Volume of n-dim \”cube\”

2. Ans: V=[Gamma(1/k)*(a/k) ]^n / Gamma(n/k +1), k>0, n=1,2,3,...
晚上再解, 我要補眠去.

2010-02-20 21:14:13 補充：
第1題為第2題特例,以下解第2題,
設所求體積為V(n)[a] (n dim with radius a), 則
(A) V(n)[a]=a^n*V(n)[1]
(B) V(1)[1]=1
(C) V(n)[a]=∫[0~a] V(n-1)[(a^k-xn^k)^(1/k)] dxn
= V(n-1)[1]∫[0~a] (a^k- x^k)^((n-1)/k) dx
(代換積分)令 x=a*t^(1/k),得
V(n)[1]=V(n-1)[1]*(1/k)∫[0~1] (1-t)^((n-1)/k)*t^(1/k -1) *dt
=(1/k)Γ(1/k)Γ[(n-1)/k + 1]/Γ(n/k +1) *V(n-1)[1] (Beta fn.)
=αΓ[(n-1)/k +1]/Γ(n/k +1)*V(n-1)[1] [α=Γ(1/k +1)=Γ(1/k) /k ]
=...=α^(n-1)Γ(1/k +1) /Γ(n/k +1) *V(1)[1]
=α^n /Γ(n/k +1)=[Γ(1/k +1)]^n /Γ(n/k +1)
故所求體積=V(n)[a]=[a*Γ(1/k +1)]^n /Γ(n/k +1)
檢驗:
(1) (n,k)=(2,1)即 x+y<=a, (x, y>0)所圍為三角形,面積=a^2/2
而V(2)[a]=[a*Γ(2)]^2/Γ(3)= a^2 /2 (合) (註:Γ(n+1)=n!)
(2) (n,k)=(3,2)即x^2+y^2+z^2<=a^2 (x,y,z>0)表第一卦限球
體積=(1/8)*(4π/3)*a^3=(π/6)a^3
而V(3)[a]=[a*Γ(3/2)]^3/Γ(5/2)=[a*(1/2)Γ(1/2)]^3/[(3/4)Γ(1/2)]
=(π/6)a^3 (註: Γ(x+1)=xΓ(x), Γ(1/2)^2=π)

圖片參考：http://imgcld.yimg.com/8/n/AE03435620/o/701002080126013873398800.jpg

or http://s585.photobucket.com/albums/ss296/mathmanliu/V_nD_cube.gif