https://bugs.kde.org/show_bug.cgi?id=409472
--- Comment #4 from David <[email protected]> --- (In reply to Christoph Feck from comment #2) > Actually, a^b with a < 0 and b not integral usually gives a complex number. > It's possible that the code just doesn't handle this. Sorry man, this isn't true either. Even thinking of the Complex set, you'd have that the cubic root of a negative Real number (that is, a complex of argument π) has 3 solutions, each 120° or 2π/3 from the others. In other words, √³(-x), x>0, has 3 solutions in C, one at π/3, one at π/3+2π/3=π, the last at 5π/3. The one at π clearly representing a negative Real number. That has to be the solution to your problem, if you consider the problem from a C prospective. Considering the Real set as well, an odd function always accepts negative arguments because of the very definition of odd function. -- You are receiving this mail because: You are watching all bug changes.
