"Bailey lattice"의 두 판 사이의 차이

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*  In the corollay above, set a=q and replace i by i-1<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}</math><br>
 
*  In the corollay above, set a=q and replace i by i-1<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}</math><br>
 
*  On LHS, we get<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+n_i+\cdots+n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br>
 
*  On LHS, we get<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+n_i+\cdots+n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br>
−
*  On RHS, we get<br><math>R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}</math><br><math>=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}</math><br> Now use the original Bailey pair,<br><math>\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}</math><br><math>\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}</math><br><math>R=\frac{1}{(q)_{\infty}}\left{[}1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right{]}</math><br>  <br>
+
*  On RHS, we get<br><math>R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}</math><br><math>=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}</math><br> Now use the original Bailey pair,<br><math>\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}</math><br><math>\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}</math><br><math>R=\frac{1}{(q)_{\infty}}\left{[}1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right{]}</math><br><math>=\frac{1}{(q)_{\infty}}\left{[}1+(-1)^{n}\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}+{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\right{]}</math><br>
 +
*  first part in the summation is<br>  <math>{q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}</math><br><math>=q^{kn^2+(k-i+1)n+n(n-1)/2}}=q^{n(2kn+2(k-i+1)+(n-1))/2</math><br><math>=q^{n((2k+1)n+2k-2i)+1)/2</math><br>
 +
*  secont part in the summation is<br><math>{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}</math><br>  <br>  <br>  <br>
  
 
 
 
 

2010년 10월 9일 (토) 11:47 판

introduction

Let \(\{\alpha_r\}, \{\beta_r\}\) be a Bailey pair relative to a and set

\(\alpha_0'=\alpha_0\), \(\alpha_n'=(1-a)a^nq^{n^2-n}(\frac{\alpha_n}{1-aq^{2n}}-\frac{aq^{2n-2}\alpha_{n-1}}{1-aq^{2n-2}})\)\(\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}\)

Then \(\{\alpha_r'\}, \{\beta_r'\}\)  is a Bailey pair relative to \(aq^{-1}\)

 

 

 

comparison with Bailey chain
  • Bailey chain
    \(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
    \(\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r\)
  • This does not change the parameter a of the Bailey pair.
  • lattice construction changes this

 

 

 

corollary

Let \(\{\alpha_r\}, \{\beta_r\}\) be the initial Bailey pair relative to a. Then the following is true \[\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_i}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}=\frac{1}{(a)_{\infty}}\left{[}\alpha_0+(1-a)\sum_{n=1}^{\infty}(\frac{a^{kn}q^{kn^2-in}\alpha_n}{1-aq^{2n}}-\frac{a^{k(n-1)+i+1}q^{k(n-1)^2+(i+2)(n-1)}\alpha_{n-1}}{1-aq^{2n-2}})\right{]}\]

 

(proof)

apply Bailey chain construction k-i times (Bailey chain)

At the (k-i)th step apply Bailey lattice

apply Bailey chain construction i-1 times again.

Then we get a Bailey pair

\(\{\alpha_r'\}, \{\beta_r'\}\)  is a Bailey pair relative to \(aq^{-1}\).

If we use the defining relation of Bailey pair to \(\{\alpha_r'\}, \{\beta_r'\}\),

\(\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}\)

and take the limit L\to\infty ■

 

Example. Do this for k=5 and i=2

 

 

application
  • the proof of Andrews-Gordon identity
  • initial Bailey pair
    \(\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
    \(\beta_{L}=\delta_{L,0}\)
  • In the corollay above, set a=q and replace i by i-1
    \(\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
  • On LHS, we get
    \(\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+n_i+\cdots+n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
  • On RHS, we get
    \(R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
    \(=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}\)
    Now use the original Bailey pair,
    \(\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}\)
    \(\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}\)
    \(R=\frac{1}{(q)_{\infty}}\left{[}1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right{]}\)
    \(=\frac{1}{(q)_{\infty}}\left{[}1+(-1)^{n}\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}+{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\right{]}\)
  • first part in the summation is
     \({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}\)
    \(=q^{kn^2+(k-i+1)n+n(n-1)/2}}=q^{n(2kn+2(k-i+1)+(n-1))/2\)
    \(=q^{n((2k+1)n+2k-2i)+1)/2\)
  • secont part in the summation is
    \({q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\)
     
     
     

 

 

 

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encyclopedia

 

 

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[[4909919|]]

 

 

articles
  • A Bailey Lattice
    • Jeremy Lovejoy, Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516

 

 

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