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Re: FindInstance what inspite ?

• To: mathgroup at smc.vnet.net
• Subject: [mg83924] Re: [mg83861] FindInstance what inspite ?
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 4 Dec 2007 04:32:18 -0500 (EST)
• References: <fire9m$roc$1@smc.vnet.net> <200712020914.EAA07009@smc.vnet.net> <200712031039.FAA18775@smc.vnet.net> <00E6DB06-7777-4F71-A011-BA1BFE1DE3FB@mimuw.edu.pl>

One correction. I wrote below:

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

Of course, while {2, 2, 2, 1, -2}  works  its negative -{2, 2, 2, 1,  
-2} does not work. Clearly, vectors v and - v would both be answers  
only if the algebraic number

  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]]
Root[#1^5 - 3*#1^3 + #1^2 + 7*#1 + 5 & , 1]


were 0, which it is not. I got confused by the fact that -{2, 2, 2, 1,  
-2} is almost the same as your proposed answer, which is also {-2, -3,  
-2, -1, 2} clearly wrong.  I am not sure if {2, 2, 2, 1, -2} is the  
answer you wanted as I have no idea how you got your wrong one.

Andrzej Kozlowski



On 3 Dec 2007, at 21:25, Andrzej Kozlowski wrote:

> *This message was transferred with a trial version of  
> CommuniGate(tm) Pro*
>
> 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

• References:
  • Re: Interpolation in 2 D, bug?
    • From: Hugh <h.g.d.goyder@cranfield.ac.uk>
  • FindInstance what inspite ?
    • From: Artur <grafix@csl.pl>