베일리 격자(Bailey lattice)
이 항목의 수학노트 원문주소==
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
\(R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
\(=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}\)
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
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
수학용어번역==
- 단어사전
- 발음사전 http://www.forvo.com/search/
- 대한수학회 수학 학술 용어집
- 한국통계학회 통계학 용어 온라인 대조표
- 남·북한수학용어비교
- 대한수학회 수학용어한글화 게시판
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/
- The Online Encyclopaedia of Mathematics
- NIST Digital Library of Mathematical Functions
- The World of Mathematical Equations
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
\(\alpha_0'=\alpha_0\), \(\alpha_n'=(1-a)a^nq^{n^2-n}(\frac{\alpha_n}{1-aq^{2n}}-\frac{aq^{2n-2}\alpha_{n-1}}{1-aq^{2n-2}})\)\(\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}\)
- 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
\(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}\)
\(\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}\)
\(\beta_{L}=\delta_{L,0}\)
\(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
\(L=\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}\)
\(R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}\)
\(=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}\)
Now use the original Bailey pair,
\(\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}\)
\(\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}\)
\(R=\frac{1}{(q)_{\infty}}\left{[}1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right{]}\)
\(=\frac{1}{(q)_{\infty}}\left{[}1+(-1)^{n}\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}+{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\right{]}\)
\((-1)^{n}\sum_{n=1}^{\infty}{q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n+n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{n(2kn+2(k-i+1)+(n-1))/2\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2\)
\((-1)^{n}\sum_{n=1}^{\infty}{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}\)
\(=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}\)
\(=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}\)
\(=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}\)
\(R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}\)
\(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}\)
\(\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}\)
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들