Abhinav Baid wrote:
>
> On 8/6/15, Waldek Hebisch <hebi...@math.uni.wroc.pl> wrote:
> > Abhinav Baid wrote:
> >>
> > Additional remarks:
> > - If there is singularity with exponential
> > part of multiplicity 1, then failure of van Hoej algorithm
> > for this exponential part applied to operator and its
> > adjount means that operator is irreducible. In such
> > case there is no need to look at other singularities
> > or even other exponnetial parts. And if our implementation
> > fails to find factor for reducible operator this is bug
> > which should be fixed.
> Yes, but how do I check this and appropriately call factor_minmult1?
> (which is removed for now). The earlier check was using
> skipped_factors but it was wrong. I think there should be a condition
> in the loop after bounds are computed, but am not sure.
1) If you get no solution from Hermite-Pade solver, or if there
is single solution, but has no common factor with f, then
van Hoej method failed for given expomential part. In fact,
if you seek for factors increasing order by one, then
once you get a single reult from Hermite-Pade solver it
will be a factor of f or relatively prime to f. If there
are multiple solution then increasing number of terms in
expansion will eventually lead to single or no solution.
2) Checking for exponential part of multiplicity 1 is easy:
you need to check that there is generalized exponents
which is not equivalent to another one.
--
Waldek Hebisch
hebi...@math.uni.wroc.pl
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