A small lamp lies in the plane of a circular ring of unit radius

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Numerical and computational mathematics · Numerical and computational mathematics

Error, Chance and the Machine, Ch Integration: error expansions, Gauss rules and adaptivity

Problem

A small lamp lies in the plane of a circular ring of unit radius, at distance ρ=12\rho = \tfrac{1}{2} from the centre, with all lengths in units of the ring radius. By the cosine rule the squared distance from the lamp to the point of the ring at angle θ\theta is 1+ρ2−2ρcos⁡θ=1.25−cos⁡θ1 + \rho^2 - 2\rho\cos\theta = 1.25 - \cos\theta, so the intensity there is proportional to f(θ)=1/(1.25−cos⁡θ)f(\theta) = 1/(1.25 - \cos\theta). The integral I=∫02πf dθ=8π/3=8.377580I = \int_0^{2\pi}f\,\mathrm{d}\theta = 8\pi/3 = 8.377580 is exact. The Fourier series of ff is f(θ)=∑kakeikθf(\theta) = \sum_ka_ke^{ik\theta} with ak=43 2−∣k∣a_k = \tfrac{4}{3}\,2^{-|k|}, from summing two geometric series. Write TN=2πN∑j=0N−1f(2πj/N)T_N = \frac{2\pi}{N}\sum_{j=0}^{N-1}f(2\pi j/N) for the trapezium rule with NN points over one period. ((a)) Compute T4T_4 and T8T_8, and their errors TN−IT_N - I. ((b)) Use the Euler-Maclaurin formula to explain why these errors are not a constant times h2h^2, and why the second error is not one quarter of the first. ((c)) Use Eq. (5.11) and the given coefficients aka_k to find TN−IT_N - I in closed form. Check it against part (a) and say what fixes the factor by which the error shrinks for each added point, in terms of the poles of ff in the complex θ\theta plane. ((d)) Find the smallest NN for which the error is below 10−1010^{-10}.

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