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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
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Luxembourg

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