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It is known, that $\phi := \frac{sqrt(5)-1}{2}$, is the number, that is hardest to approximate by rationals (cf e.g. the section properties of the golden ratio $\phi$ here: http://en.wikipedia.org/wiki/Continued_fraction#A_property_of_the_golden_ratio_.CF.86).

In the section Infinite continued fractions on the same WIKI page, one finds the following "explanation" of how the terms of an infinite continued fraction relate to its "degree of irrationality":

The larger a term is in the continued fraction, the closer the corresponding convergent is to the irrational number being approximated. Numbers like π have occasional large terms in their continued fraction, which makes them easy to approximate with rational numbers. Other numbers like e have only small terms early in their continued fraction, which makes them more difficult to approximate rationally. The golden ratio ϕ has terms equal to 1 everywhere—the smallest values possible—which makes ϕ the most difficult number to approximate rationally. In this sense, therefore, it is the "most irrational" of all irrational numbers

That explanation is somewhat fuzzy: "small terms in the beginning" and "occasionally large terms" are not mutually exclusive; so that explanation can't be used for sorting continued fractions according to decreasing irrationality in the sense of rational approximability.

Question: is it possible to compare the degree of irrationality of infinite (periodic) continued fractions on basis of their terms?

Any suggestions or pointers to the literature would be great.

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  • $\begingroup$ Usually one refers to the set of numbers having bounded digits in their continued fraction expansion as the set of "Badly Approximable" numbers (abbreviated BA), and this set is very well-studied, as this condition can be pharsed (in an analouge language) dynamically, see the works of Kleinbock and Margulis for example. The "occasionally large terms" can be thought of the first entrance time to a cuspidal neighborhood (for example, in the dynamical language). $\endgroup$ – Asaf Dec 23 '14 at 21:25
  • $\begingroup$ @Asaf thank you for the information; I will check the works of Kleinbock and Margulis. $\endgroup$ – Manfred Weis Dec 23 '14 at 22:46
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    $\begingroup$ May I suggest that my paper, "Real numbers with bounded partial quotients: a survey", might be useful? It appeared in L'Enseignement math. 38 (1992), 151-187. I can send a pdf to those who contact me by e-mail (my address is easily found with a google search). $\endgroup$ – Jeffrey Shallit Dec 24 '14 at 18:03
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A frequently used measurement is ``the largest partial quotient''. Almost all numbers (in the sense of measure theory) have unbounded partial quotients, but many interesting numbers do not (rationals and quadratic irrationals, for example). This is, in my experience, the most usable such gradation of approximability.

Another measurement is to consider the average of the first $n$ partial quotients. Again, for almost all numbers this is unbounded as $n$ increases, but this measurement goes deeper into the well-approximable numbers.

Another measurement is to consider the geometric mean of the first $n$ partial quotients. For almost all numbers, this is finite (as $n\to\infty$). Unfortunately, for almost all numbers the limit of geometric means is the same (Khinchin's constant), so this doesn't allow one to distinguish numbers very well.

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There is something called the "irrationality measure" of real number. The following brief treatment is based on mathworld, which should be consulted for more information and examples.

Let $x$ be a real number, and let $R$ be the set of positive real numbers $\mu$ for which $$0<|x-(p/q)|<q^{-\mu}$$ has (at most) finitely many solutions $p/q$ for $p$ and $q$ integers. Then the irrationality measure $\mu(x)$ is defined as the infimum of the set $R$ )if $R$ is empty, then by convention $\mu(x)=\infty$).

The irrationality measure of an irrational number $x$ can be given in terms of its simple continued fraction expansion $x=[a_0,a_1,a_2,\dots]$ and its convergents $p_n/q_n$ as $$\mu(x)=1+\limsup_{n\to\infty}{\log q_{n+1}\over\log q_n}=2+\limsup_{n\to\infty}{\log a_{n+1}\over\log q_n}$$

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