n-Sphere & n-Ball
n-Ball的体积公式：
![\[\displaystyle V_n (R) = \frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2} + 1)} R^n\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-39f98a0f19e8d4d3007ffbdd6e09fdd1_l3.png "Rendered by QuickLaTeX.com")
偶数维体积公式：
![\[\displaystyle V_{2n} (R) = \frac{\pi^n}{n!} R^{2n}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-99cf6bba9e44d0dd0a0d23dffd2a951a_l3.png "Rendered by QuickLaTeX.com")
奇数维体积公式：
![\[\displaystyle V_{2n+1} (R) = \frac{2^{n+1} \pi^n}{\left(2n+1\right)!!} R^{2n+1} = \frac{2^{2n+1} n! \pi^n}{\left(2n+1\right)!} R^{2n+1}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-1421a525b5c30a339b8e5d6417dede0b_l3.png "Rendered by QuickLaTeX.com")
n-Sphere的面积公式：
![\[\displaystyle S_n=\frac{dV_{n+1}}{dR}=\frac{nV_{n+1}}{R}={2\pi^{(n+1)/2}R^{n}\over\Gamma(\frac{n+1}{2})}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-ed6334c2aed5b41837c382a0cea5c396_l3.png "Rendered by QuickLaTeX.com")
递推公式：
![\[\displaystyle S_n= \frac{2 \pi R^2}{n-1} S_{n-2}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-58efe659f0293a6aa02c47ed1fa3e3c8_l3.png "Rendered by QuickLaTeX.com")
![\[\displaystyle V_n = \frac{2 \pi R^2}{n} V_{n-2}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-e9c62b37715f597f653805662d40a6c0_l3.png "Rendered by QuickLaTeX.com")
递推公式的推导：
![\[\begin{array}{ll} V_n (R) &\displaystyle = \int_{0}^{R} \int_{0}^{2\pi} V_{n-2}(\sqrt{R^2-r^2}) \, r \, d\theta \, dr \\ &\displaystyle = \int_{0}^{R} \int_{0}^{2\pi} C_{n-2} (R^2-r^2)^{n/2-1}\, r \, d\theta \, dr \\ &\displaystyle = 2 \pi C_{n-2} \int_{0}^{R} (R^2-r^2)^{n/2-1}\, r \, dr \\ &\displaystyle = 2 \pi C_{n-2} \left[ -\frac{1}{n}(R^2-r^2)^{n/2} \right]^{r=R}_{r=0} \\ &\displaystyle = 2 \pi C_{n-2} \frac{R^n}{n} = \frac{2 \pi R^2}{n} V_{n-2}(R). \end{array}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-cc5f459bbbc2dd1866c8373ffe80e868_l3.png "Rendered by QuickLaTeX.com")
关系：
![\[\displaystyle V_n/S_{n-1} = R/n\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-38c29c783904428b34d92eeb3b61144e_l3.png "Rendered by QuickLaTeX.com")
![\[\displaystyle S_{n+1}/V_n = 2\pi R\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-b66437061b392dbf7cc714bd44d1e159_l3.png "Rendered by QuickLaTeX.com")
一个很有趣的现象是，当n趋于无穷大的时候，单位n-Ball的体积数值随着n增大趋于0：
![\[\displaystyle \lim_{n\rightarrow\infty} V_n(1) = 0\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-50da2dc7d7b69e8d11b77c926cb0d89f_l3.png "Rendered by QuickLaTeX.com")
这件事有点反直觉，因为感觉上单位n-Ball里面应该总是能够塞进去一个单位n-Cube，但事实上这只是我们在低维空间中总结出来的错误规律。单位n-Ball的直径永远是2，而单位n-Cube的对角线长却是
![\[\sqrt{n}\]](http://zhblog.engic.org/wp-content/ql-cache/quicklatex.com-bb6b3b5bada0b88cd7d7b2b801d62d6e_l3.png "Rendered by QuickLaTeX.com")
。n<4时单位n-Cube对角线长小于n-Ball直径，n=4时单位n-Cube对角线长等于n-Ball直径，n>4时单位n-Cube对角线长大于n-Ball直径，而且随着n增大会大得越来越多。也就是说在足够高维的空间中，单位n-Ball里面根本就无法塞入一个单位n-Cube。当然，一个单位nCube里面也不可能完全塞入一个n-Sphere，毕竟单位n-Cube两个相对表面之间的最小距离只有1，而单位n-Sphere直径却是2。
插件比较
