Re: Selecting Real Roots, Again
- To: mathgroup at smc.vnet.net
- Subject: [mg67131] Re: Selecting Real Roots, Again
- From: dh <dh at metrohm.ch>
- Date: Sat, 10 Jun 2006 04:53:25 -0400 (EDT)
- References: <e6b5q5$d3c$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
formulate your requirements and let Mathematica do the work.E.g.
(excluding the trivial solution {0,0}):
eq = {.9 x^10 + .1 x^2 == y && 0 < x <= 1 && 0 < y <= 1};
Then use Reduce or, if you only need a single solution, FindInstance. E.g.
Reduce[eq,{x,y}]
FindInstance[eq,{x,y}]
Daniel
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?
>