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
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