"Quantum dilogarithm"의 두 판 사이의 차이

2010년 5월 19일 (수) 06:20 판

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

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

where C= sum of Rogers dilogarithms

[[4909919|]]

The hyperbolic volume of knots from quantum dilogarithm
- R. M. Kashaev, 1996
Remarks on the quantum dilogarithm
- V V Bazhanov and N Yu Reshetikhin, 1995 J. Phys. A: Math. Gen. 28 2217
Quantum Dilogarithm
- L.D.Fadeev and R.M.Kashaev, Mod. Phys. Lett. A. 9 (1994) p.427–434
http://ncatlab.org/nlab/show/quantum+dilogarithm

@@ 9번째 줄: / 9번째 줄: @@
 *  noncommutative geometry<br>
 * <math>uv=qvu</math><br>
+<h5 style="margin: 0px; line-height: 2em;"> </h5>
+<math>\Phi(z)=\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n</math>
+<h5 style="margin: 0px; line-height: 2em;">asymptotics</h5>
+* <math>q=e^{-t}</math> and as the t goes 0 (i.e. as q goes to 1)<br>
+<math>\sum_{n=0}^{\infty}\frac{q^{\frac{A}{2}n^2+cn}}{(q)_n}\sim\exp(\frac{C}{t})</math>
+where C= sum of Rogers dilogarithms
+*   <br>
@@ 25번째 줄: / 49번째 줄: @@
 * [[1 Fermion summation formula - quasi-particle interpretation|Boson and Fermion summation form]]<br>
+* [[asymptotic analysis of basic hypergeometric series]]<br>