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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