email@example.com wrote: > Waldek Hebisch schrieb: > > > > firstname.lastname@example.org wrote: > > > > > > Hooray, hooray. A paper has finally appeared on two-term recurrence > > > formulae for indefinite algebraic integrals: > > > > > > <http://arxiv.org/abs/1209.3758> > > > > > > I wonder though if this might be a hoax. > > > > Does not look as a hoax :). But AFAICS there is one (maybe 2-3 > > depending on how you count) theorem, which author did not state > > and a lot of examples. More precisely, author wrote: > > > > > The two-term recurrence relations have been derived by the method > > > of undetermined coefficients > > > > Of course the interesting question is why such formulas should > > exist. The answer (which the author apparently did not want > > to disclose) is that Hermite reduction method works. Fact > > that it works for increasing exponents by 1 is well-known. > > Fact that it can be used to reduce exponents by 1 is less > > known, but for example Bronstein mentions this in his thesis. > > > > Author also did not mention easy to observe fact: given > > > > R*Q*A > > > > where A is product of roots (with possibly added exponential > > factor), Q is product of powers of polynomials Q_1, ..., Q_n > > containing all radicands, such that sum of degrees of > > Q_1, ..., Q_n is m and R is a polynomial of degree at most > > m - 1, one can subtract multiple of (Q*A)' from R*Q*A and get > > a similar term with R of degree at most m - 2. So using > > the R term of degree m - 2 he still effectively has the same > > coverage as Hermite reduction. > > > > The author repeatedy writes phrases like: > > > > > To exclude integrands with confluent roots, the following > > > recurrences should be applied only if the resultant of the > > > linear polynomials does not vanish > > > > I do not know why he wants to exclude confluent roots, because > > AFAICS the formulas are equivalent to equalites between polynomials, > > so are valid for all values of parameters. When applying > > them we need to avoid division by zero, which in general > > is different condition than excluding confluent roots. > > > > Also, I find his motivation form introdution somewhat disconnected > > with rest of the article. Namely, Hermite reduction seem to > > widely used and does not eliminate form of integrals that > > the author does not like. AFAICS the main source of difficulty > > is due to logarithmic terms, which is outside of Hermite > > reduction. Minor source of difficulty is because some > > (otherwise attractive) simplifications can change branching > > pattern of the integral. Hermite reduction is of limited > > relevance for the seond problem - it gives "rational" > > approach which works without additional simplifications, but > > simplifications are typically introduced because of other > > steps. Even in contexts of rule based integration it > > may be better to keep Hermite reduction as a procedure > > istead of encoding it as set of rules. > > > > The author precomputes results of Hermite reduction for > > a few "typical" forms of integrand. If this is worth the > > effort can be decided only for an integrator as a whole, but > > probably in some cases precomputed formulas give large > > saving in compute time. > > > > Many thanks to Holmes for pointing out this interesting paper! > > In my view, the simple, trivial, answer as to why such formulae exist is > that the "method of undetermined coefficients" works: In other words, > one may insert polynomial factors and ramp up their degree until the > linear system of equations that results after differentiation becomes > soluble.
Do you have a proof that this works? It seem that if you incresase degree only on one side, then you end up with multiterm reccurences. If you increase degree on both sides you have more equations.
Hermite reduction has quite simple proof of correctness.
> Of course this could be coded as a single procedure for > arbitrary integrands of this type, but such a procedure shouldn't be > called a "formula" (nor a "rule") anymore.
> Also, there is very little > practical interest in going beyond elliptic integrals (or beyond > Lauricella FD).
Hmm, that is your opinion. Note however, that when integrating functions of form R*sqrt(P) where R is a rational function and P is of degree 3, the rules apply only if R is of degree 1. This seems to be quite serious limitation. Of course you can split R into partial fractions, but to get denominator which is power of linear factor may require extra algebraic extension (in particular may require going to complex numbers).
> By the way, what does the FriCAS (or Axiom) integrator > return when fed some arbitrary R(x,sqrt(P)) where R is rational and P is > a polynomial of degree three or four? Does it recognize the > pseudo-elliptic cases in Gradshteyn-Ryzhik 2.292?
I do not have Gradshteyn-Ryzhik handy, but the whole point of Trager algoritm is to distinguish between cases which could be done in terms of elementary functions and the ones which can not. For algebraic functions the only unimplemented part is computation of splitting field of Trager resultant. So algorithm may fail if integrability depends on relations between roots. ATM I do not have example of such behaviour (it should exist, but finding it is not easy), all I have are cases which require special functions, but the integrator can not properly analyze dependencies between roots of Trager resultant and gives up.
> It seems to me that the author wants to exclude confluent roots > precisely in order to avoid divison by zero. Apparently he expects > readers to realize that the recurrence formulae - being in essence just > equalities between polynomials - are generally valid otherwise.
Well, there is more conditions beyond confluent roots, so just excluding them is not enough -- the user still have to check for division by zero. However, the point of my remark was that repeated blurb about confluent roots is more distraction than help for the reader. Of course, the author is free to ignore my opinion...