"베일리 격자(Bailey lattice)"의 두 판 사이의 차이

2013년 2월 18일 (월) 02:49 판

개요

$\{\alpha_r\}, \{\beta_r\}$를 a에 대한 베일리 쌍이라 하고, 다음을 정의하자.

\[\alpha_0'=\alpha_0\], \[\alpha_n'=(1-a)a^nq^{n^2-n}(\frac{\alpha_n}{1-aq^{2n}}-\frac{aq^{2n-2}\alpha_{n-1}}{1-aq^{2n-2}})\] \[\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}\]

$\{\alpha_r'\}, \{\beta_r'\}$ 는 $aq^{-1}$에 대한 베일리 쌍이 된다

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

\[ \begin{aligned} R&=\frac{1}{(q)_{\infty}}\left[1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right] \\ {}&=\frac{1}{(q)_{\infty}}\left[1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right] \end{aligned} \] Now use the original Bailey pair, \[\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}\] \[\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}\] \[ \begin{aligned} R&=\frac{1}{(q)_{\infty}}\left[1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right]\\ &=\frac{1}{(q)_{\infty}}\left[1+(-1)^{n}\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}+{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\right] \end{aligned} \]

first part in the summation is

\[ \begin{aligned} (-1)^{n}\sum_{n=1}^{\infty}q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}&=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n}q^{n(n-1)/2}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n+n(n-1)/2} \\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{n(2kn+2(k-i+1)+(n-1))/2} \\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2} \end{aligned} \]

second part in the summation is

\[ \begin{aligned} (-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}\\ {}&=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})} \\ {}&=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})} \\ {}&=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2} \end{aligned} \]

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

\[\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}\]

@@ 1번째 줄: / 1번째 줄: @@
-==이 항목의 수학노트 원문주소==
-* [[베일리 격자(Bailey lattice)]]
 ==개요==
-* <math>\{\alpha_r\}, \{\beta_r\}</math>를 <em>a</em>에 대한 베일리 쌍이라 하고, 다음을 정의하자.:<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><br>
+* <math>\{\alpha_r\}, \{\beta_r\}</math>를 <em>a</em>에 대한 베일리 쌍이라 하고, 다음을 정의하자.
-* <math>\{\alpha_r'\}, \{\beta_r'\}</math>  는 <math>aq^{-1}</math>에 대한 베일리 쌍이 된다<br>
+:<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>
+* <math>\{\alpha_r'\}, \{\beta_r'\}</math>  는 <math>aq^{-1}</math>에 대한 베일리 쌍이 된다
@@ 19번째 줄: / 11번째 줄: @@
 * [[베일리 사슬(Bailey chain)]]
-* <math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math>:<math>\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r</math><br>
+:<math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math>
-*  This does not change the parameter <em>a</em> of the Bailey pair.<br>
+:<math>\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r</math>
-*  lattice construction changes this<br>
+* This does not change the parameter <em>a</em> of the Bailey pair.
+* lattice construction changes this
@@ 30번째 줄: / 21번째 줄: @@
 Let <math>\{\alpha_r\}, \{\beta_r\}</math> be the initial Bailey pair relative to a. Then the following is true :
+:<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>
-<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>
 (proof)
@@ 48번째 줄: / 38번째 줄: @@
 If we use the defining relation of Bailey pair to <math>\{\alpha_r'\}, \{\beta_r'\}</math>,
+:<math>\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}</math>
-<math>\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}</math>
+and take the limit $L\to\infty$ ■
-and take the limit L\to\infty ■
@@ 65번째 줄: / 54번째 줄: @@
 * [[앤드류스-고든 항등식(Andrews-Gordon identity)]] 의 증명<br>
 *  initial Bailey pair:<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>:<math>\beta_{L}=\delta_{L,0}</math><br>
-*  In the corollay above, set a=q and replace i by i-1:<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>
+*  In the corollay above, set a=q and replace i by i-1
+:<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>
 *  On LHS, we get:<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>
-*  On RHS, we get:<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>:<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,:<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>:<math>\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}</math>:<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>:<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>
+*  On RHS, we get
-*  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>:<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>:<math>=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2</math><br>
+:<math>
-*  secont part in the summation is:<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>:<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>:<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>:<math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}</math>:<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>:<math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}</math>:<math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}</math>:<math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}</math><br>
+\begin{aligned}
-*  by summing two parts, we get:<math>R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math><br>
+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] \\
-*  Therefore we have proved the following are equal:<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>
+{}&=\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]
-*  You can use Jacobi triple product identity to get:<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>
+\end{aligned}
+</math>
+Now use the original Bailey pair,
+:<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>
+:<math>\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}</math>
+:<math>
+\begin{aligned}
+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]
+\end{aligned}
+</math>
+*  first part in the summation is
+:<math>
+\begin{aligned}
+(-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}
+\end{aligned}
+</math>
+*  second part in the summation is
+:<math>
+\begin{aligned}
+(-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}
+\end{aligned}
+</math>
+*  by summing two parts, we get
+:<math>R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math>
+*  Therefore we have proved the following are equal
+:<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>
+*  You can use Jacobi triple product identity to get
+:<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>

"베일리 격자(Bailey lattice)"의 두 판 사이의 차이

2013년 2월 18일 (월) 02:49 판

목차

개요

comparison with Bailey chain

corollary

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리뷰논문, 에세이, 강의노트

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