"Quantum dilogarithm"의 두 판 사이의 차이

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<math>\Psi(z)=\prod_{n=0}^{\infty}(1-zq^n)=\sum_{n\geq 0}\frac{(-1)^nq^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n</math>
 
<math>\Psi(z)=\prod_{n=0}^{\infty}(1-zq^n)=\sum_{n\geq 0}\frac{(-1)^nq^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n</math>
  
<math>\Psi(z)=\exp(\frac{\operatorname{Li}_{2,q}(z)}{q-1})</math> 
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<math>\Psi(z)=\exp(\frac{\operatorname{Li}_{2,q}(z)}{q-1})</math>
  
 
 
 
 
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
  
* Qu
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* Quantum dilogarithm.<br>
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** Zudilin, W., Preprint, Bonn and Moscow (2006), 8 pages
 
* [http://dx.doi.org/10.1023/A:1007364912784 The hyperbolic volume of knots from quantum dilogarithm]<br>
 
* [http://dx.doi.org/10.1023/A:1007364912784 The hyperbolic volume of knots from quantum dilogarithm]<br>
 
** R. M. Kashaev, 1996
 
** R. M. Kashaev, 1996
 
* [http://dx.doi.org/10.1088/0305-4470/28/8/014 Remarks on the quantum dilogarithm]<br>
 
* [http://dx.doi.org/10.1088/0305-4470/28/8/014 Remarks on the quantum dilogarithm]<br>
 
** V V Bazhanov and N Yu Reshetikhin, 1995 J. Phys. A: Math. Gen. 28 2217
 
** V V Bazhanov and N Yu Reshetikhin, 1995 J. Phys. A: Math. Gen. 28 2217
*  <br>[http://dx.doi.org/10.1142/S0217732394000447 Quantum Dilogarithm]<br>
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* [http://dx.doi.org/10.1142/S0217732394000447 Quantum Dilogarithm]<br>
 
** L.D.<em style="line-height: 2em;">Fadeev</em> and R.M.<em style="line-height: 2em;">Kashaev</em>, Mod. Phys. Lett. A. 9 (1994) p.427–434
 
** L.D.<em style="line-height: 2em;">Fadeev</em> and R.M.<em style="line-height: 2em;">Kashaev</em>, Mod. Phys. Lett. A. 9 (1994) p.427–434
 
* http://ncatlab.org/nlab/show/quantum+dilogarithm
 
* http://ncatlab.org/nlab/show/quantum+dilogarithm

2010년 6월 21일 (월) 16:09 판

introduction

 

 

quantum plane
  • also called the Weyl algebra
  • noncommutative geometry
  • \(uv=qvu\)

 

 

 

 

q-integral (Jackson integral)
  • \(0<q<1\)에 대하여 다음과 같이 정의
    \(\int_0^a f(x) d_q x = a(1-q)\sum_{k=0}^{\infty}q^k f(aq^k )\)
    \(\int_0^{\infty} f(x) d_q x =(1-q)\sum_{k=-\infty}^{\infty}q^k f(aq^k )\)
  • \(q\to 1\) 이면, \(\int_0^a f(x) d_q x \to \int_0^a f(x) dx \)

 

 

 

quantum dilogarithm

\(\Psi(z)=\prod_{n=0}^{\infty}(1-zq^n)=\sum_{n\geq 0}\frac{(-1)^nq^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)

\(\Psi(z)=\exp(\frac{\operatorname{Li}_{2,q}(z)}{q-1})\)

 

\(\operatorname{Li}_{2,q}(z) = -\int_0^z{{\ln (1-t)}\over t} d_{q}t \)

\(\operatorname{Li}_2(z) = -\int_0^z{{\ln (1-t)}\over t} dt \)

 

 

asymptotics

 

  • \(q=e^{-t}\) and as the t goes 0 (i.e. as q goes to 1)

\(\sum_{n=0}^{\infty}\frac{q^{\frac{A}{2}n^2+cn}}{(q)_n}\sim\exp(\frac{C}{t})\)

where C= sum of Rogers dilogarithms

 

 

quantum 5-term relation

 

 

history

 

 

related items

 

 

encyclopedia

 

 

books

 

[[4909919|]]

 

 

articles

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

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