MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Solve given that a variable is between 0 and 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52700] Re: Solve given that a variable is between 0 and 1
  • From: "Jason Roth" <jsr123 at gmail.com>
  • Date: Thu, 9 Dec 2004 20:23:18 -0500 (EST)
  • References: <cos13d$dl2$1@smc.vnet.net><coudkj$qdt$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

OK, so you need to substitute numerical values...I see...thank you
Steve!  :-)


Steve Luttrell wrote:
> To solve algebraically you need to give a numerical value to b, and
then
> only certain values lead to simple closed-form solution. Otherwise,
you have
> to solve numerically.
>
> Algebraic example:
>
> b = 1/2;
> Solve[w - k^(1 - b)*r*((-((a*k^b - OverBar[y])/a))^
>        (1/b))^(-1 + b) == 0, k]
>
> gives
>
> {{k -> (w^2*OverBar[y]^2)/(a^2*(-r + w)^2)},
>   {k -> (w^2*OverBar[y]^2)/(a^2*(r + w)^2)}}
>
> Numerical example:
>
> b = 0.4;
> w = 1;
> a = 1;
> r = 1;
> OverBar[y] = 1;
> NSolve[w - k^(1 - b)*r*((-((a*k^b - OverBar[y])/a))^
>        (1/b))^(-1 + b) == 0, k]
>
> gives
>
> {{k -> 0.17677669529663688110021109052621}}
>
> Steve Luttrell
>
> "Jason" <jsr123 at gmail.com> wrote in message
> news:cos13d$dl2$1 at smc.vnet.net...
> > Hi all,
> >
> > I need some help with solving the following:
> >
> > \!\(Solve[
> >    w - k\^\(1 - b\)\ r\ \((\((\(-\(\(a\ k\^b -
> > y\&_\)\/a\)\))\)\^\(1\/b\))\)\
> > \^\(\(-1\) + b\) \[Equal] 0, k]\)
> >
> > given that a>0 and 0<b<1.
> >
> > How do I do this?
> >
> > Thanks MUCH!!---jason
> >


  • Prev by Date: Problem with linken general differential equations
  • Next by Date: projection of {-1,1}*{-1,1} to a ruled tetrahedral surface in rectangles
  • Previous by thread: Re: Solve given that a variable is between 0 and 1
  • Next by thread: Re: Re: Solve given that a variable is between 0 and 1