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<h5 style="line-height: 2em; margin: 0px;">Note==
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==Note==
  
 
* [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]]<br>[[twisted Chebyshev polynomials and dilogarithm identities|twisted Chebyshev polynomials and dilogarithm identities]]<br>
 
* [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]]<br>[[twisted Chebyshev polynomials and dilogarithm identities|twisted Chebyshev polynomials and dilogarithm identities]]<br>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">type of identity==
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==type of identity==
  
 
* [[Slater list|Slater's list]]
 
* [[Slater list|Slater's list]]
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<h5 style="line-height: 2em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair 1==
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==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>
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<h5 style="line-height: 2em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair 2==
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==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>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair ==
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==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>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">q-series identity==
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==q-series identity==
  
 
<math>\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}</math>
 
<math>\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}</math>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bethe type equation (cyclotomic equation)==
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==Bethe type equation (cyclotomic equation)==
  
 
Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
 
Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">dilogarithm identity==
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==dilogarithm identity==
  
 
<math>L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)</math>
 
<math>L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)</math>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">related items==
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==related items==
  
 
 
 
 
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">books==
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==books==
  
 
 
 
 
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles==
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==articles==
  
 
 
 
 

2012년 10월 28일 (일) 17:47 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

\(\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}\)

 

 

 

 

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=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1

 

\(\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2\)

\(x^3+3x^2-1=0\)

\(x, -y, -z^{-1}\)가 방정식 의 해 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0

 

 

dilogarithm identity

\(L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)\)

 

 

related items

 

 

 

books

 

4909919

 

 

articles