PRML 第1章 1.19(標準)

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$$
\begin{array}{ll}
(1.145) & \dfrac{球の体積}{立方体の体積} = \dfrac{\pi^{D/2}}{D~2^{D-1} \Gamma(D/2)}\\[1em]
(1.146) & \Gamma(x+1) \simeq (2\pi)^{1/2} e^{-x} x^{x+1/2}
\end{array}
$$

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解答

　演習問題1.18の結果より，

$$
\begin{array}{ll}
\dfrac{球の体積}{立方体の体積}
&= \dfrac{V_D(a)^D}{(2a)^D}\\[1em]
&= \dfrac{1}{2^D}V_D\\[1em]
&= \dfrac{1}{2^D} \dfrac{2\pi^{D/2}}{D\Gamma(D/2)}\\[1em]
&= \dfrac{\pi^{D/2}}{D~2^{D-1} \Gamma(D/2)}
\end{array}
$$

となる．ここで$${D\to\infty}$$の場合，(1.146)より，

$$
\begin{array}{ll}
\Gamma(D/2) &= \dfrac{2}{D}\Gamma(D/2 + 1)\\[1em]
&\simeq \dfrac{2}{D} (2\pi)^{1/2} e^{-D/2} (D/2)^{(D+1)/2}\\[1em]
&= 2 \left(\dfrac{\pi}{D}\right)^\frac12 \left(\dfrac{D}{2e}\right)^\frac{D}{2}
\end{array}
$$

が成り立つので，

$$
\begin{array}{ll}
{\displaystyle\lim_{D\to\infty}} \dfrac{\pi^{D/2}}{D~2^{D-1} \Gamma(D/2)}
&\simeq {\displaystyle\lim_{D\to\infty}} \left(\dfrac{1}{\pi D}\right)^\frac12 \left(\dfrac{\pi e}{2D}\right)^\frac{D}{2}\\[1em]
&= \left(\dfrac{2}{\pi^2 e}\right)^\frac12 {\displaystyle\lim_{D\to\infty}} \left(\dfrac{\pi e}{2D}\right)^\frac{D+1}{2}\\[1em]
&= 0
\end{array}
$$

となる．また，

$$
\begin{array}{ll}
\dfrac{中心頂点間距離}{中心側面間距離}
= \dfrac{\sqrt{a^2 + a^2 + \cdots + a^2}}{a}
= \dfrac{\sqrt{D}a}{a}
= \sqrt{D}
\end{array}
$$

となり，$${D\to\infty}$$のとき$${\infty}$$に発散する．