"Slater 83"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
imported>Pythagoras0 |
imported>Pythagoras0 잔글 (찾아 바꾸기 – “<h5 (.*)">” 문자열을 “==” 문자열로) |
||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==Note== | |
* [[Slater 86]]<br> | * [[Slater 86]]<br> | ||
| 7번째 줄: | 7번째 줄: | ||
| − | + | ==type of identity== | |
* [[Slater list|Slater's list]] | * [[Slater list|Slater's list]] | ||
| 16번째 줄: | 16번째 줄: | ||
| − | + | ==Bailey pair 1== | |
* Use the folloing<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>, <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math><br> | * Use the folloing<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>, <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math><br> | ||
| 28번째 줄: | 28번째 줄: | ||
| − | + | ==Bailey pair 2== | |
* Use the following <br><math>\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}}</math><br> | * Use the following <br><math>\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}}</math><br> | ||
| 38번째 줄: | 38번째 줄: | ||
| − | + | ==Bailey pair == | |
* Bailey pairs<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><br><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><br><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math><br> | * Bailey pairs<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><br><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><br><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math><br> | ||
| 48번째 줄: | 48번째 줄: | ||
| − | + | ==q-series identity== | |
<math>\sum_{n=0}^{\infty}\frac{q^{2n^{2}}}{ (q)_{2n}}=\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}(q^{8};q^{8})_{\infty}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}}{(q)_{\infty}}</math> | <math>\sum_{n=0}^{\infty}\frac{q^{2n^{2}}}{ (q)_{2n}}=\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}(q^{8};q^{8})_{\infty}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}}{(q)_{\infty}}</math> | ||
| 63번째 줄: | 63번째 줄: | ||
| − | + | ==Bethe type equation (cyclotomic equation)== | |
Let '''<br>'''<math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ | Let '''<br>'''<math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ | ||
| 86번째 줄: | 86번째 줄: | ||
| − | + | ==dilogarithm identity== | |
<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | <math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | ||
| 94번째 줄: | 94번째 줄: | ||
| − | + | ==related items== | |
* [[asymptotic analysis of basic hypergeometric series]]<br> | * [[asymptotic analysis of basic hypergeometric series]]<br> | ||
| 102번째 줄: | 102번째 줄: | ||
| − | + | ==books== | |
| 117번째 줄: | 117번째 줄: | ||
| − | + | ==articles== | |
* [http://arxiv.org/abs/math-ph/0406042 Hypergeometric generating function of $L$-function, Slater's identities, and quantum invariant]<br> | * [http://arxiv.org/abs/math-ph/0406042 Hypergeometric generating function of $L$-function, Slater's identities, and quantum invariant]<br> | ||
2012년 10월 28일 (일) 16:47 판
Note
type of identity
Bailey pair 1
- Use the folloing
\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\) - Specialize
\(x=q^2, y=-q, z\to\infty\). - Bailey pair
\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)
\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)
Bailey pair 2
- Use the following
\(\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\) - Specialize
\(a=q,c=-q,d=\infty\) - Bailey pair
\(\alpha_{0}=1\), \(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}\)
Bailey pair
- Bailey pairs
\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)
\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)
\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)
\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)
q-series identity
\(\sum_{n=0}^{\infty}\frac{q^{2n^{2}}}{ (q)_{2n}}=\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}(q^{8};q^{8})_{\infty}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}}{(q)_{\infty}}\)
\((q)_{2n}=(q;q^2)_{n}(q^2;q^2)_{n}\)
- Bailey's lemma
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)
Bethe type equation (cyclotomic equation)
Let
\(\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
\prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}\).
Then \(\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a\) has a unique root \(0<\mu<1\). We get
\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)
a=4,d_1=2.d_2=2,e=1
The equation becomes \((1-x^{2})^{2}=x^{4}\).
\(x^2=\frac{1}{2}\)
\(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)
dilogarithm identity
\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)
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=
articles
- Hypergeometric generating function of $L$-function, Slater's identities, and quantum invariant
- Kazuhiro Hikami, Anatol N. Kirillov, 2004
- http://www.ams.org/mathscinet
- [1]http://www.zentralblatt-math.org/zmath/en/
- [2]http://arxiv.org/
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html
- http://dx.doi.org/