How to find the inverse of elliptic function of first kind in term of angle of integration $\varphi$? This link say that Jacobi amplitude $\varphi = \text{am}(u)$ gives the value of angle $\varphi$. How do I expand the Jacobi amplitude in series so that I can numerically calculate the angle $\varphi$ as function of $u$.
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There's Fourier expansion:
$$\operatorname{am} (u,k)= \frac{\pi u}{2K}+\sum_{n=1}^{\infty} \frac{\sin \frac{n\pi u}{K}}{n\cosh \frac{n\pi K'}{K}} $$
where $K\equiv K(k)=F \left( \frac{\pi}{2},k \right)$ and $K'\equiv K(\sqrt{1-k^2})$.
By the ways, its better to use symbolic software to ease your work.
For example, in Mathematica
JacobiAmplitude[u,m]
for $\operatorname{am} (u,\sqrt{m})$
Ng Chung Tak
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this seems related to this – Mula Ko Saag Jul 18 '16 at 07:22
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As you have shown, the inverse of $\text F(x,y)$ with respect to $x$ is $\text{am}(x,y)$, but what if I wanted to solve for the other argument, $y$? – Тyma Gaidash Oct 25 '21 at 23:40
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Applying InverseSeries. for 22.10(ii), 22.16.7, 22.16.8, etc. – Ng Chung Tak Oct 26 '21 at 06:46