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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.

Michael Hardy
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Vladimir Vargas
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  • You don't need to write $e<<1$; you can write $e\ll 1$. I changed it. ${}\qquad{}$ – Michael Hardy Apr 03 '15 at 16:58
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    Someone will come along shortly to explain the specific details, but the short version is that when $e$ is small then you're dealing with a contraction mapping in the neighborhood of the fixed-point and so the recurrence moves steadily towards that fixed-point solution of $E=M+e\sin E$. – Steven Stadnicki Apr 03 '15 at 17:01

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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 $p$.

Robert Israel
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