"Slater 83"의 두 판 사이의 차이

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==Note==
  
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* [[Slater 86]]
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==type of identity==
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* [[Slater list|Slater's list]]
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* 
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==Bailey pair 1==
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*  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>
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*  Specialize<math>x=q^2, y=-q, z\to\infty</math>.
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*  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==
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*  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>
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*  Specialize<math>a=q,c=-q,d=\infty</math>
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*  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 ==
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*  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==
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<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>
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<math>(q)_{2n}=(q;q^2)_{n}(q^2;q^2)_{n}</math>
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* [[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>
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* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
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** http://www.research.att.com/~njas/sequences/?q=
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==Bethe type equation (cyclotomic equation)==
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Let ''''''<math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
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\prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>.
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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
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<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
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The equation  becomes <math>(1-x^{2})^{2}=x^{4}</math>.
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<math>x^2=\frac{1}{2}</math>
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<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==
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<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math>
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==articles==
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* [http://arxiv.org/abs/math-ph/0406042 Hypergeometric generating function of <math>L</math>-function, Slater's identities, and quantum invariant]
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** Kazuhiro Hikami, Anatol N. Kirillov, 2004
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[[분류:개인노트]]
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[[분류:math and physics]]
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[[분류:migrate]]

2020년 12월 28일 (월) 05:24 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

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