"Bailey pair and lemma"의 두 판 사이의 차이

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<h5 style="BACKGROUND-POSITION: 0px 100%; FONT-SIZE: 1.16em; MARGIN: 0px; COLOR: rgb(34,61,103); LINE-HEIGHT: 2em; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif;">Bailey lemma</h5>
 
<h5 style="BACKGROUND-POSITION: 0px 100%; FONT-SIZE: 1.16em; MARGIN: 0px; COLOR: rgb(34,61,103); LINE-HEIGHT: 2em; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif;">Bailey lemma</h5>
  
the sequence <math>\{\alpha_r\}, \{\beta_r\}</math> 
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If the sequence <math>\{\alpha_r\}, \{\beta_r\}</math>, <math>\{\delta_r\}, \{\gamma_r\}</math> satisfy the following
  
<math>\beta_L=\sum_{r=0}^{L}{\alpha_r}{u_{L-r}v_{L+r}}</math>
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<math>\beta_L=\sum_{r=0}^{L}{\alpha_r}{u_{L-r}v_{L+r}}</math>, <math>\gamma_L=\sum_{r=L}^{\infty}{\delta_r}{u_{r-L}v_{r+L}}</math>
  
<math>\gamma_L=\sum_{r=L}^{\infty}{\delta_r}{u_{r-L}v_{r+L}}</math> 
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then,
  
 
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<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math> 
  
(theorem)
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(corollay)
  
<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math> 
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Choose the following
  
 
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<math>u_{n}=\frac{1}{(q)_n}</math> ,<math>v_{n}=\frac{1}{(x)_n}</math>,<math>\delta_r=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>
  
 
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Then
 
 
(corollay)
 
  
 
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 <math>\gamma_n=\prod()\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
  
 
 
 
 

2010년 6월 20일 (일) 21:45 판

introduction
  •  q-Pfaff-Sallschutz sum

 

 

Bailey lemma

If the sequence \(\{\alpha_r\}, \{\beta_r\}\), \(\{\delta_r\}, \{\gamma_r\}\) satisfy the following

\(\beta_L=\sum_{r=0}^{L}{\alpha_r}{u_{L-r}v_{L+r}}\), \(\gamma_L=\sum_{r=L}^{\infty}{\delta_r}{u_{r-L}v_{r+L}}\)

then,

\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\) 

(corollay)

Choose the following

\(u_{n}=\frac{1}{(q)_n}\) ,\(v_{n}=\frac{1}{(x)_n}\),\(\delta_r=\frac{(y)_n(z)_n x^n}{y^n z^n}\)

Then

 \(\gamma_n=\prod()\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

 

 

Bailey pair
  • the sequence \(\{\alpha_r\}, \{\beta_r\}\) satisfying the following is called a Bailey pair
    \(\beta_L=\sum_{r=0}^{L}\frac{\alpha_r}{(q)_{L-r}(aq)_{L+r}}\)
  • conjugate Bailey pair  \(\{\delta_r\}, \{\gamma_r\}\)
    \(\gamma_L=\sum_{r=L}^{\infty}\frac{\delta_r}{(q)_{r-L}(aq)_{r+L}}\)

 

 

 

 

 

 

Bailey chain
  • we derive a new Bailey chain from a known Bailey pair
    \(\alpha^\prime_n= \frac{(\rho_1;q)_n(\rho_2;q)_n(aq/\rho_1\rho_2)^n}{(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\alpha_n\)
    \(\beta^\prime_n = \sum_{j\ge0}\frac{(\rho_1;q)_j(\rho_2;q)_j(aq/\rho_1\rho_2;q)_{n-j}(aq/\rho_1\rho_2)^j}{(q;q)_{n-j}(aq/\rho_1;q)_n(aq/\rho_2;q)_n}\beta_j\)
  • corollary. by taking  \(\rho_1,\rho_2\to \infty\) , we get 
    \(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
    \(\beta^\prime_n = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_j\)

 

 

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