Mathematics - 2024-12-02

Jakobian

01:32

in fact, and this is recent news

1

A: Even covers and collectionwise normal spaces

As implied by Tyrone, the claim that a fully normal space is strongly collectionwise normal is true. See, for example, the result in H. J. Cohen's paper Sur un problème de M. Dieudonné (translated into English and paraphrased). Proposition [1, Thm. 2]. A topological space $X$ is strongly collect...

turns out theorem 4.8 is in fact false

which is the one I was asking about (well, parts of proofs of)

so it seems like it was a good idea to ask it on mathoverflow

They were claiming that strong collectionwise normality is equivalent to every even open cover being normal

the latter is called somewhat normal space, and a normal space is somewhat normal iff evenly paracompact

but there is a normal evenly compact space which isn't strongly collectionwise normal

Ben Steffan

01:47

"somewhat normal" lol

Jakobian

I know, lol

Thomas Finley

04:11

Show that $\ell_2$ is a complete metric space.

Let $\{x^{(k)}\}$ be a Cauchy sequence in $\ell_p,$ where $x^{(k)}=(x_1^{(k)},x_2^{(k)},...\}.$

This means that, for $\epsilon^{\frac 1p}>0$ there exists $K\in \Bbb N$ such that, $d(x^{(r)},x^{(q)})\lt\epsilon,\forall r,q\geq K.$

Hence, $$d(x^{(r)},x^{(q)})=\sum_{i=1}^{\infty}|x_i^{(r)}-x_i^{(q)}|^p\lt\epsilon\tag 1$$

This implies, $|x_i^{(r)}-x_i^{(q)}|<\epsilon^{1/p}=\epsilon',$ for all $r,q\geq K,i\in \Bbb N.$

As, $\epsilon'$ is arbitrary, so, $\{x_i^{(k)}\}$ is a Cauchy sequence in $\Bbb C,$ which in turn implies that, $\{x^{(k)}_i\}$ is a convergent sequence in $\Bbb C$ such that $…

@Jakobian Can you please check if my proof is valid or not?

Thomas Finley

04:33

@copper.hat It would be really great if you consider checking it as well!

(Thanks in advance!)

copper.hat

04:50

@ThomasFinley is there something that you don't follow? also, you start with $l_2$ then jump to $l_p$. when i read proofs i stop there.

leslie townes

05:22

oh, but the best is yet to come

Lukas Heger

@leslie what do you mean?

leslie townes

i was being sarcastic. (if a written proof of something about ell_2 immediately jumps to ell_p without comment, the best is probably not yet to come)

Lukas Heger

oh I see

maybe in the end there's a limit argument that lets $p$ approach $2$

leslie townes

i stopped reading at a mandate to do something, followed by "This implies,"

Lukas Heger

@ThomasFinley I don't think your step "letting $q \to \infty$" is justified. You can't just interchange a limit and an infinite sum without further argument

the shortest proof for completeness of $\ell_p$ I know uses Fatou's lemma

not sure if you have that at your disposal

one potato two potato

05:57

study a bit of arithmetic hyperbolic 3-manifold theory and found it quite heavy

study mainly because of curiosity but the theory is quite expensive

Soham Saha

Hi everyone. Is this statement true: “If a function $f(x)$ is differentiable on $(a,b)$, and has positive derivative at $x=c$ in $(a,b)$, then there’s a neighbourhood around $c$ where $f(x)$ is increasing”?

Thomas Finley

06:13

@copper.hat actually, I tried to prove it for $\ell_p$ in general.

@LukasHeger no I don't know about Fatou's lemma

leslie townes

soham: no

soham: try to convince yourself e.g. that any function f(x) on R that is differentiable for nonzero x and satisfies x - x^2 <= f(x) <= x + x^2 will be differentiable on all of R and satisfy f'(0) = 1. then convince yourself that you can draw really wiggly shit between the graphs of x - x^2 and x + x^2 that will not be monotone on any interval containing 0

there aren't simple counterexamples because if you add in enough further regularity hypotheses (beyond "differentiable on [an interval]") that only simple functions can satisfy, the implication might actually hold

but it doesn't hold at the stated level of generality

Sine of the Time

Maybe something like x/2+x^2sin(1/x)

leslie townes

yeah you can also just write down examples that do it

i was trying to sketch a kind of geometrical picture of why the hypotheses "clearly" can't be enough

as a shoutout to ted who would have wanted that if he were here

if all you want is an answer this is probably a 20x duplicate on main

Soham Saha

06:34

@leslietownes Thanks for the explaining the general idea. I was trying to come up with an example, but hadn’t succeeded yet. I think I will try for myself again before searching on main

leslie townes

imagine just drawing a squiggle that ping-pongs back and forth between those two graphs forever (necessitating increasing frequency) as you approach 0

you always have enough room to ping-pong back and forth because one of the graphs is strictly above the other one, and you won't ruin differentiability at 0 because of the remark above about how anything that's sandwiched between those graphs will be differentiable at 0

nothing special about x - x^2 and x + x^2 or 0 here, i mean, two parabolas, one upward facing another downward facing, tangent to one another at a point where their common tangent has a nonzero slope

any function sandwiched between such things will be forced to be differentiable at that tangent point and have the same slope

so as long as you ensure that your ping-ponging elsewhere is differentiable, you're in clover

Soham Saha

@SineoftheTime Already tried that, limit of derivative doesn’t exist at $x\rightarrow0$

@leslietownes Yes, I think I’ve got the idea now. Thanks again

@leslietownes This statement was in the introduction of Counterexamples in Calculus. Pretty cool book I think

Sine of the Time

@SohamSaha so?

leslie townes

cool idea for a book

sanya

Hu

How do I find out if f(x)=cosx² is continuous or not for all x belongs to R

Soham Saha

06:46

@SineoftheTime I think I am missing something… If the derivative doesn’t converge at $x\rightarrow 0$, then how do we get derivative at $x=0$?

leslie townes

typically one would resolve this by [proving and then] applying general theorems about operations on functions that preserve their continuity

and the proved or assumed known continuity of t -> cos(t) would certainly enter into it

sanya

Im not sure how to prove this @leslietownes if it was about a point then thats easy by proving that the limit exists nd is equal to the functional value

But for an interval how

Sine of the Time

@SohamSaha maybe you're confusing the first derivative withe the second derivative? At x=0, the derivative is 1/2

Soumik Mukherjee

@sanya do you know the epsilon delta way of proving continuity?

leslie townes

sanya as a threshold matter (although it does not affect the result) i'm not sure if f(x) is (cos(x))^2 or cos(x^2) but either way the approach would be essentially the same. it absolutely does not matter whether one fixes a point or not. continuity is a 'local' phenomenon, f is continuous on R iff it is continuous at x for all x in R

so i'm not sure i understand the difference between proving continuity at a point and proving continuity at an interval. the approach should be nearly identical

and if both are hard then both are hard, but it shouldn't be that one seems easy and the other seems hard

sanya

06:54

@leslietownes its cos(x²) so I just take a variable point h which is in R and prove that the limit exists and is equal to the functional value?

Soham Saha

@SineoftheTime First derivative is -cos(1/x)+2xsin(1/x)+0.5 right?

sanya

@SoumikMukherjee no:/

leslie townes

@sanya yes

Soham Saha

@SineoftheTime which oscillates between -0.5 and 1.5 near 0?

@SineoftheTime Sorry, I will be back in just a minute…

Sine of the Time

$f'(0)=\lim_{x\to 0 }\frac{x/2-x^2\sin(1/x)}x=1/2$

Soham Saha

07:14

@SineOfTheTime Ah yes, got it now. So f’(a) doesn’t always depend on the limit of f’(x) near x=a… We are defining the function as x/2+x^2sin(1/x) at x not equal to 0, and 0 at x=0, and then simply taking derivative using first principle right?

@leslietownes and @SineoftheTime would it be OK if I ping you here in the future for other doubts?

leslie townes

soham fine with me although i am here often enough that i might see it without the ping :)

Soham Saha

@leslietownes Thanks a lot :)

leslietownes I think this matches your description

leslie townes

07:30

yes

Sine of the Time

@SohamSaha ok for me

Soham Saha

@SineoftheTime Thanks :)

Soumik Mukherjee

How to render mathjax on tab?

This is my nth time asking how to render mathjax question 🫠

n>1

Soham Saha

07:52

@SoumikMukherjee Go to the installing on mobile browsers section here

Copy the code

Delete the chat link from your search bar, and paste the code instead

Your browser might decide to delete the “javascript:” at the beginning, so need to retype it

Press enter

Oh, and I think that their “Bookmark” idea works too, but never tried it

Soumik Mukherjee

09:01

@SohamSaha where? Can you send a screenshot?

Soham Saha

09:14

Link: math.ucla.edu/~robjohn/math/mathjax.html

Copy all that code in the link

And paste it as shown below:

Then check if the “javascript:” at the starting of the code has also got copied or not. If it has not got copied, you need to append it the beginning. The press enter

Soumik Mukherjee

12:21

@SohamSaha how to get to this address bar?

psie

Let $\mu$ be a complex measure. Under what condition does it hold that $$\mu(A\setminus B)=\mu(A)-\mu(B)?$$

Of course, $B\subset A$. It seems to me that it always holds. Not like if $\mu$ were a positive measure, where it'd only hold if $\mu(B)<\infty$. But since complex measures do not assume infinite values, we can always subtract $\mu(B)$ form $\mu(A)=\mu(A\setminus B)+\mu(B)$, which is just countable additivity. Let me know if you agree or not.

Thorgott

you probably want $A,B$ to be measurable, yes?

psie

yeah, true

Soumik Mukherjee

13:17

@SohamSaha it's redirecting me to google

Soham Saha

@SoumikMukherjee You’re on chrome right? On tab?

Soumik Mukherjee

Yes

Soham Saha

Open MSE chat in google chrome app (not google app), and take a screenshot

Soumik Mukherjee

I pasted at the address bar, but it's not working

Soham Saha

After pasting did you check if there was a “javascript” at the beginning?

Chrome doesn’t allow you to paste “javascript”

Soumik Mukherjee

13:23

Yes I checked, it was not there, I added javascript: at the beginning

Soham Saha

Then pressed enter, and still nothing happened?

Strange…

Could you share a link to the google search page where you’re redirected to?

Soumik Mukherjee

Yes

It's too long, can't send

Soham Saha

Paste it here: pastebin.com and send the pastebin share link

I am on ipad and the mathjax method works fine… Don’t know why it doesn’t work for you

Soumik Mukherjee

pastebin.com/1QAvfLH7

Soham Saha

Umm, link shows 404

Soumik Mukherjee

13:33

pastebin.com/jC1simdZ

Try this one

Soham Saha

Nope

Try copying it here part by part

Wait, you can just share a screenshot of the google search page

Soumik Mukherjee

Sine of the Time

bros fishing

btw, I also tried and get the same as Soumik

Soham Saha

I don’t know why but it works all the time for me. I’m on ipad

Soumik Mukherjee

@SineoftheTime XD

Soham Saha

13:46

Hmm doesn’t work on Android it seems

@SoumikMukherjee Did you try the bookmark method mentioned in the link (above the code part)?

Soumik Mukherjee

No, I will try

User1865345

14:01

@SohamSaha Check Bartle where he discusses about increasing at a point, if I remember correctly.

He discusses the example @SineoftheTime mentioned.

Soham Saha

14:16

@User1865345 Got it, thanks :)

User1865345

👍🏻

Sine of the Time

that's also mentioned in Folland if I recall correctly

User1865345

Yeh. Examples like these have been doing rounds for quite sometimes. Seen such types at Kaczor too.

Xander Henderson

14:39

I've been trying to get students to schedule times to take their finals for three weeks. Exams need to happen in the next couple of days, and I have a student who just emailed me, asking for classroom space for her final at one of our more remote campuses. Like... maybe I can find a room? I wish you had asked earlier...

User1865345

You did your part informing them quite early. It was imperative on them to place such request not at 11th hour.

psie

16:14

If $F\in NBV$, meaning it is in $BV$ (the space of complex-valued bounded variation functions), $F(-\infty)=0$ and $F$ is right continuous, then why is $F'\in L^1(m)$ where $m$ is Lebesgue measure? I'm reading Folland by the way.

I know that $F'$ exists and is equal $m$-a.e. to $G'$, where $G(x)=F(x+)$. But I don't understand why its integrable.

I suspect it has something to do with the fact that we can write $F=(F_1^+-F_1^-)+i(F_2^+-F_2^-)$ where $F_j^\pm$ are bounded and increasing, but I'm not 100% sure.

Nah, this is not it. Since we are talking about $F'$, not $F$.

I think I understand why. I had to go back a couple of pages...in the section entitled "Differentiation on Euclidean space". Great.

ModularMindset

17:04

hello

Jakobian

@psie it should mainly be about being a function of bounded variation

@psie Why not? If $F$ is increasing and bounded, would that imply $F'$ is Lebesgue integrable?

Shouldn't it be that $\int_x^y F' \leq F(y)-F(x)$ once we show that $F'\geq 0$ is measurable

we know that $F'$ exists a.e.

so it boils down to showing two facts, that $F'$ is measurable, and that $\int F' \leq F(\infty) < \infty$

that $F'$ is measurable should follow from the proof that $F$ is a.e. differentiable, I think?

leslie townes

yes

Jakobian

Hi Leslie

good to see you

Claudio

17:22

If $A \subseteq \mathbb{R}^n$ is an open set and $\phi : A \to \mathbb{R}^n $, is it trivial that $m(\phi(A)) = \displaystyle\int_E|detJ\phi|d\mathbf{x}$?

Jakobian

Its not trivial, but it follows from a similar equality for linear maps

you should look in a textbook for this one, in my opinion

Claudio

my professor wrote this in class: $$ m(\phi(A)) = \int_{\phi(A)} 1d\mathbf{x} = \int_A |detJ\phi(\mathbf{x})|d\mathbf{x} $$ but I couldn't see the connection

@Jakobian oh ok maybe I'll take a look at my book or at Folland

Sine of the Time

@Claudio did you already see the change of variable theorem for multiple integrals?

psie

Well, my reasoning goes like this @Jakobian, maybe you're saying the same thing. Every $F\in NBV$ gives rise to a unique complex Borel measure $\mu_F$ such that $F(x)=\mu_F((-\infty,x])$ (this measure is regular by the way). Now consider the below theorem; let $d\mu_F=d\lambda+f\,dm$ be its Lebesgue-Radon-Nikodym decomposition, where $f\in L^1(m)$.

Applying Theorem 3.22 to the sets $(x-r,x]$ and $(x,x+r]$, which "shrink nicely" to $x$, we obtain for a.e. $x$, $$f(x)=\lim_{r\to 0}\frac{\mu_F((x,x+r])}{r}=\lim_{r\to 0}\frac{F(x+r)-F(x)}{r}$$and similarly for $(x-r,x]$. So $f=F'$ $m$-a.e.

Jakobian

well, clearly I'm not saying the same thing

but whatever works, works

Claudio

17:28

@SineoftheTime yeah we did today and when I asked her if it could be proven why the determinant was there (during the lecture she only wrote down an intuitive visual proof), and she wrote the above-mentioned line on the chalkboard

psie

alright :)

leslie townes

@Jakobian you too

Jakobian

@Claudio the determinant is there because you first show it for linear functions

@leslietownes :)

Claudio

@Jakobian Yeah I've found Folland's proof, it will take me some time to digest it

Sine of the Time

@Claudio do you like Folland?

ModularMindset

17:31

hi

Sine of the Time

hi

ModularMindset

how do i avoid getting banned from sports betting apps? Do I have to temper my winnings to avoid them detecting advanced betting strategies?

Claudio

@SineoftheTime it's a bit complicated for me, but it is pretty clear indeed

Sine of the Time

@ModularMindset don't install the betting app :D

Xander Henderson

@ModularMindset "Advanced betting strategies"?! Ha!

You aren't going to get banned. You are exactly the kind of rube they want on their apps.

Sine of the Time

17:35

Folland is enjoyable to read, expecially if you've already seen some material

Xander Henderson

@SineoftheTime ARE YOU HIGH?!

Sine of the Time

No I'm tall

ModularMindset

@XanderHenderson thanks - I am relieved that I can continue siphoning money from the house

Xander Henderson

Uh huh. Sure.

Sine of the Time

I'm talking about Folland Advanced Calculus, not Real Analysis

Xander Henderson

17:37

@SineoftheTime Oh, I have that one, too. It's... okay... I guess? Still: ARE YOU HIGH?!

Sine of the Time

why? :(

Jakobian

If $f:X\to Y$ is a closed continuous surjection and $K\subseteq Y$ is closed, $\overline{G}\subseteq f^{-1}(Y\setminus K)$ then can we show that $\overline{Y\setminus f(X\setminus G)}\subseteq Y\setminus K$?

Xander Henderson

@SineoftheTime I'm not a huge fan of Folland. Too terse. No pedagogy.

Though I have vague memory of Folland maybe having a proof of the Riemann rearrangement theorem, which is perhaps my favorite undergraduate analysis result, so if I am remembering that correctly, Folland scores a minor vicory.

Jakobian

I am reading a proof that monotonically normal spaces are preserved under closed continuous surjections, and failing to see one of the properties

Xander Henderson

Most books at that level seem to omit it.

@Jakobian Have you tried looking at it with only your left eye?

Sine of the Time

17:40

Well, I've not studied the whole book, but I've liked that he sometimes presents the topic in an "informal" way to give intuition

Jakobian

the issue is that interiors and images don't have much in common

unless your map is open which I don't have

Xander Henderson

@SineoftheTime If it is the book I am thinking of, the theorem is at the top of the page, on the right-hand side when the book is opened flat, in the chapter on sequences and series. But I haven't looked at that book since I took undergraduate analysis in 2007, so I might be thinking of Dangello and the other guy's book.

Sine of the Time

correct

page 298

leslie townes

waay too much insight into how xander's memory works

mine's the same way. "oh, just look in that book. you know? with the red cover? its like 80% into the book, middle third of the page, on the right."

i'm a big electronic over paper person, but this is something we lose when it's all just text on a screen

Xander Henderson

@leslietownes Heh.

Jakobian

18:09

@Jakobian From $\overline{G}\subseteq f^{-1}(Y\setminus K)$ it follows $f(\overline{G}) = \overline{f(G)}\subseteq Y\setminus K$ so $K\subseteq Y\setminus \overline{f(G)} = \text{int}(Y\setminus f(G))$. But $Y\setminus f(G)\subseteq f(X\setminus G)$ so $K\subseteq\text{int}(f(X\setminus G))$.

The inequality being equivalent to $\overline{Y\setminus f(X\setminus G)}\subseteq Y\setminus K$ by taking complements

the properties I used is that $f$ is a closed surjection

maybe one can summarize this somehow in terms of properties of closed surjections... I don't really see it, whatever

Silly Goose

19:11

How does one show that the Yang-Baxter equation and Braid equation are equivalent?

I have been mucking around a bit but can't really find a way to show that if $R$ solves one then $PR$ solves the other, where $P(x \otimes x') = x' \otimes x$

Jakobian

Before I thought that the proof that linearly ordered spaces are normal is pretty hard. But now, after reading the proof that linearly ordered spaces are monotonically normal, this method is actually quite easy to understand

Silly Goose

19:24

okay actually i got it

Daniel Adams

20:22

Has there been a big drop in activity on this site since LLM became a thing?

leslie townes

i dunno, it seems approximately as busy as it was, to me

Ben Steffan

in fact, there has been an increase in users seeking explanations for the garbage that LLMs spit out in response to their math questions :^)

leslie townes

most of the people who would go to an LLM for answers and not notice anything wrong probably wouldn't have been going to MSE in the first place

MSE is like the masters degree level of cheating on your homework

i.e. for lazy people who think somewhat critically about what machines have vomited at them

think_meaning_buildß

eeww

There's been a huge drop here:

leslie townes

well, if anyone wants to know why software stopped working, that's why

think_meaning_buildß

20:36

Yeah, there's a new species of animal roaming the net.

psie

20:48

I'm stuck with a small detail in this proposition, which isn't really proved in the book. Consider the statement if $F=\int_{-\infty}^xF'(t)\,dt$, then the unique complex Borel measure that $F$ induces, $\mu_F$, is absolutely continuous with respect to Lebesgue measure $m$.

I believe one has to show $$F(x)=\int_{-\infty}^xF'(t)\,dt\implies\mu_F(A)=\int_AF'(t)\,dt$$for all Borel sets $B$. By Caratheodory's extension theorem, it suffices to show this on the algebra of h-intervals, that is, sets of the form $(a,b]$, $(a,\infty)$ and $\varnothing$ where $-\infty\leq a<b<\infty$.

$A=\varnothing$ is obvious. For $A=(a,b]$, we have that $\mu_F((a,b])=\mu_F((-\infty,b])-\mu_F((-\infty,a])=F(b)-F(a)=\int_a^b F'(t)\,dt$. And for $A=(a,\infty)$, I'm stuck. We have $\mu_F((a,\infty))=\lim_{n\to\infty}\mu_F((-\infty,n])-\mu_F((-\infty,a])$, but is $\lim_{n\to\infty}\mu_F((-\infty,n])=\int_{-\infty}^\infty F'(t)\,dt$?

leslie townes

@psie just barely peeking into this, why isn't the equality on the right hand side of the implication that you say that you 'ha[ve] to show' the definition of mu_F? or what is the definition of mu_F?

oh, never mind

i don't render latex which may explain why i am confused about some of this

Ben Steffan

let your browser do it for you :)

leslie townes

there isn't any time

psie

@leslietownes no worries :) I don't think there is any explicit definition of $\mu_F$, it's just "induced" by $F$ and satisfies $F(x)=\mu_F((-\infty,x])$

leslie townes

that last equation sounds like it could work as a putative definition of mu_F

psie

21:01

@leslietownes do you see how, if I can show that right-hand side of the implication, then $\mu_F$ is absolutely continuous with respect to $m$?

Absolutely continuous in this case means that the Lebesgue-Radon-Nikodym decomposition is just $d\mu_F=F'\,dm$, which is the right-hand side of the implication.

leslie townes

absolutely continuous with respect to m should have been given separate definition outside of this case, which is if m assigns a set measure zero, then so should the supposedly absolutely continuous measure do that

this kind of thing is at least equivalent to absolute continuity as folland would/should have defined it

for finitely-but-not-necessarily-countably additive measures you need to be a little more careful but that is gilding a lily

psie

ok, I think a measure nu that can be written nu(E) = int_E f(x) dmu(x) will be absolutely continuous with respect to mu.

leslie townes

this is the prototype of it and from the radon nikodym style of result the only way you get those

Binky

21:16

$y'+y^2=1 , y(0)=0$

Pls help

Its a Riccati's equation

Sine of the Time

huh?

Isn't it separable?

Binky

Yes

Sine of the Time

what's difficult about it? It seems straightforward

leslie townes

psie be aware that it wouldn't surprise me if some resources use that as a definition of absolute continuity. all of this stuff is in flux

psie

@leslietownes I see :)

I agree though, the definition of absolute continuity is what you said

leslie townes

21:21

well, i don't agree. there are "definitions" of absolute continuity, of which that one is maybe the most common in the setting of countably additive measures

:)

mo-_-

@Binky $-\int{\frac{1}{{y}^{2}-1}}{dy}=\int{1}{dx}$

leslie townes

it's common enough in the countably additive setting that i would be surprised to see a textbook do otherwise. we can provisionally say it is 'the' definition

Sine of the Time

you're missing a minus sign

leslie townes

as long as we remind ourselves that we're swirling in an indifferent cosmos filled with other notions and things

mo-_-

I wrote - instead of + ... at the start

sorry

Binky

21:25

Is the solution tanh(x)?

Sine of the Time

I'd write it as $(1-y^2)^{-1}$ since its antiderivative is well known

mo-_-

No

I don't think so, wait

@SineoftheTime Right

Just Google It and you will find the steps if necessary @Binky

I mean int (1-y^2)^-1

Binky

I mean the final solution

Sine of the Time

did you try to subsitute in the equation and see if it verifies it?

Binky

Yes

mo-_-

21:33

$-\frac{\ln\left(\frac{y-1}{y+1}\right)}{2}=x+C$

Can you take It from here ?

Binky

I'm already done

mo-_-

What solution do you get?

Binky

tanh(x)

mo-_-

Correct

$\tanh(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$

Binky

Yes

mo-_-

21:44

@Binky can you suggest why its true?

Binky

Its for definition

@mo-_- i.imgur.com/eFk6YqJ.jpeg

Can you explain to me what that graph means?

Pls

It comes from the exercise above

mo-_-

?

Binky

I got it from wolfram

mo-_-

The system exhibits a decreasing trend, suggesting that $y'(x)$ decreases as $y(x)$ approaches the limiting value.

Binky

Okay...?

Yes but what does it mean

mo-_-

21:59

⚆ _ ⚆

@SineoftheTime Can you help by any chance?

Sine of the Time

what's the problem?

mo-_-

He sent a graph and wants to know what it means

It would basically be the plot solution but in graphic form

Sine of the Time

yeah it's the plot of the solution

mo-_-

Yes I think he wants to know what it means (I didn't quite understand)

Binky

Wait

this is the graph of y(x)=tanh(x)

What is the input of the other graph

mo-_-

22:09

ಠ╭╮ಠ

Sine of the Time

@Binky look at the "x axis" and "y axis"

Binky

Yes

mo-_-

Yes but there was y' and y

Binky

They are different

Why does the x disappear

Sine of the Time

it's the plot of $y'=1-y^2$

mo-_-

22:13

Binky

@SineoftheTime ah

Clear

Thank you all

Sine of the Time

@Binky what are you studying?

mo-_-

chat.stackexchange.com/transcript/message/66722248#66722248 Can someone explain me step 2 -> 3 pls :(

Binky

@SineoftheTime I'm combining a little bit of everything

mo-_-

Are you preparing for JEE?

Binky

22:22

What

No

mo-_-

@Binky @SineoftheTime What do you frequent?

Binky

I attend a university that offers me many opportunities to put what I study into practice.

mo-_-

Why are you writing like this?

Binky

To be professional

mo-_-

(ʘᗩʘ')

Binky

22:31

@mo-_- Does this seem strange to you?

mo-_-

@Binky Hmm...

Sorry :(

psie

23:21

It's a bit unclear with the domain of the function. Every absolutely continuous function is said to be uniformly continuous, but on Wikipedia only over compact sets. Is it not true for a function with domain $\mathbb R$?

Xander Henderson

2

A: Can you use "biject" as a verb?

It should be OK. One reason is that the meaning is immediately clear. The other reason is that it is used in several research papers, indicating that it is a common phrase. From Schwede's Global algebraic K-theory: Since every countably infinite set bijects with the set $\omega$ ... There are o...

Gross...

I think I need to go take a shower after reading that.

Ben Steffan

it's a usage that, uh, exists

I'm pretty sure I've heard Schwede use this in person

Xander Henderson

@BenSteffan That doesn't make it any less gross.

Ben Steffan

no, certainly not

I don't need to "buy ject" anyways; I still have ject at home

Thorgott

23:38

it really isn't that weird, we use "inject" all the time

or "surject"

Xander Henderson

The verb "to inject" predates the mathematical notion of an "injection".

And I don't really like the use of "to inject" in mathematics to describe an injective map. In my opinion (which is obviously the only one that matters), "to inject" is as bad as "to biject".

Jakobian

@psie whats your definition of absolutely continuous on $\mathbb{R}$

Xander Henderson

"To surject" is in the same terrible category.

@Jakobian Like, continuous, man, but like really, really continuous. Like, whoa! It's continuity is, like, absolute, man!

Thorgott

I don't see what you find so objectionable about the verbs

Xander Henderson

They're gross!

They taste bad in my mouth.

And they make my ears bleed.

psie

23:45

@Jakobian A function $F:\mathbb R\to\mathbb C$ is absolutely continuous if for every $\epsilon>0$ there exists $\delta>0$ such that for any finite set of disjoint intervals $(a_1,b_1),\ldots,(a_N,b_N)$, $$\sum_1^N (b_j-a_j)<\delta\implies\sum_1^N |F(b_j)-F(a_j)|<\epsilon.$$

Jakobian

Then here you go, for $N = 1$ you have uniform continuity

psie

ok, I was puzzled by wiki's parentheses "(over a compact interval)"

I mean, it's true for compact intervals, I just took it as meaning it is only true for compact intervals

Jakobian

I think some people might be using something that some people call "locally absolutely continuous" as definition of absolute continuity

psie

ok

Jakobian

I don't know, the article is a mess anyway

psie

23:54

pretty messy :)

Transcript for

$Mathematics$ Mathematics