每日彙整: 2017-09-23

樹莓派、樹莓派之學習、樹莓派之教育

GoPiGo 小汽車︰格點圖像算術《投影幾何》【五‧線性代數】《導引八》觀察者《運動‧I 》

⊙OIAB⊕OIAB
OOOOOOOIAB
IOIABIIOBA
AOABIAABOI
BOBIABBAIO

220px-Rubik's_cube.svg

220px-NegativeOne3Root.svg

220px-Clock_group.svg

170px-Cyclic_group.svg

200px-Sixteenth_stellation_of_icosahedron

那麼要如何『了解』上面那個『體』的『加法表』與『乘法表』呢？通常人們會自然的把 \bigoplus 看成『加』，將 \bigodot 想為 『乘』。然而『數學』的一般『抽象結構』是由『規則』所『定義』的，很多講的是某個『集合』內之『元素』所具有的『性質』，以及『運算』所滿足的『定律』。這與有沒有人們所『熟悉的』類似結構無關，而且那些『元素』也未必得是個『數』的啊！這或許就是『抽象數學』之所以『困難』的原因。雖然從『純粹』的『邏輯推理』能夠得到『結論』，只不過要是缺乏『經驗性』，人們通常『感覺』不實在、不具體、而且也不安心。就讓我們試著給這個『體』一的比較容易『理解』之結構的『再現』︰設想將 O, I, A, B 表現在『複數平面』上，其中 O 是『原點』，而 I, A, B 位在『單位圓』之上，定義如下
O \equiv_{rp} \ 0 + i 0 = 0
I \equiv_{rp} \ 1 + i 0 = 1
A \equiv_{rp} \ - \frac{1}{2} + i \frac{\sqrt{3}}{2}
B \equiv_{rp} \ - \frac{1}{2} - i \frac{\sqrt{3}}{2}

X \bigoplus Y \equiv_{rp} \ -(X + Y), if X \neq Y
X \bigoplus Y \equiv_{rp} \ (X - Y) = 0, if X = Y
X \bigodot Y \equiv_{rp} \ X \cdot Y

\because I \bigoplus A \equiv_{rp} \ - \left[ 1 + \left( - \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \right]
= - \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = B

A \bigodot A \equiv_{rp} \ {\left( - \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)}^2
= \frac{1}{4} - i \frac{\sqrt{3}}{2} - \frac{3}{4} = B

\therefore (I \bigoplus A) \bigoplus B = B \bigoplus B
= I \bigoplus A \bigoplus (A \bigodot A) = O

如果將它用複數改寫成 1 + A + A^2 = 0，這不就是『上上篇』裡的『三次方程式』 x^3 - 1 = (x - 1)(x^2 + x + 1) = 0 的『解』 \omega 的嗎？再徵之以『相量』的『向量加法』和『旋轉乘法』，這個四個元素的『體』之『喻義』也許可以『想像』的了。假使人們對於『抽象思考』一再重複的『練習』，那麼『邏輯推演』也將會是『經驗』中的了！就像俗語說的︰熟能生巧；『抽象的』也就成了『直覺』上的了！！

─── 《【Sonic π】電路學之補充《四》無窮小算術‧下上》

 

先生，就算知道『透視』之形式可以表示為︰

\left( \begin{array}{cc} z^{'} \\ 1 \end{array} \right) = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{cc} z \\ 1 \end{array} \right)

但那可是『整個複數面』 z 對應到『整個複數面』 z^{'} 的啊！難到說不是『任一視線』，祇該『應合』一『投影點』嗎？

 

然而即使『固定』 z_1,z_2 ，可『選擇』的 \alpha, \beta 可多了︰

z^{'} = \frac{\alpha \cdot \beta \cdot z \cdot (z_2-z_1) }{(\alpha - \beta) z + (\beta \cdot z_2 - \alpha \cdot z_1)}}

。更別說『不共線』之 z^{'}_1, z^{'}_2 的『情況』哩？？

在此反問『透視』條件下，上圖 z^{'} 與 z 的『關係』是什麼呢 ？！  z \ = \ i \cdot \frac{2 z^{'}}{z^{'} - {z^{'}}^{*}} 能不能表示一維『投影線』耶！？

特先請讀者觀此『神奇』也◎

 

Moebius Transformations Revealed

“Möbius Transformations Revealed is a wonderful video clarifying a deep topic. This is amazing work by Douglas Arnold and Jonathan Rogness of the University of Minnesota.”
— Edward Tufte        

Möbius Transformations Revealed is a short video by Douglas Arnold and Jonathan Rogness which depicts the beauty of Möbius transformations and shows how moving to a higher dimension reveals their essential unity. It was one of the winners in the 2007 Science and Engineering Visualization Challenge and was featured along with the other winning entries in the September 28, 2007 issue of journal Science. The video, which was first released on YouTube in June 2007, has been watched there by nearly two million viewers and classified as a “Top Favorite of All Time“ first in the Film & Animation category and later in the Education category. It was selected for inclusion in the MathFilm 2008 DVD, published by Springer.

From this web page you can:

plane view projection 3D view big sphere