"함수 다이로그 항등식(functional dilogarithm identity)"의 두 판 사이의 차이

2013년 7월 15일 (월) 01:41 판

개요

로저스 다이로그 함수 (Rogers' dilogarithm)가 만족시키는 두 함수 항등식의 일반화
- 2항 관계식, 반사공식(오일러) $0\leq x \leq 1$ 일 때, \[L(x)+L(1-x)=L(1)\]
- 5항 관계식 (5-term relation) $0\leq x,y\leq 1$ 일 때,

\[L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy} \right)=3L(1)\]

클러스터 대수(cluster algebra) 를 이용하여 일반화됨
n 변수로 구성된 $(n^2+3n)/2$ 항 관계식을 찾을 수 있음

2항 관계식

$S=\left\{x,\frac{1}{x}\right\}$라 두면,

\[\sum_{a\in S}L(\frac{1}{1+a})=L\left(\frac{1}{\frac{1}{x}+1}\right)+L\left(\frac{1}{x+1}\right)=L(1)\]

5항 관계식

$S=\left\{x,y,\frac{x+1}{y},\frac{y+1}{x},\frac{x+y+1}{x y}\right\}$ 이면,

\[\sum_{a\in S}L(\frac{1}{1+a})=L\left(\frac{1}{\frac{x+1}{y}+1}\right)+L\left(\frac{1}{\frac{y+1}{x}+1}\right)+L\left(\frac{1}{\frac{x+y+1}{x y}+1}\right)+L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)=2L(1)\]

9항 관계식

$S$를 다음과 같이 두자

\[ S=\left\{x,y,z,\frac{x z+x+z+1}{y},\frac{x y+x z+x+y^2+y z+2 y+z+1}{x y z},\\ \frac{x z+x+y+z+1}{x y},\frac{x z+x+y+z+1}{y z},\frac{y+1}{x},\frac{y+1}{z}\right\}\]

다이로그 함수에 대하여 다음이 성립한다

\[\sum_{a\in S}L(\frac{1}{1+a})=3L(1)\]

14항 관계식

\[ \left\{x,z,\frac{(x+1) (z+1)}{y},\frac{z+1}{w},\frac{x z+x+y+z+1}{x y}, \frac{(w+z+1) (x z+x+y+z+1)}{w y z}\\,\frac{(y+z+1) (w (x+y+1)+x z+x+y+z+1)}{w x y z}, \frac{w (x+y+1)+x z+x+y+z+1}{y z},\\ \frac{w y+w+y+z+1}{w z},\frac{(x+y+1) (w y+w+y+z+1)}{x y z}, \frac{(w+1) (y+1)}{z},\frac{y+1}{x},w,y\right\} \] \[\sum_{a\in S}L(\frac{1}{1+a})=4L(1)\]

역사

매스매티카 파일 및 계산 리소스

https://docs.google.com/leaf?id=0B8XXo8Tve1cxMzA0M2NkMzMtYTFiNy00N2YwLTlmYzktYWI2YTYwMDMyOTQz&sort=name&layout=list&num=50

관련논문

Nakanishi, Tomoki. 2011. “Dilogarithm Identities for Conformal Field Theories and Cluster Algebras: Simply Laced Case.” Nagoya Mathematical Journal 202 (June): 23–43. doi:10.1215/00277630-1260432.
Chapoton, Frédéric. 2005. “Functional Identities for the Rogers Dilogarithm Associated to Cluster Y-Systems.” Bulletin of the London Mathematical Society 37 (5) (October 1): 755 -760. doi:10.1112/S0024609305004510.
Algebraic Dilogarithm Identities ,Basil Gordon and Richard J. Mcintosh, 1997
L.J. Rogers, On Function Sum Theorems Connected with the Series Formula Proc. London Math. Soc. (1907) s2-4(1): 169-189 doi:10.1112/plms/s2-4.1.169

@@ 25번째 줄: / 25번째 줄: @@
 ==9항 관계식==
 * $S$를 다음과 같이 두자
-:<math>S=\left\{x,y,z,\frac{x z+x+z+1}{y},\frac{x y+x z+x+y^2+y z+2 y+z+1}{x y z},\frac{x z+x+y+z+1}{x y},\frac{x z+x+y+z+1}{y z},\frac{y+1}{x},\frac{y+1}{z}\right\}</math>
+:<math>
+S=\left\{x,y,z,\frac{x z+x+z+1}{y},\frac{x y+x z+x+y^2+y z+2 y+z+1}{x y z},\\
+\frac{x z+x+y+z+1}{x y},\frac{x z+x+y+z+1}{y z},\frac{y+1}{x},\frac{y+1}{z}\right\}</math>
 * 다이로그 함수에 대하여 다음이 성립한다
 :<math>\sum_{a\in S}L(\frac{1}{1+a})=3L(1)</math>
@@ 32번째 줄: / 34번째 줄: @@
 ==14항 관계식==
-:<math>\left\{x,z,\frac{(x+1) (z+1)}{y},\frac{z+1}{w},\frac{x z+x+y+z+1}{x y},\frac{(w+z+1) (x z+x+y+z+1)}{w y z},\frac{(y+z+1) (w (x+y+1)+x z+x+y+z+1)}{w x y z}, \frac{w (x+y+1)+x z+x+y+z+1}{y z},\frac{w y+w+y+z+1}{w z},\frac{(x+y+1) (w y+w+y+z+1)}{x y z},\frac{(w+1) (y+1)}{z},\frac{y+1}{x},w,y\right\}</math>
+:<math>
+\left\{x,z,\frac{(x+1) (z+1)}{y},\frac{z+1}{w},\frac{x z+x+y+z+1}{x y},
+\frac{(w+z+1) (x z+x+y+z+1)}{w y z}\\,\frac{(y+z+1) (w (x+y+1)+x z+x+y+z+1)}{w x y z}, \frac{w (x+y+1)+x z+x+y+z+1}{y z},\\
+\frac{w y+w+y+z+1}{w z},\frac{(x+y+1) (w y+w+y+z+1)}{x y z},
+\frac{(w+1) (y+1)}{z},\frac{y+1}{x},w,y\right\}
+</math>
 :<math>\sum_{a\in S}L(\frac{1}{1+a})=4L(1)</math>