"Bailey pair and lemma"의 두 판 사이의 차이

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<h5 style="margin: 0px; line-height: 2em;">specialization</h5>
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*  Choose the following (in the following, x=aq to get a Bailey pair relative to a)<br><math>u_{n}=\frac{1}{(q)_n}</math> ,<math>v_{n}=\frac{1}{(x)_n}</math>,<br>
 
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*  There is a conjugate Bailey pair<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math><br>  <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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*  If we apply Bailey lemma to the above conjugate pair, we get<br><math>\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}</math><br>
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*  If we apply Bailey lemma to the above conjugate pair, we get<br>  <br>
  
 
 
 
 

2011년 11월 12일 (토) 06:44 판

examples of Bailey pair

 

 

 

 

 

 

  • If we apply Bailey lemma to the above conjugate pair, we get
     

 

 

examples
  • Conjugate Bailey pair (\(x=q,y\to\infty, z\to\infty\))
    \(\delta_n=q^{n^2}\)
    \(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)
  • Bailey pair
    \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{1}{2}n}+q^{-\frac{1}{2}n})\)
    \(\beta_n=\frac{1}{(q)_{n}}\)
  • we get the Rogers-Ramanujan identity(Slater 18)
    \(\sum_{n=0}^{\infty}\frac{q^{n^2}}{ (q)_{n}}=\frac{(q^{3};q^{5})_{\infty}(q^{2};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}}\)

 

 

Bailey chain

 

 

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