Question

我在C ++中遇到了自适应梯形规则算法的问题 - 基本上，无论指定的容差如何，我都得到了相同的精确近似值。对于大公差，递归应该很早就停止（因为abs（粗细）将小于3.0 *大容差和递归的minLevel大约为5）。

但是，无论选择公差，此功能的作用都是最大次数。我在哪里搞砸了？编辑：也许我的助手功能有问题？

double trap_rule(double a, double b, double (*f)(double),double tolerance, int count)
{
    double coarse = coarse_helper(a,b, f); //getting the coarse and fine approximations from the helper functions
    double fine = fine_helper(a,b,f);

    if ((abs(coarse - fine) <= 3.0*tolerance) && (count >= minLevel))
        //return fine if |c-f| <3*tol, (fine is "good") and if count above
        //required minimum level
    {
        return fine;
    }
    else if (count >= maxLevel)
        //maxLevel is the maximum number of recursion we can go through
    {
        return fine;

    }
    else
    {
        //if none of these conditions are satisfied, split [a,b] into [a,c] and [c,b] performing trap_rule
        //on these intervals -- using recursion to adaptively approach a tolerable |coarse-fine| level
        //here, (a+b)/2 = c
        ++count;
        return  (trap_rule(a, (a+b)/2.0, f, tolerance/2.0, count) + trap_rule((a+b)/2.0, b, f, tolerance/2.0, count));
    }
}


EDIT: Helper and test functions:

//test function
double function_1(double a)
{
    return pow(a,2);
}

//"true" integral for comparison and tolerances




//helper functions

double coarse_helper(double a, double b, double (*f)(double))
{
    return 0.5*(b - a)*(f(a) + f(b)); //by definition of coarse approx
}


double fine_helper(double a, double b, double (*f)(double))
{
    double c = (a+b)/2.0;
    return 0.25*(b - a)*(f(a) + 2*f(c) + f(b)); //by definition of fine approx
}

double helper(double a, double b, double (*f)(double x), double tol)
{
    return trap_rule(a, b, f, tol, 1);
}

这是main（）中的内容：

std::cout << "First we approximate the integral of f(x) = x^2 on [0,2]" << std::endl;

    std::cout << "Enter a: ";
    std::cin >> a;

    std::cout << "Enter b: ";
    std::cin >> b;

    true_value1 = analytic_first(a,b);

    for (int i = 0; i<=8; i++)
    {
        result1 [i] = helper(a, b, function_1, tolerance[i]);
        error1 [i] = fabs(true_value1 - result1 [i]);
    }

    std::cout << "(Approximate integral of x^2, tolerance, error )" << std::endl;
    for (int i = 0; i<=8; i++)
    {
        std::cout << "(" << result1 [i] << "," << tolerance[i] << "," << error1[i] << ")" << std::endl;
    }

Answer 1

我发现与你的建议正好相反 - 算法仅在minLevel步骤后终止 - 原因是由于你使用了abs，而不是公差测试中fabs。 abs正在将其参数转换为int，因此任何小于1的错误都会四舍五入为零。

在abs到位的情况下，我从一个非常类似的程序中得到了这个输出：

(0.333496,0.001,0.00016276)
(0.333496,0.0001,0.00016276)
(0.333496,1e-05,0.00016276)
(0.333496,1e-06,0.00016276)
(0.333496,1e-07,0.00016276)
(0.333496,1e-08,0.00016276)
(0.333496,1e-09,0.00016276)
(0.333496,1e-10,0.00016276)
(0.333496,1e-11,0.00016276)

替换为fabs我明白了：

(0.333496,0.001,0.00016276)
(0.333374,0.0001,4.06901e-05)
(0.333336,1e-05,2.54313e-06)
(0.333334,1e-06,6.35783e-07)
(0.333333,1e-07,3.97364e-08)
(0.333333,1e-08,9.93411e-09)
(0.333333,1e-09,6.20882e-10)
(0.333333,1e-10,3.88051e-11)
(0.333333,1e-11,9.7013e-12)

自适应正交（C ++）

1 个答案: