Mathematics - 2020-08-16

Thorgott

12:03 AM

if two functions are equal a.e. then one is measurable iff the other is

Hawk

12:17 AM

do i just rename g = inf sup f_n and f? they are equal everywhere.

Thorgott

ye

Hawk

4:25 AM

actually i just realized why it is true.

its because the domain of f, call it E, can be partitioned into E = E - D \cup D. Since m(D) = 0, it means D is measurable. hence it suffices to consider only E -D.

ratatuy

5:58 AM

HI everyone, what is the theme

Artificial Stupidity

6:44 AM

@robjohn Thank you very very much for checking it. :)

Rajesh Dachiraju

8:04 AM

0

Q: Approximate convergence

$\psi:\Omega\to\mathbb{R}$ is continuous. $\Omega$ is a bounded open subset of $\mathbb{R}^m$. $f_n\in H^k(\Omega)$, $k>\frac{m}{2}$. It is given that $$\limsup\limits_{n\to\infty}(f_n(x)-\psi(x)) = \epsilon\forall x\in D,$$ $D$ is a countable dense subset of $\Omega$. It is known that $$\lim\lim...

Soheil

11:53 AM

Hi

I really enjoy doing math. I want to study math at university But I heard It is really hard. So I doubt whether being a math major or not. What is your idea?

Tobias Kildetoft

@Soheil If you enjoy doing math, then studying it is a fine idea

All subjects are "hard" at university level

Soheil

Thank you. @TobiasKildetoft

Tobias Kildetoft

Mainly you should make sure you get an idea of what studying math actually involves. Take a look at some of the topics and how they are presented. For example, there will be a lot more proofs and a lot more abstraction than you have probably encountered previously

That is probably the main thing people see and drop out, because they expected it to be just more of whatever they had seen before.

Soheil

Isn't real analysis or topology something that blowing my mind?

lol

Tobias Kildetoft

Not sure if it is blowing your mind (how would I know ? :) )

Soheil

12:04 PM

Do you had any difficulties with any specific branch of math?

Tobias Kildetoft

Not as such. I had a harder time in courses on the topics I found less interesting. But that just meant I did not take advanced classes in those

Soheil

Oh interesting

Threnody

1:29 PM

Why is an injective homomorphism from $G$ to $S_G$ necessarily an isomorphism?

I'm trying to understand Cayley's Theorem using group actions

All the proofs I found seem to stop short in proving that $phi$ is surjective so I feel like I'm missing something obvious

Tobias Kildetoft

@Threnody It is not. That is not what Cayley's theorem says

Threnody

@TobiasKildetoft I'm not saying that's what it says, but what it seems to require...? I'm a bit lost

Tobias Kildetoft

The theorem says that the group is isomorphism to a subgroup of that symmetric group

Threnody

Yes!

But an isomorphism is a bijective homomorphism

Tobias Kildetoft

Which precisely means that there is an injective homomorphism to it

Threnody

1:32 PM

Uh...

Is an isomorphism an injective or bijective homomorphism?

Tobias Kildetoft

bijective

Threnody

OOoooh!

Tobias Kildetoft

But the theorem does not say that the group is isomorphic to that symmetric group

Threnody

I see I see!

Just a subset (which happens to be a group)

Thank you!

Thorgott

the point is that an injective homomorphism is an isomorphism onto its image

Threnody

1:35 PM

mhm... and the image is not necessarily the entirely stated codomain..

Thorgott

yeah

but the image (of any homomorphism in general) is a subgroup of the codomain

Threnody

any? so not just the left translation... hmm..

Thorgott

yeah, you can try proving that, it's not complicated

Threnody

thank you :)

ronak jain

Hi everyone

Someone please try the above question

robjohn

2:11 PM

@ronakjain I get A

ronak jain

@robjohn but what is the procedure ?

robjohn

How many numbers can start with a 2? How many can start with a 3? How many can start with a 4? There is nothing special about the first digit.

Sebastiano

2:39 PM

Hi, and good afternoon from the Sicily everybody. Is there somebody that can give an answer for this question? math.stackexchange.com/questions/3781647/…

4

Q: Physical meaning of the dot product of a vector and its laplacian

What is the physical meaning of $$\boldsymbol{A}\cdot (\nabla^2\boldsymbol{A})$$ where $\boldsymbol{A}$ is a vector field in 3D space? What does it show?

vector-spaces vector-analysis physics inner-product-space laplacian

Thank you very much and excuse me for this intrusion.

Knight

3:27 PM

@anakhro Hola

anakhro

@Knight hi! How are you?

Knight

I’m in the same state when you made these comments.

Never mind

anakhro

No longer? that was fast

Knight

:) didn’t get the meaning of your reply

It was on 28th March

and I have saved it in my phone. It gives me strength sometimes

Can you please tell me what do you think “Time present and time past both are contained in time future” mean?

anakhro

Nothing very formal, that's for sure. It seems to capture some idea of "what happens in the past and present affects the future (in some manner)".

Knight

3:36 PM

Okay.

anakhro

Does that sort of jive with your understanding of the phrase?

Knight

I thought the poet was alluding to something.

anakhro

I think they most definitely are, but what you cannot say for sure.

Certainly to me, it sounds like "what happens in the past and present affects the future", but it could be more than that.

Knight

But the next lines makes everything more difficult to understand

“And time future contained in time past \\ If all time is eternally present \\ All time is unredeemable”

anakhro

Well here, let's have fun in misconstruing the poet's words as if he were writing mathematical propositions!

past, present $\subseteq$ future

and future $\subseteq$ past

So obviously then past $=$ future

But we are unsure about whether future $\subseteq$ present, so we cannot actually say "all time is eternally present", otherwise past (which is also future) $=$ present.

So we are not quite sure if all time is unredeemable.

Does the poet ever say anything more about the relation of present and past/future time?

Knight

3:50 PM

Don’t you think that just now you have done an amazing and transcendental job? My God!

what an explanation!

no, he talks about time no more

anakhro

No, I think I did a very painful thing of trying to apply mathematical notation where it is definitely not needed. :)

Knight

But what it means that all time is unredeemable?

anakhro

Hmmm, that's not clear to me.

geocalc33

you can't get time back

anakhro

Something is unredeemable when you can't save it or can't make it better.

But it's saying "all time being present forever" $\implies$ "unable to redeem any time".

But I'd think that would make more sense if we said all time was past.

But all time present? I feel like that things in the present and future are those things which are redeemable.

geocalc33

3:58 PM

hi @anakhro

anakhro

hi @geocalc33

geocalc33

@anakhro how long will you be here?

anakhro

I think the average lifespan is >70 so probably can bet I will be around for >40 years.

2

Knight

Present is contained in past

Present is contained in future

“If all time is eternally present, al time is unredeemable” is quite hard to understand.

It doesn’t make any ordinary sense.

anakhro

Poetry is often not supposed to be made sense of (at least, in any analytic sense). Sometimes you just go with the feeling that it conveys to you, rather than any particular logical statements you can deduce from it.

geocalc33

4:12 PM

Time is money

if time is money, then space is...?

Edward Evans

bare dolla

Thorgott

rate of income

this is Fourier duality

Edward Evans

nah man it's bare dolla

phat stacks

Knight

4:29 PM

Thorgott, I know it is a weird request, but can I give you a solution for it’s validation? (Of course, I got a question whose solution I have done)

Or if you’re busy, can you please suggest where can I do it?

Edward Evans

Just ask; don't ask to ask.

Knight

:)

Oops sorry

I'm writing the solution in TexStudio and will upload the it here as an image (because the solution is long)

Knight

5:22 PM

Question

Aug 13 at 15:40, by Knight

Let $f$ be a polynomial with integer coefficients. Define $a_1 = f(0);~~a_2= f(a_1) = f(f(0))$ and $a_n = f(a_{n-1})$ for $n \geq 3$. If there exists a natural number $k \geq 3$ such that $a_k = 0$, then prove that either $a_1 = 0$ or $a_2= 0$

And here is my solution in PDF format

I want someone to validate the solution (if you want to be pedantic @CalvinKhor it's okay lol)

Leaky Nun

@Knight $\Bbb Z$ is \mathbb{Z}

well know theorem -> well known theorem

"if ... implies ...": remove "if" or change the "implies" into "then"

"greater than/equal to" -> "greater than or equal to"

"less than/equal to" -> "less than or equal to"

anakhro

I don't know if he wanted that kind of correction, Leaky.

Leaky Nun

chnce -> chance

@anakhro neither do I

but where's the harm

Knight

mathbb and implies are not working in my studio

Anakhro is back again

Leaky Nun

5:38 PM

172

Q: \mathbb{Z} yields undefined control sequence error

I am using WinEdt 6 for compiling my TeX documents. I am getting an error with contains the line $\mathbb{Z}$ showing that it is undefined control sequence. How should I rectify it? Should I include any math packages or something. I am a beginner in using TeX.

math-mode errors symbols

tldr: you need \usepackage{amssymb} at the beginning of your file

Knight

okay

FruDe

5:49 PM

$\Huge{Hello}$

Alessandro Codenotti

6:22 PM

@EdwardEvans that's advanced AG

Edward Evans

bare dolla

or phat stacks

$\hat{p}$ stacks

anakhro

7:15 PM

I think he's just referring to "stacks".

As in, algebraic stacks.

Stack (mathematics)

In mathematics a stack or 2-sheaf is, roughly speaking, a sheaf that takes values in categories rather than sets. Stacks are used to formalise some of the main constructions of descent theory, and to construct fine moduli stacks when fine moduli spaces do not exist. Descent theory is concerned with generalisations of situations where isomorphic, compatible geometrical objects (such as vector bundles on topological spaces) can be "glued together" within a restriction of the topological basis. In a more general set-up the restrictions are replaced with pullbacks; fibred categories then make a good...

Hawk

Can someone explain to me why in this question math.stackexchange.com/questions/104854/… all the answers gave f(E_{11}) as just f(E_{11})? Isn't f(E_{11}) = 1?

anakhro

Only if you normalize it.

f is only (1) linear, (2) satisfies f(xy) = f(yx).

Hawk

oh because we don't have an explict def of f here. I keep thinking that it is already the trace.

anakhro

Note that $\alpha\text{tr}$ satisfies these properties.

for any scalar $\alpha,$ that is.

Ted Shifrin

8:15 PM

Howdy, lurking demonic @Alessandro.

anakhro

hovers a finger over the report flag.

Ted Shifrin

Howdy, demonic anakhro.

anakhro

hides his pointy tail.

What have you been up to lately, Ted?

Ted Shifrin

A boring life trying not to get sick ... And you?

I was supposed to have a math lesson with an old student of mine who's finishing his Ph.D. in topology, but his internet has been out for a few days. I guess I was going to do curvature of bundles and characteristic classes, but I'm not sure.

Balarka Sen

8:36 PM

Hi

Ted Shifrin

Hi, a @Balarka!

Leaky Nun

@BalarkaSen @TedShifrin is there a triangulation of the torus and a triangulation of the sphere such that their 1-simplices are "isomorphic"?

Ted Shifrin

What does that mean?

anakhro

8:51 PM

Maybe he means something like 1-skeleton?

Leaky Nun

i.e. is there a triangulation of torus such that I can remove the faces and make new faces and end up with the sphere

Ted Shifrin

So you mean identical $1$-skeleta?

Leaky Nun

yeah

I can do this with CW complexes

just take the 8 shape

Ted Shifrin

So can you ever get the Euler characteristic to work out if you have just $E=E'$?

anakhro

I have just been studying some analysis stuff for september. Not very exciting. There is a course on characteristic classes here next semester but I am thinking it might be a little dry having just done some reading on it last year. But I guess I get exercise sets and motivation to do them...

Ted Shifrin

8:54 PM

Seems like $3F=2E$ and something similar with $V$ may make you trouble.

There is so much that course could be or could not be, @anakhro.

Leaky Nun

@TedShifrin very philosophical

anakhro

I think it's just covering Milnor & Stasheff's book, Ted.

Ted Shifrin

Oh, meh.

@Leaky: Am I off base or on base with that last remark about counting?

Leaky Nun

I don't think there can be any relation with V right

a vertex can be shared by any number of edges

and any number of faces

Ted Shifrin

Right. But you have $F=F'$ from what I said.

Leaky Nun

8:58 PM

oh but the two numbers must be the same?

Ted Shifrin

I see.

Well $V-E+F=0$ and $V'-E+F=2$ says $V'=V+2$.

Leaky Nun

hmm

this is a bit disappointing XD

Ted Shifrin

Actually, you don't mean literal triangulations, I bet.

Leaky Nun

I mean a simplicial complex homeomorphic to them

Ted Shifrin

Right. Triangulation is far more rigid.

Leaky Nun

9:00 PM

I think both senses are used?

at least wiki says that a triangulation is a homeomorphic with a simplicial complex

I don't know how the word is used outside

Ted Shifrin

Triangulations are subject to technical conditions. Like an edge can't be shared by more than two faces.

Leaky Nun

that's a consequence of it being a manifold right

Ted Shifrin

(That's where I got my numerical thing before.)

No, I can give all sorts of non-triangulation $\Delta$-complex structures on the torus. I always made my diff geo students triangulate a torus (and it takes something like 20+ faces to get it).

Maybe that's wrong. 20+ edges. 9 faces, maybe.

The easy ways have overlaps that contradict triangulation.

Leaky Nun

well I know a triangulation with 9 vertices and 27 edges and 18 faces

Ted Shifrin

Oh, 18 faces. Yeah, I was doing squares, not triangles.

Anyhow, that homework was always fun to grade.

Leaky Nun

9:04 PM

but how can an edge be shared by 3 faces?

lol

how can it still be locally homeomorphic to R^2

Ted Shifrin

Because of identifications.

Just split the usual square into two triangles. That's a legitimate $\Delta$-complex structure for the torus.

Leaky Nun

aha, your simplicial complexes don't have "extensionality"

i.e. a simplex isn't determined by its vertices

Ted Shifrin

Right. Evidently that's not the case with your question, either. You're wanting to change the faces without changing the edges.

Leaky Nun

yeah

wait

well the faces can still be determined by their vertices right

Ted Shifrin

I dunno.

Then I don't know how you end up with two more vertices and the same number of faces.

Leaky Nun

9:10 PM

yeah then it's impossible

so if I have a simplicial complex satisfying extensionality and is homeomorphic to a 2-manifold, do I automatically have 3E=2F?

Ted Shifrin

I don't know.

mick

@AlgebraicGeometryStudent yes !!

Balarka Sen

10:00 PM

@LeakyNun I don't know. Your example shows you can embed a graph in both S^2 and T^2 such that their respective complements are union of simply connected regions, so I don't see any genuine topological restriction.

Seems like if it's impossible it should be because of combinatorics. I bet there are examples, though.

Leaky Nun

@BalarkaSen what's wrong with Ted's 3E=2F argument?

Balarka Sen

Did he solve it? I didn't read it

If he solved it, cool.

Leaky Nun

well if the identity 3E=2F is true, then F is determined by E

so V and E together determine the euler characteristic

Balarka Sen

OK

I am not really thinking about it

Leaky Nun

sure

Balarka Sen

10:04 PM

3E = 2F is not a topological restriction, it's an artifact of you demanding triangular faces, is all

Leaky Nun

I guess

@BalarkaSen and in the chess world, Hikaru scored many a victory against Magnus

Balarka Sen

Yeah

@LeakyNun Did he win the set? I actually haven't been following

Leaky Nun

neither have I actually

Balarka Sen

I knew yesterday he was crushing Magnus or something

Threnody

10:45 PM

https://sites.math.washington.edu/~sullivan/4027s_au12.pdf

This article says that the order of the group of symmetries of the tetrahedron is 24.

My notes say it's 12.

I'm left confused as a result...

I know how to find the order of the orbit for a tetrahedron, it's 4.

I don't know how to find the order of the stabiliser however... regardless, my notes say it's 3, the linked notes say it's 6.

Who is right?

Ted Shifrin

Both are right.

Depends whether you count only proper symmetries (orientation-preserving) or include improper ones (reflections included).

Threnody

Huh... so it all depends on the intended meaning of 'symmetries of the tetrahedron'

Ted Shifrin

Yes.

Both sources should make it clear whether they're following one of my descriptions or the other.

For example, in my algebra book when I discuss the symmetries of the regular polyhedra, I explicitly say "We consider only the proper symmetries (i.e., symmetries arising from the rotations of $\Bbb R^3$."

Threnody

I see...

In my notes, the order of the stabiliser is noted to be 3. (So proper only)
Is my reasoning correct? Fix a vertex. Then you have 3 more. We need to find out how many ways we can permute them with each other. 6 ways, but half of them are 'improper', so 3.

Ted Shifrin

Yes. You have the rotations of the triangle on the face opposite your vertex (rotating about the axis through the vertex, of course).

Threnody

10:52 PM

So we have 3 permutations/bijections which fix our chosen vertex, hence they're all in the stabiliser of that vertex...

Ted Shifrin

And that's all there is.

Say rotations.

Threnody

Thank you :)

Yes.. that's all there is :D

mick

11:14 PM

0

Q: Uniqueness criterion for tetration based on signs of derivatives?

Consider the fractional iterations of the expontential function denoted $$ \exp^{[t]}(x) $$ $$ \exp^{[0]}(x) = x $$ $$ \exp^{[t]}(x) = \exp(\exp^{[t-1]}(x) $$ $$ \exp^{[l + m]} = \exp^{[l]}(\exp^{[m]}(x))$$ Where $t,l,m,x$ are real and $t > 0$. My friend Tommy proposed once a condition for an inf...

real-analysis calculus derivatives tetration smooth-functions

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$Mathematics$ Mathematics