A note on Bailey and WP-Bailey Pairs

We begin by recalling some standard notations and terminology. Let a, q be complex numbers with 0 < |q| < 1. Then the q − shifted factorial is defined by $\begin{matrix} {(a; q)}_{n} = (1 - a) (1 - aq) ... (1 - a q^{n - 1}) if n > 0, \\ {(a; q)}_{0} = 1 \end{matrix}$ \matrix{{{{(a;q)}_n} = (1 - a)(1 - aq)...(1 - a{q^{n - 1}}) {\kern 1pt} {\rm if}{\kern 1pt} n > 0,} \cr {{{(a;q)}_0} = 1}} and ${(a; q)}_{\infty} = \prod_{r = 0}^{\infty} (1 - a q^{r}) .$ {(a;q)_\infty} = \prod\limits_{r = 0}^\infty (1 - a{q^r}).

For the sake of brevity, we often write ${(a_{1}; q)}_{n} {(a_{2}; q)}_{n} ... (a_{r}; q)_{n} = (a_{1}, a_{2}, a_{3}, ..., a_{r}; q)_{n} .$ {({a_1};q)_n}{({a_2};q)_n}...({a_r};q{)_n} = ({a_1},{a_2},{a_3},...,{a_r};q{)_n}.

The basic hypergeometric series is defined by $_{r} Φ_{s} = [\begin{array}{l} a_{1}, a_{2}, ..., a_{r}; q; z \\ b_{1}, b_{2}, ..., b_{s} \end{array}] = \sum_{n = 0}^{\infty} \frac{{(a_{1}, a_{2}, ..., a_{r}; q)}_{n} z^{n}}{{(q, b_{1}, b_{2}, ..., b_{s}; q)}_{n}} {{(- 1)}^{n} q^{n (n - 1) / 2}}^{1 + s - r} .$ {\,_r}{\Phi_s} = \left[ {\matrix{{{a_1},{a_2},...,{a_r};q;z} \hfill \cr {{b_1},{b_2},...,{b_s}} \hfill \cr}} \right] = \sum\limits_{n = 0}^\infty {{{{({a_1},{a_2},...,{a_r};q)}_n}{z^n}} \over {{{(q,{b_1},{b_2},...,{b_s};q)}_n}}}{\left\{{{{(- 1)}^n}{q^{n(n - 1)/2}}} \right\}^{1 + s - r}}.

In an attempt to clarify Rogers second proof [3] of the Rogers-Ramanujan identities, Bailey [1] made the following simple but very useful observation,

If (1.1) $β_{n} = \sum_{r = 0}^{n} α_{r} u_{n - r} v_{n + r}$ {\beta_n} = \sum\limits_{r = 0}^n {\alpha_r}{u_{n - r}}{v_{n + r}} and (1.2) $\begin{array}{l} γ_{n} & = \sum_{r = n}^{\infty} δ_{r} u_{r - n} v_{r + n} \\ = \sum_{r = 0}^{\infty} δ_{r + n} u_{r} v_{r + 2 n} \end{array}$ \matrix{{{\gamma_n}} \hfill & {= \sum\limits_{r = n}^\infty {\delta_r}{u_{r - n}}{v_{r + n}}} \hfill \cr {} \hfill & {= \sum\limits_{r = 0}^\infty {\delta_{r + n}}{u_r}{v_{r + 2n}}} \hfill} then (1.3) $\sum_{n = 0}^{\infty} α_{n} γ_{n} = \sum_{n = 0}^{\infty} β_{n} δ_{n} .$ \sum\limits_{n = 0}^\infty {\alpha_n}{\gamma_n} = \sum\limits_{n = 0}^\infty {\beta_n}{\delta_n}.

The proof is straightforword and merely requires an interchange of sums. Of course, in the above transform, suitable convergence conditions need to be imposed to make the definition of γ_n and interchange of sums meaningful.

In application of the transform, Bailey chose $u_{r} = \frac{1}{{(q; q)}_{r}}$ {u_r} = {1 \over {{{(q;q)}_r}}} , $v_{r} = \frac{1}{{(aq; q)}_{r}}$ {v_r} = {1 \over {{{(aq;q)}_r}}} and with this choice equations (1.1) and (1.2) became (1.4) $β_{n} = \sum_{r = 0}^{n} \frac{α_{r}}{{(q; q)}_{n - r} {(aq; q)}_{n + r}}$ {\beta_n} = \sum\limits_{r = 0}^n {{{\alpha_r}} \over {{{(q;q)}_{n - r}}{{(aq;q)}_{n + r}}}} and (1.5) $γ_{n} = \sum_{r = 0}^{\infty} \frac{δ_{r + n}}{{(q; q)}_{r} {(aq; q)}_{r + 2 n}}$ {\gamma_n} = \sum\limits_{r = 0}^\infty {{{\delta_{r + n}}} \over {{{(q;q)}_r}{{(aq;q)}_{r + 2n}}}} respectively.

A pair of sequence 〈α_n, β_n〉 that satisfies (1.4) is called a Bailey pair relative to the parameter a. Similarly, the pair of sequence 〈γ_n, δ_n〉 which satisfies (1.5) is called conjugate Bailey pair relative to a. For these Bailey and conjugate Bailey pairs we have (1.6) $\sum_{n = 0}^{\infty} α_{n} γ_{n} = \sum_{n = 0}^{\infty} β_{n} δ_{n},$ \sum\limits_{n = 0}^\infty {\alpha_n}{\gamma_n} = \sum\limits_{n = 0}^\infty {\beta_n}{\delta_n}, provided series involved are convergent.

Theorems Involving Bailey Pairs

(i) Choosing $δ_{r} = (ρ_{1}, ρ_{2}; q)_{r} {(\frac{aq}{ρ_{1} ρ_{2}})}^{r}$ {\delta_r} = ({\rho_1},{\rho_2};q{)_r}{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)^r} in (1.5) and using the summation formula [2, App. II(II.8) p. 236] we have, (2.1) $γ_{n} = \frac{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty} {(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty} {(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} .$ {\gamma_n} = {{{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}{{({\rho_1},{\rho_2};q)}_n}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}.

Putting these values of γ_n and δ_n in (1.6) we obtain the following theorem.

Theorem 1

If 〈α_n, β_n〉 is a Bailey pair satisfying (1.4) then(2.2) $\sum_{n = 0}^{\infty} \frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} α_{n} = \frac{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty}} \sum_{n = 0}^{\infty} {(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n} β_{n} .$ \sum\limits_{n = 0}^\infty {{{{({\rho_1},{\rho_2};q)}_n}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\alpha_n} = {{{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}}}\sum\limits_{n = 0}^\infty {({\rho_1},{\rho_2};q)_n}{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)^n}{\beta_n}.

Taking ρ₁, ρ₂ → ∞ in (2.2) we obtain the following theorem,

Theorem 2

If 〈α_n, β_n〉 is a Bailey pair then(2.3) $\sum_{n = 0}^{\infty} q^{n^{2}} a^{n} α_{n} = (aq; q)_{\infty} \sum_{n = 0}^{\infty} q^{n^{2}} a^{n} β_{n} .$ \sum\limits_{n = 0}^\infty {q^{{n^2}}}{a^n}{\alpha_n} = (aq;q{)_\infty}\sum\limits_{n = 0}^\infty {q^{{n^2}}}{a^n}{\beta_n}.

(ii) Choosing $δ_{r} = (ρ_{1}, ρ_{2}; q)_{r} {(\frac{a}{ρ_{1} ρ_{2}})}^{r}$ {\delta_r} = ({\rho_1},{\rho_2};q{)_r}{\left({{a \over {{\rho_1}{\rho_2}}}} \right)^r} in (1.5) and using the summation formula [4, (1.4) p. 771] we have, (2.4) $γ_{n} = \frac{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty}}{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty}} {\frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{a}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} + \frac{(1 - ρ_{1}) (1 - ρ_{2})}{(ρ_{1} ρ_{2} - a)} \frac{{(ρ_{1} q, ρ_{2} q; q)}_{n} {(\frac{a}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}}} .$ {\gamma_n} = {{{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}} \over {{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}}}\left\{{{{{{({\rho_1},{\rho_2};q)}_n}{{\left({{a \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}} + {{(1 - {\rho_1})(1 - {\rho_2})} \over {({\rho_1}{\rho_2} - a)}}{{{{({\rho_1}q,{\rho_2}q;q)}_n}{{\left({{a \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}} \right\}.

Putting these values of γ_n and δ_n in (1.6) we have following theorem.

Theorem 3

If 〈α_n, β_n〉 is a Bailey pair then(2.5) $\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{a}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} α_{n} + \frac{(1 - ρ_{1}) (1 - ρ_{2}) a}{(ρ_{1} ρ_{2} - a)} \sum_{n = 0}^{\infty} \frac{{(ρ_{1} q, ρ_{2} q; q)}_{n} {(\frac{a}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} α_{n} \\ = \frac{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} \sum_{n = 0}^{\infty} {(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{a}{ρ_{1} ρ_{2}})}^{n} β_{n} . \end{matrix}$ \matrix{{\sum\limits_{n = 0}^\infty {{{{\left({{\rho_1},{\rho_2};q} \right)}_n}{{\left({{a \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\alpha_n} + {{(1 - {\rho_1})(1 - {\rho_2})a} \over {({\rho_1}{\rho_2} - a)}}\sum\limits_{n = 0}^\infty {{{{({\rho_1}q,{\rho_2}q;q)}_n}{{\left({{a \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\alpha_n}} \cr {\quad \quad = {{{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}\sum\limits_{n = 0}^\infty {{({\rho_1},{\rho_2};q)}_n}{{\left({{a \over {{\rho_1}{\rho_2}}}} \right)}^n}{\beta_n}.}}

Taking ρ₁, ρ₂ → ∞ in (2.5) we have following theorem,

Theorem 4

If 〈α_n, β_n〉 is a Bailey pair then(2.6) $\sum_{n = 0}^{\infty} q^{n (n - 1)} a^{n} α_{n} + a \sum_{n = 0}^{\infty} q^{n (n + 1)} a^{n} α_{n} = (aq; q)_{\infty} \sum_{n = 0}^{\infty} q^{n (n - 1)} a^{n} β_{n}$ \sum\limits_{n = 0}^\infty {q^{n(n - 1)}}{a^n}{\alpha_n} + a\sum\limits_{n = 0}^\infty {q^{n(n + 1)}}{a^n}{\alpha_n} = (aq;q{)_\infty}\sum\limits_{n = 0}^\infty {q^{n(n - 1)}}{a^n}{\beta_n}

(iii) Taking $δ_{r} = \frac{{(ρ_{1}, ρ_{2}; q)}_{r} {(\frac{aq}{ρ_{1} ρ_{2}}; q)}_{N - r}}{{(q; q)}_{N - r}} {(\frac{aq}{ρ_{1} ρ_{2}})}^{r}$ {\delta_r} = {{{{({\rho_1},{\rho_2};q)}_r}{{\left({{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_{N - r}}} \over {{{(q;q)}_{N - r}}}}{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)^r} in (1.5) and using the summation formula [2, App. II (II. 12), p. 237] we obtain (2.7) $γ_{n} = \frac{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{N} {(ρ_{1}, ρ_{2}; q)}_{n} {(q^{- N}; q)}_{n} {(- \frac{a q^{1 + N}}{ρ_{1} ρ_{2}})}^{n}}{{(q, aq; q)}_{N} q^{n (n - 1) / 2} {(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, a q^{1 + N}; q)}_{n}} .$ {\gamma_n} = {{{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_N}{{({\rho_1},{\rho_2};q)}_n}{{({q^{- N}};q)}_n}{{\left({- {{a{q^{1 + N}}} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{(q,aq;q)}_N}{q^{n(n - 1)/2}}{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},a{q^{1 + N}};q} \right)}_n}}}.

Putting these values of γ_n and δ_n in (1.6) we have,

Theorem 5

If 〈α_n, β_n〉 is a Bailey pair then(2.8) $\sum_{n = 0}^{N} \frac{{(ρ_{1}, ρ_{2}, q^{- N}; q)}_{n} {(- \frac{a q^{1 + N}}{ρ_{1} ρ_{2}})}^{n} α_{n}}{q^{n (n - 1) / 2} {(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, a q^{1 + N}; q)}_{n}} = \frac{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{N}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{N}} \sum_{n = 0}^{N} \frac{{(ρ_{1}, ρ_{2}, q^{- N}; q)}_{n} q^{n}}{{(\frac{ρ_{1} ρ_{2}}{a} q^{- N}; q)}_{n}} β_{n} .$ \sum\limits_{n = 0}^N {{{{({\rho_1},{\rho_2},{q^{- N}};q)}_n}{{\left({- {{a{q^{1 + N}}} \over {{\rho_1}{\rho_2}}}} \right)}^n}{\alpha_n}} \over {{q^{n(n - 1)/2}}{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},a{q^{1 + N}};q} \right)}_n}}} = {{{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_N}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_N}}}\sum\limits_{n = 0}^N {{{{({\rho_1},{\rho_2},{q^{- N}};q)}_n}{q^n}} \over {{{\left({{{{\rho_1}{\rho_2}} \over a}{q^{- N}};q} \right)}_n}}}{\beta_n}.

For N → ∞, (2.8) yields (2.2).

WP-Bailey Pairs and Related Theorems

In this section we have established certain theorems involving WP-Bailey pairs.

A WP-Bailey is a pair of sequences {α_n(a,k), β_n(a,k)} satisfying α₀(a,k) = β₀(a,k) = 1 and (3.1) $\begin{array}{l} β_{n} (a, k) & = \sum_{r = 0}^{n} \frac{{(\frac{k}{a}; q)}_{n - r} {(k; q)}_{n + r}}{{(q; q)}_{n - r} {(aq; q)}_{n + r}} α_{r} (a, k) \\ = \frac{{(\frac{k}{a}, k; q)}_{n}}{{(q, aq; q)}_{n}} \sum_{r = 0}^{n} \frac{{(q^{- n}, k q^{n}; q)}_{r}}{{(\frac{a q^{1 - n}}{k}, a q^{1 + n}; q)}_{r}} {(\frac{aq}{k})}^{r} α_{r} (a, k) . \end{array}$ \matrix{{{\beta_n}(a,k)} \hfill & {= \sum\limits_{r = 0}^n {{{{\left({{k \over a};q} \right)}_{n - r}}{{(k;q)}_{n + r}}} \over {{{(q;q)}_{n - r}}{{(aq;q)}_{n + r}}}}{\alpha_r}(a,k)} \hfill \cr {} \hfill & {= {{{{\left({{k \over a},k;q} \right)}_n}} \over {{{(q,aq;q)}_n}}}\sum\limits_{r = 0}^n {{{{({q^{- n}},k{q^n};q)}_r}} \over {{{\left({{{a{q^{1 - n}}} \over k},a{q^{1 + n}};q} \right)}_r}}}{{\left({{{aq} \over k}} \right)}^r}{\alpha_r}(a,k).} \hfill}

The corresponding WP-conjugate Bailey pair 〈γ_n(a,k), δ_n(a,k)〉 is given by, (3.2) $\begin{array}{l} γ_{n} (a, k) & = \sum_{r = 0}^{\infty} \frac{{(\frac{k}{a}; q)}_{r} {(k; q)}_{r + 2 n}}{{(q; q)}_{r} {(aq; q)}_{r + 2 n}} δ_{r + n} (a, k) \\ = \frac{{(k; q)}_{2 n}}{{(aq; q)}_{2 n}} \sum_{r = 0}^{\infty} \frac{{(\frac{k}{a}; q)}_{r} {(k q^{2 n}; q)}_{r}}{{(q; q)}_{r} {(a q^{1 + 2 n}; q)}_{r}} δ_{r + n} (a, k) . \end{array}$ \matrix{{{\gamma_n}(a,k)} \hfill & {= \sum\limits_{r = 0}^\infty {{{{\left({{k \over a};q} \right)}_r}{{(k;q)}_{r + 2n}}} \over {{{(q;q)}_r}{{(aq;q)}_{r + 2n}}}}{\delta_{r + n}}(a,k)} \hfill \cr {} \hfill & {= {{{{\left({k;q} \right)}_{2n}}} \over {{{(aq;q)}_{2n}}}}\sum\limits_{r = 0}^\infty {{{{\left({{k \over a};q} \right)}_r}{{\left({k{q^{2n}};q} \right)}_r}} \over {{{\left({q;q} \right)}_r}{{(a{q^{1 + 2n}};q)}_r}}}{\delta_{r + n}}(a,k).} \hfill}

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair and 〈γ_n(a,k), δ_n(a,k)〉 is a WP-Conjugate Bailey pair then Bailey lemma gives, (3.3) $\sum_{n = 0}^{\infty} α_{n} (a, k) γ_{n} (a, k) = \sum_{n = 0}^{\infty} β_{n} (a, k) δ_{n} (a, k),$ \sum\limits_{n = 0}^\infty {\alpha_n}(a,k){\gamma_n}(a,k) = \sum\limits_{n = 0}^\infty {\beta_n}(a,k){\delta_n}(a,k), provided series involved in (3.2) and (3.3) are convergent.

(i) Choosing $δ_{r} (a, k) = {(\frac{a^{2} q}{k^{2}})}^{r}$ {\delta_r}(a,k) = {\left({{{{a^2}q} \over {{k^2}}}} \right)^r} in (3.2) and using the summation formula [2, App. II(II.8), p. 236] we find, (3.4) $γ_{n} (a, k) = \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} \frac{{(k; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} {(\frac{a^{2} q}{k^{2}})}^{n} .$ {\gamma_n}(a,k) = {{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}{{{{(k;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}{\left({{{{a^2}q} \over {{k^2}}}} \right)^n}.

Putting these values of 〈γ_n(a,k), δ_n(a,k)〉 in (3.3) we have following theorem.

Theorem 6

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.5) $\frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} \sum_{n = 0}^{\infty} \frac{{(k; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} {(\frac{a^{2} q}{k^{2}})}^{n} α_{n} (a, k) = \sum_{n = 0}^{\infty} {(\frac{a^{2} q}{k^{2}})}^{n} β_{n} (a, k) .$ {{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}\sum\limits_{n = 0}^\infty {{{{(k;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}{\left({{{{a^2}q} \over {{k^2}}}} \right)^n}{\alpha_n}(a,k) = \sum\limits_{n = 0}^\infty {\left({{{{a^2}q} \over {{k^2}}}} \right)^n}{\beta_n}(a,k).

(ii) Putting $δ_{r} (a, k) = {(\frac{a^{2}}{k^{2}})}^{r}$ {\delta_r}(a,k) = {\left({{{{a^2}} \over {{k^2}}}} \right)^r} in (3.2) and making use of the summation formula [4, (1.4), p.771] we get, (3.6) $γ_{n} (a, k) = \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} \frac{{(k; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} (\frac{k}{k + a}) (1 + a q^{2 n}) {(\frac{a^{2}}{k^{2}})}^{n} .$ {\gamma_n}(a,k) = {{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}{{{{(k;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}\left({{k \over {k + a}}} \right)(1 + a{q^{2n}}){\left({{{{a^2}} \over {{k^2}}}} \right)^n}.

Substituting these values 〈γ_n(a,k), δ_n(a,k)〉 in (3.3) we get the following theorem.

Theorem 7

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.7) $\begin{array}{l} \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} (\frac{k}{k + a}) \sum_{n = 0}^{\infty} \frac{{(k; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} α_{n} (a, k) {(\frac{a^{2}}{k^{2}})}^{n} \\ + \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} (\frac{ka}{k + a}) \sum_{n = 0}^{\infty} \frac{{(k; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} {(\frac{a^{2} q^{2}}{k^{2}})}^{n} α_{n} (a, k) \\ = \sum_{n = 0}^{\infty} β_{n} (a, k) {(\frac{a^{2}}{k^{2}})}^{n} . \end{array}$ \matrix{{{{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}\left({{k \over {k + a}}} \right)\sum\limits_{n = 0}^\infty {{{{(k;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}{\alpha_n}(a,k){{\left({{{{a^2}} \over {{k^2}}}} \right)}^n}} \hfill \cr {+ {{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}\left({{{ka} \over {k + a}}} \right)\sum\limits_{n = 0}^\infty {{{{(k;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}{{\left({{{{a^2}{q^2}} \over {{k^2}}}} \right)}^n}{\alpha_n}(a,k)} \hfill \cr {= \sum\limits_{n = 0}^\infty {\beta_n}(a,k){{\left({{{{a^2}} \over {{k^2}}}} \right)}^n}.} \hfill}

(iii) Taking $δ_{r} = \frac{{(\frac{a^{2} q}{k^{2}}; q)}_{N - r} {(\frac{a^{2} q}{k^{2}})}^{r}}{{(q; q)}_{N - r}}$ {\delta_r} = {{{{\left({{{{a^2}q} \over {{k^2}}};q} \right)}_{N - r}}{{\left({{{{a^2}q} \over {{k^2}}}} \right)}^r}} \over {{{(q;q)}_{N - r}}}} in (3.2) we find, (3.8) $γ_{n} (a, k) = \frac{{(k; q)}_{2 n}}{{(aq; q)}_{2 n}} \frac{{(\frac{a^{2} q}{k^{2}}; q)}_{N - n}}{{(q; q)}_{N - n}} {(\frac{a^{2} q}{k^{2}})}^{n}_{3} Φ_{2} [\begin{array}{l} k q^{2 n}, \frac{k}{a}, q^{- (N - n)}; q; q \\ a q^{1 + 2 n}, \frac{k^{2}}{a^{2}} q^{- (N - n)} \end{array}] .$ {\gamma_n}(a,k) = {{{{(k;q)}_{2n}}} \over {{{(aq;q)}_{2n}}}}{{{{\left({{{{a^2}q} \over {{k^2}}};q} \right)}_{N - n}}} \over {{{(q;q)}_{N - n}}}}{\left({{{{a^2}q} \over {{k^2}}}} \right)^n}{_3}{\Phi_2}\left[ {\matrix{{k{q^{2n}},{k \over a},{q^{- (N - n)}};q;q} \hfill \cr {a{q^{1 + 2n}},{{{k^2}} \over {{a^2}}}{q^{- (N - n)}}} \hfill \cr}} \right].

Now summing ₃Φ₂ series in (3.8) by making use of [2, App. II(II.12), p. 237] we get, (3.9) $γ_{n} (a, k) = \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{N}}{{(q, aq; q)}_{N}} \frac{{(aq; q)}_{2 n}}{{(\frac{a^{2} q}{k}; q)}_{2 n}} \frac{{(q^{- N}, \frac{a^{2} q^{1 + N}}{k}; q)}_{n} {(\frac{aq}{k})}^{n}}{{(a q^{1 + N}, \frac{k}{a} q^{- N}; q)}_{n}} .$ {\gamma_n}(a,k) = {{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_N}} \over {{{\left({q,aq;q} \right)}_N}}}{{{{(aq;q)}_{2n}}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}}}{{{{\left({{q^{- N}},{{{a^2}{q^{1 + N}}} \over k};q} \right)}_n}{{\left({{{aq} \over k}} \right)}^n}} \over {{{\left({a{q^{1 + N}},{k \over a}{q^{- N}};q} \right)}_n}}}.

Putting these values of 〈γ_n(a,k), δ_n(a,k)〉 in (3.3) we have following theorem.

Theorem 8

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.10) $\begin{array}{l} \sum_{n = 0}^{N} \frac{{(k; q)}_{2 n} {(q^{- N}, \frac{a^{2}}{k} q^{1 + N}; q)}_{n} {(\frac{aq}{k})}^{n}}{{(\frac{a^{2} q}{k}; q)}_{2 n} {(\frac{k q^{- N}}{a}, a q^{1 + N}; q)}_{n}} α_{n} (a, k) \\ = \frac{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{N}}{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{N}} \sum_{n = 0}^{N} \frac{{(q^{- N}; q)}_{n} q^{n}}{{(\frac{k^{2} q^{- N}}{a^{2}}; q)}_{n}} β_{n} (a, k) . \end{array}$ \matrix{{\,\,\,\sum\limits_{n = 0}^N {{{{(k;q)}_{2n}}{{\left({{q^{- N}},{{{a^2}} \over k}{q^{1 + N}};q} \right)}_n}{{\left({{{aq} \over k}} \right)}^n}} \over {{{\left({{{{a^2}q} \over k};q} \right)}_{2n}}{{\left({{{k{q^{- N}}} \over a},a{q^{1 + N}};q} \right)}_n}}}{\alpha_n}(a,k)} \hfill \cr {= {{{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_N}} \over {{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_N}}}\sum\limits_{n = 0}^N {{{{({q^{- N}};q)}_n}{q^n}} \over {{{\left({{{{k^2}{q^{- N}}} \over {{a^2}}};q} \right)}_n}}}{\beta_n}(a,k).} \hfill}

(iv) Choosing $δ_{r} (a, k) = \frac{(1 - k q^{2 r})}{(1 - k)} \frac{{(ρ_{1}, ρ_{2}; q)}_{r} {(\frac{aq}{ρ_{1} ρ_{2}})}^{r}}{{(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{r}}$ {\delta_r}(a,k) = {{(1 - k{q^{2r}})} \over {(1 - k)}}{{{{({\rho_1},{\rho_2};q)}_r}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^r}} \over {{{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}};q} \right)}_r}}} in (3.2) we find, (3.11) $\begin{array}{l} γ_{n} (a, k) & = \frac{{(k; q)}_{2 n} {(ρ_{1}, ρ_{2}; q)}_{n} (1 - k q^{2 n}) {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(aq; q)}_{2 n} {(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{n} (1 - k)} \\ ._{6} Φ_{5} [\begin{array}{l} k q^{2 n}, q^{n + 1} \sqrt{k}, - q^{n + 1} \sqrt{k}, ρ_{1} q^{n}, ρ_{2} q^{n}, \frac{k}{a}; q; \frac{aq}{ρ_{1} ρ_{2}} \\ q^{n} \sqrt{k}, - q^{n} \sqrt{k}, \frac{k q^{n + 1}}{ρ_{1}}, \frac{k q^{n + 1}}{ρ_{2}}, a q^{1 + 2 n} \end{array}] . \end{array}$ \matrix{{{\gamma _n}(a,k)} \hfill & {= {{{{(k;q)}_{2n}}{{({\rho _1},{\rho _2};q)}_n}(1 - k{q^{2n}}){{\left({{{aq} \over {{\rho _1}{\rho _2}}}} \right)}^n}} \over {{{(aq;q)}_{2n}}{{\left({{{kq} \over {{\rho _1}}},{{kq} \over {{\rho _2}}};q} \right)}_n}(1 - k)}}} \hfill \cr {} \hfill & {{{.}_6}{\Phi _5}\left[ {\matrix{{k{q^{2n}},{q^{n + 1}}\sqrt k, - {q^{n + 1}}\sqrt k,{\rho _1}{q^n},{\rho _2}{q^n},{k \over a};q;{{aq} \over {{\rho _1}{\rho _2}}}} \hfill \cr {{q^n}\sqrt k, - {q^n}\sqrt k,{{k{q^{n + 1}}} \over {{\rho _1}}},{{k{q^{n + 1}}} \over {{\rho _2}}},a{q^{1 + 2n}}} \hfill \cr}} \right].} \hfill}

Summing the ₆Φ₅ series in (3.11) by using [2, App. II (II 20), p. 238] we find, (3.12) $γ_{n} (a, k) = \frac{{(kq, \frac{kq}{ρ_{1} ρ_{2}}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty} {(ρ_{1}, ρ_{2}; q)}_{n}}{{(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty} {(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n} .$ {\gamma_n}(a,k) = {{{{\left({kq,{{kq} \over {{\rho_1}{\rho_2}}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}{{({\rho_1},{\rho_2};q)}_n}} \over {{{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)^n}.

Putting these values of 〈γ_n(a,k), δ_n(a,k)〉 in (3.3) we get following theorem.

Theorem 9

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.13) $\begin{array}{r} \frac{{(kq, \frac{kq}{ρ_{1} ρ_{2}}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty}}{{(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty}} \sum_{n = 0}^{\infty} \frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} α_{n} (a, k) \\ = \sum_{n = 0}^{\infty} (\frac{1 - k q^{2 n}}{1 - k}) \frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{n}} β_{n} (a, k) . \end{array}$ \matrix{\hfill {{{{{\left({kq,{{kq} \over {{\rho_1}{\rho_2}}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}} \over {{{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}}}\sum\limits_{n = 0}^\infty {{{{({\rho_1},{\rho_2};q)}_n}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\alpha_n}(a,k)} \cr \hfill {= \sum\limits_{n = 0}^\infty \left({{{1 - k{q^{2n}}} \over {1 - k}}} \right){{{{({\rho_1},{\rho_2};q)}_n}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}};q} \right)}_n}}}{\beta_n}(a,k).}}

As ρ₁, ρ₂ → ∞, (3.13) yields following theorem.

Theorem 10

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.14) $\frac{{(kq; q)}_{\infty}}{{(aq; q)}_{\infty}} \sum_{n = 0}^{\infty} q^{n^{2}} a^{n} α_{n} (a, k) = \sum_{n = 0}^{\infty} (\frac{1 - k q^{2 n}}{1 - k}) q^{n^{2}} a^{n} β_{n} (a, k) .$ {{{{(kq;q)}_\infty}} \over {{{(aq;q)}_\infty}}}\sum\limits_{n = 0}^\infty {q^{{n^2}}}{a^n}{\alpha_n}(a,k) = \sum\limits_{n = 0}^\infty \left({{{1 - k{q^{2n}}} \over {1 - k}}} \right){q^{{n^2}}}{a^n}{\beta_n}(a,k).

(v) Taking $δ_{r} (a, k) = \frac{{(ρ_{1}, ρ_{2}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N}; q)}_{r} {(\frac{1}{k}; q)}_{- N - r}}{{(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, \frac{ρ_{1} ρ_{2}}{a} q^{- N}; q)}_{r} {(q; q)}_{N - r}} {(\frac{q^{- 2 N}}{k})}^{r}$ {\delta_r}(a,k) = {{{{\left({{\rho_1},{\rho_2},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N}};q} \right)}_r}{{\left({{1 \over k};q} \right)}_{- N - r}}} \over {{{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},{{{\rho_1}{\rho_2}} \over a}{q^{- N}};q} \right)}_r}{{(q;q)}_{N - r}}}}{\left({{{{q^{- 2N}}} \over k}} \right)^r} in (3.2) we get (3.15) $\begin{array}{l} γ_{n} (a, k) & = \frac{{(k; q)}_{2 n} (1 - k q^{2 n}) {(ρ_{1}, ρ_{2}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N}; q)}_{n} {(q^{- N}; q)}_{n} q^{n}}{{(aq; q)}_{2 n} (1 - k) {(\frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, \frac{ρ_{1} ρ_{2}}{a} q^{- N}; q)}_{n} {(k q^{1 + N}; q)}_{n}} \frac{q^{N (N + 1) / 2} k^{N}}{{(q, kq; q)}_{N}} \\ ._{8} Φ_{7} [\begin{array}{l} k q^{2 n}, q^{n + 1} \sqrt{k}, - q^{n + 1} \sqrt{k}, ρ_{1} q^{n}, ρ_{2} q^{n}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N + n}, \frac{k}{a}, q^{- (N - n)}; q; q \\ q^{n} \sqrt{k}, - q^{n} \sqrt{k}, \frac{k}{ρ_{1}} q^{1 + n}, \frac{k}{ρ_{2}} q^{1 + n}, \frac{ρ_{1} ρ_{2}}{a} q^{- (N - n)}, a q^{1 + 2 n}, k q^{1 + N + n} \end{array}] . \end{array}$ \matrix{{{\gamma_n}(a,k)} \hfill & {= {{{{(k;q)}_{2n}}(1 - k{q^{2n}}){{\left({{\rho_1},{\rho_2},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N}};q} \right)}_n}{{({q^{- N}};q)}_n}{q^n}} \over {{{(aq;q)}_{2n}}(1 - k){{\left({{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},{{{\rho_1}{\rho_2}} \over a}{q^{- N}};q} \right)}_n}{{(k{q^{1 + N}};q)}_n}}}{{{q^{N(N + 1)/2}}{k^N}} \over {{{(q,kq;q)}_N}}}} \hfill \cr {} \hfill & {{._8}{\Phi_7}\left[ {\matrix{{k{q^{2n}},{q^{n + 1}}\sqrt k, - {q^{n + 1}}\sqrt k,{\rho_1}{q^n},{\rho_2}{q^n},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N + n}},{k \over a},{q^{- (N - n)}};q;q} \hfill \cr {{q^n}\sqrt k, - {q^n}\sqrt k,{k \over {{\rho_1}}}{q^{1 + n}},{k \over {{\rho_2}}}{q^{1 + n}},{{{\rho_1}{\rho_2}} \over a}{q^{- (N - n)}},a{q^{1 + 2n}},k{q^{1 + N + n}}} \hfill \cr}} \right].} \hfill}

Summing the ₈Φ₇ series in (3.15) by using [2, App.II (II.22), p. 238] we have (3.16) $\begin{array}{l} γ_{n} (a, k) & = \frac{{(\frac{kq}{ρ_{1} ρ_{2}}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{N} q^{N (N + 1) / 2} k^{N}}{{(q, \frac{aq}{ρ_{1} ρ_{2}}, \frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{N} {(aq; q)}_{N}} \\ . \frac{{(ρ_{1}, ρ_{2}, \frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N}, q^{- N}; q)}_{n} {(\frac{aq}{k})}^{n}}{{(a q^{1 + N}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, \frac{ρ_{1} ρ_{2}}{k} q^{- N}, \frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{n}} . \end{array}$ \matrix{{{\gamma_n}(a,k)} \hfill & {= {{{{\left({{{kq} \over {{\rho_1}{\rho_2}}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_N}{q^{N(N + 1)/2}}{k^N}} \over {{{\left({q,{{aq} \over {{\rho_1}{\rho_2}}},{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}};q} \right)}_N}{{(aq;q)}_N}}}} \hfill \cr {} \hfill & {\,\,\,.{{{{\left({{\rho_1},{\rho_2},{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N}},{q^{- N}};q} \right)}_n}{{\left({{{aq} \over k}} \right)}^n}} \over {{{\left({a{q^{1 + N}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},{{{\rho_1}{\rho_2}} \over k}{q^{- N}},{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}};q} \right)}_n}}}.} \hfill}

Putting these values of 〈γ_n(a,k), δ_n(a,k)〉 in (3.3) we obtain the following theorem.

Theorem 11

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.17) $\begin{array}{l} \frac{{(kq, \frac{kq}{ρ_{1} ρ_{2}}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{N}}{{(aq, \frac{aq}{ρ_{1} ρ_{2}}, \frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}; q)}_{N}} \sum_{n = 0}^{N} \frac{{(ρ_{1}, ρ_{2}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N}, q^{- N}; q)}_{n} {(\frac{aq}{k})}^{n}}{{(a q^{1 + N}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, \frac{ρ_{1} ρ_{2}}{k} q^{- N}; q)}_{n}} \\ = \sum_{n = 0}^{\infty} (\frac{1 - k q^{2 n}}{1 - k}) \frac{{(ρ_{1}, ρ_{2}, \frac{ak}{ρ_{1} ρ_{2}} q^{1 + N}, q^{- N}; q)}_{n} q^{n}}{{(k q^{1 + N}, \frac{kq}{ρ_{1}}, \frac{kq}{ρ_{2}}, \frac{ρ_{1} ρ_{2}}{a} q^{- N}; q)}_{n}} β_{n} (a, k) . \end{array}$ \matrix{{\,\,\,\,\,{{{{\left({kq,{{kq} \over {{\rho_1}{\rho_2}}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_N}} \over {{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}},{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}};q} \right)}_N}}}\sum\limits_{n = 0}^N {{{{\left({{\rho_1},{\rho_2},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N}},{q^{- N}};q} \right)}_n}{{\left({{{aq} \over k}} \right)}^n}} \over {{{\left({a{q^{1 + N}},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},{{{\rho_1}{\rho_2}} \over k}{q^{- N}};q} \right)}_n}}}} \hfill \cr {= \sum\limits_{n = 0}^\infty \left({{{1 - k{q^{2n}}} \over {1 - k}}} \right){{{{\left({{\rho_1},{\rho_2},{{ak} \over {{\rho_1}{\rho_2}}}{q^{1 + N}},{q^{- N}};q} \right)}_n}{q^n}} \over {{{\left({k{q^{1 + N}},{{kq} \over {{\rho_1}}},{{kq} \over {{\rho_2}}},{{{\rho_1}{\rho_2}} \over a}{q^{- N}};q} \right)}_n}}}{\beta_n}(a,k).} \hfill}

As ρ₂ → ∞, (3.17) yields the following theorem.

Theorem 12

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair then(3.18) $\begin{array}{l} \frac{{(kq, \frac{aq}{ρ_{1}}; q)}_{N}}{{(aq, \frac{kq}{ρ_{1}}; q)}_{N}} \sum_{n = 0}^{N} \frac{{(ρ_{1}, q^{- N}; q)}_{n} {(\frac{a q^{1 + N}}{ρ_{1}})}^{n} α_{n} (a, k)}{{(\frac{aq}{ρ_{1}}, a q^{1 + N}; q)}_{n}} \\ = \sum_{n = 0}^{N} (\frac{1 - k q^{2 n}}{1 - k}) \frac{{(ρ_{1}, q^{- N}; q)}_{n}}{{(\frac{kq}{ρ_{1}}, k q^{1 + N}; q)}_{n}} {(\frac{a}{ρ_{1}} q^{1 + N})}^{n} β_{n} (a, k) . \end{array}$ \matrix{{\,\,\,\,\,{{{{\left({kq,{{aq} \over {{\rho_1}}};q} \right)}_N}} \over {{{\left({aq,{{kq} \over {{\rho_1}}};q} \right)}_N}}}\sum\limits_{n = 0}^N {{{{\left({{\rho_1},{q^{- N}};q} \right)}_n}{{\left({{{a{q^{1 + N}}} \over {{\rho_1}}}} \right)}^n}{\alpha_n}(a,k)} \over {{{\left({{{aq} \over {{\rho_1}}},a{q^{1 + N}};q} \right)}_n}}}} \hfill \cr {= \sum\limits_{n = 0}^N \left({{{1 - k{q^{2n}}} \over {1 - k}}} \right){{{{({\rho_1},{q^{- N}};q)}_n}} \over {{{\left({{{kq} \over {{\rho_1}}},k{q^{1 + N}};q} \right)}_n}}}{{\left({{a \over {{\rho_1}}}{q^{1 + N}}} \right)}^n}{\beta_n}(a,k).} \hfill}

Bailey Pairs

In this section we give numerous Bailey pairs deducible from certain summation formulas.

(i) Choosing $α_{r} = \frac{q^{r (r + 1) / 2} {(a, q \sqrt{a}, - q \sqrt{a}; q)}_{r} q^{- \frac{3}{2} r}}{{(q, \sqrt{a}, - \sqrt{a}; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a,q\sqrt a, - q\sqrt a;q)}_r}{q^{- {3 \over 2}r}}} \over {{{(q,\sqrt a, - \sqrt a;q)}_r}}} in (1.4) and using the summation formula [5, (4.1), p.76] we get, (4.1) $β_{n} = \frac{1}{2} [\frac{{(- q^{- \frac{1}{2}}; q)}_{n} (1 + \sqrt{a})}{{(q, \sqrt{aq}, - \sqrt{a}; q)}_{n}} + \frac{{(- q^{- \frac{1}{2}}; q)}_{n} (1 - \sqrt{a})}{{(q, - \sqrt{aq}, \sqrt{a}; q)}_{n}}] .$ {\beta_n} = {1 \over 2}\left[ {{{{{(- {q^{- {1 \over 2}}};q)}_n}(1 + \sqrt a)} \over {{{(q,\sqrt {aq}, - \sqrt a;q)}_n}}} + {{{{(- {q^{- {1 \over 2}}};q)}_n}(1 - \sqrt a)} \over {{{(q, - \sqrt {aq},\sqrt a;q)}_n}}}} \right].α_n and β_n given in (4.1) form a Bailey pair.

(ii) Taking $α_{r} = \frac{q^{r (r + 1) / 2} {(a, q \sqrt{a}; q)}_{r} q^{- r}}{{(q, \sqrt{a}; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a,q\sqrt a;q)}_r}{q^{- r}}} \over {{{(q,\sqrt a;q)}_r}}} in (1.4) and using the summation formula [5, (4.2), p.76] we get, (4.2) $β_{n} = \frac{1 + \sqrt{a}}{2} \frac{{(- 1; q)}_{n}}{{(q, \sqrt{aq}, - \sqrt{aq}; q)}_{n}} + \frac{1 - \sqrt{a}}{2} \frac{{(- 1; q)}_{n}}{{(q, \sqrt{a}, - q \sqrt{a}; q)}_{n}} .$ {\beta_n} = {{1 + \sqrt a} \over 2}{{{{(- 1;q)}_n}} \over {{{(q,\sqrt {aq}, - \sqrt {aq};q)}_n}}} + {{1 - \sqrt a} \over 2}{{{{(- 1;q)}_n}} \over {{{(q,\sqrt a, - q\sqrt a;q)}_n}}}.α_n and β_n given in (4.2) form a Bailey pair.

(iii) Taking $α_{r} = \frac{q^{r (r + 1) / 2} {(a; q)}_{r} q^{- \frac{1}{2} r}}{{(q; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a;q)}_r}{q^{- {1 \over 2}r}}} \over {{{(q;q)}_r}}} in (1.4) and making use of the summation formula [5, (4.3), p.76] we find, (4.3) $β_{n} = \frac{1 + \sqrt{a}}{2} \frac{{(- q^{\frac{1}{2}}; q)}_{n}}{{(q, - \sqrt{aq}, q \sqrt{a}; q)}_{n}} + \frac{1 - \sqrt{a}}{2} \frac{{(- q^{\frac{1}{2}}; q)}_{n}}{{(q, \sqrt{aq}, - q \sqrt{a}; q)}_{n}} .$ {\beta_n} = {{1 + \sqrt a} \over 2}{{{{(- {q^{{1 \over 2}}};q)}_n}} \over {{{(q, - \sqrt {aq},q\sqrt a;q)}_n}}} + {{1 - \sqrt a} \over 2}{{{{(- {q^{{1 \over 2}}};q)}_n}} \over {{{(q,\sqrt {aq}, - q\sqrt a;q)}_n}}}.α_n and β_n given in (4.3) form a Bailey pair.

(iv) Choosing $α_{r} = \frac{q^{\frac{1}{2} r^{2}} {(a; q)}_{r} (1 - a q^{2 r})}{{(q; q)}_{r} (1 - a)}$ {\alpha_r} = {{{q^{{1 \over 2}{r^2}}}{{(a;q)}_r}(1 - a{q^{2r}})} \over {{{(q;q)}_r}(1 - a)}} in (1.4) and using the summation formula [5, (4.5), p.77] we find, (4.4) $β_{n} = \frac{1 + \sqrt{a}}{2 \sqrt{a}} \frac{{(- q^{- \frac{1}{2}}; q)}_{n}}{{(q, \sqrt{aq}, - q \sqrt{a}; q)}_{n}} - \frac{1 - \sqrt{a}}{2 \sqrt{a}} \frac{{(- q^{- \frac{1}{2}}; q)}_{n}}{{(q, - \sqrt{aq}, q \sqrt{a}; q)}_{n}} .$ {\beta_n} = {{1 + \sqrt a} \over {2\sqrt a}}{{{{(- {q^{- {1 \over 2}}};q)}_n}} \over {{{(q,\sqrt {aq}, - q\sqrt a;q)}_n}}} - {{1 - \sqrt a} \over {2\sqrt a}}{{{{(- {q^{- {1 \over 2}}};q)}_n}} \over {{{(q, - \sqrt {aq},q\sqrt a;q)}_n}}}.α_n and β_n given in (4.4) form a Bailey pair.

(v) Taking $α_{r} = \frac{q^{r (r + 1) / 2} {(a, q \sqrt{a}; q)}_{r}}{{(q, \sqrt{a}; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a,q\sqrt a;q)}_r}} \over {{{(q,\sqrt a;q)}_r}}} in (1.4) and using the summation formula [5, (4.6), p.77] we find, (4.5) $β_{n} = \frac{1 + \sqrt{a}}{2 \sqrt{a}} \frac{{(- 1; q)}_{n}}{{(q, \sqrt{aq}, - \sqrt{aq}; q)}_{n}} - \frac{1 - \sqrt{a}}{2 \sqrt{a}} \frac{{(- 1; q)}_{n}}{{(q, \sqrt{a}, - q \sqrt{a}; q)}_{n}} .$ {\beta_n} = {{1 + \sqrt a} \over {2\sqrt a}}{{{{(- 1;q)}_n}} \over {{{(q,\sqrt {aq}, - \sqrt {aq};q)}_n}}} - {{1 - \sqrt a} \over {2\sqrt a}}{{{{(- 1;q)}_n}} \over {{{(q,\sqrt a, - q\sqrt a;q)}_n}}}.α_n and β_n given in (4.5) form a Bailey pair.

(vi) Choosing $α_{r} = \frac{q^{r (r + 1) / 2} {(a; q)}_{r} q^{\frac{1}{2} r}}{{(q; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a;q)}_r}{q^{{1 \over 2}r}}} \over {{{(q;q)}_r}}} in (1.4) and making use of the summation formula [5, (4.7), p.77] we find, (4.6) $β_{n} = \frac{1 + \sqrt{a}}{2 \sqrt{a}} \frac{{(- q^{\frac{1}{2}}; q)}_{n}}{{(q, - \sqrt{aq}, q \sqrt{a}; q)}_{n}} - \frac{1 - \sqrt{a}}{2 \sqrt{a}} \frac{{(- q^{\frac{1}{2}}; q)}_{n}}{{(q, \sqrt{aq}, - q \sqrt{a}; q)}_{n}} .$ {\beta_n} = {{1 + \sqrt a} \over {2\sqrt a}}{{{{(- {q^{{1 \over 2}}};q)}_n}} \over {{{(q, - \sqrt {aq},q\sqrt a;q)}_n}}} - {{1 - \sqrt a} \over {2\sqrt a}}{{{{(- {q^{{1 \over 2}}};q)}_n}} \over {{{(q,\sqrt {aq}, - q\sqrt a;q)}_n}}}.α_n and β_n given in (4.6) form a Bailey pair.

(vii) Taking $α_{r} = \frac{q^{r (r + 1) / 2} {(a, - q \sqrt{a}, b; q)}_{r} {(- \frac{\sqrt{a}}{b})}^{r}}{{(q, - \sqrt{a}, \frac{aq}{b}; q)}_{r}}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a, - q\sqrt a,b;q)}_r}{{\left({- {{\sqrt a} \over b}} \right)}^r}} \over {{{\left({q, - \sqrt a,{{aq} \over b};q} \right)}_r}}} in (1.4) and using the summation formula [2, App. II (II.14), p. 237] we get (4.7) $β_{n} = \frac{{(\frac{q \sqrt{a}}{b}; q)}_{n}}{{(q, q \sqrt{a}, \frac{aq}{b}; q)}_{n}} .$ {\beta_n} = {{{{\left({{{q\sqrt a} \over b};q} \right)}_n}} \over {{{\left({q,q\sqrt a,{{aq} \over b};q} \right)}_n}}}. 〈α_n, β_n〉 given in (4.7) form a Bailey pair.

(viii) Taking $α_{r} = \frac{q^{r (r + 1) / 2} {(a, b, c; q)}_{r} (1 - a q^{2 r}) {(- \frac{a}{bc})}^{r}}{{(q, \frac{aq}{b}, \frac{aq}{c}; q)}_{r} (1 - a)}$ {\alpha_r} = {{{q^{r(r + 1)/2}}{{(a,b,c;q)}_r}(1 - a{q^{2r}}){{\left({- {a \over {bc}}} \right)}^r}} \over {{{\left({q,{{aq} \over b},{{aq} \over c};q} \right)}_r}(1 - a)}} in (1.4) and using the summation formula [2, App. II (II.21), p. 238] we get (4.8) $β_{n} = \frac{{(\frac{aq}{bc}; q)}_{n}}{{(q, \frac{aq}{b}, \frac{aq}{c}; q)}_{n}} .$ {\beta_n} = {{{{\left({{{aq} \over {bc}};q} \right)}_n}} \over {{{\left({q,{{aq} \over b},{{aq} \over c};q} \right)}_n}}}. 〈α_n, β_n〉 given in (4.8) form a Bailey pair.

WP-Bailey Pairs

In this section we give certain WP-Bailey pairs out of which some are known and some are new.

(i) Taking $α_{r} (a, k) = \frac{{(a, b; q)}_{r} (1 - a q^{2 r})}{{(q, \frac{aq}{b}; q)}_{r} (1 - a)} {(\frac{1}{b})}^{r}$ {\alpha_r}(a,k) = {{{{(a,b;q)}_r}(1 - a{q^{2r}})} \over {{{\left({q,{{aq} \over b};q} \right)}_r}(1 - a)}}{\left({{1 \over b}} \right)^r} in (3.1) and summing the series by making use of [2, App. II (II.21), p. 238] we get, (5.1) $β_{n} (a, k) = \frac{{(k, \frac{kb}{a}; q)}_{n}}{{(q, \frac{aq}{b}; q)}_{n} b^{n}} .$ {\beta_n}(a,k) = {{{{\left({k,{{kb} \over a};q} \right)}_n}} \over {{{\left({q,{{aq} \over b};q} \right)}_n}{b^n}}}. 〈α_n(a,k), β_n(a,k)〉 given in (5.1) form a WP-Bailey pair.

(ii) Choosing $α_{r} (a, k) = \frac{{(a, b, c, \frac{a^{2} q}{bck}; q)}_{r} (1 - a q^{2 r})}{{(q, \frac{aq}{b}, \frac{aq}{c}, \frac{bck}{a}; q)}_{r} (1 - a)} {(\frac{k}{a})}^{r}$ {\alpha_r}(a,k) = {{{{\left({a,b,c,{{{a^2}q} \over {bck}};q} \right)}_r}(1 - a{q^{2r}})} \over {{{\left({q,{{aq} \over b},{{aq} \over c},{{bck} \over a};q} \right)}_r}(1 - a)}}{\left({{k \over a}} \right)^r} in (3.1) and using the summation formula [2, App. II (II.22), p. 238] we get, (5.2) $β_{n} (a, k) = \frac{{(k, \frac{aq}{bc}, \frac{kb}{a}, \frac{kc}{a}; q)}_{n}}{{(q, \frac{aq}{b}, \frac{aq}{c}, \frac{kbc}{a}; q)}_{n}} .$ {\beta_n}(a,k) = {{{{\left({k,{{aq} \over {bc}},{{kb} \over a},{{kc} \over a};q} \right)}_n}} \over {{{\left({q,{{aq} \over b},{{aq} \over c},{{kbc} \over a};q} \right)}_n}}}. 〈α_n(a,k), β_n(a,k)〉 given in (5.2) form a WP-Bailey pair.

(iii) Again, choosing $α_{n} (a, k) = \frac{{(a, \frac{a}{k}; q)}_{n} (1 - a q^{2 n})}{{(q, kq; q)}_{n} (1 - a)} {(\frac{k}{a})}^{n}$ {\alpha_n}(a,k) = {{{{\left({a,{a \over k};q} \right)}_n}(1 - a{q^{2n}})} \over {{{\left({q,kq;q} \right)}_n}(1 - a)}}{\left({{k \over a}} \right)^n} in (3.1) and summing the series by making use of [2, App. II (II.21), p. 238] we get, (5.3) $β_{n} (a, k) = {\begin{array}{l} 1, n = 0 . \\ 0, n \geq 1 \end{array}$ {\beta_n}(a,k) = \left\{{\matrix{{1, n = 0} \hfill \cr {0, n \ge 1.} \hfill \cr}} \right.

So, 〈α_n(a,k), β_n(a,k)〉 given in (5.3) form a WP-Bailey pair.

(iv) If we take α_r(a,k) = δ_r,0 in (3.1) we find, (5.4) $β_{n} (a, k) = \frac{{(k, \frac{k}{a}; q)}_{n}}{{(q, aq; q)}_{n}} .$ {\beta_n}(a,k) = {{{{\left({k,{k \over a};q} \right)}_n}} \over {{{\left({q,aq;q} \right)}_n}}}. 〈α_n(a,k), β_n(a,k)〉 given in (5.4) also form a WP-Bailey pair.

(v) In the summation formula [5, (1.3), p.71] if we take $c = \frac{a}{k} q^{1 / 2}$ c = {a \over k}{q^{1/2}} we get, (5.5) $\begin{array}{l} _{4} Φ_{3} [\begin{array}{l} a, \frac{a}{k} q^{\frac{1}{2}}, k q^{n}, q^{- n}; q; q \\ k q^{\frac{1}{2}}, \frac{a}{k} q^{1 - n}, a q^{1 + n} \end{array}] & = \frac{1 + \sqrt{a}}{2} \frac{{(aq, \sqrt{q}; q)}_{n} {(\frac{k}{\sqrt{a}}, k \sqrt{\frac{q}{a}}; q)}_{n}}{{(k q^{\frac{1}{2}}, \frac{k}{a}; q)}_{n} {(\sqrt{aq}, q \sqrt{a}; q)}_{n}} \\ + \frac{1 - \sqrt{a}}{2} \frac{{(aq, \sqrt{q}; q)}_{n} {(- \frac{k}{\sqrt{a}}, - k \sqrt{\frac{q}{a}}; q)}_{n}}{{(k q^{\frac{1}{2}}, \frac{k}{a}; q)}_{n} {(- \sqrt{aq}, - q \sqrt{a}; q)}_{n}} . \end{array}$ \matrix{{{\,_4}{\Phi_3}\left[ {\matrix{{a,{a \over k}{q^{{1 \over 2}}},k{q^n},{q^{- n}};q;q} \hfill \cr {k{q^{{1 \over 2}}},{a \over k}{q^{1 - n}},a{q^{1 + n}}} \hfill \cr}} \right]} \hfill & {= {{1 + \sqrt a} \over 2}{{{{(aq,\sqrt q;q)}_n}{{\left({{k \over {\sqrt a}},k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({k{q^{{1 \over 2}}},{k \over a};q} \right)}_n}{{(\sqrt {aq},q\sqrt a;q)}_n}}}} \hfill \cr {} \hfill & {+ {{1 - \sqrt a} \over 2}{{{{(aq,\sqrt q;q)}_n}{{\left({- {k \over {\sqrt a}}, - k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({k{q^{{1 \over 2}}},{k \over a};q} \right)}_n}{{(- \sqrt {aq}, - q\sqrt a;q)}_n}}}.} \hfill}

Now, choosing $α_{r} (a, k) = \frac{{(a, \frac{a q^{\frac{1}{2}}}{k}; q)}_{n}}{{(q, k q^{\frac{1}{2}}; q)}_{n}} {(\frac{k}{a})}^{n}$ {\alpha_r}(a,k) = {{{{\left({a,{{a{q^{{1 \over 2}}}} \over k};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}};q} \right)}_n}}}{\left({{k \over a}} \right)^n} in (3.1) and summing the series by using (5.5) we get (5.6) $\begin{array}{l} β_{n} (a, k) & = \frac{1 + \sqrt{a}}{2} \frac{{(k, q^{\frac{1}{2}}, \frac{k}{\sqrt{a}}, k \sqrt{\frac{q}{a}}; q)}_{n}}{{(q, k q^{\frac{1}{2}}, q \sqrt{a}, \sqrt{aq}; q)}_{n}} \\ + \frac{1 - \sqrt{a}}{2} \frac{{(k, q^{\frac{1}{2}}, - \frac{k}{\sqrt{a}}, - k \sqrt{\frac{q}{a}}; q)}_{n}}{{(q, k q^{\frac{1}{2}}, - q \sqrt{a}, - \sqrt{aq}; q)}_{n}} . \end{array}$ \matrix{{{\beta_n}(a,k)} \hfill & {= {{1 + \sqrt a} \over 2}{{{{\left({k,{q^{{1 \over 2}}},{k \over {\sqrt a}},k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}},q\sqrt a,\sqrt {aq};q} \right)}_n}}}} \hfill \cr {} \hfill & {+ {{1 - \sqrt a} \over 2}{{{{\left({k,{q^{{1 \over 2}}}, - {k \over {\sqrt a}}, - k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}}, - q\sqrt a, - \sqrt {aq};q} \right)}_n}}}.} \hfill} 〈α_n(a,k), β_n(a,k)〉 given in (5.6) form a WP-Bailey pair.

(vi) Again, taking $c = \frac{a}{k} q^{\frac{1}{2}}$ c = {a \over k}{q^{{1 \over 2}}} in the summation formula [5, (4.4), p.77] we get, (5.7) $\begin{array}{l} _{4} Φ_{3} [\begin{array}{l} a, \frac{a}{k} q^{\frac{1}{2}}, k q^{n}, q^{- n}; q; q^{2} \\ k q^{\frac{1}{2}}, \frac{a}{k} q^{1 - n}, a q^{1 + n} \end{array}] & = \frac{1 + \sqrt{a}}{2 \sqrt{a}} \frac{{(aq, \sqrt{q}; q)}_{n} {(\frac{k}{\sqrt{a}}, k \sqrt{\frac{q}{a}}; q)}_{n}}{{(k q^{\frac{1}{2}}, \frac{k q^{\frac{1}{2}}}{a}; q)}_{n} {(\sqrt{aq}, q \sqrt{a}; q)}_{n}} \\ + \frac{1 - \sqrt{a}}{2 \sqrt{a}} \frac{{(aq, \sqrt{a}; q)}_{n} {(- \frac{k}{\sqrt{a}}, - k \sqrt{\frac{q}{a}}; q)}_{n}}{{(k q^{\frac{1}{2}}, \frac{k q^{\frac{1}{2}}}{a}; q)}_{n} {(- \sqrt{aq}, - q \sqrt{a}; q)}_{n}} . \end{array}$ \matrix{{{\,_4}{\Phi_3}\left[ {\matrix{{a,{a \over k}{q^{{1 \over 2}}},k{q^n},{q^{- n}};q;{q^2}} \hfill \cr {k{q^{{1 \over 2}}},{a \over k}{q^{1 - n}},a{q^{1 + n}}} \hfill \cr}} \right]} \hfill & {= {{1 + \sqrt a} \over {2\sqrt a}}{{{{(aq,\sqrt q;q)}_n}{{\left({{k \over {\sqrt a}},k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({k{q^{{1 \over 2}}},{{k{q^{{1 \over 2}}}} \over a};q} \right)}_n}{{(\sqrt {aq},q\sqrt a;q)}_n}}}} \hfill \cr {} \hfill & {+ {{1 - \sqrt a} \over {2\sqrt a}}{{{{(aq,\sqrt a;q)}_n}{{\left({- {k \over {\sqrt a}}, - k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({k{q^{{1 \over 2}}},{{k{q^{{1 \over 2}}}} \over a};q} \right)}_n}{{(- \sqrt {aq}, - q\sqrt a;q)}_n}}}.} \hfill}

Now, choosing $α_{n} (a, k) = \frac{{(a, \frac{a q^{\frac{1}{2}}}{k}; q)}_{n}}{{(q, k q^{\frac{1}{2}}; q)}_{n}} {(\frac{kq}{a})}^{n}$ {\alpha_n}(a,k) = {{{{\left({a,{{a{q^{{1 \over 2}}}} \over k};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}};q} \right)}_n}}}{\left({{{kq} \over a}} \right)^n} in (3.1) and using (5.7) we get (5.8) $\begin{array}{l} β_{n} (a, k) & = \frac{1 + \sqrt{a}}{2 \sqrt{a}} \frac{{(k, q^{\frac{1}{2}}, \frac{k}{\sqrt{a}}, k \sqrt{\frac{q}{a}}; q)}_{n}}{{(q, k q^{\frac{1}{2}}, q \sqrt{a}, \sqrt{aq}; q)}_{n}} \\ - \frac{1 - \sqrt{a}}{2 \sqrt{a}} \frac{{(k, q^{\frac{1}{2}}, - \frac{k}{\sqrt{a}}, - k \sqrt{\frac{q}{a}}; q)}_{n}}{{(q, k q^{\frac{1}{2}}, - q \sqrt{a}, - \sqrt{aq}; q)}_{n}} . \end{array}$ \matrix{{{\beta_n}(a,k)} \hfill & {= {{1 + \sqrt a} \over {2\sqrt a}}{{{{\left({k,{q^{{1 \over 2}}},{k \over {\sqrt a}},k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}},q\sqrt a,\sqrt {aq};q} \right)}_n}}}} \hfill \cr {} \hfill & {- {{1 - \sqrt a} \over {2\sqrt a}}{{{{\left({k,{q^{{1 \over 2}}}, - {k \over {\sqrt a}}, - k\sqrt {{q \over a}};q} \right)}_n}} \over {{{\left({q,k{q^{{1 \over 2}}}, - q\sqrt a, - \sqrt {aq};q} \right)}_n}}}.} \hfill} 〈α_n(a,k), β_n(a,k)〉 given in (5.8) form a WP-Bailey pair. Bailey pairs given in (5.6) and (5.8) are believed to be new.

Bailey Chain

If 〈α_n, β_n〉 is a Bailey pair i.e. $β_{n} = \sum_{r = 0}^{n} \frac{α_{r}}{{(q; q)}_{r} {(aq; q)}_{r}}$ {\beta_n} = \sum\nolimits_{r = 0}^n {{{\alpha_r}} \over {{{(q;q)}_r}{{(aq;q)}_r}}} , then so 〈 $α_{n}^{'}$ \alpha_n^{'} , $β_{n}^{'}$ \beta_n^{'} 〉 is also a Bailey pair, where (6.1) $α_{n}^{'} = \frac{{(ρ_{1}, ρ_{2}; q)}_{n} {(\frac{aq}{ρ_{1} ρ_{2}})}^{n}}{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} α_{n} .$ {\alpha_{n'}} = {{{{({\rho_1},{\rho_2};q)}_n}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^n}} \over {{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{\alpha_n}. and (6.2) $\begin{array}{l} β_{n}^{'} & = \sum_{r = 0}^{\infty} \frac{{(ρ_{1}, ρ_{2}; q)}_{r} {(\frac{aq}{ρ_{1} ρ_{2}}; q)}_{n - r} {(\frac{aq}{ρ_{1} ρ_{2}})}^{r} β_{r}}{{(q; q)}_{n - r} {(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} \\ = \frac{{(\frac{aq}{ρ_{1} ρ_{2}}; q)}_{n}}{{(q, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} \sum_{r = 0}^{\infty} \frac{{(ρ_{1}, ρ_{2}; q)}_{r} {(q^{- n}; q)}_{r} q^{r}}{{(\frac{ρ_{1} ρ_{2}}{a} q^{- n}; q)}_{r}} β_{r} . \end{array}$ \matrix{{{\beta_{n'}}} \hfill & {= \sum\limits_{r = 0}^\infty {{{{({\rho_1},{\rho_2};q)}_r}{{\left({{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_{n - r}}{{\left({{{aq} \over {{\rho_1}{\rho_2}}}} \right)}^r}{\beta_r}} \over {{{(q;q)}_{n - r}}{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}} \hfill \cr {} \hfill & {= {{{{\left({{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_n}} \over {{{\left({q,{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}\sum\limits_{r = 0}^\infty {{{{({\rho_1},{\rho_2};q)}_r}{{\left({{q^{- n}};q} \right)}_r}{q^r}} \over {{{\left({{{{\rho_1}{\rho_2}} \over a}{q^{- n}};q} \right)}_r}}}{\beta_r}.} \hfill}

Thus, we find that if one Bailey pair 〈α_n, β_n〉 is known then a new Bailey pair 〈 $α_{n}^{'}$ \alpha_n^{'} , $β_{n}^{'}$ \beta_n^{'} 〉 can be constructed as shown above in (6.2). Repeating this process we can have infinite number of Bailey pairs if one initial pair is known. These Bailey pairs so constructed from an initial Bailey pair form a chain called Bailey chain.

WP-Bailey Tree

Andrews proved following two theorems for constructing WP-Bailey pairs from a initial known Bailey pair.

Theorem 13

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) , $β_{n}^{'} (a, k)$ \beta_n^{'}(a,k) 〉 given by, (7.1) $\begin{matrix} α_{n}^{'} (a, k) = \frac{{(b, c; q)}_{n}}{{(\frac{aq}{b}, \frac{aq}{c}; q)}_{n}} {(\frac{k}{m})}^{n} α_{n} (a, m), \\ β_{n}^{'} (a, k) = \frac{{(\frac{mq}{b}, \frac{mq}{c}; q)}_{n}}{{(\frac{aq}{b}, \frac{aq}{c}; q)}_{n}} \sum_{r = 0}^{n} \frac{1 - m q^{2 r}}{1 - m} \frac{{(b, c; q)}_{r} {(\frac{k}{m}; q)}_{n - r} {(k; q)}_{n + r}}{{(\frac{mq}{b}, \frac{mq}{c}; q)}_{r} {(q; q)}_{n - r} {(mq; q)}_{m + r}} {(\frac{k}{m})}^{r} β_{r} (a, m), \end{matrix}$ \matrix{{{\alpha_n^{'}}(a,k) = {{{{(b,c;q)}_n}} \over {{{\left({{{aq} \over b},{{aq} \over c};q} \right)}_n}}}{{\left({{k \over m}} \right)}^n}{\alpha_n}(a,m),} \cr {{\beta_n^{'}}(a,k) = {{{{\left({{{mq} \over b},{{mq} \over c};q} \right)}_n}} \over {{{\left({{{aq} \over b},{{aq} \over c};q} \right)}_n}}}\sum\limits_{r = 0}^n {{1 - m{q^{2r}}} \over {1 - m}}{{{{(b,c;q)}_r}{{\left({{k \over m};q} \right)}_{n - r}}{{(k;q)}_{n + r}}} \over {{{\left({{{mq} \over b},{{mq} \over c};q} \right)}_r}{{(q;q)}_{n - r}}{{(mq;q)}_{m + r}}}}{{\left({{k \over m}} \right)}^r}{\beta_r}(a,m),}}

[6, Theorem (2.1)]

where $m = \frac{bck}{aq}$ m = {{bck} \over {aq}} .

Theorem 14

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) , $β_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) 〉 given by, (7.2) $\begin{matrix} α_{n}^{'} (a, k) = \frac{{(m; q)}_{2 n}}{{(k; q)}_{2 m}} {(\frac{k}{m})}^{n} α_{n} (a, m), \\ β_{n}^{'} (a, k) = \sum_{r = 0}^{n} \frac{{(\frac{k}{m}; q)}_{n - r}}{{(q; q)}_{n - r}} {(\frac{k}{m})}^{r} β_{r} (a, m), \end{matrix}$ \matrix{{{\alpha_n^{'}}(a,k) = {{{{(m;q)}_{2n}}} \over {{{(k;q)}_{2m}}}}{{\left({{k \over m}} \right)}^n}{\alpha_n}(a,m),} \cr {{\beta_n^{'}}(a,k) = \sum\limits_{r = 0}^n {{{{\left({{k \over m};q} \right)}_{n - r}}} \over {{{\left({q;q} \right)}_{n - r}}}}{{\left({{k \over m}} \right)}^r}{\beta_r}(a,m),}}

[6, Theorem (2.2)]

where $m = \frac{a^{2} q}{k}$ m = {{{a^2}q} \over k} .

From these two theorems, each WP-Bailey pair gives rise to a binary tree of WP-Bailey pairs. Andrews coined this the WP-Bailey tree. The following four theorems due to Warnaar give additional branches to the Bailey tree.

Theorem 15

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) , $β_{n}^{'} (a, k)$ \beta_n^{'}(a,k) 〉 given by, (7.3) $\begin{matrix} α_{n}^{'} (a, k) = \frac{1 - σ k^{\frac{1}{2}}}{1 - σ k^{\frac{1}{2}} q^{n}} \frac{1 + σ m^{\frac{1}{2}} q^{n}}{1 + σ m^{\frac{1}{2}}} \frac{{(m; q)}_{2 n}}{{(k; q)}_{2 m}} {(\frac{k}{m})}^{n} α_{n} (a, m), \\ β_{n}^{'} (a, k) = \frac{1 - σ k^{\frac{1}{2}}}{1 - σ k^{\frac{1}{2}} q^{n}} \sum_{r = 0}^{n} \frac{1 + σ m^{\frac{1}{2}} q^{n}}{1 + σ m^{\frac{1}{2}}} \frac{{(\frac{k}{m}; q)}_{n - r}}{{(q; q)}_{n - r}} {(\frac{k}{m})}^{r} β_{r} (a, m), \end{matrix}$ \matrix{{\alpha_n^{'}(a,k) = {{1 - \sigma {k^{{1 \over 2}}}} \over {1 - \sigma {k^{{1 \over 2}}}{q^n}}}{{1 + \sigma {m^{{1 \over 2}}}{q^n}} \over {1 + \sigma {m^{{1 \over 2}}}}}{{{{(m;q)}_{2n}}} \over {{{(k;q)}_{2m}}}}{{\left({{k \over m}} \right)}^n}{\alpha_n}(a,m),} \cr {\beta_n^{'}(a,k) = {{1 - \sigma {k^{{1 \over 2}}}} \over {1 - \sigma {k^{{1 \over 2}}}{q^n}}}\sum\limits_{r = 0}^n {{1 + \sigma {m^{{1 \over 2}}}{q^n}} \over {1 + \sigma {m^{{1 \over 2}}}}}{{{{\left({{k \over m};q} \right)}_{n - r}}} \over {{{\left({q;q} \right)}_{n - r}}}}{{\left({{k \over m}} \right)}^r}{\beta_r}(a,m),}}

[6, Theorem (2.3)]

where $m = \frac{a^{2}}{k}$ m = {{{a^2}} \over k} and σ = {−1, 1}.

The freedom in the choice of σ simply reflects that the above expressions are invariant under the simultaneous negation of $k^{\frac{1}{2}}$ {k^{{1 \over 2}}} , $m^{\frac{1}{2}}$ {m^{{1 \over 2}}} and σ.

Theorem 16

If 〈α_n(a,k), β_n(a,k)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) , $β_{n}^{'} (a, k)$ \beta_n^{'}(a,k) 〉 given by, (7.4) $\begin{matrix} α_{n}^{'} (a^{2}, k; q^{2}) = α_{n} (a, m; q), \\ β_{n}^{'} (a^{2}, k; q^{2}) = \frac{{(- mq; q)}_{2 n}}{{(- aq; q)}_{2 n}} \sum_{r = 0}^{n} \frac{(1 - m q^{2 r})}{(1 - m)} \frac{{(\frac{k}{m^{2}}; q^{2})}_{n - r} {(k; q^{2})}_{n + r} {(\frac{m}{a})}^{n - r}}{{(q^{2}; q^{2})}_{n - r} {(m^{2} q^{2}; q^{2})}_{n + r}} β_{r} (a, m; q), \end{matrix}$ \matrix{{\alpha_n^{'}({a^2},k;{q^2}) = {\alpha_n}(a,m;q),} \cr {\beta_n^{'}({a^2},k;{q^2}) = {{{{(- mq;q)}_{2n}}} \over {{{(- aq;q)}_{2n}}}}\sum\limits_{r = 0}^n {{(1 - m{q^{2r}})} \over {(1 - m)}}{{{{\left({{k \over {{m^2}}};{q^2}} \right)}_{n - r}}{{(k;{q^2})}_{n + r}}{{\left({{m \over a}} \right)}^{n - r}}} \over {{{({q^2};{q^2})}_{n - r}}{{({m^2}{q^2};{q^2})}_{n + r}}}}{\beta_r}(a,m;q),}}

[6, Theorem (2.4)]

where $m = \frac{k}{aq}$ m = {k \over {aq}} .

Theorem 17

If 〈α_n(a,k;q), β_n(a,k;q)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k; q)$ \alpha_n^{'}(a,k;q) , $β_{n}^{'} (a, k; q)$ \beta_n^{'}(a,k;q) 〉 given by, (7.5) $\begin{matrix} α_{n}^{'} (a^{2}, k; q^{2}) = q^{- n} \frac{1 + a q^{2 n}}{1 + a} α_{n} (a, m; q), \\ β_{n}^{'} (a^{2}, k; q^{2}) = q^{- n} \frac{{(- mq; q)}_{2 n}}{{(- a; q)}_{2 n}} \sum_{r = 0}^{n} \frac{(1 - m q^{2 r})}{(1 - m)} \frac{{(\frac{k}{m^{2}}; q^{2})}_{n - r} {(k; q^{2})}_{n + r} {(\frac{m}{a})}^{n - r}}{{(q^{2}; q^{2})}_{n - r} {(m^{2} q^{2}; q^{2})}_{n + r}} β_{r} (a, m; q), \end{matrix}$ \matrix{{\alpha_n^{'}({a^2},k;{q^2}) = {q^{- n}}{{1 + a{q^{2n}}} \over {1 + a}}{\alpha_n}(a,m;q),} \cr {\beta_n^{'}({a^2},k;{q^2}) = {q^{- n}}{{{{(- mq;q)}_{2n}}} \over {{{(- a;q)}_{2n}}}}\sum\limits_{r = 0}^n {{(1 - m{q^{2r}})} \over {(1 - m)}}{{{{\left({{k \over {{m^2}}};{q^2}} \right)}_{n - r}}{{(k;{q^2})}_{n + r}}{{\left({{m \over a}} \right)}^{n - r}}} \over {{{({q^2};{q^2})}_{n - r}}{{({m^2}{q^2};{q^2})}_{n + r}}}}{\beta_r}(a,m;q),}}

[6, Theorem (2.5)]

where $m = \frac{k}{a}$ m = {k \over a} .

Theorem 18

If 〈α_n(a,k;q), β_n(a,k;q)〉 is a WP-Bailey pair, then so is the pair 〈 $α_{n}^{'} (a, k; q)$ \alpha_n^{'}(a,k;q) , $β_{n}^{'} (a, k; q)$ \beta_n^{'}(a,k;q) 〉 given by, (7.6) $\begin{matrix} α_{2 n}^{'} (a, k; q) = α_{n} (a, m; q^{2}), α_{2 n + 1}^{'} (a, k; q) = 0, \\ β_{n}^{'} (a, k; q) = \frac{{(mq; q^{2})}_{n}}{{(aq; q)}_{n}} \sum_{r = 0}^{[\frac{n}{2}]} \frac{(1 - m q^{2 r})}{(1 - m)} \frac{{(\frac{k}{m}; q)}_{n - r} {(k; q)}_{n + 2 r} {(- \frac{k}{a})}^{n - 2 r}}{{(q; q)}_{n - r} {(mq; q)}_{n + 2 r}} β_{r} (a, m; q^{2}), \end{matrix}$ \matrix{{{\alpha_{2n}^{'}}(a,k;q) = {\alpha_n}(a,m;{q^2}), {\alpha_{2n + 1}^{'}}(a,k;q) = 0,} \cr {{\beta_n^{'}}(a,k;q) = {{{{(mq;{q^2})}_n}} \over {{{(aq;q)}_n}}}\sum\limits_{r = 0}^{[{n \over 2}]} {{(1 - m{q^{2r}})} \over {(1 - m)}}{{{{\left({{k \over m};q} \right)}_{n - r}}{{(k;q)}_{n + 2r}}{{\left({- {k \over a}} \right)}^{n - 2r}}} \over {{{(q;q)}_{n - r}}{{(mq;q)}_{n + 2r}}}}{\beta_r}(a,m;{q^2}),}}

[6, Theorem (2.6)]

where $m = \frac{k}{a}$ m = {k \over a} .

Applications

In this section we shall establish certain transformation formulas by making use of the results established in previous sections.

(a) Substituting the Bailey pairs given in (4.1) in (2.2) we get, (8.1) $\begin{array}{l} \frac{{(\frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{\infty}}{{(aq, \frac{aq}{ρ_{1} ρ_{2}}; q)}_{\infty}} & _{5} Φ_{5} [\begin{array}{l} a, q \sqrt{a}, - q \sqrt{a}, ρ_{1}, ρ_{2}; q; - \frac{a q^{\frac{1}{2}}}{ρ_{1} ρ_{2}} \\ \sqrt{a}, - \sqrt{a}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, 0 \end{array}] \\ = \frac{1 + \sqrt{a}}{2}_{3} Φ_{2} [\begin{array}{l} ρ_{1}, ρ_{2}, - q^{- \frac{1}{2}}; q; \frac{aq}{ρ_{1} ρ_{2}} \\ \sqrt{aq}, - \sqrt{a} \end{array}] \\ + \frac{1 - \sqrt{a}}{2}_{3} Φ_{2} [\begin{array}{l} ρ_{1}, ρ_{2}, - q^{- \frac{1}{2}}; q; \frac{aq}{ρ_{1} ρ_{2}} \\ - \sqrt{aq}, \sqrt{a} \end{array}], | \frac{a q^{\frac{1}{2}}}{ρ_{1} ρ_{2}} | < 1 . \end{array}$ \matrix{{{{{{\left({{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_\infty}} \over {{{\left({aq,{{aq} \over {{\rho_1}{\rho_2}}};q} \right)}_\infty}}}} \hfill & {{\,_5}{\Phi_5}\left[ {\matrix{{a,q\sqrt a, - q\sqrt a,{\rho_1},{\rho_2};q; - {{a{q^{{1 \over 2}}}} \over {{\rho_1}{\rho_2}}}} \hfill \cr {\sqrt a, - \sqrt a,{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},0} \hfill \cr}} \right]} \hfill \cr {} \hfill & {= {{1 + \sqrt a} \over 2}{_3}{\Phi_2}\left[ {\matrix{{{\rho_1},{\rho_2}, - {q^{- {1 \over 2}}};q;{{aq} \over {{\rho_1}{\rho_2}}}} \hfill \cr {\sqrt {aq}, - \sqrt a} \hfill \cr}} \right]} \hfill \cr {} \hfill & {+ {{1 - \sqrt a} \over 2}{_3}{\Phi_2}\left[ {\matrix{{{\rho_1},{\rho_2}, - {q^{- {1 \over 2}}};q;{{aq} \over {{\rho_1}{\rho_2}}}} \hfill \cr {- \sqrt {aq},\sqrt a} \hfill \cr}} \right], \left| {{{a{q^{{1 \over 2}}}} \over {{\rho_1}{\rho_2}}}} \right| < 1.} \hfill}

Similar transformations can be established by putting Bailey pairs given in (4.1), (4.2), (4.3), (4.4), (4.5), (4.6), (4.7) and (4.8) in any one of the results given in (2.2), (2.3), (2.5), (2.6) and (2.8).

(b) If we put WP-Bailey pair given in (5.1) and (3.5) we get, (8.2) $\begin{array}{l} \frac{{(\frac{a^{2} q}{k}, \frac{aq}{k}; q)}_{\infty}}{{(aq, \frac{a^{2} q}{k^{2}}; q)}_{\infty}} & _{8} Φ_{7} [\begin{array}{l} a, q \sqrt{a}, - q \sqrt{a}, b, \sqrt{k}, - \sqrt{k}, \sqrt{kq}, - \sqrt{kq}; q; - \frac{a^{2} q}{b k^{2}} \\ \sqrt{a}, - \sqrt{a}, \frac{aq}{b}, \frac{aq}{\sqrt{k}}, - \frac{aq}{\sqrt{k}}, a \sqrt{\frac{q}{k}}, - a \sqrt{\frac{q}{k}} \end{array}] \\ =_{2} Φ_{1} [\begin{array}{l} k, \frac{kb}{a}; q; \frac{a^{2} q}{b k^{2}} \\ \frac{aq}{b} \end{array}] . \end{array}$ \matrix{{{{{{\left({{{{a^2}q} \over k},{{aq} \over k};q} \right)}_\infty}} \over {{{\left({aq,{{{a^2}q} \over {{k^2}}};q} \right)}_\infty}}}\,} \hfill & {\,\,\,\,{\,_8}{\Phi_7}\left[ {\matrix{{a,q\sqrt a, - q\sqrt a,b,\sqrt k, - \sqrt k,\sqrt {kq}, - \sqrt {kq};q; - {{{a^2}q} \over {b{k^2}}}} \hfill \cr {\sqrt a, - \sqrt a,{{aq} \over b},{{aq} \over {\sqrt k}}, - {{aq} \over {\sqrt k}},a\sqrt {{q \over k}}, - a\sqrt {{q \over k}}} \hfill \cr}} \right]} \hfill \cr {} \hfill & {{=_2}{\Phi_1}\left[ {\matrix{{k,{{kb} \over a};q;{{{a^2}q} \over {b{k^2}}}} \hfill \cr {{{aq} \over b}} \hfill \cr}} \right].} \hfill}

Similar transformations can be established by substituting WP-Bailey pairs given in (5.2), (5.3), (5.4), (5.6) and (5.8) in any one of the results given in (3.5), (3.7), (3.10), (3.13), (3.14), (3.17) and (3.18).

(c) Replacing b, c by ρ₁ and ρ₂ in (7.1) respectively and then putting the values of α_n(a, m), β_n(a, m) from (5.1) we get new Bailey pairs. (8.3) $\begin{matrix} α_{n}^{'} (a, k) = \frac{{(a, q \sqrt{a}, - q \sqrt{a}, b, ρ_{1}, ρ_{2}; q)}_{n}}{{(q, \sqrt{a}, - \sqrt{a}, \frac{aq}{b}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} {(\frac{k}{mb})}^{n}, \\ β_{n}^{'} (a, k) = \frac{{(k, \frac{k}{m}, \frac{mq}{ρ_{1}}, \frac{mq}{ρ_{2}}; q)}_{n}}{{(q, mq, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}; q)}_{n}} \sum_{r = 0}^{n} \frac{1 - m q^{2 r}}{1 - m} \frac{{(m, \frac{mb}{a}, ρ_{1}, ρ_{2}, k q^{n}, q^{- n}; q)}_{r} {(\frac{q}{b})}^{r}}{{(q, \frac{aq}{b}, \frac{mq}{ρ_{1}}, \frac{mq}{ρ_{2}}, \frac{m}{k} q^{1 - n}, m q^{1 + n}; q)}_{r}}, \end{matrix}$ \matrix{{\alpha_n^{'}(a,k) = {{{{\left({a,q\sqrt a, - q\sqrt a,b,{\rho_1},{\rho_2};q} \right)}_n}} \over {{{\left({q,\sqrt a, - \sqrt a,{{aq} \over b},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}{{\left({{k \over {mb}}} \right)}^n},} \cr {\beta_n^{'}(a,k) = {{{{\left({k,{k \over m},{{mq} \over {{\rho_1}}},{{mq} \over {{\rho_2}}};q} \right)}_n}} \over {{{\left({q,mq,{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}};q} \right)}_n}}}\sum\limits_{r = 0}^n {{1 - m{q^{2r}}} \over {1 - m}}{{{{\left({m,{{mb} \over a},{\rho_1},{\rho_2},k{q^n},{q^{- n}};q} \right)}_r}{{\left({{q \over b}} \right)}^r}} \over {{{\left({q,{{aq} \over b},{{mq} \over {{\rho_1}}},{{mq} \over {{\rho_2}}},{m \over k}{q^{1 - n}},m{q^{1 + n}};q} \right)}_r}}},}} where $m = \frac{k ρ_{1} ρ_{2}}{aq}$ m = {{k{\rho_1}{\rho_2}} \over {aq}} .

Now, putting these values of $α_{n}^{'} (a, k)$ \alpha_n^{'}(a,k) and $β_{n}^{'} (a, k)$ \beta_n^{'}(a,k) given in (8.3) in (3.1) we get, (8.4) $\begin{array}{l} \frac{{(\frac{k}{m}, \frac{mq}{ρ_{1}}, \frac{mq}{ρ_{2}}, aq; q)}_{n}}{{(mq, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, \frac{k}{a}; q)}_{n}} & _{8} Φ_{7} [\begin{array}{l} m, q \sqrt{m}, - q \sqrt{m}, \frac{mb}{a}, ρ_{1}, ρ_{2}, k q^{n}, q^{- n}; q; \frac{q}{b} \\ \sqrt{m}, - \sqrt{m}, \frac{aq}{b}, \frac{mq}{ρ_{1}}, \frac{mq}{ρ_{2}}, \frac{m}{k} q^{1 - n}, m q^{1 + n} \end{array}] \\ =_{8} Φ_{7} [\begin{array}{l} a, q \sqrt{a}, - q \sqrt{a}, b, ρ_{1}, ρ_{2}, k q^{n}, q^{- n}; q; \frac{aq}{mb} \\ \sqrt{a}, - \sqrt{a}, \frac{aq}{b}, \frac{aq}{ρ_{1}}, \frac{aq}{ρ_{2}}, \frac{a}{k} q^{1 - n}, a q^{1 + n} \end{array}] . \end{array}$ \matrix{{{{{{\left({{k \over m},{{mq} \over {{\rho_1}}},{{mq} \over {{\rho_2}}},aq;q} \right)}_n}} \over {{{\left({mq,{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},{k \over a};q} \right)}_n}}}} \hfill & {\,\,\,\,{\,_8}{\Phi_7}\left[ {\matrix{{m,q\sqrt m, - q\sqrt m,{{mb} \over a},{\rho_1},{\rho_2},k{q^n},{q^{- n}};q;{q \over b}} \hfill \cr {\sqrt m, - \sqrt m,{{aq} \over b},{{mq} \over {{\rho_1}}},{{mq} \over {{\rho_2}}},{m \over k}{q^{1 - n}},m{q^{1 + n}}} \hfill \cr}} \right]} \hfill \cr {} \hfill & {= {\,_8}{\Phi_7}\left[ {\matrix{{a,q\sqrt a, - q\sqrt a,b,{\rho_1},{\rho_2},k{q^n},{q^{- n}};q;{{aq} \over {mb}}} \hfill \cr {\sqrt a, - \sqrt a,{{aq} \over b},{{aq} \over {{\rho_1}}},{{aq} \over {{\rho_2}}},{a \over k}{q^{1 - n}},a{q^{1 + n}}} \hfill \cr}} \right].} \hfill} where $m = \frac{k ρ_{1} ρ_{2}}{aq}$ m = {{k{\rho_1}{\rho_2}} \over {aq}} .

Putting the WP-Bailey pairs given in (5.1), (5.2), (5.3), (5.4), (5.5), (5.6), (5.7) and (5.8) in any one of the results given in (7.1), (7.2), (7.3), (7.4), (7.5) and (7.6) one finds new WP-Bailey pairs, on substituting these new Bailey pairs in (3.1) we get transformations similar to (8.4).

Conclusions

In this paper, certain transformation formulas involving q-hypergeometric series have been obtained by making use of theorems, Bailey Pairs and WP-Bailey Pairs established herein. From these transformation formulas q-series identities can be deduced which may have partition theoretic interpretations. Results of this paper are quite useful and we hope that these results will form the base of further research in the subject.

eISSN:: 2444-8656
Język:: Angielski

Częstotliwość wydawania:: Volume Open
Dziedziny czasopisma:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

Kanał RSS czasopisma

A note on Bailey and WP-Bailey Pairs

Data publikacji: 20 sie 2020

Zakres stron: 143 - 156

Otrzymano: 20 wrz 2019

Przyjęty: 18 sty 2020

DOI: https://doi.org/10.2478/amns.2020.2.00021

Słowa kluczoweBailey pair, WP-Bailey pair, transformation formula, summation formula, basic(q-) hypergeometric series

© 2020 Satya Prakash Singh et al., published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Słowa kluczowe
Bailey pair, WP-Bailey pair, transformation formula, summation formula, basic(q-) hypergeometric series