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20:00
Well, you're talking about students but you never mentioned who is your audience. I'd say, there is some expectations from a mathematics student, more than just, say, physics student.
when it comes to teaching math*
@Jakobian My first comment to you was about students learning calculus for the first time.
At that level (in the US, at least), there really is no distinction between a physics student, a math student, and an anthropology student.
It is all one audience.
you mean the calculations class?
@Jakobian I don't know what that means.
the class where they calculate rather than being given a theory
@Jakobian I don't teach any such class.
Nor do I know of anyone who does (except, possibly, some of the waste-of-time business calculus classes?).
20:03
Waste of time?
I mean a class, well, lets call it analysis, where they are going more in-depth with integrals
The standard calculus curriculum in the US starts with limits (epsilon-delta is usually discussed, though minimized; the calculus of limits is given more time), uses limits to define continuity, the derivative, and to work out properties of continuous and differentiable functions.
for example, in my analysis class, the teacher introduced us Riemann-Stieltjes integrals
Then the Riemann integral is introduced.
In the typical US curriculum, most of what is taught in such a calculus class is justified, if not completely rigorously proved. The exceptions are (1) epsilon-delta arguments are typically downplayed significantly (and left for real analysis), and (2) the extreme value theorem is typically left unjustified, as proof requires some slightly nuanced understanding of the real number system.
Students are not necessarily supposed to internalize the proofs, but they are supposed to be familiar with them.
There is more than just "plug-and-chug".
@user726941 There are a number of institutions which have a "business calculus" class which is supposed to cover limits, differentiation, integration, linear ODE, and multivariable calculus in one semester. These are typically very rote classes with a lot of drill-and-kill, and very little conceptual understanding. I think that they are largely a waste of time for students.
But, like, that just my opinion, man.
maybe I'm overestimating capabilities that students have...
20:10
Ok, thanks for clarifying that @XanderHenderson
And in a more advanced setting, I don't really see what problem an HK integral solves. You can integrate more functions... so what? Where, in the "real world", are people actually using HK integrals, and not Lebesgue integrals?
@Jakobian Which is another reason why my question about whether or not you had ever taught the class was a good-faith effort to understand your background. If you had answered "yes", the next question would have been about the level of students you have worked with.
25 mins ago, by Koro
Can I say that $f_1=f_2$ in S(R)?
@XanderHenderson I mean could mention things that I know of that gauge integrals enable us to do, and what do they make simpler, but they won't be very convincing
I can imagine an analysis class with bright and dedicated students where one might discuss the HK integral (after digging into the Riemann integral in detail, since students in such a class are supposed to be familiar with the Riemann integral, and it is valuable to build on what they already know), as "bonus topic".
@Jakobian Who is "us" in that context?
What is S(R), @Koro? Schwarz space?
20:16
I lack the linguistic knowledge about English to answer that question
@anak Hi :). Yes, it is Schwartz space.
I don't mean you or me
(never forget the t there)
This guy is different from Schwarz from Cauchy Schwarz Bunyakovsky inequality :).
is it even important who am I talking about though? I think the important part is that its possible
@Koro Doesn't Plancherel give you an automorphism of $L^2$ and then injectivity follows?
Since you are just asking if the Fourier transform is injective on S.
20:19
@Jakobian Well, as an example, people in physics get a lot of milage out of the Lebesgue integral. Lebesgue integrals are "nice" in the sense that the space of integrable functions is a handy-dandy vector space (among other things). The Lebesgue integral gives you a nice theory of $L^p$ spaces, and useful integral transforms (such as the Laplace and Fourier transform).
@anak Well, the theorem statement that I am using is $\|f\|_2=\|\hat f\|_2$ for $f\in L^2$.
Lebesgue integrals generalize very nicely to higher dimensional spaces, complex spaces, manifolds, and a host of other "nonstandard" spaces.
But is it important enough to put at the beginning of an introductory course @Jakobian
and I know that the map $f\mapsto \hat f$ is a homeomorphism.
I don't think people in physics think of Lebesgue integrals at all. Unless dealing with people in mathematical physics
20:20
@Jakobian Well, at that level, people in physics just need to know how to produce a number when they see $\int f$.
@user726941 its not important, no
So it is of no relevance at all whether that $\int$ denotes a Riemann integral, a Lebesgue integral, an HK integral, or something else entirely.
@Koro You mean you don't interpret Plancherel as saying that $\hat\cdot\colon L^2\to L^2$ is an isometry?
They just need a number.
@anak no :(
20:21
But on the more theoretical side, the Lebesgue integral is the useful tool.
In physics its more about the results than the way to get there
@Koro What part are you missing of that statement?
@Jakobian Depends.
@anak injectivity.
But, again, the deep, dark heart of physics is mostly built on the Lebesgue integral.
20:22
I have surjectivity as follows: hat of hat of f (x)= f(-x)
@anak This is not something I can just eyeball, but you do know that $d(g^{-1})=-g^{-1}dg\,g^{-1}$. Maybe that’s germane.
For injectivity, let's say $\hat f=0$ for some f in S(R). I must show that f=0.
Thats the way it felt in my physics class. The lecturer didn't care much about the mathematical setting, as long as it deals with things, or gives a "convincing argument", we are fine. Fourier transform? Okay, as long as it gives us results.
By PLR, I get $\|f\|_2=0$ so $f$ is 0 a.e.
In any event, I'll reiterate my main thesis: the HK integral is more complicated than the Riemann integral, and doesn't solve any problems of relevance to introductory calculus students. As such, there doesn't seem to be a strong reason to teach it at that level.
20:24
How do I get $f=0$ everywhere?
@Koro Why do you need $f=0$ everywhere?
@Koro continuity
But what about to Rudin-level analysis students? The Stieltjes integral shows up in probability and statistics, I guess.
Or, when we talk about equality of f,g in S(R), do we mean $f=g$ a.e.?
@Koro if two continuous functions are a.e. equal, they are plainly equal
20:26
@TedShifrin Sure, but Jakobian has been arguing that we should teach the Henstock-Kurzweil integral.
@Jakobian I think you're right. Schwartz functions are continuous!
Infact, they are infinitely differentiable.
@Koro Smooth, baby!
If not then the definition of Schwartz space won't make sense.
I thought he meant to analysis students. Remember that calculus in Europe is analysis.
well, more like, replace it with Riemann one, than teach it
I'm arguing that if theres a reason to teach Riemann integral, I don't see a reason to not teach HK integral
20:27
@XanderHenderson infinitely smooth :).
yes, to analysis students
I still disagree with you, though.
As one with 50 years teaching experience at various levels.
@Koro I was taught that "smooth" means $C^{\infty}$, so "infinitely smooth" is redundant. :P
@Koro If $\lambda(A) = 0$, then $\mathbb{R}\setminus A$ is dense in $\mathbb{R}$. Hence...
the cats are still fighting... why??
since yesterday
what could be the dispute?
20:31
Is $C^3$ more than enough for practical purposes
Has the AoPS tried to contact you to teach some more courses @TedShifrin
Just asking for a friend
@Koro territory I guess?
@Jakobian yes, but it's been more than 24 hrs!!
cats do things out of instinct, so it has to be something simple like this
aren't they supposed to settle the matter within like 1 hr?
20:33
also I understand your question about rabies now, Koro
its because there's a lot of dogs in India, isn't it?
@TedShifrin Hmmm, juxtaposition here is composition?
@Jakobian India is endemic for rabies.
because of dogs? I heard you even have a law against feeding them?
mostly due to dogs.
but you see cats and dogs also get into fight sometimes, so cat may also carry it.
also I've noticed, there's a sort of culture of feeding the poor in India?
20:36
More stray dogs than any other in the world
@Koro yeah that makes sense
don't get me wrong, I'm not actively searching for this information about India, I just encounter it
And the most deaths from rabies
it's becoming a problem in India.
US is almost free of rabies.
@TedShifrin math.stackexchange.com/questions/4797221/… I hope I re-tagged this question correctly
the worst that can happen is a group of dogs attacking a person when the person is alone.
there have been cases when the group kills the person.
20:47
Rabies kills slowly
once it happened to a retired doctor inside a university during morning walk.
the person was mauled to ... by the dogs.
In rugby mauling is a regular part of the game
@Jakobian I'm not aware of any such prevailing law here.
@anak I’m writing the naive formula with matrices. You can fix it up.
@user726941 No reason they would. I pursued them in the first place and then quit after 2 years.
@TedShifrin thanks for responding professor
20:55
@Jakobian Too fancy. It’s just multivariable calculus. Other than the fancy word immersion.
21:16
@Jakobian SkyNet has Hunter-Killer integrals?
21:29
I had rabies shots a long time ago.
I was bitten by a dog in MX.
21:40
Are we being rabid here?
my hw is making my head numb
david: that's how you know it's working
its really boring analysis problems
21:55
Usually numb heads come from banging them hard on the desk for not knowing how to do the "boring" problems.
there are no boring analysis problems, only boring analysis solutions
Oh, numbskulls, not numb heads.
22:42
Suppose $f$ is a power series with radius of convergence $R$. Is it then correct that $g(x)=f(x+a)$ is another power series for some real or complex $a$? Does $R$ change? I know this holds for entire functions and that $R$ does not change, but if $R$ is finite, I assume $R$ changes under translation. I'm unsure.
sunny: you're basically asking what the definition of 'power series' is, but under any sensible definition the answer would be yes, with the same radius of convergence (the center of the interval of convergence will change but its width will not)
@DavidRaveh I've never had homework doing that. Is this a class in head numbing?
@sunny "Suppose $f$ is a power series"
I don't like that you write that and treat it like a function later as well
@sunny The radius of convergence of the power series for $f(x+a)$ is between $R-|a|$ and $R+|a|$
@robjohn its just an overly-tedious collection of "prove this sequence converges to 0"
22:50
@leslietownes No?
you should hear what those "prove this sequence converges to 0" exercises say about you
ted: no as to which part? translation of 'power series' not a power series? or translation a power series, but 'R' not the same?
You made it seem like the roc stays the same under translation? Or did I misread?
I'd argue that power series is more than just a function. It has a center, it has its coefficients
@robjohn so the roc does change?
Consider $f(x)=\frac1{1+x}$. Compute the power series for $f(x+1/2)$ and $f(x-1/2)$, and compute the ROCs.
22:53
@sunny not if you are willing to change the center of power series
ted: doesn't it?
Think of examples, sunny. Where $R\ne\infty$.
ok
your question is just so ambiguous that robjohn interpreted it in a totally different way from leslie...
i guess amend my answer to say it also depends not just on what 'power series' but on what 'R" means
22:54
Roc
Robjohn and I agree
i, like jacobian is pointing out, am assuming that you're defining 'R' for a 'power series' such that it wouldn't change under translation
@TedShifrin Does that signal the end of the world?
$\sum a_n(x-a)^n$ is also a power series
we don't have to have $a = 0$
Probably. When leslie and I agree, it does.
Ted. Would you be able to elaborate on your comment about the torsion = 0 being a condition on the connection and not on the bracket?
22:56
if R is something like "the sup of |x| over the set of x for which the series converges" then that will definitely change under translation, but i guess i would regard that as an unusual way of defining the 'radius of convergence'
It’s the torsion of a connection, Lemon.
Every power series is given locally as a power series. If you have power series of the form $\sum a_nx^n$ and don't shift it too much, so that $0$ is still in the domain of the translation, then the translated power series can also be represented as $\sum b_nx^n$ in some smaller disk
Oh, apologies, leslie. I was thinking of translating the center but not the function.
I was stoopid.
It is the end of the world!
just in time for halloween
i'm going as ted in his multivariable math youtube videos
That ugly?
23:02
when we define "radius of convergence", we presume that we are already dealing with a power series, that is an expression of the form $\sum a_n(x-a)^n$
where the sequence $a_n$, as well as the center $a$ is fixed
there's a difference between "is power series" and "is represented as power series"
and a power series defines a function but its more than just a function
that said, I really dislike the question
@robjohn @sunny Here's the confusion. You're talking about $f(x+a)$. But you did not specify at what point you expand the series (i.e., the "center"). Is it changing by translating the same way? Or is it staying the same?
Hmm, I'm unsure. The reason for the question was simply the last sentence that I read in this passage.
"an power series"? Oy.
Yes, this book is dubious.
They are thinking what robjohn and I thought, not what leslie and jakobian thought. But it's totally sloppy.
They are thinking of translating the center, keeping the function the same — or of translating the function, keeping the center the same.
Since they're not discussing power series centered at other points, I'm assuming they want you to think about the function $1/(x-1/2)$, expand it as a power series with center $x=0$, and ask what its radius of convergence is.
23:19
alright
well, I'm not really "team leslie", I considered both possibilities but dismissed the question as too ambiguous
OK, so on this one you sit higher atop the mountain than the rest of us.
I see it differently
you were trying to answer the question, I was trying to deconstruct it
23:54
In mathematics, the regulated integral is a definition of integration for regulated functions, which are defined to be uniform limits of step functions. The use of the regulated integral instead of the Riemann integral has been advocated by Nicolas Bourbaki and Jean Dieudonné. == Definition == === Definition on step functions === Let [a, b] be a fixed closed, bounded interval in the real line R. A real-valued function φ : [a, b] → R is called a step function if there exists a finite partition Π = { a = t...
this makes me think of Ito integral
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