The lemma can be understood intuitively as saying that the integral of a product of f(x) with a rapidly varying (n-> oo) periodic function is simply the integral of f(x) times the average of the periodic function within its period interval.
Proof of the lemma is not right. From 5:00 on it should be, integrating by parts: Goal + Integral(a,b): f(x)h(nx)dx = Goal + 1/n*(H(nb)f(b) - H(na)f(a)) - 1/n*(Integral(a,b): f'(x)H(nx)dx = Goal as n-> infinity.
Better justification for the integral of f(x)h(nx) going to zero: The integral of h(x) is zero over any interval of length p. The integral of h(nx) is zero over any interval of length p/n. Therefore over any integral from a to b, we are just adding zero for every period of h(nx) that can fit in the interval, plus whatever is left over. That leftover bit is bounded above by h_max since h is bounded, and has a max width of the period p/n, therefore a max area of h_max*p/n. Since h_max and p are finite, the whole thing will go to zero as n grows.
For sufficently large n, we will have 0 < pi/n < 1, and for any x in (0, pi/n), we have: 1/(1 + |sin(nx)|)² = 1/(1 + sin(nx))² So the derivative from below will be equal to: -2ncos(nx)/(1 + sin(nx))², which approaches 2n. For any x in (pi/n, 1) that is sufficently close to pi/n, we will have: 1/(1 + |sin(nx)|)² = 1/(1 - sin(nx))² And therefore, the derivative from above will be: 2ncos(nx)/(1 - sin(nx))³, which approaches -2n. So the answer is clearly no, this function will have many points at which its derivative doesn't exist. Considering these points will always be finitely many though, I think we may still apply a similar lemma here.
@@Debg91Taking the limit results in the Lemma. The Lemma only cares if g(x) is smooth, not what happens to g(nx) when n approaches infinity Also note that g(nx) is still smooth for variable n as you can simply rewrite it with t=nx to be g(t), which we already know is smooth
@@Debg91 If we look at the proof of the lemma, we only use "smoothness" when we rewrite h(nx) as h'(nx) inside an integral, and we then we only care about the global boundedness of h'(x) on R. We don't actually need smoothness here so far as I can tell, I really don't know why Michael has put such a restrictive condition here. Since the derivative g'(x) is periodic, continuous and bounded on its domain, we can define it to be 0 on the points I've discussed, so that it is periodic, bounded and piecewise continuous on R. Note that we only need to do this for g'(x), we won't need to take limits of anything. We can then proceed with the proof as before and all will be fine.
@@Debg91 Based on the proof of the lemma, it looks like the function only needs to be piecewise differentiable and continuous for the lemma to hold. If the lemma strictly required smoothness, then it would not apply in this scenario
@@ve4rexe One could still save this by introducing an antiderivative H of h and then integrate by parts (with an extra assumption: for f to be C^1 on [a,b]). The way h is designed, H would be p-periodic and therefore bounded... That is one way to prove the usual version of Riemann-Lebesgue (the one with circular functions) for C^1 functions defined on [a,b]
@@frankstengel6203 bro im gonna be very honest with you i haven’t learnt any of this i am in 11th grade i simply know basic calculus and you were using the anti-linear chain rule wrong 💀 also he has assumed that h(0) is zero but even if one doesn’t assume that the stuff he said is mostly right as n tends to infinity that whole non-goal integral tends to zero so could you elucidate on what you’re trying to save here?
@@ve4rexe what you wrote in your first comment is exactly what I wrote in my first one. I am under the impression that at around 5:18, one sees something that boils down to writing: h(x)=\frac{1}{n}\int_0^x h’(nt)dt Which is quite puzzling since it conflicts with chain derivation/substitution in integrals…
I noticed the same thing. He never used the fact that the integral of h was 0, and his “proof” could have been done on g as well as h, so it was obviously wrong that way. I think your fix is the best way to finish the proof
tan is not continuous except if you restrict the domain to just one period, which may not be the case when the input is nx for x in [a,b]. so it cannot really be considered “smooth” as michael has written as one of the hypotheses in the lemma. though the confusion understandably stems from his being a bit fast and loose with those hypotheses
The lemma can be understood intuitively as saying that the integral of a product of f(x) with a rapidly varying (n-> oo) periodic function is simply the integral of f(x) times the average of the periodic function within its period interval.
Proof of the lemma is not right. From 5:00 on it should be, integrating by parts:
Goal + Integral(a,b): f(x)h(nx)dx = Goal + 1/n*(H(nb)f(b) - H(na)f(a)) - 1/n*(Integral(a,b): f'(x)H(nx)dx = Goal as n-> infinity.
Better justification for the integral of f(x)h(nx) going to zero:
The integral of h(x) is zero over any interval of length p. The integral of h(nx) is zero over any interval of length p/n. Therefore over any integral from a to b, we are just adding zero for every period of h(nx) that can fit in the interval, plus whatever is left over. That leftover bit is bounded above by h_max since h is bounded, and has a max width of the period p/n, therefore a max area of h_max*p/n. Since h_max and p are finite, the whole thing will go to zero as n grows.
Is 1/(1+|sin(nx)|)^2 smooth at all?
For sufficently large n, we will have 0 < pi/n < 1, and for any x in (0, pi/n), we have:
1/(1 + |sin(nx)|)² = 1/(1 + sin(nx))²
So the derivative from below will be equal to:
-2ncos(nx)/(1 + sin(nx))², which approaches 2n.
For any x in (pi/n, 1) that is sufficently close to pi/n, we will have:
1/(1 + |sin(nx)|)² = 1/(1 - sin(nx))²
And therefore, the derivative from above will be:
2ncos(nx)/(1 - sin(nx))³, which approaches -2n.
So the answer is clearly no, this function will have many points at which its derivative doesn't exist. Considering these points will always be finitely many though, I think we may still apply a similar lemma here.
@@Balequalm I agree the function behaves sufficiently well for a fixed *n* but will it still hold when we take the limit?
@@Debg91Taking the limit results in the Lemma. The Lemma only cares if g(x) is smooth, not what happens to g(nx) when n approaches infinity
Also note that g(nx) is still smooth for variable n as you can simply rewrite it with t=nx to be g(t), which we already know is smooth
@@Debg91 If we look at the proof of the lemma, we only use "smoothness" when we rewrite h(nx) as h'(nx) inside an integral, and we then we only care about the global boundedness of h'(x) on R. We don't actually need smoothness here so far as I can tell, I really don't know why Michael has put such a restrictive condition here.
Since the derivative g'(x) is periodic, continuous and bounded on its domain, we can define it to be 0 on the points I've discussed, so that it is periodic, bounded and piecewise continuous on R. Note that we only need to do this for g'(x), we won't need to take limits of anything. We can then proceed with the proof as before and all will be fine.
@@Debg91 Based on the proof of the lemma, it looks like the function only needs to be piecewise differentiable and continuous for the lemma to hold. If the lemma strictly required smoothness, then it would not apply in this scenario
Today there is no " that's good place to stop", there is just "that's good" (2:23).
5:18 I have a doubt: as far as I can tell \int_0^x n h'(nt) dt=h(nx)-h(0), so one should multiply by n and not divide by n...
n*int(hprime(nt)) = n*(h(nx)/n) = h(nx)
@@ve4rexe One could still save this by introducing an antiderivative H of h and then integrate by parts (with an extra assumption: for f to be C^1 on [a,b]). The way h is designed, H would be p-periodic and therefore bounded... That is one way to prove the usual version of Riemann-Lebesgue (the one with circular functions) for C^1 functions defined on [a,b]
@@frankstengel6203 bro im gonna be very honest with you i haven’t learnt any of this i am in 11th grade i simply know basic calculus and you were using the anti-linear chain rule wrong 💀 also he has assumed that h(0) is zero but even if one doesn’t assume that the stuff he said is mostly right as n tends to infinity that whole non-goal integral tends to zero so could you elucidate on what you’re trying to save here?
@@ve4rexe what you wrote in your first comment is exactly what I wrote in my first one. I am under the impression that at around 5:18, one sees something that boils down to writing:
h(x)=\frac{1}{n}\int_0^x h’(nt)dt
Which is quite puzzling since it conflicts with chain derivation/substitution in integrals…
I noticed the same thing. He never used the fact that the integral of h was 0, and his “proof” could have been done on g as well as h, so it was obviously wrong that way. I think your fix is the best way to finish the proof
Non-trivial calculus aggregations in the limit of the unexpected. Love it!
That lemma can be generalized to a double integral (seems like not many ppl know this), see MSE 4566609. The main theme is equidistribution.
Something's off about thi proof , you never used that the integral of h is 0. So the proof could've worked with any constant "offset"
You should've used integration by parts. Integral of h over a period is 0 ensures that its antiderivative is also periodic..
6:34 How do we know that h is bounded on the interval? Tan is periodic, integrable, and differentiable but it still has no upper bound on its period
It could be that Tan is hinted at being excluded as it is not really nice @ 0:30 and so not simple enough for the topic explained today?
tan is not continuous except if you restrict the domain to just one period, which may not be the case when the input is nx for x in [a,b]. so it cannot really be considered “smooth” as michael has written as one of the hypotheses in the lemma. though the confusion understandably stems from his being a bit fast and loose with those hypotheses
thumbnail is wrong
It's the same integral though.
thats just a ploy of his, for more views
@@Ijkbeautyso it's clickbait? That's not appreciated.
infinity
Mike used to explain clearly; getting worse
The proof is fishy. He does not explain at all why the second term in h(x) has to be like that.
It would be interesting to see some problems where the final answer is the problem has no solutions. Just something different
third!