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All square roots can be expressed as a regular Continued Fraction. For example: $$\newcommand{\contfrac}{\raise{-0.5ex}\mathop{\Large\mathrm{K}}}\sqrt{2}+1=2+\contfrac_{n = 0}^\infty \frac{1}{2}$$

It is easy to see that comparable regular Continued Fractions for higher order roots are impossible. But at least some higher roots can actually be expressed as Continued Fractions, following simple patterns. For example:

$$4\sqrt[3]{4}-1=7+\contfrac_{n = 0}^\infty \frac{-4\cdot\frac{3n+2}{n+1}}{7} =7+\cfrac{-4\cdot\cfrac{2}{1}}{7+\cfrac{-4\cdot\cfrac{5}{2}}{7+\cfrac{-4\cdot\cfrac{8}{3}}{7+\cfrac{-4\cdot\cfrac{11}{4}}{7+\ddots}}}}$$ $$6\sqrt[5]{6}-1=11+\contfrac_{n = 0}^\infty \frac{-6\cdot\frac{5n+4}{n+1}}{11}$$

$$(1+p)\sqrt[p]{p+1}-1=1+2p+\contfrac_{n = 0}^\infty \frac{-(1+p)\cdot\frac{(n+1)p-1}{n+1}}{1+2p}$$ I have proven these CF's by applying Euler's Differential Method for Continued Fractions.

The question remains: is there a way to express an arbitrary higher order root like $\sqrt[3]{5}$ as a "patterned" Continued Fraction?

metamorphy
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Paul vdVeen
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    Every real number is expressible as a continued fraction, so... yes? Whether they are "nice" continued fractions is another matter. – Jakob Streipel Nov 07 '21 at 23:57
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    Per this answer, very little is known about the continued fractions of algebraic numbers of degree $d \geq 3$. – Jakob Streipel Nov 08 '21 at 00:16
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    The above comments apply to simple continued fractions, that is, continued fractions where the numerators are all $1$, but OP is clearly interested in general continued fractions, with arbitrary numerators. . – Gerry Myerson Nov 08 '21 at 02:24
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    Several examples (of cube roots of integers and suchlike, with patterned generalized continued fractions) are given at https://en.wikipedia.org/wiki/Generalized_continued_fraction#Roots_of_positive_numbers (and it is indicated that such exist for all $\root n\of m$). – Gerry Myerson Nov 08 '21 at 03:05
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    This is very interesting! In General Method for Extracting Roots Manny Sardina explains this method. Indeed, this proofs that every higher root can be expressed in a "patterned" Continued Fraction. Which is a result, not widely known, I guess. The debate if these "Patterned" Continued Fractions show aesthetic qualities is a different matter. – Paul vdVeen Nov 08 '21 at 16:27
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    Let me encourage you to work up a summary of the Wikipedia essay and/or the Sardina essay to post here as an answer. – Gerry Myerson Nov 09 '21 at 06:42
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    An application of some historical interest is the approximation of a musical scale with "equal temperament", because of the underlying twelfth root of two. It is mentioned in this MathOverflow Question in connection with an algorithm by J.B. Rosser. – hardmath Nov 19 '21 at 23:03

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You can definitely express higher order roots as Continued Fractions. However my little experience with the General Continued Fractions proved most of them to be trash because almost always they converge to the target with a snail's pace.

For higher order roots as Continued Fractions, using the Simple Continued Fraction approach is the most efficient and also the simplest way. It's like every discovered coefficient takes up at least a decimal place in the result. As an example, when I test $2190^\frac{1}{7}$ in my code, only 50 SCF coefficients converge to 72 correct decimal places.

However it is kind of a little convoluted and also a long topic to explain here since they involve calculations like translating, inverting and finding the positive root's lower bound of $n^{th}$ order polynomials... $n$ being the higher order that you mention.

If you would like to know more then perhaps an earlier post of mine may be of your interest.

Redu
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