藉著引進『單位尺』 $\ddagger$ 度量平面，

然後思考什麼『變』了，什麼『不變』的呢？

pi@raspberrypi:~  \left( \begin{array}{cc} A & 0 \\ C & D \end{array} \right) l \to l^{'}\left( \begin{array}{cc} \frac{b}{a} & 0 \\ \frac{1- \frac{1}{k}}{a} & \frac{1}{k}  \end{array} \right)\overline{AB}\ddaggera=1, \ \left( \begin{array}{cc} \frac{b}{a} & 0 \\ \frac{1- \frac{1}{k}}{a} & \frac{1}{k}  \end{array} \right) \Rightarrow \ \left( \begin{array}{cc} b & 0 \\ 1- \frac{1}{k} & \frac{1}{k}  \end{array} \right) = \left( \begin{array}{cc} b & 0 \\ 0 & 1 \end{array} \right)   \  \left( \begin{array}{cc} 1 & 0 \\ 1- \frac{1}{k} & \frac{1}{k}  \end{array} \right) l=1BAsll^{'}A^{'}sa,b(x_{\Box}, t_{\Box})(x_{\bigcirc}, t_{\bigcirc})v\begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix} = G_v \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix} = \begin{pmatrix} 1 & -v \\0 & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}(0, t_{\Box})x_{\bigcirc} = - v \cdot t_{\Box}t_{\bigcirc} = t_{\Box}v(x_{\Box}^1, t_{\Box})(x_{\Box}^2, t_{\Box})(x_{\Box}^1 - v \cdot t_{\Box}, t_{\Box})(x_{\Box}^2 - v \cdot t_{\Box}, t_{\Box})\begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix} = G_{-v} \begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix} G_{-v} = {G_v}^{-1} = \begin{pmatrix} 1 & v \\0 & 1 \end{pmatrix}\begin{pmatrix} 1 & v \\0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & - v \\0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & - v \\0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & v \\0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}(x_{\star}, t_{\star})w\begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix} = G_v \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix} = \begin{pmatrix} 1 & -v \\0 & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}\begin{bmatrix} x_{\star} \\ t_{\star} \end{bmatrix} = G_w \begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix} = \begin{pmatrix} 1 & -w \\0 & 1 \end{pmatrix} \begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix}=  \begin{pmatrix} 1 & -w \\0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & -v \\0 & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}=  \begin{pmatrix} 1 & -(w+v) \\0 & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}=  G_{(w+v)} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}\begin{bmatrix} x_{\bigcirc} \\ t_{\bigcirc} \end{bmatrix} =  L_v \begin{bmatrix} x \\ t \end{bmatrix} = \frac{1}{\sqrt{1 - {(\frac{v}{c})}^2}}  \begin{pmatrix} 1 & -v \\ -\frac{v}{c^2} & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}-\frac{v}{c^2} L_{- v} = L_{v}^{-1}L_w \cdot L_v = L_{w \bigoplus v} = \frac{1}{\sqrt{1 - {[\frac{(w+v)/c}{1+(wv/c^2)}]}^2}}  \begin{pmatrix} 1 & -[\frac{(w+v)}{1+(wv/c^2)}] \\ -\frac{1}{c^2} {[\frac{(w+v)}{1+(wv/c^2)}]}^2 & 1 \end{pmatrix} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}\neq L_{(w + v)} \begin{bmatrix} x_{\Box} \\ t_{\Box} \end{bmatrix}w \bigoplus v = \frac{w + v}{1 + (w v / c^2)}L_{w \bigoplus v}  = \frac{1}{\sqrt{1 - {(\frac{(w \bigoplus v)}{c})}^2}}  \begin{pmatrix} 1 & -(w \bigoplus v) \\ -\frac{(w \bigoplus v)}{c^2} & 1 \end{pmatrix} $ 這又能有什麼『不對』的嗎？因是之故，『狹義相對論』所帶來的『困惑』遠勝於『運動之不可能性』，反倒以為『運動』果真能是這種『現象』的嘛！！


果真『阿基里斯』 Achilles 追得上『烏龜』，然而這件事又怎麽『可能』用著『數理邏輯』來『證明』的呢？假使說『大自然』果是『如是』，我們卻『不要』『如是』的『相信』，那麼我們『該用』什麼『道理』來『論辯』的呢？？偉大的科學家『愛因斯坦』說︰我會為親愛的上帝感到遺憾。不管怎樣理論都還是正確的。



誰趕的上芝諾的烏龜呢？


─── 摘自《【Sonic π】電聲學之電路學《四》之《 !!!! 》下》

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 III 》

2017-08-25 懸鉤子

承上篇，在此『單點透視』

※ 此處 $x,y,z$ 都是本地座標系，皆如是賦值︰

$\frac{\overline{C^{''}A^{''}}}{\overline{C^{''}B^{''}}} = \frac{\overline{C^{''}A^{''}}}{\overline{C^{''}A{''}} - \overline{A^{''}B^{''}}} = \frac{z}{z -1}, \ z =_{df} \frac{\overline{C^{''}A{''}}}{\overline{A{''}B{''}}}$ 。

之特定安排下，任意『變換矩陣』

$\left( \begin{array}{cc} y \\ 1 \end{array} \right) \ {\overset {P}{\doublebarwedge }} \ \left( \begin{array}{cc} 1 & 0 \\ 1 - \frac{1}{k} & \frac{1}{k} \end{array} \right) \left( \begin{array}{cc} x \\ 1 \end{array} \right)$

$\left( \begin{array}{cc} x \\ 1 \end{array} \right) \ {\overset {P}{\doublebarwedge }} \ { \left( \begin{array}{cc} 1 & 0 \\ 1 - \frac{1}{k} & \frac{1}{k} \end{array} \right)}^{-1} \left( \begin{array}{cc} y \\ 1 \end{array} \right)$

，都是『下三角』形式

$\left( \begin{array}{cc} A & 0 \\ C & D \end{array} \right)$ 。

故而皆是『可交換矩陣』也。

Commuting matrices

In linear algebra, two matrices $A$ and $B$ are said to commute if $AB=BA$ and equivalently, their commutator $[A,B]= AB-BA$ is zero. A set of matrices $A_{1},\ldots ,A_{k}$ is said to commute if they commute pairwise, meaning that every pair of matrices in the set commute with each other.

但是我們又怎麼知道這不是來自『本地座標系』的『單位長度』之不同呢？？且把『』尺寸引入，建立

度量空間

在數學中，度量空間是個具有距離函數的集合，該距離函數定義集合內所有元素間之距離。此一距離函數被稱為集合上的度量。

度量空間中最符合人們對於現實直觀理解的為三維歐幾里得空間。事實上，「度量」的概念即是歐幾里得距離四個周知的性質之推廣。歐幾里得度量定義了兩點間之距離為連接這兩點的直線段之長度。此外，亦存在其他的度量空間，如橢圓幾何與雙曲幾何，而在球體上以角度量測之距離亦為一度量。特殊相對論使用雙曲幾何的雙曲面模型，作為速度之度量空間。

度量空間還能導出開集與閉集之類的拓撲性質，這導致了對更抽象的拓撲空間之研究。

吧！！假設

$x \overline{AB} = x \cdot a \ddagger = x^{'} \ddagger$ ，

$y \overline{A^{'}B{'}} = y \cdot b \ddagger = y^{'} \ddagger$ 。

且用矩陣代數計算看看哩︰

pi@raspberrypi:~ l, l^{'}PlXPXl^{'}X^{'}l^{'}Y^{'}PYlYf_{P} \colon : [ \overline{AC} ]  \mapsto [ \overline{A^{'} C^{'}}]ll^{'}A, B\frac{\overline{CA}}{\overline{CB}} = \frac{\overline{PA}}{\overline{PB}}  \cdot \frac{\sin (\angle{APC})}{\sin(\angle{BPC})}A^{'}, B^{'}\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B{'}}} = \frac{\overline{PA^{'}}}{\overline{PB^{'}}}  \cdot \frac{\sin (\angle{APC})}{\sin(\angle{BPC})}\frac{\frac{\overline{CA}}{\overline{CB}}}{\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B{'}}} } = \frac{\frac{\overline{PA}}{\overline{PB}}}{\frac{\overline{PA^{'}}}{\overline{PB^{'}}}}\frac{\overline{CA}}{\overline{CB}}\frac{\overline{CA}}{\overline{CB}} = \frac{\overline{CA}}{\overline{CA} - \overline{AB}} = \frac{x}{x -1}, \ x =_{df} \frac{\overline{CA}}{\overline{AB}} \frac{\overline{CA}}{\overline{CB}} = \frac{\overline{CB} + \overline{BA}}{\overline{CB}} = \frac{y+1}{y}, \ y =_{df} \frac{\overline{CB}}{\overline{BA}} f_P (f_P (X)) = X\frac{x}{x-1} \Longrightarrow  \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} = \frac{x}{x-1} \frac{y+1}{y} \Longrightarrow  \frac{\frac{y+1}{y} +1}{\frac{y+1}{y}}  = \frac{2 y +1}{y+1}(\lambda x. ((\lambda y. (x \ y)) \ \Box)  \ \bigcirc)\beta ((\lambda y. ( \bigcirc \ y))  \ \Box )\beta(\lambda x. ( (x \ \Box )   \ \bigcirc)\beta(\bigcirc \ \Box )\beta$ 化約』不管是用著怎麽樣的『步驟次序』，都一定能夠得到『相同結果』的嗎？？
─── 摘自《光的世界︰矩陣光學六壬》

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 II 》

2017-08-24 懸鉤子

如何『轉變』 Transform 想法呢？比方求 $7^{819} \mod 10 = ?$ 。假使先計算 $7^{819}$ 再除以 $10$ ，恐怕使用『全球計算機網路』，還得計算很久哩！要是『轉換』觀點，借著『模算數』之『定理』︰

$(m \times n) \mod k = \left[ (m \mod k) \times (n \mod k) \right] \mod k$ 。

那麼知道

$7 = 7 \mod 10, \ 7^2 = 49 \mod 10 =9, \ 7^3=343 \mod 10 = 3, \ 7^4=2401 \mod 10 =1, \ \cdots$

的人就可以利用

$7^{819} \mod 10 = \left[ {(7^4)}^{204} \times (7^3) \right] \mod 10 = (1 \times 3) \mod 10 = 3$

得知也！！

此所以天下『觀物』與『物觀』之觀點，尚得能有『形變』 Transformation 之『心』乎？？

旅夜書懷

細草微風岸，
危檣獨夜舟。

星垂平野闊，
月湧大江流。

名豈文章著？
官應老病休。

飄飄何所似？
天地一沙鷗。

杜甫

在《一個奇想！！》一文中，提及『計算機種子』 Lick 一張標題為『Members and Affiliates of the Intergalactic Computer Network』給工作同仁的備忘錄︰
“imagined as an electronic commons open to all, ‘the main and essential medium of informational interaction for governments, institutions, corporations, and individuals.’”

許下了『銀河際網路』的願景，開啟了今天的『網際網路』！！

古人有『玉有十德』之說『仁、知、義、禮、樂、忠、信、天、地、德』，而且都是源自『大道』，真可說是善於『觀物』取象，意在象外的了。有人認為『玉有十德』實屬『牽強附會』之說，也許他有些不明白，人類所『珍惜的價值』其實都是一種『信念』，於是才用著各種『象徵』來『表意』，即使是『流行』與『時尚』所代表的『意義』，或者『整形』和『美容』所追求之『目的』，歸根究底『作法相似』，不過『取向不同』罷了！『觀』之一事，確實是『行』之不易的啊！祇就『觀人』這事而言，無怪乎，連『孔老夫子』都只能說︰『始吾於人也，聽其言而信其行。今吾於人也，聽其言而觀其行。於予與改是！』的吧！！

過去東方一代宗師『陳寅恪』認為『對對子』，包含了『微觀』與『宏觀』的『文化』，於是『大學聯考』出了『一道怪題』︰以『孫行者』為上聯要求對下聯。而西方思想種子『Lick』能夠『由微知顯』所以會生『銀河際網路』的『一個奇想』。因此我們可以知道『觀』的重要性，由於『錯覺』與『謬觀』也可能發生，如何校之以『合理性』就更顯『必要』的了。『易經‧師』卦有『彖』曰：

師，眾也，貞正也，能以眾正，可以王矣。剛中而應，行險而順，以此毒天下，而民從之，吉又何咎矣。

這個『毒天下』之『毒』應當怎麼『解釋』的呢？如果依據《説文解字》：毒，厚也。害人之艸，往往而生。从屮，从毒。『毒草』果真能『以毒攻毒』談『生民』的嗎？為何又『民從之』？難道是以『苦毒為樂』的嗎？？

赫林錯視
平行之不平行

加斯特羅圖形
相同卻不同

赫曼方格
不存在能存在嗎？

弗雷澤圖形
同心還是不同心！

─── 摘自《【Sonic π】電聲學之電路學《四》之《 !!! 》》

如是『』尺寸無涉？祇靠幾何概念！

※ 此處 $x,y,z$ 都是本地座標系，皆如是賦值︰

又將通往哪裡矣！！

如果 $f$ 將 $l$ 投射到 $l^{'}$ ，且 $g$ 將 $l^{'}$ 投射到 $l^{''}$

$\frac{\frac{\overline{CA}}{\overline{CB}}}{\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B{'}}} } = \frac{\frac{\overline{PA}}{\overline{PB}}}{\frac{\overline{PA^{'}}}{\overline{PB^{'}}}} = \frac{1}{k_{ll^{'}}}$

$\frac{\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B^{'}}}}{\frac{\overline{C^{''}A^{''}}}{\overline{C^{''}B{''}}} } = \frac{\frac{\overline{PA^{'}}}{\overline{PB^{'}}}}{\frac{\overline{PA^{''}}}{\overline{PB^{''}}}} = \frac{1}{k_{l^{'}l^{''}}}$

此時若講

$\frac{\frac{\overline{CA}}{\overline{CB}}}{\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B{'}}} } \cdot \frac{\frac{\overline{C^{'}A^{'}}}{\overline{C^{'}B^{'}}}}{\frac{\overline{C^{''}A^{''}}}{\overline{C^{''}B{''}}} } = \frac{\frac{\overline{CA}}{\overline{CB}}}{\frac{\overline{C^{''}A^{''}}}{\overline{C^{''}B{''}}} } =\frac{1}{k_{ll^{'}}} \cdot \frac{1}{k_{l^{'}l^{''}}}$ 。

，當下是否可證『 f∘g 』必滿足 $l \to l^{''}$ 組合耶◎

Function composition

In mathematics, function composition is the pointwise application of one function to the result of another to produce a third function. For instance, the functions $f : X \to Y$ and $g : Y \to Z$ can be composed to yield a function which maps $x$ in $X$ to $g (f (x))$ in $Z$ . Intuitively, if $z$ is a function of $y$ , and $y$ is a function of $x$ , then $z$ is a function of $x$ . The resulting composite function is denoted $g \circ f : X \to Z$ , defined by $(g \circ f)(x) = g (f (x))$ for all $x$ in $X$ .^{[note 1]} The notation $g \circ f$ is read as “ $g$ circle $f$ “, or “ $g$ round $f$ “, or “ $g$ composed with $f$ “, “ $g$ after $f$ “, “ $g$ following $f$ “, or “ $g$ of $f$ “, or “ $g$ on $f$ “. Intuitively, composing two functions is a chaining process in which the output of the inner function becomes the input of the outer function.

The composition of functions is a special case of the composition of relations, so all properties of the latter are true of composition of functions.^[1] The composition of functions has some additional properties.

Concrete example for the composition of two functions.

Composition monoids

Suppose one has two (or more) functions $f : X \to X,$ $g : X \to X$ having the same domain and codomain; these are often called transformations. Then one can form chains of transformations composed together, such as $f \circ f \circ g \circ f$ . Such chains have the algebraic structure of a monoid, called a transformation monoid or (much more seldom) composition monoid. In general, transformation monoids can have remarkably complicated structure. One particular notable example is the de Rham curve. The set of all functions $f : X \to X$ is called the full transformation semigroup^[3] or symmetric semigroup^[4] on $X$ . (One can actually define two semigroups depending how one defines the semigroup operation as the left or right composition of functions.^[5])

If the transformation are bijective (and thus invertible), then the set of all possible combinations of these functions forms a transformation group; and one says that the group is generated by these functions. A fundamental result in group theory, Cayley’s theorem, essentially says that any group is in fact just a subgroup of a permutation group (up to isomorphism).^[6]

The set of all bijective functions $f : X \to X$ (called permutations) forms a group with respect to the composition operator. This is the symmetric group, also sometimes called the composition group.

In the symmetric semigroup (of all transformations) one also finds a weaker, non-unique notion of inverse (called a pseudoinverse) because the symmetric semigroup is a regular semigroup.^[7]

The similarity that transforms triangle EFA into triangle ATB is the composition of a homothety $H$ and a rotation $R$ , of which the common centre is S. For example, the image of A under the rotation $R$ is U, which may be written $R$ (A) = U. And $H$ (U) = B means that the mapping $H$ transforms U into B. Thus $H (R$ (A)) = $(H \circ R)$ (A) = B.

※ 參考

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]: k1, k2 = symbols('k1, k2')

In [4]: f = Matrix(([1,0],[1 - 1/k1 , 1/k1]))

In [5]: f
Out[5]: 
⎡  1     0 ⎤
⎢          ⎥
⎢    1   1 ⎥
⎢1 - ──  ──⎥
⎣    k₁  k₁⎦

In [6]: g = Matrix(([1,0],[1 - 1/k2 , 1/k2]))

In [7]: g
Out[7]: 
⎡  1     0 ⎤
⎢          ⎥
⎢    1   1 ⎥
⎢1 - ──  ──⎥
⎣    k₂  k₂⎦

In [8]: g * f
Out[8]: 
⎡       1           0  ⎤
⎢                      ⎥
⎢        1             ⎥
⎢    1 - ──            ⎥
⎢        k₁   1     1  ⎥
⎢1 + ────── - ──  ─────⎥
⎣      k₂     k₂  k₁⋅k₂⎦

In [9]: (g * f)[1,0].simplify()
Out[9]: 
      1  
1 - ─────
    k₁⋅k₂

In [10]: f * g
Out[10]: 
⎡       1           0  ⎤
⎢                      ⎥
⎢        1             ⎥
⎢    1 - ──            ⎥
⎢        k₂   1     1  ⎥
⎢1 + ────── - ──  ─────⎥
⎣      k₁     k₁  k₁⋅k₂⎦

In [11]: (f * g)[1,0].simplify()
Out[11]: 
      1  
1 - ─────
    k₁⋅k₂

In [12]:

Triangular matrix

In the mathematical discipline of linear algebra, a triangular matrix is a special kind of square matrix. A square matrix is called lower triangular if all the entries above the main diagonal are zero. Similarly, a square matrix is called upper triangular if all the entries below the main diagonal are zero. A triangular matrix is one that is either lower triangular or upper triangular. A matrix that is both upper and lower triangular is called a diagonal matrix.

Because matrix equations with triangular matrices are easier to solve, they are very important in numerical analysis. By the LU decomposition algorithm, an invertible matrix may be written as the product of a lower triangular matrix L and an upper triangular matrix U if and only if all its leading principal minors are non-zero.

Binary lower unitriangular Toeplitz matrices, multiplied using F₂ operations
They form the Cayley table of Z₄ and correspond to powers of the 4-bit Gray code permutation.

Simultaneous triangularisability

A set of matrices $A_{1},\ldots ,A_{k}$ are said to be simultaneously triangularisable if there is a basis under which they are all upper triangular; equivalently, if they are upper triangularizable by a single similarity matrix P. Such a set of matrices is more easily understood by considering the algebra of matrices it generates, namely all polynomials in the $A_{i},$ denoted $K[A_{1},\ldots ,A_{k}].$ Simultaneous triangularizability means that this algebra is conjugate into the Lie subalgebra of upper triangular matrices, and is equivalent to this algebra being a Lie subalgebra of a Borel subalgebra.

The basic result is that (over an algebraically closed field), the commuting matrices $A,B$ or more generally $A_{1},\ldots ,A_{k}$ are simultaneously triangularizable. This can be proven by first showing that commuting matrices have a common eigenvector, and then inducting on dimension as before. This was proven by Frobenius, starting in 1878 for a commuting pair, as discussed at commuting matrices. As for a single matrix, over the complex numbers these can be triangularized by unitary matrices.

The fact that commuting matrices have a common eigenvector can be interpreted as a result of Hilbert’s Nullstellensatz: commuting matrices form a commutative algebra $K[A_{1},\ldots ,A_{k}]$ over $K[x_{1},\ldots ,x_{k}]$ which can be interpreted as a variety in k-dimensional affine space, and the existence of a (common) eigenvalue (and hence a common eigenvector) corresponds to this variety having a point (being non-empty), which is the content of the (weak) Nullstellensatz. In algebraic terms, these operators correspond to an algebra representation of the polynomial algebra in k variables.

This is generalized by Lie’s theorem, which shows that any representation of a solvable Lie algebra is simultaneously upper triangularizable, the case of commuting matrices being the abelian Lie algebra case, abelian being a fortiori solvable.

More generally and precisely, a set of matrices $A_{1},\ldots ,A_{k}$ is simultaneously triangularisable if and only if the matrix $p(A_{1},\ldots ,A_{k})[A_{i},A_{j}]$ is nilpotent for all polynomials p in k non-commuting variables, where $[A_{i},A_{j}]$ is the commutator; note that for commuting $A_{i}$ the commutator vanishes so this holds. This was proven in (Drazin, Dungey & Gruenberg 1951); a brief proof is given in (Prasolov 1994, pp. 178–179). One direction is clear: if the matrices are simultaneously triangularisable, then $[A_{i},A_{j}]$ is strictly upper triangularizable (hence nilpotent), which is preserved by multiplication by any $A_{k}$ or combination thereof – it will still have 0s on the diagonal in the triangularizing basis.

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 I 》

2017-08-23 懸鉤子

『』尺有所短，『』寸有所長◎

※ 此處 $x,y,z$ 都是本地座標系，皆如是賦值︰

若問是否存在 $k_{ll'} <0$ 之『透視』情況呢？

知道『孟氏定理』者

Menelaus’ theorem

Menelaus’ theorem, named for Menelaus of Alexandria, is a proposition about triangles in plane geometry. Given a triangle ABC, and a transversal line that crosses BC, AC and AB at points D, E and F respectively, with D, E, and F distinct from A, B and C, then

{\frac {AF}{FB}}\times {\frac {BD}{DC}}\times {\frac {CE}{EA}}=-1.

or simply

AF\times BD\times CE=-FB\times DC\times EA.

This equation uses signed lengths of segments, in other words the length AB is taken to be positive or negative according to whether A is to the left or right of B in some fixed orientation of the line. For example, AF/FB is defined as having positive value when F is between A and B and negative otherwise.

The converse is also true: If points D, E and F are chosen on BC, AC and AB respectively so that

{\frac {AF}{FB}}\times {\frac {BD}{DC}}\times {\frac {CE}{EA}}=-1,

then D, E and F are collinear. The converse is often included as part of the theorem.

The theorem is very similar to Ceva’s theorem in that their equations differ only in sign.

Menelaus’ theorem, case 1: line DEF passes inside triangle ABC

情況1：直線LMN穿過三角形ABC

證明

如情況一，設 $\angle ANM=\alpha$ ， $\angle AMN=\beta$ ， $\angle MLC=\gamma$ ，則在 $\triangle AMN$ 中由正弦定理，有

{\frac {AN}{AM}}={\frac {\sin \beta }{\sin \alpha }},

同理，因對頂角相等在 $\triangle NBL$ 和 $\triangle CLM$ 中有

{\frac {BL}{BN}}={\frac {\sin \alpha }{\sin \gamma }},

及

{\frac {CM}{CL}}={\frac {\sin \gamma }{\sin \beta }}.

三式相乘即得

{\frac {AN}{AM}}\cdot {\frac {BL}{BN}}\cdot {\frac {CM}{CL}}={\frac {\sin \beta }{\sin \alpha }}\cdot {\frac {\sin \alpha }{\sin \gamma }}\cdot {\frac {\sin \gamma }{\sin \beta }}=1,

即

{\frac {AN}{NB}}\cdot {\frac {BL}{LC}}\cdot {\frac {CM}{MA}}=1.

或可假借構圖也！

※ 這裡

$A \ {\overset {P}{\doublebarwedge }} \ A^{'}$ 、 $B \ {\overset {P}{\doublebarwedge }} \ B^{'}$ 、 $C \ {\overset {P}{\doublebarwedge }} \ C^{'}$

線 $L$ 與線 $L'$ 相交於 $X$ 點。

且由該定理推知固有『 $k = -1$ 』之組態矣？！

其實從落於線段 $\overline{AB}$ 之外的 $C$ 點，被 $P$ 投影至線段 $\overline{A^{'}B{'}}$ 之內的 $C^{'}$ 點，依『賦值約定』亦可知也！？

當是時

$\frac{\frac{x}{x-1}}{\frac{y}{y-1}} = -1, \ \Rightarrow y = \frac{x}{2x -1}$ 。

它的矩陣表現為

$\left( \begin{array}{cc} y \\ 1 \end{array} \right) \ {\overset {P}{\doublebarwedge }} \ \left( \begin{array}{cc} 1 & 0 \\ 1 - (-1) & -1 \end{array} \right) \left( \begin{array}{cc} x \\ 1 \end{array} \right)$

$= \left( \begin{array}{cc} 1 & 0 \\ 2 & -1 \end{array} \right) \left( \begin{array}{cc} x \\ 1 \end{array} \right)$ 。

為什麼特別談這個矩陣呢？？難道不是將 $k=-1$ 代入

一般『透視關係式』就得到了嘛！！

但思

$\frac{x}{x-1} \ \equiv_{df} \left( \begin{array}{cc} 1 & 0 \\ 1 & -1 \end{array} \right)$

雖然是『冪等』的︰

Idempotence

Idempotence (UK: /ˌɪdɛmˈpoʊtns/;^[1] US: /ˌaɪdəmˈpoʊtəns/ EYE-dəm-POH-təns)^[2] is the property of certain operations in mathematics and computer science, that can be applied multiple times without changing the result beyond the initial application. The concept of idempotence arises in a number of places in abstract algebra (in particular, in the theory of projectors and closure operators) and functional programming (in which it is connected to the property of referential transparency).

The term was introduced by Benjamin Peirce^[3] in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means “(the quality of having) the same power”, from idem + potence (same + power).

There are several meanings of idempotence, depending on what the concept is applied to:

A unary operation (or function) is idempotent if, whenever it is applied twice to any value, it gives the same result as if it were applied once; i.e., ƒ(ƒ(x)) ≡ ƒ(x). For example, the absolute value function, where abs(abs(x)) ≡ abs(x), is idempotent.
Given a binary operation, an idempotent element (or simply an “idempotent”) for the operation is a value for which the operation, when given that value for both of its operands, gives that value as the result. For example, the number 1 is an idempotent of multiplication: 1 × 1 = 1.
A binary operation is called idempotent if all elements are idempotent elements with respect to the operation. In other words, whenever it is applied to two equal values, it gives that value as the result. For example, the function giving the maximum value of two equal values is idempotent: max(x, x) ≡ x.

Other examples

In Boolean algebra, both the logical and and the logical or operations are idempotent. This implies that every element of Boolean algebra is idempotent with respect to both of these operations. Specifically, $x \wedge x = x$ and $x \vee x = x$ for all $x$ . In linear algebra, projections are idempotent. In fact, the projections of a vector space are exactly the idempotent elements of the ring of linear transformations of the vector space. After fixing a basis, it can be shown that the matrix of a projection with respect to this basis is an idempotent matrix. An idempotent semiring (also sometimes called a dioid) is a semiring whose addition (not multiplication) is idempotent. If both operations of the semiring are idempotent, then the semiring is called doubly idempotent.^[8]

卻無法用那個一般式得到哩？？！！

故而

$\left( \begin{array}{cc} 1 & 0 \\ 2 & -1 \end{array} \right)$

實在有趣吧！！？？

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]: m = Matrix(([1,0],[2,-1]))

In [4]: m
Out[4]: 
⎡1  0 ⎤
⎢     ⎥
⎣2  -1⎦

In [5]: m*m
Out[5]: 
⎡1  0⎤
⎢    ⎥
⎣0  1⎦

In [6]: m.eigenvals()
Out[6]: {-1: 1, 1: 1}

In [7]: m.eigenvects()
Out[7]: 
⎡⎛-1, 1, ⎡⎡0⎤⎤⎞, ⎛1, 1, ⎡⎡1⎤⎤⎞⎤
⎢⎜       ⎢⎢ ⎥⎥⎟  ⎜      ⎢⎢ ⎥⎥⎟⎥
⎣⎝       ⎣⎣1⎦⎦⎠  ⎝      ⎣⎣1⎦⎦⎠⎦

In [8]:

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 》

2017-08-22 懸鉤子

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。

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
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:

∠ 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 \sim △ 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 $△ ABC \sim △ DEF$ . 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◎

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

FreeSandal

每月彙整: 2017 年 8 月

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 IV 》

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 III 》

Commuting matrices

度量空間

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 II 》

Function composition

Composition monoids

Triangular matrix

Simultaneous triangularisability

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 I 》

Menelaus’ theorem

證明

Idempotence

Other examples

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 》

Parallel postulate

速讀

起源

原理

Similar triangles

Cross-Ratio

Lemma

Remark

Proof of Lemma

Theorem 1

輕。鬆。學。部落客

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