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MathGroup Archive 2006

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Re: Is -1^(2/5) really undefined in R?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69094] Re: [mg69075] Is -1^(2/5) really undefined in R?
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Wed, 30 Aug 2006 06:32:36 -0400 (EDT)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <200608290847.EAA00784@smc.vnet.net>
  • Reply-to: murray at math.umass.edu
  • Sender: owner-wri-mathgroup at wolfram.com

The difficulty is, of course, exactly what the meaning of z^(m/n) is 
when m and n are integers and z is negative.

By the same kind of reasoning as Ben gave,

   (-1)^(2/5) = ( (-1)^(1/5) )^2 = (-1)^2 = 1.

[Because (-1)^5 = -1, it seems correct to say that (-1)^(1/5) = -1.]

So that's consistent with what Ben calculated.  (I inserted missing 
parentheses in Ben's calculation.)

But is it correct (as was asked)?

Any real value of (-1)^(2/5) must also be a complex value (with zero 
imaginary part).  So let's see what the complex values are.  The 
definition of the multi-valued power z^a is, in Mathematica syntax, the 
set of all complex numbers of the form

   Exp[a ( Log[Abs[z]] + I (Arg[z] + 2n Pi) )]

where n can be any integer.

Let's evaluate this in Mathematica, noting from further calculations 
that additional values of n outside the range {0,1,2,3} just give the 
same values again:

   a = 2/5; z = -1;
   vals = Table[
                 Exp[a ( Log[Abs[z] ] + I (Arg[z] + 2 n Pi) )],
                 {n, 0, 3}
                ] // ComplexExpand
{-1/4 + Sqrt[5]/4 + (I/2)*Sqrt[(5 + Sqrt[5])/2],
  -1/4 - Sqrt[5]/4 - (I/2)*Sqrt[(5 - Sqrt[5])/2], 1,
  -1/4 - Sqrt[5]/4 + (I/2)*Sqrt[(5 - Sqrt[5])/2]}

(I actually converted the output to InputForm to make the result display 
"linearly".)

And if you either take N of these values or plot them -- most easily 
using David Park's Cardano3`ComplexGraphics` routines

    ComplexGraphics[{PointSize[0.025], ComplexPoint /@ vals}];

-- you see that you have 4 distinct values.

Moreover, exactly one of these 4 values is real, namely, 1.  So it IS 
correct to say that (-1)^(2/5) = 1 -- provided you're looking only for 
real roots.

And Mathematica seems to agree:

   <<Miscellaneous`RealOnly`

   (-1)^(2/5)
1


Ben wrote:
> Is -1^(2/5) really undefined in R?
> 
> Mathematica seems to think so, I guess since it looks like a negative square root, but
> 
> (-1)^(2/5) = ((-1)^2)^(1/5) = 1^(1/5) = 1
> 
> Is this correct mathematically?
> 
> cheers,
> 
> BC
> 
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305


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