The following question/topic is an extension of a question I asked a while ago: Click here
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 $w_0 =1; w_{n+1} =1 + \frac{ \displaystyle 1}{ \displaystyle 1+ w_n(1+w_n)}$).