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

**Follow-Ups**:**Re: Re: Is -1^(2/5) really undefined in R?***From:*János <janos.lobb@yale.edu>

**References**:**Is -1^(2/5) really undefined in R?***From:*Ben <ben.carbery@spam.me>