Mathematics - 2023-10-13

David Raveh

01:52

If I have two functions $f,g: X\to\mathbb{R}$ such that $f$ and $g$ are closed maps. Is $f+g$ a closed map as well? this would make one of my hw problems much simpler

Dannyu NDos

Counterexample: $\sinh x + \cosh x = \exp x$.

It's worthy to note that $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is open but not closed.

David Raveh

dang. I am trying to show that $h(x,y)=x+y^2$ over reals is quotient map, and this was my idea of how to do it. Any thoughts?

Dannyu NDos

It seems you already know that all closed surjective maps are quotient maps. Hint: All open surjective maps are quotient maps as well.

Though, there are quotient maps that are neither open nor closed.

David Raveh

I began by trying open maps, but was running into trouble because $y^2$ appears to ruin it

Dannyu NDos

Oh

I examined the fibers of $h$, and the fibers are $F_x = \{(x, x-y^2) : y \in \mathbb{R}\}$.

David Raveh

02:13

I don't know what fibers are

Dannyu NDos

A fiber is the inverse image of a singleton set.

That said, the saturated open sets in the domain of $h$ are precisely $\bigcup_{x \in U} F_x$ for an arbitrary open set $U \subset \mathbb{R}$, which $h$ maps to precisely $U$.

So the quotient space induced by $h$ agrees with the usual $\mathbb{R}$. Case closed.

David Raveh

You use terminology that I am unfamiliar with.

Why are you looking at the inverse images of singleton sets?

Dannyu NDos

I looked at fibers because, a function induces a quotient space if and only if the fibers cover the entire domain.

To elaborate, every surjective function "carries" the topology of the domain to endow the codomain with another topology, resulting in the quotient space. In your case, the codomain of $h$, namely $\mathbb{R}$, already carries a topology, so an extra step to check whether the topologies agree is required.

Ted Shifrin

03:04

@DannyuNDos Where did you get this? If I set $x+y^2=c$, then h^{-1}(c) = \{(x,y): x=c-y^2\}$.

Oh, you reversed coordinates and changed letters ... or something.

Dannyu NDos

Sorry; my fault.

copper.hat

03:49

If $A$ is a (bounded) Hermitian operator in a complex Hilbert space then $\|A\| = \sup_{\|x\| \le 1} | \langle x, Ax \rangle |$. Is this true in a real Hilbert space as well?

Hey Joe.

Ted Shifrin

@copper.hat Sure.

copper.hat

It is easy to show in the complex case, but I'm stuck on the reals.

My brain has gone a bit fuzzy the last few weeks.

Ted Shifrin

There’s still the spectral theorem, no?

Koro

Quotienting a point in a space means doing nothing to the space, right?

X, p is in X, then X is homeo. to X/{p}

hmm

copper.hat

@TedShifrin That was my first shot, but all versions of the spectral theorem i have seen are over the complex numbers.

Not sure why I did not stumble on this earlier math.stackexchange.com/q/638216/27978

Urrgh, it is simple, silly me. No need for the spectral theorem.

If I don't do problems for a few weeks my knowledge evaporates away.

Koro

05:25

I don't see the point of graded cohomology rings.

Why shouldn't one take the usual and (better) cohomology rings?

Please explain 🙏

@Thorgott when you're available.

I say this because I don't see how it's obvious to recover H^k from graded ring H^\ast

leslie townes

huh? isn't it the kth graded component of H^dot? how are you defining cohomology as a graded ring?

Koro

hmm that seems to be some heavy terminology from algebra.

leslie townes

i didn't understand your question from earlier either so maybe i'm just confused. i haven't studied any of this in 20 years

awaiting thorgott

Koro

I am referring to Hatcher's definition on page. no. 212

@leslietownes oh which one?

leslie townes

i don't understand what you mean by the 'usual and better cohomology rings.' the usual cohomology as i understand it is naturally a graded ring.

if you think i'm going to find and open hatcher, you are mistaken

Koro

05:38

@leslietownes The thing is Hatcher didn't use the word graded component. So I thought 'graded component' is from the jungles of algebra.

Lemme write out the definition from Hatcher here to clear things up:

leslie townes

whatever. the stuff of pure degree k.

in general stuff in the ring is going to be formal sums of things of different degrees.

but the stuff that's just of one fixed degree.

i should be 1000% clear i do not want to go through anything in hatcher with you, i'm just thinking out loud.

this is all to provide thorgott with more context when he eventually reads this.

i am 20 years too late to be intentionally using terms with any fixed meaning, put "graded component" and "pure degree" in scare quotes.

Koro

A is said to be a graded ring if $A=\Oplus A_i$, where A_i are additive subgroups such that the multiplication takes $A_i\times A_j$ to $A_{i+j}$. If x in A belongs to A_i, we say |x|=i.

leslie townes

right, so here H^dot is the sum of the H^k's. no individual one of which is a ring, as far as i know, in any natural way. the cup product makes the conglomeration of them into a ring.

conglomeration again not being some intentional reference to a high powered algebra word, but my own vague phrasing.

Koro

yeah, so suppose one is given H^* (graded ring). Then how do they get H^k out of it?

(i.e., only it is told that H^* is a graded ring, that's it.)

leslie townes

it's the stuff in H^dot that has degree k.

not everything in H^dot has a degree, of course. but the stuff that does oughta form a group, and it oughta be H^k

Koro

05:44

ooh

leslie townes

i think the grading structure genuinely does encode maybe more information than you think it does. the particulars might not even be spelled out in hatcher for all i know.

Koro

degree of x in A is said to be i if x is in A_i.

as per Hatcher's definition.

leslie townes

i do agree that if you forget entirely about the grading, then maybe i don't know how you recover one of the H^k summands from the graded direct sum.

but i think the grading does allow you to do that.

Koro

@leslietownes it should, that was, I think, the whole point of coming up with this compact notation.

leslie townes

so you consider those things up to graded isomorphism, and not just abstract ring isomorphism.

okay was this what was going on earlier? you had two things that were isomorphic as additive groups, or maybe even as abstract rings, but had a different grading?

or were they also non isomorphic as rings. i wasn't following it too closely.

Koro

05:48

but I don't understand yet how it does that (but your def. hints to some other one which I find helping. You say that H^k is the set of all k degree elements in H^*although it seems we mean different things by degree but I think they ought to be identical somehow)

leslie townes

we're almost literally at the 20th anniversary of my farewell to algebraic topology. when is yours? :)

Koro

@leslietownes it was about two spaces being still not homotopic.

leslie townes

i think i walked in after the actual problem was posed and thinking it was just about a comparison of cohomology rings, when maybe it wasn't.

Koro

despite how it looks like from my messages, I don't want to bid farewell to AT. :)

in one of my exams it was asked to show that RP^n is not homotopic to RP^n-1 \vee S^n.

Unknown x

@TedShifrin Consider $\psi_{1,n}$, As the domain of the function is $U_1$(Right semi circle). So, $x\in (0.1]$ and $y\in (-1,1)$. We have no issue with continuity. Right? $x\neq0$. Also tangent to the semi circle makes an angle from $(0,\pi)$. So, $\psi_{1,n}(U_1)\subseteq (0,pi).$ Right?Then why do we need more angles?

Koro

05:56

I answered: if they were then their universal covers would be homotopic too. That of RP^n, RP^n-1\vee S^n are respectively S^n and S^{n-1}\vee S^n\vee S^n.

and clearly their homology groups don't agree on n-1 level.

so contradiction.

But I think they were expecting some cohomology related answer.

Also it seems intuitively clear to me why universal cover of RP^n-1\vee S^n is what I wrote above. But not sure how to prove it.

leslie townes

does cup length distinguish the graded cohomology rings of those spaces? i thought this fed directly into what ted was saying about the difference in graded structure.

i guess maybe only in the n=3 case.

if you bid adieu to algebraic topology you could just have this be one of life's mysteries.

Koro

but I suppose that can be discounted by the fact that sphere is known to be homeomorphic to a non spherical potato but still no one writes a proof of it? This much handwaving should be allowed so my answer should be correct to that level of handwaving.

@leslietownes cup length?

@leslietownes :(

not for other n's?

leslie townes

i meant that the discussion above that i saw was only about RP^3 and some pdf on the rutgers website.

definitely not saying generally that whatever it is doesn't hold except for n = 3. my mind isn't capable of forming that thought.

Koro

okay, noted. thanks.

Ted Shifrin

@Unknownx The tangent vector doesn't stay in the right semicircle for the whole curve. That's the whole issue. Look even at any circle. Draw more complicated curves. Draw the unit tangent and follow it along the curves.

Unknown x

06:05

@TedShifrin But here domain is $U_1$. Right?

Ted Shifrin

For $\psi_1$, sure. The point is that as you go around your curve, you will need to progress to other $\psi_i$. I don't think you get the idea at all :(

We're trying to measure the angle from the unit tangent vector to the positive x-axis.

What does that angle do as you go around a circle?

Unknown x

@TedShifrin Around the circle, it increases.

Ted Shifrin

Be more specific.

Unknown x

the image will be $\mathbb R$

Ted Shifrin

You go once around the circle.

Koro

06:08

say, one downloads some folder from icloud drive and keeps it in the icloud drive folder. Then, will the downloaded folder stay after expiry of icloud subscription?

Ted Shifrin

Why should it?

Do you get to keep your furniture in your apartment after you're no longer renting it?

Koro

because we downloaded it.

Ted Shifrin

Download it to your PC, then.

I assume the icloud drive folder is stuff on the icloud.

leslie townes

i'm not a mac guy but it seems like there might, just might, be a distinction between a local folder labeled "icloud drive" or "icloud downloads" or whatever, and whatever you pay for when you have an icloud folder in a cloud.

Ted Shifrin

I don't actually pay for space on the cloud, so I am not an expert.

I use the cloud only to sync photos and calendars.

Soumik Mukherjee

06:10

Hello

I am trying to prove Lusternik-Schnirelmann theorem using open subsets instead of closed subsets. For $n=1$ say we have $S^1= A_1 \cup A_2$, where $A_1$ and $A_2$ are open subsets. They can't be disjoint as that will make $S^1$ disconnected. So let $x \in A_1 \cap A_2$. Now $-x \in A_1$ or $-x \in A_2$. That completes the proof.

Koro

Ted Shifrin

That is on the iCloud, no?

Soumik Mukherjee

Is it possible to upgrade this argument to higher dimensions?

Koro

so say I download the download folders (I have done this already). What happens after expiry of the subscription?

Unknown x

@TedShifrin At $(0,-1)$. tangent vector makes an angle $0$ with $x $ axis. Right?

Ted Shifrin

06:12

If you put things on your own computer, then I'm pretty sure they don't magically vanish.

@Unknownx Sure.

Koro

@TedShifrin two things here 1) the folder is not downloaded, i.e., it is still in the 'cloud', 2) downloaded. 2) is confusing here.

Ted Shifrin

Look on your hard drive, not on the iCloud drive.

Koro

If I move the folder to say the 'downloads' under the favourites, then it starts uploading it again to icloud.

hence eating up more icloud space

@TedShifrin as I said, two situations here.

Ted Shifrin

@copper.hat I was going to suggest you consult with leslie. This is his bailiwick.

Unknown x

@TedShifrin At $(1,0)$------> $\pi/2$, $(0,1)$------> $\pi$, $(-1,0)$------> $-\pi/2$, $(0,-1)$------> $2\pi$

Ted Shifrin

06:14

As often happens with you, Koro, we go around and around in circles and get nowhere.

Unknown x

And so on.

Ted Shifrin

@Unknownx ... Absolutely correct. But you move from $U_2$ to $U_3$ to $U_4$ to $U_1$ to $U_2$. NO "and so on." You go once around the curve.

Now if you draw a more complicated simple closed curve, you'll find that the angle meanders up and down.

Unknown x

@TedShifrin okay.

Ted Shifrin

Oh, we didn't quite finish, Unknownx. What happens when you come back to your starting point?

Unknown x

@TedShifrin it increases from $0$ to $2\pi$

Ted Shifrin

06:22

No. You were at $2\pi$ at the point $(0,-1)$. Now you do the last quarter circle and return to $(1,0)$. What does $\theta$ do?

Unknown x

Last quarter circle means? $U_1$?

Ted Shifrin

No, you're on the circle, telling me what the tangent vector does.

You didn't complete the trip around the curve. You said and so on, but I mistakenly said you were done.

Unknown x

@TedShifrin it increases from $0$ to $2\pi$. Right?

Ted Shifrin

AGH.

You told me it started at $\pi/2$ and that it had gotten to $2\pi$ by the time you got to $(0,-1)$. You have to finish going around the circle.

Unknown x

@TedShifrin I started from $(0,-1)$. That is why told like that.

Ted Shifrin

06:30

Oh, ok. What if we started at $(1,0)$? What would $\theta$ do?

It would start, say, at $\pi/2$ (it might be $\pi/2 + 28\pi$ if you were being difficult).

Unknown x

@TedShifrin If it would start at $(1,0)$, it increases from $\pi/2$ to $2\pi+\pi/2$

Are we using that $2k\pi$ to remove that extra term in the returning?

@TedShifrin If it would start from $\pi/2 + 28\pi$. Then it end to the starting point making an angle $\pi/2 + 30\pi$

Am I correct?

Koro

the folder will stay. nvm, The question was not relevant here anyways.

Ted Shifrin

06:49

@Unknownx Yes, both are correct.

Now trace through how you use the $\psi$s to track $\theta$.

Unknown x

@TedShifrin Case 1 for $U_1$ it traces from $pi/2$ to $pi$, if it starts from $(1,0)$.

Should I start from $(0,-1)$?

@TedShifrin If it starts from (1,0), How do we trace continuously after(0,1)? That is why I said it traces from $pi/2$ to $pi$

I am confused, the other part of the semicircle.

$S^1$ is a unit circle. Right?

Soumik Mukherjee

07:35

@Unknownx yes

one potato two potato

08:09

I heard that CLT in probability is a very important result. However the statement is about the convergence in distribution, which may be the weakest convergence. So in probability, convergent in distribution is 'usually' enough? or is there a way to improve a convergence? (not by changing the given RV).

it's a vague question.

Sine of the Time

08:41

In the proof of the Cauchy's Integral formula, why Cauchy theorem hold? I've this doubt bacause in the Cauchy theorem the domain is simply connected, whereas the domain in the proof of the formula is similar to an annulus and so it's only connected. en.wikipedia.org/wiki/Cauchy%27s_integral_formula

s.harp

The caucyh theorem tells you that if you can deform the contour of integration continuously then the integral doesn't change

the formulation with "simply connected domain" etc is just way of guaranteeing you can always deform to the constant path

Sine of the Time

makes sense, it's like considering the ray of the small circle approaching zero?

leslie townes

i haven't clicked through, but wikipedia might not be the best source for this. there are a lot of slightly different things called "cauchy's integral formula" (differing mainly in their hypotheses) and lots of ways to prove such formulas (depending on what one 'knows already' or is allowed to take as given). one can't expect different pages within a resource like wikipedia to maintain textbook-style consistency, which you need to understand "the proof" [i.e., any given proof] of the formula.

it wouldn't even surprise me to find that some crummy book out there has presented these ideas in an incoherent or logically circular way.

sharp's comment gets at the heart of it, whatever your specific hypotheses are.

Sine of the Time

I've seen the proof done in this way in three different sources, and they refer always to Cauchy theorem

and sometimes,the author does a cut and the domain becomes simply connected

leslie townes

these are the sorts of choices that come from how an author has chosen to set things up pedagogically, to deduce more complex results from previously proved ones. there's no inherent "simply connected domains only" restriction to the ideas behind the proof.

Sine of the Time

08:54

ok thank you

leslie townes

again, it's quite possible that someone has written up a set of notes where they don't put their results in the right order and they're using something they haven't proved yet. but it isn't inherently suspicious to me to hear of someone using something called "cauchy's integral formula" in evaluating integrals around the boundary of an annulus.

Sine of the Time

our professor used the cauchy theorem to proof the cauchy integral formula actually, and the idea is basically $\oint_{|z|=R}=\oint_{|z|=r}$ where $r<R$ and this holds because of the cauchy theorem

Unknown x

11:12

@TedShifrin I found the tangent direction like this for $\psi_i$.

Pedro

12:22

Maybe there's a probabilist in the chat: I have a uniform distribution into $n$ boxes that fills box $i$ with $k_i$ balls with probability $1/n$. Thus, after say $N$ trials, I would expect about $k_iN/k$ balls in the $i$th box where $k = \sum_i k_i$. It looks like if I let $p_i = k_i/k$ and give box $i$ a probability of $p_i^{-1}/ \sum p_i^{-1}$, I end up with a uniform distribution where each box gets filled with one ball. That is, after $N$ trials, I should expect $N/n$ balls per box.

I am aware of the inverse sampling from a uniform distribution, i.e. that $F_X(X)$ is uniformly distributed for any continuous $X$. Perhaps it is something along these lines?

Does this process have a name?

user 726941

I would call it "The first question from Pedro in many years"

Welcome back pal

Pedro

Well, I never left the site!

user 726941

True true.

Lucky Chouhan

@Pedro As you are a moderator sir.

Hello @Pedro ! Are you a university student?

Pedro

12:39

@LuckyChouhan No. Why do you ask?

nickbros123

13:21

Let's say I have two fields of characteristic zero, now, the identity element , 0, is this the same for these two fields

shintuku

13:35

any element of a non-finitely generated ideal is nevertheless the sum of finitely many elements of the parent ring, right?

one potato two potato

The reason I don't like physics or that kind of probability problem is that I kinda lack literacy. It's hard for me to formulate situations described in words in mathematical form.

Jakobian

@shintuku itself?

shintuku

@Jakobian i didn't understand, what's the subject of 'itself'?

Jakobian

I think you mean... if $x\in I$ and $I$ is generated by $x_i$, then $x$ is a finite $A$-linear combination of $x_i$?

shintuku

@Jakobian right, necessarily right?

even if $I$ is non-f.g.

@Jakobian no, that can't be the statement

Thorgott

13:47

what Jakobian says is correct

shintuku

nvm yes it can be, since the statement $x$ is a finite $A$-linear combination of $x_i$ does not imply all $x_i$ terms must be stated

ok thanks

nvmmmmmmmmmmmmmmmm

thanks for the help

shintuku

14:07

can an arbitrary ideal of a non-Noetherian ring $R[X]$ be stated: $\langle r_{L_i} + X^{L_i}\rangle_{i \in \mathbb N; L_i \in \mathbb Z}$?

no, consider $\langle r_aX^a + r_bX^b \rangle$

Jakobian

14:53

@Thorgott I want to prove that if $B$ is a flat $A$-algebra, $A$ is Dedekind domain and $f\in B$ is not a zero divisor in $B/\mathfrak{m}B$ for every maximal ideal $\mathfrak{m}$ of $A$, then $B/fB$ is a flat $A$-module

To do thisI want to prove its torsion free

I have something like $ax = fy$ for $x, y\in B$ and $a\in A$, but I don't see what to do without assuming, say, maximal ideals of $A$ are principal

My def. of Dedekind domain is a Noetherian integral domain such that all localizations by prime ideals are PID's

I don't have anything else, and my experience with Dedekind domains is null

Lukas Heger

@Jakobian a module $M$ is flat iff $M_{\mathfrak p}$ is flat over $A_{\mathfrak{p}}$ for all $\mathfrak{p}$, maybe you can use that to reduce to the local case

Lukas Heger

15:14

In the local setting after having localized everything at $\mathfrak{m}$, suppose that $p$ generates $\mathfrak{m}$. A module over $A$ is flat iff it is $p$-torsion free. So suppose that $px=fy$ for $x,y \in B$. Then $fy \in \mathfrak{m}B$, this implies that $y \in \mathfrak{m}B$, because $f$ is not a zero divisor in $B/\mathfrak{m}B$. Thus $y \in \mathfrak{m}B=pB$, write $y=pz$ for $z \in B$.
Using the equation $px=fy$, we get $px=fpz$, but because $B$ is flat, i.e. torsion-free, we can cancel $p$ and get $x=fz$

@Jakobian then we're done

Jakobian

15:25

ah

thank you

Thorgott

(this statement is also true without assuming $A$ is Dedekind, but it's incredibly annoying to prove)

Jakobian

I was overthinking it

Lukas Heger

@Thorgott without the Dedekind assumption, I've only ever seen it with some finiteness assumption on $A \to B$ or Noetherian assumptions or something like that

Thorgott

oh right, no finiteness assumptions on $B$ here

needs finite presentation in general

Lukas Heger

essentially of finite presentation is probably enough, but yeah

Thorgott

15:31

essentially is the assumption in the local case

that's the stacks lemma you linked me a couple days ago :P

Lukas Heger

oh okay

Thorgott

I gave up on the proof for the non-Noetherian case though

Lukas Heger

hmm

Thorgott

they write an essentially f.p. local homomorphism as colimit of essentially f.p. local homomorphisms of noetherian rings (thats still ok) and then they argue if the colimit is flat, the maps in the system are eventually flat (this made me quit)

Lukas Heger

I think there's a stacks lemma on that :P

can't find it though

Thorgott

15:44

yeah there is

it rests on a lemma in the Tor section that relies on another lemma in the Tor section that relies on another etc pp

Lukas Heger

classic stacks

Thorgott

i recently learned that you can look up metadata on stacks like how many previous result a given result (implicitly) references and vice versa

oh, that flatness lemma only relies on a total of 48 preliminary results

Lucky Chouhan

16:06

@Pedro just like that

Ted Shifrin

@Pedro Greetings, stranger!

Lucky Chouhan

@TedShifrin is coolest person here

Ted Shifrin

Ha ha ha.

Lucky Chouhan

@TedShifrin how are you doing sir?

Thomas Finley

If $f$ is assumed to be a continuous function on $\mathbb{R}$, then if $A$ is bounded then $f(A)$ is bounded.

How to prove this?

A is bounded but it seems $A$ might not be closed

So, I don't think I can say, $A$ is a compact set from Heine-Borel Theorem

Thorgott

16:19

perhaps $A$ is contained in some compact set?

Thomas Finley

@Thorgott Yeah, $A\subset \Bbb R$ and $\Bbb R$ is compact

But are you saying this suffices as a reason to claim $A$ is compact?

I might be getting you wrong.

Sourav Ghosh

16:35

@ThomasFinley $cl(A) $ is compact.

$f(cl(A)) $ is compact (implies bounded).

Thorgott

@ThomasFinley very wrong, rethink this

Thomas Finley

@Thorgott I am sorry, what a stupid thing I wrote

@SouravGhosh How is f(cl(A)) is compact? Is it a standard theorem

Sourav Ghosh

@ThomasFinley Yes. "Continuous image of a compact set is compact ".

Thomas Finley

@SouravGhosh ok, but how do you know whether $f$ is continuous on $cl(A)$ just from the fact that $f$ is continuous on $\Bbb R$ and $cl(A)\subset \Bbb R$ ? Sorry, if I sound stupid

Ok, wait if $c\in cl(A)$ is an isolated point then $f$ is continuous on $c$ automatically. If, $c\in cl(A)$ is a cluster point then, then $c\in R$ is also a cluster point and so, if $(x_n)$ is a sequence converging to $c$ then $(x_n)$ is also in $\Bbb R$ then, $f(x_n)$ converges to $f(c)$ as $f$ is continuous at $c\in\Bbb R.$ This means, $f$ is continuous at $c\in cl(A)$

I got it, @SouravGhosh

Thanks!

Shashaank

17:17

Hello!

integral from 0 to x/x-2 dx is divergent right?

since the function is not defined at x=2?

Jakobian

@ThomasFinley If $f:X\to Y$ is continuous and $A\subseteq X$, then $f\restriction_A:A\to Y$ is continuous, since if $U\subseteq Y$ is open then $(f\restriction_A)^{-1}[U] = f^{-1}[U]\cap A$ which is an open subset of $A$.

Sine of the Time

@Shashaank you have to consider the limit or the asymptotic behaviour.

Shashaank

@SineoftheTime I didn't get u

I am asking about the integral

Sine of the Time

consider $\int_0^1 \frac{dx}{x}$ and $\int_0^1\frac{dx}{\sqrt x}$

Shashaank

I can split the integral as from 0 to 2 and from 2 to 5?

Sine of the Time

17:26

not being defined does not imply divergence

Shashaank

@SineoftheTime I can take limit x-->0+ right?

@SineoftheTime Oh

Sine of the Time

your integral is $\int_0^5\frac{x}{x-2} dx$?

Shashaank

yes

I can compute the integral but is it convergent or divergent?

Sine of the Time

you have to study what happen in a neighbourhood of $2$

Jakobian

its divergent. Like Sine says, $\frac{x}{x-2}\sim \frac{1}{x-2}$ in a neighbourhood of $0$, which doesn't have a finite integral there

Shashaank

17:28

Yes

That is because I can't define integral from 0 to 2 x/x-2 right?

@SineoftheTime here the first one is divergent and the second one is convergent right?

Sine of the Time

yes

Shashaank

Thank you @SineoftheTime @Jakobian

Sine of the Time

@Shashaank try substituting $x-2=y$ and the integrand has the behaviour of $1/y$ near $0$ and so it diverges

Shashaank

@SineoftheTime Oh Thank you :-)

Sine of the Time

:)

Cotton Headed Ninnymuggins

18:43

Are there infinitely many Mersenne Primes?

user 726941

8

Q: Are there infinitely many Mersenne primes?

known facts : $1.$ There are infinitely many Mersenne numbers : $M_p=2^p-1$ $2.$ Every Mersenne number greater than $7$ is of the form : $6k\cdot p +1$ , where $k$ is an odd number $3.$ There are infinitely many prime numbers of the form $6n+1$ , where $n$ is an odd number $4.$ If $p$ is prime n...

Jakobian

20:17

@Thorgott what's a DVR?

Lukas Heger

@Jakobian a local PID

Jakobian

ah. What's it short for?

oh, discrete valuation ring

Lukas Heger

right

Thorgott

indeed

sunny

21:28

I'm reading in a book on ODEs by Adkins and Davidson that a function is of exponential type of order $a$ if $|f(t)|\leq Ke^{at}.$ Then it is claimed the set of these functions is a vector space, since it is closed under addition and scalar multiplication. Closed under addition and scalar multiplication...is this the definition of a vector space or that of subspace of a vector space? In case the latter, which vector space is the set of functions of exponential type a subspace of?

Thorgott

if anything, the space of functions from whatever your domain is to whatever your codomain is (which is a real vector space)

sunny

21:48

ok, makes sense, thanks

Jakobian

@sunny All functions $f:\mathbb{R}\to\mathbb{R}$ already have a structure of a vector space, so yes, we are looking at the one inherited from this one as a subspace

I guess... more like what Thorgott said with domain and codomain getting more arbitrary

sunny

@Jakobian if say our codomain is $\mathbb C$, isn't our vector space then a vector space over $\mathbb C$?

Jakobian

@sunny yes

which is also a vector space over $\mathbb{R}$, I suppose

sunny

22:12

👍

冥王 Hades

23:02

The superiority complex when you use extremely basic high school math concept to simplify a "complex" programming problem for someone else

Jakobian

@冥王Hades yes?

leslie townes

you just reflect A across BC to A' and draw the circle and see that your 'complex' programming is really just a 30 degree angle

ahh, that sweet, sweet rush of superiority

冥王 Hades

Something like that yes

I don't wanna become addicted to coffee

Jakobian

You can't be

2 small starbucks coffee's already give you a lot more caffeine than necessary, I heard

冥王 Hades

I need a large usually

twice a day

Jakobian

23:09

I heard thats unhealthy

for your heart

leslie townes

i can think of less healthy things.

Jakobian

since even coke has a lot less caffeine than that

to break the limit of 400 mg of coffee you need like 10 cans of coke

and I think, large starbucks coffee already had like, more than 200?

leslie townes

yeah, two coffees of the right size would get you there.

冥王 Hades

On top of that I hardly sleep too, especially if I lose too many matches in one night

leslie townes

400 mg is an arbitrary line. some of us just need that to get to normal.

hades: you should keep them in a book, or, alternatively, a box.

Jakobian

23:13

@leslietownes I mean. The "normal" probably comes from drug abuse anyway

冥王 Hades

Reminds me of how I hid inside a box at one point to avoid someone

Jakobian

are you snake from metal gear

冥王 Hades

I have played Metal Gear, which snake exactly?

Metal Gear 3?

Jakobian

23:43

whichever

Transcript for

$Mathematics$ Mathematics