"Bailey pair and lemma"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
33번째 줄: 33번째 줄:
 
Hence by Bailey's lemma,
 
Hence by Bailey's lemma,
  
<math>\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}</math> 
+
<math>\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}</math>
  
 
 
 
 
131번째 줄: 131번째 줄:
 
* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]<br>
 
* [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]<br>
 
**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475<br>
 
**  Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475<br>
*  On two theorems <br>
+
*  On two theorems of combinatory analysis and some allied identities <br>
 
* http://www.ams.org/mathscinet<br>
 
* http://www.ams.org/mathscinet<br>
  

2010년 6월 23일 (수) 18:35 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

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

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

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

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 
    \(\alpha^\prime_n= a^nq^{n^2}\alpha_n\)
    \(\beta^\prime_n = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_j\)

 

 

history

 

 

 

related items

 

 

encyclopedia

 

 

books

[[4909919|]]

 

 

articles

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links