Continued Fractions of Algebraic Numbers of Higher Degree

Definition 5.4 (Algebraic Number)   An algebraic number is a root of a polynomial $f\in\mathbb{Q}[x]$ .

Open Problem 5.4   Give a simple description of the complete continued fractions expansion of the algebraic number $\sqrt[3]{2}$ . It begins

$$[ 1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14,$$

$$3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, \ldots]$$

The author does not see a pattern, and the $534$ reduces his confidence that he will. Lang and Trotter (see [#!langtrotter1!#]) analyzed many terms of the continued fraction of $\sqrt[3]{2}$ statistically, and their work suggests that $\sqrt[3]{2}$ has an ``unusual'' continued fraction; later work in [#!langtrotter2!#] suggests that maybe it does not.

Khintchine (see [#!khintchine!#, pg. 59])

No properties of the representing continued fractions, analogous to those which have just been proved, are known for algebraic numbers of higher degree [as of $1963$ ]. [...] It is of interest to point out that up till the present time no continued fraction development of an algebraic number of higher degree than the second is known [emphasis added]. It is not even known if such a development has bounded elements. Generally speaking the problems associated with the continued fraction expansion of algebraic numbers of degree higher than the second are extremely difficult and virtually unstudied.

Richard Guy (see [#!guy:unsolved!#, pg. 260])

Is there an algebraic number of degree greater than two whose simple continued fraction has unbounded partial quotients? Does every such number have unbounded partial quotients?

Baum and Sweet [#!baum_sweet!#] answered the analogue of Richard Guy's question but with algebraic numbers replaced by elements of a field $K$ other than  $\mathbb {Q}$ . (The field $K$ is $\mathbb{F}_2((1/x))$ , the field of Laurent series in the variable $1/x$ over the finite field with two elements. An element of $K$ is a polynomial in $x$ plus a formal power series in $1/x$ .) They found an $\alpha$ of degree three over $K$ whose continued fraction has all terms of bounded degree, and other elements of various degrees greater than $2$ over $K$ whose continued fractions have terms of unbounded degree.