|
|
楼主 |
青青子衿
发表于 2023-12-12 20:20
本帖最后由 青青子衿 于 2023-12-14 08:53 编辑 青青子衿 发表于 2023-11-19 19:25
\begin{align*}
f(E,q)&=\dfrac{q\frac{\mathrm{d}x(q)}{\mathrm{d}q}}{2y(q) + a_1x(q) + a_3}=\dfrac{q\frac{\mathrm{d}x(q)}{\mathrm{d}q}}{2y(q)+1}\\
&=q \prod _{k=1}^{+\infty} \left(1-q^k\right)^2 \left(1-q^{11 k}\right)^2\\
\end{align*}
\begin{align*}
\vartheta_1(z,\mathscr{q})
&=2\mathscr{q}^{1/4}\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}\sin((2n+1)z)\\
&=2\mathscr{q}^{1/4}\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}i\frac{e^{-i(2n+1)z}-e^{i(2n+1)z}}{2}\\
&=i\mathscr{q}^{1/4}\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{-i(2n+1)z}-\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
&=i\mathscr{q}^{1/4}\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{i(-2n-1)z}-\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
&=i\mathscr{q}^{1/4}\sum_{n=-\infty}^{0}(-1)^{-n}\mathscr{q}^{n(n-1)}e^{i(2n-1)z}-\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
&=i\mathscr{q}^{1/4}\sum_{n=-\infty}^{0}(-1)^{n}\mathscr{q}^{n(n-1)}e^{i(2n-1)z}-\sum_{n=0}^{+\infty}(-1)^n\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
&=i\mathscr{q}^{1/4}\sum_{n=-\infty}^{-1}(-1)^{n+1}\mathscr{q}^{n(n+1)}e^{i(2n+1)z}+\sum_{n=0}^{+\infty}(-1)^{n+1}\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
&=i\mathscr{q}^{1/4}\sum_{n=-\infty}^{+\infty}(-1)^{n+1}\mathscr{q}^{n(n+1)}e^{i(2n+1)z}
\end{align*}
\begin{align*}
\vartheta_1(z,\mathscr{q})
&=i\mathscr{q}^{1/4}\sum_{n=-\infty}^{+\infty}(-1)^{n+1}\mathscr{q}^{n(n+1)}e^{i(2n+1)z}\\
\vartheta_1(z,\mathscr{q})
&=2\mathscr{q}^{1/4}\sin(z)\prod_{k=1}^{+\infty}(1-\mathscr{q}^{2k})(1-2\cos(2z)\mathscr{q}^{2k}+\mathscr{q}^{4k})\\
&=-i\mathscr{q}^{1/4}e^{iz}\prod_{k=1}^{+\infty}(1-\mathscr{q}^{2k})(1-\mathscr{q}^{2k}e^{2iz})(1-\mathscr{q}^{2k-2}e^{-2iz})
\end{align*}
\begin{align*}
\vartheta_1(-g\pi\tau,q^{N/2})&=\vartheta_1(-g\pi\tau,\mathscr{q}^N)\\
&=-2q^{N/8}\sin(g\pi\tau)\prod_{k=1}^{+\infty\,\,}(1-q^{kN})(1-2\cos(2g\pi\tau)q^{kN}+q^{2kN})\\
&=-iq^{-g^2/(2N^2)}E_{g,N}(\tau)\eta(N\tau)
\\
\\
E_{g,N}\left(\tau\right)&=q^{b(g,N)}\prod_{k=1}^{\infty}\left[\left(1-q^{\left(k-1\right)N+g}\right)\left(1-q^{kN-g}\right)\right]\\
q&=\exp(2\pi\,\!i\tau)\\
B(p)&=p^2-p+\frac{1}{6}\\
b(g,N)&=\frac{N}{2}\cdot B\left(\frac{g}{N}\right)=\frac{g^2}{2N}-\frac{g}{2}+\frac{N}{12}\\
\eta(\tau)&=q^{1/24}\prod_{k=1}^{\infty}\left(1-q^{k}\right)
\end{align*}
\begin{align*}
\\
&\begin{split}
x&=\frac{E_{4,11}^{2}E_{2,11}}{E_{1,11}^{3}}-\frac{E_{3,11}^{2}E_{4,11}}{E_{2,11}^{3}}+\frac{E_{1,11}^{2}E_{5,11}}{E_{3,11}^{3}}-\frac{E_{5,11}^{2}E_{3,11}}{E_{4,11}^{3}}+\frac{E_{2,11}^{2}E_{1,11}}{E_{5,11}^{3}}\\
y&=\frac{E_{2,11}^{4}}{E_{4,11}^{3}E_{1,11}}-\frac{E_{4,11}^{4}}{E_{3,11}^{3}E_{2,11}}-\frac{E_{5,11}^{4}}{E_{1,11}^{3}E_{3,11}}+\frac{E_{3,11}^{4}}{E_{5,11}^{3}E_{4,11}}-\frac{E_{1,11}^{4}}{E_{2,11}^{3}E_{5,11}}-2\\
\\
y^2 + y &= x^3 − x^2 − 10x − 20\\
\\
E_{g,N}\left(\tau\right)&=q^{g^2/(2N)-g/2+N/12}\prod_{k=1}^{\infty}\left[\left(1-q^{\left(k-1\right)N+g}\right)\left(1-q^{kN-g}\right)\right]\\
q&=\exp(2\pi\,\!i\tau)\\
\\
E_{g,N}(\tau)&=-iq^{g^2/(2N^2)}\dfrac{\vartheta_1(g\pi\tau,q^{N/2})}{\eta(N\tau)}\\
\vartheta_1(g\pi\tau,q^{N/2})
&=2q^{N/8}\sin(g\pi\tau)\prod_{k=1}^{+\infty\,\,}\Big[(1-q^{kN})\big(1-2\cos(2g\pi\tau)q^{kN}+q^{2kN}\big)\Big]\\
\eta(N\tau)&=q^{N/24}\prod_{k=1}^{\infty}\left(1-q^{kN}\right)\\
\\
x(\tau)&=\frac{q^{3/2}\,\vartheta _1^2\left(4 \pi \tau ,q^{11/2}\right) \vartheta _1\left(2 \pi \tau ,q^{11/2}\right)}{\vartheta _1^3\left(\pi \tau ,q^{11/2}\right)}-\frac{q\,\vartheta _1^2\left(3 \pi \tau ,q^{11/2}\right) \vartheta _1\left(4 \pi \tau ,q^{11/2}\right)}{\vartheta _1^3\left(2 \pi \tau ,q^{11/2}\right)}\\
&\qquad\quad+\frac{\vartheta _1^2\left(\pi \tau ,q^{11/2}\right) \vartheta _1\left(5 \pi \tau ,q^{11/2}\right)}{\vartheta _1^3\left(3 \pi \tau ,q^{11/2}\right)}
-\frac{q^{1/2}\vartheta _1^2\left(5 \pi \tau ,q^{11/2}\right)\vartheta _1\left(3 \pi \tau ,q^{11/2}\right) }{\vartheta _1^3\left(4 \pi \tau ,q^{11/2}\right)}\\
&\qquad\qquad\quad+\frac{\vartheta _1^2\left(2 \pi \tau ,q^{11/2}\right)\vartheta _1\left(\pi \tau ,q^{11/2}\right) }{q^3 \vartheta _1^3\left(5 \pi \tau ,q^{11/2}\right)}\\
y(\tau)&=\frac{\vartheta _1^4\left(2 \pi \tau ,q^{11/2}\right)}{q^{3/2}\,\vartheta _1^3\left(4 \pi \tau ,q^{11/2}\right)\vartheta _1\left(\pi \tau ,q^{11/2}\right) }-\frac{q^{3/2} \vartheta _1^4\left(4 \pi \tau ,q^{11/2}\right)}{\vartheta _1^3\left(3 \pi \tau ,q^{11/2}\right)\vartheta _1\left(2 \pi \tau ,q^{11/2}\right) }\\
&\qquad-\frac{q^4\,\vartheta _1^4\left(5 \pi \tau ,q^{11/2}\right)}{\vartheta _1^3 \left(\pi \tau ,q^{11/2}\right)\vartheta _1\left(3 \pi \tau ,q^{11/2}\right)}+\frac{\vartheta _1^4\left(3 \pi \tau ,q^{11/2}\right)}{q^{5/2}\,\vartheta _1^3\left(5 \pi \tau ,q^{11/2}\right)\vartheta _1\left(4 \pi \tau ,q^{11/2}\right) }\\
&\qquad\qquad-\frac{\vartheta _1^4\left(\pi \tau ,q^{11/2}\right)}{q^{3/2}\,\vartheta _1^3\left(2 \pi \tau ,q^{11/2}\right) \vartheta _1\left(5 \pi \tau ,q^{11/2}\right)}-2\\
\\
x_1&=x\left(\tau=\frac{-3+i \sqrt{35}}{22}\right)\\
&=3+2\sqrt{5}+i\sqrt{7}\,(7+3 \sqrt{5}) \approx7.47214 + 36.2685 i\\
\\
x_2&=x\left(\tau=\frac{-25+i \sqrt{35}}{66}\right)\\
&=3-2 \sqrt{5}-i \sqrt{7}(7-3 \sqrt{5})\approx-1.47214 - 0.77202 i\\
\\
y_1&=y\left(\tau=\frac{-3+i \sqrt{35}}{22}\right)\\
&=53+24 \sqrt{5}-i \sqrt{7}(37+17 \sqrt{5})\approx106.666 - 198.466i\\
\\
y_2&=y\left(\tau=\frac{-25+i \sqrt{35}}{66}\right)\\
&=53-24 \sqrt{5}-i \sqrt{7} \left(17 \sqrt{5}-37\right)\approx-0.665631 - 2.68056 i\\
\\
x_1\oplus\,\!x_2&=\frac{-16-2 i\sqrt{35}}{9}\approx-1.77778 - 1.31468 i\\
y_1\oplus\,\!y_2&=\frac{-40+4 i \sqrt{35}}{27}\approx-1.48148 + 0.876456 i
\end{split}
\end{align*}
|
|
