多维数据集的根，C ++

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我想问一个很短的问题，它是这样的:在C ++中找到一个数字(正负)的立方根时，如何将输出限制为仅实数解? 我目前正在编写一个程序，用卡尔达诺公式求解一个三次方，而我使用的中间变量之一随机输出了复数和实方根-我只需要实根.

I would like to ask a very short question, and it is as follows: in finding the cube root of a number (both neg. and pos.) in C++, how does one restrict the output to real solutions only? I am currently writing a program to solve a cubic with Cardano's formula, and one of the intermediate variables I am using randomly outputs the complex and real cube roots - and I only need the real roots.

(例如，在评估-0.0127378的立方根时，三个根将为0.11677095 + 0.202253218i，-0.2335419、0.11677095-0.202253218i-我希望忽略将其替换为以后公式的复杂根)

(E.g. in evaluating the cube root of -0.0127378, the three roots would be 0.11677095+0.202253218i, −0.2335419, 0.11677095−0.202253218i - I wish to ignore the complex ones for substitution into a later formula)

谢谢！

解决了！ :)我在使用SPrime和TPrime的绝对值的幂后创建了一个signum函数并对其进行了调整，因此现在它仅继承了真实的多维数据集根.

Solved it! :) I created a signum function and tweaked the sign after taking the power of the absolute value of SPrime and TPrime, so now it carries forward only the real cube root.

/* ... */ #include <iostream> #include <cmath> #include <complex> #include <cstdio> #include <cassert> using namespace std; int signum(std::complex<double> z) { if (z.real() < 0 || z.imag() < 0) return -1; else if (z.real() >= 0 || z.imag() >= 0) return 1; } // POST: The function is intended to solve a cubic equation with coefficients a, b, c and d., such that // ax^3 + bx^2 + cx + d = 0. If there exist infinitely many solutions, we output -1, i.e. if a=b=c=d=0 // (trivial solution). void solve(std::complex<double> a, std::complex<double> b, std::complex<double> c, std::complex<double> d, std::complex<double>& x1, std::complex<double>& x2, std::complex<double>& x3) { complex<double> i = complex<double> (0, 1.0); // Consider implementing Cardano's method for obtaining the solution of a degree 3 polynomial, as suggested // We must hence define the discriminant D of such an equation through complex doubles Q and R std::complex<double> Q; Q = (3.0*a*c - pow(b, 2)) / (9.0*pow(a, 2)); cout << "Q=" << Q << endl; std::complex<double> R; R = (9.0*a*b*c - 27.0*d*pow(a, 2) - 2.0*pow(b, 3)) / (54.0*pow(a, 3)); cout << "R=" << R << endl; std::complex<double> D; D = pow(Q, 3) + pow(R, 2); // Possible types of output for discriminant if (abs(D) < 0.0) { cout << "The cubic has three distinct, real roots." << endl; } else if (abs(D) == 0.0) { cout << "The cubic has three real roots, at least two of which are equal." << endl; } else if (abs(D) > 0.0) { cout << "The cubic has one real root and two complex conjugate roots." << endl; } // Defining two further complex double variables S and T, which are required to obtain the final solution for x1, x2 and x3 std::complex<double> S; std::complex<double> SPrime; SPrime = R+sqrt(Q*Q*Q + R*R); cout << "SPrime=" << SPrime << endl; if (signum(SPrime) == -1) { S = (-1)*pow(abs(SPrime), 0.3333333333333); } else if (signum(SPrime) == 1) { S = pow(abs(SPrime), 0.3333333333333); } cout << "S=" << S << endl; std::complex<double> T; std::complex<double> TPrime; TPrime = (R-sqrt(Q*Q*Q + R*R)); if (signum(TPrime) == -1) { T = (-1)*pow(abs(TPrime), 0.3333333333333); } else if (signum(TPrime) == 1) { T = pow(abs(TPrime), 0.3333333333333); } cout << "T=" << T << endl; cout << "TPrime= " << TPrime << endl; // Expressions for the solutions x1 = S + T - (b/(3.0*a)); x2 = (-0.5)*(S + T) - (b/(3.0*a)) + (sqrt(3.0)*0.5)*(S - T)*i; x3 = conj(x2); if (abs(x1) < 0.000000000001) { x1 = 0; } } // Driver code int main () { // Taking user input for a, b, c and d std::complex<double> a, b, c, d, x1, x2, x3; cout << "Please enter the coefficients of the polynomial in successive order." << endl; cin >> a >> b >> c >> d; solve (a, b, c, d, x1, x2, x3); cout << x1 << ", " << x2 << ", " << x3 << "." << endl; return 0; }

推荐答案

您所说的问题可以轻松解决(对于实数，-x的立方根与x的立方根相反):

The problem as you're stating it can be solved trivially (with real numbers the cubic root of -x is the opposite of the cubic root of x):

double cuberoot(double x) { if (x < 0) { return -pow(-x, 1.0/3.0); } else if (x > 0) { return pow(x, 1.0/3.0); } else { return 0; } }

如果输入改为一般的复数z，而您正在寻找最真实的"(主)立方根，则可以使用复数pow版本将相同的推理应用于z或取决于实部的符号:

If the input is instead in general complex z and you're looking for the "most real" (principal) cubic root the same reasoning can be applied using complex pow version to either z or -z depending on the sign of the real part:

std::complex<double> cuberoot(std::complex<double> z) { if (z.real() < 0) { return -pow(-z, 1.0/3.0); } else { return pow(z, 1.0/3.0); } }