Golden number, diagonals of polygons and continued fractions

Question

It is well know that, for a pentagon with unit side, the diagonal $\delta$ is such that $$ \delta : 1=1:(\delta-1) $$ so that its length is the positive solution of the equation $x^2-x-1=0$.

i.e. the Golden Number $\delta=\Phi$, that, from this equation, can be written as the continued fraction $$ \Phi=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}} $$ All this is very nice and it is interesting to see if can be extended to polygons with $n>5$ sides. Obviously for $n>5$ we have two or more diagonals and, for the shorter diagonal $\delta_n$ and for $n$ odd, it is show (https://www.jstor.org/stable/2691048) that its length satisfies an algebraic equation of degree $(n-1)/2$.

E.G. for an heptagon the equation is $$x^3-x^2-2x+1=0$$ whose solutions are cubic irrationals that, as far as i know, cannot be represented with a continued fraction extracted from the equation. So my question is: there is some way to express this diagonal with a continued (but non periodic) fraction? . More in general: there is some general result about the possible representation of diagonals of regular polygons with continued fractions?

Answer 1

(Edited Oct 2024 to incorporate data from this new post.)

If we allow $q$-continued fractions, then there is other ways to express diagonals of $p$-gons for prime $p=4m+1$, with the pentagon simply being the simplest case.

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I. For p = 5

Let $q = -e^{-\pi\sqrt{5}},$ and $\phi = \dfrac{\sin\frac{2\pi}5}{\sin\frac{\pi}5} = \dfrac{1+\sqrt5}2$ where $\phi$ is $\frac{5-1}4=\color{blue}1$ diagonal of a regular $5$-gon with unit side length. Then,

$$\phi^{1/4} = \cfrac{2^{-1/4}\,(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}$$

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II. For p = 13

Let $q = -e^{-\pi\sqrt{13}},$ and $u = \dfrac{\sin\frac{3\pi}{13}}{\sin\frac{\pi}{13}}+\dfrac{\sin\frac{4\pi}{13}}{\sin\frac{\pi}{13}}-\dfrac{\sin\frac{5\pi}{13}}{\sin\frac{\pi}{13}}= \dfrac{1+\sqrt{13}}2$ where the three addends are $\frac{13-1}4=\color{blue}3$ diagonals of a regular $13$-gon with unit side length. Then,

$$ \left(u+1\right)^{1/4} = \cfrac{2^{-1/4}\,(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}\quad$$

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III. For p = 17

Let $q = -e^{-\pi\sqrt{17}},$ and $v = -\dfrac{\sin\frac{2\pi}{17}}{\sin\frac{\pi}{17}}+\dfrac{\sin\frac{5\pi}{17}}{\sin\frac{\pi}{17}}-\dfrac{\sin\frac{7\pi}{17}}{\sin\frac{\pi}{17}}+\dfrac{\sin\frac{8\pi}{17}}{\sin\frac{\pi}{17}}= \dfrac{1+\sqrt{17}}2$ where the four addends are $\frac{17-1}4=\color{blue}4$ diagonals of a regular $17$-gon with unit side length. Then,

$$ \left(\frac{\sqrt{v}+1}{\sqrt{v}-1} \right)^{1/4} = \cfrac{2^{-1/4}\,(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}\quad$$

and so on. But one has to deal with more diagonals and square roots (especially when dealing with the $17$th root of unity, like in my answer to an old post).