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For $n\geq1$, the largest solution to this lovely equation is a local extremum on a function related to the Fibonacci sequence:

$$\sum_{k=1}^{n} k{(-1)^{k}} \cdot \frac{\sin(\frac{k\pi}{x} )}{3+2\cos(\frac{k\pi}{x} )} = 0$$

For $n=1$, the largest solution is $1$. For $n=2$, the largest solution is:

$$\frac{\pi}{2\arctan(\frac{\sqrt{4\sqrt{10}-5}}{3})} $$

Can it be proven that for any integer $n\geq2$, the solution to the above equation can be expressed as

$$\frac{\pi}{2\arctan(A)} $$

where $A$ is algebraic? If so, what are some minimal polynomials for $A$ when $n\geq3$?

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    $\begingroup$ Beauty sure is in the eye of the beholder. $\endgroup$ Mar 12, 2021 at 22:52
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    $\begingroup$ where could I read about this equation and its relation to the fibonacci sequence? $\endgroup$
    – Dabed
    Mar 13, 2021 at 2:04

1 Answer 1

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Yes, you can write it in terms of an algebraic number as you desire. Both $\sin\frac{k\pi}{x}$ and $\cos\frac{k\pi}{x}$ can be written as polynomials in $\cos\frac{\pi}{x},\sin\frac{\pi}{x}$, which can in turn be written as rational functions in $\tan\frac{\pi}{2x}$. It follows that for any solution $x$ of your equation, $A=\tan\frac{\pi}{2x}$ will be algebraic. But since $x=\frac{\pi}{2\arctan A}$, this is your desired result.

From this method you can easily figure out a polynomial of which $A$ is a root of, but I doubt there is an easy way to find its minimal polynomial.

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