"An analogue of Rogers-Ramanujan continued fraction"의 두 판 사이의 차이
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| 10번째 줄: | 10번째 줄: | ||
* http://pythagoras0.springnote.com/pages/3004578<br> | * http://pythagoras0.springnote.com/pages/3004578<br> | ||
| − | <math>R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n) | + | <math>R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n)}</math> |
<math>H(q)=R(q), G(q)=R(1)</math> | <math>H(q)=R(q), G(q)=R(1)</math> | ||
| 17번째 줄: | 17번째 줄: | ||
(정리) | (정리) | ||
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<math>R(z)=R(zq)+zqR(zq^2)</math> | <math>R(z)=R(zq)+zqR(zq^2)</math> | ||
2010년 3월 19일 (금) 21:29 판
- A=2 (2,5) minimal model
\(\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\)
\(\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\) - A=1/2 (3,5) minimal model
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})+o(t^5)\)
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})+o(t^5)\)
recurrence relation
\(R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n)}\)
\(H(q)=R(q), G(q)=R(1)\)
(정리)
\(R(z)=R(zq)+zqR(zq^2)\)