Bootstrap percolation

수학노트
Pythagoras0 (토론 | 기여)님의 2020년 11월 16일 (월) 08:50 판
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introduction

  • one of important question in 2d percolation is the calculation of power-law exponent for boostrap percolation
  • this is related to the theory of partitions without k-gaps  

 

bootstrap percolation

 

 

partitions without k-gaps

  • partitions without k-gaps (or k-sequences)
  • p_k(n) is the number of partitions of n that do not contain any sequence of consecutive integers of length k. p_2 (7) = 8.
  • examples: partition of 7 {{7},{6,1},{5,2},{5,1,1},{4,3},{4,2,1},{4,1,1,1},{3,3,1},{3,2,2},{3,2,1,1},{3,1,1,1,1},{2,2,2,1},{2,2,1,1,1},{2,1,1,1,1,1},{1,1,1,1,1,1,1}} 7, 6 + 1, 5 + 2, 5 + 1 + 1, 4 + 1 + 1 + 1, 3 + 3 + 1, 3 + 1 + 1 + 1 + 1, and 1 + 1 + 1 + 1 + 1 + 1 + 1. so there are 8 partitions without 2-gaps
  • Anderew's result
    • generating function for partitions without k-gaps\(G_2(q)=1+\sum_{n=1}^{\infty}\frac{q^n\prod_{j=1}^{n-1}(1-q^j+q^{2j})}{(q;q)_n}\)A116931
  1. (*define a gap as 'b' *) b := 2 G[b_, x_] :=  Sum[x^k*Product[1 + x^(b*j)/(1 - x^j), {j, 1, k - 1}]/(1 - x^k), {k,    1, 30}] Series[G[b, x], {x, 0, 20}] Table[SeriesCoefficient[%, n], {n, 0, 20}]

 

 

q-series from percolation

  • definition\(P_k(q)=(q;q)_{\infty}G_k(q)\)
  • Andrews and Zagier expression of \(P_k(q)\)
  • result of [HLR04] if \(q=e^{-t}\) and  \(t\sim 0\)\(P_k(q) \sim \frac{-\lambda_k}{1-q}\) as \(q \to 1\)

 

 

Andrews' conjecture on asymptotics

  • asymptotics of P_2(q) is known \(q=e^{-t}\) 으로 두면 \(t\sim 0\) 일 때,\(P_2(q) \sim \sqrt\frac{2\pi}{t}\exp(-\frac{\pi^2}{18t})\)
  • conjecture\(P_k(q) \sim \sqrt\frac{2\pi}{t}\exp(-\frac{\lambda_k}{t})\) where \(\lambda_k=\frac{\pi^2}{3k(k+1)}\)

 

 

tricky integrals

  • Henrik Eriksson: A Tricky Integral\(f_1(x)=1-x\)\(f_2(x)=\frac{1-x+\sqrt{(1-x)(1+3x)}}{2}\)
  • \(\lambda_k=\frac{\pi^2}{3k(k+1)}\)
  • \(\lambda_2=\frac{\pi^2}{18}\)

 

 

relevance to dedekind eta function

  • Dedekind eta function (데데킨트 에타함수)\(q=e^{-t}\) 으로 두면 \(t\sim 0\) 일 때,\(\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n\geq 1}^{\infty}\frac{(-1)^nq^{n(n+1)/2}}{(q)_n}\sim \sqrt\frac{2\pi}{t}\exp(-\frac{\pi^2}{6t})\) 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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