Clausenの公式はもう雲の上の公式ではない1(副題:まずは対称二乗作用素について調べる)

[1]
\begin{eqnarray} \left\{ \begin{array}{l} u_{ij}=y_{i}y_{j}\\ u_{ij}^{'}=y_{i}^{'}y_{j}+y_{i}y_{j}^{'}\\ u_{ij}^{''}=y_{i}^{''}y_{j}+2y_{i}^{'}y_{j}^{'}+y_{i}y_{j}^{''}\\ u_{ij}^{'''}=y_{i}^{'''}y_{j}+3y_{i}^{''}y_{j}^{'}+3y_{i}^{'}y_{j}^{''}+y_{i}y_{j}^{'''} \end{array} \right. \end{eqnarray}
[2]
\begin{eqnarray} u_{ij}^{'''}&=&-(p_{1}y_{i}^{'}+p_{2}y_{i})^{'}y_{j}-3(p_{1}y_{i}^{'}+p_{2}y_{i})y_{j}^{'}-3y_{i}^{'}(p_{1}y_{j}^{'}+p_{2}y_{j})-y_{i}(p_{1}y_{j}^{'}+p_{2}y_{j})^{'}\\ &=&-\{p_{1}y_{i}^{''}+(p_{1}^{'}+p_{2})y_{i}^{'}+p_{2}^{'}y_{i}\}y_{j}-3(p_{1}y_{i}^{'}+p_{2}y_{i})y_{j}^{'}-3y_{i}^{'}(p_{1}y_{j}^{'}+p_{2}y_{j})-y_{i}\{p_{1}y_{j}^{''}+(p_{1}^{'}+p_{2})y_{j}^{'}+p_{2}^{'}y_{j}\}\\ &=&-\{p_{1}y_{i}^{''}y_{j}+(p_{1}^{'}+p_{2})y_{i}^{'}y_{j}+p_{2}^{'}y_{i}y_{j}+3p_{1}y_{i}^{'}y_{j}^{'}+3p_{2}y_{i}y_{j}^{'}+3p_{1}y_{i}^{'}y_{j}^{'}+3p_{2}y_{i}^{'}y_{j}+p_{1}y_{i}y_{j}^{''}+(p_{1}^{'}+p_{2})y_{i}y_{j}^{'}+p_{2}^{'}y_{i}y_{j}\}\\ &=&-\{p_{1}(y_{i}^{''}y_{j}+y_{i}y_{j}^{''})+3p_{2}(y_{i}^{'}y_{j}+y_{i}y_{j}^{'})+6p_{1}y_{i}^{'}y_{j}^{'}+(p_{1}^{'}+p_{2})(y_{i}^{'}y_{j}+y_{i}y_{j}^{'})+2p_{2}^{'}u_{ij}\}\\ &=&-\{p_{1}u_{ij}^{''}+4p_{1}y_{i}^{'}y_{j}^{'}+(p_{1}^{'}+4p_{2})u_{ij}^{'}+2p_{2}^{'}u_{ij}\}\\ &=&-\{3p_{1}u_{ij}^{''}-2p_{1}(y_{i}^{''}y_{j}+y_{i}y_{j}^{''})+(p_{1}^{'}+4p_{2})u_{ij}^{'}+2p_{2}^{'}u_{ij}\}\\ &=&-\{3p_{1}u_{ij}^{''}+2p_{1}^{2}u_{ij}^{'}+4p_{1}p_{2}u_{ij}+(p_{1}^{'}+4p_{2})u_{ij}^{'}+2p_{2}^{'}u_{ij}\}\\ &=&-\{3p_{1}u_{ij}^{''}+(p_{1}^{'}+2p_{1}^{2}+4p_{2})u_{ij}^{'}+(2p_{2}^{'}+4p_{1}p_{2})u_{ij}\} \end{eqnarray}
[3]以上の計算により下記の微分方程式を得る。
\begin{equation} u_{ij}^{'''}+3p_{1}u_{ij}^{''}+(p_{1}^{'}+2p_{1}^{2}+4p_{2})u_{ij}^{'}+(2p_{2}^{'}+4p_{1}p_{2})u_{ij}=0 \end{equation}

[1]まず定理1を参照する。元の微分演算子$L$側での指数方程式は以下の様に与えられる。
\begin{eqnarray} \left\{ \begin{array}{l} I(\rho)=(\rho)_{2}+\sum_{k=1}^{2}q_{k}(\rho)_{2-k}\\ q_{k}(z)\coloneqq(z-a)^{k}p_{k}(z) \end{array} \right. \end{eqnarray}
ただし$q_{k}(a)\coloneqq q_{k}$とした。
[2]次の様な記号を定める。
\begin{eqnarray} \left\{ \begin{array}{l} A(z)=3p_{1}(z)\\ B(z)=p_{1}^{'}(z)+2p_{1}^{2}(z)+4p_{2}(z)\\ C(z)=2p_{2}^{'}(z)+4p_{1}(z)p_{2}(z) \end{array} \right. \end{eqnarray}
[3]
\begin{eqnarray} \left\{ \begin{array}{l} A(z)=\frac{3q_{1}(z)}{z-a}\therefore \lim_{z\rightarrow a}(z-a)A(z)=3q_{1}(a)\\ B(z)=\frac{(z-a)q_{1}^{'}-q_{1}(z)+2q_{1}^{2}(z)}{(z-a)^{2}}+\frac{4q_{2}(z)}{(z-a)^{2}}\therefore \lim_{z\rightarrow a}(z-a)^{2}B(z)=2q_{1}^{2}-q_{1}+4q_{2}\\ C(z)=\frac{2(z-a)q_{2}^{'}(z)-4q_{2}(z)}{(z-a)^{3}}+\frac{4q_{1}(z)q_{2}(z)}{(z-a)^{3}}\therefore \lim_{z\rightarrow a}(z-a)^{3}C(z)=-4q_{2}+4q_{1}q_{2} \end{array} \right. \end{eqnarray}
[4]
\begin{eqnarray} I(\rho)&=&\rho(\rho-1)+q_{1}\rho+q_{2}\\ &=&\rho^{2}-(1-q_{1})\rho+\rho_{2}\\ &=&(\rho-\rho_{1})(\rho-\rho_{2})\\ &=&\rho^{2}-(\rho_{1}+\rho_{2})\rho+\rho_{1}\rho_{2} \end{eqnarray}
なので
\begin{eqnarray} \left\{ \begin{array}{l} q_{1}=1-\rho_{1}-\rho_{2}\\ \rho_{1}\rho_{2}=q_{2} \end{array} \right. \end{eqnarray}
[5]
\begin{align} &I_{\mathrm{Sym}}(\rho)\\ &=(\rho)_{3}+3q_{1}(\rho)_{2}+(2q_{1}^{2}-q_{1}+4q_{2})(\rho)_{1}-4q_{2}+4q_{1}q_{2}\\ &=\rho(\rho-1)(\rho-2)+3(1-\rho_{1}-\rho_{2})\rho(\rho-1)+\{2(1-\rho_{1}-\rho_{2})^{2}-(1-\rho_{1}-\rho_{2})+4\rho_{1}\rho_{2}\}\rho-4\rho_{1}\rho_{2}+4\rho_{1}\rho_{2}(1-\rho_{1}-\rho_{2})\\ &=\rho^{3}-3\rho^{2}+2\rho\\ &+3(1-\rho_{1}-\rho_{2})\rho^{2}-3(1-\rho_{1}-\rho_{2})\rho\\ &+\{(1-\rho_{1}-\rho_{2})(1-2\rho_{1}-2\rho_{2})+4\rho_{1}\rho_{2}\}\rho\\ &-4\rho_{1}\rho_{2}(\rho_{1}+\rho_{2})\\ &=\rho^{3}\\ &-3(\rho_{1}+\rho_{2})\rho^{2}\\ &-\{-2+2(1-\rho_{1}-\rho_{2})(1+\rho_{1}+\rho_{2})-4\rho_{1}\rho_{2}\}\rho\\ &-4\rho_{1}\rho_{2}(\rho_{1}+\rho_{2})\\ &=\rho^{3}\\ &-3(\rho_{1}+\rho_{2})\rho^{2}\\ &+2\{(\rho_{1}+\rho_{2})^{2}+2\rho_{1}\rho_{2}\}\rho\\ &-4\rho_{1}\rho_{2}(\rho_{1}+\rho_{2})\\ &=\rho^{3}-3(\rho_{1}+\rho_{2})\rho^{2}+2(\rho_{1}^{2}+\rho_{2}^{2}+4\rho_{1}\rho_{2})\rho-4\rho_{1}\rho_{2}(\rho_{1}+\rho_{2})\\ \end{align}
[6]
\begin{eqnarray} \left\{ \begin{array}{l} 2\rho_{1}+2\rho_{2}+(\rho_{1}+\rho_{2})=3(\rho_{1}+\rho_{2})\\ (2\rho_{1})(2\rho_{2})+(\rho_{1}+\rho_{2})(2\rho_{1}+2\rho_{2})=2(\rho_{1}^{2}+\rho_{2}^{2}+4\rho_{1}\rho_{2})\\ (2\rho_{1})(2\rho_{2})(\rho_{1}+\rho_{2})=4\rho_{1}\rho_{2}(\rho_{1}+\rho_{2}) \end{array} \right. \end{eqnarray}
[7]以上の計算により以下の結果を得る。
\begin{equation} I_{\mathrm{Sym}}(q)=(\rho-2\rho_{1})(\rho-2\rho_{2})\{\rho-(\rho_{1}+\rho_{2})\} \end{equation}

[1]$\displaystyle A_{n}=\frac{\prod_{k=1}^{P}(a_{k})_{n}}{n!\prod_{k=1}^{Q}(b_{k})_{n}}$の様に定める。すると以下の式が成り立つ。
\begin{eqnarray} A_{n}n\prod_{k=1}^{Q}(b_{k}+n-1)=A_{n-1}\prod_{k=1}^{P}(a_{k}+n-1) \end{eqnarray}
[2]すると以下の式を得る。
\begin{equation} \frac{1}{z}\theta\prod_{k=1}^{Q}(b_{k}+\theta-1)A_{n}x^{n}=\prod_{k=1}^{P}(a_{k}+\theta)A_{n-1}x^{n-1} \end{equation}
[3]ゆえに以下の式を得る。
\begin{equation} \{\theta\prod_{k=1}^{Q}(b_{k}+\theta-1)-z\prod_{k=1}^{P}(a_{k}+\theta)\}{}_{P}F_{Q}(\vb*{a}_{P};\vb*{b}_{Q};z)=0 \end{equation}

[1]
\begin{align} &\theta\prod_{k=1}^{2}(b_{k}+\theta-1)-z\prod_{k=1}^{3}(a_{k}+\theta)\\ &=z\frac{d}{dz}(z\frac{d}{dz}+b_{1}-1)(z\frac{d}{dz}+b_{2}-1)-z(a_{1}+z\frac{d}{dz})(a_{2}+z\frac{d}{dz})(a_{3}+z\frac{d}{dz})\\ &=z\frac{d}{dz}\{z^{2}\frac{d^{2}}{dz^{2}}+(b_{1}+b_{2}-1)z\frac{d}{dz}+(b_{1}-1)(b_{2}-1)\}-z(a_{1}+z\frac{d}{dz})\{z^{2}\frac{d^{2}}{dz^{2}}+(a_{2}+a_{3}+1)z\frac{d}{dz}+a_{2}a_{3}\}\\ &=z^{3}\frac{d^{3}}{dz^{3}}+(b_{1}+b_{2}+1)z^{2}\frac{d^{2}}{dz^{2}}+\{(b_{1}+b_{2}-1)+(b_{1}-1)(b_{2}-1)\}z\frac{d}{dz}\\ &-z\{z^{3}\frac{d^{3}}{dz^{3}}+(a_{1}+a_{2}+a_{3}+3)z^{2}\frac{d^{2}}{dz^{2}}+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3}+1)z\frac{d}{dz}+a_{1}a_{2}a_{3}\}\\ &=z^{3}\frac{d^{3}}{dz^{3}}+(b_{1}+b_{2}+1)z^{2}\frac{d^{2}}{dz^{2}}+b_{1}b_{2}z\frac{d}{dz}\\ &-z\{z^{3}\frac{d^{3}}{dz^{3}}+(a_{1}+a_{2}+a_{3}+3)z^{2}\frac{d^{2}}{dz^{2}}+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3}+1)z\frac{d}{dz}+a_{1}a_{2}a_{3}\} \end{align}
[2]ゆえに
\begin{equation} [z^{2}(1-z)\frac{d^{3}}{dz^{3}}+\{(b_{1}+b_{2}+1)-(a_{1}+a_{2}+a_{3}+3)z\}z\frac{d^{2}}{dz^{2}}+\{b_{1}b_{2}-(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3}+1)z\}\frac{d}{dz}-a_{1}a_{2}a_{3}]y=0 \end{equation}

[1-1]$z=0$は確定特異点
\begin{eqnarray} \left\{ \begin{array}{l} \lim_{z\rightarrow 0}zp_{1}(z)=b_{1}+b_{2}+1\\ \lim_{z\rightarrow 0}z^{2}p_{2}(z)=b_{1}b_{2}\\ \lim_{z\rightarrow 0}z^{3}p_{3}(z)=0 \end{array} \right. \end{eqnarray}
[1-2]また指数は下記の様に与えられる。
\begin{eqnarray} \rho(\rho-1)(\rho-2)+(b_{1}+b_{2}+1)\rho(\rho-1)+b_{1}b_{2}\rho&=&\rho\{(\rho-1)(\rho-2)+(b_{1}+b_{2}+1)(\rho-1)+b_{1}b_{2}\}\\ &=&\rho\{\rho^{2}+(b_{1}+b_{2}-2)\rho+b_{1}b_{2}-(b_{1}+b_{2})+1\}\\ &=&\rho(\rho+b_{1}-1)(\rho+b_{2}-1) \end{eqnarray}
なのでその指数は$\rho=0,1-b_{1},1-b_{2}$
[2-1]$z=1$は確定特異点
\begin{eqnarray} \left\{ \begin{array}{l} \lim_{z\rightarrow 1}(z-1)p_{1}(z)=a_{1}+a_{2}+a_{3}-(b_{1}+b_{2})+2\\ \lim_{z\rightarrow 1}(z-1)^{2}p_{2}(z)=0\\ \lim_{z\rightarrow 1}(z-1)^{3}p_{3}(z)=0 \end{array} \right. \end{eqnarray}
[2-2]その指数は以下の様に書ける。
\begin{align} &\rho(\rho-1)(\rho-2)+\{a_{1}+a_{2}+a_{3}-(b_{1}+b_{2})+2\}\rho(\rho-1)\\ &=\rho(\rho-1)\{\rho+a_{1}+a_{2}+a_{3}-(b_{1}+b_{2})\} \end{align}
なので$\rho=0,1,b_{1}+b_{2}-(a_{1}+a_{2}+a_{3})$
[3-1]$z=\infty$は確定特異点
\begin{eqnarray} \left\{ \begin{array}{l} \frac{d}{dz}=-w^{2}\frac{d}{dw}\\ \frac{d^{2}}{dz^{2}}=w^{4}\frac{d^{2}}{dw^{2}}+2w^{3}\frac{d}{dw}\\ \frac{d^{3}}{dz^{3}}=-w^{6}\frac{d^{3}}{dw^{3}}-6w^{5}\frac{d^{2}}{dw^{2}}-6w^{4}\frac{d}{dw}\\ p_{1}(\frac{1}{w})=\frac{(b_{1}+b_{2}+1)w^{2}-(a_{1}+a_{2}+a_{3}+3)w}{w-1}\\ p_{2}(\frac{1}{w})=\frac{b_{1}b_{2}w^{3}-(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3}+1)w^{2}}{w-1}\\ p_{3}(\frac{1}{w})=-\frac{a_{1}a_{2}a_{3}w^{3}}{w-1} \end{array} \right. \end{eqnarray}
これを代入すると以下の微分方程式を得る。
\begin{align} &\frac{d^{3}}{dz^{3}}+p_{1}\frac{d^{2}}{dz^{2}}+p_{2}\frac{d}{dz}+p_{3}\\ &=-w^{6}\frac{d^{3}}{dw^{3}}-6w^{5}\frac{d^{2}}{dw^{2}}-6w^{4}\frac{d}{dw}+p_{1}(w^{4}\frac{d^{2}}{dw^{2}}+2w^{3}\frac{d}{dw})-w^{2}p_{2}\frac{d}{dw}+p_{3}\\ &=-w^{6}\frac{d^{3}}{dw^{3}}+(w^{4}p_{1}-6w^{5})\frac{d^{2}}{dw^{2}}+(2w^{3}p_{1}-w^{2}p_{2}-6w^{4})\frac{d}{dw}+p_{3}\\ &=-w^{6}\{\frac{d^{3}}{dw^{3}}+(-\frac{p_{1}}{w^{2}}+\frac{6}{w})\frac{d^{2}}{dw^{2}}+(-\frac{2p_{1}}{w^{3}}+\frac{p_{2}}{w^{4}}+\frac{6}{w^{2}})\frac{d}{dw}-\frac{p_{3}}{w^{6}}\} \end{align}
全体を$-w^{6}$で割り
\begin{eqnarray} \left\{ \begin{array}{l} \lim_{w\rightarrow 0}w(-\frac{p_{1}}{w^{2}}+\frac{6}{w})=3-(a_{1}+a_{2}+a_{3})\\ \lim_{w\rightarrow 0}w^{2}(-\frac{2p_{1}}{w^{3}}+\frac{p_{2}}{w^{4}}+\frac{6}{w^{2}})=-2(a_{1}+a_{2}+a_{3})+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3})+1\\ \lim_{w\rightarrow 0}w^{3}(-\frac{p_{3}}{w^{6}})=-a_{1}a_{2}a_{3} \end{array} \right. \end{eqnarray}
[3-2]
\begin{align} &\rho(\rho-1)(\rho-2)+\{3-(a_{1}+a_{2}+a_{3})\}\rho(\rho-1)+\{-2(a_{1}+a_{2}+a_{3})+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3})+1\}\rho-a_{1}a_{2}a_{3}\\ &=\rho^{3}-3\rho^{2}+2\rho\\ &+\{3-(a_{1}+a_{2}+a_{3})\}\rho^{2}-\{3-(a_{1}+a_{2}+a_{3})\}\rho\\ &+\{-2(a_{1}+a_{2}+a_{3})+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}+a_{1}+a_{2}+a_{3})+1\}\rho\\ &-a_{1}a_{2}a_{3}\\ &=\rho^{3}\\ &-(a_{1}+a_{2}+a_{3})\rho^{2}\\ &+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})\rho\\ &-a_{1}a_{2}a_{3}\\ &=(\rho-a_{1})(\rho-a_{2})(\rho-a_{3}) \end{align}
なので$z=\infty$での指数は$a_{1},a_{2},a_{3}$