3 ms·

I see what you're getting at now. Let's check the conditions for integrability. Your function is bounded (everywhere, but let's restrict to [0,1]). But I'm n

by mturmon 13y ago

I see what you're getting at now. Let's check the conditions for integrability. Your function is bounded (everywhere, but let's restrict to [0,1]). But I'm not sure it is continuous due to the limit process. If you can't show it's continuous almost everywhere (as noted by @clintonc) then it is not Riemann-integrable, therefore its integral will not exist. That would be the answer to (1). If you go through some manipulations (i.e., (2)) and get an expression that seems correct, but the underlying function is not actually Riemann-integrable (as in (1)), then I'd suggest your manipulations have fooled you. So, do you know the answers to these questions?

Strilanc 13y 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 13y ago
  The 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 13y ago
    Bingo. Very clear.

by mturmon 13y ago

Strilanc 13y 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 13y ago
The 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 13y ago
  Bingo. Very clear.