"Slater 37"의 두 판 사이의 차이

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==Note==
 
==Note==
  
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* [[Slater list|Slater's list]]
 
* [[Slater list|Slater's list]]
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*  I(17)<br>
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*  I(17)
  
 
 
 
 
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==Bailey pair 1==
 
==Bailey pair 1==
  
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*  Use the folloing<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math><br>
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*  Use the folloing<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
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*  Specialize<br><math>x=q^{3}, y=-q, z\to\infty</math>.<br>
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*  Specialize<math>x=q^{3}, y=-q, z\to\infty</math>.
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*  Bailey pair<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><br><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math><br>
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*  Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>
  
 
 
 
 
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==Bailey pair 2==
 
==Bailey pair 2==
  
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*  Use the following '''[Slater52-1] '''(4.2)<br>  <br>
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*  Use the following '''[Slater52-1] '''(4.2)  
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*  Specialize<br><math>a=q^{2},d=q^2,e=q</math><br>
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*  Specialize<math>a=q^{2},d=q^2,e=q</math>
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*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><br><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math><br>
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*  Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
  
 
 
 
 
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==Bailey pair ==
 
==Bailey pair ==
  
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*  Bailey pairs<br>  <br><math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><br><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math><br>  <br><math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><br><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math><br>
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*  Bailey pairs  <math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math>  <math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math>
  
 
 
 
 
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<math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\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=1}^{\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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* [[Bailey pair and lemma|Bailey's lemma]]<br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<br><math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math><br>
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* [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math>
  
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 
** http://www.research.att.com/~njas/sequences/?q=
 
** http://www.research.att.com/~njas/sequences/?q=
  
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==related items==
 
==related items==
  
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* [[asymptotic analysis of basic hypergeometric series]]<br>
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* [[asymptotic analysis of basic hypergeometric series]]
  
 
 
 
 
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* [[2010년 books and articles]]<br>
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* [[2010년 books and articles]]
 
* http://gigapedia.info/1/
 
* http://gigapedia.info/1/
 
* http://gigapedia.info/1/
 
* http://gigapedia.info/1/
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==articles==
 
==articles==
  
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* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]<br>
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* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]
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**  McLaughlin, 2008<br>
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**  McLaughlin, 2008
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* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>
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* [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]
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**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167<br>
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**  Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167
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* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]<br>
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* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]
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**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475<br>
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**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475
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*   <br>
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*   
 
* http://www.ams.org/mathscinet
 
* http://www.ams.org/mathscinet
 
* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
 
* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
 
* [http://arxiv.org/ ]http://arxiv.org/
 
* [http://arxiv.org/ ]http://arxiv.org/
 
* http://pythagoras0.springnote.com/
 
* http://pythagoras0.springnote.com/
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* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
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* http://dx.doi.org/
 
* http://dx.doi.org/
 
[[분류:개인노트]]
 
[[분류:개인노트]]
 
[[분류:math and physics]]
 
[[분류:math and physics]]
 
[[분류:migrate]]
 
[[분류:migrate]]

2020년 11월 13일 (금) 23:41 판

Note

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type of identity

 

 

Bailey pair 1

  • Use the folloing\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\),  \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)
  • Specialize\(x=q^{3}, y=-q, z\to\infty\).
  • Bailey pair\(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)\(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\)

 

 

Bailey pair 2

  • Use the following [Slater52-1] (4.2)  
  • Specialize\(a=q^{2},d=q^2,e=q\)
  • Bailey pair\(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)

 

 

Bailey pair 

  • Bailey pairs  \(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)\(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\)  \(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)\(\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)

 

 

q-series identity

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

  • Bailey's lemma\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\) (오일러의 오각수정리(pentagonal number theorem) was used to verify this)\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)

 

 

Bethe type equation (cyclotomic equation)

Let \(\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}\).

Then \(\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a\)  has a unique root \(0<\mu<1\). We get

\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)

 

a=1,d=1,e=1

The equation  becomes \(1-x=x\).

\(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)

 

 

dilogarithm identity

\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)

 

 

related items

 

 

books

 


 

 

articles

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