# Representing the continued fractions of non-square-root numbers ($\sqrt[a]{x}$ where $a>2$) NOT in simplest form

In that question, I asked why the SIMPLE continued fraction of square roots are periodic and why this doesn't work for cubic roots, 4th roots, etc. I received many substantial answers which answered the question and as a result, I selected the best answer (I felt) from the 4 answers given. However, I feel like I only gave this question one approach: in SIMPLEST form... $$\sqrt[n]{x} = a_1+ \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \cfrac{1}{a_5 + \cfrac{1}{a_6 + \cfrac{1}{a_7 + \cdots}}}}}}$$

The above continued fraction is in simplest form (it is always $\dfrac{1}{a+\cdots}$) However, this is not the only way to represent continued fractions.

I am going to use the following notation:

$\sqrt{5} = 2+ \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4 + \cdots}}}}$

$\sqrt5 = [2;\overline{4}]_1$

Where the subscript $_1$ represents the numerator of each term in the continued fraction

Similarly: $\sqrt7 = [2;\overline{1,1,1,4}]_1$

...etc

To showcase this "notation"... and to show it does work for square roots, look at the following:

Take the identity: $$\sqrt x = 1 + \frac{x-1}{1+\sqrt x }$$

Replacing $\sqrt x$ on the right hand side with $1 + \frac{x-1}{1+\sqrt x }$ gives us:

$\sqrt{x} = 1+ \cfrac{x-1}{2 + \cfrac{x-1}{2 + \cfrac{x-1}{2 + \cfrac{x-1}{2 + \cdots}}}}$

$\sqrt{x} = [1;\overline{2}]_{x-1}$

Ok... enough with square roots does this work with other roots? I am not sure... but there must be a way to write cubic roots, etc. without having to use $1$ in the numerator of each term in the continued fraction...

Now, $\pi$ isn't algebraic, but this is quite amazing:

$$\pi = 3+ \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \cfrac{7^2}{6 + \cdots}}}}$$ $$\pi = [3;\overline{6}]_{(2n-1)^2}$$

Tl;Dr?

Is there a way to represent continued fractions of non-square root numbers? Not in simplest form?

Kind Regards

Joshua Lochner

a) You can't write $\sqrt[3]{x}$ as $\sqrt[3]{x}= a + \frac{b + cx}{d + e \sqrt[3]{x}}$ with constants $a,b,c,d,e$ because then $\sqrt[3]{x}$ would not satisfy an algebraic equation of degree 3.

b) Starting from $a^3-b^3 =(a-b)(a^2+ab + b^2)$ you can write $\sqrt[3]{2}=1 + \frac{ \displaystyle1}{ \displaystyle 1+ \sqrt[3]{2}(1+\sqrt[3]{2})}$.

And you can of course iterate it, yielding in a next step (putting $w=\sqrt[3]{2}$):

$w =1 + \frac{ \displaystyle1}{ \displaystyle 1+ (1 + \frac{ \displaystyle1}{ \displaystyle 1+ w(1+w)})(1+(1 + \frac{ \displaystyle1}{ \displaystyle 1+ w(1+w)}))}$.

You will obtain not a contiuned fraction (which is a unbifurcated chain of nested fractions) but a tree-like fraction which bifurcates at each iteration thus yielding an infinite binary tree.

(Evaluating until a certain depth will be equivalent with calculating the recursion $w_0 =1; w_{n+1} =1 + \frac{ \displaystyle 1}{ \displaystyle 1+ w_n(1+w_n)}$).

• This reminds me of an article of Domingo Gomez "The Fifth Arithmetical Operation, Number Revolution" which has been online years ago. Perhaps it is still online anywhere or at the webarchive. There he developed the concept of the generalized fractal continued fraction, however having a tree in the denominator but also in the numerator. I can't say much about it, but might be interesting to the OP. Commented Dec 19, 2016 at 5:59
• A question that arises for me: I was from time to time trying to generalize the matrix-concept for continued fractions to 3x3-matrices for third roots, (and larger sized for algebraic numbers of higher orders), but didn't succeed so far. Might this ansatz of yours be a/the key to a 3x3-matrix expression? Commented Dec 19, 2016 at 6:03

Old question, but for completeness sake:

Yes, it can be done. Please look for "General Method for Extracting Roots" by Manny Sardina. He develops it in some detail. The formula he gives is somewhat involved to write down directly.

The core involves recasting $y^{\frac{m}{n}}$ as an almost integer root $(a^n + b)^\frac{m}{n}$ by finding the appropriate $a$ and $b$. Then the first term is $a^m$ and the first numerator is $bm$. Then the sequence of denominator terms are

$$na^{n-m},~ 2a^m,~ 3na^{n-m},~ 2a^m,\cdots, (2i+1)na^{n-m}, 2a^m, \cdots$$ and the sequence of numerators is also an interleaving of two separate indexed functions with a clear pattern.

$$b(n-m),~ b(n+m),~ b(2n-m),~ b(2n+m), \cdots$$.

In practice it's best probably to just apply the exponent first and then extract the root so that $m=1$ and there are no nasty negative numerators for the first few terms.