Cube root of a negative number

Question

I\'m trying to find the cube root of a negative number but I get a NaN. Any help?

System.out.println(Math.pow(-8, 1.0 / 3.0));

Answer 1

The Java documentation for Math.pow states:

If the first argument is finite and less than zero [...] [and] if the second argument is finite and not an integer, then the result is NaN.

You could use Math.cbrt to get the cube root:

double result = Math.cbrt(-8.0);

Answer 2

http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Math.html#cbrt(double)

System.out.println(Math.cbrt(-8));

Answer 3

Remember that mathematically, there are 3 cube-roots. Assuming you want the root that is real, you should do this:

x = 8;  //  Your value

if (x > 0)
    System.out.println(Math.pow(x, 1.0 / 3.0));
else
    System.out.println(-Math.pow(-x, 1.0 / 3.0));

EDIT : As the other answers mention, there is Math.cbrt(x). (which I didn't know existed)

The reason why pow returns NaN with a negative base and non-integral power is that powering is usually done by angle-magnitude in the complex plane.

• For positive real numbers, the angle is zero, so the answer will still be positive and real.
• For negative real numbers, the angle is 180 degrees, which (after multiplying by a non-integral power) will always produce a complex number - hence a NaN.

Answer 4

From http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html:

If the first argument is finite and less than zero

• if the second argument is a finite even integer, the result is equal to the result of raising the absolute value of the first argument to the power of the second argument
• if the second argument is a finite odd integer, the result is equal to the negative of the result of raising the absolute value of the first argument to the power of the second argument
• if the second argument is finite and not an integer, then the result is NaN.