Re: Solvedoesn't

*To*: mathgroup at smc.vnet.net*Subject*: [mg90642] Re: Solvedoesn't*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Thu, 17 Jul 2008 05:35:35 -0400 (EDT)

On 7/16/08 at 6:29 AM, fc3a501 at uni-hamburg.de (Hauke Reddmann) wrote: >This is extremely annoying. I have a very large (this seems to be >critical) equation system. It's partly linear, partly horror, but >already after I solve the linear part, a certain solution (which is >obviously one since substituting it in makes all equations identical >zero) can't be found anymore. The solution has some surds (this also >seems critical, the ones without are found) but the equations are >strictly polynomic. >Faked Example (the real one only PM-ed, since it's a bit oversized): >x-y=1 >x^2+y^2=3 >x^3-y^3=4 >and after y was eliminated, it doesn't >find x. (=golden mean here) After fixing things to reflect correct syntax, Solve seems to work just fine for the example you posted. That is, In[5]:= eqns = {x - y == 1, x^2 + y^2 == 3, x^3 - y^3 == 4}= ; In[6]:= Solve[eqns, {x, y}] Out[6]= {{x -> (1/2)*(1 - Sqrt[5]), y -> (1/2)*(-1 - Sqrt[5])}, {x -> (1/2)*(1 + Sqrt[5]), y -> (1/2)*(-1 + Sqrt[5])}} Also, the first two equations are sufficient. That is, In[7]:= Solve[Most@eqns, {x, y}] Out[7]= {{x -> (1/2)*(1 - Sqrt[5]), y -> (1/2)*(-1 - Sqrt[5])}, {x -> (1/2)*(1 + Sqrt[5]), y -> (1/2)*(-1 + Sqrt[5])}} =46inally, I think Reduce is preferred when the number of equations differs from the number of variables. And I like the output I see better, i.e., In[8]:= Reduce[And @@ eqns, {x, y}] Out[8]= (x == (1/2)*(1 - Sqrt[5]) || x == (1/2)*(1 + Sqrt[5])) && y == x - 1