4 ms·

Pretty cool that you can determine the exact probability distribution for the digits of fraction expansion of a random real number: lim_{n->inf} Pr[k_n = k] =

Pretty cool that you can determine the exact probability distribution for the digits of fraction expansion of a random real number: lim_{n->inf} Pr[k_n = k] = -log_2(1 - 1/(k+1)^2) And this was determine already in 1929! I think fraction expansions was all the rage back then. https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin_distribution https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin_distribut...

svat 4y ago
Yes! Not only that, here's something I found mind-blowing: for almost all real numbers, the nth root of the nth convergent's denominator has as limit the same value, and that value is e^(pi^2/12ln2). Am typing from phone so can't write it down here properly, but some details at an old blog post of mine: https://shreevatsa.wordpress.com/2010/04/30/some-incredibly-beautiful-things/ https://shreevatsa.wordpress.com/2010/04/30/some-incredibly-...
- madcaptenor 4y ago
  there's also Khinchin's constant - for almost all real numbers, the geometric mean of the first n continued fraction coefficients approaches a constant K as ngoes to infinity. K is about 2.68 (the geometric mean of the Gauss-Kuzmin distribution). I like to pair this with a more trivial fact - for almost all real numbers, the arithmetic mean of the first n digits of the decimal expansion, as n goes to infinity, approaches 4.5.
  - tialaramex 4y ago
    > for almost all real numbers, the arithmetic mean of the first n digits of the decimal expansion, as n goes to infinity, approaches 4.5. Is this just because Almost All Reals are Normal, and so in every base they have an equal distribution of all digits ?
    - madcaptenor 4y ago
      That’s exactly it.
  - cperciva 4y ago
    Here's an even more trivial fact: For almost all real numbers, the geometric mean of the first N digits of the decimal expansion, as N goes to infinity, approaches zero. (Really really fast!)
    - madcaptenor 4y ago
      and a less trivial one: the arithmetic mean of the continued fraction coefficients goes to infinity.
arutar 4y ago
Results like this are a lot of fun, and like many a reasonable number of cool results in number theory, are "surprisingly easy". There's a proof only using two ideas: 1) Birkhoff ergodic theorem, which states for a "nice" dynamical system, the probability that certain events occur can be described explicitly by an invariant distribution (see [1]), and 2) Continued fractions have an associated "nice" dynamical system (the Gauss map) which has an explicit probability distribution that is not too challenging to compute. Of course, writing this argument out takes a bit of work [2]. In fact, the argument is structured in the exact same way as the fact that uniformly randomly chosen numbers in [0,1] are normal (i.e. the digit frequencies in a base-b expansion are all 1/b). However, proving such results about _specific_ numbers is notoriously hard [3]. As far as I am aware, there has not been a single irrational algebraic number proven to be normal. Normality of well-known constants like pi and e is also an open problem! I would not be surprised if proving distributional results for continued fraction expansions of pi is also very hard. [1]: https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorem... [2]: http://www.geometrie.tugraz.at/karpenkov/cf2011/cf2011s_7.pdf http://www.geometrie.tugraz.at/karpenkov/cf2011/cf2011s_7.pd... [3]: https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples https://en.wikipedia.org/wiki/Normal_number#Properties_and_e...
- joshuamorton 4y ago
  > However, proving such results about _specific_ numbers is notoriously hard [3]. As far as I am aware, there has not been a single irrational algebraic number proven to be normal. Normality of well-known constants like pi and e is also an open problem! I would not be surprised if proving distributional results for continued fraction expansions of pi is also very hard. Is it known if algebraic numbers can be normal? I'm not a mathematician, but almost all numbers are normal, and almost all numbers are non-algebraic (or even non-computable!). Something akin to "most of the non-algebraics and non-computables are normal, and none of the algebraics are normal" is feasible, right? It contradicts the commonly held idea that e or sqrt(2) or pi are normal, but we don't even have a (non-constructuve) proof that there exist irrational algebraic normals, do we?
  - 4y ago

svat 4y ago

Yes! Not only that, here's something I found mind-blowing: for almost all real numbers, the nth root of the nth convergent's denominator has as limit the same value, and that value is e^(pi^2/12ln2). Am typing from phone so can't write it down here properly, but some details at an old blog post of mine: https://shreevatsa.wordpress.com/2010/04/30/some-incredibly-beautiful-things/ https://shreevatsa.wordpress.com/2010/04/30/some-incredibly-...

madcaptenor 4y ago
there's also Khinchin's constant - for almost all real numbers, the geometric mean of the first n continued fraction coefficients approaches a constant K as ngoes to infinity. K is about 2.68 (the geometric mean of the Gauss-Kuzmin distribution). I like to pair this with a more trivial fact - for almost all real numbers, the arithmetic mean of the first n digits of the decimal expansion, as n goes to infinity, approaches 4.5.
- tialaramex 4y ago
  > for almost all real numbers, the arithmetic mean of the first n digits of the decimal expansion, as n goes to infinity, approaches 4.5. Is this just because Almost All Reals are Normal, and so in every base they have an equal distribution of all digits ?
  - madcaptenor 4y ago
    That’s exactly it.
- cperciva 4y ago
  Here's an even more trivial fact: For almost all real numbers, the geometric mean of the first N digits of the decimal expansion, as N goes to infinity, approaches zero. (Really really fast!)
  - madcaptenor 4y ago
    and a less trivial one: the arithmetic mean of the continued fraction coefficients goes to infinity.

arutar 4y ago

Results like this are a lot of fun, and like many a reasonable number of cool results in number theory, are "surprisingly easy". There's a proof only using two ideas: 1) Birkhoff ergodic theorem, which states for a "nice" dynamical system, the probability that certain events occur can be described explicitly by an invariant distribution (see [1]), and 2) Continued fractions have an associated "nice" dynamical system (the Gauss map) which has an explicit probability distribution that is not too challenging to compute. Of course, writing this argument out takes a bit of work [2]. In fact, the argument is structured in the exact same way as the fact that uniformly randomly chosen numbers in [0,1] are normal (i.e. the digit frequencies in a base-b expansion are all 1/b). However, proving such results about _specific_ numbers is notoriously hard [3]. As far as I am aware, there has not been a single irrational algebraic number proven to be normal. Normality of well-known constants like pi and e is also an open problem! I would not be surprised if proving distributional results for continued fraction expansions of pi is also very hard. [1]: https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorem... [2]: http://www.geometrie.tugraz.at/karpenkov/cf2011/cf2011s_7.pdf http://www.geometrie.tugraz.at/karpenkov/cf2011/cf2011s_7.pd... [3]: https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples https://en.wikipedia.org/wiki/Normal_number#Properties_and_e...

joshuamorton 4y ago
> However, proving such results about _specific_ numbers is notoriously hard [3]. As far as I am aware, there has not been a single irrational algebraic number proven to be normal. Normality of well-known constants like pi and e is also an open problem! I would not be surprised if proving distributional results for continued fraction expansions of pi is also very hard. Is it known if algebraic numbers can be normal? I'm not a mathematician, but almost all numbers are normal, and almost all numbers are non-algebraic (or even non-computable!). Something akin to "most of the non-algebraics and non-computables are normal, and none of the algebraics are normal" is feasible, right? It contradicts the commonly held idea that e or sqrt(2) or pi are normal, but we don't even have a (non-constructuve) proof that there exist irrational algebraic normals, do we?
- 4y ago