next up previous
Next: About this document ... Up: Lecture 19: Continued Fractions Previous: Periodic Continued Fractions

What About Higher Degree?

Definition 3.1   An algebraic number is a root of a polynomial $ f\in\mathbb{Q}[x]$.



Open Problem: 1 What is the continued fraction expansion of the algebraic number $ \sqrt[3]{2}$?

? contfrac(2^(1/3))
%5 = [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, 
 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1, 1, 1, 3, 7, 2, 2, 9, 4, 1, 3, 7, 
 6, 1, 1, 2, 2, 9, 3, 1, 1, 69, 4, 4, 5, 12, 1, 1, 5, 15, 1, 4, 1, 1, 
 1, 1, 1, 89, 1, 22, 186, 5, 2, 4, 3, 3, 1, \ldots]
I sure don't see a pattern, and that $ 534$ strips me of any confidence that I ever will. One could at least try to analyze the first few terms of the continued fraction statistically (see Lang and Trotter, 1972).






Khintchine (1963), page 59:

No properties of the representing continued fractions, analogous to those which have just been proved, are known for algebraic numbers of higher degree. [...] 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. 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 Unsolved Problems in Number Theory (1994), page 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?


next up previous
Next: About this document ... Up: Lecture 19: Continued Fractions Previous: Periodic Continued Fractions
William A Stein 2001-10-25