       • To: mathgroup at smc.vnet.net
• Subject: [mg72123] Re: radical question again
• From: dh <dh at metrohm.ch>
• Date: Wed, 13 Dec 2006 06:39:04 -0500 (EST)
• Organization: hispeed.ch
• References: <elgm4j\$ha9\$1@smc.vnet.net>

```Hi Dimitris,
I can not fully answer your question, but here is a hint:
In Root notation we can indicate the number of the root. The first root
must, by definition, be real for a 3.order poly. In Radical notation
we have no such notation.  As long as we have a symbolic parameter,
radical notation may not be able to indicate the real root. This may
explain why Cubics->True does not work as soon as you restrict the
domain to real.
It may also prevent radical notation if you want to pick out the real
root, having a symbolic parameter.
Daniel

dimitris wrote:
> Before proceeding to the post let me thank Andrzej and Daniel
> for their responses to my previous question appeared here
>
>
> So consider again the following equation
>
> req = (2 - x)^2 - 4*Sqrt[1 - x]*Sqrt[1 - c*x] == 0;
>
> with 0<x<1 and 0<c<1.
>
> A quick solution is provided by
>
> Reduce[{req, 0 < x < 1, 0 < c < 1}, {x}, Reals]
> 0 < c < 1 && x == Root[-16 + 16*c + (24 - 16*c)*#1 - 8*#1^2 + #1^3 & ,
> 1]
>
> Here comes my first question:
>
> Why the following setting returns the same output?
>
> Reduce[{req,0<x<1,0<c<1},{x},Reals,Cubics->True(*the default is
> False*)]
> 0 < c < 1 && x == Root[-16 + 16*c + (24 - 16*c)*#1 - 8*#1^2 + #1^3 & ,
> 1]
>
> The following plots verify the obtained solution by Reduce
>
> f[c_] = x /. ToRules[Last[%]];
> Plot[f[c], {c, 0, 1}]
> Plot[Chop[First[req] /. x -> f[c]], {c, 0, 1}]
>
> Let
>
> 8/3 - (8 - 48*c)/(6*(-17 + 45*c + 3*Sqrt*Sqrt[11 - 62*c + 107*c^2 -
> 64*c^3])^(1/3)) +
>   (2/3)*(-17 + 45*c + 3*Sqrt*Sqrt[11 - 62*c + 107*c^2 -
> 64*c^3])^(1/3)
>
> with
>
> g[c] == x /. DeleteCases[Solve[req, x], {x -> 0}][]
> True
>
> Now
>
> Block[{Message}, Plot[g[c], {c, 0, 1}, PlotPoints -> 100]]
> Show[GraphicsArray[Block[{\$DisplayFunction = Identity},
>     (Plot[Chop[#1[g[c]]], {c, 0, 1}, Frame -> True, Axes -> False,
> PlotPoints -> 100] & ) /@ {Re, Im}]], ImageSize -> 500]
>
> The graphs can be justified due the choise of principal cube root by
> Mathematica.
> In the case of negative numbers, it turns out that it is not a real
> number.
> (see relevant descussion here:
> )
>
> And here comes my second question:
>
> With the Root object we obtain a closed form root which is valid for
> 0<x,c<1.
> Converting it to radical we lost that and we must consider in more
> detail the range
> of c.
>
> Is it possible to get a radical expression which is equivalent to the
> Root object?
> If yes how?
>
> Thanks
> Dimitris
>

```

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