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Topic: 136 theorems on 29 pages
Replies: 20   Last Post: Nov 19, 2012 4:55 PM

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clicliclic@freenet.de

Posts: 978
Registered: 4/26/08
Re: 136 theorems on 29 pages
Posted: Sep 26, 2012 5:25 PM
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Waldek Hebisch schrieb:
>
> clicliclic@freenet.de wrote:

> >
> > 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.


You are overlooking that there are three sides to the equation: the
given integral, the transformed integral, and the integrated term. Extra
"numerator" polynomials are inserted into each: the coefficients of that
in the given integrand are assumed to be given, those of the other two
assumed to be unknown. For a recurrence formula, the polynomials in the
two integrands should have the same degree, and the highest admissible
degree of that in the integrated term will obviously correlate with
that. When the degrees are ramped up simultaneously, the numbers of
unknowns rises twice as fast as the number of knowns, and a solution
must eventually become feasible. This (fairly obvious) argument could no
doubt be formalized into a mathematical proof, but I doubt the author
would be interested in this.

> >
> > [...] 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).


Let's say, this summarizes my personal experience in science and
engineering. I agree with the author that the x^1 and x^3 terms in an
elliptic radicand are best annihilated right away to simplify the
problem. If the result involves elliptic integrals of the third kind,
Pi(x,n,k), the rational cofactor has to be split into partial fractions
anyway in order to extract the parameters n, which may therefore come in
complex conjugate pairs, etc.

>
> I do not have Gradshteyn-Ryzhik handy, [...]


One of the pseudo-elliptic examples (here taken from Goursat's Cours
d'analyse) is given on page 50 of the booklet mentioned.

>
> 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...


I expect your opinion will be appreciated by the author. As far as I can
see, excluding additional confluences at any level of degeneracy is a
sufficient condition (though not a necessary condition) to prevent zero
division: Some confluences play no role (e.g. 4*d*f = e^2 doesn't affect
8.1.1, 8.1.2, 8.1.3 in the paper at all) whereas others play a selective
role (e.g. 4*a*c = b^2 affects 8.1.1 and 8.1.3 in the paper but doesn't
affect 8.1.2). While simply excluding confluent roots then seems to be
enough indeed, it also does no harm since those integrands are dealt
with separately anyway.

Martin.



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