i++) { 37 double xi = x_data[i], yi = y_data[i]; // 第i个数据点 38 double error = yi - exp(ae * xi * xi + be * xi + ce); 39 Vector3d J; // 雅可比矩阵 40 J[0] = -xi * xi * exp(ae * xi * xi + be * xi + ce); // de/da 41 J[1] = -xi * exp(ae * xi * xi + be * xi + ce); // de/db 42 J[2] = -exp(ae * xi * xi + be * xi + ce); // de/dc 43 44 H += inv_sigma * inv_sigma * J * J.transpose();//构造Hx=b 45 b += -inv_sigma * inv_sigma * error * J; 46 47 cost += error * error;//计算最小二乘结果，看是否是最小的 48 } 49 50 // 求解线性方程 Hx=b 51 Vector3d dx = H.ldlt().solve(b); 52 if (isnan(dx[0])) { 53 cout << "result is nan!" << endl; 54 break; 55 } 56 57 if (iter > 0 && cost >= lastCost) { 58 cout << "cost: " << cost << ">= last cost: " << lastCost << ", break." << endl; 59 break; 60 } 61 62 ae += dx[0];//更新迭代量，继续迭代寻找最优值 63 be += dx[1]; 64 ce += dx[2]; 65 66 lastCost = cost; 67 68 cout << "total cost: " << cost << ", \t\tupdate: " << dx.transpose() << 69 "\t\testimated params: " << ae << "," << be << "," << ce << endl; 70 } 71 72 chrono::steady_clock::time_point t2 = chrono::steady_clock::now(); 73 chrono::duration<double> time_used = chrono::duration_cast<chrono::duration<double>>(t2 - t1);//给出优化用时 74 cout << "solve time cost = " << time_used.count() << " seconds. " << endl; 75 76 cout << "estimated abc = " << ae << ", " << be << ", " << ce << endl; 77 waitKey(0); 78 return 0; 79 80 }
原文：https://www.cnblogs.com/fuzhuoxin/p/12678498.html