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The function is continuous. The cumulative effect of the sum after the nth item are bounded by a [-2^-n, 2^-n] offset, and varies very slowly as x is changed (t
by Strilanc 12y ago
The function is continuous. The cumulative effect of the sum after the nth item are bounded by a [-2^-n, 2^-n] offset, and varies very slowly as x is changed (the argument to cos is divided by increasingly huge factors). I haven't done an explicit epsilon-delta proof, but I'm confident I could make one work by using those two facts.
So I don't expect to be able to show lack of Riemann-integrable-ness via that route. ... I just don't think it has an antiderivative.
... wait, this makes no sense. I've made a mistake because if the function is continuous then it must have an antiderivative by the first fundamental theorem of calculus. I will have to actually do the epsilon-delta proof and find out where I made the tricky mistake. Probably something to do with that infinity...
- SamReidHughes 12y agoThe series (of continuous functions) converges uniformly, which means that it's Reimann integrable. And you can compute its integral as a sum of the integrals inside the sum (which you showed upthread). Also, as hinted upthread, just change f(x) = - sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) to f(x) = - sum[n = 0 to infinity](10^n / pi (cos((1/20)^n pi x) - 1)) That gives you an actual formula for the antiderivative that converges at every point.
- mturmon 12y agoBingo. Very clear.