An analogue of Rogers-Ramanujan continued fraction

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general theory of hypergeometric series and continued fraction

\[R(z)=\sum_{n=0}^{\infty}\frac{z^{an}q^{bn^2}}{(q)_n}\] \[R(z)=R(zq^{1/a})+z^{a}q^{b}R(zq^{2b/a})\] i.e.


examples

\[\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})\] \[2\prod_{n=1}^{\infty}(1+q^n)=\sum_{n\geq 0}^{\infty}\frac{q^{n(n-1)/2}}{(q)_n}\sim \sqrt{2}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\] \[\prod_{n=1}^{\infty}(1+q^n)=\sum_{n\geq 0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\]


Rogers-Ramanujan

  • A=2 (2,5) minimal model

\[\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\] \[\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\]

  • recusrion \(R(z)=R(zq)+zqR(zq^2)\)
  • if \(z=q^{n}\), \(R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\)

\[\frac{R(q^{n+1})}{R(q^n)}=\cfrac{1}{1+q^{n+1}\cfrac{R(q^{n+2})}{R(q^{n+1})}}\] \[\frac{H(q)}{G(q)}=\cfrac{R(q)}{R(1)} = \cfrac{1}{1+q\cfrac{R(q^2)}{R(q)}}=\cfrac{1}{1+\cfrac{q}{1+q^2\cfrac{R(q^3)}{R(q^2)}}}=\cdots\]

  • By repeating this, we get a continued fraction

\[\frac{H(q)}{G(q)} = \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}\]


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

\[R(z)=\sum_{n\geq 0}\frac{z^nq^{n^2/4}}{(1-q)\cdots(1-q^n)}\] \[H(q)=R(q^{1/2})=\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n} \sim \frac{2}{\sqrt{5+\sqrt{5}}}\exp(\frac{\pi^2}{10t}+\frac{t}{40})+o(t^5)\] \[G(q)=R(1)=\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\sim \frac{2}{\sqrt{5-\sqrt{5}}}\exp(\frac{\pi^2}{10t}-\frac{t}{40})+o(t^5)\]


(theorem) Let \(a=1,b=1/4\) \[R(z)=R(zq)+zq^{1/4}R(zq^{1/2})\] If \(z=q^{n/4}\), \[R(q^{\frac{n}{4}})=R(q^{\frac{n+4}{4}})+q^{\frac{n+1}{4}}R(q^{\frac{n+2}{4}})\] \[\frac{R(q^{\frac{n+2}{4}})}{R(q^{\frac{n}{4}})}=\cfrac{1}{q^{\frac{n+1}{4}}+\cfrac{R(q^{\frac{n+4}{4}})}{R(q^{\frac{n+2}{4}})}}\]


(proof) \[R(z)=R(zq)+zqR(zq^2)\] \[R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\]

■


(cor)

\[\frac{R(q^{\frac{n+2}{4}})}{R(q^{\frac{n}{4}})}=\cfrac{1}{q^{\frac{n+1}{4}}+\cfrac{R(q^{\frac{n+4}{4}})}{R(q^{\frac{n+2}{4}})}}\] \[\frac{H(q)}{G(q)}=\cfrac{R(q^{1/2})}{R(1)} = \cfrac{1}{q^{1/4}+\cfrac{R(q)}{R(q^{1/2})}}=\cfrac{1}{q^{1/4}+\cfrac{1}{q^{3/4}+\cfrac{R(q^{3/2})}{R(q)}}}\] \[\frac{1}{q^{1/4}+\frac{1}{q^{3/4}+\frac{1}{q^{5/4}+\frac{1}{\frac{1}{q^{9/4}}+\cdots}}}}\]


modular function

\[q^{1/40}H(q)=q^{1/40}R(q^{1/2})=q^{1/40}\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}+\frac{n}{2}}} {(q;q)_n}\] \[q^{-1/40}G(q)=q^{-1/40}R(1)=q^{-1/40}\sum_{n=0}^\infty \frac {q^{\frac{n^2}{4}}} {(q;q)_n}\]


modular function and continued fraction

\[q^{1/20}\frac{H(q)}{G(q)}=\cfrac{q^{1/20}}{q^{1/4}+\cfrac{1}{q^{3/4}+\cfrac{R(q^{3/2})}{R(q)}}}\]


related items



computational resources


references

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