"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: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">introduction</h5>
 
<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">introduction</h5>
 
*   q-Pfaff-Sallschutz sum<br>
 
 
 
 
  
 
 
 
 
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* [[Slater list|Slater's list]]<br>
 
* [[Slater list|Slater's list]]<br>
 
 
 
 
 
 
 
<h5 style="margin: 0px; line-height: 2em;">how to obtain and check Bailey pair?</h5>
 
 
*  various complicated q-series identities<br>
 
 
 
 
 
 
 
 
<h5 style="margin: 0px; line-height: 2em;">why do we care about Bailey pair?</h5>
 
 
*  When we have a Bailey pair, we can produce q-series identities<br>
 
**  (1) Bailey lemma gives an identity involving q-series<br>
 
**  (2) using the definition of Bailey pair<br><math>\beta_L=\sum_{r=0}^{L}\frac{\alpha_r}{(q)_{L-r}(aq)_{L+r}}</math><br>
 
  
 
 
 
 
142번째 줄: 120번째 줄:
 
* [[q-analogue of summation formulas]]<br>
 
* [[q-analogue of summation formulas]]<br>
 
* [[Rogers-Ramanujan continued fraction]]<br>
 
* [[Rogers-Ramanujan continued fraction]]<br>
 
 
 
 
 
 
 
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* http://en.wikipedia.org/wiki/Bailey_pair
 
* http://en.wikipedia.org/wiki/Wilfrid_Norman_Bailey
 
* http://en.wikipedia.org/wiki/
 
* http://www.scholarpedia.org/
 
* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
 
 
 
 
 
 
 
 
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* [[2010년 books and articles]]<br>
 
* http://gigapedia.info/1/
 
* http://gigapedia.info/1/
 
* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
 
 
[[4909919|]]
 
 
 
 
 
 
 
 
 
 
 
<h5 style="margin: 0px; line-height: 2em;">expositions</h5>
 
 
* [http://arxiv.org/abs/0910.2062v2 50 Years of Bailey's lemma] S. Ole Warnaar, 2009<br>
 
 
 
 
  
 
 
 
 
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<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">articles</h5>
  
* [http://arxiv.org/abs/math.QA/9909044 A generalization of the q-Saalschutz sum and the Burge transform]<br>
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*  <br> A. Schilling, S.O. Warnaar [http://arxiv.org/abs/math.QA/9909044 A generalization of the q-Saalschutz sum and the Burge transform], 2009<br>
**  A. Schilling, S.O. Warnaar, 2009<br>
 
 
* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]<br>
 
* [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities]<br>
 
**  Mc Laughlin, 2008<br>
 
**  Mc Laughlin, 2008<br>
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* [http://plms.oxfordjournals.org/cgi/reprint/s2-50/1/1.pdf Identities of Rogers-Ramanujan type]<br>
 
* [http://plms.oxfordjournals.org/cgi/reprint/s2-50/1/1.pdf Identities of Rogers-Ramanujan type]<br>
 
**  Bailey, 1944<br>
 
**  Bailey, 1944<br>
*  On two theorems of combinatory analysis and some allied identities <br>
 
* http://www.ams.org/mathscinet<br>
 
 
* http://www.zentralblatt-math.org/zmath/en/
 
* http://pythagoras0.springnote.com/
 
* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html][http://www.ams.org/mathscinet ]
 
* http://front.math.ucdavis.edu/search?a=&t=&c=&n=40&s=Listings&q=
 
* http://www.ams.org/mathscinet/search/publications.html?pg4=AUCN&s4=&co4=AND&pg5=TI&s5=&co5=AND&pg6=PC&s6=&co6=AND&pg7=ALLF&co7=AND&Submit=Search&dr=all&yrop=eq&arg3=&yearRangeFirst=&yearRangeSecond=&pg8=ET&s8=All&s7=
 
* http://dx.doi.org/10.1112/plms/s2-53.6.460
 
 
 
 
 
 
 
 
<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">question and answers(Math Overflow)</h5>
 
 
* http://mathoverflow.net/search?q=
 
* http://mathoverflow.net/search?q=
 
 
 
 
 
 
 
 
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*  구글 블로그 검색<br>
 
** http://blogsearch.google.com/blogsearch?q=
 
** http://blogsearch.google.com/blogsearch?q=
 
 
 
 
 
 
 
 
<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">experts on the field</h5>
 
 
* http://arxiv.org/
 
 
 
 
 
 
 
 
<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">links</h5>
 
 
* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
 
* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
 
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 
* http://functions.wolfram.com/
 
*
 

2011년 11월 12일 (토) 05:48 판

introduction

 

 

 

examples of Bailey pair

 

 

Bailey lemma
  • Bailey lemma involved a Bailey pair and a conjugate Bailey pair
  • 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,
    \(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)

 

 

specialization
  • Choose the following (in the following, x=aq to get a Bailey pair relative to a)
    \(u_{n}=\frac{1}{(q)_n}\) ,\(v_{n}=\frac{1}{(x)_n}\),
  • There is a conjugate Bailey pair
    \(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\)
     \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

 

(proof)

By the basic analogue of Gauss' theorem 

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

[[q-analogue of summation formulas|]]

Also note that \((a)_{n+r}=(a)_{r}(aq^{r})_{n}\) and \((a)_{\infty}=(a)_{r}(aq^{r})_{\infty}\)

Put \(a=yq^{r},b=zq^{r},c=xq^{2r}\). 

Then we get (*)

\(A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(xq^{r}/y)_{\infty}(xq^{r}/z)_{\infty}}{(xq^{2r})_{\infty}(x/(yz))_{\infty}}=B\)

From the left hand side,

\(A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}}{(y)_{r}(z)_{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n+r}\)

Now let

\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\) so that

\(A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}\)

From the right hand side of (*), we get

\(B=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}\)

Therefore,

\(A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}=B\)

By simplifying the above equation, we obtain

\(\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{1}{(x/y)_{r}(x/z)_{r}}\frac{(y)_{r}(z)_{r}x^{r}}{y^{r}z^{r}}\)

\(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)  with \(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\) gives a conjugate Bailey pair

(a different notation

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

is also used sometimes) ■

  • If we apply Bailey lemma to the above conjugate pair, we get
    \(\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}\)

 

 

examples
  • Conjugate Bailey pair (\(x=q,y\to\infty, z\to\infty\))
    \(\delta_n=q^{n^2}\)
    \(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)
  • Bailey pair
    \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{1}{2}n}+q^{-\frac{1}{2}n})\)
    \(\beta_n=\frac{1}{(q)_{n}}\)
  • we get the Rogers-Ramanujan identity(Slater 18)
    \(\sum_{n=0}^{\infty}\frac{q^{n^2}}{ (q)_{n}}=\frac{(q^{3};q^{5})_{\infty}(q^{2};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}}\)

 

 

Bailey chain

 

 

history

 

 

related items

 

 

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