"베일리 격자(Bailey lattice)"의 두 판 사이의 차이

수학노트
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1번째 줄: 1번째 줄:
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 수학노트 원문주소</h5>
  
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* [[베일리 격자(Bailey lattice)]]
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<h5>개요</h5>
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Let <math>\{\alpha_r\}, \{\beta_r\}</math> be a Bailey pair relative to a and set
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<math>\alpha_0'=\alpha_0</math>, <math>\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}})</math><math>\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}</math>
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Then <math>\{\alpha_r'\}, \{\beta_r'\}</math>  is a Bailey pair relative to <math>aq^{-1}</math>
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<h5 style="line-height: 2em; margin: 0px;">comparison with Bailey chain</h5>
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* [[베일리 사슬(Bailey chain)]]
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* <math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math><br><math>\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r</math><br>
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*  This does not change the parameter <em>a</em> of the Bailey pair.<br>
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*  lattice construction changes this<br>
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<h5 style="line-height: 2em; margin: 0px;">corollary</h5>
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Let <math>\{\alpha_r\}, \{\beta_r\}</math> be the initial Bailey pair relative to a. Then the following is true :
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<math>\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{]}</math>
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(proof)
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apply Bailey chain construction k-i times  [[베일리 사슬(Bailey chain)]]
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At the (k-i)th step apply Bailey lattice
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apply Bailey chain construction i-1 times again.
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Then we get a Bailey pair
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<math>\{\alpha_r'\}, \{\beta_r'\}</math>  is a Bailey pair relative to <math>aq^{-1}</math>.
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If we use the defining relation of Bailey pair to <math>\{\alpha_r'\}, \{\beta_r'\}</math>,
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<math>\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}</math>
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and take the limit L\to\infty ■
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Example. Do this for k=5 and i=2
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<h5 style="line-height: 2em; margin: 0px;">응용</h5>
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* [[앤드류스-고든 항등식(Andrews-Gordon identity)]] 의 증명<br>
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*  initial Bailey pair<br><math>\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}}</math><br><math>\beta_{L}=\delta_{L,0}</math><br>
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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-1}\geq0}\frac{q^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^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>
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*  On LHS, we get<br><math>L=\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br>
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*  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>
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*  first part in the summation is<br>  <math>(-1)^{n}\sum_{n=1}^{\infty}{q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n+n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{n(2kn+2(k-i+1)+(n-1))/2</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2</math><br>
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*  secont part in the summation is<br><math>(-1)^{n}\sum_{n=1}^{\infty}{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}</math><br><math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}</math><br><math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}</math><br>
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*  by summing two parts, we get<br><math>R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math><br>
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*  Therefore we have proved the following are equal<br><math>\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math><br>
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*  You can use Jacobi triple product identity to get<br><math>\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}</math><br>
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<h5>역사</h5>
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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* [[수학사연표 (역사)|수학사연표]]
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<h5>메모</h5>
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* Math Overflow http://mathoverflow.net/search?q=
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<h5>관련된 항목들</h5>
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* [[앤드류스-고든 항등식(Andrews-Gordon identity)]]
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">수학용어번역</h5>
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*  단어사전<br>
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** http://translate.google.com/#en|ko|
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** http://ko.wiktionary.org/wiki/
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* 발음사전 http://www.forvo.com/search/
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
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* [http://www.kss.or.kr/pds/sec/dic.aspx 한국통계학회 통계학 용어 온라인 대조표]
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
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<h5>사전 형태의 자료</h5>
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* http://ko.wikipedia.org/wiki/
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* http://en.wikipedia.org/wiki/
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* [http://eom.springer.de/default.htm The Online Encyclopaedia of Mathematics]
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* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
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* [http://eqworld.ipmnet.ru/ The World of Mathematical Equations]
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<h5>리뷰논문, 에세이, 강의노트</h5>
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<h5>관련논문</h5>
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* Jeremy Lovejoy [http://www.liafa.jussieu.fr/%7Elovejoy/lattice.pdf A Bailey Lattice], Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516
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*  David Bressoud, The Bailey lattice, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.<br>
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*  A. Agarwal, G.E. Andrews, and D. Bressoud, The Bailey Lattice  J. Indian Math. Soc. 51 (1987), 57-73.<br>
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* http://www.jstor.org/action/doBasicSearch?Query=
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* http://www.ams.org/mathscinet
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* http://dx.doi.org/
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<h5>관련도서</h5>
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*  도서내검색<br>
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** http://books.google.com/books?q=
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** http://book.daum.net/search/contentSearch.do?query=

2011년 11월 12일 (토) 06:33 판

이 항목의 수학노트 원문주소

 

 

개요

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
  • 베일리 사슬(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  베일리 사슬(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

 

 

응용
  • 앤드류스-고든 항등식(Andrews-Gordon identity) 의 증명
  • 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
    \(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
  • On LHS, we get
    \(L=\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
  • 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
     \((-1)^{n}\sum_{n=1}^{\infty}{q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n+n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{n(2kn+2(k-i+1)+(n-1))/2\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2\)
  • secont part in the summation is
    \((-1)^{n}\sum_{n=1}^{\infty}{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}\)
    \(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}\)
    \(=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}\)
    \(=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}\)
  • by summing two parts, we get
    \(R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}\)
  • Therefore we have proved the following are equal
    \(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}\)
  • You can use Jacobi triple product identity to get
    \(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}\)

 

 

 

역사

 

 

 

메모

 

 

 

관련된 항목들

 

 

수학용어번역

 

 

 

사전 형태의 자료

 

 

리뷰논문, 에세이, 강의노트

 

 

 

관련논문
  • Jeremy Lovejoy A Bailey Lattice, Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516
  • David Bressoud, The Bailey lattice, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.
  • A. Agarwal, G.E. Andrews, and D. Bressoud, The Bailey Lattice  J. Indian Math. Soc. 51 (1987), 57-73.

 

 

관련도서
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