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Gaussの三角数定理のq級数による証明

非負整数 $n$ を用いて, $N = (\binom{n + 1}{2}) = \frac{n (n + 1)}{2}$ と表されるような $N$ を三角数という. 全ての自然数が3つの三角数の和で表されることはGaussの三角数定理と呼ばれており, Gauss, Legendreによって示されているが, ここでは $q$ 級数による証明を与える.

F. H. Jackson(1921)

$\begin{aligned} \sum_{0 \leq n} \frac{1 - a q^{2 n}}{1 - a} \frac{(a, y; q)_{n}}{(q, a q / y; q)_{n}} q^{n^{2}} a^{n} y^{- n} & = \frac{(a q; q)_{\infty}}{(a q / y; q)_{\infty}} \end{aligned}$

$\begin{aligned} f (a) & := \sum_{0 \leq n} \frac{1 - a q^{2 n}}{1 - a} \frac{(a, y; q)_{n}}{(q, a q / y; q)_{n}} q^{n^{2}} a^{n} y^{- n} \\ = \sum_{0 \leq n} \frac{(1 - q^{n}) + q^{n} (1 - a q^{n})}{1 - a} \frac{(a, y; q)_{n}}{(q, a q / y; q)_{n}} q^{n^{2}} a^{n} y^{- n} \\ = \sum_{0 < n} \frac{1}{1 - a} \frac{(a, y; q)_{n}}{(q; q)_{n - 1} (a q / y; q)_{n}} q^{n^{2}} a^{n} y^{- n} + \sum_{0 \leq n} \frac{1}{1 - a} \frac{(a; q)_{n + 1} (y; q)_{n}}{(q, a q / y; q)_{n}} q^{n^{2} + n} a^{n} y^{- n} \\ = \sum_{0 \leq n} \frac{1}{1 - a} \frac{(a; q)_{n + 1} (y; q)_{n + 1}}{(q; q)_{n} (a q / y; q)_{n + 1}} q^{n^{2} + 2 n + 1} a^{n + 1} y^{- n - 1} + \sum_{0 \leq n} \frac{1}{1 - a} \frac{(a; q)_{n + 1} (y; q)_{n}}{(q, a q / y; q)_{n}} q^{n^{2} + n} a^{n} y^{- n} \\ = \sum_{0 \leq n} \frac{1}{1 - a} \frac{(a; q)_{n + 1} (y; q)_{n}}{(q; q)_{n} (a q / y; q)_{n + 1}} q^{n^{2} + n} a^{n} y^{- n} ((1 - y q^{n}) q^{n + 1} a y^{- 1} + (1 - \frac{a q^{n + 1}}{y})) \\ = \frac{1 - a q}{1 - a q / y} \sum_{0 \leq n} \frac{1 - a q^{2 n + 1}}{1 - a q} \frac{(a q, y; q)_{n}}{(q, a q^{2} / y; q)_{n}} q^{n^{2}} (a q)^{n} y^{- n} \\ = \frac{1 - a q}{1 - a q / y} f (a q) \end{aligned}$
よって, これを繰り返すと,
$\begin{aligned} f (a) & = lim_{N \to \infty} \frac{(a q; q)_{N}}{(a q / y; q)_{N}} f (a q^{N}) \\ = \frac{(a q; q)_{\infty}}{(a q / y)_{\infty}} f (0) \\ = \frac{(a q; q)_{\infty}}{(a q / y)_{\infty}} \end{aligned}$
となって補題が示される.

$α_{0} = β_{0} = 1$ として, $n \geq 1$ に対して,
$\begin{array}{r} α_{n} = (1 - a q^{2 n}) \sum_{j = 0}^{n} \frac{(a q; q)_{n + j - 1} (- 1)^{n - j} q^{(\binom{n - j}{2})}}{(q; q)_{n - j}} β_{j} \end{array}$
とするとき,
$\begin{aligned} \sum_{0 \leq n} (y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})} β_{n} & = \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n} \frac{(y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})}}{(a q / y; q)_{n}} α_{n} \end{aligned}$
が成り立つ.

$\begin{aligned} \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n} \frac{(y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})}}{(a q / y; q)_{n}} α_{n} \\ = \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n} \frac{(y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})} (1 - a q^{2 n})}{(a q / y; q)_{n}} \sum_{j = 0}^{n} \frac{(a q; q)_{n + j - 1} (- 1)^{n - j} q^{(\binom{n - j}{2})}}{(q; q)_{n - j}} β_{j} \\ = \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n, j} \frac{(y; q)_{n + j} a^{n + j} y^{- n - j} q^{(\binom{n + j + 1}{2})} (1 - a q^{2 n + 2 j}) (a q; q)_{n + 2 j - 1} (- 1)^{j} q^{(\binom{n}{2})}}{(a q / y; q)_{n + j} (q; q)_{n}} β_{j} (n \mapsto n + j) \\ = \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n} \frac{(y; q)_{j} a^{j} y^{- j} q^{(\binom{j + 1}{2})} (a q; q)_{2 j} (- 1)^{j}}{(a q / y; q)_{j}} β_{j} \sum_{0 \leq n} \frac{1 - a q^{2 n + 2 j}}{1 - a q^{2 j}} \frac{(a q^{2 j}, y q^{j}; q)_{n}}{(q, a q^{j + 1} / y; q)_{n}} a^{n} y^{- n} q^{n^{2} + n j} \end{aligned}$
ここで, 補題1より,
$\begin{aligned} \sum_{0 \leq n} \frac{1 - a q^{2 n + 2 j}}{1 - a q^{2 j}} \frac{(a q^{2 j}, y q^{j}; q)_{n}}{(q, a q^{j + 1} / y; q)_{n}} a^{n} y^{- n} q^{n^{2} + n j} & = \frac{(a q^{2 j + 1}; q)_{\infty}}{(a q^{j + 1} / y; q)_{\infty}} \end{aligned}$
だから, これを代入することによって補題を得る.

D. Shanks(1958)

非負整数 $n$ に対して,
$\begin{aligned} \frac{(q^{2}; q^{2})_{n}}{(q; q^{2})_{n}} \sum_{j = 0}^{n} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} & = \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \end{aligned}$
が成り立つ.

$\begin{aligned} S_{n} & := \frac{(q^{2}; q^{2})_{n}}{(q; q^{2})_{n}} \sum_{j = 0}^{n} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} \\ T_{n} & := \frac{(q^{2}; q^{2})_{n}}{(q; q^{2})_{n}} \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} \end{aligned}$
とすると,
$\begin{aligned} S_{n} - T_{n} & = \frac{(q^{2}; q^{2})_{n}}{(q; q^{2})_{n}} (\sum_{j = 0}^{n} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} - \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)}) \\ = q^{- (\binom{2 n + 1}{2})} \end{aligned}$
であり,
$\begin{aligned} T_{n} - S_{n - 1} & = \frac{(q^{2}; q^{2})_{n}}{(q; q^{2})_{n}} \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} - \frac{(q^{2}; q^{2})_{n - 1}}{(q; q^{2})_{n - 1}} \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- (n - 1) (2 j + 1)} \\ = \frac{(q^{2}; q^{2})_{n - 1}}{(q; q^{2})_{n}} \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} ((1 - q^{2 n}) - q^{2 j + 1} (1 - q^{2 n - 1})) \\ = \frac{(q^{2}; q^{2})_{n - 1}}{(q; q^{2})_{n}} \sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} ((1 - q^{2 j + 1}) - q^{2 n} (1 - q^{2 j})) \\ = \frac{(q^{2}; q^{2})_{n - 1}}{(q; q^{2})_{n}} (\sum_{j = 0}^{n - 1} \frac{(q; q^{2})_{j + 1}}{(q^{2}; q^{2})_{j}} q^{- n (2 j + 1)} - \sum_{j = 1}^{n - 1} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j - 1}} q^{- n (2 j - 1)}) \\ = \frac{(q^{2}; q^{2})_{n - 1}}{(q; q^{2})_{n}} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{n - 1}} q^{- n (2 n - 1)} \\ = q^{- (\binom{2 n}{2})} \end{aligned}$
よって, これらより,
$\begin{array}{r} S_{n} - S_{n - 1} = q^{- (\binom{2 n + 1}{2})} + q^{- (\binom{2 n}{2})} \end{array}$
である. $S_{0} = 1$ より, 補題が従う.

Andrews(1986)

$\begin{aligned} {(\sum_{0 \leq n} q^{(\binom{n + 1}{2})})}^{3} & = \sum_{0 \leq n} \frac{q^{2 n^{2} + 2 n} (1 + q^{2 n + 1})}{1 - q^{2 n + 1}} \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \end{aligned}$

補題2
$\begin{aligned} \sum_{0 \leq n} (y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})} β_{n} & = \frac{(a q / y; q)_{\infty}}{(a q; q)_{\infty}} \sum_{0 \leq n} \frac{(y; q)_{n} (- a)^{n} y^{- n} q^{(\binom{n + 1}{2})}}{(a q / y; q)_{n}} α_{n} \end{aligned}$
において, $q$ を $q^{2}$ に置き換えて, $a = q^{2}, y = q$
$\begin{array}{r} β_{j} := \frac{(- 1)^{j} (q; q^{2})_{j} q^{- j^{2} - j}}{(q^{2}, q^{3}; q^{2})_{j}} \end{array}$
とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(q; q^{2})_{n}^{2}}{(q^{2}, q^{3}; q^{2})_{n}} q^{n} & = \frac{(q^{3}; q^{2})_{\infty}}{(q^{4}; q^{2})_{\infty}} \sum_{0 \leq n} (- 1)^{n} \frac{1 - q}{1 - q^{2 n + 1}} q^{n^{2} + 2 n} α_{n} \end{aligned}$
を得る. 左辺は Heineの和公式により,
$\begin{aligned} \sum_{0 \leq n} \frac{(q; q^{2})_{n}^{2}}{(q^{2}, q^{3}; q^{2})_{n}} q^{n} & = \frac{(q^{2}; q^{2})_{\infty}^{2}}{(q, q^{3}; q^{2})_{\infty}} \end{aligned}$
である. 一方,
$\begin{aligned} α_{n} & = \frac{(- 1)^{n} (1 - q^{4 n + 2})}{1 - q^{2}} q^{n^{2} - n} \sum_{j = 0}^{n} \frac{(q^{2}; q^{2})_{n + j} (q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{2}, q^{3}; q^{2})_{j}} \\ = \frac{(- 1)^{n} (1 - q^{4 n + 2}) (q^{2}; q^{2})_{n}}{1 - q^{2}} q^{n^{2} - n} \sum_{j = 0}^{n} \frac{(q^{2 n + 2}, q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{2}, q^{3}; q^{2})_{j}} \end{aligned}$
ここで, $q$ 二項定理より
$\begin{aligned} (q^{2 n + 2}; q^{2})_{j} & = \sum_{k = 0}^{j} \frac{(q^{2}; q^{2})_{j}}{(q^{2}; q^{2})_{k} (q^{2}; q^{2})_{j - k}} (- 1)^{k} q^{2 n k + k^{2} + k} \end{aligned}$
であるから,
$\begin{aligned} \sum_{j = 0}^{n} \frac{(q^{2 n + 2}, q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{2}, q^{3}; q^{2})_{j}} \\ = \sum_{j = 0}^{n} \frac{(q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{3}; q^{2})_{j}} \sum_{k = 0}^{j} \frac{(- 1)^{k} q^{2 n k + k^{2} + k}}{(q^{2}; q^{2})_{k} (q^{2}; q^{2})_{j - k}} \\ = \sum_{j = 0}^{n} \sum_{k = 0}^{n} \frac{(- 1)^{k} (q; q^{2})_{j + k} q^{- 2 n j + k^{2} + k}}{(q^{2}; q^{2})_{n - j - k} (q^{3}; q^{2})_{j + k} (q^{2}; q^{2})_{k} (q^{2}; q^{2})_{j}} \\ = \sum_{j = 0}^{n} \frac{(q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{2}, q^{3}; q^{2})_{j}} \sum_{k = 0}^{n} \frac{(q^{2 j + 1}, q^{2 j - 2 n}; q^{2})_{k}}{(q^{2 j + 3}, q^{2}; q^{2})_{k}} q^{2 k (n - j + 1)} \end{aligned}$
ここで, $q$ -Vandermondeの和公式より,
$\begin{aligned} \sum_{k = 0}^{n} \frac{(q^{2 j + 1}, q^{2 j - 2 n}; q^{2})_{k}}{(q^{2 j + 3}, q^{2}; q^{2})_{k}} q^{2 k (n - j + 1)} & = \frac{(q^{2}; q^{2})_{n - j}}{(q^{2 j + 3}; q^{2})_{n - j}} \\ = \frac{(q^{2}; q^{2})_{n - j} (q^{3}; q^{2})_{j}}{(q^{3}; q^{2})_{n}} \end{aligned}$
だから,
$\begin{aligned} \sum_{j = 0}^{n} \frac{(q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{n - j} (q^{2}, q^{3}; q^{2})_{j}} \sum_{k = 0}^{n} \frac{(q^{2 j + 1}, q^{2 j - 2 n}; q^{2})_{k}}{(q^{2 j + 3}, q^{2}; q^{2})_{k}} q^{2 k (n - j + 1)} \\ = \frac{(q^{2}; q^{2})_{n}}{(q^{3}; q^{2})_{n}} \sum_{j = 0}^{n} \frac{(q; q^{2})_{j} q^{- 2 n j}}{(q^{2}; q^{2})_{j}} \end{aligned}$
ここで, 補題3を用いれば,
$\begin{aligned} α_{n} & = \frac{(- 1)^{n} q^{n^{2} - n} (1 - q^{4 n + 2})}{1 - q^{2}} \frac{(q^{2}; q^{2})_{n}}{(q^{3}; q^{2})_{n}} \sum_{j = 0}^{n} \frac{(q; q^{2})_{j}}{(q^{2}; q^{2})_{j}} q^{- 2 n j} \\ = \frac{(- 1)^{n} q^{n^{2}} (1 + q^{2 n + 1})}{1 + q} \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \end{aligned}$
を得る. よってこれより,
$\begin{aligned} \frac{(q^{2}; q^{2})_{\infty}^{2}}{(q, q^{3}; q^{2})_{\infty}} & = \frac{(q^{3}; q^{2})_{\infty}}{(q^{4}; q^{2})_{\infty}} \sum_{0 \leq n} (- 1)^{n} \frac{1 - q}{1 - q^{2 n + 1}} q^{n^{2} + 2 n} α_{n} \\ = \frac{(1 - q) (q; q^{2})_{\infty}}{(q^{2}; q^{2})_{\infty}} \sum_{0 \leq n} \frac{1 + q^{2 n + 1}}{1 - q^{2 n + 1}} q^{2 n^{2} + 2 n} \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \end{aligned}$
より,
$\begin{aligned} {(\frac{(q^{2}; q^{2})_{\infty}}{(q; q^{2})_{\infty}})}^{3} & = \sum_{0 \leq n} \frac{1 + q^{2 n + 1}}{1 - q^{2 n + 1}} q^{2 n^{2} + 2 n} \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \end{aligned}$
最後に補題3において $q \mapsto q^{- 1}$ としてから $n \to \infty$ として得られる式(これはJacobiの三重積からも示せる)
$\begin{aligned} \sum_{0 \leq n} q^{(\binom{n + 1}{2})} & = \frac{(q^{2}; q^{2})_{\infty}}{(q; q^{2})_{\infty}} \end{aligned}$
を用いることによって定理を得る.

$\begin{aligned} \sum_{0 \leq n} \frac{q^{2 n^{2} + 2 n} (1 + q^{2 n + 1})}{1 - q^{2 n + 1}} \sum_{j = 0}^{2 n} q^{- (\binom{j + 1}{2})} \\ = \sum_{0 \leq n} q^{n} \frac{1 + q^{2 n + 1}}{1 - q^{2 n + 1}} \sum_{j = 0}^{2 n} q^{\frac{j (4 n + 1 - j)}{2}} \end{aligned}$
と書き換えれば, 全ての自然数 $n$ に対し, $q^{n}$ の係数が $1$ 以上であることが分かる. よって, 以下を得る.