"Q-초기하급수의 점근 급수"의 두 판 사이의 차이
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| 8번째 줄: | 8번째 줄: | ||
* <math>a>0,x>0,b\in\mathbb{R}</math>라 두자 | * <math>a>0,x>0,b\in\mathbb{R}</math>라 두자 | ||
| − | * z>0는 방정식 <math>1- | + | * z>0는 방정식 <math>1-x=zx^{a}</math> 의 해라 하자. |
* 다음 근사식이 성립함 '''[McIntosh1995]'''<br><math>\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\})</math> 또는<br> | * 다음 근사식이 성립함 '''[McIntosh1995]'''<br><math>\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\})</math> 또는<br> | ||
| − | <math>\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 | + | <math>\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)}{\log q})</math> |
| 18번째 줄: | 18번째 줄: | ||
<h5>예</h5> | <h5>예</h5> | ||
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| + | * A=1/2 (3,5) minimal model<br><math>\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})+o(t^5)</math><br><math>\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})+o(t^5)</math><br> | ||
* A=1 (3,4) minimal model<br><math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math><br><math>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})</math><br><math>\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})</math><br> | * A=1 (3,4) minimal model<br><math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math><br><math>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})</math><br><math>\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})</math><br> | ||
* A=2 (2,5) minimal model<br><math>\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})+o(1)</math><br><math>\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})+o(1)</math><br> | * A=2 (2,5) minimal model<br><math>\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})+o(1)</math><br><math>\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})+o(1)</math><br> | ||
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2011년 10월 14일 (금) 15:36 판
이 항목의 수학노트 원문주소
개요
- \(a>0,x>0,b\in\mathbb{R}\)라 두자
- z>0는 방정식 \(1-x=zx^{a}\) 의 해라 하자.
- 다음 근사식이 성립함 [McIntosh1995]
\(\sum_{n=0}^{\infty}\frac{z^nq^{\frac{a}{2}n^2+bn}}{(q)_n}\sim \frac{x^b}{\sqrt[[:틀:X+a(1-x)]]} \exp (-\frac{1}{\log q}\{\operatorname{Li}_2(zx^{a})+\frac{a}{2}\log^2 x\})\) 또는
\(\sum_{n=0}^{\infty}\frac{z^nq^{\frac{a}{2}n^2+bn}}{(q)_n}\sim \frac{x^b}{\sqrt[[:틀:X+a(1-x)]]} \exp (-\frac{L(1-x)}{\log q})\)
예
- A=1/2 (3,5) minimal model
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})+o(t^5)\)
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})+o(t^5)\)
- 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
\(\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\)
\(\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\)
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/
- The Online Encyclopaedia of Mathematics
- NIST Digital Library of Mathematical Functions
- The World of Mathematical Equations
리뷰논문, 에세이, 강의노트
관련논문
- [McIntosh1995]Some Asymptotic Formulae for q-Hypergeometric Series Richard J. McIntosh, Journal of the London Mathematical Society 1995 51(1):120-136
- http://www.jstor.org/action/doBasicSearch?Query=
- http://www.ams.org/mathscinet
- http://dx.doi.org/