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<h5 style="line-height: 2em; margin: 0px;">Note</h5>
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<h5 style="line-height: 2em; margin: 0px;">Note==
  
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* [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]]<br>[[twisted Chebyshev polynomials and dilogarithm identities|]]<br>
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* [[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>
 
*  Loxton & Lewin<br><math>x, -y, -z^{-1}</math>가 방정식 <math>x^3+3x^2-1=0</math>의 해라고 하자.<br><math>3L(x^3)-9L(x^2)-9L(x)+7L(1)=0</math><br><math>3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0</math><br><math>3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0</math><br>
 
*  Loxton & Lewin<br><math>x, -y, -z^{-1}</math>가 방정식 <math>x^3+3x^2-1=0</math>의 해라고 하자.<br><math>3L(x^3)-9L(x^2)-9L(x)+7L(1)=0</math><br><math>3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0</math><br><math>3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0</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%;">type of identity</h5>
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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==
  
 
* [[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</h5>
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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==
  
 
*  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</h5>
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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==
  
 
*  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 </h5>
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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 ==
  
 
*  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</h5>
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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==
  
 
<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)</h5>
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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)==
  
 
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</h5>
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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==
  
 
<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</h5>
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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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<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</h5>
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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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* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
 
* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
  
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[[4909919|]]
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[[4909919|4909919]]
  
 
 
 
 
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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</h5>
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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==
  
 
 
 
 

2012년 10월 28일 (일) 14:42 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==  
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