Neusis IV

 



Tridecagon Neusis


let \(f(x)\) be \(13^{th}\) roots of unity 

$$f^x=f(x)$$

for all \(x\in\mathbb{Z}\), \(f(x)\) forms a regular heptagon on unit circle in complex plane and satisfies

$$f(x)^{13}=1$$

which factors into

$$\Big(f(x) - 1\Big)\Big(f(x)^{12} + f(x)^{11} + \cdots + f(x)^2 + f(x)^1 + 1\Big)=0$$

which we can obtain from (where \(f(x)\neq1\) 

$$f(12x) + f(11x) + \cdots + f(x) = -1$$

let

$$p(x)=f(x) + f(3x) + f(9x) $$

we have 

$$\begin{array}{lllll}p(x)&=p(3x)&=p(9x)\\ \\p(2x)&=p(6x)&=p(5x)\\ \\p(4x)&=p(12x)&=p(10x)\\ \\p(8x)&=p(11x)&=p(7x)\\ \\p(13x)&=3\end{array}$$

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