エンタメ解説

MZVで遊ぼう

MZV,級数,積分

この著者は初心者として投稿しています。間違いや考慮が足りていない点が含まれている可能性が高いです。見つけたらコメント欄で優しく指摘してあげましょう。

あいさつ

んちゃ！
今回はMZVで色々遊びます。
正直後半の議論は怪しいかも()疑いながら読むくらいが丁度いいと思います。

基本事項の復習
調和積・シャッフル積
一般化
応用
最後に

表示

$H_{N} : = \sum_{n = 1}^{N} \frac{1}{n}$
$H_{M} (k_{1}, k_{2}, . . ., k_{N}) : = \sum_{M > n_{1} > n_{2} > \dots > n_{N} \geq 1} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}}$
$S (X, Y) : = {(◻_{0}, x_{1}, ◻_{1}, . . ., x_{| X |}, ◻_{| X |}) | ◻_{k} = {\begin{cases} \emptyset \\ y_{k_{1}}, y_{k_{2}}, . . ., y_{k_{j_{k}}} (k_{1} < k_{2} < \dots < k_{j_{k}}) \end{cases}, m (◻_{k}) = {\begin{cases} 0 (◻_{k} = \emptyset) \\ k_{1} \end{cases}, M (◻_{k}) = {\begin{cases} | Y | (◻_{k} = \emptyset) \\ k_{j_{k}} \end{cases}, ◻_{0} \cup ◻_{1} \cup \dots ◻_{| X |} = Y}$

基本事項の復習

MZV

整数 $k_{1}, k_{2}, . . ., k_{N} \in N (1 < k_{1})$ に対して以下の様な級数を定め、この級数をMultiple Zeta Valueと呼ぶ。
$\begin{array}{r} ζ (k_{1}, k_{2}, . . ., k_{N}) : = \sum_{n_{1} > n_{2} > \dots n_{N} \geq 1} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} \end{array}$
なお $w t : = k_{1} + k_{2} + \dots + k_{N}$ を重み。
$N$ を深さという。

深さ $N$ ,重み $w t$ の異なるMZVの個数を $d (N, w t)$ とおくと下記の様になる。
$d (N, w t) = (\begin{matrix} w t - 2 \\ N - 1 \end{matrix})$
また、重み $w t$ の異なるMZVの個数を $d (w t)$ とすると次の様に書ける。
$d (w t) = 2^{w t - 2}$

[1]
$(1)^{k} : = \overset{k 個}{\overset{⏞}{1 + 1 + 1 \dots + 1}}$ の様に定める。すると以下の様に $w t - N + 1$ は書ける。
$w t - N - 1 = \overset{N 個}{\overset{⏞}{(1)^{k_{1} - 2} + (1)^{k_{2} - 1} + \dots + (1)^{k_{N} - 1}}}$
[2]右辺は $1$ と $+$ がそれぞれ $w t - N - 1$ , $N - 1$ 個ありそれらを並べる個数分だけ表し方がある。
そしてその表し方の分だけ異なる $M Z V$ を取る事が出来るので
$d (N, w t) = (\begin{matrix} w t - 2 \\ N - 1 \end{matrix})$
[3]
$\begin{array}{rcl} d (w t) & = & \sum_{N = 1}^{w t - 1} (\begin{array}{c} w t - 2 \\ N - 1 \end{array}) \\ = & (1 + 1)^{w t - 2} \\ = & 2^{w t - 2} \end{array}$

$ζ (k_{1}, k_{2}, . . ., k_{N})$ は $k_{1}, k_{2}, . . ., k_{N} \in N (1 < k_{1})$ に対して絶対収束する。

[1]
$\begin{array}{rcl} ζ (k_{1}, k_{2}, . . ., k_{N}) & = & \sum_{n_{1} > n_{2} > \dots n_{N} \geq 1} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} \\ < & \sum_{n_{1} > n_{2}} \frac{1}{n_{1}^{2}} (\sum_{n_{3} = 1}^{n_{2}} \frac{1}{n_{3}})^{N - 1} \\ = & \sum_{n_{1} > n_{2}} \frac{1}{n_{1}^{2}} (H_{n_{2}})^{N - 1} \end{array}$
[2]
$\begin{array}{rcl} 0 & < & H_{n} - \log n \\ = & \sum_{k = 1}^{n} \frac{1}{k} - \sum_{k = 1}^{n} \int_{k}^{k + 1} \frac{1}{x} d x \\ = & \sum_{k = 1}^{n} \int_{0}^{1} (\frac{1}{k} - \frac{1}{x + k}) d x \\ = & \sum_{k = 1}^{n} \int_{0}^{1} (\frac{x}{k (x + k)}) d x \\ < & \sum_{k = 1}^{n} \frac{1}{k^{2}} \int_{0}^{1} x d x \\ = & \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k^{2}} < \frac{1}{2} ζ (2) < 1 \end{array}$
ゆえに $H_{n} < 1 + \log n$
[3]下記の様な $(0, \infty)$ 上で定義された関数を考える。
$\begin{array}{r} \forall r \in (0, 1] : f (x) : = \frac{1 + \log x}{x^{r}} \end{array}$
すると
$\begin{array}{rcl} lim_{x \to \infty} f (x) & = & lim_{x \to \infty} \frac{1 + \log x}{x^{r}} \\ = & lim_{x \to \infty} \frac{\log x^{r}}{r x^{r}} \\ = & \frac{1}{r} lim_{y \to \infty} \frac{\log y}{y} \\ = & 0 \end{array}$
ゆえに、 $r = \frac{1}{2 (N - 1)}$ とすると
$\begin{array}{rcl} ζ (k_{1}, k_{2}, . . ., k_{N}) & < & \sum_{n_{1} > n_{2}} \frac{1}{n_{1}^{2}} (H_{n_{2}})^{N - 1} \\ = \sum_{n_{1} > n_{2}} \frac{o (n_{2}^{\frac{1}{2}})}{n_{1}^{2}} & < & + \infty \end{array}$

多重対数関数

整数 $k_{1}, k_{2}, . . ., k_{N} \in N (1 < k_{1})$ に対して以下の様な級数を定め、この級数を多重対数関数と呼ぶ。
$L i_{k_{1}, k_{2}, . . ., k_{N}} (x_{1}, x_{2}, . . ., x_{N}) = \sum_{n_{1} > n_{2} > \dots > n_{N}} \frac{x_{1}^{n_{1}} x_{1}^{n_{2}} \dots x_{N}^{n_{N}}}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} (| x_{k} | < 1)$

Bernoulli数

下記の母関数により定まる数列 ${B_{n}}_{n \in N_{0}}$ をBernouli数と呼ぶ。
$\frac{x}{e^{x} - 1} = \sum_{n = 0}^{\infty} B_{n} \frac{x^{n}}{n!}$

Bernouli数は ${B_{n}}_{n \in N_{0}}$ 下記の性質を満たす。
$\begin{array}{r} {\begin{cases} B_{0} = 1 \\ B_{1} = - \frac{1}{2} \\ \frac{1}{(2 N + 1)!} - \frac{1}{2 (2 N)!} + \sum_{n = 1}^{N} B_{2 n} \frac{1}{(2 N - 2 n + 1)! (2 n)!} = 0 \end{cases} \end{array}$

[1]
$\begin{array}{rcl} x & = & \sum_{m = 1}^{\infty} \sum_{n = 0}^{\infty} B_{n} \frac{x^{m + n}}{m! n!} \\ = & \sum_{N = 1}^{\infty} x^{N} \sum_{n = 0}^{N - 1} B_{n} \frac{1}{(N - n)! n!} \end{array}$
[2]係数比較により下記の式が得られる。
$\begin{array}{r} {\begin{cases} B_{0} = 1 \\ \sum_{n = 0}^{N - 1} B_{n} \frac{1}{(N - n)! n!} = 0 \end{cases} \end{array}$
[3]また
$\begin{array}{rcl} \frac{x}{e^{x} - 1} + \frac{x}{2} & = & \frac{x}{2} \frac{e^{x} + 1}{e^{x} - 1} \\ = & \frac{x}{2} \frac{e^{\frac{x}{2}} + e^{- \frac{x}{2}}}{e^{\frac{x}{2}} - e^{- \frac{x}{2}}} \\ = & \frac{x}{2} \coth \frac{x}{2} \\ = & \frac{- x}{2} \coth \frac{- x}{2} \end{array}$
より $\frac{x}{2} \coth \frac{x}{2}$ は遇関数である事が分かったので
$\begin{array}{r} {\begin{cases} B_{0} = 1 \\ B_{1} = - \frac{1}{2} \\ \frac{1}{(2 N + 1)!} - \frac{1}{2 (2 N)!} + \sum_{n = 1}^{N} B_{2 n} \frac{1}{(2 N - 2 n + 1)! (2 n)!} = 0 \end{cases} \end{array}$

多項式 $f (x)$ が $x = a$ で $m$ 重根を持つ為の必要十分条件は下記の様に与えられる。
『 $f^{(k)} (a) = 0 (k = 0, 1, 2, . . ., m - 1)$ かつ $f^{(m)} (a) = 0$ 』

[ $\Rightarrow$ ] $f^{(k)} (a) = 0 (k = 0, 1, 2, . . ., m - 1)$ かつ $f^{(m)} (a) = 0$ が成り立つとする。
(i) $f (a) = 0$ および因数定理より $\exists g_{0} (x) \in C [x] s . t . f (x) = (x - a) g_{0} (x)$ が成り立つ。
(ii) $f^{(1)} (x) = (x - a) g_{0}^{(1)} (x) + g_{0} (x)$ なので、 $f^{(1)} (a) = g_{0} (a)$ が得られる。(i)同様に因数定理から $\exists g_{1} (x) \in C [x] s . t . g_{0} (x) = (x - a) g_{1} (x)$ が得られる。
(iii)これを帰納的に繰り返し $\exists g_{m - 1} (x) \in C [x] s . t . f (x) = (x - a)^{m} g_{m - 1} (x)$ が得られる。
(iv)最後に $f^{(m)} (x) = \sum_{k = 0}^{m - 1} (\begin{matrix} m \\ k \end{matrix}) \frac{m!}{(m - k)!} (x - a)^{m - k} g_{m - 1}^{(m - k)} (x) + g_{m - 1} (x)$ なので、仮定より $f^{(m)} (a) = g_{m - 1} (a) \neq 0$ が得られる。この事から因数定理より $g_{m - 1} (x)$ は $x - a$ で割る事は出来ない事が分かる。
(v)以上をまとめると、 $\exists g_{m - 1} (x) \in C [x] s . t . f (x) = (x - a)^{m} g_{m - 1} (x)$ が得られ $f (x)$ が $x = a$ で $m$ 重根を持つ事が示された。
[ $\Leftarrow$ ] $f (x)$ が $x = a$ で $m$ 重根を持つとすると $\exists g (x) \in C [x] (g (a) \neq 0) s . t . f (x) = (x - a)^{m} g (x)$ が成り立つ。
ゆえに下記の式が得られるので必要性も示せた。
$\begin{array}{r} {\begin{cases} f^{(k)} (x) = \sum_{l = 0}^{k} (\begin{array}{c} k \\ l \end{array}) (x - a)^{m - l} g^{(l)} (x) \Rightarrow f^{(k)} (a) = 0 \\ f^{(m)} (x) = \sum_{k = 0}^{m - 1} (\begin{array}{c} m \\ k \end{array}) \frac{m!}{(m - k)!} (x - a)^{m - k} g^{(m - k)} (x) + g (x) \Rightarrow f^{(m)} (a) = g (a) \neq 0 \end{cases} \end{array}$

$\sin π x = π x \prod_{n = 1}^{\infty} (1 - \frac{x^{2}}{n^{2}})$

厳密な証明ではありません。
[1] $\sin π x$ は $x = n \in Z$ で1位の零点を持つ。
$\begin{array}{r} {\begin{cases} \sin π n = 0 (n \in Z) \\ π \cos π n = (- 1)^{n} π \neq 0 (n \in Z) \end{cases} \end{array}$
補題2より $x = n \in Z$ で一位の零点を持つ事が分かる。👈本当は補題2が適用できるのか確認する必要がある。
[2][1]から
$\begin{array}{rcl} \sin π x & = & lim_{N \to \infty} C_{N} \prod_{n = - N}^{N} (x - n) \\ = & x lim_{N \to \infty} C_{N} \prod_{n = 1}^{N} (x^{2} - n^{2}) \end{array}$
実際にはこの乗積が $N \to \infty$ で発散しない様にしないといけない。
それらを考慮すると
$\sin π x = x lim_{N \to \infty} {C_{N} \prod_{n = 1}^{N} n^{2}} \prod_{n = 1}^{N} (\frac{x^{2}}{n^{2}} - 1)$
両辺を $x$ で割り $x \to 0$ の極限を取ると下記の式が得られるので $C_{N} = \frac{(- 1)^{N}}{\prod_{n = 1}^{N} n^{2}}$
$1 = lim_{N \to \infty} {C_{N} \prod_{n = 1}^{N} n^{2}} (- 1)^{N}$
これを用いると
$\sin π x = x \prod_{n = 1}^{\infty} (1 - \frac{x^{2}}{n^{2}})$

$\forall n \in N : ζ (2 n) = \frac{(- 1)^{n - 1} 2^{2 n - 1}}{(2 n)!} B_{2 n} π^{2 n}$

[1]
$\begin{array}{r} x \coth \frac{x}{2} = \sum_{n = 1}^{\infty} B_{2 n} \frac{1}{(2 n)!} x^{2 n} \end{array}$
[2]
$2 x \coth x = \sum_{n = 1}^{\infty} B_{2 n} \frac{2^{2 n}}{(2 n)!} x^{2 n}$
[3]
$\begin{array}{rcl} 2 i x \coth (i x) & = & 2 i x \frac{\cos x}{2 i \sin x} \\ = & x \cot x \\ = & \sum_{n = 1}^{\infty} (- 1)^{n} B_{2 n} \frac{2^{2 n}}{(2 n)!} x^{2 n} \end{array}$
[4] $c o t x$ は定理3より $x = π n \in Z$ で一位の極を持つ。その留数を求めると下記の様に計算できる。👈ただし収束性に関して十分に注意せよ。
$\begin{array}{rcl} lim_{x \to π n} (x - π n) \cot x & = & lim_{x \to π n} (x - π n) \frac{\cos x}{\sin x} \\ = & lim_{x \to π n} \frac{x - π n}{\sin (x - π n) \cos π n} \cos x \\ = & 1 \end{array}$
[5]
$\begin{array}{rcl} \cot x & = & \frac{1}{x} + lim_{N \to \infty} \sum_{n = 1}^{N} (\frac{1}{x + π n} + \frac{1}{x - π n}) \\ = & \frac{1}{x} + 2 \sum_{n = 1}^{\infty} \frac{x}{x^{2} - π^{2} n^{2}} \end{array}$
[6]
$\begin{array}{rcl} x \cot x & = & 1 + 2 \sum_{n = 1}^{\infty} \frac{x^{2}}{x^{2} - π^{2} n^{2}} \\ = & 1 - \frac{2}{π^{2}} \sum_{n = 1}^{\infty} \frac{x^{2}}{n^{2}} \sum_{m = 0}^{\infty} \frac{x^{2 m}}{π^{2 m} n^{2 m}} \\ = & 1 - \sum_{m = 0}^{\infty} \frac{2 x^{2 m + 2}}{π^{2 m + 2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2 m + 2}} \\ = & 1 - \sum_{m = 0}^{\infty} \frac{2 x^{2 m + 2}}{π^{2 m + 2}} ζ (2 m + 2) \\ = & 1 - \sum_{m = 1}^{\infty} \frac{2 x^{2 m}}{π^{2 m}} ζ (2 m) \end{array}$
[7]最後に $x^{2 n}$ の係数を比較して証明完了

$\forall k_{1}, k_{2} \in N (k_{1}, k_{2} > 1) : ζ (k_{1}) ζ (k_{2}) = ζ (k_{1}, k_{2}) + ζ (k_{2}, k_{1}) + ζ (k_{1} + k_{2})$

$\begin{array}{rcl} ζ (k_{1}) ζ (k_{2}) & = & \sum_{n_{1}, n_{2} \geq 1} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}}} \\ = & (\sum_{n_{1} > n_{2} \geq 1} + \sum_{n_{2} > n_{1} \geq 1} + \sum_{n_{1} = n_{2} \geq 1}) \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}}} \\ = & ζ (k_{1}, k_{2}) + ζ (k_{2}, k_{1}) + ζ (k_{1} + k_{2}) \end{array}$

$ζ (3) = ζ (2, 1)$

$\begin{array}{rcl} ζ (2, 1) & = & \sum_{n_{1} > n_{2} \geq 1} \frac{1}{n_{1}^{2} n_{2}} \\ = & \sum_{n_{1}, n_{2} \geq 1} \frac{1}{(n_{1} + n_{2})^{2} n_{2}} \\ = & \sum_{n_{1}, n_{2} \geq 1} \frac{1}{n_{1} (n_{1} + n_{2})} (\frac{1}{n_{2}} - \frac{1}{n_{1} + n_{2}}) \\ = & \sum_{n_{1}, n_{2} \geq 1} \frac{1}{n_{1} n_{2} (n_{1} + n_{2})} - ζ (2, 1) \\ = & \sum_{n_{1}, n_{2} \geq 1} \frac{1}{n_{1}^{2}} (\frac{1}{n_{2}} - \frac{1}{n_{1} + n_{2}}) - ζ (2, 1) \\ = & \sum_{n_{1} \geq 1} \frac{1}{n_{1}^{2}} \sum_{n_{2} = 1}^{n_{1}} \frac{1}{n_{3}} - ζ (2, 1) \\ = & ζ (3) \end{array}$

反復積分

$G (a_{1}, a_{2}, . . ., a_{N}; x) : = \int_{0}^{x} d t_{1} \frac{1}{t_{1} - a_{1}} G (a_{2}, a_{3}, . . ., a_{N}; t_{1})$

$(- 1)^{N} G ({0}^{k_{1} - 1}, 1, {0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; x) = L i_{k_{1}, k_{2}, . . ., k_{N}} (\overset{N}{\overset{⏞}{x, x, . . ., x}})$
特に $x = 1$ とすると、以下の結果が得られる。
$(- 1)^{N} G ({0}^{k_{1} - 1}, 1, {0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; 1) = ζ (k_{1}, k_{2}, . . ., k_{N})$

以下 ${0}^{n} = (\overset{n 個}{\overset{⏞}{0, 0, . . ., 0}})$ の様に定める。
そして次の計算を行っていく。
$\begin{array}{r} (- 1)^{N} G ({0}^{k_{1} - 1}, 1, {0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; x) = (- 1)^{N} \overset{k_{1} - 1 個}{\overset{⏞}{\int_{0}^{x} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 2}} d x_{k - 1}}} \int_{0}^{x_{k - 1}} \frac{d x_{k}}{x_{k} - 1} G ({0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; x_{k}) \end{array}$
[1] $N = 1$ の場合
$\begin{array}{rcl} - G ({0}^{k_{1} - 1}, 1; x) & = & - \overset{k_{1} - 1 個}{\overset{⏞}{\int_{0}^{x} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 2}} d x_{k - 1}}} \int_{0}^{x_{k - 1}} \frac{d x_{k}}{x_{k} - 1} \\ = & \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{k_{1}}} \\ = & L i_{k_{1}} (x) \end{array}$
[2]そこで $1, 2, . . ., N - 1$ まで $(- 1)^{N - 1} G ({0}^{k_{1} - 1}, 1, {0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N - 1} - 1}, 1; x) = L i_{k_{1}, k_{2}, . . ., k_{N - 1}} (x)$ が成り立つとする。
[3]
$\begin{array}{rcl} (- 1)^{N} G ({0}^{k_{1} - 1}, 1, {0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; x) & = & - \overset{k_{1} - 1 個}{\overset{⏞}{\int_{0}^{x} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 2}} d x_{k - 1}}} \int_{0}^{x_{k - 1}} \frac{d x_{k}}{x_{k} - 1} (- 1)^{N - 1} G ({0}^{k_{2} - 1}, 0, . . ., {0}^{k_{N} - 1}, 1; x_{k}) \\ = & - \overset{k_{1} - 1 個}{\overset{⏞}{\int_{0}^{x} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 2}} d x_{k - 1}}} \int_{0}^{x_{k - 1}} \frac{d x_{k}}{x_{k} - 1} (- 1)^{N - 1} L i_{k_{2}, k_{3}, . . ., k_{N}} (x_{k}) \\ = & \sum_{n_{2}, n_{3}, \dots, n_{N} \geq 1} \frac{1}{(n_{2} + n_{3} + \dots + n_{N})^{k_{2}} (n_{2} + n_{3} + \dots + n_{N})^{k_{3}} \dots n_{N}^{k_{N}}} (- 1) \overset{k_{1} - 1 個}{\overset{⏞}{\int_{0}^{x} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 2}} d x_{k - 1}}} \int_{0}^{x_{k - 1}} \frac{x_{k}^{n_{2} + n_{3} + \dots + n_{N}} d x_{k}}{x_{k} - 1} \\ = & \sum_{n_{1}, n_{2}, . . ., n_{N} \geq 1} \frac{x^{n_{1} + n_{2} + \dots + n_{N}}}{(n_{1} + n_{2} + \dots + n_{N})^{k_{1}} (n_{2} + n_{3} + \dots + n_{N})^{k_{2}} \dots n_{N}^{k_{N}}} \\ = & L i_{k_{1}, k_{2}, . . ., k_{N}} (\overset{N}{\overset{⏞}{x, x, . . ., x}}) \end{array}$

調和積・シャッフル積

$h = Q < x, y >$ を $Q$ 係数 $x, y$ 変数の非可換多項式環。
また、 $h^{1} = Q + h y, h^{0} = Q + x h y$ とする。
そして $Q -$ 線形関数 $Z : h^{0} \to R$ を次の様に定める。
$\begin{array}{r} {\begin{cases} Z (1) = 1 \\ Z (x^{k_{1} - 1} y x^{k_{2} - 1} y \dots x^{k_{N} - 1} y) = ζ (k_{1}, k_{2}, . . ., k_{N}) \end{cases} \end{array}$

👆今までと違って、MZVを代数的に取り扱える様にする意図があります。
ちなみにMZVの重さ、深さはそれぞれ単項式 $x^{k_{1} - 1} y x^{k_{2} - 1} y \dots x^{k_{N} - 1} y$ 、 $y$ の次数が対応する様になります。

$x y, x^{2} y^{2}, x y x^{5} y x y \in h^{0}$ によって定まる $M Z V$ を求めてください。

[1] $Z (x y) = ζ (2)$
[2] $Z (x^{2} y^{2}) = ζ (3, 1, 1)$
[3] $Z (x y x^{5} y x y) = ζ (2, 6, 2)$

$h^{1} = Q + Q {\sum_{j} z_{k_{1 j}} z_{k_{2 j}} \dots z_{k_{l j}} y | z_{k_{i j}} = x^{k_{i j} - 1} y, k_{i j} \in N}$ の様に書ける。

有理数か終端が $y$ である様な $x, y$ を並べた項の $Q -$ 線形結合全体が $h^{1}$ である事を考えれば明らか。

以下の計算規則
$ζ (k) ζ (l) = ζ (k, l) + ζ (l, k) + ζ (k + l)$
を一般化して取り扱うために、下記の演算を導入します。
発散する場合も同一の規則に従うので $h^{1}$ に積 $*$ を導入します。

調和積

調和積 $* : h^{1} \times h^{1} \to h^{1}$ を下記の様に定める。
$\forall w, w_{1}, w_{2}, w_{3}, z_{k}, z_{l} \in h^{1} : {\begin{cases} w * 1 = 1 * w = w \\ z_{k} w_{1} * z_{l} w_{2} = z_{k} (w_{1} * z_{l} w_{2}) + z_{l} (z_{k} w_{1} * w_{2}) + z_{k} z_{l} (w_{1} * w_{2}) \\ (w_{1} + w_{2}) * w_{3} = w_{1} * w_{3} + w_{1} * w_{3} \\ w_{1} * (w_{2} + w_{3}) = w_{1} * w_{2} + w_{1} * w_{3} \end{cases}$

調和積 $*$ は可換

帰納法を用いて示す。
まず、語 $w_{1}, w_{2}$ の $y$ に関する次数を $d_{1}, d_{2}$ とする。
[1] $(d_{1}, d_{2}) = (1, 1)$ の場合:
この場合は $w_{1} = z_{k_{1}}, w_{2} = z_{l_{1}}$ となる様な自然数 $k_{1}, l_{1}$ を取る事が出来る。
これを用いて計算を行う。
$\begin{array}{rcl} w_{1} * w_{2} & = & (z_{k_{1}} 1 * z_{l_{1}} 1) \\ = & z_{k_{1}} (1 * z_{l_{1}} 1) + z_{l_{1}} (z_{k_{1}} 1 * 1) + z_{k_{1} + l_{1}} (1 * 1) \\ = & z_{k_{1}} z_{l_{1}} + z_{l_{1}} z_{k_{1}} + z_{k_{1} + l_{1}} \\ = & z_{l_{1}} z_{k_{1}} + z_{k_{1}} z_{l_{1}} + z_{l_{1} + k_{1}} \\ = & z_{l_{1}} (1 * z_{k_{1}}) + z_{k_{1}} (z_{l_{1}} 1 * 1) + z_{l_{1} + k_{1}} (1 * 1) \\ = & w_{2} * w_{1} \end{array}$
[2] $(d_{1}, d_{2}) \neq (1, 1)$ の場合
(A) $d_{1}$ を固定し $d_{2} + 1$ の場合を考える。
$\begin{array}{r} {\begin{cases} w_{1} = z_{k_{1}} z_{k_{2}} \dots z_{k_{d_{2}}} \\ w_{2} = z_{l_{1}} z_{l_{2}} \dots z_{l_{d_{2}}} z_{l_{d_{2} + 1}} \end{cases} \end{array}$
$\begin{array}{rcl} w_{1} * w_{2} & = & z_{k_{1}} (z_{k_{2}} \dots z_{k_{d_{2}}} * z_{l_{1}} z_{l_{2}} \dots z_{l_{d_{2}}} z_{l_{d_{2} + 1}}) + z_{l_{1}} (z_{k_{1}} z_{k_{2}} \dots z_{k_{d_{2}}} * z_{l_{2}} \dots z_{l_{d_{2}}} z_{l_{d_{2} + 1}}) + (可換な項) \\ = & z_{k_{1}} z_{k_{2}} (z_{k_{3}} \dots z_{k_{d_{2}}} * z_{l_{1}} z_{l_{2}} \dots z_{l_{d_{2}}} z_{l_{d_{2} + 1}}) + z_{l_{1}} z_{l_{2}} (z_{k_{1}} z_{k_{2}} \dots z_{k_{d_{2}}} * z_{l_{3}} \dots z_{l_{d_{2}}} z_{l_{d_{2} + 1}} + (可換な項) \\ = & z_{1} z_{2} + z_{2} z_{1} + (可換な項) \\ = & z_{2} z_{1} + z_{1} z_{2} + (可換な項) \\ = & w_{2} * w_{1} \end{array}$
(B) $d_{2}$ を固定し $d_{1} + 1$ の場合も同様の計算をすればいい。
[3]最終的に帰納法の仮定より証明完了。

調和積 $*$ は結合法則を持つ

帰納法を用いて示す。
まず、語 $w_{1}, w_{2}, w_{3}$ の $y$ に関する次数を $d_{1}, d_{2}, d_{3}$ とする。
[1] $(d_{1}, d_{2}, d_{3}) = (1, 1, 1)$ の場合:
$\begin{array}{rcl} (w_{k} * w_{l}) * w_{m} & = & (z_{k} * z_{l}) * z_{m} \\ = & {z_{k} z_{l} + z_{l} z_{k} + z_{k + l}} * z_{m} \\ = & z_{k} z_{l} * z_{m} 1 + z_{l} z_{k} * z_{m} 1 + z_{k + l} * z_{m} \\ = & z_{k} (z_{l} * z_{m} 1) + z_{m} (z_{k} z_{l} * 1) + z_{k + m} (z_{l} * 1) + z_{l} (z_{k} * z_{m} 1) + z_{m} (z_{l} z_{k} * 1) + z_{l + m} z_{k} + z_{k + l} z_{m} + z_{m} z_{k + l} + z_{k + l + m} \\ = & z_{k} z_{l} z_{m} + z_{k} z_{m} z_{l} + z_{k} z_{l + m} + z_{m} z_{k} z_{l} + z_{k + m} z_{l} + z_{l} z_{k} z_{m} + z_{l} z_{m} z_{k} + z_{l} z_{k + m} + z_{m} z_{l} z_{k} + z_{l + m} z_{k} + z_{k + l} z_{m} + z_{m} z_{k + l} + z_{k + l + m} \end{array}$
$\begin{array}{rcl} w_{k} * (w_{l} * w_{m}) & = & z_{k} * (z_{l} * z_{m}) \\ = & z_{k} * (z_{l} z_{m} + z_{m} z_{l} + z_{l + m}) \\ = & z_{k} 1 * z_{l} z_{m} + z_{k} 1 * z_{m} z_{l} + z_{k} 1 * z_{l + m} 1 \\ = & z_{k} (1 * z_{l} z_{m}) + z_{l} (z_{k} 1 * z_{m}) + z_{k + l} (1 * z_{m}) + z_{k} (1 * z_{m} z_{l}) + z_{m} (z_{k} 1 * z_{l}) + z_{k + m} (1 * z_{l}) + z_{k} (1 * z_{l + m} 1) + z_{l + m} (z_{k} 1 * 1) + z_{k + l + m} (1 * 1) \\ = & z_{k} z_{l} z_{m} + z_{l} z_{k} z_{m} + z_{l} z_{m} z_{k} + z_{l} z_{k + m} + z_{k + l} z_{m} + z_{k} z_{m} z_{l} + z_{m} z_{k} z_{l} + z_{m} z_{l} z_{k} + z_{m} z_{k + l} + z_{k + m} z_{l} + z_{k} z_{l + m} + z_{l + m} z_{k} + z_{k + l + m} \end{array}$
[2] $(d_{1}, d_{2}, d_{3}) \neq (1, 1, 1)$ まで成り立つと仮定
$\begin{array}{r} {\begin{cases} w_{1} = z_{k_{1}} w_{1}^{^{'}} \\ w_{2} = z_{l_{1}} w_{2}^{^{'}} \\ w_{3} = z_{m_{1}} w_{3}^{^{'}} \end{cases} \end{array}$
$d_{2}, d_{3}$ を固定して $d_{1} + 1$ とした場合
$\begin{array}{rcl} (w_{1} * w_{2}) * w_{3} & = & {z_{k_{1}} (w_{1}^{^{'}} * z_{l_{1}} w_{2}^{^{'}}) + z_{l_{1}} (z_{k_{1}} w_{1}^{^{'}} * w_{2}^{^{'}}) + z_{k_{1} + l_{1}} w_{1}^{^{'}} * w_{2}^{^{'}}} * z_{m_{1}} w_{3}^{^{'}} \\ = & z_{k_{1}} {(w_{1}^{^{'}} * z_{l_{1}} w_{2}^{^{'}}) * z_{m_{1}} w_{3}^{^{'}}} + z_{l_{1}} (z_{k_{1}} w_{1}^{^{'}} * w_{2}^{^{'}}) * z_{m_{1}} w_{3}^{^{'}} + z_{k_{1} + l_{1}} {(w_{1}^{^{'}} * w_{2}^{^{'}}) * z_{m_{1}} w_{3}^{^{'}}} 👈 赤色部分は結合法則が成り立つ事が帰納法の仮定で既に分かっている。以下結合法則が成立している部分は省略する。 \\ = & z_{l_{1}} (z_{k_{1}} w_{1}^{^{'}} * w_{2}^{^{'}}) * z_{m_{1}} w_{3}^{^{'}} + (結合律が成り立つ事が分かっている部分) \\ = & z_{l_{1}} z_{l_{2}} (z_{k_{1}} w_{1}^{^{'}} * w_{2}^{^{″}}) * z_{m_{1}} w_{3}^{^{'}} + (結合律が成り立つ事が分かっている部分) \\ = & z_{l_{1}} z_{l_{2}} \dots z_{l_{d_{2}}} (z_{k_{1}} w_{1}^{^{'}} * 1) * z_{m_{1}} w_{3}^{^{'}} + (結合律が成り立つ事が分かっている部分) \\ = & z_{l_{1}} z_{l_{2}} \dots z_{l_{d_{2}}} {z_{k_{1}} w_{1}^{^{'}} * (1 * z_{m_{1}} w_{3}^{^{'}})} + (結合律が成り立つ事が分かっている部分) 👈 出てくるすべての項が結合律を満たす項で出来ている事が分かる。後は逆向きに構成しなおしていけばいい。 \\ = & w_{1} * (w_{2} * w_{3}) \end{array}$
他の $d_{2}, d_{3}$ に対しても同様に計算すればいい。

調和積 $* : h^{1} \times h^{1} \to h^{1}$ の演算が入っている $h^{1}$ を $h_{*}^{1}$ とする。
また、部分集合 $h^{0} (\subset h^{1})$ に対して同様の調和積を継承させその様な $h^{0}$ を $h_{*}^{0}$ とする。
このとき、 $* -$ 積に関して $Z : h_{*}^{0} \to R$ は同型射になる。

[1]
$\begin{array}{r} {\begin{cases} w_{1} = z_{k_{1}} \\ w_{2} = z_{l_{1}} \end{cases} \end{array}$
$\begin{array}{rcl} Z (w_{1} * w_{2}) & = & Z (z_{k_{1}} z_{l_{1}} + z_{l_{1}} z_{k_{1}} + z_{k_{1} + l_{1}}) \\ = & Z (z_{k_{1}} z_{l_{1}}) + Z (z_{l_{1}} z_{k_{1}}) + Z (z_{k_{1} + l_{1}}) \\ = & ζ (k_{1}, l_{1}) + ζ (l_{1}, k_{1}) + ζ (k_{1} + l_{1}) \\ = & ζ (k_{1}) ζ (l_{1}) \\ = & Z (k_{1}) Z (l_{1}) \end{array}$
[2]
$\begin{array}{r} {\begin{cases} w_{1} = z_{k_{1}} z_{k_{2}} \dots z_{k_{M}} \\ w_{2} = z_{l_{1}} z_{l_{2}} \dots z_{l_{N}} \end{cases} \end{array}$
$\begin{array}{rcl} Z (w_{1} * w_{2}) & = & Z (z_{k_{1}} (z_{k_{2}} \dots z_{k_{M}} * z_{l_{1}} z_{l_{2}} \dots z_{l_{N}})) + Z (z_{l_{1}} (z_{k_{2}} \dots z_{k_{M}} * z_{l_{1}} z_{l_{2}} \dots z_{l_{N}})) + Z (z_{k_{1}} z_{l_{1}} (z_{k_{2}} \dots z_{k_{M}} * z_{l_{2}} \dots z_{l_{N}})) 👈 Z の線形性を用いている。 \end{array}$
[3]次に
$\begin{array}{r} {\begin{cases} S (X, Y) : = {(◻_{0} x_{1} ◻_{1} \dots ◻_{| X - 1 |} x_{N} ◻_{| X |}) | ◻_{k} = {\begin{matrix} \emptyset \\ y_{k_{1}}, y_{k_{2}}, . . ., y_{k_{j_{k}}} (k_{1} < k_{2} < \dots < k_{j_{k}}) \end{matrix}, m (◻_{k}) = {\begin{matrix} 0 (◻_{k} = \emptyset) \\ k_{1} (o t h e r w i s e) \end{matrix}, M (◻_{k}) = {\begin{matrix} | Y | (◻_{k} = \emptyset) \\ k_{j_{k}} (o t h e r w i s e) \end{matrix}, M (◻_{k}) < m (◻_{k + 1}), ◻_{0} \cup ◻_{1} \dots ◻_{| X |} = Y} \\ T (S (X, Y)) : = {z_{1} ⋄_{1} z_{2} ⋄_{2} \dots ⋄_{| Z | - 1} z_{| Z |} | ⋄_{k} = {\begin{matrix} > \\ \geq \end{matrix} ただし、 x_{k} ⋄_{k} x_{k + 1} または y_{k} ⋄ y_{k + 1} ならば ⋄ =>} \\ {\overset{―}{M}}_{k} = (m_{k}, m_{k + 1}, . . ., m_{M}) \\ {\overset{―}{N}}_{k} = (n_{k}, n_{k + 1}, . . ., n_{N}) \end{cases} \end{array}$
の様に定める。
$\begin{array}{rcl} Z (w_{1}) Z (w_{2}) & = & \sum_{m_{1} > m_{2} > \dots > m_{M} \geq 1} \frac{1}{m_{1}^{k_{1}} m_{2}^{k_{2}} \dots m_{M}^{k_{M}}} \sum_{n_{1} > n_{2} > \dots > n_{N} \geq 1} \frac{1}{n_{1}^{l_{1}} n_{2}^{l_{2}} \dots n_{N}^{l_{N}}} \\ = & (\sum_{m_{1} > T (S (M_{2}, N_{1}))} + \sum_{n_{1} \geq T (S (M_{1}, N_{2}))} + \sum_{m_{1} = n_{1} \geq T (S (M_{2}, N_{2}))}) \frac{1}{m_{1}^{k_{1}} m_{2}^{k_{2}} \dots m_{M}^{k_{M}} n_{1}^{l_{1}} n_{2}^{l_{2}} \dots n_{N}^{l_{N}}} \\ = & Z (z_{k_{1}} w_{1}^{^{'}} * w_{2}) + Z (z_{l_{1}} w_{1} * w_{2}^{^{'}}) + Z (z_{k_{1} + l_{1}} w_{1}^{^{'}} * w_{2}^{^{'}}) \\ = & Z (w_{1} * w_{2}) \end{array}$

$ζ (k, l) = {\begin{cases} ζ (3, {1}^{k - 2}) (l = 1) \\ ζ (2, {1}^{l - 1}, 2, {1}^{k - 2}) (l = 2, 3, . . .) \end{cases}$

$\begin{array}{rcl} ζ (k, l) & = & G ({0}^{k - 1}, 1, {0}^{l - 1}, 1; 1) \\ = & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{x_{2}} \dots \int_{0}^{x_{k - 1}} \frac{d x_{k}}{1 - x_{k}} \int_{0}^{x_{k}} \frac{d x_{k + 1}}{x_{k + 1}} \int_{0}^{x_{k + 1}} \frac{d x_{k + 2}}{x_{k + 2}} \dots \int_{0}^{x^{k + l - 1}} \frac{d x_{k + l}}{1 - x_{k + l}} \\ = & \int_{0}^{1} \frac{d x_{k + l}}{1 - x_{k + l}} \int_{x_{k + l}}^{1} \frac{d x_{k + l - 1}}{x_{k + l - 1}} \dots \int_{x_{k + 1}}^{1} \frac{d x_{k}}{1 - x_{k}} \int_{x_{k}}^{1} \frac{d x_{k - 1}}{x_{k - 1}} \int_{x_{k - 1}}^{1} \frac{d x_{k - 2}}{x_{k - 2}} \dots \int_{x_{2}}^{1} \frac{d x_{1}}{x_{1}} \\ = & \int_{0}^{1} \frac{d y_{k + l}}{y_{k + l}} \int_{0}^{y_{k + l}} \frac{d y_{k + l - 1}}{1 - y_{k + l - 1}} \dots \int_{0}^{y_{k + 1}} \frac{d y_{k}}{y_{k}} \int_{0}^{y_{k}} \frac{d y_{k - 1}}{1 - y_{k - 1}} \int_{0}^{y_{k - 1}} \frac{d y_{k - 2}}{1 - y_{k - 2}} \dots \int_{0}^{y_{2}} \frac{d y_{1}}{1 - y_{1}} \\ = & G (0, {1}^{l - 1}, 0, {1}^{k - 1}) \\ = & {\begin{matrix} ζ (3, {1}^{k - 2}) (l = 1) \\ ζ (2, {1}^{l - 1}, 2, {1}^{k - 2}) (l = 2, 3, . . .) \end{matrix} \end{array}$

$ζ (2)^{2} = 2 ζ (2, 2) + 4 ζ (3, 1)$

[1]
$\begin{array}{rcl} ζ (2)^{2} & = & G (0, 1; 1)^{2} \\ = & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{1 - x_{2}} \int_{0}^{1} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{4}}{1 - x_{4}} \\ = & \int_{1 \geq x_{1} \geq x_{2} \geq 0 \land 1 \geq x_{3} \geq x_{4} \geq 0} \frac{d x_{1}}{x_{1}} \frac{d x_{2}}{1 - x_{2}} \frac{d x_{3}}{x_{3}} \frac{d x_{4}}{1 - x_{4}} \end{array}$
[2]
$\begin{array}{r} {\begin{cases} S (X, Y) : = {(◻_{0}, x_{1}, ◻_{1}, . . ., x_{| X |}, ◻_{| X |}) | ◻_{k} = {\begin{matrix} \emptyset \\ y_{k_{1}}, y_{k_{2}}, . . ., y_{k_{j_{k}}} (k_{1} < k_{2} < \dots < k_{j_{k}}) \end{matrix}, m (◻_{k}) = {\begin{matrix} 0 (◻_{k} = \emptyset) \\ k_{1} \end{matrix}, M (◻_{k}) = {\begin{matrix} | Y | (◻_{k} = \emptyset) \\ k_{j_{k}} \end{matrix}, ◻_{0} \cup ◻_{1} \cup \dots ◻_{| X |} = Y} \\ T (S (X, Y)) : = {1 \geq z_{1} \geq z_{2}, . . . \geq z_{| Z |} \geq 0 | (z_{1}, z_{2}, . . ., z_{| Z |}) \in S (X, Y)} \end{cases} \end{array}$
の様に置く。
特に $X = (x_{1}, x_{2}), Y = (x_{3}, x_{4})$ とすると
$\begin{array}{rcl} T (S (X, Y)) = & { & 1 \geq x_{3} \geq x_{4} \geq x_{1} \geq x_{2} \geq 0 \\ , & 1 \geq x_{1} \geq x_{3} \geq x_{4} \geq x_{2} \geq 0 \\ , & 1 \geq x_{1} \geq x_{2} \geq x_{3} \geq x_{4} \geq 0 \\ , & 1 \geq x_{3} \geq x_{1} \geq x_{4} \geq x_{2} \geq 0 \\ , & 1 \geq x_{3} \geq x_{1} \geq x_{2} \geq x_{4} \geq 0 \\ , & 1 \geq x_{1} \geq x_{3} \geq x_{2} \geq x_{4} \geq 0} \end{array}$
以下 $\int_{v \in T (S (X, Y))} = \sum_{v \in T (S (X, Y))} \int_{v}$ の様に記号を定めると
$\begin{array}{rcl} ζ (2)^{2} & = & \int_{v \in T (S (X, Y))} \frac{d x_{1}}{x_{1}} \frac{d x_{2}}{1 - x_{2}} \frac{d x_{3}}{x_{3}} \frac{d x_{4}}{1 - x_{4}} \\ = & \int_{0}^{1} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{4}}{1 - x_{4}} \int_{0}^{x_{4}} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{1 - x_{2}} \\ + & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{4}}{1 - x_{4}} \int_{0}^{x_{4}} \frac{d x_{2}}{1 - x_{2}} \\ + & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{1 - x_{2}} \int_{0}^{x_{2}} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{4}}{1 - x_{4}} \\ + & \int_{0}^{1} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{4}}{1 - x_{4}} \int_{0}^{x_{4}} \frac{d x_{2}}{1 - x_{2}} \\ + & \int_{0}^{1} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{1 - x_{2}} \int_{0}^{x_{2}} \frac{d x_{4}}{1 - x_{4}} \\ + & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{3}}{x_{3}} \int_{0}^{x_{3}} \frac{d x_{2}}{1 - x_{2}} \int_{0}^{x_{2}} \frac{d x_{4}}{1 - x_{4}} \\ = & ζ (2, 2) + ζ (3, 1) + ζ (2, 2) + ζ (3, 1) + ζ (3, 1) + ζ (3, 1) \\ = & 2 ζ (2, 2) + 4 ζ (3, 1) \end{array}$

上記計算は調和積の場合において不等式を $\geq$ だけに制限したものと完全に一致する事が分かると思います。この計算を一般化し代数的に取り扱えるようにするために下記のシャッフル積を導入致します。

シャッフル積

二項演算 $⌣ : h \times h \to h$ を下記の様に定めシャッフル積と呼ぶ。
$\begin{array}{r} {\begin{cases} \forall w \in h : w ⌣ 1 = 1 ⌣ w = w \\ \forall u_{i} \in {x, y}, w_{i} \in h (i = 1, 2) : (u_{1} w_{1}) ⌣ (u_{2} w_{2}) = u_{1} (w_{1} ⌣ u_{2} w_{2}) + u_{2} (u_{1} w_{1} ⌣ w_{2}) \end{cases} \end{array}$

👆定理14において $u_{1} = x_{1}$ は, $u_{2} = x_{3}$ としてみたら意味が分かると思います。

シャッフル積 $⌣$ は結合法則、可換性を満たす。
また、 $⌣ -$ 積に関して $Z : h_{⌣}^{0} \to R$ は同型射になる。

証明は、調和積と同じなので省略。

シャッフル積を用いて下記の公式を証明せよ。
$ζ (2)^{2} = 2 ζ (2, 2) + 4 ζ (3, 1)$

$\begin{array}{rcl} z_{2} ⌣ z_{2} & = & x y ⌣ x y \\ = & x (y ⌣ x y) + x (x y ⌣ y) \\ = & x {y (1 ⌣ x y) + x (y 1 ⌣ y)} + x {x (y ⌣ y 1) + y (x y ⌣ 1)} \\ = & x y x y + 2 x^{2} y^{2} + 2 x^{2} y^{2} + x y x y \\ = & 2 x y x y + 4 x^{2} y^{2} \end{array}$
$\begin{array}{rcl} Z (z_{2} ⌣ z_{2}) & = & 2 Z (x y x y) + 4 Z (x^{2} y^{2}) \\ = & 2 ζ (2, 2) + 4 ζ (3, 1) \\ = & Z (z_{2})^{2} \\ = & ζ (2)^{2} \end{array}$

有限複シャッフル関係式

$\begin{array}{r} \forall w_{1}, w_{2} \in h^{0} : Z (w_{1} * w_{2}) = Z (w_{1} ⌣ w_{2}) \end{array}$

これは $Z$ が $* -$ 積そして $⌣ -$ 積に対して同型なので
$\begin{array}{rcl} Z (w_{1} * w_{2}) & = & Z (w_{1}) Z (w_{2}) \\ = & Z (w_{1} ⌣ z_{2}) \end{array}$

$ζ (3, 1) = \frac{π^{4}}{360}$

[1]
$\begin{array}{rcl} z_{2} * z_{2} & = & z_{2} 1 * z_{2} 1 \\ = & z_{2} (1 * z_{2} 1) + z_{2} (z_{2} 1 * 1) + z_{4} (1 * 1) \\ = & 2 z_{2} z_{2} + z_{4} \\ = & 2 x y x y + x^{3} y \end{array}$
[2]
$z_{2} ⌣ z_{2} = 2 x y x y + 4 x^{2} y^{2}$
[3][1],[2]を用いると
$\begin{array}{rcl} ζ (2)^{2} & = & Z (z_{2} * z_{2}) \\ = & Z (2 x y x y + x^{3} y) \\ = & 2 Z (x y x y) + Z (x^{3} y) \\ = & 2 ζ (2, 2) + ζ (4) \\ = & 2 ζ (2, 2) + 4 ζ (3, 1) \\ = & Z (2 x y x y + 4 x^{2} y^{2}) \\ = & Z (z_{2} ⌣ z_{2}) \end{array}$
[4]以上の計算より
$\begin{array}{rcl} ζ (3, 1) & = & \frac{1}{4} ζ (4) \\ = & - \frac{1}{4} \frac{1}{3} B_{4} π^{4} \\ = & \frac{1}{12} \frac{1}{30} π^{4} \\ = & \frac{π^{4}}{360} \end{array}$

一般化

上記の理論は別に $M Z V$ だけに制限する必要はありません。
調和積は総和と不等式の性質。
シャッフル積は積分と不等式の性質から来ている。
そこで下記の様な $M Z V$ を一般化した級数を考えてみよう。

pochhammer記号

下記の様な記号を定義する。
$\begin{array}{r} \forall n, N \in N : \forall α = (\begin{array}{c} α_{1} \\ α_{2} \\ ⋮ \\ α_{N} \end{array}) \in R^{N} : (α)_{n} = \prod_{k = 1}^{N} \prod_{l = 1}^{n} (α_{k} + l - 1) \end{array}$

一般化

M Z V

ただし、 $α_{i} = (\begin{matrix} α_{11} \\ α_{12} \\ ⋮ \\ α_{i n_{i}} \end{matrix}) \in R^{s_{i}}, β_{i} = (\begin{matrix} β_{i 1} \\ β_{i 2} \\ ⋮ \\ β_{i m_{i}} \end{matrix}) \in R^{t_{i}} (i = 1, 2, . . ., N)$ はそれぞれ任意の有限次元実ベクトル、 $K, L, N, s_{1}, s_{2}, . . ., s_{K}, t_{1}, t_{2}, . . ., t_{L}, k_{1}, k_{2}, . . ., k_{N}$ は自然数。ただし、 $k_{1} > 1$ とする。
$\begin{array}{r} ζ ((\begin{array}{c} α_{1} & α_{2} & \dots & α_{N} \\ β_{1} & β_{2} & \dots & β_{N} \end{array}); k_{1}, k_{2}, . . ., k_{N}) = \sum_{n_{1} > n_{2} > \dots > n_{N} \geq 1} \frac{(α_{1})_{n_{1} - 1} (α_{2})_{n_{2} - 1} \dots (α_{N})_{n_{N} - 1}}{(β_{1})_{n_{1} - 1} (β_{2})_{n_{2} - 1} \dots (β_{N})_{n_{N} - 1}} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} \end{array}$

$\begin{array}{r} \forall N \in N : \forall α \in R : \exists C \in R s . t . (α)_{N} = C (N + α - 1)^{N + α - \frac{1}{2}} e^{- N + O (\frac{1}{N})} \end{array}$

[1]
$f (x) = \log (x + α - 1)$ とおく。するとEuler-Maclaurinの定理より
$\begin{array}{rcl} \log (α)_{N} & = & \sum_{n = 1}^{N} \log (n + α - 1) \\ = & \int_{1}^{N} \log (x + α - 1) + \frac{f (N) + f (1)}{2} + \frac{f^{^{'}} (N) + f^{^{'}} (1)}{12} - \frac{f^{(3)} (N) - f^{(3)} (1)}{720} + \\ = & (N + α - 1) \log (N + α - 1) - N + 1 + \frac{\log (N + α - 1) + \log α}{2} + \frac{\frac{1}{N + α - 1} - \frac{1}{α}}{12} - \frac{\frac{2}{(N + α - 1)^{3}} - \frac{1}{α^{3}}}{720} + \dots \\ = & \log C + \log (N + α - 1)^{N + α - 1} - N + \frac{1}{2} \log (N + α - 1) + O (\frac{1}{N}) \\ = & \log C (N + α - 1)^{N + α - \frac{1}{2}} e^{- N} + O (\frac{1}{N}) \end{array}$
[2]
$(α)_{N} = C (N + α - 1)^{N + α - \frac{1}{2}} e^{- N + O (\frac{1}{N})}$

一般化MZVが絶対収束するための十分条件は
$\begin{array}{r} \forall i \in {1, 2, . . ., N} : s_{i} - t_{j} \geq \sum_{j = 1}^{s_{i}} \log | N + α_{i j} - 1 | - \sum_{j = 1}^{t_{i}} \log | N + β_{i j} - 1 | \end{array}$

[1] $| \frac{(α_{1})_{n_{1}} (α_{2})_{n_{2}} \dots (α_{N})_{n_{N}}}{(β_{1})_{n_{1}} (β_{2})_{n_{2}} \dots (β_{L})_{n_{N}}} |$ が $n_{1}, n_{2}, . . ., n_{N}$ が $o (n_{i}^{r}) (0 \leq r < 1, i = 1, 2, . . ., N)$ となる条件を探す。
$\begin{array}{rcl} | \frac{(α_{1})_{n_{1}} (α_{2})_{n_{2}} \dots (α_{N})_{n_{N}}}{(β_{1})_{n_{1}} (β_{2})_{n_{2}} \dots (β_{L})_{n_{N}}} | & = & | \prod_{i = 1}^{N} \frac{\prod_{j = 1}^{s_{i}} (a_{i j})_{n_{i}}}{\prod_{j = 1}^{t_{i}} (b_{i j})_{n_{i}}} | \\ = & | \prod_{i = 1}^{N} \frac{\prod_{j = 1}^{s_{i}} C (N + α_{i j} - 1)^{n_{i} + α_{i j} - \frac{1}{2}} e^{- n_{i}}}{\prod_{j = 1}^{t_{i}} C (N + β_{i j} - 1)^{n_{i} + β_{i j} - \frac{1}{2}} e^{- n_{i}}} | \\ = & C^{\sum_{i}^{N} (s_{i} - t_{i})} e^{- \sum_{i = 1}^{N} (s_{i} - t_{i}) n_{i}} \prod_{i = 1}^{N} \frac{\prod_{j = 1}^{s_{i}} | (N + α_{i j} - 1)^{n_{i} + α_{i j} - \frac{1}{2}} |}{\prod_{j = 1}^{t_{i}} | (N + β_{i j} - 1)^{n_{i} + β_{i j} - \frac{1}{2}} |} \\ = & C^{\sum_{i}^{N} (s_{i} - t_{i})} e^{- \sum_{i = 1}^{N} (s_{i} - t_{i}) n_{i}} \prod_{i = 1}^{N} \frac{\prod_{j = 1}^{s_{i}} | N + α_{i j} - 1 |^{n_{i} + R e (α_{i j}) - \frac{1}{2}}}{\prod_{j = 1}^{t_{i}} | N + β_{i j} - 1 |^{n_{i} + R e (β_{i j}) - \frac{1}{2}}} \end{array}$
[2]
$\begin{array}{rcl} \log \prod_{i = 1}^{N} \frac{\prod_{j = 1}^{s_{i}} | N + α_{i j} - 1 |^{n_{i} + R e (α_{i j}) - \frac{1}{2}}}{\prod_{j = 1}^{t_{i}} | N + β_{i j} - 1 |^{n_{i} + R e (β_{i j}) - \frac{1}{2}}} & \sim & \sum_{i = 1}^{N} \sum_{j = 1}^{s_{i}} {n_{i} + R e (α_{i j} - \frac{1}{2})} \log | N + α_{i j} - 1 | - \sum_{i = 1}^{N} \sum_{j = 1}^{t_{i}} {n_{i} + R e (β_{i j} - \frac{1}{2})} \log | N + β_{i j} - 1 | \end{array}$
[3] $n_{1}, n_{2}, . . ., n_{N}$ に依存する部分だけが問題なのでその部分だけを抜き出して考えると
$\sum_{i = 1}^{N} n_{i} {- (s_{i} - t_{i}) + \sum_{j = 1}^{s_{i}} \log | N + α_{i j} - 1 | - \sum_{j = 1}^{t_{i}} \log | N + β_{i j} - 1 |} \leq 0$
[4]上記不等式を整理すると与えられた不等式が得られる。
$\begin{array}{r} s_{i} - t_{j} \geq \sum_{j = 1}^{s_{i}} \log | N + α_{i j} - 1 | - \sum_{j = 1}^{t_{i}} \log | N + β_{i j} - 1 | \end{array}$
[5]以上の計算結果を用いると、 $| \frac{(α_{1})_{n_{1} - 1} (α_{2})_{n_{2} - 1} \dots (α_{N})_{n_{N} - 1}}{(β_{1})_{n_{1} - 1} (β_{2})_{n_{2} - 1} \dots (β_{L})_{n_{N} - 1}} |$ は上に有界なので $\exists M \in R (0 < M) s . t . | \frac{(α_{1})_{n_{1} - 1} (α_{2})_{n_{2} - 1} \dots (α_{N})_{n_{N} - 1}}{(β_{1})_{n_{1} - 1} (β_{2})_{n_{2} - 1} \dots (β_{L})_{n_{N} - 1}} | < M$ が成り立つ。これを用いて計算すると一般化されたMZVは絶対収束する事が示された。
$\begin{array}{rcl} | ζ ((\begin{array}{c} α_{1} & α_{2} & \dots & α_{K} \\ β_{1} & β_{2} & \dots & β_{L} \end{array}); k_{1}, k_{2}, . . ., k_{N}) | & \leq & \sum_{n_{1} > n_{2} > \dots > n_{N} \geq 1} | \frac{(α_{1})_{n_{1} - 1} (α_{2})_{n_{2} - 1} \dots (α_{N})_{n_{N}} - 1}{(β_{1})_{n_{1} - 1} (β_{2})_{n_{2} - 1} \dots (β_{L})_{n_{N} - 1}} | \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} \\ \leq & M \sum_{n_{1} > n_{2} > \dots > n_{N} \geq 1} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}} \dots n_{N}^{k_{N}}} \\ < & M ζ (2, \overset{N - 1 個}{\overset{⏞}{1, 1, . . ., 1}}) \end{array}$

連結積(by YANA)

ベクトル $α = (\begin{matrix} α_{1} \\ α_{2} \\ ⋮ \\ α_{m} \end{matrix}) \in R^{m}, β = (\begin{matrix} β_{1} \\ β_{2} \\ ⋮ \\ β_{m} \end{matrix}) \in R^{m}$ に対して以下の様な積を考える。
$\begin{array}{r} α ♯ β : = (\begin{array}{c} α_{1} \\ α_{2} \\ ⋮ \\ α_{m} \\ β_{1} \\ β_{2} \\ ⋮ \\ β_{n} \end{array}) \in R^{m + n} \end{array}$

$\forall N \in N : \forall n_{i} \in N : \forall α_{i} \in R^{n_{i}} (i = 1, 2, . . ., N) : ♯_{i = 1}^{N} α_{i} : = (♯_{i = 1}^{N - 1} α_{i}) ♯ α_{N}$

下記の式を証明せよ。
$ζ^{2} ((\begin{matrix} α_{1} \\ β_{1} \end{matrix}), 2) = 2 ζ ((\begin{matrix} α_{1} & α_{1} \\ β_{1} & β_{1} \end{matrix}), 2, 2) + ζ ((\begin{matrix} α_{1} ♯ α_{1} \\ β_{1} ♯ β_{1} \end{matrix}), 4)$

$\begin{array}{rcl} ζ^{2} ((\begin{array}{c} α_{1} \\ β_{1} \end{array}), 2) & = & \sum_{n_{1} \geq 1} \frac{(α_{1})_{n_{1} - 1}}{(β_{1})_{n_{1} - 1}} \frac{1}{n_{1}^{2}} \sum_{n_{1} \geq 1} \frac{(α_{1})_{n_{2} - 1}}{(β_{1})_{n_{2} - 1}} \frac{1}{n_{2}^{2}} \\ = & (\sum_{n_{1} > n_{2} \geq 1} + \sum_{n_{2} > n_{1} \geq 1} + \sum_{n_{1} = n_{2} = 1}) \frac{(α_{1})_{n_{1} - 1}}{(β_{1})_{n_{1} - 1}} \frac{1}{n_{1}^{2}} \frac{(α_{1})_{n_{2} - 1}}{(β_{1})_{n_{2} - 1}} \frac{1}{n_{2}^{2}} \\ = & 2 ζ ((\begin{array}{c} α_{1} & α_{1} \\ β_{1} & β_{1} \end{array}), 2, 2) + ζ ((\begin{array}{c} α_{1} ♯ α_{1} \\ β_{1} ♯ β_{1} \end{array}), 4) \end{array}$

$ζ ((\begin{matrix} α_{1} \\ β_{1} \end{matrix}), 2) ζ ((\begin{matrix} α_{2} \\ β_{2} \end{matrix}), 2) = ζ ((\begin{matrix} α_{1} & α_{2} \\ β_{1} & β_{2} \end{matrix}), 2, 2) + ζ ((\begin{matrix} α_{2} & α_{1} \\ β_{2} & β_{1} \end{matrix}), 2, 2) + ζ ((\begin{matrix} α_{1} ♯ α_{2} \\ β_{1} ♯ β_{2} \end{matrix}), 4)$

$\begin{array}{rcl} ζ ((\begin{array}{c} α_{1} \\ β_{1} \end{array}), 2) ζ ((\begin{array}{c} α_{2} \\ β_{2} \end{array}), 2) & = & \sum_{n_{1} \geq 1} \frac{(α_{1})_{n_{1} - 1}}{(β_{1})_{n_{1} - 1}} \frac{1}{n_{1}^{2}} \sum_{n_{1} \geq 1} \frac{(α_{2})_{n_{2} - 1}}{(β_{2})_{n_{2} - 1}} \frac{1}{n_{2}^{2}} \\ = & (\sum_{n_{1} > n_{2} \geq 1} + \sum_{n_{2} > n_{1} \geq 1} + \sum_{n_{2} = n_{1} \geq 1}) \frac{(α_{1})_{n_{1} - 1}}{(β_{1})_{n_{1} - 1}} \frac{1}{n_{1}^{2}} \frac{(α_{2})_{n_{2} - 1}}{(β_{2})_{n_{2} - 1}} \frac{1}{n_{2}^{2}} \\ = & ζ ((\begin{array}{c} α_{1} & α_{2} \\ β_{1} & β_{2} \end{array}), 2, 2) + ζ ((\begin{array}{c} α_{2} & α_{1} \\ β_{2} & β_{1} \end{array}), 2, 2) + ζ ((\begin{array}{c} α_{1} ♯ α_{2} \\ β_{1} ♯ β_{2} \end{array}), 4) \end{array}$

$I$ は連続体濃度を持つ添え字集合とする。そして次の様な非可換環を考える。 $y : = Q + Q {\prod_{i \in I} x_{α_{i}, β_{i}}^{k_{i} - 1} y | k_{i} \in N, α_{i} \in R^{s_{i}} / ≃_{s_{i}}, β_{i} \in R^{t_{i}} / ≃_{t_{i}}}$ を考える。
またその部分集合 $y^{1} : = Q + h y$ 、 $h^{0} (α, β) : = Q + x_{α, β} h y$ を定める。
ただし、 $≃$ は次の様な同値関係
$x ≃_{n} y \overset{d e f}{\equiv} σ : R^{n} \to R^{n} が存在して σ (y) = x}$

$Q -$ 線形写像 $Y : y \to R$ を次の様に定める。
$Y (\prod_{i \in {1, 2, . . ., N}} x_{α_{i}, β_{i}}^{k_{i} - 1} y) = ζ ((\begin{matrix} α_{1} & α_{2} & \dots & α_{N} \\ β_{1} & β_{2} & \dots & β_{N} \end{matrix}), k_{1}, k_{2}, . . ., k_{N})$

調和積

二項演算 $* y^{1} \times y^{1} \to y^{1}$ を下記の様に定める。
$\begin{array}{r} {\begin{cases} \forall w \in h^{1} : 1 w = w = w 1 \\ \forall z_{α, β}^{k_{1}} = x_{α, β}^{k_{1} - 1} y, z_{γ, δ}^{k_{2}} = x_{γ, δ}^{k_{2} - 1}, Z, W \in y^{1} : z_{α, β}^{k_{1}} Z * z_{γ, δ}^{k_{2}} W = z_{α, β}^{k_{1}} (Z * z_{γ, δ}^{k_{2}} W) + z_{γ, δ}^{k_{2}} (z_{α, β}^{k_{1}} Z * W) + z_{α ♯ γ, β ♯ δ}^{k_{1} + k_{2}} Z * W \end{cases} \end{array}$

上記の意味を具体例で確認しよう。

実験

今定義した調和積を最も簡単な場合に適用してみよう。
すると次の様な計算が出来る。

\begin{array}{rcl} x_{α, β}^{k_{1} - 1} y * x_{β, γ}^{k_{2} - 1} y & = & x_{α, β}^{k_{1} - 1} y x_{β, γ}^{k_{2} - 1} y + x_{β, γ}^{k_{2} - 1} y x_{α, β}^{k_{1} - 1} y + x_{α ♯ γ, β ♯ δ}^{k_{1} + k_{2} - 1} y \end{array}

これは次の計算に対応している事が分かる。

\begin{array}{rcl} ζ ((\begin{array}{c} α \\ β \end{array}), k_{1}) ζ ((\begin{array}{c} γ \\ δ \end{array}), k_{2}) & = & \sum_{n_{1} \geq 1} \sum_{n_{1} \geq 1} \frac{(α)_{n_{1}} (γ)_{n_{2}}}{(β)_{n_{1}} (δ)_{n_{2}}} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}}} \\ = & (\sum_{n_{1} > n_{2} \geq 1} + \sum_{n_{2} > n_{1} \geq 1} + \sum_{n_{2} = n_{1} \geq 1}) \frac{(α)_{n_{1}} (γ)_{n_{2}}}{(β)_{n_{1}} (δ)_{n_{2}}} \frac{1}{n_{1}^{k_{1}} n_{2}^{k_{2}}} \\ = & ζ ((\begin{array}{c} α & γ \\ β & δ \end{array}), k_{1}, k_{2}) + ζ ((\begin{array}{c} γ & α \\ δ & β \end{array}), k_{2}, k_{1}) + ζ ((\begin{array}{c} α ♯ γ \\ β ♯ δ \end{array}), k_{1} + k_{2}) \end{array}

調和積は可換、結合律を満たす。また、 $Y$ は $* -$ 積に関して同型射を成す。

[1]調和積は可換
まず一般化されたMZVの定義より $ζ ((\begin{matrix} α ♯ γ \\ β ♯ δ \end{matrix}), k) = ζ ((\begin{matrix} β ♯ α \\ δ ♯ β \end{matrix}), k)$ である事に注意する。
つまり $♯$ は可換ではないが、一般化されたMZVでは可換に振る舞うという事である。
以上の事を用いる。
[1A]以下 $z_{1}, z_{2} \in y$ の $y$ の次数 $d_{1}, d_{2}$ により帰納法を用いて示す。
$(d_{1}, d_{2}) = (1, 1)$ の場合： $z_{1} = x_{α, β}^{k_{1} - 1} y, z_{2} = x_{β, γ}^{k_{2} - 1} y$ と書けるので以下の様に計算できる。
$\begin{array}{rcl} x_{α, β}^{k_{1} - 1} y * x_{β, γ}^{k_{2} - 1} y & = & x_{α, β}^{k_{1} - 1} y x_{β, γ}^{k_{2} - 1} y + x_{β, γ}^{k_{2} - 1} y x_{α, β}^{k_{1} - 1} y + x_{α ♯ γ, β ♯ δ}^{k_{1} + k_{2} - 1} y \\ = & x_{β, γ}^{k_{2} - 1} y x_{α, β}^{k_{1} - 1} y + x_{α, β}^{k_{1} - 1} y x_{β, γ}^{k_{2} - 1} y + x_{γ ♯ α, δ ♯ β}^{k_{1} + k_{2} - 1} y \\ = & x_{β, γ}^{k_{2} - 1} y * x_{α, β}^{k_{1} - 1} y \end{array}$
[1-B]
$(d_{1}, d_{2})$ の場合まで成り立つと仮定する。そして $(d_{1} + 1, d_{2})$ でも成り立つ事を示す。
まず、 $z_{1} = \prod_{i = 1}^{d_{1} + 1} z_{α_{i}, β_{i}}^{k_{2} - 1}, z_{2} = \prod_{i = 1}^{d_{2}} z_{γ_{i}, δ_{i}}^{k_{2} - 1}$ の様に置く。
すると下記の様に計算できる。
$\begin{array}{rcl} z_{1} * z_{2} & = & z_{α_{1}, β_{1}}^{k_{1} - 1} (\prod_{i = 2}^{d_{1} + 1} z_{α_{i}, β_{i}}^{k_{2} - 1}, z_{2}) + z_{γ_{1}, δ_{1}}^{k_{1} - 1} (z_{1} * \prod_{i = 2}^{d_{2}} z_{γ_{i}, δ_{i}}^{k_{2} - 1}) + (可換な項) \\ = & z_{α_{1}, β_{1}}^{k_{1} - 1} z_{α_{2}, β_{2}}^{k_{2} - 1} (\prod_{i = 3}^{d_{1} + 1} z_{α_{i}, β_{i}}^{k_{2} - 1}, z_{2}) + z_{γ_{1}, δ_{1}}^{k_{1} - 1} z_{γ_{2}, δ_{2}}^{k_{2} - 1} (z_{1} * \prod_{i = 3}^{d_{2}} x_{γ_{i}, δ_{i}}^{k_{i} - 1} y) + (可換な項) \\ = & \dots \\ = & z_{1} z_{2} + z_{2} z_{1} + (可換な項) \end{array}$
[1-C] $(d_{1}, d_{2} + 1)$ の場合も同様の計算を行えばよいので可換性に関しては証明完了。
[2]結合律：
[2-A] $z_{α_{1}, β_{1}}^{k_{1}}, z_{α_{2}, β_{2}}^{k_{2}}, z_{α_{3}, β_{3}}^{k_{3}}$
$\begin{array}{rcl} z_{α_{1}, β_{1}}^{k_{1}} * (z_{α_{2}, β_{2}}^{k_{2}} * z_{α_{3}, β_{3}}^{k_{3}}) & = & z_{α_{1}, β_{1}}^{k_{1}} * (z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}}) \\ = & z_{α_{1}, β_{1}}^{k_{1}} 1 * z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{1}, β_{1}}^{k_{1}} 1 * z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{1} ♯ α_{2} ♯ α_{3}, β_{1} ♯ β_{2} ♯ β_{3}}^{k_{1} + k_{2} + k_{3}} \\ = & z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{2}, β_{2}}^{k_{2}} (z_{α_{1}, β_{1}}^{k_{1}} 1 * z_{α_{3}, β_{3}}^{k_{3}}) + z_{α_{1} ♯ α_{2}, β_{1} ♯ β_{2}}^{k_{1} + k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} (z_{α_{1}, β_{1}}^{k_{1}} 1 * z_{α_{2}, β_{2}}^{k_{2}}) + z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{1} ♯ α_{2} ♯ α_{3}, β_{1} ♯ β_{2} ♯ β_{3}}^{k_{1} + k_{2} + k_{3}} \\ = & z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} + z_{α_{1} ♯ α_{2}, β_{1} ♯ β_{2}}^{k_{1} + k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1} ♯ α_{2}, β_{1} ♯ β_{2}}^{k_{1} + k_{2}} + z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{1} ♯ α_{2} ♯ α_{3}, β_{1} ♯ β_{2} ♯ β_{3}}^{k_{1} + k_{2} + k_{3}} \end{array}$
$\begin{array}{rcl} (z_{α_{1}, β_{1}}^{k_{1}} * z_{α_{2} * β_{2}}^{k_{2}}) * z_{α_{3}, β_{3}}^{k_{3}} & = & (z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{1} ♯ α_{2}, β_{1} ♯ α_{2}}^{k_{1} + k_{2}}) * z_{α_{3}, β_{3}}^{k_{3}} 1 \\ = & z_{α_{1}, β_{1}}^{k_{1}} (z_{α_{2} * β_{2}}^{k_{2}} * z_{α_{3}, β_{3}}^{k_{3}} 1) + z_{α_{3}, β_{3}}^{k_{3}} (z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} * 1) + z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{2}, β_{2}}^{k_{2}} (z_{α_{1}, β_{1}}^{k_{1}} * z_{α_{3}, β_{3}}^{k_{3}} 1) + z_{α_{3}, β_{3}}^{k_{3}} (z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} * 1) + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{2}, β_{2}}^{k_{1}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} z_{α_{1}, β_{1}} + z_{α_{1} ♯ α_{2}, β_{1} ♯ α_{2}}^{k_{1} + k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1} ♯ α_{2}, β_{1} ♯ α_{2}}^{k_{1} + k_{2}} + z_{α_{1} ♯ α_{2} ♯ α_{3}, β_{1} ♯ β_{2} ♯ β_{3}}^{k_{1} + k_{2} + k_{3}} \\ = & z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1}, β_{1}}^{k_{1}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} z_{α_{2}, β_{2}}^{k_{2}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1} ♯ α_{3}, β_{1} ♯ β_{3}}^{k_{1} + k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{2}, β_{2}}^{k_{2}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{2} ♯ α_{3}, β_{2} ♯ β_{3}}^{k_{2} + k_{3}} z_{α_{1}, β_{1}}^{k_{1}} + z_{α_{1} ♯ α_{2}, β_{1} ♯ α_{2}}^{k_{1} + k_{2}} z_{α_{3}, β_{3}}^{k_{3}} + z_{α_{3}, β_{3}}^{k_{3}} z_{α_{1} ♯ α_{2}, β_{1} ♯ α_{2}}^{k_{1} + k_{2}} + z_{α_{1} ♯ α_{2} ♯ α_{3}, β_{1} ♯ β_{2} ♯ β_{3}}^{k_{1} + k_{2} + k_{3}} \end{array}$
[2-B] $(d_{1}, d_{2}, d_{3}) = (1, 1, 1)$ の場合は正しい事が分かったので、 $(d_{1}, d_{2}, d_{3}) \neq (1, 1, 1)$ まで正しいと仮定して通常のMZV同様の計算を行えばよい。
[3]まず、 $z_{α_{1}, β_{1}}^{k_{1}}, z_{α_{2}, β_{2}}^{k_{2}}, z_{α_{3}, β_{3}}^{k_{3}}$ の場合は上記実験結果から成り立つ事が分かる。
後はMZV同様帰納法を用いて証明できるので省略する。

区間[0,1]で積分可能な関数を変数に持つ関数
$\begin{array}{r} G (f_{1}, f_{2}, . . ., f_{N}; x) = \int_{0}^{x} d t f_{1} (t) G (f_{2}, . . ., f_{N}; t) \end{array}$

$\begin{array}{r} G (f_{1}, f_{2}, . . ., f_{N}; x) = G (f_{N} \circ (1 - x), f_{N - 1} \circ (1 - x), . . ., f_{1} \circ (1 - x); x) \end{array}$

$\begin{array}{rcl} G (f_{1}, f_{2}, . . ., f_{N}; x) & = & \int_{0}^{x} d x_{1} f_{1} (x_{1}) \int_{0}^{x_{1}} d x_{2} f_{2} (x_{2}) \dots \int_{0}^{x_{N - 1}} d x_{N - 1} f_{1} (x_{N}) d x_{N} \\ = & \int_{0}^{1} d x_{N} f (x_{N}) \int_{x_{N}}^{1} d x_{N - 1} f_{N - 1} (x_{N - 1}) \dots \int_{x_{2}}^{1} d x_{1} f_{1} (x_{1}) \\ = & \int_{0}^{1} d y_{N} f (1 - y_{N}) \int_{0}^{y_{N}} d y_{N - 1} f_{N - 1} (1 - y_{N - 1}) \dots \int_{0}^{y_{2}} d y_{1} f_{1} (1 - y_{1}) \\ = & G (f_{N} \circ (1 - x), f_{N - 1} \circ (1 - x), . . ., f_{1} \circ (1 - x); x) \end{array}$

$\sum_{n = 0}^{\infty} \frac{(\begin{matrix} 2 n \\ n \end{matrix})}{2^{2 n}} \frac{1}{(n + 1)^{2}} = 4 (1 - \log 2)$

$\begin{array}{rcl} G (\frac{1}{x}, \frac{1}{\sqrt{1 - x}}; x) & = & \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} \frac{d x_{2}}{\sqrt{1 - x_{2}}} \\ = & \sum_{n = 0}^{\infty} \frac{(\begin{array}{c} 2 n \\ n \end{array})}{2^{n}} \int_{0}^{1} \frac{d x_{1}}{x_{1}} \int_{0}^{x_{1}} d x_{2} x_{2}^{n} \\ = & \sum_{n = 0}^{\infty} \frac{(\begin{array}{c} 2 n \\ n \end{array})}{2^{2 n}} \frac{1}{(n + 1)^{2}} \\ = & G (\frac{1}{\sqrt{x}}, \frac{1}{1 - x}; x) \\ = & \int_{0}^{1} \frac{d x_{2}}{\sqrt{x_{2}}} \int_{0}^{x_{2}} \frac{d x_{1}}{1 - x_{1}} \\ = & \sum_{n = 0}^{\infty} \frac{1}{n + 1} \int_{0}^{1} d x_{2} x_{2}^{n + \frac{1}{2}} \\ = & \sum_{n = 0}^{\infty} \frac{1}{(n + 1) (n + \frac{3}{2})} \\ = & \sum_{n = 1}^{\infty} \frac{1}{n (n + \frac{1}{2})} \\ = & 2 \sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{2}{2 n + 1}) \\ = & 2 lim_{N \to \infty} {2 + 2 H_{N} - 2 \log N - 2 H_{2 N} + 2 \log (2 N) - \frac{2}{2 N + 1} + 2 \log N - 2 \log (2 N)} \\ = & 4 (1 - \log 2) \end{array}$

$f_{i} (x) =_{s_{i} + 1} F_{t_{i}} (α_{i 1}, α_{i 2}, . . ., α_{i s_{i}}, 1; β_{i 1}, β_{i 2}, . . ., β_{i s_{i}}; x)$ とすると
$\begin{array}{r} G ({\frac{1}{x}}^{k_{1} - 1}, f_{1}, . . ., {\frac{1}{x}}^{k_{N} - 1}, f_{N}; 1) = ζ ((\begin{array}{c} α_{1} & α_{2} & \dots & α_{N} \\ β_{1} & β_{2} & \dots & β_{N} \end{array}); k_{1}, k_{2}, . . ., k_{N}) \end{array}$

MZVの場合と同様の計算を行うだけなので省略。

シャッフル積

シャッフル積 $⌣ : y \times y \to y$ を下記の様に定める。
$\begin{array}{r} {\begin{cases} \forall w \in y : w ⌣ 1 = w = 1 ⌣ w \\ \forall u_{1}, u_{2} \in {x_{α, β} | α, β \in \cup_{k = 1}^{\infty} R^{k}} \cup {y} : Z, W \in y : u_{1} Z ⌣ u_{2} W = u_{1} (Z ⌣ u_{2} W) + u_{2} (u_{1} Z ⌣ W) \end{cases} \end{array}$

シャッフル積に関しても通常のMZV同様、結合律、可換律を満たし、そして $Y$ は $⌣ -$ 積に関して同型となる。

MZVと同様なので省略。

応用

$\sum_{n = 0}^{\infty} \frac{x^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})} = \frac{4}{4 - x} + \frac{1}{2 x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \arctan \sqrt{\frac{x}{4 - x}} (| x | < 4)$

$\begin{array}{rcl} \sum_{n = 0}^{\infty} \frac{x^{n}}{(\begin{array}{c} 2 n \\ n \end{array})} & = & \sum_{n = 0}^{\infty} x^{n} \frac{Γ (n + 1) Γ (n + 1)}{Γ (2 n + 1)} \\ = & 1 + \sum_{n = 1}^{\infty} x^{n} n B (n, n + 1) \\ = & 1 + \sum_{n = 0}^{\infty} x^{n} n \int_{0}^{1} t^{n - 1} (1 - t)^{n} d t \\ = & 1 + \int_{0}^{1} \frac{x t (1 - t) d t}{t {1 - x t (1 - t)}^{2}} \\ = & 1 + x \int_{0}^{1} \frac{(1 - t) d t}{{x (t - \frac{1}{2})^{2} + 1 - \frac{x}{4}}^{2}} \\ = & 1 + \frac{1}{x} \int_{0}^{1} \frac{(1 - t) d t}{{(t - \frac{1}{2})^{2} + \frac{4 - x}{4 x}}^{2}} \\ = & 1 + \frac{1}{x} (\frac{4 x}{4 - x})^{2} \sqrt{\frac{4 - x}{4 x}} \int_{- \sqrt{\frac{x}{4 - x}}}^{\sqrt{\frac{x}{4 - x}}} \frac{\frac{1}{2} - \sqrt{\frac{4 - x}{x}} u}{(u^{2} + 1)^{2}} d u (u = \tan θ) \\ = & 1 + \frac{1}{x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \int_{0}^{\arctan \sqrt{\frac{x}{4 - x}}} \cos^{2} θ d θ \\ = & 1 + \frac{1}{x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \frac{1}{2} {\arctan \sqrt{\frac{x}{4 - x}} + \frac{1}{2} \sin (2 \arctan \sqrt{\frac{x}{4 - x}})} \\ = & 1 + \frac{1}{x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \frac{1}{2} {\arctan \sqrt{\frac{x}{4 - x}} + \frac{\sqrt{\frac{x}{4 - x}}}{1 + \frac{x}{4 - x}}} \\ = & 1 + \frac{1}{2 x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \arctan \sqrt{\frac{x}{4 - x}} + \frac{\sqrt{x (4 - x)}}{8 x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \\ = & 1 + \frac{1}{2 x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \arctan \sqrt{\frac{x}{4 - x}} + \frac{x}{4 - x} \\ = & \frac{4}{4 - x} + \frac{1}{2 x} (\frac{4 x}{4 - x})^{\frac{3}{2}} \arctan \sqrt{\frac{x}{4 - x}} \end{array}$

$\begin{array}{r} \sum_{n = 1}^{\infty} \frac{x^{n}}{n (\begin{array}{c} 2 n \\ n \end{array})} = \sqrt{\frac{4 x}{4 - x}} \arctan \sqrt{\frac{x}{4 - x}} \end{array}$

$\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{x^{n}}{n (\begin{array}{c} 2 n \\ n \end{array})} & = & \sum_{n = 1}^{\infty} x^{n} \frac{Γ (n) Γ (n + 1)}{Γ (2 n + 1)} \\ = & \sum_{n = 1}^{\infty} x^{n} \int_{0}^{1} t^{n - 1} (1 - t)^{n} d t \\ = & \int_{0}^{1} \frac{x t (1 - t) d t}{t {1 - x t (1 - t)}} \\ = & \int_{0}^{1} \frac{x (1 - t) d t}{{x (t - \frac{1}{2})^{2} + \frac{4 - x}{4}}} \\ = & \int_{0}^{1} \frac{(1 - t) d t}{{(t - \frac{1}{2})^{2} + \frac{4 - x}{4 x}}} \\ = & \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{(\frac{1}{2} - u) d u}{(u^{2} + \frac{4 - x}{4 x})} \\ = & \frac{1}{2} \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{d u}{u^{2} + \frac{4 - x}{4 x}} \\ = & \sqrt{\frac{4 x}{4 - x}} \int_{0}^{\arctan \sqrt{\frac{x}{4 - x}}} d θ \\ = & \sqrt{\frac{4 x}{4 - x}} \arctan \sqrt{\frac{x}{4 - x}} \end{array}$

定理25を用いると $x = 1$ とした時
$\sum_{n = 1}^{\infty} \frac{1}{n (\begin{matrix} 2 n \\ n \end{matrix})} = \frac{π}{3 \sqrt{3}}$
これを用いると
$\begin{array}{r} \frac{π^{2}}{27} = \sum_{n = 1}^{\infty} \frac{1}{n^{2} {(\begin{array}{c} 2 n \\ n \end{array})}^{2}} + 2 \sum_{n_{1} > n_{2} \geq 1} \frac{1}{n_{1} n_{2} (\begin{array}{c} 2 n_{1} \\ n_{1} \end{array}) (\begin{array}{c} 2 n_{2} \\ n_{2} \end{array})} \end{array}$

$ζ ((\begin{matrix} α_{1} ♯ α_{2} \\ β_{1} ♯ β_{2} \end{matrix}), k_{1} + k_{2}) = 2 {ζ ((\begin{matrix} α_{1} & α_{2} \\ β_{1} & β_{2} \end{matrix}), k_{1}, k_{2}) + {ζ ((\begin{matrix} α_{2} & α_{1} \\ β_{2} & β_{1} \end{matrix}), k_{2}, k_{1})}$

[1]調和積
$\begin{array}{rcl} x_{α_{1}, β_{1}}^{k_{1} - 1} y * x_{α_{2}, β_{2}}^{k_{2} - 1} y & = & x_{α_{1}, β_{1}}^{k_{1} - 1} y x_{α_{2}, β_{2}}^{k_{2} - 1} y + x_{α_{2}, β_{2}}^{k_{2} - 1} y x_{α_{1}, β_{1}}^{k_{1} - 1} y + x_{α_{1} ♯ α_{1}, β_{1} ♯ β_{1}}^{k_{1} + k_{2} - 1} y \end{array}$
[2]シャッフル積
$\begin{array}{rcl} x_{α_{1}, β_{1}}^{k_{1} - 1} y ⌣ x_{α_{2}, β_{2}}^{k_{2} - 1} y & = & x_{α_{1}, β}^{k_{1} - 1} (y 1 ⌣ x_{α_{2}, β_{2}}^{k_{2} - 1} y) + x_{α_{2}, β_{2}}^{k_{2} - 1} (x_{α_{1}, β_{1}}^{k_{1} - 1} y ⌣ y 1) \\ = & x_{α_{1}, β_{1}}^{k_{1} - 1} y (1 ⌣ x_{α_{2}, β_{2}}^{k_{2} - 1} y) + x_{α_{1}, β_{1}}^{k_{1} - 1} x_{α_{2}, β_{2}}^{k_{2} - 1} (y 1 ⌣ y 1) + x_{α_{2}, β_{2}}^{k_{2} - 1} x_{α_{1}, β_{1}}^{k_{1} - 1} (y ⌣ y 1) + x_{α_{2}, β_{2}}^{k_{2} - 1} y (x_{α_{1}, β_{1}} y ⌣ 1) \\ = & x_{α_{1}, β_{1}}^{k_{1} - 1} y x_{α_{2}, β_{2}}^{k_{2} - 1} y + 2 x_{α_{1}, β_{1}}^{k_{1} - 1} x_{α_{2}, β_{2}}^{k_{2} - 1} y^{2} + 2 x_{α_{2}, β_{2}}^{k_{2} - 1} x_{α_{1}, β_{1}}^{k_{1} - 1} y^{2} + x_{α_{2}, β_{2}}^{k_{2} - 1} y x_{α_{1}, β_{1}}^{k_{1} - 1} y \end{array}$
[3] $Y (x_{α_{1}, β_{1}}^{k_{1} - 1} y * x_{α_{2}, β_{2}}^{k_{2} - 1} y) = Y (x_{α_{1}, β_{1}}^{k_{1} - 1} y ⌣ x_{α_{2}, β_{2}}^{k_{2} - 1} y)$ を用いると
$\begin{array}{r} ζ ((\begin{array}{c} α_{1} ♯ α_{2} \\ β_{1} ♯ β_{2} \end{array}), k_{1} + k_{2}) = 2 {ζ ((\begin{array}{c} α_{1} & α_{2} \\ β_{1} & β_{2} \end{array}), k_{1}, k_{2}) + {ζ ((\begin{array}{c} α_{2} & α_{1} \\ β_{2} & β_{1} \end{array}), k_{2}, k_{1})} \end{array}$

最後に

今回は前半部分ではMZVで色々遊び、
後半部分では、MZVの一般化を試みました。
本記事で最も重要な事は不等式の組み合わせを代数的に取り扱えるように調和積・シャッフル積を考えたことです。
この調和積・シャッフル積の構造を壊さない様にパラメータを増やせばいくらでも一般化出来ると思います。
それではばいちゃ！

投稿日：15日前

更新日：14日前