The Physics Language – Page 16

August 18, 2020March 15, 2023

202008181129 Homework 1 (Q1)

Solve for $z\in \mathbb{C}$ in the following equations:

(a) $z^4+z^3+z^2+z+1=0$ ,

(b) $3z^3+29z^2+497z-169=0$ .

Attempts.

(a) Take notice that $z\neq 1$ . Try and see having both sides of the equation multiplied by $(1-z)$ ,

$\begin{aligned} 0 & = (1-z)(z^4+z^3+z^2+z+1) \\ 0 & = 1-z^5 \\ z^5 & = 1 = 1(1+0\,\mathrm{i}) \\ z^5 & = e^{\mathrm{i}(2n\pi)}\qquad \qquad \textrm{where }n=0,1,2,3,4. \\ \end{aligned}$

$\therefore z=e^{\mathrm{i}(\frac{2n\pi}{5})}$ .

( $n=0$ is rejected for $z\neq 1=e^{\mathrm{i}(\frac{2(0)\pi}{5})}$ .)

In polar expression $z=e^{\mathrm{i\frac{2\pi}{5}}},\, e^{\mathrm{i\frac{4\pi}{5}}},\, e^{\mathrm{i\frac{6\pi}{5}}},\, e^{\mathrm{i\frac{8\pi}{5}}}$ .

In trigonometric expression

$z=\mathrm{cis}(\frac{2n\pi}{5})=\mathrm{cis}(\frac{2\pi}{5}),\, \mathrm{cis}(\frac{4\pi}{5}),\, \mathrm{cis}(\frac{6\pi}{5}),\, \mathrm{cis}(\frac{8\pi}{5})$

I.e.,

$\begin{aligned} z^1 & = \cos 72^\circ +\mathrm{i}\sin 72^\circ = \mathrm{cis\,} 72^\circ \\ z^2 & = \cos 144^\circ +\mathrm{i}\sin 144^\circ = \mathrm{cis\,} 144^\circ \\ z^3 & = \cos 216^\circ +\mathrm{i}\sin 216^\circ = \mathrm{cis\,} 216^\circ \\ z^4 & = \cos 288^\circ +\mathrm{i}\sin 288^\circ = \mathrm{cis\,} 288^\circ \\ \end{aligned}$

(b)Let $f(z)=3z^3+29z^2+497z-169=0$ . Then $(3z-1)$ is a factor, because $\frac{1}{3}$ is a zero (i.e., $f(\frac{1}{3}) = \frac{1}{9}+\frac{29}{9}+\frac{497}{3}-169=0$ ).

$\begin{aligned} f(z) & =(3z-1)(z^2+10z+169) \\ & = (3z-1)g(z) \\ \end{aligned}$

When $g(z)=0$ , $z=\displaystyle{\frac{-10\pm \sqrt{100-4(169)}}{2}}=-5\pm 12\,\mathrm{i}$ .

The solution to $f(z)=0$ gives $z_1=\frac{1}{3}$ , $z_2=-5+12\,\mathrm{i}$ , and $z_3=-5-12\,\mathrm{i}$ .

August 15, 2020July 25, 2022

202008151451 Arithmetic 0

Multiplication Table of the Set of Numbers One to Ten

$\begin{array}{|c|c|c|c|c|} \hline 1\quad &\quad 2\quad &\quad 3\quad &\quad 4\quad &\quad 5 \\ \hline\hline 1\times 1=1 & 2\times 1=2 & 3\times 1=3 & 4\times 1=4 & 5\times 1=5 \\ 1\times 2=2 & 2\times 2=4 & 3\times 2=6 & 4\times 2=8 & 5\times 2=10 \\ 1\times 3=3 & 2\times 3=6 & 3\times 3=9 & 4\times 3=12 & 5\times 3=15 \\ 1\times 4=4 & 2\times 4=8 & 3\times 4=12 & 4\times 4=16 & 5\times 4=20 \\ 1\times 5=5 & 2\times 5=10 & 3\times 5=15 & 4\times 5=20 & 5\times 5=25 \\ 1\times 6=6 & 2\times 6=12 & 3\times 6=18 & 4\times 6=24 & 5\times 6=30 \\ 1\times 7=7 & 2\times 7=14 & 3\times 7=21 & 4\times 7=28 & 5\times 7=35 \\ 1\times 8=8 & 2\times 8=16 & 3\times 8=24 & 4\times 8=32 & 5\times 8=40 \\ 1\times 9=9 & 2\times 9=18 & 3\times 9=27 & 4\times 9=36 & 5\times 9=45 \\ 1\times 10=10 & 2\times 10=20 & 3\times 10=30 & 4\times 10=40 & 5\times 10=50 \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|c|} \hline 6&7&8&9&10 \\ \hline\hline 6\times 1=6 & 7\times 1=7 & 8\times 1=8 & 9\times 1=9 & 10\times 1=10\\ 6\times 2=12 & 7\times 2=14 & 8\times 2=16 & 9\times 2=18 & 10\times 2=20\\ 6\times 3=18 & 7\times 3=21 & 8\times 3=24 & 9\times 3=27 & 10\times 3=30\\ 6\times 4=24 & 7\times 4=28 & 8\times 4=32 & 9\times 4=36 & 10\times 4=40 \\ 6\times 5=30 & 7\times 5=35 & 8\times 5=40 & 9\times 5=45 & 10\times 5=50 \\ 6\times 6=36 & 7\times 6=42 & 8\times 6=48 & 9\times 6=54 & 10\times 6=60 \\ 6\times 7=42 & 7\times 7=49 & 8\times 7=56 & 9\times 7=63 & 10\times 7=70\\ 6\times 8=48 & 7\times 8=56 & 8\times 8=64 & 9\times 8=72 & 10\times 8=80\\ 6\times 9=54 & 7\times 9=63 & 8\times 9=72 & 9\times 9=81 & 10\times 9=90\\ 6\times 10=60 & 7\times 10=70 & 8\times 10=80 & 9\times 10=90 & 10\times 10=100\\ \hline \end{array}$

Figures One to Ten

$\begin{array}{cc} \hline \hline \\\\ \bigcirc \qquad \qquad \qquad \hfill & 1 \\\\ \hline \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & 9\\\\ \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+8\\\\ \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+7 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =3+6\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =4+5\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =5+4\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =6+3 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =7+2\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =8+1\\\\ \hline \hline \end{array}$

$\begin{array}{cc} \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill & 2 \\\\ \bigcirc\qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc \hfill & =1+1\\\\ \hline \\\\ \noindent \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \hfill & 8\\\\ \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+7 \\\\ \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+6\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =3+5\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =4+4 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =5+3\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =6+2\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =7+1\\\\ \hline\hline \\ \end{array}$

$\begin{array}{cc} \noindent \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \hfill & 3\\\\ \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =1+2 \\\\ \bigcirc\enspace \bigcirc\qquad\qquad \qquad \qquad \qquad \qquad\qquad \qquad \hfill \bigcirc\hfill & =2+1 \\\\ \hline \\\\ \noindent \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill & 7\\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+6\\\\ \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+5\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =3+4 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =4+3 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =5+2 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =6+1\\\\ \hline \hline \\\\ \end{array}$

$\begin{array}{cc} \noindent \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\hfill & 4\\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+3\\\\ \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =2+2 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \hfill \bigcirc\hfill & =3+1\\\\ \hline \\\\ \noindent \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill & 6\\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+5\\\\ \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+4\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =3+3\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =4+2\\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =5+1\\\\ \hline \hline \end{array}$

$\begin{array}{cc} \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill & 5 \\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+4 \\\\ \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+3 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =3+2 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =4+1 \\\\ \hline \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill & 5 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill & =4+1 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill & =3+2 \\\\ \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =2+3 \\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill & =1+4 \\\\ \hline \hline \end{array}$

$\begin{array}{cc} \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill &10 \\\\ \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=1+9 \\\\ \bigcirc\enspace \bigcirc\qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=2+8 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=3+7 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=4+6 \end{array}$

$\begin{array}{cc} \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=5+5 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=6+4 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\enspace \bigcirc\hfill &=7+3 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\enspace \bigcirc\hfill &=8+2 \\\\ \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc\enspace \bigcirc \qquad \qquad \qquad \qquad \qquad \qquad \hfill \bigcirc\hfill &=9+1 \\\\ \hline \hline \end{array}$

Ex. 01

$\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline + & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \hline 1 & & & & & & & & & & \\ \hline 2 & & & & & & & & & & \\ \hline 3 & & & & & & & & & & \\ \hline 4 & & & & & & & & & \bullet & \\ \hline 5 & & & & & & & & & & \\ \hline 6 & & & & & & & & & & \\ \hline 7 & & & & & & & & & & \\ \hline 8 & & & & & & & & & & \\ \hline 9 & & & & & & & & & & \\ \hline 10 & & & & & & & & & & \\ \hline \end{array}$

$4+9=\,\, ?$

$\begin{array}{ccccccccccc} & & & & & & \bigcirc &\bigcirc &\bigcirc &\bigcirc & 4 \\\\ + & \bigcirc &\bigcirc &\bigcirc &\bigcirc &\bigcirc &\bigcirc &\bigcirc &\bigcirc & \bigcirc & 9 \\\\ \hline \\ &&&&&&&&&&\\ \hline \hline \end{array}$

$4+9=4+(6+3)=4+6+3=\,\, ?$

$\begin{array}{cccccccccccc} &&&&&&&\bigcirc&\bigcirc&\bigcirc&\bigcirc& 4 \\\\ + & & \bigotimes & \bigotimes & \bigotimes & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & 6 \\\\ \hline \\ & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & 10 \\\\ \hline \hline \end{array}$

$(4+6)+3=10+3=13$

Ex. 02

$\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline + & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \hline 1 & & & & & & & & & & \\ \hline 2 & & & & & & & & & & \\ \hline 3 & & & & & & & & & & \\ \hline 4 & & & & & & & & & & \\ \hline 5 & & & & & & & & & & \\ \hline 6 & & & & & & & & & & \\ \hline 7 & & & & & & & & & & \\ \hline 8 & & & & & \bullet & & & & & \\ \hline 9 & & & & & & & & & & \\ \hline 10 & & & & & & & & & & \\ \hline \end{array}$

$8+5=\,\, ?$

$\begin{array}{ccccccccccc} & & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc &\bigcirc & \bigcirc &\bigcirc & 8 \\\\ + & & & & & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & 5 \\\\ \hline \\ &&&&&&&&&&\\ \hline \hline \end{array}$

$8+5=8+(2+3)=8+2+3=\,\, ?$

$\begin{array}{cccccccccccc} &&& \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & 8 \\\\ + &&&&&& \bigotimes & \bigotimes & \bigotimes & \bigcirc & \bigcirc & 2 \\\\ \hline \\ & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc &10 \\\\ \hline\hline \end{array}$

$(8+2)+3=10+3=13$

April 24, 2020January 12, 2023

202004241907 Homework 1 (Q2)

(a) State the Mean Value Theorem and the Taylor’s Theorem.

(b) Find an $O(h^4)$ and an $O(h^5)$ approximation to $\cos h$ . Compare those approximate values to the actual value $0.9950042$ when $h=0.1$ . Correct answers in this question to seven decimal places.

Solution.

(a)

Theorem. (Mean-Value Theorem) Suppose $f\in C[a,b]$ , and $f'(x)$ exists on $(a,b)$ . For every $x\in [a,b]$ and some $x_0\in [a,b]$ ,

$f(x)=f(x_0)+f'(\eta (x))(x-x_0)$

where $\eta (x)$ is between $x_0$ and $x$ .

Theorem. (Taylor’s Theorem) Suppose $f\in C^n[a,b]$ , and $f^{(n+1)}$ exists on $(a,b)$ . For every $x\in (a,b)$ and some $x_0\in [a,b]$ , there exists a number $\eta (x)$ between $x_0$ and $x$ with $f(x)=P_n(x)+R_n(x)$ , where

$P_n(x)=f(x_0)+f'(x_0)(x-x_0)+ \displaystyle{\frac{f''(x_0)}{2!}}(x-x_0)^2+\cdots + \displaystyle{\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n}$ , and

$R_n(x)=\displaystyle{\frac{f^{(n+1)}(\eta (x))}{(n+1)!}}(x-x_0)^{n+1}$ .

(b) Let $f(x)=\cos x$ . Then we have $f'(x)=-\sin x$ , $f''(x)=-\cos x$ , $f'''(x)=\sin x$ , $f^{(4)}(x)=\cos x$ , $f^{(5)}(x)=-\sin x$ , etc. By Taylor expansion of $\cos x$ at $x_0$ , we have

$\cos x=\cos x_0+(-\sin x_0)(x-x_0)+\displaystyle{\frac{-\cos x_0}{2!}}(x-x_0)^2+\displaystyle{\frac{\sin x_0}{3!}}(x-x_0)^3+\displaystyle{\frac{\cos x_0}{4!}}(x-x_0)^4+\displaystyle{\frac{-\sin x_0}{5!}}(x-x_0)^5+\cdots$

Now letting $x_0=0$ , we obtain further that

$\cos x=1-\displaystyle{\frac{x^2}{2!}}+\displaystyle{\frac{x^4}{4!}}+\cdots =\displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}}$

Thus, the $O(h^4)$ approximation to $\cos h$ is in the form of

$\cos h=1-\displaystyle{\frac{h^2}{2!}}+O(h^4)$

where $O(h^4)$ is the higher-order terms $\displaystyle{\sum_{n=2}^\infty \frac{(-1)^n}{(2n)!}h^{2n}}$ .

And the $O(h^5)$ approximation to $\cos h$ is in the form of

$\cos h=1-\displaystyle{\frac{h^2}{2!}}+\displaystyle{\frac{h^4}{4!}}+O(h^5)$

where $O(h^5)$ is the higher-order terms $\displaystyle{\sum_{n=3}^\infty \frac{(-1)^n}{(2n)!}h^{2n}}$ .

Then, when $h=0.1$ , $\cos h$ in the $O(h^4)$ approximation would be

$\cos 0.1=1-\displaystyle{\frac{0.1^2}{2!}}=0.995$ ;

whereas in the $O(h^5)$ approximation,

$\cos 0.1=1-\displaystyle{\frac{0.1^2}{2!}}+\displaystyle{\frac{0.1^4}{4!}}=0.995004166\approx 0.9950042\enspace \textrm{(corr. to 7 d.p.)}$ .

Given the actual value $\cos h=0.9950042$ , the $O(h^4)$ approximation has an absolute error of $|0.9950042-0.995|=0.0000042$ and a relative error of $\displaystyle{\frac{|0.9950042-0.995|}{|0.9950042|}}=4.2210877\times 10^{-6}$ . And the $O(h^5)$ approximation has an absolute error of $|0.9950042-0.9950042|=0$ and thus a relative error of zero.

Remark. The $0$ error is owing to the correction of 7 decimal places in both the direct computation of $\cos 0.1$ and its $O(h^5)$ approximation.

April 24, 2020October 24, 2022

202004241649 Problem 2, Ch. 1 Sec. 1

Prove the following equalities:

(a) $|ab|=|a|\cdot |b|$ ;

(b) $|a|^2=a^2$ ;

(c) $\displaystyle{\bigg|\frac{a}{b}\bigg|=\frac{|a|}{|b|}}$ (where $b\neq 0$ );

(d) $\sqrt{a^2}=|a|$ .

Proof.

(a) (proof by cases)

i. When $a,b\geqslant 0$ :

$|ab|=ab=|a||b|$ .

ii. When $a,b< 0$ :

$|ab|=ab=(-a)(-b)=|a||b|$ .

iii. When $a\geqslant 0$ and $b<0$ :

$|ab|=-ab=a(-b)=|a||b|$ .

iv. When $a<0$ and $b\geqslant 0$ :

$|ab|=-ab=(-a)b=|a||b|$ .

QED

(b) (proof by induction)

By making a stronger claim

$P(n)$ : For any positive integer $n$ , $|a^n|=|a|^n$ .

Proof.

The trivial cases $n=0$ and $n=1$ are evident.

Consider the case $n=2$ ,

$\begin{aligned} |a^2| & = |a||a| \qquad \textrm{(by equality (a))} \\ & = |a|^2 \end{aligned}$

$P(2)$ is true.

Assume now that $P(n)$ is true,

$\begin{aligned} P(n+1): \qquad |a^{n+1}|& = |a^n\cdot a| \\ & = |a^n||a| \qquad \textrm{(by equality (a))}\\ & = |a|^n|a| \qquad \textrm{(by the assumption }P(n)\textrm{ is true)} \\ & = |a|^{n+1} \end{aligned}$

it can be seen that $P(n+1)$ is also true.

From the fact that $P(2)$ is true and by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$ .

It follows that $|a|^2=|a^2|=a^2$ holds.

QED

(c) (direct proof)

$\because |a|=\displaystyle{\bigg| \frac{a}{b}\cdot b\bigg|} =\bigg|\displaystyle{\frac{a}{b}}\bigg| |b|$ ,

where the second equality sign is due to equality (a),

$\therefore \displaystyle{\bigg| \frac{a}{b} \bigg| = \frac{|a|}{|b|}}$ .

QED

(d) (proof by definition)

The absolute value of a real number $a$ , denoted by $|a|$ , is defined by

$|a|= \begin{cases} x & \textrm{if } x\geqslant 0, \\ -x & \textrm{if }x<0 \end{cases}$

For any non-negative real number $a$ , the symbol $\sqrt{a}$ denotes the non-negative square root of $a$ .

QED

April 24, 2020July 25, 2022

202004241533 Problem 1, Ch. 1 Sec. 1

Prove that if $a$ and $b$ are real numbers then

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

Proof.

As $|a|$ is non-negative and $-|a|$ non-positive, one has

Eq. (1):

$-|a|\leqslant a\leqslant |a|$

Eq. (2):

$-|b|\leqslant b\leqslant |b|$

Combining Eq. (1) and Eq. (2),

$-(|a|+|b|)\leqslant a+b\leqslant |a|+|b|$ ,

or, Eq. (3): (the triangle inequality)

$|a+b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a-b|$ , one gets

Eq. (4):

$|a-b|=|a+(-b)|\leqslant |a|+|-b|=|a|+|b|$ ,

or, Eq. (4)’:

$|a-b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a|=|(a-b)+b|$ , one gets

Eq. (5):

$|a|=|(a-b)+b|\leqslant |a-b|+|b|$ ,

or, Eq. (5)’:

$|a|-|b|\leqslant |a-b|$ .

Applying the triangle inequality to $|b|=|(b-a)+a|$ , one gets

Eq. (6):

$|b|=|(b-a)+a|\leqslant |b-a|+|a|$ ,

Roughwork.

or, Eq. (6)’:

$-|a-b| \leqslant |a|-|b|$

Combining Eq. (5)’ and Eq. (6)’:

$-|a-b| \leqslant |a|-|b| \leqslant |a-b|$ ,

or, Eq. (7):

$||a|-|b||\leqslant |a-b|$ .

Combining Eq. (4)’ and Eq. (7), one obtains readily

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

QED

April 24, 2020February 5, 2023

202004240713 Homework 1 (Q3)

Suppose $G$ be a finite group of even order. Prove that there exists an element $a\in G$ such that $a\neq e$ and $a^2=e$ .

Attempts.

Try negating the statement by the following:

$\forall\, a\in G$ , either $a=e$ or $a^2\neq e$ ,

then looking for contradiction to the assumption that $G$ should be a finite group of even order.

CASE 1: Would it be possible that $a=e$ ?

Given so, the inverse and the identity of $G$ would be $e$ . And the set $G$ would have contained $e$ only, i.e., the singleton set $G=\{ e\}$ . This contradicts with the assumption that its order be even.

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .

If $a^2\neq e$ , then $a\neq a^{-1}$ for all $a\in G$ . It would then become an ill-posited negation, because the identity element $e$ is one of all $a\in G$ , and both statements $e^2\neq e$ and $e\neq e^{-1}$ are abhorrent to the basic axioms of a group. So perhaps to my discretion $e$ should be precluded from $a$ ‘s, and understood as one ( $e=e^{-1}$ ) member apart from the other members.

Proceeding to observe that if $a\neq a^{-1}$ for all $a\in G$ , I may then construct a pair of $a_i$ and $a_i^{-1}$ for $i\in \{ 1,2,\dots ,n:n\in\mathbb{Z}^+\}$ , provided from the definition of a group that the inverse of each element must exist and, as it follows, that it must be unique. That is, if $a_5=a_{27}^{-1}$ , I may relabel $a_{27}$ as $a_5^{-1}$ . By such a construction I can guarantee that all elements, except the identity $e$ , are now in pairs.

One therefore, by counting in total how many elements there are in group $G$ , will get an odd number $(2n+1)$ , the $2n$ counted from the pairs, and the single $1$ the identity $e$ itself alone counts.

This contradicts with the assumption that $|G|$ be of even parity.

From the negation of statement arises contradiction, the original statement is therefore proven by contradiction.

Remark.

I could not prove otherwise than loosely to so naive myself. Not until I had survived mathematical rigor.

April 24, 2020October 24, 2022

202004240630 Homework 1 (Q2)

Call $G$ a group and $e$ the identity of $G$ . If the order of $G$ is $2$ , then $a^2=e$ for all $a\in G$ and $G$ is abelian (Prove it yourself). Show that

$G$ is abelian if $a^2=e$ for all $a\in G$ .

Give an example of such a group whose order is greater than $2$ .

Solution.

Recall that a group $G$ is said to be abelian if $xy=yx$ for all $x,y\in G$ .

For any $a,b\in G$ , there is $ab=(bb^{-1})ab(a^{-1}a)$ , because a group satisfies, first, the inverse axiom $xx^{-1}=x^{-1}x=e$ under the notation $x^{-1}$ being the inverse of $x$ , and secondly, the identity axiom $ex=xe=x$ .

Thence by (associativity axiom) $x*(y*z)=(x*y)*z$ I may change brackets and obtain as follows:

$\begin{aligned} ab & =(bb^{-1})ab(a^{-1}a) \\ & =b(b^{-1}a)(ba^{-1})a \\ & =b(b^{-1}a)(b^{-1}a)^{-1}a \end{aligned}$

Provided that $a^2=e$ for all $a\in G$ , one may deduce $a=a^{-1}$ , $b=b^{-1}$ , etc. Thus,

$ab=b(b^{-1}a)(b^{-1}a)a$

In addition $(b^{-1}a)(b^{-1}a)=(b^{-1}a)^2=e$ , hence $ab=ba$ , and $G$ abelian.

The product group $\mathbb{Z}_2\times \mathbb{Z}_2$ is an example of one abelian group whose order is greater than $2$ and the order of whose elements is $2$ .

April 24, 2020October 24, 2022

202004240606 Homework 1 (Q1)

Let $G=\{ a,b,c,d\}$ and the Cayley table of $G$ be

$\begin{aligned} * \quad | &\textrm{} \quad a &\textrm{}\quad b \quad &\textrm{}\quad c &\textrm{}\quad d \\ \hline a \quad | &\quad a &\quad b \quad &\quad c &\quad d \\ b \quad | &\quad b &\quad a \quad &\quad c &\quad d \\ c \quad | &\quad c &\quad b \quad &\quad a &\quad d \\ d \quad | &\quad d &\quad d \quad &\quad b &\quad c \end{aligned}$

Is $(G,*)$ a group?

Attempts.

I recall that a group $(G,*)$ is a set $G$ altogether with an operation $*:G\times G\rightarrow G$ , $(x,y)\mapsto x*y$ such that the following axioms be satisfied:

i. (Associativity) $(x*y)*z=x*(y*z)$ ;

ii. (Identity) There exists $e\in G$ such that $e*x=x*e=x$ for all $x\in G$ ; and

iii. (Inverse) For each $a\in G$ , there exists $b\in G$ such that $a*b=b*a=e$ .

Let me check them one by one.

i. For $x,y,z\in \{ a,b,c,d \}$ , it suffices to check $4\times 4\times 4=64$ operations. And I found that the following operation failed axiom (i):

$d*(c*b)=d*b=d$ but $(d*c)*b=b*b=a$ .

That said, $G$ under the operation $*$ is not a group.

ii. By inspection, the identity of $G$ exists, and that is $a$ because $a*x=x*a=x$ for all $x\in \{ a,b,c,d\}$ . Axiom (ii) is thus satisfied.

iii. There exists $d\in G$ , such that for all $x^i\in\{ a,b,c,d\}=G$ ,

$d*x^i=x^i*d\neq a$ .

The claim above is validated by simply checking the identity found to be $a$ in part ii. cannot be found in the entries, and by the fact that $a$ is unique.

Axioms i. and iii. having been failed, I may conclude that $G$ under the operation $*$ cannot form a group.

April 23, 2020October 14, 2023

202004231606 Exercise 1, Section 1.1

Determine whether the vectors emanating from the origin and terminating at the following pair of points are parallel.

(a) $(3,1,2)$ and $(6,4,2)$

(b) $(-3,1,7)$ and $(9,-3,-21)$

(c) $(5,-6,7)$ and $(-5,6,-7)$

(d) $(2,0,-5)$ and $(5,0,-2)$

Background.

Two nonzero vectors $x$ and $y$ are called parallel if $y=tx$ for some nonzero real number $t$ . (Thus nonzero vectors having the same or opposite directions are parallel.)

_{Text on pg.3}

Solution.

(a) Let $x=(3,1,2)$ and $y=(6,4,2)$ . Apparently $\nexists\, t\in \mathbb{R}$ such that $y=tx$ . For otherwise $(6,4,2)=t(3,1,2)$ , the system of equations

$\begin{aligned} 6 & = 3t \\ 4 & = 1t \\ 2 & = 2t \\ \end{aligned}$

is inconsistent. They are not parallel.

(b) Let $x=(-3,1,7)$ and $y=(9,-3,-21)$ , then $y=-3x$ . The vectors $x$ and $y$ are in opposite direction and the magnitude of $y$ is three times that of $x$ . They are parallel.

(c) Let $x=(5,-6,7)$ and $y=(-5,6,-7)$ . Observe that they are in equal magnitude but in opposite direction, i.e., $y=-x$ . They are also parallel.

(d) Let $x=(2,0,-5)$ and $y=(5,0,-2)$ . Assume $t\in\mathbb{R}$ s.t. $y=tx$ , i.e.,

$\begin{aligned} 5 & = 2t \\ 0 & = 0t \\ -2 & = 5t \\ \end{aligned}$

no way will the first and the third lines agree. They are nonparallel.

November 19, 2019October 25, 2022

201911190336 Homework 1, Differential Geometry

Let $\boldsymbol{\alpha} :I\rightarrow \mathbb{R}^3$ be a regular parametrized curve (not necessarily by arc length) and let $\beta :J\rightarrow \mathbb{R}^3$ be a reparametrization of $\boldsymbol{\alpha}$ by the arc length $s=s(t)$ measured from $t_0\in I$ .

Let also $t=t(s)$ be the inverse function of $s$ and denote the derivative of $\boldsymbol{\alpha}$ wrt $t$ by $\boldsymbol{\alpha}'$ . Prove that

i. $\mathrm{d}t/ \mathrm{d}s=1/\|\boldsymbol{\alpha}'\|$ and $\mathrm{d}^2t/ \mathrm{d}s^2=-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4$ ;

ii. The curvature of $\boldsymbol{\alpha}$ at $t\in I$ is $\kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}$ ; and

iii. The torsion of $\boldsymbol{\alpha}$ at $t\in I$ is $\tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}$ .

Solution.

i. By definition $\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\boldsymbol{\alpha}'$ .

Using chain rule,

$\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}$ .

After taking the norm, as a consequence of natural parametrization

(i.e., $\bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\bigg| =1$ ),

we have

$\| \boldsymbol{\alpha}'\|=\bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}} \bigg| \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}} \bigg|= \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}\bigg|$ .

Hence $\mathrm{d}t/\mathrm{d}s=1/ \| \boldsymbol{\alpha}'\|$ .

That said,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\|\boldsymbol{\alpha}' \|}}\bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ \end{aligned}$ .

But,

$\begin{aligned} & \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ & =-\displaystyle{\frac{1}{2}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\Big( \langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle +\langle \boldsymbol{\alpha}'',\boldsymbol{\alpha}' \rangle \Big) \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ \end{aligned}$ .

Thus,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =-\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & =-\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & = \displaystyle{-\frac{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle}{\big( \langle \boldsymbol{\alpha}', \boldsymbol{\alpha}'\rangle^{1/2}\big)^4}} \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4 \\ \end{aligned}$ .

ii.

$\boldsymbol{\alpha}'= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s'$ .

Besides,

$\boldsymbol{\alpha}''=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\dot{\boldsymbol{\alpha}}s')=\dot{\boldsymbol{\alpha}}\displaystyle{\frac{\mathrm{d}s'}{\mathrm{d}t}}+s'\displaystyle{\frac{\mathrm{d}\dot{\boldsymbol{\alpha}}}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}}$ .

Then,

$\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\big( \dot{\boldsymbol{\alpha}}s' \big) \times \big( \dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}} \big) =(s')^3\langle\dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle =\|\boldsymbol{\alpha}'\|^3 \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle$ .

(For $\mathrm{d}s/\mathrm{d}t=\| \boldsymbol{\alpha}'\|$ of part i. is used.)

It follows that

$\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\|\boldsymbol{\alpha}\|^3 \| \dot{\boldsymbol{\alpha}} \|\| \ddot{\boldsymbol{\alpha}}\|\sin \measuredangle (\dot{\boldsymbol{\alpha}},\ddot{\boldsymbol{\alpha}})$ .

Note that $\dot{\boldsymbol{\alpha}}=\mathbf{t}$ and $\ddot{\boldsymbol{\alpha}}=\dot{t}$ are orthogonal,

$\| \boldsymbol{\alpha}\|=1$ and $\| \ddot{\boldsymbol{\alpha}}\|=\| \dot{\mathbf{t}}\|=\| \kappa\|$ .

We obtain

$\kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}$ .

iii.

First,

$\dot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}=\boldsymbol{\alpha}'\dot{t}$ ;

secondly,

$\ddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}(\boldsymbol{\alpha}'\dot{t})=\boldsymbol{\alpha}'\ddot{t}+\bigg( \displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\boldsymbol{\alpha}' \bigg)\dot{t}=\boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2$ ;

thirdly,

$\dddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) =\boldsymbol{\alpha}'\dddot{t}+\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}''2\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3=\boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3$ .

Compute $\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle$ as follows:

$\begin{aligned} \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle & = \big( \boldsymbol{\alpha}'\dot{t} \big) \wedge \big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big) \\ & =\big( \boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2\big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big)\\ \end{aligned}$

Now that the cross product $\boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}''$ is orthogonal to both $\boldsymbol{\alpha}'$ and $\boldsymbol{\alpha}''$ , we can ignore the dot product among them and what remains is

$(\boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2)\cdot \boldsymbol{\alpha}'''\dot{t}^3$ , or,

$\dot{t}^6\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle$ .

Using the result of part i., substitute $1/\| \boldsymbol{\alpha}'\|$ for $\dot{t}$ , we obtain

$\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\|\boldsymbol{\alpha}'\|^6}}$ .

Using the formula for curvature in part ii. and put it into

$\tau =\displaystyle{\frac{\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}},\dddot{\boldsymbol{\alpha}} \rangle}{\kappa^2}}=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\kappa^2 \|\boldsymbol{\alpha}'\|^6}}$ ,

i.e.,

$\tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}$ .

CASE 1: Would it be possible that ?

CASE 2: If for all elements in the group , their order cannot ever be .

CASE 1: Would it be possible that $a=e$ ?

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .