C中的中心差分近似值

Question

我在为决赛学习时遇到了这个问题，而且我似乎无法使其发挥作用。这个问题本身如下所示。任何关于如何解决这个问题的帮助将不胜感激。

下面是我解决类似问题的代码。我希望它可以作为一个基础上解决这一问题

#include <stdio.h> 
    #include <stdlib.h> 
    #include <cmath> 
    using namespace std; 

     double f(double x) 
     { 
      return (cos(x)); 
     } 


    /*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 
    Numerical Differentiation Formulae (n-th derivative) 
    ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/ 
    void dnf_dxn (int n, double x, double h, double& fd, double& cd) 
    { 

    if(n==1) 
     // Approximation to the 1st Derivative of f at x 
     { 
     // 1st order forward differencing 
     fd = (f(x+h) - f(x))/h; 

     // 2nd order centered differencing 
     cd = (f(x+h) - f(x-h))/(2*h); 

     } 

    else if(n==2) 
     // Approximation to the 2nd Derivative of f at x 
     { 
     // 1st order forward differencing 
     fd = (f(x+2*h) - 2*f(x+h) + f(x))/(h*h); 

     // 2nd order centered differencing 
     cd = (f(x+h) - 2*f(x) + f(x-h))/(h*h); 

     } 

    else if(n==3) 
     // Approximation to the 3rd Derivative of f at x 
     { 
     // 1st order forward differencing 
     fd = (f(x+3*h) - 3*f(x+2*h) + 3*f(x+h) - f(x))/(h*h*h); 

     // 2nd order centered differencing 
     cd = (f(x+2*h) - 2*f(x+h) + 2*f(x-h) - f(x-2*h))/(2*h*h*h); 

     } 

    else 
     { 
     printf("Only derivatives of orders 1, 2 and 3 are implemented. \n"); 
     getchar(); 
     exit(1); 
     } 

    } 



    /*:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 
     NUM_DIFF  M A I N  P R O G R A M 
    ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/ 
    int main() 
    { 

    printf("\n Numerical Differentiation of f(x)=cos(x) at x=1 \n \n"); 
    printf(" Derivative of order 1, 2 and 3 using forward  \n"); 
    printf(" and centered difference approximations (h=0.01): \n \n"); 


    double x = 0.5; 
    double h = 0.01; 
    int n; 
    double fd, cd, exact, cd_error, fd_error; 
    double true_fx = - sin(x); 
    double true_fxx = - cos(x); 
    double true_fxxx = sin(x); 


    printf("Derivative Stepsize Differencing  Result Abs Error \n"); 

    for(n=1; n<4; n++) 
     { 

     dnf_dxn (n, x, h, fd, cd); 

     if(n==1) 
      { exact = true_fx; } 
     else if(n==2) 
      { exact = true_fxx; } 
     else 
      { exact = true_fxxx; } 

     fd_error = abs(exact - fd); 
     cd_error = abs(exact - cd); 
     printf("  %i  %4.2f  Forward  %10.7f %10.3e \n", 
        n, h, fd, fd_error); 
     printf("      Centered %10.7f %10.3e \n", 
         cd, cd_error); 

     } 

    printf("\n \n <Press the RETURN key to exit num_diff.cpp> \n \n"); 
    getchar(); 

    } 

下面是实际的问题：

Answer 1

嗯，一个，你说你在x = 1计算这些差异，但你实际上在x = 0.5进行计算。

仅仅说“它不起作用”是不够的。你在期待什么，而你又得到了什么？