Question

我正在尝试实现一个利用先前函数评估的梯形规则，以避免冗余计算。两件事：a）计算结果没有收敛，我有点不确定为什么。我将发布数学背后的原因，为什么我认为算法应该在需要时产生收敛，并且b）do while循环终止于n = 8并且我也无法想出那个;它应该运行直到n> 128？（n是子间隔的数量）我的代码如下。提前谢谢！

void NestedTrap(int n) //Trapezoidal with reuse of function evaluations
{
    double a,b; //interval end points
    double x[n+1]; //equally spaced nodes
    double c[n]; //midpoints
    double T; //Initial integral evaluation
    double T2; //Evaluation with reuse of previous function evaluations
    double h, h2; //step sizes for T and T2
    double temp1, temp2;

    std::cout <<"Enter interval end points (lesser first; enter 999 for pi & 999.2 for pi/2 & 999.4 for pi/4): ";
    std::cin >> a >> b;

    if (b == 999)
    {
        b = M_PI;
    }
    if (a == 999)
    {
        a = M_PI;
    }
    if (b == 999.4)
    {
        b = M_PI/4;
    }
    if (a == 999.4)
    {
        a = M_PI/4;
    }
    if (b == 999.2)
    {
        b = M_PI/2;
    }
    if (a == 999.2)
    {
        a = M_PI/2;
    }

    h = (b-a)/n;
    T = 0;
    temp1 = 0;
    temp2 = 0;

    for (int i=0; i<=n; i++)
    {
        x[i] = 0;
    }


    for (int i=0; i<n; i++)
    {
        x[i+1] = x[i] + h;
    }

    for (int i=1; i<n; i++)
    {
        temp1 += I1(x[i]);
    }
    T = (h/2)*exp(x[0]) + (h/2)*exp(x[n]) + (h*temp1);
    std::cout << "T_" << n <<": " << T << std::endl;

    do
    {
    temp2 = 0;
    n = 2*n;
    h2 = (b-a)/(n);

    for (int i=0; i<n; i++)
    {
        c[i] = 0;
    }
    for (int i=1; i<=n; i++)
    {
        c[i] = a + h2*(i-0.5);
        //std::cout << c[i] << std::endl;
    }

    for (int i=0; i<n; i++)
    {
        temp2 += exp(c[i]);
    }

    T2 = (T/2) + h2*temp2;
    std::cout << "T_" << n <<": " << T2 << std::endl;

    T = T2;
    }   while (n <= 128);
}

Answer 1

您可以在此处创建大小为n的数组

double x[n+1]; //equally spaced nodes
double c[n]; //midpoints

（请注意，这不是有效的c ++）

然后你在这里增加n：

n = 2*n;

然后你在这里写过数组的结尾：

for (int i=0; i<n; i++)
{
    c[i] = 0;
}

导致未定义的行为（可能会覆盖其他一些变量）

梯形法c ++的数值积分

1 个答案: