Newton-Raphson is used to solve Kepler's equation for a highly eccentric orbit with
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Computational Physics, Ch Where a function vanishes, and what lies between the points
Problem
Newton-Raphson is used to solve Kepler's equation E−esinE=M for a highly eccentric orbit with e=0.90, all angles in radians. Near its root E∗≈2.4rad the derivative is f′(E∗)=1−ecosE∗=1.66, and the three quantities entering f(E)=E−esinE−M (namely E, esinE and M) are each of order unity in magnitude. Every real number is stored in IEEE 754 binary64, for which ϵmach=2.22×10−16 and each stored quantity carries a relative rounding error no larger than 21ϵmach.Someone sets the stopping criterion ∣f(E)∣<10−19, reasoning: "a smaller number in the tolerance always buys more accurate digits, so asking for 10−19 costs nothing and only helps."
((a)) Using ∣f(E)∣≈∣f′(E∗)∣∣E−E∗∣, state the error in E that a criterion ∣f(E)∣<10−19 would leave, if it could ever be satisfied.
((b)) Explain, from the size of the quantities entering f and the machine epsilon, why the computed residual can never be driven that low, and say what the iteration does instead once it has run past a coded iteration cap.
((c)) Give a residual-type stopping criterion at a defensible tolerance that the arithmetic can actually meet, and state the error in E it leaves.