ラマヌジャン・佐藤級数を理解したい(その6、公式集1)

　ある
$$z=z(\tau ),x=x(\tau )$$
が存在して
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[a,b,c]x^n$$
が成り立つ。

$$\begin{array}{rl}{j}_X(\tau )& =j_Y(\tau )+2a+\frac{a^2+4c}{j_Y(\tau )}\\ j_Y(\tau )& =\frac{1}{x}-a-cx\end{array}$$
および
$$\begin{array}{rl}{s}_X(n)& =s_n[a,b,c](\genfrac{}{}{0ex}{}{2n}{n})\\ s_Y(n)& =t_n[a,b,c]\end{array}$$
とおくと
$$\sqrt{x(1-ax-c{x}^2)}{z}^2=\sum _{n=0}^{\mathrm{\infty }}\frac{s_X(n)}{j_X(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_Y(n)}{j_Y(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j_X-4a-16c{j}_X^{-1}}\sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2n}{n})s_n\frac{n+\frac{1}{2}+\lambda }{j_X(\tau {)}^{n+\frac{1}{2}}}\\ & =\sqrt{j_Y+2a+(a^2+4c)j_Y^{-1}}\sum _{n=0}^{\mathrm{\infty }}{t}_n\frac{n+\frac{1}{2}+\mu }{j_Y(\tau {)}^{n+\frac{1}{2}}}\end{array}$$

$$z=\sqrt[4]{E_4(\tau )},x=\frac{1}{864}(1-\sqrt{\frac{1728-j(\tau )}{j(\tau )}})$$
とおくと
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[432,60,0]x^n$$
が成り立つ。

$$j^{\ast }(\tau )=432\frac{\sqrt{j(\tau )}+\sqrt{j(\tau )-1728}}{\sqrt{j(\tau )}-\sqrt{j(\tau )-1728}}$$
および
$$\begin{array}{rlrl}{s}_{1A}(n)& =s_n[432,60,0](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{6n}{3n})(\genfrac{}{}{0ex}{}{3n}{n})(\genfrac{}{}{0ex}{}{2n}{n})\\ s_{1B}(n)& =t_n[432,60,0]& & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{6k}{3k})(\genfrac{}{}{0ex}{}{3k}{k})(\genfrac{}{}{0ex}{}{2k}{k})(\genfrac{}{}{0ex}{}{n+k}{n-k})(-432{)}^{n-k}\end{array}$$
とおくと
$$\frac{\eta (\tau {)}^{12}}{E_4(\tau )}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{1A}(n)}{j(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{1B}(n)}{j^{\ast }(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j(\tau )-1728}\sum _{n=0}^{\mathrm{\infty }}{s}_{1A}(n)\frac{n+\frac{1}{2}+\lambda }{j(\tau {)}^{n+\frac{1}{2}}}\\ & =\frac{j^{\ast }(\tau )+432}{\sqrt{j^{\ast }(\tau )}}\sum _{n=0}^{\mathrm{\infty }}{s}_{1B}(n)\frac{n+\frac{1}{2}+\mu }{j^{\ast }(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは$|j(\tau )|>1728,|j^{\ast }(\tau )|>432$において収束する。

$$\begin{array}{rl}z& =\sqrt{{\theta }_2(2\tau {)}^4+{\theta }_3(2\tau {)}^4}={(\frac{\eta (\tau {)}^{16}}{\eta (2\tau {)}^8}+64\frac{\eta (2\tau {)}^{16}}{\eta (\tau {)}^8})}^{\frac{1}{4}}\\ x& =\frac{\eta (2\tau {)}^{24}}{\eta (\tau {)}^{24}+64\eta (2\tau {)}^{24}}\end{array}$$
とおくと
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[64,12,0]x^n$$
が成り立つ。

$$\begin{array}{rl}{j}_{2A}(\tau )& ={(\sqrt{j_{2B}(\tau )}+\frac{64}{\sqrt{j_{2B}(\tau )}})}^2\\ j_{2B}(\tau )& ={\left(\frac{\eta (\tau )}{\eta (2\tau )}\right)}^{24}\end{array}$$
および
$$\begin{array}{rlrl}{s}_{2A}(n)& =s_n[64,12,0](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{4n}{2n}){(\genfrac{}{}{0ex}{}{2n}{n})}^2\\ s_{2B}(n)& =t_n[64,12,0]& & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{4k}{2k}){(\genfrac{}{}{0ex}{}{2k}{k})}^2(\genfrac{}{}{0ex}{}{n+k}{n-k})(-64{)}^{n-k}\end{array}$$
とおくと
$$\frac{(\eta (\tau )\eta (2\tau ){)}^8}{\sqrt{\eta (\tau {)}^{24}+64\eta (2\tau {)}^{24}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{2A}(n)}{j_{2A}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{2B}(n)}{j_{2B}(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j_{2A}(\tau )-256}\sum _{n=0}^{\mathrm{\infty }}{s}_{2A}(n)\frac{n+\frac{1}{2}+\lambda }{j_{2A}(\tau {)}^{n+\frac{1}{2}}}\\ & =\frac{j_{2B}(\tau )+64}{\sqrt{j_{2B}(\tau )}}\sum _{n=0}^{\mathrm{\infty }}{s}_{2B}(n)\frac{n+\frac{1}{2}+\mu }{j_{2B}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは$|j_{2A}(\tau )|>256,|j_{2B}(\tau )|>64$において収束する。

$$\begin{array}{rl}z& =a(\tau )={(\frac{\eta (\tau {)}^9}{\eta (3\tau {)}^3}+27\frac{\eta (3\tau {)}^9}{\eta (\tau {)}^3})}^{\frac{1}{3}}\\ x& =\frac{\eta (3\tau {)}^{12}}{\eta (\tau {)}^{12}+27\eta (3\tau {)}^{12}}\end{array}$$
とおくと
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[27,6,0]x^n$$
が成り立つ。

$$\begin{array}{rl}{j}_{3A}(\tau )& ={(\sqrt{j_{3B}(\tau )}+\frac{27}{\sqrt{j_{3B}(\tau )}})}^2\\ j_{3B}(\tau )& ={\left(\frac{\eta (\tau )}{\eta (3\tau )}\right)}^{12}\end{array}$$
および
$$\begin{array}{rlrl}{s}_{3A}(n)& =s_n[27,6,0](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{3n}{n}){(\genfrac{}{}{0ex}{}{2n}{n})}^2\\ s_{3B}(n)& =t_n[27,6,0]& & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{3k}{k}){(\genfrac{}{}{0ex}{}{2k}{k})}^2(\genfrac{}{}{0ex}{}{n+k}{n-k})(-27{)}^{n-k}\end{array}$$
とおくと
$$\frac{(\eta (\tau )\eta (3\tau ){)}^4}{\sqrt[3]{\eta (\tau {)}^{12}+27\eta (3\tau {)}^{12}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{3A}(n)}{j_{3A}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{3B}(n)}{j_{3B}(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j_{3A}(\tau )-108}\sum _{n=0}^{\mathrm{\infty }}{s}_{3A}(n)\frac{n+\frac{1}{2}+\lambda }{j_{3A}(\tau {)}^{n+\frac{1}{2}}}\\ & =\frac{j_{3B}(\tau )+27}{\sqrt{j_{3B}(\tau )}}\sum _{n=0}^{\mathrm{\infty }}{s}_{3B}(n)\frac{n+\frac{1}{2}+\mu }{j_{2B}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは$|j_{3A}(\tau )|>108,|j_{3B}(\tau )|>27$において収束する。

$$\begin{array}{rl}z& ={\theta }_3(2\tau {)}^2=\frac{\eta (2\tau {)}^{10}}{(\eta (\tau )\eta (4\tau ){)}^4}\\ x& =\frac{\eta (4\tau {)}^8}{\eta (\tau {)}^8+16\eta (4\tau {)}^8}={\left(\frac{\eta (\tau )\eta (4\tau {)}^2}{\eta (\tau {)}^3}\right)}^8\end{array}$$
とおくと
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[16,4,0]x^n$$
が成り立つ。

$$\begin{array}{rl}{j}_{4A}(\tau )& ={(\sqrt{j_{4B}(\tau )}+\frac{16}{\sqrt{j_{4B}(\tau )}})}^2={\left(e^{-\frac{\pi i}{24}}\frac{\eta (\frac{2\tau +1}{2})}{\eta (2\tau )}\right)}^{24}\\ j_{4B}(\tau )& ={\left(\frac{\eta (\tau )}{\eta (4\tau )}\right)}^8\end{array}$$
および
$$\begin{array}{rlrl}{s}_{4A}(n)& =s_n[16,4,0](\genfrac{}{}{0ex}{}{2n}{n})& & ={(\genfrac{}{}{0ex}{}{2n}{n})}^3\\ s_{4B}(n)& =t_n[16,4,0]& & =\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{2k}{k})}^3(\genfrac{}{}{0ex}{}{n+k}{n-k})(-16{)}^{n-k}\end{array}$$
とおくと
$${\left(\frac{\eta (\tau )\eta (4\tau )}{\eta (2\tau )}\right)}^4=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{4A}(n)}{j_{4A}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{4B}(n)}{j_{4B}(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j_{4A}(\tau )-64}\sum _{n=0}^{\mathrm{\infty }}{s}_{4A}(n)\frac{n+\frac{1}{2}+\lambda }{j_{4A}(\tau {)}^{n+\frac{1}{2}}}\\ & =\frac{j_{4B}(\tau )+16}{\sqrt{j_{4B}(\tau )}}\sum _{n=0}^{\mathrm{\infty }}{s}_{4B}(n)\frac{n+\frac{1}{2}+\mu }{j_{4B}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは$|j_{4A}(\tau )|>64,|j_{4B}(\tau )|>16$において収束する。

　$q=e^{2\pi i\tau }$とし
$$\begin{array}{rl}z& =\prod _{n=1}^{\mathrm{\infty }}\frac{(1-q^n{)}^2}{((1-q^{5n-4})(1-q^{5n-1}){)}^5}\\ x& ={\left(q^{\frac{1}{5}}\prod _{n=1}^{\mathrm{\infty }}\frac{(1-q^{5n-4})(1-q^{5n-1})}{(1-q^{5n-3})(1-q^{5n-2})}\right)}^5\end{array}$$
とおくと
$$z=\sum _{n=0}^{\mathrm{\infty }}{s}_n[11,3,1]x^n$$
が成り立つ。

$$\begin{array}{rl}{j}_{5A}(\tau )& =j_{5B}(\tau )+22+\frac{125}{j_{5B}(\tau )}\\ j_{5B}(\tau )& ={\left(\frac{\eta (\tau )}{\eta (5\tau )}\right)}^6\end{array}$$
および
$$\begin{array}{rlrl}{s}_{5A}(n)& =s_n[11,3,1](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{2n}{n})\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2(\genfrac{}{}{0ex}{}{n+k}{k})\\ s_{5B}(n)& =t_n[11,3,1]& & =\sum _{k=0}^n(-1{)}^{n-k}{(\genfrac{}{}{0ex}{}{n}{k})}^3(\genfrac{}{}{0ex}{}{4n-5k}{3n})\end{array}$$
とおくと
$$(\eta (\tau )\eta (5\tau ){)}^2=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{5A}(n)}{j_{5A}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{5B}(n)}{j_{5B}(\tau {)}^{n+\frac{1}{2}}}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{j_{5A}-44-16j_{5A}^{-1}}\sum _{n=0}^{\mathrm{\infty }}{s}_{5A}(n)\frac{n+\frac{1}{2}+\lambda }{j_{5A}(\tau {)}^{n+\frac{1}{2}}}\\ & =\sqrt{j_{5B}+22+125j_{5B}^{-1}}\sum _{n=0}^{\mathrm{\infty }}{s}_{5B}(n)\frac{n+\frac{1}{2}+\mu }{j_{5B}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは$|j_{5A}(\tau )|>2(11+5\sqrt{5})=44.36\dots ,|j_{5B}(\tau )|>5\sqrt{5}=11.18\dots$において収束する。

$$\begin{array}{rlrl}{z}_B& =\frac{[1{]}^6[6]}{[2{]}^3[3{]}^2}& x_B& =\frac{[2][6{]}^5}{[1{]}^5[3]}\\ z_C& =\frac{[2{]}^6[3]}{[1{]}^3[6{]}^2}& x_C& ={\left(\frac{[1][6{]}^2}{[2{]}^2[3]}\right)}^4\\ z_D& =\frac{[2][3{]}^6}{[1{]}^2[6{]}^3}& x_D& ={\left(\frac{[1][6{]}^3}{[2][3{]}^3}\right)}^3\end{array}$$
とおくと
$$\begin{array}{rl}{z}_B& =\sum _{n=0}^{\mathrm{\infty }}{s}_n[-17,-6,-72]x_B^n\\ z_C& =\sum _{n=0}^{\mathrm{\infty }}{s}_n[10,3,-9]x_C^n\\ z_D& =\sum _{n=0}^{\mathrm{\infty }}{s}_n[7,2,8]x_D^n\end{array}$$
が成り立つ。

$$\begin{array}{rl}{j}_{6A}(\tau )& ={(\sqrt{j_{6B}(\tau )}-\frac{1}{\sqrt{j_{6B}(\tau )}})}^2={(\sqrt{j_{6C}(\tau )}+\frac{8}{\sqrt{j_{6C}(\tau )}})}^2={(\sqrt{j_{6D}(\tau )}+\frac{9}{\sqrt{j_{6D}(\tau )}})}^2-4\\ j_{6B}(\tau )& ={\left(\frac{[2][3]}{[1][6]}\right)}^{12}\\ j_{6C}(\tau )& ={\left(\frac{[1][3]}{[2][6]}\right)}^6\\ j_{6D}(\tau )& ={\left(\frac{[1][2]}{[3][6]}\right)}^4\\ j_{6E}(\tau )& ={\left(\frac{[2][3{]}^3}{[1][6{]}^3}\right)}^3=\frac{1}{x_D}\end{array}$$
および
$$\begin{array}{rlrl}{\alpha }_B(n)& =s_n[-17,-6,-72](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{2n}{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-8{)}^{n-k}\sum _{j=0}^k{(\genfrac{}{}{0ex}{}{k}{j})}^3\\ {\alpha }_C(n)& =s_n[10,3,-9](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{2n}{n})\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2(\genfrac{}{}{0ex}{}{2k}{k})\\ {\alpha }_D(n)& =s_n[7,2,8](\genfrac{}{}{0ex}{}{2n}{n})& & =(\genfrac{}{}{0ex}{}{2n}{n})\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^3\\ s_{6B}(n)& =t_n[-17,-6,-72]& & =\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2{(\genfrac{}{}{0ex}{}{n+k}{k})}^2\\ s_{6C}(n)& =t_n[10,3,-9]& & =(-1{)}^n\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2(\genfrac{}{}{0ex}{}{2n-2k}{n-k})(\genfrac{}{}{0ex}{}{2k}{k})\\ s_{6D}(n)& =t_n[7,2,8]& & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+k}{n})(\genfrac{}{}{0ex}{}{n}{3k})(\genfrac{}{}{0ex}{}{3k}{2k})(\genfrac{}{}{0ex}{}{2k}{k})(-3{)}^{n-3k}\end{array}$$
とおくと
$$\begin{array}{rl}[1][2][3][6]& =\sum _{n=0}^{\mathrm{\infty }}\frac{{\alpha }_B(n)}{(j_{6A}(\tau )-32{)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\alpha }_C(n)}{(j_{6A}(\tau )+4{)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\alpha }_D(n)}{(j_{6A}(\tau ){)}^{n+\frac{1}{2}}}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{s_{6B}(n)}{j_{6B}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{6C}(n)}{j_{6C}(\tau {)}^{n+\frac{1}{2}}}=\sum _{n=0}^{\mathrm{\infty }}\frac{s_{6D}(n)}{j_{6D}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
が成り立つ。

$$\begin{array}{rl}\frac{1}{2\pi \mathrm{Im}(\tau )}& =\sqrt{\frac{(j_{6A}(\tau )+4)j_{6A}(\tau )}{j_{6A}(\tau )-32}}\sum _{n=0}^{\mathrm{\infty }}{\alpha }_B(n)\frac{n+\frac{1}{2}+{\lambda }_B}{(j_{6A}(\tau )-32{)}^{n+\frac{1}{2}}}\\ & =\sqrt{\frac{j_{6A}(\tau )(j_{6A}(\tau )-32)}{j_{6A}(\tau )+4}}\sum _{n=0}^{\mathrm{\infty }}{\alpha }_C(n)\frac{n+\frac{1}{2}+{\lambda }_C}{(j_{6A}(\tau )+4{)}^{n+\frac{1}{2}}}\\ & =\sqrt{\frac{(j_{6A}(\tau )-32)(j_{6A}(\tau )+4)}{j_{6A}(\tau )}}\sum _{n=0}^{\mathrm{\infty }}{\alpha }_D(n)\frac{n+\frac{1}{2}+{\lambda }_D}{(j_{6A}(\tau ){)}^{n+\frac{1}{2}}}\\ & =\sqrt{j_{6A}(\tau )-32}\sum _{n=0}^{\mathrm{\infty }}{s}_{6B}(n)\frac{n+\frac{1}{2}+{\mu }_B}{j_{6B}(\tau {)}^{n+\frac{1}{2}}}\\ & =\sqrt{j_{6A}(\tau )+4}\sum _{n=0}^{\mathrm{\infty }}{s}_{6C}(n)\frac{n+\frac{1}{2}+{\mu }_C}{j_{6C}(\tau {)}^{n+\frac{1}{2}}}\\ & =\sqrt{j_{6A}(\tau )}\sum _{n=0}^{\mathrm{\infty }}{s}_{6D}(n)\frac{n+\frac{1}{2}+{\mu }_D}{j_{6D}(\tau {)}^{n+\frac{1}{2}}}\end{array}$$
これらは
$$\begin{array}{rlrl}|j_{6A}(\tau )-32|& >36& |j_{6B}(\tau )|& >17+12\sqrt{2}=33.97\dots \\ |j_{6A}(\tau )+4|& >36& |j_{6C}(\tau )|& >16\\ |j_{6A}(\tau )|& >32& |j_{6D}(\tau )|& >9\end{array}$$
において収束する。