Re: solving with inverse functions/inexact coefficients
- To: mathgroup at smc.vnet.net
- Subject: [mg67403] Re: [mg67397] solving with inverse functions/inexact coefficients
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Thu, 22 Jun 2006 06:21:18 -0400 (EDT)
- Reply-to: hanlonr at cox.net
- Sender: owner-wri-mathgroup at wolfram.com
eqns={F2 == Exp[a* NU]*F1,F1 == F2/2* Exp[a* NU], F2/2+NU == 15};
soln=Reduce[eqns,{F1,F2,NU},Reals]//Simplify
(F1 == 0 && F2 == 0 && NU == 15) || (F2 + 2*NU == 30 && (Sqrt[2]*(30*a - Log[2]))/a ==
2*F1 && ((Sqrt[2]*Sqrt[F1^2] == F2 && (30*a > Log[2] || a < 0)) ||
(F2 + Sqrt[2]*Sqrt[F1^2] == 0 && 0 < a && 30*a < Log[2])))
soln2=Solve[soln[[2]]/.a->76/100,{F1,F2,NU}][[1]]
{NU -> (25*Log[2])/38, F2 -> (5/19)*(114 - 5*Log[2]),
F1 -> (5/38)*(114*Sqrt[2] - 5*Sqrt[2]*Log[2])}
%//N
{NU -> 0.45601788194733245, F2 -> 29.087964236105336, F1 -> 20.56829676226186}
Bob Hanlon
---- Dave Spence <dspence at ics.mq.edu.au> wrote:
> I'm trying to solve a set of simultaneous equations describing a laser cavity. A much simplified set of equations in Real unknowns F1, F2 and NU looks like this:
>
> F2 == Exp[0.76 NU] F1, F1 == 0.5 F2 Exp[0.76 NU], 0.5 F2 + NU == 15
>
> They can be solved by hand very easily with two solutions, {F1=F2=0 && NU = 15} and {NU=0.46, F1=14.5, F2=29.1}
>
> Now Solve cannot solve these equations unless I change both instances of 0.76 to integers. Reduce seems to hang (or at least take too long to return and answer). Reduce can solve the problem quickly if I change both instances of 0.76 to 0.75! Reduce gives the message "Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result"
>
> I guess this is just a difficulty with the way mathematica attempts to solve problem. There are references in the manuals to difficulties solving multiple simultaneous equations that all invoke inverse functions. Are there any options I can set to enable mathematica to quickly solve these equations?
>
> I note that while these equations can be manipulated by hand into a form that can be solved by mathematica, by full set of equations are far more complex and can't easily be edited by hand.
>