Bailey pair and lemma

Choose the following (in the following, x=aq to get a Bailey pair relative to a)
\(u_{n}=\frac{1}{(q)_n}\) ,\(v_{n}=\frac{1}{(x)_n}\),
There is a conjugate Bailey pair
\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\)
\(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

If we apply Bailey lemma to the above conjugate pair, we get
\(\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}\)

Conjugate Bailey pair (\(x=q,y\to\infty, z\to\infty\))
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)
Bailey pair
\(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{1}{2}n}+q^{-\frac{1}{2}n})\)
\(\beta_n=\frac{1}{(q)_{n}}\)
we get the Rogers-Ramanujan identity(Slater 18)
\(\sum_{n=0}^{\infty}\frac{q^{n^2}}{ (q)_{n}}=\frac{(q^{3};q^{5})_{\infty}(q^{2};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}}\)

Multiple series Rogers-Ramanujan type identities.
- George E. Andrews, Pacific J. Math. Volume 114, Number 2 (1984), 267-283.
Special values of the dilogarithm function
- J. H. Loxton, 1984
Wilfrid Norman Bailey
- Slater, L. J. (1962), Journal of the London Mathematical Society. Second Series 37: 504–512
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
Identities of Rogers-Ramanujan type
- Bailey, 1944

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