Kepler and the contraction mapping theorem

We’ll make this statement more precise and give a historically important application.Definitions and theoremA function f on a metric space X is a contraction if there exists a constant q with 0 ≤ q < 1 such that for any pair of points x and y in X,d( f(x),  f(y) ) ≤ q d(x, y)where d is the metric on X.A point x is a fixed point of a function f if f(x) = x.Banach’s fixed point theorem, also known as the contraction mapping theorem, says that every contraction on a complete metric space has a fixed point..The proof is constructive: start with any point in the space and repeatedly apply the contraction..The sequence of iterates will converge to the fixed point.Application: Kepler’s equationKepler’s equation in for an object in an elliptical orbit saysM + e sin E = Ewhere M is the mean anomaly, e is the eccentricity, and E is the eccentric anomaly..These “anomalies” are parameters that describe the location of an object in orbit..Kepler solved for E given M and e using the contraction mapping theorem, though he didn’t call it that.Kepler speculated that it is not possible to solve for E in closed form—he was right—and used a couple iterations [1] off(E) = M + e sin Eto find an approximate fixed point..Since the mean anomaly is a good approximation for the eccentric anomaly, M makes a good starting point for the iteration..The iteration will converge from any starting point, as we will show below, but you’ll get a useful answer sooner starting from a good approximation.Proof of convergenceKepler came up with his idea for finding E around 1620, and Banach stated his fixed point theorem three centuries later..Kepler had the idea of Banach’s theorem, but he didn’t have a rigorous formulation of the theorem or a proof.In modern terminology, the real line is a complete metric space and so we only need to prove that the function f above is a contraction..By the mean value theorem, it suffices to show that the absolute value of its derivative is less than 1..That is, we can use an upper bound on |f ‘| as the q in the definition of contraction.Nowf ‘ (E) = e cos Eand so|f ‘(E)| ≤ efor all E..If our object is in an elliptical orbit, e < 1 and so we have a contraction.ExampleThe following example comes from [2], though the author uses Newton’s method to solve Kepler’s equation..This is more efficient, but anachronistic.Consider a satellite on a geocentric orbit with eccentricity e = 0.37255.. More details

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