# Cube root of two $\sqrt2$ 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 $\sqrt{2}$?

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

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

So you get:

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

Thanks for the help.

• As far as I am aware very little is known about the continued fraction of any algebraic number with degree greater than $2$. However your computations can be simplified by working with polynomials, starting with the fact that $\root3\of2$ is a root of $x^3-2$. I have an exposition using this very example here starting at page 80. – David May 3 '14 at 13:21
• @David Yeh, that was mentioned in our number theory class. Thanks for the link to the ps. It's really helpful (+1). Its pretty much an answer, that polynomial method and the identity that comes after it, for the more complicated way. Thanks for the post. – snulty May 3 '14 at 17:43
• There are rules to derive CF for higher roots. For more info look at this Wikipedia page: en.wikipedia.org/wiki/Generalized_continued_fraction – Stefan Gruenwald Sep 9 '17 at 3:34

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 $$\sqrt{2}$$.

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 $$1$$: $$\sqrt{2}=1+\cdots$$

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 $$3$$: $$\sqrt{2}=1+\frac 1{3+}\cdots$$ 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.

• So just to see if I understand a bit why this works. $(1,0,0,-2)$ is representing $1x^3+0x^2+0x-2$. Then $A$ and $B$ are some kind of linear maps on this polynomial ring/vector space. $A$ looks like $p\mapsto p+p’+p’’/2!+p’’’/3!$, which is a kind of Taylor expansion, while B I suppose can be seen like the map $p(x)->x^3*p(1/x)$. I have no idea why this works though. Is there some intuition for it? It seems really cool. – snulty Oct 26 '19 at 10:37
• @snulty: I added explanation in answer, thank'you – Fabio Lucchini Oct 26 '19 at 11:09
• @FabioLucchini Can this method be used to bound the continued fraction coefficients of $\sqrt{2}$? – Franklin Pezzuti Dyer Nov 10 '19 at 18:49
• @franklin I am not sure the ciefficients are bounded in the first place. – Oscar Lanzi Nov 10 '19 at 21:57
• @OscarLanzi No, they’re probably not bounded, but I meant bounding them using something like big-O notation (showing that they have sub-quadratic growth, or sub cubic growth, or sub-exponential growth, or something like that). – Franklin Pezzuti Dyer Nov 10 '19 at 22:28