Re: Problem to evaluate cube root of a negative cube nember where a real value is expected
- To: mathgroup at smc.vnet.net
- Subject: [mg28068] Re: Problem to evaluate cube root of a negative cube nember where a real value is expected
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Fri, 30 Mar 2001 04:12:21 -0500 (EST)
- References: <99urns$5nv@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Gary, There's a standard package: <<Miscellaneous`RealOnly` (-8)^(1/3) -2 -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Gary" <garylga at magix.com.sg> wrote in message news:99urns$5nv at smc.vnet.net... > 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 > > > ______________________________________ > Gary Lee Guanan (garylga at magix.com.sg) > Director - Business Development > IQExplorers.com Pte Ltd > Tel 874-1345/6 > ========================C/o Address============================ > Incubation Centre, School of Computing(SoC), NUS. > S15 #01-09, 1 Science Drive 2 (along Lower Kent Ridge Road), S117543 > =============================================================== > >