Talk:Pi (mathematical constant)

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Revision as of 07:31, 3 May 2007 by imported>Catherine Woodgold (→‎zeta(n) for n odd: "might have" seemed probable?)
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Article Checklist for "Pi (mathematical constant)"
Workgroup category or categories Mathematics Workgroup [Categories OK]
Article status Developing article: beyond a stub, but incomplete
Underlinked article? Yes
Basic cleanup done? Yes
Checklist last edited by --AlekStos 09:53, 26 March 2007 (CDT)

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[Image:http://commons.wikimedia.org/wiki/Image:Pi-symbol.svg]

what's the policy on using wikimedia images/media?

--Chief Mike 04:24, 25 January 2007 (CST)

Algebraically independent?

It says "lthough π itself is transcendental, it is an open problem whether many other transcendental numbers are algebraically independent from π. " in "Relation to other constants". If this means that it's not known whether there exist many transcendental numbers which are algebraically independent from π, then it sounds wrong to me. Maybe it means that for many other known transcendental numbers, it is an open problem whether those numbers are algebraically independent from π. If so, it should be reworded to clarify. --Catherine Woodgold 17:32, 2 May 2007 (CDT)

It can't be the first, by a simple countability argument. Greg Woodhouse 17:35, 2 May 2007 (CDT)
That's what I thought. Thanks for rewording it. --Catherine Woodgold 20:27, 2 May 2007 (CDT)

zeta(n) for n odd

It's remarked in the current version that it "may seem probable" that zeta(n) is a rational multiple of pi^n when n>1 is odd; I don't feel this is correct.

First off "may" should be "might". But more importantly, if zeta(3) were a rational multiple of pi^3 for example, then presumably the rational multiple wouldn't involve a ridiculously large numerator and denominator. I just did a quick computation (by computer) that verified, however, that zeta(3)/pi^3 doesn't equal any rational number whose numerator or denominator has less than 500 digits. I'm sure this sort of computation has been done before and to more precision. So I doubt many number theorists believe zeta(n)/pi^n is rational for n odd. - Greg Martin 02:02, 3 May 2007 (CDT)

Indeed, that's why many number theorists believe; the point is that it might come as a surprise given the pattern of the even-indexed zeta numbers. Anyway, I'd rather see this section deleted (if not the entire article). Fredrik Johansson 02:27, 3 May 2007 (CDT)
Possibly "might have seemed probable" but that makes it sound as if we're sure it isn't. I like this article but maybe it goes on just a little too long about calculation records. --Catherine Woodgold 07:31, 3 May 2007 (CDT)