Re: Selecting Real Roots, Again
- To: mathgroup at smc.vnet.net
- Subject: [mg67132] Re: [mg67104] Selecting Real Roots, Again
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 10 Jun 2006 04:53:26 -0400 (EDT)
- Reply-to: hanlonr at cox.net
- Sender: owner-wri-mathgroup at wolfram.com
allSoln=x/.Solve[{.9*x^10 + .1*x^2 ==y},x]; In this case, you can just use a "typical" value of y (say .5) to select soln=Select[allSoln, (FreeQ[(#/.y->.5),Complex]&& 0<=(#/.y->.5)<=1)&] {Root[0.9*#1^10 + 0.1*#1^2 - 1.*y & , 2]} Plot[soln,{y,0,1}]; Bob Hanlon ---- DOD <dcodea at gmail.com> 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? > -- Bob Hanlon hanlonr at cox.net