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『』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!

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:

床前月光
疑是地上霜
舉頭望
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}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}

(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\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)\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☆

 

 

 

 

 

 

 

 

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‧III 》

假設讀者已經知道『複數

Complex number

A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit (which satisfies the equation i2 = −1).[1] In this expression, a is called the real part of the complex number, and b is called the imaginary part. If  {\displaystyle z=a+bi}, then we write {\displaystyle \operatorname {Re} (z)=a,} and  {\displaystyle \operatorname {Im} (z)=b.}

Complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. The complex number a + bi can be identified with the point (a, b) in the complex plane. A complex number whose real part is zero is said to be purely imaginary, whereas a complex number whose imaginary part is zero is a real number. In this way, the complex numbers are a field extension of the ordinary real numbers, in order to solve problems that cannot be solved with real numbers alone.

As well as their use within mathematics, complex numbers have practical applications in many fields, including physics, chemistry, biology, economics, electrical engineering, and statistics. The Italian mathematician Gerolamo Cardano is the first person known to have introduced complex numbers. He called them “fictitious” during his attempts to find solutions to cubic equations in the 16th century.[2]

A complex number can be visually represented as a pair of numbers (a, b) forming a vector on a diagram called an Argand diagram, representing the complex plane. “Re” is the real axis, “Im” is the imaginary axis, and i is the imaginary unit which satisfies i2 = −1.

 

是什麼?熟悉它的

運算

通過形式上應用代數結合律交換律分配律,再加上等式i ² = −1,定義複數的加法、減法、乘法和除法:

法則。此處只是簡介如何利用『複數平面』以及『複數運算』表達『平面幾何』的概念。接續文本中祇借 z 起頭的符號 z, z^{'}, z_{\Box}, z^{'}_{\Box}, \cdots 代表『複數』,其餘符號除非特別指定外,都是『實數』也。

這個『複數』 z 有兩種表示法

『笛卡爾座標系形式』 z = x + y*i ,方便表達

【複數加法‧平移】

z + z_t = ( x+y*i ) + ( x_t + y_t * i )

= (x + x_t) + (y+y_t)*i

『極座標形式』 z = r \cdot e^{i \cdot \theta} ,容易展現

【複數乘法‧旋轉】

e^{i \cdot \phi} \cdot r \cdot e^{i \cdot \theta} = r \cdot e^{i \cdot (\phi + \theta)}

若能深入了解

【共軛複數】 z^{*} \ {\equiv}_{df} \ x - y*i = r \cdot e^{- i \cdot \theta}

,將可用它表達許多概念哩!

‧【實部】 re(z) = x = \frac{z + z*}{2}

‧【虛部】 im(z) = y = \frac{z - z*}{2}

‧【長度】 Abs(z) = |z| = \sqrt{x^2+y^2} = \sqrt{z \cdot z^{*}}

‧……

舉例說 z_a \ {\equiv}_{rp} \ \vec{a}, z_b \ {\equiv}_{rp} \ \vec{b} ,那麼

z_a \cdot {z_b}^{*} = \left( r_a \cdot e^{i \cdot {\theta}_a} \right) \left( r_b \cdot e^{- i \cdot {\theta}_b} \right) = r_a \cdot r_b \cdot e^{i({\theta}_a - {\theta}_b)}

= r_a \cdot r_b \cdot \cos({\theta}_a - {\theta}_b) + r_a \cdot r_b \cdot \sin({\theta}_a - {\theta}_b) * i

可以看成統和了『內積外積』︰

Dot product

In mathematics, the dot product or scalar product[note 1] is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used and often called inner product (or rarely projection product); see also inner product space.

Algebraically, the dot product is the sum of the products of the corresponding entries of the two sequences of numbers. Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them. These definitions are equivalent when using Cartesian coordinates. In modern geometry, Euclidean spaces are often defined by using vector spaces. In this case, the dot product is used for defining lengths (the length of a vector is the square root of the dot product of the vector by itself) and angles (the cosine of the angle of two vectors is the quotient of their dot product by the product of their lengths).

The name “dot product” is derived from the centered dot ” · ” that is often used to designate this operation; the alternative name “scalar product” emphasizes that the result is a scalar, rather than a vector, which is the case for the vector product in three-dimensional space.

Scalar projection

Cross product

In mathematics and vector algebra, the cross product or vector product (occasionally directed area product to emphasize the geometric significance) is a binary operation on two vectors in three-dimensional space (R3) and is denoted by the symbol ×. Given two linearly independent vectors a and b, the cross product, a × b, is a vector that is perpendicular to both a and b and thus normal to the plane containing them. It has many applications in mathematics, physics, engineering, and computer programming. It should not be confused with dot product (projection product).

If two vectors have the same direction (or have the exact opposite direction from one another, i.e. are not linearly independent) or if either one has zero length, then their cross product is zero. More generally, the magnitude of the product equals the area of a parallelogram with the vectors for sides; in particular, the magnitude of the product of two perpendicular vectors is the product of their lengths. The cross product is anticommutative (i.e., a × b = −(b × a)) and is distributive over addition (i.e., a × (b + c) = a × b + a × c). The space R3 together with the cross product is an algebra over the real numbers, which is neither commutative nor associative, but is a Lie algebra with the cross product being the Lie bracket.

Like the dot product, it depends on the metric of Euclidean space, but unlike the dot product, it also depends on a choice of orientation or “handedness“. The product can be generalized in various ways; it can be made independent of orientation by changing the result to pseudovector, or in arbitrary dimensions the exterior product of vectors can be used with a bivector or two-form result. Also, using the orientation and metric structure just as for the traditional 3-dimensional cross product, one can in n dimensions take the product of n − 1 vectors to produce a vector perpendicular to all of them. But if the product is limited to non-trivial binary products with vector results, it exists only in three and seven dimensions.[1] If one adds the further requirement that the product be uniquely defined, then only the 3-dimensional cross product qualifies. (See § Generalizations, below, for other dimensions.)

The cross-product in respect to a right-handed coordinate system

 

,借之探討幾何上『垂直‧平行』矣。

雖然『SymPy』 的複數運算感覺有點陽春,依舊值得嚐鮮乎!◎

pi@raspberrypi:~ $ ipython3
Python 3.4.2 (default, Oct 19 2014, 13:31:11) 
Type "copyright", "credits" or "license" for more information.

IPython 2.3.0 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.

In [1]: from sympy import *

In [2]: init_printing()

In [3]: z, z1, z2 = symbols('z,z1,z2')

In [4]: x,y,x1,y1,x2,y2,t = symbols('x,y,x1,y1,x2,y2,t', real=True)

In [5]: z = x + y*I

In [6]: z1 = x1 + y1*I

In [7]: z2 = x2 + y2*I

In [8]: z
Out[8]: x + ⅈ⋅y

In [9]: conjugate(z)
Out[9]: x - ⅈ⋅y

In [10]: Abs(z)
Out[10]: 
   _________
  ╱  2    2 
╲╱  x  + y  

In [11]: arg(z)
Out[11]: arg(x + ⅈ⋅y)

In [12]: z1 + z2
Out[12]: x₁ + x₂ + ⅈ⋅y₁ + ⅈ⋅y₂

In [13]: re(z1 + z2)
Out[13]: x₁ + x₂

In [14]: im(z1 + z2)
Out[14]: y₁ + y₂

In [15]: z1 * z2
Out[15]: (x₁ + ⅈ⋅y₁)⋅(x₂ + ⅈ⋅y₂)

In [16]: re(z1 * z2)
Out[16]: x₁⋅x₂ - y₁⋅y₂

In [17]: im(z1 * z2)
Out[17]: x₁⋅y₂ + x₂⋅y₁

In [18]: exp(t*I)
Out[18]: 
 ⅈ⋅t
ℯ   

In [19]: exp(t*I).expand(complex=True)
Out[19]: ⅈ⋅sin(t) + cos(t)

In [20]: re(exp(t*I) *z)
Out[20]: x⋅cos(t) - y⋅sin(t)

In [21]: im(exp(t*I) *z)
Out[21]: x⋅sin(t) + y⋅cos(t)

In [22]: 

 

 

 

 

 

 

 

 

 

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‧II 》

《老子》‧第二章

天下皆知美之為美,斯惡已;皆知善之為善,斯不善已。故有無相生,難易相成,長短相形,高下相傾,音聲相和,前後相隨。

是以聖人處無為之事,行不言之教。萬物作焉而不辭,生而不有,為而不恃,功成而弗居。夫唯弗居,是以不去。

何 謂『矛盾』的呢? 『此』且『非此』,『彼』又『非彼』一類的『概念』。若問『善』與『惡』是『矛盾』的嗎?假使人世間有『不善不惡』之事或『既善且惡』之行,當然『善‧ 惡』就不可說是邏輯的『矛盾』。即使沒有『又善又惡』的事件,那『善‧惡』的概念是否就可窮盡一切現象哩!要是存在『無記』之『非善非惡 』的作為,那『善‧惡』依然不可說是邏輯的『矛盾』!!

 

讀者莫要以為『透視』在『二維』平面上的表達就一定比『三維』空間裡容易?如果說 \overline{z_1 z_2} 是一條『投影線』︰

 

代數上講可用 (1 - \lambda) z_1 + \lambda z_2, \ \lambda \in Real 表示也。此時平面上之『任一點』 z 與這條『投影線』可有多種關係︰

‧I︰ z = z_1z = z_2

‧II︰ z 不在『投影線』上,但在 \overline{Pz_1}\overline{Pz_2} 『視線』上。

‧III︰ 非 I 且 II , z 在『投影線』上。

‧IV:︰ 非 I、II、III 之『投影線』外一點。

 

此所以前一篇寫來總覺拗口為難矣︰

【相 依三點】 z_1 = x_1 + y_1 \cdot i , \ z_2 = x_2 + y_2 \cdot i, \ z_3 = x_3 + y_3 \cdot i , \ det_{12}, det_{13}, det_{23} \neq 0

如果 a \cdot z_1 + b \cdot z_2 = - c \cdot z_3 ,當下

\right) \left( \begin{array}{cc} a \\ b \end{array} \right) = - c  {\left( \begin{array}{cc} x_1 & x_2 \\ y_1 & y_2 \end{array} \right)}^{-1}  \left( \begin{array}{cc} x_3 \\ y_3 \end{array} \right)

豈能『不相依』耶!◎

 

假設 c = -1 ,上式不過是說平面裡『任一點』 z ,都可以用『任選兩個』 z_1, z_2 『線性獨立』向量所構成 z = \alpha \cdot z_1 + \beta \cdot z_2 。若講這 (\alpha, \beta) 一般也稱作『廣義座標』︰

Generalized coordinates

In analytical mechanics, specifically the study of the rigid body dynamics of multibody systems, the term generalized coordinates refers to the parameters that describe the configuration of the system relative to some reference configuration. These parameters must uniquely define the configuration of the system relative to the reference configuration.[1] This is done assuming that this can be done with a single chart. The generalized velocities are the time derivatives of the generalized coordinates of the system.

An example of a generalized coordinate is the angle that locates a point moving on a circle. The adjective “generalized” distinguishes these parameters from the traditional use of the term coordinate to refer to Cartesian coordinates: for example, describing the location of the point on the circle using x and y coordinates.

Although there may be many choices for generalized coordinates for a physical system, parameters which are convenient are usually selected for the specification of the configuration of the system and which make the solution of its equations of motion easier. If these parameters are independent of one another, the number of independent generalized coordinates is defined by the number of degrees of freedom of the system.[2][3]

 

首要『二維』『自由度』不變也。

此時如果

\alpha + \beta \neq 1

z 怎麼可能在 \overline{z_1 z_2} 線上耶??

然而 \overline{Pz} 『視線』 \lambda \cdot z = \lambda \cdot \alpha \cdot z_1 + \lambda \cdot \beta \cdot z_2 終有

\lambda \cdot \alpha + \lambda \cdot \beta = 1 時乎!!

解析者務請小心也★☆

俗話說︰不以規矩,不足以成方圓。在西方文明史中,這個僅使用『直尺』與『圓規』來『作圖』的傳統實在是淵源流長的,據聞它是來自於古希臘的『數學課題』,而且只准用『圓規』和『直尺』,並且祇能在『有限次』的『操作』裡,去解決各種不同的平面幾何『作圖題』。還有人講,它還有『抽象』的『限制條件』︰

一、直尺沒有刻度,但可以是『無限長』,而且只能夠使用直尺之固定的一側。它的『作用』可以將『已有』的『兩個點』連接在一起,成為一條『』。

二、圓規所畫的圓能夠達半徑長『無限寬』,但是上面也不可以刻有『角度』。它的主要製圖『作用』是『複製』已有的『長度』或是產生與其它之『』或者『』彼此之間的『交點』。

這引發了『古希臘三大難題
一、【化圓為方】
求一個正方形的邊長,使它的面積與一個已知的圓相等。
二、【三等分角】
求一角,使它的角度是一個已知角度的三分之一。
三、【倍立方】
求一立方體的邊長,使它的體積是一個已知立方體的二倍。

經過了上千年的努力,人們開始思考這些『問題』的『可行性』之『條件』,假使我們按造它規定所說的『條件』,這些都是『不可能』被『作出來』的了。這又是為什麼的呢?因為假使只使用『』與『』,如果從『解析幾何』的觀點來看,『尺規作圖』 Compass-and-straightedge construction 最多也只能夠是多個『二次方程式』之『疊套a_1 + b_1 \sqrt{a_2 + b_2 \sqrt{a_3 + b_3 \sqrt{\dots}}} 的罷了!若是有『不滿足』這個『條件』的『構圖』,難到它是有『可能』的『作出圖』的嗎??

比方說『化圓為方』 的方程式像是 x^2 = \pi r^2,不要說 \sqrt{\pi} 畫不畫的出來,那個 \pi 難道就畫的出來的嗎?如果思考一個『單位圓』 ── 直徑等於一,圓周長等於 \pi ── 的『內接』以及『外切』的正四邊形、八邊形、十六邊形等等『倍邊數』逼近的『正多邊形』,這個圓的『周長\pi  總是介於這些內外的『正多邊形』之『周長』之間,就算那些內外切『正多邊形』都可以借助於『半角公式\sin\frac{\theta}{2}=\pm\sqrt\frac{1-\cos\theta}{2} 表達成『二次方程式』的『疊套』形式,恐怕也得要『無限』之『尺規步驟』才能得出這個 \pi 的吧!

雖然我們已經談過了『三次方程式』的求解以及一般方程式『二次方根』解的個數是『偶數』的,如果我們直接計算 \sqrt[3]{r} = a + b \sqrt{Q}, \ Q \geq 0 會得到什麼的呢?
r = {(a + b \sqrt{Q})}^3
= a^3 + 3 a^2 b \sqrt{Q} + 3 a b^2 Q + b^3 Q \sqrt{Q}
= (a^3 + 3 a b^2 Q) + (3 a^2 b + b^3 Q) \sqrt{Q}
\therefore  3 a^2 b + b^3 Q = 0
,所以 3 a^2 + b^2 Q = 0,因為 a^2, b^2 > 0,所以 Q < 0,這就產生了假設『矛盾』的了,因此『立方根』終究是無法表達成『二次方根』的吧!!

假使我們從『三角函數』的『三倍角公式』來看
\cos \theta = 4 \cos^3 {\frac{\theta}{3}} -3 \cos {\frac{\theta}{3}}
,這又是一個『三次方程式』的啊!無怪乎,它也是『幾何作圖』之『不可能』的吧!!

110px-Cyrkiel_RB1

400px-Basic-construction-demo

220px-Straight_Square_Inscribed_in_a_Circle_240px

220px-Regular_Hexagon_Inscribed_in_a_Circle_240px

Pentagon_construct

220px-Pentagon

Squaring_the_circle.svg
化圓為方

Bisection_construction

250px-Doubling_the_cube_problem
倍立方

300px-Doubling_the_cube.svg

Trisecting_angles_three.svg
三等分角

── 摘自《【Sonic π】電路學之補充《四》無窮小算術‧中下下‧中

 

 

 

 

 

 

 

 

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‧I 》

LOST 對話錄

人的存在久遠矣,就像物種的存在一樣,哪有什麼同不同、做不做的事呢?如果存在有道理,那它若不普遍似乎比較神奇的吧!

 誰說人們超越了芝諾,人們果真聽明白了他的話嗎??有窮與無窮並不是人給的條件,而只是認知之不足的啊。就像諸神的時代已經遠離,人們怎麼還不知道如何過日子哩!!

我親愛的普羅米修斯火種是不夠用的,哪怕你認為把光明帶給世界仍舊徒然??因為人類根本無法承受那光亮照明的呢!!

䁗奧思,你為什麼這麼說呢?你明知道這火種既不屬於你也不屬於你的孿生兄弟奧德。它從無物反思自身存有迸發出的大霹靂之火而來 。只要還有時間,自然存在機會阿!當虛空歸寂反噬萬有之時,你曾經歷千百萬次,終究無法回想起是吧!!…

 

即使只有『一維視覺』,如果能作『二維運動』,不知『平面國』的科學家會應用什麼樣的數學『描述所見』呢?由於典章早就流失 ,父老傳說幾成神話!與其迷失於想像國度的『奧秘之文』,不如回首人類歷史上的『神奇之數』── 複數 ──

220px-Algebra_by_Rafael_Bombelli

150px-Leonhard_Euler_2
e^{i \pi} + 1 = 0

 

200px-Euler's_formula.svg

 

Complex_conjugate_picture.svg

220px-ImaginaryUnit5.svg

一五七二年義大利數學家拉斐爾‧邦貝利 Rafael Bombelli 是文藝復興時期歐洲著名的工程師,也是一個卓越的數學家,出版了《代數學》 L’Algebra 一書,他在書中討論了『負數的平方根\sqrt{- a}, \ a>0,這在歐洲產生了廣泛影響力。

一六三七年笛卡爾在他的著作《幾何學》 La Géométrie 書中創造了『虛數』imaginary numbers 一詞,說明這種『真實上並不存在的數字』。

瑞士大數學家和物理學家李昂哈德‧尤拉 Leonhard Euler 傳說年輕時曾研讀神學,一生虔誠篤信上帝,並不能容忍任何詆毀上帝的言論在他面前發表。一回,德尼‧狄德羅 Denis Diderot ── 法國啟蒙思想家、唯物主義哲學家、無神論者和作家,百科全書派的代表 ── 造訪葉卡捷琳娜二世的宮廷,尤拉挑戰狄德羅說︰『先生,e^{i \pi} + 1 = 0,所以上帝存在,請回答!』。作者以為這或許只是個『杜撰』。然而尤拉是位多產的作家,一生著作有六十到八十巨冊。一七八三年九月十八日,晚餐後,尤拉邊喝著茶邊和小孫女玩耍,突然間,煙斗從他手中掉了下來。他說了聲:『我的煙斗』,將彎腰去撿,就再也沒有站起來了,他祇是抱著頭說了一句:『我死了』。法國哲學家馬奎斯‧孔多塞 marquis de Condorcet 講︰..il cessa de calculer et de vivre,『尤拉停止了計算和生命』!!

一七九七年挪威‧丹麥數學家卡斯帕爾‧韋塞爾 Caspar Wessel 在『Royal Danish Academy of Sciences and Letters』上發表了『Om directionens analytiske betegning』,提出了『複數平面』,研究了複數的幾何意義,由於是用『丹麥文』寫成的,幾乎沒有引起任何重視。一八零六年法國業餘數學家讓-羅貝爾‧阿爾岡 Jean-Robert Argand 與一八三一年德國著名大數學家约翰‧卡爾‧弗里德里希‧高斯 Johann Karl Friedrich Gauß 都再次『重新發現』同一結果!!

虛數軸和實數軸構成的平面稱作複數平面,複平面上每一點對應著一個複數。 

那麽要怎樣理解『複數z = x + i \ y 的呢?如果說『複數』起源於『方程式』的『求解』,比方說 x^2 + 1 = 0, \ x = \pm i,這定義了『i = \sqrt{-1}』,但是它的『意義』依然晦澀。即使說從『複數平面』的每一個『(x, y) 都對應著一個『複數z = x + i \ y 可能還是不清楚『i』的意思到底是什麼?假使再從『複數』的『加法上看』︰

假使 z_1 = x_1 + i \ y_1z_2 = x_2 + i \ y_2

那麼 z_1 + z_2 = (x_1 + x_2) + i \ (y_1 + y_2)

這是一種類似『向量』的加法,是否『i』的意義就藏在其中的呢?

positive_negative_rotation

imaginary_rotation

220px-90-Degree_Rotations_in_the_Complex_Plane

一九九八年美國新罕布希爾大學 University of New Hampshire 的
Paul J. Nahin 教授寫了一本『An Imaginary Tale: the Story of the Square Root of −1』的書,指出韋塞爾當初所講的『幾何意義』就是︰

i = \sqrt{-1} = 1 \ \angle 90^{\circ}

也就是說『i』就是『逆時鐘旋轉九十度』的『運算子』!

假使從複數的『極座標』表示法來看複數的『乘法』︰

假 使 z_1 = r \cdot e^{i \ \theta}, \ z_2 = \alpha \cdot e^{i \ \beta},那麼 z_1 \cdot z_2 = \alpha \cdot r \cdot e^{i \ (\theta +\beta)}

就可以解釋成 Z1 『向量』被『逆時鐘旋轉』了『β』角度,它的『長度』被『縮放』了『α』倍!!

複數果真不是簡單的『』啊!也難怪它是『完備的』的喔!!

─── 摘自《【Sonic π】電聲學補充《二》

 

品嚐一下『代數法』料理『透視』的味道,或可借其『神奇』搭條通往『線性代數』之『橋樑』乎?

此處略講一些『 □ ○ ☆ 術語』作個開始︰

 

【投影線 點對應】 z_p = x+y \cdot i , {z^{'}}_p = \alpha x + \alpha y \cdot i = \alpha z_p

也可說『線性相依』︰

\alpha z_p + (-1) {z^{'}}_p = 0 。可得

det \left| \begin{array}{cc} x & \alpha x \\ y & \alpha y \end{array} \right| = 0 也。

若說兩點間非『線性相依』,則知必是『相異兩點』矣。此時稱作『線性獨立』。

 

【獨立兩點】 z_1 = x_1 + y_1 \cdot i , \ z_2 = x_2 + y_2 \cdot i , \ det_{12} = det \left| \begin{array}{cc} x_1 & x_2 \\ y_1 & y_2 \end{array} \right| \neq  0

假設

a \cdot z_1 + b \cdot z_2 = 0 ,那麼

\Rightarrow \ a = 0, \ b=0

這從『聯立方程式』求解可知乎?

\left( \begin{array}{cc} x_1 & x_2 \\ y_1 & y_2  \end{array} \right) \left( \begin{array}{cc} a \\ b \end{array} \right) =0

 

【相依三點】 z_1 = x_1 + y_1 \cdot i , \ z_2 = x_2 + y_2 \cdot i, \ z_3 = x_3 + y_3 \cdot i , \ det_{12}, det_{13}, det_{23} \neq 0

如果 a \cdot z_1 + b \cdot z_2 = - c \cdot z_3 ,當下

\right) \left( \begin{array}{cc} a \\ b \end{array} \right) = - c  {\left( \begin{array}{cc} x_1 & x_2 \\ y_1 & y_2 \end{array} \right)}^{-1}  \left( \begin{array}{cc} x_3 \\ y_3 \end{array} \right)

 

豈能『不相依』耶!◎

 

 

 

 

 

 

 

 

 

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 V 》

無奈這個『一維鏡頭』只有『單點』之象觀,無論『視野』 A,B 之『角度』大小︰

Angle of view

In photography, angle of view (AOV)[1] describes the angular extent of a given scene that is imaged by a camera. It is used interchangeably with the more general term field of view.

425px-Angle_of_view.svg

A camera’s angle of view can be measured horizontally, vertically, or diagonally.

It is important to distinguish the angle of view from the angle of coverage, which describes the angle range that a lens can image. Typically the image circle produced by a lens is large enough to cover the film or sensor completely, possibly including some vignetting toward the edge. If the angle of coverage of the lens does not fill the sensor, the image circle will be visible, typically with strong vignetting toward the edge, and the effective angle of view will be limited to the angle of coverage.

A camera’s angle of view depends not only on the lens, but also on the sensor. Digital sensors are usually smaller than 35mm film, and this causes the lens to have a narrower angle of view than with 35mm film, by a constant factor for each sensor (called the crop factor). In everyday digital cameras, the crop factor can range from around 1 (professional digital SLRs), to 1.6 (consumer SLR), to 2 (Micro Four Thirds ILC) to 4 (enthusiast compact cameras) to 6 (most compact cameras). So a standard 50mm lens for 35mm photography acts like a 50mm standard “film” lens even on a professional digital SLR, but would act closer to an 80mm lens (1.6 x 50mm) on many mid-market DSLRs, and the 40 degree angle of view of a standard 50mm lens on a film camera is equivalent to a 28 – 35mm lens on many digital SLRs.

Angle_of_View_F_V_Chambers_1916

In 1916, Northey showed how to calculate the angle of view using ordinary carpenter’s tools.[2] The angle that he labels as the angle of view is the half-angle or “the angle that a straight line would take from the extreme outside of the field of view to the center of the lens;” he notes that manufacturers of lenses use twice this angle.

Camera_focal_length_distance_house_animation

In this simulation, adjusting the angle of view and distance of the camera while keeping the object in frame results in vastly differing images. At distances approaching infinity, the light rays are nearly parallel to each other, resulting in a “flattened” image. At low distances and high angles of view objects appear “foreshortened”.

Derivation of the angle-of-view formula

Consider a rectilinear lens in a camera used to photograph an object at a distance  S_{1}, and forming an image that just barely fits in the dimension,  d, of the frame (the film or image sensor). Treat the lens as if it were a pinhole at distance  S_{2} from the image plane (technically, the center of perspective of a rectilinear lens is at the center of its entrance pupil):[7]

Lens_angle_of_view.svg

Now  \alpha /2 is the angle between the optical axis of the lens and the ray joining its optical center to the edge of the film. Here  \alpha is defined to be the angle-of-view, since it is the angle enclosing the largest object whose image can fit on the film. We want to find the relationship between:

the angle  \alpha
the “opposite” side of the right triangle, d/2 (half the film-format dimension)
the “adjacent” side,  S_{2} (distance from the lens to the image plane)

Using basic trigonometry, we find:

  \tan(\alpha /2)={\frac {d/2}{S_{2}}}.

which we can solve for α, giving:

\alpha =2\arctan {\frac {d}{2S_{2}}}

To project a sharp image of distant objects, S_{2} needs to be equal to the focal lengthF, which is attained by setting the lens for infinity focus. Then the angle of view is given by:

\alpha =2\arctan {\frac {d}{2f}} where  f=F

Note that the angle of view varies slightly when the focus is not at infinity (See breathing (lens)), given by  S_{2}={\frac {S_{1}f}{S_{1}-f}} rearranging the lens equation.

Macro photography

For macro photography, we cannot neglect the difference between S_{2} and  F. From the thin lens formula,

{\frac {1}{F}}={\frac {1}{S_{1}}}+{\frac {1}{S_{2}}}.

From the definition of magnificationm=S_{2}/S_{1}, we can substitute  S_{1} and with some algebra find:

S_{2}=F\cdot (1+m)

Defining  f=S_{2} as the “effective focal length”, we get the formula presented above:

\alpha =2\arctan {\frac {d}{2f}} where f=F\cdot (1+m).

A second effect which comes into play in macro photography is lens asymmetry (an asymmetric lens is a lens where the aperture appears to have different dimensions when viewed from the front and from the back). The lens asymmetry causes an offset between the nodal plane and pupil positions. The effect can be quantified using the ratio (P) between apparent exit pupil diameter and entrance pupil diameter. The full formula for angle of view now becomes:[5]

\alpha =2\arctan {\frac {d}{2F\cdot (1+m/P)}}

─── 摘自《光的世界︰矩陣光學六辛

 

難免受限於

笛沙格定理

笛沙格(Desargues)定理說明:在射影空間中,有六點A,B,C,a,b,c。Aa,Bb,Cc共點若且唯若AB∩ab,BC∩bc,CA∩ca共線。

射影幾何的對偶性來看,笛沙格定理是自對偶的。

笛沙格(Desargues)定理說明:在射影空間中,有六點A,B,C,a,b,c。Aa,Bb,Cc共點若且唯若AB∩ab,BC∩bc,CA∩ca共線。

射影幾何的對偶性來看,笛沙格定理是自對偶的。

 

也。

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Desargues’ Theorem

Let A1B1C1 and A2B2C2 be two triangles. Consider two conditions:

  1. Lines A1A2, B2B2, C1C2 joining the corresponding vertices are concurrent.
  2. Points ab, bc, ca of intersection of the (extended) sides A1B1 and A2B2, B1C1 and B2C2, C1A1 and C2A2, respectively, are collinear.

Desargues’ Theorem claims that 1. implies 2. It’s dual asserts that 1. follows from 2. In particular, the dual to Desragues’ theorem coincides with its converse.

(In the applet below each of the triangles as well as each of the vertices is draggable.)

Created with GeoGebra

Two triangles that satisfy the first condition are said to be perspecitve from a point. Two triangles that satisfy the second condition are said to be perspecitve from a line. Desargues‘ theorem thus claims that two triangles perspective from a point are perspective from a line. Its dual asserts that two triangles perspective from a line are also perspective from a point.

Monge’s theorem can be derived from that of Desargues and in fact is the latter in disguise. The existence of an orthic axis of a triangle is also an immediate consequence of Desargues’ Theorem. In a somewhat disguised form Desargues’ Theorem establishes a relationship between a triangle and a cevian triangle of a point not on a triangle itself.

Curiously, Desragues’ theorem admits an intuitive proof if considered as a statement in the 3-dimensional space, but is not as easy in the 2-dimensional case, where it is often taken as an axiom.

Following is the proof (kindly supplied by Hubert Shutrick) that adopts the 3-dimensional perspective. (The proof fails in some exceptional configurations, e.g., when A1, A2, B1, and B2 are collinear. These are simple enough to be treated individually and will not be considered here. We simply restrict the theorem to two triangles in general position, i.e., assuming that A1≠A2, B1≠B2, C1≠C2 and that, similarly, no two corresponding side lines coincide.)

 

聽聞此恰是『平面國』之『點投派』將『投影線』『抽象公設化』的時候,所遭遇『四元 □ ○ 完備』困難哩◎

 Complete Quadrangle

CompleteQuadrangle

If the four points making up a quadrilateral are joined pairwise by six distinct lines, a figure known as a complete quadrangle results. A complete quadrangle is therefore a set of four points, no three collinear, and the six lines which join them. Note that a complete quadrilateral is different from a complete quadrangle.

The midpoints of the sides of any complete quadrangle and the three diagonal points all lie on a conic known as the nine-point conic. If it is an orthocentric quadrilateral, the conic reduces to a circle.

Complete Quadrilateral

CompleteQuadrilateral

The figure determined by four lines, no three of which are concurrent, and their six points of intersection (Johnson 1929, pp. 61-62). Note that this figure is different from a complete quadrangle. A complete quadrilateral has three diagonals (compared to two for an ordinary quadrilateral). The midpoints of the diagonals of a complete quadrilateral are collinear on a line M (Johnson 1929, pp. 152-153).

A theorem due to Steiner (Mention 1862ab, Johnson 1929, Steiner 1971) states that in a complete quadrilateral, the bisectors of angles are concurrent at 16 points which are the incenters and excenters of the four triangles. Furthermore, these points are the intersections of two sets of four circles each of which is a member of a conjugate coaxal system. The axes of these systems intersect at the point common to the circumcircles of the quadrilateral.

Newton proved that, if a conic section is inscribed in a complete quadrilateral, then its center lies on M (Wells 1991). In addition, the orthocenters of the four triangles formed by a complete quadrilateral lie on a line which is perpendicular to M. Plücker proved that the circles having the three diagonals as diameters have two common points which lie on the line joining the four triangles’ orthocenters (Wells 1991).