Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.symbolic.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
clicliclic@freenet.de

Posts: 997
Registered: 4/26/08
Re: FriCAS failure on Goursat(?) pseudo-elliptics
Posted: Sep 29, 2012 4:21 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


Waldek Hebisch schrieb:
>
> clicliclic@freenet.de wrote:

> >
> > clicliclic@freenet.de schrieb:

> > >
> > > Waldek Hebisch schrieb:

> > > >
> > > > clicliclic@freenet.de wrote:

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

> > > >
> > > > AFAICS this example is wrong -- FriCAS result indicates that
> > > > there is no elementary integral in this case and I see no
> > > > reasons why it should have one. OTOH, if one modifies
> > > > the example to have
> > > >
> > > > f/sqrt((1 - x^2)*(1 - k^4*x^2))
> > > >
> > > > under integral, then FriCAS can handle it for particular f-s.
> > > >

> > >
> > > Thanks for trying. My Gradshteyn-Ryzhik has three such integrands and
> > > adds more detail (without specifying the source). In particular, the
> > > substitution z*x = SQRT(x*(1-x)*(1-k^2*x)) is said to turn the
> > > integrand into a rational function. I will look into this claim. The
> > > other two examples are the same integrand f(x)/sqrt(x*(1-x)*(1-k^2*x))
> > > with f(x) + f((1-k^2*x)/(k^2*(1-x))) = 0 and with
> > > f(x) + f((1-x)/(1-k^2*x)) = 0.
> > >

> >
> > If you made no mistake, FriCAS is in trouble here: I have confirmed that
> > the first example (the one cited by Hardy) is indeed elementary, using
> > the rational factor f(x) = (k^2*x^2 - 1)/((a*k^2*x + b)*(b*x + a)) for
> > which f(x) + f(1/(k^2*x)) = 0 as required.

>
> I made a mistake: I took the irrational factor to be
> sqrt(x*(1 - x)*(1 - k^2*x^2)). Goursat example is clearly
> based on symmetry under x -> 1/(k^2*x) map, but that factor
> is not symmetric under such map, so has no reason to have
> elementary integral. When irrational factor is changed to
> sqrt((1-x^2)*(1 - k^2*x^2))) we get required symmetry, and
> it works fine. I did not notice that replacing k^2*x^2 by
> k^2*x also gives symmetry (and that this is what text
> actually contained).
>
> To make things clear: with sqrt(x*(1 - x)*(1 - k^2*x)) FriCAS
> can do the integrals.
>


Excellent. So a computation of the "splitting field of the Trager
resultant" is not needed here. Goursat was clearly an authority in
pseudo-elliptic integrals, as shown by this 1887 paper of his:

<http://www.numdam.org/item?id=BSMF_1887__15__106_1>

Martin.



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.