Mathematics - 2021-06-18 (page 2 of 2)

Thorgott

3:03 PM

@wklm No guarantee, but I think it's one of those you can solve by contour integration

wklm

thanks @Thorgott I'll google that!

epsilon-emperor

3:25 PM

@robjohn Sure, but I don't know the meaning of "sharp up to polynomial approximations"

Where can I read more about this?

robjohn

3:58 PM

@wklm have you tried completing the square in the exponent?

@epsilon-emperor I am not sure what that exactly means

Ron Gordon

4:46 PM

@TedShifrin Hi Ted, not to put you on the spot, but out of curiosity what did I do to annoy you? (To be fair I just came across your post from 2 months ago so if you do not remember I understand.)

robjohn

4:58 PM

@RonGordon The trail I find leads to this answer and attendant comments, but that does not look to be annoying, so I have no answer.

copper.hat

i love intrigue when i am not part of it

robjohn

@copper.hat enjoy the intrigue until Ted comes back and answers the question.

AMDG

Frustration. Frustration is my name.

I would like to attempt to ask a question here since my attempt to ask in basic mathematics failed. It's about integration, and yes, it is about my on-going search regarding circular functions. I just recently got an idea that I haven't tried yet.

The idea involves some form of integration which I am uncertain of how to properly describe or notate as follows: you have a function $f(x)$ and a (possibly unrelated) function $g(x)$ for which you want to find the indefinite integral of. You can find the indefinite integral of $f(x)$, but not $g(x)$ for whatever reason.

Given $f(x)$, I can find the indefinite integral of $g(x)$ by allowing its domain of integration to vary and taking some infinitesimal difference between two different integrations of $f(x)$ and summing these together by limiting the domain to be the region between the real intersection(s) of $f(x)$ and $g(x)$ as $f(x)$ moves up or down, effectively generalizing $f$ to be a function $f(x,y)$ of some kind.

How would I even begin to describe this, assuming my point has come across here?

robjohn

5:22 PM

An example would be useful in understanding what you are trying to do.

AMDG

Alright, I'll make a demo in desmos.

desmos.com/calculator/u6tfzhi5rx

desmos.com/calculator/dme1yyvjun

Forgot to square the left side.

The red lines define the domain of integration for $f$ and $g$. We know the integral of $f$ but not $g$. Given this changing domain, what is the integral of $f$ here for all values of $n$ between [0, 1], and given that integral, given some infinitesimal difference $\Delta$, and computing the integral of $f$ for all values of $n$ between $[0 + \Delta, 1 + \Delta]$, what is the difference of the two areas for all values of $x$?

I'm strugglin' here to describe this precisely, but do you get the general idea of what I'm going for?

AMDG

5:44 PM

I'll call the former integral of $f$, $f_1$, and the latter, $f_2$. What I want is an integral: $$\int_{-x-\Delta}^{x+\Delta}f_{2}\left(x\right)-\int_{-x}^{x}f_{1}\left(x\right)+\int_{-1}^{1}f\left(x\right)\ =\int_{-x}^{x}g\left(x\right)$$

Ted Shifrin

@RonGordon What post from 2 months ago? I don't remember interacting with you for many years. I do remember frustration years ago when I was engaged in "hinting teaching" with some OP and you just posted one of your flawless solutions. But I've been on the losing side of this battle as long as I've been here (which is now 8 years or so, I guess).

AMDG

6:26 PM

Ok, I made something much clearer on mathb.in . Please let me know if it is clear enough or not. The formatting might not be up to par. mathb.in/58462

Ron Gordon

7:07 PM

Hi Ted, this one: https://chat.stackexchange.com/transcript/message/57627991#57627991
I am sorry I seem to be making a big deal about this; it was just jarring to see that someone like you - whose work here I admire - might be put off by something I did. I think I understand that it was part of a discussion going back many many years at this point. Thanks for understanding.

@TedShifrin (You did nothing wrong and please feel free to criticize me however I may deserve it. I just felt it was better to ask you than to pretend that such things don't bother me.)

@robjohn Yeah looks like I made a mountain out of a molehill as is my tendency and blew what Ted said out of proportion. Sigh.

Yorch

I didn't know the legend @RonGordon came to these earthly corners of the web

Ron Gordon

@Yorch Didn't realize I was anything like a legend, but I very much appreciate the sentiment!

copper.hat

7:28 PM

this is a good place for making mountains out of molehills. we even have legal support.

Larry

@robjohn Thank you @robjohn !

schn

7:48 PM

Maybe a very basic question, but are continuous functions bounded?

hyper-neutrino

@schn $f(x)=x$

schn

@hyper-neutrino Obviously not :)

Vrouvrou

Hello i have a question if $A\cap B=\emptyset $ and $O\subset A\cup B$ can we say that O is on A or on B ?

schn

@hyper-neutrino I guess my intended question was more along the lines; if $f\in C^m$ for some positive integer $m$, does $f^{(m)}(t)$ exist for some $t\in D_f$, where $D_f$ is the domain of $f$?

what guarantees that $-\infty<f^{(m)}(t)< \infty$?

hyper-neutrino

I'm not exactly sure I understand, but what about $f(x)=e^x$?

any number of derivatives of that is still unbounded

robjohn

8:02 PM

@Vrouvrou Let $A$ be the odd integers and $B$ be the even integers. The integers is in the union, but not in one or the other of $A$ or $B$

you could even let $O=\{1,2\}$

Yorch

you could also let $A=\{1\}$ and $B=\{2\}$

oh nvm

robjohn

there are just too many counterexamples

hyper-neutrino

$A={0,1},B={2},O={1,2}$ would work :P and is probably a minimum working example? I think?

Yorch

I don't understand what the question is to be honest

what does it mean for a set to be on another?

hyper-neutrino

good point. i think i read that as "in" subconsciously, lol

Vrouvrou

8:07 PM

is there a proof in algebra ?

hyper-neutrino

@Vrouvrou ?

first of all what does "on" mean

and second of all why ask for an algebra proof to a set theory problem?

Thorgott

@schn if that isn't the very definition of C^m, I don't know what is

Ted Shifrin

@RonGordon Yes, this was specifically in reference to what I said earlier. After 7+ years I no longer remember the specific posts, of course, but it was your (and others', of course) posting picture-perfect solutions while I was trying to get the OP to figure things out for him/herself. It was annoying. It doesn't make either of us a horrible person that we have different philosophies about this site. Sorry I upset you.

@copper.hat Our legal support is too busy training his daughter to swear.

schn

@hyper-neutrino@Thorgott Thanks for the replies. Here is the context of my question. I have a function $f\in C^m$ for some positive integer $m$. I'd like to make a Taylor expansion of this function around $t$, so $f(t-a)=\sum_{l=0}^m \frac{f^{(l)}(t)}{l!} (-a)^l+o((a)^m)$. Obviously this expansion is only possible if $f$ and all its $m$ derivatives are defined at $t$. Does this follow from $f\in C^m$?

hyper-neutrino

oh

you just need it to be defined

yeah then that's true. keep in mind that a function does not need to be bounded to be defined at every point

what is $C^m$ though...?

schn

8:23 PM

@hyper-neutrino With $f\in C^m$ I meant that $f$ is $m$ times continuously differentiable. For the record then, just because $f\in C^m$ does not mean that $f,...,f^{m}$ are all defined at $t$ (where the Taylor expansion is centered), right?

Vrouvrou

@hyper-neutrino on that is subset

hyper-neutrino

Oh, yeah, sorry it's been a while since I did this so I forgot a continuous function doesn't always have a continuous derivative

leslie townes

she hasn't said the f word in several days.

hyper-neutrino

isn't the definition of $f\in C^m$ that $f,f',\dots,f^{(m-1)}$ are all continuous and therefore defined at all points (specifically $t$)?

schn

@hyper-neutrino I'm unsure about the last part, but according to Wikipedia it seems to be $f',..,f^{m}$ for $f\in C^m$

Smoothness

In mathematical analysis, the smoothness of a function is a property measured by the number of continuous derivatives it has over some domain. At the very minimum, a function could be considered "smooth" if it is differentiable everywhere (hence continuous). At the other end, it might also possess derivatives of all orders in its domain, in which case it is said to be infinitely differentiable and referred to as a C-infinity function (or C ∞ {\displaystyle C^{\infty }} function). == Differentiability... ==

hyper-neutrino

8:29 PM

or sorry i meant up to $f^{(m)}$ now that i think about it more

leslie townes

that "at the very minimum" a function could be considered smooth if it's differentiable is a weird remark. maybe you'd use that in a high school class but not with any precise meaning in mind. to most, smooth means C^infty.

amWhy

8:56 PM

@robjohn Hey, Ted, I see that I upvoted your meta post, way back when! I would do so again, today.

Ted Shifrin

@leslietownes Mazltov!

I think it's been an inexorable slide downhill, @AmWhy, fueled plenty by COVID.

amWhy

^^^ I did not ping you, @Ted, in my last comment!

Ted Shifrin

Yes, but I wanted to respond nevertheless.

But thanks for the upvote :)

amWhy

Oh, @Ted, I just meant that I should have pinged you, but failed to have pinged you. Thanks for catching it! I think the sentiments you expressed in that post are precisely why you do such good work in this chat, and on site. I adore the Socratic Method!

Ted Shifrin

Oh :D

I noticed that this young fellow Andrade had responded recently to that meta post. I feel like a proud papa, having watched him mature from very, very basic questions in geometry/analysis to posting good answers these days himself. So my methods have worked out for a few OPs. :)

amWhy

9:06 PM

@Ted Have you ever been taught by profs, or taught yourself, using the Moore Method?

My first graduate course was taught via the Moore Method

Ted Shifrin

No, I haven't. I'm way too impatient. I did try to make my classroom as conversation-based as possible — and it worked pretty well in point set topology, actually, having the undergraduates steer the proofs and try to come up with examples, and then I would write the final version on the board. But the class was good enough that we still covered the material.

I also used the classroom one time on Tuesdays and Thursdays for actual office hours for the class, so it was almost like having extra time. Ironically, the two math ed doctoral students complained to me that — despite what they preach — they wished I'd go back to a traditional lecture format.

amWhy

@TedShifrin Hah! ;D

Ted Shifrin

Of course, I understand what was going on. That class had a bunch of amazingly bright and quick undergraduates, and so the math ed grad students just couldn't quite keep up. I encouraged them to take more part in the "discovery" portion, but I think they were too shy/embarrassed to do so.

A good object lesson, though, for all the math educators spouting bull about how we should be teaching math by the discovery method.

amWhy

@TedShifrin I think they call that "flipping" or something similar, where students are assigned reading, a video lecture, and then, when their in class, they work in groups, or individually, on exercises, with the teacher/professor supervising and mingling, and available.

Ted Shifrin

It takes engaged teachers and students for that to work.

The flipping is what's prevalent now. That wasn't a flipped class at all. However, had I not retired pre-COVID, I probably would have used my YouTube lectures to flip my multivariable math classroom these years. I know that a lot of people using my textbook at various places have asked permission to do so (not that they needed my permission).

amWhy

9:14 PM

@TedShifrin Wow! Tip of the hat, t'you. !

Ted Shifrin

Just serendipity that my students made the videos to start with. I never instigated that, but I think they really helped out a lot of people.

leslie townes

you should sue them for copyright infringement if you didn't instigate it.

are they monetizing the videos by running ads for dietary supplements over them? i could think of a few more claims.

let me think this out

Ted Shifrin

Who's "them," leslie?

amWhy

9:31 PM

@TedShifrin Indeed. As a pre-college student, I was lucky and had engaged teachers, all along. And by fourth grade, took honors or gifted courses in math and English, So for the most part students were engaged. I really feel blessed, because I came of age at a time where my teachers were challenging, and the culture was challenging, previous conceptions of "males innately more talented in math, girls should train to be school teachers, nurses, or secretaries, until marriage."

Ted Shifrin

I went to public schools in MA for grades 9-12 and, despite one of the "best" school systems, there were truly only a handful of gifted and good teachers. The math teachers were in fact terrible. The woman who taught me geometry did have a Ph.D., but she punished me for asking (sincere) questions.

Then she stopped teaching the calculus class, saying she wanted to teach students who needed her; those students she rewarded by saying things like, "You're not an honors class, so I don't expect you to understand this."

My fabulous teachers were 2 English teachers, one history teacher, one science teacher (despite one truly horrendous physics teacher), and a French teacher.

amWhy

@TedShifrin Yikes!! There are so many nightmare students who literally fear math, and some users who went on to defy the limits of their teachers' conceptions. But for women (girls), expectations to succeed in math when out the window, save for a few trend setters, who were, for me, remarkably, male teachers!

Ted Shifrin

I would like to believe that some (of course not all) of my women students prospered because of my encouragement and pushing them to be great. But I know more than a few who did not.

amWhy

@TedShifrin It doesn't happen with all girls/women. Many come to college, and even this site, without any confidence in math, having decided "I guess I'm just not good at math." Partly because generally our culture seems to think "you either are good at math, or you're not", and partly because girls often learn better through cooperation, and not competition, at least earlier on.

@TedShifrin My favorite teachers and profs, across the board, were enthusiastic, encouraging but also challenging. I have no doubt that you had all those characteristics. Kudos!

Ted Shifrin

Yes, the whole competition thing is a big deal. So is accepting praise, sometimes. One of my most brilliant woman students, who ultimately finished her Ph.D. at Chicago, was mortified when I bragged to the whole class when I returned the first homework that she had given a solution to a challenge problem far more clever than anything I had thought of and that I was so impressed. I learned to be very low-key about being impressed by her.

Another woman student, who was brilliant, tried to outdo every previous student in the challenge problems. She wanted a higher score than all the males. She sadly had some depression issues and bailed out of her doctorate. I hope she will get things together and be happy.

amWhy

9:49 PM

@TedShifrin Wow, you had good intuition. I think over time, there'll be more similarities between genders. Woman are particularly sensitive to suffering the "Imposter Complex"... and are more apt to see their success as luck, and their failures as deserved. I think we're still at a point where men succeed in part because they seem to believe their success is deserved, and their failures the fault of others or bad luck. These are absolutely, overgeneralizations, but it does seem to reflect

How some women see themselves, and also, how some men see themselves. I really do not believe it is rooted in genetics, but culture. But now we need another SE site: Nature.vs.Nurture.SE!

Ted Shifrin

I observed in 15 years of teaching the Spivak Calculus with Theory course that most of the women students didn't enjoy the struggle as much as the guys. That said, they were better disciplined students, but didn't react well to the frustration. The guys were more willing to shrug it off and say, Oh well. I had one superb woman student one year who was driving herself nuts wanting to be perfect. She didn't want to go on to the second semester.

I told her this: I'll make a deal with you. You put in half the time on homework (so quitting before the frustration gets to her) and I guarantee you'll still be an A student. She did and she was.

I did have some perfectionist guys, but they didn't seem to be in duress about it — or perhaps I'm not remembering the cases now.

@AmWhy Why did you (and others) vote to close this after I had answered it? The question was not great, but there is good math content in the comments and answer.

leslie townes

welcome to PSQville.

population, you

amWhy

@TedShifrin It wasn't closed because you answered it. But the question was one sentence long, with no supplementary context provided by the asker. There have been no delete votes. And nor did you receive any downvotes. Yes, as @leslie suggests, it was a bare PSQ.

Ted Shifrin

I guess I am not so suspicious when things are ostensibly at the graduate level. The student wasn't sophisticated and certainly didn't make it clear what his/her level of knowledge was, but I don't see the point of closing it once it has a good answer.

If it's someone asking for a true/false quiz, the level of the course is pretty steep.

leslie townes

do closed questions with accepted answers go away in the future? i was under the impression they just couldn't be answered anymore.

Ted Shifrin

10:00 PM

Do they show up in searches, etc.?

I don't see the point of closing once there's a good answer.

amWhy

@leslietownes @Ted They show up in searches, and they just can't be answered anymore.

Ted Shifrin

If it's at a suitably advanced level. Of course, at the level of elementary and high school, precalculus, etc., I don't agree with what I just said.

I guess I just don't get the point here in this case.

I guess if I hadn't answered it and the OP hadn't tried to make any progress, it might make sense.

amWhy

@TedShifrin It's just tricky to grant what looks like a privilege for the advanced student, and penalty for the elementary student, and it is not always clear which is which.

Ted Shifrin

Well, people who don't recognize graduate-level differential geometry shouldn't be casting close votes.

If someone posts a poor question in functional analysis, I'm not going to be the one voting to close it.

amWhy

Advanced students, after all, sometimes plagiarized their theses! They are not neccessarily more honest.

Ted Shifrin

10:04 PM

Oh good grief.

I resign and leave this to my attorney to handle in the future.

amWhy

@TedShifrin If you want to stop this discussion, fine. I do not want to argue with you.

@TedShifrin Hah! ;P

@TedShifrin I'll have my attorney contact yours. ;D

Hi there, @JoséCarlosSantos !

@TedShifrin Can we agree to disagree? with a handshake?

Yorch

handshakes not allowed here without vaccination passport

amWhy

10:27 PM

@Yorch Hah! Can I give a "shoulder bump" if wearing a mask? Or an "elbow bump" wearing a mask? ;D

Yorch

sure thing :)

Ted Shifrin

10:43 PM

Well, I'm safely vaccinated, so I'll shake hands.

amWhy

@TedShifrin Me too! :-)

copper.hat

you can get a qr code version of your covid vax record in california

thankfully my daughter has finally managed to get scheduled for vax.

Ted Shifrin

Oh really, @copper? I just took a photo of my card. Should I google this?

copper.hat

myvaccinerecord.cdph.ca.gov

you don't need it, just if you want a qr code instead.

i just took a photo of my card

i don't trust computers

Ted Shifrin

well, it doesn't hurt to have multiple accesses

No response yet from the website.

copper.hat

10:50 PM

i agree. i am a belt & braces sort of fellow

i did mine a short while ago, the text arrived more or less immediately.

we have crappy service

Ted Shifrin

Mine didn't. I was sure I used phone, not email. At least the CDC texts came to my phone.

copper.hat

I guess you could try the other. I did not remember, but must have picked the relevant one.

Ted Shifrin

Oh, I forgot to check the checkbox. Dope.

copper.hat

damn lawyers

Ted Shifrin

Hmm, no, I think I must have checked it to get to the submit button. Who knows.

copper.hat

10:53 PM

good luck!

unrelated: i keep getting scam calls from people purporting to be from social security. i reply with, that's odd, i work for social security administration, what department are you calling from and surprisingly they hang up!

MphLee

Hi, someone can help me with this question?

1

Q: Functors making functions natural trasformations and vice-versa.

I apologize in advance if this is naive. In this answer Conjugation in a groupoid it is said that given a groupoid $\mathcal G$, and an arbitrary function $\mu:\mathcal G_0\to \bigcup_{x\in \mathcal G_0}\mathcal G(x,x)$ s.t. $\mu_x\in \mathcal G(x,x)$ we can define an endofunctor $F_\mu\in[\mathc...

amWhy

@copper.hat Love that idea!!

MphLee

If someone is interested there is a 50rep bounty on it.

Ted Shifrin

Nope, no luck. I tried the email and got an email response saying it didn't match their records. But no texts.

copper.hat

oh, well.

Ted Shifrin

10:57 PM

Our functorial people are asleep and not here currently, @MphLee.

copper.hat

it bothers me that those scammers are preying on old folks.

i am not an abstract sort of guy

plenty of nonsense

MphLee

Oh, hahah xd ok @TedShifrin thank you anyways. It's pretty late here (EU time)... I was betting on US ppl...
Even functor ppl need abstract nonsense dreams sometimes

copper.hat

:-)

Thorgott

maybe I'm missing something, but what does the thing you describe have to do with groupoids?

MphLee

@Thorgott The main example, the motivating example which I linked in the question, appears naturally in the context of groupoids, because it is some kind of generalized conjugation. At first sight, I'd say one can extend the argument to category as long as at least one object, or all of them, have at least one non-trivial automorphism.

Thorgott

11:12 PM

I don't see a need for non-triviality, the construction goes through as long as you pick an automorphism at each object

Ted Shifrin

Functor Thor has arrived.

MphLee

ok, but then if you just chose the identity for each object you're just defining the identity endofunctor... pretty boring isn't it

Thorgott

that option already existed for groupoids too

this construction just describes all functors naturally isomorphic to the identity functor

MphLee

Sure. I'm sure I don't understading where youre getting at. I'm missing something probably.

Thorgott

functors naturally isomorphic to the identity functor that act as the identity on objects*

this is tautological, of course, but perhaps a cleaner phrasing

MphLee

11:18 PM

Yes. Ok. Because $\mu$'s components are isomorphisms (by definition).

What puzzles me is the fact that before I define the functor, \mu can't be a natural transformation, because it has not a functor to be natural with...
And the component on object of the functor needed can't be defined if we haven't made a choice of \mu. When we do that F becomes functorial, and only then \mu become natural.
I see a kind of circularity here.

Thorgott

well, since the $\mu$'s are automorphisms, they could only constitute a natural transformation between functors that act as identity on objects

the point is that for each choice of automorphisms on each object, there is one and only one functor acting as the identity on objects such that the automorphisms chosen define a natural isomorphism from it to the identity functor

MphLee

Is this a theorem, and exercise in some book, a basic result? Something so trivial is not even worth mentioning. Has this a name?
That was the spirit of Q1, in my question.

Thorgott

I think it's "trivial". The diagram defining naturality immediately forces the functor to have the form given in your post and it's not hard to see that that formula indeed defines a functor.

MphLee

Sure, I agree. I know that I'm defining a functor. And I proved the naturality. It's kinda trivial. But I see some circularity. Probably because... in CT what is mu before I define the functor? It cannot be a natural transformation....because there is no functor yet.

Ted Shifrin

@copper It finally showed up. I think their website is slammed. Unfortunately, my record is only half there.

MphLee

11:27 PM

When you say for each choice there is only...
I'd like to make this precise. Few comments above you stated "
functors naturally isomorphic to the identity functor that act as the identity on objects" I'd hope to phrase this a something regarding objects of the slice category over the identity functor...
I' not sure how: that was question Q2.

@Thorgott I'm sorry If my question was unclear. I'm not good at explaining myself in english. But it is harder when I don't get feed back on my questions.

hahah I remember back in the days when a misplaced comma would result in bad comments, downvotes and votes for closures.

Thorgott

I guess you could phrase this all as follows, though I'm not sure if it's helpful: you can consider the indiscrete category $\mathcal{A}$ with underlying set $\prod_{x\in\mathcal{C}}\mathrm{Aut}(x)$; then the construction described in your post yields a functor $\mathcal{A}\rightarrow[\mathcal{C},\mathcal{C}]/1_F$, which is an embedding whose image precisely consists of those endofunctors over $1_F$ acting as the identity on objects

though that's more or less just windowdressing, I don't think there's really anything "deeper" going on (of course, I could be mistaken)

MphLee

So, basically, that's literally a function assigning to every choice function and endofunctor. By your previous remark, if I understand it, the image of the functor should be exactly the slice category of Automorphisms of the category over the identity functor.

Thorgott

no, I corrected that part earlier

the functors constructed act as identity on objects

MphLee

11:44 PM

ok. Sure, I forgot to write it.

Well, thank you very much. That already something interesting I believe. I believe there should be something deeper btw.

Btw, so I'll do this as an exercise: I need to prove that it is an embedding, and to prove that the image is the one you stated, that should be the easy part.

Btw, the deep part I believe is going on I'm referring to is when we stop to consider aoutomorphisms and extend this to isos in general using a and endofunction $f:{\rm Ob}\to{\rm Ob}$. Like in the case of turning cardinality into an endofunctor of Sets, or when we make a non-canonical choice of a basis for every vec. space, making us able to define and endofunctor that is not the identity on the objects.

Oh @Thorgott If you want you can post this as an answer to Q2. I believe You completely answered that here.
If you can give me some cent on question Q3, that is now basically something like an adjoint situation to Q2, I'll award you the bounty.

Transcript for

$Mathematics$ Mathematics