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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...

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closed as too broad by Andrés E. Caicedo, Alexey Ustinov, Stopple, Ryan Budney, David Hansen Nov 29 '15 at 5:44

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ "Pattern" seems like a vague concept to me. $\endgroup$ – Todd Trimble Nov 23 '15 at 14:19
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    $\begingroup$ 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. $\endgroup$ – Stanley Yao Xiao Nov 23 '15 at 14:19
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    $\begingroup$ @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. $\endgroup$ – john mangual Nov 23 '15 at 14:20
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    $\begingroup$ oeis.org/A002945 $\endgroup$ – Douglas Zare Nov 23 '15 at 14:29
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    $\begingroup$ 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 Nov 23 '15 at 15:18
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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 + , ...]
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  • $\begingroup$ 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 $2$. $\endgroup$ – Robert Israel Nov 24 '15 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 $\mathbb{Z}[i]$. $\endgroup$ – Kevin Casto Nov 24 '15 at 19:26
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    $\begingroup$ You are right. $\sqrt[3]{2}$ has a nice continued fraction when the coefficients can be in $\mathbb Z[\sqrt[3]{2}]$ ... $\endgroup$ – Gerald Edgar Nov 28 '15 at 20:11
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(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.

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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.

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