Talk:Khinchin's constant

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Untitled[edit]

it has not been proven for any specific real number.

That surely ought to be rephrased, or else I could specify a number by a continued fraction, which worked. Charles Matthews 14:05, 19 Mar 2005 (UTC)

I see the problem; but I don't think this is the solution: the present text leaves open the possibility that there is a number which
  • is of some interest in any other context,
  • has a known continued fraction and
  • does satisfy Khinchin's condition. Septentrionalis PMAnderson 13:26, 6 August 2007 (UTC)[reply]

Algorithm[edit]

Why does it take Mathematica 5 so damn long to calculate the constant to 10000 decimal place, while it can calculate it to 2000 places in 0. seconds

Khinchin's constant is very expensive to calculate even for just a few thousand digits. It seems that Mathematica has the 5000 first digits or so pre-computed. Fredrik Johansson 11:03, 19 November 2006 (UTC)[reply]

Euler-Mascheroni constant[edit]

How is that rational numbers do not posses this property, yet the Euler-Mascheroni constant does? 121.216.136.243 (talk) 19:57, 16 May 2008 (UTC)[reply]

Nevermind, I think the wording threw me off. Maybe the second sentence of that paragraph should come first? 121.216.136.243 (talk) 20:00, 16 May 2008 (UTC)[reply]

New Relevant Result?[edit]

I just learned (from [1]) that there has been a recent advance on this problem, but I have not accessed the paper (and might not be best placed to evaluate it in any case).

Title: A Khinchin Sequence Author(s): Thomas Wieting Journal: Proc. Amer. Math. Soc. 136 (2008), 815-824. —Preceding unsigned comment added by Josephorourke (talkcontribs) 19:18, 31 December 2009 (UTC)[reply]

Irrational / Transcendental ?[edit]

Is it known if Khinchin's constant is transcendental, or at least irrational? 71.225.89.166 (talk) 08:26, 21 January 2012 (UTC)[reply]

Since it's not known to be transcendental, how can we be sure that no solutions of quadratic equations with rational coefficients have Khinchin's property (as claimed in the article)? The geometric mean of their coefficients always converges to an n-th root of an integer m, and for any given integers n>0 and m, one can construct a quadratic irrational whose coefficients mean converges to the n-th root of m. --Roentgenium111 (talk) 11:32, 15 July 2014 (UTC)[reply]
I agree with Roentgenium111, and I have removed the claim on general quadratic algebraic numbers. /195.249.119.154 (talk) 13:41, 1 August 2014 (UTC)[reply]

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