"Slater 83"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 3명의 중간 판 10개는 보이지 않습니다) | |||
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| − | + | ==Note== | |
| − | * [[Slater 86]] | + | * [[Slater 86]] |
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| − | + | ==type of identity== | |
* [[Slater list|Slater's list]] | * [[Slater list|Slater's list]] | ||
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| − | + | ==Bailey pair 1== | |
| − | * Use the folloing | + | * Use the folloing<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> |
| − | * Specialize | + | * Specialize<math>x=q^2, y=-q, z\to\infty</math>. |
| − | * Bailey pair | + | * Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math> |
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| − | + | ==Bailey pair 2== | |
| − | * Use the | + | * Use the following <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> |
| − | * Specialize | + | * Specialize<math>a=q,c=-q,d=\infty</math> |
| − | * Bailey pair | + | * Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\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}}</math> |
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| − | + | ==Bailey pair == | |
| − | * Bailey pairs | + | * Bailey pairs<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math> |
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| − | + | ==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> | ||
| 54번째 줄: | 54번째 줄: | ||
<math>(q)_{2n}=(q;q^2)_{n}(q^2;q^2)_{n}</math> | <math>(q)_{2n}=(q;q^2)_{n}(q^2;q^2)_{n}</math> | ||
| − | * [[Bailey pair and lemma|Bailey's lemma]] | + | * [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math> |
| − | * [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] | + | * [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] |
** http://www.research.att.com/~njas/sequences/?q= | ** http://www.research.att.com/~njas/sequences/?q= | ||
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| − | + | ==Bethe type equation (cyclotomic equation)== | |
| − | + | Let ''''''<math>\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}</math>. | \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>. | ||
| − | + | Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math> has a unique root <math>0<\mu<1</math>. We get | |
<math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | <math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | ||
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a=4,d_1=2.d_2=2,e=1 | a=4,d_1=2.d_2=2,e=1 | ||
| − | + | The equation becomes <math>(1-x^{2})^{2}=x^{4}</math>. | |
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| + | <math>x^2=\frac{1}{2}</math> | ||
<math>4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2</math> | <math>4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2</math> | ||
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| − | + | ==dilogarithm identity== | |
<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | <math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | ||
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| − | + | ==articles== | |
| − | + | * [http://arxiv.org/abs/math-ph/0406042 Hypergeometric generating function of <math>L</math>-function, Slater's identities, and quantum invariant] | |
| + | ** Kazuhiro Hikami, Anatol N. Kirillov, 2004 | ||
| − | + | [[분류:개인노트]] | |
| − | + | [[분류:math and physics]] | |
| − | + | [[분류:migrate]] | |
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2020년 12월 28일 (월) 04:24 기준 최신판
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
\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)
articles
- Hypergeometric generating function of \(L\)-function, Slater's identities, and quantum invariant
- Kazuhiro Hikami, Anatol N. Kirillov, 2004