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Also fun is the Dirichlet function[1], which is discontinuous at every point yet is integrable everywhere. There's also a similar function that is discontinuous at all rational numbers but continuous at all irrational numbers[2]. [1] http://en.wikipedia.org/wiki/Dirichlet_function [2] http://en.wikipedia.org/wiki/Thomae%27s_function
Also fun is the Dirichlet function[1], which is discontinuous at every point yet is integrable everywhere. There's also a similar function that is discontinuous at all rational numbers but continuous at all irrational numbers[2]. [1] http://en.wikipedia.org/wiki/Dirichlet_function [2] http://en.wikipedia.org/wiki/Thomae%27s_function
I believe that is only true when you consider certain types of integration. E.g. True for Lebesgue, false for Riemann
Earlier quoted context omitted.
Consider this function: f(x) = sum[n = 0 to infinity]((1/2)^n sin((1/20)^n pi x)) It converges because each successive term is bounded by [-1/2, +1/2]. But if you integrate it you get: f(x) = sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) (Well... you get that if you play fast and loose about swapping the order of the integral and the sum) Which diverges because the input to the cosine function limits to 0 as n…
What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take the limit as n -> infinity. This is not meaningful. You can only do that with a definite integral (under conditions). Using your reasoning, even a function like exp(x) would not integrate, because the terms in the integral of its series expansion would all go…
1. Can you give me any function whose derivative is the function I specified? (i.e. I give you permission to force all the integration constants to zero)
2. Can you compute the area under the curve from `x1` to `x2` of the function I specified?
3. Do you agree that you can do (2) but not (1)?
Earlier quoted context omitted.
What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take the limit as n -> infinity. This is not meaningful. You can only do that with a definite integral (under conditions). Using your reasoning, even a function like exp(x) would not integrate, because the terms in the integral of its series expansion would all go…
I mentioned that I was playing fast and loose when swapping them. But nevermind that exact process, let's be a bit more direct. 1. Can you give me any function whose derivative is the function I specified? (i.e. I give you permission to force all the integration constants to zero) 2. Can you compute the area under the curve from `x1` to `x2` of the function I specified? 3. Do you agree that you can do (2) but not (1)…
Earlier quoted context omitted.
I am quite surprised it does not sound like white noise. Anyway, I played the output of that script through some filters, and this sounds nice: play -c 2 weierstrass.wav gain -12 loudness chorus 0.5 0.9 50 0.4 0.25 2 -t 60 0.32 0.4 2.3 -t 40 0.3 0.3 1.3 -s reverb gain 12
Ah, it's easy to see why it won't sound like white noise. The standard function is a Fourier series - a sum of sine waves. It's a power series where the b parameter must be an odd positive integer value 7 or greater. If we want the n = 0 case to map to 20 Hz (minimum of human hearing), and the maximum frequency we can hear is 20kHz (valid for children) then the highest term we can hear is the solution to 20000/7^n =…
Earlier quoted context omitted.
What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take the limit as n -> infinity. This is not meaningful. You can only do that with a definite integral (under conditions). Using your reasoning, even a function like exp(x) would not integrate, because the terms in the integral of its series expansion would all go…
I mentioned that I was playing fast and loose when swapping them. But nevermind that exact process, let's be a bit more direct. 1. Can you give me any function whose derivative is the function I specified? (i.e. I give you permission to force all the integration constants to zero) 2. Can you compute the area under the curve from `x1` to `x2` of the function I specified? 3. Do you agree that you can do (2) but not (1)…
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?
Earlier quoted context omitted.
I mentioned that I was playing fast and loose when swapping them. But nevermind that exact process, let's be a bit more direct. 1. Can you give me any function whose derivative is the function I specified? (i.e. I give you permission to force all the integration constants to zero) 2. Can you compute the area under the curve from `x1` to `x2` of the function I specified? 3. Do you agree that you can do (2) but not (1)…
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 manipu…
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...
Earlier quoted context omitted.
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 manipu…
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.…
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.
Earlier quoted context omitted.
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.…
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 antid…