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針小棒大な証明

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あいさつ

んちゃ！
今回は証明の難易度に対して得られるものが余りにも少ない針小棒大な証明を紹介するのだ。

$n\in\mathbb{N}\setminus\{1\}$に対して以下の式が成り立つ。
\begin{equation} (\prod_{i=1}^{n}\int_{0}^{1}dx_{i})\frac{1}{1-\prod_{i=1}^{n}x_{i}}=\zeta(n) \end{equation}

半径$1$の$n-1$次元球面の面積

半径$1$の$n-1$次元球面の面積を$\omega_{n}$とすると下記の様に書ける。
\begin{equation} \omega_{n}=\frac{2\pi^{\frac{n}{2}}}{\Gamma{(\frac{n}{2})}} \end{equation}

証明を見る

\begin{eqnarray} (\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{n}&=&(\prod_{i=1}^{n}\int_{-\infty}^{\infty}dx_{i})e^{\sum_{i=1}^{n}x_{i}^{2}}\\ &=&\omega_{n}\int_{0}^{\infty}r^{n-1}e^{-r^{2}}dr\quad(r^{2}=s)\\ &=&\frac{\omega_{n}}{2}\int_{0}^{\infty}s^{\frac{n}{2}-1}e^{-s}ds\\ &=&\frac{\Gamma{(\frac{n}{2})}}{2}\omega_{n}\\ &=&\pi^{\frac{n}{2}} \end{eqnarray}
より、
\begin{equation} \omega_{n}=\frac{2\pi^{\frac{n}{2}}}{\Gamma{(\frac{n}{2})}} \end{equation}

$n(\geq 2)$次元空間の極形式は以下の様に与えられる。
\begin{eqnarray} \left\{ \begin{array}{l} x_{1}=r\cos{\theta_{1}}\\ x_{2}=r\sin{\theta_{1}}\cos{\theta_{2}}\\ \vdots\\ x_{i}=r\sin{\theta_{1}}\cdots\sin{\theta_{i-1}}\cos{\theta_{i}}\\ \vdots\\ x_{n-1}=r\sin{\theta_{1}}\cdots\sin{\theta_{n-2}}\cos{\theta_{n-1}}\\ x_{n}=r\sin{\theta_{1}}\cdots\sin{\theta_{n-2}}\sin{\theta_{n-1}} \end{array} \right. \end{eqnarray}
ただし、$0\leq \theta_{i}\leq \pi\quad(1\leq i\leq n-2),0\leq \theta_{n-1}\lt 2\pi$

証明を見る

これは帰納法によって証明できる。
[1]$n=2,3$の場合は明らか
[2]$1,2,...,n$まで成立すると仮定する。
[3]そこで$n+1$次元の場合を考える。まず$x_{1}$軸と$\vb*{x}$のなす角を$\theta_{1}$とおくと、$\vb*{x}$の$x_{1}$方向の成分は$r\cos{\theta_{1}}$。
また、\vb*{x}から$x_{1}$軸に平行な成分を取り除いた$\vb*{x}^{'}=\vb*{x}-r\cos{\theta_{1}}\vb*{e}_{1}$は$\vb*{e}_{2},...,\vb*{e}_{n+1}$によって張られる座標$x_{2}\cdots x_{n+1}$で表せる。帰納法の仮定より、$|\vb*{x}^{'}|=r\sin{\theta_{1}}$を用いると
\begin{eqnarray} \left\{ \begin{array}{l} x_{2}=r\sin{\theta_{1}}\cos{\theta_{2}}\\ \vdots\\ x_{i}=r\sin{\theta_{1}}\cdots\sin{\theta_{i-1}}\cos{\theta_{i}}\\ \vdots\\ x_{n}=r\sin{\theta_{1}}\cdots\sin{\theta_{n-1}}\cos{\theta_{n}}\\ x_{n+1}=r\sin{\theta_{1}}\cdots\sin{\theta_{n-1}}\sin{\theta_{n}} \end{array} \right. \end{eqnarray}
が得られるので、すべての$n\in\mathbb{N}\setminus\{1\}$で所要の極形式が成り立つ事が示された。

極形式のJacobian

\begin{equation} J_{n}=\begin{vmatrix}\frac{\partial x_{1}}{\partial r}&\frac{\partial x_{1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{1}}{\partial \theta_{n-1}}\\ \frac{\partial x_{2}}{\partial r}&\frac{\partial x_{2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{2}}{\partial \theta_{n-1}}\\ \cdots&\cdots&\cdots&\cdots\\ \frac{\partial x_{n}}{\partial r}&\frac{\partial x_{n}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{n}}{\partial \theta_{n-1}}\end{vmatrix} \end{equation}
とおくと、以下の漸化式が成り立つ。
\begin{equation} J_{n}=r\sin{\theta_{1}}\sin{\theta_{2}}\cdots\sin{\theta_{n-2}}J_{n-1} \end{equation}

証明を見る

第$n$列に対して余因子行列展開を行う。
すると下記の式を得る。
\begin{equation} J_{n}=-\frac{\partial x_{n-1}}{\partial \theta_{n-1}}\begin{vmatrix}\frac{\partial x_{1}}{\partial r}&\frac{\partial x_{1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{1}}{\partial \theta_{n-2}}\\ \frac{\partial x_{2}}{\partial r}&\frac{\partial x_{2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{2}}{\partial \theta_{n-2}}\\ \cdots&\cdots&\cdots&\cdots\\ \frac{\partial x_{n-2}}{\partial r}&\frac{\partial x_{n-2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{n-2}}{\partial \theta_{n-2}}\\ \frac{\partial x_{n}}{\partial r}&\frac{\partial x_{n}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{n}}{\partial \theta_{n-2}}\end{vmatrix}+\frac{\partial x_{n}}{\partial \theta_{n-1}}\begin{vmatrix}\frac{\partial x_{1}}{\partial r}&\frac{\partial x_{1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{1}}{\partial \theta_{n-1}}\\ \frac{\partial x_{2}}{\partial r}&\frac{\partial x_{2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{2}}{\partial \theta_{n-1}}\\ \cdots&\cdots&\cdots&\cdots\\ \frac{\partial x_{n-1}}{\partial r}&\frac{\partial x_{n-1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{n-1}}{\partial \theta_{n-2}}\end{vmatrix} \end{equation}
また
\begin{eqnarray} \left\{ \begin{array}{l} \frac{\partial x_{n}}{\partial r}=\sin{\theta_{1}}\cdots\sin{\theta_{n-1}}\\ \frac{\partial x_{n-1}}{\partial r}=\sin{\theta_{1}}\cdots\cos{\theta_{n-1}}\\ \frac{\partial x_{n}}{\partial \theta_{i}}=r\sin{\theta_{1}}\cdots\sin{\theta_{i-1}}\cos{\theta_{i}}\sin{\theta_{i+1}}\cdots\sin{\theta_{n-1}}\\ \frac{\partial x_{n-1}}{\partial \theta_{i}}=r\sin{\theta_{1}}\cdots\sin{\theta_{i-1}}\cos{\theta_{i}}\sin{\theta_{i+1}}\cdots\cos{\theta_{n-1}} \end{array} \right. \end{eqnarray}
より
\begin{eqnarray} \left\{ \begin{array}{l} \frac{\partial x_{n}}{\partial r}=\frac{\sin{\theta_{n-1}}}{\cos{\theta_{n-1}}}\frac{\partial x_{n-1}}{\partial r}\\ \frac{\partial x_{n}}{\partial \theta_{i}}=\frac{\sin{\theta_{n-1}}}{\cos{\theta_{n-1}}}\frac{\partial x_{n-1}}{\partial \theta_{i}} \end{array} \right. \end{eqnarray}
ゆえに、
\begin{eqnarray} J_{n}&=&(-\frac{\partial x_{n-1}}{\partial \theta_{n-1}}\frac{\sin{\theta_{n-1}}}{\cos{\theta_{n-1}}}+\frac{\partial x_{n}}{\partial \theta_{n-1}})\begin{vmatrix}\frac{\partial x_{1}}{\partial r}&\frac{\partial x_{1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{1}}{\partial \theta_{n-2}}\\ \frac{\partial x_{2}}{\partial r}&\frac{\partial x_{2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{2}}{\partial \theta_{n-2}}\\ \cdots&\cdots&\cdots&\cdots\\ \frac{\partial x_{n-1}}{\partial r}&\frac{\partial x_{n-1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{n-1}}{\partial \theta_{n-2}}\end{vmatrix}\\ &=&(-\frac{\partial x_{n-1}}{\partial \theta_{n-1}}\sin{\theta_{n-1}}+\frac{\partial x_{n}}{\partial \theta_{n-1}}\cos{\theta_{n-1}})\begin{vmatrix}\frac{\partial x_{1}}{\partial r}&\frac{\partial x_{1}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{1}}{\partial \theta_{n-2}}\\ \frac{\partial x_{2}}{\partial r}&\frac{\partial x_{2}}{\partial \theta_{1}}&\cdots&\frac{\partial x_{2}}{\partial \theta_{n-2}}\\ \cdots&\cdots&\cdots&\cdots\\ \frac{1}{\cos{\theta_{n-1}}}\frac{\partial x_{n-1}}{\partial r}&\frac{1}{\cos{\theta_{1}}}\frac{\partial x_{n-1}}{\partial \theta_{1}}&\cdots&\frac{1}{\cos{\theta_{n-1}}}\frac{\partial x_{n-1}}{\theta_{n-2}}\end{vmatrix}\\ &=&r\sin{\theta_{1}}\sin{\theta_{2}}\cdots\sin{\theta_{n-2}}J_{n-1} \end{eqnarray}

Jacobian

\begin{eqnarray} \left\{ \begin{array}{l} J_{2}=r\\ J_{n}=r^{n-1}\sin^{n-2}{\theta_{1}}\sin^{n-3}{\theta_{2}}\cdots\sin{\theta_{n-2}} \end{array} \right. \end{eqnarray}

Beta関数

\begin{equation} B(x,y)=2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}{\theta}\cos^{2y-1}{\theta}d\theta \end{equation}

\begin{eqnarray} B(x,y)&=&\int_{0}^{1}t^{x-1}(1-x)^{t-1}dx\quad(x=\sin^{2}{\theta})\\ &=&2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}{\theta}\cos^{2y-1}{\theta}d\theta \end{eqnarray}

極形式による$\zeta$値の表示

\begin{equation} \zeta{(n)}=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{r^{n-1}e^{-r}}{1-e^{-r}} \end{equation}

証明を見る

\begin{eqnarray} \zeta(n)&=&(\prod_{i=1}^{n}\int_{0}^{1} dx_{i})\frac{1}{1-\prod_{i=1}^{n}x_{i}}\quad(y_{i}=-\log{x_{i}})\\ &=&(\prod_{i=1}^{n}\int_{0}^{\infty} dy_{i})\frac{\prod_{i=1}^{n}e^{-y_{i}}}{1-\prod_{i=1}^{n}e^{-y_{i}}}\quad(y_{i}=z_{i}^{2})\\ &=&2^{n}(\prod_{i=1}^{n}\int_{0}^{\infty} dz_{i})\frac{(\prod_{i=1}^{n}z_{i})e^{-\sum_{i=1}^{n}z_{i}^{2}}}{1-e^{-\sum_{i=1}^{n}z_{i}^{2}}}\\ &=&2^{n}\int_{0}^{\infty}dr\int_{0}^{\frac{\pi}{2}}d\theta_{1}\cdots\int_{0}^{\frac{\pi}{2}}d\theta_{n-1}\frac{e^{-r^{2}}}{1-e^{-r^{2}}}r^{n}\cos{\theta_{1}}\sin^{n-1}{\theta_{1}}\cos{\theta_{2}}\sin^{n-2}{\theta_{2}}\cdots\cos{\theta_{n-1}}\sin{\theta_{n-1}}r^{n-1}\sin^{n-2}{\theta_{1}}\sin^{n-3}{\theta_{2}}\cdots\sin{\theta_{n-2}}\\ &=&2^{n}\int_{0}^{\infty}dr\int_{0}^{\frac{\pi}{2}}d\theta_{1}\cdots\int_{0}^{\frac{\pi}{2}}d\theta_{n-1}\frac{e^{-r^{2}}}{1-e^{-r^{2}}}r^{2n-1}\cos{\theta_{1}}\sin^{2n-3}{\theta_{1}}\cos{\theta_{2}}\sin^{2n-5}{\theta_{2}}\cdots\cos{\theta_{n-2}}\sin^{3}{\theta_{n-2}}\cos{\theta_{n-1}}\sin{\theta_{1}}\\ &=&2B(1,n-1)B(1,n-2)\cdots B(1,1)\int_{0}^{\infty}dr\frac{r^{2n-1}e^{-r^{2}}}{1-e^{-r^{2}}}\\ &=&\frac{2}{(n-1)!}\int_{0}^{\infty}dr\frac{r^{2n-1}e^{-r^{2}}}{1-e^{-r^{2}}}\quad(r^{2}=s)\\ &=&\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{r^{n-1}e^{-r}}{1-e^{-r}} \end{eqnarray}