27
$\begingroup$

Is there any pattern to the continued fraction of $\sqrt[3]{2}$ ? Wolfram Alpha returns for cube root of 2:

$\sqrt[3]{2}=$ [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, ...]

So the answer is likely no. There certainly won't be any repeating pattern such as:

$$ 1+\sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots }}}$$

But I was hoping maybe for a pattern like we might find for the number $e = 2.718\dots$:

$$ e = [2; 1, \color{blue}{2}, 1, 1, \color{blue}{4}, 1, 1, \color{blue}{6}, 1, 1, \color{blue}{8}, 1, 1, \color{blue}{10}, 1, 1, \color{blue}{12}, 1, 1, \color{blue}{14}, 1, 1, \color{blue}{16}, 1, 1, 1, ...]$$

So in fact $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ has a nicer continued fraction pattern than $\sqrt[3]{2}$.

You can actually derive the continued fraction of $e$ (Henry Cohn) - but I haven't worked through it yet.


Proposal: Tree-Like Continued Fraction

If we try to derive a continued fraction for cube roots let's see how we get in trouble:

$$ \sqrt[3]{2} \approx 1 $$

That's really lousy guess but let's see how far we are off:

$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \tag{$\square$}$$

so let's backtrack we actually needed two pieces of information:

$$ \sqrt[3]{2} \approx 1 \hspace{0.25in} \textbf{and} \hspace{0.25in} \sqrt[3]{4} \approx 1 $$

The cube root of 4 is still not quite $2 = \sqrt[3]{8}$. Now let's try:

$$ \sqrt[3]{4} - 1 = \frac{3}{1 + \sqrt[3]{4} + \sqrt[3]{16}} = \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}} \tag{$\Delta$} $$

And then run all of this back into the equation we started from:

$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} = \cfrac{1}{1 + \left(1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \right) + \left( 1 + \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}}\right)}$$

Let's make it look a little bit cleaner but the same:

$$ \sqrt[3]{2} = 1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} = 1 + \cfrac{1}{3 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{3}{1 + 2\sqrt[3]{2}+ \sqrt[3]{4}}}$$

Using the $\square$ and $\Delta$ pattern we can get a regular infinite pattern this way.


Clarification

I am being repeatedly asked to clarify the question what do I mean by "pattern" the truth is I don't know. To this end I make two points:

  • the are transcedental numbers with clear patterns if their continued fraction such as $e$. We find that every third digit is an even number starting with $2$ and increasing by $+2$. Every other digit is $1$.

  • I have already proposed a "tree-like" generalized continued fraction that may have recursive properties similar to what repeated patterns havefor numberes like $\sqrt{n}$

  • This paper by Yann Bugeaud states continued fraction digits definitely cannot have any pattern:

Let $\alpha = [0; a_1, a_2,...]$ be an algebraic number of degree at least three. One of our criteria implies that the sequence of partial quotients $(a_n)_{n≥1}$ of $\alpha$ cannot be generated by a finite automaton, and that the complexity function of $(a_n)_{n≥1}$ cannot increase too slowly.

Personally it's hard for me to fathom a number that doesn't satisfy any pattern. There must be some pattern. My question is then what could we try instead?

Lastly, if the cubic case is this intractable, I might shift my attention to the quadratic case where more results are known...

$\endgroup$
13
  • 6
    $\begingroup$ "Pattern" seems like a vague concept to me. $\endgroup$
    – Todd Trimble
    Commented Nov 23, 2015 at 14:19
  • 9
    $\begingroup$ It is a folklore conjecture that all algebraic numbers of degree $d > 2$ are unexceptional, i.e. they look like a typical real number from Khintchine's theorem. $\endgroup$ Commented Nov 23, 2015 at 14:19
  • 3
    $\begingroup$ @ToddTrimble the question is deliberately vague. the continued fraction digits of $\sqrt[3]{2}$ can't repeat. so we need to find some other notion of pattern. $\endgroup$ Commented Nov 23, 2015 at 14:20
  • 5
    $\begingroup$ oeis.org/A002945 $\endgroup$ Commented Nov 23, 2015 at 14:29
  • 4
    $\begingroup$ So then, I guess you are saying that doesn't qualify as a 'yes' in your view. Which leaves one still to wonder what might plausibly qualify... (I don't mean to be argumentative; I'm honestly unsure what the parameters of the problem are). $\endgroup$
    – Todd Trimble
    Commented Nov 23, 2015 at 15:18

3 Answers 3

15
$\begingroup$

To my knowledge, much more general question is open:

OPEN DIOPHANTINE PROBLEMS p. 15

Essentially nothing is known about the continued fraction expansion of a real algebraic number of degree ≥ 3; one does not know the answer to any of the following two questions.

Question 2.9. Does there exist a real algebraic number of degree ≥ 3 with bounded partial quotients?

Question 2.10 . Does there exist a real algebraic number of degree ≥ 3 with unbounded partial quotients?

According to Wolfram Alpha there might be chance for the Hurwitz expansion of $\sqrt{i}$ which starts

 [1 + i; -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + 2i, 2 + 2i, -2 + , ...]
$\endgroup$
3
  • 1
    $\begingroup$ You get algebraic numbers of degree $2$ from periodic simple continued fractions, where the elements are positive integers. Here they are not integers, so it's not surprising that the result is not of degree $2$. $\endgroup$ Commented Nov 24, 2015 at 1:15
  • $\begingroup$ Yeah, I would think here, the point is that it's degree two over $\mathbb{Z}[i]$, so you should get periodic partial quotients when your coefficients can be in $\mathbb{Z}[i]$. $\endgroup$ Commented Nov 24, 2015 at 19:26
  • 2
    $\begingroup$ You are right. $\sqrt[3]{2}$ has a nice continued fraction when the coefficients can be in $\mathbb Z[\sqrt[3]{2}]$ ... $\endgroup$ Commented Nov 28, 2015 at 20:11
10
$\begingroup$

(I have a comment regarding your Tree-Like Continued Fraction generalization, except I only have enough reputation to write answers.)

Given algebraic $x_0$, an instance of your generalization can be found by searching for some function who has a fixed point at $x_0$ and is the inverse of a polynomial.

Suppose a polynomial $p(x)$ has a root at $x_0$. Following your example, if $x_0 = \sqrt[3]{2}-1$, then we can pick the minimal polynomial $p(x) = x^3 + 3x^2 + 3x -1$. Set $p(x) = 0$ to write $\frac{1}{x} = x^2 + 3x + 3$. Thus, $x_0$ is a fixed point of $f(x)$:

$$ f(x) = \frac{1}{x^2+3x+3} $$

The Tree-like continued fraction can then be directly read off of $f(x)$. Going back to the general case: if $p(x)=\sum_{i=0}^{n}a_ix^i$, then $$ f(x) = \frac{-a_0}{\sum_{i=1}^{n}a_ix^{i-1}} $$

We make no restriction on whether $|f'(x_0)|<1$.

You have thought of this implicitly already, but I feel that this is a more explicit way of writing down what are you looking for.

$\endgroup$
0
6
$\begingroup$

This is the Hermite Problem: https://en.wikipedia.org/wiki/Hermite%27s_problem

The Jacobi-Perron algorithm attempts to solve this.

Mittal and Gupta have defined Bifurcating Continued Fractions- http://arxiv.org/abs/math/0002227 http://arxiv.org/abs/math/0008060

Specifically, they say the pair $2^{1/3}$ and $2^{2/3}$ has the expansion: $1,\overline{1,2} $ and $ \overline{1,0}$

Lehmer talks of Ternary Continued Fractions - https://oeis.org/A000962/a000962.pdf

How it all fits together I have no idea.

$\endgroup$
1

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