Mathematics - 2024-12-05

pie

00:10

Now I got ads lol

copper.hat

00:39

@psie Yes.

Xander Henderson

01:04

@copper.hat Your into BDSM, aren't you? "Liking" Folland. Kinky.

Sine of the Time

bro has beef with Folland

Xander Henderson

@SineoftheTime No, I don't like Folland enough to have a meal with him.

Sine of the Time

so no duck with Folland? :'(

copper.hat

01:24

@XanderHenderson no, the bdsm is Rudin's functional analysis

psie

@copper.hat essentially you are taking the measure $\sigma$ defined on the Borel sigma algebra on $S^{n-1}$ and extending it to $\mathbb R^n$ by zeroing it outside $S^{n-1}$

copper.hat

@psie Yep, hence the singularity

Xander Henderson

@copper.hat Ah, true enough.

Folland is kind of 50 Shades of Grey in comparison.

copper.hat

my real porn is Rockafellar's convex analysis. i often look at stuff there, don't really understand what is going on, but i keep staring anyway. and then i feel guilty.

psie

ok, I'm somewhat hesitant to accept that as legit, but I don't see how else you'd show mutual singularity,

leslie townes

01:28

psie you might work out whatever details are troubling you in the case n = 2 with the unit circle in the plane

copper.hat

or $n=1$ for less excitement.

leslie townes

many of the more technical questions you are asking always have like 10 parameters that can be given specific values without destroying the essential point

copper, psie needs a little more excitement than that

Xander Henderson

@copper.hat I feel that way about Haran's book on the "infinite prime".

psie

@leslietownes yeah, it's kind of nerdy of me :) but I'll look into the plane, see if I find something convincing

copper.hat

01:57

i'm still searching for the other even primes

leslie townes

copper i'm hard at work on the triple prime conjecture

Xander Henderson

@copper.hat I found two.

leslie townes

boo

copper.hat

i have a short proof of the $n$ prime conjecture for small values of $n$.

Soumik Mukherjee

07:41

Where is the fact that R is Noetherian used in this proof?

Thorgott

08:30

very first sentence

"Noetherian" is equivalent to "every non-empty set of ideals possesses a maximal element"

mo-_-

09:23

Hello

Monty

09:51

History was made, bitcoin hits 100k

Soumik Mukherjee

10:02

@Thorgott Oh right, thanks

Jakobian

12:09

@mo-_- hey

Soumik Mukherjee

Hello

Sine of the Time

hi all

Soumik Mukherjee

12:33

What do you think, who'll win?

Jakobian

me

Soumik Mukherjee

:)

mo-_-

12:52

hi

Do you think this diagram is good?

Lucky Chouhan

13:13

Hi @SoumikMukherjee

Soumik Mukherjee

Hi

Lucky Chouhan

@SoumikMukherjee I just read your this question on Academia SE and its answers. Has this issue been resolved now?

Soumik Mukherjee

I didn't pursue that any further

Binky

14:27

Do you know if there is a theorem called Riemann-Fourier?

leslie townes

there is if someone decides to give a theorem that name :)

the most common use of riemann's name in fourier analysis is in connection with a result like this en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma (really a family of results, not just the stated thing about L^1(R^n))

Sine of the Time

@Binky in which context?

leslie townes

and you will see books and papers cite "the riemann-lebesgue lemma" without separately introducing it and giving it that name. it might be reasonable to expect an audience familiar with fourier analysis to know (approximately, up to some mild differences in hypotheses) what is meant

i don't think "riemann-fourier" has a similar status. if you asked me what that was i'd have no idea

Sine of the Time

yeah never heard about that

Binky

treccani.it/enciclopedia/…

Here

Xander Henderson

14:42

@Binky That calls it a formula, not a theorem...

Though I've never heard of either.

leslie townes

OK. i have never seen anything like that given that name in english

but nothing stops anybody from saying "by [made up name] we mean ..." and with that kind of context, basically anything goes

riemann certainly did work in fourier analysis :)

mo-_-

2

Q: Inverse Laplace transform via the table formulas

In my inverse Laplace table there is this inversion "formula": $(1) \frac{1}{s-a} \rightarrow e^{at}$ I understand that $\mathcal{L}^{-1}[\frac{1}{s+4}] = \frac{1}{2}\sin(2t)$ But why can I not do the following: $\mathcal{L}^{-1}[\frac{1}{s+4}] = \mathcal{L}^{-1}[\frac{1}{s-(-4)}]$ Using $(...

Sine of the Time

That's an encyclopedia, I don't know if it's reliable for scientific terminology

leslie townes

15:11

a google search for "riemann-fourier formula" (with quotes) returns that one encyclopedia page and two documents at the university of trento. so, maybe an example of one author's idiosyncratic terminology :)

in the specific context of those documents, though, the author makes clear what they are talking about

and does not presume the name to speak for itself

psie

15:24

> Theorem If $F$ and $G$ are in $NBV$ [meaning they are in $BV$, $G(-\infty)=F(-\infty)=0$ and are right continuous] and at least one of them is continuous, then for $-\infty<a<b<\infty$, $$\int_{(a,b]}F\,dG+\int_{(a,b]}G\,dF=F(b)G(b)-F(a)G(a).\tag1$$

> Proof. $F$ and $G$ are linear combinations of increasing functions in $NBV$, so a simple calculation shows that it suffices to assume $F$ and $G$ increasing. ...

Question I'm trying to understand the "simple calculation". To this end, I'm trying to show that if $(1)$ holds for increasing functions $F,G$, it holds for $F,G\in NBV$. I've been trying to work backwards by writing out $F=(F_1^+-F_1^-)+i(F_2^+-F_2^-)$ in the left-hand side of $(1)$ and similarly for $G$, but I'm getting stuck at combining $F_1^+$ with $G_1^+$, since these are being integrated with respect to different measures. Is there a quicker way to see the "simple calculation"?

Actually, I think I might be on the right track. Let me write something down on a paper.

Hmm, no. It doesn't work. Because if I write out $F=(F_1^+-F_1^-)+i(F_2^+-F_2^-)$ and combine the integral of $F_1^+$ with the integral of $G_1^+$, then these integrals are still with respect to $\mu_G$ and $\mu_F$ respectively, whereas I'd want them to be with respect to $\mu_{G_1^+}$ and $\mu_{F_1^+}$ :(

Jakobian

15:53

@psie well first of all, assume $F = aF_0$ instead, and then $F = F_1+F_2$, where the equality works for $F_0, F_1, F_2$. Why work hard, when you can work smart

tell us what happens when you do that

psie

@Jakobian alright :) is $a$ here the endpoint of the interval $(a,b]$? and what is $F_0$?

Jakobian

@psie $a$ is a complex number. Name it something else if it clashes with your notation

$F_0$ is a function in NBV for which the theorem holds

psie

@Jakobian hmm, but you're still working backwards. You're saying that if the equality holds for NBV functions, then it holds for linear combination of NBV functions. But the proof goes; we show the equality for increasing functions, then it holds for NBV functions.

Jakobian

no

you just don't understand what I am saying

when someone says "without loss of generality" or "it suffices to assume" what do you think that means

it means that we are working out of the natural order (as far as string of implications goes, not necessarily natural for a human being), but the proof will be equally valid

If you want to show that $P$ implies $R$ and you have shown that $Q$ implies $R$ first, and then $P$ implies $Q$, then of course that means $P$ implies $R$

it doesn't matter that you have proven $Q$ implies $R$ before you did that $P$ implies $Q$

psie

16:11

@Jakobian ok, but you're saying here "where the equality works for $F_0,F_1,F_2$" and then later that $F_0$ is a function in $NBV$ (I assume also $F_1,F_2\in NBV$). How do we know the equality works for $NBV$ functions?

Jakobian

here, to be concrete, we can take $P$ to be our assumptions, $Q$ to be that the statement holds for monotone functions $F$, and $R$ to be that it holds for NBV functions $F$

@psie this is exhausting... of course I don't know that. I am assuming that this concrete function $F_0$, that it works for it

this will prove that the set of NBV functions for which the theorem holds is closed under linear combinations

and since you know that it does for monotone ones, it will hold for everyone single one of them

psie

@Jakobian so the proof shows $Q\implies R$ and you're asking me to show $P\implies Q$?

Jakobian

No, the proof that you are reading shows that $P$ implies $Q$, and what you need to prove is that $Q$ implies $R$

psie

ok 👍

Jakobian

and to do that, show that linear combinations of functions for which the theorem holds, the theorem still holds for them. And to do that, you need to consider multiples of functions for which it holds, and sums of functions for which it holds

psie

16:40

@Jakobian alright, I'll tell you what happens. Let $F_0, G_0$ be decreasing or increasing, and $c<0$ or $c\geq0$ respectively. Thus $F=cF_0, G=cG_0$ are increasing and the theorem holds (provided one of them is continuous). Then $$\int_{(a,b]} F\,dG+\int_{(a,b]} G\,dF=c\left(\int_{(a,b]} F_0\,dG+\int_{(a,b]} G_0\,dG\right).$$ This is where I'm stuck. Not only might $F_0,G_0$ not be increasing anymore, but we are also not integrating with respect to $dG_0$ and $dF_0$, which confuses me.

EDIT: there's a typo within the parenthesis (second occurence of $dG$ should be $dF$).

Jakobian

16:58

One function at a time, not both of them

also, $c$ is supposed to be a complex number

psie

@Jakobian but then we can not speak of increasing functions anymore? Recall, we are assuming $Q$.

Jakobian

You can first show that the theorem holds for NBV functions $F$ and increasing functions $G$, and then show that it holds where both are NBV by repeating the same argument with linearity for $G$ this time

Suppose that theorem holds for $F_0$ and increasing function $G$, and then show that it holds for $F = c\cdot F_0$ and $G$

psie

Ok, that's what you mean by one function at a time, ok.

Hmm. I still feel like I'm running into the same issue with the integration with respect to $dF$, whereas we'd want it to be with respect to $dF_0$, i.e. $$\int_{(a,b]} F\,dG+\int_{(a,b]} G\,dF=c\int_{(a,b]} F_0\,dG+\int_{(a,b]} G\,dF.$$We want here the $dF$ to be $dF_0$, since the theorem holds for $F_0,G$.

Also, by assuming $F_0$ (which I assume is NBV) and increasing $G$, aren't we assuming more than $Q$?

Jakobian

17:29

@psie $dF = c\cdot dF_0$

@psie I don't know what you mean here, actually

we aren't using the assumption $Q$ here

psie

@Jakobian well, $Q$ is the assumption that the theorem holds for increasing functions, this is what we can use. So we need to apply the equality to increasing functions. In my most recent equation, $F_0$ is NBV and $G$ is increasing. This is not a pair we've shown the equality for yet.

ModularMindset

$$ \zeta(s)= I(s)-K(s) $$

Is that circular?

psie

17:44

@Jakobian ok, so I guess $G\,dF=cG\,dF_0$ is just true by the usual measure theoretic arguments? It's the definition for indicator function $G$, then for simple functions by linearity, and so on...

Jakobian

17:56

@psie No, you are misunderstanding. What we are trying to show is an implication, that if it holds for $F_0$ and $G$, then it holds for $F$ and $G$

So we are not using the assumption anywhere

@psie the definition of $dF$ is, I presume, integral over $\mu_F$. And since $\mu_F = c\cdot \mu_{F_0}$ is obvious from looking at the generating set i.e. half-open intervals, what you really want is that integration with respect to $\mu_F$ is the same as integrating with respect to $\mu_{F_0}$ and multiplying the end result by $c$

for the last one, if you haven't yet, you can show that $d(c\cdot \mu) = c\cdot d\mu$ for any measure $\mu$

and indeed one way would be to go by the standard "measure-theoretic induction" i.e. simple functions, and so on

psie

hmm, this "simple calculation" turned out to be quite involved :(

leslie townes

folland is assuming that the reader is either capable of doing this or not troubled by not doing it

its sometimes helpful for basic results of this nature to think, if they failed, what would a counterexample look like. could there be a counterexample with strictly increasing F, G. etc. you wind up in the same place as those 'without loss of generality' reductions

Jakobian

@psie it is indeed a simple calculation. At the end of the day you're just showing that a given expression is linear with respect to both variables

The two facts I mentioned, and symmetry, provided that the theorem holds for $(F_0, G)$ then it holds for $(cF_0, G)$, and provided it holds for $(F_k, G)$ then it holds for $(F_1+F_2, G)$, suffices

leslie townes

or just look at the proof for slightly more regular F and G that you might find in a calculus book. there are essential ideas behind all of this stuff that go beyond any particular definitional setup

you can certainly go over the proofs in any textbook and find variations in how the author does or does not rely on the reader to fill in details. it's none of my business, but i'm not sure it's particularly beneficial to focus on any one text like this

Jakobian

Assuming that it holds for all pairs which are monotone functions $F, G$ you can first use those to say it holds when $F$ is NBV and $G$ is monotone, and after, that it holds when $F$ and $G$ are NBV

leslie townes

18:12

the main lesson i get out of the last several weeks is that folland is more than happy to leave certain details to the reader, and even at times say things that aren't literally true with the understanding that a reader can correct infelicities in statements

imvho one of terence tao's analysis books is slightly better at presenting this stuff around the FTC

but it's all kind of the same

Jakobian

Being honest, those are the type of things I wish you would try to, at least partially, come up with and solve on your own

psie

@Jakobian that's the step I think we've been skipping; assuming it holds for all monotone functions $F,G$ and showing it holds for $F$ in NBV and $G$ monotone.

Jakobian

There's no steps

Doing math is not following instructions

Its building a robot in random order

nickbros123

is it reasonable for a course to do, say, till van kampen theorem from munkres topology, in a course in general topology (a semester long course)

Jakobian

@nickbros123 yes

nickbros123

18:22

I see. Thank you

Jakobian

So I haven't been skipping any steps, I just didn't get to the point where I am building that particular part of the proof that you wish it was already there

Xander Henderson

@nickbros123 Why not?

Frieren

On my book, there's written that if for each $b \in f(A)$ there is only one $a \in A$ such that $f(a)=b$ then $f: A \to B$ is one-to-one. I tried to prove this implication, can someone check the proof please? Let $x,y \in A$ such that $f(x)=f(y)$. By hypothesis, there exists a unique $a \in A$ such that $f(a)=f(x)$. But $f(x)=f(y)$, so there exists a unique $a \in A$ such that $f(a)=f(x)$ and $f(a)=f(y)$. By uniqueness of $a$, we have $x=a=y$ and this proves that $f$ is one-to-one.

nickbros123

@XanderHenderson I dont know, thats why I am asking. I havent had a course in topology yet.

my university is known to try to cram a lot of things in a course

Xander Henderson

@nickbros123 Okay, but then what is the reason to ask? A course in topology will cover whatever the instructor decides it should. If you are working out of Munkres, an intro course would likely have to leave out at least half the book, but anything in the text is likely fair game.

@Frieren What is your definition of one-to-one? The first sentence you write could be the definition...

Frieren

18:31

@XanderHenderson: The definition I have is that a function $f:A \to B$ is one-to-one if for each $x,y \in A, f(x)=f(y) \implies x=y$.

nickbros123

@XanderHenderson The reason I am asking this is, I have put algebraic topology as a possible thing to study, in the application for summer programs. Now even though most professors dealing with summer programs know that the students know jack shit, I still want to make sure that I can actually handle the thing, in the summer

Ben Steffan

@nickbros123 yes

Xander Henderson

@Frieren In that case, your proof is fine, but I find it a bit much.

Ben Steffan

we usually do van Kampen & covering theory

(in the general topology course)

Sine of the Time

My topology course did not cover Van Kampen also

Frieren

18:32

@XanderHenderson: Thanks for help. You mean that it is too verbose?

Xander Henderson

@Frieren Yes, and there are parts of it that seem redundant.

Ben Steffan

should have just said that it's obvious :^)

Sine of the Time

that's clear

Frieren

@XanderHenderson: I see. Usually, when I write here I sacrifice brevity in favor to showing understanding. If I had to rewrite it, I would write this: "Let $x,y \in A$ such that $f(x)=f(y)$. By hypothesis, there exists a unique $a \in A$ such that $f(a)=f(x)=f(y)$. By uniqueness of $a$, we have $x=y$".

nickbros123

@BenSteffan my concern was that (from past experience) I wont be able to absorb the entire thing to a good extent. A good chunk of solvable groups and nilpotent groups, upper / lower central series stuff, I have very barebones knowledge- I couldnt solve Dummit and foote past the 4th chapter- all because they tried to teach us 6 chapters of group theory in a semester. so im also kinda asking from traumatic experience :)

Xander Henderson

18:37

Suppose that $f$ has the property that for each $b \in f(A)$, there is a unique $a \in A$ such that $f(a) = b$, and further suppose that $f(x) = f(x') = y$. Then $f^{-1}(y) \subseteq \{x,x'\}$ and, by hypothesis, this is a singleton set. Therefore $x=x'$.

There is no reason to introduce $a$. You already have an $x$ and a $y$.

If you prefer, suppose that $f(x) = f(x') = y$. By hypothesis, there is a unique $x \in A$ such that $f(x)= y$, hence $x = x'$.

Frieren

@XanderHenderson: Got it, thanks! Another question incoming.

While talking about the fact that the zero product property does not hold for the canonical product of a matrix and a vector, I was asking myself: which is the correct way of phrasing this when talking about where it does not hold? More explicitly: given $n\in\mathbb{N}\setminus\{0,1\}$, a field $\mathbb{K}$, a vector space $V$ on $\mathbb{K}$ and a matrix $A \in \text{Mat}_{n \times n}(\mathbb{K})$, should I say that the zero product property does not hold on $\text{Mat}_{n \times n}(\mathbb{K}) \times V$? Or just that it does not hold on $\text{Mat}_{n \times n}(\mathbb{K})$?

Xander Henderson

What is the zero product property?

Frieren

In the case of real numbers $a$ and $b$, it is the one that says $ab = 0$ if and only if $a=0$ or $b=0$.

Xander Henderson

Okay, but that is a property of multiplication over the reals. It very much does not hold in general.

And the multiplication of a matrix by a vector is a very different kind of operation than multiplication in the reals.

Ben Steffan

@nickbros123 point-set topology should not give you much trouble: at this level it is easy (forgive the word)

a first course in point-set topology largely consists of setting up a number of different properties a space can have, proving a few basic results, and studying how they interact when applicable

it is a subject that is broad but not very deep

as for fundamental groups, there's only so much material there to cover

There's also a point to be made that learning what to "absorb to a good extent" and what to maybe gloss over is a vital skill you should acquire as early as possible

nickbros123

18:51

@BenSteffan I dont think thats a skill I possess :(

Ben Steffan

@nickbros123 emphasis on acquire

but for this in particular, you also have the benefit that there are people around here you can tell you what parts of point-set topology are relevant to algebraic topology and which less so :)

nickbros123

right, that exists :)

Jakobian

@BenSteffan General topology is mainly the study of examples of spaces and properties concerning those spaces. But while this is a lot of it, I disagree its all of it. So I wouldn't say its "not very deep", rather, "most of it is not very deep"

Ben Steffan

@Jakobian I'm mostly referring to what you would learn in a first course

I don't really know enough about the field as a whole to say much apart from that :)

Jakobian

I don't know how "deep" is "deep" either

Xander Henderson

18:57

@Jakobian Anything over 20 feet.

nickbros123

@BenSteffan but at a fundamental level though I haven't really figured out (or maybe gone deep enough in a topic to figure out) which topic I will end up putting most of my energy into. So, though I have a summer program in alg.topology coming up, for which I have to prepare myself, generally speaking I get FOMO skipping / glossing over stuff. ATM, im like a dog chasing cars, I wouldn't know what I'd do if I caught one :)

anyways, good night y'all

ModularMindset

19:39

Let $X$ be a smooth, open, geodesically convex $n$-dimensional manifold with regular polytope boundary, and let $V = \{v_i\}$ be a finite set of vertices in $\partial X$. A block is defined as $\mathcal{B} := X \cup \partial X$, where $\partial X$ denotes the boundary (here boundary is not being used in the traditional sense).

Should I not be using $\partial X$ here?

leslie townes

people are free to make whatever notational choices they want, but it certainly sometimes helps if you do not make choices that are at odds with common usage

Thorgott

people can make whatever notational choices they want, until I disagree

ModularMindset

haha

leslie townes

at least if it actually matters, i wouldn't worry too much about using something in a nonstandard way if it isn't going to come up a lot

if i get halfway through reading something, only to find that a word that hasn't been given a definition doesn't quite mean what i would think it did without an express definition, i usually stop reading

ModularMindset

@leslietownes I use it as a foundation for defining what I call a completion

leslie townes

19:48

i would pay particular attention to definitions and provide definitions in that case

the context matters more than what word you would use to abbreviate it

Thorgott

attention is for losers, make claims first, worry about details later

6

leslie townes

something that people familiar with manifolds often expect about notions ascribed to manifolds is proof that the thing being defined is intrinsic to the manifold and not e.g. dependent on how the thing is embedded in a euclidean space

AFAIK this is already highly nontrivial with the usual notion of 'boundary' for manifolds

you might take the level of care that textbooks use for that as an example

ModularMindset

20:09

@leslietownes The surfaces only embed as surfaces of revolution with constant Gaussian curvature $K$ under certain conditions. I think I understand what you're saying.

Anyway what I'm taking away from this convo is that I should try to define the whole setup intrinsically?

leslie townes

well, just that if someone were to read a laundry list of definitions relating to a "manifold" they would probably be relieved to see an author distinguish between concepts that might really only be a property of an embedding and concepts that are intrinsic to the manifold

and when an author doesn't do that they might think that any departure from familiar concepts was word salad instead of something meaningful

or wonder if the author was even aware of any working definition of "manifold"

and maybe "manifold" isn't the right setting. i don't know

Nasser

20:40

Math experts, any idea why Maple says 5=c1*infinity has solution c=0? what logic do you think lead to this?

restart;
eq:=5=c1*infinity;
eliminate([eq],[c1])

gives {c1 = 0}

I also include screen shot below.

isn't 0*infinity supposed to be undefined? so 5=undefined? Maple's eliminate is described here

I found site doing google, that says "SHORT ANSWER: 0 x ∞ = anything you like" So if this is true, then may be this is why Maple said 0*infinity =5 ?

Ben Steffan

does that article explicitly talk about Maple?/

Because $0 \cdot \infty$ is not anything you like: it is undefined, and generally left that way

at most you'll find people sometimes define that $0 \cdot \infty = 0$

you will not find anybody put $0 \cdot \infty = 5$, like, ever

Sine of the Time

20:55

if $f(x)\to 0$ and $g(x)\to \infty$ then $f(x)g(x)\to \text{anything you like}$

Xander Henderson

Does this work in ChatJax...?: $\raisebox{\depth}{\rotatebox[origin=c]{90}{B}}$

No. :(

Oh, well.

Nasser

@BenSteffan no, that web page does not talk about Maple. It just said x*infinity is anything you want.

@SineoftheTime Ok, so Maple is correct then by saying solution to 5=c1*infinity is c1=0, right?

Sine of the Time

I'd interpret it this way: if $f\to c_1$ and $g\to \infty$ and $fg\to 5$, then the only "hope" is that $c_1=0$

Ben Steffan

@Nasser do not trust webpages that say $0 \cdot \infty$ is anything you want :)

Nasser

@BenSteffan Ok, but how else then to explain why Maple gives c=0 as solution to 5=c1*infinity?

Ben Steffan

21:09

@Nasser maple isn't mathematics

it's a computer program

it has its own rules and conventions

if there's a full spec for maple, I'd look there

Nasser

@BenSteffan Maple is commercial software so it is closed source. But one can print on the screen many of its functions source code. I put link to the eliminate web page. If you think this is a bug in Maple, I will report it then.

Ben Steffan

otherwise you'd need to ask somebody with enough familiarity with the program. I don't know anybody in this chatroom who would fulfill this criterion

@Nasser I saw the link to the help page

closed source does not mean it does not have a spec

but I would not be surprised in the least if it doesn't

I have no idea whether it is a bug

I'm sure there's some kind of maple forum where you could ask, no?

Nasser

@BenSteffan I am asking if the math is correct. Regardless of the system used.

Ben Steffan

as I pointed out above, this question is not well-posed

there's no generally agreed upon meaning of the expression $0 \cdot \infty$, so asking whether it equals some value is not a mathematical question

again, I find positing $5 = 0 \cdot \infty$ bizarre, but it is a definition I could technically make

whether this is the intended semantics in maple or not is impossible to say for an outsider

Nasser

@BenSteffan the equation 5=c1*infinity, easily comes up in many places. It is when we have A=c * f(x) and then take the limit of f(x) as x->infinity. This can generate infinity. So we end up with A=c*infinity. Where A happens to be a number and c is not known.

Ben Steffan

21:21

sure, but why are you telling me this

Nasser

@BenSteffan because you said "this question is not well-posed" so was trying to explain how it came up. But any way, thanks for the feedback. bye.

GroveRover

21:50

@Nasser infinity doesn't belong to reals so you can't do operations with it. You may have been using it informally in calculus classes, but in formal mathematics it does not exist. When you write c * inf = sign(c) inf it's a shorthand notation for lim x to inf c f(x) = sgn(x) inf for every f. The infinite on the RHS side is not a number but is defined using epsilons and deltas.

Ben Steffan

It does exist in formal mathematics, even if it is not an element of the real line.

As for extending operations to it, things like $c \cdot \infty = \infty$ when $c > 0$ are fairly benign and commonplace

GroveRover

@BenSteffan As I said it is a shorthand notation. There is no definition of infinity by itself that prescinds from the use of limits

Xander Henderson

@GroveRover This is a wild overstatement

Ben Steffan

I'm telling you that it is not. Look up, say, the extended real line.

Sine of the Time

that's not true

GroveRover

21:59

(actually I think there is someone attempting to do it, something called nonstandard analysis, but well, that's non standard)

Ben Steffan

...

Sine of the Time

that's another thing

plus nonstandard analysis is already a thing

GroveRover

@BenSteffan yes - and you can't extend operations to it

The restricted operations are the same you get with limits

Xander Henderson

@GroveRover It is fascinating to me that you are popping off about how infinity isn't a thing in formal mathematics, yet you can't seem to pin down what non-standard analysis is.

Are you familiar with the surreal numbers?

Ben Steffan

@GroveRover I literally just told you you can.

to some extent.

Xander Henderson

22:02

or the hyperreal numbers?

or, g-d forbid, wheel theory?

Ben Steffan

ducks

Xander Henderson

geese

GroveRover

Right I may have put this the wrong way

Ben Steffan

it has been 0 days since wheel theory has been mentioned in this chatroom

Xander Henderson

@BenSteffan Sorry to break the streak.

I did wince when I wrote it, if that helps.

And I think that I need to go take a shower now, to wash it off.

GroveRover

22:03

There are ways to formally define infinity, but not in the sense he meant

Ben Steffan

@XanderHenderson what kind of show? standup comedy?

@GroveRover But this is coming from maple?

And maple has some concept of infinity like that, and allows you to operate on it, apparently?

That's why this discussion came about in the first place

Xander Henderson

The real question is "Double ewe tee eff is Maple doing?!"

GroveRover

@BenSteffan Maybe I'm tired, but reading his concerns I understood that he was thinking about infinity the way you do in an informal high school class. I was trying to state that those are short-hand notations for theorems, and infinity is not well-defined in standard analysis

This way of thinking leads you to a straightforward interpretation: if lim x to inf f(x)g(x) = 5 and lim x to inf g(x) = inf, then necessarily lim x to inf f(x) = 0

Ben Steffan

If they had a solid understanding of the concept of $\infty$ then the discussion would probably have been somewhat shorter

GroveRover

@BenSteffan That's what I was trying to express, and I overstated my argument

Ben Steffan

22:13

but in the end the thing they were asking about is just strange, independent of that

That's alright, that's alright :)

happens to the best of us

Thorgott

I don't know $\infty$, only $\infty-$

Ben Steffan

oh, on that note: I learned today that apparently you can proof straightening/unstraightening model-independently if you already know it for anima

more precisely, apparently if you could produce the yoneda embedding in some other way, then you could have a nicer proof

Thorgott

huh, that's weird

I thought the case of anima was much simpler

Ben Steffan

22:29

the model independent proof is still a lot of work

apparently there's a paper out there that does it this way, for $(\infty, n)$-categories

Thorgott

I have this saved lol

Ben Steffan

...this is the exact paper I pulled up lol

Moser-Rasekh-Rovelli

I thought it was this one but apparently not

Thorgott

yeah, stuff like this makes me feel like I'll never understand anything

Ben Steffan

yeah :'(

I do not understand how people do $\infty$-category theory at the research level

this is just absurd

Thorgott

neither do I, but with the caveat I'm supposed to be writing a thesis in the subject

Ben Steffan

22:43

relative to me at least you seem to understand a lot :)

not that that necessarily means much but

what's your thesis' topic?

Thorgott

@BenSteffan thanks :)

@BenSteffan I'm supposed to motivate that spectra can be modeled by fully dualizable symmetric monoidal $(\infty,\infty)$-categories or something along these lines

Ben Steffan

:o

Thorgott

practically speaking, I'm currently struggling to derive an explicit formula for the left adjoint to the inclusion $\mathbf{Spc}\rightarrow\mathbf{Cat}_{(\infty,\infty)}$

Ben Steffan

@Thorgott that sounds... very scary

Thorgott

and I am very scared

it's supposed to fix some old ideas of Quinn (later elaborated on by Laures and McClure) on how to construct bordism spectra

Ben Steffan

22:53

ah, what else

but $(\infty, \infty)$-categories??

Thorgott

yeah, that's what you need for Lurie's bordism categories in full generality (apparently; I still don't know how to construct the bordism categories)

Ben Steffan

bordism is so strange

Thorgott

anyway, here's my (tentative) fun proposition of the day (some details in the proof still not fleshed out): if $K\times\Delta^1\rightarrow\mathbf{Cat}_{\infty}$ is a natural transformation of diagrams of $\infty$-categories that is an object-wise left adjoint and s.t. the transfer maps are compatible with the corresponding right adjoints up to equivalence, then the induced map on limits is again a left adjoint

Ben Steffan

huh

I guess that makes sense

how much time do you have left on the thesis?

Thorgott

ideally I'll finish this semester

Ben Steffan

23:00

oh, ok

that's not a lot :/

Thorgott

yeah, but hopefully I'll manage somehow

Ben Steffan

I believe in you :)

but at the same time I do not envy you over the end-of-thesis stress

leslie townes

i wrote my thesis largely over a summer and then spent the fall trying to get my committee to look at it, which was a different kind of stress

Claudio

Regarding uniform convergence of a series of functions, I know I can use the following result

In the case where $g_k(x) = g_k$, namely a standard numerical series, the above condition should give me a much easier criterion, but I can't see how

if $g_k$ converges, this means that $g_{n+1} \to 0 \text{ as } n \to +\infty$

leslie townes

i don't know what you mean by "easier." the dependence on x goes away. it should resemble the cauchy criterion for the convergence of the sequence of partial sums

it implies that. it means (i.e. is equivalent to) slightly more than that, i.e. convergence of the sequence of partial sums

Claudio

23:14

Yeah but I can't just say $$|\sum_{k \ge n+1} g_k| \to 0 \text{ as } n \to \infty$$

@leslietownes I see so this just becomes Cauchy's criterion

leslie townes

there's certainly a good exercise in there, is the convergence of the series equivalent to the sequence of tail sums going to zero, but yes, if you just literally interpret the above condition with constant sequences you get the cauchy criterion

the reason i wouldn't use 'easier' in this context is, the convergence of a series (even a numerical series) can be very subtle to determine even if the condition is expressible in relatively simpler terms than some other conditions

Claudio

this is why I asked that btw

@leslietownes yeah I thought it'd be easier :p turns out it isn't hahah

leslie townes

in that example the more relevant thing might be that for any given value of x only finitely many terms will enter into the sum

at least, that's what jumps out at me as i look at it

it's also a nonnegative sequence so the sum (if it exists) is the sup

Claudio

yeah but at each value of $k$ x varies in a completely different interval so I'm kind of stuck

leslie townes

maybe try a more general problem where E_k is any family of disjoint subsets of R and c_k is any sequence and g_k is defined by g_k(x) = c_k for x in E_k and 0 otherwise

before assessing uniform convergence, can you identify whether the sequence has a pointwise limit? that is the natural first step

Claudio

23:26

it should be 0 if my previous calculations arent wrong

if I fix an $x \in (0,1]$ I can always find a $\bar{k}$ s.t. $x>1/\bar{k}$ $\forall k\ge \bar{k}$

also the series does not converge totally in [0,1]

Sine of the Time

Are you sure the pointwise limit is $0$?

leslie townes

if 1/3 < x < 1/2, what is sum g_k(x)

Claudio

@leslietownes wait I'm not getting the idea: trying to estimate the sup in each $E_k$

leslie townes

if 1/3 < x < 1/2, for which value(s) of k is g_k(x) nonzero

Claudio

k=2

leslie townes

23:34

i agree

Claudio

yeah I might've got it wrong

for every $x \in (0,1)$ I can find a $g_k$ which is nonzero

leslie townes

one thing that might be throwing you off is that the pointwise limit of sum g_k(x) has its most natural expression as a 'piecewise function' i.e. the formula for it is not any simpler than the formula for the kth term in the sum

Claudio

I'm kind of confused

I can't figure out the pointwise limit hahahah

leslie townes

what is sum g_k(x) if x is in (1/(k+1), 1/k]

Claudio

1/(k+1)

and this does not converge

Sine of the Time

23:44

I suggest drawing the intervals $]\frac{1}{k+1},\frac 1k]$

and to note that are disjoint

leslie townes

well you've identified the pointwise limit

Claudio

@SineoftheTime that much I can see for now, but what I'm trying to say is that If I have no pointwise conv in (0,1]

Sine of the Time

so every $x$ is in only one of these intervals

and the sum is reduced to one term

Claudio

to only one nonzero term right

I see what you were saying by you've identified the pointwise limit @leslietownes

I have pointwise convergence in [0,1] then

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