二重q级数与西塔函数齐次和

青青子衿

\begin{align*}
\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}q^{m^{2}+mn+\frac{\ell+1}{4}n^{2}}&=\vartheta _2(0,q) \vartheta _2(0,q^{\ell})+\vartheta _3(0,q) \vartheta _3(0,q^{\ell})\\
&\qquad\qquad\qquad\qquad\qquad\quad\big(\ell=3,7,11\big)\qquad\\
\vartheta _2(0,q)&=2q^{1/4}\prod_{k=1}^{+\infty}\frac{(1-q^{4k})^2}{1-q^{2k}}\\
&=2q^{1/4}\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k}\right)^2\\
\vartheta _3(0,q)&=\prod_{k=1}^{+\infty}\frac{(1-q^{2k})^5}{(1-q^{k})^2(1-q^{4k})^2}\\
&=\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k-1}\right)^2\\
\eta(\tau)=\eta\langle\,q\,\rangle&={\mathfrak{q}}^{1/24}\prod\limits_{k=1}^{+\infty\>\>}\left(1-\mathfrak{q}^{k}\right)={q}^{1/12}\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{2k}\right)
\end{align*}

\begin{align*}
\vartheta_2(0,e^{\pi\,\!i\tau})&=\frac{2\eta^2(2\tau)}{\eta(\tau)}\\
\vartheta_3(0,e^{\pi\,\!i\tau})&=\frac{\eta^5(\tau)}{\eta^2(\frac{\tau}{2})\eta^2(2\tau)}\\
\vartheta_4(0,e^{\pi\,\!i\tau})&=\frac{\eta^2(\frac{\tau}{2})}{\eta(\tau)}\\
\lambda(\tau)&=\dfrac{\vartheta_2^4(0,e^{\pi\,\!i\tau})}{\vartheta_3^4(0,e^{\pi\,\!i\tau})}\\
&=\left(\frac{2\eta^2(\frac{\tau}{2})\eta^4(2\tau)}{\eta^6(\tau)}\right)^4
\end{align*}

1. FullSimplify[
2. Series[EllipticTheta[3, 0, q] EllipticTheta[3, 0, q^11] +
3.    EllipticTheta[2, 0, q] EllipticTheta[2, 0, q^11], {q, 0, 20}],
4. Assumptions -> q > 0]
5. Series[Sum[q^(m^2 + m*n + 3 n^2), {m, -10, 10}, {n, -10, 10}], {q, 0,
6.   20}]

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kuing.cjhb.site/forum.php?mod=redirect&goto=findpost&ptid=8440&pid=53857
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msp.org/pjm/2008/235-1/pjm-v235-n1-p01-p.pdf
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[Tom Weston]
swc-math.github.io/notes/files/01Weston1.pdf
Hypergeometric transformation formulas of degrees 3, 7, 11 and 23
Shaun Cooper, Massey University
Jinqi Ge, Massey University
Dongxi Ye, Sun Yat-Sen University
sciencedirect.com/science/article/pii/S0022247X14007045
【超几何变换】【戴德金伊塔函数】

青青子衿

\begin{align*}
\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}q^{m^{2}+mn+n^{2}}
=\frac{4}{3} \prod _{k=1}^{+\infty} \frac{\left(1-q^{2 k}\right)^6 \left(1-q^{3 k}\right)}{\left(1-q^k\right)^3 \left(1-q^{6 k}\right)^2}-\frac{1}{3} \prod _{k=1}^{+\infty} \frac{\left(1-q^k\right)^6 \left(1-q^{6 k}\right)}{\left(1-q^{2 k}\right)^3 \left(1-q^{3 k}\right)^2}
\end{align*}
(Ogasawara 2018)
\begin{align*}
\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}q^{m^{2}+mn+3n^{2}}&=\dfrac{4\eta^2\langle\,q^2\rangle\eta^2\langle\,q^{22}\rangle}{\eta\langle\,q\,\rangle\eta\langle\,q^{11}\rangle}+\frac{\eta^5\langle\,q\,\rangle\eta^5\langle\,q^{11}\rangle}{\eta^2\langle\,q^{1/2}\rangle\eta^2\langle\,q^2\rangle\eta^2\langle\,q^{11/2}\rangle\eta^2\langle\,q^{22}\rangle}\\
\eta\langle\,q\,\rangle&\,\colon\!={q}^{1/12}\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{2k}\right)
\end{align*}

\begin{align*}
\dfrac{\mathrm{d}\eta(\tau)}{\mathrm{d}\tau} &= \frac{i \pi}{12} \eta(\tau) E_{2}\!\left(\tau\right)\\
\dfrac{\mathrm{d}\eta\langle\,q\,\rangle}{\mathrm{d}q} = \left.\dfrac{\mathrm{d}\eta(\tau)}{\mathrm{d}\tau}\middle/\dfrac{\mathrm{d}q(\tau)}{\mathrm{d}\tau}\right.&=\frac{i\pi\eta(\tau) E_{2}\!\left(\tau\right)}{12i\pi\,q}
\\
&=\frac{\eta\langle\,q\,\rangle\,\!E_{2}\langle\,q\,\rangle}{12q} \\
\frac{12q\frac{\mathrm{d}}{\mathrm{d}q}\left[q^{1/12}\prod\limits_{k=1}^{+\infty} \left(1-q^{2 k}\right)\right]}{q^{1/12}\prod \limits_{k=1}^{+\infty} \left(1-q^{2 k}\right)}&=1-24 \sum _{k=1}^{+\infty} \frac{k q^{2 k}}{1-q^{2 k}}
\end{align*}

青青子衿

青青子衿 发表于 2023-11-19 01:12
\begin{align*}
\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}q^{m^{2}+mn+3n^{2}}&=\vartheta _2(0,q) \vartheta _2(0,q^{11})+\vartheta _3(0,q) \vartheta _3(0,q^{11})
\end{align*}

\begin{align*}
\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{k}\right)\left(1-q^{11k}\right)=\dfrac{1}{2}\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}(-1)^{n}q^{\frac{(2m-1)^{2}+(2m-1)n+3n^{2}-1}{2}}
\end{align*}


Bruno Schoeneberg formula
sciencedirect.com/science/article/pii/S138572586750029X



\begin{align*}
\prod\limits_{k=1}^{+\infty\>\>}\left(1-q^{k}\right)\left(1-q^{11k}\right)&=\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}(-1)^{m+n}q^{\frac{(6 m - 1)^2 + 11 (6 n - 1)^2 - 12}{24}}\\
&=\sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}(-1)^{m+n}q^{\frac{m (3 m - 1) + 11 n (3 n - 1)}{2}}\\

\end{align*}