"Slater 31"의 두 판 사이의 차이

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<h5>Note</h5>
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==Note==
  
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* [[Rogers-Selberg identities]]<br><math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math><br>  <br>  <br>
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* [[Rogers-Selberg identities]]<math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math>
  
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<h5>type of identity</h5>
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<math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math>
  
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* [[Slater list|Slater's list]]
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* G(1)<br>
 
  
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==Bailey pair 2==
  
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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 (conjugate Bailey pair)</h5>
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*  Use the following <math>\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}</math>
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*  Specialize<math>a=q,d=-q^{\frac{3}{2}},e=\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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* Use the following<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}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{\delta_{n}}{(x/y)_{n}(x/z)_{n}}</math><br>
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*  Specialize<br><math>x=q, y=\to\infty, z\to\infty</math>.<br>
 
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*  Bailey pair<br><math>\delta_n=q^{n^2}</math><br><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</math><br>
 
  
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==Bailey pair ==
  
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*  Bailey pairs<math>\delta_n=q^{n^2}</math><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</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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<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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*  Use the following <br><math>\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}</math><br>
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==q-series identity==
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*  Specialize<br><math>a=q,d=-q^{\frac{3}{2}},e=\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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*   <br> Bailey pairs<br><math>\delta_n=q^{n^2}</math><br><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</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>  <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^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math>
 
<math>\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</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^{n^2}}{(q)_{n}}</math><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math>
  
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
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<math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math>
 
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* [[Bailey pair and lemma|Bailey's lemma]]<br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><br><math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math><br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math><br>
 
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
 
 
** 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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To apply the above result, we rewrite the equation in the suitable form.
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<math>(-q;q)_{2n+1}=(-q)_{2n}(1+q^{2n+1})=\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})</math> ([[useful techniques in q-series]])
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<math>\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{(q^{2};q^{2})_{n}\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}(q;q^2)_{n}}{(q^2;q^4)_{n}(q^4;q^4)_{n}(1+q^{2n+1})}</math>
  
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a=1,d=1,e=1
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a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1
  
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The equation  becomes <math>1-x=x</math>.
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The equation  becomes <math>(1-x^4)(1-x^4)=x^4(1-x^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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This is factorized into <math>(-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)</math>
  
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So <math>\mu^2=\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</math>
  
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<math>4(\frac{1}{4}L(1-\alpha^2)+\frac{1}{4}L(1-\alpha^2)-\frac{1}{2}L(1-\alpha))=\frac{1}{14}(\frac{2}{3}\pi^2)</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%;">dilogarithm identity</h5>
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<math>2L(1-\alpha^2)-2L(1-\alpha)=\frac{\pi^2}{21}</math>
  
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<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math>
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<math>7L(\alpha^2)-7L(\alpha)+L(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%;">related items</h5>
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* [[asymptotic analysis of basic hypergeometric series]]<br>
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==dilogarithm identity==
  
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<math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> where <math>\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</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%;">books</h5>
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* [[2010년 books and articles]]<br>
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==related items==
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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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* [[asymptotic analysis of basic hypergeometric series]]
  
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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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*   <br>
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[[분류:개인노트]]
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* http://www.ams.org/mathscinet
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[[분류:math and physics]]
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* http://www.zentralblatt-math.org/zmath/en/
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[[분류:migrate]]
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* http://arxiv.org/
 
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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/
 

2020년 12월 28일 (월) 05:06 기준 최신판

Note

  • Rogers-Selberg identities\(C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)


\(C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)



Bailey pair 2

  • Use the following \(\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\)
  • Specialize\(a=q,d=-q^{\frac{3}{2}},e=\infty\)
  • 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\(\delta_n=q^{n^2}\)\(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)


q-series identity

\(\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)

  • Bailey's lemma\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{n^2}}{(q)_{n}}\)\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\)



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

\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)

To apply the above result, we rewrite the equation in the suitable form.

\((-q;q)_{2n+1}=(-q)_{2n}(1+q^{2n+1})=\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})\) (useful techniques in q-series)

\(\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{(q^{2};q^{2})_{n}\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}(q;q^2)_{n}}{(q^2;q^4)_{n}(q^4;q^4)_{n}(1+q^{2n+1})}\)

a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1

The equation becomes \((1-x^4)(1-x^4)=x^4(1-x^2)\).

This is factorized into \((-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)\)

So \(\mu^2=\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots\)

\(4(\frac{1}{4}L(1-\alpha^2)+\frac{1}{4}L(1-\alpha^2)-\frac{1}{2}L(1-\alpha))=\frac{1}{14}(\frac{2}{3}\pi^2)\)

\(2L(1-\alpha^2)-2L(1-\alpha)=\frac{\pi^2}{21}\)

\(7L(\alpha^2)-7L(\alpha)+L(1)=0\)




dilogarithm identity

\(7L(\alpha^2)-7L(\alpha)+L(1)=0\) where \(\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots\)




related items

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