Kepler's equation is $E-e\sin E = M$, where $e,M$ are constants.
My teacher of celestial mechanics told me that if $e\ll 1$, I should take a first aproximation $E_1=M$, then a second aproximation $E_2=M+e\sin E_1,\ldots,$ then a $n$-th aproximation $E_n=M+e\sin E_{n-1}$, until it converges, why does it work?
Thanks, I know nothing about these recurrence relations.
OK, Steven, I've come along...
If you define the function $f(E) = M + e \sin(E)$, your equation becomes the fixed-point equation $E = f(E)$. Now $f'(E) = e \cos(E)$, so if $|e| < 1$ and $E$ is real, $|f'(E)| \le |e| < 1$. This means $f$ is a contraction mapping: $|f(x) - f(y)| \le |e| |x - y|$. If $p$ is a fixed point and $E_0$ is any initial "guess" that does not satisfy the equation, this contraction property says that $E_1 = f(E_0)$ is closer to $p$ than $E_0$ is, by at worst a factor of $|e|$. Continuing with $E_2 = f(E_1)$, $E_3 = f(E_2)$, etc., we get $$\eqalign{|E_1 - p| &\le |e| |E_0 - p|\cr |E_2 - p| &\le |e| |E_1 - p| \le |e|^2 |E_0 - p|\cr |E_3 - p| &\le |e| |E_2 - p| \le |e|^3 |E_0 - p|\cr \ldots}$$ Since $|e|^n \to 0$ as $n \to \infty$, we conclude that the sequence $E_0, E_1, E_2, \ldots$ converges to $p$.