Re: Problem to evaluate cube root of a negative cube nember where a real value is expected
- To: mathgroup at smc.vnet.net
- Subject: [mg28065] Re: [mg28042] Problem to evaluate cube root of a negative cube nember where a real value is expected
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Fri, 30 Mar 2001 04:12:18 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Actually what you are doing is not simplifying (-8)^(1/3) at all, because in
Mathematica's notation (-8)^(1/3) is definitely not -2. You can check it as
follows:
In[14]:=
(-8)^(1/3)==-2//Simplify
Out[14]=
False
In fact you can see exactly what (-8)^(1/3) is in terms of radicals:
In[15]:=
RootReduce[(-8)^(1/3)]
Out[15]=
1 + I Sqrt[3]
The point is this. There are three complex third roots of -8. The one
denoted by (-8)^(1/3) is by convention (at least in mathematics a little
beyond High School) taken to be the principal value, which is exactly what
Mathematica does, and it is 1 + I Sqrt[3].
The easiest way to obtain all the roots is to enter them as root objects,
that is in the form Root[#^3+8&,1], Root[#^3+8&,2], Root[#^3+8&,3].
Mathematica orders the roots of a polynomial in such a way that the real
ones come first. Thus in this case we get:
In[16]:=
Table[Root[#^3+8,i],{i,3}]
Out[16]=
{-2, 1 - I Sqrt[3], 1 + I Sqrt[3]}
--
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
on 3/29/01 9:24 AM, Gary at garylga at magix.com.sg wrote:
> Hi,
>
> Does anyone know a simplier way to simplify (-8)^(1/3)=(-2) other than what
> I did below because no complex answer is expected in the solution.
>
> In[161]:=
> p=(-8)^(1/3)
> q=Abs[p]
> Level[p,3]
> r=Extract[Level[p,3],2]
> (* if p is real, then p should be as below *)
> q*r
>
> Out[161]=
> \!\(2\ \((\(-1\))\)\^\(1/3\)\)
>
> Out[162]=
> 2
>
> Out[163]=
> \!\({2, \(-1\), 1\/3, \((\(-1\))\)\^\(1/3\)}\)
>
> Out[164]=
> -1
>
> Out[165]=
> -2
>
>
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