Slater 37

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imported>Pythagoras0님의 2012년 10월 28일 (일) 14:42 판 (찾아 바꾸기 – “</h5>” 문자열을 “==” 문자열로)
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Note==
  • not checked
   
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
    \(x=q^{3}, y=-q, z\to\infty\).
  • Bailey pair
    \(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)
    \(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\)
   
Bailey pair 2==
  • Use the following [Slater52-1] (4.2)
     
  • Specialize
    \(a=q^{2},d=q^2,e=q\)
  • Bailey pair
    \(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)
    \(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)
   
Bailey pair ==
  • Bailey pairs
     
    \(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)
    \(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\)
     
    \(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)
    \(\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)
   
q-series identity== \(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)
  • Bailey's lemma
    \(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
    \(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\) (오일러의 오각수정리(pentagonal number theorem) was used to verify this)
    \(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)
   
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=1,d=1,e=1 The equation  becomes \(1-x=x\). \(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)    
dilogarithm identity== \(L(\frac{1}{2})=\frac{1}{12}\pi^2\)    
related items==    
books==   4909919    
articles==
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