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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
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