furliinotes

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

Proof | Hendecagon construction by Neusis for any complex number \(z\) : $$z + z^*\in\mathbb{R}$$ $$z z^*\in\mathbb{R}^{+}$$ $$\dfrac{\sqrt{\overset{\phantom{0}}{zz^*}}}{z^*}\in\mathbb{S}^{1}$$ $$z - z^*\in\Big\{ix\mid x\in\mathbb{R}\Big\}$$ let \(f(x)\) be \(11^{th}\) 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 $$f(x)^{11}=1$$ which factors into $$\Big(f(x) - 1\Big)\Big(f(x)^{10} + f(x)^9 + \cdots + f(x)^2 + f(x)^1 + 1\Big)=0$$ which we can obtain from (where \(f(x)\neq1\)  $$f(10x) + f(9x) + f(8x) + f(7x) + f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$ let $$p(x)=f(x) + f(4x) + f(5x) + f(9x)+f(3x)$$ we have  $$\begin{array}{lllll}p(x)&=p(4x)&=p(5x)&=p(9x)&=p(3x)\\ \\p(2x)&=p(8x)&=p(10x)&=p(7x)&=p(6x)\\ \\p(11x)&=5\end{array}$$ $$\Big(p(x)\Big)^*=p(2x)$$ $$p(x) + p(2x) = -1$$ which shows that \(p(x),p(2x)\) lies on line \(z+z^*=-1\) ...