我在 C++ 中的自适应梯形规则算法存在问题——基本上，无论指定的容差如何，我都会得到相同的精确近似值。对于较大的公差，递归应该很早就停止（因为 abs(coarse-fine) 将小于 3.0*large tolerance 并且递归的 minLevel 约为 5）。
但是，此函数的作用是运行最大次数，而不管公差的选择如何。我哪里搞砸了？编辑：也许我的辅助函数有问题？
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);
}
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;
}