Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$

Question

Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$

Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.

For $x=11$, it's $[363; \mathbf{11}, 544, ...]$. The pattern continues (I have only checked up to $21$, but I assume that it's true) that the second term in the continued fraction expansion is $x$.

I have two questions:

1. Why does this pattern occur and how do we prove that it occurs?
2. Is there a similar pattern with later terms?

I haven't found an answer to 2, and I've had no luck as of yet for 1. All I have done so far is simplify it to $x\sqrt[3]{x^3+1}+\sqrt[3]{(x^3+1)^2}+x^2$.

Thanks in advance.

Answer 1

Let $$\sqrt[3]{x^3+1} - x = a$$ Then $$a^3 + 3a^2x + 3ax^2 = 1$$ This shows that $a < \frac{1}{3x^2}$. Now we have everything we need to work out the continued fraction:

$$\begin{align*} \frac{1}{a} &= 3x^2 + a(3x + a)\\ &=3x^2 + \frac{1}{\frac{1}{a}\frac{1}{3x+a}}\\ &=3x^2 + \frac{1}{\frac{3x^2 + 3ax + a^2}{3x+a}}\\ &=3x^2 + \frac{1}{x + a\frac{2x + a}{3x+a}}\\ \end{align*}$$

Using the upper bound $a<\frac{1}{3x^2}$, you can show that $a\frac{2x + a}{3x+a} < 1$, which proves the relationship you observed.

You can continue this process to find more continued fraction terms. $$a\frac{2x + a}{3x+a} = \frac{1}{\frac{1}{a}\frac{3x+a}{2x+a}}$$ Again, replace $\frac{1}{a} = 3x^2 + 3ax + a^2$ and separate into term independent of $a$ and term proportionate to $a$

Answer 2

Partial answer: the provided numerical examples strongly suggest that the continued fraction expansion would be of the form $[3x^2,x,\ldots]$. So we only need to show that for $x>0$ we have $$ 3x^2+\frac{1}{x+1} < \frac{1}{\sqrt[3]{x^3+1}-x} < 3x^2+\frac{1}{x} $$ A few steps of algebraic manipulations on the right-side inequality give us $$ \begin{array}{rl} \dfrac{1}{\sqrt[3]{x^3+1}-x} &< \ 3x^2+\dfrac{1}{x} \\ x &< \ \left(\sqrt[3]{x^3+1}-x\right)\left(3x^3+1 \right) \\ \dfrac{3x^4+2x}{3x^3+1} &< \ \sqrt[3]{x^3+1} \\ \left(\dfrac{3x^4+2x}{3x^3+1}\right)^3 &< \ x^3+1 \\ x^3+1-\dfrac{2}{3(3x^3+1)^2}-\dfrac{1}{3(3x^3+1)^3} &< \ x^3+1 \\ \end{array}$$ which is evidently true (last step courtesy of WolframAlpha). A completely similar derivation shows that the left-side inequality also holds, and we're done with (1).

I'll admit, I have no idea how to approach (2) though.