自适应正交 (C++）

我在C++的自适应梯形规则算法中遇到了问题 - 基本上，无论指定的公差如何，我都会得到相同的精确近似值。对于大容差，递归应该很早就停止（因为 abs（粗细）将小于 3.0*大容差，最小递归级别约为 5）。

但是，此函数的作用是运行最大次数，而不考虑选择公差。我哪里搞砸了？编辑：也许我的帮助程序函数有问题？

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);
}

以下是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;
    }