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MathGroup Archive 2006

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Re: Selecting Real Roots, Again

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67183] Re: Selecting Real Roots, Again
  • From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 11 Jun 2006 02:18:59 -0400 (EDT)
  • References: <e6b5q5$d3c$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

DOD wrote:
> I've read the many posts already here about how to get mathematica to
> select real roots for you, but I have a slightly(very slighty, I
> thought) different problem ,and I don't know how to get mathematica to
> do what I want.
>
> I want to get the solution for a polynomial of the following form:
> d x^n + (1-d) x^2 =y
>
> so for example, I do
> Solve[.9 x^10 + .1 x^2 ==y,x]
> and I get a whole bunch of solution,  very good.  For my purposes, y
> lives in the [0,1], as must the solution.  So I can see, by hand, which
> root I want; exactly one root is both real, and has solutions in my
> inverval.  So I want to tell mathematica to:
>
> A: look at only solutions x* that are real over y in [0,1]
>
> and
>
> B: of those solutions, give the one x* that itself lies is [0,1].
>
> So, when I try to do something from reading previous posts, I cannot
> get it to work:
> In[24]:=
> Select[Solve[.9 x^10 + .1 x^2 ==y,x],Element[x/.#,Reals]&]
>
> Out[24]=
> {}
> or perhaps
> In[41]:=
> Select[Solve[.9 x^10 + .1 x^2
> ==y,x],Assuming[Element[y,Reals],Eement[x/.#,Reals]]&]
> Out[41]=
> {}
>
>
>  So How to I tell mathematica to do this?

What about using Reduce [1]? For instance,

Reduce[(9/10)*x^(10) + (1/10)*x^(2) == y && 0 <= y <= 1,
  x, Reals]

Out[3]=
0 <= y <= 1 &&
  (x == -Sqrt[Root[-10*y + #1 + 9*#1^5 & , 1]] ||
   x == Sqrt[Root[-10*y + #1 + 9*#1^5 & , 1]])

Regards,
/Jean-Marc

[1]: http://documents.wolfram.com/mathematica/functions/Reduce


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