I am trying to solve Kepler's equation ( $ 0 = E - e·\sin(E) + M $ ) with an iterative method but due to the elliptic orbit of the body for some typical values of e and M which causes eccentric anomaly, I prefer to use Secant Method instead of Newton Method. However, I cannot decide on the initial points. For Newton, I use the initial point: $$ E_0=M + \frac{e\sin(M) }{1 - \sin(M + e) + \sin(M)} $$
However, what could be a good choice for the second initial point? I guess close values may work but I am not sure.
Not using the secant method.
I always think that it is much better to use Newton method provided that we are able to generate a good starting point. One thing I use to do is to build the $[1,m]$ Padé approximant of function $f(x)$ around a given point $x_0$.
These approximants write $$f(x) \sim \frac {f(x_0)+a^{(m)}(x-x_0) } {1+\sum_{i=1}^m b_i^{(m)} (x-x_0)^i}$$ where all coefficients are defined in terms of the function and high-order derivative value at $x_0$. I find this more efficient than series reversion which are often suggested (have a look here)
Then the estimate $$x_{(m)}=x_0-\frac {f(x_0)}{a^{(m)}}$$ For small values of $m$, this correspond to Newton method, original Halley method or original Householder method.
For Kepler's equation $$M=E-e \sin(E)$$ using $x_0=0$ this would give the following estimates $$E_{(0)}=E_{(1)}=\frac M{1-e}$$ $$E_{(2)}=\frac {6(1-e)^2 M} {e M^2+6(1-e)^3}$$ $$E_{(3)}=\frac{e M^3+6 (1-e)^3 M}{2 e (1-e) M^2+6 (1-e)^4}$$
Suppose $e=0.1$ and $M=0.5678$. Converted to decimals, the above formulae (and the next ones) give
$$\left( \begin{array}{cc} m & E_{(m)} \\ 0 & 0.63088888888888888889 \\ 1 & 0.63088888888888888889 \\ 2 & 0.62627278539121795082 \\ 3 & 0.62630631527966334097 \\ 4 & 0.62642926189145170922 \\ 5 & 0.62642749747544870702 \\ 6 & 0.62642515169269243491 \\ 7 & 0.62642520885980656763 \\ \dots & \cdots \\ \infty & 0.62642525037762133960 \end{array} \right)$$