Slater 83

Note

Slater 86

type of identity

Slater's list

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

$\sum_{n=0}^{\infty}\frac{q^{2n^{2}}}{ (q)_{2n}}=\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}(q^{8};q^{8})_{\infty}(q^{6};q^{16})_{\infty}(q^{10};q^{16})_{\infty}}{(q)_{\infty}}$

$(q)_{2n}=(q;q^2)_{n}(q^2;q^2)_{n}$

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

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=4,d_1=2.d_2=2,e=1

The equation becomes $(1-x^{2})^{2}=x^{4}$.

$x^2=\frac{1}{2}$

$4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2$

dilogarithm identity

$L(\frac{1}{2})=\frac{1}{12}\pi^2$

related items

asymptotic analysis of basic hypergeometric series

books

[[4909919|]]

articles

Hypergeometric generating function of $L$-function, Slater's identities, and quantum invariant
- Kazuhiro Hikami, Anatol N. Kirillov, 2004
http://www.ams.org/mathscinet
[1]http://www.zentralblatt-math.org/zmath/en/
[2]http://arxiv.org/
http://pythagoras0.springnote.com/
http://math.berkeley.edu/~reb/papers/index.html
http://dx.doi.org/