[NA]Lab6:正交多项式拟合

任务概述

给定函数f(x)在m个采样点处的值f(xi)以及每个点的权重wi，求曲线拟合的正交多项式Pn(x)满足最小二乘误差||err||2=∑mi=1wi[f(xi)−Pn(xi)]2<TOL。
函数接口定义：

int OPA( double (*f)(double t), int m, double x[], double w[], double c[], double *eps );

裁判程序样例：

#include <stdio.h>#include <math.h>#define MAX_m 200#define MAX_n 5double f1(double x){    return sin(x);}double f2(double x){    return exp(x);}int OPA( double (*f)(double t), int m, double x[], double w[], double c[], double *eps );void print_results( int n, double c[], double eps){       int i;    printf("%d\n", n);    for (i=0; i<=n; i++)        printf("%12.4e ", c[i]);    printf("\n");    printf("error = %9.2e\n", eps);    printf("\n");}int main(){    int m, i, n;    double x[MAX_m], w[MAX_m], c[MAX_n+1], eps;    m = 90;    for (i=0; i<m; i++) {        x[i] = 3.1415926535897932 * (double)(i+1) / 180.0;        w[i] = 1.0;    }    eps = 0.001;    n = OPA(f1, m, x, w, c, &eps);    print_results(n, c, eps);    m = 200;    for (i=0; i<m; i++) {        x[i] = 0.01*(double)i;        w[i] = 1.0;    }    eps = 0.001;    n = OPA(f2, m, x, w, c, &eps);    print_results(n, c, eps);    return 0;}/* Your function will be put here */

输出样例：

3 -2.5301e-03   1.0287e+00  -7.2279e-02  -1.1287e-01 error =  6.33e-054  1.0025e+00   9.6180e-01   6.2900e-01   7.0907e-03   1.1792e-01 error =  1.62e-04

算法思路