每日彙整: 2017-08-31

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 VI‧IV 》

人們將如何『預約』未來，又怎麼『翻譯』過去呢？

Li Bai

A Quiet Night Thought

In front of my bed, there is bright moonlight.
It appears to be frost on the ground.
I lift my head and gaze at the August Moon,
I lower my head and think of my hometown.

 

Contemplation

Moon twilight approaches, coating the ground through the window,
Resembles a touch of frost,
Moon at the window,
Taking me back to where I am from.

李白

静夜思

床前明月光
疑是地上霜
舉頭望明月
低頭思故鄉

假使將李白的《靜夜思》翻譯成英文，藉由『中英對照』，是否更能『理解』原作之『意境』呢？還是會少了點『詩』的『味道』？？或許這個『利弊得失』就落在︰

『文化』之『盲點』，常顯現在『意義』的『忽略』之中。

『人文』之『偏見』，普遍藏於『字詞』之『情感』之內。

故而同一『內容』的多種『語言文本』，也許可見那『通常之所不見』。

或許萬象『錯綜複雜』，難得『意義確解』哩！

n4x26s-panamaeclipsebloodmoonbinary1598272
四月四日之連環四月蝕

讓我們趁著『血月』全蝕之餘威，乘著『想像之翼』走進詩人的『情感世界』︰

李白

静夜思

床前明月光
疑是地上霜
舉頭望明月
低頭思故鄉

Li Bai

A Quiet Night Thought

In front of my bed, there is bright moonlight.
It appears to be frost on the ground.
I lift my head and gaze at the August Moon,
I lower my head and think of my hometown.

李白的《靜夜思》在述說些什麼呢？既然已寫了【床前明月光】，作者難道會不知道『月將圓』嗎？為何又寫【疑是地上霜】 的呢！也許人們『知道』一事終將發生，與『感受』那事現實之刻，有巨大的差別，故而『生疑』，總以為時間還很多，事情也還沒發生。所謂『春雪秋霜』，這個 『秋之明月』，又能意指什麼的呢？？在此『中秋賞月』佳節，不是萬戶笙簫？早是團欒樂夥之時？？任誰只消雙手推開，就能見那『窗前月』的了吧！又怎會俳佪 『床前』的呢！！是想『入夢』？又或者『難眠』？？眼巴巴的，竟捱到了『月上中天』，不得不『舉頭望』明月！！或許此刻正『萬籟俱寂 』，夜深人靜，方敢 『思念』，卻不能『想家』，又不忍『思親』 ，祇得說︰【低頭思故鄉】，觸景偶然『思憶』起了『所來之處』 ……

由於一首《静夜思》卻有多個『版本』，據考證，現今『最早』的文本是宋朝郭茂倩《樂府詩集》作：

床前看月光
疑是地上霜
舉頭望山月
低頭思故鄉

又因為這是『詩仙』『李白』流傳最廣泛的一首詩，所以引起學術界許多議論質疑，那位清代文人『蘅塘退士』『孫洙』所編纂的《唐詩三百首》之收錄版本是否是『原作』？如果 Li Bai 不『仙人指路』，那麼我們能夠知道『答案』嗎？或許『流傳的早』就代表『抄寫的對』！！又或許還有尚未發現之『更早版本』？？要是說已知︰李白的《静夜思》創作于唐玄宗開元十四年九月十五日之『揚州旅舍』，時年李白二十六歲。在同時同地所作的還有一首《秋夕旅懷》

涼風度秋海，吹我鄉思飛。
連山去無際，流水何時歸。
目極浮雲色，心斷明月暉。
芳草歇柔豔，白露催寒衣。
夢長銀漢落，覺罷天星稀。
含悲想舊國，泣下誰能揮。

。之後在一個月明星稀的夜晚，詩人抬望天空一輪皓月，思鄉之情油然而生，寫下了這首傳誦古今，中外皆知的名詩。如此我們會怎麼『判斷』這個『版本』的問題呢？駐留於『可能』與『不可能』之間的到底是什麼ㄚ！！

─── 摘自《如何閱讀 W!o+ 之傳真？？》

 

若思假借『幾何』與『代數』談及同一平面之『論述』，應當駕輕就熟的吧？？且讓我們試著使用『比』、『交比』講講所謂『同義反復』也！！

《Cut The Knot》網站上有一篇文章介紹了『交比』︰

Cross-Ratio

The cross-ratio is a surprising and a fundamental concept that plays a key role in projective geometry. In the spirit of duality, cross-ratio is defined for two sets of objects: 4 collinear points and 4 concurrent lines.

The cross-ratio (ABCD) of four collinear points A, B, C, D is defined as the “double ratio”:

(1)

(ABCD) = CA/CB : DA/DB,

where all the segments are thought to be signed. The cross-ratio obviously does not depend on the selected direction of the line ABCD, but does depend on the relative position of the points and the order in which they are listed.

The cross-ratio (abcd) of four (coplanar and) concurrent lines is defined as another double ratio, now of sines:

(abcd) = sin(cMa)/sin(cMb) : sin(dMa)/sin(dMb),

where angles are also considered signed (in a natural way.) If points A, B, C, D are chosen on four lines a, b, c, d concurrent at M, then we often write (abcd) = M(ABCD). The fact that the four points (lines) are grouped into two pairs of points (lines) is reflected in another popular notation: (AB; CD) and (ab; cd).

The relationship between the two definitions is established by the following

Lemma

Let A, B, C, D be the points of intersection of 4 concurrent lines a, b, c, d by another straight line. Then (ABCD) = (abcd).

Remark

When the lines a, b, c, d are defined by the points, as above, it is often convenient to write (abcd) = M(ABCD).

Proof of Lemma

Consider 4 triangles CMA, CMB, DMA, and DMB and represent their areas in two different ways:

  Area(CMA): h·CA/2 = MC·MA·sin(CMA)/2
  Area(CMB): h·CB/2 = MC·MB·sin(CMB)/2
  Area(DMA): h·DA/2 = MD·MA·sin(DMA)/2
  Area(DMB): h·DB/2 = MD·MB·sin(DMB)/2,

where h is the length of the common altitude of the four triangles from vertex M. The required identity now follows immediately.

The lemma helps explain the significance of the cross-ratio in projective geometry.

Theorem 1

The cross-ratio of collinear points does not change under central (and, trivially, parallel) projections.

Indeed, from Lemma, (ABCD) = (abcd) = (A’B’C’D’).

It’s worth noting that central projection does not, in general, preserve either the distance or the ratio of two distances.

A permutation of the points may or may not change the cross-ratio. If any two pairs of points are swaped simultaneously, the cross-ratio does not change, e.g., (ABCD) = (BADC) = (DCBA). Wherever it changes, there are only five possible values. If (ABCD) = m, the possible values are 1-m, 1/m, (m-1)/m, 1/(1-m), m/(m-1).

If (ABCD) = 1, then either A = B or C = D. A more important case is where (ABCD) = -1. If (ABCD) = -1 then the points C and D are called harmonic conjugates of each other with respect to the pair A and B. A and B are then also harmonic conjugates with respect to C and D. Each of the pairs is said to divide the other’s segment harmonically. There exists a straight edge only construction of harmonic conjugates. The four lines through an arbitrary point M and four conjugate points are called a harmonic bundle.

One of the four points may lie at infinity. On such occasions, it is useful to consider the limit when a finite point moves to infinity along the common line of the four. The limit is quite simple. For example, if D = , then (ABC) = CA/CB.

The theorem has been established by Pappus in the seventh book of his Mathematical Collections. It was further developed by Desargues starting with 1639 [Wells, p. 41].

或 許可當作探索『幾何意義』的起點。首先為什麼『交比』又稱為『雙比』 double ratio 呢？因為 (ABCD) \equiv (A, B ; C, D) =_{df} \frac{CA}{CB} : \frac{DA}{DB} 是『比之比』。這建議了『比』之『定義』可由三『共線點』(A, B; C) = \frac{CA}{CB} 給出，因此『交比』 (A, B; C, D) 就是 \frac{(A, B ; C)}{(A, B ; D)} 的了。

那 麼這個『比』 (A, B ; C) 有什麼『幾何意義』嗎？事實上它可以用來『確定』直線 \overline {AB} 上任意點 C 的『位置』。甚至賦予『座標數值』 ── AC 、BC 同向取正，反向取負 ── 。如果與『共點』 M 線結合起來︰

可得『邊』、『角』對應關係︰

\triangle MAC 之面積 I = \frac{1}{2} CA \cdot Mh = \frac{1}{2} MA \cdot \sin(\angle AMC) \cdot MC 。

\triangle MBC 之面積 II = \frac{1}{2} CB \cdot Mh = \frac{1}{2} MB \cdot \sin(\angle BMC) \cdot MC 。

\frac{I}{II} = \frac{CA}{CB} = \frac{MA}{MB} \cdot \frac{\sin(\angle AMC) }{\sin(\angle BMC)} 。

故知對固定之『共點』 M 與『定點』 A, B 而言，『邊角比』

\frac{CA}{CB} : \frac{\sin(\angle AMC) }{\sin(\angle BMC)} = \frac{MA}{MB} 和 C 點的『位置無關』也。

─── 摘自《GoPiGo 小汽車︰格點圖像算術《投影幾何》【四‧平面國】《壬》》

 

如果說三條『視線』共點 M \ or \ P ，假設其為『原點』有何妨乎 ？畢竟『幾何』無此依賴耶！

故而

z= l \cdot e^{i \cdot \theta}, \ z_1= l_1 \cdot e^{i \cdot {\theta}_1}, \ z_2= l_2 \cdot e^{i \cdot {\theta}_2}

在 (0,0) 共點也。

今知 z, z_1, z_2 『三點共線』，故知恰有一『實數』 \lambda ，使得

z = (1-\lambda) \cdot z_1 + \lambda \cdot z_2 。所以

\therefore \ z-z_1 = \lambda (z_2-z_1)

z-z_2 = (\lambda -1)(z_2-z_1) 。當然

\frac{z-z_1}{z-z_2} = \frac{\lambda}{\lambda -1} ，自然推出

\Rightarrow \ \frac{|z-z_1|}{|z-z_2|} = \frac{\overline{AC}}{\overline{BC}} = \left| \frac{\lambda}{\lambda -1} \right| 吧！

但是這也等於三角形 \Delta p-z-z_1 與 \Delta p-z-z_2 之面積比，可用

\frac{\Delta p-z-z_1}{\Delta p-z-z_2} = \frac{im(z \cdot {z_1}^{*})}{im(z \cdot {z_2}^{*})} 來計算︰

\frac{im(z \cdot {z_1}^{*})}{im(z \cdot {z_2}^{*})} = \frac{im(l \cdot l_1 \cdot e^{i(\theta - {\theta}_1)})}{im(l \cdot l_2 \cdot e^{i(\theta - {\theta}_2)})} = \frac{l_1}{l_2} \cdot \frac{sin(\theta - {\theta}_1)}{sin(\theta - {\theta}_2)} 。果然

(z_1,z_2;z) = \frac{\overline{z_1 z}}{\overline{z_2z}} = \left| \frac{\lambda}{\lambda -1} \right| = \frac{l_1}{l_2} \cdot \frac{sin(\theta - {\theta}_1)}{sin(\theta - {\theta}_2)} 也。

此時若說有另一點 z^{'} 形成『四點共線』，那麼 z^{'} 當可表示為 (1-{\lambda}^{'}) z_1 + {\lambda}^{'} z_2, \ {\lambda}^{'} \neq \lambda 吧。這可由 z^{'} -z = ({\lambda}^{'} - \lambda)(z_2 - z_1) 知 \overline{z^{'}z} \parallel \overline{z_2 z_1} 而確認哩。所以

\therefore (z_1,z_2;z^{'}) = \frac{\overline{z_1 z^{'}}}{\overline{z_2 z^{'}}} = \left| \frac{{\lambda}^{'}}{{\lambda}^{'} -1} \right| = \frac{l_1}{l_2} \cdot \frac{sin({\theta}^{'} - {\theta}_1)}{sin({\theta}^{'} - {\theta}_2)} 。如是『交比』

(z_1,z_2;z,z^{'}) = \frac{(z_1,z_2;z)}{(z_1,z_2;z^{'})}

= \frac{\overline{z_1 z}}{\overline{z_2z}} \cdot \frac{\overline{z_2 z^{'}}}{\overline{z_1 z^{'}}} = \left| \frac{\lambda}{\lambda -1} \right| \cdot \left| \frac{{\lambda}^{'}-1}{{\lambda}^{'}} \right| = \frac{sin(\theta - {\theta}_1)}{sin(\theta - {\theta}_2)} \cdot \frac{sin({\theta}^{'} - {\theta}_2)}{sin({\theta}^{'} - {\theta}_1)}

『邊』、『角』分立左右且『相等』呦◎

縱是『同義反覆』，假使改弦易轍︰

\frac{\Delta p-z-z_1}{\Delta p-z-z_2} = \frac{im(z \cdot {z_1}^{*})}{im(z \cdot {z_2}^{*})}

= \frac{\frac{1}{2} (z {z_1}^{*} - z^{*} z_1)}{\frac{1}{2} (z {z_2}^{*} - z^{*} z_2)} = \frac{z {z_1}^{*} - z^{*} z_1}{z {z_2}^{*} - z^{*} z_2}

，怕那『交比性質』不易見勒☆