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Re: FindInstance what inspite ?


On 3 Dec 2007, at 19:39, Artur wrote:

> Who have idea what function uses inspite FindInstance in procedure?
> \!\(FindInstance[Chop[N[Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 + #1\^5
> &, 2] \
> + Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 + #1\^5 &, 3], 500]] == a +  
> b\
>                Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 + #1\^5 &,
>                     1] + c\ Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 +
> #1\^5 \
> &, 1]^2 + d\
>            Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 + #1\^5 &, 1]^3 + e\
>                    Root[\(-1\) - 2\ #1 - 2\ #1\^2 - #1\^3 + #1\^5 &,
>                   1]^4 && a != 0, {a, b, c, d, e}, Integers]\)
> And anser is empty set {}
> Good answer  is {a,b,c,d,e}={-2,-3,-2,-1,2}
> Who know how I can realize that procedure in Mathematica ?
>
> Best wishes
> Artur
>
>
>


First of all your answer is incorrect. You can check it with  
Mathematica:

I FullSimplify[
    Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 2] +
        Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 3] ==
      {-2, -3, -2, -1, 2} .
        Table[Root[#1^5 - #1^3 - 2*#1^2 - 2*#1 - 1 & ,
              1]^i, {i, 0, 4}]]
  False

The correct answer is {2, 2, 2, 1, -2} or, if you prefer  
{-2,-2,-2,-1,2}, whic differs from yours in the second place.

Neither FindInstance not Reduce or any other general algorithm based  
on polynomial algebra (or algebraic geometry) will work because they  
all work over the real or compelex number fields and you are looking  
for integer solutions. But in Mathematica 6 you can find this answer  
as follows:

ToNumberField[Root[-1 - 2*#1 - 2*#1^2 - #1^3 +
            #1^5 & , 2] +
      Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 3],
    Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 1]]

  AlgebraicNumber[Root[#1^5 - #1^3 - 2*#1^2 - 2*#1 -
1 & , 1], {2, 2, 2, 1, -2}]

You can check that is correct:

  FullSimplify[
    Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 2] +
        Root[-1 - 2*#1 - 2*#1^2 - #1^3 + #1^5 & , 3] ==
      {2, 2, 2, 1, -2} .
        Table[Root[#1^5 - #1^3 - 2*#1^2 - 2*#1 - 1 & ,
              1]^i, {i, 0, 4}]]
True

Best regards
Andrzej Kozlowski


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