Finding real roots of negative numbers (for example, $\sqrt[3]{-8}$)

Question

Say I want to quickly calculate $\sqrt[3]{-8}$, to which the most obvious solution is $-2$.

When I input $\sqrt[3]{-8}$ or Power[-8, 3^-1], Mathematica gives the result $2 (-1)^{1/3}$. Not what I want.

When I input Power[-8, 3^-1] // N or Power[-8., 3^-1], Mathematica gives the result $1. + 1.73205i$. While technically correct, this is messy and not always useful.

How can I get the real cube root of a negative number? Or more generally, how can I get a list of all valid roots?

Answer 1

Mathematica 9 introduces two new functions, CubeRoot and Surd, that give real-valued roots:

In[1]:= CubeRoot[-8]
Out[1]= -2

In[2]:= Surd[-32, 5]
Out[2]= -2

You can use these to plot real roots:

Plot[CubeRoot[x], {x, -3, 3}]

Note that these functions are undefined for complex numbers:

In[5]:= CubeRoot[1 + I]

CubeRoot::preal: The parameter 1+I should be real valued. >>

Out[5]= Surd[1 + I, 3]

and the typeset form has a small "tail" at the end of the overbar to visually distinguish them from the usual roots:

Answer 2

In general, a typical root of a negative number is complex, so you need to get rid of most roots. A nice approach would be Root, e.g.

Root[ x^3 + 8, #] & /@ Range[3]

{-2, 1 - I Sqrt[3], 1 + I Sqrt[3]}    

To get only real roots you can do :

Select[Root[ x^3 + 8, #] & /@ Range[3], Re[#] == # &]

{-2}

This is a handy approach when you have roots of lower orders.

However you'd rather use Reduce or Solve for higher order roots, in this case it works like this :

Reduce[ x^3 + 8 == 0, x]

x == -2 || x == 1 - I Sqrt[3] || x == 1 + I Sqrt[3]

Solve[ x^3 + 8 == 0, x]

{{x -> -2}, {x -> 2 (-1)^(1/3)}, {x -> -2 (-1)^(2/3)}}  

To get only real roots one can use for example : Reduce[x^3 + 8 == 0, x, Reals] or Solve[x^3 + 8 == 0, x, Reals]. They do almost the same, but their outputs are a bit different, respectively : in the boolean form and in the form of rules.

As a more appropriate example of what you want to do I could choose this one : (-3)^(1/7). Mathematica treats variables (in general) as complex. So one gets seven roots and there is the only one real.

Solve[ x^7 + 3 == 0, x, Reals]

{{x -> -3^(1/7)}}

To get the full output one can do this :

points = {Re @ #, Im @ #} & /@ Last @@@ Solve[x^7 + 3 == 0, x]

Absolute values of the roots are the same, so they are found on the circle of a given radius (== 3^(1/7)) :

{ Equal @@ #, radius = #[[1]] } & @ Simplify @ (Norm /@ points)

 {True, 3^(1/7)} 

To visualize the structure of the output one makes use of ContourPlot of real and imaginary parts of the function (x + I y)^7 + 3 (we write the function explicitly in the complex form since we make plots in real domains of x and y ) :

GraphicsRow[{
    ContourPlot[ Re[(x + I y)^7 + 3], {x, -2, 2}, {y, -2, 2}, PlotPoints -> 25, MaxRecursion -> 5, 
                 Epilog -> {Darker@Green, Thick, Line[{{0, 0}, #}] & /@ points, 
                 Gray, Dashed, Circle[{0, 0}, radius], PointSize[0.02], Blue, Point[points]}], 

    ContourPlot[ Im[(x + I y)^7 + 3], {x, -2, 2}, {y, -2, 2}, PlotPoints -> 25, MaxRecursion -> 5, 
                 Epilog -> {Darker@Green, Thick, Line[{{0, 0}, #}] & /@ points, 
                 Gray, Dashed, Circle[{0, 0}, radius], PointSize[0.02], Blue, Point[points]}]}]

Clarification

• The blue points --- roots
• lengths of the green lines --- absolute values of the roots
• the dashed circle --- a set of all complex numbers z such that Abs[z] == radius

The only one real root lies on the line y == 0.

Answer 3

You may use a function, which gives you the "real Power":

rprule=(b_?Negative)^Rational[m_,n_?OddQ]:>(-(-b)^(1/n))^m;
Attributes[realPower]={Listable, NumericFunction,OneIdentity}  (* same as Power *)
realPower[b_?Negative, Rational[m_, n_?OddQ]] := (-(-b)^(1/n))^m;
realPower[x_,y_]:=Power[x,y];
realPower[x_]:=x//.rprule;

Then you'll get:

realPower[{8^(1/3),(-8)^(1/3)]

{2,-2}

The "two arguments" form is needed eg. in plots:

Plot[realPower[x, 1/3], {x, -2, 2}] 

Answer 4

In general, to get a list of all the cube roots of -8 (or the $m$ roots of any number $n$), you can use either the the Roots or Solve or Reduce functions.

Roots[x^3 == -8, x]
(* Out[1]=  x == 2 || x == 2 (-1)^(2/3) || x == -2 (-1)^(1/3) *)

Reduce and Solve are perhaps more flexible because you can specify the domain that you want or leave it out for all the solutions.

Reduce[x^3 == 8, x]
(* Out[2]= x == 2 || x == -1 - I Sqrt[3] || x == -1 + I Sqrt[3] *)

Reduce[x^3 == 8, x, Reals]
(* Out[3]= x == 2

Solve[x^3 == 8, x]
(* Out[4]= {{x -> 2}, {x -> -2 (-1)^(1/3)}, {x -> 2 (-1)^(2/3)}} *)

Solve[x^3 == 8, x, Reals]
(* Out[5]= {{x -> 2}} *)

Note that the structure of the output returned for Reduce and Solve are different.

Answer 5

I quite prefer realPower[x_, r_] := Sign[x]*Abs[x]^r myself. (A similar thing is done in the old package Miscellaneous`RealOnly`.) realPower[-8, 1/3] yields -2, as expected. To get all the real roots, Artes's solution is best.

Answer 6

Let us introduce a rule:

cubicRootRule = a_^(1/3) /; Negative[a] -> -(Abs[a])^(1/3);

To check it let us form a list with cubic roots of various numbers, positive and negative:

 lst = Table[(RandomInteger[{-16, 16}])^(1/3), {20}]

(* {(-11)^(1/3), 10^(1/3), (-10)^(1/3), 6^(1/3), (-5)^(1/3),  
  2 (-1)^(1/3), (-10)^(1/3), (-3)^(1/3), 2^(1/3), (-15)^(1/3),
  3^( 1/3), (-7)^(1/3), (-7)^(1/3), 15^(1/3), (-5)^(1/3),  
   2 (-1)^(1/3), 13^(1/3), (-13)^(1/3), 2 (-2)^(1/3), 2^(2/3) 3^(1/3)} *)

and apply the rule to the list:

lst /. cubicRootRule

(* {-11^(1/3), 10^(1/3), -10^(1/3), 6^(
 1/3), -5^(1/3), -2, -10^(1/3), -3^(1/3), 2^(1/3), -15^(1/3), 3^(
 1/3), -7^(1/3), -7^(1/3), 15^(1/3), -5^(1/3), -2, 13^(
 1/3), -13^(1/3), -2 2^(1/3), 2^(2/3) 3^(1/3)} *)

That's it?