Neusis IV

 



Tridecagon Neusis

https://drive.google.com/file/d/1pg8vfu3726R26nc6E7f-i3D4VIgXgH1T/view?usp=drive_link



Comments

Post a Comment

Popular posts from this blog

Geometry

p1 p2 p3 Unit Vector (grade 1) $$e_1\quad,e_2\quad,e_3\quad,e_4\quad\dots$$ depending on the dimension used Unit Bivector (grade 2) 2D : $$\quad e_1e_2$$ 3D : $$\quad e_1e_2,\quad e_1e_3,\quad e_2e_3$$ 4D : $$\quad e_1e_2,\quad e_1e_3,\quad e_1e_4,\quad e_2e_3,\quad e_2e_4,\quad e_3e_4$$ etc. depending on dimension used Unit Trivector (grade 3) 3D : $$\quad e_1e_2e_3$$ 4D : $$\quad e_1e_2e_3,\quad e_1e_2e_4,\quad e_1e_3e_4,\quad e_2e_3e_4$$ etc. depending on dimension used 2D Setup : $$\overset{\text{scalar}}{{1}}\quad\quad\overset{\text{vector}}{\overbrace{e_1\quad\quad e_2}}\quad\quad \overset{\text{bivector}}{\overbrace{e_1e_2}}$$ $$a=a_1e_1 + a_2e_2$$ $$b=b_1e_1 + b_2e_2$$ Addition $$\begin{array}{ll}a+b&=(a_1+b_1)e_1+(a_2+b_2)e_2\end{array}$$ Multiplication (Product) Warning : not commutative $$\begin{array}{ll}ab&=(a_1b_1 + a_2b_2) + (a_1b_2 - a_2b_1)e_1e_2\end{array}$$ $$\begin{array}{ll}ab &=(a_1e_1 + a_2e_2)(b_1e_1 + b_2e_2) \\ \\ &=a_1e_1(b_1e_1 + b_2e_2) + a_...
Read more

Neusis II

  Proof | Heptagon Construction by Neusis let \(f(x)\) be \(7^{th}\) 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 $$f(x)^7=1$$ which factors into $$\Big(f(x) - 1\Big)\Big(f(x)^6 + f(x)^5 + f(x)^4 + f(x)^3 + f(x)^2 + f(x)^1 + 1\Big)=0$$ which we can obtain from (where \(f(x)\neq1\)) $$f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$ let $$p(x)=f(x) + f(2x) + f(4x)$$ we can find that $$\begin{array}{llll}p(x)&=p(2x)&=p(4x)\\ \\p(3x)&=p(6x)&=p(5x)\\ \\p(7x)&=3\end{array}$$ $$\begin{array}{ll}\Big(p(x)\Big)^*&=f(x)^*+f(2x)^*+f(4x)^*\\ \\&=f(6x)+f(5x)+f(3x)\\ \\&=p(3x)\end{array}$$ $$p(x) + p(3x) = -1$$ which shows that \(p(x), p(3x)\) lies on line \(z+z^*=-1\), and $$\begin{array}{llllllll}p(x)\cdot p(3x) &=& &f(3x) &f(6x) &(5x)&=p(x)+p(3x)+3&=2\\ \\ &&f(x)&f(4x)&1&f(6x)\\ \\ &&f(2x)&f(5x)&f(x...
Read more

Neusis III

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 \(z+z^*=-1\) ...
Read more