"An analogue of Rogers-Ramanujan continued fraction"의 두 판 사이의 차이
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<math>R(q^{\frac{n}{4}}})=R(q^{\frac{n+4}{4}})+q^{\frac{n+1}{4}}}R(q^{\frac{n+2}{4}}})</math> | <math>R(q^{\frac{n}{4}}})=R(q^{\frac{n+4}{4}})+q^{\frac{n+1}{4}}}R(q^{\frac{n+2}{4}}})</math> | ||
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| + | <math>R(z)=R(zq)+zqR(zq^2)</math> | ||
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| + | 로부터 <math>R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})</math> | ||
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| + | <math>\frac{H(q)}{G(q)}=\cfrac{R(q^{1/2})}{R(1)} = \cfrac{1}{q^{1/4}+q\cfrac{R(q)}{R(q^{1/2})}}=\cfrac{1}{1+\cfrac{q}{1+q^2\cfrac{R(q^3)}{R(q^2)}}}=\cdots</math> | ||
2010년 3월 25일 (목) 09:14 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.
\(\frac{H(q)}{G(q)} = \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\)
analogue
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.
\(R(z)=R(zq)+zq^{1/4}R(zq^{1/2})\)
\(z=q^{n/4}\)
\(R(q^{\frac{n}{4}}})=R(q^{\frac{n+4}{4}})+q^{\frac{n+1}{4}}}R(q^{\frac{n+2}{4}}})\)
(proof)
\(R(z)=R(zq)+zqR(zq^2)\)
로부터 \(R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\)
■
(cor)
\(\frac{H(q)}{G(q)}=\cfrac{R(q^{1/2})}{R(1)} = \cfrac{1}{q^{1/4}+q\cfrac{R(q)}{R(q^{1/2})}}=\cfrac{1}{1+\cfrac{q}{1+q^2\cfrac{R(q^3)}{R(q^2)}}}=\cdots\)