Find the inverse of an equation reminiscent of Kepler's equation

Question

Context

For an eccentricity, $e$, Kepler's Equation is given as \begin{align*} M &= E - e\, \sin {\left(E\right)} \end{align*} For a historical treatment of Kepler's equation and how to invert it, see [1] and references therein.

Question

Let $E$ be such that $E = [-\pi, \pi]$ and $M$ be such that $M = [-\pi, \pi]$. I have a form reminiscent of Kepler's Equation, which is \begin{align*} M &=E - \, \sin{\left( E \right)} \, \left(- \dfrac{2}{3 } \,\cos^2{\left(\dfrac{E}{2}\right)} - 1\right) \end{align*} How can I invert the above to find E(M)?

Bibliography

[1] Colwell, P. Solving Kepler's Equation over Three Centuries. Richmond, VA: Willmann-Bell, 1993.

Answer 1

Here is a closed form from:

Numerical inverse of a function: $\frac34 x - \frac 12\sin(2x) + \frac 1{16} \sin(4x)$.

$$M=E - \, \sin{\left( E \right)} \, \left(- \dfrac{2}{3 } \,\cos^2{\left(\dfrac{E}{2}\right)} - 1\right)\mathop\iff^{E=2\sin^{-1}(x)} M=2\sin^{-1}(x)+\frac23x(5-2x^2)\sqrt{1-x^2}$$

Therefore we use $\int_0^x (1-t^2)^r=\frac12\text B_{x^2}\left(\frac12,r+1\right)$ with the incomplete beta function $\text B_z(a,b)$, convert to beta regularized $\text I_z(a,b)$, and use the haversine hav$(x)$

$$M=\frac{16}3\int_0^{\sin\left(\frac E2\right)}(1-t^2)^\frac32dt=\frac83 \text B_{\sin^2\left(\frac E2\right)}\left(\frac12,\frac52\right)=\pi \text I_{\text{hav}(E)}\left(\frac12,\frac52\right) $$

Finally, we use inverse beta regularized $\text I^{-1}_s(a,b)$ and inverse haversine hav$^{-1}(x)$:

$$\boxed{E=\text{hav}^{-1}\left(\text I^{-1}_{\frac M\pi -2n}\left(\frac12,\frac52\right)\right)+2\pi n, \text{hav}^{-1}\left(\text I^{-1}_{\frac M\pi -(2n+1) }\left(\frac52,\frac1 2\right)\right)+(2n+1)\pi,n\in\Bbb Z}$$

Shown is the $0\le M\le \pi$ case:

$\text I^{-1}_s\left(\frac52,\frac12\right)$ reduces the problem to a student T distribution quantile of $v=5$ degrees of freedom.

Answer 2

There are different methods of proofs that the function of Kepler's equation doesn't have partial inverses (over a complex domain without isolated points) that are elementary functions.

$$M=E-\sin(E)\left(-\frac{2}{3}\cos^2\left(\frac{E}{2}\right)-1\right)$$ $$E-\sin(E)\left(-\frac{2}{3}\cos^2\left(\frac{E}{2}\right)-1\right)-M$$ $$-\frac{1}{12}i(e^{iE})^4-\frac{2}{3}i(e^{iE})^3+(E-M)(e^{iE})^2+\frac{2}{3}ie^{iE}+\frac{1}{12}i=0$$ $E\to \frac{t}{I}$: $$-\frac{1}{12}i(e^t)^4-\frac{2}{3}i(e^t)^3+(-M-it)(e^t)^2+\frac{2}{3}ie^t+\frac{1}{12}i=0$$

The function on the left-hand side of the latter equation is an algebraic function in dependence of both $t$ and $e^t$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions.

The equation is also an algebraic equation in dependence of both $t$ and $e^t$. Lin proved, assuming Schanuel's conjecture is true, that such kind of equations don't have solutions except $0$ that are elementary numbers.

The latter equation also shows that the equation cannot be solved in terms of elementary functions and Lambert W or Generalized Lambert W either.

How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?