每日彙整: 2017-09-10

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】《導引七‧變換組合 VII‧A 》

既已體會了『複數』 z 之神奇。這裡說說藉著它的『絕對值』 |z| ──   原點到點 z 的距離 ── 之性質︰

‧ |z_1 \cdot z_2| = |z_1| \cdot |z_2|

‧ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}

，在不含混模糊情況下，簡化『幾何表述』也。比方講

z_1, z_2 兩點『線段長度』 \overline{z_1 z_2} 記作 z_2 - z_1, \because |z_2-z_1| = \overline{z_1 z_2} 。

依樣畫葫蘆︰

‧ \overline{z_1 z_2} \times \overline{{z_1}^{'} {z_2}^{'}} \Rightarrow (z_2 - z_1)({z_2}^{'} - {z_1}^{'})

‧ \frac{\overline{z_1 z_2}}{\overline{{z_1}^{'} {z_2}^{'}}} \Rightarrow \frac{z_2 - z_1}{{z_2}^{'} - {z_1}^{'}} 。

如是亦能啟發解讀『代數式』之『幾何意義』矣。

且再舉『透視』的『特徵平行四邊形』 p Z_{{\infty}^{'}} Z_x {Z^{'}}_{\infty} 推導為例︰

 

上圖假設『透視中心』 p 是『原點』，

\therefore z_{\Box}^{\bigcirc} - p = z_{\Box}^{\bigcirc} - 0 = z_{\Box}^{\bigcirc} 。

如果用 z^{'} = P(z) 表示這個『透視函數』，那麼

P(Z_x) = Z_x

P(\infty) = {Z^{'}}_{\infty}

 \infty = P(Z_{{\infty}^{'}}) 。

幾何意指

‧ Z_x 是 l 、l^{'} 兩線交點。

‧ Z_{{\infty}^{'}} \ \parallel \ l^{'}

‧ {Z^{'}}_{\infty} \ \parallel \ l 。

 

因三角形 \Delta p Z_{{\infty}^{'}} z 和 \Delta z' Z_x z 相似，故

\frac{z^{'}-z}{z} = \frac{Z_x - z}{z - Z_{{\infty}^{'}}}

左右兩邊加一自得

\Rightarrow \frac{z^{'}}{z} = \frac{(Z_x - z) + (z - Z_{{\infty}^{'}})} {z - Z_{{\infty}^{'}}} = \frac{Z_x - Z_{{\infty}^{'}}} {z - Z_{{\infty}^{'}}} = \frac{{{Z^{'}}_{\infty}}}{z - Z_{{\infty}^{'}}} 哩◎

 

假使 p 不是『原點』，可得

z^{'} - p = \frac{({{Z^{'}}_{\infty}} - p)(z-p)}{(z-p) - (Z_{{\infty}^{'}} - p)}

\Rightarrow z^{'} = \frac{{{Z^{'}}_{\infty}} z \ - \ p({{Z^{'}}_{\infty}} + Z_{{\infty}^{'}} - p)}{z - Z_{{\infty}^{'}}} = \frac{{{Z^{'}}_{\infty}} z \ - \ p \cdot Z_x}{z - Z_{{\infty}^{'}}}

可知它的兩個『定點』 z^{'} = z 就是 p 和 Z_x ，而且 p + Z_x = {{Z^{'}}_{\infty}} + Z_{{\infty}^{'}}, \ \because Z_x - p = ({{Z^{'}}_{\infty}} -p) + (Z_{{\infty}^{'}}-p)

也能清楚明白

Fixed points

Every non-identity Möbius transformation has two fixed points  \gamma _{1},\gamma _{2} on the Riemann sphere. Note that the fixed points are counted here with multiplicity; the parabolic transformations are those where the fixed points coincide. Either or both of these fixed points may be the point at infinity.

Determining the fixed points

The fixed points of the transformation

f(z)={\frac {az+b}{cz+d}}

are obtained by solving the fixed point equation f(γ) = γ. For c ≠ 0, this has two roots obtained by expanding this equation to

  c\gamma ^{2}-(a-d)\gamma -b=0\ ,

and applying the quadratic formula. The roots are

\gamma _{{1,2}}={\frac {(a-d)\pm {\sqrt {(a-d)^{2}+4bc}}}{2c}}={\frac {(a-d)\pm {\sqrt {(a+d)^{2}-4(ad-bc)}}}{2c}}.

Note that for parabolic transformations, which satisfy (a+d)2 = 4(ad−bc), the fixed points coincide. Note also that the discriminant is

(a-d)^{2}+4cb=(a-d)^{2}+4ad-4=(a+d)^{2}-4=\operatorname {tr}^{2}{\mathfrak {H}}-4.

When c = 0, the quadratic equation degenerates into a linear equation. This corresponds to the situation that one of the fixed points is the point at infinity. When a ≠ d the second fixed point is finite and is given by

\gamma =-{\frac {b}{a-d}}.

In this case the transformation will be a simple transformation composed of translations, rotations, and dilations:

z\mapsto \alpha z+\beta .\,

If c = 0 and a = d, then both fixed points are at infinity, and the Möbius transformation corresponds to a pure translation:

  z\mapsto z+\beta .

 

意指吧◎