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 Φ 》

假使說歐幾里得

歐幾里得（希臘文：Ευκλειδης，前325年—前265年），有時被稱為亞歷山卓的歐幾里得，以便區別於墨伽拉的歐幾里得，古希臘數學家，被稱為「幾何學之父」。他活躍於托勒密一世（公元前323年－公元前283年）時期的亞歷山卓，也是亞歷山太學派的成員。他在著作《幾何原本》中提出五大公設，成為歐洲數學的基礎 。[1][2][3]歐幾里得也寫過一些關於透視、圓錐曲線、球面幾何學及數論的作品。歐幾里得幾何被廣泛的認為是數學領域的經典之作。

 

未曾好好選擇『幾何原理』之『公設』，恐怕沒人相信。如果講他對『第五公設』──  平行 ── 之表述出人意料︰

Parallel postulate

In geometry, the parallel postulate, also called Euclid‘s fifth postulate because it is the fifth postulate in Euclid’s Elements, is a distinctive axiom in Euclidean geometry. It states that, in two-dimensional geometry:

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

This postulate does not specifically talk about parallel lines,[1] it is only a postulate related to parallelism. Euclid gave the definition of parallel lines in Book I, Definition 23[2] just before the five postulates.[3]

Euclidean geometry is the study of geometry that satisfies all of Euclid’s axioms, including the parallel postulate. A geometry where the parallel postulate does not hold is known as a non-Euclidean geometry. Geometry that is independent of Euclid’s fifth postulate (i.e., only assumes the modern equivalent of the first four postulates) is known as absolute geometry (or, in other places known as neutral geometry).

If the sum of the interior angles α and β is less than 180°, the two straight lines, produced indefinitely, meet on that side.

 

難到是想統整『透視』耶？也許就有爭議了！

反思『科學精神』總只能是一再再地『實驗』

人類發現『規 律』的天性，使得人世間總有一些不傳的『技藝』。這一些未經『科學』解釋之『經驗』，正是理性『思辨』的『種子 』 ！若非冥冥中自有『系統』、『環境』、『法則』 …… 等等促成之『作用』，那一種『經驗』又怎可能的呢？於是『思辨』不得不落入『無可思議』的乎！！

一般默念或念出聲音

一般閱讀方式──念出聲或默念

眼腦直映

速讀──眼腦直映

Eye-exercise-for-speed-reading_thumb

速讀的眼睛訓練

有人說人可以『速讀』的也？？維基百科詞條講︰

速讀

速讀是快速而有效的閱讀，是一種在不影響理解和記憶的情況下，提升閱讀速率的閱讀方法。用到的方法常包括有各種心理學技巧，如組塊化（chunking）、去除默讀（subvocalization）等。 其中速讀的原理有兩個：擴大視野及眼腦直映。

  1. 擴大視野藉由視野的擴大，眼睛可閱讀的字數增加，當然速度自然倍數成長。

     

    擴大視野練習

  2. 眼腦直映是一種眼睛看到，頭腦直接反應的一種直覺化反應的過程。

那麼這樣的『如人飲水』之事！！到底是人人有之『共性』或者說『殊性 』之論的耶？？即使從其始初

起源

速讀起源[1]於第二次世界大戰時， 美國國防部為使軍官能有效且迅速地判斷空中快速行進的戰機標誌和各式型體，發明速視儀裝置，快速閃示視覺刺激，藉由閃示訓練，軍官可以在1/500秒內辨 別極小的飛機圖像。閱讀學家的後續研究發現受過訓練的普通人平均可在一分鐘內閱讀2萬個英文單字，未受過訓練的人平均可閱讀200個單字，可見人類在閱讀 上具有相當大的潛力。 戰後美國西北大學視聽教育中心繼續研究速讀方法，哈佛大學首先開辦第一期速讀訓練班，而後速讀訓練班在各地大、中小學校普及，至今美國80%以上的高等院 校都開設有速讀課程，且有專門研究和教授速讀的速讀學院，可授予學習者博士學位。1964年速讀傳入台灣逐漸推廣。

以及歸結

原理

一 般人看書時是：「眼睛看」→「嘴巴念或默念」→「頭腦想與記憶」。 嘴巴念或默念是使閱讀速度無法大量增進的主因之一。眼睛視線所及常是一個面而非一個點，如看風景、照片、電影等圖像時，人能看到整個畫面，並不需要從上到 下或從左到右依序地看，也不用嘴巴念或默念。速讀的原理便在於改變原本一行一行逐字念出或默念的習慣，養成「眼睛看」→「頭腦想與記憶」的閱讀習慣(眼腦 直映)，則看書可一眼看一行以上，或將多行文字以形成一個畫面的方式，看整個頁面。[2]

來講，任何一個複雜系統的『表面行為』與其『內在機制』之關係可能也是『複雜』的吧！更不要說任『一個人』還可以『學習』、『改變』、 …… 增益『其所不能』的矣！！

若 問『速描』一張『圖像』，用著『來不及想』那樣的速度，其所『反映』的『本徵』要點，是屬於『描者』，還是屬於『被描者』 、或是屬於『描』與『被描』所構成的『系統』耶？？？假使以『量子』量測來看，大概得是『系統』的吧！！！如果用『經典』力學觀察，可能『最大機率』依舊 是『被描者』的哩？？？

如是說來，假使『小樹林系統』是人自己『 Top Down 』從上往下用人之『邏輯』建構的『人為系統』。那麼擁有『生命』與『智慧 』的『自然系統』，是否可以『判定 』其是來自『上→下』，還是『下→上』，或是『上↑↓下』的嗎？

─── 摘自《W!o+ 的《小伶鼬工坊演義》︰小樹林系統之速描》

 

檢證『真理』的吧！！所以他特重『 SAS 』反而無『 SSS 』哩！！

就像我們所知，『幾何事實』之『單點透視』可如是呈現︰

\left( \begin{array}{cc} v \\ 1 \end{array} \right) = \ \left( \begin{array}{cc} \frac{u}{(1- \frac{1}{k}) u + \frac{1}{k}} \\ 1 \end{array} \right) \ {\overset {P}{\doublebarwedge }} \ \left( \begin{array}{cc} 1 & 0 \\ 1 - \frac{1}{k} & \frac{1}{k} \end{array} \right) \left( \begin{array}{cc} u \\ 1 \end{array} \right)

 

。或他早知

若是 L_{\Phi} \parallel L \parallel L^{'} ，滿足下述關係也︰

I = \frac{\overline{CA}}{\overline{CB}} = \frac{\overline{PA}}{\overline{PB}} \cdot \frac{\sin( \angle {\theta}_{AB} + \angle {\theta}_{BC})} {\sin( \angle {\theta}_{BC})}

II = \frac{\overline{DA}}{\overline{DB}} = \frac{\overline{PA}}{\overline{PB}} \cdot \frac{\sin( \angle {\theta}_{AB} + \angle {\theta}_{BD})} {\sin( \angle {\theta}_{BD})}

III = \frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B^{'}}} = \frac{\overline{PA^{'}}}{\overline{PB^{'}}} \cdot \frac{\sin( \angle {\theta}_{AB} + \angle {\theta}_{BC})} {\sin( \angle {\theta}_{BC})}

 

 

且知此『比』實通『相似』與『全等』關係之關鍵呢★？

Similar triangles

In geometry two triangles, △ABC and △A′B′C′, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional.[1] It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. This is known as the AAA similarity theorem.[2] Due to this theorem, several authors simplify the definition of similar triangles to only require that the corresponding three angles are congruent.[3]

There are several statements each of which is necessary and sufficient for two triangles to be similar:

  • The triangles have two congruent angles,[4] which in Euclidean geometry implies that all their angles are congruent.[5] That is:
If ∠BAC is equal in measure to ∠B′A′C′, and ∠ABC is equal in measure to ∠A′B′C′, then this implies that ∠ACB is equal in measure to ∠A′C′B′ and the triangles are similar.
  • All the corresponding sides have lengths in the same ratio:[6]
AB/A′B′ = BC/B′C′ = AC/A′C′. This is equivalent to saying that one triangle (or its mirror image) is an enlargement of the other.
  • Two sides have lengths in the same ratio, and the angles included between these sides have the same measure.[7] For instance:
AB/A′B′ = BC/B′C′ and ∠ABC is equal in measure to ∠A′B′C′.

This is known as the SAS similarity criterion.[8]

When two triangles △ABC and △A′B′C′ are similar, one writes[9]:p. 22

△ABC ∼ △A′B′C′.

There are several elementary results concerning similar triangles in Euclidean geometry:[10]

  • Any two equilateral triangles are similar.
  • Two triangles, both similar to a third triangle, are similar to each other (transitivity of similarity of triangles).
  • Corresponding altitudes of similar triangles have the same ratio as the corresponding sides.
  • Two right triangles are similar if the hypotenuse and one other side have lengths in the same ratio.[11]

Given a triangle △ABC and a line segment DE one can, with ruler and compass, find a point F such that △ABC ∼ △DEF. The statement that the point F satisfying this condition exists is Wallis’s postulate[12] and is logically equivalent to Euclid’s parallel postulate.[13] In hyperbolic geometry (where Wallis’s postulate is false) similar triangles are congruent.

In the axiomatic treatment of Euclidean geometry given by G.D. Birkhoff (see Birkhoff’s axioms) the SAS similarity criterion given above was used to replace both Euclid’s Parallel Postulate and the SAS axiom which enabled the dramatic shortening of Hilbert’s axioms.[8]

 

能解所謂『交比』之『不變性』哩☆！

《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。

此乃『交比不變性』之基礎吧！

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

 

進而探究 \frac{x}{x-1} 『對合』矣◎

{\left( \begin{array}{cc} 1 & 0 \\ 1 & -1 \end{array} \right) } {\left( \begin{array}{cc} 1 & 0 \\ 1 & -1 \end{array} \right) } = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) }