Is there a mathematical method to simplify this infinitely recursive function?
I have tried to approximate it, but for work with the ranges I am considering, precision becomes an important factor.
the recursion is as follows
$f_i = a + b \ \sin( f_{i-1} )$ with $f_0 = a$
For example, 3 stage recursion:
$a + b \ \sin( a + b \ \sin( a + b \ \sin( a ) ) )$
Variable ranges if it helps for an approximation
$a \in [-2\pi , 2\pi]$ and $b \in [0,1]$
P.s. I apologize for the poor formatting. This is my first time in StackExchange.
If your goal is to solve for $E$ Kepler's equation $$E-e\sin(E)=M$$ there should not be any problem using high-order iterative methods. The only problem is to have a reasonable starting guess.
The solution being bounded $M < E <M+e$, expand as a series around $E=M$ and use series reversion to get $$E_0=M+\frac{e \sin (M)}{1-e \cos (M)}-\frac{e^3 \sin ^3(M)}{2 (1-e \cos (M))^3 }+\cdots$$ Now, to find the zero of function $$f(E)=E-e\sin(E)-M$$ use Newton, Halley or Householder method with this guess $E_0$.
In fact; if I properly remember, there are explicit solutions in terms of Lambert hyperfunction writing the equation in the exponential form $$E \exp\Bigg[\log \left(1-e\frac{ \sin (E)}{E}\right)\Bigg]=M$$