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MathGroup Archive 2001

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Re: Re: Solve[] for equations?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31960] Re: [mg31946] Re: [mg31928] Solve[] for equations?
  • From: "Fred Simons" <f.h.simons at tue.nl>
  • Date: Fri, 14 Dec 2001 04:21:24 -0500 (EST)
  • References: <200112130608.BAA01031@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Only a few remarks:


> The concept may be simple but the practice is not quite so. You can see
> that as follows.
> Here are your equations.
>
> eq1 = Rac == R1(R2 + R3 + R4)/(R1 + R2 + R3 + R4)
> eq2 = Rad == R2(R1 + R3 + R4)/(R1 + R2 + R3 + R4)
> eq3 = Rbc == R3(R1 + R2 + R4)/(R1 + R2 + R3 + R4)
> eq4 = Rbd == R4(R1 + R2 + R3)/(R1 + R2 + R3 + R4)
>
> we ask Mathematica to eliminate all the variables except one (say R4).
>
> eq5 = Eliminate[{eq1, eq2, eq3, eq4}, {R1, R2, R3}];
>
> You have to wait a bit for this to work (Mathematica 4.1).
>
> If you want to see the 4th degree equation in R4 that you get you can
> evaluate:
>
> eq5 /. Equal[x_, y_] :> Collect[x - y, R4] == 0
>
> What you see is a fourth degree equation with symbolic coefficients
> which is far from simple. Mathematica can actually "solve" it with:
>
> Solve[eq5, R4]
>
> You have to wait quite a while and then you will see something
> phenomenally complicated and in my opinion essentially useless (and in
> addition there is basically no way of checking its correctness).
>

The result is simpler when we force Solve not to solve cubics and quartics:

SetOptions[Roots, Cubics -> False, Quartics -> False];
Solve[eq5, R4]

But still the result is rather useless.

> If you like that you can use the same approach to get R1,R2 and R3 or
> much  better, you can just use the symmetry of your equations to find
> out the other answers (to get R3 just replace Rbd by Rbc and vice versa).
>

It is not so simple. Indeed, in the above way we can find the four values
for R4. Similarly, or by cyclic permutation, we can find the four values for
R1, for R2 and for R3 in symbolic form. But it is not clear how this four
times four values for each of the unknowns have to be combined for finding a
solution for the set of equations. Some testing with numerical values shows
that it is unlikely that a solution of the set of equations indeed can be
expressed as a combination of the results found in this way. Maybe that
explains that the Solve command in Mathematica is unable to solve the
equations within 14 hours, while in the above way it finds formula for each
of the unknowns within 2 minutes.

Fred Simons
Eindhoven University of Technology




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