"슬레이터 목록 (Slater's list)"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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==개요== | ==개요== | ||
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* [[자코비 삼중곱(Jacobi triple product)]] | * [[자코비 삼중곱(Jacobi triple product)]] | ||
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==관련논문== | ==관련논문== | ||
| − | + | * McLaughlin, Sills, Zimmer [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities], 2008 | |
* '''[Slater52]'''Slater, L. J.[http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>Proc. London Math. Soc.<br>1952s2-54: 147–167<br> | * '''[Slater52]'''Slater, L. J.[http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type]<br>Proc. London Math. Soc.<br>1952s2-54: 147–167<br> | ||
* '''[Slater51]'''Slater, L. J. [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]Proc. London Math. Soc. 1951 s2-53: 460-475<br> | * '''[Slater51]'''Slater, L. J. [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]Proc. London Math. Soc. 1951 s2-53: 460-475<br> | ||
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2013년 5월 13일 (월) 02:30 판
개요
주요 항등식
- [Slater51] (1.3)
- [Slater51] (2.1)
- [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
- [Slater51] (4.2)\[\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\]
- [Slater51] (4.3)
Group B
- [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
- B(1)
- B(2)
Group E
Group H
- [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
슬레이터 목록
- 슬레이터 1\[\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}})=\sum_{n=-\infty}^\infty(-1)^nq^{n(3n-1)/2}\]
- 슬레이터 2\[\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\]
- 슬레이터 8\[\sum_{n=0}^{\infty}\frac{(q^2;q^2)_{n}q^{n(n+1)/2}}{ (q)_{n}^2}=\frac{(-q)_{\infty}}{(q^2;q^4)_{\infty}}\]
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
리뷰논문, 에세이, 강의노트
관련논문
- McLaughlin, Sills, Zimmer Rogers-Ramanujan-Slater Type identities, 2008
- [Slater52]Slater, L. J.Further identities of the Rogers-Ramanujan type
Proc. London Math. Soc.
1952s2-54: 147–167 - [Slater51]Slater, L. J. A New Proof of Rogers's Transformations of Infinite SeriesProc. London Math. Soc. 1951 s2-53: 460-475