Weierstrass椭圆函数化雅可比

青青子衿

\begin{align*}
\sqrt[3]{4}\,\wp\!\left(\tfrac{u}{\sqrt[3]{4}};0,1\right)&=\frac{\sqrt{3}\left(1+\text{cn}\left(\sqrt[4]{3}\,u,\frac{\sqrt{6}-\sqrt{2}}{4}\right)\right)}
{1-\text{cn}\left(\sqrt[4]{3}\,u,\frac{\sqrt{6}-\sqrt{2}}{4}\right)}+1\\

\sqrt[3]{4}\,\wp\!\left(\tfrac{u}{\sqrt[3]{4}};0,-1\right)&=\frac{\sqrt{3}\left(1+\text{cn}\left(\sqrt[4]{3}\,u,\frac{\sqrt{6}+\sqrt{2}}{4}\right)\right)}
{1-\text{cn}\left(\sqrt[4]{3}\,u,\frac{\sqrt{6}+\sqrt{2}}{4}\right)}-1\\
\end{align*}

青青子衿

\begin{align*}
S(u)&=\sqrt[3]{4}\,\wp\!\left(\frac{u}{\sqrt[3]{4}}-\frac{\Gamma \left(\frac{1}{3}\right)^3}{2 \sqrt{3} \pi };0,-1\right)
+\sqrt[3]{4}\,\wp\!\left(\frac{u}{\sqrt[3]{4}};0,-1\right)\\

&=\left(\frac{2 \sqrt[4]{27} \operatorname{sn}\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\operatorname{dn}\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right) +\left(1-\operatorname{cn}\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\right)^2}

{\sqrt{3} \operatorname{sn}^2\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)-\left(1-\operatorname{cn}\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\right)^2}\right)^2\\

\end{align*}

青青子衿

\begin{align*}
&\qquad\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}\\
&\qquad\alpha_1>\alpha_2>\alpha_3\\
\sigma_1&=\alpha_1+\alpha_2+\alpha_3=0\\
\wp_{\rm{R}}(u)&=4\wp\left(u;{\small{-}}\tfrac{\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3}{4},\tfrac{\alpha_1\alpha_2\alpha_3}{16}\right)\\
&=\frac{\alpha_1-\alpha_3}{\operatorname{sn}^2\left(\frac{\sqrt{\alpha_1-\alpha_3}}{2} \,u\,,\sqrt{\frac{\alpha_2-\alpha_3}{\alpha_1-\alpha_3}}\,\right)}+\alpha_3\\
\\
u&=\int_{\wp_{\rm{R}}(u)}^{+\infty}\frac{\mathrm{d}t}{\sqrt{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)}}\\
\\

&\qquad\alpha_1,\alpha_3\in\mathbb{C},\,\alpha_2\in\mathbb{R}\\

\sigma_1&=\alpha_1+\alpha_2+\alpha_3=0\\
\wp_{\rm{C}}(u)&=4\wp\left(u;{\small{-}}\tfrac{\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3}{4},\tfrac{\alpha_1\alpha_2\alpha_3}{16}\right)\\

&=\tfrac{\sqrt{(\alpha_2-\alpha_1)(\alpha_2-\alpha_3)}\operatorname{cn}^2\!\Big(\frac{(\alpha_2-\alpha_1)^{1/4}(\alpha_2-\alpha_3)^{1/4}}{2}u,k_\alpha\>\!\Big)}{\operatorname{sn}^2\!\Big(\frac{(\alpha_2-\alpha_1)^{1/4}(\alpha_2-\alpha_3)^{1/4}}{2}u,k_\alpha\>\!\Big)\operatorname{dn}^2\!\Big(\frac{(\alpha_2-\alpha_1)^{1/4}(\alpha_2-\alpha_3)^{1/4}}{2}u,k_\alpha\>\!\Big)}+\alpha_2\\
k_\alpha&=\sqrt{\frac{1}{2}-\frac{3\alpha_2}{4 \sqrt{(\alpha_2-\alpha_1)(\alpha_2-\alpha_3)}}}\\
\\
u&=\int_{\wp_{\rm{C}}(u)}^{+\infty}\frac{\mathrm{d}t}{\sqrt{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)}}\\
\end{align*}

\begin{align*}
{\large\int}_{x}^{+\infty}\frac{\mathrm{d}t}{\sqrt{t^{3}-30t+56}}
&=
\frac{\sqrt{\small{6(2-\sqrt{2}\>\!)}}\,}{3}{\large\int}_{0}^{\sqrt{\frac{6+3\sqrt{2}}{x+2+3\sqrt{2}}}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-(\!\sqrt{\small{2(\sqrt{2}-1\>\>\!\!)}}\>\>\!\!)^2t^{2})}}\\
&=\frac{1}{i\sqrt{2}}{\large\int}_{\frac{18-4 x+x^2}{2 (4-x)}}^{+\infty}\frac{\mathrm{d}t}{\sqrt{t^{3}-30t+56}}\\
\\
{\large\int}_{\frac{6+3\sqrt{2}-(2+3\sqrt{2}\>\!)X^2}{X^{2}}}^{+\infty}\frac{\mathrm{d}t}{\sqrt{t^{3}-30t+56}}
&=\frac{\sqrt{\small{6(2-\sqrt{2}\>\!)}}\,}{3}{\large\int}_0^{X}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-(\!\sqrt{\small{2(\sqrt{2}-1\>\>\!\!)}}\>\>\!\!)^2t^{2})}}\\
&=\frac{\sqrt{\small{6(2-\sqrt{2}\>\!)}}\,}{3}\cdot\frac{1}{i\sqrt{2}}{\large\int}_{0}^{\frac{\sqrt{2}iX\sqrt{1-X^{2}}}{1-\sqrt{2}X^{2}}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-(\!\sqrt{\small{2(\sqrt{2}-1\>\>\!\!)}}\>\>\!\!)^2t^{2})}}\\
&=\frac{1}{i\sqrt{2}}{\large\int}_{-\frac{6+3\sqrt{2}-2(4+3\sqrt{2}\>\!)X^{2}+8X^{4}}{2X^{2}-2X^{4}}}^{+\infty}\frac{\mathrm{d}t}{\sqrt{t^{3}-30t+56}}\\

\end{align*}

青青子衿

青青子衿 发表于 2023-8-23 13:44
\begin{align*}
&\qquad\alpha_1,\alpha_3\in\mathbb{C},\,\alpha_2\in\mathbb{R}\\

\sigma_1&=\alpha_1+\alpha_2+\alpha_3=0\\
\wp_{\rm{C}}(u)&=4\wp\left(u;{\small{-}}\tfrac{\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3}{4},\tfrac{\alpha_1\alpha_2\alpha_3}{16}\right)\\
\\
u&=\int_{\wp_{\rm{C}}(u)}^{+\infty}\frac{\mathrm{d}t}{\sqrt{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)}}\\
\end{align*}

\begin{align*}
\left(\frac{\partial\left(4 \wp \left(u;\raise{0.5pt}\scriptsize{-\tfrac{A}{4},-\tfrac{B}{4^2}}\right)\right)}{\partial u}\right)^2&=\big(4 \wp \left(u;\raise{0.5pt}\scriptsize{-\tfrac{A}{4},-\tfrac{B}{4^2}}\right)\big)^3+A\cdot\big(4 \wp \left(u;\raise{0.5pt}\scriptsize{-\tfrac{A}{4},-\tfrac{B}{4^2}}\right)\big)+B
\\
\left(\frac{\partial\left(\wp (u;g_2,g_3)\right)}{\partial u}\right)^2&=4\cdot\wp (u;g_2,g_3)^3-g_2\cdot\wp (u;g_2,g_3)-g_3
\end{align*}

1. D[WeierstrassP[u, {-(A/4), -(B/4^2)}], u]^2 /. {A -> 322,
2.     B -> 1453} /. u -> 0.4 // N
3. 4 X^3 + A/4 X + B/4^2 /.
4.     X -> WeierstrassP[u, {-(A/4), -(B/4^2)}] /. {A -> 322,
5.     B -> 1453} /. u -> 0.4 // N

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Four Lectures on Weierstrass Elliptic Function and Applications in Classical and Quantum Mechanics
arxiv.org/pdf/1706.07371
Fukushima, T. (2013). Recursive computation of derivatives of elliptic functions and of incomplete elliptic integrals.
sci-hub.ru/https://doi.org/10.1016/j.amc.2013.06.008
Elliptic constructions of hyperkaehler metrics III: Gravitons and Poncelet polygons
arxiv.org/pdf/0712.3601

1. f[z_?NumericQ] :=
2.   InverseWeierstrassP[{(4 WeierstrassP[z, {-(3/4), -(5/16)}])/4,
3.     -Sqrt[
4.       4 ((4 WeierstrassP[z, {-(3/4), -(5/16)}])/4)^3 +
5.        3/4 ((4 WeierstrassP[z, {-(3/4), -(5/16)}])/4) + 5/16]}, {-(3/
6.      4), -(5/16)}];
7. g[z_?NumericQ] :=
8.   InverseWeierstrassP[(4 WeierstrassP[z, {-(3/4), -(5/16)}])/
9.    4, {-(3/4), -(5/16)}];
10. h[z_?NumericQ] :=
11.   NIntegrate[1/
12.    Sqrt[t^3 + 3 t + 5], {t,
13.     4 WeierstrassP[z, {-3/4, -5/16}], +Infinity}];
14. hs[z_?NumericQ] :=
15.   1/2 NIntegrate[1/
16.     Sqrt[u^3 + 3 u + 5], {u,
17.      WeierstrassP[z, {-4*3, -4*5}], +Infinity}];
19. Plot[{f[z], g[z], h[z], hs[z]}, {z, -10, 10},
20. Frame -> True, FrameLabel -> {{"f(z)", None}, {"z", "my integral"}},
21. GridLines -> Automatic, GridLinesStyle -> LightGray,
22. PlotStyle -> {{Green}, {Blue, Dotted}, {Red, Dashed}, {Brown,
23.     DotDashed}}, BaseStyle -> 12, ImageSize -> Large]

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\begin{align*}
\zeta \left(z;g_2,g_3\right)
&=\frac{1}{z}-\int_0^z \left(\wp \left(t;g_2,g_3\right)-\frac{1}{t^2}\right) \mathrm{d}t\\
&=\frac{1}{z}-\int_0^z \frac{\wp \left(t;g_2,g_3\right)-\frac{1}{t^2}}{\wp'\!\left(t;g_2,g_3\right)} \mathrm{d}(\wp\left(t;g_2,g_3\right))\\
&=\frac{1}{z}-\int_0^{\wp\left(z;g_2,g_3\right)}\frac{u-\frac{1}{[\wp^{-1}\left(u;g_2,g_3\right)]^2}}{\sqrt{4u^3-g_2u-g_3}}\mathrm{d}u\\

\end{align*}

青青子衿

青青子衿 发表于 2023-7-6 01:04
\begin{align*}
S(u)&=\sqrt[3]{4}\,\wp\!\left(\frac{u}{\sqrt[3]{4}}-\frac{\Gamma \left(\frac{1}{3}\right)^3}{2 \sqrt{3} \pi };0,-1\right)
+\sqrt[3]{4}\,\wp\!\left(\frac{u}{\sqrt[3]{4}};0,-1\right)\\

&=\left(\frac{2 \sqrt[4]{27} \operatorname{sn}\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\operatorname{dn}\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right) +\left(1-\operatorname{cn}\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\right)^2}

{\sqrt{3} \operatorname{sn}^2\!\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)-\left(1-\operatorname{cn}\left(\sqrt[4]{3} u,\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\right)^2}\right)^2\\

\end{align*}

\begin{align*}
\operatorname{sl}(x)
&=\operatorname{sn}(x,i)=\frac{\operatorname{sn}(\sqrt{2}x,\frac{\sqrt{2}}{2})}{\sqrt{2}\operatorname{dn}(\sqrt{2}x,\frac{\sqrt{2}}{2})}=
\frac{\operatorname{sn}(\sqrt{2}x,\sqrt{2})}{\operatorname{cn}(\sqrt{2}x,\sqrt{2})}\\

\operatorname{sl}(x)&
=\frac{1}{\sqrt{\wp(x;4,0)}}=-i\,\wp^{1/2}\left(x+\tfrac{(1+i) \Gamma^2(\frac{1}{4})}{4 \sqrt{2 \pi }};4,0\right)\\
&=\frac{1-i}{\sqrt{2\wp(\frac{1+i}{\sqrt{2}}x;-4,0)}}
=\left(\frac{2\wp(\frac{x}{\sqrt{2}};-4,0)}{1+\wp^2(\frac{x}{\sqrt{2}};-4,0)}\right)^{1/2}\\
&=\frac{1-i}{2 \sqrt{\wp ((1+i) x;-1,0)}}=
2\left(\frac{\wp (x;-1,0)}{1+4 \wp^2(x;-1,0)}\right)^{1/2}\\
&=\frac{1}{\sqrt{2 \wp({\scriptsize{\sqrt{2}}\,}x;1,0)}}=- i\sqrt{2}\,\wp^{1/2}\left(\sqrt{2} \left(x+\tfrac{(1+i)\Gamma^2(\frac{1}{4})}{4 \sqrt{2 \pi }}\right);1,0\right)
\end{align*}

1. -Sqrt[2] I*Sqrt[
2.   WeierstrassP[
3.    Sqrt[2] (x + (1 + I) Gamma[1/4]^2/(4 Sqrt[2 \[Pi]])), {1, 0}]] /.
4. x -> 0.1346
5. 1/Sqrt[2 WeierstrassP[Sqrt[2] x, {1, 0}]] /. x -> 0.1346
7. -I Sqrt[WeierstrassP[
8.    x + (1 + I) Gamma[1/4]^2/(4 Sqrt[2 \[Pi]]), {4, 0}]] /.
9. x -> 0.1346
10. 1/Sqrt[WeierstrassP[x, {4, 0}]] /. x -> 0.1346

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\begin{gather*}
\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1-t^{3}}}
=\frac{1}{\sqrt[4]{3}}\operatorname{sn}^{-1}\left(\sqrt[4]{3}\tfrac{\big(2+(1+\sqrt{3})x\big)\sqrt{1+x+x^{2}}-\big(2-(\sqrt{3}-1)x\big)\sqrt{1-x}}{2+2\sqrt{3}+2x+(2+\sqrt{3})x^{2}},\tfrac{\sqrt{2}+\sqrt{6}}{4}\right)\\
\\
-\left(2+\sqrt{3}+\frac{\sqrt[4]{3}}{\sqrt{2}}\left(3+\sqrt{3}\right)\right)\leqslant\,\!x<1\\
\\

\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1-t^{3}}}
=\frac{1}{\sqrt[4]{3}}\operatorname{sn}^{-1}\left(\tfrac{\frac{(1+i)(\sqrt{3}-i)3^{1/4}}{4\sqrt{2}}\Big(\sqrt{1+{\scriptsize{\frac{1+i\sqrt{3}}{2}}}x}+\sqrt{(1-x)(1+{\scriptsize{\frac{1-i\sqrt{3}}{2}}}x)}\Big)}{1-\frac{1+i\sqrt{3}}{4}x},\tfrac{\sqrt{3}-i}{2}\right)
\end{gather*}

\begin{align*}
\operatorname{sl}_3(x)
&=1+\frac{\frac{-3+i\sqrt{3}}{2}}{{\operatorname{sn}}^{2}\left(\frac{3^{1/4}(1+i)(\sqrt{3}-i)^{2}}{8\sqrt{2}}x+\frac{2K(\frac{\sqrt{3}-i}{2})}{\sqrt{3}},\frac{\sqrt{3}+i}{2}\right)}\\
&=1-\frac{1-{\operatorname{cn}}\left(\sqrt[4]{3}x-\frac{2K(\frac{\sqrt{2}+\sqrt{6}}{4})}{3},\frac{\sqrt{2}+\sqrt{6}}{4}\right)}{1+{\operatorname{cn}}\left(\sqrt[4]{3}x-\frac{2K(\frac{\sqrt{2}+\sqrt{6}}{4})}{3},\frac{\sqrt{2}+\sqrt{6}}{4}\right)}\sqrt{3}\\

\operatorname{sl}_3(x)&=\wp \left(\frac{i}{2}\left(x+\frac{\Gamma \left(\frac{1}{3}\right)^3}{2^{1/3}3^{1/2} \pi }\right);0,4\right)\\
&=-\wp \left(\frac{1}{2}\left(x+\frac{\Gamma \left(\frac{1}{3}\right)^3}{2^{1/3}3^{1/2} \pi }\right);0,-4\right)\\
&=-4^{1/3} \wp \left(\frac{2^{1/3}i}{2} \left( x+\frac{\Gamma \left(\frac{1}{3}\right)^3}{2^{1/3}3^{1/2}\pi}\right);0,-1\right)\\
&=4^{1/3} \wp \left(\frac{2^{1/3}i}{2} \left( x+\frac{\Gamma \left(\frac{1}{3}\right)^3}{2^{1/3}3^{1/2}\pi}\right);0,1\right)
\end{align*}


\begin{align*}
\frac{1-\operatorname{cn}\left(z,k\right)}{1+\operatorname{cn}\left(z,k
\right)}=
\frac{{\operatorname{sn}}^{2}\left(\frac{z}{2},k\right){\operatorname{dn}}^{2}
\left(\frac{z}{2},k\right)}{{\operatorname{cn}}^{2}\left(\frac{z}{2},k\right)}
\end{align*}