n維角錐及球體積

多維空間角錐

(X>0, X1 + X2 + .... + Xn < U 之體積 U^n / n!)

傳遞引擎:
Y=0 to Z 則 (Z-Y)^k dY = -Z^(k+1) (1-Y/Z)^k d(1-Y/Z) = -Z^(k+1)/(k+1) d (1-Y/Z)^(k+1)
積分得 Z^(k+1)/(k+1)

開始證明:
當X2-Xn已知 => X1 < U- (X2 + .... + Xn) => dX1 => 積分得 U- (X2 + .... + Xn)

當X3-Xn已知 => X2 < U- (X3 + .... + Xn) => (U- (X2 + .... + Xn)) dX2
可令 Z = U- (X3 + .... + Xn), Y=X2, K=1 -> 積分得(U- (X3 + .... + Xn))^2/2

當X4-Xn已知 => X3 < U- (X4 + .... + Xn) => (U- (X3 + .... + Xn))^2/2 dX3
可令 Z = U- (X3 + .... + Xn), Y=X3, K=2 -> 原式變 [(Z-Y)^2 dY ]/2
-> 積分得 [U- (X4 + .... + Xn)]^3/3!

類推至dX(n-1) => (U-Xn)^(n-1)/(n-1)!
最後dXn積完 => 體積 U^n / n!

多維空間球體 (n-sphere)

(X>0, X1^2 + X2^2 + .... + Xn^2 < R^2 之體積再乘上2^n可得球體體積)

背景:

令F(k)為: 1U從0到1積分U^k/[(1-U^2)^0.5] dU
F(0)=1/(1-U^2)^0.5 dU =>令U=cos(t),t從pi/2到0 =>原式 -sin(t)/sin(t) dt = -dt積分為pi/2
F(1)=U/(1-U^2)^0.5 dU = -d [(1-U^2)^0.5] U從0到1積分後為 1

F(k+1) = U^(k+1)/[(1-U^2)^0.5] dU 如下:
當k>0時 d[U^k (1-U^2)^0.5] /dU = k U^(k-1) (1-U^2)^0.5 - U^(k+1)/(1-U^2)^0.5
= [k U^(k-1) (1-U^2) - U^(k+1)]/(1-U^2)^0.5 = [k U^(k-1) -(k+1) U^(k+1)]/(1-U^2)^0.5
又當k>0時U從0到1之 d[U^k (1-U^2)^0.5]積分為1x0-0x1=0
則 U^(k+1)/(1-U^2)^0.5 dU = k^2/[(k+1)k] [ U^(k-1)]/(1-U^2)^0.5 dU ]
若k為奇數 [k(k-2)...3x1]^2/[(k+1)k.....2x1] [ 1/(1-U^2)^0.5 dU ]
F(k+1) = (pi/2) [k(k-2)...3x1]^2/(k+1)!
若k為偶數 [k(k-2)...4x2]^2/[(k+1)k.....3x2] [ U/(1-U^2)^0.5 dU ]
F(k+1) = [k(k-2)...4x2]^2/(k+1)!

F(k)xF(k+1)
不論k為奇數或偶數,F(k)xF(k+1)
= (pi/2) [(k-1)(k-3)...]^2/k! [k(k-2)...]^2/(k+1)!
= (pi/2) (k!)^2/((k+1)(k!)^2) = pi/2/(k+1)
(則F(k+1)= pi/2/(k+1)/F(k) => F(2)= pi/4)

F(1)......F(2h)= (pi/2)^h / [(2h)(2h-2)...x2] = (pi^0.5/2)^(2h) / h!
F(1)......F(2h)F(2h+1)
= (pi/2)^h / [(2h)(2h-2)...x2] x [2h(2h-2)...4x2]^2/(2h+1)!
= (pi/2)^h [2h(2h-2)..4x2]/(2h+1)! = (pi)^h h!/(2h+1)!

傳遞引擎:

Y=0 to Z 則 (Z^2-Y^2)^(k/2) dY = Z^(k+1) (1-(Y/Z)^2)^(k/2) d(Y/Z)
令U=(1-(Y/Z)^2)^0.5 U從1到0,原式成為 Z^(k+1) U^k d (1-U^2)^0.5
U從0到1: Z^(k+1) U^(k+1)/[(1-U^2)^0.5] dU = Z^(k+1) F(k+1)

開始證明:

當X2-Xn已知 => X1 < (R^2- (X2^2 + .... + Xn^2))^0.5 => dX1
=> 積分得 (R^2- (X2^2 + .... + Xn^2))^0.5 F(1)

當X3-Xn已知 => X2 < (R^2- (X3^2 + .... + Xn^2))^0.5
=> (R^2- (X2^2 + .... + Xn^2))^0.5 dX2
可令 Z = (R^2- (X3^2 + .... + Xn^2))^0.5, Y=X2, K=1
原式成為 (Z^2 - Y^2)^0.5 dY
-> 積分得 (R^2- (X3^2 + .... + Xn^2))^(0.5x2) F(1)F(2)

當X4-Xn已知 => X3 < (R^2- (X4^2 + .... + Xn^2))^0.5
=> F(2) (R^2- (X3^2 + .... + Xn^2))^(0.5x2) dX3
可令 Z = (R^2- (X4^2 + .... + Xn^2))^0.5, Y=X3, K=2
原式成為 (Z^2 - Y^2)^(0.5x2) dY
-> 積分得 (R^2- (X3^2 + .... + Xn^2))^(0.5x3) F(1)F(2)F(3)

當Xn已知 => X(n-1) < (R^2- Xn^2)^0.5
=> [F(1)F(2)...F(n-2)] (R^2- X(n-1)^2- Xn^2)^(0.5x(n-2)) dX(n-1)
可令 Z = (R^2- Xn^2)^0.5, Y=X(n-1), K=(n-2)
原式成為 (Z^2 - Y^2)^(0.5x(n-2)) dY
-> 積分得 (R^2- Xn^2)^(0.5x(n-1)) [F(1)F(2)...F(n-1)]

最後dXn積完 => R^n [F(1)F(2)...F(n)]
總體積 R^n 2^n [F(1)F(2)...F(n)]
若n為奇數 R^n (2^n/n!) [pi^((n-1)/2) x ((n-1)/2)!]
= R^n 2 (2pi)^((n-1)/2) /[n(n-2)....3x1] 係數 2, 4pi/3, 8pi^2/15,..
若n為偶數 R^n pi^(n/2) / (n/2)! 係數 pi, pi^2/2, pi^3/6,..