"An analogue of Rogers-Ramanujan continued fraction"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
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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)=R(zq^{1/a})+z^{a}q^{b}R(zq^{2b/a})</math> i.e. | ||
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| − | + | ==examples== | |
| − | + | * [[asymptotic analysis of basic hypergeometric series|asymptotic analysis of basic hypergeometric series]] | |
| + | :<math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math> | ||
| + | :<math>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})</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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| − | + | ==Rogers-Ramanujan== | |
| + | * A=2 (2,5) minimal model | ||
| + | :<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> | ||
| + | :<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> | ||
| + | * recusrion <math>R(z)=R(zq)+zqR(zq^2)</math> | ||
| + | * if <math>z=q^{n}</math>, <math>R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})</math> | ||
| + | :<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> | ||
| + | :<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> | ||
| + | * 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> | ||
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| − | <math>\sum_{n | + | ==analogue== |
| + | * A=1/2 (3,5) minimal model | ||
| + | :<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> | ||
| + | :<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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| − | <math>\ | + | ==recurrence relation== |
| + | * {{수학노트|url=로저스-라마누잔_항등식}} | ||
| + | * {{수학노트|url=로저스-라마누잔_연분수}} | ||
| + | :<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^{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> | ||
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(theorem) | (theorem) | ||
| + | Let <math>a=1,b=1/4</math> | ||
| + | :<math>R(z)=R(zq)+zq^{1/4}R(zq^{1/2})</math> | ||
| + | If <math>z=q^{n/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> | ||
| + | :<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) | ||
| − | + | :<math>R(z)=R(zq)+zqR(zq^2)</math> | |
| − | <math>R(z)=R(zq)+zqR(zq^2)</math> | + | :<math>R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})</math> |
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| − | <math>R(q^n)=R(q^{n+1})+q^{n+1}R(q^{n+2})</math> | ||
■ | ■ | ||
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(cor) | (cor) | ||
| − | <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> |
| − | + | :<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> | + | :<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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| − | <math>\frac{1}{q^{1/4}+\frac{1}{q^{3/4}+\frac{1}{q^{5/4}+\frac{1}{\frac{1}{q^{9/4}}+\cdots}}} | ||
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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}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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| − | + | ==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> | ||
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| − | + | ==related items== | |
| − | + | * [[rank 2 case]] | |
| + | * [[asymptotic analysis of basic hypergeometric series]] | ||
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| − | + | ==computational resources== | |
| + | * https://docs.google.com/file/d/0B8XXo8Tve1cxOHRUNER3RGJiMEk/edit | ||
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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] | + | * [http://www.ams.org/bull/2005-42-02/S0273-0979-05-01047-5/home.html#References Continued fractions and modular functions] |
| − | ** W. Duke, | + | ** W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162 |
| − | + | * [http://dx.doi.org/10.1006/aima.1994.1077 Diagonalization of Certain Integral Operators] | |
| − | * [http://dx.doi.org/10.1006/aima.1994.1077 Diagonalization of Certain Integral Operators] | + | ** Ismail M. E. H. and Zhang R. Advances in Mathematics Volume 109, Issue 1, November 1994, Pages 1-33 |
| − | ** | + | ** [http://dx.doi.org/10.1016/0377-0427%2895%2900263-4 http://dx.doi.org/10.1016/0377-0427(95)00263-4] |
| + | [[분류:개인노트]] | ||
| + | [[분류:q-series]] | ||
| + | [[분류:migrate]] | ||
2020년 11월 16일 (월) 07:09 기준 최신판
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
\[q^{1/20}\frac{H(q)}{G(q)}=\cfrac{q^{1/20}}{q^{1/4}+\cfrac{1}{q^{3/4}+\cfrac{R(q^{3/2})}{R(q)}}}\]
computational resources
references
- Continued fractions and modular functions
- W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162
- Diagonalization of Certain Integral Operators
- Ismail M. E. H. and Zhang R. Advances in Mathematics Volume 109, Issue 1, November 1994, Pages 1-33
- http://dx.doi.org/10.1016/0377-0427(95)00263-4