総和の積と代数
あいさつ
んちゃ！
今までも何回か同じ様な事を書いてきましたけど、可能な限り調和積とシャッフル積を一般化します。
級数と積の研究1
引数の数は可変長で複素数列を引数に持つ関数$S$を以下の様に定める。
\begin{equation}
S_{N}(\{a_{1n}\},\{a_{2n}\},...,\{a_{rn}\})\coloneqq\sum_{0\lt n_{1}\lt n_{2}\lt\cdots\lt n_{r}\leq N}\prod_{k=1}^{r}a_{kn_{k}}
\end{equation}
次の様な記号を定める。
\begin{equation}
\tilde{S}_{N}(\{a_{1n}\},\{a_{2n}\},...,\{a_{rn}\},\{b_{1n}\},\{b_{2n}\},\cdots\{b_{sn}\})\coloneqq
S_{N}(\{a_{1n}\},\{a_{2n}\},...,\{a_{rn}\})S_{N}(\{b_{1n}\},\{b_{2n}\},...,\{b_{sn}\})
\end{equation}
すると以下の式が成り立つ。
\begin{align}
&\tilde{S}_{N}(\{a_{1n}\},\{a_{2n}\},...,\{a_{rn}\},\{b_{1n}\},\{b_{2n}\},\cdots\{b_{sn}\})\\
&=\sum_{0\lt m_{r}\lt n_{s}}b_{sn_{s}}\tilde{S}_{m_{r}}(\{a_{1n}\},\{a_{2n}\},...,\{a_{rn}\},\{b_{1n}\},\{b_{2n}\},\cdots\{b_{s-1n}\})\\
&+\sum_{0\lt n_{s}\lt m_{r}}a_{rm_{r}}\tilde{S}_{n_{s}}(\{a_{1n}\},\{a_{2n}\},...,\{a_{r-1n}\},\{b_{1n}\},\{b_{2n}\},\cdots\{b_{sn}\})\\
&+\sum_{0\lt m_{r}=n_{s}}a_{rm_{r}}b_{sn_{s}}\tilde{S}_{m_{r}}(\{a_{1n}\},\{a_{2n}\},...,\{a_{r-1n}\},\{b_{1n}\},\{b_{2n}\},\cdots\{b_{s-1n}\})
\end{align}
ただ計算するだけ。
アルファベット$\Sigma$および単語$\Sigma^{\ast}$を以下の様に定める。
\begin{eqnarray}
\left\{
\begin{array}{l}
\Sigma_{1}\coloneqq\{1\}\sqcup\{x_{1},x_{2},...,x_{r}\}\\
\Sigma_{2}\coloneqq\{y_{1}\diamondsuit y_{2}\diamondsuit\cdots\diamondsuit y_{n}|n\in\mathbb{N},y_{1},y_{2},...,y_{n}\in\Sigma_{1}\}/\sim\\
y_{1}\diamondsuit y_{2}\diamondsuit\cdots\diamondsuit y_{n}\sim y_{1}^{'}\diamondsuit y_{2}^{'}\diamondsuit\cdots\diamondsuit y_{n}^{'}\overset{def}\Leftrightarrow\exists\sigma\in\mathfrak{S}_{n}\ s.t.\ y_{\sigma(1)}^{'}\diamondsuit y_{\sigma(2)}^{'}\diamondsuit \cdots\diamondsuit y_{\sigma(n)}^{'}=y_{1}\diamondsuit y_{2}\diamondsuit\cdots\diamondsuit y_{n}\\
\Sigma\coloneqq\Sigma_{2}\\
\Sigma^{\ast}\coloneqq\{y_{1}y_{2}\cdots y_{n}|n\in\mathbb{N},y_{1},y_{2},...,y_{n}\in\Sigma\}
\end{array}
\right.
\end{eqnarray}
$\mathbb{Q}$-係数ベクトル空間:$\mathfrak{H}\coloneqq\mathbb{Q}<\Sigma^{\ast}>\coloneqq\bigoplus_{z\in\Sigma^{\ast}}\mathbb{Q}z$を定義する。この時、$\mathbb{Q}$-双対線形写像:$\ast:\mathfrak{H}\times \mathfrak{H}\rightarrow\mathfrak{H}$を次の様に定める。
- \begin{align} \forall w\in\Sigma:1\ast w=w\ast 1=w \end{align}
- \begin{align} &\forall a_{1},a_{2}\in\Sigma_{1}:\forall w_{1},w_{2}\in\Sigma:\\ &(w_{1}a_{1})\ast(w_{2}a_{2})=(w_{1}\ast(w_{2}a_{2}))a_{1}+((w_{1}a_{1})\ast w_{2})a_{2}+(w_{1}\ast w_{2})(a_{1}\diamondsuit a_{2})\\ \end{align}
調和積$\ast$は結合律を満たす。
自然数$M\in\mathbb{N}$そして数列$\{a_{mn}\}_{m,n\in\mathbb{N}}$をパラメータに持つ$\mathbb{Q}$-線形写像$H_{M}(\{a_{mn}\}_{m,n\in\mathbb{N}};\ast):\mathfrak{H}\rightarrow\mathbb{C}$を以下の様に定める。
\begin{align}
&\forall w\in\Sigma^{\ast}:\exists N\in\mathbb{N}\ s.t.\ \exists w_{1},w_{2},...,w_{N}\in\Sigma\ s.t. \exists x_{f_{k}(1)},x_{f_{k}(2)},...,x_{f_{k}(m_{k})}\ s.t. w_{k}=x_{f_{k}(1)}\diamondsuit x_{f_{k}(2)}\diamondsuit \cdots \diamondsuit x_{f_{k}(m_{k})}\\
&H_{M}(\{a_{mn}\}_{m,n\in\mathbb{N}};w)\coloneqq\sum_{0\lt n_{1}\lt n_{2}\lt \cdots\lt n_{N}\leq M}\prod_{k=1}^{N}\prod_{l=1}^{m_{k}}a_{f_{k}(l)n_{k}}
\end{align}
自然数$M\in\mathbb{N}$そして数列$\{a_{mn}\}_{m,n\in\mathbb{N}}$に対して以下の式が成り立つ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\forall w\in\Sigma^{\ast}:\exists x_{k_{1}},x_{k_{2}},...,x_{k_{m}}\in\Sigma\ s.t.\ w=x_{k_{1}}x_{k_{2}}\cdots x_{k_{m}}: H_{M}(w)=S(\{a_{k_{1}n}\},\{a_{k_{2}n}\},...,\{a_{k_{m}n}\})=\sum_{0\lt n_{1}\lt n_{2}\lt\cdots\lt n_{m}\leq M}a_{k_{1}n_{1}}a_{k_{2}n_{2}}\cdots a_{k_{m}n_{m}}\\
\forall w_{1},w_{2}\in\mathfrak{H}:H_{M}(\{a_{mn}\}_{m,n\in\mathbb{N}}; w_{1}\ast w_{2})=H_{M}(\{a_{mn}\}_{m,n\in\mathbb{N}}; w_{1})H_{M}(\{a_{mn}\}_{m,n\in\mathbb{N}}; w_{2})
\end{array}
\right.
\end{eqnarray}
$\Sigma_{1}=\{1,x\}$とする。
この時以下の計算を行え。
- $x\ast x\ast x$を求めよ。
- $S_{N}(\{a_{n}\})^{3}$を求めよ。
[1-1]
\begin{eqnarray}
x\ast x&=& (1x)\ast(1x)\\
&=&\{1\ast (1x)\}x+\{(1x)\ast 1\}x+(1\ast 1)(x\diamondsuit x)\\
&=&2xx+x\diamondsuit x
\end{eqnarray}
[1-2]
\begin{eqnarray}
(x\ast x)\ast x&=&(2xx+x\diamondsuit x)\ast x\\
&=&2(xx)\ast x+(x\diamondsuit x)\ast x
\end{eqnarray}
[1-3]
\begin{eqnarray}
(xx)\ast x&=&(xx)\ast(1x)\\
&=&\{x\ast(1x)\}x+\{(xx)\ast1\}x+(x\ast1)(x\diamondsuit x)\\
&=&(2xx+x\diamondsuit x)x+xxx+x(x\diamondsuit x)\\
&=&3xxx+(x\diamondsuit x)x+x(x\diamondsuit x)
\end{eqnarray}
[1-4]
\begin{eqnarray}
(x\diamondsuit x)\ast x&=&[\{1(x\diamondsuit x)\}\ast(1x)]\\
&=&\{1\ast(1x)\}(x\diamondsuit x)+[\{1(x\diamondsuit x)\}\ast 1]x+(1\ast 1)\{(x\diamondsuit x)\diamondsuit x\}\\
&=&x(x\diamondsuit x)+(x\diamondsuit x)x+x\diamondsuit x\diamondsuit x
\end{eqnarray}
[1-5]
\begin{equation}
(x\ast x)\ast x=6xxx+3x(x\diamondsuit x)+3(x\diamondsuit x)x+x\diamondsuit x\diamondsuit x
\end{equation}
[2]
\begin{eqnarray}
S_{N}(\{a_{n}\})^{3}&=&H_{N}(\{a_{n}\},x\ast x\ast x)\\
&=&6H_{N}(\{a_{n}\};xxx)+3H_{N}(\{a_{n}\};x(x\diamondsuit x))+3H_{N}(\{a_{n}\};(x\diamondsuit x)x)+H_{N}(\{a_{n}\};x\diamondsuit x\diamondsuit x)\\
&=&6S_{N}(\{a_{n}\},\{a_{n}\},\{a_{n}\})+3S_{N}(\{a_{n}\},\{a_{n}^{2}\})+3S_{N}(\{a_{n}^{2}\},\{a_{n}\})+S_{N}(\{a_{n}^{3}\})
\end{eqnarray}
$\Sigma_{1}=\{1,x,y\}$とする。
この時以下の計算を行え。
- $x\ast y\ast y$
- $S_{N}(\{a_{n}\})S_{N}(\{b_{n}\})^{2}$
[1-1]
\begin{eqnarray}
y\ast y=2yy+y\diamondsuit y
\end{eqnarray}
[1-2]
\begin{eqnarray}
x\ast (y\ast y)&=&x\ast (2yy+y\diamondsuit y)\\
&=&2x\ast (yy)+x\ast (y\diamondsuit y)
\end{eqnarray}
[1-3]
\begin{eqnarray}
x\ast(yy)&=&(1x)\ast(yy)\\
&=&\{1\ast(yy)\}x+\{(1x)\ast y\}y+(1\ast y)(x\diamondsuit y)\\
&=&yyx+(x\ast y)y+y(x\diamondsuit y)
\end{eqnarray}
\begin{eqnarray}
x\ast y&=&(1x)\ast(1y)\\
&=&\{1\ast(1y)\}x+\{(1x)\ast 1\}y+(1\ast 1)(x\diamondsuit y)\\
&=&yx+xy+x\diamondsuit y
\end{eqnarray}
\begin{eqnarray}
x\ast(yy)=yyx+yxy+xyy+(x\diamondsuit y)y+y(x\diamond y)
\end{eqnarray}
[1-5]
\begin{eqnarray}
x\ast(y\diamondsuit y)&=&(1x)\ast\{1(y\diamondsuit y)\}\\
&=&[1\ast\{1(y\diamondsuit y)\}]x+\{(1x)\ast1\}(y\diamondsuit y)+(1\ast1)(x\diamondsuit y\diamondsuit y)\\
&=&(y\diamondsuit y)x+x(y\diamondsuit y)+x\diamondsuit y\diamondsuit y
\end{eqnarray}
[1-6]
\begin{equation}
x\ast (y\ast y)=2yyx+2yxy+2xyy+2y(x\diamondsuit y)+(y\diamondsuit y)x+x(y\diamondsuit y)+x\diamondsuit y\diamondsuit y
\end{equation}
[1-7]
\begin{align}
&S_{N}(\{a_{n}\})S_{N}(\{b_{n}\})^{2}\\
&=2\{S_{N}(\{b_{n}\},\{b_{n}\},\{a_{n}\})+S(\{b_{n}\},\{a_{n}\},\{b_{n}\})+S_{N}(\{a_{n}\},\{b_{n}\},\{b_{n}\})+S_{N}(\{a_{n}b_{n}\},\{b_{n}\})+S_{N}(\{b_{n}\},\{a_{n}b_{n}\})\}\\
&+S_{N}(\{b_{n}^{2}\},\{a_{n}\})+S_{N}(\{a_{n}\},\{b_{n}^{2}\})+S_{N}(\{a_{n}b_{n}^{2}\})
\end{align}
問題2で$a_{n}\coloneqq \frac{1}{n^{2}},b_{n}=\frac{1}{2^{n}}$の様に置く。すると次の結果を得る。
\begin{eqnarray}
\frac{\pi^{2}}{6}&=&2\{\sum_{0\lt n_{1}\lt n_{2}\lt n_{3}}(\frac{1}{2^{n_{1}+n_{2}}n_{3}^{2}}+\frac{1}{2^{n_{1}+n_{3}}n_{2}^{2}}+\frac{1}{2^{n_{2}+n_{3}}n_{1}^{2}})+\sum_{0\lt n_{1}\lt n_{2}}(\frac{1}{2^{n_{1}+n_{2}}n_{1}^{2}}+\frac{1}{2^{n_{1}+n_{2}}n_{2}^{2}})\}+\sum_{0\lt n_{1}\lt n_{2}}(\frac{1}{2^{2n_{1}}n_{2}^{2}}+\frac{1}{2^{2n_{2}}n_{1}^{2}})+\sum_{0\lt n_{1}}\frac{1}{2^{2n_{1}}n_{1}^{2}}\\
&=&\sum_{0\lt n_{1}\lt n_{2}\lt n_{3}}\frac{1}{2^{n_{1}+n_{2}+n_{3}-1}}(\frac{2^{n_{1}}}{n_{1}^{2}}+\frac{2^{n_{2}}}{n_{2}^{2}}+\frac{2^{n_{3}}}{n_{3}^{2}})+\sum_{0\lt n_{1}\lt n_{2}}\frac{1}{2^{n_{1}+n_{2}}}(\frac{1}{n_{1}^{2}}+\frac{1}{n_{2}^{2}}+\frac{2^{n_{1}}}{2^{n_{2}}n_{1}^{2}}+\frac{2^{n_{2}}}{2^{n_{1}}n_{2}^{2}})+\sum_{0\lt n_{1}}\frac{1}{2^{2n_{1}}n_{1}^{2}}
\end{eqnarray}
\begin{equation} \frac{\pi^{2}}{6}=\sum_{0\lt n_{1}\lt n_{2}\lt n_{3}}\frac{1}{2^{n_{1}+n_{2}+n_{3}-1}}(\frac{2^{n_{1}}}{n_{1}^{2}}+\frac{2^{n_{2}}}{n_{2}^{2}}+\frac{2^{n_{3}}}{n_{3}^{2}})+\sum_{0\lt n_{1}\lt n_{2}}\frac{1}{2^{n_{1}+n_{2}}}(\frac{1}{n_{1}^{2}}+\frac{1}{n_{2}^{2}}+\frac{2^{n_{1}}}{2^{n_{2}}n_{1}^{2}}+\frac{2^{n_{2}}}{2^{n_{1}}n_{2}^{2}})+\sum_{0\lt n_{1}}\frac{1}{2^{2n_{1}}n_{1}^{2}} \end{equation}
関数列$\{f_{n}\}\subset C([0,1])$に対して以下の様に定める。
\begin{equation}
G(f_{1},f_{2},...,f_{n})(x)\coloneqq\int_{0}^{x}dx_{1}f_{1}(x_{1})\int_{0}^{x_{1}}dx_{2}f_{2}(x_{2})\cdots\int_{0}^{x_{n-1}}dx_{n}f_{n}(x_{n})
\end{equation}
$G(f_{1},f_{2})G(f_{2},f_{4})$について計算しその性質について調べよ。
\begin{eqnarray} G(f_{1},f_{2})G(f_{3},f_{4})(x)&=&\int_{0\leq x_{1}\lt x_{2}\leq x}dx_{1}dx_{2}f_{1}(x_{1})f_{2}(x_{2})\int_{0\leq x_{3}\lt x_{4}\leq x}dx_{3}dx_{4}f_{3}(x_{3})f_{4}(x_{4})\\ &=&(\int_{0\leq x_{1}\lt x_{3}\lt x_{2}\lt x_{4}\leq x}+\int_{0\leq x_{3}\lt x_{1}\lt x_{2}\lt x_{4}\leq x}+\int_{0\leq x_1\lt x_{2}\lt x_{3}\lt x_{4}\leq x}+\int_{0\leq x_{1}\lt x_{3}\lt x_{4}\leq x_{2}\leq x}+\int_{0\leq x_{3}\lt x_{1}\lt x_{4}\leq x_{2}\leq x}+\int_{0\leq x_{3}\lt x_{4}\lt x_{1}\leq x_{2}\leq x})dx_{1}dx_{2}dx_{3}dx_{4}f_{1}(x_{1})f_{2}(x_{2})f_{3}(x_{3})f_{4}(x_{4})\\ &=&G(f_{1},f_{3},f_{2},f_{4})+G(f_{3},f_{1},f_{2},f_{4})+G(f_{1},f_{2},f_{3},f_{4})+G(f_{1},f_{3},f_{4},f_{2})+G(f_{3},f_{1},f_{4},f_{2})+G(f_{3},f_{4},f_{1},f_{2}) \end{eqnarray}
関数列$f_{1},f_{2},...,f_{r},g_{1},g_{2},...,g_{s}$に対して以下の様な記号を定める。
\begin{equation}
\tilde{G}(f_{1},f_{2},...,f_{r};g_{1},g_{2},...,g_{s})(x)\coloneqq G(f_{1},f_{2},...,f_{r})(x)G(g_{1},g_{2},...,g_{s})(x)
\end{equation}
すると以下の式が成り立つ。
\begin{equation}
\tilde{G}(f_{1},f_{2},...,f_{r};g_{1},g_{2},...,g_{s})(x)=\int_{0}^{x}dx_{r}\tilde{G}(f_{1},f_{2},...,f_{r-1};g_{1},g_{2},...,g_{s})(x_{r})f_{r}(x_{r})+\int_{0}^{x}dy_{s}\tilde{G}(f_{1},f_{2},...,f_{r};g_{1},g_{2},...,g_{s-1})(y_{s})g_{s}(y_{s})
\end{equation}
\begin{eqnarray} \tilde{G}(f_{1},f_{2},...,f_{r};g_{1},g_{2},...,g_{s})(x)&=&G(f_{1},f_{2},...,f_{r})(x)G(g_{1},g_{2},...,g_{s})(x)\\ &=&\int_{0\leq x_{1}\lt x_{2}\lt\cdots\lt x_{r}\leq x}dx_{1}dx_{2}\cdots dx_{r}\int_{0\leq y_{1}\lt y_{2}\lt \cdots\lt y_{s}\leq x}dy_{1}dy_{2}\cdots dy_{s}f_{1}(x_{1})f_{2}(x_{2})\cdots f_{r}(x_{r})g_{1}(y_{1})g_{2}(y_{2})\cdots g_{s}(y_{s})\\ &=&\int_{0}^{x}dx_{r}\tilde{G}(f_{1},f_{2},...,f_{r-1};g_{1},g_{2},...,g_{s})(x_{r})f_{r}(x_{r})+\int_{0}^{x}dy_{s}\tilde{G}(f_{1},f_{2},...,f_{r};g_{1},g_{2},...,g_{s-1})(y_{s})g_{s}(y_{s}) \end{eqnarray}
アルファベット$\Sigma$および単語$\Sigma^{\backepsilon}$を以下の様に定める。
\begin{eqnarray}
\left\{
\begin{array}{l}
\Sigma_{1}\coloneqq\{1\}\sqcup\{x_{1},x_{2},...,x_{r}\}\\
\Sigma^{\backepsilon}\coloneqq\{y_{1}y_{2}\cdots y_{n}|n\in\mathbb{N},y_{1},y_{2},...,y_{n}\in\Sigma_{1}\}
\end{array}
\right.
\end{eqnarray}
$\mathbb{Q}$-係数ベクトル空間:$\mathfrak{H}^{\backepsilon}\coloneqq\mathbb{Q}<\Sigma^{\backepsilon}>\coloneqq\bigoplus_{z\in\Sigma^{\backepsilon}}\mathbb{Q}z$を定義する。この時、$\mathbb{Q}$-双対線形写像:$\backepsilon:\mathfrak{H}^{\backepsilon}\times \mathfrak{H}^{\backepsilon}\rightarrow\mathfrak{H}^{\backepsilon}$を次の様に定める。
- \begin{align} \forall w\in\Sigma^{\backepsilon}:1\backepsilon w=w\backepsilon 1=w \end{align}
- \begin{align} &\forall a_{1},a_{2}\in\Sigma_{1}:\forall w_{1},w_{2}\in\Sigma^{\backepsilon}:\\ &(w_{1}a_{1})\backepsilon(w_{2}a_{2})=(w_{1}\backepsilon(w_{2}a_{2}))a_{1}+((w_{1}a_{1})\backepsilon w_{2})a_{2} \end{align}
シャッフル積$\backepsilon$は結合律を満たす。
関数列$\{f_{n}\}_{n\in\mathbb{N}}$をパラメータに持つ$\mathbb{Q}$-線形写像$I(\{f_{n}\}_{n\in\mathbb{N}};\ast):\mathfrak{H}^{\backepsilon}\rightarrow\mathbb{C}$を以下の様に定める。
\begin{align}
&\forall w\in\Sigma^{\backepsilon}:\exists N\in\mathbb{N}\ s.t.\ \exists x_{k_{1}},x_{k_{2}},...,x_{k_{m}}\in \Sigma_{1}\ s.t.\\
&I(\{f_{n}\}_{n\in\mathbb{N}};w)(x)\coloneqq \int_{0}^{x}dx_{1}f_{k_{1}}(x_{1})\int_{0}^{x_{1}}dx_{2}f_{k_{2}}(x_{2})\cdots\int_{0}^{x_{m-1}}dx_{m}f_{k_{m}}(x_{m})=G(f_{k_{1}},f_{k_{2}},...,f_{k_{m}})(x)
\end{align}
以下の式が成り立つ。
\begin{equation}
\forall w_{1},w_{2}\in\mathfrak{H}^{\backepsilon}:I(\{f_{n}\}_{n\in\mathbb{N}};w_{1}\backepsilon w_{2})(x)=I(\{f_{n}\}_{n\in\mathbb{N}};w_{1})(x)I(\{f_{n}\}_{n\in\mathbb{N}};w_{2})(x)
\end{equation}
$\Sigma_{1}\coloneqq \{1,x,y\}$とする。
この時以下の計算を行え。
- $x\backepsilon x\backepsilon y$を求めよ。
- $\tilde{G}(f,f;g)(x)\coloneqq G(f,f)(x)G(g)(x)$を求めよ。
[1-1]
\begin{eqnarray}
x\backepsilon x&=&(1x)\backepsilon (1x)\\
&=&\{1\backepsilon(1x)\}x+\{(1x)\backepsilon1\}x\\
&=&2xx
\end{eqnarray}
[1-2]
\begin{eqnarray}
(x\backepsilon x)\backepsilon y&=&2(xx)\backepsilon (1y)\\
&=&2\{x\backepsilon(1y)\}x+2\{(xx)\backepsilon 1\}y\\
&=&2(x\backepsilon y)x+2xxy\\
&=&2yxx+2xyx+2xxy
\end{eqnarray}
[2]
\begin{eqnarray}
\tilde{G}(f,f;g)(x)&=&G(f,f)(x)G(g)(x)\\
&=&2\int_{0\leq x_{1}\lt x_{2}\lt x_{3}\leq x}dx_{1}dx_{2}dx_{3}\{g(x_{1})f(x_{2})f(x_{3})+f(x_{1})g(x_{2})f(x_{3})+f(x_{1})f(x_{2})g(x_{3})\}
\end{eqnarray}
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