[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 
#include 

#define MAX_m 200
#define MAX_n 5

double 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; i3.1415926535897932 * (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; i0.01*(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-05

4
  1.0025e+00   9.6180e-01   6.2900e-01   7.0907e-03   1.1792e-01 
error =  1.62e-04

算法思路