Evaluating N[(-1)^(1/3)] returns 0.5 + 0.866i in my v12.0 notebook. Of course, the equation x^3 = -1 has three distinct solutions for x. Is there a global (or local) option that will force Mathematica to output the real solution in favor of the complex ones? In this case, simply -1?
Further to this, if my notebook defaults to evaluating (-1)^(1/3) to a complex number, can I assume that Mathematica is doing the same when plotting complicated expressions with cube roots? Namely that some areas of the plot may appear imaginary when in fact there is a real solution available?
Thanks
Surd[]? – J. M.'s missing motivation Feb 27 '21 at 20:21Surd[(-1),3] would give the desired result here, but for a complicated expression where some terms contain a cube root, it would be very cumbersome to have to manually call Surd[ ] for the terms which happen to be cube roots.
– subzero366 Feb 27 '21 at 20:50Exp[2 I π/3]if need be. – J. M.'s missing motivation Feb 27 '21 at 21:00$f(x)=(- x^3 + 5 x + (-2 + \sqrt{-27 x^2 + 10 x^3 - 48 x + 115}))^{1/3}$
The question is, for what values of x is this function real? If one simply plots the function, it will appear to be real on the interval (0,1.9). The term under the square root however, is real on (0,1.9) AND when $x>3$. Suppose $x = 4$, then the function reduces to:
$f(4)=(\sqrt{131}-46)^{1/3}$
– subzero366 Feb 28 '21 at 17:55As I am examining many plots visually that contain this term, it is important to get Mathematica to display the real branch (when $x>3$) in my plots.
– subzero366 Feb 28 '21 at 17:55Power[w, 1/3]intoSurd[w, 3], if so. – J. M.'s missing motivation Feb 28 '21 at 17:59