[Date Index]
[Thread Index]
[Author Index]
Re: Re: Solve[] for equations?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg31961] Re: [mg31946] Re: [mg31928] Solve[] for equations?
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Fri, 14 Dec 2001 04:21:26 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
You are of course completely right. To use this approach one would need
to be able to prove that the system does have solutions. If it does than
they will be found in this way. But indeed they may very well not exist
at all.
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
On Friday, December 14, 2001, at 12:14 AM, Fred Simons wrote:
> Only a few remarks:
>
> From Andrzej Kozlowsky's message:
>
>> 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
>
>
>
>
Prev by Date:
**Re: numerical approximation to the diffusion equation in Mathematica?**
Next by Date:
**Re: symbolic calculations with complex numbers**
Previous by thread:
**Re: Re: Solve[] for equations?**
Next by thread:
**Re: Re: Re: Solve[] for equations?**
| |