Is there any reason to always solve the Kepler equation for the eccentric anomaly, $E$, instead of the more meaningful (at least to me) true anomaly, $\theta$?Solving for the eccentric anomaly usually would mean calculating the true anomaly from its solution and maybe even the initial guess of the eccentric anomaly from an initial guess of the true anomaly. So this would require more calculations, however the Kepler equation expressed in the eccentric anomaly is simpler and thus less expensive to calculate. When I search for solving Kepler's equation I only have some approaches which use the eccentric anomaly, however I can imagine that when the error tolerance of the solution is not very tight or the initial guess is good it might be cheaper to solve for the true anomaly. Or are there other reasons for solving for the eccentric anomaly?The relevant equations, using Newton's method, are:$$E = 2 \tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan\frac{\theta}{2}\right),$$$$M = E - e \sin{E},$$where $M$ is the mean anomaly and $e$ is the eccentricity. Such that using Newton's method results in$$E_{n+1} = E_n - \frac{E_n - e \sin{E_n} - M}{1 - e \cos{E_n}}.$$And for the true anomaly,$$M = 2\tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan{\frac{\theta}{2}}\right)-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}},$$$$\theta_{n+1} = \theta_n - \left(1 + e \cos{\theta_n}\right)^2 \frac{2\tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan{\frac{\theta_n}{2}}\right)-\frac{e\sqrt{1-e^2}\sin{\theta_n}}{1+e\cos{\theta_n}} - M}{\sqrt{(1 - e^2)^3}}.$$
Solving Kepler equation for true or eccentric anomaly
newton method
The most robust way of answering this is to benchmark it. Failingthat, there are several things to note (roughly in order ofimportance).First, the most cheap floating-point operations on a modern CPU areaddition and multiplication (both are equally fast; same as fusedmultiply-adds when available). Division is much slower (by a factor of~20), trigonometric functions are also slower (~200), square roots andlogs similar. In fact, sometimes (depends on architecture) trigfunctions, roots, logs are implemented in a library. The bestreference for this is usually the optimization guide published by thewhichever company made your CPU.  So as a rule of thumb you want tominimize special functions first, then divisions, then multiplicationsand additions.So by this measure the second formula is much worse: the there is oneinverse trig function, one square root (it would be wasteful toimplement your formula as written, with three square roots), and onesincos (since they are expensive it is better to evaluate sincos onceand then write things like $\tan\frac\theta2$ in terms of those). Yourfirst formula involves the other special functions, apart from sincos,only once at the end.Second, you should not stop at Newton's method. For example, if youcompare Newton's method with Halley's method,$$ \frac{e \sin (x)-e x \cos (x)+M}{1-e \cos (x)}, \qquadx-\frac{(e \cos (x)-1) (e \sin (x)+M-x)}{e^2+e (M-x) \sin (x)-2 e \cos  (x)+1}, $$both evaluate the expensive trigonometric functions at the samearguments, but with a little more algebra Halley's method can convergefaster; the extra algebra might not outweigh the savings ofevaluating trigonometric fewer times. So Halley's method (and otheriterative methods) also need to be checked.Third, you can precompute some things. For example, if you start byreducing the argument to the range $0<E<2\pi$, you can experimentally,in advance, find the maximum number of iterations taken by themethod. Since the number of iterations is fairly small (depending onprecision) and mispredicted branches can be expensive, it may alsomake sense to unroll the iteration loop by hand to a fixed sufficientnumber of iterations. Unrolled code can also more easily benefit fromvectorization.Fourth, it is difficult to intuitively predict which optimizationswill do the best. And since this is just one fairly simple equation,it is probably best to benchmark many different approaches and findthe best. When doing this, also consider what optimization flags yourcompiler supports for floating-point arithmetic (some of them areinteresting and important, many people know about -ffast-math, butit actually decomposes into different helpful or harmful optimizationflags; here is gcc'slist for example). Another thing to do is to look at the assemblyoutput of your compiler to see which CPU instructions it actually endsup using.Fifth, if you need to solve this equation many times for the sameeccentricity $e$, it is possible to rewrite the problem. If youconsider the function $E = E(M)$, on the range $[0,2\pi]$, given bythe solutions of the equation for fixed known $e$, you can approximatethe function $E(M)$ using, for example, Chebyshev series, which takesonly a small number of evaluations of $E(M)$ (which can be done withany root-finding method). Once you have a sufficiently closeapproximation, which might be a 20-term Chebyshev series or somethinglike that, you can evaluate that later without needing to solve theequation again.