文献あり

対称両側Bailey変換

以下2つの命題はAndrews-Warnaarの2007年の論文において用いられている.

対称両側Bailey変換

$\begin{aligned} β_{n} & = \sum_{k = - n}^{n} α_{k} u_{n - k} v_{n + k} \\ γ_{n} & = \sum_{k = | n |}^{\infty} δ_{k} u_{k - n} u_{k + n} \end{aligned}$
とするとき,
$\begin{aligned} \sum_{n \in Z} α_{n} γ_{n} & = \sum_{0 \leq n} β_{n} δ_{n} \end{aligned}$
が成り立つ.

$\begin{aligned} \sum_{0 \leq n} β_{n} δ_{n} & = \sum_{0 \leq n} δ_{n} \sum_{k = - n}^{n} α_{k} u_{n - k} v_{n + k} \\ = \sum_{k \in Z} α_{k} \sum_{n = | k |}^{\infty} δ_{k} u_{n - k} v_{n + k} \\ = \sum_{k \in Z} α_{k} δ_{k} \end{aligned}$

全く同様に以下が示される.

非対称両側Bailey変換

$\begin{aligned} β_{n} & = \sum_{k = - n - 1}^{n} α_{k} u_{n - k} v_{n + k + 1} \\ γ_{n} & = \sum_{k = max {n, - n - 1}} δ_{k} u_{k - n} v_{k + n + 1} \end{aligned}$
とするとき,
$\begin{aligned} \sum_{n \in Z} α_{n} γ_{n} & = \sum_{0 \leq n} β_{n} δ_{n} \end{aligned}$
が成り立つ.

Andrews-Warnaar(2007)

対称両側Bailey変換において, $u_{n} = v_{n} = \frac{1}{(q^{2}; q^{2})_{n}}$
$\begin{aligned} δ_{n} & = \frac{(q^{2}; q^{2})_{2 n}}{(- q; q)_{2 n + 1}} q^{n} \end{aligned}$
のとき,
$\begin{aligned} γ_{n} & = q^{- n^{2}} \sum_{k = | n |}^{\infty} q^{k^{2} + k} \end{aligned}$
が成り立つ.

$m = | n |$ として,
$\begin{array}{r} \sum_{k = m}^{\infty} \frac{(q^{2}; q^{2})_{2 k} q^{k}}{(- q; q)_{2 k + 1} (q^{2}; q^{2})_{k - m} (q^{2}; q^{2})_{k + m}} = q^{- m^{2}} \sum_{k = m}^{\infty} q^{k^{2} + k} \end{array}$
を示せば良い.
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, - a, a q \\ a^{2}, - a q^{2} \end{array}; q^{2}; q] & = \frac{(- q; q)_{\infty}}{(- a q; q)_{\infty}} \sum_{0 \leq n} a^{n} q^{n^{2}} \end{aligned}$
が示されれば, そこで $a = q^{2 m + 1}$ とすればよい. Jainによる変換公式
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, b, - b \\ b^{2}, a x \end{array}; - x] & = \frac{(x; q)_{\infty}}{(a x; q)_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a, a q \\ b^{2} q \end{array}; q^{2}; x^{2}] \end{aligned}$
において, $a \mapsto a q, b = a, x = - q$ とすると,
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, - a, a q \\ a^{2}, - a q^{2} \end{array}; q^{2}; q] & = \frac{(- q^{2}; q^{2})_{\infty}}{(- a q; q^{2})_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a q, a q^{3} \\ a^{2} q^{2} \end{array}; q^{4}; q^{2}] \end{aligned}$
を得る. $a \mapsto a / q, q^{2} \mapsto q$ として,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, a q \\ a^{2} \end{array}; q^{2}; q] & = \frac{(- q; q)_{\infty}}{(- a; q)_{\infty}} \sum_{0 \leq n} a^{n} q^{\frac{1}{2} n (n - 1)} \end{aligned}$
を示せば良い. ここで, Heineの変換公式
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, b \\ c \end{array}; x] & = \frac{(c / b, b x; q)_{\infty}}{(c, x; q)_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a b x / c, b \\ b x \end{array}; \frac{c}{b}] \end{aligned}$
より,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, a q \\ a^{2} \end{array}; q^{2}; q] & = \frac{(a, a q; q^{2})_{\infty}}{(a^{2}, q; q^{2})_{\infty}}_{2} ϕ_{1} [\begin{array}{c} q^{2}, a \\ a q \end{array}; q^{2}; a] \\ = \frac{(- q; q)_{\infty}}{(- a; q)_{\infty}} \sum_{0 \leq n} \frac{(a; q^{2})_{n}}{(a q; q^{2})_{n}} a^{n} \end{aligned}$
ここで, Rogers-Fineの恒等式より,
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q^{2})_{n}}{(a q; q^{2})_{n}} a^{n} & = \sum_{0 \leq n} a^{2 n} q^{2 n^{2} - n} (1 + a q^{2 n}) \\ = \sum_{0 \leq n} a^{n} q^{\frac{1}{2} n (n - 1)} \end{aligned}$
となるので示すべきことが得られる.

Andrews-Warnaar(2007)

対称両側Bailey変換において, $u_{n} = v_{n} = \frac{1}{(q^{2}; q^{2})_{n}}$
$\begin{aligned} δ_{n} & = (q; q)_{2 n} q^{n} \end{aligned}$
のとき,
$\begin{aligned} γ_{n} & = q^{- 2 n^{2}} \sum_{k = 2 | n |}^{\infty} q^{\frac{1}{2} k (k + 1)} \end{aligned}$

$m = | n |$ として,
$\begin{aligned} \sum_{k = m}^{\infty} \frac{(q; q)_{2 k} q^{k}}{(q^{2}; q^{2})_{k - m} (q^{2}; q^{2})_{k + m}} & = q^{- 2 k^{2}} \sum_{k = 2 m}^{\infty} q^{\frac{1}{2} k (k + 1)} \end{aligned}$
を示せば良い. これは定理2の証明における等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q^{2})_{n}}{(a q; q^{2})_{n}} a^{n} & = \sum_{0 \leq n} a^{n} q^{\frac{1}{2} n (n - 1)} \end{aligned}$
において $a = q^{2 m + 1}$ とすると示される.

定理3, 定理4

$\begin{aligned} β_{n} & = \sum_{k = - n}^{n} \frac{α_{k}}{(q^{2}; q^{2})_{n - k} (q^{2}; q^{2})_{n + k}} \end{aligned}$
のとき,
$\begin{aligned} \sum_{0 \leq n} \frac{(q^{2}; q^{2})_{2 n} q^{n} β_{n}}{(- q; q)_{2 n + 1}} & = \sum_{0 \leq n} q^{n^{2} + n} \sum_{k = - n}^{n} α_{k} q^{- k^{2}} \\ \sum_{0 \leq n} (q; q)_{2 n} q^{n} β_{n} & = \sum_{0 \leq n} q^{\frac{1}{2} n (n + 1)} \sum_{k = - ⌊ \frac{n}{2} ⌋}^{⌊ \frac{n}{2} ⌋} α_{k} q^{- 2 k^{2}} \end{aligned}$

Andrews-Warnaar(2007)

非対称両側Bailey変換において, $u_{n} = v_{n} = \frac{1}{(q^{2}; q^{2})_{n}}$
$\begin{aligned} δ_{n} & = \frac{(q^{2}; q^{2})_{2 n + 1}}{(- q; q)_{2 n + 2}} q^{n} \end{aligned}$
のとき,
$\begin{aligned} γ_{n} & = q^{- n^{2} - n} \sum_{k = max {n, - n - 1}}^{\infty} q^{k^{2} + 2 k} \end{aligned}$
が成り立つ.

$m = max {n, - n - 1}$ として,
$\begin{aligned} \sum_{k = m}^{\infty} \frac{(q; q)_{2 k + 1} q^{k}}{(- q; q)_{2 k + 2} (q^{2}; q^{2})_{k - m} (q^{2}; q^{2})_{k + m + 1}} & = q^{- m^{2} - m} \sum_{k = m}^{\infty} q^{k^{2} + 2 k} \end{aligned}$
を示せば良い. それは定理2の証明における等式
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, - a, a q \\ a^{2}, - a q^{2} \end{array}; q^{2}; q] & = \frac{(- q; q)_{\infty}}{(- a q; q)_{\infty}} \sum_{0 \leq n} a^{n} q^{n^{2}} \end{aligned}$
において, $a = q^{2 m + 2}$ とすれば良い.

Andrews-Warnaar(2007)

対称両側Bailey変換において, $u_{n} = v_{n} = \frac{1}{(q^{2}; q^{2})_{n}}$
$\begin{aligned} δ_{n} & = (q; q)_{2 n + 1} q^{n} \end{aligned}$
のとき,
$\begin{aligned} γ_{n} & = q^{- 2 n^{2} - 2 n} \sum_{k = 2 max {n, - n - 1}}^{\infty} q^{\frac{1}{2} k (k + 3)} \end{aligned}$

$m = max {n, - n - 1}$ として,
$\begin{aligned} \sum_{k = m}^{\infty} \frac{(q; q)_{2 k + 1} q^{k}}{(q^{2}; q^{2})_{k - m} (q^{2}; q^{2})_{k + m + 1}} & = q^{- 2 k^{2} - 2 k} \sum_{k = 2 m}^{\infty} q^{\frac{1}{2} k (k + 3)} \end{aligned}$
を示せば良い. これは定理2の証明における等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q^{2})_{n}}{(a q; q^{2})_{n}} a^{n} & = \sum_{0 \leq n} a^{n} q^{\frac{1}{2} n (n - 1)} \end{aligned}$
において $a = q^{2 m + 2}$ とすると示される.

定理5, 定理6

$\begin{aligned} β_{n} & = \sum_{k = - n - 1}^{n} \frac{α_{k}}{(q^{2}; q^{2})_{n - k} (q^{2}; q^{2})_{n + k + 1}} \end{aligned}$
のとき,
$\begin{aligned} \sum_{0 \leq n} \frac{(q^{2}; q^{2})_{2 n + 1} q^{n} β_{n}}{(- q; q)_{2 n + 2}} & = \sum_{0 \leq n} q^{n^{2} + 2 n} \sum_{k = - n - 1}^{n} α_{k} q^{- k^{2} - k} \\ \sum_{0 \leq n} (q; q)_{2 n + 1} q^{n} β_{n} & = \sum_{0 \leq n} q^{\frac{1}{2} n (n + 3)} \sum_{k = - ⌊ \frac{n}{2} ⌋ - 1}^{⌊ \frac{n}{2} ⌋} α_{k} q^{- 2 k^{2} - 2 k} \end{aligned}$