"An analogue of Rogers-Ramanujan continued fraction"의 두 판 사이의 차이

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<h5>general theory of hypergeometric series and continued fraction</h5>
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==general theory of hypergeometric series and continued fraction==
  
 
<math>R(z)=\sum_{n=0}^{\infty}\frac{z^{an}q^{bn^2}}{(q)_n}</math>
 
<math>R(z)=\sum_{n=0}^{\infty}\frac{z^{an}q^{bn^2}}{(q)_n}</math>
  
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<math>R(z)=R(zq^{1/a})+z^{a}q^{b}R(zq^{2b/a})</math> i.e.
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<math>R(z)=R(zq^{1/a})+z^{a}q^{b}R(zq^{2b/a})</math> i.e.
  
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<h5>examples</h5>
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==examples==
  
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* [[asymptotic analysis of basic hypergeometric series|asymptotic analysis of basic hypergeometric series]]<br>
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* [[asymptotic analysis of basic hypergeometric series|asymptotic analysis of basic hypergeometric series]]<br>
  
 
<math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math>
 
<math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math>
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<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>
 
<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>
  
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<h5>Rogers-Ramanujan</h5>
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==Rogers-Ramanujan==
  
 
A=2 (2,5) minimal model
 
A=2 (2,5) minimal model
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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>
 
<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>
 
<math>R(z)=R(zq)+zqR(zq^2)</math>
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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>
 
<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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By repeating this, we get a continued fraction 
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By repeating this, we get a continued fraction
  
 
<math>\frac{H(q)}{G(q)} = \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}</math>
 
<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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<h5 style="margin: 0px; line-height: 2em;">analogue</h5>
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==analogue==
  
 
A=1/2 (3,5) minimal model
 
A=1/2 (3,5) minimal model
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<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>
 
<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>
  
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<h5 style="margin: 0px; line-height: 2em;">recurrence relation</h5>
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==recurrence relation==
  
 
* http://pythagoras0.springnote.com/pages/3004578<br>
 
* http://pythagoras0.springnote.com/pages/3004578<br>
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<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>
 
<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>
  
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(theorem)
 
(theorem)
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<math>\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}})}}</math>
 
<math>\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}})}}</math>
  
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(proof)
 
(proof)
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(cor)
 
(cor)
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<math>\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)}}}</math>
 
<math>\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)}}}</math>
  
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<math>\frac{1}{q^{1/4}+\frac{1}{q^{3/4}+\frac{1}{q^{5/4}+\frac{1}{\frac{1}{q^{9/4}}+\cdots}}}}</math>
 
<math>\frac{1}{q^{1/4}+\frac{1}{q^{3/4}+\frac{1}{q^{5/4}+\frac{1}{\frac{1}{q^{9/4}}+\cdots}}}}</math>
  
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<h5 style="margin: 0px; line-height: 2em;">modular function</h5>
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==modular function==
  
 
<math>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}</math>
 
<math>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}</math>
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<math>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}</math>
 
<math>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}</math>
  
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<h5 style="margin: 0px; line-height: 2em;">modular function and continued fraction</h5>
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==modular function and continued fraction==
  
 
<math>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)}}}</math>
 
<math>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)}}}</math>
  
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<h5 style="margin: 0px; line-height: 2em;">related items</h5>
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==related items==
  
 
* [[rank 2 case]]<br>
 
* [[rank 2 case]]<br>
 
* [[asymptotic analysis of basic hypergeometric series]]<br>
 
* [[asymptotic analysis of basic hypergeometric series]]<br>
  
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<h5>computational resources</h5>
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==computational resources==
  
 
* [[5399075/attachments/4950801|an_analogue_of_Rogers-Ramanujan_continued_fraction.nb]]
 
* [[5399075/attachments/4950801|an_analogue_of_Rogers-Ramanujan_continued_fraction.nb]]
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* [[mathematica and experimental mathematics]]<br>
 
* [[mathematica and experimental mathematics]]<br>
  
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<h5 style="margin: 0px; line-height: 2em;">references</h5>
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==references==
  
 
* [http://www.ams.org/bull/2005-42-02/S0273-0979-05-01047-5/home.html#References Continued fractions and modular functions]<br>
 
* [http://www.ams.org/bull/2005-42-02/S0273-0979-05-01047-5/home.html#References Continued fractions and modular functions]<br>
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** W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162
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** W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162
 
* [http://dx.doi.org/10.1016/0377-0427%2895%2900263-4 http://dx.doi.org/10.1016/0377-0427(95)00263-4]
 
* [http://dx.doi.org/10.1016/0377-0427%2895%2900263-4 http://dx.doi.org/10.1016/0377-0427(95)00263-4]
 
* [http://dx.doi.org/10.1006/aima.1994.1077 Diagonalization of Certain Integral Operators]<br>
 
* [http://dx.doi.org/10.1006/aima.1994.1077 Diagonalization of Certain Integral Operators]<br>
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**  Ismail M. E. H. and Zhang R. Advances in Mathematics Volume 109, Issue 1, November 1994, Pages 1-33<br>  <br>
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**  Ismail M. E. H. and Zhang R. Advances in Mathematics Volume 109, Issue 1, November 1994, Pages 1-33<br> <br>

2012년 10월 27일 (토) 14:43 판

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

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

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


\(R(z)=R(zq)+zqR(zq^2)\)

\(z=q^{n}\)

\(R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\)

\(\frac{R(q^{n+1})}{R(q^n)}=\cfrac{1}{1+q^{n+1}\cfrac{R(q^{n+2})}{R(q^{n+1})}}\)

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

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)

\(a=1,b=1/4\)

\(R(z)=R(zq)+zq^{1/4}R(zq^{1/2})\)

\(z=q^{n/4}\)

\(R(q^{\frac{n}{4}}})=R(q^{\frac{n+4}{4}})+q^{\frac{n+1}{4}}}R(q^{\frac{n+2}{4}}})\)

\(\frac{R(q^{\frac{n+2}{4}})}{R(q^{\frac{n}{4}})}=\cfrac{1}{q^{\frac{n+1}{4}}+\cfrac{R(q^{\frac{n+4}{4}})}{R(q^{\frac{n+2}{4}})}}\)


(proof)

\(R(z)=R(zq)+zqR(zq^2)\)

\(R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})\)

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(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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