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>