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MathGroup Archive 2004

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Re: Selecting cubic roots in functional form

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50290] Re: [mg50282] Selecting cubic roots in functional form
  • From: DrBob <drbob at bigfoot.com>
  • Date: Wed, 25 Aug 2004 03:35:57 -0400 (EDT)
  • References: <200408241022.GAA06697@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

This seems to work:

Clear[preSolve, qSolve]
preSolve[c_] = Solve[2000c*(5c + 7q) - 4(c + 8q)*(12q + 5c)^2 == 0, q];
qSolve[c_Real] := Max@Cases[Chop[q /. preSolve[c]], _Real]
qSolve[c_] := qSolve@N@c
Plot[qSolve@c, {c, 2, 4}]

Bobby

On Tue, 24 Aug 2004 06:22:27 -0400 (EDT), Carol Ting <suomesta at yahoo.com> wrote:

> Hello, list,
>
> I have an equation
>
> 2000c*(5c+7q)-4(c+8q)*(12q+5c)^2==0
>
> and I need to find q as a function of c, where both c and q are
> positive quantities.
>
> I use InequalitySolve and find that when c<3.11665 the 3rd root gives
> the positive function I need, and when c>3.11665 I should pick the
> first root instead.  So I need to pick different roots in different
> regions of c and combine them into a piecewise continuous function.
>
> Trouble is, I have a whole series of equations generated by another
> function and it is impossible for me to pick the roots one by one.  Is
> there a way to make Mathematica do this automatically?
>
> Thanks in advance!
>
> Carol
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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