"Slater 92"의 두 판 사이의 차이

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<h5 style="line-height: 2em; margin: 0px;">Note</h5>
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==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]][[twisted Chebyshev polynomials and dilogarithm identities|twisted Chebyshev polynomials and dilogarithm identities]]
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*  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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*  Loxton & Lewin<math>x, -y, -z^{-1}</math>가 방정식 <math>x^3+3x^2-1=0</math>의 해라고 하자.<math>3L(x^3)-9L(x^2)-9L(x)+7L(1)=0</math><math>3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0</math><math>3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0</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%;">type of identity</h5>
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==type of identity==
  
 
* [[Slater list|Slater's list]]
 
* [[Slater list|Slater's list]]
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*  B(3)<br>
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*  B(3)
  
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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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<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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* 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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*  Specialize<br><math>x=q^2, y=-q, z\to\infty</math>.<br>
 
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*  Bailey pair<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>
 
  
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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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<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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==Bailey pair ==
  
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*  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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*  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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*  Specialize<br><math>a=q,c=-q,d=\infty</math><br>
 
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*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><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><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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==q-series identity==
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*  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>
 
  
 
<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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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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* [http://www.research.att.com/%7Enjas/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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<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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==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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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>.
  
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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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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=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1
 
a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1
  
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<math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math>
 
<math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math>
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<math>x^3+3x^2-1=0</math>
 
<math>x^3+3x^2-1=0</math>
  
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<math>x, -y, -z^{-1}</math>가 방정식 의 해 [http://www.wolframalpha.com/input/?i=x%5E3%2B3x%5E2-1%3D0 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0]
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<math>x, -y, -z^{-1}</math>가 방정식 의 해 [http://www.wolframalpha.com/input/?i=x%5E3%2B3x%5E2-1%3D0 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0]
  
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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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==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%;">books</h5>
 
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* [[2010년 books and articles]]<br>
 
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* http://gigapedia.info/1/
 
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* http://gigapedia.info/1/
 
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* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
 
  
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[[4909919|]]
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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%;">articles</h5>
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* http://www.ams.org/mathscinet
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[[분류:개인노트]]
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* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
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[[분류:math and physics]]
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* [http://arxiv.org/ ]http://arxiv.org/
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[[분류:migrate]]
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* http://pythagoras0.springnote.com/
 
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* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
 
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* http://dx.doi.org/
 
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2020년 12월 28일 (월) 04:02 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

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





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

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