"Bailey pair and lemma"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
16번째 줄: 16번째 줄:
  
 
<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math> 
 
<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math> 
 +
 +
 
 +
 +
 
  
 
(corollay)
 
(corollay)
21번째 줄: 25번째 줄:
 
Choose the following
 
Choose the following
  
−
<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>
+
<math>u_{n}=\frac{1}{(q)_n}</math> ,<math>v_{n}=\frac{1}{(x)_n}</math>,<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>
 +
 
 +
Then by the basic analogue of Gauss's theorem (q-Gauss sum)
 +
 
 +
 <math>\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}</math>
 +
 
 +
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> 
  
−
Then
+
 
  
−
 <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) 22:26 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
원본 주소 "https://wiki.mathnt.net/index.php?title=Bailey_pair_and_lemma&oldid=37731"