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Re: Selecting Real Roots, Again


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?

Mathematica can't determine whether the Root objects returned by Solve 
are real without knowing something about y, so your Select condition 
will not work. A better approach would be to use Reduce, where you can 
specify the conditions you are interested in:

In[21]:= Reduce[
  d x^10 + (1 - d) x^2 == y && 0 <= y <= 1 && 0 <= x <= 1 && 0 <= d <= 1, x]

Out[21]= (y == 0 && 0 <= d <= 1 &&
    x == Root[d #1^10 + (1 - d) #1^2 &, 1]) || (0 < y <= 1 && 0 <= d <= 
1 &&
    x == Root[d #1^10 + (1 - d) #1^2 - y &, 2])

So, the root you are looking for is the second one returned above (the 
first is only defined for y==0). For example,

Plot[ Root[ .9 #^10 + .1 #^2 - y&, 2], {y,0,1}]

will plot the dependence of x on y for d==.9.

Carl Woll
Wolfram Research


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