"Bootstrap percolation"의 두 판 사이의 차이

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*  calculation of power-law exponent for boostrap percolation<br>
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* http://mathworld.wolfram.com/BootstrapPercolation.html<br>
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*  Andrews' conjecture<br>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">integrals</h5>
  
 
<math>\lambda_k=\frac{\pi^2}{3k(k+1)}</math>
 
<math>\lambda_k=\frac{\pi^2}{3k(k+1)}</math>
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Henrik Eriksson: [http://www.math.ubc.ca/~holroyd/integral.pdf A Tricky Integral]
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">relevance to dedekind eta function</h5>
  
 
 
 
 
  
http://mathworld.wolfram.com/BootstrapPercolation.html
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*  Dedekind eta function ([http://pythagoras0.springnote.com/pages/3325777 데데킨트 에타함수])<br><math>q=e^{-t}</math> 으로 두면 <math>t\sim 0</math> 일 때,<br><math>\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n\geq 1}^{\infty}\frac{(-1)^nq^{n(n+1)/2}}{(q)_n}\sim \sqrt\frac{2\pi}{t}\exp(-\frac{\pi^2}{6t})=\sqrt{\frac{2\pi}{t}}\exp(-\frac{(2\pi)^2}{24t})</math><br> more generally, <math>q=\exp(\frac{2\pi ih}{k})e^{-t}</math>  and  <math>t\to 0</math> implies<br><math>\sqrt{\frac{t}{2\pi}}\exp({\frac{\pi^2}{6k^2t}})\eta(\frac{h}{k}+i\frac{t}{2\pi})\sim  \frac{\exp\left(\pi i (\frac{h}{12k}-s(h,k)\right)}{\sqrt{k}}</math><br>
  
 
 
 
 
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* [[asymptotic analysis of basic hypergeometric series]]<br>
  
 
 
 
 
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2010년 3월 15일 (월) 13:18 판

introduction

 

  • calculation of power-law exponent for boostrap percolation

 

  • Andrews' conjecture

 

 

integrals

\(\lambda_k=\frac{\pi^2}{3k(k+1)}\)

Henrik Eriksson: A Tricky Integral

 

 

 

relevance to dedekind eta function

 

  • Dedekind eta function (데데킨트 에타함수)
    \(q=e^{-t}\) 으로 두면 \(t\sim 0\) 일 때,
    \(\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n\geq 1}^{\infty}\frac{(-1)^nq^{n(n+1)/2}}{(q)_n}\sim \sqrt\frac{2\pi}{t}\exp(-\frac{\pi^2}{6t})=\sqrt{\frac{2\pi}{t}}\exp(-\frac{(2\pi)^2}{24t})\)
    more generally, \(q=\exp(\frac{2\pi ih}{k})e^{-t}\)  and  \(t\to 0\) implies
    \(\sqrt{\frac{t}{2\pi}}\exp({\frac{\pi^2}{6k^2t}})\eta(\frac{h}{k}+i\frac{t}{2\pi})\sim \frac{\exp\left(\pi i (\frac{h}{12k}-s(h,k)\right)}{\sqrt{k}}\)

 

 

 

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[[4909919|]]

 

 

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question and answers(Math Overflow)

 

 

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