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Re: Question on Solve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112740] Re: Question on Solve
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 29 Sep 2010 04:12:13 -0400 (EDT)

I don't think one can fault Solve since it is not supposed to return complete solutions anyway. I don't have anymore Mathematica 5.2 installed but Mathematica 7 returns a very complex output to your second example and produces a warning that it may not give solutions for all Solve variables. The output is too complicated for me to try to check if it is correct (in some sense anyway) or not.

However, here is something that does worry me a little more:

eqs == {(-3*Sqrt[3]*v1 + v2 + 17*v3 + (8*I)*Sqrt[3]*v3 -
       9*Sqrt[3]*v4 + 9*v5 - Sqrt[3]*v6 - 3*Sqrt[3]*v7 + 9*v8)/16 ====
    0, (5*v1 - Sqrt[3]*v2 + 3*Sqrt[3]*v3 + 11*v4 + (8*I)*Sqrt[3]*v4 -
       5*Sqrt[3]*v5 + 3*v6 - 3*v7 - Sqrt[3]*v8)/16 ====
    0, (-(Sqrt[3]*v1) + 3*v2 + 3*v3 + 5*Sqrt[3]*v4 +
       11*v5 + (8*I)*Sqrt[3]*v5 - 3*Sqrt[3]*v6 - Sqrt[3]*v7 - 5*v8)/
     16 ==== 0, (-9*v1 - 3*Sqrt[3]*v2 + Sqrt[3]*v3 + 9*v4 +
       9*Sqrt[3]*v5 + 17*v6 + (8*I)*Sqrt[3]*v6 - v7 - 3*Sqrt[3]*v8)/
     16 ==== 0, (9*v1 - Sqrt[3]*v2 + 3*Sqrt[3]*v3 - 9*v4 +
       3*Sqrt[3]*v5 - v6 + 17*v7 + (8*I)*Sqrt[3]*v7 - 9*Sqrt[3]*v8)/
     16 ==== 0, (-5*Sqrt[3]*v1 + 3*v2 + 3*v3 + Sqrt[3]*v4 - 5*v5 +
       Sqrt[3]*v6 + 3*Sqrt[3]*v7 + 11*v8 + (8*I)*Sqrt[3]*v8)/16 ====
    0, ((11 + (8*I)*Sqrt[3])*v1 - 3*Sqrt[3]*v2 + Sqrt[3]*v3 + 5*v4 +
       Sqrt[3]*v5 - 3*v6 + 3*v7 + 5*Sqrt[3]*v8)/16 ====
    0, (9*Sqrt[3]*v1 + (17 + (8*I)*Sqrt[3])*v2 + v3 + 3*Sqrt[3]*v4 +
       9*v5 + 3*Sqrt[3]*v6 + Sqrt[3]*v7 + 9*v8)/16 ==== 0};
vars == {v1, v2, v3, v4, v5, v6, v7, v8};



 Reduce[eqs, vars]

 v1 ==== 0 && v2 ==== 0 && v3 ==== 0 && v4 ==== 0 && v5 ==== 0 &&
   v6 ==== 0 && v7 ==== 0 && v8 ==== 0


However, let's just add the condition that not all vi's are 0 and we get:

Reduce[And @@ eqs &&  ! And @@ Thread[vars ==== 0], vars]

v2 ==== I*v1 && v3 ==== I*v1 && v4 ==== -v1 &&
   v5 ==== (-I)*v1 && v6 ==== v1 && v7 ==== -v1 &&
   v8 ==== (-I)*v1 && v1 !== 0


Note that your particular solution is a special case of this one. The gener=
al solution should be the disjunction of the trivial solution, that Reduce =
returned and this solutions. It seems to me that this is exactly what Reduc=
e ought to have returned. Of course Solve could not return this since it ca=
n't return inequalities.

Andrzej Kozlowski

On 28 Sep 2010, at 12:04, carlos at colorado.edu wrote:

> I have noticed some erratic behavior of Solve,
> illustrated in the following two examples.
> Suppose I have 8 homogenous linear equations
>
> eqs=={(9*v1-3*Sqrt[3]*v2+v3+v4-(8*I)*Sqrt[3]*v4-
> 9*Sqrt[3]*v5+9*v6-Sqrt[3]*v7-3*Sqrt[3]*v8)/16====0,
> (-(Sqrt[3]*v1)+5*v2-Sqrt[3]*v3+3*Sqrt[3]*v4-5*v5-
> (8*I)*Sqrt[3]*v5-5*Sqrt[3]*v6+3*v7-3*v8)/16====0,
> (-5*v1-Sqrt[3]*v2+3*v3+3*v4+5*Sqrt[3]*v5-5*v6-
> (8*I)*Sqrt[3]*v6-3*Sqrt[3]*v7-Sqrt[3]*v8)/16====0,
> (-3*Sqrt[3]*v1-9*v2-3*Sqrt[3]*v3+Sqrt[3]*v4+9*v5+
> 9*Sqrt[3]*v6+v7-(8*I)*Sqrt[3]*v7-v8)/16====0,
> (-9*Sqrt[3]*v1+9*v2-Sqrt[3]*v3+3*Sqrt[3]*v4-9*v5+
> 3*Sqrt[3]*v6-v7+v8-(8*I)*Sqrt[3]*v8)/16====0,
> ((-5-(8*I)*Sqrt[3])*v1-5*Sqrt[3]*v2+3*v3+3*v4+
> Sqrt[3]*v5-5*v6+Sqrt[3]*v7+3*Sqrt[3]*v8)/16====0,
> (5*Sqrt[3]*v1+(-5-(8*I)*Sqrt[3])*v2-3*Sqrt[3]*v3+
> Sqrt[3]*v4+5*v5+Sqrt[3]*v6-3*v7+3*v8)/16====0,
> (9*v1+9*Sqrt[3]*v2+v3-(8*I)*Sqrt[3]*v3+v4+
> 3*Sqrt[3]*v5+9*v6+3*Sqrt[3]*v7+Sqrt[3]*v8)/16====0};
>
> The variables are
> v=={v1,v2,v3,v4,v5,v6,v7,v8};
> Then
> sol==Solve[eqs,v]; Print[sol];
>
> gives the parametric solution
> {{v4->-v3,v5->v2,v6->(-2*I)*v2-v3,
> v7->-3*v2+(2*I)*v3,v8->-3*v2+(2*I)*v3,v1->(2*I)*v2+v3}};
> which is correct.
>
> Change the above equations to
>
> eqs=={(-3*Sqrt[3]*v1+v2+17*v3+(8*I)*Sqrt[3]*v3-
> 9*Sqrt[3]*v4+9*v5-Sqrt[3]*v6-3*Sqrt[3]*v7+9*v8)/16====0,
> (5*v1-Sqrt[3]*v2+3*Sqrt[3]*v3+11*v4+(8*I)*Sqrt[3]*v4-
> 5*Sqrt[3]*v5+3*v6-3*v7-Sqrt[3]*v8)/16====0,
> (-(Sqrt[3]*v1)+3*v2+3*v3+5*Sqrt[3]*v4+11*v5+
> (8*I)*Sqrt[3]*v5-3*Sqrt[3]*v6-Sqrt[3]*v7-5*v8)/16====0,
> (-9*v1-3*Sqrt[3]*v2+Sqrt[3]*v3+9*v4+9*Sqrt[3]*v5+
> 17*v6+(8*I)*Sqrt[3]*v6-v7-3*Sqrt[3]*v8)/16====0,
> (9*v1-Sqrt[3]*v2+3*Sqrt[3]*v3-9*v4+3*Sqrt[3]*v5-v6+
> 17*v7+(8*I)*Sqrt[3]*v7-9*Sqrt[3]*v8)/16====0,
> (-5*Sqrt[3]*v1+3*v2+3*v3+Sqrt[3]*v4-5*v5+Sqrt[3]*v6+
> 3*Sqrt[3]*v7+11*v8+(8*I)*Sqrt[3]*v8)/16====0,
> ((11+(8*I)*Sqrt[3])*v1-3*Sqrt[3]*v2+Sqrt[3]*v3+5*v4+
> Sqrt[3]*v5-3*v6+3*v7+5*Sqrt[3]*v8)/16====0,
> (9*Sqrt[3]*v1+(17+(8*I)*Sqrt[3])*v2+v3+3*Sqrt[3]*v4+
> 9*v5+3*Sqrt[3]*v6+Sqrt[3]*v7+9*v8)/16====0};
> v=={v1,v2,v3,v4,v5,v6,v7,v8};
>
> sol==Solve[eqs,v]; Print[sol];
>
> and I get only the trivial solution
> {{v1->0,v2->0,v3->0,v4->0,v5->0,v6->0,v7->0,v8->0}};
>
> But the system has an infinity of nontrivial solutions,
> for example (this one was obtained with another method)
>
> vsol=={v1->-I,v2->1,v3->1,v4->I,v5->-1,v6->-I,v7->I,v8->-1};
> Print[Simplify[eqs/.vsol]];
> gives {True,True,True,True,True,True,True,True};
>
> These 2 examples are extracted from several thousands
> similar ones.  Solve returns the trivial solution in
> about 1/3 of the instances.  Mathematica version used
> is 5.2 under Mac OS 10.5.9.
>
> Question: what do I have to do to get the correct
> parametric answers in all cases?
>


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