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<div id="content">
<h1 class="title">PRML 第1章 演習 1.11-1.20</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">PRML 第1章 演習 1.11-1.20</a>
<ul>
<li><a href="#sec-1-1"><span class="done DONE">DONE</span> 1.11 対数尤度関数からサンプル平均とサンプル分散を求める</a></li>
<li><a href="#sec-1-2"><span class="done DONE">DONE</span> 1.12 [www] ガウス分布の下での2つのデータ点の積の平均値</a></li>
<li><a href="#sec-1-3"><span class="todo TODO">TODO</span> 1.13 ガウス分布の分散の推定値</a></li>
<li><a href="#sec-1-4"><span class="done DONE">DONE</span> 1.14 D次元多項式の2次の項の独立パラメータの数</a></li>
<li><a href="#sec-1-5"><span class="done DONE">DONE</span> 1.15 [www] M次のD次元多項式の独立なパラメータの数</a></li>
<li><a href="#sec-1-6"><span class="todo TODO">TODO</span> 1.16 M次までのすべての項における独立なパラメータの数</a></li>
<li><a href="#sec-1-7"><span class="done DONE">DONE</span> 1.17 [www] ガンマ関数と階乗</a></li>
<li><a href="#sec-1-8"><span class="todo TODO">TODO</span> 1.18 [www] D次元単位球の表面積と体積</a></li>
<li><a href="#sec-1-9"><span class="todo TODO">TODO</span> 1.19 D次元超立方体の体積</a></li>
<li><a href="#sec-1-10"><span class="todo TODO">TODO</span> 1.20 [www] 高次元ガウス分布の振る舞い</a></li>
</ul>
</li>
</ul>
</div>
</div>
<div id="outline-container-sec-1" class="outline-2">
<h2 id="sec-1">PRML 第1章 演習 1.11-1.20</h2>
<div class="outline-text-2" id="text-1">
</div><div id="outline-container-sec-1-1" class="outline-3">
<h3 id="sec-1-1"><span class="done DONE">DONE</span> 1.11 対数尤度関数からサンプル平均とサンプル分散を求める</h3>
<div class="outline-text-3" id="text-1-1">
<p>
(1.54) 対数尤度関数<br />
</p>
\begin{align*}
\ln p(x|μ,σ^2) = - \frac{1}{2σ^2} \sum_{n=1}^N (x_n - μ)^2
- \frac{N}{2} \ln σ^2 - \frac{N}{2} \ln (2π)
\end{align*}
\begin{align*}
\frac{∂}{∂μ} \ln p(x|μ,σ^2) ∝ & \frac{∂}{∂μ} \sum_{n=1}^N (x_n - μ)^2 \\
= & \sum_{n=1}^N \frac{∂}{∂μ} (x_n - μ)^2 \\
= & -2μ \sum_{n=1}^N (x_n - μ) \\
= & -2μ [\sum_{n=1}^N x_n - Nμ] \\
0 = & \sum_{n=1}^N μ_{ML} (x_n - μ_{ML}) \\
μ_{ML} = & 0, \frac{1}{N} \sum_{n=1}^N x_n & \text{(1.55)}
\end{align*}
\begin{align*}
\frac{∂}{∂(σ^2)} \ln p(x|μ,σ^2)
= & [\frac{∂}{∂(σ^2)} -\frac{1}{2σ^2}] \sum_{n=1}^N (x_n - μ)^2
- \frac{N}{2} \frac{∂}{∂(σ^2)} \ln σ^2 \\
= & \frac{1}{2σ^4} \sum_{n=1}^N (x_n - μ)^2 - \frac{N}{2σ^2} \\
0 = & \frac{1}{2σ_{ML}^4} \sum_{n=1}^N (x_n - μ_{ML})^2 - \frac{N}{2σ_{ML}^2} \\
\frac{N}{2σ_{ML}^2} = & \frac{1}{2σ_{ML}^4} \sum_{n=1}^N (x_n - μ_{ML})^2 \\
N σ_{ML}^2 = & \sum_{n=1}^N (x_n - μ_{ML})^2 \\
σ_{ML}^2 = & \frac{1}{N} \sum_{n=1}^N (x_n - μ_{ML})^2 & \text{(1.56)}
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-2" class="outline-3">
<h3 id="sec-1-2"><span class="done DONE">DONE</span> 1.12 [www] ガウス分布の下での2つのデータ点の積の平均値</h3>
<div class="outline-text-3" id="text-1-2">
\begin{align*}
E[x] = & ∫_{-∞}^∞ N(x|μ,σ^2)x dx = & μ & \text{(1.49)} \\
E[x^2] = & ∫_{-∞}^∞ N(x|μ,σ^2)x^2 dx = & μ^2 + σ^2 & \text{(1.50)}
\end{align*}
<p>
\(n = m\)の場合、\(x_n\)と\(x_m\)は同じ試行の結果なので、\(E[x_n x_m]\)は\(E[x^2]\)と同じ。<br />
</p>
\begin{align*}
E[x_n x_m] = μ^2 + σ^2
\end{align*}
<p>
\(n ≠ m\)の場合、\(x_n\)と\(x_m\)は独立した試行の結果なので、<br />
</p>
\begin{align*}
E[x_n x_m] = E[x_n] E[x_m]
\end{align*}
<p>
\(E[x_n]\)、\(E[x_m]\)は\(E[x]\)と同じ。<br />
</p>
\begin{align*}
E[x_n x_m] = E[x]^2 = μ^2
\end{align*}
<p>
まとめると<br />
</p>
\begin{align*}
E[x_n x_m] = μ^2 + δ_{nm}σ^2
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-3" class="outline-3">
<h3 id="sec-1-3"><span class="todo TODO">TODO</span> 1.13 ガウス分布の分散の推定値</h3>
<div class="outline-text-3" id="text-1-3">
<p>
ガウス分布の分散の推定量<br />
</p>
\begin{align*}
σ_{ML}^2 = \frac{1}{N} \sum_{n=1}^N (x_n - μ_{ML})^2 & \text{(1.56)}
\end{align*}
<p>
平均の最尤推定値 μ<sub>ML</sub> を真の値 μ で置き換える。<br />
</p>
\begin{align*}
σ_{ML}^2 = \frac{1}{N} \sum_{n=1}^N (x_n - μ)^2
\end{align*}
<p>
期待値をとる。<br />
</p>
\begin{align*}
E[σ_{ML}^2] = & ∫ p(x|μ,σ^2) σ_{ML}^2 dx \\
= & \frac{1}{N} ∫ [\prod_{n=1}^N N(x_n|μ,σ^2)] [\sum_{n=1}^N (x_n - μ)^2] dx
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-4" class="outline-3">
<h3 id="sec-1-4"><span class="done DONE">DONE</span> 1.14 D次元多項式の2次の項の独立パラメータの数</h3>
<div class="outline-text-3" id="text-1-4">
<p>
任意の\(w_{ij}\)に対して、次の方程式を満たす\(w^S_{ij}\)、\(w^A_{ij}\)が存在するか否かを問うている。<br />
</p>
\begin{align*}
w_{ij} = & w^S_{ij} + w^A_{ij} \\
w^S_{ij} = & w^S_{ji} \\
w^A_{ij} = & -w^A_{ji} \\
\end{align*}
<p>
\(w^S_{ij}\)、\(w^A_{ij}\)を以下のように定義すると、上記の条件を満たす。<br />
</p>
\begin{align*}
w^S_{ij} = & (w_{ij} + w_{ji})/2 \\
w^A_{ij} = & (w_{ij} - w_{ji})/2 \\
\end{align*}
\begin{align*}
& \sum_{i=1}^D \sum_{j=1}^D w_{ij} x_i x_j \\
= & \sum_{i=1}^D \sum_{j=1}^D (w^S_{ij} + w^A_{ij}) x_i x_j \\
= & \sum_{i=1}^D \sum_{j=1}^D w^S_{ij} x_i x_j
+ \sum_{i=1}^D \sum_{j=1}^D w^A_{ij} x_i x_j \\
= & \sum_{i=1}^D \sum_{j=1}^D w^S_{ij} x_i x_j \\
\end{align*}
<p>
\(w^S_{ij}\)は\(D^2\)個のパラメータを含むが、<br />
\(w^S_{ij} = w^S_{ji}\)という\(D(D-1)/2\)個の独立な方程式を満たさなければならないので、<br />
独立なパラメータの数は<br />
</p>
\begin{align*}
& D^2 - D(D-1)/2 \\
= & D^2/2 + D/2 \\
= & D(D+1)/2 \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-5" class="outline-3">
<h3 id="sec-1-5"><span class="done DONE">DONE</span> 1.15 [www] M次のD次元多項式の独立なパラメータの数</h3>
<div class="outline-text-3" id="text-1-5">
<p>
D次元の多項式のM次の項<br />
</p>
\begin{align*}
\sum_{i_1=1}^D \sum_{i_2=1}^D ... \sum_{i_M=1}^D
w_{i_1 i_2 ... i_M} x_{i_1} x_{i_2} ... x_{i_M} & \text{(1.133)}
\end{align*}
</div>
<div id="outline-container-sec-1-5-1" class="outline-4">
<h4 id="sec-1-5-1"><span class="done DONE">DONE</span> (1.134)の証明</h4>
<div class="outline-text-4" id="text-1-5-1">
<p>
以下の式が成り立つことを証明する。<br />
</p>
\begin{align*}
\sum_{i_1=1}^D \sum_{i_2=1}^{i_1} ... \sum_{i_M=1}^{i_{M-1}}
\tilde{w}_{i_1 i_2 ... i_M} x_{i_1} x_{i_2} ... x_{i_M} & \text{(1.134)}
\end{align*}
<p>
\(x_{i_1} x_{i_2} ... x_{i_M}\)の掛け算の順序を並べ替えたものは等しい。<br />
そこで、常に添字が減少する順序に並べ替えることにする。<br />
並べ替えた添字を\(j_1, j_2, ..., j_M\)とすると、<br />
</p>
\begin{align*}
j_1 ≧ j_2, j_2 ≧ j_3, ..., j_{M-1} ≧ j_M
\end{align*}
<p>
が成り立つ。<br />
この条件を満たすようにすべての項の和をとったものが(1.134)である。<br />
</p>
</div>
</div>
<div id="outline-container-sec-1-5-2" class="outline-4">
<h4 id="sec-1-5-2"><span class="done DONE">DONE</span> (1.135)の証明</h4>
<div class="outline-text-4" id="text-1-5-2">
\begin{align*}
n(D,M) = \sum_{i=1}^D n(i,M-1) & \text{(1.135)}
\end{align*}
<p>
(1.134)は冗長性のない和なので、<br />
その項の数がすなわち独立なパラメータの数である。<br />
項の数は各項を1で置き換えることにより得られる。<br />
</p>
\begin{align*}
n(D,M) = & \sum_{i_1=1}^D \sum_{i_2=1}^{i_1} ... \sum_{i_M=1}^{i_{M-1}} 1 \\
n(i,M-1) = & \sum_{i_1=1}^i \sum_{i_2=1}^{i_1} ... \sum_{i_{(M-1)}=1}^{i_{(M-1)-1}} 1 \\
\end{align*}
<p>
(1.135)の右辺から出発し、上の式を用いて\(n(i,M-1)\)を展開し、<br />
適当に添字を付け替えれば、(1.135)の左辺が得られる。<br />
</p>
\begin{align*}
& \sum_{i=1}^D n(i,M-1) \\
= & \sum_{i=1}^D \sum_{i_1=1}^i \sum_{i_2=1}^{i_1} ... \sum_{i_{M-1}=1}^{i_{(M-1)-1}} 1 \\
= & \sum_{i_1=1}^D \sum_{i_2=1}^{i_1} \sum_{i_3=1}^{i_2} ... \sum_{i_M=1}^{i_{M-1}} 1
& i→i_1, i_n→i_{n+1} \\
= & n(D,M) \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-5-3" class="outline-4">
<h4 id="sec-1-5-3"><span class="done DONE">DONE</span> (1.136)の証明</h4>
<div class="outline-text-4" id="text-1-5-3">
\begin{align*}
\sum_{i=1}^D \frac{(i+M-2)!}{(i-1)!(M-1)!} = \frac{(D+M-1)!}{(D-1)!M!} & \text{(1.136)}
\end{align*}
<p>
\(D=1\)と任意の\(M\)の場合に成り立つことを示す。<br />
</p>
\begin{align*}
左辺 = & \sum_{i=1}^D \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \sum_{i=1}^1 \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \frac{(1+M-2)!}{(1-1)!(M-1)!} \\
= & \frac{(M-1)!}{0!(M-1)!} \\
= & 1 \\
\end{align*}
\begin{align*}
右辺 = & \frac{(D+M-1)!}{(D-1)!M!} \\
= & \frac{(1+M-1)!}{(1-1)!M!} \\
= & \frac{M!}{0!M!} \\
= & 1 \\
\end{align*}
<p>
\(D=d\)と任意の\(M\)の場合に成り立つという仮定のもとで、<br />
\(D=d+1\)と任意の\(M\)の場合に成り立つことを示す。<br />
</p>
\begin{align*}
左辺 = & \sum_{i=1}^D \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \sum_{i=1}^{d+1} \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \frac{(d+1+M-2)!}{(d+1-1)!(M-1)!} + \sum_{i=1}^d \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \frac{(d+1+M-2)!}{(d+1-1)!(M-1)!} + \frac{(d+M-1)!}{(d-1)!M!}
& \text{帰納法の仮定より} \\
= & \frac{(d+M-1)!}{d!(M-1)!} + \frac{(d+M-1)!}{(d-1)!M!} \\
= & \frac{(d+M-1)!M}{d!M!} + \frac{(d+M-1)!d}{d!M!} \\
= & \frac{(d+M-1)!(M+d)}{d!M!} \\
= & \frac{(d+M)!}{d!M!} \\
\end{align*}
\begin{align*}
右辺 = & \frac{(D+M-1)!}{(D-1)!M!} \\
= & \frac{(d+1+M-1)!}{(d+1-1)!M!} \\
= & \frac{(d+M)!}{d!M!} \\
\end{align*}
<p>
よって、数学的帰納法により、任意の\(D\)、\(M\)について(1.136)が成り立つ。<br />
</p>
</div>
</div>
<div id="outline-container-sec-1-5-4" class="outline-4">
<h4 id="sec-1-5-4"><span class="done DONE">DONE</span> (1.137)の証明</h4>
<div class="outline-text-4" id="text-1-5-4">
\begin{align*}
n(D,M) = \frac{(D+M-1)!}{(D-1)!M!} (1.137)
\end{align*}
<p>
\(M=2\)と任意の\(D≧1\)の場合に成り立つことを示す。<br />
</p>
\begin{align*}
左辺 = & n(D,M) \\
= & n(D,2) \\
= & \sum_{i_1=1}^D \sum_{i_2=1}^{i_1} 1 \\
= & \sum_{i_1=1}^D i_1 \\
= & \frac{(D+1)D}{2} \\
\end{align*}
\begin{align*}
右辺 = & \frac{(D+M-1)!}{(D-1)!M!} \\
= & \frac{(D+2-1)!}{(D-1)!2!} \\
= & \frac{(D+1)!}{(D-1)!2} \\
= & \frac{(D+1)D!D}{D!2} \\
= & \frac{(D+1)D}{2} \\
\end{align*}
<p>
\(M-1\)次で成り立っているという仮定のもとで、<br />
\(M\)次の場合に成り立つことを示す。<br />
</p>
\begin{align*}
左辺 = & n(D,M) \\
= & \sum_{i=1}^D n(i,M-1) & \text{(1.135)より} \\
= & \sum_{i=1}^D \frac{(i+(M-1)-1)!}{(i-1)!(M-1)!} & \text{帰納法の仮定より} \\
= & \sum_{i=1}^D \frac{(i+M-2)!}{(i-1)!(M-1)!} \\
= & \frac{(D+M-1)!}{(D-1)!M!} & \text{(1.136)より} \\
\end{align*}
<p>
よって、数学的帰納法により、任意の\(D\)、\(M\)について(1.137)が成り立つ。<br />
</p>
</div>
</div>
</div>
<div id="outline-container-sec-1-6" class="outline-3">
<h3 id="sec-1-6"><span class="todo TODO">TODO</span> 1.16 M次までのすべての項における独立なパラメータの数</h3>
<div class="outline-text-3" id="text-1-6">
</div><div id="outline-container-sec-1-6-1" class="outline-4">
<h4 id="sec-1-6-1"><span class="done DONE">DONE</span> </h4>
<div class="outline-text-4" id="text-1-6-1">
<p>
\(M\)次までのすべての項における独立パラメータの数<br />
</p>
\begin{align*}
N(D,M) = & \sum_{m=0}^M n(D,m) & \text{(1.138)}
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-6-2" class="outline-4">
<h4 id="sec-1-6-2"><span class="done DONE">DONE</span> \(N(D,M) = (D+M)!/D!M!\) (1.139)の証明</h4>
<div class="outline-text-4" id="text-1-6-2">
<p>
\(M=0\)と任意の\(D≧1\)の場合に成り立つことを示す。<br />
</p>
\begin{align*}
左辺 = & N(D,0) \\
= & \sum_{m=0}^0 n(D,m) \\
= & n(D,0) \\
= & (D+0-1)!/(D-1)!0! & \text{(1.137)より} \\
= & 1 \\
右辺 = & (D+0)!/D!0! \\
= & 1 \\
\end{align*}
<p>
\(M\)次で成り立つという仮定のもとで\(M+1\)次の場合に成り立つことを示す。<br />
</p>
\begin{align*}
& N(D,M+1) \\
= & \sum_{m=0}^{M+1} n(D,m) \\
= & n(D,M+1) + \sum_{m=0}^M n(D,m) \\
= & n(D,M+1) + N(D,M) \\
= & n(D,M+1) + \frac{(D+M)!}{D!M!} & \text{帰納法の仮定より} \\
= & \frac{(D+(M+1)-1)!}{(D-1)!(M+1)!} + \frac{(D+M)!}{D!M!} & \text{(1.137)より} \\
= & \frac{D(D+M)!}{D!(M+1)!} + \frac{(D+M)!(M+1)}{D!(M+1)!} \\
= & \frac{(D+(M+1))!}{D!(M+1)!} \\
\end{align*}
<p>
よって、数学的帰納法により、任意の\(D\)、\(M\)について(1.139)が成り立つ。<br />
</p>
</div>
</div>
<div id="outline-container-sec-1-6-3" class="outline-4">
<h4 id="sec-1-6-3"><span class="todo TODO">TODO</span> </h4>
<div class="outline-text-4" id="text-1-6-3">
\begin{align*}
D≫M のとき O(N(D,M)) = & D^M \\
M≫D のとき O(N(D,M)) = & M^D \\
\end{align*}
\begin{align*}
N(D,M) = & (D+M)!/D!M! \\
= & \prod_{i=D+M}^{} \\
\end{align*}
<p>
\(D≫M\)のとき<br />
</p>
\begin{align*}
N(D,M) ≒ & (D+M)^(D+M)e^{-(D+M)}/D^De^{-D}M! \\
= & (D+M)^(D+M)e^{-M}/D^DM! \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-6-4" class="outline-4">
<h4 id="sec-1-6-4"><span class="done DONE">DONE</span> </h4>
<div class="outline-text-4" id="text-1-6-4">
<p>
>>> import math<br />
>>> math.factorial(10+3)/math.factorial(10)/math.factorial(3)<br />
286L<br />
>>> math.factorial(100+3)/math.factorial(100)/math.factorial(3)<br />
176851L<br />
</p>
<p>
10<sup>3</sup>=1000<br />
100<sup>3</sup>=1000000<br />
</p>
</div>
</div>
</div>
<div id="outline-container-sec-1-7" class="outline-3">
<h3 id="sec-1-7"><span class="done DONE">DONE</span> 1.17 [www] ガンマ関数と階乗</h3>
<div class="outline-text-3" id="text-1-7">
<p>
ガンマ関数<br />
</p>
\begin{align*}
Γ(x) = ∫_0^∞ u^{x-1}e^{-u} du
\end{align*}
<p>
部分積分の公式<br />
</p>
\begin{align*}
∫_a^b f(x)g'(x)dx = [f(x)g(x)]_a^b - ∫_a^b f'(x)g(x)dx
\end{align*}
\begin{align*}
Γ(x+1) = & ∫_0^∞ u^xe^{-u} du \\
= & -[u^xe^{-u}]_0^∞ - x ∫_0^∞ u^{x-1}e^{-u} du \\
= & x Γ(x)
\end{align*}
\begin{align*}
Γ(1) = & ∫_0^∞ e^{-u} du \\
= & -[e^{-u}]_0^∞ \\
= & -(e^{-∞} - e^0) \\
= & -(0 - 1) \\
= & 1
\end{align*}
\begin{align*}
Γ(x+1) = & x Γ(x) \\
Γ(1) = & 1
\end{align*}
<p>
これは \(x!\) の定義そのものである。<br />
</p>
</div>
</div>
<div id="outline-container-sec-1-8" class="outline-3">
<h3 id="sec-1-8"><span class="todo TODO">TODO</span> 1.18 [www] D次元単位球の表面積と体積</h3>
<div class="outline-text-3" id="text-1-8">
\begin{align*}
\prod_{i=1}^D ∫_{-∞}^∞ e^{-x_i^2} dx_i = S_D ∫_0^∞ e^{-r^2}r^{D-1} dr & \text{(1.142)}
\end{align*}
</div>
<div id="outline-container-sec-1-8-1" class="outline-4">
<h4 id="sec-1-8-1"><span class="done DONE">DONE</span> \(S_D = \frac{2π^{D/2}}{Γ(D/2)}\) の証明</h4>
<div class="outline-text-4" id="text-1-8-1">
\begin{align*}
∫_{-∞}^∞ exp(-x^2/2σ^2)dx = (2πσ^2)^{1/2} & \text{(1.126)}
\end{align*}
<p>
\(σ^2 = 1/2\)を代入すると<br />
</p>
\begin{align*}
∫_{-∞}^∞ exp(-x^2)dx = π^{1/2}
\end{align*}
<p>
(1.142)左辺<br />
</p>
\begin{align*}
\prod_{i=1}^D ∫_{-∞}^∞ e^{-x_i^2} dx_i = & \prod_{i=1}^D π^{1/2} \\
= & π^{D/2}
\end{align*}
<p>
(1.142)右辺の積分で、\(r^2=u\)とおくと\(du=2rdr\)<br />
</p>
\begin{align*}
∫_0^∞ e^{-r^2}r^{D-1} dr = & ∫_0^∞ \frac{e^{-u}r^{D-1}}{2r} du \\
= & ∫_0^∞ \frac{e^{-u}r^{D-2}}{2} du \\
= & \frac{1}{2} ∫_0^∞ e^{-u}u^{D/2-1} du \\
= & \frac{Γ(D/2)}{2}
\end{align*}
<p>
これらを(1.142)に代入すると<br />
</p>
\begin{align*}
π^{D/2} = & S_D \frac{Γ(D/2)}{2} \\
S_D = & \frac{2π^{D/2}}{Γ(D/2)}
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-8-2" class="outline-4">
<h4 id="sec-1-8-2"><span class="todo TODO">TODO</span> \(V_D = \frac{S_D}{D}\) の証明</h4>
<div class="outline-text-4" id="text-1-8-2">
\begin{align*}
V_D = & S_D ∫_0^1 r^{D-1} dr \\
= & S_D \left[ \frac{r^D}{D} \right]_0^1 \\
= & \frac{S_D}{D} \\
\end{align*}
<p>
:TODO: $ V<sub>D</sub> = & S<sub>D</sub> ∫<sub>0</sub><sup>1</sup> r<sup>D-1</sup> dr $ これはなぜ？<br />
</p>
</div>
</div>
<div id="outline-container-sec-1-8-3" class="outline-4">
<h4 id="sec-1-8-3"><span class="done DONE">DONE</span> \(D=2\)および\(D=3\)の場合</h4>
<div class="outline-text-4" id="text-1-8-3">
<p>
\(D=2\)のとき<br />
</p>
\begin{align*}
S_D = & 2π^{D/2}/Γ(D/2) \\
= & 2π^{2/2}/Γ(2/2) \\
= & 2π
\end{align*}
\begin{align*}
V_D = & S_D/D \\
= & 2π/2 \\
= & π
\end{align*}
<p>
\(D=3\)のとき<br />
</p>
\begin{align*}
S_D = & 2π^{D/2}/Γ(D/2) \\
= & 2π^{3/2}/Γ(3/2) \\
= & 2π^{3/2}/\sqrt{π}/2 \\
= & 2π^{3/2}π^{-1/2}/2 \\
= & 4π
\end{align*}
\begin{align*}
V_D = & S_D/D \\
= & 4π/3
\end{align*}
</div>
</div>
</div>
<div id="outline-container-sec-1-9" class="outline-3">
<h3 id="sec-1-9"><span class="todo TODO">TODO</span> 1.19 D次元超立方体の体積</h3>
<div class="outline-text-3" id="text-1-9">
</div><div id="outline-container-sec-1-9-1" class="outline-4">
<h4 id="sec-1-9-1"><span class="done DONE">DONE</span> </h4>
<div class="outline-text-4" id="text-1-9-1">
\begin{align*}
\text{立方体の体積} = & \prod_{i=1}^D ∫_{-a}^{a} dx_i \\
= & (2a)^D
\end{align*}
\begin{align*}
\frac{ \text{球の体積} }{ \text{立方体の体積} }
= & \frac{ 2π^{D/2}a^D }{ DΓ(D/2)(2a)^D } \\
= & \frac{ 2π^{D/2} }{ DΓ(D/2)2^D } \\
= & \frac{ π^{D/2} }{ D2^{D-1}Γ(D/2) } \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-9-2" class="outline-4">
<h4 id="sec-1-9-2"><span class="todo TODO">TODO</span> </h4>
<div class="outline-text-4" id="text-1-9-2">
<p>
スターリングの公式<br />
\(x≫1\)のとき<br />
</p>
\begin{align*}
Γ(x+1) ≒ (2π)^{1/2}e^{-x}x^{x+1/2}
\end{align*}
<p>
\(D→∞\)のとき<br />
</p>
\begin{align*}
& \frac{ π^{D/2} }{ D2^{D-1}Γ(D/2) } \\
= & \frac{ π^{D/2} }{ D2^{D-1}(2π)^{1/2}e^{-D/2}(D/2)^{D/2+1/2} } \\
= & \frac{ π^{D/2} }{ D2^{D-1}(2π)^{1/2}e^{-D/2}(D/2)^{(D+1)/2} } \\
= & \frac{ π^{D/2}e^{D/2}2^{(D+1)/2} }{ D2^{D-1}(2π)^{1/2}D^{(D+1)/2} } \\
= & \frac{ π^{D/2}e^{D/2}2^{(D+1)/2} }{ 2^{D-1}(2π)^{1/2}D^{(D+3)/2} } \\
= & \frac{ π^{(D-1)/2}e^{D/2}2^{-D/2+1} }{ D^{(D+3)/2} } \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-9-3" class="outline-4">
<h4 id="sec-1-9-3"><span class="done DONE">DONE</span> </h4>
<div class="outline-text-4" id="text-1-9-3">
<p>
立方体の中心から1つの頂点までの距離<br />
</p>
\begin{align*}
d_1 = & a \\
d_D = & \sqrt{d_{D-1}^2 + a^2} \\
d_D = & \sqrt{Da^2} \\
d_D = & a\sqrt{D} \\
\end{align*}
<p>
中心から側面までの距離で割った比<br />
</p>
\begin{align*}
d_D/a = & \sqrt{D}
\end{align*}
<p>
\(D→∞\)のとき\(d_D/a→∞\)。<br />
</p>
</div>
</div>
</div>
<div id="outline-container-sec-1-10" class="outline-3">
<h3 id="sec-1-10"><span class="todo TODO">TODO</span> 1.20 [www] 高次元ガウス分布の振る舞い</h3>
<div class="outline-text-3" id="text-1-10">
<p>
D次元ガウス分布<br />
</p>
\begin{align*}
p(x) = & 1/(2πσ^2)^{D/2} exp(-||x||^2/2σ^2) & \text{(1.147)}
\end{align*}
\begin{align*}
p(r) = & S_Dr^{D-1}/(2πσ^2)^{D/2} exp(-r^2/2σ^2) & \text{(1.148)}
\end{align*}
<p>
(1.147)を極座標に変換する。<br />
</p>
</div>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="creator"><a href="http://www.gnu.org/software/emacs/">Emacs</a> 24.4.4 (<a href="http://orgmode.org">Org</a> mode 8.2.10)</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
</html>