Kleptog

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Hypertranscendental function

Hi, Kleptog! Thank you for the interesting articles about "hypertranscendental" numbers / functions.

I did manage to download the article from the Japanese math journal, but had a problem with the file from the site at Gottingen. Did you manage to download that one? I got 22 Mbytes downloaded, but then I couldn't open the file (I got an Acroread message "file is damaged"). So I can only check part of what you wrote.

Anyway, I think the definition of a hypertranscendental function should be that it's not expressible as the solution of an ODE with coefficients that are rational functions in the independent variable. For example, the Bessel functions, which are the solutions of

$x^{2}y^{\prime \prime }+xy^{\prime }+(x^{2}-k^{2})y=0\,$

are not hypertranscendental functions (but the coefficients in the ODE are not integers – they're polynomials in x).

I'm also curious why you mention a boundary condition in the article about hypertranscendental functions. I can see where the boundary condition is important in the definition of a hypertranscendental number. But since we're talking about single-valued functions of a single variable, all a different boundary condition (say y(0) = k) is going to do is translate the function by adding in a constant term, right?

Anyway, thanks for the articles. I want to make a couple of minor corrections, but thought I'd check with you first. Have a great day! DavidCBryant 15:48, 3 May 2007 (UTC)

In a sense I think you're right. It all depends on the definition of "algebraic differential equation". I was not able to find a definition that would tell me if your equation for the bessel functions is indeed such an equation. I think the "coefficients in the integers" is to stop people sticking numbers like pi en e in there. There is a wikipedia article for "differential algebraic equations" but I think they are something else. So I think you're right, but I can't verify it yet (though I've requested a book from the library which might have it).

As for the constant term, yes, all it's going to do is shift the solution by a constant, but if I set the constant to be "pi" or any other transcendeantal number, then suddenly formerly transcendental numbers become solutions. So it has to be an algebraic constant. You could possibly require it to be zero, but I can't prove it. An n-th order differential equation has n initial conditions and I can't be sure the effects are trivial. Kleptog 09:33, 4 May 2007 (UTC)

Thanks for the reply. I'll dig through the paper from the Japanese journal a little more – they have an appendix about the "algebraic differential equations". I'm pretty sure what they're talking about is rational functions where the coefficients in the two polynomials have to be algebraic numbers (which really boils down to integer coefficients in the polynomials). They also mention a class of meromorphic functions (which fits with the rational function coefficients, since each such coefficient would have a finite number of poles). I'll read it to be sure, though, and get back to you. Good luck with the library! DavidCBryant 10:44, 4 May 2007 (UTC)

Aha, if that the case then it's all clear. It also means that arctan, arccos and arcsin are also normal, as well as the bessel functions and most others . Rational is a subset of meromorphic so I think it's right. That certainly clears things up. Kleptog 11:54, 4 May 2007 (UTC)

Ways to improve Recognition (family law)

Hi, I'm Robvanvee. Kleptog, thanks for creating Recognition (family law)!

I've just tagged the page, using our page curation tools, as having some issues to fix. Just needs more references. Otherwise, good work!

The tags can be removed by you or another editor once the issues they mention are addressed. If you have questions, you can leave a comment on my talk page. Or, for more editing help, talk to the volunteers at the Teahouse. Robvanvee 13:57, 26 November 2016 (UTC)