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From: Artur on 4 Mar 2010 05:25
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

From: dh on 5 Mar 2010 04:30
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.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh(a)metrohm.com>
Internet:<http://www.metrohm.com>


From: Daniel Lichtblau on 5 Mar 2010 04:30
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

Can be done using numerical approximation. One might then need to
provide proof that the given rationals are in fact correct (or not; I
make no attempt to check results).

There might be something already built in to Mathematica for this, but I
did not find anything in a brief search. So here is some code. It is
based on lattice reduction. One first finds an integer relation between
target and a given set of values. One then simply divides through by the
multiplier used for the target, in order to get rational multipliers for
the values.

findRelation[target_, vals_, prec_] := Module[
{nvals, lat, redlat, relation},
nvals = Round[10^prec*N[Append[vals, -target], prec]];
lat = Transpose[Join[IdentityMatrix[Length[nvals]], {nvals}]];
redlat = LatticeReduce[lat];
relation = Most[First[redlat]];
Most[relation]/Last[relation]
]

Here are your examples:

In[93]:= vals = {Pi^2, 1, Catalan};
findRelation[Zeta[2, 5/4], vals, 50]
findRelation[Zeta[2, 5/2], vals, 50]

Out[94]= {1, -16, 8}

Out[95]= {1/2, -(40/9), 0}


Daniel Lichtblau
Wolfram Research

From: Daniel Lichtblau on 5 Mar 2010 04:31
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

Here is an approach that involves much less code than what I last sent
(in the tehcnical sense that "none" is much less than "some").

(1) Go to
http://www.wolframalpha.com

(2) Enter
zeta(2,5/2)
or
zeta(2,5/4)

Results for teh first include a pane
Exact result:
pi^2/2-40/9

Results for the second have a pane
Alternate form:
8 C-16+pi^2

That Wolfram|Alpha is one clever gal.

Daniel Lichtblau
Wolfram Research

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