Slater 37

Note

• not checked

 

 

type of identity

• Slater's list
• I(17)

 

 

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)\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)

• The On-Line Encyclopedia of Integer Sequences
  • http://www.research.att.com/~njas/sequences/?q=

 

 

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

• asymptotic analysis of basic hypergeometric series

 

 

==articles==  

• Rogers-Ramanujan-Slater Type identities
  • McLaughlin, 2008
• Further identities of the Rogers-Ramanujan type
  • Slater, L. J. (1952),  Proceedings of the London Mathematical Society. Second Series 54: 147–167
• A New Proof of Rogers's Transformations of Infinite Series
  • Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475