Cube root of two $\sqrt[3]2$ continued fraction

I know there is a nice way of getting the continued fraction expansion of quadratic irrationals mainly because they recur after a point, and if they recur after a point they are quadratic irrationals. When constructing the expansion you can multiply by conjugates (kind of), e.g.

$\sqrt 3 =1+\sqrt 3 -1 = 1+\frac {1}{\frac {\sqrt 3 +1}{2}} $

Where you use $(\sqrt 3 - 1)(\sqrt 3 +1)=2$.

Are there identities that would help with the construction for $ \sqrt[3]{2} $?

One I thought was useful in the first step to get [1; 3,...] was

$ (\sqrt[3]{2}-1)( \sqrt[3]{4} + \sqrt[3]{2}+1 )=1$,

So you get:

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

Thanks for the help.

Solution 1:

Starting from the column vector $(1,0,0,-2)$, consider the following steps:

Step a) Repeat multiplication by the matrix $A$ $$A=\begin{bmatrix} 1&0&0&0\\ 3&1&0&0\\ 3&2&1&0\\ 1&1&1&1 \end{bmatrix}$$ while the coefficients of the resulting vector have different signs.

Step b) Reverse the coefficients of the vector, or equivalently multiply by $$B=\begin{bmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0 \end{bmatrix}$$

Then the number of times you multiply by $A$ in step a gives the partial quotients of continued fraction of $\sqrt[3]{2}$.

For, starting from $(1,0,0,-2)$, successive multiplication by $A$ gives: \begin{align} (1,0,0,-2) &\xrightarrow A\color{red}{(1,3,3,-1)}\\ &\xrightarrow A(1,6,12,6) \end{align} hence in step a we multiply by $A$ one time only, because $(1,6,12,6)$ have positive coefficients only, hence the first partial quotient is $1$: $$\sqrt[3]{2}=1+\cdots$$

Apply step b to $(1,3,3,-1)$ we get $(-1,3,3,1)$. Then applying step a to $(-1,3,3,1)$, successive multiplication by $A$ gives: \begin{align} (-1,3,3,1) &\xrightarrow A(-1,0,6,6)\\ &\xrightarrow A(-1,-3,3,11)\\ &\xrightarrow A\color{red}{(-1,-6,-6,10)}\\ &\xrightarrow A(-1,-9,-21,-3)\\ \end{align} hence the second partial quotient is $3$: $$\sqrt[3]{2}=1+\frac 1{3+}\cdots$$ and so on...

This algorithm holds for every algebraic number of third degree which is the only positive root of it minimal polynomial. For higher degree the matrix $A$ is enlarged as in the Tartaglia-Pascal triangle; for example for fourth degree: $$A=\begin{bmatrix} 1&0&0&0&0\\ 4&1&0&0&0\\ 6&3&1&0&0\\ 4&3&2&1&0\\ 1&1&1&1&1 \end{bmatrix}$$

For the intuition behind this algorithm. Then vector $(1,0,0,-2)$ corresponds to the polynomial $p(x)=x^3-2$. Multyplication by $A$ corresponds to $p(x)\mapsto p(x+1)$, while revesing in step b corresponds to $p(x)\mapsto x^3p(1/x)$. Finally Descartes's signs rule provide the stopping criterion in step a.