finding cube root in C++?

Question

Strange things happen when i try to find the cube root of a number.

The following code returns me undefined. In cmd : -1.#IND

cout<<pow(( double )(20.0*(-3.2) + 30.0),( double )1/3)

While this one works perfectly fine. In cmd : 4.93242414866094

cout<<pow(( double )(20.0*4.5 + 30.0),( double )1/3)

From mathematical way it must work since we can have the cube root from a negative number. Pow is from Visual C++ 2010 math.h library. Any ideas?

Answer 1

pow(x, y) from <cmath> does NOT work if x is negative and y is non-integral.

This is a limitation of std::pow, as documented in the C standard and on cppreference:

Error handling

• Errors are reported as specified in math_errhandling
• If base is finite and negative and exp is finite and non-integer, a domain error occurs and a range error may occur.
• If base is zero and exp is zero, a domain error may occur.
• If base is zero and exp is negative, a domain error or a pole error may occur.

There are a couple ways around this limitation:

• Cube-rooting is the same as taking something to the 1/3 power, so you could do std::pow(x, 1/3.).

• In C++11, you can use std::cbrt. C++11 introduced both square-root and cube-root functions, but no generic n-th root function that overcomes the limitations of std::pow.

Answer 2

The power 1/3 is a special case. In general, non-integral powers of negative numbers are complex. It wouldn't be practical for pow to check for special cases like integer roots, and besides, 1/3 as a double is not exactly 1/3!

I don't know about the visual C++ pow, but my man page says under errors:

EDOM The argument x is negative and y is not an integral value. This would result in a complex number.

You'll have to use a more specialized cube root function if you want cube roots of negative numbers - or cut corners and take absolute value, then take cube root, then multiply the sign back on.

Note that depending on context, a negative number x to the 1/3 power is not necessarily the negative cube root you're expecting. It could just as easily be the first complex root, x^(1/3) * e^(pi*i/3). This is the convention mathematica uses; it's also reasonable to just say it's undefined.