Slater 31

Note

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

type of identity

Slater's list
G(1)

Bailey pair 1 (conjugate Bailey pair)

Use the following
\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{\delta_{n}}{(x/y)_{n}(x/z)_{n}}\)
Specialize
\(x=q, y=\to\infty, z\to\infty\).
Bailey pair
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)

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
\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)
\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)
\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)
\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)

q-series identity

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

\(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)

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

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

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

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