"An analogue of Rogers-Ramanujan continued fraction"의 두 판 사이의 차이
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| − | + | <h5>Rogers-Ramanujan</h5> | |
| − | * 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 | + | |
| + | A=2 (2,5) minimal model | ||
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| + | <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> | ||
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| + | <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> | ||
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| + | <math>R(z)=R(zq)+zqR(zq^2)</math> | ||
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| + | 이 정리로부터 <math>R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})</math> | ||
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| + | 즉 <math>\frac{R(q^{n+1})}{R(q^n)}=\cfrac{1}{1+q^{n+1}\cfrac{R(q^{n+2})}{R(q^{n+1})}}</math>를 얻는다. | ||
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| + | <math>\frac{H(q)}{G(q)}=\cfrac{R(q)}{R(1)} = \cfrac{1}{1+q\cfrac{R(q^2)}{R(q)}}=\cfrac{1}{1+\cfrac{q}{1+q^2\cfrac{R(q^3)}{R(q^2)}}}=\cdots</math> | ||
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| + | 이를 반복하면, 다음을 얻는다. | ||
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| + | <math>\frac{H(q)}{G(q)} = \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}</math> | ||
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| + | * <br> A=1/2 (3,5) minimal model<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><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> | ||
| 12번째 줄: | 41번째 줄: | ||
<math>R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n)}</math> | <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)</math> | + | <math>H(q)=R(q^{1/2})=\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> |
| − | + | <math>G(q)=R(1)=\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> | |
2010년 3월 25일 (목) 08:56 판
Rogers-Ramanujan
A=2 (2,5) minimal model
\(\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\)
\(\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\)
\(R(z)=R(zq)+zqR(zq^2)\)
이 정리로부터 \(R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\)
즉 \(\frac{R(q^{n+1})}{R(q^n)}=\cfrac{1}{1+q^{n+1}\cfrac{R(q^{n+2})}{R(q^{n+1})}}\)를 얻는다.
\(\frac{H(q)}{G(q)}=\cfrac{R(q)}{R(1)} = \cfrac{1}{1+q\cfrac{R(q^2)}{R(q)}}=\cfrac{1}{1+\cfrac{q}{1+q^2\cfrac{R(q^3)}{R(q^2)}}}=\cdots\)
이를 반복하면, 다음을 얻는다.
\(\frac{H(q)}{G(q)} = \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\)
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A=1/2 (3,5) minimal model
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})+o(t^5)\)
\(\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})+o(t^5)\)
recurrence relation
\(R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n)}\)
\(H(q)=R(q^{1/2})=\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})+o(t^5)\)
\(G(q)=R(1)=\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})+o(t^5)\)
(정리)
\(a=1,b=1/4\)
\(R(z)=R(zq^{1/a})+z^{a}q^{b}R(zq^{2b/a})\) i.e.
\(R(z)=R(zq)+zq^{1/4}R(zq^{1/2})\)