6

Edit: as multiple users have pointed out, the premise of my question assumes some canonical representation of real numbers as infinite nested radicals. There does not seem to be any such representation.

Khinchin's constant is the peculiar number $K$ such that for almost any real number $x$, if we write out $x$'s continued fraction representation $$x = a_0+\frac1{a_1+\frac1{\ddots}}$$ Then we have $$\lim_{n\to\infty}\sqrt[n]{a_1a_2\dots a_n} = K$$ My question begins with the fact that any real number $x$ may be written as $$x = b_0+\sqrt{b_1 + \sqrt{b_2+\dots}}$$ And, given the similarity between continued fractions and nested radicals as iterated function systems/contractions, I would think there must be some number $S$ and non-trivial function $f$ such that for almost all $x$ we have $$\lim_{n \to\infty}f(b_0,b_1 \dots b_n) = S$$ Where $f$ is probably defined independent of $b_0$.

I nervously tag this post because I know Khinchin relied on it in the proof for his constant.

Descartes Before the Horse
  • 7,386
  • 4
    Given a positive irrational $x$, the sequence $(a_0, a_1, ...)$ giving the contiue fraction representation exists and is unique. How do you define, in general, the seuqnce $(b_0, b_1, ...)$? Does it exist and is it unique? – D. Thomine Dec 14 '19 at 00:29
  • 3
    Similar to D. Thomine's point, there is a canonical continued fraction, which associates each real $x$ with a unique continued fraction. You would need to canonically define the nested radical. – Jam Dec 14 '19 at 00:31
  • 3
    You'd need to be careful about how you do it because $\sqrt{2}=\sqrt{1+\sqrt{1+\sqrt{0+\sqrt{0+\ldots}}}}=\sqrt{1+\sqrt{0+\sqrt{1+\sqrt{0+\ldots}}}}$ and so on. So the nested radicals, $(1,1,\bar{0}),\ (1,0,1,\bar{0}),\ldots$ all non-uniquely specify the same value. – Jam Dec 14 '19 at 00:41

0 Answers0