Q-초기하급수의 점근 급수

개요

• \(a>0,z>0,b\in\mathbb{R}\)라 두자
• $x>0$는 방정식 \(1-x=zx^{a}\) 의 해라 하자.
• 다음 근사식이 성립함 [McIntosh1995] \[\sum_{n=0}^{\infty}\frac{z^nq^{\frac{a}{2}n^2+bn}}{(q)_n}\sim \frac{x^b}{\sqrt{x+a(1-x)}} \exp (-\frac{1}{\log q}\{\operatorname{Li}_2(zx^{a})+\frac{a}{2}\log^2 x\})\] 또는 \[\sum_{n=0}^{\infty}\frac{z^nq^{\frac{a}{2}n^2+bn}}{(q)_n}\sim \frac{x^b}{\sqrt{x+a(1-x)}} \exp (\frac{L(1-x)}{t})\]   이 때, \(q=e^{-t}\).

 

 

예

• A=1/2 (3,5) minimal model\[\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})\]

\[\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})\]

• A=1 (3,4) minimal model\[\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})\]\[2\prod_{n=1}^{\infty}(1+q^n)=\sum_{n\geq 0}^{\infty}\frac{q^{n(n-1)/2}}{(q)_n}\sim \sqrt{2}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\]\[\prod_{n=1}^{\infty}(1+q^n)=\sum_{n\geq 0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\]
  
• A=2 (2,5) minimal model 로저스-라마누잔 항등식

\[\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})\] \[\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})\]

 

 

 

역사

 

• http://www.google.com/search?hl=en&tbs=tl:1&q=
• 수학사 연표

 

 

메모

 

• Math Overflow http://mathoverflow.net/search?q=

 

 

관련된 항목들

• 로저스 다이로그 함수 (Rogers' dilogarithm)

 

 

매스매티카 파일 및 계산 리소스

• https://docs.google.com/file/d/0B8XXo8Tve1cxODk1ZjBiYWEtYjMyOS00MDdmLTg1ZjItMTJhOTA0MzZmYmY5/edit

 

 

 

관련논문

• [McIntosh1995]Some Asymptotic Formulae for q-Hypergeometric Series Richard J. McIntosh, Journal of the London Mathematical Society 1995 51(1):120-136