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MathGroup Archive 2006

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Re: radical equation once more!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72058] Re: radical equation once more!
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 10 Dec 2006 04:49:07 -0500 (EST)
  • References: <421128.34967.qm@web30214.mail.mud.yahoo.com>

On 10 Dec 2006, at 02:54, dimitris anagnostou wrote:


> ---------------------------------------------------------------------- 
> ----------------------
>
> Before proceeding to the post let me thank Andrzej and Daniel
> for their responses to my previous question appeared here
>
> http://groups.google.com/group/comp.soft-sys.math.mathematica/ 
> browse_thread/thread/78d57749388d7384/e072ef4f6aa4f9c0? 
> hl=en#e072ef4f6aa4f9c0
>
> 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]


Because in the presence of inequalities Reduce will not use the  
Cardano and Tartaglia formulas no matter what setting fro the options  
cubic and Quartics you use. The main tool used in such cases is  
CylindricalDecomposition and that defines having a function that will  
be real for all values of the parameter c. Only using Root[....,1]  
will give this kind of function. In fact this is how (probably)  
mathematica gets this answer. Let

In[1]:=
req1 = (2 - x)^4 - 16*(1 - x)*(1 - c*x) == 0;

In[2]:=
CylindricalDecomposition[req1 && 0 < c < 1 && 0 < x < 1,
   {c, x}]

Out[2]=
0 < c < 1 && x == Root[#1^3 - 8*#1^2 - 16*c*#1 + 24*#1 +
       16*c - 16 & , 1]


>
> And here comes my second question:
>
> With the Root object, somehow, 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?
>

In general the answer is: no.

Andrzej Kozlowski


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