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$\def\F\{\operatorname F}\def\E{\operatorname E}$

Elliptic F$(x,m)$ appears in elliptic E$(x,m) =\int_0^x\sqrt{1-m\sin^2(x)}dx$ parameter $m$ transformations. Luckily the DLMF has reciprocal/imaginary modulus transformations and Gauss/Landen transformations, but these both use $\E(x,k)= \int_0^x\sqrt{1-k^2\sin^2(x)} dx$. The following formulas work around $z=0$:

Reciprocal parameter transformation:

$$\E(z,m)=\sqrt m\E\left(\sin^{-1}\left(\sqrt m\sin(z)\right),\frac1m\right)+\frac{1-m}{\sqrt m}\F\left(\sin^{-1}(\sqrt m\sin(z),\frac1m\right)\tag1$$

Complementary parameter/imaginary argument transformation:

$$\E(z,m)=i\E(i\tanh^{-1}(\sin(z)),1-m)-i\F(i\tanh^{-1}(\sin(z)),1-m)+\sin(z)\sqrt{\frac{1-m\sin^2(z)}{\cos^2(z)}}$$

Negative parameter transform from ResearchGate using Elliptic $\E(z)$ and complex conjugate $\bar z$

$$\E(z,m)\mathop=^{m<1}\sqrt{1-m}\left(\E\left(\frac m{1-m}\right)-\E\left(\frac\pi 2-z,\frac m{1-m}\right)\right)\mathop=^{m>1}-\overline{\sqrt{1-m}\left(\E\left(\frac m{1-m}\right)-\E\left(\frac\pi 2-z,\frac m{1-m}\right)\right)}\tag2$$

Ascending Landen transformation:

$$\E(z,m)=(1+\sqrt m)\E\left(\frac12(\sin^{-1}(\sqrt m\sin(z))+z),\frac{2\sqrt m}{m+1}\right)+ (1-\sqrt m)\F\left(\frac12(\sin^{-1}(\sqrt m\sin(z))+z),\frac{2\sqrt m}{m+1}\right)-\sqrt m\sin(z)$$

Descending Landen transformation:

$$\E(z,m)=\frac{\sqrt{1-m}+1}2\E\left(\tan^{-1}(\sqrt{1-m}\tan(z))+z,\left(\frac{\sqrt{1-m}-1}{\sqrt{1-m}+1}\right)^2\right)-\sqrt{1-m}\F(z,m)+\frac{m\cos(z)\sin(z)}{2\sqrt{1-m\sin^2(z)}}$$

These formulas answer most of the original question, but they use $\frac{m\cos(z)\sin(z)}{2\sqrt{1-m\sin^2(z)}}$ and other complicated trigonometric expressions.

What are other simpler transformations for $\E(z,m)$, perhaps without trigonometric expressions like in $(1),(2)$?

Тyma Gaidash
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    Do you know what is a nome ? – Barackouda Mar 18 '23 at 13:10
  • generally the update should be in the question itself, unless your self-answer completely answers the question to your satisfaction – kodlu Mar 19 '23 at 14:17
  • Hi, it could be difficult to find sources that uses Wolfram's notation given all the theory of Elliptic integrals at least since Cayley was written with $k$ instead of $m$ but for some unknown reason Wolfram uses a different notation. – Bertrand87 Mar 20 '23 at 01:21
  • However, knowing that $m$ is just $k^2$ in the standard notation, we have the following:

    $$ E(\varphi, im) = \int_{0}^{\varphi} \sqrt{1-im\sin^2\varphi} = \int_{0}^{\varphi} \sqrt{1-(\sqrt{im})^2\sin^2\varphi}=\int_{0}^{\varphi} \sqrt{1-(e^{\frac{i\pi}{4}}\sqrt{m})^2\sin^2\varphi}$$

    So $k = e^{\frac{i\pi}{4}}\sqrt{m}$ in the standard notation.

     – Bertrand87 Mar 20 '23 at 01:22
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    But note that

    $$ k = \sqrt{m}e^{\frac{i\pi}{4}} = \frac{\sqrt{m}}{\sqrt{2}}+\frac{i\sqrt{m}}{\sqrt{2}}$$

    This is a non--purely imaginary modulus. All the transoformations in the theory for imaginary modulus are made with purely imaginary modulus. i.e. modulus of the form $i\bar{k}$ with $\bar{k}\in \mathbb{R} $

     – Bertrand87 Mar 20 '23 at 01:32
  • @Bertrand87 $k\to-i k$ gives a formula for $\text E(z,k)$ in terms of $\text E\left(z,-\frac{i k}{\sqrt{1-k^2}}\right)$, so there is a formula for $\text E\left(z,\sqrt m e^\frac{\pi i}4\right)$ in terms of $\text E\left(z,\frac{e^{-\frac{\pi i}4}\sqrt m}{\sqrt{1-i m}}\right)$. From there, a formula for $\text E(z,m)$ is possible. Is this right? – Тyma Gaidash Mar 20 '23 at 02:15
  • @TymaGaidash the first transformation assumes that $ik$ has a real part equal to zero (ie is purely imaginary) while $\sqrt{m}e^{\frac{\pi i}{4}}$ does not. We should find a transofrmation for an imaginary modulus that is not purely imaginary, but I does not know one. – Bertrand87 Mar 20 '23 at 03:47
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    If you only want the equivalent formulas for the imaginary modulus with Wolfram's notation note that if $m = k^2$

    $$ \int_{0}^{\varphi} \sqrt{1-(ik)^2\sin \varphi}d\varphi =\int_{0}^{\varphi} \sqrt{1+k^2\sin \varphi} d\varphi = \int_{0}^{\varphi} \sqrt{1-(-m)\sin \varphi}d\varphi = E(\varphi,-m) $$

    So the transformation of "negative modulus" in Mathematica is equivalent to the transformation of imaginary modulus in the standard notation.

     – Bertrand87 Mar 20 '23 at 04:05
  • @Bertrand87 Thanks for the help. The “negative modulus” term helped find a good reference for a significantly simpler transformation. Maybe you know about simpler forms of the other transformations in the question. – Тyma Gaidash Mar 20 '23 at 15:24

1 Answers1

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(This is not an answer but an extended comment)

Note for example, the following transformation (using Mathematica notation). Suppose that $0<m<1$, then

$$E(\varphi,-m) = \sqrt{1+m}E\left(\varphi,\frac{ m}{m+1}\right)- \frac{m\sin(\varphi)\cos(\varphi)}{\sqrt{1+m\cos^2\varphi}}$$

For convenience put $\displaystyle \varphi = \frac{\pi}{2}$ and $\displaystyle m=\frac{1}{2}$

$$E\left(\frac{\pi}{2},-\frac{1}{2}\right) =\sqrt{\frac{3}{2}}E\left(\frac{\pi}{2},\frac{1}{3}\right)$$

So, in Mathematica notation we are transforming a negative modulus to a positive modulus while using the standard notation we are transforming a purely imaginary modulus to a positive one:

$$E\left(\frac{\pi}{2},\frac{i}{\sqrt{2}}\right) = \sqrt{\frac{3}{2}}E\left(\frac{\pi}{2},\frac{1}{\sqrt{3}}\right) $$

My advice is to completely ignore the Wolfram's notation for the theory of elliptic integrals and use it only for computation. They also put a warning on this issue in MathWorld.

Bertrand87
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