clicliclic@freenet.de wrote:
>
> "Nasser M. Abbasi" schrieb:
> >
> > On 4/1/2018 2:37 PM, clicliclic@freenet.de wrote:
> > >
> > > Here is another "moderately-sized square-root pseudo-elliptic integral
> > > on which FriCAS is still known to fail".
> > >
> > > Being confronted with:
> > >
> > > setSimplifyDenomsFlag(true)
> > >
> > > integrate((x^2 + 2*x - 3)/((x^4 - 8*x^3 + 94*x^2 + 552*x + 657)
> > > *sqrt(x^3 - 15*x - 22)), x)
> > >
> > > the current web-interface version 1.3.2 responds with:
> > >
> > >>> Error detected within library code:
> > > Not integrable (provided residues have no relations)
> > >
> > > Why can't FriCAS establish non-integrability here?
> > >
> >
> > FYI, same error shows up in Fricas 1.3.3 on Linux.
> >
> So this failure should be related to limitations of results from
> torsionIfCan() when q(z) (whatever this square-free thingy represents:
> can a polynomial be doubly-transitive?) has degree > two. The
> implementation of torsionIfCan() resides in another source file
> (presumably pfo.spad).
>
> I think NOTI should be improved to inform the user why FriCAS cannot
> handle his integral.
I am not sure what more FriCAS could say. The message is:
Not integrable (provided residues have no relations)
Should the message say that 'relations' above mean linear
dependence over rationals? Core problem here is that
FriCAS has kind of "virtual integral", namely log-like
terms (divisors) and we want to convert virtal thing into
a real one. For single term it is done by 'torsionIfCan()',
it decides if the term is integrable (that is corresponds to
log of algebraic function) or not.
However, there is potential possiblity that two (or more)
nonintegrable terms combine into integrable one. Trager
proved that when residues are linearly independent over
rationals then the sum is integrable if and only if
each term is integrable. So, if we know that residues
are linearly independent over rationals, then after finding
a nonintegrable term we can declare the integral as
nonelementary. If we _know_ linear relation betwen
residues we can use it to combine terms and get smaller
number of terms.
So the problem here is to determine structure of linear
space over ratinals spanned by residues. FriCAS knows
that residues are roots of Trager double resultant.
However, the only known method to determine linear
dependencies between roots seem to be construction of
the splitting field. This is rather heavy process
involving several factorizations of polynomials
with coefficients in algebraic extensions.
One problem is that sum of roots gives one of coefficients
of polynomial. So if coefficients are rational we
get at least one relation. Paradoxicaly, the
most troblesome case is when sum of roots is 0
(the trace0 case). For polynomial of degree 2 in
such case we can note that one root, call it a1, is
irrational (because polynomial is irreducible) and
the second root a2 satisfies a2 = -a1. Clearly,
there can be no more linear relations over rationals.
Doubly transitive above is about Galois group of q(z).
If Galois group is doubly transitive, then one can
take a shortcut and combine two terms. If such
combination is integrable one can gets whole interal.
If not in doubly transitve case the function is
not integrable. Theoreticaly assuming that Galois group
is doubly transitive is nice condtion, but computing
Galois group seem to be almost as hard as computing
splitting space.
> The deadly failure reported recently in "elementarily integrable or
> not?" is most likely related to torsion too.
Well, the limitation above exist even if torsionIfCan()
works correctly. The failure from the other post most
likely means a bug in torsionIfCan().
--
Waldek Hebisch
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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