       Re: How to instruct Math to take a certain (e.g. real)

• To: mathgroup at smc.vnet.net
• Subject: [mg105051] Re: [mg105018] How to instruct Math to take a certain (e.g. real)
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Thu, 19 Nov 2009 05:23:19 -0500 (EST)
• References: <200911181157.GAA04186@smc.vnet.net>

```Each cube root on the left has 3 possible values (except 0^(1/3), so the
sum has nine possible values. Only three are zero.

We can replace the terms with primary roots, after substituting y->0:

Here's the method applied to each term separately:

eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3);
Cases[eq,
Power[x_, Rational[1, n_Integer]] :>
Root[#^n - x /. y -> 0 &, 1], Infinity]

{-2, 2, 0}

And here it is, applied to the equation:

eq /. Power[x_, Rational[1, n_Integer]] :> Root[#^n - x /. y -> 0 &, 1]

True

Bobby

On Wed, 18 Nov 2009 05:57:41 -0600, 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:= eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3);
> eq /. y -> 0
>
> Out= False
>
> just because the expression
>
> In:= eq[] /. y -> 0
>
> Out= 2 + 2 (-1)^(1/3)
>
> Mathematica does not interpret as zero. And if I ask it to give a
> In:= 2 + 2 (-1.)^(1/3)
>
> Out= 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
>

--
DrMajorBob at yahoo.com

```

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