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Re: [Axiom-developer] The Gosper algorithm for sum of hypergeometric fun


From: Francois Maltey
Subject: Re: [Axiom-developer] The Gosper algorithm for sum of hypergeometric functions.
Date: 27 Nov 2006 19:24:18 +0100
User-agent: Gnus/5.09 (Gnus v5.9.0) Emacs/21.4

Thanks a lot Martin and Waldek,

In fact I'm looking for testing with axiom the sum that my students do.
I hope that axiom computes right (as the two M....) the most possible series.
By example students search the series bellow. 

Are they all hypergeometrics ? 
this means that u(n+1) / u(n) is an alg├ębraic fraction
as (n-...)(n-...) - - - (n-...)/(n-...) - - - (n-...)

I believe yes but the cos/cosh/sin/sinh. 
In this case the standard way with complex, combine and expand is right.
The very last one can be find by expand sum ... + sum ...
The only real problem is about the malicious sum ((2^((-1)^n), n)
for n even and n odd.

Am I right ?

Theses exercices have a closed form so the Gosper algorithm gives the 
solution as a hypergeometric serie.

Am I right (again) ?

How can I use the GosperPackage : 
I don't know how to use the signature for the command 

    GosperMethod (Q, V, ()-> V) -> Union (Q, "failed")
            ++ GospersMethod(b, n, new) returns a rational function
            ++ \spad{rf(n)} such that \spad{a(n) * rf(n)} is the indefinite
            ++ sum of \spad{a(n)}
            ++ with respect to upward difference on \spad{n}, i.e.
            ++ \spad{a(n+1) * rf(n+1) - a(n) * rf(n) = a(n)},

The package is :

GosperSummationMethod(E, V, R, P, Q): Exports == Impl where
    E: OrderedAbelianMonoidSup
    V: OrderedSet
    R: IntegralDomain
    P: PolynomialCategory(R, E, V)
    Q: Join(RetractableTo Fraction Integer, Field with
              (coerce: P -> %; numer : % -> P; denom : % -> P))

Then can I recognize << with my eyes >> the hypergeometric serie ?

The easiest hypergeometric serie are 0F0 (x)=exp x and 1F0 (a, x)=(1-x)^(-a).

For others functions must I cut the hypergeometric serie in order to recognize
atan, exp and so. Is it possible ? 

The main problem is perhaps to simplify (n+1)!/n!, 
but an expand/rewrite function can do it, no ? 

Martin, I'm sorry but I don't understand what your package do,
And I don't find your package on my silver axiom.

My aim is to force axiom to compute theses exercices about series.

Thanks a lot if you accept to drive me inside theses algorithms.

sum (n^2 * x^n,                         n=0..%plusInfinity) -- YES
sum (n^3 * x^n,                         n=0..%plusInfinity) -- YES

sum (x^n / (2*n-1),                     n=1..%plusInfinity) -- no

sum (1 / (n+1) / (n+3),                 n=0..%plusInfinity) -- YES
sum (x^n / (n+1) / (n+3),               n=0..%plusInfinity) -- no

sum (1 / (4*n^2-1),                     n=0..%plusInfinity) -- YES
sum ((-1)^n * x^(2*n+1) / (4*n^2-1),    n=0..%plusInfinity) -- no
sum ((-1)^n * x^(2*n+1) / (4*n^2-1),    n=0..%plusInfinity) -- no 

sum (x^n / (4*n-1),                     n=1..%plusInfinity) -- no
sum ((n+3) * x^n / (2*n+1),             n=0..%plusInfinity) -- no

sum (x^n * exp (n*a),                   n=0..%plusInfinity) -- YES
sum (x^n * cosh (n*a),                  n=0..%plusInfinity) -- no
sum (x^n * exp (n*a) / n,               n=0..%plusInfinity) -- no
sum (x^n * cosh (n*a) / n,              n=0..%plusInfinity) -- no

sum (x^n * sin (n*a)^2 / 2^n,           n=0..%plusInfinity) -- YES with exp %i
sum (x^n * (n^2+1) / (n+1),             n=0..%plusInfinity) -- no

sum (x^n / factorial n,                 n=0..%plusInfinity) -- no
sum (x^n / factorial (2*n),             n=0..%plusInfinity) -- no
sum (x^n * sin (n*a)^2 / factorial (n), n=0..%plusInfinity) -- no 
sum (x^n * n^5 / factorial n,           n=0..%plusInfinity) -- no
sum (x^(3*n) / factorial (3*n),         n=0..%plusInfinity) -- no
sum (x^n * binomial (n+1, 2*n),         n=0..%plusInfinity) -- no
sum ((-1)^n * binomial (n, 2*n) / 4^n,  n=0..%plusInfinity) -- no

sum (1 / (n^2-1),                       n=0..%plusInfinity) -- YES
sum (x^n / (n^2-1),                     n=0..%plusInfinity) -- no
sum (1 / (3*n+1) / (6*n+5),             n=0..%plusInfinity) -- no

sum (1 / n / (n-1),                     n=0..%plusInfinity) -- YES
sum (x^n / n / (n-1),                   n=0..%plusInfinity) -- no

sum ((4*n+1) / (2*n^2+n-1),             n=0..%plusInfinity) -- no
sum ((4*n+1) * x^n / (2*n^2+n-1),       n=0..%plusInfinity) -- no

sum ((2^((-1)^n),                       n=0..%plusInfinity) -- no
sum (a^n * (1+b^n) / n,                 n=0..%plusInfinity) -- no




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