Cube root of a number, C ++

Question

I would like to ask a very short question, and it looks like this: when finding the cube root of a number (both negative and positive) in C ++, how to limit the output to only real solutions? I am currently writing a program to solve a Cardano formula cube, and one of the intermediate variables that I use randomly outputs the complex and real roots of the cube, and I only need real roots.

(For example, when evaluating the cubic root of -0.0127378, the three roots will be 0.11677095 + 0.202253218i, -0.2335419, 0.11677095-0.202253218i. I want to ignore the difficult ones to replace with a later formula)

Thank!

EDITOR: Solved! :) I created the signum function and changed the sign after I took the absolute values ​​of SPrime and TPrime, so now it only carries the real cubic 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 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;
}

Answer 1

The problem you are claiming can be solved trivially (with real numbers, the cubic root -x is opposite to the cubic root 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;
    }
 }

If the input is generally complicated z, and you are looking for the “real” (main) cubic root, then using the complex version powyou can use the same reasoning as zor -zdepending 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);
    }
 }

Answer 2

Problems with your code:

• , , .

• abs(D) . D==0, , .

• , S*T=-Q. , u=T^3 0==u^2 - 2*R*u - Q^3 (u-R)^2 = D = R^2+Q^3

  rtD = sqrt(D);
  T = cuberoot( R + (abs(R+rtD)>=abs(R-rtD)) ? rtD : -rtD );
  S = (abs(T)<epsilon) ? 0 : -Q/T;
  

  - abs(R)<=abs(T)^3 abs(D)<=abs(T)^6 abs(Q)<=2^(1/3)*abs(T)^2,

  abs(S)=abs(Q/T) <= 2^(1/3)*abs(T)
  

  S=-Q/T , , R, Q , , T. , double epsilon=1e-150 .

---

:

T . , ,

std::complex<double> cuberoot(std::complex<double> z) {
    double r=abs(z), phi=arg(z);
    double k = round(2*phi/pi); 
    // closest multiple of pi/2
    // an equivalent angle is (phi-k*pi/2) - k*3*pi/2
    return std::polar( pow(r,1.0/3), (phi-k*pi/2)/3 - k*pi/2 );
}

abs(phi-k*pi/2)<=pi/4, , , , pi/12=15°. cuberoot(i) -i, cuberoot(-1) -1, 60 ° (60 ° -90 °)/3-90 ° = -100 ° ..