nth root implementation

Question

I am working on a way to calculate the n^th root of a number. However, I am having problems with the n^th root of negative numbers.

Most people s

Answer 1

What are you trying to do? Unless you're planning to fully and properly handle complex numbers you cannot take the nth root of a negative number.

For example, while (-8)^(1/3) has a principal branch of -2, the only branches of (-4)^(1/2) are 2i and -2i.

To handle this properly you need to transform the number into its polar form and then take the required root in that form.

So -8 is the complex number 8*exp(i*pi). The 1/3 roots of that are 2*exp(i*pi/3), 2*exp(i*pi), and 2*exp[i*(-pi)/3]. Then you can use de Moivre' formula to compute the roots in the form a + bi.

Answer 2

I'm not too sure about the exact code, but add an extra if statement to clarify between odd and even roots. something along the lines of

public static double root(double num, double root) {
    if (num < 0) {
        if(root%2==1) {
            return -Math.pow(Math.abs(num), (1 / root));
        }
    }
    return Math.pow(num, 1.0 / root);
}

Not entirely sure if this will work with your other code, but I hope it can help

Answer 3

(num) ^ (1/root) is similar to exp( (1/root) * log(num) ), so you can do it like:

public static double root(double num, double root)
{
    return Math.pow(Math.E, Math.log(num)/root);
} 

Answer 4

Either use one of the many complex number packages available on the Internet, or convert your number to a rectangular position on the Argand plane, rotate it the appropriate angle as given by the root, then read it out as a (real, imaginary) pair.

Answer 5

You could do if(num < 0){ return Math.abs(Math.pow(num, 1 / root)) } Then just use ' + "i"' whenever stating the value. Or use the absolute value for equations and later factor in the positive/negative and i when needed. That's what worked for me.

Answer 6

System.out.println( Math.pow(10, Math.log10(Number)/root));