Slater 31
Note
- 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>
<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>
Bailey pair 2
- 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>
- Specialize<math>a=q,d=-q^{\frac{3}{2}},e=\infty</math>
- 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>
Bailey pair
- 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>
q-series identity
<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>
- 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>
Bethe type equation (cyclotomic equation)
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>.
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>
To apply the above result, we rewrite the equation in the suitable form.
<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)
<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>
a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1
The equation becomes <math>(1-x^4)(1-x^4)=x^4(1-x^2)</math>.
This is factorized into <math>(-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)</math>
So <math>\mu^2=\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</math>
<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>
<math>2L(1-\alpha^2)-2L(1-\alpha)=\frac{\pi^2}{21}</math>
<math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math>
dilogarithm identity
<math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> where <math>\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</math>