[math-fun] Riemann integration question
I've got another question arising from the honors calculus course I teach. Suppose f is continuous on [a,b] with I = int_a^b f(x) dx, and for every epsilon > 0 let delta(epsilon) be the largest delta > 0 such that every Riemann sum arising from a partition of [a,b] with mesh less than delta differs from I by less than epsilon. Is it true that (leaving aside the case where f is constant) delta(epsilon) goes to zero like epsilon^2, in the sense that delta(epsilon)/epsilon^2 is bounded above and below by constants as epsilon goes to zero? Jim Propp
I don't think this is true, but haven't worked out the details. My idea for a counterexample f(x): Start with a baseline of 0, and add triangular spikes in stages. The first stage puts spikes at x=0 and x=1; later stages put spikes in between all previous spikes. The number of spikes (roughly) doubles at each stage. The heights gradually drift down, say as 1/stage. The widths drop off by a factor of (say) 2 or 3 each time, enough so the spikes don't overlap previous stages, and leave some room for following stages. This f should be continuous, since the spike height goes to 0; and its Riemann integral should converge, for the same reason. But the convergence will be very slow as delta shrinks, since an unlucky placement of the points that sample the function, at the spike tips, will seriously overestimate the integral. Rich ---- Quoting James Propp <jpropp@cs.uml.edu>:
I've got another question arising from the honors calculus course I teach.
Suppose f is continuous on [a,b] with I = int_a^b f(x) dx, and for every epsilon > 0 let delta(epsilon) be the largest delta > 0 such that every Riemann sum arising from a partition of [a,b] with mesh less than delta differs from I by less than epsilon.
Is it true that (leaving aside the case where f is constant) delta(epsilon) goes to zero like epsilon^2, in the sense that delta(epsilon)/epsilon^2 is bounded above and below by constants as epsilon goes to zero?
Jim Propp
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Exactly. A closely related object that seems easier to visualize: given a continuous function f, let Mf_epsilon(t) be the maximum of f in the interval of width epsilon centered at t, and let mf_epsilon(t) be the minimum in this interval. Claim: given any continous monotone function delta(epsilon) with delta(0) = 0, you can find a continous function on [0,1] where the integral of Mf_epsilon - mf_epsilon is greater than delta(epsilon) Proof: Construct the graph of a function f as the intersection of a decreasing sequence of constant-width bands, where successive bands have width say oscillate up and down at rapidly-increasing frequencies 1/epsilon_i and have width 2*delta(epsilon_i). Note that f is he uniform limit of a sequence continuous functions, so it is continuous. If the frequencies increase quickly enough, then the region below Mf_epsilon_i and mf_epsilon_i contains a band of width delta(epsilon_i). Similarly, the upper and lower Riemann sums differ by more than delta(epsilon_i). Bill On Jan 30, 2010, at 3:20 PM, rcs@xmission.com wrote:
I don't think this is true, but haven't worked out the details. My idea for a counterexample f(x): Start with a baseline of 0, and add triangular spikes in stages. The first stage puts spikes at x=0 and x=1; later stages put spikes in between all previous spikes. The number of spikes (roughly) doubles at each stage. The heights gradually drift down, say as 1/stage. The widths drop off by a factor of (say) 2 or 3 each time, enough so the spikes don't overlap previous stages, and leave some room for following stages. This f should be continuous, since the spike height goes to 0; and its Riemann integral should converge, for the same reason. But the convergence will be very slow as delta shrinks, since an unlucky placement of the points that sample the function, at the spike tips, will seriously overestimate the integral.
Rich
---- Quoting James Propp <jpropp@cs.uml.edu>:
I've got another question arising from the honors calculus course I teach.
Suppose f is continuous on [a,b] with I = int_a^b f(x) dx, and for every epsilon > 0 let delta(epsilon) be the largest delta > 0 such that every Riemann sum arising from a partition of [a,b] with mesh less than delta differs from I by less than epsilon.
Is it true that (leaving aside the case where f is constant) delta(epsilon) goes to zero like epsilon^2, in the sense that delta(epsilon)/epsilon^2 is bounded above and below by constants as epsilon goes to zero?
Jim Propp
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