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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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