Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The continued fraction $$[1;1,2,3,4,5,\dots]=1+\cfrac{1}{1+\cfrac{1}{2+\cdots}}, $$ for instance, is known explicitly as a ratio of Bessel function values and is (I believe - SS) known to be transcendental. Similarly, $[1,2,2^2,2^3,2^4,2^5,\dots] $ is surely transcendental (and likely related to Liouville numbers). Are there any explicitly known algebraic numbers with unbounded continued fraction coefficients? (Of course, such a number must have degree greater than $2$, by all the usual theorems on continued fraction representations of quadratics.) Any references would be greatly appreciated.

share|improve this question
2  
XL: I've revised the question substantially; in particular, the bits about 'computable' unbounded coefficients are rather out of place; CF coefficients for any algebraic number are certainly computable (there are known algorithms) but they're also largely 'patternless' so far as anyone knows (again, at least for degree $\gt 2$). AFAIK even the simpler question of unboundedness is open, and as that's the most central question here I've revised your Q to be about that specifically. –  Steven Stadnicki Jul 31 at 0:57
3  
To the best of my knowledge, there is no algebraic number known to have unbounded partial quotients, and there is no algebraic number of degree exceeding 2 known to have bounded partial quotients. –  Gerry Myerson Jul 31 at 2:36
1  
Every algebraic number is "explicitly known", and it would be very surprising if any algebraic number of degree $> 2$ had bounded coefficients, but of course there are no proofs and there is no reasonable prospect of a proof in the near future. –  Robert Israel Jul 31 at 2:43
2  
I think you'll find, XL, that you can only notify one person per comment. The second @ is ignored. And, yes, everyone believes $\root3\of2$ has unbounded partial quotients, and if it were easy to prove it, it would have been done long ago. But some things that are "certainly true" are highly resistant to proof. –  Gerry Myerson Jul 31 at 5:33
2  
Another example of non-algebraic number is $\sqrt{e}$ which has a particular nice CF. The numbers you get are 1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1, 1, 25, 1, 1, 29, 1, 1, 33, 1, 1, 37, 1, 1, 41, 1, 1, 45, 1, 1, 49, 1, 1 ... –  Per Alexandersson Jul 31 at 10:29

3 Answers 3

up vote 26 down vote accepted

As you indicate, real algebraic numbers of degree $\leq 2$ have this property in view of Lagrange's classical result characterizing them by the eventual periodicicty of the continued fractions expansion. It may be useful to know (if you don't already) that $\alpha \in \mathbb{R}$ having bounded continued fractions coefficients is equivalent to the sharpness of Dirichlet's approximation theorem: $|\alpha - p/q| > cq^{-2}$ for all but finitely many $p/q \in \mathbb{Q}$.

For algebraic numbers this means that the exponent $2+\varepsilon$ in Roth's theorem can be reduced to $2$. For quadratic irrationalities this holds with the uniform constant $c = 1/\sqrt{5} - \epsilon$; google "Lagrange spectrum." As far as I know it is widely believed that this may not happen for any algebraic number of degree $> 2$, although Serge Lang has suggested that a milder improvement in Roth's theorem, whereby $q^{2+\varepsilon}$ gets replaced with $(q\log{q})^2$, is always possible. (This is wide open; there is an analogous statement known to hold true in Nevanlinna theory). Certainly no algebraic number of degree $> 2$ is known to have unbounded partial fractions coefficients, but worse, it does not appear to be even known that there exists an algebraic number having this property.

A reference where this appears explicitly in print is p. 366 of Hindry and Silverman's Diophantine Geometry: An Introduction.


Added. There is also a rather interesting variant of this question for $\mathbb{Z}[i]$-continued fractions expansions of complex algebraic numbers.

As is typical in diophantine analysis, both Roth's theorem and the continued fractions algorithm extend to the relative setting over number fields other than $\mathbb{Q}$; and to a large extent, so does the relationship between the two. To be concrete, consider rational approximations over the Gaussian field $\mathbb{Q}(i)$. The relative Roth theorem over $\mathbb{Q}(i)$ states: If $\alpha \in \mathbb{C}$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many pairs $p,q \in \mathbb{Z}[i]$ in the Gaussian lattice satisfying $|\alpha - p/q| < |q|^{-2-\varepsilon}$. [For the general statement over any number field see Thm. 6.2.3 of Bombieri and Gubler's Heights in Diophantine Geometry.]

Likewise, Hurwitz has attached to a complex number a canonical continued fractions expansion with entries in $\mathbb{Z}[i]$, by using the same algorithm as over $\mathbb{Q}$, but with the nearest rounding to the Gaussian lattice $\mathbb{Z}[i]$ (whereas over $\mathbb{Q}$ the conventional choice of rounding involves the floor function $\lfloor \cdot \rfloor$ instead of the nearest rounding to $\mathbb{Z}$). The trichotomy "rational-quadratic-higher degree" would appear to extend to the relative setting over this (or any other) number field $\mathbb{Q}(i)$: the $\alpha \in \mathbb{C}$ with terminating expansion are of course precisely the numbers in $\mathbb{Q}(i)$, and Hurwitz has shown that his expansion is eventually periodic if and only if $[\mathbb{Q}(\alpha,i):\mathbb{Q}(i)] \leq 2$.

We can ask, then, the same questions for Hurwitz's complex continued fractions. Rather suprisingly, there are algebraic numbers whose $\mathbb{Z}[i]$-continued fractions expansion has bounded coefficients, and whose relative degree over $\mathbb{Q}(i)$ is $> 2$! One such number, due to D. Hensley [1, Ch. 5] in 2006, is $\sqrt{2} + i\sqrt{5}$, of relative degree four. More generally, W. Bosma and D. Gruenewald [3] have shown that a complex number has this property if the square of its modulus is a rational integer which is not a norm from $\mathbb{Z}[i]$; algebraic such examples thus include $\sqrt[m]{2} + i\sqrt{n - \sqrt[m]{4}}$ for all $n \equiv 3 \mod{4}$ and $m$.

(On the other hand, to my knowledge no particular algebraic number has been proven to have unbounded coefficients in its Hurwitz $\mathbb{Z}[i]$-continued fractions expansion.)

But now there is something even more curious about the implication of such examples for Roth's theorem over $\mathbb{Q}(i)$ (diophantine approximations by Gaussian numbers). I realize this just now after looking up Hensley's very interesting paper [2] (from 2006), and I wonder why this point isn't brought up in the literature on complex continued fractions.

While the convergents $p_n/q_n \in \mathbb{Q}(i)$ in Hurwitz's expansion no longer exhaust all good $\mathbb{Q}(i)$-rational approximations to $\alpha \in \mathbb{C}$ (as they do in the case over $\mathbb{Q}$), Theorem 1 in [2] shows that up to a multiplicative constant, they still give the best $\mathbb{Q}(i)$-rational approximations. As a result, for those numbers (algebraic of arbitrarily high relative degree over $\mathbb{Q}(i)$), the exponent $2+\varepsilon$ in Roth's theorem rel. $\mathbb{Q}(i)$ (stated above) can be reduced to $2$, and this is moreover effective:

Consequence: If $\alpha \in \mathbb{C}$ has $|\alpha|^2 \in \mathbb{Q}$ a rational number not a norm from $\mathbb{Q}(i)$ [e.g., the rel. degree-four algebraic example $\sqrt{2} + i\sqrt{5}$], then there is an effective $c(\alpha) > 0$ such that $|\alpha - \beta| > cH(\beta)^{-2}$ for all $\beta \in \mathbb{Q}(i)$. (Here, $H(\cdot)$ is the absolute multiplicative height on $\bar{\mathbb{Q}}$; for rational or imaginary quadratic integers $n$, it coincides with $|n|$)

Consider now Khintchine's principle according to which an algebraic number of degree $> 2$ should be generic. As almost every complex number $x \in \mathbb{C}$ has infinitely many $\mathbb{Q}(i)$-rational approximants $\beta \in \mathbb{Q}(i)$ with $|x - \beta| < 1/(H(\beta)^2\log{H(\beta)})$, the numbers $\alpha$ of the above form are not generic in this sense. As they contain algebraic numbers of arbitrarily high (though necessarily even) rel. degree over $\mathbb{Q}(i)$, Khintchine's principle would appear to fail in the relative setting over $\mathbb{Q}(i)$!

Nevertheless I think that the story is more interesting, as the above examples still resemble quadratic irrationalities in some vague sense. Perhaps, in the relative setting of diophantine approximations over a number field $K$, we could salvage Khintchine's principle and the above trichotomy by enlarging the class of special algebraic numbers -- which a priori contain all numbers of rel. degree $\leq 2$ over $K$.

Allow me to record a few

Problems. Might the numbers of degree $\leq 2$ over $\mathbb{Q}(i)$ (whose Hurwitz expansions are eventually periodic) and the algebraic numbers mentioned in the above "Consequence" (whose Hurwitz expansions are aperiodic yet have bounded coefficients) exhaust all the algebraic numbers for which the exponent $2+\varepsilon$ in Roth's theorem rel $\mathbb{Q}(i)$ may be reduced to $2$? Should we expect that all algebraic numbers not of this shape ought to satisfy Khintchine's principle rel $\mathbb{Q}(i)$? What are the special algebraic numbers in diophantine approximations rel a general given number field $K$? Finally, the same problem can be considered about $p$-adic (and $S$-adic) $K$-rational approximations to algebraic numbers; I do not know if this has been done even for $K = \mathbb{Q}$.

References:

[1] D. Hensley: Continued Fractions (World Scientific, Singapore, 2006).

[2] D. Hensley: The Hurwitz complex continued fraction (2006): http://mosaic.math.tamu.edu/~dhensley/SanAntonioShort.pdf

[3] W. Bosma, D. Gruenewald: Complex numbers with bounded partial quotients, J. Aust. Math. Soc., vol. 93 (2012), pp. 9--20.

share|improve this answer
    
thank you Vesselin for your answer. –  XL _at_China Jul 31 at 5:58
    
It is interesting for quadric irrational numbers and other irrational algebraic or other computable irrational numbers to have such a different property,moreover,transcendental number with such a property is easy to find ,but such a algebraic number's existence is difficult to be proved.Possibly we need new idea? –  XL _at_China Jul 31 at 6:13
2  
There is a basic trichotomy in diophantine approximations whereby algebraic numbers are divided into rationals, quadratic irrationalities, and numbers of degree $> 2$. It is a general principle that a number of the last type should have all the metric properties of a generic (in the a.e. sense) real number. For instance, it is easily seen that almost all reals are badly approximable to order $2+\varepsilon$, and correspondingly, we have Roth's deep theorem. –  Vesselin Dimitrov Jul 31 at 6:39
1  
Lang's conjecture I mentioned is founded on this principle: for almost all $x \in \mathbb{R}$, the inequality $|x - p/q| < (q\log{q})^{-2}$ has finitely many solutions. Clearly, almost all real numbers have unbounded partial fractions coefficients, so by the same principle we expect that so does every algebraic number of degree $> 2$. –  Vesselin Dimitrov Jul 31 at 6:40
    
I have gotten a lot of knowledge from your answer and comment,you know I have know very little a bit of Diophantine approximations,hope to have your more comment.thank you very much –  XL _at_China Jul 31 at 6:42

My survey paper, Real numbers with bounded partial quotients, discusses this question a bit. It is in L'Enseignement Math. 38 (1992), 151-187. In particular it seems that the first person to raise the unboundedness of the partial quotients of an algebraic number of degree > 2 was perhaps Khintchine in 1949. Some weak results on the growth of the partial quotients of algebraic numbers are summarized in section 4 of my survey.

share|improve this answer
    
Thank you, Jeffrey,I will find your survey and read it. –  XL _at_China Jul 31 at 11:37
    
If you contact me by e-mail (use google) I'll send you a pdf. –  Jeffrey Shallit Aug 1 at 0:24
    
Thank you very much,Jeffrey.I will send you an email. –  XL _at_China Aug 1 at 0:42
    
I have sent an email through my gmail to your email at waterloo U,since I have not found your gmail.Thank you again. –  XL _at_China Aug 1 at 1:03

I have one thought on this. But it would take some work to carry out.
(*) Perron (Die Lehre von den Ketterbrüchen) in Kap. 11 evaluates certain continued fractions with polynomial entries in terms of hypergeometric functions.
(**) There is an algorithm to determine when a hypergeometric function is an algebraic function.
Maybe these can be combined to get an example of an algebraic number with polynomial (and thus unbounded) denominators.

added: Examples of these two parts...

(*) The Perron method evaluates the (non-simple) continued fraction $$ b_0+\frac{a_1}{b_1+\frac{a_2}{b_2}+\ddots}\qquad\text{with}\qquad a_n=18n^2-33n, b_n=3n-2 $$ as $$ \frac{-\frac{9}{2}}{{}_2F_1\left(-\frac{5}{6},1;\frac{1}{2};\frac{1}{3}\right)} $$ Here, $a_n$ and $b_n$ are polynomials in $n$ which is what I meant by "polynomial entries" above.

(**) Unexpectedly, $$ {}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right) $$ is an algebraic function of $z$, see LINK

So, the (somewhat remote) hope would be that a case of continued fraction with $a_n=1$ evaluates to something in terms of hypergeometric functions on the list of known algebraic cases.

share|improve this answer
2  
I think in most cases the continued fractions for algebraic functions will not be simple continued fractions. For example: $$(x+1)^{1/3} = 1 + x/(3+x/(1+2x/(9+5x/\ldots)))$$ –  Robert Israel Aug 1 at 22:56
3  
The @ in a comment on an answer only notifies people who have posted, edited, or commented on that answer. In particular, it didn't notify me, I was just pasing through and noticed it. –  Gerry Myerson Aug 2 at 5:08
2  
Continued fractions with polynomial coefficients violate the Gauss-Kuzmin statistics (cf. page 33 in Shallit's survey), so they are not generic, and if we are to believe Khintchine's principle, no algebraic number of degree $> 2$ is expected to have polynomial continued fractions coefficients. So it would be extremely surprising (and very remarkable) if you could construct an example in this way. Anyway, the determination of the algebraic hypergeometric functions is a beautiful classical result (see en.wikipedia.org/wiki/Schwarz's_list) [continued.] –  Vesselin Dimitrov Aug 2 at 7:17
1  
@XL_at_China: No, this has not been proved for neither $\pi$ nor any algebraic number, it is only expected to be true. –  Vesselin Dimitrov Aug 3 at 18:43
1  
@XL_at_China: Roth's theorem provides a major evidence, as do some of the many (and far-reaching) analogies with results in complex analysis (Nevanlinna's theory of the distribution of values of meromorphic maps). Anyway, it is just a conjecture. It could very well not be true, especially after the examples over $\mathbb{Q}(i)$ to which I have referred above. Many things remain mysterious regarding algebraic numbers and transcendence questions. –  Vesselin Dimitrov Aug 3 at 20:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.