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Henriciの三重積公式

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前の記事( 超幾何級数の積 , 超幾何級数の積2 )において, 超幾何級数の積を再び超幾何級数で表すタイプの公式を示したが, それらは全て2つの積に関するものだった. 今回は超幾何級数の3つの積が再び超幾何級数になる以下のような公式を示す.

Henrici(1987)

$ω := e^{\frac{2 π i}{3}}$ とするとき,
$\begin{aligned} _{0} F_{1} [\begin{array}{c} - \\ a \end{array}; x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω^{2} x] \\ =_{2} F_{7} [\begin{array}{c} \frac{a}{2} - \frac{1}{4}, \frac{a}{2} + \frac{1}{4} \\ a, \frac{a}{3}, \frac{a + 1}{3}, \frac{a + 2}{3}, \frac{2 a - 1}{3}, \frac{2 a}{3}, \frac{2 a + 1}{3} \end{array}; {(\frac{4}{9} x)}^{3}] \end{aligned}$
が成り立つ.

$_{0} F_{1}$ はBessel関数として古くから研究されてきたが, 上の等式が示されたのは1987年と比較的最近である. 超幾何級数が何らかの累乗になっている場合や指数関数で書ける場合を除いて, パラメータが入った超幾何級数の3つの積が再び超幾何級数で表されるというタイプの結果は他に見たことが無い.

超幾何級数が何らかの累乗になっている場合として, 例えば $_{1} F_{0}$ の場合,
$\begin{array}{r} _{1} F_{0} [\begin{array}{c} a \\ - \end{array}; x] = (1 - x)^{- a} \end{array}$
であるので,
$\begin{aligned} _{1} F_{0} [\begin{array}{c} a \\ - \end{array}; x]_{1} F_{0} [\begin{array}{c} a \\ - \end{array}; ω x]_{1} F_{0} [\begin{array}{c} a \\ - \end{array}; ω^{2} x] & = (1 - x^{3})^{- a} \\ =_{1} F_{0} [\begin{array}{c} a \\ - \end{array}; x^{3}] \end{aligned}$
と表される.

定理の証明の前にいくつか補題を用意する. Appellの超幾何関数を
$\begin{array}{r} _{} F_{2} [\begin{array}{c} a, b, c \\ d, e \end{array}; x, y] := \sum_{0 \leq n, m} \frac{(a)_{n + m} (b)_{n} (c)_{m}}{n! m! (d)_{n} (e)_{m}} x^{n} y^{m} \\ _{} F_{4} [\begin{array}{c} a, b \\ c, d \end{array}; x, y] := \sum_{0 \leq n, m} \frac{(a, b)_{n + m}}{n! m! (c)_{n} (d)_{m}} x^{n} y^{m} \end{array}$
とする.

二次変換公式

$\begin{aligned} _{} F_{4} [\begin{array}{c} a, a + \frac{1}{2} - b \\ c, b + \frac{1}{2} \end{array}; x, y^{2}] & = (1 + y)^{- 2 a}_{} F_{2} [\begin{array}{c} a, a + \frac{1}{2} - b, b \\ c, 2 b \end{array}; \frac{x}{(1 + y)^{2}}, \frac{4 y}{(1 + y)^{2}}] \end{aligned}$

$\begin{aligned} _{} F_{4} [\begin{array}{c} a, a + \frac{1}{2} - b \\ c, b + \frac{1}{2} \end{array}; x, y^{2}] & = \sum_{0 \leq n, m} \frac{{(a, a + \frac{1}{2} - b)}_{n + m}}{n! m! (c)_{n} {(b + \frac{1}{2})}_{m}} x^{n} y^{2 m} \\ = \sum_{0 \leq n} \frac{{(a, a + \frac{1}{2} - b)}_{n}}{n! (c)_{n}} x^{n}_{2} F_{1} [\begin{array}{c} a + n, a + \frac{1}{2} - b + n \\ b + \frac{1}{2} \end{array}; y^{2}] \end{aligned}$
ここで, $_{2} F_{1}$ の二次変換公式
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, a + \frac{1}{2} - b \\ b + \frac{1}{2} \end{array}; y^{2}] & = \frac{1}{(1 + y)^{2 a}}_{2} F_{1} [\begin{array}{c} a, b \\ 2 b \end{array}; \frac{4 y}{(1 + y)^{2}}] \end{aligned}$
において, $a \mapsto a + n$ として,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a + n, a + \frac{1}{2} - b + n \\ b + \frac{1}{2} \end{array}; y^{2}] & = \frac{1}{(1 + y)^{2 a + 2 n}}_{2} F_{1} [\begin{array}{c} a + n, b \\ 2 b \end{array}; \frac{4 y}{(1 + y)^{2}}] \end{aligned}$
を得る. よって,
$\begin{aligned} _{} F_{4} [\begin{array}{c} a, a + \frac{1}{2} - b \\ c, b + \frac{1}{2} \end{array}; x, y^{2}] & = (1 + y)^{- 2 a} \sum_{0 \leq n} \frac{{(a, a + \frac{1}{2} - b)}_{n}}{n! (c)_{n}} {(\frac{x}{(1 + y)^{2}})}^{n}_{2} F_{1} [\begin{array}{c} a + n, b \\ 2 b \end{array}; \frac{4 y}{(1 + y)^{2}}] \\ = (1 + y)^{- 2 a} \sum_{0 \leq n} \frac{(a)_{n + m} {(a + \frac{1}{2} - b)}_{n} (b)_{m}}{n! m! (c)_{n} (2 b)_{m}} {(\frac{x}{(1 + y)^{2}})}^{n} {(\frac{4 y}{(1 + y)^{2}})}^{m} \end{aligned}$
が得られる.

$a = - n$ が負整数のとき,
$\begin{aligned} _{} F_{2} [\begin{array}{c} a, b, c \\ d, e \end{array}; 1, x] & = \frac{(d - b)_{n}}{(d)_{n}} \sum_{0 \leq m} \frac{(a, c, 1 + a - d)_{m}}{m! (e, 1 + a + b - d)_{m}} x^{m} \end{aligned}$

Gaussの超幾何定理より,
$\begin{aligned} _{} F_{2} [\begin{array}{c} a, b, c \\ d, e \end{array}; 1, x] & = \sum_{0 \leq n, m} \frac{(a)_{n + m} (b)_{n} (c)_{m}}{n! m! (d)_{n} (e)_{m}} x^{m} \\ = \sum_{0 \leq m} \frac{(a)_{n + m} (b)_{n} (c)_{m}}{n! m! (d)_{n} (e)_{m}} x^{m}_{2} F_{1} [\begin{array}{c} a + m, b \\ e \end{array}; 1] \\ = \sum_{0 \leq m} \frac{(a, c)_{m}}{m! (e)_{m}} x^{m}_{2} F_{1} [\begin{array}{c} a + m, b \\ d \end{array}; 1] \\ = \sum_{0 \leq m} \frac{(a, c)_{m}}{m! (e)_{m}} x^{m} \frac{Γ (d) Γ (d - a - b - m)}{Γ (d - a - m) Γ (d - b)} \\ = \frac{(d - b)_{n}}{(d)_{n}} \sum_{0 \leq m} \frac{(a, c, 1 + a - d)_{m}}{m! (e, 1 + a + b - d)_{m}} x^{m} \end{aligned}$
となって示される.

$n = 3 k$ のとき,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} - n, a - \frac{1}{2}, 1 - n - a \\ 2 a - 1, 2 - 2 a - 2 n \end{array}; 4] & = \frac{(3 k)! {(a - \frac{1}{2})}_{2 k}}{4^{k} k! (a)_{k} {(a - \frac{1}{2})}_{3 k}} \end{aligned}$
が成り立つ.

Baileyによる3次変換公式
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 3 a, b, 3 a - b + \frac{1}{2} \\ 2 b, 6 a - 2 b + 1 \end{array}; 4 x] & = (1 - x)^{- 3 a}_{3} F_{2} [\begin{array}{c} a, a + \frac{1}{3}, a + \frac{2}{3} \\ b + \frac{1}{2}, 3 a - b + 1 \end{array}; \frac{27 x^{2}}{4 (1 - x)^{3}}] \end{aligned}$
において, $a = - n$ とすると, 右辺は
$\begin{array}{r} (1 - x)^{3 n}_{3} F_{2} [\begin{array}{c} - n, - n + \frac{1}{3}, - n + \frac{2}{3} \\ b + \frac{1}{2}, 1 - 3 n - b \end{array}; \frac{27 x^{2}}{4 (1 - x)^{3}}] \end{array}$
となる. ここで, $x \to 1$ とすると, 超幾何級数の定数項だけが残り, それは
$\begin{array}{r} \frac{(3 n)! (b)_{2 n}}{4^{n} n! {(b + \frac{1}{2})}_{n} (b)_{3 n}} \end{array}$
に一致する. よって,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} - 3 n, b, - 3 n - b + \frac{1}{2} \\ 2 b, 1 - 6 n - 2 b \end{array}; 4] & = \frac{(3 n)! (b)_{2 n}}{4^{n} n! {(b + \frac{1}{2})}_{n} (b)_{3 n}} \end{aligned}$
を得る. $b \mapsto a - \frac{1}{2}, n \mapsto k$ とすれば, 示すべき等式になる.

以下は, Karlsson-Srivastavaによる1990年の証明である.

定理1の証明

まず, 左辺の $3 k + 1, 3 k + 2$ 次の係数は $0$ である. 左辺を展開すると,
$\begin{aligned} _{0} F_{1} [\begin{array}{c} - \\ a \end{array}; x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω^{2} x] \\ = \sum_{0 \leq n} x^{n} \sum_{0 \leq i, j} \frac{ω^{i} ω^{2 j}}{i! (a)_{i} j! (a)_{j} (n - i - j)! (a)_{n - i - j}} \\ = \sum_{0 \leq n} \frac{x^{n}}{n! (a)_{n}} \sum_{0 \leq i, j} \frac{(- n, 1 - n - a)_{i + j}}{i! (a)_{i} j! (a)_{j}} ω^{i} ω^{2 j} \\ = \sum_{0 \leq n} \frac{x^{n}}{n! (a)_{n}}_{} F_{4} [\begin{array}{c} - n, 1 - n - a \\ a, a \end{array}; ω, ω^{2}] \end{aligned}$
となる. ここで, 補題2, 補題3, 補題4を用いると, $n = 3 k$ のとき,
$\begin{aligned} _{} F_{4} [\begin{array}{c} - n, 1 - n - a \\ a, a \end{array}; ω, ω^{2}] & = ω^{n}_{} F_{2} [\begin{array}{c} - n, 1 - n - a, a - \frac{1}{2} \\ a, 2 a - 1 \end{array}; 1, 4] \\ = \frac{4^{n} {(a - \frac{1}{2})}_{n}}{(2 a - 1)_{n}}_{3} F_{2} [\begin{array}{c} - n, a - \frac{1}{2}, 1 - n - a \\ 2 a - 1, 2 - 2 a - 2 n \end{array}; 4] \\ = \frac{2^{4 k} (3 k)! {(a - \frac{1}{2})}_{2 k}}{(2 a - 1)_{3 k} k! (a)_{k}} \end{aligned}$
であるから, これを代入して定理を得る.

この等式は
$\begin{aligned} _{0} F_{1} [\begin{array}{c} - \\ a \end{array}; x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω x]_{0} F_{1} [\begin{array}{c} - \\ a \end{array}; ω^{2} x] & = \sum_{0 \leq n} \frac{{(2 a - 1)}_{2 n}}{n! (a)_{n} (a)_{2 n} (a, 2 a - 1)_{3 n}} x^{3 n} \end{aligned}$
と書くこともできる. 例として, $a = 1$ の場合,
$\begin{aligned} \sum_{0 \leq n} \frac{x^{n}}{n!^{2}} \sum_{0 \leq m} \frac{(ω x)^{m}}{m!^{2}} \sum_{0 \leq l} \frac{(ω^{2} x)^{l}}{l!^{2}} & = \sum_{0 \leq n} \frac{(4 n)!}{n!^{2} (2 n)! (3 n)!^{2}} x^{3 n} \end{aligned}$
が成り立つ.