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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?

Emilio Novati
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    "there is some way to express this diagonal with a continued (but non periodic) fraction?" Every real number has a way to do it. Some numbers have clearer and simpler patterns than others. – Arthur May 14 '23 at 11:02
  • Yes. I know how find the continued fraction if I can write out the number in some other way, but here I'm searching the converse: a fraction from which I can write the digits of a number that is defined by an equation that I dont want or cannot solve algebrically. – Emilio Novati May 14 '23 at 13:50
  • See E. Bombieri and A. J. van der Poorten, Continued fractions of algebraic numbers, in Computational Algebra and Number Theory, Kluwer, 1995, pp. 137–152. Also, P. Shiu, ‘Computation of continued fractions without input values’, Math. Comp. 64 (1995), 1307–1317. – Gerry Myerson Aug 27 '23 at 08:39

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If you allow $q$-continued fractions, then there is another way to express the golden ratio $\Phi$. Let $q = -e^{-\pi\sqrt{5}},$ and $\Phi = 1+2\cos\big(\frac{2\pi}5\big).$ Then,

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

Since $p = 5$ involves a constructible polygon, the next prime is $p=17$.

Let $q = -e^{-\pi\sqrt{17}},$ and $x = 2\cos\big(\frac{2\pi}{17}\big)+2\cos\big(\frac{8\pi}{17}\big).$ Then,

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

Or alternatively,

$$\small{\left(\frac{(1+\sqrt{17})^{3/2}}{2\sqrt2}+\frac{(5x-3)\sqrt2+(x-1)\sqrt{34}}{\sqrt{17-\sqrt{17}}}\right)^{1/4} = \cfrac{(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}}$$

We could express the continued fraction purely in terms of $2\cos\big(\frac{2\pi}{17}\big)$, but that would entail MORE nested square roots (which is expected when dealing with the $17$th root of unity, like in my answer to this post).

Tito Piezas III
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