→ ANUBISANKH:?? 推 140.112.7.59 03/24
作者 黃彥傑。
黃彥傑聲明:完全放棄本網頁之版權。
本畫面之網址是:
http://mavang2.tripod.com/square/index.txt
歡迎來信 mailto:zyeks@excite.com
我是黃彥傑,畢業於民國76年,
於民國72年至76年就讀於中山大學電機系,校長是李煥先生,李煥先生曾經當過行政院長,
別人( 釋迦牟尼佛、愛因斯坦)說的法都不夠究竟!只有我黃彥傑現在準備要說的才是究竟
圓滿!
我將要解說一切世界、宇宙的真實結構,含一切物質世界、精神世界!
能究竟圓滿解釋一切的'統一哲學'!
我將這個統一哲學稱作'squares theory'('正方形理論'),
正方形理論中含有'圖示'、'敘述'兩部份,
正方形理論有一點類似太極,正方形理論之基本道理是binary(二元的),
二元之兩者是彼此complementary的(彼此相反的,彼此補足的),
比如,羅漢、菩薩兩者是binary(二元的),兩者是彼此complementary的(彼此相反的,彼此補
足的),
另外,二元之兩者彼此'必定'是處於平等ranking(地位,名次,果位)!
因此,羅漢、菩薩兩者彼此'必定'是處於平等ranking(地位,名次,果位)!
squares theory認定說:
'一切菩薩平等!這個菩薩與那個菩薩之分別是妄分別!
這個菩薩與那個菩薩同樣是平等的菩薩果位ranking!'
'一切羅漢平等!這個羅漢與那個羅漢之分別是妄分別!
這個羅漢與那個羅漢同樣是平等的羅漢果位ranking!'
我再簡述一次:binary之兩者是彼此complementary的,但是處於平等ranking!
每一個square當然有四邊,我規定每一邊的邊長(之相對值)一定是1!
因此每一個square面積都是1,
一切square都是以兩種'小矩形'來指出一個'物質'或'精神'體
在特定之'時'(time)、'處'(site)的'法',
我規定用黑白兩色來代表兩種小矩形,
'法'只有兩種,即是黑白兩種小矩形,
當然,這兩種法(或黑白兩種小矩形)之兩者是彼此complementary的,
一個square內之小矩形的'種類'雖然只有兩種,
但是一個square內之黑白小矩形之'數目'極多,
黑色小矩形之面積總合一定是1/2,
白色小矩形之面積總合一定是1/2,
上述之
'時'是一個'物質'或'精神'體之週期的開始之時至結束之時,
上述之
'處'是一個'物質'或'精神'體之體係的底端之處至頂端之處,(或 體積)
square可用笛卡兒座標系來標示其'時'(time)、'處'(site),
我規定
square之最左下角必須置於笛卡兒座標系之(0,0)點,
square之最下邊必須與笛卡兒座標系之+X axis 重疊,在+X axis的(0,0)點至(1,0)點,
square之最左邊必須與笛卡兒座標系之+Y axis 重疊,在+Y axis的(0,0)點至(0,1)點,
我規定
+X axis是t axis,t是time,from 0 to 1,
+Y axis是s axis,s是site,from 0 to 1,
square四邊之意義-
最左邊:t=0,time是在一個週期的開始之時,而s,site則是from 0 to 1,此邊是from(0,0)點to(0,1)
點,
最右邊:t=1,time是在一個週期的結束之時,而s,site則是from 0 to 1,此邊是from(1,0)點to(1,1)
點,
最下邊:s=0,site是在一個體係的底端之處,而t,time則是from 0 to 1,此邊是from(0,0)點to(1,0)
點,
最上邊:s=1,site是在一個體係的頂端之處,而t,time則是from 0 to 1,此邊是from(0,1)點to(1,1)
點,
注意:s和t之'相對值'都是from 0 to 1,可是'實際值'視實際之週期與體係而定,
X_1 square之圖示:見[圖示x1]
Y_1 square之圖示:見[圖示y1]
[圖示x1]:
http://mavang2.tripod.com/square/x1.bmp
Y-axis
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[圖示y1]:
http://mavang2.tripod.com/square/y1.bmp
Y-axis
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Higher ranking rule:
X_(n+1)=X_n.Y_n. (式子A);
Y_(n+1)=Y_n+X_n+ (式子B);
for n=1 to positive infinity ;
狄摩根定律:
(.)'=(+)
(+)'=(.)
(X_n)'=(Y_n)
(Y_n)'=(X_n)
[X_(n+1)]'=[X_n.Y_n.]'= Y_n+X_n+ = Y_(n+1) ,
[Y_(n+1)]'=[Y_n+X_n+]'= X_n.Y_n. = X_(n+1) ,
比如:
X_2=X_1.Y_1. ,
Y_2=Y_1+X_1+ ,
X_3=X_2.Y_2. ,
Y_3=Y_2+X_2+ ,
以此類推,所以,這是recursive(遞歸的,反覆的,循環的)規則,
二元之兩者彼此'必定'是處於平等ranking(地位,名次,果位),
比如X_2和Y_2是二元之兩者,是彼此complementary的(彼此相反的,彼此補足的),
彼此'必定'是處於平等ranking(地位,名次,果位),
可是(n+1)和n比較時,
X_(n+1)和Y_(n+1)是上級,higher ranking,
X_n 和Y_n 是下級,lower ranking,
上級與下級是不平等的,
上級,higher ranking的果位比較高,
下級,lower ranking的果位比較低,
現在我要解釋
{
X_2=X_1.Y_1.
}
之意義,請先看X_2 square之幾個例子之圖示,
這幾個例子之圖示在下面的[圖示x2],[圖示x2-2],[圖示x2-3]中,
[圖示x2]:
http://mavang2.tripod.com/square/x2.bmp
Y-axis
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|________________________________|________________________________|_X-axis
(0,0) (1/2,0) (1,0)
[圖示x2-2]:
http://mavang2.tripod.com/square/x2-2.bmp
Y-axis
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|________________________________|________________________________|_X-axis
(0,0) (1/2,0) (1,0)
[圖示x2-3]:
http://mavang2.tripod.com/square/x2-3.bmp
Y-axis
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|________________|________________|________________|________________|_X-axis
(0,0) (1/4,0) (1/2,0) (3/4,0) (1,0)
在上面的[圖示x2]、[圖示x2-2]、[圖示x2-3]的X_2 square之中,其
黑色矩形當然是來自(全黑的)X_1 square,
白色矩形當然是來自(全白的)Y_1 square,
X_2 square都必須合於下述兩個規矩:
第一,X_2 square之中,
每個黑色矩形,其time不管分佈在甚麼時段,其site分佈規矩是from 0 to 1而不間斷,
每個白色矩形,其time不管分佈在甚麼時段,其site分佈規矩是from 0 to 1而不間斷,
第二,X_2 square之中,
黑色矩形面積總合一定是1/2,
白色矩形面積總合一定是1/2,
只要是合於上述兩個規矩的square,便都是代表
{
X_2=X_1.Y_1.
}
之X_2 square!
[圖示x2]、[圖示x2-2]、[圖示x2-3]的三種X_2 square都是平等的X_2 ranking!
現在我要解釋
{
Y_2=Y_1+X_1+
}
之意義,請先看Y_2 square之幾個例子之圖示,
這幾個例子之圖示在下面的[圖示y2],[圖示y2-2],[圖示y2-3]中,
[圖示y2]:
http://mavang2.tripod.com/square/y2.bmp
Y-axis
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[圖示y2-2]:
http://mavang2.tripod.com/square/y2-2.bmp
Y-axis
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[圖示y2-3]:
http://mavang2.tripod.com/square/y2-3.bmp
Y-axis
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(0,0) (1,0)
在上面的[圖示y2]、[圖示y2-2]、[圖示y2-3]的Y_2 square之中,其
黑色矩形當然是來自(全黑的)X_1 square,
白色矩形當然是來自(全白的)Y_1 square,
Y_2 square都必須合於下述兩個規矩:
第一,Y_2 square之中,
每個黑色矩形,其site不管分佈在甚麼地段,其time分佈規矩是from 0 to 1而不間斷,
每個白色矩形,其site不管分佈在甚麼地段,其time分佈規矩是from 0 to 1而不間斷,
第二,Y_2 square之中,
黑色矩形面積總合一定是1/2,
白色矩形面積總合一定是1/2,
只要是合於上述兩個規矩的square,便都是代表
{
Y_2=Y_1+X_1+
}
之Y_2 square!
[圖示y2]、[圖示y2-2]、[圖示y2-3]的三種Y_2 square都是平等的Y_2 ranking!
現在我要說如何看懂square所代表之法,
以[圖示y2]Y_2 square作例子,
若設定
X_1 square是'貪嗔癡',
Y_1 square是'戒定慧',
則
Y_2 square之下面白色矩形代表:
在一天的24小時中,持續於下半身作'戒定慧'之法,
Y_2 square之上面黑色矩形代表:
在一天的24小時中,持續於上半身作'貪嗔癡'之法,
佛教說佛陀有法身佛、報身佛、化身佛三種,
Y_2 square代表某一種佛陀,
至於Y_2 square是否為報身佛,我則不清楚,
X_2 square之圖示:見[圖示x2],
Y_2 square之圖示:見[圖示y2],
[圖示x2]:
http://mavang2.tripod.com/square/x2.bmp
Y-axis
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|________________________________|________________________________|_X-axis
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[圖示y2]:
http://mavang2.tripod.com/square/y2.bmp
Y-axis
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X_3 square之圖示:見[圖示x3]
Y_3 square之圖示:見[圖示y3]
[圖示x3]:
http://mavang2.tripod.com/square/x3.bmp
Y-axis
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|________________|________________|________________________________|_X-axis
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[圖示y3]:
http://mavang2.tripod.com/square/y3.bmp
Y-axis
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下述之四個圖示步驟可用來表示complementary的兩者之間的干係,
我要以X_3 square[圖示x3]和Y_3 square[圖示y3]作例子,
當然,X_3 square和Y_3 square兩者是彼此complementary的,
第一.
本來是X_3 square
見[圖示x3]
[圖示x3]:
http://mavang2.tripod.com/square/x3.bmp
Y-axis
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第二.
將X_3 square
翻轉或旋轉:
旋轉角度:
270度:
見[圖示c-1]
[圖示c-1]:
http://mavang2.tripod.com/square/c-1.bmp
Y-axis
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第三.
將c-1 square
翻轉或旋轉:
水平翻轉:
見[圖示c-2]
[圖示c-2]:
http://mavang2.tripod.com/square/c-2.bmp
Y-axis
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第四.
將c-2 square
色彩對換:
見[圖示c-3]
[圖示c-3]:
http://mavang2.tripod.com/square/c-3.bmp
Y-axis
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[圖示c-3]即是Y_3 square!
由於
(X_3)'=(Y_3) ,
(Y_3)'=(X_3) ,
故上述之四個圖示步驟,亦能使第一.步驟中的Y_3 square,於第四.步驟中形成X_3 square!
因果律:
若因是X_n,則果是Y_n,
若因是Y_n,則果是X_n,
已知:羅漢、菩薩兩者是X_2、Y_2,則
X_2:菩薩,
Y_2:羅漢,
X_3:魔,
Y_3:辟支佛,
X_4:眾生,(不可思議解脫三昧)(請參考 維摩詰所說經)
Y_4:佛陀,
已知:羅漢、菩薩兩者是X_3、Y_3,則
X_3:菩薩,
Y_3:羅漢,
X_4:魔,
Y_4:辟支佛,
X_5:眾生,
Y_5:佛陀,
羅漢可以當作是Y_2,或Y_3,或Y_4,或Y_5,以此類推,
於是辟支佛便是Y_3,或Y_4,或Y_5,或Y_6,以此類推,
X_n和Y_n不可對換,因X_n是(..),Y_n是(++),
菩薩、魔、眾生 都是X_n,不是Y_n,
羅漢、辟支佛、佛陀都是Y_n,不是X_n,
佛陀並不是法界中的最高尊者!
在佛陀之上還有更高ranking(地位,名次,果位)的上級尊者,比如菩提心普作王,
菩提心普作王是'遍作王續'的'遍作王尊者',
菩提心普作王之果位是'法界遍行地',佛陀眾生不二或不相屬,
當然菩提心普作王之上還有更高ranking(地位,名次,果位)的上級尊者,
法界的真相是這樣的:越是上級,higher ranking,越是正法!
上面我說一個square內,
黑色小矩形之面積總合一定是1/2,
白色小矩形之面積總合一定是1/2,
可是X_1,Y_1看起來似乎不是如此,
事實上,
X_1=X_0.Y_0. ,
Y_1=Y_0+X_0+ ,
X_0=X_-1.Y_-1. ,
Y_0=Y_-1+X_-1+ ,
以此類推, 到 負無限大 ;
X_0和Y_0以及其下級,都不再有黑白兩種小矩形,
於是,改顏色便可以了,比如
令
X_-100 是全綠色,
Y_-100 是全紅色,
X_1仍然是全黑色,
Y_1仍然是全白色,
便OK!
--
[圖示x2],[圖示x2-2],[圖示x2-3]都可以代表X_2,可是只有[圖示x2]能被稱做X_2之'典型正
方形',
[圖示x2-2]、[圖示x2-3]都不是X_2之'典型正方形',
每個X_n只有一個'典型正方形',每個Y_n只有一個'典型正方形',
for n=negative infinity to positive infinity.
對應於典型正方形,我要定義'Z函數'(Z function),Z函數之'定義域'(domain)是D,
D={(x,y)|0小於等於x小於1 and 0小於等於y小於1 and x is real number and y is real
number.}
Z函數之'值'(value)只能是-1/2或+1/2!
因Z函數之定義域是D,
故Z函數沒有值,若0大於x,或1小於等於x,或0大於y,或1小於等於y,
Z_X_n是一個'Z函數',
If n=1,Z_X_n(x,y)=Z_X_1(x,y)=-1/2,for ALL points in D,
Z_Y_n是一個'Z函數',
If n=1,Z_Y_n(x,y)=Z_Y_1(x,y)=+1/2,for ALL points in D,
因X_1 square是全黑的正方形,而且Z_X_1(x,y)=-1/2,故
每個X_n之典型正方形內之黑色矩形區域內之ALL points之Z函數之值,都是-1/2,
每個Y_n之典型正方形內之黑色矩形區域內之ALL points之Z函數之值,都是-1/2,
for n=2 to positive infinity.
因Y_1 square是全白的正方形,而且Z_Y_1(x,y)=+1/2,故
每個X_n之典型正方形內之白色矩形區域內之ALL points之Z函數之值,都是+1/2,
每個Y_n之典型正方形內之白色矩形區域內之ALL points之Z函數之值,都是+1/2,
for n=2 to positive infinity.
因X_(n+1)=X_n.Y_n.
故 Z_X_(n+1)(x,y)=Z_X_n(2x,y)+Z_Y_n(2x-1,y),
If n=1 and 0小於等於x小於1/2 and
0小於等於y小於1,Z_X_(n+1)(x,y)=Z_X_n(2x,y)=Z_X_1(2x,y)= -1/2.
If n=1 and 1/2小於等於x小於1 and
0小於等於y小於1,Z_X_(n+1)(x,y)=Z_Y_n(2x-1,y)=Z_Y_1(2x-1,y)=+1/2.
因Y_(n+1)=Y_n+X_n+
故 Z_Y_(n+1)(x,y)=Z_Y_n(x,2y)+Z_X_n(x,2y-1),
If n=1 and 0小於等於y小於1/2 and
0小於等於x小於1,Z_Y_(n+1)(x,y)=Z_Y_n(x,2y)=Z_Y_1(x,2y)= +1/2.
If n=1 and 1/2小於等於y小於1 and
0小於等於x小於1,Z_Y_(n+1)(x,y)=Z_X_n(x,2y-1)=Z_X_1(x,2y-1)=-1/2.
接著,
因
X_(n+2)=X_(n+1).Y_(n+1).
故
Z_X_(n+2)(x,y)=
Z_X_(n+1)(2x,y)+Z_Y_(n+1)(2x-1,y)=
[Z_X_n(4x,y)+Z_Y_n(4x-1,y)]+[Z_Y_n(2x-1,2y)+Z_X_n(2x-1,2y-1)],
令n=1,X_(n+2)=X_3,[圖示x3]是X_3之典型正方形,
(1).
If 0小於等於x小於1/4 and 0小於等於y小於1,then 0小於等於4x小於1 and 0小於等於y小
於1,
Z_X_3(x,y)=Z_X_(n+2)(x,y)=Z_X_n(4x,y)=Z_X_1(4x,y)=-1/2.
在[圖示x3]內,
{(x,y)|0小於等於x小於1/4 and 0小於等於y小於1}
是從左邊開始算起第一個黑色矩形區域,
此區域是黑色矩形區域,故此區域內之ALL points之Z函數之值,都是-1/2,
(2).
If 1/4小於等於x小於1/2 and 0小於等於y小於1,then 0小於等於4x-1小於1 and 0小於等於y
小於1,
Z_X_3(x,y)=Z_X_(n+2)(x,y)=Z_Y_n(4x-1,y)=Z_Y_1(4x-1,y)=+1/2.
在[圖示x3]內,
{(x,y)|1/4小於等於x小於1/2 and 0小於等於y小於1}
是從左邊開始算起第一個白色矩形區域,
此區域是白色矩形區域,故此區域內之ALL points之Z函數之值,都是+1/2,
(3).
If 1/2小於等於x小於1 and 0小於等於y小於1/2,then 0小於等於2x-1小於1 and 0小於等於2y
小於1,
Z_X_3(x,y)=Z_X_(n+2)(x,y)=Z_Y_n(2x-1,2y)=Z_Y_1(2x-1,2y)=+1/2.
在[圖示x3]內,
{(x,y)|1/2小於等於x小於1 and 0小於等於y小於1/2}
是從左邊開始算起第二個白色矩形區域,
此區域是白色矩形區域,故此區域內之ALL points之Z函數之值,都是+1/2,
(4).
If 1/2小於等於x小於1 and 1/2小於等於y小於1,then 0小於等於2x-1小於1 and 0小於等於2y-1
小於1,
Z_X_3(x,y)=Z_X_(n+2)(x,y)=Z_X_n(2x-1,2y-1)=Z_X_1(2x-1,2y-1)=-1/2.
在[圖示x3]內,
{(x,y)|1/2小於等於x小於1 and 1/2小於等於y小於1}
是從左邊開始算起第二個黑色矩形區域,
此區域是黑色矩形區域,故此區域內之ALL points之Z函數之值,都是-1/2,
接著,
因
Y_(n+2)=Y_(n+1)+X_(n+1)+
故
Z_Y_(n+2)(x,y)=
Z_Y_(n+1)(x,2y) +Z_X_(n+1)(x,2y-1)=
[Z_Y_n(x,4y)+Z_X_n(x,4y-1)]+[Z_X_n(2x,2y-1)+Z_Y_n(2x-1,2y-1)],
令n=1,Y_(n+2)=Y_3,[圖示y3]是Y_3之典型正方形,
(1).
If 0小於等於y小於1/4 and 0小於等於x小於1,then 0小於等於4y小於1 and 0小於等於x小
於1,
Z_Y_3(x,y)=Z_Y_(n+2)(x,y)=Z_Y_n(x,4y)=Z_Y_1(x,4y)=+1/2.
在[圖示y3]內,
{(x,y)|0小於等於y小於1/4 and 0小於等於x小於1}
是從下邊開始算起第一個白色矩形區域,
此區域是白色矩形區域,故此區域內之ALL points之Z函數之值,都是+1/2,
(2).
If 1/4小於等於y小於1/2 and 0小於等於x小於1,then 0小於等於4y-1小於1 and 0小於等於x
小於1,
Z_Y_3(x,y)=Z_Y_(n+2)(x,y)=Z_X_n(x,4y-1)=Z_X_1(x,4y-1)=-1/2.
在[圖示y3]內,
{(x,y)|1/4小於等於y小於1/2 and 0小於等於x小於1}
是從下邊開始算起第一個黑色矩形區域,
此區域是黑色矩形區域,故此區域內之ALL points之Z函數之值,都是-1/2,
(3).
If 1/2小於等於y小於1 and 0小於等於x小於1/2,then 0小於等於2y-1小於1 and 0小於等於2x
小於1,
Z_Y_3(x,y)=Z_Y_(n+2)(x,y)=Z_X_n(2x,2y-1)=Z_X_1(2x,2y-1)=-1/2.
在[圖示y3]內,
{(x,y)|1/2小於等於y小於1 and 0小於等於x小於1/2}
是從下邊開始算起第二個黑色矩形區域,
此區域是黑色矩形區域,故此區域內之ALL points之Z函數之值,都是-1/2,
(4).
If 1/2小於等於y小於1 and 1/2小於等於x小於1,then 0小於等於2y-1小於1 and 0小於等於2x-1
小於1,
Z_Y_3(x,y)=Z_Y_(n+2)(x,y)=Z_Y_n(2x-1,2y-1)=Z_Y_1(2x-1,2y-1)=+1/2.
在[圖示y3]內,
{(x,y)|1/2小於等於y小於1 and 1/2小於等於x小於1}
是從下邊開始算起第二個白色矩形區域,
此區域是白色矩形區域,故此區域內之ALL points之Z函數之值,都是+1/2,
--
上面有說到典型正方形與Z函數,
現在我要說任何一個正方形與Z函數,
以X_2來說,
[圖示x2],[圖示x2-2],[圖示x2-3]都可以代表X_2,可是只有[圖示x2]能被稱做X_2之'典型正
方形',
對應於'典型正方形'(typical square),我們可以定義'Z函數',如下所說:
因X_(n+1)=X_n.Y_n.
故 Z_X_(n+1)(x,y)=Z_X_n(2x,y)+Z_Y_n(2x-1,y),
因Y_(n+1)=Y_n+X_n+
故 Z_Y_(n+1)(x,y)=Z_Y_n(x,2y)+Z_X_n(x,2y-1),
而對應於可以代表X_2之任何一個正方形,恰好有一個Z函數存在,
已知
Z_X_n(x,y)是X_n之Z函數,Z_Y_n(x,y)是Y_n之Z函數,(x,y)是笛卡兒座標系內的點,
Z函數之'定義域'(domain)是D,
D={(x,y)|0小於等於x小於1 and 0小於等於y小於1 and x is real number and y is real
number.}
令n=1,
Z_X_n(x,y)=Z_X_1(x,y)=-1/2,for ALL points in D,
Z_Y_n(x,y)=Z_Y_1(x,y)=+1/2,for ALL points in D,
(甲).
X_(n+1)=[X_n.Y_n.]
Z_X_(n+1)(x,y)是X_(n+1)之Z函數,
我要以Z_X_n(x,y)和Z_Y_n(x,y)來表示Z_X_(n+1)(x,y),
t_i是在X axis之時間,i是其索引,
t_i is real number and 0小於等於t_i小於等於1 for all i,
where i is integer and 1小於等於i小於等於m,
3小於等於m,
t_1=0,t_m=1,
t_1小於t_2,
t_2小於t_3,
t_3小於t_4,
...etc...
t_(m-1)小於t_m,
(一).
若m是奇數正數,
Z_X_(n+1)(x,y)只可以是下列二種函數中的某一種:
第一種函數:
if
Z_X_(n+1)(x,y)=Z_X_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1,
then
Z_X_(n+1)(x,y)=Z_X_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_3)/(t_4-t_3),y),
where t_3小於等於x小於t_4 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_5)/(t_6-t_5),y),
where t_5小於等於x小於t_6 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_(m-2))/(t_(m-1)-t_(m-2)),y),
where t_(m-2)小於等於x小於t_(m-1) and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_2)/(t_3-t_2),y),
where t_2小於等於x小於t_3 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_4)/(t_5-t_4),y),
where t_4小於等於x小於t_5 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_6)/(t_7-t_6),y),
where t_6小於等於x小於t_7 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_(m-1))/(t_(m-0)-t_(m-1)),y),
where t_(m-1)小於等於x小於t_(m-0) and 0小於等於y小於1;
and
1*{ [t_2-t_1]+[t_4-t_3]+[t_6-t_5]...etc...+[t_(m-1)-t_(m-2)] }=1/2;
and
1*{ [t_3-t_2]+[t_5-t_4]+[t_7-t_6]...etc...+[t_(m-0)-t_(m-1)] }=1/2.
第二種函數:
if
Z_X_(n+1)(x,y)=Z_Y_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1,
then
Z_X_(n+1)(x,y)=Z_Y_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_3)/(t_4-t_3),y),
where t_3小於等於x小於t_4 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_5)/(t_6-t_5),y),
where t_5小於等於x小於t_6 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_(m-2))/(t_(m-1)-t_(m-2)),y),
where t_(m-2)小於等於x小於t_(m-1) and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_2)/(t_3-t_2),y),
where t_2小於等於x小於t_3 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_4)/(t_5-t_4),y),
where t_4小於等於x小於t_5 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_6)/(t_7-t_6),y),
where t_6小於等於x小於t_7 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_(m-1))/(t_(m-0)-t_(m-1)),y),
where t_(m-1)小於等於x小於t_(m-0) and 0小於等於y小於1;
and
1*{ [t_2-t_1]+[t_4-t_3]+[t_6-t_5]...etc...+[t_(m-1)-t_(m-2)] }=1/2;
and
1*{ [t_3-t_2]+[t_5-t_4]+[t_7-t_6]...etc...+[t_(m-0)-t_(m-1)] }=1/2.
(二).
若m是偶數正數,
Z_X_(n+1)(x,y)只可以是下列二種函數中的某一種:
第一種函數:
if
Z_X_(n+1)(x,y)=Z_X_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1,
then
Z_X_(n+1)(x,y)=Z_X_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_3)/(t_4-t_3),y),
where t_3小於等於x小於t_4 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_5)/(t_6-t_5),y),
where t_5小於等於x小於t_6 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_(m-1))/(t_(m-0)-t_(m-1)),y),
where t_(m-1)小於等於x小於t_(m-0) and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_2)/(t_3-t_2),y),
where t_2小於等於x小於t_3 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_4)/(t_5-t_4),y),
where t_4小於等於x小於t_5 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_6)/(t_7-t_6),y),
where t_6小於等於x小於t_7 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_(m-2))/(t_(m-1)-t_(m-2)),y),
where t_(m-2)小於等於x小於t_(m-1) and 0小於等於y小於1;
and
1*{ [t_2-t_1]+[t_4-t_3]+[t_6-t_5]...etc...+[t_(m-0)-t_(m-1)] }=1/2;
and
1*{ [t_3-t_2]+[t_5-t_4]+[t_7-t_6]...etc...+[t_(m-1)-t_(m-2)] }=1/2.
第二種函數:
if
Z_X_(n+1)(x,y)=Z_Y_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1,
then
Z_X_(n+1)(x,y)=Z_Y_n((x-t_1)/(t_2-t_1),y),
where t_1小於等於x小於t_2 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_3)/(t_4-t_3),y),
where t_3小於等於x小於t_4 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_5)/(t_6-t_5),y),
where t_5小於等於x小於t_6 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_Y_n((x-t_(m-1))/(t_(m-0)-t_(m-1)),y),
where t_(m-1)小於等於x小於t_(m-0) and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_2)/(t_3-t_2),y),
where t_2小於等於x小於t_3 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_4)/(t_5-t_4),y),
where t_4小於等於x小於t_5 and 0小於等於y小於1;
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_6)/(t_7-t_6),y),
where t_6小於等於x小於t_7 and 0小於等於y小於1;
...etc...
and
Z_X_(n+1)(x,y)=Z_X_n((x-t_(m-2))/(t_(m-1)-t_(m-2)),y),
where t_(m-2)小於等於x小於t_(m-1) and 0小於等於y小於1;
and
1*{ [t_2-t_1]+[t_4-t_3]+[t_6-t_5]...etc...+[t_(m-0)-t_(m-1)] }=1/2;
and
1*{ [t_3-t_2]+[t_5-t_4]+[t_7-t_6]...etc...+[t_(m-1)-t_(m-2)] }=1/2.
(乙).
Y_(n+1)=[Y_n+X_n+]
Z_Y_(n+1)(x,y)是Y_(n+1)之Z函數,
我要以Z_Y_n(x,y)和Z_X_n(x,y)來表示Z_Y_(n+1)(x,y),
s_i是在Y axis之體積,i是其索引,
s_i is real number and 0小於等於s_i小於等於1 for all i,
where i is integer and 1小於等於i小於等於m,
3小於等於m,
s_1=0,s_m=1,
s_1小於s_2,
s_2小於s_3,
s_3小於s_4,
...etc...
s_(m-1)小於s_m,
(一).
若m是奇數正數,
Z_Y_(n+1)(x,y)只可以是下列二種函數中的某一種:
第一種函數:
if
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1,
then
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_3)/(s_4-s_3)),
where s_3小於等於y小於s_4 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_5)/(s_6-s_5)),
where s_5小於等於y小於s_6 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_(m-2))/(s_(m-1)-s_(m-2))),
where s_(m-2)小於等於y小於s_(m-1) and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_2)/(s_3-s_2)),
where s_2小於等於y小於s_3 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_4)/(s_5-s_4)),
where s_4小於等於y小於s_5 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_6)/(s_7-s_6)),
where s_6小於等於y小於s_7 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_(m-1))/(s_(m-0)-s_(m-1))),
where s_(m-1)小於等於y小於s_(m-0) and 0小於等於x小於1;
and
1*{ [s_2-s_1]+[s_4-s_3]+[s_6-s_5]...etc...+[s_(m-1)-s_(m-2)] }=1/2;
and
1*{ [s_3-s_2]+[s_5-s_4]+[s_7-s_6]...etc...+[s_(m-0)-s_(m-1)] }=1/2.
第二種函數:
if
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1,
then
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_3)/(s_4-s_3)),
where s_3小於等於y小於s_4 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_5)/(s_6-s_5)),
where s_5小於等於y小於s_6 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_(m-2))/(s_(m-1)-s_(m-2))),
where s_(m-2)小於等於y小於s_(m-1) and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_2)/(s_3-s_2)),
where s_2小於等於y小於s_3 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_4)/(s_5-s_4)),
where s_4小於等於y小於s_5 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_6)/(s_7-s_6)),
where s_6小於等於y小於s_7 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_(m-1))/(s_(m-0)-s_(m-1))),
where s_(m-1)小於等於y小於s_(m-0) and 0小於等於x小於1;
and
1*{ [s_2-s_1]+[s_4-s_3]+[s_6-s_5]...etc...+[s_(m-1)-s_(m-2)] }=1/2;
and
1*{ [s_3-s_2]+[s_5-s_4]+[s_7-s_6]...etc...+[s_(m-0)-s_(m-1)] }=1/2.
(二).
若m是偶數正數,
Z_Y_(n+1)(x,y)只可以是下列二種函數中的某一種:
第一種函數:
if
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1,
then
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_3)/(s_4-s_3)),
where s_3小於等於y小於s_4 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_5)/(s_6-s_5)),
where s_5小於等於y小於s_6 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_(m-1))/(s_(m-0)-s_(m-1))),
where s_(m-1)小於等於y小於s_(m-0) and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_2)/(s_3-s_2)),
where s_2小於等於y小於s_3 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_4)/(s_5-s_4)),
where s_4小於等於y小於s_5 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_6)/(s_7-s_6)),
where s_6小於等於y小於s_7 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_(m-2))/(s_(m-1)-s_(m-2))),
where s_(m-2)小於等於y小於s_(m-1) and 0小於等於x小於1;
and
1*{ [s_2-s_1]+[s_4-s_3]+[s_6-s_5]...etc...+[s_(m-0)-s_(m-1)] }=1/2;
and
1*{ [s_3-s_2]+[s_5-s_4]+[s_7-s_6]...etc...+[s_(m-1)-s_(m-2)] }=1/2.
第二種函數:
if
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1,
then
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_1)/(s_2-s_1)),
where s_1小於等於y小於s_2 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_3)/(s_4-s_3)),
where s_3小於等於y小於s_4 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_5)/(s_6-s_5)),
where s_5小於等於y小於s_6 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_X_n(x,(y-s_(m-1))/(s_(m-0)-s_(m-1))),
where s_(m-1)小於等於y小於s_(m-0) and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_2)/(s_3-s_2)),
where s_2小於等於y小於s_3 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_4)/(s_5-s_4)),
where s_4小於等於y小於s_5 and 0小於等於x小於1;
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_6)/(s_7-s_6)),
where s_6小於等於y小於s_7 and 0小於等於x小於1;
...etc...
and
Z_Y_(n+1)(x,y)=Z_Y_n(x,(y-s_(m-2))/(s_(m-1)-s_(m-2))),
where s_(m-2)小於等於y小於s_(m-1) and 0小於等於x小於1;
and
1*{ [s_2-s_1]+[s_4-s_3]+[s_6-s_5]...etc...+[s_(m-0)-s_(m-1)] }=1/2;
and
1*{ [s_3-s_2]+[s_5-s_4]+[s_7-s_6]...etc...+[s_(m-1)-s_(m-2)] }=1/2.
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