초구면 좌표계(Hyperspherical coordinate)

정의1

실삼각함수가 $\cos,\sin:\mathbb{R}\to \mathbb{R}$이고 

$2$차원 구면 좌표변환 $\Phi_2 :\mathbb{R}^2\to \mathbb{R}^2$와 $\Phi_2$의 성분함수 $\Phi_{2,1},\Phi_{2,2}:\mathbb{R}^2\to \mathbb{R}$가 

모든 $(\rho,\varphi_1)\in \mathbb{R}^2$에 대해 $\Phi_2(\rho, \varphi_1)= (\rho\cdot \cos\varphi_1, \rho\cdot \sin\varphi_1) = (\Phi_{2,1}(\rho,\varphi_1),\Phi_{2,2}(\rho,\varphi_1))$일때

$k\ge 2$인 모든 $k\in $ $\mathbb{Z}^+$에 대해

$k$차원 구면 좌표변환 $\Phi_k :\mathbb{R}^k\to \mathbb{R}^k$와 $\Phi_k$의 성분함수 $\Phi_{k,1},\cdots ,\Phi_{k,k}:\mathbb{R}^k\to \mathbb{R}$가 귀납적으로 정의되어 

모든 $(\rho,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^k$가 $\Phi_k(\rho,\varphi_{k-1},\cdots, \varphi_{1}) = (\Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_{1}),\cdots , \Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_{1}))$이면 

$k+1$차원 구면 좌표변환 $\Phi_{k+1} :\mathbb{R}^{k+1}\to \mathbb{R}^{k+1}$과 $\Phi_{k+1}$의 성분함수 $\Phi_{k+1,1},\cdots ,\Phi_{k+1,k},\Phi_{k+1,k+1}:\mathbb{R}^{k+1}\to \mathbb{R}$을

모든 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^{k+1}$에 대해 

$\begin{align*}\Phi_{k+1}(\rho,\varphi_k, \varphi_{k-1},\cdots ,\varphi_1) & = (\sin \varphi_k\cdot \Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_{1}),\cdots , \sin\varphi_k\cdot \Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_{1}), \rho\cdot \cos\varphi_k) \\[0.5em] & = (\Phi_{k+1,1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1),\cdots,\Phi_{k+1,k}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1),\Phi_{k+1,k+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)) \text{ 로 정의한다.} \end{align*}$

$2$차원 구면 좌표변환 $\Phi_2$는 모든 $(\rho,\theta)\in \mathbb{R}^2$에 대해

$\Phi_2(\rho, \theta)= (\Phi_{2,1}(\rho,\theta),\Phi_{2,2}(\rho,\theta)) = (\rho\cdot \cos\theta, \rho\cdot \sin\theta)$이므로 $\Phi_2$를 극 좌표변환이라 하고

$3$차원 구면 좌표변환 $\Phi_3$은 모든 $(\rho,\varphi,\theta)\in \mathbb{R}^3$에 대해

$\begin{align*} \Phi_3(\rho, \varphi,\theta) &= (\Phi_{3,1}(\rho,\varphi,\theta),\Phi_{3,2}(\rho,\varphi,\theta),\Phi_{3,3}(\rho,\varphi,\theta)) \\[0.5em] & = (\sin\varphi\cdot \Phi_{2,1}(\rho,\theta),\sin\varphi\cdot \Phi_{2,1}(\rho,\theta), \rho\cdot \cos\varphi) \\[0.5em] & = (\sin\varphi\cdot \rho\cdot \cos\theta , \sin\varphi\cdot \rho\cdot \sin\theta, \rho\cdot \cos\varphi) \\[0.5em] & = (\rho\cdot \sin\varphi\cdot \cos\theta , \rho\cdot \sin\varphi\cdot \sin\theta, \rho\cdot \cos\varphi) \text{ 이므로} \end{align*}$  $\phantom{ \displaystyle \sum_{i=1}^n}$

$\Phi_3$을 구면 좌표변환이라 한다.

 

 

 

정리1

실삼각함수가 $\cos,\sin:\mathbb{R}\to \mathbb{R}$이고 $n \ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해

실수체 $(\mathbb{R},+,\cdot,0,1)$위의 $n$-순서쌍 벡터공간 $(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$의 표준순서기저가 $\beta_n$이고

$(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$위의 점곱 내적공간의 노름공간 $(\mathbb{R}^n,\lVert\cdot\rVert_n)$의 노름거리공간이 $(\mathbb{R}^n,d_n)$일때 

$n$차원 구면 좌표변환 $\Phi_n :\mathbb{R}^n\to \mathbb{R}^n$과 $\Phi_n$의 성분함수 $\Phi_{n,1},\cdots ,\Phi_{n,n}:\mathbb{R}^n\to \mathbb{R}$에 대해 다음이 성립한다.

1. $\Phi_n$은 $\mathbb{R}^n$에서 연속미분가능하다.

2. 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해 $\dfrac{\partial \Phi_{n}}{\partial \rho}(\rho,\varphi_{n-1},\cdots,\varphi_1) = \Phi_{n}(1,\varphi_{n-1},\cdots,\varphi_1)$이다.

3. 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해 $\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1) = \rho\cdot_n \Phi_n(1,\varphi_{n-1},\cdots,\varphi_1)$이다.

4. $\Phi_n$은 $\mathbb{R}^n$에서 미분가능하다.

5. 임의의 $(\rho,\varphi_n, \varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^{n+1}$에서 $\Phi_{n+1}$의 도함수가 $D\Phi_{n+1}(\rho,\varphi_n,\varphi_{n-1},\cdots,\varphi_1) : \mathbb{R}^{n+1}\to \mathbb{R}^{n+1}$이고

$(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에서 $\Phi_{n}$의 도함수가 $D\Phi_{n}(\rho,\varphi_{n-1},\cdots,\varphi_1) : \mathbb{R}^{n}\to \mathbb{R}^{n}$일때

$\Phi_n$의 야코비행렬이 $A_{n} = [D\Phi_{n}(\rho,\varphi_{n-1},\cdots,\varphi_1)]_{\beta_{n}}\in M_{n\times n}(\mathbb{R})$이고

$\Phi_{n+1}$의 야코비행렬이 $A_{n+1} = [D\Phi_{n+1}(\rho,\varphi_n,\varphi_{n-1},\cdots,\varphi_1)]_{\beta_{n+1}}\in M_{(n+1)\times (n+1)}(\mathbb{R})$이면 

$A_{n+1} = \begin{bmatrix} \sin \varphi_n\cdot (A_n)_{1,1} & \rho\cdot \cos \varphi_n\cdot (A_n)_{1,1} & \sin \varphi_n\cdot (A_n)_{1,2} & \cdots & \sin \varphi_n\cdot (A_n)_{1,n} \\ \vdots & \vdots & \vdots &\ddots &\vdots \\ \sin \varphi_n\cdot (A_n)_{n,1} & \rho\cdot \cos \varphi_n\cdot (A_n)_{n,1} & \sin \varphi_n\cdot (A_n)_{n,2} & \cdots & \sin \varphi_n\cdot (A_n)_{n,n} \\ \cos \varphi_n & -\rho\cdot \sin \varphi_n & 0 & \cdots & 0 \end{bmatrix} \text{ 이고}$

$A_{n+1}$의 행렬식은 $\det(A_{n+1}) = (-1)^n \cdot \rho\cdot (\sin \varphi_n)^{n-1}\cdot \det(A_n)$이다.

6. $\mathbb{R}^n$에서 $\Phi_{n}$의 야코비안 $J\Phi_n : \mathbb{R}^n\to \mathbb{R}$은 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해

$J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1) = (-1)^{\frac{n^2 - n-2}{2}}\cdot \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0$이다.

증명

거리공간 정리로 $(\mathbb{R}^n,d_n)$은 $n$차원 유클리드 거리공간이고 거리공간 정리로 $\mathbb{R}^n$은 $(\mathbb{R}^n,d_n)$에서 열린집합이다.

1, 2, 3

$n\ge 2$인 $n\in \mathbb{Z}^+$에 대한 귀납법으로 증명한다.

$n = 2$이면

극 좌표변환 정리로 $\Phi_2$는 $\mathbb{R}^2$에서 연속미분가능하여 모든 $(\rho,\varphi_1)\in \mathbb{R}^2$에 대해 편도함수 정리와 야코비행렬 정리로

$\dfrac{\partial \Phi_2}{\partial \rho}(\rho,\varphi_1) = \left (\dfrac{\partial \Phi_{2,1}}{\partial \rho}(\rho,\varphi_1), \dfrac{\partial \Phi_{2,2}}{\partial \rho}(\rho,\varphi_1) \right) = (\cos\varphi_1,\sin\varphi_1) = (\Phi_{2,1}(1,\varphi_1),\Phi_{2,2}(1,\varphi_1)) = \Phi_2(1,\varphi_1) \text{ 이고}$

$\Phi_2(\rho,\varphi_1) = (\rho\cdot\cos\varphi_1,\rho\cdot \sin \varphi_1) = \rho\cdot_2 (\cos\varphi_1,\sin\varphi_1) = \rho\cdot_2 \Phi_{2}(1,\varphi_1)$이다.

$k\ge 2$인 모든 $k\in \mathbb{Z}^+$에 대해 $\Phi_k$가 $\mathbb{R}^k$에서 연속미분가능하고 모든 $(\rho,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^k$에 대해 

$\dfrac{\partial \Phi_{k}}{\partial \rho}(\rho,\varphi_{k-1},\cdots,\varphi_1) = \Phi_{k}(1,\varphi_{k-1},\cdots,\varphi_1)$이고 $\Phi_k(\rho,\varphi_{k-1},\cdots,\varphi_1) = \rho\cdot_k \Phi_k(1,\varphi_{k-1},\cdots,\varphi_1)$이면

연속미분 정리로 $\Phi_k$의 성분함수 $\Phi_{k,1},\cdots,\Phi_{k,k}$는 $\mathbb{R}^k$에서 연속미분가능하여

편도함수 정리로 모든 $i=1,2,\cdots,k$에 대해 $\dfrac{\partial \Phi_{k,i}}{\partial \rho}(\rho,\varphi_{k-1},\cdots,\varphi_1) = \Phi_{k,i}(1,\varphi_{k-1},\cdots,\varphi_1)$이고

삼각함수 정리와 미분연속성과 연속함수 정리와 연속미분 정리로 $\cos,\sin$은 $\mathbb{R}$에서 연속미분가능하므로

연속미분 정리로 모든 $(\rho,\varphi) = (\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^{k+1}$에 대해 

$p_1(\rho,\varphi) = \rho$이고 $p_2(\rho,\varphi) = \varphi_k$인 함수 $p_1,p_2 : \mathbb{R}^{k+1}\to \mathbb{R}$가 $\mathbb{R}^{k+1}$에서 연속미분가능하고 

$q(\rho,\varphi) = (\rho,\varphi_{k-1},\cdots, \varphi_1)$인 함수 $q : \mathbb{R}^{k+1}\to \mathbb{R}^k$가 $\mathbb{R}^{k+1}$에서 연속미분가능하여

$k+1$차원 구면 좌표변환의 정의로 모든 $i= 1,2,\cdots,k$에 대해 

$\Phi_{k+1,i}(\rho,\varphi) = \sin \varphi_k \cdot \Phi_{k,i}(\rho,\varphi_{k-1},\cdots,\varphi_1) = \sin (p_2(\rho,\varphi)) \cdot \Phi_{k,i}(q(\rho,\varphi))  = (\sin\circ \,p_2)(\rho,\varphi)\cdot (\Phi_{k,i}\circ q)(\rho,\varphi)\text{ 이고}$

$\Phi_{k+1,k+1}(\rho,\varphi) = \rho\cdot \cos \varphi_k = p_1(\rho,\varphi)\cdot (\cos \circ \, p_2)(\rho,\varphi)$임에 따라

연속미분 정리와 연속미분 정리로 $\Phi_{k+1}$의 성분함수 $\Phi_{k+1,1},\cdots,\Phi_{k+1,k},\Phi_{k+1,k+1}$은 $\mathbb{R}^{k+1}$에서 연속미분가능하므로

연속미분 정리로 $\Phi_{k+1}$은 $\mathbb{R}^{k+1}$에서 연속미분가능하여 편미분 정리와 편미분 정리와 편미분 정리와 삼각함수 정리로 

$\begin{align*} \dfrac{\partial \Phi_{k+1,i}}{\partial \rho}(\rho,\varphi) & = \dfrac{\partial (\sin\circ \,p_2)}{\partial \rho}(\rho,\varphi) \cdot (\Phi_{k,i}\circ q)(\rho,\varphi) + (\sin\circ \,p_2)(\rho,\varphi) \cdot \dfrac{\partial (\Phi_{k,i}\circ q)}{\partial \rho}(\rho,\varphi) \\[0.5em] & = \sin'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \rho}(\rho,\varphi) \cdot \Phi_{k,i}(q(\rho,\varphi)) + \sin(p_2(\rho,\varphi))\cdot \dfrac{\partial \Phi_{k,i}}{\partial \rho}(q(\rho,\varphi))\cdot \dfrac{\partial p_1}{\partial \rho}(\rho,\varphi) \\[0.5em] & = \sin'(p_2(\rho,\varphi)) \cdot 0\cdot \Phi_{k,i}(q(\rho,\varphi)) + \sin \varphi_{k}\cdot \dfrac{\partial \Phi_{k,i}}{\partial \rho}(q(\rho,\varphi))\cdot 1 \\[0.5em] & =\sin \varphi_{k}\cdot \dfrac{\partial \Phi_{k,i}}{\partial \rho}(\rho,\varphi_{k-1},\cdots,\varphi_1) \\[0.5em] & = \sin\varphi_{k} \cdot \Phi_{k,i}(1,\varphi_{k-1},\cdots,\varphi_1) \\[0.5em] & = \Phi_{k+1,i}(1,\varphi) \text{ 이고} \end{align*}$

$\begin{align*} \dfrac{\partial \Phi_{k+1,k+1}}{\partial \rho}(\rho,\varphi) & = \dfrac{\partial p_1}{\partial \rho}(\rho,\varphi)\cdot (\cos \circ\, p_2)(\rho,\varphi) + p_1(\rho,\varphi)\cdot \dfrac{\partial (\cos\circ\, p_2)}{\partial \rho}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial p_1}{\partial \rho}(\rho,\varphi)\cdot \cos(p_2(\rho,\varphi)) + \rho\cdot \cos'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \rho}(\rho,\varphi) \\[0.5em] & = 1\cdot \cos \varphi_{k} + \rho \cdot \cos'(p_2(\rho,\varphi))\cdot 0 \\[0.5em] & = \cos\varphi_{k} \\[0.5em] & = \Phi_{k+1,k+1}(1,\varphi) \text{ 이므로} \end{align*}$

편도함수 정리로

$\begin{align*}\dfrac{\partial \Phi_{k+1}}{\partial \rho}(\rho,\varphi) & = \left (\dfrac{\partial \Phi_{k+1,1}}{\partial \rho}(\rho,\varphi),\cdots, \dfrac{\partial \Phi_{k+1,k}}{\partial \rho}(\rho,\varphi) ,\dfrac{\partial \Phi_{k+1,k+1}}{\partial \rho}(\rho,\varphi) \right) \\[0.5em] & = (\Phi_{k+1,1}(1,\varphi),\cdots,\Phi_{k+1,k}(1,\varphi),\Phi_{k+1,k+1}(1,\varphi)) \\[0.5em]& = \Phi_{k+1}(1,\varphi) \text{ 이고} \end{align*}$

$k+1$-데카르트곱의 정의로

$\begin{align*} \Phi_{k+1}(\rho,\varphi) & = (\sin \varphi_k\cdot \Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_{1}),\cdots , \sin\varphi_k\cdot \Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_{1}), \rho\cdot \cos\varphi_k) \\[0.5em] & = ((\sin \varphi_k\cdot \Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_{1}),\cdots , \sin\varphi_k\cdot \Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_{1})), \rho\cdot \cos\varphi_k) \\[0.5em] & = (\sin \varphi_k\cdot_k ( \Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_{1}),\cdots , \Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_{1})), \rho\cdot \cos\varphi_k) \\[0.5em] & = (\sin \varphi_k\cdot_k \Phi_{k}(\rho,\varphi_{k-1},\cdots,\varphi_{1}), \rho\cdot \cos\varphi_k) \\[0.5em] & = (\sin \varphi_k \cdot_k (\rho \cdot_k \Phi_{k}(1,\varphi_{k-1},\cdots,\varphi_{1})), \rho\cdot \cos\varphi_k) \\[0.5em] & = ( \rho \cdot_k (\sin \varphi_k \cdot_k \Phi_{k}(1,\varphi_{k-1},\cdots,\varphi_{1})), \rho\cdot \cos\varphi_k) \\[0.5em] & = \rho \cdot_{k+1} ( \sin \varphi_k \cdot_k \Phi_{k}(1,\varphi_{k-1},\cdots,\varphi_{1}), \cos\varphi_k) \\[0.5em] & = \rho \cdot_{k+1} \Phi_{k+1}(1,\varphi) \text{ 이다.} \end{align*}$

따라서 $n\ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해 $\Phi_n$은 $\mathbb{R}^n$에서 연속미분가능하고 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해

$\dfrac{\partial \Phi_{n}}{\partial \rho}(\rho,\varphi_{n-1},\cdots,\varphi_1) = \Phi_{n}(1,\varphi_{n-1},\cdots,\varphi_1)$이고 $\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1) = \rho\cdot_n \Phi_n(1,\varphi_{n-1},\cdots,\varphi_1)$이다.

4.

1번과 연속미분의 정의와 미분 정리로 $\Phi_n$은 $\mathbb{R}^n$에서 미분가능하다.

5.

연속미분 정리로 모든 $(\rho,\varphi) = (\rho,\varphi_n,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^{n+1}$와 모든 $r = 2,\cdots,n,n+1$에 대해 

$p_1(\rho,\varphi) = \rho$이고 $p_r(\rho,\varphi) = \varphi_{n+2-r}$인 함수 $p_1,p_r : \mathbb{R}^{n+1}\to \mathbb{R}$은 $\mathbb{R}^{n+1}$에서 연속미분가능하고 

$q(\rho,\varphi) = (\rho,\varphi_{n-1},\cdots, \varphi_1) = (p_1(\rho,\varphi),p_3(\rho,\varphi),\cdots,p_{n+1}(\rho,\varphi))$인

함수 $q : \mathbb{R}^{n+1}\to \mathbb{R}^{n}$는 $\mathbb{R}^{n+1}$에서 연속미분가능하여 모든 $i = 1,2,\cdots, n$에 대해

$\Phi_{n+1,i}(\rho,\varphi)  = \sin\varphi_{n}\cdot \Phi_{n,i}(\rho,\varphi_{n-1},\cdots,\varphi_1) = (\sin\circ \,p_2)(\rho,\varphi) \cdot (\Phi_{n,i}\circ q)(\rho,\varphi)$이고

$\Phi_{n+1,n+1}(\rho,\varphi) = \rho\cdot \cos \varphi_{n} = p_1(\rho,\varphi)\cdot (\cos \circ \,p_2)(\rho,\varphi)$이므로

1, 2, 3번과 야코비행렬 정리와 편미분 정리와 편미분 정리와 편미분 정리와 삼각함수 정리로 

$\begin{align*}([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{i,1} & = \dfrac{\partial \Phi_{n+1,i}}{\partial \rho}(\rho,\varphi) \\[0.5em] & = \Phi_{n+1,i}(1,\varphi) \\[0.5em] & = \sin\varphi_{n} \cdot \Phi_{n,i}(1,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \sin\varphi_n\cdot \dfrac{\partial \Phi_{n,i}}{\partial \rho}(\rho,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \sin\varphi_n\cdot ([D\Phi_{n}(\rho,\varphi_{n-1},\cdots,\varphi_1)]_{\beta_n})_{i,1} \text{ 이고} \end{align*}$

$([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{n+1,1}  = \dfrac{\partial \Phi_{n+1,n+1}}{\partial \rho}(\rho,\varphi)  = \Phi_{n+1,n+1}(1,\varphi) = \cos \varphi_n$이고 

$\begin{align*} ([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{i,2} & = \dfrac{\partial \Phi_{n+1,i}}{\partial \varphi_{n}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial (\sin\circ \,p_2)}{\partial \varphi_{n}}(\rho,\varphi) \cdot (\Phi_{n,i}\circ q)(\rho,\varphi) + (\sin\circ \,p_2)(\rho,\varphi) \cdot \dfrac{\partial (\Phi_{n,i}\circ q)}{\partial \varphi_{n}}(\rho,\varphi) \\[0.5em] & = \sin'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \varphi_{n}}(\rho,\varphi) \cdot \Phi_{n,i}(q(\rho,\varphi)) + \sin(p_2(\rho,\varphi))\cdot 0 \\[0.5em] & = \cos(p_2(\rho,\varphi))\cdot 1 \cdot \Phi_{n,i}(q(\rho,\varphi)) \\[0.5em] & = \cos \varphi_{n}\cdot \Phi_{n,i}(\rho,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \cos \varphi_{n}\cdot \rho \cdot \Phi_{n,i}(1,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \rho\cdot \cos \varphi_n \cdot \dfrac{\partial \Phi_{n,i}}{\partial \rho}(\rho,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \rho\cdot \cos \varphi_{n}\cdot ([D\Phi_{n}(\rho,\varphi_{n-1},\cdots,\varphi_1)]_{\beta_{n}})_{i,1}  \text{ 이고} \end{align*}$

$\begin{align*} ([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{n+1,2} & = \dfrac{\partial \Phi_{n+1,n+1}}{\partial \varphi_{n}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial p_1}{\partial \varphi_{n}}(\rho,\varphi)\cdot (\cos \circ\, p_2)(\rho,\varphi) + p_1(\rho,\varphi)\cdot \dfrac{\partial (\cos\circ\, p_2)}{\partial \varphi_{n}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial p_1}{\partial \varphi_{n}}(\rho,\varphi)\cdot \cos(p_2(\rho,\varphi)) + \rho\cdot \cos'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \varphi_{n}}(\rho,\varphi) \\[0.5em] & = 0\cdot \cos(p_2(\rho,\varphi)) + \rho \cdot (-\sin(p_2(\rho,\varphi))\cdot 1 \\[0.5em] & = - \rho\cdot \sin \varphi_{n} \text{ 이고} \end{align*}$

모든 $j = 2,\cdots,n$에 대해 $1 = n+1 - n \le n+1 - j\le n+1 - 2 = n-1 < n$이므로 

$\begin{align*} \dfrac{\partial (\Phi_{n,i}\circ q)}{\partial \varphi_{n+1-j}}(\rho,\varphi) & = \dfrac{\partial \Phi_{n,i}}{\partial \rho}(q(\rho,\varphi)) \cdot \dfrac{\partial p_1}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\ & \qquad +\; \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n-1}}(q(\rho,\varphi)) \cdot \dfrac{\partial p_3}{\partial \varphi_{n+1-j}}(\rho,\varphi) +\cdots +\dfrac{\partial \Phi_{n,i}}{\partial \varphi_{1}}(q(\rho,\varphi)) \cdot \dfrac{\partial p_{n+1}}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial \Phi_{n,i}}{\partial \rho}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\rho) \\ & \qquad +\; \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n-1}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{n+2-3}) +\cdots +\dfrac{\partial \Phi_{n,i}}{\partial \varphi_{1}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{n+2 - (n+1)}) \\[0.5em] & = \dfrac{\partial \Phi_{n,i}}{\partial \rho}(q(\rho,\varphi)) \cdot 0 \\ & \qquad +\; \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n-1}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{n-1}) +\cdots +\dfrac{\partial \Phi_{n,i}}{\partial \varphi_{1}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{1}) \\[0.5em] & = \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n+1-j}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{n+1-j}) \\[0.5em] & = \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n+1-j}}(q(\rho,\varphi)) \cdot \dfrac{\partial}{\partial \varphi_{n+1-j}}(\varphi_{n+2- (j+1)}) \\[0.5em] & = \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n+1-j}}(q(\rho,\varphi)) \cdot \dfrac{\partial p_{j+1}}{\partial \varphi_{n+1-j}}(\rho,\varphi) \text{ 임에 따라} \end{align*}$

$\begin{align*} ([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{i,j+1} & = \dfrac{\partial \Phi_{n+1,i}}{\partial \varphi_{n+1 - j}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial (\sin\circ \,p_2)}{\partial \varphi_{n+1-j}}(\rho,\varphi) \cdot (\Phi_{n,i}\circ q)(\rho,\varphi) + (\sin\circ \,p_2)(\rho,\varphi) \cdot \dfrac{\partial (\Phi_{n,i}\circ q)}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = \sin'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \varphi_{n+1-j}}(\rho,\varphi) \cdot \Phi_{n,i}(q(\rho,\varphi)) + \sin(p_2(\rho,\varphi))\cdot \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n+1-j}}(q(\rho,\varphi)) \cdot \dfrac{\partial p_{j+1}}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = \sin'(p_2(\rho,\varphi))\cdot 0 \cdot \Phi_{n,i}(q(\rho,\varphi)) + \sin(p_2(\rho,\varphi))\cdot \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n+1-j}}(q(\rho,\varphi))\cdot 1 \\[0.5em] & = \sin \varphi_{n} \cdot \dfrac{\partial \Phi_{n,i}}{\partial \varphi_{n-(j-1)}}(\rho,\varphi_{n-1},\cdots,\varphi_1) \\[0.5em] & = \sin \varphi_{n} \cdot ([D\Phi_{n}(\rho,\varphi_{n-1},\cdots,\varphi_1)]_{\beta_{n}})_{i,j} \text{ 이고} \end{align*}$

$\begin{align*} ([D\Phi_{n+1}(\rho,\varphi)]_{\beta_{n+1}})_{n+1,j+1} & = \dfrac{\partial \Phi_{n+1,n+1}}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial p_1}{\partial \varphi_{n+1-j}}(\rho,\varphi)\cdot (\cos \circ\, p_2)(\rho,\varphi) + p_1(\rho,\varphi)\cdot \dfrac{\partial (\cos\circ\, p_2)}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = \dfrac{\partial p_1}{\partial \varphi_{n+1-j}}(\rho,\varphi)\cdot \cos(p_2(\rho,\varphi)) + \rho\cdot \cos'(p_2(\rho,\varphi))\cdot \dfrac{\partial p_2}{\partial \varphi_{n+1-j}}(\rho,\varphi) \\[0.5em] & = 0\cdot \cos(p_2(\rho,\varphi)) + \rho \cdot \cos'(p_2(\rho,\varphi))\cdot 0 \\[0.5em] & = 0 \text{ 이 되어} \end{align*}$

$A_{n+1} = \begin{bmatrix} \sin \varphi_n\cdot (A_n)_{1,1} & \rho\cdot \cos \varphi_n\cdot (A_n)_{1,1} & \sin \varphi_n\cdot (A_n)_{1,2} & \cdots & \sin \varphi_n\cdot (A_n)_{1,n} \\ \vdots & \vdots & \vdots &\ddots &\vdots \\ \sin \varphi_n\cdot (A_n)_{n,1} & \rho\cdot \cos \varphi_n\cdot (A_n)_{n,1} & \sin \varphi_n\cdot (A_n)_{n,2} & \cdots & \sin \varphi_n\cdot (A_n)_{n,n} \\ \cos \varphi_n & -\rho\cdot \sin \varphi_n & 0 & \cdots & 0 \end{bmatrix} \text{ 이다.}$

행렬식 정리와 행렬식 정리와 피타고라스 항등식으로

$\begin{align*} \det(A_{n+1}) & = (-1)^{n+1 +1}\cdot \cos\varphi_n \cdot \det(\begin{bmatrix} \rho\cdot \cos \varphi_n\cdot (A_n)_{1,1} & \sin \varphi_n\cdot (A_n)_{1,2} & \cdots & \sin \varphi_n\cdot (A_n)_{1,n} \\ \vdots & \vdots &\ddots &\vdots \\ \rho\cdot \cos \varphi_n\cdot (A_n)_{n,1} & \sin \varphi_n\cdot (A_n)_{n,2} & \cdots & \sin \varphi_n\cdot (A_n)_{n,n} \end{bmatrix}) \\[0.5em] & \qquad +\; (-1)^{n+1 +2}\cdot (-\rho\cdot \sin\varphi_n) \cdot \det( \begin{bmatrix} \sin \varphi_n\cdot (A_n)_{1,1} & \sin \varphi_n\cdot (A_n)_{1,2} & \cdots & \sin \varphi_n\cdot (A_n)_{1,n} \\ \vdots & \vdots &\ddots &\vdots \\ \sin \varphi_n\cdot (A_n)_{n,1} & \sin \varphi_n\cdot (A_n)_{n,2} & \cdots & \sin \varphi_n\cdot (A_n)_{n,n} \end{bmatrix}) \\[0.5em] & = (-1)^{n+2} \cdot \rho\cdot (\cos\varphi_n)^2 \cdot (\sin\varphi_n)^{n-1}\cdot \det(\begin{bmatrix} (A_n)_{1,1} & (A_n)_{1,2} & \cdots & (A_n)_{1,n} \\ \vdots & \vdots &\ddots &\vdots \\ (A_n)_{n,1} & (A_n)_{n,2} & \cdots & (A_n)_{n,n} \end{bmatrix}) \\[0.5em] & \qquad +\; (-1)^{n+ 3}\cdot (-\rho)\cdot (\sin\varphi_n)^{n+1}\cdot \det(\begin{bmatrix} (A_n)_{1,1} & (A_n)_{1,2} & \cdots & (A_n)_{1,n} \\ \vdots & \vdots &\ddots &\vdots \\ (A_n)_{n,1} & (A_n)_{n,2} & \cdots & (A_n)_{n,n} \end{bmatrix}) \\[0.5em] & = (-1)^n\cdot \rho\cdot (\cos\varphi_n)^2 \cdot (\sin\varphi_n)^{n-1}\cdot \det(A_n) + (-1)^{n}\cdot \rho\cdot (\sin\varphi_n)^{n+1}\cdot\det(A_n) \\[0.5em] & = (-1)^n\cdot \rho \cdot (\sin\varphi_n)^{n-1}\cdot \det(A_n) \cdot ((\cos\varphi_n)^2 + (\sin\varphi_n)^2 ) \\[0.5em] & = (-1)^n\cdot \rho \cdot (\sin\varphi_n)^{n-1}\cdot \det(A_n) \text{ 이다.} \end{align*}$

6.

$n\ge 2$인 $n\in \mathbb{Z}^+$에 대한 귀납법으로 증명한다.

$n = 2$이면 극 좌표변환 정리로 모든 $(\rho,\varphi_1)\in \mathbb{R}^2$에 대해

$J\Phi_2(\rho,\varphi_1) = \rho = \rho \cdot (\sin\varphi_1)^0 = (-1)^0\cdot \rho \cdot (\sin\varphi_1)^0 = (-1)^{\frac{4-2-2}{2}}\cdot \rho \cdot(\sin\varphi_1)^0 = (-1)^{\frac{2^2 - 2 -2}{2}}\cdot \rho \cdot (\sin\varphi_1)^0 \text{ 이다.}$

$k \ge 2$인 모든 $k\in \mathbb{Z}^+$에 대해 모든 $(\rho,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^{k}$가 

$J\Phi_k(\rho,\varphi_{k-1},\cdots,\varphi_1) = (-1)^{\frac{k^2 - k-2}{2}}\cdot \rho^{k-1}\cdot (\sin \varphi_{k-1})^{k-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0$이면 

$\dfrac{(k+1)^2 - (k+1) -2}{2}= \dfrac{k^2 + 2\cdot k +1 - k-1 -2}{2}  = \dfrac{k^2-k -2 + 2\cdot k}{2} = \dfrac{k^2 - k-2 }{2} + k$이므로

모든 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^{k+1}$에 대해 야코비안의 정의와 5번으로 

$\begin{align*} J\Phi_{n+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) & = \det([D\Phi_{k+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)]_{\beta_{k+1}}) \\[0.5em] & = (-1)^k\cdot \rho \cdot (\sin\varphi_k)^{k-1}\cdot \det([D\Phi_{k}(\rho,\varphi_{k-1},\cdots,\varphi_1)]_{\beta_{k}}) \\[0.5em] & = (-1)^k\cdot \rho \cdot (\sin\varphi_k)^{k-1}\cdot (-1)^{\frac{k^2 - k-2}{2}}\cdot \rho^{k-1}\cdot (\sin \varphi_{k-1})^{k-2} \cdot \,\cdots\,\cdot (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 \\[0.5em] & = (-1)^{\frac{k^2 - k-2}{2} + k}\cdot \rho^{k}\cdot(\sin\varphi_k)^{k-1}\cdot (\sin \varphi_{k-1})^{k-2} \cdot \,\cdots\,\cdot (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 \\[0.5em] & = (-1)^{\frac{(k+1)^2 - (k+1)-2}{2} }\cdot \rho^{k}\cdot(\sin\varphi_k)^{k-1}\cdot (\sin \varphi_{k-1})^{k-2} \cdot \,\cdots\,\cdot (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 \text{ 이다.} \end{align*}$

 

 

 

정리2

실삼각함수가 $\cos,\sin:\mathbb{R}\to \mathbb{R}$이고 $n \ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해 $n$차원 유클리드 거리공간이 $(\mathbb{R}^n,d_n)$일때

$I_1 = [0,\infty)$이고 $I_n = [0, 2\cdot \pi]$이고 $1 < i < n$인 모든 $i =1,2,\cdots,n$에 대해 $I_i = [0,\pi]$이면

$n$차원 구면 좌표변환 $\Phi_n :\mathbb{R}^n\to \mathbb{R}^n$과 $\Phi_n$의 성분함수 $\Phi_{n,1},\cdots ,\Phi_{n,n}:\mathbb{R}^n\to \mathbb{R}$에 대해 다음이 성립한다.

1. 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해 

$(\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 + (\Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 +\cdots + (\Phi_{n,n}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 = \rho^2$이다.

2. $\displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) \subseteq \Omega_n \subseteq \prod_{i=1}^nI_i$인 어떤 $\Omega_n \subseteq \mathbb{R}^n$이 존재하여 $\Phi_n$의 제한함수 $\Phi_n|_{\Omega_n} : \Omega_n\to \mathbb{R}^n$은 전단사이다.

3. $\mathbb{R}^n$에서 $\Phi_{n}$의 야코비안이 $J\Phi_n : \mathbb{R}^n\to \mathbb{R}$일때

모든 $\displaystyle (\rho,\varphi_{n-1},\cdots,\varphi_1)\in \prod_{i=1}^n I_i$에 대해

$|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| = \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 \ge 0$이고

모든 $\displaystyle (\rho,\varphi_{n-1},\cdots,\varphi_1)\in \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)$에 대해

$|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| = \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 > 0$이다.

4. 임의의 $n$차원 조르당영역 $E\subseteq \displaystyle \prod_{i=1}^nI_i$에 대해 $\Phi_n(E)$는 $n$차원 조르당영역이다.

이때 $\Phi_n(E)\subseteq S$인 임의의 $S \subseteq \mathbb{R}^n$에 대해 함수 $f : S \to \mathbb{R}$가 $\Phi_n(E)$에서 리만적분가능하고

모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in E$에 대해 $f_{\Phi_n}(\rho,\varphi_{n-1},\cdots,\varphi_1) = f(\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1))$인

함수 $f_{\Phi_n} : E\to \mathbb{R}$이 $E$에서 리만적분가능하면

$\displaystyle \int_{\Phi_n(E)}f(x) \operatorname{d}\!x = \int_{E} f(\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1))\cdot |J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| \operatorname{d}(\rho,\varphi_{n-1},\cdots,\varphi_1)$이다.

증명

실수체 $(\mathbb{R},+,\cdot,0,1)$위의 $n$-순서쌍 벡터공간이 $(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$일때

거리공간 정리로 $(\mathbb{R}^n,d_n)$은 $(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$위의 점곱 내적공간의 노름공간 $(\mathbb{R}^n,\lVert\cdot\rVert_n)$의 노름거리공간이다.

거리공간 정리로 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_1) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,\infty)) = (0,\infty)$와 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_n) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,2\cdot\pi]) = (0,2\cdot \pi)$가 성립하고

$1 < i < n$인 모든 $i =1,2,\cdots,n$에 대해 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_i) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,\pi]) = (0,\pi)$이므로 

거리공간 정리와 위상공간 정리와 곱공간 정리로

$\displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) = \prod_{i=1}^n \underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_i) = (0,\infty)\times (0,\pi)\times\cdots \times (0,\pi)\times (0,2\cdot \pi)$이다.

1.

$n\ge 2$인 $n\in \mathbb{Z}^+$에 대한 귀납법으로 증명한다.

$n = 2$이면 모든 $(\rho,\varphi_1)\in \mathbb{R}^2$에 대해 피타고라스 항등식으로

$(\Phi_{2,1}(\rho,\varphi_1))^2 + (\Phi_{2,2}(\rho,\varphi_1))^2 = (\rho\cdot \cos\varphi_1)^2 + (\rho\cdot\sin\varphi_1)^2 = \rho^2\cdot ((\cos\varphi_1)^2 + (\sin\varphi_1)^2) = \rho^2$이다.

$k\ge 2$인 모든 $k\in \mathbb{Z}^+$에 대해

모든 $(\rho,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^k$가 $(\Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2 +\cdots + (\Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2 = \rho^2$이면 

모든 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)\in \mathbb{R}^{k+1}$에 대해 피타고라스 항등식으로

$\begin{align*} \rho^2 & = \rho^2\cdot ((\sin\varphi_k)^2 + (\cos\varphi_k)^2) \\[0.5em] & = (\sin\varphi_k)^2\cdot \rho^2 + \rho^2\cdot (\cos\varphi_k)^2 \\[0.5em] & =(\sin\varphi_k)^2\cdot ((\Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2 +\cdots + (\Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2) + (\rho \cdot \cos\varphi_k)^2 \\[0.5em] & = (\sin\varphi_k\cdot \Phi_{k,1}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2 +\cdots + (\sin\varphi_k\cdot\Phi_{k,k}(\rho,\varphi_{k-1},\cdots,\varphi_1))^2 + (\rho \cdot \cos\varphi_k)^2 \\[0.5em] & = (\Phi_{k+1,1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1))^2 +\cdots + (\Phi_{k+1,k}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1))^2 + (\Phi_{k+1,k+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1))^2 \text{ 이다.} \end{align*}$

따라서 $n\ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$가 

$(\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 + (\Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 +\cdots + (\Phi_{n,n}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 = \rho^2$이다.

2.

$n\ge 2$인 $n\in \mathbb{Z}^+$에 대한 귀납법으로 증명한다.

$n = 2$이면 극 좌표변환 정리로 $\displaystyle (0,\infty)\times (0,2\cdot \pi) \subseteq \Omega_2 \subseteq [0,\infty)\times [0,2\cdot \pi]$인

어떤 $\Omega_2\subseteq \mathbb{R}^2$가 존재하여 $\Phi_2$의 제한함수 $\Phi_2|_{\Omega_2} : \Omega_2\to \mathbb{R}^2$는 전단사이다.

$k\ge 2$인 모든 $k\in \mathbb{Z}^+$에 대해 조건을 만족하는 어떤 $\Omega_k\subseteq \mathbb{R}^k$가 존재하면 

$P = \{ (\rho,\varphi_k, \varphi_{k-1},\cdots, \varphi_1)\in \mathbb{R}^{k+1} : \rho\in (0,\infty)\text{이고 } \varphi_k\in (0,\pi) \text{이고 }(1,\varphi_{k-1},\cdots,\varphi_1) \in \Omega_k\}$이고

$A = \{ (\rho,\varphi_k,0,\cdots,0) : (\rho,\varphi_k)\in (0,\infty)\times \{ 0,\pi\} \}$일때 $\Omega_{k+1} = P\cup A\cup \{ \vec{0}_{k+1}\} \subseteq \mathbb{R}^{k+1}$은 

$(0,\infty) \times (0,\pi)\times \cdots \times (0,\pi)\times (0,2\cdot \pi) \subseteq P \subseteq \Omega_{k+1} \subseteq [0,\infty)\times [0,\pi]\times \cdots\times [0,\pi]\times [0,2\cdot\pi]$이다.

임의의 $(x_1,\cdots,x_k,x_{k+1})\in \mathbb{R}^{k+1}$에 대해 $\rho = \sqrt{x_1^2 + \cdots + x_k^2 + x_{k+1}^2} \in [0,\infty)$일때

$\rho = 0$이면 $x_1 = \cdots= x_k= x_{k+1} = 0$이므로 $\vec{0}_{k+1}\in \Omega_{k+1}$임에 따라 제한함수의 정의와 위 정리로

$\Phi_{k+1}|_{\Omega_{k+1}}(\vec{0}_{k+1}) = \Phi_{k+1}(\vec{0}_{k+1}) = 0\cdot_{k+1} \Phi_{k+1}(1,0,0,\cdots,0) = \vec{0}_{k+1} = (x_1,\cdots,x_k,x_{k+1})$이다.

$\rho > 0$이면 절댓값 정리로 $|x_{k+1}| = \sqrt{x_{k+1}^2}\le \sqrt{x_1^2 + \cdots + x_k^2 + x_{k+1}^2} = \rho$이므로

$-1\le \dfrac{x_{k+1}}{\rho} \le 1$이 되어 삼각함수 정리로 $\cos \varphi_k = \dfrac{x_{k+1}}{\rho}$인 $\varphi_k\in [0,\pi]$가 존재하고

$\sin\varphi_k = 0$일때 삼각함수 정리로 $\varphi_k\in \{ 0,\pi\}$가 되어 삼각함수 정리로 $\cos \varphi_k \in \{ -1,1\}$이고 

$x_{k+1}^2 = (\rho\cdot \cos\varphi_k)^2 = \rho^2\cdot (\cos\varphi_k)^2 = \rho^2  = x_1^2 +\cdots + x_k^2 + x_{k+1}^2$임에 따라 $x_1 = \cdots= x_k= 0$이고

$(\rho,\varphi_k,0,\cdots,0)\in A\subseteq \Omega_{k+1}$이므로 제한함수의 정의와 구면 좌표변환의 정의와 $k+1$-데카르트곱의 정의로

$\begin{align*}\Phi_{k+1}|_{\Omega_{k+1}}(\rho,\varphi_k,0,\cdots,0) & =\Phi_{k+1}(\rho,\varphi_k,0,\cdots,0) \\[0.5em] & = (\sin\varphi_k\cdot_k \Phi_k(\rho,0,\cdots,0), \rho\cdot \cos\varphi_k) \\[0.5em] & = (0\cdot_k \Phi_k(\rho,0,\cdots,0),x_{k+1}) \\[0.5em] & = (0,\cdots,0,x_{k+1}) \\[0.5em] & = (x_1,\cdots,x_k,x_{k+1}) \text{ 이다.} \end{align*}$

$\sin\varphi_k \ne 0$일때 삼각함수 정리로 $\varphi_k\notin \{ 0,\pi\}$이므로 $\varphi_k\in (0,\pi)$이고 $\dfrac{1}{\rho\cdot \sin\varphi_k}\cdot_k (x_1,\cdots,x_k)\in \mathbb{R}^k$이므로 

전사의 정의로 $\Phi_k|_{\Omega_k}(r,\varphi_{k-1},\cdots,\varphi_1) = \dfrac{1}{\rho\cdot \sin\varphi_k}\cdot_k (x_1,\cdots,x_k)$인 $(r,\varphi_{k-1},\cdots,\varphi_1)\in \Omega_k$가 존재하여

$r\in [0,\infty)$이고 제한함수의 정의와 1번과 피타고라스 항등식으로

$\begin{align*} r^2 & = (\Phi_{k,1}(r,\varphi_{k-1},\cdots,\varphi_1))^2 + \cdots + (\Phi_{k,k}(r,\varphi_{k-1},\cdots,\varphi_1))^2 \\[0.5em] & = \dfrac{1}{\rho^2 \cdot (\sin\varphi_k)^2} \cdot (x_1^2 +\cdots + x_k^2) \\[0.5em] & = \dfrac{1}{\rho^2\cdot (\sin\varphi_k)^2} \cdot (\rho^2 - x_{k+1}^2) \\[0.5em] & = \dfrac{1}{\rho^2 \cdot (\sin\varphi_k)^2} \cdot (\rho^2 - \rho^2 \cdot (\cos \varphi_k)^2) \\[0.5em] & = \dfrac{1}{(\sin\varphi_k)^2} \cdot (1 - (\cos \varphi_k)^2) \\[0.5em] & = \dfrac{1}{(\sin\varphi_k)^2}\cdot (\sin \varphi_k)^2 \\[0.5em] & = 1 \text{ 임에 따라}\end{align*}$

$r = 1$이고 $(1,\varphi_{k-1},\cdots,\varphi_1) = (r,\varphi_{k-1},\cdots,\varphi_1)\in \Omega_k$이므로 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1)\in P\subseteq \Omega_{k+1}$이고

위 정리와 구면 좌표변환의 정의와 $k+1$-데카르트곱의 정의로

$\begin{align*} \Phi_{k+1}|_{\Omega_{k+1}}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) & =\Phi_{k+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) \\[0.5em] & = (\sin\varphi_k\cdot_k \Phi_k(\rho,\varphi_{k-1},\cdots,\varphi_1), \rho\cdot \cos\varphi_k) \\[0.5em] & = (\sin\varphi_k\cdot_k (\rho\cdot_k \Phi_k(1,\varphi_{k-1},\cdots,\varphi_1)),x_{k+1}) \\[0.5em] & = ((\rho\cdot \sin\varphi_k\cdot \tfrac{1}{\rho \cdot \sin\varphi_k})\cdot_k (x_1,\cdots,x_k) ,x_{k+1} ) \\[0.5em] & = ((x_1,\cdots,x_k),x_{k+1}) \\[0.5em] & = (x_1,\cdots,x_k,x_{k+1}) \text{ 이 되어} \end{align*}$

$\Phi_{k+1}|_{\Omega_{k+1}}$은 전사이다.

임의의 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1), (r,\theta_k,\theta_{k-1},\cdots,\theta_1)\in \Omega_{k+1} = P\cup A\cup \{ \vec{0}_{k+1}\}$에 대해

$\Phi_{k+1}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) = \Phi_{k+1}|_{\Omega_{k+1}}(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) = \Phi_{k+1}|_{\Omega_{k+1}}(r,\theta_k,\theta_{k-1},\cdots,\theta_1) = \Phi_{k+1}(r,\theta_k,\theta_{k-1},\cdots,\theta_1)\text{ 이면}$

$\rho,r\in [0,\infty)$이므로 1번으로 $\rho^2 = r^2$임에 따라 절댓값 정리로 $\rho = |\rho| = \sqrt{\rho^2} = \sqrt{r^2} = |r| = r$이고 

$\rho = r= 0$일때 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1), (r,\theta_k,\theta_{k-1},\cdots,\theta_1)\in \{ \vec{0}_{k+1}\}$이 되어

$(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) =\vec{0}_{k+1} = (r,\theta_k,\theta_{k-1},\cdots,\theta_1)$이다.

$\rho = r > 0$일때 

$(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1), (r,\theta_k,\theta_{k-1},\cdots,\theta_1)\in P\cup A$이고 구면 좌표변환의 정의로

$(\sin\varphi_k\cdot_k \Phi_k(\rho,\varphi_{k-1},\cdots,\varphi_1),\rho\cdot \cos\varphi_k) = (\sin\varphi_k\cdot_k\Phi_{k}(r,\theta_{k-1},\cdots,\theta_1), r\cdot \cos\theta_k)$이므로

$\cos \varphi_k = \cos \theta_k$가 되어 $\varphi_k,\theta_k \in [0,\pi]$임에 따라 삼각함수 정리로 $\varphi_k = \theta_k$이고 

$\sin \varphi_k = \sin \theta_k = 0$이면

삼각함수 정리로 $\varphi_k = \theta_k \in \{ 0,\pi\}$이므로 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1), (r,\theta_k,\theta_{k-1},\cdots,\theta_1)\in A$가 되어 

$(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) = (\rho,\varphi_k,0,\cdots,0) = (r,\theta_k,0,\cdots,0) = (r,\theta_k,\theta_{k-1},\cdots,\theta_1)$이고 

$\sin \varphi_k = \sin \theta_k \ne 0$이면 삼각함수 정리로 $\varphi_k = \theta_k \notin \{ 0,\pi\}$이므로 $\varphi_k = \theta_k \in (0,\pi)$가 되어

$(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1), (r,\theta_k,\theta_{k-1},\cdots,\theta_1)\in P$임에 따라 $(1,\varphi_{k-1},\cdots,\varphi_1), (1,\theta_{k-1},\cdots,\theta_1)\in \Omega_k$이고 위 정리로

$\rho\cdot_k \Phi_k(1,\varphi_{k-1},\cdots,\varphi_1)= \Phi_k(\rho,\varphi_{k-1},\cdots,\varphi_1) = \Phi_{k}(r,\theta_{k-1},\cdots,\theta_1) = r\cdot_k \Phi_k(1,\theta_{k-1},\cdots,\theta_1)$이므로 

$\Phi_k|_{\Omega_k}(1,\varphi_{k-1},\cdots,\varphi_1)= \Phi_k(1,\varphi_{k-1},\cdots,\varphi_1)= \Phi_k(1,\theta_{k-1},\cdots,\theta_1) = \Phi_k|_{\Omega_k}(1,\theta_{k-1},\cdots,\theta_1)$이 되어 

단사의 정의로 $(1,\varphi_{k-1},\cdots,\varphi_1)= (1,\theta_{k-1},\cdots,\theta_1)$이고 

순서쌍의 상등으로 $(\rho,\varphi_k,\varphi_{k-1},\cdots,\varphi_1) = (r,\theta_k,\theta_{k-1},\cdots,\theta_1)$임에 따라 $\Phi_{k+1}|_{\Omega_{k+1}}$은 단사이다. 

따라서 $n\ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해 $\displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) \subseteq \Omega_n \subseteq \prod_{i=1}^nI_i$인 $\Omega_n \subseteq \mathbb{R}^n$이 존재하여 $\Phi_n|_{\Omega_n}$은 전단사이다.

3.

모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^n$에 대해 $(\sin \varphi_1)^0 = 1$이고 위 정리로

$\begin{align*}|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| & = |(-1)^{\frac{n^2 - n-2}{2}}\cdot \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0| \\[0.5em] &= |\rho^{n-1}\cdot (\sin\varphi_{n-1})^{n-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0| \text{ 이므로} \end{align*}$

삼각함수 정리로 모든 $\theta\in [0,\pi]$에 대해 $0\le \sin \theta$이고 모든 $\theta\in (0,\pi)$에 대해 $0 < \sin \theta$임에 따라 

모든 $\displaystyle (\rho,\varphi_{n-1},\cdots,\varphi_1)\in \prod_{i=1}^n I_i = [0,\infty)\times [0,\pi]\times \cdots\times [0,\pi] \times [0,2\cdot \pi]$에 대해

$|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| = \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 \ge 0$이고

모든 $\displaystyle (\rho,\varphi_{n-1},\cdots,\varphi_1)\in \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) = (0,\infty)\times (0,\pi)\times \cdots \times (0,\pi)\times (0,2\cdot \pi)$에 대해

$|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| = \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 > 0$이다.

4.

내부 정리와 2번으로 $\displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)} \subseteq \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) \subseteq \Omega_n$이므로 $\Phi_n|_{\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)}}: \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)} \to \mathbb{R}^n$은 단사이다.

거리공간 정리로 $\mathbb{R}^n$은 $(\mathbb{R}^n,d_n)$에서 열린집합이고 폐포의 정의로 $\underset{(\mathbb{R}^n,d_n)}{\operatorname{cl}(E)}\subseteq \mathbb{R}^n$이므로

3번으로 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)} \subseteq \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\displaystyle \prod_{i=1}^n I_i)$에 대해 

$|J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| = \rho^{n-1}\cdot (\sin \varphi_{n-1})^{n-2} \cdot \,\cdots\,\cdot  (\sin\varphi_2)^1 \cdot (\sin \varphi_1)^0 > 0$이 되어

거리공간 정리와 폐포 정리로 $\underset{(\mathbb{R}^n,d_n)}{\operatorname{cl}(\emptyset)} = \emptyset$이고 부피 정리로 $\emptyset$의 부피가 $0$임에 따라

위 정리와 조르당영역 정리로 $\Phi_n(E), \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)})=\Phi_n({\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)}}\setminus \emptyset)$은 조르당영역이다.

내부 정리로 $\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)}\subseteq E$이고 조르당영역 정리로 $\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)}$는 조르당영역이므로

$f$가 $\Phi_n(E)$에서 리만적분가능하고 $f_{\Phi_n}$이 $E$에서 리만적분가능함에 따라

적분 정리로 $f$는 $\Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)})$에서 리만적분가능하고 $f_{\Phi_n}$은 $\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}(E)}$에서 리만적분가능하여

적분 정리로 $\displaystyle \int_{\Phi_n(E)}f(x) \operatorname{d}\!x = \int_{E} f(\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1))\cdot |J\Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)| \operatorname{d}(\rho,\varphi_{n-1},\cdots,\varphi_1)$이다.

 

 

 

정리3

실삼각함수가 $\cos,\sin:\mathbb{R}\to \mathbb{R}$이고 $n \ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해

실수체 $(\mathbb{R},+,\cdot,0,1)$위의 $n$-순서쌍 벡터공간이 $(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$이고

$(\mathbb{R}^n,+_n,\cdot_n,\vec{0}_n)$위의 점곱 내적공간의 노름공간 $(\mathbb{R}^n,\lVert\cdot\rVert_n)$의 노름거리공간이 $(\mathbb{R}^n,d_n)$일때 

임의의 $r \in (0,\infty)$에 대해 $I_1 = [0,r]$이고 $I_n = [0, 2\cdot \pi]$이고

$1 < i < n$인 모든 $i =1,2,\cdots,n$에 대해 $I_i = [0,\pi]$이면

$n$차원 구면 좌표변환 $\Phi_n :\mathbb{R}^n\to \mathbb{R}^n$과 $\Phi_n$의 성분함수 $\Phi_{n,1},\cdots ,\Phi_{n,n}:\mathbb{R}^n\to \mathbb{R}$에 대해 다음이 성립한다.

1. $B[r] = \{ x\in \mathbb{R}^n : \lVert x\rVert_n \le r \}$일때 $\displaystyle \Phi_n(\prod_{i=1}^n I_i) = B[r]$이다.

2. $B(r) = \{ x\in \mathbb{R}^n : \lVert x\rVert_n < r \}$일때 $Z_1 = [0,r]$이고 $Z_2 = \{ 0\}$이고

$i > 2$인 모든 $i=1,2,\cdots, n$에 대해 $Z_i = [-r,r]$이면 $\displaystyle \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)) = B(r)\setminus \prod_{i=1}^n Z_i$이다.

3. 임의의 $x_0\in \mathbb{R}^n$에 대해 $(\mathbb{R}^n,d_n)$에서 열린공과 닫힌공 $\underset{(\mathbb{R}^n,d_n)}{B(x_0,r)},\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}$은 $n$차원 조르당영역이고 

감마함수 $\Gamma : (0,\infty)\to \mathbb{R}$에 대해 $\underset{(\mathbb{R}^n,d_n)}{B(x_0,r)},\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}$의 부피는 $vol(\underset{(\mathbb{R}^n,d_n)}{B(x_0,r)}) = \dfrac{2\cdot \pi^\frac{n}{2}\cdot r^n}{n\cdot \Gamma(\frac{n}{2})}= vol(\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]})$이다.

증명

거리공간 정리로 $(\mathbb{R}^n,d_n)$은 $n$차원 유클리드 거리공간이고 거리공간 정리로 $\mathbb{R}^n$은 $(\mathbb{R}^n,d_n)$에서 열린집합이다.

거리공간 정리로 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_1) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,r]) = (0,r)$과 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_n) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,2\cdot\pi]) = (0,2\cdot \pi)$가 성립하고

$1 < i < n$인 모든 $i =1,2,\cdots,n$에 대해 $\underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_i) = \underset{(\mathbb{R},d_1)}{\operatorname{int}}([0,\pi]) = (0,\pi)$이므로 

거리공간 정리와 위상공간 정리와 곱공간 정리로

$\displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i) = \prod_{i=1}^n \underset{(\mathbb{R},d_1)}{\operatorname{int}}(I_i) = (0,r)\times (0,\pi)\times\cdots \times (0,\pi)\times (0,2\cdot \pi)$이다.

1.

모든 $x \in \Phi_n (\displaystyle \prod_{i=1}^n I_i)$에 대해 $x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)$인 $(\rho,\varphi_{n-1},\cdots,\varphi_1) \in \displaystyle \prod_{i=1}^n I_i$가 존재하여

위 정리로 $\lVert x\rVert_n = \sqrt{(\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 + \cdots+ (\Phi_{n,n}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2} = \sqrt{\rho^2} = |\rho| = \rho\le r$이므로 

$x\in B[r]$임에 따라 $\Phi_n (\displaystyle \prod_{i=1}^n I_i)\subseteq B[r]$이다.

위 정리로 어떤 $\Omega_n \subseteq [0,\infty)\times [0,\pi]\times \cdots\times [0,\pi]\times [0,2\cdot\pi] \subseteq \mathbb{R}^n$이 존재하여

$\Phi_n$의 제한함수 $\Phi_n|_{\Omega_n} : \Omega_n\to \mathbb{R}^n$은 전사이므로 

모든 $x\in B[r] \subseteq \mathbb{R}^n$에 대해 $x = \Phi_n|_{\Omega_n}(\rho,\varphi_{n-1},\cdots,\varphi_1)$인 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \Omega_n$가 존재하고

제한함수의 정의로 $x = \Phi_n|_{\Omega_n}(\rho,\varphi_{n-1},\cdots,\varphi_1) = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)$이므로 위 정리로

$\rho = |\rho|= \sqrt{\rho^2} = \sqrt{(\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2 + \cdots+ (\Phi_{n,n}(\rho,\varphi_{n-1},\cdots,\varphi_1))^2} = \lVert x\rVert_n \le r$임에 따라 

$\rho\in [0,r]$이고 $(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \displaystyle \prod_{i=1}^n I_i$가 되어 $x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_1)\in \Phi_n (\displaystyle \prod_{i=1}^n I_i)$이므로

$B[r]\subseteq \Phi_n (\displaystyle \prod_{i=1}^n I_i)$이고 집합 정리로 $\displaystyle \Phi_n(\prod_{i=1}^n I_i) = B[r]$이다.

2.

$n = 2$이면 극 좌표변환 정리로 $\Phi_2((0,r)\times (0,2\cdot \pi)) = B(r)\setminus ([0,r]\times \{ 0\})$이다.

$n\ge 3$이면 위 정리와 구면 좌표변환의 정의로 모든 $(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in \mathbb{R}^n$에 대해 

$\begin{align*} \Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) & = \rho\cdot \Phi_{n,1}(1,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) \\[0.5em] & = \rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \Phi_{2,1}(1,\varphi_1) \\[0.5em] & = \rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \cos \varphi_1 \text{ 이고}\end{align*}$

$\begin{align*} \Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) & = \rho\cdot \Phi_{n,2}(1,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) \\[0.5em] & = \rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \Phi_{2,2}(1,\varphi_1) \\[0.5em] & = \rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \sin \varphi_1 \text{ 이다.}\end{align*}$

모든 $x\in \displaystyle \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)) = \Phi_n((0,r)\times (0,\pi)\times \cdots \times (0,\pi)\times (0,2\cdot\pi))$에 대해

$x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)$인 $(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\displaystyle \prod_{i=1}^n I_i)$이 존재하여 

위 정리로 $\lVert x\rVert_n = \sqrt{\rho^2} = |\rho| = \rho < r$이므로 $x\in B(r)$이고

$x = (x_1,x_2,x_3,\cdots,x_n)\in \displaystyle \prod_{i=1}^n Z_i = [0,r] \times \{ 0\} \times [-r,r]\times \cdots \times [-r,r]$이라고 가정하면 

$\rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \sin \varphi_1 = \Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) = x_2 = 0$이므로 

$\rho > 0$이고 $1 < i < n$인 모든 $i=1,2,\cdots,n$에 대해 $\varphi_i\in (0,\pi)$임에 따라 삼각함수 정리로 $\sin \varphi_i > 0$이 되어 

$\sin\varphi_1 = 0$이고 삼각함수 정리로 $\dfrac{\varphi_1}{\pi}\in \mathbb{Z}$인데

$0 < \varphi_1 < 2\cdot \pi$이므로 $\varphi_1 = \pi$이고 삼각함수 정리와 삼각함수 정리로 $\cos \varphi_1 = \cos \pi = -\cos 0 = -1$임에 따라

$0\le x_1 = \Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) = \rho \cdot \sin\varphi_{n-1}\cdot\,\cdots\,\cdot \sin\varphi_2\cdot \cos \varphi_1 < 0$이 되어 모순이므로 

$x \notin \displaystyle \prod_{i=1}^n Z_i $이고 $x\in B(r)\setminus \displaystyle \prod_{i=1}^n Z_i$가 되어 $\displaystyle \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)) \subseteq B(r)\setminus \prod_{i=1}^n Z_i$이다.

모든 $x\in B(r) \setminus \displaystyle \prod_{i=1}^n Z_i \subseteq B[r] = \Phi_n(\prod_{i=1}^nI_i) = \Phi_n([0,r]\times [0,\pi]\times \cdots \times [0,\pi]\times [0,2\cdot \pi])$에 대해 

$x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)$인 $(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in \displaystyle \prod_{i=1}^n I_i$가 존재하여 위 정리로 $\rho = \sqrt{\rho^2}= \lVert x\rVert_n < r$이고 

$\rho = 0$이라고 가정하면

위 정리로 $x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) = \rho\cdot_n \Phi_n(1,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) = \vec{0}_n\in \displaystyle \prod_{i=1}^n Z_i$임에 따라 모순이므로

$\rho > 0$이 되어 $\rho \in (0,r)$이고 

$1 < i < n$인 임의의 $i=1,2,\cdots,n$에 대해 $\varphi_i\in \{ 0,\pi\}$라고 가정하면

삼각함수 정리로 $\sin\varphi_i = 0$이므로 $\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_{2},\varphi_1) =0 =\Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)$이고

삼각함수 정리와 구면 좌표변환의 정의로

$\Phi_{n,3}(\rho,\varphi_{n-1},\cdots,\varphi_{2},\varphi_1) , \cdots ,\Phi_{n,n}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in [-\rho,\rho]\subseteq [-r,r]$이 되어

$x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) \in \displaystyle \prod_{i=1}^n Z_i$임에 따라 모순이므로 $\varphi_i \in (0,\pi)$이고 

$\varphi_1 \in \{ 0,2\cdot \pi\}$라고 가정하면 삼각함수 정리와 삼각함수 정리로 $0 < \sin\varphi_i \le 1$이고

삼각함수 정리와 삼각함수 정리로 $\cos\varphi_1= 1 = \cos0 = \cos(2\cdot \pi)$와 $\sin\varphi_1= 0 = \sin 0 = \sin (2\cdot \pi)$가 성립하여 

$\Phi_{n,1}(\rho,\varphi_{n-1},\cdots,\varphi_{2},\varphi_1) \in [0,\rho]\subseteq [0,r]$이고 $\Phi_{n,2}(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) = 0$임에 따라 

$x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1) \in \displaystyle \prod_{i=1}^n Z_i$가 되어 모순이므로 $\varphi_1\in (0,2\cdot\pi)$이고 

$(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in \displaystyle \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)$이 되어 $x = \Phi_n(\rho,\varphi_{n-1},\cdots,\varphi_2,\varphi_1)\in \displaystyle \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i))$이므로 

$B(r) \setminus \displaystyle \prod_{i=1}^n Z_i \subseteq \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^nI_i))$임에 따라 집합 정리로 $\displaystyle \Phi_n(\underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\prod_{i=1}^n I_i)) = B(r)\setminus \prod_{i=1}^n Z_i$이다.

3.

$\mathbb{R}^n$에서 $\Phi_{n}$의 야코비안이 $J\Phi_n : \mathbb{R}^n\to \mathbb{R}$일때

위 정리와 야코비안 정리로 $J\Phi_n$의 절댓값 $|J\Phi_n| : \mathbb{R}^n\to \mathbb{R}$은 $(\mathbb{R}^n,d_n)$에서 $(\mathbb{R},d_1)$로의 연속함수이므로 

위 정리와 연속함수 정리로 모든 $(\varphi_{n-1},\cdots,\varphi_1)\in \mathbb{R}^{n-1}$에 대해

$\gamma(\varphi_{n-1},\cdots,\varphi_1) = |J\Phi_n(1,\varphi_{n-1},\cdots,\varphi_1)| = (\sin\varphi_{n-1})^{n-2}\cdot \,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0$인

함수 $\gamma : \mathbb{R}^{n-1}\to \mathbb{R}$는 $(\mathbb{R}^{n-1},d_{n-1})$에서 $(\mathbb{R},d_1)$로의 연속함수이다.

$n\ge 2$인 $n\in \mathbb{Z}^+$에 대한 귀납법으로 

$\dfrac{2\cdot \pi^\tfrac{n}{2}}{\Gamma(\tfrac{n}{2})} = \displaystyle \int_{\overset{n}{\underset{i=2}{\prod}} I_i} (\sin\varphi_{n-1})^{n-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\varphi_{n-1},\cdots,\varphi_2,\varphi_1)$임을 증명한다.

$n = 2$이면 적분 정리와 상자 정리와 감마함수 정리로

$\displaystyle \int_{[0,2\cdot\pi]} (\sin \varphi_1)^0\operatorname{d}\!\varphi_1 = \int_{[0,2\cdot \pi]} 1\operatorname{d}\!\varphi_1 = 2\cdot \pi -0 = \dfrac{2\cdot \pi^1}{0!} = \dfrac{2\cdot \pi^{\tfrac{2}{2}}}{\Gamma(\tfrac{2}{2})} $이다.

$k\ge 2$인 모든 $k\in \mathbb{Z}^+$에 대해 조건이 성립할때 감마함수 정리와 푸비니 정리와 적분 정리로 

$\begin{align*} \dfrac{2\cdot \pi^\tfrac{k+1}{2}}{\Gamma(\tfrac{k+1}{2})} & = \dfrac{2\cdot \pi^\tfrac{k}{2} \cdot \pi^\tfrac{1}{2}}{\Gamma(\tfrac{k+1}{2})}\cdot \dfrac{\Gamma(\tfrac{k}{2})}{\Gamma(\tfrac{k}{2})} \\[0.5em] & = \sqrt{\pi}\cdot \dfrac{\Gamma(\tfrac{k}{2})}{\Gamma(\tfrac{k+1}{2})} \cdot \dfrac{2\cdot \pi^\tfrac{k}{2}}{\Gamma(\tfrac{k}{2})} \\[0.5em] & = \left ( \int_0^\pi (\sin\varphi_k)^{k-1} \operatorname{d}\!\varphi_k \right ) \cdot \left (\int_{\overset{k+1}{\underset{i=3}{\prod}} I_i} (\sin\varphi_{k-1})^{k-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\varphi_{k-1},\cdots,\varphi_2,\varphi_1) \right ) \\[0.5em] & = \left ( \int_0^\pi (\sin\varphi_k)^{k-1} \operatorname{d}\!\varphi_k \right ) \cdot \left (\int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi} (\sin\varphi_{k-1})^{k-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\! \varphi_1 \operatorname{d}\!\varphi_2 \cdots \operatorname{d}\!\varphi_{k-1} \right ) \\[0.5em] & = \int_0^\pi (\sin\varphi_k)^{k-1} \cdot \int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi} (\sin\varphi_{k-1})^{k-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\! \varphi_1 \operatorname{d}\!\varphi_2 \cdots \operatorname{d}\!\varphi_{k-1} \operatorname{d}\!\varphi_k \\[0.5em] & = \int_0^\pi \int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi}(\sin\varphi_k)^{k-1}\cdot (\sin\varphi_{k-1})^{k-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\! \varphi_1 \operatorname{d}\!\varphi_2 \cdots \operatorname{d}\!\varphi_{k-1} \operatorname{d}\!\varphi_{k} \\[0.5em] & = \int_{\overset{k+1}{\underset{i=2}{\prod}} I_i} (\sin\varphi_k)^{k-1}\cdot (\sin\varphi_{k-1})^{k-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\varphi_k,\varphi_{k-1},\cdots,\varphi_2,\varphi_1) \text{ 이다.} \end{align*}$

따라서 $n\ge 2$인 모든 $n\in \mathbb{Z}^+$에 대해

$\dfrac{2\cdot \pi^\tfrac{n}{2}}{\Gamma(\tfrac{n}{2})} = \displaystyle \int_{\overset{n}{\underset{i=2}{\prod}} I_i} (\sin\varphi_{n-1})^{n-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\varphi_{n-1},\cdots,\varphi_2,\varphi_1)$이므로 

$n -1 \ge 1 > 0$임에 따라 1번과 위 정리와 적분 정리와 푸비니 정리와 적분 정리와 적분 정리로

$\begin{align*} vol(B[r]) & = \int_{B[r]} 1\operatorname{d}\!x \\[0.5em] & = \int_{\Phi_n(\overset{n}{\underset{i=1}{\prod}} I_i)} 1\operatorname{d}\!x \\[0.5em] & = \int_{\overset{n}{\underset{i=1}{\prod}} I_i} \rho^{n-1} \cdot (\sin \varphi_{n-1})^{n-2}\cdot\,\cdots\, \cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\rho,\varphi_{n-1},\cdots, \varphi_2,\varphi_1) \\[0.5em] & = \int_0^r \int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi} \rho^{n-1} \cdot (\sin \varphi_{n-1})^{n-2}\cdot\,\cdots\, \cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\!\varphi_1\operatorname{d}\!\varphi_2\cdots \operatorname{d}\!\varphi_{n-1}\operatorname{d}\!\rho \\[0.5em] & = \int_0^r \rho^{n-1} \cdot \int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi} (\sin \varphi_{n-1})^{n-2}\cdot\,\cdots\, \cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\!\varphi_1\operatorname{d}\!\varphi_2\cdots \operatorname{d}\!\varphi_{n-1}\operatorname{d}\!\rho \\[0.5em] & = \left (\int_0^r \rho^{n-1} \operatorname{d}\!\rho \right ) \cdot \left ( \int_0^\pi \cdots \int_0^\pi \int_0^{2\cdot \pi} \rho^{n-1} \cdot (\sin \varphi_{n-1})^{n-2}\cdot\,\cdots\, \cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}\!\varphi_1\operatorname{d}\!\varphi_2\cdots \operatorname{d}\!\varphi_{n-1} \right ) \\[0.5em] & = \left (\int_0^r \rho^{n-1} \operatorname{d}\!\rho \right ) \cdot \left ( \int_{\overset{n}{\underset{i=2}{\prod}} I_i} (\sin\varphi_{n-1})^{n-2}\cdot\,\cdots\,\cdot (\sin\varphi_2)^1\cdot (\sin\varphi_1)^0 \operatorname{d}(\varphi_{n-1},\cdots,\varphi_2,\varphi_1) \right ) \\[0.5em] & = \dfrac{r^n -0^n}{n} \cdot \dfrac{2\cdot \pi^\tfrac{n}{2}}{\Gamma(\tfrac{n}{2})} \\[0.5em] & = \dfrac{2\cdot \pi^\frac{n}{2}\cdot r^n}{n\cdot \Gamma(\frac{n}{2})} \text{ 이다.} \end{align*}$

모든 $x\in B[r]$에 대해 $d_n(x_0,x_0 +_n x) = d_n(x_0+_n x,x_0) = \lVert x_0 +_n x - x_0\rVert_n = \lVert x\rVert_n \le r$이므로

닫힌공의 정의로 $x_0 +_n x\in \underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}$이 되어 $\{ x_0 +_n x : x\in B[r]\} \subseteq \underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}$이고 

모든 $y \in \underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}$에 대해 닫힌공의 정의로 $\lVert y - x_0 \rVert_n = d_n(y,x_0) = d_n(x_0,y)\le r$이므로 $y - x_0 \in B[r]$가 되어 

$y  =x_0 - x_0 +_n y = x_0 +_n (y- x_0)\in \{ x_0 +_n x: x\in B[r]\}$이고 $\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}\subseteq \{ x_0+_n x: x\in B[r]\}$임에 따라

집합 정리와 평행이동 정리로 $\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]} = \{ x_0 +_n x : x\in B[r]\}$은 $n$차원 조르당영역이고

$vol(\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}) = vol(\{ x_0 +_n x : x\in B[r]\}) = vol(B[r]) = \dfrac{2\cdot \pi^\frac{n}{2}\cdot r^n}{n\cdot \Gamma(\frac{n}{2})}$이므로

노름공간 정리와 조르당영역 정리로 $\underset{(\mathbb{R}^n,d_n)}{B(x_0,r)} = \underset{(\mathbb{R}^n,d_n)}{\operatorname{int}}(\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]})$은 $n$차원 조르당영역이고 

$vol(\underset{(\mathbb{R}^n,d_n)}{B(x_0,r)}) = vol(\underset{(\mathbb{R}^n,d_n)}{B[x_0,r]}) = \dfrac{2\cdot \pi^\frac{n}{2}\cdot r^n}{n\cdot \Gamma(\frac{n}{2})}$이다.

 

 

 

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정의의 링크 : 

https://openknowledgevl.tistory.com/144#def번호

번호는 해당 정의 옆에 붙어있는 작은 숫자입니다.

 

정리의 링크 : 

https://openknowledgevl.tistory.com/144#thm번호

번호는 해당 정리 옆에 붙어있는 작은 숫자입니다.

 

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출처(저자 - 제목 - ISBN13)

William R. Wade - Introduction to Analysis - 9780132296380