"Bailey lattice"의 두 판 사이의 차이
| 63번째 줄: | 63번째 줄: | ||
* the proof of [[Andrews-Gordon identity]]<br> | * the proof of [[Andrews-Gordon identity]]<br> | ||
* 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> | * 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> | ||
| − | * In the corollay above, set a=q and replace i by i-1<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2 | + | * In the corollay above, set a=q and replace i by i-1<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^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> |
| − | + | * On LHS, we get<br><math>\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+n_i+\cdots+n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}</math><br> | |
| − | + | * On RHS, we get<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-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> <br> | |
2010년 10월 9일 (토) 10:36 판
introduction
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
- 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 (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
application
- the proof of Andrews-Gordon identity
- 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
\(\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\) - On LHS, we get
\(\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{q^{n_1^2+\cdots+n_{k}^2+n_i+\cdots+n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\) - On RHS, we get
\(\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
\(=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}\)
history
encyclopedia
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
[[4909919|]]
articles
- A Bailey Lattice
- Jeremy Lovejoy, Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516
- The Bailey lattice
- David Bressoud, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.
- David Bressoud, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.
- The Bailey Lattice
- A. Agarwal, G.E. Andrews, and D. Bressoud, J. Indian Math. Soc. 51 (1987), 57-73.
- A. Agarwal, G.E. Andrews, and D. Bressoud, J. Indian Math. Soc. 51 (1987), 57-73.
- http://www.ams.org/mathscinet
- [1]http://www.zentralblatt-math.org/zmath/en/
- http://arxiv.org/
- http://www.pdf-search.org/line-height: 2em;">
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html
- http://dx.doi.org/
question and answers(Math Overflow)
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experts on the field