"Quantum dilogarithm"의 두 판 사이의 차이

2010년 5월 19일 (수) 06:33 판

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

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

\(0<q<1\)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, \(\int_0^a f(x) d_q x \to \int_0^a f(x) dx \)

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

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

@@ 17번째 줄: / 17번째 줄: @@
 <h5 style="margin: 0px; line-height: 2em;">quantum dilogarithm</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>
+<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>\operatorname{Li}_2(z) = -\int_0^z{{\ln (1-t)}\over t} dt </math>
+<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">정의</h5>
+* <math>0<q<1</math>에 대하여 다음과 같이 정의<br><math>\int_0^a f(x) d_q x = a(1-q)\sum_{k=0}^{\infty}q^k f(aq^k )</math><br><math>\int_0^{\infty} f(x) d_q x =(1-q)\sum_{k=-\infty}^{\infty}q^k f(aq^k )</math><br>
+* <math>q\to 1</math> 이면, <math>\int_0^a f(x) d_q x \to  \int_0^a f(x) dx </math><br>