Bailey pair and lemma

수학노트
http://bomber0.myid.net/ (토론)님의 2010년 6월 23일 (수) 22:05 판
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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

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

Then by the basic analogue of Gauss's theorem (q-Gauss sum)

(Recall \(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}}{(c ,q)_{n}(q ,q)_{n}}(\frac{c}{ab})^{n}=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}\) )

 

 \(\gamma_n=\prod{{x/y,x/z;q}\choose {x,x/yz;}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

Hence by Bailey's lemma,

\(\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\prod{{x/y,x/z;q}\choose {x,x/yz;}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}\)

 

 

 

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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