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<h1>位相圧力 と Rényi エントロピーの関係<a class="headerlink" href="#renyi" title="Permalink to this headline"></a></h1>
<p><a class="reference internal" href="dynamical-renyi-entropies.html#dynamical-renyi-entropy"><em>Rényi エントロピー</em></a> <img class="math" src="_images/math/ee05acc2882ef23b33730ab49cfb8704bbdf1757.png" alt="K(\beta)"/>自然不変測度によって <img class="math" src="_images/math/fc97ef67268cd4e91bacdf12b8901d7036c9a056.png" alt="N"/>-シリンダー と対応付けられた
確率分布によって定義されていた.
ここで,逆温度 <img class="math" src="_images/math/0615acc3725de21025457e7d6f7694dab8e2f758.png" alt="q"/> の Gibbs 測度 <img class="math" src="_images/math/50ab8f8663319322f3b506ab268e96efd7537855.png" alt="\mu_q"/> について同様に定義した
Rényi エントロピー を <img class="math" src="_images/math/d92e13359f78618914fe17e2c596b75454b46107.png" alt="K(\beta, q)"/> と書く.
<img class="math" src="_images/math/26520f2333a3d48eb4929b7517dea54293f36605.png" alt="q=1"/> の Gibbs 測度である SRB 測度は自然不変測度と同一である
(<a class="reference internal" href="gibbs-measures-and-srb-measures.html#gibbs-measures-and-srb-measures"><em>Gibbs 測度 と SRB 測度</em></a> を参照) ので,
<img class="math" src="_images/math/bd5ff4891c67426e840d187b6e9eb303b1453afd.png" alt="K(\beta, 1) = K(\beta)"/> である.</p>
<p>Rényi エントロピー <img class="math" src="_images/math/d92e13359f78618914fe17e2c596b75454b46107.png" alt="K(\beta, q)"/> を具体的に書くと</p>
<div class="math">
<p><img src="_images/math/0da1bfa9004f162bdf77027ec674022df1055bf4.png" alt="K(\beta, q) = \limooN \ooomb \ln \sum_j \pob{\Pseq}" /></p>
</div><p>となる.ただし確率分布は Gibbs 測度 <img class="math" src="_images/math/50ab8f8663319322f3b506ab268e96efd7537855.png" alt="\mu_q"/></p>
<div class="math">
<p><img src="_images/math/d7544320e5a686f92c112b96c58fb4703948734f.png" alt="\Pseq = \mu_q (J_j^{(N)})" /></p>
</div><p>の形で対応している.</p>
<p>また, <img class="math" src="_images/math/587deaa115238bac53970a4d6e0a68e303834080.png" alt="N \to \infty"/> では</p>
<div class="math">
<p><img src="_images/math/665fd33379a39f7f704c0d961364c5b484fccfbc.png" alt="\Pseq = \frac{\pob[q]{l_j^{(N)}}}{\Ztop_N}" /></p>
</div><div class="math">
<p><img src="_images/math/112d3afc0210fb00a6cb4eebc4402383f575adfe.png" alt="\Ztop_N = \sum_j \pob[q]{l_j^{(N)}} = \exp(N \Ftop(q))" /></p>
</div><p>である.ここで, <img class="math" src="_images/math/e43cdf3b77639f88b338a5c3c0479f714768a97f.png" alt="\sum_j \pob{\Pseq}"/> を二通りの方法で
(二つの自由エネルギーを用いて) 計算する.
<img class="math" src="_images/math/918ff908410611d99921408ee31472e36f5f0a41.png" alt="\Ftop(q)"/> を使えば,  <img class="math" src="_images/math/587deaa115238bac53970a4d6e0a68e303834080.png" alt="N \to \infty"/> で,</p>
<div class="math" id="equation-sum-pob-pseq-ftop">
<p><span class="eqno">(1)</span><img src="_images/math/2da037835dfcea228a4d1416ad39b982499ce1d3.png" alt="\sum_j \pob{\Pseq}
&amp;\sim \exp(- N \beta \Ftop(q)) \sum_j \pob[q \beta]{l_j^{(N)}} \\
&amp;= \exp(- N \beta \Ftop(q)) \exp(N \Ftop(q \beta)) \\
&amp;= \exp( N [\Ftop(q \beta) - \beta \Ftop(q)] )" /></p>
</div><p>となる.二番目の変形で, <img class="math" src="_images/math/d67a8fd1b0a2697a0cc94f522b04ca40dede70df.png" alt="q \beta"/> を逆温度とみなした位相圧力
が出てきている.
次に, <img class="math" src="_images/math/d92e13359f78618914fe17e2c596b75454b46107.png" alt="K(\beta, q)"/> を使えば,  <img class="math" src="_images/math/587deaa115238bac53970a4d6e0a68e303834080.png" alt="N \to \infty"/> で,</p>
<div class="math">
<p><img src="_images/math/ada6da49a7782dceec8f3fada274f2e30ceeff21.png" alt="\sum_j \pob{\Pseq} = \exp (N (1-\beta) K(\beta, q))" /></p>
</div><p>となる.以上の二式を合わせて <img class="math" src="_images/math/30d938049e2d8bdbff5c53527041fc1dac96cb49.png" alt="\ln"/> を施せば</p>
<div class="math">
<p><img src="_images/math/c4319ed80b7057c9ed08550e778f23acaa59f47c.png" alt="N (1-\beta) K(\beta, q)
= N [\Ftop(q \beta) - \beta \Ftop(q)] + \text{const.}" /></p>
</div><p>より (右辺の定数は式 <a href="#equation-sum-pob-pseq-ftop">(1)</a><img class="math" src="_images/math/e55156a4008b0944ad00d5bc71bc5aa6315aabb7.png" alt="\sim"/> より),
<img class="math" src="_images/math/587deaa115238bac53970a4d6e0a68e303834080.png" alt="N \to \infty"/> で,</p>
<div class="math">
<p><img src="_images/math/fca639cd4d5ff57fff4f968f33b1a91c54fb8f93.png" alt="K(\beta, q) = \ooomb \, [\Ftop(q \beta) - \beta \Ftop(q)]" /></p>
</div><p>を得る. <img class="math" src="_images/math/26520f2333a3d48eb4929b7517dea54293f36605.png" alt="q=1"/> を代入すれば,</p>
<div class="math" id="equation-k_beta_1">
<p><span class="eqno">(2)</span><img src="_images/math/907067c70f83204a83884de104b7937d577eb04a.png" alt="K(\beta) = \ooomb \, [\Ftop(\beta) - \beta \Ftop(1)]" /></p>
</div><p>である. <img class="math" src="_images/math/bd0841c3936bc875fefe9192ca5209033f156466.png" alt="\beta=1"/> について計算するために <img class="math" src="_images/math/705ac32118e62bc7b95348085ac331b2965c4588.png" alt="\Ftop(\beta)"/>
<img class="math" src="_images/math/bd0841c3936bc875fefe9192ca5209033f156466.png" alt="\beta=1"/> まわりで展開する (<img class="math" src="_images/math/705ac32118e62bc7b95348085ac331b2965c4588.png" alt="\Ftop(\beta)"/> が滑らか
であることを仮定):</p>
<div class="math">
<p><img src="_images/math/04e82585f6835e3648031796f30b4a1c88d9d9b3.png" alt="\Ftop(\beta) = \Ftop(1) + (\beta - 1) \Ftop'(\beta) + O((\beta - 1)^2)" /></p>
</div><p>これを式 <a href="#equation-k_beta_1">(2)</a> に代入して計算すれば,</p>
<div class="math">
<p><img src="_images/math/9055dd8d206607befd59ce4d5ebc037cda345627.png" alt="K(\beta)
&amp;= \ooomb \, [(1 - \beta) \Ftop(1) - (1 - \beta) \Ftop'(\beta)]
   + O((\beta - 1)) \\
&amp;\xrightarrow{\beta \to 1} \Ftop(1) - \Ftop'(\beta)" /></p>
</div><p>を得る.ここで出てきた <img class="math" src="_images/math/4b8c4efb9f0f7f5af6646f9a2508ad2649d2418a.png" alt="\Ftop'(\beta)"/> を計算しよう.</p>
<div class="math">
<p><img src="_images/math/22677a36fb23b236f7bbedc1f867c4457c480619.png" alt="\Ftop(\beta)
= \limooN \ln \sum_j \exp \left( - \beta N E_N(\xk[0][j]) \right)" /></p>
</div><p>を微分すれば,</p>
<div class="math">
<p><img src="_images/math/c0c251bda6fe31161a413f5f43713338abae336e.png" alt="\Ftop'(\beta)
&amp;= \limooN \frac{\sum_j \left(- N E_N(\xk[0][j]) \right)
                        \exp \left( - \beta N E_N(\xk[0][j]) \right)}
                {\sum_{j'} \exp \left( - \beta N E_N(\xk[0][j']) \right)} \\
&amp;= - \lim_{N \to \infty} \sum_j E_N(\xk[0][j]) \Pseq \\
&amp;= - \BraAvg{E}(\beta)" /></p>
</div><p><img class="math" src="_images/math/bcaa07b62b35c70593a6c70697d4815930e0dc50.png" alt="\BraAvg{E}(1)"/> は局所拡大率の SRB 測度 (すなわち自然不変測度)
に関する平均なので,これは Lyapunov 指数 <img class="math" src="_images/math/ce4588fd900d02afcbd260bc07f54cce49a7dc4a.png" alt="\lambda"/> である.
また, <img class="math" src="_images/math/5b5ee4b858ec5224e789b29e9fbfa0d46a92df20.png" alt="\Ftop(1) = - \kappa"/> (<a class="reference internal" href="topological-pressure.html#escape-rate"><em>流出率</em></a> を参照)を
用いて,</p>
<div class="math">
<p><img src="_images/math/94aae784064cdacb2751773f74135f6e498bc616.png" alt="K(1) = \lambda - \kappa" /></p>
</div><p>を得る.これは, KS エントロピー, 流出率,そして Lyapunov 指数
の関係を表している.</p>
</div>


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