C++ 截断错误？

Question

我正在学习 C++，在尝试使用公式计算电流时遇到了这个问题。

我得到:0.628818 答案应该是:

f=200 Hz

R=15 Ohms

C=0.0001 (100µF)

L=0.01476 (14.76mH)

E = 15 V

Answer: I = 0.816918A (calculated)

下面是我的代码:

#include <iostream>
#include <cmath>

int main()
{
    const double PI = 3.14159;
    double r = 15;
    double f = 200;
    double c = 0.0001;
    double l = 0.01476;
    double e = 15;

    double ans = e / std::sqrt(std::pow(r, 2) + (std::pow(2 * PI*f*l - (1.0 / 2.0 * PI*f*c), 2)));

    std::cout << "I = " << ans << "A" << std::endl;
}

我已阅读有关截断错误的信息并尝试使用 1.0/2.0，但似乎也不起作用。

Answer 1

截断误差是指仅使用无穷级数的前 N ​​项来估计一个值。所以你的问题的答案是“否”。但是，您可能会发现以下内容有些有趣......

#include <iostream>
#include <cmath>  
#include <iomanip>

using namespace std;

template<typename T>
T fsqr(T x) { return x * x; }

// Numerically stable and non-blowuppy way to calculate
// sqrt(a*a+b*b)
template<typename T>
T pythag(T a, T b) {
    T absA = fabs(a);
    T absB = fabs(b);
    if (absA > absB)
    {
        return absA*sqrt(1.0 + fsqr(absB / absA));
    } else if (0 == absB) {
        return 0;
    } else {
        return absB*sqrt(1.0 + fsqr(absA / absB));
    }
}
int main () {

double e, r, f, l, c, ans;

const double PI = 3.14159265358972384626433832795028841971693993751058209749445923078164062862089986280348253421170;
cout << "Insert value for resistance: " << endl;
cin >> r ;

cout << "Insert value for frequency: " << endl;
cin >> f;

cout << "Insert value for capacitance: " << endl;
cin >> c;

cout << "Insert value for inductance: " << endl;
cin >> l;

cout << "Insert value for electromotive force (voltage): " << endl;
cin >> e;

ans = e / pythag(r, 2*PI*f*l - (1/(2*PI*f*c)) );

cout << "I = " << ans << "A" << endl;

system("pause");

return 0;
}

关于 PI 开个玩笑。