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MathGroup Archive 2001

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Re Solve for Equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32104] Re Solve for Equations
  • From: "Alan Mason" <swt at austin.rr.com>
  • Date: Mon, 24 Dec 2001 23:44:35 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 12/12/01 5:51:22 AM, djvango at sandia.gov writes:

>I have a system of 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)
>
>I'd like to get R1, R2, R3, and R4 in terms of Rac, Rad, Rbc, and Rbd.
> Four
>equations, four unknowns -- it seems simple in concept so I figured
Mathematica
>could do it easily.  I thought that something along the lines of:
>
>Solve[{eqs},{R1, R2, R3, R4}] or
>Solve[{eqs}, R1, {R2, R3, R4}]
>
>would have worked, but these give {}.  They only permutation of commands
>that seemed to do anything was something like
>
>Solve[eq1, R1]
>
>but one can do that easily by hand so what's the point.
>
>Is there some other command I should be using besides Solve?   Is there
>some
>mathematical reason that I don't recognize as to why this won't work?
> I
>would appreciate any insights.
>
 Hello,
This problem is nicely susceptible to mathematical analysis with bare hands.
I replace your Rac, etc., with t1, t2, t3, t4.
It is much simpler to work with "reduced" variables rho_i and tau_i defined
by
rho_i = Ri/s, tau_i = ti/s where s = R1 + ... R4. The motivation for this is
the observation that the equations are homogeneous of degree one.  Thus, to
find the structure of the solution set we need only make a change of scale
and suppose that s = 1 (more precisely, we have sum of rho_i's = 1).  Then
everything is quite simple and the equations decouple.

In detail, tau_i = rho_i (1 - rho_i), i = 1, ..., 4.
Solve quadratic eqn to get

rho_i = (1 +- Sqrt[1-4 tau_i])/2, where +- is "plus or minus", call it
sgn_i; there are 16 choices of these signs.

To find s more explicitly, sum over i, using sum rho_i = 1 to get
-2 = Sum[sgn_i Sqrt[1 - 4 tau_i], {i, 1, 4}]
This is equivalent to the hard part that Solve was working on.  Now solve
this for the tau_i and get s from s = ti/tau_i.  There will be 16
(complex-valued) branches from the choices for the sgn_i, and three
independent tau's (in suitable coordinate charts -- the equations define a
projective variety).

This treatment extends to the analogous equations with N R's and N t's;
basically, they all decouple into independent quadratic equations
parameterized by s, which can be found as above.

Alan






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