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青青子衿
发表于 2023-5-15 15:46
本帖最后由 青青子衿 于 2023-5-15 20:58 编辑
\begin{align*}
\left.\partial_z\vartheta_1(z,q)\right|_{z=0}&=2 \sqrt[4]{q}\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right)^3\\
\vartheta_2(0,q)&=2 \sqrt[4]{q}\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k}\right)^2\\
\vartheta_3(0,q)&=\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k-1}\right)^2\\
\vartheta_4(0,q)&=\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1-q^{2 k-1}\right)^2\\
\sqrt{k}&=\dfrac{\vartheta_2(0,q)}{\vartheta_3(0,q)}=2\sqrt[4]{q}\cdot\left[\dfrac{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k})}{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k-1})}\right]^2\\
\sqrt{k'}&=\dfrac{\vartheta_4(0,q)}{\vartheta_3(0,q)}=\left[\dfrac{\prod\limits_{i=1}^{+\infty\>\>}(1-q^{2k-1})}{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k-1})}\right]^2\\
k^2+k'^2&=1
\end{align*}
- 2 q^(1/4) Product[(1 - q^(2 k))^3, {k, 1, +Infinity}] /. q -> 0.3
- EllipticThetaPrime[1, 0, 0.3]
- 2 q^(1/4) Product[(1 - q^(2 k)) (1 + q^(2 k))^2, {k,
- 1, +Infinity}] /. q -> 0.3
- EllipticTheta[2, 0, 0.3]
- Product[(1 - q^(2 k)) (1 + q^(2 k - 1))^2, {k, 1, +Infinity}] /.
- q -> 0.3
- EllipticTheta[3, 0, 0.3]
- Product[(1 - q^(2 k)) (1 - q^(2 k - 1))^2, {k, 1, +Infinity}] /.
- q -> 0.3
- EllipticTheta[4, 0, 0.3]
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Czhang271828
发表于 2022-1-30 11:58
