Re: Re: Is -1^(2/5) really undefined in R?
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- Subject: [mg69147] Re: [mg69094] Re: [mg69075] Is -1^(2/5) really undefined in R?
- From: János <janos.lobb at yale.edu>
- Date: Thu, 31 Aug 2006 04:39:00 -0400 (EDT)
- References: <200608290847.EAA00784@smc.vnet.net> <200608301032.GAA27866@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I do not want to split electrons here, but for me the question in
this case is this:
What has higher precedence: Sign/Times ? or Power ?
I vote for Power :)
According to my vote -1^x is not the same as (-1)^x , so Ben's second
equation:
>> (-1^2)^(1/5) = 1^(1/5)
is not correct.
According to Mathematica:
In[13]:=
-1^2
Out[13]=
-1
János
On Aug 30, 2006, at 6:32 AM, Murray Eisenberg wrote:
> 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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corpse.
(S. Lem: His Master Voice)
- References:
- Is -1^(2/5) really undefined in R?
- From: Ben <ben.carbery@spam.me>
- Re: Is -1^(2/5) really undefined in R?
- From: Murray Eisenberg <murray@math.umass.edu>
- Is -1^(2/5) really undefined in R?