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,..

cholesterol and LDL lowering

使用statin降低之總膽固醇和LDL有相當高之線性相關性
單位: mmol/L
資料來源: Lancet. 2010 November 13; 376(9753): 1670–1681.
Supplementary webappendix: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2988224/bin/mmc1.pdf

jacobian 二維座標轉換積分

前言

vector a x b : vector a sweeps in direction of vector b creates the area
0=(ai+bj)x(ai+bj)=ab (i x j + j x i) => i x j = - j x i
vector(a,b) moved as vector (c,d) painted green area
green area=blue-yellow=orange-pink=ad-bc

類 jacobian 二維座標轉換積分

for U(x,y), V(x,y)
dU = Ux dx + Uy dy
dV = Vx dx + Vy dy

Assume:
U(x,y)=U, V(x,y)=V cross at (x0,y0)
U(x,y)=U+dU, V(x,y)=V cross at (x0+dx1,y0+dy1)
U(x,y)=U, V(x,y)=V+dV cross at (x0+dx2,y0+dy2)

Ux dx1 + Uy dy1= dU
Vx dx1 + Vy dy1= 0
Ux dx2 + Uy dy2= 0
Vx dx2 + Vy dy2= dV

let S= UxVy - VxUy
(dx1,dy1)= (Vy/S dU , -Vx/S dU)
(dx2,dy2)= (Uy/S dV ,  Ux/S dV)

Area(面積) = ad - bc = (dx1 dy2) - (dy1 dx2)
= (Vy/S dU) x (Ux/S dV) - (-Vx/S dU) x (Uy/S dV) = (UxVy - VxUy)/S^2 dU dV = 1/S dU dV

for S= UxVy - VxUy , 1/S is jacobian directly (直接就是jacobian)

Conclusion:

When U moves dU, and V moves dV, the area in U-V plane is dU dV
which projected to area of 1/(UxVy - VxUy) dU dV in X-Y plane

座標轉換(一次聯立方程求解)與行列式(面積,體積) Cramer's rule

2D 面積比求新座標 (二元一次聯立方程求解)

以原座標體系向量A(Ax,Ay),B(Bx,By)做為新座標體系單位向量
預將原座標體系之向量C(Cx,Cy)轉換成新座標(Da,Db)
(Cx,Cy)= C = DaA + DbB = (DaAx + DbBx , DaAy + DbBy)
AxDa + BxDb = Cx
AyDa + ByDb = Cy
座標轉換等於求取上面二元一次聯立方程Da,Db解

由圖中可見
C沿著平行於B的方向交A之處即為A上分量DaA
C沿著平行於A的方向交B之處即為B上分量DbB
A和C圍成黃色平行四邊形,與同高之A與DbB圍成平行四邊形相等面積
而A和B圍成藍色平行四邊形
預知Db=DbB/B可由 (A x DbB)/(A x B) =(A x C)/(A x B)=黃色面積/藍色面積
而面積可表為二向量放入形成之行列式
Db=(A x C)/(A x B)=Δb/Δ
|Ax Cx|    |Ax Bx|
|Ay Cy| /  |Ay By|
同理Da=紅色面積/藍色面積= (C x B)/(A x B)=Δa/Δ
|Cx Bx|    |Ax Bx|
|Cy By| /  |Ay By|
基本上
分母則為所有新座標軸形成之行列式
分子就是以待轉換向量代位分母行列式中欲求參數所屬之新座標軸向量

3D體積比求新座標 (三元一次聯立方程求解)

以原座標體系向量A(Ax,Ay,Az),B(Bx,By,Bz),C(Cx,Cy,Cz)做為新座標體系單位向量
預將原座標體系之向量D(Dx,Dy,Dz)轉換成新座標(Ea,Eb,Ec)
AxEa + BxEb + CxEc = Dx
AyEa + ByEb + CyEc = Dy
AzEa + BzEb + CzEc = Dz
座標轉換等於求取上面三元一次聯立方程Ea,Eb,Ec解

由圖中可見
D沿著平行於AB平面的方向交C之處即為C上分量EcC
AB平面和D圍成紅色平行體,與同高之AB平面與EcC圍成平行體相等體積
預知Ec=EcAB/CAB可由 (A x B x EcC)/(AxBxC) =(AxBxD)/(AxBxC)=紅色體積/藍色體積
而體積可表為三向量放入形成之行列式
Ec=(AxBxD)/(AxBxC)=Δc/Δ
|Ax Bx Dx|    |Ax Bx Cx|
|Ay By Dy| /  |Ay By Cy|
|Az Bz Dz|    |Az Bz Cz|
同理可得Ea,Eb

物資流動圖

物資流動圖

維持勞動力及資本存量之成本包括薪資,借貸利率及資本折舊(衰減)成本

Straitified utility function of consumer

Utility means conception of brains to stimuli of various products
via external sensory receptors(visual, auditory, olfactory, taste, .....)
and internal receptors to sense body composition(sugar, osmolality, effective circulating volume...)
There are plenty of neural conection to integrate these stimuli to a single aggregate utility.

Assume:
Product P1,P2,.......,Pm with consuming amount Q1,Q2,.......,Qm
substitutible functional pre-utilities: food(S1), cloth(S2), house(S3), .... , recreation(Sn)

each product Pj contributes Si with Aij x Qj

Matrix Aij
A11 A12..A1j...A1m
A21 A22..A2j...A2m
.....
Ai1 Ai2..Aij...Aim
.....
An1 An2..Anj...Anm

Then, consuming matrix Q, pre-utility matrix S, Matrix Aij, have relation
A x Q = S

apply Cobb-Douglas function to S , then
aggregate U = c S1^k1 S2^K2 ..... Sn^Kn

furthermores, input-output model for producing sectors may be applied to
neural "utility sectors" to deal with various intermediate "utility products"

Cobb–Douglas生產函數最佳淨產出

Cobb–Douglas生產函數F = c K^a L^b
令借貸利率i, 資本折舊(自然衰減)率j,單位勞工薪資W
資本存量成本包括利息成本iK和資本折舊(自然衰減)成本jK
勞工成本WL

淨產出獲利Y = 總產出 - 總成本 =  f - (i+j)K - WL
dY = dF - (i+j)dK - W dL
而dF = Fk dK + Fl dl (Fk表F對K之偏微分,Fl表F對L之偏微分)
帶入則 dY = (Fk-(i+j)) dK + (Fl-W) dL
故若要使Y達極值, 需找出K,L使 Fk=(i+j);Fl=W 二者皆要成立

Cobb–Douglas生產函數之 Fk = (a/K)F , Fl = (b/L)F
K=(a/(i+j))F, L=(b/W)F 代回 F = c K^a L^b
F= c (a/(i+j))^a (b/W)^b F^(a+b)
得極值下之F
F= c^(1/(1-a-b)) (a/(i+j))^(a/(1-a-b)) (b/W)^(b/(1-a-b))
K=(a/(i+j))F= c^(1/(1-a-b)) (a/(i+j))^((1-b)/(1-a-b)) (b/W)^(b/(1-a-b))
L=(b/W)F= c^(1/(1-a-b)) (a/(i+j))^(a/(1-a-b)) (b/W)^((1-a)/(1-a-b))


此時
資金成本 aF 其中利率成本 [ai/(i+j)] F 資本折舊成本 [aj/(i+j)] F
勞工工資成本 bF
總成本 (a+b)F
生產單位可獲取利潤(淨產出)Y = F- (a+b)F = (1-a-b)F

那此F為最大或最小值呢?
a+b>1時該值為同成本下F最大,但不同成本下F最小值(Y,K,L之3D圖呈馬鞍形)
a+b<1時該值為F為最大值,即最佳狀況

當市場需求超越或不足此F時, (成本之物資不一定從自產成品來)
等生產(F)線須滿足: Fk dK + Fl dL
等成本線須滿足: (i+j)dK - W dL
二者相切之處: (a/K)/(i+j)=(b/L)/W 令其=r
=> K= [a/(i+j)]r, L= [b/W]r, 總成本=(a+b)r ,
F= c [a/(i+j)]^a [b/W]^b r^(a+b) = c [a/(i+j)]^a [b/W]^b (a+b)^(-(a+b)) (總成本)^(a+b)

此時 r= c^(1/(a+b)) [a/(i+j)]^(a/(a+b)) [b/W]^(b/(a+b)) F^(1/(a+b)) 
令 s=c [a/(i+j)]^a [b/W]^b => r=  (sF)^(1/(a+b)) 
資金成本 a(sF)^(1/(a+b))  其中利率成本 [ai/(i+j)] (sF)^(1/(a+b)) 資本折舊成本 [aj/(i+j)] (sF)^(1/(a+b)) 勞工工資成本 b(sF)^(1/(a+b)) 總成本 (a+b)(sF)^(1/(a+b))
生產單位可獲取利潤(淨產出)Y = F- (sF)^(1/(a+b))


dY/d(成本) = dF/d(成本) -1 = dF/d[(a+b)r] = c [a/(i+j)]^a [b/W]^b r^(a+b-1) -1
故a+b<1 (Y''<0) 曲線凹向下,Y有最大值,a+b>1(Y''>0)曲線凹向上,
當a+b=1 (Y''=0) Y為直線

橫軸為r, F=c'r^(a+b), 成本=(a+b)r

當a+b>1, r越大成本佔率越低,生產單位純益佔率越高
當a+b<1, r越大成本佔率越高,生產單位純益佔率越低
當a+b=1, r越大,成本佔率及生產單位純益佔率固定

當a+b=1則利率i=aF/K-j,K太大會讓實質利率i變負的

需求為波動時

當各時間點競爭市場需求為波動時,
假定Q=cK^0.5 L^0.5, Q(t)=Q0 (cos(ht))^2,(Q在0-1之間波動,平均0.5)
C = rK + wL,
單位K(元),L(人),r(貨/元-月),w(貨/人-月),C(貨/月),Q0(貨/月),c(貨/月-(元-人)^0.5), h(/月)
預選定長期(T->∞)不動之K使平均成本t=0~T,(∫(C dt))/T 最低
L(t)=(Q/c)^2/K= (Q0/c)^2 (cos(ht))^4/K
又 (T->∞)則(∫[(cos(ht))^4 d(ht)])/(hT) = 3/8
(∫(C dt))/T= (∫(rK + wL)dt)/T = rK + w/K (Q0/c)^2(∫[(cos(ht))^4 dt])/T = rK + 3/8 (Q0/c)^2 w/K
d( rK + 3/8 w/K )/dK= r - 3/8 (Q0/c)^2 w/K^2 =0 則 K= (Q0/c) (3w/(8r))^0.5

在非波動下,(Q=0 則 K=0),(Q=0.5Q0 則 K=0.25Q0/r),(Q=Q0 則 K=0.5Q0/r),
同樣平均Q=0.5Q0下,波動K/非波動K=((3w/(8r))^0.5/c)/(0.25/r)=(6rw/c^2)^0.5
rw> c^2/6則波動K>非波動K,rw < c^2/6則波動K<非波動K

非波動下 AC=rK+wL=0.5Q0
波動下AC= rK+ 3/8 (Q0/c)^2 w/K = (Q0/c)(3wr/8)^0.5 + (Q0/c)(3rw/8)^0.5 = 0.5 (Q0/c)(6rw)^0.5
rw < c^2/6則波動總成本下降,生產者有盈餘

經濟筆記

(一)梭羅模型-1

令生產函數 f=cK^a L^b, 儲蓄(投入資本)比例s 資本衰減率 j
K'(t) = sf - jK = scK^a L^b - jK
生產盈餘Y=(1-s)cK^a L^b

最大K值時 scK^a L^b = jK ; K= (sc/j L^b)^(1/(1-a))
生產盈餘Y=(1-s)c K^a L^b = (1-s)c(sc/j L^b)^(a/(1-a)) L^b
= (1-s)s^(a/(1-a)) c^(1/(1-a)) L^(b/(1-a)) /j^(a/(1-a))
當Y'(s)=0 => -f + (1-s)[a/(1-a)]f/s =0
=> (1-s)a/(1-a)/s=1 => s=a
儲蓄率為a時有最大盈餘Y= (1-a)a^(a/(1-a)) c^(1/(1-a)) L^(b/(1-a))/j^(a/(1-a))

K(t)隨時間之變化:
dK/dt +jK = scK^a L^b
=> (1-a)scL^b = (1-a)K^(-a)dK/dt + (1-a)jK^(1-a) = d(K^(1-a))/dt + (1-a)jK^(1-a)
令K(t)^(1-a)=G(t) =>  dG/dt + (1-a)iG = (1-a)scL^b
d(G-scL^b/j)/dt + (1-a)j(G-scL^b/j)=0
=> G(t)= Xe^((a-1)jt) + scL^b/j (X為常數)
K(t)=G(t)^(1/(1-a))= (Xe^((a-1)jt) + scL^b/j)^(1/(1-a))
又t=0時K0=(X + scL^b/j)^(1/(1-a)) => X= K0^(1-a)-scL^b/j
K(t)= (K0^(1-a)e^((a-1)jt) + scL^b/j(1-e^((a-1)jt)))^(1/(1-a))
f(t)=  cL^b (K0^(1-a)e^((a-1)jt) + scL^b/j(1-e^((a-1)jt)))^(a/(1-a))
(若K0接近0,當a>0.5時a/(1-a)>1此時f(t)存在凹凸轉折點t=ln(a/(1-a))/[(1-a)j]使f''(t)=0
於該處達最大成長f'(t))

當a<1,t->無限大 => K= (scL^b/j)^(1/(1-a))
預找出s=1下達最佳K(該時s=a)之時間t
=> (Xe^((a-1)jt) + cL^b/j)^(1/(1-a))=(acL^b/j)^(1/(1-a))
=> e^((a-1)jt) =(a-1)cL^b/j / (K0^(1-a)-acL^b/j) = (1-a)/(a-(j/(cL^b))K0^(1-a))
t= [ln(a-(j/(cL^b))K0^(1-a))-ln(1-a)]/(1-a)


(二)梭羅模型-2

當生產函數f未知 儲蓄(投入資本)比例S 資本衰減率j
K'(t) = Sf - jK 生產盈餘Y=(1-S)f

K達最大時K'=0 => Sf - jK =0 =>  d (lnS + lnf - lnK)= d lnj =0
=> d(lnS) = (dlnK/dlnf - 1) d(lnf)
當Y'=0 => d (ln(1-S) + lnf)=0 => [-S/(1-S) dlnS] + d(lnf) =0
=> d(lnS) = (1/S -1) d(lnf)

合併兩式 S=d(lnf)/d(lnK)  (d(lnf)/d(lnK)為邊際資本生產彈性)
=> 當 S=K/f df/dK 時, 生產盈餘Y可得最大值

(帶入f=cK^a L^b則可得 S=a)

(三)梭羅模型與人口及技術成長

若勞力成長率g, L=L0 e^(gt)
K(t)=(X e^((a-1)jt) + sc/(j+bg/(1-a)) L0^b e^(bgt))^(1/(1-a))

又若技術成長率n, C=C0 e^(nt)
Y=CK^aL^b=(C0L0^b)e^[(n+bg)t] K^a
dK/dt=sY-jK=(sC0L0^b)e^[(n+bg)t] K^a-jK
dK^(1-a)/dt=(1-a)K^(-a) dK/dt=(s(1-a)C0L0^b)e^[(n+bg)t]-(1-a)jK^(1-a)
令U=(s(1-a)C0L0^b),V=(1-a)j
dK^(1-a)/dt+VK^(1-a)=U e^[(n+bg)t]
d[K^(1-a)-we^[(n+bg)t]/dt+V[K^(1-a)-we^[(n+bg)t]
=dK^(1-a)/dt+VK^(1-a)+[-w(n+bg+V)]e^[(n+bg)t]
=U e^[(n+bg)t]-w(n+bg+V)e^[(n+bg)t]=[U-w(n+bg+V)]e^[(n+bg)t]

當w=U/(n+bg+V)=(sC0L0^b)/(j+(n+bg)/(1-a))時,
d[K^(1-a)-we^[(n+bg)t]/dt+V[K^(1-a)-we^[(n+bg)t]=0
[K^(1-a)-we^[(n+bg)t]=Xe^(-Vt) (X為常數)
則K(t)=[Xe^(-Vt)+we^[(n+bg)t]]^(1/(1-a))
K(t)=[Xe^((a-1)jt)+(sC0)/(j+(n+bg)/(1-a)) L0^b e^[(n+bg)t]]^(1/(1-a))
(以(n+bg)及C0取代無技術成長時bg及C即可)

Yk=(aC0L0^b)e^[(n+bg)t] K^(a-1)=a/[X1 e^(-(n+bg+(1-a)j)t)+s/(j+(n+bg)/(1-a))]
當t非常大且a+b=1時,Yk=a/[s(j+(n+bg)/(1-a))]=a/[s(j+n/b+g)]

圖化算式:

人口成長率g的固定規模報酬函數:

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