Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

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

Search the Archive

equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31993] equations
  • From: "Fred Simons" <f.h.simons at tue.nl>
  • Date: Sun, 16 Dec 2001 03:44:28 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Andrzej and Daniel,

I think I formulated my comment a little bit too short; the postings of
Daniel and mine are not mutually contradictory, as Andrzej wrote.  As Daniel
did, I drew the conclusion that the set of equations must have four
solutions. I my posting I referred to another problem. Knowing the general
values of the unknowns does not imply that we know the general solution. Let
me explain that more clearly.

We do have (symbolic) expressions for the four possible values of each of
the four unknowns R1, R2, R3 and R4:

R1 may have one of the values v11, v12, v13, v14,
R2 may have one of the values v21, v22, v23, v24,
R3 may have one of the values v31, v32, v33, v34,
R4 may have one of the values v41, v42, v43, v44.

Nevertheless, we have not yet found any solution of the system of equations.
Such a solution much have the form

{R1->value1, R2->value2, R3->value3, R4->value4},

where we expect that value(i) will be one of the four values for R(i).

The testing I did was the following. For a given set of parameters we can
solve the set of equations and therefore we can combine the symbolic values
in a unique way such that say

{R1 -> v11, R2 -> v24, R3 -> v31, R4 -> v44}
{R1 -> v12, R2 -> v22, R3 -> v32,  R4-> v42}
{R1 -> v13, R2 -> v21, R3 -> v34, R4 -> v43}
{R1 -> v14, R2 -> v23, R3 -> v33, R4 -> v41}

is the solution of the set of equations for that choice of the parameters.
However, this solution is definitely NOT the general solution (i.e. for all
choices of the parameters). It is easy to find values of the parameters for
which substituting in the equations results in one or more outcomes False.
So the general solution must be more complicated (a combination of roots of
a polynomial of degree 16, in the worst case?).  Maybe this explains why I
did not manage to solve the set of equations using Solve within 14 hours,
while finding the values v11, ..., v44 is pretty fast.

By the way, playing with this example I also found that

Solve[ Eliminate[eqs, {R1, R2, R3}], R4] is done within 3 seconds, while

Solve[eqs, R4, {R1, R2, R3}]

takes about 2 minutes. I thought these two commands were equivalent.


Fred Simons
Eindhoven University of Technology








  • Prev by Date: Re: equations
  • Next by Date: Re: Solve[] for equations?
  • Previous by thread: Re: equations
  • Next by thread: Re: equations