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Re: What inspite FindInstance ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107997] Re: What inspite FindInstance ?
  • From: dh <dh at metrohm.com>
  • Date: Fri, 5 Mar 2010 04:30:06 -0500 (EST)
  • References: <hmo1q7$pr6$1@smc.vnet.net>

Hi Artur,
you can get all possibilities using Reduce.E.g.
Reduce[Zeta[2, 5/4] - a1/a2 Pi^2 - b1/b2 - c1/c2  Catalan == 0, {a1,
   a2, b1, b2, c1, c2}, Integers, Backsubstitution -> True]
This list different cases. You get a shorter but nested output of you 
eliminate: Backsubstitution.
Daniel


On 04.03.2010 11:25, Artur wrote:
> Dear Mathematica Gurus,
>
> Mathematical problem is following:
> Find rational numbers a,b,c such that
> (Pi^2)*a+b+c*Catalan==Zeta[2,5/k] for some k
> e.g.
> FindInstance[
>   Zeta[2, 5/4] -a Pi^2 - b - c Catalan == 0, {a, b, c}, Rationals]
> give answer
> FindInstance::nsmet: The methods available to FindInstance are
> insufficient to find the requested instances or prove they do not exist.>>
>
>
> What inspite FindInstance? (I know that we can do 6 loops (3
> Denominators and 3 Numerators) but we have to have luck to give good
> range of loops..
>
> Good answer for my example is {a,b,c}={1,-16,8}but in general case these
> a,b,c will be rationals (not integers)
> e.g. (Pi^2)*a+b+c*Catalan==Zeta[2,5/2]  we have {a,b,c}={1/2,-40/9,0}
> but this last case Mathematica deduced autmathically if we execute :
> Zeta[2,5/2]
> first one none.
>
> Best wishes
> Artur
>


-- 

Daniel Huber
Metrohm Ltd.
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CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh at metrohm.com>
Internet:<http://www.metrohm.com>



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