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