Re: How to instruct Math to take a certain (e.g. real)
- To: mathgroup at smc.vnet.net
- Subject: [mg105059] Re: [mg105018] How to instruct Math to take a certain (e.g. real)
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Thu, 19 Nov 2009 05:24:49 -0500 (EST)
- Reply-to: hanlonr at cox.net
Your equation does not have a solution
eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3);
Reduce[eq, y]
False
Solve[eq, y]
{}
Apparently, you are looking for the solution of a different equation
f[x_] := Sign[x]*Abs[x]^(1/3)
eq2 = f[-8 + y] + f[8 + y] == f[y];
sol = {Reduce[eq2, y, Reals] // ToRadicals // ToRules}
{{y -> 0}, {y -> -12*Sqrt[3/7]}, {y -> 12*Sqrt[3/7]}}
eq2 /. sol // FullSimplify
{True,True,True}
eq /. sol
{False,False,False}
There is a legacy RealOnly add-on package at
http : // library.wolfram.com/infocenter/MathSource/6771/
Bob Hanlon
---- Alexei Boulbitch <Alexei.Boulbitch at iee.lu> wrote:
=============
Dear Community,
I came to a problem, that I cannot check the solution y=0 of equation
In[345]:= eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3);
eq /. y -> 0
Out[346]= False
just because the expression
In[347]:= eq[[1]] /. y -> 0
Out[347]= 2 + 2 (-1)^(1/3)
Mathematica does not interpret as zero. And if I ask it to give a
numerical answer
In[348]:= 2 + 2 (-1.)^(1/3)
Out[348]= 3.+ 1.73205 \[ImaginaryI]
It returns the complex root out of the three possible.
My question is the following:
1) How should I instruct Mathematica to take a certain root that I want
of say, (-1)^(1/3)?
2) I think there is a general possibility instruct Mathematica that all
calculations should be done on reals only. Is it right?
Thank you, Alexei
--
Alexei Boulbitch, Dr., habil.
Senior Scientist
IEE S.A.
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 Contern
Luxembourg
Phone: +352 2454 2566
Fax: +352 2454 3566
Website: www.iee.lu
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